Business Statistics A Decision Making Approach 10th Edition Groebner Test Bank by Ace Test Banks

TEST ITEM FILE LINDA DAWSON University of Washington Tacoma

B USINESS S TATISTICS A D ECISION -M AKING A PPROACH TENTH EDITION

David F. Groebner Boise State University

Patrick W. Shannon Boise State University

Phillip C. Fry Boise State University

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ISBN-13: 978-0-13-450725-5 ISBN-10: 0-13-450725-8

Business Statistics, 9e (Groebner/Shannon/Fry) Chapter 1 The Where, Why, and How of Data Collection 1) Statistics is a discipline that involves tools and techniques used to describe data and draw conclusions. Answer: TRUE Diff: 1 Keywords: descriptive statistics Section: 1-1 What Is Business Statistics? Outcome: none 2) In this course, the term business statistics refers to the set of tools and techniques that are used to convert information into meaningful data. Answer: FALSE Diff: 1 Keywords: descriptive statistics and/or inferential statistics Section: 1-1 What Is Business Statistics? Outcome: none 3) Descriptive statistics allow a decision maker to reach a conclusion about a population based on a subset from the population. Answer: FALSE Diff: 2 Keywords: descriptive statistics and/or inferential statistics Section: 1-1 What Is Business Statistics? Outcome: none 4) An accountant has recently prepared a report for a client that contains a variety of graphs and charts. In doing so, she has used descriptive statistical methods. Answer: TRUE Diff: 1 Keywords: descriptive statistics Section: 1-1 What Is Business Statistics? Outcome: none 5) Descriptive statistical tools include graphs, charts, and numerical measures. Answer: TRUE Diff: 1 Keywords: descriptive statistics Section: 1-1 What Is Business Statistics? Outcome: none 6) A histogram is an example of a numerical measure. Answer: FALSE Diff: 1 Keywords: descriptive statistics Section: 1-1 What Is Business Statistics? Outcome: none 1-1 Copyright © 2018 Pearson Education, Inc.

7) Companies frequently use charts and graphs in their regular communications with stockholders and investors; this shows the use of descriptive statistics. Answer: TRUE Diff: 1 Keywords: descriptive statistics Section: 1-1 What Is Business Statistics? Outcome: none 8) A manufacturing manager has developed a table that shows the average production volume each day for the past three weeks. The average production level is an example of a numerical measure. Answer: TRUE Diff: 1 Keywords: descriptive statistics Section: 1-1 What Is Business Statistics? Outcome: none 9) An accountant who recently examined 200 accounts from a company's total of 4,000 accounts in an effort to estimate the percentage of all accounts that have incorrect journal entries is using descriptive statistical analysis to reach the conclusion. Answer: FALSE Diff: 2 Keywords: descriptive statistics Section: 1-1 What Is Business Statistics? Outcome: none 10) The editor of a local newspaper is interested in determining the percentage of subscribers who read the paper's editorials. The statistical technique that he would use is called estimation. Answer: TRUE Diff: 2 Keywords: descriptive statistics Section: 1-1 What Is Business Statistics? Outcome: none 11) Hypothesis testing and estimation are two statistical tools that are used to draw inferences about a large data set based on a subset of the data. Answer: TRUE Diff: 1 Keywords: inferential statistics Section: 1-1 What Is Business Statistics? Outcome: none 12) Another term for the arithmetic average is the mean. Answer: TRUE Diff: 1 Keywords: descriptive statistics, mean Section: 1-1 What Is Business Statistics? Outcome: none

13) Statistical inference would be used as the primary statistical tool by a quality control manager who wishes to estimate the average weight of her company's products. Answer: TRUE Diff: 2 Keywords: inferential statistics Section: 1-1 What Is Business Statistics? Outcome: none 14) A light bulb manufacturer wants to advertise the average life of its light bulbs so it tests a subset of light bulbs. This is an example of inferential statistics. Answer: TRUE Diff: 2 Keywords: inferential statistics Section: 1-1 What Is Business Statistics? Outcome: none 15) A sales manager has five salespeople. The following are the number of units sold by the five salespeople during the past week: {5, 13, 6, 2, 4}. Based on the data, the mean number of units sold was 6 units. Answer: TRUE Diff: 2 Keywords: descriptive statistics, mean Section: 1-1 What Is Business Statistics? Outcome: none 16) Some of the most common methods of collecting data include experiments, telephone surveys, mail questionnaires, direct observations, and personal interviews. Answer: TRUE Diff: 1 Keywords: data collection Section: 1-2 Procedures for Collecting Data Outcome: 1 17) An experiment is a process that generates data as its outcome. Answer: TRUE Diff: 1 Keywords: data collection Section: 1-2 Procedures for Collecting Data Outcome: 1 18) Experimental design is a plan for performing an experiment where the effects of one or more factors on the variable of interest are measured. Answer: TRUE Diff: 2 Keywords: data collection, experiments Section: 1-2 Procedures for Collecting Data Outcome: 1

19) Typically, it is possible to include a larger number of questions in a phone survey than in a mail survey since it takes less time to complete the survey over the phone. Answer: FALSE Diff: 2 Keywords: data collection Section: 1-2 Procedures for Collecting Data Outcome: 1 20) An Internet-based or emailed survey is not an alternative method of data collection. Answer: FALSE Diff: 1 Keywords: data collection Section: 1-2 Procedures for Collecting Data Outcome: 1 21) An open-end question requires respondents to choose from a short list of choices Answer: FALSE Diff: 1 Keywords: data collection Section: 1-2 Procedures for Collecting Data Outcome: 1 22) A short survey with closed-end questions is likely to have a better response rate than a long survey with open-ended questions. Answer: TRUE Diff: 1 Keywords: data collection Section: 1-2 Procedures for Collecting Data Outcome: 1 23) The Cranston Company recently met with a group of its customers to ask questions about the service and products provided by the company. The data collected in this process would be an example of data collected through direct observation. Answer: FALSE Diff: 1 Keywords: data collection Section: 1-2 Procedures for Collecting Data Outcome: 1 24) The Georgia Company, a pharmaceutical company, recently conducted a study in which 20 people were given a new drug and 20 other people were given a placebo. The objective was to determine whether there was a difference in pain relief between those using the new drug versus those using the placebo. The data collection used here is an example of an experiment. Answer: TRUE Diff: 2 Keywords: data collection Section: 1-2 Procedures for Collecting Data Outcome: 1 1-4 Copyright © 2018 Pearson Education, Inc.

25) When comparing experiments, surveys, and direct observation as methods of data collection, the method that would typically be the least expensive is surveys. Answer: TRUE Diff: 2 Keywords: data collection Section: 1-2 Procedures for Collecting Data Outcome: 1 26) Assuming that you are planning to collect data using an experiment, it will be very important to establish an appropriate survey design. Answer: FALSE Diff: 1 Keywords: data collection Section: 1-2 Procedures for Collecting Data Outcome: 1 27) Mail questionnaires typically generate poor response rates. Answer: TRUE Diff: 1 Keywords: data collection Section: 1-2 Procedures for Collecting Data Outcome: 1 28) In an unstructured interview the questions are scripted. Answer: FALSE Diff: 1 Keywords: data collection, interviews Section: 1-2 Procedures for Collecting Data Outcome: 1 29) One way to improve the response rate for a survey is to administer the surveys directly to the respondents. Answer: TRUE Diff: 1 Keywords: data collection Section: 1-2 Procedures for Collecting Data Outcome: 1 30) On a survey, the questions pertaining to the background of the respondent (age, gender, etc.) are referred to as demographic questions. Answer: TRUE Diff: 2 Keywords: data collection Section: 1-2 Procedures for Collecting Data Outcome: 1

31) When an interviewer asks a specified series of questions in the course of a personal interview, he/she is conducting an unstructured interview. Answer: FALSE Diff: 2 Keywords: data collection Section: 1-2 Procedures for Collecting Data Outcome: 1 32) The marketing division of a company is interested in determining whether increased advertising will increase sales in three of its target cities. Three levels of advertising are used and the sales are recorded for the month immediately following the ads. In this case, the data are considered to have been collected using an experiment. Answer: TRUE Diff: 2 Keywords: data collection, survey Section: 1-2 Procedures for Collecting Data Outcome: 1 33) Data collected using open-end questions is generally easier to analyze than data collected from closed-end questions. Answer: FALSE Diff: 1 Keywords: data collection, survey Section: 1-2 Procedures for Collecting Data Outcome: 1 34) One of the advantages of data check sheets is that as the data are being recorded, they are also being displayed in a useful format. Answer: TRUE Diff: 2 Keywords: data collection Section: 1-2 Procedures for Collecting Data Outcome: 1 35) The primary purpose of performing a pre-test when developing a telephone or mail survey is to make sure that the respondents can understand the questions and are able to provide meaningful responses. Answer: TRUE Diff: 2 Keywords: data collection, protest Section: 1-2 Procedures for Collecting Data Outcome: 1

36) Close-end questions provide the greatest opportunity to obtain ideas and thoughts on the part of those surveyed but the resulting data are more difficult to analyze. Answer: FALSE Diff: 1 Keywords: data collection, survey Section: 1-2 Procedures for Collecting Data Outcome: 1 37) Questions on a written survey dealing with the characteristics of the respondent (age, income, etc.) are referred to as categorical questions. Answer: FALSE Diff: 2 Keywords: data collection Section: 1-2 Procedures for Collecting Data Outcome: 1 38) Open-end questions are typically included in a survey when the objective is to provide the maximum opportunity for the respondent to express his or her opinion. Answer: TRUE Diff: 1 Keywords: data collection Section: 1-2 Procedures for Collecting Data Outcome: 1 39) The method of data collection called direct observation is always associated with gathering data from people. Answer: FALSE Diff: 1 Keywords: data collection Section: 1-2 Procedures for Collecting Data Outcome: 1 40) Data gathered from a structured interview is generally easier to analyze than data collected from an unstructured interview. Answer: TRUE Diff: 1 Keywords: data collection, structured interview Section: 1-2 Procedures for Collecting Data Outcome: 1 41) When a survey is done you can always assume that non-respondents would have answered the same way as those who did respond. Answer: FALSE Diff: 2 Keywords: data collection, nonresponse bias Section: 1-2 Procedures for Collecting Data Outcome: 1

42) When a company scans the bar codes on its products in an effort to count the number of products that remain in inventory, the company is collecting data through experimentation. Answer: FALSE Diff: 2 Keywords: data collection, UPC Section: 1-2 Procedures for Collecting Data Outcome: 1 43) Data collected on the Internet can generally be considered accurate since the data must go through a screening process before they can be placed on the Internet. Answer: FALSE Diff: 1 Keywords: data collection Section: 1-2 Procedures for Collecting Data Outcome: 1 44) It is possible for an interviewer to interject bias into the data collection project by the way he or she asks the questions. Answer: TRUE Diff: 1 Keywords: data collection, bias Section: 1-2 Procedures for Collecting Data Outcome: 1 45) When people fail to respond to a survey, the data collection process may suffer from nonresponse bias. Answer: TRUE Diff: 1 Keywords: data collection Section: 1-2 Procedures for Collecting Data Outcome: 1 46) Selection bias occurs when the respondent decides which of the questions on the survey to answer. Answer: FALSE Diff: 2 Keywords: data collection, bias Section: 1-2 Procedures for Collecting Data Outcome: 1 47) Recently, an analyst in a company's marketing department surveyed customers regarding how often they buy a particular product. One customer indicated that she purchased the product 17 times in the last six months, but the analyst recorded the response as 71 times. This is an example of observer bias. Answer: FALSE Diff: 2 Keywords: data collection, bias Section: 1-2 Procedures for Collecting Data Outcome: 1

48) When the United States conducts a census that counts all people in the country, this is an example of using a sample. Answer: FALSE Diff: 1 Keywords: population, sample Section: 1-3 Populations, Samples and Sampling Techniques Outcome: 2 49) When the marketing manager for a large company surveys a portion of the total customers of his company, he is using a sample from the population. Answer: TRUE Diff: 1 Keywords: sample, population Section: 1-3 Populations, Samples and Sampling Techniques Outcome: 2 50) A census is an enumeration of the entire sample of items selected from the population of interest. Answer: FALSE Diff: 2 Keywords: sample Section: 1-3 Populations, Samples and Sampling Techniques Outcome: 2 51) A sample is selected from a population in cases where selecting data from the entire population is either very difficult or very expensive. Answer: TRUE Diff: 1 Keywords: sample Section: 1-3 Populations, Samples and Sampling Techniques Outcome: 2 52) A parameter is the boundary on the population of interest. Answer: FALSE Diff: 1 Keywords: parameter, population Section: 1-3 Populations, Samples and Sampling Techniques Outcome: 3 53) Population parameters are descriptive numerical measures, such as an average, that describe the entire population. Answer: TRUE Diff: 1 Keywords: parameter Section: 1-3 Populations, Samples and Sampling Techniques Outcome: 3

54) Statistics are measures computed from the entire population of data. Answer: FALSE Diff: 1 Keywords: statistics Section: 1-3 Populations, Samples and Sampling Techniques Outcome: 2 55) When the production manager selects a sample of items that have been produced on her production line and computes the proportion of those items that are defective, the proportion is referred to as a statistic. Answer: TRUE Diff: 2 Keywords: statistics, proportion Section: 1-3 Populations, Samples and Sampling Techniques Outcome: 3 56) The First National Bank mailed out a survey to all 3,456 savings account customers. A total of 568 surveys were returned. Values computed from the returned surveys would constitute parameters since all 568 customers were surveyed. Answer: FALSE Diff: 2 Keywords: parameter, statistic Section: 1-3 Populations, Samples and Sampling Techniques Outcome: 3 57) If an analyst computes statistics from a sample, the sample is by definition a statistical sample. Answer: FALSE Diff: 2 Keywords: sample, statistic Section: 1-3 Populations, Samples and Sampling Techniques Outcome: 3 58) When university newspaper reporters take a poll of students by standing outside of the nearest pub to survey students about the university offering upgraded food options, the sampling method used is called a random sample. Answer: FALSE Diff: 2 Keywords: convenience sample Section: 1-3 Populations, Samples and Sampling Techniques Outcome: 3 59) A pharmaceutical company conducts a study where 50 patients are given a drug. They find that 10 percent of patients experience nausea as a side effect. This 10 percent is an example of a parameter. Answer: FALSE Diff: 2 Keywords: parameter, statistic Section: 1-3 Populations, Samples and Sampling Techniques Outcome: 3 1-10 Copyright © 2018 Pearson Education, Inc.

60) It is possible for a nonstatistical sample to yield statistics that have values closer to the corresponding parameter than will a statistical sample. Answer: TRUE Diff: 3 Keywords: nonstatistical sample Section: 1-3 Populations, Samples and Sampling Techniques Outcome: 3 61) One of the most common statistical sampling techniques is convenience sampling. Answer: FALSE Diff: 2 Keywords: convenience sample Section: 1-3 Populations, Samples and Sampling Techniques Outcome: 3 62) Possibly the most frequently used nonstatistical sampling procedure is the simple random sample. Answer: FALSE Diff: 1 Keywords: simple random sample Section: 1-3 Populations, Samples and Sampling Techniques Outcome: 3 63) A common underpinning of all statistical sampling techniques is the concept of random selection. Answer: TRUE Diff: 1 Keywords: random sample Section: 1-3 Populations, Samples and Sampling Techniques Outcome: 3 64) Simple random sampling involves selecting members of the population in such a way that all members are equally likely to be chosen. Answer: TRUE Diff: 1 Keywords: sampling techniques Section: 1-3 Populations, Samples and Sampling Techniques Outcome: 3 65) When stratified random sampling is employed, the population is divided into homogeneous subgroups called strata. Answer: TRUE Diff: 2 Keywords: stratified random sample Section: 1-3 Populations, Samples and Sampling Techniques Outcome: 3

66) In election years, the polls that are conducted by such companies as Gallup and Harris typically employ stratified random sampling to reduce the number of people that will need to be surveyed. Answer: TRUE Diff: 2 Keywords: stratified random sample Section: 1-3 Populations, Samples and Sampling Techniques Outcome: 3 67) If a state agency wishes to conduct on-site surveys of small businesses throughout the state, cluster sampling could potentially be used to reduce the geographical area over which the surveys would need to be conducted. Answer: TRUE Diff: 2 Keywords: cluster sample Section: 1-3 Populations, Samples and Sampling Techniques Outcome: 3 68) Cluster sampling is the same thing as stratified random sampling. Answer: FALSE Diff: 1 Keywords: sampling techniques Section: 1-3 Populations, Samples and Sampling Techniques Outcome: 3 69) When a small sample is used, a stratified random sample is more likely to provide the desired information than a simple random sample. Answer: TRUE Diff: 2 Keywords: sampling techniques Section: 1-3 Populations, Samples and Sampling Techniques Outcome: 3 70) Suppose a professor collects survey data by passing out surveys in his/her classes, where the population of interest is defined as all students enrolled at that university. This is an example of nonstatistical sampling technique. Answer: TRUE Diff: 2 Keywords: sampling techniques Section: 1-3 Populations, Samples and Sampling Techniques Outcome: 3 71) One of the reasons that managers prefer statistical sampling to nonstatistical sampling is that statistical sampling is generally easier to perform and less expensive. Answer: FALSE Diff: 2 Keywords: statistical sample Section: 1-3 Populations, Samples and Sampling Techniques Outcome: 3 1-12 Copyright © 2018 Pearson Education, Inc.

72) A market research firm that surveys customers in a shopping mall by asking various people to respond to a short survey about a new product is performing convenience sampling. Answer: TRUE Diff: 2 Keywords: convenience sample Section: 1-3 Populations, Samples and Sampling Techniques Outcome: 3 73) If a population is very large, it may be better to select a sample from the population than to try to obtain a census in an effort to reduce measurement error. Answer: TRUE Diff: 2 Keywords: measurement error, data collection Section: 1-3 Populations, Samples and Sampling Techniques Outcome: 3 74) The sales data for a company measured in the week following an increased ad campaign would be considered cross-sectional data. Answer: TRUE Diff: 1 Keywords: data type, cross-sectional Section: 1-4 Data Types and Data Measurement Levels Outcome: 4 75) When students are asked to list their age and the percentage of their college expenses that they pay for themselves, the type of data being collected is quantitative. Answer: TRUE Diff: 2 Keywords: data type, quantitative Section: 1-4 Data Types and Data Measurement Levels Outcome: 4 76) It is possible for the same survey questionnaire to yield both quantitative and qualitative data. Answer: TRUE Diff: 1 Keywords: data type, quantitative, qualitative Section: 1-4 Data Types and Data Measurement Levels Outcome: 4 77) Sales data measured each week for the past twenty weeks are examples of time-series data. Answer: TRUE Diff: 1 Keywords: data type, time series Section: 1-4 Data Types and Data Measurement Levels Outcome: 4

78) Recording vehicle type as sedan, minivan, pick-up truck, etc. is an example of qualitative data. Answer: TRUE Diff: 1 Keywords: data type Section: 1-4 Data Types and Data Measurement Levels Outcome: 4 79) When customers return a product to a store and the store asks the customer to indicate the reason that the merchandise was returned, the resulting data are quantitative since multiple people will be providing the data. Answer: FALSE Diff: 2 Keywords: data type, qualitative Section: 1-4 Data Types and Data Measurement Levels Outcome: 4 80) Nominal data is the highest level of data. Answer: FALSE Diff: 1 Keywords: measurement levels Section: 1-4 Data Types and Data Measurement Levels Outcome: 4 81) At the end of the school term, students are asked to rate the course and instructor by indicating on a scale of 1-5 how well they liked the course. The data generated from this question are examples of ordinal data. Answer: TRUE Diff: 2 Keywords: data type, ordinal Section: 1-4 Data Types and Data Measurement Levels Outcome: 4 82) On a survey, amount of education is recorded as some high school, high school graduate, some college, college graduate, etc. This is an example of ordinal data. Answer: TRUE Diff: 1 Keywords: measurement levels Section: 1-4 Data Types and Data Measurement Levels Outcome: 4 83) A variable, i.e., the length of time it takes for an employee to complete an assembly procedure at an automotive plant, is a ratio level variable. Answer: TRUE Diff: 2 Keywords: levels of measurement, ratio Section: 1-4 Data Types and Data Measurement Levels Outcome: 4

84) A variable that has all the properties of an interval variable, but also has a true zero, is a ratio level variable. Answer: TRUE Diff: 2 Keywords: levels of measurement, ratio Section: 1-4 Data Types and Data Measurement Levels Outcome: 4 85) Cross-sectional data is a set of data values observed at successive points in time. Answer: FALSE Diff: 1 Keywords: data type, time series Section: 1-4 Data Types and Data Measurement Levels Outcome: 4 86) Data collected on marital status (married, divorced, single, other) would be an ordinal level variable. Answer: FALSE Diff: 2 Keywords: data type, ordinal Section: 1-4 Data Types and Data Measurement Levels Outcome: 4 87) Recently, a bank manager pulled a sample of customer accounts and recorded data for two variables, checking account balance and total number of transactions during the previous 30 days. The data collected would be considered time-series data. Answer: FALSE Diff: 2 Keywords: data type, time series Section: 1-4 Data Types and Data Measurement Levels Outcome: 4 88) Flavors of ice cream (chocolate, vanilla, strawberry, etc.) are an example of nominal data. Answer: TRUE Diff: 1 Keywords: measurement levels Section: 1-4 Data Types and Data Measurement Levels Outcome: 4 89) A major fast-food chain has installed a device that measures the temperature of the hamburgers on the grill. These data are stored in a computer file. If you were to analyze these data, you would be working with ordinal level data. Answer: TRUE Diff: 2 Keywords: levels of measurement, ordinal Section: 1-4 Data Types and Data Measurement Levels Outcome: 4

90) The difference between interval data and ratio data is that interval data has a natural zero. Answer: FALSE Diff: 2 Keywords: measurement levels Section: 1-4 Data Types and Data Measurement Levels Outcome: 4 91) If you have an ordinal variable, it is possible to precisely measure the magnitude of the difference between the possible values of the variable. Answer: FALSE Diff: 3 Keywords: levels of measurement, ordinal Section: 1-4 Data Types and Data Measurement Levels Outcome: 4 92) A cell phone service provider has 14,000 customers. Recently, the sales department selected a random sample of 400 customer accounts and recorded the number of minutes of long distance time used during the previous billing period. The data for this variable is considered to be nominal since the values are based on sample data. Answer: FALSE Diff: 2 Keywords: levels of measurement, nominal Section: 1-4 Data Types and Data Measurement Levels Outcome: 4 93) A cell phone service provider has 14,000 customers. Recently, the sales department selected a random sample of 400 customer accounts and recorded the number of minutes of long distance time used during the previous billing period. The company analyst used Excel to sort these values in order from high to low. She then assigned the highest value a rank of 1, the next highest value a rank of 2, and so forth. These ranks would be considered to be ordinal data. Answer: TRUE Diff: 2 Keywords: levels of measurement, ordinal Section: 1-4 Data Types and Data Measurement Levels Outcome: 4 94) A survey conducted by a local real estate agency asked respondents to indicate whether they preferred natural gas, electric, or oil furnaces for heating their home. The data collected for this variable would be of ordinal level. Answer: FALSE Diff: 2 Keywords: levels of measurement, ordinal Section: 1-4 Data Types and Data Measurement Levels Outcome: 4

95) A small engine repair shop tracks the number of customers who call each day. This variable is a timeseries variable and also ratio level. Answer: TRUE Diff: 2 Keywords: levels of measurement, time series, ratio Section: 1-4 Data Types and Data Measurement Levels Outcome: 4 96) The use of charts and graphs is an example of: A) descriptive statistics. B) inferential statistics. C) estimation. D) hypothesis testing. Answer: A Diff: 1 Keywords: descriptive statistics Section: 1-1 What Is Business Statistics? Outcome: none 97) When an administrator at a local hospital prepares a series of charts and graphs pertaining to the patients that have stayed at the hospital during the past month, she is using which general category of statistical analysis? A) Quantitative statistics B) Inferential statistics C) Descriptive statistics D) Random sampling Answer: C Diff: 2 Keywords: descriptive statistics Section: 1-1 What Is Business Statistics? Outcome: none 98) Which of the following is an example of graphs used to describe data? A) Histograms B) Bar charts C) Both A and B are correct. D) None of the above. Answer: C Diff: 1 Keywords: descriptive statistics, graphs Section: 1-1 What Is Business Statistics? Outcome: none

99) When a marketing manager surveys a few of the customers for the purpose of drawing a conclusion about the entire list of customers, she is applying: A) inferential statistics. B) descriptive statistics. C) quantitative models. D) numerical measures. Answer: A Diff: 1 Keywords: inferential statistics Section: 1-1 What Is Business Statistics? Outcome: none 100) When the park ranger at Yellowstone National Park reports the average length of time that visitors spend in the park, he is using: A) graphical tools. B) numerical measures. C) statistical charts. D) histograms or bar charts. Answer: B Diff: 2 Keywords: descriptive statistics Section: 1-1 What Is Business Statistics? Outcome: none 101) A car manufacturer stated in its advertising that the gas mileage for its hybrids will be greater than 40 mpg on average. A consumer agency tested a sampling of the hybrids under a variety of conditions. Based on these tests, the agency concluded that the manufacturer was justified in making this claim. The process described is an example of: A) descriptive statistics. B) hypothesis testing. C) statistical inference. D) Both B and C are correct. Answer: D Diff: 2 Keywords: descriptive statistics, inferential statistics, hypothesis testing Section: 1-1 What Is Business Statistics? Outcome: none

102) A consumer products company is considering introducing a new product nationally. To help make the decision, it first conducts a test market by selling the product for a few months in one city. This is an example of: A) descriptive statistics. B) charts and graphs. C) estimation. D) hypothesis testing. Answer: C Diff: 2 Keywords: inferential statistics Section: 1-1 What Is Business Statistics? Outcome: none 103) The Biltmore Hotel manager is getting ready to make a presentation that she hopes will justify adding additional staff. As part of the presentation, she has constructed charts and graphs. The general type of statistical analysis she is using is: A) hypothesis testing. B) estimation. C) inferential statistics. D) descriptive statistics. Answer: D Diff: 2 Keywords: descriptive statistics, graphs Section: 1-1 What Is Business Statistics? Outcome: none 104) Estimation and hypothesis testing are categories of: A) inferential statistics. B) descriptive statistics. C) numerical measurement. D) statistical charts. Answer: A Diff: 1 Keywords: inferential statistics Section: 1-1 What Is Business Statistics? Outcome: none 105) A political poll that is used to indicate the percentage of voters who will vote for a particular candidate makes use of which of the following? A) Hypothesis testing B) Numerical analysis C) Estimation D) Both B and C Answer: D Diff: 2 Keywords: inferential statistics Section: 1-1 What Is Business Statistics? Outcome: none 1-19 Copyright © 2018 Pearson Education, Inc.

106) The company that makes a new weight loss pill claims that people who use this pill according to instructions will lose an average of 20 pounds during a four-month period. They say the claim is based on a study of 300 people. Which of the following statistical methods was most likely used to arrive at the company's conclusion? A) Estimation B) Hypothesis testing C) Histograms D) Bar charts Answer: B Diff: 2 Keywords: inferential statistics Section: 1-1 What Is Business Statistics? Outcome: none 107) When the California Highway Patrol states that a study of drivers on a rural highway shows that the average speed is between 62.5 mph and 64.5 mph, they are most likely basing this statement on: A) descriptive statistics. B) estimation. C) hypothesis testing. D) graphical analysis. Answer: B Diff: 2 Keywords: inferential statistics Section: 1-1 What Is Business Statistics? Outcome: none 108) The summaries of data, which may be in forms of tabular, graphical, or numerical, are referred to as: A) inferential statistics. B) descriptive statistics. C) statistical inference. D) report generation. Answer: B Diff: 1 Keywords: descriptive statistics Section: 1-1 What Is Business Statistics? Outcome: none

109) Based on a survey of 400 students in a university in which 20 percent indicated that they were business majors. The university student newspaper reported that "20 percent of all the students at the university are business majors." This report is an example of: A) a sample. B) a population. C) statistical inference. D) descriptive statistics. Answer: C Diff: 2 Keywords: inferential statistics Section: 1-1 What Is Business Statistics? Outcome: none 110) A company is interested in determining which of several advertising layouts is most effective at generating additional sales. The data collection tool that would most likely be used in this situation is: A) telephone survey. B) mail questionnaire. C) experiment. D) observation. Answer: C Diff: 2 Keywords: data collection Section: 1-2 Procedures for Collecting Data Outcome: 1 111) A company conducted a survey of its employees to determine their level of satisfaction with various company policies. The data collected from this survey are: A) primary data. B) secondary data. C) experimental data. D) census data. Answer: A Diff: 2 Keywords: data collection, primary data Section: 1-2 Procedures for Collecting Data Outcome: 1

112) The Dalton Company has recently made a decision to build a new plant in Denver. In making this decision it used data supplied by the U.S. Census Bureau. For the Dalton Company, these data are examples of: A) primary data. B) secondary data. C) reliable data. D) experimental data. Answer: B Diff: 1 Keywords: data collection, secondary data Section: 1-2 Procedures for Collecting Data Outcome: 1 113) An Internet service provider wants to determine its level of customer satisfaction. The best data collection method to obtain the results most quickly is: A) experiment. B) telephone survey. C) mailed survey. D) personal interview. Answer: B Diff: 2 Keywords: data collection Section: 1-2 Procedures for Collecting Data Outcome: 1 114) A tire manufacturing company is interested in obtaining data on stopping distances for each of the three main tread types made by the company. The data collection method that would be most likely used in this case would be: A) telephone survey. B) written questionnaire. C) demographic surveying. D) experiments. Answer: D Diff: 2 Keywords: data collection, experimental Section: 1-2 Procedures for Collecting Data Outcome: 1 115) Which of the following data collection methods is most likely to generate the largest nonresponse? A) Mail surveys B) Direct observation C) Telephone surveys D) Personal interviews Answer: A Diff: 2 Keywords: data collection, mail surveys Section: 1-2 Procedures for Collecting Data Outcome: 1 1-22 Copyright © 2018 Pearson Education, Inc.

116) In developing and conducting a survey, what is the purpose of the pre-test phase? A) To make sure that the cost of developing the survey instrument is not too great B) To generate initial data for analysis C) To catch any problems with the questionnaire before it is fully administered D) To make sure that the respondents like the issues being addressed by the survey Answer: C Diff: 2 Keywords: data collection, survey pre-test Section: 1-2 Procedures for Collecting Data Outcome: 1 117) For which data collection method is it most important to have a polished-looking survey form? A) Telephone survey B) Written questionnaire C) Experimental design D) Personal interview Answer: B Diff: 1 Keywords: data collection, questionnaire Section: 1-2 Procedures for Collecting Data Outcome: 1 118) Which of the following types of questions provide the respondent with the greatest choice in responding to a question? A) Open-end questions B) Close-end questions C) Multiple choice questions D) True/false questions Answer: A Diff: 1 Keywords: data collection, open-ended Section: 1-2 Procedures for Collecting Data Outcome: 1 119) A consumer products company wants to interview customers regarding a new product. If it wishes to adhere to a predetermined pattern of questions in the interview, which of the following would likely be used? A) Structured interview B) Open-end questioning C) Unstructured interview D) Written questionnaire Answer: A Diff: 2 Keywords: data collection, structured interview Section: 1-2 Procedures for Collecting Data Outcome: 1

120) A marketing researcher decides to perform in-depth interviews for ten individuals chosen from a group of twenty five consumers who taste tasted some new food products. In conducting a personal interview, what problem can result if the interviewer is arbitrarily decides who will be interviewed? A) Nonresponses B) Missing data C) Bias D) Poor response rate Answer: C Diff: 2 Keywords: data collection, bias Section: 1-2 Procedures for Collecting Data Outcome: 1 121) One of the major challenges for developing a good written questionnaire or telephone survey instrument is that: A) nonresponses are too high. B) there will always be missed data. C) bias cannot be controlled. D) wording can influence responses. Answer: D Diff: 3 Keywords: data collection Section: 1-2 Procedures for Collecting Data Outcome: 1 122) When an accounting auditor randomly selects 20 accounts from all the accounts to check for accuracy, she has selected: A) a personal observation. B) a sample from the population. C) a census. D) a convenience sample. Answer: B Diff: 1 Keywords: data collection, sample Section: 1-3 Populations, Samples and Sampling Techniques Outcome: 2 123) An Internet service provider has the capability of tracking the time that each of its customers spends connected to the Internet during a month. These data would constitute: A) a simple random sample. B) a convenience sample. C) a cluster sample. D) a population. Answer: D Diff: 2 Keywords: data collection, population Section: 1-3 Populations, Samples and Sampling Techniques Outcome: 2 1-24 Copyright © 2018 Pearson Education, Inc.

124) A professor hands out survey forms during her classes, where the population is all students attending the college. This is an example of: A) a convenience sample. B) a simple random sample. C) a stratified sample. D) a cluster sample. Answer: A Diff: 1 Keywords: convenience sampling Section: 1-3 Populations, Samples and Sampling Techniques Outcome: 3 125) Another term used for statistical sampling is: A) probability sampling. B) convenience sampling. C) ratio sampling. D) numerical sampling. Answer: A Diff: 1 Keywords: probability sample Section: 1-3 Populations, Samples and Sampling Techniques Outcome: 3 126) A retail store is interested in determining what its customers are looking for specifically when they go to the store. To collect the necessary data, interviewers stand near the store's entrance and survey every 10th customer. This type of sampling is called: A) systematic random sampling. B) ratio sampling. C) convenience sampling. D) stratified sampling. Answer: A Diff: 2 Keywords: systematic random sampling Section: 1-3 Populations, Samples and Sampling Techniques Outcome: 3

127) The Polson Pole and Fence Company recently did a quality check on the length of fence posts. To do this, each of the 400 posts in inventory was numbered. Numbers from 1 to 400 were placed in a bowl. Twenty numbers were selected from the bowl without looking. These 20 poles were the ones selected for the study. This type of sampling is called: A) cluster sampling. B) simple random sampling. C) nonstatistical. D) convenience sampling. Answer: B Diff: 2 Keywords: simple random samples Section: 1-3 Populations, Samples and Sampling Techniques Outcome: 3 128) In order to determine a parameter (such as a mean) of a population you would need to conduct a: A) population. B) random sample. C) census. D) statistic. Answer: C Diff: 2 Keywords: population, parameter Section: 1-3 Populations, Samples and Sampling Techniques Outcome: 3 129) If a systematic random sample is to be selected of size 100 from a population with 5,000 items, the first item selected from the ordered population will be: A) randomly selected between 1 and 100. B) randomly selected between 1 and 50. C) any randomly selected value between 1 and 5,000. D) item 50. Answer: B Diff: 2 Keywords: systematic sample Section: 1-3 Populations, Samples and Sampling Techniques Outcome: 3

130) If a stratified random sample is to be conducted, which of the following is true? A) The population will be broken down into subgroups called strata. B) Each subgroup should contain items that are homogeneous with respect to the characteristic of interest. C) If effective, the total required sample size should be less than that which would be needed if a simple random sample were selected. D) All of the above. Answer: D Diff: 2 Keywords: stratified sample Section: 1-3 Populations, Samples and Sampling Techniques Outcome: 3 131) A food warehouse manager plans to conduct a check on damaged packages. The warehouse covers a large area and products are spread out over the entire building. Assuming that no products are more likely to have damaged packages than any other, what statistical sampling method would be used to reduce the time and effort required to do the study? A) Convenience sampling B) Stratified random sampling C) Cluster random sampling D) Systematic random sampling Answer: C Diff: 3 Keywords: cluster sample Section: 1-3 Populations, Samples and Sampling Techniques Outcome: 3 132) Some stores and restaurants have "tell us what you think" cards available for customers. Assuming that angry customers are more likely to take the time to fill these out, this is an example of: A) simple random sampling. B) stratified sampling. C) cluster sampling. D) nonstatistical sampling. Answer: D Diff: 2 Keywords: sampling methods Section: 1-3 Populations, Samples and Sampling Techniques Outcome: 3

133) The mayor of a large U.S. city is interested in addressing complaints from many property owners regarding recent property assessments. Many people feel that they are being overtaxed and that their assessments are too high. To study this issue, the mayor plans to hire consultants to randomly select homes in the city and have these homes independently assessed for value. However, she is concerned that the cost of sampling will be very high since the city is spread out over a wide geographical area. To potentially reduce the cost of sampling, which of the following statistical sampling techniques should be applied? A) Cluster sampling B) Ratio sampling C) Simple random sampling D) Stratified random sampling Answer: A Diff: 3 Keywords: sampling methods Section: 1-3 Populations, Samples and Sampling Techniques Outcome: 3 134) The human resources department at a major high tech company plans to conduct an employee satisfaction study by sampling 100 employees from the 3,000 total employees. They plan to use systematic random sampling since the employee file is in alphabetic order. The first employee selected in the study should be: A) the 30th employee. B) employee 1 to 30 randomly selected. C) employee 1 to 100 randomly selected. D) the first employee. Answer: B Diff: 2 Keywords: systematic random sample Section: 1-3 Populations, Samples and Sampling Techniques Outcome: 3 135) A value computed from a population is called: A) a statistic. B) a real number. C) a parameter. D) a point estimate. Answer: C Diff: 1 Keywords: statistic Section: 1-3 Populations, Samples and Sampling Techniques Outcome: 2

136) Which of the following statements is true? A) Random samples are easier to select than nonstatistical samples. B) Nonstatistical samples can provide useful data. C) Stratified random sampling involves breaking the population down into geographic subgroups. D) Systematic sampling is an example of nonstatistical sampling. Answer: B Diff: 2 Keywords: nonstatistical sampling Section: 1-3 Populations, Samples and Sampling Techniques Outcome: 3 137) In Excel, what procedure is used to select random numbers? A) The random numbers function B) Click on the Data tab, then click on Data Analysis, then click on Sampling C) Click on the Data tab, then click on Data Analysis, then click on Random Number Generation D) Random numbers are not available in Excel. Answer: C Diff: 2 Keywords: Excel, random number generation Section: 1-3 Populations, Samples and Sampling Techniques Outcome: 3 138) A sampling plan that requires a person to interview 100 people as they exit a department store would most likely be: A) a simple random sample. B) a convenience sample. C) a systematic random sample. D) a stratified sample. Answer: B Diff: 2 Keywords: convenience samples Section: 1-3 Populations, Samples and Sampling Techniques Outcome: 3 139) Which one of the following is NOT statistical sampling? A) Simple random sample B) Stratified random sampling C) Cluster sampling D) Convenience sampling Answer: D Diff: 3 Keywords: statistical and nonstatistical sampling Section: 1-3 Populations, Samples and Sampling Techniques Outcome: 3

140) When a survey uses the responses strongly disagree, disagree, neutral, agree, strongly agree, this is an example of: A) nominal data. B) ordinal data. C) interval data. D) ratio data. Answer: B Diff: 2 Keywords: measurement levels Section: 1-4 Data Types and Data Measurement Levels Outcome: 4 141) General Electric Corporation tracks employee turnover annually. It currently has a data set that contains turnover for the past 20 years. What type of data does it have? A) Time-series data B) Cross-sectional data C) Nominal data D) Ordinal data Answer: A Diff: 2 Keywords: data types, time series Section: 1-4 Data Types and Data Measurement Levels Outcome: 4 142) The human resources department at a major high tech company recently conducted an employee satisfaction survey of 100 of its 3,000 employees. Data were collected on such variables as age, gender, marital status, current salary, level of overall satisfaction on a scale from 1 to 5, number of years with the company, and job title. Which of the following best describes the overall data set that was generated from the study? A) Cross-sectional data B) Time-series data C) Nominal data D) Quantitative data Answer: A Diff: 2 Keywords: data types, cross-sectional Section: 1-4 Data Types and Data Measurement Levels Outcome: 4

143) The human resources department at a major high tech company recently conducted an employee satisfaction survey of 100 of its 3,000 employees. Data were collected on such variables as age, gender, marital status, current salary, level of overall satisfaction on a scale from 1 to 5, number of years with the company, and job title. Considering the age variable where employees were asked to list their age at their last birthday, which of the following best describes the level of data measurement for that variable? A) Interval level B) Nominal level C) Ratio level D) Cross-sectional data Answer: C Diff: 2 Keywords: data measurement levels, ratio Section: 1-4 Data Types and Data Measurement Levels Outcome: 4 144) The high temperature is recorded each day for a period of 1 year. This is an example of: A) nominal data. B) ordinal data. C) time-series data. D) cross-sectional data. Answer: C Diff: 1 Keywords: data type, time series Section: 1-4 Data Types and Data Measurement Levels Outcome: 4 145) The human resources department at a major high tech company recently conducted an employee satisfaction survey of 100 of its 3000 employees. Data were collected on such variables as age, gender, marital status, current salary, level of overall satisfaction on a scale from 1 to 5, number of years with the company, and job title. Which of the variables listed are considered to be ratio level data? A) Age and years with the company B) Gender and marital status C) Job title D) None of the variables is ratio level. Answer: A Diff: 2 Keywords: data measurement levels, ratio Section: 1-4 Data Types and Data Measurement Levels Outcome: 4

146) The human resources department at a major high tech company recently conducted an employee satisfaction survey of 100 of its 3,000 employees. Data were collected on such variables as age, gender, marital status, current salary, level of overall satisfaction on a scale from 1 to 5, number of years with the company, and job title. Which of the variables would be classified as nominal level data? A) Age and years with the company B) Overall satisfaction C) Gender, marital status, and job title D) Age and gender Answer: C Diff: 2 Keywords: data measurement levels, nominal Section: 1-4 Data Types and Data Measurement Levels Outcome: 4 147) The human resources department at a major high tech company recently conducted an employee satisfaction survey of 100 of its 3,000 employees. Data were collected on such variables as age, gender, marital status, current salary, level of overall satisfaction on a scale from 1 to 5, number of years with the company, and job title. Which of the variables would be considered to be qualitative data? A) Gender, marital status, job satisfaction, and job title B) Age C) Years with the company D) All variables listed are qualitative. Answer: A Diff: 2 Keywords: data types, qualitative data Section: 1-4 Data Types and Data Measurement Levels Outcome: 4 148) Weekly sales for Amazon.com would be classified as which of the following? A) Cross-sectional data B) Time-series data C) Nominal data D) Ordinal data Answer: B Diff: 1 Keywords: data types, time series Section: 1-4 Data Types and Data Measurement Levels Outcome: 4

149) A college data base includes the number of people who are enrolled in each class the college offers. This is an example of: A) nominal data. B) ordinal data. C) interval data. D) ratio data. Answer: D Diff: 2 Keywords: measurement levels Section: 1-4 Data Types and Data Measurement Levels Outcome: 4 150) When data are organized into levels, the highest data level is: A) interval level data. B) nominal level data. C) ordinal level data. D) ratio level data. Answer: D Diff: 1 Keywords: data measurement levels, ratio Section: 1-4 Data Types and Data Measurement Levels Outcome: 4 151) Data collected at a fixed point in time are: A) time-series data. B) approximate time-series data. C) cross-sectional data. D) panel data. Answer: C Diff: 2 Keywords: time-series data, cross-sectional data Section: 1-4 Data Types and Data Measurement Levels Outcome: 4 152) What is meant by the term statistical inference? Answer: A statistical inference is a conclusion reached about a population value (parameter) based upon an analysis of data in a statistical sample from the population. The idea is that we can study the sample data and then draw a conclusion (inference) about what the entire population looks like with respect to the measure of interest. There are two main categories of statistical inference: estimation and hypothesis testing. Diff: 2 Keywords: inferential statistics Section: 1-1 What Is Business Statistics? Outcome: none

153) Discuss the differences and similarities between statistical estimation and statistical hypothesis testing. Answer: Both estimation and hypothesis testing fall under the main category of statistical procedures called inferential statistics. In both cases, we are attempting to better understand the population of interest by examining the data in a sample from the population. With estimation, we begin with a goal of estimating a population value such as a population mean. We don't have a preconception about what that value is. Instead, we look at a corresponding value for the sample (i.e., sample mean) and use that as our "best" guide to what the population value is. With hypothesis testing, we begin with a claim or idea (hypothesis) about what the population value is and then we use the corresponding sample value to either support or refute the claim or idea. Diff: 2 Keywords: inferential statistics Section: 1-1 What Is Business Statistics? Outcome: none 154) Explain what an experimental design is. Answer: When an experiment is conducted with an experimental design the factors that may affect the variable of interest are controlled so their effect can be observed. Each factor has several predefined levels and the combinations of the various factor levels are tested. Diff: 2 Keywords: data collection, experimental design Section: 1-2 Procedures for Collecting Data Outcome: 1

155) Discuss the steps involved in developing and carrying out a written survey. Answer: While there are several possible approaches to developing and conducting a written survey, there are several steps that most approaches would include. These are: Define the issue — You need a clear understanding of what it is that you wish to learn from the survey. Define the population of interest — You must determine who the people are that are to be surveyed. You need to be specific about this since those surveyed will hopefully represent the views of the entire population. Design the survey instrument — You need to develop the questions that will get at the answers to your research questions. A written survey must be a reasonable length and the questions need to be wellwritten and clear. You need to be careful not to interject bias in the way the question is written. The survey design should be clear and easy for the respondents to use. Pre-test — You should always pre-test the survey with a small subgroup from the population. Use their feedback to make constructive improvements in the survey and to make sure that you have included the right questions given your original objective. Determine the sample size and sampling method — Whenever possible you want to survey a large number of people, but your sampling budget will limit you. It will be necessary to determine how many are needed for statistical validity and for meeting objectives. You also need to specify how the sample is to be selected from the population—statistical or nonstatistical—and specifically what sampling method to use. Select the sample and administer the survey — The survey can be administered in several ways—through the mail, on the Internet, or in person. Diff: 2 Keywords: data collection, survey Section: 1-2 Procedures for Collecting Data Outcome: 1 156) What are the advantages and disadvantages of open-end questions in either a written survey or a personal interview? Answer: Open-end questions allow the respondent to provide a broad range of input and not be restricted to a defined set of response options. This is an advantage since it is possible to tap into the true feelings of the respondent and to obtain responses that might not have been anticipated when the survey was developed. The disadvantage is that analysis of the responses to open-end questions is difficult. The analyst must somehow code the potentially broad range of responses into categories. This takes time and often requires interpretation that might interject bias into the analysis. Diff: 2 Keywords: data collection, survey, interview Section: 1-2 Procedures for Collecting Data Outcome: 1

157) Discuss the two major types of descriptive statistics. Answer: Descriptive statistics includes methods for summarizing a data set. The two major types of descriptive statistics are (1) charts and graphs, and (2) summary numerical measures. Charts and graphs include "picture" type tools such as histograms, bar charts and pie charts that visually summarize a data set. Numerical summary measures are numbers such as the average that describe a characteristic of the data set. Diff: 1 Keywords: descriptive statistics Section: 1-1 What Is Business Statistics? Outcome: none 158) Is there ever a reason why we might prefer to work with a sample rather than with an entire population? Discuss. Answer: A sample is a subset of a population and might not be a perfect representation of the population. As a result, it is possible that when our objective is to know characteristics of the population, the values we get from the sample might be misleading. Thus, in the general case, we would prefer to have access to all the data in the population. However, there are reasons why we might be better off with a sample in some instances. For example, if the population is very large, the time and cost of collecting data from the entire population might make the project unfeasible. In addition, when a great number of measurements are required for a large population, measurement and/or coding errors could be introduced that would give us incorrect information about the population. In addition, if the measurement process requires that we destroy, or otherwise modify, the population values, it would not make sense to deal with the entire population. In these cases, a sample might well be preferable to a census of the population. Diff: 2 Keywords: sample, population Section: 1-3 Populations, Samples and Sampling Techniques Outcome: 2 159) What is the underlying common element of all statistical sampling techniques? Answer: The basis for all statistical sampling techniques is that the items selected in the sample are chosen at random from the population. This provides the potential to perform an objective analysis of the data and reach objective conclusions about the population based on the sample. Diff: 1 Keywords: random sampling Section: 1-3 Populations, Samples and Sampling Techniques Outcome: 3

160) Explain the difference between stratified random sampling and cluster random sampling. Answer: Both techniques are considered statistical sampling techniques since the elements to be included in the sample are randomly selected. Likewise, both techniques have as their objective reducing the cost of sampling as compared to a simple random sample. However, the two techniques are fundamentally different. With stratified random sampling, the population is divided into homogeneous subgroups (strata) with the intent of having the elements in a particular subgroup be as much alike as possible, thereby making it possible to know about all items in the subgroup by examining only a few of the subgroup items. Thus, the overall sample size, which is the sum of the samples from each stratum, could be smaller than the sample size needed if the sampling was done using a simple random sampling approach. Cluster sampling is used when the population values are spread out over a relatively wide geographical area and data collection would be costly due to movement among the population values. With cluster sampling, the population is divided into groups called clusters that are usually defined by geographical boundaries. Each cluster is intended to be as heterogeneous as the population as a whole. Then a random selection of clusters is selected and the sampled items are all selected from only those clusters. This potentially reduces the travel costs. Diff: 3 Keywords: sampling methods, stratified, cluster Section: 1-3 Populations, Samples and Sampling Techniques Outcome: 3 161) Open the data file provided with the text called AirlinePassengers. Indicate the level of data measurement for each variable in the data set. Answer: Airline - nominal Day of Week - ordinal Time of Day - nominal Male/Female - nominal Business/Pleasure - nominal Length of Trip- ordinal Hours to Destination - ratio Children < 10 years - nominal Pieces of Luggage - ratio Pieces Carried On - ratio Times Flown - ratio Diff: 2 Keywords: data measurement levels Section: 1-4 Data Types and Data Measurement Levels Outcome: 4

162) Open the data file provided with the text called Computer Use. Indicate the level of data measurement for each variable in the data set. Answer: Major - nominal Comp. Use - ordinal Comp. Hours - ratio Labs - ordinal Benefit - nominal Higher Tuition - nominal Enrollment Decision - nominal Price to Pay - ratio Diff: 2 Keywords: data measurement levels Section: 1-4 Data Types and Data Measurement Levels Outcome: 4 163) Of the following techniques for collecting data, which one is generally considered the least costly to implement: experiments, telephone surveys, mail questionnaires, direct observation, personal interview? Answer: In most instances, the mail questionnaire is the least costly method for data collection. Even though response rates are notoriously low, the cost of sending surveys in the mail is very low on a per item basis. However, there are many instances where mail surveys can't be used to collect the required data. For instance, if we are interested in the impact of greater drying heat in the press when plywood is manufactured, we can't very well send a survey to a piece of plywood asking for information about its strength and quality! An experiment would be required. In other instances, we need to use observation to get accurate data since people might be tempted to reply with the desired response rather than the truth. For instance, a question like, "Do you wear your seatbelt when driving in your car?" might not provide accurate data in a written survey. Observation might provide better data on driver behavior. Diff: 1 Keywords: data collection Section: 1-2 Procedures for Collecting Data Outcome: 1 164) What are the major categories of statistical tools that will be covered in this course? Answer: Business statistics can be divided into two main categories: descriptive statistics and inferential statistics. Probability is a link between the two and is a major part of the statistical inference section. Diff: 1 Keywords: descriptive, inferential statistics Section: 1-1 What Is Business Statistics? Outcome: none

165) In a survey, what is meant by demographic questions and why might we want to include demographic questions in survey? Answer: Demographic questions are questions that pertain to the respondent such as age, gender, education level, etc. The purpose of demographic questions is to be able to group responses to the central survey questions by category of the demographic questions. For instance, we might group responses to a question about product satisfaction by male/female to see whether males and females have different views about the product. Diff: 2 Keywords: data collection, demographics Section: 1-2 Procedures for Collecting Data Outcome: 1 166) As a member of the student council at your university, you have been assigned the task of conducting a phone survey of undergraduate students to determine satisfaction with the campus food service. Explain how you would go about selecting a simple random sample. Answer: We need to obtain a frame (list) of the population that might be available through the registrar's office. We would assign each student a number from 1 through the number of students. Assuming that this list contains a contact phone number for each student, we could select a simple random sample by using Excel (or a random numbers table) to select the desired sample size. We would probably select extra students since some would be unreachable or not use campus food service. We should establish a call-back procedure to reduce sampling bias. Diff: 2 Keywords: simple, random, sample, bias Section: 1-3 Populations, Samples and Sampling Techniques Outcome: 3 167) Explain the difference between a stratified random sample and cluster random sample. Answer: First, both sampling techniques are examples of statistical sampling procedures. However, they are very different in purpose. In stratified random sampling, the population is broken down into homogeneous groups called strata. The idea is that the elements in each strata are as much alike as possible so that the required sample size from each strata is reduced. Then the total sample size selected from all strata will hopefully be less than the required sample size that would be needed if stratification were not used. In cluster sampling, the population is divided into geographical subgroups. The hope is that each subgroup is a mirror image of the population as a whole. Then a few of the clusters are randomly selected and all sampling of individual items is taken from only the selected clusters. The objective is to reduce the cost of sampling by reducing the physical area that must be covered in the sample. Diff: 3 Keywords: stratified, random, cluster, sample Section: 1-3 Populations, Samples and Sampling Techniques Outcome: 3

Business Statistics, 9e (Groebner/Shannon/Fry) Chapter 2 Graphs, Charts and Tables—Describing Your Data 1) For the same data, a graph of a relative frequency distribution will look exactly the same as a graph of the frequency distribution. Answer: TRUE Diff: 1 Keywords: graph, relative, frequency, distribution Section: 2-1 Frequency Distributions and Histograms Outcome: 1 2) When choosing class boundaries for a frequency distribution, classes such as 60-70, 70-80, 80-90 would be acceptable. Answer: FALSE Diff: 1 Keywords: frequency distribution, classes Section: 2-1 Frequency Distributions and Histograms Outcome: 1 3) Recently a survey was conducted in which customers of a large insurance company were asked to indicate the number of speeding tickets they had received in the past three years. The data in this case would most likely be analyzed using a frequency distribution with the data grouped into classes such as 0-2, 3-5, 6-8, etc. Answer: FALSE Diff: 2 Keywords: frequency, distribution, classes Section: 2-1 Frequency Distributions and Histograms Outcome: 1 4) Recently a survey was conducted in which customers of a large insurance company were asked to indicate the number of speeding tickets they had received in the past three years. The minimum value in the data was zero and the largest was six tickets. If you wished to illustrate the proportion of people who had three or fewer tickets, you would most likely construct a cumulative relative frequency distribution. Answer: TRUE Diff: 2 Keywords: cumulative, relative, frequency distribution Section: 2-1 Frequency Distributions and Histograms Outcome: 1 5) Frequency distributions are specifically for analyzing discrete data. Answer: FALSE Diff: 1 Keywords: frequency, distribution, discrete Section: 2-1 Frequency Distributions and Histograms Outcome: 1

6) When developing a frequency distribution, the following bin classifications follow the definition of appropriate intervals of a variable: 0 to < 10 10 to < 20 20 to < 25 Answer: FALSE Diff: 1 Keywords: frequency distribution, classes Section: 2-1 Frequency Distributions and Histograms Outcome: 1 7) It is often a good idea to convert frequency distributions to relative frequency distributions when you wish to compare two distributions with different amounts of data. Answer: TRUE Diff: 1 Keywords: relative, frequency, distribution Section: 2-1 Frequency Distributions and Histograms Outcome: 1 8) In a report describing the number of people in the family of each of the 400 employees at a manufacturing company, the frequency count at the value 3 was 220. This means that the relative frequency at the 3 level is .44. Answer: FALSE Diff: 1 Keywords: relative, frequency Section: 2-1 Frequency Distributions and Histograms Outcome: 1 9) One way to develop a frequency distribution using Excel is to use the Frequency function. Answer: TRUE Diff: 2 Keywords: frequency, distribution, Excel Section: 2-1 Frequency Distributions and Histograms Outcome: 1 10) There is no difference between cumulative frequency and relative frequency. Answer: FALSE Diff: 2 Keywords: frequency Section: 2-1 Frequency Distributions and Histograms Outcome: 1

11) A study of 2000 Verizon cellphone customers listed the annual incomes of the customers as well as other variables. The lowest income was $25,000 and the highest income was $145,000. To develop a frequency distribution with 6 classes, the smallest value that the class width can be is 20K. Answer: TRUE Diff: 2 Keywords: frequency, distribution, class, width Section: 2-1 Frequency Distributions and Histograms Outcome: 1 12) A cumulative frequency distribution shows the percentage of observations for the variable of interest with values less than or equal to the upper limit of each class. Answer: FALSE Diff: 2 Keywords: cumulative, frequency, distribution Section: 2-1 Frequency Distributions and Histograms Outcome: 1 13) In constructing a frequency distribution for the savings account balances for customers at a bank, the following class boundaries might be acceptable if the minimum balance is $5.00 and the maximum balance is $18,700: $0.00 - $5,000 $5,000 - 10,000 $10,000 - $15,000 $15,000 - $20,000 Answer: FALSE Diff: 1 Keywords: frequency, distribution, class, boundary Section: 2-1 Frequency Distributions and Histograms Outcome: 1 14) The appropriate number of classes should generally be between 5 and 20. Answer: TRUE Diff: 2 Keywords: frequency distribution, classes Section: 2-1 Frequency Distributions and Histograms Outcome: 1 15) Once you have determined the class width using the formula, high-low divided by the number of classes, it is appropriate to round to the nearest integer to make the analysis easier. Answer: FALSE Diff: 2 Keywords: class, width, formula Section: 2-1 Frequency Distributions and Histograms Outcome: 1

16) There is no hard-and-fast rule regarding the number of classes that must be used when establishing a frequency distribution for a continuous variable. Answer: TRUE Diff: 1 Keywords: class, frequency, distribution, continuous Section: 2-1 Frequency Distributions and Histograms Outcome: 1 17) The upper and lower limits of each class in a frequency distribution are also referred to as the data array. Answer: FALSE Diff: 2 Keywords: class, frequency, distribution, array Section: 2-1 Frequency Distributions and Histograms Outcome: 1 18) The following class limits would be acceptable for developing a frequency distribution on income: $0 < $5,000 $5001 < $10,000 $10,001 < $20,000 Over $20,000 Answer: FALSE Diff: 2 Keywords: class, limit, frequency, distribution Section: 2-1 Frequency Distributions and Histograms Outcome: 1 19) A histogram can be created for discrete or continuous data. Answer: TRUE Diff: 1 Keywords: histogram Section: 2-1 Frequency Distributions and Histograms Outcome: 1 20) In a recent study at First National Bank, a frequency count was made for the variable marital status for the bank's 10,000 customers. It would also be appropriate to develop a histogram for this variable to show how marital status is distributed. Answer: FALSE Diff: 2 Keywords: frequency, histogram, distribution Section: 2-1 Frequency Distributions and Histograms Outcome: 2

21) After developing a frequency distribution for a quantitative variable, a histogram can be developed with the horizontal axis representing the values of the variable and the vertical axis representing the frequency of occurrence in each class or group. Answer: TRUE Diff: 2 Keywords: frequency, distribution, histogram, class Section: 2-1 Frequency Distributions and Histograms Outcome: 2 22) A histogram can be constructed for data that are either quantitative or qualitative. Answer: FALSE Diff: 1 Keywords: histogram, quantitative, qualitative Section: 2-1 Frequency Distributions and Histograms Outcome: 2 23) In a recent study of retail daily sales by stores at a mall in Kansas, the minimum daily sales was $700 and the maximum was $51,000. If you wish to construct a frequency distribution with 10 classes, the minimum class width would be $5,100. Answer: FALSE Diff: 2 Keywords: frequency, distribution, class, width Section: 2-1 Frequency Distributions and Histograms Outcome: 1 24) Consider a situation in which both a frequency distribution and a relative frequency distribution have been developed for the same quantitative variable. If histograms are constructed from each distribution, the graphs will appear to have the same shape. Answer: TRUE Diff: 2 Keywords: relative, frequency, distribution, histogram, quantitative Section: 2-1 Frequency Distributions and Histograms Outcome: 2 25) When a histogram is constructed for discrete numerical data, there should be spaces between the bars of the histogram. Answer: FALSE Diff: 2 Keywords: histogram Section: 2-1 Frequency Distributions and Histograms Outcome: 2

26) When the Histogram tool in Excel is used to construct a frequency distribution and histogram, the default histogram is in the proper format and will require only that you add appropriate labels. Answer: FALSE Diff: 2 Keywords: histogram, Excel, frequency, distribution Section: 2-1 Frequency Distributions and Histograms Outcome: 2 27) When using the Histogram tool in Excel to construct a frequency distribution and histogram, the bins represent the upper class limits. Answer: TRUE Diff: 2 Keywords: histogram, Excel, frequency, bin, class Section: 2-1 Frequency Distributions and Histograms Outcome: 2 28) When using the Histogram tool in Excel to construct a frequency distribution and histogram, if the first bin value is 10 and the second bin value is 20, the frequency count for the second class will include all values from 10 up to, but not including, 20. Answer: FALSE Diff: 3 Keywords: histogram, Excel, bin, frequency Section: 2-1 Frequency Distributions and Histograms Outcome: 2 29) If you wish to construct a graph of a relative frequency distribution, you would most likely construct an ogive. Answer: FALSE Diff: 2 Keywords: ogive, relative, frequency, distribution Section: 2-1 Frequency Distributions and Histograms Outcome: 2 30) A joint frequency distribution is used to describe the number of occurrences where two observations in a data set have the same value. Answer: FALSE Diff: 1 Keywords: joint, frequency, distribution Section: 2-1 Frequency Distributions and Histograms Outcome: 3 31) A joint frequency distribution can be constructed for either quantitative or qualitative data. Answer: TRUE Diff: 2 Keywords: joint, frequency, distribution Section: 2-1 Frequency Distributions and Histograms Outcome: 3

32) If a manager is interested in analyzing the relationship between the age of customers and the dollar volume of business that is done in the store, a relative frequency distribution would be most appropriate. Answer: FALSE Diff: 2 Keywords: relative, frequency, distribution Section: 2-1 Frequency Distributions and Histograms Outcome: 1 33) A recent study of students at the university contained data on year in school and student age. An appropriate tool for analyzing the relationship between these two variables would be a joint frequency distribution. Answer: TRUE Diff: 2 Keywords: joint, frequency, distribution Section: 2-1 Frequency Distributions and Histograms Outcome: 3 34) A histogram can be used to display a joint frequency distribution between two quantitative variables. Answer: FALSE Diff: 3 Keywords: histogram, joint, frequency, distribution Section: 2-1 Frequency Distributions and Histograms Outcome: 3 35) In Excel, joint frequency distributions can be generated using the Pivot Table feature under the Data tab. Answer: TRUE Diff: 2 Keywords: Excel, joint, distribution, pivot Section: 2-1 Frequency Distributions and Histograms Outcome: 3 36) An ogive is a graph that shows cumulative relative frequency. Answer: TRUE Diff: 1 Keywords: ogive Section: 2-1 Frequency Distributions and Histograms Outcome: 2 37) If you have constructed a joint frequency distribution manually and now wish to convert it to a joint relative distribution, the proper method is to divide each cell frequency by the cell's row total. Answer: FALSE Diff: 2 Keywords: joint, relative, frequency, distribution Section: 2-1 Frequency Distributions and Histograms Outcome: 3

38) Another name for a joint frequency distribution is a cross-tabulation table. Answer: TRUE Diff: 1 Keywords: joint, frequency, distribution, cross-tabulation Section: 2-1 Frequency Distributions and Histograms Outcome: 3 39) Two separate frequency distributions for two variables provide the same information as one joint frequency distribution involving the same two variables. Answer: FALSE Diff: 3 Keywords: joint frequency distribution Section: 2-1 Frequency Distributions and Histograms Outcome: 3 40) In Excel a joint frequency distribution table can be created using a tool called PivotTable. Answer: TRUE Diff: 2 Keywords: Excel, joint, frequency, distribution, pivot Section: 2-1 Frequency Distributions and Histograms Outcome: 3 41) An ogive is a graph of a joint frequency distribution. Answer: FALSE Diff: 1 Keywords: ogive, joint, frequency, distribution Section: 2-1 Frequency Distributions and Histograms Outcome: 3 42) A histogram is an effective tool for graphically describing a joint frequency distribution. Answer: FALSE Diff: 2 Keywords: histogram, joint, frequency, distribution Section: 2-1 Frequency Distributions and Histograms Outcome: 3 43) In a study involving car owners, one question asked the owner for the number of miles driven last year. A second question asked the owner for the age of the vehicle. A joint frequency distribution would be useful for determining whether newer cars tend to be driven more miles than older cars. Answer: TRUE Diff: 2 Keywords: joint, frequency, distribution Section: 2-1 Frequency Distributions and Histograms Outcome: 3

44) In a study involving car owners, one question asked the owner for the number of miles driven last year. A second question asked the owner for the age of the vehicle. A histogram would be useful for analyzing the relationship between miles driven and the age of the vehicle. Answer: FALSE Diff: 3 Keywords: histogram, joint, frequency, distribution Section: 2-1 Frequency Distributions and Histograms Outcome: 3 45) In constructing a histogram for a joint frequency distribution, the histogram will have the most meaning for the decision maker if there are no gaps between the bars on the histogram. Answer: FALSE Diff: 2 Keywords: histogram, gap, bar, joint, frequency, distribution Section: 2-1 Frequency Distributions and Histograms Outcome: 3 46) A bar chart is the same as a histogram. Answer: FALSE Diff: 1 Keywords: bar, chart, histogram Section: 2-2 Bar Charts, Pie Charts and Stem and Leaf Diagrams Outcome: 4 47) Histograms cannot have gaps between the bars, whereas bar charts can have gaps. Answer: TRUE Diff: 1 Keywords: histogram, gap, bar, chart Section: 2-2 Bar Charts, Pie Charts and Stem and Leaf Diagrams Outcome: 4 48) The regional sales manager for a medical supply company recently collected data on the reasons why customers returned the merchandise for a refund. She actually formed a frequency distribution for this variable. It would now be acceptable to construct a bar chart to graphically display the results. Answer: TRUE Diff: 2 Keywords: frequency, distribution, bar, chart Section: 2-2 Bar Charts, Pie Charts and Stem and Leaf Diagrams Outcome: 4 49) Bar charts can typically be formed with the bars vertical or horizontal without adversely affecting the interpretation. Answer: TRUE Diff: 2 Keywords: bar, chart, horizontal, vertical Section: 2-2 Bar Charts, Pie Charts and Stem and Leaf Diagrams Outcome: 4

50) Bar charts can show either frequency or percentage. Answer: TRUE Diff: 1 Keywords: bar chart Section: 2-2 Bar Charts, Pie Charts and Stem and Leaf Diagrams Outcome: 4 51) A tire store manager has collected data showing the number of tires of each brand sold during the past month. A bar chart might be effective in graphically illustrating which brands tend to sell best at this store. Answer: TRUE Diff: 2 Keywords: bar chart Section: 2-2 Bar Charts, Pie Charts and Stem and Leaf Diagrams Outcome: 4 52) A pie chart is almost always constructed when the variable of interest is qualitative. Answer: FALSE Diff: 2 Keywords: pie, chart, qualitative Section: 2-2 Bar Charts, Pie Charts and Stem and Leaf Diagrams Outcome: 4 53) In situations involving two or more variables, both histograms and bar charts can be used for multiple variables on the same graph. Answer: FALSE Diff: 3 Keywords: multiple, variable, histogram, bar chart Section: 2-2 Bar Charts, Pie Charts and Stem and Leaf Diagrams Outcome: 4 54) The Wilson company monitors customer complaints and organizes these complaints into six distinct categories. Over the past year, the company has received 534 complaints. One possible graphical method for representing these data would be a histogram. Answer: FALSE Diff: 2 Keywords: histogram, category Section: 2-2 Bar Charts, Pie Charts and Stem and Leaf Diagrams Outcome: 4 55) The difference between bar charts and histograms is that bar charts always show percentage while histograms always show frequency. Answer: FALSE Diff: 1 Keywords: bar, chart, histogram, percentage Section: 2-2 Bar Charts, Pie Charts and Stem and Leaf Diagrams Outcome: 4

56) When developing a bar chart, it is usually preferable to organize the bars in order from high to low. Answer: FALSE Diff: 2 Keywords: bar, chart Section: 2-2 Bar Charts, Pie Charts and Stem and Leaf Diagrams Outcome: 4 57) A stem and leaf diagram is most similar to a bar chart. Answer: FALSE Diff: 1 Keywords: stem, leaf, diagram Section: 2-2 Bar Charts, Pie Charts and Stem and Leaf Diagrams Outcome: 5 58) One of the differences between a stem and leaf diagram and a histogram is that even for variables involving a large number of different values, the stem and leaf diagram shows the individual data values whereas the histogram requires you to group the data and lose the individual values. Answer: TRUE Diff: 2 Keywords: stem, leaf, histogram Section: 2-2 Bar Charts, Pie Charts and Stem and Leaf Diagrams Outcome: 5 59) A study was recently conducted in which makers of toothpaste tracked sales for the month at different stores in a market area. The variable of interest was the number of units sold. The numbers ranged from 1,200 to 22,700. In this case, the stems in a stem and leaf diagram might be values such as 1 and 22 while the leaves would be 200 and 700. Answer: TRUE Diff: 2 Keywords: stem, leaf Section: 2-2 Bar Charts, Pie Charts and Stem and Leaf Diagrams Outcome: 5 60) In constructing a stem and leaf diagram, the stem and leaves are defined using a similar method to the definition of the bins of a histogram. Answer: TRUE Diff: 1 Keywords: stem, leaf Section: 2-2 Bar Charts, Pie Charts and Stem and Leaf Diagrams Outcome: 5

61) A stem and leaf diagram is more appropriate for graphically displaying a joint frequency distribution than is a histogram since the stems can be used to display one variable while the leaves can be used to display the second variable. Answer: FALSE Diff: 2 Keywords: stem, leaf, joint, frequency Section: 2-2 Bar Charts, Pie Charts and Stem and Leaf Diagrams Outcome: 5 62) If the Viking Sales Company plans to display the sales for each of its six major products for the year 2001, an effective chart to do this would be a histogram. Answer: FALSE Diff: 2 Keywords: histogram, chart Section: 2-2 Bar Charts, Pie Charts and Stem and Leaf Diagrams Outcome: 4 63) In preparing a line chart, the horizontal axis shows time and the vertical axis shows the value of the variable of interest. Answer: TRUE Diff: 1 Keywords: line, chart, axis Section: 2-3 Line Charts and Scatter Diagrams Outcome: 6 64) In a scatter plot the points should always be connected with a line. Answer: FALSE Diff: 1 Keywords: scatter, plot, line, chart Section: 2-3 Line Charts and Scatter Diagrams Outcome: 7 65) A university recently collected data for a sample of 200 business majors. One variable collected was the number of credits left to be taken before graduation. This variable could effectively be displayed using a line chart. Answer: FALSE Diff: 2 Keywords: line, chart Section: 2-3 Line Charts and Scatter Diagrams Outcome: 6 66) A scatter diagram is a line graph without the points connected by a line. Answer: FALSE Diff: 1 Keywords: scatter, diagram, line Section: 2-3 Line Charts and Scatter Diagrams Outcome: 7

67) A major insurance company believes that for drivers between 16 years of age and 60 years of age, the number of accidents per year tends to decrease as age increases. If this is the case, a scatter diagram should show a negative relationship between the two variables. Answer: TRUE Diff: 2 Keywords: scatter, diagram Section: 2-3 Line Charts and Scatter Diagrams Outcome: 7 68) Sawyer & Company is a law firm in Dallas, Texas. Recently, the administrative manager prepared a report for the managing partners that showed the number of court cases handled by the firm monthly over the past three years. It was appropriate for her to use a line chart in this case. Answer: TRUE Diff: 2 Keywords: line, chart Section: 2-3 Line Charts and Scatter Diagrams Outcome: 6 69) Sawyer & Company is a law firm in Dallas, Texas. Recently, the administrative manager prepared a report for the managing partners that showed the number of court cases handled by the firm monthly over the past three years. One of the objectives of graphing these data might have been to identify a trend in the number of court cases. Answer: TRUE Diff: 1 Keywords: trend, graph Section: 2-3 Line Charts and Scatter Diagrams Outcome: 6 70) To show the relationship between amount of rainfall and the number of car accidents, the best type of graph to use is a scatter diagram. Answer: TRUE Diff: 1 Keywords: scatter diagram Section: 2-3 Line Charts and Scatter Diagrams Outcome: 7 71) The J.B. Hanson Company is interested in analyzing the relationship between end-of-the-week inventory levels and sales for the same week. The graph that most likely would be used to show this relationship is a histogram. Answer: FALSE Diff: 2 Keywords: scatter, relationship, histogram Section: 2-3 Line Charts and Scatter Diagrams Outcome: 7

72) A study at State University involved an analysis of students' GPAs and the number of hours that they work at jobs off-campus. An appropriate graph to display the relationship between these two variables might be a scatter diagram. Answer: TRUE Diff: 2 Keywords: scatter, diagram, relationship Section: 2-3 Line Charts and Scatter Diagrams Outcome: 7 73) When developing a scatter diagram, it is appropriate to connect the points on the graph with straight lines or the lines can be omitted. Answer: FALSE Diff: 2 Keywords: scatter, diagram, connect Section: 2-3 Line Charts and Scatter Diagrams Outcome: 7 74) Scatter diagrams can be used for either quantitative or qualitative data. Answer: FALSE Diff: 1 Keywords: scatter diagram, data Section: 2-3 Line Charts and Scatter Diagrams Outcome: 7 75) If two variables are graphed on the same line chart, two separate scales are always required. Answer: FALSE Diff: 2 Keywords: variable, graph, scale Section: 2-3 Line Charts and Scatter Diagrams Outcome: 6 76) If a scatter diagram shows points that are reasonably aligned and are sloping downward from left to right, this implies that there is a negative linear relationship between the two variables. Answer: TRUE Diff: 2 Keywords: scatter, diagram, linear, relationship Section: 2-3 Line Charts and Scatter Diagrams Outcome: 7 77) In constructing and analyzing a scatter diagram, it is important to focus on the data points of interest by establishing appropriate scales for both the x and y axes. Otherwise, a nonlinear relationship can appear to be linear. Answer: TRUE Diff: 2 Keywords: scatter, diagram, relationship, nonlinear Section: 2-3 Line Charts and Scatter Diagrams Outcome: 7

78) A scatter diagram can show whether a pair of variables has a strong or weak relationship, and also whether it is linear or curved. Answer: TRUE Diff: 2 Keywords: scatter, diagram Section: 2-3 Line Charts and Scatter Diagrams Outcome: 7 79) Roscoe and Associates makes computer software for use in the telecommunications industry. Recently, managers at the company collected data for the year 2001 on three variables: total dollars spent on research and development, total sales dollars, and total employee salaries. To graphically present these three variables, the managers would be justified in using a line chart with all three variables plotted. Answer: FALSE Diff: 3 Keywords: line, chart Section: 2-3 Line Charts and Scatter Diagrams Outcome: 6 80) On a scatter diagram, the independent variable should be placed on the horizontal axis and the dependent variable should be placed on the vertical axis. Answer: TRUE Diff: 2 Keywords: scatter, diagram, independent Section: 2-3 Line Charts and Scatter Diagrams Outcome: 7 81) In analyzing a single quantitative variable, you will generally choose to use a scatter diagram if the variable is measured over time and a histogram if the variable is cross-sectional. Answer: FALSE Diff: 2 Keywords: scatter, diagram, histogram Section: 2-3 Line Charts and Scatter Diagrams Outcome: 7 82) A histogram is most commonly used to analyze which of the following? A) Nominal level data B) Quantitative data C) Time-series data D) Ordinal data Answer: B Diff: 2 Keywords: histogram, quantitative, data Section: 2-1 Frequency Distributions and Histograms Outcome: 2

83) The Maple Grove Hotel manager has collected data on the number of rooms occupied each evening for the past 700 nights. The fewest rooms occupied during that period was 11 and the most was the capacity, 430. Based on this information, which of the following would be reasonable class limits for the first class if the manager wishes to use 8 classes to develop a frequency distribution? A) 0 to 40 B) 10 to < 65 C) 11 to 19 D) 0 to 52.38 Answer: B Diff: 2 Keywords: histogram, class, limit Section: 2-1 Frequency Distributions and Histograms Outcome: 2 84) Recently a study of fans attending the New York Mets baseball games was conducted and 500 fans were surveyed. In forming a frequency distribution of the number of miles fans traveled from home to the stadium, it was found that 247 fans traveled between 0 and 5 miles. Based on this information what was the relative frequency for this class? A) 0.247 B) 0.30 C) 0.494 D) Can't be determined without more information. Answer: C Diff: 2 Keywords: relative, frequency, distribution Section: 2-1 Frequency Distributions and Histograms Outcome: 1 85) Frequency distributions can be formed from which of the following types of data? A) Both discrete and continuous B) Discrete only C) Continuous only D) Only qualitative data Answer: A Diff: 2 Keywords: frequency, distribution, data Section: 2-1 Frequency Distributions and Histograms Outcome: 1

86) A common rule of thumb for determining how many classes to use when developing a frequency distribution with classes is: A) between 5 and 20 classes. B) no fewer than 6 classes. C) equal to 0.25 times the number of data values. D) at least 10 classes. Answer: A Diff: 1 Keywords: frequency, distribution, classes Section: 2-1 Frequency Distributions and Histograms Outcome: 1 87) Which of the following is an acceptable format for setting up class boundaries for a frequency distribution? A) 20 to under 40 B) 20 to 40 C) 200 to 299.99 D) All of the above. Answer: D Diff: 2 Keywords: class, boundaries, frequency, distribution Section: 2-1 Frequency Distributions and Histograms Outcome: 1 88) Which of the following is not considered desirable when constructing a frequency distribution for continuous data? A) Open-ended classes B) Mutually exclusive classes C) Equal-width classes D) All-inclusive classes Answer: A Diff: 1 Keywords: frequency, distribution, continuous, classes Section: 2-1 Frequency Distributions and Histograms Outcome: 1

89) Many Walmart stores have automotive departments where customers can buy tires, have their vehicles serviced, and obtain other automotive services. Recently, the manager at an Ohio Walmart collected data on the time customers had to wait to get the desired automotive service. Of the 500 cars in the sample, the shortest time any customer spent waiting was 3 minutes and the longest time was 183 minutes. Assuming that the manager wishes to develop a frequency distribution with 9 classes, which of the following would be an appropriate class width for each class? A) 10.50 B) 19.99 C) 20.00 D) 3 to 23 Answer: C Diff: 2 Keywords: class, width, frequency, distribution Section: 2-1 Frequency Distributions and Histograms Outcome: 1 90) A histogram is used to display which of the following characteristics for a quantitative variable? A) The approximate center of the data B) The spread in the data C) The shape of the distribution D) All of the above. Answer: D Diff: 2 Keywords: histogram, display, quantitative, variable Section: 2-1 Frequency Distributions and Histograms Outcome: 2 91) When using Excel's Histogram option under the Data Analysis tool, the term bins refers to: A) the mid-point of each class. B) the column where the data are located. C) the upper limits of each class. D) the lower limits of each class. Answer: C Diff: 2 Keywords: Excel, histogram, bin, class Section: 2-1 Frequency Distributions and Histograms Outcome: 2

92) In forming the classes for a frequency distribution and histogram, suppose there were a number of empty classes. You should: A) increase the class width. B) decrease the class width. C) keep the current class width. D) use an ogive instead. Answer: B Diff: 2 Keywords: histogram, Excel, gaps Section: 2-1 Frequency Distributions and Histograms Outcome: 2 93) A frequency histogram should be computed from which type of data? A) Quantitative data B) Categorical data C) Nominal level data D) Ordinal data Answer: A Diff: 2 Keywords: frequency, histogram, data, quantitative Section: 2-1 Frequency Distributions and Histograms Outcome: 2 94) Which of the following is a reason for constructing a joint frequency distribution? A) To determine the trend between the two variables B) To measure the spread between the two variables C) To help analyze the relationship between the two variables D) To show the average of two variables Answer: C Diff: 1 Keywords: joint, frequency, distribution Section: 2-1 Frequency Distributions and Histograms Outcome: 3 95) Joint frequency distributions are used to display: A) the histograms of two variables analyzed simultaneously. B) the number of occurrences at each of the possible joint occurrences of two variables. C) the cumulative distribution of a variable with two possible outcomes. D) the relative frequency of two variables. Answer: B Diff: 2 Keywords: joint, frequency, distribution Section: 2-1 Frequency Distributions and Histograms Outcome: 3

96) A study was recently done in which the brand preference for breakfast cereal was analyzed against the gender of the shopper. The study consisted of 200 male shoppers and 300 female shoppers. Three different cereal brands were considered: A, B, and C. A total of 250 female shoppers preferred brand A, 25 female shoppers preferred brand C. The number of female shoppers that preferred brand B was: A) 25. B) 100. C) 75. D) 50. Answer: A Diff: 1 Keywords: joint, frequency, distribution Section: 2-1 Frequency Distributions and Histograms Outcome: 3 97) A study was recently done in which the brand preference for breakfast cereal was analyzed against the gender of the shopper. The study consisted of 200 male shoppers and 300 female shoppers. Three different cereal brands were considered: A, B, and C. A total of 250 female shoppers preferred brand A, 25 female shoppers preferred brand C. A total of 100 shoppers preferred brand B. The number of male shoppers that preferred brand B was: A) 25. B) 100. C) 75. D) 50. Answer: C Diff: 2 Keywords: joint, frequency, distribution Section: 2-1 Frequency Distributions and Histograms Outcome: 3 98) The undergraduate students at your university are classified as freshmen, sophomores, juniors, or seniors. A recent study of undergraduates asked the students to indicate the number of credits they were registered for this term. The responses were 3, 6, 9, 12, 15, and 18. The number of cells in a joint frequency distribution for the two variables, class standing, and credit hours is: A) 4. B) 10. C) 24. D) None of the above. Answer: C Diff: 1 Keywords: joint, frequency, distribution, cells Section: 2-1 Frequency Distributions and Histograms Outcome: 3

99) Which of the following CANNOT be shown effectively with a histogram? A) A frequency distribution B) A joint frequency distribution C) A relative frequency distribution D) The center, shape and spread of a distribution Answer: B Diff: 2 Keywords: histogram, frequency, distribution Section: 2-1 Frequency Distributions and Histograms Outcome: 2 100) Which of the following is NOT true of a bar chart? A) It is used for numerical data. B) The bars can be either horizontal or vertical. C) It can show either frequency or relative frequency. D) It is used for categorical data. Answer: A Diff: 2 Keywords: bar chart Section: 2-2 Bar Charts, Pie Charts and Stem and Leaf Diagrams Outcome: 4 101) One characteristic of a bar chart is: A) the bars can be displayed either vertically or horizontally. B) there can be no gaps between the bars. C) it is used to display the distribution of a continuous variable. D) it shows cumulative frequency. Answer: A Diff: 1 Keywords: bar, chart Section: 2-2 Bar Charts, Pie Charts and Stem and Leaf Diagrams Outcome: 4 102) A bar chart is most likely used to display which of the following? A) A continuous variable B) A nominal level variable C) An ordinal level variable D) Either B or C Answer: D Diff: 2 Keywords: bar, chart, variable Section: 2-2 Bar Charts, Pie Charts and Stem and Leaf Diagrams Outcome: 4

103) A study was recently conducted by the regional electric and gas company. Data were collected for three customer categories showing the dollar amount of natural gas and the dollar amount of electricity consumed during the year. Which of the following graphs would most likely be used to display both sets of data together? A) Pie chart B) Bar chart C) Line chart D) Histogram Answer: B Diff: 2 Keywords: bar, chart, display Section: 2-2 Bar Charts, Pie Charts and Stem and Leaf Diagrams Outcome: 4 104) One of the key differences between a bar chart and histogram is: A) the histogram contains gaps between the bars and the bar chart does not. B) a bar chart is used to display a categorical variable and a histogram is used to display the distribution of a quantitative variable. C) the histogram shows relative frequency while the bar chart shows frequency. D) the bar chart must be vertical while the histogram must be horizontal. Answer: B Diff: 2 Keywords: bar, chart, histogram, variable Section: 2-2 Bar Charts, Pie Charts and Stem and Leaf Diagrams Outcome: 4 105) The city counsel has just voted to pass the city's budget for next year. If you were writing a report describing the budget so the citizens could understand how the total tax dollars will be spent, which of the following graphs might be most appropriate? A) Pie chart B) Scatter diagram C) Histogram D) Ogive Answer: A Diff: 2 Keywords: pie, chart Section: 2-2 Bar Charts, Pie Charts and Stem and Leaf Diagrams Outcome: 4

106) Which of the following is a false statement? A) A bar chart is usually constructed so that gaps exist between the bars. B) The bars on a bar chart can be different colors. C) A histogram is usually constructed without gaps between the bars. D) A bar chart and histogram can typically be used interchangeably. Answer: D Diff: 2 Keywords: bar, chart, histogram Section: 2-2 Bar Charts, Pie Charts and Stem and Leaf Diagrams Outcome: 4 107) At a manufacturing plant workers are divided into 4 different teams that rotate shifts. The number of units produced by each team is recorded. The best type of chart to display the data is a: A) pie chart. B) histogram. C) ogive. D) line chart. Answer: A Diff: 2 Keywords: pie, chart Section: 2-2 Bar Charts, Pie Charts and Stem and Leaf Diagrams Outcome: 4 108) A homeowners association consists of 20 homes. The family in each home is considered an automatic member of the association. Recently, one of the homes fell into a state of disrepair. A survey was conducted of the homeowners both on the same street as the house in question and on the second street. At issue was whether legal action should be brought against the homeowner with the problem house. There are 8 homes on the same street as the problem house and 6 of these called for legal action. The percentage of houses on the second street that favored legal action is 50 percent. Which type of chart might be most effective for conveying the information about percentage of residents favoring legal action by street? A) Histogram B) Stem and leaf diagram C) Bar chart D) Pie chart Answer: C Diff: 2 Keywords: bar, chart Section: 2-2 Bar Charts, Pie Charts and Stem and Leaf Diagrams Outcome: 4

109) Which of the following is a key difference between a bar chart and a histogram? A) A bar chart typically has gaps between the bars while a histogram has no gaps. B) A bar chart is developed to analyze a continuous variable, while a histogram is used to analyze discrete variables. C) Both A and B are correct. D) There is actually no real difference between a bar chart and a histogram. Answer: A Diff: 2 Keywords: bar, chart, histogram Section: 2-2 Bar Charts, Pie Charts and Stem and Leaf Diagrams Outcome: 4 110) The Grangeville Power Company has four classifications for its customers. For each customer classification, the company tracks the total amount of electricity used during the year. Which of the following types of graphs would be most appropriate to use? A) A horizontal bar chart B) A vertical bar chart C) Both A and B would be appropriate. D) A histogram Answer: C Diff: 2 Keywords: bar, chart, horizontal, vertical Section: 2-2 Bar Charts, Pie Charts and Stem and Leaf Diagrams Outcome: 4 111) Which of the following is true about the difference between stem and leaf diagrams and histograms? A) There is no difference. B) The stem and leaf diagram shows more information by showing the individual values. C) The histogram shows the shape center and spread of the distribution while the stem and leaf does not. D) The stem and leaf diagram shows less information than a histogram. Answer: B Diff: 2 Keywords: stem and leaf, histogram Section: 2-2 Bar Charts, Pie Charts and Stem and Leaf Diagrams Outcome: 5

112) Ace Hardware Store collected data on the sales volume for five locations for the past four months. The data are input into an Excel spreadsheet where columns A through E represents the location names and column F represents the sales for each day for the past four months at that location. In order to construct a scatter diagram that displays the percent of total sales by location for each day for the past six months, the required steps in Excel will be to: 1) Determine the total sales by location. 2) Input the daily sales for each location 3) Divide the daily sales for a particular location by the total sales for that location. What would be the correct order of these steps? A) 1) 2) 3). B) 2) 1) 3). C) 2) 3) 1). D) 1) 3) 2). Answer: B Diff: 3 Keywords: Excel, bar, chart Section: 2-3 Line Charts, Scatter Diagrams, and Pareto Charts Outcome: 6 113) A bar chart possesses which of the following? A) Capability of displaying the distribution for a quantitative variable B) The option of displaying the data in scatter diagram form C) The option for displaying two or more variables on the same chart D) An easy method for displaying the general shape of a continuous variable Answer: C Diff: 2 Keywords: bar, chart Section: 2-2 Bar Charts, Pie Charts and Stem and Leaf Diagrams Outcome: 4 114) The Canyon Water Company collects data on the number of gallons of water consumed during a month for each customer. The production manager has divided the usage into 6 classes. To display these data effectively, she could use which of the following types of graphs to convey information about the water usage? A) A stem and leaf diagram B) A bar chart C) A histogram D) Either a histogram or a pie chart Answer: D Diff: 2 Keywords: histogram, pie, chart Section: 2-2 Bar Charts, Pie Charts and Stem and Leaf Diagrams Outcome: 4

115) A stem and leaf diagram is used to: A) display the distribution of a quantitative variable. B) show the joint relationship between two variables. C) graph a joint frequency distribution. D) show relative cumulative frequency. Answer: A Diff: 2 Keywords: stem, leaf, quantitative Section: 2-2 Bar Charts, Pie Charts and Stem and Leaf Diagrams Outcome: 5 116) A stem and leaf diagram is an alternative to using: A) a pie chart. B) a bar chart. C) a histogram. D) an ogive. Answer: C Diff: 2 Keywords: stem, leaf, histogram Section: 2-2 Bar Charts, Pie Charts and Stem and Leaf Diagrams Outcome: 5 117) One of the advantages that a stem and leaf diagram has over a histogram is: A) the detail of the data is preserved. B) it shows the general distribution of a quantitative variable. C) it can be used with nominal data. D) There are no advantages. Answer: A Diff: 2 Keywords: stem, leaf, histogram Section: 2-2 Bar Charts, Pie Charts and Stem and Leaf Diagrams Outcome: 5 118) If two variables show a positive linear relationship in a scatter diagram: A) most of the data values will plot in the lower left-hand quadrant. B) most of the data values will plot in the lower left-hand and upper right-hand quadrants. C) most of the data values will cluster close to the x and y axes. D) the data will cluster in the center of the graph. Answer: B Diff: 2 Keywords: scatter, diagram, linear, relationship Section: 2-3 Line Charts and Scatter Diagrams Outcome: 7

119) To show how the price of a stock has changed over the last 3 months, the best type of chart to use is: A) a pie chart. B) a histogram. C) a line chart. D) a bar chart. Answer: C Diff: 1 Keywords: line, chart, trend Section: 2-3 Line Charts and Scatter Diagrams Outcome: 6 120) A line chart is most appropriate for: A) cross-sectional data. B) nominal level data. C) ordinal level data. D) time-series data. Answer: D Diff: 1 Keywords: line, chart, time, series Section: 2-3 Line Charts and Scatter Diagrams Outcome: 6 121) The managers at Harris Pizza in Boston have tracked the tips received by their drivers along with the total bill to the customer. An appropriate graph for analyzing the relationship between these two variables is: A) a scatter diagram. B) a line chart. C) a histogram. D) a pie chart. Answer: A Diff: 2 Keywords: scatter, diagram, relationship Section: 2-3 Line Charts and Scatter Diagrams Outcome: 7 122) A scatter diagram can be used to do which of the following? A) Determine the trend in a variable B) Analyze the relationship between two variables C) Describe the basic distribution for a quantitative variable D) Show the percentage of a variable that is associated with each category into which that variable has been divided Answer: B Diff: 2 Keywords: scatter, diagram, relationship Section: 2-3 Line Charts and Scatter Diagrams Outcome: 7

123) In constructing a scatter diagram: A) the independent variable should be on the vertical axis. B) the independent variable should be on the horizontal axis. C) the dependent variable should be on the horizontal axis. D) It does not matter which variable goes on which axis. Answer: B Diff: 2 Keywords: scatter, diagram, independent, dependent Section: 2-3 Line Charts and Scatter Diagrams Outcome: 7 124) Which of the following questions CANNOT be answered using a scatter diagram? A) Is there a causal relationship between the 2 variables? B) Is there a curved or linear relation between the 2 variables? C) Is there a weak or strong relation between the 2 variables? D) Is there a positive or negative relation between the 2 variables? Answer: A Diff: 2 Keywords: scatter diagram Section: 2-3 Line Charts and Scatter Diagrams Outcome: 7 125) The sales manager at Western Furniture Company tracked data on the number of customers who came into the store each day and the total dollar volume of sales at the store during the same day. She is considering putting together a report for top management and wishes to show the relationship between these two variables. Which of the following graphs would likely be most useful? She has a sample of 36 days worth of data. A) Scatter diagram B) Bar chart C) Frequency histogram D) Pie chart Answer: A Diff: 2 Keywords: scatter, diagram, relationship Section: 2-3 Line Charts and Scatter Diagrams Outcome: 7

126) Consider the following chart. Which of the following statements is most correct?

A) There is a negative linear relationship between the two variables. B) There is a positive linear relationship between the two variables. C) There is a perfect linear relationship between the two variables. D) There is no apparent relationship between the two variables. Answer: B Diff: 2 Keywords: scatter, diagram, linear, relationship Section: 2-3 Line Charts and Scatter Diagrams Outcome: 7

127) Consider the following chart. Which of the following statements is most correct?

A) The values for the dependent variable are determined by the values for the independent variable. B) The values in a scatter plot should be connected by a straight line. C) The variable on the horizontal axis should be the independent variable. D) A scatter plot like this one shows the trend in the data over time. Answer: C Diff: 2 Keywords: scatter, plot, independent Section: 2-3 Line Charts and Scatter Diagrams Outcome: 7 128) The Fitness Center manager has collected data on the number of visits to the club each week for the past 8 weeks. These data are shown as follows. Which of the following statements is most correct? Week 1 1415

Week 2 1623

Week 3 1934

Week 4 1879

Week 5 2102

Week 6 2156

Week 7 2511

Week 8 2499

A) The proper graph for displaying these data is a pie chart. B) There has been a gradual downward trend in these data. C) A frequency histogram should be developed to help identify the trend in these data. D) The data lend themselves to a line chart. Answer: D Diff: 2 Keywords: line, chart, trend Section: 2-3 Line Charts and Scatter Diagrams Outcome: 6

129) A professor wants to know if the amount of time her students spent working on a statistics assignment relates to the grade the student receives. She surveyed 10 students and recorded the data below. Which of the following statements is most correct? #1

# 10

5 90

3 80

3.5 80

1 60

4.5 90

1 70

3 75

4 85

2 70

2.5 75

A) A histogram will illustrate whether a linear relationship exists between the number of hours studied and the grade received. B) A scatter diagram would be useful for displaying a relationship between the number of hours studied and the grade received. C) A line chart for these data will show a trend between the student number and the grade received. D) None of the above Answer: B Diff: 2 Keywords: line, chart, trend Section: 2-3 Line Charts and Scatter Diagrams Outcome: 6 130) Assuming you have data for a variable with 2,000 values, using the 2k ≥ n guideline, what is the least number of groups that should be used in developing a grouped data frequency distribution? A) 9 B) 11 C) 13 D) 12 Answer: B Diff: 1 Keywords: descriptive statistics Section: 2-1 Frequency Distributions and Histograms Outcome: 1 131) In creating a frequency distribution for numerical data, describe the steps in choosing the classes. Answer: The steps are (1) determine the number of classes, (2) determine the class width, (3) determine the class boundaries, and (4) count how many values are in each class. The number of classes should generally be between 5 and 20, where the more data you have means more classes. To determine the class width, the classes must span the entire range of the data set from the smallest value to the largest value. So an initial estimate of class width is to divide the range by the number of classes. W= This width usually needs to be adjusted up to a "round" number. To determine the class boundaries there are several considerations. The classes should be of equal width, have no gaps between classes (all inclusive), and no overlaps (mutually exclusive). Finally the number of observation in each class is tallied. Diff: 2 Keywords: classes, class width 2-31 Copyright © 2018 Pearson Education, Inc.

Section: 2-1 Frequency Distributions and Histograms Outcome: 2 132) Explain what information can be conveyed by a frequency histogram. Answer: A frequency histogram can be used to convey information about three different characteristics of a quantitative variable. First, the histogram can give us an idea of where the center of the data falls. The histogram can show the spread in the variable. Finally, the histogram can show the shape of the distribution. Diff: 2 Keywords: frequency, histogram, quantitative Section: 2-1 Frequency Distributions and Histograms Outcome: 2 133) Explain why it is appropriate to connect the points on a line graph, but the points on a scatter plot should not be connected. Answer: A line chart is used to display data that are measured in sequence over time. The points are in a defined order. Connecting the points serves to illustrate any trend that may be present in the data. A scatter plot is used to show the relationship between two variables. The XY data points are plotted in a two-dimensional space. The order that the XY points are recorded is of no consequence, so connecting the points would be meaningless. Diff: 3 Keywords: line, scatter, connect, points Section: 2-3 Line Charts and Scatter Diagrams Outcome: 7 134) Why should a histogram contain no gaps between the bars but a bar chart may have gaps? Answer: A histogram is used to convey the distribution of a quantitative variable. The horizontal axis represents the range of possible values for the variable of interest. The bars are joined together to form the distribution. A bar chart is used to illustrate a qualitative variable. By using gaps between the bars, the differences between the categories are more easily seen. Diff: 2 Keywords: histogram, bar, chart, gaps Section: 2-2 Bar Charts, Pie Charts and Stem and Leaf Diagrams Outcome: 4

135) Discuss the steps that you would use to manually construct a histogram for the salaries of the 1,124 employees in the Ferris Steel Company. Answer: We might begin by sorting the data from low to high. Since salaries would likely be somewhat continuous, there would likely be lots of different possible salaries represented in the 1,124 data points. Thus, we would need to group the data into classes. We need to determine how many classes we want to use. The rule of thumb is somewhere between 5-20 depending on the data. Once the number of classes is determined, we next compute the class width using the formula: W= We can round this class width up to a nice round number for ease of interpretation. Next, we form the classes making sure that they are mutually exclusive and all inclusive. In addition, they should be the same width, if possible. Once the classes are determined, the next step is to form a frequency distribution by counting how many of the salaries fall in each class. A histogram can be formed directly from the frequency distribution by placing the variable (salaries) on the horizontal axis. The vertical axis will represent the frequency in each class. Bars are drawn with width corresponding to the class limits and height corresponding to frequency in each class, no gaps are allowed except in cases where the frequency is equal to zero. Label the histogram appropriately. Diff: 3 Keywords: histogram, construction, class, width Section: 2-1 Frequency Distributions and Histograms Outcome: 2 136) Suppose you are given the following data. 99 93 109

107 121 104

89 100 99

94 115 96

119 88 97

99 93 93

87 83 103

116 112 89

97 109 108

101 116 94

111 99 93

99 94 98

If you wish to have a histogram with five classes, what should the first class limits be? Answer: The first class limits can be determined by first determining the width for each class. Given that we want 5 classes, the class width is determined by: W=

= 7.60

However, we could round the 7.60 up to 8 for ease of interpretation. Then the first class could be set up as 83 and under 91 or 83 to 90.99 or 83 < 91. Diff: 2 Keywords: histogram, class, limits, width Section: 2-1 Frequency Distributions and Histograms Outcome: 2

137) Suppose that you have a data set of 512 observations and the data values range from 36 to 187. What classes would you choose for this data set? Explain why you would choose these values. Answer: For n=512 values, 9 classes would be about the right number of classes based on the 2k ≥ n rule. Then the class width should be about (187-36)/9 = 16.8, which should be rounded to 20. So the classes should start at 20 or 30 to be low enough to contain the lowest value. So one way to do the classes is: 20 to < 40 40 to < 60 60 to < 80 80 to < 100 100 to < 120 120 to < 140 140 to < 160 160 to < 180 180 to < 200 This is 9 classes, all inclusive, mutually exclusive, equal width, and contains all the data values. Diff: 2 Keywords: classes Section: 2-1 Frequency Distributions and Histograms Outcome: 1 138) Explain why a relative frequency histogram is sometimes preferable to a regular frequency histogram. Answer: Relative frequencies are used when we are interested in comparing two or more distributions when the number of data values in the distributions differs. For instance, suppose we have two frequency distributions on salaries, one for college graduates and one for non-college graduates. There are 100 college graduates in the data set and 1,000 non-college graduates. In each distribution, the first class limits are $0 through $15,000. If there are 5 college graduates in the first class and 50 non-college graduates in the first class, it may appear that non-college graduates are much more heavily represented in the low salary bracket. However, if we convert these to relative frequencies, it turns out that the relative frequency is 0.05 for both groups. Diff: 2 Keywords: relative, frequency, histogram Section: 2-1 Frequency Distributions and Histograms Outcome: 2 139) A company has 400 employees. The manager of human resources has recorded the annual salary and wages for each employee. These value range from $17,500 to $67,800. Provide an example of a stem and a leaf that could be developed to describe the distribution of the salary and wage data. Answer: Given these data, the stems could be the thousands (e.g. 17 and 67) and the leaves could be the hundreds (e.g. 5 and 8). Diff: 2 Keywords: stem, leaf Section: 2-2 Bar Charts, Pie Charts and Stem and Leaf Diagrams Outcome: 5

Business Statistics, 9e (Groebner/Shannon/Fry) Chapter 3 Describing Data Using Numerical Measures 1) If after graphing the data for a quantitative variable of interest, you notice that the distribution is highly skewed in the positive direction, the measure of central location that would likely provide the best assessment of the center would be the median. Answer: TRUE Diff: 2 Keywords: skew, median, center Section: 3-1 Measures of Center and Location Outcome: 1 2) A statistic is just another name for a parameter. Answer: FALSE Diff: 1 Keywords: statistic, parameter Section: 3-1 Measures of Center and Location Outcome: 1 3) The owner of a local gasoline station has kept track of the number of gallons of regular unleaded sold at his station every day since he purchased the station. This morning, he computed the mean number of gallons. This value would be considered a statistic. Answer: FALSE Diff: 2 Keywords: statistic, parameter Section: 3-1 Measures of Center and Location Outcome: 1 4) The Parks and Recreation manager for the city of Detroit recently submitted a report to the city council in which he indicated that a random sample of 500 park users indicated that the average number of visits per month was 4.56. This value should be viewed as a statistic by the city council. Answer: TRUE Diff: 2 Keywords: statistic, parameter, average Section: 3-1 Measures of Center and Location Outcome: 1 5) A statistic is a value that describes a population characteristic while a parameter is computed from a sample. Answer: FALSE Diff: 2 Keywords: statistic, parameter Section: 3-1 Measures of Center and Location Outcome: 1

6) The symbol μ is used to represent the sample mean. Answer: FALSE Diff: 1 Keywords: measure, spread, mean Section: 3-1 Measures of Center and Location Outcome: 1 7) A marketing manager for the Verizon cell phone company selected a random sample of 250 customers from the list of 10,556 total customers in a particular region. The mean monthly bill for the last six months based on the sample data is $65.70. The manager should realize that the mean bill for all 10,556 customers will actually be much higher than the sample mean of $65.70. Answer: FALSE Diff: 2 Keywords: sample, mean Section: 3-1 Measures of Center and Location Outcome: 1 8) You are given the following data: 23

If these data were considered to be a population and you computed the mean, you would get the same answer as if these data were considered to be a sample from a larger population. Answer: TRUE Diff: 2 Keywords: population, mean, sample Section: 3-1 Measures of Center and Location Outcome: 1 9) You are given the following data: 23

Assuming that the data reflect a sample from a larger population, the sample mean is 30.00. Answer: TRUE Diff: 1 Keywords: sample, mean, population Section: 3-1 Measures of Center and Location Outcome: 1

10) You are given the following data: 23

Assuming that the data reflect the population of interest, the mean of the population is 36.00. Answer: FALSE Diff: 1 Keywords: mean, population Section: 3-1 Measures of Center and Location Outcome: 1 11) Data are considered to be right-skewed when the mean lies to the right of the median. Answer: TRUE Diff: 1 Keywords: skewed, skew, mean, median Section: 3-1 Measures of Center and Location Outcome: 1 12) The sample mean is an estimate of μ and may be either higher or lower than μ depending on the sample. Answer: TRUE Diff: 2 Keywords: mean, sample, population Section: 3-1 Measures of Center and Location Outcome: 1 13) A lakeside neighborhood in Washington is home to a couple of successful baseball and football players in Seattle. If news articles usually report the median household income in that neighborhood rather than the mean income, this would be because income is a left-skewed distribution. Answer: FALSE Diff: 2 Keywords: mean, median, skewed Section: 3-1 Measures of Center and Location Outcome: 1 14) You are given the following data: 9

Assuming that these data reflect the population of interest, these data can be considered symmetric. Answer: FALSE Diff: 2 Keywords: mean, median, symmetric Section: 3-1 Measures of Center and Location Outcome: 1

15) You are given the following data: 23

Assuming that these data are a sample selected from a larger population, the median value for these sample data is 25.5. Answer: FALSE Diff: 2 Keywords: median, sample Section: 3-1 Measures of Center and Location Outcome: 1 16) A distribution is said to be symmetric when the sample mean and the population mean are equal. Answer: FALSE Diff: 2 Keywords: mean, sample, population, symmetric Section: 3-1 Measures of Center and Location Outcome: 1 17) In a recent study of the sales prices of houses in a Midwestern city, the mean sales price has been reported to be $167,811 while the median sales price was $155,600. From this information, you can determine that the data involved in the study are left-skewed. Answer: FALSE Diff: 2 Keywords: mean, median, skew, skewed Section: 3-1 Measures of Center and Location Outcome: 1 18) One of the primary advantages of using the median as a measure of the center for a set of data is that the median is not affected by extreme values in the data. Answer: TRUE Diff: 1 Keywords: median, center, extreme Section: 3-1 Measures of Center and Location Outcome: 1

19) Suppose a study of houses that have sold recently in your community showed the following frequency distribution for the number of bedrooms: Bedrooms 1 2 3 4 5

Frequency 1 18 140 57 11

Based on this information, the mode for the data is 140. Answer: FALSE Diff: 2 Keywords: mode, frequency Section: 3-1 Measures of Center and Location Outcome: 1 20) Suppose a study of houses that have sold recently in your community showed the following frequency distribution for the number of bedrooms: Bedrooms 1 2 3 4 5

Frequency 1 18 140 57 11

Based on this information the mean number of bedrooms in houses that sold is approximately 3.26. Answer: TRUE Diff: 2 Keywords: mean, weighted Section: 3-1 Measures of Center and Location Outcome: 1

21) Suppose a study of houses that have sold recently in your community showed the following frequency distribution for the number of bedrooms: Bedrooms 1 2 3 4 5

Frequency 1 18 140 57 11

Based on this information, the median number of bedrooms in houses sold is 3.20. Answer: FALSE Diff: 2 Keywords: median Section: 3-1 Measures of Center and Location Outcome: 1 22) Suppose a study of houses that have sold recently in your community showed the following frequency distribution for the number of bedrooms: Bedrooms 1 2 3 4 5

Frequency 1 18 140 57 11

Based on this information, it is possible to determine that the distribution of bedrooms in homes sold is right-skewed. Answer: TRUE Diff: 3 Keywords: mean, median, skew, skewed Section: 3-1 Measures of Center and Location Outcome: 1 23) A data set in which the mean, median, and mode are all equal is said to be a skewed distribution. Answer: FALSE Diff: 1 Keywords: mean, median, mode, symmetric, skewed Section: 3-1 Measures of Center and Location Outcome: 1

24) First Pacific Bank has determined that the mean checking account balance for all its customers is currently $743.50. Based on this, it is fair to say that about half the customers have balances exceeding $743.50. Answer: FALSE Diff: 2 Keywords: mean, median Section: 3-1 Measures of Center and Location Outcome: 1 25) When analyzing annual incomes of adults in a market area, the marketing manager's report indicated that the 90th percentile is $123,400. That means that 90 percent of the adult incomes in the market area fall at or below $123,400. Answer: TRUE Diff: 1 Keywords: percentile Section: 3-1 Measures of Center and Location Outcome: 1 26) When the median of a data set is 110 and the mean is 127, the percentile associated with the mean must be higher than 50 percent. Answer: TRUE Diff: 2 Keywords: mean, median, percentile Section: 3-1 Measures of Center and Location Outcome: 1 27) The second quartile for a set of data will have the same value as the 50th percentile only when the data are symmetric. Answer: FALSE Diff: 2 Keywords: quartile, percentile, symmetric Section: 3-1 Measures of Center and Location Outcome: 1 28) If a set of data has 1,500 values, the 30th percentile value will correspond to the 450th value in the data when the data have been arranged in numerical order. Answer: TRUE Diff: 3 Keywords: percentile, location Section: 3-1 Measures of Center and Location Outcome: 1

29) If a set of data has 540 values, the 3rd quartile corresponds to approximately the 135th value when the data have been arranged in numerical order. Answer: FALSE Diff: 3 Keywords: quartile, percentile, location Section: 3-1 Measures of Center and Location Outcome: 1 30) A set of data is considered to be symmetric if the 3rd quartile is three times larger than the 1st quartile. Answer: FALSE Diff: 2 Keywords: quartile, percentile, symmetric Section: 3-1 Measures of Center and Location Outcome: 1 31) For a symmetric distribution, if the median value is 150 and the second quartile is 40, the fourth quartile must also be 40. Answer: FALSE Diff: 2 Keywords: mean, median, quartile Section: 3-1 Measures of Center and Location Outcome: 1 32) Recently an article in a newspaper stated that 75 percent of the households in the state had incomes of $20,200 or below. Given this input, it is certain that mean household income is less than $20,200. Answer: FALSE Diff: 3 Keywords: mean, percentile Section: 3-1 Measures of Center and Location Outcome: 1 33) It is possible for a set of data to have multiple modes as well as multiple medians, but there can be only one mean. Answer: FALSE Diff: 2 Keywords: mean, median, mode Section: 3-1 Measures of Center and Location Outcome: 1 34) A box and whisker plot shows where the mean value falls relative to the median for a variable. Answer: FALSE Diff: 1 Keywords: box, whisker, mean, median Section: 3-1 Measures of Center and Location Outcome: 2

35) The right and left edges of the box in a box and whisker plot represent the 3rd and 1st quartiles, respectively. Answer: TRUE Diff: 1 Keywords: quartile, box, whisker, edge Section: 3-1 Measures of Center and Location Outcome: 2 36) A recent study involving a sample of 3,000 vehicles in California showed the following statistics related to the number of miles driven per day: Q1 = 12, Q2 = 45, and Q3 = 56. Based on these data, we know that the distribution is skewed. Answer: TRUE Diff: 3 Keywords: quartile, skew, skewed, median Section: 3-1 Measures of Center and Location Outcome: 1 37) A recent study involving a sample of 3,000 vehicles in California showed the following statistics related to the number of miles driven per day: Q1 = 12, Q2 = 45, and Q3 = 56. Based on these data, if a box and whisker plot is developed, a value of 110 is an outlier. Answer: FALSE Diff: 3 Keywords: box, whisker, limit Section: 3-1 Measures of Center and Location Outcome: 2 38) A recent study involving a sample of 3,000 vehicles in California showed the following statistics related to the number of miles driven per day: Q1 = 12, Q2 = 45, and Q3 = 56. Based on these data, if a box and whisker plot is developed, the upper limit value is 122 miles. Answer: TRUE Diff: 3 Keywords: box, whisker, limit Section: 3-1 Measures of Center and Location Outcome: 2 39) In drawing a box and whisker plot the upper limit length of the whiskers is 1.5(Q3-Q1). Answer: TRUE Diff: 2 Keywords: box, whisker, limit Section: 3-1 Measures of Center and Location Outcome: 2

40) When surveyed, a sample of 1,250 patients at a regional hospital provided interviewers with the following summary statistics pertaining to the hospital charges: Minimum = $278.00 Q1 = $1,245 Q2 = $3,567 Q3= $4,702. Based on these data, if you were to construct a box and whisker plot, the value corresponding to the right-hand edge of the box would be $4,702. Answer: TRUE Diff: 2 Keywords: box, whisker, edge Section: 3-1 Measures of Center and Location Outcome: 2 41) When surveyed, a sample of 1,250 patients at a regional hospital provided interviewers with the following summary statistics pertaining to the hospital charges: Minimum = $278.00 Q1 = $1,245 Q2 = $3,567 Q3= $4,702. Based on these data, if you were to construct a box and whisker plot, the value $278 would be considered an outlier. Answer: FALSE Diff: 3 Keywords: box, whisker, outlier Section: 3-1 Measures of Center and Location Outcome: 2 42) When surveyed, a sample of 1,250 patients at a regional hospital provided interviewers with the following summary statistics pertaining to the hospital charges: Minimum = $278.00 Q1 = $1,245 Q2 = $3,567 Q3= $4,702. Based on these data, the distribution is seen to be symmetric. Answer: FALSE Diff: 2 Keywords: median, quartile, symmetric Section: 3-1 Measures of Center and Location Outcome: 1 43) A dairy farm in Wisconsin bottles milk in one gallon containers. At a recent meeting, the production manager asked top management for a new filling machine that he argued would assure that all containers had exactly one gallon of milk. Based on sound statistical principles, the top management group should conclude that the production manager could have merit to his argument. Answer: FALSE Diff: 2 Keywords: variation Section: 3-2 Measures of Variation Outcome: 3 44) The range is an ideal measure of variation since it is not sensitive to extreme values in the data. Answer: FALSE Diff: 1 Keywords: range, variation, sensitive Section: 3-2 Measures of Variation Outcome: 3 3-10 Copyright © 2018 Pearson Education, Inc.

45) When a variance is calculated for a data set, the resulting value is the same regardless of whether the data set is treated as a population or a sample. Answer: FALSE Diff: 1 Keywords: variance, population, sample Section: 3-2 Measures of Variation Outcome: 3 46) The Good-Guys Car Dealership has tracked the number of used cars sold at its downtown dealership. Consider the following data as representing the population of cars sold in each of the 8 weeks that the dealership has been open. 3

The population range is 9. Answer: TRUE Diff: 1 Keywords: range, population Section: 3-2 Measures of Variation Outcome: 3 47) The Good-Guys Car Dealership has tracked the number of used cars sold at its downtown dealership. Consider the following data as representing the population of cars sold in each of the 8 weeks that the dealership has been open. 3

The population variance is approximately 9.43. Answer: FALSE Diff: 2 Keywords: population, variance Section: 3-2 Measures of Variation Outcome: 3 48) The Good-Guys Car Dealership has tracked the number of used cars sold at its downtown dealership. Consider the following data as representing the population of cars sold in each of the 8 weeks that the dealership has been open. 3

The population standard deviation is approximately 2.87 cars. Answer: TRUE Diff: 2 Keywords: population, standard, deviation Section: 3-2 Measures of Variation Outcome: 3

49) One of the reasons that the standard deviation is preferred as a measure of variation over the variance is that the standard deviation is measured in the original units. Answer: TRUE Diff: 1 Keywords: standard, deviation, variation, units Section: 3-2 Measures of Variation Outcome: 3 50) The interquartile range is the difference between the mean and the median. Answer: FALSE Diff: 1 Keywords: interquartile range, median, mean Section: 3-2 Measures of Variation Outcome: 3 51) A store manager tracks the number of customer complaints each week. The following data reflect a random sample of ten weeks. 11

The range for these data is 8. Answer: FALSE Diff: 1 Keywords: range, variation Section: 3-2 Measures of Variation Outcome: 3 52) A store manager tracks the number of customer complaints each week. The following data reflect a random sample of ten weeks. 11

The variance for these data is approximately 27.78. Answer: TRUE Diff: 2 Keywords: sample, variance Section: 3-2 Measures of Variation Outcome: 3

53) A store manager tracks the number of customer complaints each week. The following data reflect a random sample of ten weeks. 11

The standard deviation for these data is approximately 27.78. Answer: FALSE Diff: 2 Keywords: standard, deviation, sample Section: 3-2 Measures of Variation Outcome: 3 54) The interquartile range contains the middle 50 percent of a data set. Answer: TRUE Diff: 2 Keywords: interquartile range Section: 3-2 Measures of Variation Outcome: 3 55) For a given set of data, if the data are treated as a population, the calculated standard deviation will be less than it would be had the data been treated as a sample. Answer: TRUE Diff: 2 Keywords: sample, standard, deviation, population Section: 3-2 Measures of Variation Outcome: 3 56) If a population standard deviation is computed to be 345, it will almost always be the case that a standard deviation computed from a random sample from that population will be larger than 345. Answer: TRUE Diff: 2 Keywords: population, sample, standard, deviation Section: 3-2 Measures of Variation Outcome: 3 57) The advantage of using the interquartile range as a measure of variation is that it utilizes all the data in its computation. Answer: FALSE Diff: 2 Keywords: variation, interquartile, range Section: 3-2 Measures of Variation Outcome: 3

58) Suppose the standard deviation for a given sample is known to be 20. If the data in the sample are doubled, the standard deviation will be significantly greater. Answer: FALSE Diff: 3 Keywords: standard, deviation Section: 3-2 Measures of Variation Outcome: 3 59) Populations with larger means will also have larger standard deviations since the data will be more spread out for populations with larger means. Answer: FALSE Diff: 2 Keywords: mean, standard deviation Section: 3-3 Using the Mean and Standard Deviation Together Outcome: 4 60) In comparing two distributions with the same mean, the coefficient of variation is the only way to assess which distribution has the greatest relative variability. Answer: FALSE Diff: 1 Keywords: mean, coefficient, variation Section: 3-3 Using the Mean and Standard Deviation Together Outcome: 4 61) Consider a situation involving two populations where population 1 is known to have a higher coefficient of variation than population 2. In this situation, we know that population 1 has a higher standard deviation than population 2. Answer: FALSE Diff: 2 Keywords: coefficient, variation, standard deviation Section: 3-3 Using the Mean and Standard Deviation Together Outcome: 4 62) Acme Taxi has two taxi cabs. The manager tracks the daily revenue for each cab. Over the past 20 days, Cab A has averaged $76.00 per night with a standard deviation equal to $11.00. Cab B has averaged $200.00 per night with a standard deviation of $18.00. Based on this information, Cab B has the greatest relative variation. Answer: FALSE Diff: 3 Keywords: standard deviation, relative variation, coefficient of variation Section: 3-3 Using the Mean and Standard Deviation Together Outcome: 4

63) Acme Taxi has two taxi cabs. The manager tracks the daily revenue for each cab. Over the past 20 days, Cab A has averaged $76.00 per night with a standard deviation equal to $11.00. Cab B has averaged $200.00 per night with a standard deviation of $18.00. Based on this information, the coefficient of variation for Cab B is 9 percent. Answer: TRUE Diff: 2 Keywords: standard deviation, relative variation, coefficient of variation Section: 3-3 Using the Mean and Standard Deviation Together Outcome: 4 64) Based on the empirical rule we can assume that all bell-shaped distributions have approximately 95 percent of the values within ± 2 standard deviations of the mean. Answer: TRUE Diff: 1 Keywords: empirical rule Section: 3-3 Using the Mean and Standard Deviation Together Outcome: 5 65) Suppose a distribution has a mean of 80 and standard deviation of 10. It is found that 84 percent of the values in the data set lie between 70 and 90. This implies that the distribution is not bell-shaped. Answer: TRUE Diff: 2 Keywords: mean, standard deviation, empirical rule Section: 3-3 Using the Mean and Standard Deviation Together Outcome: 5 66) The credit card balances for customers at State Bank and Trust has a mean equal to $800 and a standard deviation equal to $60.00. Kevin Smith's balance is $1,352. Based on this, his standardized value is 9.20. Answer: TRUE Diff: 2 Keywords: mean, standard deviation, z score, standardized value Section: 3-3 Using the Mean and Standard Deviation Together Outcome: 4 67) Based on the empirical rule we can expect about 95 percent of the values in bell-shaped distributions to be within ± one standard deviation of the mean. Answer: FALSE Diff: 1 Keywords: mean, standard deviation, empirical rule Section: 3-3 Using the Mean and Standard Deviation Together Outcome: 5

68) A major automobile maker has two models of sedans. The first model has been shown to get an average of 27 mpg on the highway with a standard deviation equal to 5 mpg. The second model gets 33 mpg on average with a standard deviation of 8 mpg. Based on this information the first car model is relatively more variable than the second car model. Answer: FALSE Diff: 2 Keywords: mean, standard deviation, coefficient of variation, relative variation Section: 3-3 Using the Mean and Standard Deviation Together Outcome: 4 69) The distribution of bankcard balances for customers is highly right-skewed with a mean of $1,100 and a standard deviation equal to $250. Based on this information, approximately 68 percent of the customers will have bank balances between $850 and $1,350. Answer: FALSE Diff: 2 Keywords: skewed, mean, standard deviation, Tchebysheff Section: 3-3 Using the Mean and Standard Deviation Together Outcome: 5 70) The distribution of dollars paid for car insurance by car owners in a major east coast city is bellshaped with a mean equal to $750 every six months and a standard deviation equal to $100. Based on this information we should use Tchebysheff's theorem to determine the conservative percentage of car owners that will pay between $550 and $950 for car insurance. Answer: FALSE Diff: 2 Keywords: mean, standard deviation, Tchebysheff Section: 3-3 Using the Mean and Standard Deviation Together Outcome: 5 71) A population measure, such as the population mean, is called a: A) statistic. B) parameter. C) prime number. D) sample value. Answer: B Diff: 1 Keywords: population, parameter Section: 3-1 Measures of Center and Location Outcome: 1

72) If a business manager selected a sample of customers and computed the mean income for this sample of customers, she has computed: A) a statistic. B) an ordinal value. C) a nominal value. D) a parameter. Answer: A Diff: 1 Keywords: sample, statistic Section: 3-1 Measures of Center and Location Outcome: 1 73) Which of the following statements is true? A) The mean of a population will always be larger than the population standard deviation. B) The mean of the population will generally be larger than the mean of the sample selected from that population. C) The population mean and a sample mean for a sample selected from that population will usually be different values. D) The population mean and sample mean will always be identical. Answer: C Diff: 2 Keywords: population, sample, mean Section: 3-1 Measures of Center and Location Outcome: 1 74) The most frequently used measure of central tendency is: A) median. B) mean. C) mode. D) middle value. Answer: B Diff: 1 Keywords: central tendency, mean Section: 3-1 Measures of Center and Location Outcome: 1

75) Consider the following sample data: 25

For these data the sample mean is: A) 8 B) 10 C) 3 D) 12 Answer: B Diff: 1 Keywords: sample, mean Section: 3-1 Measures of Center and Location Outcome: 1 76) Consider the following sample data: 25

For these data the median is: A) 7.5 B) 3.5 C) 10 D) None of the above Answer: A Diff: 2 Keywords: sample, median Section: 3-1 Measures of Center and Location Outcome: 1 77) A small company has 7 employees. The numbers of years these employees have worked for this company are shown as follows: 4

Based upon this information, the mean number of years that employees have been with this company is: A) 16 B) C) 8.40 D) 10 Answer: D Diff: 1 Keywords: population, mean Section: 3-1 Measures of Center and Location Outcome: 1

78) A small company has 7 employees. The numbers of years these employees have worked for this company are shown as follows: 4

Based upon this information, the median number of years that employees have been with this company is: A) 9 years. B) 16 years. C) 10 years. D) 14 years. Answer: A Diff: 2 Keywords: population, median Section: 3-1 Measures of Center and Location Outcome: 1 79) A small company has 7 employees. The numbers of years these employees have worked for this company are shown as follows: 4

Based upon this information, the mode number of years that employees have been with this company is: A) 16 B) 2 C) 9 D) 10 Answer: A Diff: 1 Keywords: population, mode Section: 3-1 Measures of Center and Location Outcome: 1 80) A sample of people who have attended a college football game at your university has a mean = 3.2 members in their family. The mode number of family members is 2 and the median number is 2.0. Based on this information: A) the population mean exceeds 3.2. B) the distribution is bell-shaped. C) the distribution is right-skewed. D) the distribution is left-skewed. Answer: C Diff: 2 Keywords: skewed, mean, median Section: 3-1 Measures of Center and Location Outcome: 1

81) A major retail store has studied customer behavior and found that the distribution of time customers spend in a store per visit is symmetric with a mean equal to 17.3 minutes. Based on this information, which of the following is true? A) The distribution is right-skewed. B) The median is to the right of the mean. C) The median is approximately 17.3 minutes. D) The median is to the left of the mean. Answer: C Diff: 1 Keywords: symmetric, mean, median Section: 3-1 Measures of Center and Location Outcome: 1 82) A large retail company gives an employment screening test to all prospective employees. Franklin Gilman recently took the test and it was reported back to him that his score placed him at the 80th percentile. Therefore: A) 80 people who took the test scored below Franklin. B) Franklin scored as high or higher than 80 percent of the people who took the test. C) Franklin was in the bottom 20 percent of those that have taken the test. D) Franklin's score has a z-score of 80. Answer: B Diff: 2 Keywords: percentile Section: 3-1 Measures of Center and Location Outcome: 1 83) A large retail company gives an employment screening test to all prospective employees. If a prospective employee receives a report saying that she scored at the 40th percentile: A) she scored above the median. B) she scored better than 40 percent of people who took the test. C) she scored in the top 40 percent of people who took the test. D) her z-score was a 40. Answer: B Diff: 2 Keywords: percentile, median Section: 3-1 Measures of Center and Location Outcome: 1

84) If a data set has 740 values that have been sorted from low to high, which value in the data set will be the 20th percentile? A) The average of the 148th and 149th values B) The 20th value C) The 148th value D) None of the above Answer: A Diff: 2 Keywords: percentile, location Section: 3-1 Measures of Center and Location Outcome: 1 85) If a data set has 1,133 sorted values, what value corresponds to the 3rd quartile? A) The 250th value B) The 850th value C) The 760th value D) The 849th value Answer: B Diff: 2 Keywords: percentile, quartile, value Section: 3-1 Measures of Center and Location Outcome: 1 86) At a sawmill in Oregon, a process improvement team measured the diameters for a sample of 1,500 logs. The following summary statistics were computed:

Given this information, the boundaries on the box in a box and whisker plot are: A) 8.9 in and 15.6 in. B) 13.5 in ± 1.5 (Q3-Q1). C) 14.2 in ± 1.5 (Q3-Q1). D) 8.9 in and 14.2 in. Answer: A Diff: 2 Keywords: box, whisker, plot, boundary Section: 3-1 Measures of Center and Location Outcome: 2

87) At a sawmill in Oregon, a process improvement team measured the diameters for a sample of 1,500 logs. The following summary statistics were computed:

Given this information, in a box and whisker plot, which of these four values will NOT appear? A) 8.9 in. B) 13.5 in. C) 15.6 in. D) 14.2 in. Answer: D Diff: 2 Keywords: box, whisker, plot Section: 3-1 Measures of Center and Location Outcome: 2 88) At a sawmill in Oregon, a process improvement team measured the diameters for a sample of 1,500 logs. The following summary statistics were computed:

Given this information, which of the following statements is correct? A) The distribution of log diameters is symmetric. B) A log that is over 20 inches in diameter can be considered an outlier. C) The distribution of log diameters is right-skewed. D) The distribution is left-skewed. Answer: C Diff: 2 Keywords: skewed, mean, median, quartile Section: 3-1 Measures of Center and Location Outcome: 1 89) At a sawmill in Oregon, a process improvement team measured the diameters for a sample of 1,500 logs. The following summary statistics were computed:

Given this information, for a box and whisker plot which of the following statements is appropriate? A) Seventy-five percent of the trees in the sample have values between 8.9 in. and 15.6 in. B) Virtually all of the data should fall between 0 in. and 25.65 in. C) No tree will have a diameter of more than 22.3 in. D) Fifty percent of the trees will have diameters between 13.5 in. and 15.6 in. Answer: B Diff: 3 Keywords: box, whisker, outlier Section: 3-1 Measures of Center and Location Outcome: 2 3-22 Copyright © 2018 Pearson Education, Inc.

90) If a distribution for a quantitative variable is thought to be nearly symmetric with very little variation, and a box and whisker plot is created for this distribution, which of the following is true? A) The box will be quite wide but the whisker will be very short. B) The left and right-hand edges of the box will be approximately equal distance from the median. C) The whiskers should be about half as long as the box is wide. D) The upper whisker will be much longer than the lower whisker. Answer: B Diff: 2 Keywords: box, whisker, symmetric, median Section: 3-1 Measures of Center and Location Outcome: 2 91) The box and whisker plot CANNOT be used to identify: A) skewedness. B) centerness. C) outliers. D) symmetry. Answer: B Diff: 3 Keywords: box, whisker, outlier Section: 3-1 Measures of Center and Location Outcome: 2 92) For ordinal data, ________ is the preferred measure of central location. A) the mean B) the median C) the percentile D) the quartile Answer: B Diff: 3 Keywords: mean, median, percentile, quartile Section: 3-1 Measures of Center and Location Outcome: 1 93) Which of the following is the most frequently used measure of variation? A) The range B) The standard deviation C) The variance D) The mode Answer: B Diff: 1 Keywords: variation, standard deviation Section: 3-2 Measures of Variation Outcome: 3

94) Which of the following measures is not affected by extreme values in the data? A) The mean B) The median C) The range D) The standard deviation Answer: B Diff: 2 Keywords: extreme, insensitive, median Section: 3-2 Measures of Variation Outcome: 3 95) The following data reflect the number of customers who test drove new cars each day for a sample of 20 days at the Redfield Ford Dealership. 5 9 5 6

7 7 6 3

2 10 4 4

9 4 0 14

4 7 7 6

Given these data, what is the range? A) 14 B) 1 C) Approximately 3.08 D) 5.95 Answer: A Diff: 1 Keywords: range, sample Section: 3-2 Measures of Variation Outcome: 3 96) The following data reflect the number of customers who test drove new cars each day for a sample of 20 days at the Redfield Ford Dealership. 5 9 5 6

7 7 6 3

2 10 4 4

9 4 0 14

4 7 7 6

Given these data, what is the variance? A) 0.69 B) Approximately 3.08 C) Approximately 9.52 D) Approximately 181 Answer: C Diff: 2 Keywords: variance, sample Section: 3-2 Measures of Variation Outcome: 3 3-24 Copyright © 2018 Pearson Education, Inc.

97) The following data reflect the number of customers who test drove new cars each day for a sample of 20 days at the Redfield Ford Dealership. 5 9 5 6

7 7 6 3

2 10 4 4

9 4 0 14

4 7 7 6

Given these data, what is the interquartile range? A) 3 B) 7 C) 4 D) 14 Answer: A Diff: 3 Keywords: interquartile range, sample Section: 3-2 Measures of Variation Outcome: 3 98) The advantage of using the interquartile range versus the range as a measure of variation is: A) it is easier to compute. B) it utilizes all the data in its computation. C) it gives a value that is closer to the true variation. D) it is less affected by extremes in the data. Answer: D Diff: 1 Keywords: range, variation, interquartile range Section: 3-2 Measures of Variation Outcome: 3 99) The following data reflect the number of customers who return merchandise for a refund on Monday. Note these data reflect the population of all 10 Mondays for which data are available. 40 46

12 13

17 22

25 16

9 7

Based on these data, what is the standard deviation? A) 13.03 B) 12.36 C) 39 D) 152.8 Answer: B Diff: 2 Keywords: population, standard deviation Section: 3-2 Measures of Variation Outcome: 3 3-25 Copyright © 2018 Pearson Education, Inc.

100) The following data reflect the number of customers who return merchandise for a refund on Monday. Note these data reflect the population of all 10 Mondays for which data are available. 40 46

12 13

17 22

25 16

9 7

Assume that this same exact pattern of data were replicated for the next ten days. How would this affect the standard deviation for the new population with 20 items? A) The standard deviation would be doubled. B) The standard deviation would be cut in half. C) The standard deviation would not be changed. D) There is no way of knowing the exact impact without knowing how the mean is changed. Answer: C Diff: 3 Keywords: standard deviation, population Section: 3-2 Measures of Variation Outcome: 3 101) In order to compute the mean and standard deviation, the level of data measurement should be: A) ratio or interval. B) qualitative. C) nominal. D) ordinal. Answer: A Diff: 1 Keywords: mean, standard deviation, ratio, interval Section: 3-2 Measures of Variation Outcome: 3 102) Consider the following data, which represent the number of miles that employees commute from home to work each day. There are two samples: one for males and one for females. Males: 13 5

Females: 15 6

Which of the following statements is true? A) The female distribution is more variable since the range for the females is greater than for the males. B) Females in the sample commute farther on average than do males. C) The males in the sample commute farther on average than the females. D) Males and females on average commute the same distance. Answer: C Diff: 2 Keywords: range, mean Section: 3-2 Measures of Variation Outcome: 3 3-26 Copyright © 2018 Pearson Education, Inc.

103) Consider the following data, which represent the number of miles that employees commute from home to work each day. There are two samples: one for males and one for females. Males: 13 5

Females: 15 6

The coefficient of variation of commute miles for the males is: A) approximately 76 percent. B) about 7.8. C) approximately 61.5. D) about 67 percent. Answer: A Diff: 2 Keywords: coefficient of variation, mean, standard deviation Section: 3-3 Using the Mean and Standard Deviation Together Outcome: 4 104) Consider the following data, which represent the number of miles that employees commute from home to work each day. There are two samples: one for males and one for females. Males: 13 5

Females: 15 6

Which of the following statements is true? A) Females have the larger mean. B) The coefficient of variation is larger for females than for males. C) The coefficient of variation is larger for males than for females. D) Females have the larger range. Answer: B Diff: 2 Keywords: sample, coefficient of variation, relative variability Section: 3-3 Using the Mean and Standard Deviation Together Outcome: 4

105) If the age distribution of customers at a major retail chain is thought to be bell-shaped with a mean equal to 43 years and a standard deviation equal to 7 years, the percentage of customers between the ages of 29 and 57 years is: A) approximately 81.5. B) approximately 68. C) at least 75. D) approximately 95. Answer: D Diff: 1 Keywords: empirical rule, distribution Section: 3-3 Using the Mean and Standard Deviation Together Outcome: 5 106) Under what circumstances is it necessary to use the coefficient of variation to compare relative variability between two or more distributions? A) When the means of the distributions are equal B) When the means of the distributions are not equal C) When the standard deviations of the distributions are not equal D) When the standard deviations of the distributions are equal Answer: B Diff: 2 Keywords: coefficient of variation, mean, standard deviation Section: 3-3 Using the Mean and Standard Deviation Together Outcome: 4 107) In the annual report, a major food chain stated that the distribution of daily sales at its Detroit stores is known to be bell-shaped, and that 95 percent of all daily sales fell between $19,200 and $36,400. Based on this information, what were the mean sales? A) Around $20,000 B) Close to $30,000 C) Approximately $27,800 D) Can't be determined without more information. Answer: C Diff: 2 Keywords: empirical rule, mean, standard deviation Section: 3-3 Using the Mean and Standard Deviation Together Outcome: 5

108) The number of days that homes stay on the market before they sell in Seattle is bell-shaped with a mean equal to 8 days. Further, 95 percent of all homes are on the market between 2 and 14 days. Based on this information, what is the standard deviation for the number of days that houses stay on the market in Seattle? A) 12 B) 3 C) 9 D) 3 Answer: D Diff: 2 Keywords: empirical rule, mean, standard deviation Section: 3-3 Using the Mean and Standard Deviation Together Outcome: 5 109) Incomes in a particular market area are known to be right-skewed with a mean equal to $33,100. In a report issued recently, a manager stated that at least 89 percent of all incomes are in the range of $26,700 to $39,500, and this was based on Tchebysheff's theorem. Given these facts, what is the standard deviation for the incomes in this market area? A) Approximately $6,400 B) Approximately $3,200 C) Approximately $2,133 D) Approximately $4266 Answer: C Diff: 3 Keywords: standard deviation, mean, Tchebysheff Section: 3-3 Using the Mean and Standard Deviation Together Outcome: 5 110) A distribution has a coefficient of variation of 65 percent and mean of 74. What is the value of the standard deviation? A) 0.65 B) 4810 C) 113.8 D) 48.1 Answer: D Diff: 2 Keywords: mean, standard deviation, coefficient of variation Section: 3-3 Using the Mean and Standard Deviation Together Outcome: 4

111) The asking price for homes on the real estate market in Baltimore has a mean value of $286,455 and a standard deviation of $11,200. Four homes are listed by one real estate company with the following prices: Home 1: Home 2: Home 3: Home 4:

$456,900 $306,000 $266,910 $201,456

Based upon this information, which house has a standardized value that is relatively closest to zero? A) Home 1 B) Home 2 C) Home 3 D) Home 2 and home 3 Answer: D Diff: 2 Keywords: mean standard deviation, standardized value, z score Section: 3-3 Using the Mean and Standard Deviation Together Outcome: 4 112) The asking price for homes on the real estate market in Baltimore has a mean value of $286,455 and a standard deviation of $11,200. The mean and standard deviation in asking price for homes in Denver are $188,468 and $8,230, respectively. Recently, one home sold in each city where the asking price for each home was $193,000. Based on these data, which of the following conclusions can be made? A) The two homes have approximately the same standardized values. B) The distribution of asking prices in the two cities is bell-shaped. C) The house in Baltimore is relatively farther from the mean than the house in Denver. D) The asking prices of homes in Denver is less variable than those in Baltimore. Answer: C Diff: 2 Keywords: mean, standard deviation, standardized value, z score Section: 3-3 Using the Mean and Standard Deviation Together Outcome: 4 113) The asking price for homes on the real estate market in Baltimore has a mean value of $286,455 and a standard deviation of $11,200. The mean and standard deviation in asking price for homes in Denver are $188,468 and $8,230, respectively. Recently, one home sold in each city where the asking price for each home was $193,000. Assuming that both distributions are bell-shaped, which of the following statements is true? A) The Baltimore home has the higher standard z-value. B) The coefficient of variation for Denver is less than for Baltimore. C) The Denver home has a higher standard z-value. D) Both cities have the same coefficient of variation. Answer: C Diff: 2 Keywords: mean, standard deviation, standardized value, z score Section: 3-3 Using the Mean and Standard Deviation Together Outcome: 4 3-30 Copyright © 2018 Pearson Education, Inc.

114) A report on spending by adults on recreation stated the following: At least 75 percent of the people in the survey spend between $750 and $1,250 per year. The report also said that at least 88 percent spend between $625 and $1,375 per year. Given this information, which of the following is most apt to be true? A) The standard deviation is approximately $125. B) The distribution of spending on recreation can be assumed to be bell-shaped. C) The standard deviation is approximately $187.5. D) The standard deviation is approximately $250. Answer: A Diff: 3 Keywords: standard deviation, Tchebysheff Section: 3-3 Using the Mean and Standard Deviation Together Outcome: 5 115) The distribution of the actual weight of potato chips in a 16 ounce sack is thought to be bell-shaped with a mean equal to 16 ounces and a standard deviation equal to 0.45 ounces. Based on this, between what two limits could we expect 95 percent of all sacks to weigh? A) 14 to 18 ounces B) 15.10 to 16.90 ounces C) 15.55 to 16.45 ounces D) 14.65 to 17.35 ounces Answer: B Diff: 2 Keywords: standard deviation, empirical rule Section: 3-3 Using the Mean and Standard Deviation Together Outcome: 5 116) A recent study in the restaurant business determined that the mean tips for male waiters per hour of work are $6.78 with a standard deviation of $2.11. The mean tips per hour for female waiters are $7.86 with a standard deviation of $2.20. Based on this information, which of the following statements do we know to be true? A) The distribution of tips for both males and females is right-skewed. B) The variation in tips received by females is more variable than males. C) The median tips for females exceeds that of males. D) On a relative basis, males have more variation in tips per hour than do females. Answer: D Diff: 2 Keywords: mean, standard deviation, coefficient of variation Section: 3-3 Using the Mean and Standard Deviation Together Outcome: 4

117) Data was collected on the number of television sets in a household, and it was found that the mean was 3.5 and the standard deviation was 0.75. Based on these sample data, what is the standardized value corresponding to 5 televisions? A) -2.00 B) 1.5 C) 2.00 D) 1.125 Answer: C Diff: 2 Keywords: mean, standard deviation, sample, z score, standardized value Section: 3-3 Using the Mean and Standard Deviation Together Outcome: 4 118) Portfolio A of a collection of stocks is considered more risky than portfolio B if: A) portfolio A has a higher mean than portfolio B. B) portfolio A has a higher variance than portfolio B. C) portfolio A has a higher standard deviation. D) portfolio A has a higher coefficient of variation than portfolio B. Answer: D Diff: 3 Keywords: mean, standard deviation, coefficient of variation Section: 3-3 Using the Mean and Standard Deviation Together Outcome: 4 119) When comparing data measured by substantially different scales, we must use: A) standardized data values. B) standardized data scales. C) standardized data variations. D) standardized data scores. Answer: A Diff: 2 Keywords: z-score, standardized value Section: 3-3 Using the Mean and Standard Deviation Together Outcome: 4 120) The Empirical Rule states that for a bell-shaped distribution, approximately 95 percent of data should lie within: A) one standard deviation from either side of the mean. B) two standard deviations from either side of the mean. C) three standard deviations from either side of the mean. D) four standard deviations from either side of the mean. Answer: B Diff: 2 Keywords: mean, standard deviation, bell-shaped, empirical rule Section: 3-3 Using the Mean and Standard Deviation Together Outcome: 5

121) A professor wishes to develop a numerical method for giving grades. He intends to base the grade on homework, two midterms, a project, and a final examination. He wishes the final exam to have the largest influence on the grade. He wants the project to have 10%, each midterm to have 20%, and the homework to have 10% of the influence of the semester grade. Determine the weights the professor should use to produce a weighted average for grading purposes. A) Instrument Final Project Midterm 1 Midterm 2 Homework Weight 40 10 20 20 10 B) Instrument Final Weight 50

Project 10

Midterm 1 Midterm 2 Homework 20 20 10

C) Instrument Final Weight 40

Project 15

Midterm 1 Midterm 2 Homework 15 15 15

D) Instrument Final Weight 30

Project 10

Midterm 1 Midterm 2 Homework 20 20 10

Answer: A Diff: 3 Keywords: measure of variation Section: 3-2 Measures of Variation Outcome: 1 122) A professor wishes to develop a numerical method for giving grades. He intends to base the grade on homework, two midterms, a project, and a final examination. He wishes the final exam to have the largest influence on the grade. He wants the project to have 10%, each midterm to have 20%, and the homework to have 10% of the influence of the semester grade. For a student with the following grades during the quarter, calculate a weighted average for the course: Instrument Percentage Grade

Final 64

Project 98

Midterm 1 Modterm 2 Homework 67 63 89

A) 68.50 B) 73.30 C) 68.30 D) 70.30 Answer: D Diff: 1 Keywords: measure of center and location Section: 3-1 Measures of Center and Location Outcome: 1

123) Todd Lindsey & Associates, a commercial real estate company located in Boston, owns six office buildings in the Boston area that it leases to businesses. The lease price per square foot differs by building due to location and building amenities. Currently, all six buildings are fully leased at the prices shown here. Price per Number of Square Foot Square Feet Building 1 $ 75 125,000 Building 2 $ 85 37,500 Building 3 $ 90 77,500 Building 4 $ 45 35,000 Building 5 $ 55 60,000 Building 6 $110 130,000 Compute the weighted average (mean) price per square foot for these buildings. A) 86.25 B) 83.25 C) 80.15 D) 86.15 Answer: B Diff: 1 Keywords: measure of center and location Section: 3-1 Measures of Center and Location Outcome: 1

124) Each year, Business Week publishes information and rankings of master of business administration (MBA) programs. The data file MBA Analysis contains data on several variables for eight reputable MBA programs as presented in the October 2, 2000, issue of Business Week. The variables include pre- and postMBA salary, percentage salary increase, undergraduate GPA, average Graduate Management Admission Test (GMAT) score, annual tuition, and expected annual student cost. Compute the mean and median for each of the variables in the database. A) Mean Median Pre-MBA Salary 39077 43337.63 Post-MBA Salary 82203 98902 Percentage Increase in Salary 116 123 Undergraduate GPA 3.46 3.40 GMAT Score 635.00 631.13 Annual Tuition 13163.50 15967.50 Expected Annual Student Cost 23169.5 27980.75 B) Pre-MBA Salary Post-MBA Salary Percentage Increase in Salary Undergraduate GPA GMAT Score Annual Tuition Expected Annual Student Cost

Mean 46667.63 98902 113 3.40 661.16 15967.50 28980.75

Median 39077 84403 116 3.56 635.00 19163.70 23179.5

Mean 43337.63 98902 123 3.40 631.13 15967.50 27980.75

Median 39077 82203 116 3.46 635.00 13163.50 23169.5

Mean 39077 84403 116 3.56 635.00 19163.70 23179.5

Median 46667.63 98902 113 3.40 661.16 15967.50 28980.75

C) Pre-MBA Salary Post-MBA Salary Percentage Increase in Salary Undergraduate GPA GMAT Score Annual Tuition Expected Annual Student Cost D) Pre-MBA Salary Post-MBA Salary Percentage Increase in Salary Undergraduate GPA GMAT Score Annual Tuition Expected Annual Student Cost

Answer: C Diff: 2 Keywords: measure of center and location Section: 3-1 Measures of Center and Location Outcome: 1 125) Dynamic random-access memory (DRAM) memory chips are made from silicon wafers in manufacturing facilities through a very complex process called wafer fabs. The wafers are routed through the fab machines in an order that is referred to as a recipe. The wafers may go through the same machine several times as the chip is created. The data file DRAM Chips contains a sample of processing times, measured in fractions of hours, at a particular machine center for one chip recipe. Compute the mean processing time. A) 0.24 minutes B) 0.22 minutes C) 0.31 minutes D) 0.33 minutes Answer: D Diff: 1 Keywords: measure of center and location Section: 3-1 Measures of Center and Location Outcome: 1 126) Dynamic random-access memory (DRAM) memory chips are made from silicon wafers in manufacturing facilities through a very complex process called wafer fabs. The wafers are routed through the fab machines in an order that is referred to as a recipe. The wafers may go through the same machine several times as the chip is created. The data file DRAM Chips contains a sample of processing times, measured in fractions of hours, at a particular machine center for one chip recipe. Compute the median processing time. A) 0.31 minutes B) 0.24 minutes C) 0.21 minutes D) 0.44 minutes Answer: A Diff: 1 Keywords: measure of center and location Section: 3-1 Measures of Center and Location Outcome: 1

127) Dynamic random-access memory (DRAM) memory chips are made from silicon wafers in manufacturing facilities through a very complex process called wafer fabs. The wafers are routed through the fab machines in an order that is referred to as a recipe. The wafers may go through the same machine several times as the chip is created. The data file DRAM Chips contains a sample of processing times, measured in fractions of hours, at a particular machine center for one chip recipe. Determine what the mode processing time is. A) 0.22 B) 0.24 C) 0.33 D) 0.34 Answer: B Diff: 1 Keywords: measure of center and location Section: 3-1 Measures of Center and Location Outcome: 1 128) Dynamic random-access memory (DRAM) memory chips are made from silicon wafers in manufacturing facilities through a very complex process called wafer fabs. The wafers are routed through the fab machines in an order that is referred to as a recipe. The wafers may go through the same machine several times as the chip is created. The data file DRAM Chips contains a sample of processing times, measured in fractions of hours, at a particular machine center for one chip recipe. Calculate the 80th percentile for processing time. A) 0.40 minutes B) 0.35 minutes C) 0.45 minutes D) 0.20 minutes Answer: A Diff: 2 Keywords: measure of variation Section: 3-2 Measures of Variation Outcome: 1 129) Suppose that the distribution of grocery purchases is thought to be symmetric. If the mean purchase is $23.14, what would the median purchase be? Answer: For a symmetric distribution, the mean and median are equal. Therefore, the median purchase should also be $23.14. Diff: 1 Keywords: mean, median, symmetric, distribution Section: 3-1 Measures of Center and Location Outcome: 1

130) The AMI Company has two assembly lines in its Kansas City plant. Line A produces an average of 335 units per day with a standard deviation equal to 11 units. Line B produces an average of 145 units per day with a standard deviation equal to 8 units. Based on this information, which line is relatively more consistent? Answer: At first glance it may appear that Line B is more consistent since it has a smaller standard deviation. However, if we wish to compare relative variability when the means of two distributions are different, then we need to compute the coefficient of variation for each. The one with the smallest coefficient of variation is the more consistent. The coefficient of variation is given by: CV =

100

Then for Line A we get:

CV =

100 =

100 = 3.28%

For Line B we get:

CV =

100 =

100 = 5.52%

Thus, Line A is the more consistent of the two lines with respect to daily production output. Diff: 2 Keywords: mean standard deviation, relative variation, coefficient of variation Section: 3-1 Measures of Center and Location Outcome: 1

131) The following sample data reflect electricity bills for ten households in San Diego in March. $118.20 $76.90

$67.88 $144.56

$133.40 $127.89

$88.42 $89.34

$110.34 $129.10

Determine three measures of central tendency for these sample data. Then, based on these measures, determine whether the sample data are symmetric or skewed. Answer: The three measures of central tendency are the mean, median, and mode. These are computed as follows:

Mean:

= $108.60

Median: Arrange the data in order from low to high. Since we have an even number of values, the median is the mean of the 5th and 6th values. $67.88

$76.90

$88.42

$89.34

The median is found as: Md ≈

$110.34 $118.20 $127.89 $129.10 $133.40 $144.56 = $114.27

Mode: The mode is the value in the data that occurs most frequently. Since no value in this sample occurs more frequently than one time, there is no mode. Data are symmetric if the mean and the median are equal. Since the sample mean = $108.60 and the median equals $114.27 the data are not symmetric. Since the mean is less than median, we conclude that the sample data are left-skewed. Diff: 2 Keywords: central tendency, mean, median, mode Section: 3-1 Measures of Center and Location Outcome: 1

132) The following sample data reflect electricity bills for ten households in San Diego in March. $118.20 $76.90

$67.88 $144.56

$133.40 $127.89

$88.42 $89.34

$110.34 $129.10

Compute the range, variance, and standard deviation for these sample data. Discuss which of these three measures you would prefer to use as a measure of variation. Answer: The range is found as follows: Range = High - Low = $144.56 - $67.88 = $76.68 The sample variance is found using: S2 = We begin by determining the sample mean:

= 108.60

We then sum the deviations of the individual values from the sample mean giving:

We now divide this sum by n-1 giving: S2 =

= 692.99

Thus, the sample variance is 692.99. The sample deviation is found by taking the square root of the sample variance: S=

= $26.32

Thus, the sample standard deviation is $26.32. Although the range is far easier to compute, it contains information only from the extremes in the data. The variance is in squared units and therefore does not have any meaning in the context of money spent on electricity. The standard deviation is preferred since it uses all the data in its calculation and is expressed in the original units. Diff: 2 Keywords: mean, standard deviation, range, variance Section: 3-2 Measures of Variation Outcome: 3

133) Why is it that when we find the sample standard deviation, we divide by n-1 but when we find the population standard deviation we divide by n? Answer: The technical answer to this question is beyond the scope of the text. However, we can think of it this way. If our objective in computing the sample standard deviation is to estimate the population standard deviation, we would want an estimate that would be correct on the average. That means that if we took repeated random samples from the population and for each sample we computed the standard deviation, we would want the average of the sample standard deviations to equal the population standard deviation. If our formula for the sample standard deviation uses n-1, this occurs. Diff: 3 Keywords: sample, population, standard deviation Section: 3-2 Measures of Variation Outcome: 3 134) Explain when the empirical rule can be used to help describe data in a population or a sample, defining which data distributions cannot be explained by this rule. Answer: The empirical rule applies when the data distribution is bell-shaped. When this is the case, we know that approximately 68 percent of the data will fall with ± 1 standard deviation of the mean, approximately 95 percent will fall within ± 2 standard deviations of the mean, and virtually all of the data will fall within ± 3 standard deviations of the mean. However, if a data distribution is right or left skewed, it cannot be analyzed using these methods. Other methods have to be researched to analyze the placement of data. Diff: 2 Keywords: empirical rule, bell-shaped, standard deviation Section: 3-3 Using the Mean and Standard Deviation Together Outcome: 5 135) Explain how Tchebysheff's theorem can be used to help describe data in a population or a sample. Answer: If the data in a sample or a population are known to be bell-shaped, the empirical rule can be used to help us describe the data. However, when the sample or the population is not bell-shaped, Tchebysheff's theorem is very useful. It is a conservative theorem because it applies to any distribution. If we know the mean and standard deviation, Tchebysheff tells us that at least 75 percent of the data values will fall with ± 2 standard deviations of the mean and at least 88 percent will lie within 3 standard deviations of the mean. Keep in mind that in most instances, the percentage of observations will exceed these minimum amounts, but at least it gives us some idea of how the data are distributed without actually looking at all the data. Diff: 2 Keywords: Tchebysheff, data, population, sample Section: 3-3 Using the Mean and Standard Deviation Together Outcome: 5

136) Explain what is meant by percentiles and quartiles. Answer: Percentiles and quartiles are measures that help us understand how the data are distributed. Percentiles divide the data into 100 parts so that the same number of observations fall in each percentile. In order to construct the percentiles, the data must be arranged in order from low to high. The percentile value is determined by determining the location of the data value that corresponds to the percent of the way through the ordered data that we wish to go. For example, if we want to find the 80th percentile, we locate the value using: i=

n where P = 80 and n equal the number of values in the data set. If we have 1400 data values, then

the 80th percentile value is: i=

(1400 ) = 1,120. But since this is an integer, we would compute the 80th percentile to be

the average of the 1,120th and 1,121st values in the data. Quartiles are similar to percentages except that the ordered data are divided into four segments with an equal number of values in each. The 1st quartile corresponds to the 25th percentile, the 2nd quartile corresponds to the 50th percentile, and the 3rd quartile corresponds to the 75th percentile. Diff: 2 Keywords: percentile, quartile Section: 3-2 Measures of Variation Outcome: 3 137) Consumer products are required by law to contain at least as much as the amount printed on the package. For example a bag of potato chips that is labeled as 10 ounces should contain at least 10 ounces. Assume that the standard deviation of the packaging equipment yields a bag weight standard deviation of 0.2 ounces. Explain what average bag weight must be used to achieve at least 97.5 percent of the bags having 10 or more ounces in the bag. Assume the bag weight distribution is bell-shaped. Answer: If the average bag weight were 10 ounces this would mean only 50 percent of the bags would weigh enough and the other 50 percent would be underweight, so the average must be set higher to achieve 97.5 percent having at least 10 ounces. We want to allow only 2.5 percent of the bags be underweight. Based on the empirical rule, about 95 percent of the bags will be within 2 standard deviations of the mean. This means that 5 percent of the bags will be farther from the mean, and since the distribution is symmetrical that puts 2.5 percent in each tail. Therefore the mean needs to be 2 standard deviations above 10 to have only 2.5 percent in the lower tail below 10 ounces. Z = -2.0 =

so, μ = 10 + 2.0(0.2) = 10.4 ounces

Diff: 3 Keywords: standardized value, empirical distribution Section: 3-3 Using the Mean and Standard Deviation Together Outcome: 5

138) What is meant by the concept, standardizing the data? Explain why a decision maker may wish to compute a standardized value. Answer: The concept, standardized data, refers to the number of standard deviations a value is from the mean of the sample or population from which it was selected. The population standardized value is referred to as the z-value and is computed as follows: z= The reason that we might want to use the standardized value rather than the original value is if we are interested in comparing individual values from two or more distributions that have different means and standard deviations. By comparing z-values, we are able to determine which original value falls relatively more closely to its mean or which value is relatively more extreme compared to the other data in the sample or population. Diff: 2 Keywords: standardized value, z-score, mean, standard deviation Section: 3-3 Using the Mean and Standard Deviation Together Outcome: 4

Business Statistics, 9e (Groebner/Shannon/Fry) Chapter 4 Introduction to Probability 1) A car salesman has noted that the probability that the dealership sells a car on a Saturday morning is .30. Then the probability of the dealership not selling a car on Saturday morning is .70. Answer: TRUE Diff: 1 Keywords: probability, dependent, mutually exclusive Section: 4-1 The Basics of Probability Outcome: 1 2) A car salesman states that the probability that the dealership sells a car on a Saturday morning is .30. The method of probability assessment that he has used is most likely classical assessment. Answer: FALSE Diff: 1 Keywords: probability, assessment, relative frequency of occurrence Section: 4-1 The Basics of Probability Outcome: 1 3) If the probability of one event occurring is .40 and the probability of a second event occurring is 0.60, then the probability that both events will occur must be 1.0 since that is the maximum value a probability can be. Answer: FALSE Diff: 1 Keywords: probability, event Section: 4-1 The Basics of Probability Outcome: 1 4) If a company has the opportunity to bid on three contracts, A, B, and C, then the number of these contracts that are awarded to the company would be considered an elementary event. Answer: TRUE Diff: 2 Keywords: probability, elementary event Section: 4-1 The Basics of Probability Outcome: 1 5) If two events are mutually exclusive, it is possible for them to also be independent of each other. Answer: FALSE Diff: 2 Keywords: experiment, outcome, probability Section: 4-1 The Basics of Probability Outcome: 1

6) Suppose a single die (a 6-sided cube with sides numbered 1 through 6) is rolled once. The event of interest is defined as rolling an even number. This can be said to be an elementary event. Answer: FALSE Diff: 1 Keywords: sample space, elementary event Section: 4-1 The Basics of Probability Outcome: 1 7) In most situations, there is no difference between the events and the elementary events. Answer: FALSE Diff: 1 Keywords: event, elementary event, experiment Section: 4-1 The Basics of Probability Outcome: 1 8) A manufacturing company makes three types of products. Each time it makes a product, the item can be either good or defective and it can be either customized or standard. The events consisting of customized and defective would be considered mutually exclusive since they apply to different attributes of the product. Answer: FALSE Diff: 2 Keywords: event, mutually exclusive Section: 4-1 The Basics of Probability Outcome: 1 9) Two football teams play in the Super Bowl. The event of team A winning and the event of team B winning can be said to be mutually exclusive. Answer: TRUE Diff: 1 Keywords: event, mutually exclusive Section: 4-1 The Basics of Probability Outcome: 1 10) A product that is produced at Ramsey Manufacturing goes through three steps to be built. At step one, the components are assembled by technicians. At step two, the product is sanded, and at step three the product is painted. The product can become defective if any of these three steps is performed incorrectly. The three steps are done by different people in different locations. We let D1 = defect introduced at step 1, D2 = defect introduced at step 2, and D3 = defect introduced at step three. Based on this situation these three events would be considered to be mutually exclusive. Answer: FALSE Diff: 2 Keywords: mutually exclusive, independent event Section: 4-1 The Basics of Probability Outcome: 1

11) Suppose a player is dealt 2 cards from a standard deck of 52 playing cards. To determine the probability of having a blackjack would involve classical probability. Answer: TRUE Diff: 2 Keywords: classical probability Section: 4-1 The Basics of Probability Outcome: 1 12) In playing the game Monopoly, the probability of a player landing on Park Place would be assessed using classical probability assessment. Answer: TRUE Diff: 2 Keywords: classical, probability, assessment Section: 4-1 The Basics of Probability Outcome: 1 13) If a manager were interested in assessing the probability that a new product will be successful in a New Jersey market area, she would most likely use relative frequency of occurrence as the method for assessing the probability. Answer: FALSE Diff: 2 Keywords: probability, subjective Section: 4-1 The Basics of Probability Outcome: 1 14) Classical probability assessment is likely to be the most common method of probability assessment used in business decision making. Answer: FALSE Diff: 1 Keywords: subjective, probability, assessment Section: 4-1 The Basics of Probability Outcome: 1 15) The owners of Greg's Department Store have reason to believe that one of their employees has been stealing from the store. In an interview with the police, the owner says that she is 75 percent sure that the employee is stealing. This probability is an example of one that was assessed using classical probability. Answer: FALSE Diff: 2 Keywords: subjective, probability, assessment Section: 4-1 The Basics of Probability Outcome: 1

16) When a patient arrives at a clinic complaining of several specific symptoms, the doctor who makes the diagnosis says that he is 80 percent certain that the patient has a particular problem. It is likely that he is basing this assessment on relative frequency of occurrence. Answer: TRUE Diff: 2 Keywords: probability, relative, frequency, assessment Section: 4-1 The Basics of Probability Outcome: 1 17) If you were planning to take a small group out to dinner on a Thursday evening and you were considering whether to call ahead for a reservation, the method of probability assessment you would most likely use to assess the chances of being able to get in for dinner without having a reservation would be subjective assessment. Answer: TRUE Diff: 2 Keywords: probability, subjective, assessment Section: 4-1 The Basics of Probability Outcome: 1 18) One of the difficulties in using the relative frequency of occurrence method for assessing probabilities in business situations is getting a large enough set of examples that match the one in question. Answer: TRUE Diff: 1 Keywords: relative, frequency, probability, assessment Section: 4-1 The Basics of Probability Outcome: 1 19) Suppose a coin is flipped twice. The event of getting heads on the first toss and the event of getting heads on the second toss could be said to be mutually exclusive. Answer: FALSE Diff: 1 Keywords: event, mutually exclusive, independent Section: 4-1 The Basics of Probability Outcome: 1 20) At a potato processing plant in the state of Washington, 400 potatoes have been examined for disease. Of these, four were diseased. Based on this, the plant manager has stated that the probability of finding a diseased potato is 0.01. He is applying subjective probability to arrive at this 0.01 value. Answer: FALSE Diff: 2 Keywords: relative, frequency, probability Section: 4-1 The Basics of Probability Outcome: 1

21) A New Jersey company relies on a steady supply of power to keep its manufacturing going. Recently at a planning meeting, the general manager stated that the chance of a rolling blackout affecting production is 0.15. She most likely made this assessment using subjective probability assessment. Answer: TRUE Diff: 2 Keywords: subjective, probability, assessment Section: 4-1 The Basics of Probability Outcome: 1 22) It is correct to say that subjective probability assessments are neither right nor wrong, but are merely reflections of the state of mind of the individual making the probability assessment. Answer: TRUE Diff: 2 Keywords: subjective, probability, assessment Section: 4-1 The Basics of Probability Outcome: 1 23) A dam on a river that holds back a water reservoir begins to leak. Engineers say that there is a 10 percent chance of the dam breaking if repairs are not made. This is an example of classical probability. Answer: FALSE Diff: 2 Keywords: subjective, probability, assessment Section: 4-1 The Basics of Probability Outcome: 1 24) During the past week, of the 250 customers at the Dairy Queen who ordered a Blizzard, 50 ordered strawberry. This means that of the next five Blizzard customers, exactly one will order strawberry. Answer: FALSE Diff: 2 Keywords: relative, frequency, probability, assessment Section: 4-1 The Basics of Probability Outcome: 1 25) When a construction company bids on a contract, the events will be win or lose. The closer the probability is to 0.50, the greater the uncertainty about whether the company will win or lose the bid. Answer: TRUE Diff: 1 Keywords: event, mutually exclusive Section: 4-1 The Basics of Probability Outcome: 1 26) Sometime it is necessary to assign probabilities based on a person's belief that an outcome will occur. Answer: TRUE Diff: 1 Keywords: Subjective Probability Assessment Section: 4-1 The Basics of Probability Outcome: 1

27) Mutually exclusive means that the occurrence of event A has no effect on the probability of the occurrence of event B, and independent means the occurrence of event A prevents the occurrence of event B. Answer: FALSE Diff: 3 Keywords: independent, mutually exclusive Section: 4-1 The Basics of Probability Outcome: 1 28) A neighborhood has 15 houses for sale. Five of these houses were less than 10 years old and the others were 10 years or older. If one house is purchased at random from this neighborhood, the probability that it is 10 years or older is 0.66. Answer: TRUE Diff: 1 Keywords: probability Section: 4-2 True Rules of Probability Outcome: 2 29) If a single die is rolled (a cube where the sides are numbered 1 through 6), the probability of rolling at least a 3 is 0.33. Answer: FALSE Diff: 1 Keywords: probability Section: 4-2 True Rules of Probability Outcome: 2 30) Suppose 10 students are enrolled in a class and the probability of at least 8 showing up on a given day is 90 percent. Then the probability of 7 or fewer showing that day is 10 percent. Answer: TRUE Diff: 1 Keywords: probability, complement Section: 4-2 True Rules of Probability Outcome: 2

31) The following probability distribution was subjectively assessed for the number of sales a salesperson would make if he or she made five sales calls in one day. Sales 0 1 2 3 4 5

Probability 0.10 0.15 0.20 0.30 0.20 0.05

Given this distribution, the probability that the number of sales is less than 2 is 0.15. Answer: FALSE Diff: 1 Keywords: probability, addition Section: 4-2 True Rules of Probability Outcome: 2 32) The following probability distribution was subjectively assessed for the number of sales a salesperson would make if he or she made five sales calls in one day. Sales 0 1 2 3 4 5

Probability 0.10 0.15 0.20 0.30 0.20 0.05

Given this distribution, the probability that the number of sales is 2 or 3 is 0.50. Answer: TRUE Diff: 1 Keywords: probability, addition Section: 4-2 True Rules of Probability Outcome: 2

33) The following probability distribution was subjectively assessed for the number of sales a salesperson would make if he or she made five sales calls in one day. Sales 0 1 2 3 4 5

Probability 0.10 0.15 0.20 0.30 0.20 0.05

Given this distribution, the probability that the number of sales is more than 2 is 0.80. Answer: FALSE Diff: 1 Keywords: probability, addition Section: 4-2 True Rules of Probability Outcome: 2 34) When the salesperson makes a sale, there are three possible sales levels: large, medium, and small. The probability of a large sale is 0.20 and the chance of a medium sale is 0.60. Thus, when a sale is made, the chance of it being a small sale is 0.20. Answer: TRUE Diff: 2 Keywords: probability, addition Section: 4-2 True Rules of Probability Outcome: 2 35) Assume P(A) = 0.4 and P(B) = 0.2 and P(A and B) = 0.1, then the probability of P(A or B) = 0.7. Answer: FALSE Diff: 1 Keywords: probability, addition rule Section: 4-2 True Rules of Probability Outcome: 2 36) When the salesperson makes a sale, there are three possible sales levels: large, medium, and small. The probability of a large sale is 0.20 and the chance of a medium sale is 0.60. If a salesperson makes two sales, the probability that at least one is large is 0.36. Answer: TRUE Diff: 3 Keywords: probability, conditional Section: 4-2 True Rules of Probability Outcome: 2

37) The following probability distribution was subjectively assessed for the number of sales a salesperson would make if he or she made five sales calls in one day. Sales 0 1 2 3 4 5

Probability 0.10 0.15 0.20 0.30 0.20 0.05

When the salesperson makes a sale, there are three possible sales levels: large, medium, and small. The probability of a large sale is 0.20 and the chance of a medium sale is 0.60. The probability on a given day that the salesperson will make one sale and that it is medium is 0.09. Answer: TRUE Diff: 2 Keywords: probability, joint, marginal Section: 4-2 True Rules of Probability Outcome: 2 38) When customers come to a bank, there are three primary locations they may select to go to: teller, loan officer, or escrow department. Based on past experience, the following probability distribution applies: Location Teller Loan Officer Escrow

Probability 0.60 0.30 0.10

Seventy percent of customers are males. Thus, the probability that the next customer to enter the bank is a male who goes to the teller is 1.30. Answer: FALSE Diff: 1 Keywords: probability Section: 4-2 True Rules of Probability Outcome: 2

39) When customers come to a bank, there are three primary locations they may select to go to: teller, loan officer, or escrow department. Based on past experience, the following probability distribution applies: Location Teller Loan Officer Escrow

Probability 0.60 0.30 0.10

Seventy percent of customers are males. The probability that the next two customers to enter the bank are males and go to the Loan Officer is 0.42. Answer: FALSE Diff: 2 Keywords: probability, multiplication Section: 4-2 True Rules of Probability Outcome: 3 40) When customers come to a bank, there are three primary locations they may select to go to: teller, loan officer, or escrow department. Based on past experience, the following probability distribution applies: Location Teller Loan Officer Escrow

Probability 0.60 0.30 0.10

Seventy percent of customers are males. The probability that three consecutive customers all go to a teller is approximately 0.22. Answer: TRUE Diff: 2 Keywords: probability, multiplication Section: 4-2 True Rules of Probability Outcome: 3 41) When customers come to a bank, there are three primary locations they may select to go to: teller, loan officer, or escrow department. Based on past experience, the following probability distribution applies: Location Teller Loan Officer Escrow

Probability 0.60 0.30 0.10

Seventy percent of customers are males. The probability that the next customer will be male and will go to either the teller or the escrow department is 0.49. Answer: TRUE Diff: 3 Keywords: probability, multiplication Section: 4-2 True Rules of Probability Outcome: 3 4-10 Copyright © 2018 Pearson Education, Inc.

42) There are three general locations that a taxi can go to: the airport, downtown, and elsewhere. When a taxi driver starts in the downtown location, there is a 0.40 chance that his first call will take him to the airport and a 0.40 chance of going to another downtown location. Once a taxi is at the airport, there is a 0.80 probability that the next fare will take him downtown and a 0.20 chance of going elsewhere. The probability of a call from anywhere except downtown taking him to the airport is 0.20. Therefore, the probability that the taxi is at the airport when the third call arrives after going on shift is 0.20. Answer: TRUE Diff: 3 Keywords: probability, joint Section: 4-2 True Rules of Probability Outcome: 2 43) Assume P(A) = 0.6, P(B) = 0.7, and P(A and B) = 0.42, which means that events A and B are independent of each other. Answer: TRUE Diff: 2 Keywords: probability, independence Section: 4-2 True Rules of Probability Outcome: 2 44) The Crystal Window Company makes windows at three locations: Reno, Las Vegas, and Boise. Some windows made by the company contain a visible defect and must be replaced. Each defect costs the company $45.00. The Reno plant makes 40 percent of all windows while the Las Vegas and Boise plants split the remaining production evenly. A recent quality study shows that 8 percent of the Reno windows contain a defect, 11 percent of the Las Vegas windows contain a defect, while 4 percent of the windows made in Boise have a defect. Once the windows are made, they are shipped to a central warehouse where they are commingled and the location where they were made is lost. Based on this information, if a defective window is discovered, it was most likely made by the Las Vegas plant. Answer: TRUE Diff: 3 Keywords: probability, marginal Section: 4-2 True Rules of Probability Outcome: 2

45) The Crystal Window Company makes windows at three locations: Reno, Las Vegas, and Boise. Some windows made by the company contain a visible defect and must be replaced. Each defect costs the company $45.00. The Reno plant makes 40 percent of all windows while the Las Vegas and Boise plants split the remaining production evenly. A recent quality study shows that 8 percent of the Reno windows contain a defect, 11 percent of the Las Vegas windows contain a defect, while 4 percent of the windows made in Boise have a defect. Once the windows are made, they are shipped to a central warehouse where they are commingled and the location where they were made is lost. Based on this information the probability that a defective window was made by the Boise plant is approximately 0.16. Answer: TRUE Diff: 3 Keywords: probability, marginal Section: 4-2 True Rules of Probability Outcome: 2 46) The Crystal Window Company makes windows at three locations: Reno, Las Vegas, and Boise. Some windows made by the company contain a visible defect and must be replaced. Each defect costs the company $45.00. The Reno plant makes 40 percent of all windows while the Las Vegas and Boise plants split the remaining production evenly. A recent quality study shows that 8 percent of the Reno windows contain a defect, 11 percent of the Las Vegas windows contain a defect, while 4 percent of the windows made in Boise have a defect. Once the windows are made, they are shipped to a central warehouse where they are commingled and the location where they were made is lost. Based on this information, the percentage of the defective cost that should be allocated to the Reno plant is approximately 42 percent. Answer: TRUE Diff: 3 Keywords: probability, marginal Section: 4-2 True Rules of Probability Outcome: 2 47) The Vintage Clothing Company has three retail locations, code-named A, B, and C. The following table illustrates the percentage of total company sales at each store and also the percentage of customers at that store who make purchases with credit cards: Store A B C

Proportion of Total Sales 0.15 0.25 0.60

Proportion of Customers Using Credit 0.32 0.45 0.65

Based on this information, the probability that a customer will use a credit card is just slightly greater than 0.55. Answer: TRUE Diff: 3 Keywords: probability, joint Section: 4-2 True Rules of Probability Outcome: 2

48) The Vintage Clothing Company has three retail locations, code-named A, B, and C. The following table illustrates the percentage of total company sales at each store and also the percentage of customers at that store who make purchases with credit cards: Store A B C

Proportion of Total Sales 0.15 0.25 0.60

Proportion of Customers Using Credit 0.32 0.45 0.65

Based on this information, given that a customer has used a credit card to make the purchase, the sale was most likely made at store C. Answer: TRUE Diff: 3 Keywords: probability, joint Section: 4-2 True Rules of Probability Outcome: 2 49) The Vintage Clothing Company has three retail locations, code-named A, B, and C. The following table illustrates the percentage of total company sales at each store and also the percentage of customers at that store who make purchases with credit cards: Store A B C

Proportion of Total Sales 0.15 0.25 0.60

Proportion of Customers Using Credit 0.32 0.45 0.65

Based on this information, given that a customer has used a purchase method other than a credit card, the sale was least likely made at store A. Answer: FALSE Diff: 3 Keywords: probability, joint Section: 4-2 True Rules of Probability Outcome: 2 50) If a six-sided die is tossed two times and "4" shows up both times, the probability of "4" on the third trial is much larger than any other outcome. Answer: FALSE Diff: 2 Keywords: probability, independence Section: 4-2 True Rules of Probability Outcome: 2

51) An event is: A) the list of possible outcomes that can occur from a selection or decision. B) a collection of elementary events. C) similar to an experiment but not controlled by the decision maker. D) more frequently found in business than in other disciplines. Answer: B Diff: 1 Keywords: event Section: 4-1 The Basics of Probability Outcome: 1 52) The method of probability assessment that relies on an examination of historical data from similar situations is: A) relative frequency of occurrence. B) classical assessment. C) historical assessment. D) subjective assessment. Answer: A Diff: 2 Keywords: relative frequency, probability assessment Section: 4-1 The Basics of Probability Outcome: 1 53) The method of probability assessment that is least likely to be used by business decision makers is: A) subjective assessment. B) relative frequency of occurrence. C) classical assessment. D) None of the above is used by decision makers. Answer: C Diff: 2 Keywords: classical assessment, probability Section: 4-1 The Basics of Probability Outcome: 1 54) At gambling casinos all over the country, a popular dice game is called craps. The probability of a player winning at this game can be assessed using: A) subjective assessment. B) classical probability. C) relative frequency of occurrence. D) None of the above Answer: B Diff: 2 Keywords: classical probability assessment Section: 4-1 The Basics of Probability Outcome: 1

55) A consumer products company is planning to introduce a new product. The method that is least likely to be used to assess the probability of the product being successful is: A) classical probability assessment. B) subjective assessment. C) relative frequency of occurrence. D) elementary events. Answer: A Diff: 2 Keywords: classical, relative, assessment Section: 4-1 The Basics of Probability Outcome: 1 56) A study was recently done in which 500 people were asked to indicate their preferences for one of three products. The following table shows the breakdown of the responses by gender of the respondents.

If the people conducting the study wish to assess the probability that product A will be preferred by members of the target population, the method of assessment to be used would most likely be: A) classical probability assessment. B) subjective assessment. C) relative frequency of occurrence. D) independent events. Answer: C Diff: 2 Keywords: relative frequency, probability assessment Section: 4-1 The Basics of Probability Outcome: 1

57) A study was recently done in which 500 people were asked to indicate their preferences for one of three products. The following table shows the breakdown of the responses by gender of the respondents.

Based on these data, the probability that a person in the population will prefer product A can be assessed as: A) 0.18 B) 0.56 C) 0.286 D) 0.16 Answer: B Diff: 2 Keywords: probability, relative frequency Section: 4-1 The Basics of Probability Outcome: 1 58) A study was recently done in which 500 people were asked to indicate their preferences for one of three products. The following table shows the breakdown of the responses by gender of the respondents.

Suppose one person is randomly chosen. Based on this data, what is the probability that the person chosen is a female who prefers product C? A) 0.24 B) 0.86 C) 0.92 D) 0.31 Answer: A Diff: 2 Keywords: probability, relative frequency Section: 4-1 The Basics of Probability Outcome: 1

59) When a football team plays a game, there are three outcomes that can occur: they can win, they can lose, or they can tie. The events are: A) independent. B) mutually exclusive. C) all inclusive. D) dependent events. Answer: B Diff: 1 Keywords: mutually exclusive, event Section: 4-1 The Basics of Probability Outcome: 1 60) When a pair of dice are rolled, the outcome for each die can be said to be: A) mutually exclusive. B) mutually inclusive. C) dependent. D) independent. Answer: A Diff: 1 Keywords: dependent, event Section: 4-1 The Basics of Probability Outcome: 1 61) If two events are independent, then A) they must be mutually exclusive. B) the sum of their probabilities must be equal to one. C) their intersection must be zero. D) None of the above Answer: D Diff: 3 Keywords: mutually exclusive, independent event Section: 4-1 The Basics of Probability Outcome: 1

62) The managers of a local golf course have recently conducted a study of the types of golf balls used by golfers based on handicap. A joint frequency table for the 100 golfers covered in the survey is shown below:

Based on these data, the probability of a golfer having a handicap less than 10 is: A) 0.52 B) 0.10 C) 0.34 D) None of the above Answer: A Diff: 2 Keywords: probability, events Section: 4-2 True Rules of Probability Outcome: 2 63) The managers of a local golf course have recently conducted a study of the types of golf balls used by golfers based on handicap. A joint frequency table for the 100 golfers covered in the survey is shown below:

Based on these data, the probability that a player will use a Strata golf ball is: A) 0.15 B) 0.20 C) 0.18 D) None of the above Answer: B Diff: 2 Keywords: probability, events Section: 4-2 True Rules of Probability Outcome: 2

64) The managers of a local golf course have recently conducted a study of the types of golf balls used by golfers based on handicap. A joint frequency table for the 100 golfers covered in the survey is shown below:

If a player comes to the course using a Nike golf ball, the probability that he or she has a handicap of at least 10 is: A) 0.22 B) 0.48 C) slightly greater than 0.45 D) 0.10 Answer: C Diff: 2 Keywords: probability, conditional Section: 4-2 True Rules of Probability Outcome: 2 65) The managers of a local golf course have recently conducted a study of the types of golf balls used by golfers based on handicap. A joint frequency table for the 100 golfers covered in the survey is shown below:

Based on these data, the probability of someone using a Strata ball and having a handicap under 2 is: A) 0.05 B) 0.38 C) 0.25 D) None of the above Answer: C Diff: 2 Keywords: probability, events Section: 4-2 True Rules of Probability Outcome: 2

66) The managers of a local golf course have recently conducted a study of the types of golf balls used by golfers based on handicap. A joint frequency table for the 100 golfers covered in the survey is shown below:

Based on these data, if a player has a handicap that is 10 or more, the probability that he or she will use a Nike golf ball is: A) 0.21 B) 0.10 C) 0.45 D) 0.48 Answer: A Diff: 2 Keywords: probability, conditional Section: 4-2 True Rules of Probability Outcome: 2 67) The Anderson Lumber Company has three sawmills that produce boards of different lengths. The following table is a joint frequency distribution based on a random sample of 1,000 boards selected from the lumber inventory.

Based on these data, the probability of selecting a board from inventory that is 10 feet long is: A) 0.196 B) 0.450 C) 0.084 D) 0.170 Answer: D Diff: 1 Keywords: probability, joint frequency Section: 4-2 True Rules of Probability Outcome: 2

68) The Anderson Lumber Company has three sawmills that produce boards of different lengths. The following table is a joint frequency distribution based on a random sample of 1,000 boards selected from the lumber inventory.

Based on these data, if a board is selected that is 12 feet long, the probability that it was made at sawmill A is: A) 0.08 B) 0.41 C) 0.24 D) 0.20 Answer: B Diff: 2 Keywords: probability, joint frequency Section: 4-2 True Rules of Probability Outcome: 2 69) The Anderson Lumber Company has three sawmills that produce boards of different lengths. The following table is a joint frequency distribution based on a random sample of 1000 boards selected from the lumber inventory.

Based on these data, if three boards are selected at random, the probability that all three were made at sawmill A is: A) 0.037 B) 0.334 C) 1.00 D) 0.556 Answer: A Diff: 2 Keywords: probability, joint frequency Section: 4-2 True Rules of Probability Outcome: 2

70) Harrison Water Sports has three retail outlets: Seattle, Portland, and Phoenix. The Seattle store does 50 percent of the total sales in a year, while the Portland store does 35 percent of the total sales. Further analysis indicates that of the sales in Seattle, 20 percent are in boat accessories. The percentage of boat accessories at the Portland store is 30 and the percentage at the Phoenix store is 25. If a sales dollar is recorded as a boat accessory, the probability that the sale was made at the Portland store is: A) slightly greater than 0.43 B) 0.35 C) 0.2425 D) None of the above Answer: A Diff: 3 Keywords: probability, conditional Section: 4-2 True Rules of Probability Outcome: 2 71) Harrison Water Sports has three retail outlets: Seattle, Portland, and Phoenix. The Seattle store does 50 percent of the total sales in a year, while the Portland store does 35 percent of the total sales. Further analysis indicates that of the sales in Seattle, 20 percent are in boat accessories. The percentage of boat accessories at the Portland store is 30 and the percentage at the Phoenix store is 25. Overall, the probability that a sale by Harrison Water Sports will be for a boat accessory is: A) 0.105 B) 0.2425 C) 0.75 D) None of the above Answer: B Diff: 3 Keywords: probability, joint frequency Section: 4-2 True Rules of Probability Outcome: 2 72) Of the last 100 customers entering a computer shop, 25 have purchased a computer. If the classical probability assessment for computing probability is used, the probability that the next customer will purchase a computer is: A) 0.25 B) 0.50 C) 1.00 D) 0.75 Answer: B Diff: 3 Keywords: probability, classical probability assessment Section: 4-2 True Rules of Probability Outcome: 2

73) A special roulette wheel, which has an equal number of red and black spots, has come up red four times in a row. Assuming that the roulette wheel is fair, what concept allows a player to know that the probability the next spin of the wheel will come up black is 0.5? A) Concept of independent events B) Concept of mutually exclusive events C) Concept of dependent events D) Concept of mutually inclusive events Answer: A Diff: 1 Keywords: probability Section: 4-1 The Basics of Probability Outcome: 1 74) In a survey, respondents were asked to indicate their favorite brand of cereal (Post or Kellogg's). They were allowed only one choice. What is the probability concept that implies it is not possible for a single respondent to state both Post and Kellogg's to be the favorite cereal? A) Concept of independent events B) Concept of mutually exclusive events C) Concept of dependent events D) Concept of mutually inclusive events Answer: B Diff: 1 Keywords: probability Section: 4-1 The Basics of Probability Outcome: 1 75) What method of probability assessment would most likely be used to assess the probability that a major earthquake will occur in California in the next three years? A) Classical probability based on the ratio of the number of ways the event can occur B) Relative frequency based on previous history C) Subjective probability based on expert opinion D) Independent probability based on two unrelated outcomes Answer: C Diff: 1 Keywords: probability Section: 4-1 The Basics of Probability Outcome: 1

76) What method of probability assessment would most likely be used to assess the probability that a customer will return a purchase for a refund? A) Classical probability based on the ratio of the number of ways the event can occur B) Relative frequency based on previous history C) Subjective probability based on expert opinion D) Independent probability based on two unrelated outcomes Answer: B Diff: 1 Keywords: probability Section: 4-1 The Basics of Probability Outcome: 1 77) An inventory of appliances contains four white washers and one black washer. If a customer selects one at random, what is the probability that the black washer will be selected? A) 0.5 B) 0.4 C) 0.2 D) 0.8 Answer: C Diff: 1 Keywords: probability Section: 4-2 True Rules of Probability Outcome: 1 78) Long-time friends, Pat and Tom, agree on many things, but not the outcome of the American League pennant race and the World Series. Pat is originally from Boston, and Tom is from New York. They have a steak dinner bet on next year's race, with Pat betting on the Red Sox and Tom on the Yankees. Both are convinced they will win. What probability assessment technique is being used by the two friends? A) Subjective probability B) Classical probability C) Relative frequency probability D) Independent probability Answer: A Diff: 1 Keywords: probability Section: 4-1 The Basics of Probability Outcome: 1

79) Students who live on campus and purchase a meal plan are randomly assigned to one of three dining halls: the Commons, Northeast, and Frazier. What is the probability that the next student to purchase a meal plan will be assigned to the Commons? A) 0.66 B) 0.5 C) 0.25 D) 0.33 Answer: D Diff: 1 Keywords: probability Section: 4-2 True Rules of Probability Outcome: 1 80) The results of a census of 2,500 employees of a mid-sized company with 401(k) retirement accounts are as follows: Account Balance (to nearest $) $50,000 $50,000-$99,999 $100,000-$150,000 > $150,000

Male 635 185 515 155

Female 495 210 260 45

81) The results of a census of 2,500 employees of a mid-sized company with 401(k) retirement accounts are as follows: Account Balance (to nearest $) $50,000 $50,000-$99,999 $100,000-$150,000 > $150,000

Male 635 185 515 155

Female 495 210 260 45

Suppose researchers are going to sample employees from the company for further study. Based on the relative frequency assessment method, what is the probability that a randomly selected employee will have a 401(k) account balance less than $150,000? A) 0.9200 B) 0.0800 C) 0.6160 D) 0.4040 Answer: A Diff: 2 Keywords: probability Section: 4-2 True Rules of Probability Outcome: 1 82) The results of a census of 2,500 employees of a mid-sized company with 401(k) retirement accounts are as follows: Account Balance (to nearest $) $50,000 $50,000-$99,999 $100,000-$150,000 > $150,000

Male 635 185 515 155

Female 495 210 260 45

Suppose researchers are going to sample employees from the company for further study. Compute the probability that a randomly selected employee will be a female with an account balance between $100,000 and $150,000. A) 0.1580 B) 0.1040 C) 0.6160 D) 0.4040 Answer: B Diff: 2 Keywords: probability Section: 4-2 True Rules of Probability Outcome: 1 4-26 Copyright © 2018 Pearson Education, Inc.

83) Cross County Bicycles makes two mountain bike models that each come in three colors. The following table shows the production volumes for last week:

Based on the relative frequency assessment method, what is the probability that a manufactured item is brown? A) 0.2088 B) 0.3819 C) 0.3157 D) 0.1324 Answer: C Diff: 2 Keywords: probability Section: 4-2 True Rules of Probability Outcome: 1 84) Cross County Bicycles makes two mountain bike models that each come in three colors. The following table shows the production volumes for last week:

What is the probability that the product manufactured is a YZ-99? A) 0.2088 B) 0.3819 C) 0.3157 D) 0.1324 Answer: B Diff: 2 Keywords: probability Section: 4-2 True Rules of Probability Outcome: 1

85) Cross County Bicycles makes two mountain bike models that each come in three colors. The following table shows the production volumes for last week:

What is the joint probability that a product manufactured is a YZ-99 and brown? A) 0.2088 B) 0.3819 C) 0.3157 D) 0.1324 Answer: A Diff: 2 Keywords: probability Section: 4-2 True Rules of Probability Outcome: 1 86) Based on weather data collected in Racine, Wisconsin, on Christmas Day, the weather had the following distribution: Event Relative Frequency Clear & dry 0.20 Cloudy & dry 0.30 Rain 0.40 Snow 0.10 /Thompson_sn3t_WordExports/Thompson_sn3t_WordExports Based on these data, what is the probability that next Christmas will be dry? A) 0.45 B) 0.50 C) 0.60 D) 0.70 Answer: B Diff: 2 Keywords: probability Section: 4-2 True Rules of Probability Outcome: 1

87) Based on weather data collected in Racine, Wisconsin, on Christmas Day, the weather had the following distribution: Event Clear & dry Cloudy & dry Rain Snow

Relative Frequency 0.20 0.30 0.40 0.10

Based on the data, what is the probability that next Christmas will be rainy or cloudy and dry? A) 0.45 B) 0.50 C) 0.60 D) 0.70 Answer: D Diff: 2 Keywords: probability Section: 4-2 True Rules of Probability Outcome: 1 88) Based on weather data collected in Racine, Wisconsin, on Christmas Day, the weather had the following distribution: Event Clear & dry Cloudy & dry Rain Snow

Relative Frequency 0.20 0.30 0.40 0.10

Supposing next Christmas is dry, determine the probability that it will also be cloudy. A) 0.45 B) 0.50 C) 0.60 D) 0.70 Answer: C Diff: 2 Keywords: probability Section: 4-2 True Rules of Probability Outcome: 1

89) The Jack In The Box franchise in Bangor, Maine, has determined that the chance a customer will order a soft drink is 0.90. The probability that a customer will order a hamburger is 0.60. The probability that a customer will order french fries is 0.50. If a customer places an order, what is the probability that the order will include a soft drink and no fries if these two events are independent? A) 0.45 B) 0.50 C) 0.65 D) 0.70 Answer: A Diff: 3 Keywords: probability Section: 4-2 True Rules of Probability Outcome: 1 90) The Jack In The Box franchise in Bangor, Maine, has determined that the chance a customer will order a soft drink is 0.90. The probability that a customer will order a hamburger is 0.60. The probability that a customer will order french fries is 0.50. The restaurant has also determined that if a customer orders a hamburger, the probability the customer will also order fries is 0.80. Determine the probability that the order will include a hamburger and fries. A) 0.45 B) 0.58 C) 0.68 D) 0.48 Answer: D Diff: 3 Keywords: probability Section: 4-2 True Rules of Probability Outcome: 1 91) Ponderosa Paint and Glass carries three brands of paint. A customer wants to buy another gallon of paint to match paint she purchased at the store previously. She can't recall the brand name and does not wish to return home to find the old can of paint. So she selects two of the three brands of paint at random and buys them. What is the probability that she matched the paint brand? A) 3/2 B) 2/3 C) 1/9 D) 3/4 Answer: B Diff: 3 Keywords: probability Section: 4-2 True Rules of Probability Outcome: 1

92) Ponderosa Paint and Glass carries three brands of paint. A customer wants to buy another gallon of paint to match paint she purchased at the store previously. She can't recall the brand name and does not wish to return home to find the old can of paint. So she selects two of the three brands of paint at random and buys them. Her husband also goes to the paint store and fails to remember what brand to buy. So he also purchases two of the three brands of paint at random. Determine the probability that both the woman and her husband fail to get the correct brand of paint. (Hint: Are the husband's selections independent of his wife's selections?) A) 3/2 B) 2/3 C) 1/9 D) 3/4 Answer: C Diff: 3 Keywords: probability Section: 4-2 True Rules of Probability Outcome: 1 93) The college basketball team at West Texas State University has 10 players; 5 are seniors, 2 are juniors, and 3 are sophomores. Two players are randomly selected to serve as captains for the next game. What is the probability that both players selected are seniors? A) 0.22 B) 0.33 C) 0.50 D) 0.66 Answer: A Diff: 3 Keywords: probability Section: 4-2 True Rules of Probability Outcome: 1

94) Micron Technology has sales offices located in four cities: Dallas, Seattle, Boston, and Los Angeles. An analysis of the company's accounts receivables reveals the number of overdue invoices by days, as shown here. Days Overdue Under 30 days 30-60 days 61-90 days Over 90 days

Dallas 137 85 33 18

Seattle 122 46 27 32

Boston 198 76 55 45

Los Angeles 287 109 48 66

Assume the invoices are stored and managed from a central database. What is the probability that a randomly selected invoice from the database is from the Boston sales office? A) 0.2702 B) 0.0231 C) 0.3461 D) 0.7765 Answer: A Diff: 2 Keywords: probability Section: 4-2 True Rules of Probability Outcome: 1 95) Micron Technology has sales offices located in four cities: Dallas, Seattle, Boston, and Los Angeles. An analysis of the company's accounts receivables reveals the number of overdue invoices by days, as shown here. Days Overdue Under 30 days 30-60 days 61-90 days Over 90 days

Dallas 137 85 33 18

Seattle 122 46 27 32

Boston 198 76 55 45

Los Angeles 287 109 48 66

Assume the invoices are stored and managed from a central database. What is the probability that a randomly selected invoice from the database is between 30 and 90 days overdue? A) 0.2702 B) 0.0231 C) 0.3461 D) 0.7765 Answer: C Diff: 2 Keywords: probability Section: 4-2 True Rules of Probability Outcome: 1

96) Micron Technology has sales offices located in four cities: Dallas, Seattle, Boston, and Los Angeles. An analysis of the company's accounts receivables reveals the number of overdue invoices by days, as shown here. Days Overdue Dallas Seattle Boston Los Angeles Under 30 days 137 122 198 287 30-60 days 85 46 76 109 61-90 days 33 27 55 48 Over 90 days 18 32 45 66 Assume the invoices are stored and managed from a central database. What is the probability that a randomly selected invoice from the database is over 90 days old and from the Seattle office? A) 0.2702 B) 0.0231 C) 0.3461 D) 0.7765 Answer: B Diff: 2 Keywords: probability Section: 4-2 True Rules of Probability Outcome: 1 97) Micron Technology has sales offices located in four cities: Dallas, Seattle, Boston, and Los Angeles. An analysis of the company's accounts receivables reveals the number of overdue invoices by days, as shown here. Days Overdue Dallas Seattle Boston Los Angeles Under 30 days 137 122 198 287 30-60 days 85 46 76 109 61-90 days 33 27 55 48 Over 90 days 18 32 45 66 Assume the invoices are stored and managed from a central database. If a randomly selected invoice is from the Los Angeles office, what is the probability that it is 60 or fewer days overdue? A) 0.2702 B) 0.0231 C) 0.3461 D) 0.7765 Answer: D Diff: 2 Keywords: probability Section: 4-2 True Rules of Probability Outcome: 1

98) Three events occur with probabilities P(E1) = 0.35, P(E2) = 0.15, P(E3) = 0.40. If the event B occurs, the probability becomes P(E1|B) = 0.25, P(B) = 0.30. Calculate P(E1 and B). A) 0.575 B) 0.075 C) 0.021 D) 0.475 Answer: B Diff: 3 Keywords: probability Section: 4-2 True Rules of Probability Outcome: 1 99) Three events occur with probabilities P(E1) = 0.35, P(E2) = 0.15, P(E3) = 0.40. If the event B occurs, the probability becomes P(E1|B) = 0.25, P(B) = 0.30. Compute P(E1 or B). A) 0.575 B) 0.075 C) 0.021 D) 0.475 Answer: A Diff: 2 Keywords: probability Section: 4-2 True Rules of Probability Outcome: 1 100) Three events occur with probabilities P(E1) = 0.35, P(E2) = 0.15, P(E3) = 0.40. If the event B occurs, the probability becomes P(E1|B) = 0.25, P(B) = 0.30. Assume that E1, E2, and E3 are independent events. Calculate P(E1 and E2 and E3). A) 0.575 B) 0.075 C) 0.021 D) 0.475 Answer: C Diff: 3 Keywords: probability Section: 4-2 True Rules of Probability Outcome: 1

101) The URS construction company has submitted two bids, one to build a large hotel in London and the other to build a commercial office building in New York City. The company believes it has a 40% chance of winning the hotel bid and a 25% chance of winning the office building bid. The company also believes that winning the hotel bid is independent of winning the office building bid. What is the probability the company will win both contracts? A) 0.55 B) 0.44 C) 0.10 D) 0.75 Answer: C Diff: 2 Keywords: probability Section: 4-2 True Rules of Probability Outcome: 1 102) The URS construction company has submitted two bids, one to build a large hotel in London and the other to build a commercial office building in New York City. The company believes it has a 40% chance of winning the hotel bid and a 25% chance of winning the office building bid. The company also believes that winning the hotel bid is independent of winning the office building bid. What is the probability the company will win at least one contract? A) 0.55 B) 0.45 C) 0.10 D) 0.75 Answer: A Diff: 2 Keywords: probability Section: 4-2 True Rules of Probability Outcome: 1 103) The URS construction company has submitted two bids, one to build a large hotel in London and the other to build a commercial office building in New York City. The company believes it has a 40% chance of winning the hotel bid and a 25% chance of winning the office building bid. The company also believes that winning the hotel bid is independent of winning the office building bid. What is the probability the company will lose both contracts? A) 0.55 B) 0.45 C) 0.10 D) 0.75 Answer: B Diff: 2 Keywords: probability Section: 4-2 True Rules of Probability Outcome: 1

104) Suppose a quality manager for Dell Computers has collected the following data on the quality status of disk drives by supplier. She inspected a total of 700 disk drives.

Based on these inspection data, what is the probability of randomly selecting a disk drive from company B? A) 0.07 B) 0.28 C) 0.021 D) 0.76 Answer: B Diff: 2 Keywords: probability Section: 4-2 True Rules of Probability Outcome: 1 105) Suppose a quality manager for Dell Computers has collected the following data on the quality status of disk drives by supplier. She inspected a total of 700 disk drives.

What is the probability of a defective disk drive being received by the computer company? A) 0.07 B) 0.28 C) 0.021 D) 0.76 Answer: A Diff: 2 Keywords: probability Section: 4-2 True Rules of Probability Outcome: 1

106) Suppose a quality manager for Dell Computers has collected the following data on the quality status of disk drives by supplier. She inspected a total of 700 disk drives.

What is the probability of a defect given that company B supplied the disk drive? A) 0.077 B) 0.28 C) 0.021 D) 0.76 Answer: D Diff: 2 Keywords: probability Section: 4-2 True Rules of Probability Outcome: 1 107) Men have a reputation for not wanting to ask for directions. A Harris study conducted for Lincoln Mercury indicated that 42% of men and 61% of women would stop and ask for directions. The U.S. Census Bureau's 2012 population estimate was that for individuals 18 or over, 48.2% were men and 51.8% were women. This exercise addresses this age group. A randomly chosen driver gets lost on a road trip. Determine the probability that the driver is a woman and stops to ask for directions. A) 0.518 B) 0.420 C) 0.316 D) 0.390 Answer: C Diff: 2 Keywords: probability Section: 4-2 True Rules of Probability Outcome: 1

108) Men have a reputation for not wanting to ask for directions. A Harris study conducted for Lincoln Mercury indicated that 42% of men and 61% of women would stop and ask for directions. The U.S. Census Bureau's 2012 population estimate was that for individuals 18 or over, 48.2% were men and 51.8% were women. This exercise addresses this age group. Calculate the probability that the driver stops to ask for directions. A) 0.518 B) 0.420 C) 0.316 D) 0.390 Answer: A Diff: 2 Keywords: probability Section: 4-2 True Rules of Probability Outcome: 1 109) Men have a reputation for not wanting to ask for directions. A Harris study conducted for Lincoln Mercury indicated that 42% of men and 61% of women would stop and ask for directions. The U.S. Census Bureau's 2012 population estimate was that for individuals 18 or over, 48.2% were men and 51.8% were women. This exercise addresses this age group. Given that a driver stops to ask for directions, determine the probability that the driver was a man. A) 0.518 B) 0.420 C) 0.316 D) 0.390 Answer: D Diff: 2 Keywords: probability Section: 4-2 True Rules of Probability Outcome: 1 110) A local FedEx/Kinkos has three black-and-white copy machines and two color copiers. Based on historical data, the chances that each black-and-white copier will be down for repairs is 0.10. The color copiers are more of a problem and are down 20% of the time each. Based on this information, what is the probability that if a customer needs a color copy, both color machines will be down for repairs? A) 0.04 B) 0.96 C) 0.47 D) 0.42 Answer: A Diff: 2 Keywords: probability Section: 4-2 True Rules of Probability Outcome: 1

111) A local FedEx/Kinkos has three black-and-white copy machines and two color copiers. Based on historical data, the chances that each black-and-white copier will be down for repairs is 0.10. The color copiers are more of a problem and are down 20% of the time each. If a customer wants both a color copy and a black-and-white copy, what is the probability that the necessary machines will be available? (Assume that the color copier can also be used to make a blackand-white copy if needed.) A) 0.04 B) 0.96 C) 0.47 D) 0.42 Answer: B Diff: 2 Keywords: probability Section: 4-2 True Rules of Probability Outcome: 1 112) Hubble Construction Company has submitted a bid on a state government project that is to be funded by the federal government's stimulus money in Arizona. The price of the bid was predetermined in the bid specifications. The contract is to be awarded on the basis of a blind drawing from those who have bid. Five other companies have also submitted bids. What is the probability of the Hubble Construction Company winning the bid? A) 0.2778 B) 0.1667 C) 0.6944 D) 0.0278 Answer: B Diff: 2 Keywords: probability Section: 4-2 True Rules of Probability Outcome: 1 113) Hubble Construction Company has submitted a bid on a state government project that is to be funded by the federal government's stimulus money in Arizona. The price of the bid was predetermined in the bid specifications. The contract is to be awarded on the basis of a blind drawing from those who have bid. Five other companies have also submitted bids. Suppose that there are two contracts to be awarded by a blind draw. What is the probability of Hubble winning both contracts? Assume sampling with replacement. A) 0.2778 B) 0.1667 C) 0.6944 D) 0.0278 Answer: D Diff: 2 Keywords: probability Section: 4-2 True Rules of Probability Outcome: 1

114) Drake Marketing and Promotions has randomly surveyed 200 men who watch professional sports. The men were separated according to their educational level (college degree or not) and whether they preferred the NBA or the National Football League (NFL). The results of the survey are shown: Sports Preference NBA NFL

College Degree 40 10

No College Degree 55 95

What is the probability that a randomly selected survey participant prefers the NFL? A) 0.5250 B) 0.2000 C) 0.6050 D) 0.5880 Answer: A Diff: 1 Keywords: probability Section: 4-2 True Rules of Probability Outcome: 1 115) Drake Marketing and Promotions has randomly surveyed 200 men who watch professional sports. The men were separated according to their educational level (college degree or not) and whether they preferred the NBA or the National Football League (NFL). The results of the survey are shown: Sports Preference NBA NFL

College Degree 40 10

No College Degree 55 95

What is the probability that a randomly selected survey participant has a college degree and prefers the NBA? A) 0.5250 B) 0.2000 C) 0.6050 D) 0.5880 Answer: B Diff: 1 Keywords: probability Section: 4-2 True Rules of Probability Outcome: 1

116) Drake Marketing and Promotions has randomly surveyed 200 men who watch professional sports. The men were separated according to their educational level (college degree or not) and whether they preferred the NBA or the National Football League (NFL). The results of the survey are shown: Sports Preference NBA NFL

College Degree 40 10

No College Degree 55 95

Suppose a survey participant is randomly selected and you are told that he has a college degree. What is the probability that this man prefers the NFL? A) 0.5250 B) 0.2000 C) 0.6050 D) 0.5880 Answer: B Diff: 1 Keywords: probability Section: 4-2 True Rules of Probability Outcome: 1 117) Until the summer of 2006, the real estate market in Fresno, California, had been booming, with prices skyrocketing. Recently, a study showed the sales patterns in Fresno for single-family homes. One chart presented in the commission's report is reproduced here. It shows the number of homes sold by price range and number of days the home was on the market.

Using the relative frequency approach to probability assessment, what is the probability that a house will be on the market more than 7 days? A) 0.31 B) 0.099 C) 0.58 D) 0.48 Answer: C Diff: 1 Keywords: probability Section: 4-2 True Rules of Probability Outcome: 1

118) Vegetables from the summer harvest are currently being processed at Skone and Conners Foods, Inc. The manager has found a case of cans that has not been properly sealed. There are three lines that processed cans of this type, and the manager wants to know which line is most likely to be responsible for this mistake. Provide the manager this information. Line 1 2 3

Contribution to Total 0.40 0.35 0.25

Proportion Defective 0.05 0.10 0.07

A) Line 1 B) Line 2 C) Line 3 D) Cannot be determined from this information Answer: B Diff: 1 Keywords: probability Section: 4-2 True Rules of Probability Outcome: 1 119) Parts and Materials for the skis made by the Downhill Adventures Company are supplied by two suppliers. Supplier A's materials make up 30% of what is used, with supplier B providing the rest. Past records indicate that 15% of supplier A's materials are defective and 10% of B's are defective. Since it is impossible to tell which supplier the materials came from once they are in inventory, the manager wants to know which supplier most likely supplied the defective materials the foreman has brought to his attention. Provide the manager this information. A) Supplier A B) Supplier B C) Both are equally likely D) Cannot be determined from this information Answer: B Diff: 2 Keywords: probability Section: 4-2 True Rules of Probability Outcome: 1

120) A distributor of outdoor yard lights has four suppliers. This past season she purchased 40% of the lights from Franklin Lighting, 30% from Wilson & Sons, 20% from Evergreen Supply, and the rest from A. L. Scott. In prior years, 3% of Franklin's lights were defective, 6% of the Wilson lights were defective, 2% of Evergreen's were defective, and 8% of the Scott lights were defective. When the lights arrive at the distributor, she puts them in inventory without identifying the supplier. Suppose that a defective light string has been pulled from inventory; what is the probability that it was supplied by Franklin Lighting? A) 0.33 B) 0.45 C) 0.18 D) 0.29 Answer: D Diff: 3 Keywords: probability Section: 4-2 True Rules of Probability Outcome: 1 121) List three methods of assessing probabilities and indicate which is least likely to be used in business decision making. Answer: The three methods of probability assessment are classical assessment, relative frequency of occurrence, and subjective assessment. The classical method has very little application for business decision making since rarely are the elementary events equally likely. Diff: 2 Keywords: classical, relative frequency, subjective, probability assessment Section: 4-1 The Basics of Probability Outcome: 1 122) The accountant for a large U.S. company is interested in finding the probability that an account will have an incorrect balance due to being overstated or being understated. To find this probability, which probability rule is she likely to use? Answer: The key word here is or, which, in dealing with probability, always means addition. Thus, the addition rule will need to be used. Diff: 2 Keywords: probability, addition rule Section: 4-2 True Rules of Probability Outcome: 2

123) Assume that a standard deck of 52 playing cards is randomly shuffled and the first 2 cards are dealt to you. What is the probability that you have a blackjack? A blackjack is where one card is an ace and the other card is worth 10 points. The 10-point cards are kings, queens, jacks and 10's. Answer: It doesn't matter which card was first so there are two ways to have a blackjack. One is to get the ace first and the 10-point card second. The other way is to get the cards in the reverse order. There are 4 aces and 16 of the 10-point cards. P( Ace on 1st card) = 4/52 P(10 on 2nd card given that 1st card was Ace) = 16/51 So P(black where Ace is 1st) =(4/52)(16/51) = 0.0241 P(10 on 1st card) = 16/52 P(Ace on 2nd card given the first card was a 10) = 4/51 So P(blackjack where Ace is 2nd) = (16/52)(4/51) = 0.0241 The two methods of getting a blackjack are mutually exclusive so these probabilities can be added to find the total probability of a blackjack. P(blackjack) = 2 (0.0241) = 0.0482 Diff: 2 Keywords: conditional probability, mutually exclusive Section: 4-2 True Rules of Probability Outcome: 2 124) Explain what is meant by the term mutually exclusive events. Cite an example. Answer: Mutually exclusive events exist if the occurrence of one event precludes the occurrence of another event. For example, if a customer is selected at random for a store's master file, if the customer is a female, the same customer cannot also be a male. Thus the events female and male are mutually exclusive. Diff: 1 Keywords: mutually exclusive, event Section: 4-2 True Rules of Probability Outcome: 2 125) Explain why it is possible for two managers to assess different values for the probability that a supplier will fail to deliver a shipment on time. Answer: Since it is likely that the managers are using subjective probability assessment, their assessments reflect their state of mind about the shipment. They are basing this on any information that they have and their methods of processing the information may be different. Neither assessment can be considered wrong—they simply reflect the thinking of each decision maker. All that is required is that the decision makers process the information in a rational manner. Diff: 3 Keywords: probability, subjective assessment Section: 4-1 The Basics of Probability Outcome: 1 4-44 Copyright © 2018 Pearson Education, Inc.

Business Statistics, 9e (Groebner/Shannon/Fry) Chapter 5 Discrete Probability Distributions 1) A random variable is generated when a variable's value is determined by using classical probability. Answer: FALSE Diff: 1 Keywords: random, variable, probability Section: 5-1 Introduction to Discrete Probability Distributions Outcome: 1 2) The only two types of random variables are discrete and continuous random variables. Answer: TRUE Diff: 1 Keywords: random, discrete, continuous Section: 5-1 Introduction to Discrete Probability Distributions Outcome: 1 3) If a random variable is discrete, it means that the outcome for the random variable can take on only one of two possible values. Answer: FALSE Diff: 1 Keywords: random, discrete, variable Section: 5-1 Introduction to Discrete Probability Distributions Outcome: 1 4) The random variable, number of customers entering a store between 9 AM and noon, is an example of a discrete random variable. Answer: TRUE Diff: 1 Keywords: discrete, random, variable Section: 5-1 Introduction to Discrete Probability Distributions Outcome: 1 5) When a single value is randomly chosen from a discrete distribution, the different possible values are mutually exclusive. Answer: TRUE Diff: 2 Keywords: discrete, random, variable Section: 5-1 Introduction to Discrete Probability Distributions Outcome: 1 6) The graph of a discrete random variable looks like a histogram where the probability of each possible outcome is represented by a bar. Answer: TRUE Diff: 1 Keywords: discrete, random, variable Section: 5-1 Introduction to Discrete Probability Distributions Outcome: 1 5-1 Copyright © 2018 Pearson Education, Inc.

7) When a market research manager records the number of potential customers who were surveyed indicating that they like the product design, the random variable, number who like the design, is a discrete random variable. Answer: TRUE Diff: 1 Keywords: discrete, random, variable Section: 5-1 Introduction to Discrete Probability Distributions Outcome: 1 8) The time required to assemble two components into a finished part is recorded for each employee at the plant. The resulting random variable is an example of a continuous random variable. Answer: TRUE Diff: 1 Keywords: continuous, random, variable Section: 5-1 Introduction to Discrete Probability Distributions Outcome: 1 9) The Colbert Real Estate Agency has determined the number of home showings given by its agents is the same each day of the week. Then the variable, number of showings, is a continuous distribution. Answer: FALSE Diff: 1 Keywords: discrete, continuous, random, variable Section: 5-1 Introduction to Discrete Probability Distributions Outcome: 1 10) The Argon Construction Company has the opportunity to enter into a contract to remodel a building. The following table shows the probability distribution for the profit that could occur if it takes the contract: Profit $40,000 $60,000 $80,000 $100,000

Probability 0.25 0.30 0.30 0.15

Based on this information, the probability of profit being at least $80,000 is 0.85. Answer: TRUE Diff: 1 Keywords: discrete, random, variable Section: 5-1 Introduction to Discrete Probability Distributions Outcome: 1

11) The Argon Construction Company has the opportunity to enter into a contract to remodel a building. The following table shows the probability distribution for the profit that could occur if it takes the contract: Profit $40,000 $60,000 $80,000 $100,000

Probability 0.25 0.30 0.30 0.15

Based on this information, the expected profit for the company if it takes the contract is $67,000. Answer: TRUE Diff: 2 Keywords: mean, discrete, random, variable Section: 5-1 Introduction to Discrete Probability Distributions Outcome: 1 12) The Argon Construction Company has the opportunity to enter into a contract to remodel a building. The following table shows the probability distribution for the profit that could occur if it takes the contract: Profit $40,000 $60,000 $80,000 $100,000

Probability 0.25 0.30 0.30 0.15

Based on this information, the probability that the profit is $80,000 or less is 0.55. Answer: FALSE Diff: 2 Keywords: mean, discrete random, variable Section: 5-1 Introduction to Discrete Probability Distributions Outcome: 1 13) A probability distribution with an expected value greater than the expected value of a second probability distribution will also have a higher standard deviation. Answer: FALSE Diff: 2 Keywords: discrete, random, standard deviation, expected value Section: 5-1 Introduction to Discrete Probability Distributions Outcome: 1

14) The Argon Construction Company has the opportunity to enter into a contract to remodel a building. The following table shows the probability distribution for the profit that could occur if it takes the contract: Profit $40,000 $60,000 $80,000 $100,000

Probability 0.25 0.30 0.30 0.15

Based on this information, the probability of profit being < $80,000 is 0.55. Answer: TRUE Diff: 1 Keywords: discrete, random Section: 5-1 Introduction to Discrete Probability Distributions Outcome: 1 15) The Argon Construction Company has the opportunity to enter into a contract to remodel a building. The following table shows the probability distribution for the profit that could occur if it takes the contract: Profit $40,000 $60,000 $80,000 $100,000

Probability 0.25 0.30 0.30 0.15

Based on this information, the profit standard deviation for the company if it takes the contract is < $20,000. Answer: TRUE Diff: 2 Keywords: discrete, random, standard deviation Section: 5-1 Introduction to Discrete Probability Distributions Outcome: 1

16) The number of no-shows each day for dinner reservations at the Cottonwood Grille is a discrete random variable with the following probability distribution: No-shows 0 1 2 3 4

Probability 0.30 0.20 0.20 0.15 0.15

Based on this information, the expected number of no-shows is 1.65 customers. Answer: TRUE Diff: 2 Keywords: expected value, discrete Section: 5-1 Introduction to Discrete Probability Distributions Outcome: 1 17) The number of no-shows for dinner reservations at the Cottonwood Grille is a discrete random variable with the following probability distribution: No-shows 0 1 2 3 4

Probability 0.30 0.20 0.20 0.15 0.15

Based on this information, the most likely number of no-shows on any given day is 0 customers. Answer: TRUE Diff: 2 Keywords: discrete distribution Section: 5-1 Introduction to Discrete Probability Distributions Outcome: 1

18) The number of no-shows each day for dinner reservations at the Cottonwood Grille is a discrete random variable with the following probability distribution: No-shows 0 1 2 3 4

Probability 0.30 0.20 0.20 0.15 0.15

Based on this information, the standard deviation for the number of no-shows is about 0.36 customers. Answer: FALSE Diff: 2 Keywords: standard deviation, discrete Section: 5-1 Introduction to Discrete Probability Distributions Outcome: 1 19) The Ski Patrol at Criner Mountain Ski Resort has determined the following probability distribution for the number of skiers that are injured each weekend: Injured Skiers 0 1 2 3 4

Probability 0.05 0.15 0.40 0.30 0.10

Based on this information, the expected number of injuries per weekend is 2.25. Answer: TRUE Diff: 2 Keywords: expected value, discrete Section: 5-1 Introduction to Discrete Probability Distributions Outcome: 1

20) The Ski Patrol at Criner Mountain Ski Resort has determined the following probability distribution for the number of skiers that are injured each weekend: Injured Skiers 0 1 2 3 4

Probability 0.05 0.15 0.40 0.30 0.10

Based on this information, the standard deviation for the number of injuries per weekend is 2.25. Answer: FALSE Diff: 2 Keywords: standard deviation, expected value, discrete Section: 5-1 Introduction to Discrete Probability Distributions Outcome: 1 21) Holmstead Company owns two small engine repair stores. The expected value of the number of complaints received per month at store 1 is 4.5 complaints. Further, the expected number of complaints per month for store 1 and store 2 combined is 13.6. This means the expected number of complaints per month at store 2 must be 9.1 complaints. Answer: TRUE Diff: 2 Keywords: expected value, sum, discrete Section: 5-1 Introduction to Discrete Probability Distributions Outcome: 1 22) In a Florida town, the probability distribution for the number of legitimate emergency calls per day for the Fire Department is given as follows. Also shown is the probability distribution for the number of false alarms:

Given this information, the expected number of total calls to the fire department is 1.60 calls. Answer: TRUE Diff: 3 Keywords: expected value, discrete Section: 5-1 Introduction to Discrete Probability Distributions Outcome: 1

23) The Ace Construction Company has entered into a contract to widen a street in Boston. The possible payoffs for this project have been determined by management. The probabilities for these payoffs could be determined using a binomial distribution. Answer: FALSE Diff: 1 Keywords: probability, distribution Section: 5-2 The Binomial Probability Distribution Outcome: 2 24) A construction company has found it has a probability of 0.10 of winning each time it bids on a project. The probability of winning a given number of projects out of 12 bids could be determined with a binomial distribution. Answer: TRUE Diff: 1 Keywords: binomial, distribution Section: 5-2 The Binomial Probability Distribution Outcome: 2 25) Six managers at a company all enjoy golf. Each Saturday, four of the six get together for 18 holes of golf. They have decided to set up a schedule so that the same foursome does not play twice before all possible foursomes have played. The number of weekends that will pass before the same group would play twice is 15. Answer: TRUE Diff: 2 Keywords: combination Section: 5-2 The Binomial Probability Distribution Outcome: 2 26) Bill Price is a sales rep in northern California representing a line of athletic socks. Each day, he makes 10 sales calls. The chance of making a sale on each call is thought to be 0.30. The probability that he will make exactly two sales is approximately 0.2335. Answer: TRUE Diff: 2 Keywords: binomial formula, probability Section: 5-2 The Binomial Probability Distribution Outcome: 2 27) One of the characteristics of the binomial distribution is that the probability of success for each trial depends on whether the previous trial was a success or not. Answer: FALSE Diff: 2 Keywords: binomial distribution Section: 5-2 The Binomial Probability Distribution Outcome: 2

28) Each week American Stores receives a shipment from a supplier. The contract specifies that the maximum allowable percent defective is 5 percent. When the shipment arrives, a sample of 20 parts is randomly selected. If 2 or more of the sampled parts are defective, the shipment is rejected and returned to the supplier. Assume that a shipment arrives that actually has 4 percent defective parts and the distribution of defective parts is described by a binomial distribution. The probability that the shipment is rejected is approximately 0.19. Answer: TRUE Diff: 2 Keywords: binomial formula, probability Section: 5-2 The Binomial Probability Distribution Outcome: 2 29) Each week American Stores receives a shipment from a supplier. The contract specifies that the maximum allowable percent defective is 5 percent. When the shipment arrives, a sample of 20 parts is randomly selected. If 2 or more of the sampled parts are defective, the shipment is rejected and returned to the supplier. Assume that a shipment arrives that actually has 4 percent defective parts and the distribution of defective parts is described by a binomial distribution. The probability that the shipment is accepted is approximately 0.81 Answer: TRUE Diff: 2 Keywords: binomial, probability Section: 5-2 The Binomial Probability Distribution Outcome: 2 30) A pizza restaurant uses 7 different toppings on its pizzas. At lunch time it has a pizza buffet and makes pizzas with 2 toppings. If it wants to serve every possible combination of 2 toppings, it would need to make 14 different pizzas. Answer: FALSE Diff: 2 Keywords: combinations Section: 5-2 The Binomial Probability Distribution Outcome: 2 31) Ace Computer Manufacturer buys disk drives in lots of 5,000 units from a supplier in California. The contract calls for, at most, 3 percent of the disk drives to be defective. When a shipment arrives, a sample of n = 15 parts is selected. If zero defects are found in this sample, the shipment is accepted. If 3 or more defects are found, the shipment is rejected and sent back to the supplier. If the number of defects found is 1 or 2, a second sample of 15 parts is selected. If this sample yields 1 or fewer defects, the shipment is accepted; otherwise the shipment is rejected. Based on a binomial distribution, the probability that Ace will reject a shipment of parts that meets the contract requirements is approximately 0.0355. Answer: TRUE Diff: 3 Keywords: binomial, probability Section: 5-2 The Binomial Probability Distribution Outcome: 2

32) When using the binomial distribution, the maximum possible number of success is the number of trials. Answer: TRUE Diff: 1 Keywords: binomial, probability Section: 5-2 The Binomial Probability Distribution Outcome: 2 33) The Nationwide Motel Company has determined that 70 percent of all calls for motel reservations request nonsmoking rooms. Recently, the customer service manager for the company randomly selected 25 calls. Assuming that the distribution of calls requesting nonsmoking rooms is described by a binomial distribution, the expected number of requests for nonsmoking rooms is 14. Answer: FALSE Diff: 1 Keywords: binomial, expected value Section: 5-2 The Binomial Probability Distribution Outcome: 2 34) The Nationwide Motel Company has determined that 70 percent of all calls for motel reservations request nonsmoking rooms. Recently, the customer service manager for the company randomly selected 25 calls. Assuming that the distribution of calls requesting nonsmoking rooms is described by a binomial distribution, the standard deviation of requests for nonsmoking rooms is 5.25 customers. Answer: FALSE Diff: 2 Keywords: binomial, standard deviation Section: 5-2 The Binomial Probability Distribution Outcome: 2 35) The Nationwide Motel Company has determined that 70 percent of all calls for motel reservations request nonsmoking rooms. Recently, the customer service manager for the company randomly selected 25 calls. Assuming that the distribution of calls requesting nonsmoking rooms is described by a binomial distribution, the probability that more than 20 customers in the sample will request nonsmoking rooms is approximately 0.09. Answer: TRUE Diff: 2 Keywords: binomial, probability Section: 5-2 The Binomial Probability Distribution Outcome: 2

36) The Nationwide Motel Company has determined that 70 percent of all calls for motel reservations request nonsmoking rooms. Recently, the customer service manager for the company randomly selected 25 calls. Assuming that the distribution of calls requesting nonsmoking rooms is described by a binomial distribution, the probability that fewer than 5 customers will request smoking rooms is approximately 0.09. Answer: TRUE Diff: 2 Keywords: binomial, probability, complement Section: 5-2 The Binomial Probability Distribution Outcome: 2 37) The distribution for the number of emergency calls to a city's 911 emergency number in a one-hour time period is likely to be described by a binomial distribution. Answer: FALSE Diff: 2 Keywords: distribution, Poisson Section: 5-2 The Binomial Probability Distribution Outcome: 2 38) One difference between the binomial distribution and Poisson distribution is that the binomial's upper bound is the number of trials while the Poisson has no particular upper bound. Answer: TRUE Diff: 2 Keywords: binomial, Poisson Section: 5-2 The Binomial Probability Distribution Outcome: 2 39) The number of defects discovered in a random sample of 100 products produced at the Berdan Manufacturing Company is binomially distributed with p = .03. Based on this, the standard deviation of the number of defects per sample of size 100 is 2.91. Answer: FALSE Diff: 2 Keywords: binomial, standard deviation Section: 5-2 The Binomial Probability Distribution Outcome: 2 40) A direct marketing company believes that the probability of making a sale when a call is made to an individual's home is .02. The probability of making 2 or 3 sales in a sample of 20 calls is .0593. Answer: TRUE Diff: 2 Keywords: binomial, combination Section: 5-2 The Binomial Probability Distribution Outcome: 2

41) A company has 20 copy machines and every day there is a 5 percent chance for each machine that it will not be working that day. If the company wants to calculate the probability of, say, 2 machines not working, it should use the Poisson distribution. Answer: FALSE Diff: 2 Keywords: binomial, Poisson Section: 5-2 The Binomial Probability Distribution Outcome: 2 42) The Hawkins Company randomly samples 10 items from every large batch before the batch is packaged and shipped. According to the contract specifications, 5 percent of the items shipped can be defective. If the inspectors find 1 or fewer defects in the sample of 10, they ship the batch without further inspection. If they find 2 or more, the entire batch is inspected. Based on this sampling plan, the probability that a batch that meets the contract requirements will be shipped without further inspection is approximately .9139. Answer: TRUE Diff: 2 Keywords: binomial, combination Section: 5-2 The Binomial Probability Distribution Outcome: 2 43) The Hawkins Company randomly samples 10 items from every large batch before the batch is packaged and shipped. According to the contract specifications, 5 percent of the items shipped can be defective. If the inspectors find 1 or fewer defects in the sample of 10, they ship the batch without further inspection. If they find 2 or more, the entire batch is inspected. Based on this sampling plan, the probability that a batch that contains twice the amount of defects allowed by the contract requirements will be shipped without further inspection is approximately .3874. Answer: FALSE Diff: 3 Keywords: binomial, combination Section: 5-2 The Binomial Probability Distribution Outcome: 2 44) The Hawkins Company randomly samples 10 items from every large batch before the batch is packaged and shipped. According to the contract specifications, 5 percent of the items shipped can be defective. If the inspectors find 1 or fewer defects in the sample of 10, they ship the batch without further inspection. If they find 2 or more, the entire batch is inspected. Based on this sampling plan, the probability that a batch that meets the contract requirements will end up being 100 percent inspected is approximately .0746. Answer: FALSE Diff: 3 Keywords: binomial, combination Section: 5-2 The Binomial Probability Distribution Outcome: 2

45) The probability of the outcome changes from trial to trial in a binomial experiment. Answer: FALSE Diff: 2 Keywords: discrete, binomial Section: 5-2 The Binomial Probability Distribution Outcome: 2 46) The primary difference between the binomial distribution and the Poisson distribution is that the Poisson is used to describe a continuous random variable and the binomial is used for discrete random variables. Answer: FALSE Diff: 1 Keywords: binomial, Poisson, distribution Section: 5-3 Other Discrete Probability Distributions Outcome: 3 47) The number of calls to an Internet service provider during the hour between 6:00 and 7:00 p.m. is described by a Poisson distribution with mean equal to 15. Given this information, the expected number of calls in the first 30 minutes is 7.5 calls. Answer: TRUE Diff: 1 Keywords: Poisson, expected value Section: 5-3 Other Discrete Probability Distributions Outcome: 3 48) The number of calls to an Internet service provider during the hour between 6:00 and 7:00 p.m. is described by a Poisson distribution with mean equal to 15. Given this information, the standard deviation for the call distribution is about 3.87 calls. Answer: TRUE Diff: 2 Keywords: Poisson, standard deviation Section: 5-3 Other Discrete Probability Distributions Outcome: 3 49) The number of customers who arrive at a fast food business during a one-hour period is known to be Poisson distributed with a mean equal to 8.60. The probability that exactly 8 customers will arrive in a one-hour period is 0.1366. Answer: TRUE Diff: 1 Keywords: Poisson, probability Section: 5-3 Other Discrete Probability Distributions Outcome: 3

50) The number of customers who arrive at a fast food business during a one-hour period is known to be Poisson distributed with a mean equal to 8.60. The probability that more than 4 customers will arrive in a 30-minute period is 0.1933. Answer: FALSE Diff: 2 Keywords: Poisson, probability Section: 5-3 Other Discrete Probability Distributions Outcome: 3 51) The number of customers who arrive at a fast food business during a one-hour period is known to be Poisson distributed with a mean equal to 8.60. The probability that between 2 and 3 customers inclusively will arrive in one hour is 0.0263. Answer: TRUE Diff: 2 Keywords: Poisson, probability Section: 5-3 Other Discrete Probability Distributions Outcome: 3 52) The Brockingham Carpet Company prides itself on high quality carpets. At the end of each day, the company quality managers select 3 square yards for inspection. The quality standard requires an average of no more than 2.3 defects per square yard. The expected number of defects that the inspector will find during the inspection is 6.9. Answer: TRUE Diff: 2 Keywords: Poisson, expected value Section: 5-3 Other Discrete Probability Distributions Outcome: 3 53) The Brockingham Carpet Company prides itself on high quality carpets. At the end of each day, the company quality managers select 3 square yards for inspection. The quality standard requires an average of no more than 2.3 defects per square yard. Last night, the inspector found 8 defects in the sample of 3 square yards. The chance of finding 8 or more defects in the sample is 0.9975. Answer: FALSE Diff: 2 Keywords: Poisson, probability Section: 5-3 Other Discrete Probability Distributions Outcome: 3 54) The makers of Crustfree Bread in Boston have a quality standard that allows no more than 3 burned loaves per batch on average. Recently, the manager inspected a batch and found 5 burned loaves. She did not appear to be upset at the production meeting. This is because the chance of exactly 5 burned loaves occurring is 0.1008. Answer: TRUE Diff: 2 Keywords: Poisson Section: 5-3 Other Discrete Probability Distributions Outcome: 3

55) The makers of Crustfree Bread in Boston have a quality standard that allows for no more than 3 burned loaves per batch on average. Assuming that the average of 3 per batch is being met, the standard deviation for the number of burned loaves in 4 batches is approximately 1.73 loaves. Answer: FALSE Diff: 2 Keywords: standard deviation, Poisson Section: 5-3 Other Discrete Probability Distributions Outcome: 3 56) The primary application for the hypergeometric probability distribution is in situations where the sampling is done without replacement from a finite population. Answer: TRUE Diff: 1 Keywords: hypergeometric, sampling, replacement Section: 5-3 Other Discrete Probability Distributions Outcome: 3 57) A warehouse contains 5 parts made by the Stafford Company and 8 parts made by the Wilson Company. If an employee selects 3 of the parts from the warehouse at random, the probability that none of the 3 parts is from the Wilson Company is approximately .03496. Answer: TRUE Diff: 2 Keywords: hypergeometric, sample Section: 5-3 Other Discrete Probability Distributions Outcome: 3 58) A warehouse contains 5 parts made by the Stafford Company and 8 parts made by the Wilson Company. If an employee selects 3 of the parts from the warehouse at random, the probability that all 3 parts are from the Wilson Company is approximately .1958. Answer: TRUE Diff: 2 Keywords: hypergeometric, sample Section: 5-3 Other Discrete Probability Distributions Outcome: 3 59) The city council consists of 3 Democrats, 5 Republicans and 3 independents. Subcommittees are supposed to be randomly assigned from the council. Suppose the 5-member planning and zoning subcommittee is composed of 3 Democrats and 2 Republicans. The probability of this happening by chance alone is approximately .4545. Answer: FALSE Diff: 2 Keywords: hypergeometric, sample Section: 5-3 Other Discrete Probability Distributions Outcome: 3

60) A company has 20 cars that are available for use by company executives for official business purposes. Six of these cars are SUVs, 8 are luxury type cars, and the rest are basic sedans. Suppose the cars are randomly assigned each week. If 5 cars are put into use, the chance that none of the SUVs or luxury cars will be in the group is approximately .0004. Answer: TRUE Diff: 2 Keywords: hypergeometric sample Section: 5-3 Other Discrete Probability Distributions Outcome: 3 61) When dealing with the number of occurrences of an event over a specified interval of time or space, the appropriate probability distribution is hypergeometric. Answer: FALSE Diff: 2 Keywords: Poisson, hypergeometric Section: 5-3 Other Discrete Probability Distributions Outcome: 3 62) The following probability distribution has been assessed for the number of accidents that occur in a Midwestern city each day: Accidents 0 1 2 3 4

Probability 0.25 0.20 0.30 0.15 0.10

The probability of having less than 2 accidents on a given day is: A) 0.30 B) 0.75 C) 0.45 D) 0.25 Answer: C Diff: 1 Keywords: discrete, random variable Section: 5-1 Introduction to Discrete Probability Distributions Outcome: 1

63) The following probability distribution has been assessed for the number of accidents that occur in a Midwestern city each day: Accidents 0 1 2 3 4

Probability 0.25 0.20 0.30 0.15 0.10

This distribution is an example of: A) a uniform distribution. B) a continuous probability distribution. C) a discrete probability distribution. D) an expected value distribution. Answer: C Diff: 1 Keywords: discrete, probability distribution Section: 5-1 Introduction to Discrete Probability Distributions Outcome: 1 64) The following probability distribution has been assessed for the number of accidents that occur in a Midwestern city each day: Accidents 0 1 2 3 4

Probability 0.25 0.20 0.30 0.15 0.10

Based on this distribution, the expected number of accidents in a given day is: A) 0.30 B) 1.65 C) 2.00 D) 2.50 Answer: B Diff: 2 Keywords: expected value, discrete distribution Section: 5-1 Introduction to Discrete Probability Distributions Outcome: 1

65) The following probability distribution has been assessed for the number of accidents that occur in a Midwestern city each day: Accidents 0 1 2 3 4

Probability 0.25 0.20 0.30 0.15 0.10

Based on this probability distribution, the standard deviation in the number of accidents per day is: A) 2.0 B) 1.63 C) 2.65 D) 1.28 Answer: D Diff: 2 Keywords: standard deviation, discrete distribution Section: 5-1 Introduction to Discrete Probability Distributions Outcome: 1 66) A sales rep for a national clothing company makes 4 calls per day. Based on historical records, the following probability distribution describes the number of successful calls each day: Successful Calls 0 1 2 3 4

Probability 0.10 0.30 0.30 0.20 0.10

Based on this information, the probability that the sales rep will have a total of 2 successful calls in a twoday period is: A) 0.60 B) 0.09 C) 0.15 D) 0.06 Answer: C Diff: 3 Keywords: discrete, probability Section: 5-1 Introduction to Discrete Probability Distributions Outcome: 1

67) A sales rep for a national clothing company makes 4 calls per day. Based on historical records, the following probability distribution describes the number of successful calls each day: Successful Calls 0 1 2 3 4

Probability 0.10 0.30 0.30 0.20 0.10

The expected number of successful sales calls per day is: A) 2.00 B) 1.15 C) 1.90 D) 2.50 Answer: C Diff: 2 Keywords: expected value, discrete distribution Section: 5-1 Introduction to Discrete Probability Distributions Outcome: 1 68) A sales rep for a national clothing company makes 4 calls per day. Based on historical records, the following probability distribution describes the number of successful calls each day: Successful Calls 0 1 2 3 4

Probability 0.10 0.30 0.30 0.20 0.10

Based on the information provided, what is the probability of having at least 2 successful calls in one day? A) 0.60 B) 0.20 C) 0.30 D) 0.10 Answer: A Diff: 1 Keywords: discrete distribution Section: 5-1 Introduction to Discrete Probability Distributions Outcome: 1

69) Consider the following two probability distributions:

Which of the following is an accurate statement regarding these two distributions? A) Distribution A has a higher variance. B) Distribution B has a higher variance. C) Both distributions are positively skewed. D) Both distributions are uniform. Answer: A Diff: 2 Keywords: discrete distribution, variance Section: 5-1 Introduction to Discrete Probability Distributions Outcome: 1 70) Which of the following statements is incorrect? A) The expected value of a discrete probability distribution is the long-run average value assuming the experiment will be repeated many times. B) The standard deviation of a discrete probability distribution measures the average deviation of the random variable from the mean. C) The distribution is considered uniform if all the probabilities are equal. D) The mean of the probability distribution is equal to the square root of the variance. Answer: D Diff: 2 Keywords: probability distribution, expected value Section: 5-1 Introduction to Discrete Probability Distributions Outcome: 1 71) The probability function for random variable X is specified as:

The expected value of X is A) 0.333 B) 0.500 C) 2.000 D) 2.333 Answer: D Diff: 2 Keywords: probability distribution, expected value Section: 5-1 Introduction to Discrete Probability Distributions Outcome: 1 5-20 Copyright © 2018 Pearson Education, Inc.

72) Which of the following is not a condition of the binomial distribution? A) Two possible outcomes for each trial B) The trials are independent. C) The standard deviation is equal to the square root of the mean. D) The probability of a success remains constant from trial to trial. Answer: C Diff: 1 Keywords: binomial distribution, standard deviation Section: 5-2 The Binomial Probability Distribution Outcome: 2 73) A package delivery service claims that no more than 5 percent of all packages arrive at the address late. Assuming that the conditions for the binomial hold, if a sample of size 10 packages is randomly selected, and the 5 percent rate holds, what is the probability that exactly 2 packages in the sample arrive late? A) 0.0746 B) 0.9884 C) 0.2347 D) 0.0439 Answer: A Diff: 2 Keywords: binomial distribution, probability Section: 5-2 The Binomial Probability Distribution Outcome: 2 74) A package delivery service claims that no more than 5 percent of all packages arrive at the address late. Assuming that the conditions for the binomial hold, if a sample of size 10 packages is randomly selected and the 5 percent rate holds, what is the probability that more than 2 packages will be delivered late? A) 0.0115 B) 0.0105 C) 0.0862 D) 0.0746 Answer: A Diff: 2 Keywords: binomial distribution, probability Section: 5-2 The Binomial Probability Distribution Outcome: 2

75) The Vardon Exploration Company is getting ready to leave for South America to explore for oil. One piece of equipment requires 10 batteries that must operate for more than 2 hours. The batteries being used have a 15 percent chance of failing within 2 hours. The exploration leader plans to take 15 batteries. Assuming that the conditions of the binomial apply, the probability that the supply of batteries will contain enough good ones to operate the equipment is: A) 0.0449 B) 0.9832 C) 0.0132 D) 0.9964 Answer: B Diff: 3 Keywords: binomial distribution, probability Section: 5-2 The Binomial Probability Distribution Outcome: 2 76) The Vardon Exploration Company is getting ready to leave for South America to explore for oil. One piece of equipment requires 10 batteries that must operate for more than 2 hours. The batteries being used have a 15 percent chance of failing within 2 hours. The exploration leader plans to take 15 batteries. Assuming that the conditions of the binomial apply, the probability that the supply of batteries will not contain enough good ones to operate the equipment is: A) 0.0449 B) 0.0132 C) 0.9832 D) 0.0168 Answer: D Diff: 3 Keywords: binomial distribution, probability Section: 5-2 The Binomial Probability Distribution Outcome: 2 77) Previous research shows that 60 percent of adults who drink non-diet cola prefer Coca-Cola to Pepsi. Recently, an independent research firm questioned a random sample of 25 adult non-diet cola drinkers. That chance that 20 or more of these people will prefer Coca-Cola is: A) essentially zero. B) 0.0199. C) 0.0294. D) None of the above Answer: C Diff: 2 Keywords: binomial, probability Section: 5-2 The Binomial Probability Distribution Outcome: 2

78) Many people believe that they can tell the difference between Coke and Pepsi. Other people say that the two brands can't be distinguished. To test this, a random sample of 20 adults was selected to participate in a test. After being blindfolded, each person was given a small taste of either Coke or Pepsi and asked to indicate which brand soft drink it was. If people really can't tell the difference, the expected number of correct identifications in the sample would be: A) 10. B) 0. C) between 4 and 9. D) Can't be determined without more information. Answer: A Diff: 2 Keywords: binomial, expected value Section: 5-2 The Binomial Probability Distribution Outcome: 2 79) Many people believe that they can tell the difference between Coke and Pepsi. Other people say that the two brands can't be distinguished. To test this, a random sample of 20 adults was selected to participate in a test. After being blindfolded, each person was given a small taste of either Coke or Pepsi and asked to indicate which brand soft drink it was. If people really can't tell the difference, the probability that fewer than 6 people will guess correctly is: A) 0.0148 B) approximately 0.02 C) 0.0307 D) 0.0514 Answer: B Diff: 2 Keywords: binomial, probability Section: 5-2 The Binomial Probability Distribution Outcome: 2 80) Many people believe that they can tell the difference between Coke and Pepsi. Other people say that the two brands can't be distinguished. To test this, a random sample of 20 adults was selected to participate in a test. After being blindfolded, each person was given a small taste of either Coke or Pepsi and asked to indicate which brand soft drink it was. Suppose 14 people correctly identified the soft drink brand. Which of the following conclusions would be warranted under the circumstance? A) Since the chance of getting 14 correct is 0.0370, which is quite small, the study shows that people are not able to identify brands effectively. B) Since the probability of getting 14 or more correct is 0.0577, which is quite low, this means that people are not effective in identifying the soft drink brand. C) Since the probability of getting 14 or more correct is 0.0577, which is quite low, the conclusion could be that people are effective at identifying soft drink brands. D) The expected value for this binomial distribution is very close to 14 so this supports that people cannot tell the difference. Answer: C Diff: 3 Keywords: binomial, expected value, probability Section: 5-2 The Binomial Probability Distribution Outcome: 2 5-23 Copyright © 2018 Pearson Education, Inc.

81) Madam Helga claims to be psychic. A national TV talk personality plans to test her in a live TV broadcast. The process will entail asking Madam Helga a series of 20 independent questions with yes/no answers. The questions would be of the nature that she could not have any way of knowing the answer from prior knowledge. She will be considered psychic if she correctly answers more than a specified number (called the cut-off) of the questions. The cut-off must be set so that the chance of guessing that number or more is no greater than 5 percent. The cut-off value should be: A) 12 B) 14 C) 10 D) Can't be determined without more information. Answer: B Diff: 3 Keywords: binomial, probability, expected value Section: 5-2 The Binomial Probability Distribution Outcome: 2 82) Madam Helga claims to be psychic. A national TV talk personality plans to test her in a live TV broadcast. The process will entail asking Madam Helga a series of 20 independent questions with yes/no answers. The questions would be of the nature that she could not have any way of knowing the answer from prior knowledge. Suppose that Madam Helga correctly answered 15 of the 20 questions, which of the following would be a viable conclusion to reach? A) Because the probability of guessing 15 or more correctly is 0.0207, it is unlikely that she is guessing at the questions and may, in fact, have some special ability. B) Because the probability of getting 15 or more correct is 0.0207, it is likely that she is just guessing at the questions. C) If she were guessing, 15 is within one standard deviation of the mean and therefore she must not have any special psychic abilities. D) Because the probability of guessing exactly 15 correct is 0.0148, she must just be guessing. Answer: A Diff: 3 Keywords: binomial, probability Section: 5-2 The Binomial Probability Distribution Outcome: 2 83) Which of the following is true with respect to the binomial distribution? A) As the sample size increases, the expected value of the random variable decreases. B) The binomial distribution becomes more skewed as the sample size is increased for a given probability of success. C) The binomial distribution tends to be more symmetric as p approaches 0.5. D) In order for the binomial distribution to be skewed, the sample size must be quite large. Answer: C Diff: 2 Keywords: binomial, symmetric Section: 5-2 The Binomial Probability Distribution Outcome: 2

84) Which of the following statements is true? A) A binomial distribution with n = 20 and p = 0.05 will be right-skewed. B) A binomial distribution with n = 6 and p = 0.50 will be symmetric. C) A binomial distribution with n = 20 and p = 0.05 has an expected value equal to 1. D) A, B, and C are all true. Answer: D Diff: 2 Keywords: binomial, symmetric, expected value Section: 5-2 The Binomial Probability Distribution Outcome: 2 85) If the number of defective items selected at random from a parts inventory is considered to follow a binomial distribution with n = 50 and p = 0.10, the expected number of defective parts is: A) 5 B) approximately 2.24 C) more than 10 D) 0.5 Answer: A Diff: 1 Keywords: binomial, expected value Section: 5-2 The Binomial Probability Distribution Outcome: 2 86) If the number of defective items selected at random from a parts inventory is considered to follow a binomial distribution with n = 50 and p = 0.10, the standard deviation of the number of defective parts is: A) 5 B) 4.5 C) 45 D) about 2.12 Answer: D Diff: 2 Keywords: binomial, standard deviation Section: 5-2 The Binomial Probability Distribution Outcome: 2 87) The probability that a product is found to be defective is 0.10. If we examine 50 products, which of the following has the highest probability? A) 3 defective products are found. B) 4 defective products are found. C) 5 defective products are found. D) 6 defective products are found. Answer: C Diff: 2 Keywords: binomial, expected value Section: 5-2 The Binomial Probability Distribution Outcome: 2

88) If a study is set up in such a way that a sample of people is surveyed to determine whether they have ever used a particular product, the likely probability distribution that would describe the random variable, the number who say yes, is a: A) binomial distribution. B) Poisson distribution. C) uniform distribution. D) continuous distribution. Answer: A Diff: 2 Keywords: binomial, distribution Section: 5-3 Other Discrete Probability Distributions Outcome: 3 89) Assuming that potholes occur randomly along roads, the number of potholes per mile of road could best be described by the: A) binomial distribution. B) Poisson distribution. C) hypergeometric distribution. D) continuous distribution Answer: B Diff: 2 Keywords: Poisson distribution Section: 5-3 Other Discrete Probability Distributions Outcome: 3 90) The number of visible defects on a product container is thought to be Poisson distributed with a mean equal to 3.5. Based on this, the probability that 2 containers will contain a total of less than 2 defects is: A) 0.0223 B) 0.1359 C) 0.0073 D) 0.1850 Answer: C Diff: 2 Keywords: Poisson distribution Section: 5-3 Other Discrete Probability Distributions Outcome: 3 91) The number of visible defects on a product container is thought to be Poisson distributed with a mean equal to 3.5. Based on this, how many defects should be expected if 3 containers are inspected? A) 10.5 B) Approximately 3.24 C) Between 4 and 7 D) 3.5 Answer: A Diff: 1 Keywords: Poisson, expected value Section: 5-3 Other Discrete Probability Distributions Outcome: 3 5-26 Copyright © 2018 Pearson Education, Inc.

92) The number of customers who enter a bank is thought to be Poisson distributed with a mean equal to 10 per hour. What are the chances that no customers will arrive in a 15-minute period? A) Approximately zero B) 0.0067 C) 0.0821 D) 0.0250 Answer: C Diff: 2 Keywords: Poisson, expected value Section: 5-3 Other Discrete Probability Distributions Outcome: 3 93) The number of customers who enter a bank is thought to be Poisson distributed with a mean equal to 10 per hour. What are the chances that 2 or 3 customers will arrive in a 15-minute period? A) 0.0099 B) 0.4703 C) 0.0427 D) 0.0053 Answer: B Diff: 2 Keywords: Poisson, probability Section: 5-3 Other Discrete Probability Distributions Outcome: 3 94) If cars arrive to a service center randomly and independently at a rate of 5 per hour on average, what is the probability of 0 cars arriving in a given hour? A) 0.1755 B) 0.0067 C) 0.0000 D) 0.0500 Answer: B Diff: 2 Keywords: Poisson, probability Section: 5-3 Other Discrete Probability Distributions Outcome: 3 95) If cars arrive to a service center randomly and independently at a rate of 5 per hour on average, what is the probability that exactly 5 cars will arrive during a given hour? A) 0.1755 B) 0.6160 C) 0.1277 D) Essentially zero Answer: A Diff: 2 Keywords: Poisson, probability Section: 5-3 Other Discrete Probability Distributions Outcome: 3 5-27 Copyright © 2018 Pearson Education, Inc.

96) The number of weeds that remain living after a specific chemical has been applied averages 1.3 per square yard and follows a Poisson distribution. Based on this, what is the probability that a 1-square yard section will contain less than 4 weeds? A) 0.0324 B) 0.0998 C) Nearly 0.5000 D) 0.9569 Answer: D Diff: 2 Keywords: Poisson, probability Section: 5-3 Other Discrete Probability Distributions Outcome: 3 97) The number of weeds that remain living after a specific chemical has been applied averages 1.3 per square yard and follows a Poisson distribution. Based on this, what is the probability that a 3-square yard section will contain less than 4 weeds? A) 0.4532 B) 0.2001 C) 0.6482 D) 0.1951 Answer: A Diff: 2 Keywords: Poisson, probability Section: 5-3 Other Discrete Probability Distributions Outcome: 3 98) Which of the following statements is true with respect to a Poisson distribution? A) The Poisson distribution is symmetrical when the mean is close to 5. B) The Poisson distribution is more right-skewed for smaller values of the mean. C) The variance of the Poisson distribution is equal to the square root of the expected value. D) The Poisson distribution is an example of a continuous probability distribution. Answer: B Diff: 2 Keywords: Poisson, skewed Section: 5-3 Other Discrete Probability Distributions Outcome: 3

99) The manager of a movie theater has determined that the distribution of customers arriving at the concession stand is Poisson distributed with a standard deviation equal to 2 people per 10 minutes. What is the probability that more than 3 customers arrive during a 10-minute period? A) 0.1804 B) 0.5665 C) 0.4335 D) 0.1954 Answer: B Diff: 3 Keywords: Poisson, probability Section: 5-3 Other Discrete Probability Distributions Outcome: 3 100) The manager of a movie theater has determined that the distribution of customers arriving at the concession stand is Poisson distributed with a standard deviation equal to 2 people per 10 minutes. What is the probability that 0 customers arrive during a 10-minute period? A) 0.1353 B) 0.0183 C) 0.9817 D) Essentially 0 Answer: B Diff: 2 Keywords: Poisson, probability Section: 5-3 Other Discrete Probability Distributions Outcome: 3 101) If the standard deviation for a Poisson distribution is known to be 3, the expected value of that Poison distribution is: A) 3 B) about 1.73 C) 9 D) Can't be determined without more information. Answer: C Diff: 2 Keywords: Poisson, expected value, mean, standard deviation Section: 5-3 Other Discrete Probability Distributions Outcome: 3

102) If a distribution is considered to be Poisson with a mean equal to 11, the most frequently occurring value for the random variable will be: A) 10.5 B) 11 C) 10 and 11 D) 22 Answer: C Diff: 2 Keywords: Poisson, mean, distribution Section: 5-3 Other Discrete Probability Distributions Outcome: 3 103) The hypergeometric probability distribution is used rather than the binomial or the Poisson when: A) the sampling is performed with replacement. B) the sampling is performed without replacement from an infinite population. C) the sampling is performed without replacement from a finite population. D) the sampling is performed with replacement from a finite population. Answer: C Diff: 2 Keywords: hypergeometric, sampling, replacement Section: 5-3 Other Discrete Probability Distributions Outcome: 3 104) A small city has two taxi companies (A and B). Each taxi company has 5 taxis. A motel has told these companies that they will randomly select a taxi company when one of its customers needs a cab. This morning 3 cabs were needed. Assuming that no one individual taxi can be used more than once, what is the probability that 2 of the cabs selected will be from Company A and the other will be from B? A) 0.417 B) 0.25 C) 0.583 D) 0.5 Answer: A Diff: 2 Keywords: hypergeometric, probability Section: 5-3 Other Discrete Probability Distributions Outcome: 3

105) A local paint store carries 4 brands of paint (W, X, Y, and Z). The store has 5 cans of W, 3 cans of X, 6 cans of Y, and 15 cans of Z, all in white. It is thought that customers have no preference for one of these brands over another. If this is the case, what is the probability that the next 5 customers will select 1 can of W, X, Y and 2 cans of brand Z? A) About .23 B) Approximately .08 C) Over .30 D) 0.25 Answer: B Diff: 3 Keywords: hypergeometric, probability Section: 5-3 Other Discrete Probability Distributions Outcome: 3 106) The random variable x is the number of customers arriving at the service desk of a local car dealership over an interval of 10 minutes. It is known that the average number of arrivals in 10 minutes is 5.3. The probability that there are less than 3 arrivals in any 10 minutes is: A) .0659 B) .0948 C) .1016 D) .1239 Answer: C Diff: 2 Keywords: Poisson, distribution Section: 5-3 Other Discrete Probability Distributions Outcome: 3

107) Because of bad weather, the number of days next week that the captain of a charter fishing boat can leave port is uncertain. Let x = number of days that the boat is able to leave port per week. The following probability distribution for the variable, x, was determined based on historical data when the weather was poor: x 0 1 2 3 4 5 6 7

P(x) 0.05 0.10 0.10 0.20 0.20 0.15 0.15 0.05

Based on the probability distribution, what is the expected number of days per week the captain can leave port? A) 3.7 B) 4.5 C) 2.8 D) 1.7 Answer: A Diff: 1 Keywords: discrete, probability distribution Section: 5-1 Introduction to Discrete Probability Distributions Outcome: 1

108) The roll of a pair of dice has the following probability distribution, where the random variable is the sum of the values produced by each die: x 2 3 4 5 6 7 8 9 10 11 12

P(x) 1/36 2/36 3/36 4/36 5/36 6/36 5/36 4/36 3/36 2/36 1/36

Calculate the expected value of x. A) 6 B) 7 C) 8 D) 9 Answer: B Diff: 2 Keywords: discrete, probability distribution Section: 5-1 Introduction to Discrete Probability Distributions Outcome: 1

109) The roll of a pair of dice has the following probability distribution, where the random variable is the sum of the values produced by each die: x 2 3 4 5 6 7 8 9 10 11 12

P(x) 1/36 2/36 3/36 4/36 5/36 6/36 5/36 4/36 3/36 2/36 1/36

Calculate the variance of x. A) 5.833 B) 6.122 C) 5.666 D) 5.122 Answer: A Diff: 2 Keywords: discrete, probability distribution Section: 5-1 Introduction to Discrete Probability Distributions Outcome: 1

110) The roll of a pair of dice has the following probability distribution, where the random variable is the sum of the values produced by each die: x 2 3 4 5 6 7 8 9 10 11 12

P(x) 1/36 2/36 3/36 4/36 5/36 6/36 5/36 4/36 3/36 2/36 1/36

Calculate the standard deviation of x. A) 3.415 B) 2.333 C) 3.125 D) 2.415 Answer: D Diff: 2 Keywords: discrete, probability distribution Section: 5-1 Introduction to Discrete Probability Distributions Outcome: 1

111) The number of cars sold (x) in a day at Smitty’s Auto Sales is described by the probability distribution in the following table: x 0 1 2 3 4 5

P(x) 0.18 0.39 0.24 0.14 0.04 0.01

Calculate the expected number of cars sold in a day. A) 0.82 B) 0.39 C) 1.50 D) 1.33 Answer: C Diff: 2 Keywords: discrete, probability distribution Section: 5-1 Introduction to Discrete Probability Distributions Outcome: 1

112) The number of cars sold (x) in a day at Smitty’s Auto Sales is described by the probability distribution in the following table: x 0 1 2 3 4 5

P(x) 0.18 0.39 0.24 0.14 0.04 0.01

Compute the variance and standard deviation of the number of people in families per household. A) Variance=1.25, standard deviation=1.118 B) Variance=1.86, standard deviation=1.364 C) Variance=1.118, standard deviation=1.057 D) Variance=1.50, standard deviation=1.225 Answer: A Diff: 2 Keywords: discrete, probability distribution Section: 5-1 Introduction to Discrete Probability Distributions Outcome: 1

113) Jennings Assembly in Hartford, Connecticut, uses a component supplied by a company in Brazil. The component is expensive to carry in inventory and consequently is not always available in stock when requested. Furthermore, shipping schedules are such that the lead time for transportation of the component is not a constant. Using historical records, the manufacturing firm has developed the following probability distribution for the product's lead time. The distribution is shown here, where the random variable is the number of days between the placement of the replenishment order and the receipt of the item. x 2 3 4 5 6

P(x) 0.15 0.45 0.30 0.0.75 0.025

What is the average lead time for the component? A) 2.375 B) 2.875 C) 3.275 D) 3.375 Answer: D Diff: 2 Keywords: discrete, probability distribution Section: 5-1 Introduction to Discrete Probability Distributions Outcome: 1

114) Jennings Assembly in Hartford, Connecticut, uses a component supplied by a company in Brazil. The component is expensive to carry in inventory and consequently is not always available in stock when requested. Furthermore, shipping schedules are such that the lead time for transportation of the component is not a constant. Using historical records, the manufacturing firm has developed the following probability distribution for the product's lead time. The distribution is shown here, where the random variable is the number of days between the placement of the replenishment order and the receipt of the item. x 2 3 4 5 6

P(x) 0.15 0.45 0.30 0.0.75 0.025

What is the coefficient of variation for delivery lead time? A) 38.461% B) 27.065% C) 27.891% D) 31.772% Answer: B Diff: 2 Keywords: discrete, probability distribution Section: 5-1 Introduction to Discrete Probability Distributions Outcome: 1

115) Cramer's Bar and Grille in Dallas can seat 130 people at a time. The manager has been gathering data on the number of minutes a party of four spends in the restaurant from the moment they are seated to when they pay the check. Number of Minutes 60 70 80 90 100 110

Probability 0.05 0.15 0.20 0.45 0.10 0.05

What is the mean number of minutes for a dinner party of four? A) 65.5 B) 67.5 C) 85.5 D) 75.5 Answer: C Diff: 2 Keywords: discrete, probability distribution Section: 5-1 Introduction to Discrete Probability Distributions Outcome: 1 116) Cramer's Bar and Grille in Dallas can seat 130 people at a time. The manager has been gathering data on the number of minutes a party of four spends in the restaurant from the moment they are seated to when they pay the check. Number of Minutes 60 70 80 90 100 110

Probability 0.05 0.15 0.20 0.45 0.10 0.05

What is the variance and standard deviation? A) Variance = 164.99, standard deviation = 12.84 B) Variance = 233.75, standard deviation = 15.89 C) Variance = 128.75, standard deviation = 11.35 D) Variance = 134.75, standard deviation = 11.61 Answer: D Diff: 2 Keywords: discrete, probability distribution Section: 5-1 Introduction to Discrete Probability Distributions Outcome: 1 5-39 Copyright © 2018 Pearson Education, Inc.

117) The manager for State Bank and Trust has recently examined the credit card account balances for the customers of her bank and found that 20% have an outstanding balance at the credit card limit. Suppose the manager randomly selects 15 customers and finds 4 that have balances at the limit. Assume that the properties of the binomial distribution apply. What is the probability of finding 4 customers in a sample of 15 who have "maxed out" their credit cards? A) 0.1876 B) 0.8358 C) 0.6482 D) 0.3832 Answer: A Diff: 2 Keywords: binomial distribution, probability Section: 5-2 The Binomial Probability Distribution Outcome: 1 118) The manager for State Bank and Trust has recently examined the credit card account balances for the customers of her bank and found that 20% have an outstanding balance at the credit card limit. Suppose the manager randomly selects 15 customers and finds 4 that have balances at the limit. Assume that the properties of the binomial distribution apply. What is the probability that 4 or fewer customers in the sample will have balances at the limit of the credit card? A) 0.1876 B) 0.8358 C) 0.6482 D) 0.3832 Answer: B Diff: 2 Keywords: binomial distribution, probability Section: 5-2 The Binomial Probability Distribution Outcome: 1 119) For a binomial distribution with a sample size equal to 10 and a probability of a success equal to 0.30, what is the probability that the sample will contain exactly three successes? Use the binomial formula to determine the probability. A) 0.3277 B) 0.3288 C) 0.2668 D) 0.2577 Answer: C Diff: 2 Keywords: binomial distribution, probability Section: 5-2 The Binomial Probability Distribution Outcome: 1

120) If a binomial distribution applies with a sample size of n = 20, find the probability of 5 successes if the probability of a success is 0.40. A) 0.1246 B) 0.1286 C) 0.0746 D) 0.0866 Answer: C Diff: 2 Keywords: binomial distribution, probability Section: 5-2 The Binomial Probability Distribution Outcome: 1 121) If a binomial distribution applies with a sample size of n = 20, find the probability of at least 7 successes if the probability of a success is 0.25. A) 0.1814 B) 0.2142 C) 0.2333 D) 0.3123 Answer: B Diff: 2 Keywords: binomial distribution, probability Section: 5-2 The Binomial Probability Distribution Outcome: 1 122) If a binomial distribution applies with a sample size of n = 20, find the expected value, n = 20, p = 0.20. A) 2 B) 3 C) 4 D) 5 Answer: C Diff: 2 Keywords: binomial distribution, probability Section: 5-2 The Binomial Probability Distribution Outcome: 1 123) If a binomial distribution applies with a sample size of n = 20, find the standard deviation, n = 20, p = 0.20. A) 1.7889 B) 2.1889 C) 2.7889 D) 3.1221 Answer: A Diff: 2 Keywords: binomial distribution, probability Section: 5-2 The Binomial Probability Distribution Outcome: 1

124) Given a binomial distribution with n = 8 and p = 0.40, obtain the mean. A) 2.8 B) 3.2 C) 3.6 D) 4.2 Answer: B Diff: 1 Keywords: binomial distribution, probability Section: 5-2 The Binomial Probability Distribution Outcome: 1 125) Given a binomial distribution with n = 8 and p = 0.40, obtain the standard deviation. A) 1.921 B) 1.386 C) 1.848 D) 1.465 Answer: B Diff: 2 Keywords: binomial distribution, probability Section: 5-2 The Binomial Probability Distribution Outcome: 1 126) Given a binomial distribution with n = 8 and p = 0.40, obtain the probability that the number of successes is larger than the mean. A) 0.4059 B) 0.3882 C) 0.2582 D) 0.6070 Answer: A Diff: 2 Keywords: binomial distribution, probability Section: 5-2 The Binomial Probability Distribution Outcome: 1 127) Given a binomial distribution with n = 8 and p = 0.40, obtain the probability that the number of successes is within 2 standard deviations of the mean. A) 0.6887 B) 0.7334 C) 0.8665 D) 0.9334Answer: D Diff: 2 Keywords: binomial distribution, probability Section: 5-2 The Binomial Probability Distribution Outcome: 1

128) Magic Valley Memorial Hospital administrators have recently received an internal audit report that indicates that 15% of all patient bills contain an error of one form or another. After spending considerable effort to improve the hospital's billing process, the administrators are convinced that things have improved. They believe that the new error rate is somewhere closer to 0.05.Suppose that recently the hospital randomly sampled 10 patient bills and conducted a thorough study to determine whether an error exists. It found 3 bills with errors. Assuming that managers are correct that they have improved the error rate to 0.05, what is the probability that they would find 3 or more errors in a sample of 10 bills? A) 0.0115 B) 0.0233 C) 0.0884 D) 0.0766 Answer: A Diff: 2 Keywords: binomial distribution, probability Section: 5-2 The Binomial Probability Distribution Outcome: 1 129) Dell Computers receives large shipments of microprocessors from Intel Corp. It must try to ensure the proportion of microprocessors that are defective is small. Suppose Dell decides to test five microprocessors out of a shipment of thousands of these microprocessors. Suppose that if at least one of the microprocessors is defective, the shipment is returned. If Intel Corp.'s shipment contains 10% defective microprocessors, calculate the probability the entire shipment will be returned. A) 0.4980 B) 0.4209 C) 0.4095 D) 0.4550 Answer: C Diff: 2 Keywords: binomial distribution, probability Section: 5-2 The Binomial Probability Distribution Outcome: 1 130) Dell Computers receives large shipments of microprocessors from Intel Corp. It must try to ensure the proportion of microprocessors that are defective is small. Suppose Dell decides to test five microprocessors out of a shipment of thousands of these microprocessors. Suppose that if at least one of the microprocessors is defective, the shipment is returned. If Intel and Dell agree that Intel will not provide more than 5% defective chips, calculate the probability that the entire shipment will be returned even though only 5% are defective. A) 0.2262 B) 0.3478 C) 0.4564 D) 0.1812 Answer: A Diff: 2 Keywords: binomial distribution, probability Section: 5-2 The Binomial Probability Distribution Outcome: 1 5-43 Copyright © 2018 Pearson Education, Inc.

131) Dell Computers receives large shipments of microprocessors from Intel Corp. It must try to ensure the proportion of microprocessors that are defective is small. Suppose Dell decides to test five microprocessors out of a shipment of thousands of these microprocessors. Suppose that if at least one of the microprocessors is defective, the shipment is returned. Calculate the probability that the entire shipment will be kept by Dell even though the shipment has 10% defective microprocessors. A) 0.3995 B) 0.3979 C) 0.5905 D) 0.4550 Answer: C Diff: 2 Keywords: binomial distribution, probability Section: 5-2 The Binomial Probability Distribution Outcome: 1 132) The mean number of errors per page made by a member of the word processing pool for a large company is thought to be 1.5 with the number of errors distributed according to a Poisson distribution. If three pages are examined, what is the probability that more than 3 errors will be observed? A) 0.6577 B) 0.6969 C) 0.7324 D) 0.7860 Answer: A Diff: 2 Keywords: discrete, probability distribution Section: 5-3 Other Discrete Probability Distributions Outcome: 1 133) Arrivals to a bank automated teller machine (ATM) are distributed according to a Poisson distribution with a mean equal to three per 15 minutes.Determine the probability that in a given 15minute segment no customers will arrive at the ATM. A) 0.0124 B) 0.0281 C) 0.0314 D) 0.0498 Answer: D Diff: 2 Keywords: discrete, probability distribution Section: 5-3 Other Discrete Probability Distributions Outcome: 1

134) Arrivals to a bank automated teller machine (ATM) are distributed according to a Poisson distribution with a mean equal to three per 15 minutes. What is the probability that fewer than four customers will arrive in a 30-minute segment? A) 0.1512 B) 0.1889 C) 0.2515 D) 0.2576 Answer: A Diff: 2 Keywords: discrete, probability distribution Section: 5-3 Other Discrete Probability Distributions Outcome: 1 135) Consider a situation in which a used-car lot contains five Fords, four General Motors (GM) cars, and five Toyotas. If five cars are selected at random to be placed on a special sale, what is the probability that three are Fords and two are GMs? A) 0.09 B) 0.03 C) 0.04 D) 0.06 Answer: B Diff: 3 Keywords: discrete, probability distribution Section: 5-3 Other Discrete Probability Distributions Outcome: 1 136) A population of 10 items contains 3 that are red and 7 that are green. What is the probability that in a random sample of 3 items selected without replacement, 2 red and 1 green items are selected? A) 0.175 B) 0.086 C) 0.124 D) 0.211 Answer: A Diff: 3 Keywords: discrete, probability distribution Section: 5-3 Other Discrete Probability Distributions Outcome: 1

137) A corporation has 11 manufacturing plants. Of these, 7 are domestic and 4 are located outside the United States. Each year a performance evaluation is conducted for 4 randomly selected plants. What is the probability that a performance evaluation will include exactly 1 plant outside the United States? A) 0.4242 B) 0.3776 C) 0.3523 D) 0.4696 Answer: A Diff: 3 Keywords: discrete, probability distribution Section: 5-3 Other Discrete Probability Distributions Outcome: 1 138) A corporation has 11 manufacturing plants. Of these, 7 are domestic and 4 are located outside the United States. Each year a performance evaluation is conducted for 4 randomly selected plants. What is the probability that a performance evaluation will contain 3 plants from the United States? A) 0.4242 B) 0.3776 C) 0.3523 D) 0.4696 Answer: A Diff: 3 Keywords: discrete, probability distribution Section: 5-3 Other Discrete Probability Distributions Outcome: 1 139) A corporation has 11 manufacturing plants. Of these, 7 are domestic and 4 are located outside the United States. Each year a performance evaluation is conducted for 4 randomly selected plants. What is the probability that a performance evaluation will include 2 or more plants from outside the United States? A) 0.4242 B) 0.3776 C) 0.3523 D) 0.4696 Answer: D Diff: 3 Keywords: discrete, probability distribution Section: 5-3 Other Discrete Probability Distributions Outcome: 1

140) A new phone answering system installed by the Ohio Power Company is capable of handling five calls every 10 minutes. Prior to installing the new system, company analysts determined that the incoming calls to the system are Poisson distributed with a mean equal to two every 10 minutes. If this incoming call distribution is what the analysts think it is, what is the probability that in a 10-minute period more calls will arrive than the system can handle? A) 0.174 B) 0.0812 C) 0.0166 D) 0.0233 Answer: C Diff: 3 Keywords: discrete, probability distribution Section: 5-3 Other Discrete Probability Distributions Outcome: 1 141) The Weyerhauser Lumber Company headquartered in Tacoma, Washington, is one of the largest timber and wood product companies in the world. Weyerhauser manufactures plywood at one of its Oregon plants. Plywood contains minor imperfections that can be repaired with small "plugs." One customer will accept plywood with a maximum of 3.5 plugs per sheet on average. Suppose a shipment was sent to this customer and when the customer inspected two sheets at random, 10 plugged defects were counted. What is the probability of observing 10 or more plugged defects if in fact the 3.5 average per sheet is being satisfied? A) 0.1887 B) 0.1695 C) 0.2115 D) 0.2675 Answer: B Diff: 3 Keywords: discrete, probability distribution Section: 5-3 Other Discrete Probability Distributions Outcome: 1 142) A stock portfolio contains 20 stocks. Of these stocks, 10 are considered "large-cap" stocks, 5 are "midcap," and 5 are "small cap." The portfolio manager has been asked by his client to develop a report that highlights 7 randomly selected stocks. When she presents her report to the client, all 7 of the stocks are large-cap stocks. The client is very suspicious that the manager has not randomly selected the stocks. She believes that the chances of all 7 of the stocks being large cap must be very low. Compute the probability of all 7 being large cap. A) 0.0015 B) 0.0008 C) 0.0121 D) 0.0309 Answer: A Diff: 3 Keywords: discrete, probability distribution Section: 5-3 Other Discrete Probability Distributions Outcome: 1

143) College-Pro Painting does home interior and exterior painting. The company uses inexperienced painters that do not always do a high-quality job. It believes that its painting process can be described by a Poisson distribution with an average of 4.8 defects per 400 square feet of painting. What is the probability that a 400-square-foot painted section will have fewer than 6 blemishes? A) 0.2818 B) 0.3414 C) 0.4857 D) 0.6510 Answer: D Diff: 3 Keywords: discrete, probability distribution Section: 5-3 Other Discrete Probability Distributions Outcome: 1 144) College-Pro Painting does home interior and exterior painting. The company uses inexperienced painters that do not always do a high-quality job. It believes that its painting process can be described by a Poisson distribution with an average of 4.8 defects per 400 square feet of painting. What is the probability that six randomly sampled sections of size 400 square feet will each have 7 or fewer blemishes? A) 0.2818 B) 0.3414 C) 0.4857 D) 0.6509 Answer: C Diff: 3 Keywords: discrete, probability distribution Section: 5-3 Other Discrete Probability Distributions Outcome: 1 145) John Thurgood founded a company that translates Chinese books into English. His company is currently testing a computer-based translation service. Since Chinese symbols are difficult to translate, John assumes the computer program will make some errors, but then so do human translators. The computer error rate is supposed to be an average of 3 per 400 words of translation. Suppose John randomly selects a 1,200-word passage. Assuming that the Poisson distribution applies, if the computer error rate is actually 3 errors per 400 words, determine the probability that no errors will be found. A) 0.0001 B) 0.0141 C) 0.0415 D) 0.4557 Answer: A Diff: 3 Keywords: discrete, probability distribution Section: 5-3 Other Discrete Probability Distributions/Thompson_sn3t_WordExports Outcome: 1

146) John Thurgood founded a company that translates Chinese books into English. His company is currently testing a computer-based translation service. Since Chinese symbols are difficult to translate, John assumes the computer program will make some errors, but then so do human translators. The computer error rate is supposed to be an average of 3 per 400 words of translation. Suppose John randomly selects a 1,200-word passage. Assuming that the Poisson distribution applies, if the computer error rate is actually 3 errors per 400 words, calculate the probability that more than 14 errors will be found. A) 0.000123 B) 0.0141 C) 0.0415 D) 0.4557 Answer: C Diff: 3 Keywords: discrete, probability distribution Section: 5-3 Other Discrete Probability Distributions Outcome: 1 147) John Thurgood founded a company that translates Chinese books into English. His company is currently testing a computer-based translation service. Since Chinese symbols are difficult to translate, John assumes the computer program will make some errors, but then so do human translators. The computer error rate is supposed to be an average of 3 per 400 words of translation. Suppose John randomly selects a 1,200-word passage. Assuming that the Poisson distribution applies, if the computer error rate is actually 3 errors per 400 words, find the probability that fewer than 9 errors will be found. A) 0.000123 B) 0.0141 C) 0.0415 D) 0.4557 Answer: D Diff: 3 Keywords: discrete, probability distribution Section: 5-3 Other Discrete Probability Distributions Outcome: 1 148) Beacon Hill Trees & Shrubs currently has an inventory of 10 fruit trees, 8 pine trees, and 14 maple trees. It plans to give 4 trees away at next Saturday's lawn and garden show in the city park. The 4 winners can select which type of tree they want. Assume they select randomly. What is the probability that all 4 winners will select the same type of tree? A) 0.0058 B) 0.0218 C) 0.0355 D) 0.0709 Answer: C Diff: 3 Keywords: discrete, probability distribution Section: 5-3 Other Discrete Probability Distributions Outcome: 1

149) Beacon Hill Trees & Shrubs currently has an inventory of 10 fruit trees, 8 pine trees, and 14 maple trees. It plans to give 4 trees away at next Saturday's lawn and garden show in the city park. The 4 winners can select which type of tree they want. Assume they select randomly. What is the probability that 3 winners will select pine trees and the other tree will be a maple? A) 0.0058 B) 0.0218 C) 0.0355 D) 0.0709 Answer: B Diff: 3 Keywords: discrete, probability distribution Section: 5-3 Other Discrete Probability Distributions Outcome: 1 150) Beacon Hill Trees & Shrubs currently has an inventory of 10 fruit trees, 8 pine trees, and 14 maple trees. It plans to give 4 trees away at next Saturday's lawn and garden show in the city park. The 4 winners can select which type of tree they want. Assume they select randomly. What is the probability that no fruit trees and 2 of each of the others will be selected? A) 0.0058 B) 0.0218 C) 0.0355 D) 0.0709 Answer: D Diff: 3 Keywords: discrete, probability distribution Section: 5-3 Other Discrete Probability Distributions Outcome: 1 151) What is the difference between a discrete random variable and a continuous random variable? Answer: A discrete random variable is one that can assume only a countable number of values, while a continuous random variable can assume any value along a continuum and the possible values are uncountable. Diff: 1 Keywords: discrete, continuous, random variable Section: 5-1 Introduction to Discrete Probability Distributions Outcome: 1 152) Explain what the expected value of a discrete random variable measures. Answer: The expected value is the weighted average of the possible values for the discrete random variable where the weights are the individual probabilities associated with each possible value. The expected value represents the long-run average value for the random variable if the experiment that generates the random variable is repeated a very large number of times. Diff: 2 Keywords: expected value, discrete random variable Section: 5-1 Introduction to Discrete Probability Distributions Outcome: 1

153) Explain how to determine whether the binomial distribution can be used in a particular application. Answer: First, the random variable of interest must be discrete. Then look to see whether there are a specified number of trials, of which there are only 2 possible outcomes, each of which is conducted in an identical manner. Then determine whether it is possible to count the number of successes and the number of failures that occur when the trials are performed. Finally, check to see if the probability of a success can be assumed to remain constant from trial to trial. Diff: 2 Keywords: binomial distribution Section: 5-2 The Binomial Probability Distribution Outcome: 2 154) A small city has 2 ambulances. Emergency calls for ambulances arrive randomly with an average of 0.2 calls per hour. They are concerned about the possibility of both ambulances being busy when an additional call comes in. What is the probability of more than 2 calls in a 1-hour period? Determine the correct distribution, explain why it is the best distribution to use, and find the probability. Answer: This would be the Poisson distribution because it deals with the discrete number of calls arriving randomly over time. There is no particular upper bound as you would have in the binomial distribution because there are no yes/no type trials going on. Using the average of 0.2 we have P(0) = 0.8187 P(1) = 0.1638 P(2) = 0.0164

So P(# calls ≤ 2) = .9989

This means P(# calls > 2) = P(# calls ≥ 3) = 1 - 0.9988 = 0.0012. So it would be a very rare event to receive more than 2 calls during a 1-hour period. Diff: 2 Keywords: Poisson distribution, probability Section: 5-3 Other Discrete Probability Distributions Outcome: 3 155) Explain how to use the binomial distribution table when p, the probability of a success, exceeds 0.50. Answer: When p > 0.50, you have two options. The first option is to think in terms of failures rather than successes—for instance, if the sample size is n =10 and we originally want to find the probability of 3 successes when p = 0.70. Instead we can switch the problem around and instead of finding the probability of 3 successes, we can find the probability of 7 failures with the probability of a failure being p = 0.30. An alternative method is to use the q values at the bottom of each column (if the binomial distribution table being used has these values). Treat these as if they were the p values. Then locate the number of successes of interest in the right-hand column of the binomial table. Both methods will give the same result. Diff: 3 Keywords: binomial distribution, probability Section: 5-2 The Binomial Probability Distribution Outcome: 2

156) A company that makes chocolate chip cookies has found that the number of chips per cookie follows a Poisson distribution. What should the minimum average number of chips be to result in at least 98 percent of the cookies having more than 2 chips? Find the minimum average to nearest whole chip (i.e. choose an average that is a whole number). Answer: Here the problem is to determine the minimum average number of chips per cookie that will result in at least 98 percent of the cookies having more than 2 chips. This means that at most 2 percent of the cookies have 2 chips or less. Trying an average of 5 chips per cookie gives us: P(0 or 1 or 2) = .0067 + .0337 + .0842 = 0.1246, which is much too large. Trying an average of 6 chips per cookie gives us: P(0 or 1 or 2) = .0025 + .0149 + .0446 = 0.0620, which is also too large. Trying an average of 7 chips per cookie gives us: P(0 or 1 or 2) = .0009 + .0064 + .0223 = 0.0296, which is getting close. Trying an average of 8 chips per cookie gives us: P(0 or 1 or 2) = .0003 + .0027 + .0107 = 0.0137, which is less than 2 percent. Therefore it should use enough chips so that the average number of chips per cookie is 8, which will result in only 1.37 percent of the cookies having 2 chips or less, and 98.63 percent of the cookies having more than 2 chips. Diff: 2 Keywords: Poisson distribution Section: 5-3 Other Discrete Probability Distributions Outcome: 3

157) The binomial distribution is frequently used to help companies decide whether to accept or reject a shipment based on the results of a random sample of items from the shipment. For instance, suppose a contract calls for, at most, 10 percent of the items in a shipment to be red. To check this without looking at every item in the large shipment, a sample of n = 10 items is selected. If 1 or fewer are red, the shipment is accepted; otherwise it is rejected. Using probability, determine whether this is a "good" sampling plan. (Assume that a bad shipment is one that has 20 percent reds.) Answer: The sampling plan calls for a random sample of n =10 items with a cut-off of a = 1 red items. If 1 or fewer reds are found the shipment will be accepted; otherwise it will be rejected. The objectives are: 1. If the shipment meets the contract (no greater than 10 percent reds) we want to accept the shipment. 2. If the shipment violates the contract (more than 10 percent reds) we want to reject the shipment. The binomial distribution can be used to determine the probability of meeting these objectives. First, we find P(x ≤ = 1, n = 10, p = 0.10) from the binomial table to be 0.7361. This is the probability that we will meet the first objective. Next, we find P(x ≥ 2, n = 10, p = 0.20) from the binomial table to be 0.6242, which is the probability of meeting the second objective. We would like both probabilities to be high (close to 1.000). These are middle-of-the-road and we most likely would conclude that the plan as stated is inadequate. However, the final decision needs to be based on the costs of not meeting the desired objectives. Diff: 3 Keywords: binomial distribution, sampling plan Section: 5-2 The Binomial Probability Distribution Outcome: 2 158) The Swanson Auto Body business repaints cars that have been in an accident or which are in need of a new paint job. Its quality standards call for an average of 1.2 paint defects per door panel. Explain why there is a difference between the probability of finding exactly 1 defect when 1 door panel is inspected and finding exactly 2 defects when 2 doors are inspected. Answer: First, using the Poisson distribution, the probability of exactly 1 defect when 1 door panel is inspected is 0.3614. The probability of twice that many defects (x = 2) when 2 door panels are examined is 0.2613. The reason that these probabilities are different, even though it might seem like they should be the same, is that when the mean is changed from 1.2 to 2.4 when going from 1 to 2 door panels, the total probability (that sums to 1.0 for the probability distribution) is spread over more possible outcomes, and that means that the probability of any one individual value occurring will be lower. Diff: 3 Keywords: Poisson distribution, probability Section: 5-3 Other Discrete Probability Distributions Outcome: 3 159) Under what conditions is the binomial distribution symmetric? Answer: The binomial distribution is perfectly symmetric when p = 0.50 for any size sample. It also approaches symmetry when p is not equal to 0.50 as the sample size increases. Diff: 2 Keywords: binomial distribution, symmetric Section: 5-3 Other Discrete Probability Distributions Outcome: 2

160) The manager of a fast food store realizes that his staffing problems are a result of the variation in the number of customers that arrive at the store. If the same number of customers came each hour, she would know exactly how many servers to have working. It turns out that the Poisson distribution works well to describe the arrivals of customers in any given hour. Explain why the manager has more trouble staffing the store during those hours when the average arrival rate is higher. Answer: The Poisson distribution has a very special characteristic, the standard deviation is equal to the square root of the mean. Thus, when the mean is small, the standard deviation is also small making it easier to plan. However, in hours where the mean arrival rate is higher, the standard deviation is also higher, thereby causing staffing problems. Diff: 2 Keywords: Poisson distribution, standard deviation, mean Section: 5-3 Other Discrete Probability Distributions Outcome: 3 161) A mid-management team consists of 10 people, 6 males and 4 females. Recently top management selected 4 people from this team for promotion. It was stated that the selections were based on random selection. All 4 people selected were males. The females are upset and believe that there may have been more than random selection involved here. What probability distribution should be used to analyze this situation and what is the probability that all 4 promotions would go to males if the selections were random? Do you believe that the females have a valid complaint in this situation? Answer: The appropriate probability distribution in this case is the hypergeometric since the number of promotions given is discrete and since the sampling is without replacement from a finite population. The probability that 4 out of 4 promotions would go to males can be computed using the hypergeometric formula:

P(x) =

Where: x = 0 = number of females selected N = 10 employees X = 4 = number of females n = 4 = number of promotions granted Then we get:

P(0) =

= .0714

Therefore, there is a .0714 chance that this result would occur due to chance alone. This is a fairly small probability. Thus, we might not expect to see such a result. The females have some evidence to support a claim that something other than random selection was involved. However, there is still a .07 chance that this result would come from random selection. Diff: 2 Keywords: hypergeometric distribution, probability Section: 5-3 Other Discrete Probability Distributions Outcome: 3 5-54 Copyright © 2018 Pearson Education, Inc.

Business Statistics, 9e (Groebner/Shannon/Fry) Chapter 6 Introduction to Continuous Probability Distributions 1) The normal distribution is one of the most frequently used discrete probability distributions. Answer: FALSE Diff: 1 Keywords: normal distribution Section: 6-1 The Normal Probability Distribution Outcome: 1 2) Typically, a continuous random variable is one whose value is determined by measurement instead of counting. Answer: TRUE Diff: 1 Keywords: continuous, random, variable Section: 6-1 The Normal Probability Distribution Outcome: 3 3) The number of defects manufactured by workers in a small engine plant is an example of a discrete random variable. Answer: TRUE Diff: 1 Keywords: discrete, random, variable Section: 6-1 The Normal Probability Distribution Outcome: 3 4) One example of a difference between discrete random variables and continuous random variables is that in a discrete distribution P(x > 2) = P(x ≥ 3) while in a continuous distribution P(x > 2) is treated the same as P(x ≥ 2). Answer: TRUE Diff: 2 Keywords: discrete, continuous, random, variable Section: 6-1 The Normal Probability Distribution Outcome: 3 5) The probability distribution for a continuous random variable is represented by a probability density function that defines a curve. Answer: TRUE Diff: 1 Keywords: probability, density, continuous, variable Section: 6-1 The Normal Probability Distribution Outcome: 2

6) When graphed, the probability distribution for a discrete random variable looks like a histogram. Answer: TRUE Diff: 2 Keywords: discrete, variable, distribution Section: 6-1 The Normal Probability Distribution Outcome: 3 7) For a continuous distribution the total area under the curve is equal to 100. Answer: FALSE Diff: 1 Keywords: normal, distribution, continuous, variable Section: 6-1 The Normal Probability Distribution Outcome: 3 8) A continuous random variable approaches normality as the level of skewness increases. Answer: FALSE Diff: 2 Keywords: continuous, variable, random Section: 6-1 The Normal Probability Distribution Outcome: 3 9) If the mean, median and mode are all equal for a continuous random variable, then the random variable is normally distributed. Answer: FALSE Diff: 2 Keywords: continuous, random, variable, normal Section: 6-1 The Normal Probability Distribution Outcome: 3 10) When a single die is rolled, each of the six sides are equally likely. This is an example of a uniform distribution. Answer: TRUE Diff: 1 Keywords: uniform, random, variable Section: 6-2 Other Continuous Probability Distributions Outcome: 1 11) All symmetric distributions can be assumed normally distributed. Answer: FALSE Diff: 1 Keywords: normal, probability Section: 6-1 The Normal Probability Distribution Outcome: 1

12) The parameters of a normal distribution are the mean and the standard deviation. Answer: TRUE Diff: 1 Keywords: normal, distribution, mean, standard deviation Section: 6-1 The Normal Probability Distribution Outcome: 1 13) The actual weight of 2-pound sacks of salted peanuts is found to be normally distributed with a mean equal to 2.04 pounds and a standard deviation of 0.25 pounds. Given this information, the probability of a sack weighing more than 2.40 pounds is 0.4251. Answer: FALSE Diff: 2 Keywords: standardized normal, z-value Section: 6-1 The Normal Probability Distribution Outcome: 1 14) The standard normal distribution table provides probabilities for the area between the z-value and the population mean. Answer: TRUE Diff: 2 Keywords: z-value, standard normal distribution Section: 6-1 The Normal Probability Distribution Outcome: 1 15) The standard normal distribution has a mean of 0 and a standard deviation of 1.0. Answer: TRUE Diff: 1 Keywords: standard normal Section: 6-1 The Normal Probability Distribution Outcome: 1 16) A manufacturer advertised that the time it takes a person to assemble a bookcase has been shown to be normally distributed with a mean equal to 125 minutes with a standard deviation equal to 20 minutes. Given this information, the probability that it will take a randomly selected person more than 105 minutes is about 0.1587. Answer: FALSE Diff: 2 Keywords: z-value, standard normal, probability Section: 6-1 The Normal Probability Distribution Outcome: 2 17) The State Department of Forests has determined that annual tree growth in a particular forest area is normally distributed with a mean equal to 17 inches and a standard deviation equal to 6 inches. Based on this information, it is possible for a randomly selected tree not to have grown any during a year. Answer: TRUE Diff: 2 Keywords: z-value, standard normal, probability Section: 6-1 The Normal Probability Distribution 6-3 Copyright © 2018 Pearson Education, Inc.

Outcome: 2 18) The State Department of Forests has determined that annual tree growth in a particular forest area is normally distributed with a mean equal to 17 inches and a standard deviation equal to 6 inches. If 2 trees are randomly chosen, the probability that both trees will have grown more than 20 inches during the year is approximately .037. Answer: FALSE Diff: 3 Keywords: z-value, standard normal, probability Section: 6-1 The Normal Probability Distribution Outcome: 2 19) Watersports Rental at Flathead Lake rents jet skis and power boats for day use. Each piece of equipment has a clock that records the time that it was actually in use while rented. The company has observed over time that the distribution of time used is normally distributed with a mean of 3.6 hours and a standard deviation equal to 1.2 hours. Watersports management has decided to give a rebate to customers who use the equipment for less than 2.0 hours. Based on this information, the probability that a customer will get the rebate is 0.4082. Answer: FALSE Diff: 2 Keywords: z-value, standard normal, probability Section: 6-1 The Normal Probability Distribution Outcome: 2 20) Watersports Rental at Flathead Lake rents jet skis and power boats for day use. Each piece of equipment has a clock that records the time that it was actually in use while rented. The company has observed over time that the distribution of time used is normally distributed with a mean of 3.6 hours and a standard deviation equal to 1.2 hours. Watersports management has decided to give a rebate to customers who use the equipment for only a short amount of time. They wish to grant a rebate to no more than 10 percent of all customers. Based on the information provided, the amount of time that should be set as the cut-off between getting the rebate and not getting the rebate is approximately 2.06 hours. Answer: TRUE Diff: 2 Keywords: z-value, standard normal, probability Section: 6-1 The Normal Probability Distribution Outcome: 2 21) The Varden Packaging Company has a contract to fill 50 gallon barrels with gasoline for use by the U.S. Army. The machine that Varden uses has an adjustable device that allows the average fill per barrel to be adjusted as desired. However, the actual distribution of fill volume from the machine is known to be normally distributed with a standard deviation equal to 0.5 gallons. The contract that Varden has with the military calls for no more than 2 percent of all barrels to contain less than 49.2 gallons of gasoline. In order to meet this requirement, Varden should set the mean fill to approximately 49.92 gallons. Answer: FALSE Diff: 3 Keywords: z-value, standard normal, probability Section: 6-1 The Normal Probability Distribution 6-4 Copyright © 2018 Pearson Education, Inc.

Outcome: 2 22) The Varden Packaging Company has a contract to fill 50-gallon barrels with gasoline for use by the U.S. Army. The machine that Varden uses has an adjustable device that allows the average fill per barrel to be adjusted as desired. However, the actual distribution of fill volume from the machine is known to be normally distributed with a standard deviation equal to 0.5 gallons. The contract that Varden has with the military calls for no more than 2 percent of all barrels to contain less than 49.2 gallons of gasoline. In order to meet this requirement, Varden should set the mean fill to approximately 50.225 gallons. Answer: TRUE Diff: 3 Keywords: z-value, standard normal, probability Section: 6-1 The Normal Probability Distribution Outcome: 2 23) The Varden Packaging Company has a contract to fill 50-gallon barrels with gasoline for use by the U.S. Army. The machine that Varden uses has an adjustable device that allows the average fill per barrel to be adjusted as desired. However, the actual distribution of fill volume from the machine is known to be normally distributed with a standard deviation equal to 0.5 gallons. The contract that Varden has with the military calls for no more than 2 percent of all barrels to contain less than 49.2 gallons of gasoline. Suppose Varden managers are unwilling to set the mean fill at any level higher than 50 gallons. Given that, in order to meet the requirements, they will need to increase the standard deviation of fill volume. Answer: FALSE Diff: 3 Keywords: z-value, standard normal, probability Section: 6-1 The Normal Probability Distribution Outcome: 2 24) A seafood shop sells salmon fillets where the weight of each fillet is normally distributed with a mean of 1.6 pounds and a standard deviation of 0.3 pounds. They want to classify the largest fillets as extra large and charge a higher price for them. If they want the largest 15 percent of the fillets to be classified as extra large, the minimum weight for an extra large fillet should be 1.91 pounds. Answer: TRUE Diff: 3 Keywords: z-value, standard normal, probability Section: 6-1 The Normal Probability Distribution Outcome: 2 25) A seafood shop sells salmon fillets where the weight of each fillet is normally distributed with a mean of 1.6 pounds and a standard deviation of 0.3 pounds. Based on this information we can conclude that 90 percent of the fillets weight more than 1.0 pound. Answer: FALSE Diff: 2 Keywords: z-value, standard normal, probability Section: 6-1 The Normal Probability Distribution Outcome: 2

26) The miles per gallon for hybrid vehicles on city streets have been determined to be normally distributed with a mean of 33.2 mpg and a variance of 16. Based on this information, the probability that if three randomly selected vehicles are monitored and that two of the three will exceed the 35 mpg is slightly greater than 0.18. Answer: FALSE Diff: 3 Keywords: z-value, standard normal, probability Section: 6-1 The Normal Probability Distribution Outcome: 2 27) Any normal distribution can be converted to a standard normal distribution. Answer: TRUE Diff: 2 Keywords: standard normal, probability Section: 6-1 The Normal Probability Distribution Outcome: 2 28) For a normal distribution, the probability of a value being between a positive z-value and its population mean is the same as that of a value being between a negative z-value and its population mean. Answer: TRUE Diff: 3 Keywords: z-value, standard normal, probability Section: 6-1 The Normal Probability Distribution Outcome: 2 29) Suppose the time it takes for a customer to be served at a fast-food chain business is thought to be uniformly distributed between 3 and 8 minutes, then the probability that it will take exactly 5 minutes is 0.20. Answer: FALSE Diff: 2 Keywords: uniform, probability Section: 6-2 Other Continuous Probability Distributions Outcome: 4 30) One of the basic differences between a uniform probability distribution and a normal probability distribution is that the uniform is symmetrical but the normal is skewed depending on the value of the standard deviation. Answer: FALSE Diff: 1 Keywords: uniform, normal, distribution, skewed Section: 6-2 Other Continuous Probability Distributions Outcome: 4

31) Suppose the time it takes for a customer to be served at a fast-food chain business is thought to be uniformly distributed between 3 and 8 minutes, then the probability that a customer is served in less than 3 minutes is 0. Answer: TRUE Diff: 1 Keywords: uniform distribution Section: 6-2 Other Continuous Probability Distributions Outcome: 4 32) If the time it takes for a customer to be served at a fast-food chain business is thought to be uniformly distributed between 3 and 8 minutes, then the probability that the time it takes for a randomly selected customer to be served will be less than 5 minutes is 0.40. Answer: TRUE Diff: 2 Keywords: uniform, probability, distribution Section: 6-2 Other Continuous Probability Distributions Outcome: 4 33) If a uniform distribution and normal distribution both have the same mean and the same range, the normal distribution will have a larger standard deviation than the uniform distribution Answer: FALSE Diff: 2 Keywords: uniform distribution, normal distribution Section: 6-2 Other Continuous Probability Distributions Outcome: 4 34) It has been determined the weight of bricks made by the Dillenger Stone Company is uniformly distributed between 1 and 1.5 pounds. Based on this information, the probability that two randomly selected bricks will each weigh more than 1.3 pounds is 0.16. Answer: TRUE Diff: 3 Keywords: uniform distribution, probability Section: 6-2 Other Continuous Probability Distributions Outcome: 4 35) The amount of drying time for the paint applied to a plastic component part is thought to be uniformly distributed between 30 and 60 minutes. Currently, the automated process selects the part from the drying bin after the part has been there for 50 minutes. Based on this, the probability that a part selected will not be dry is approximately 0.33. Answer: TRUE Diff: 2 Keywords: uniform distribution, probability Section: 6-2 Other Continuous Probability Distributions Outcome: 4

36) The amount of drying time for the paint applied to a plastic component part is thought to be uniformly distributed between 30 and 60 minutes. Currently, the automated process selects the part from the drying bin after the part has been there for 50 minutes. The probability that none of three parts picked are still wet when they are selected is approximately 0.04. Answer: FALSE Diff: 3 Keywords: uniform distribution, probability Section: 6-2 Other Continuous Probability Distributions Outcome: 4 37) An assembly process takes between 20 and 40 minutes to complete with the distribution of time thought to be uniformly distributed. Based on this, the percentage of assemblies that require less than 25 minutes is 0.05. Answer: FALSE Diff: 2 Keywords: uniform distribution, probability Section: 6-2 Other Continuous Probability Distributions Outcome: 4 38) Service time for customers at a drive-through coffee shop has been shown to be uniformly distributed between 2 and 10 minutes. Customers will complain when service time exceeds 7.5 minutes. Based on this information, the probability of getting a complaint based on service time is 0.3125. Answer: TRUE Diff: 2 Keywords: uniform distribution, probability Section: 6-2 Other Continuous Probability Distributions Outcome: 4 39) An electronics repair shop has determined that the time between failures for a particular electronic component part is exponentially distributed with a mean time between failures of 200 hours. Based on this information, the probability that a part will fail in the first 20 hours is approximately 0.095. Answer: TRUE Diff: 3 Keywords: exponential distribution, probability Section: 6-2 Other Continuous Probability Distributions Outcome: 5 40) An electronics repair shop has determined that the time between failures for a particular electronic component part is exponentially distributed with a mean time between failures of 200 hours. Based on this information, the probability that a part will fail between 20 and 100 hours is approximately 0.30. Answer: TRUE Diff: 3 Keywords: exponential distribution, probability Section: 6-2 Other Continuous Probability Distributions Outcome: 5

41) An electronics repair shop has determined that the time between failures for a particular electronic component part is exponentially distributed with a mean time between failures of 200 hours. Based on this information, the probability that a part will not fail in the first 200 hours is 0.50. Answer: FALSE Diff: 2 Keywords: exponential distribution, probability Section: 6-2 Other Continuous Probability Distributions Outcome: 4 42) A study of cars arriving at a parking structure at the local airport shows that the time between arrivals is 1.2 minutes and is exponentially distributed. Based on this information, the mean number of cars arriving per minute is about 0.83. Answer: TRUE Diff: 2 Keywords: exponential distribution, lambda Section: 6-2 Other Continuous Probability Distributions Outcome: 4 43) A study of cars arriving at a parking structure at the local airport shows that the time between arrivals is 1.2 minutes and is exponentially distributed. The probability that more than 2 minutes will elapse between the arrivals of cars is about 0.81. Answer: FALSE Diff: 3 Keywords: exponential distribution, probability Section: 6-2 Other Continuous Probability Distributions Outcome: 4 44) Which of the following probability distributions can be used to describe the distribution for a continuous random variable? A) Normal distribution B) Binomial distribution C) Poisson distribution D) Hypergeometric Answer: A Diff: 1 Keywords: normal, distribution Section: 6-1 The Normal Probability Distribution Outcome: 1

45) Which of the following is not a characteristic of the normal distribution? A) Symmetric B) Mean = median = mode C) Bell-shaped D) Equal probabilities at all values of x Answer: D Diff: 2 Keywords: normal, uniform, distribution Section: 6-1 The Normal Probability Distribution Outcome: 1 46) Which of the following probability distributions could be used to describe the distribution for a continuous random variable? A) Exponential distribution B) Normal distribution C) Uniform distribution D) All of the above Answer: D Diff: 1 Keywords: normal, exponential, uniform, continuous, distribution Section: 6-1 The Normal Probability Distribution Outcome: 1 47) Assuming that the change in daily closing prices for stocks on the New York Stock Exchange is a random variable that is normally distributed with a mean of $.35 and a standard deviation of $.33. Based on this information, what is the probability that a randomly selected stock will close up $.75 or more? A) 0.3869 B) 0.1131 C) 0.7100 D) 0.8869 Answer: B Diff: 2 Keywords: normal, probability, z-value Section: 6-1 The Normal Probability Distribution Outcome: 2

48) Assuming that the change in daily closing prices for stocks on the New York Stock Exchange is a random variable that is normally distributed with a mean of $0.35 and a standard deviation of $0.33. Based on this information, what is the probability that a randomly selected stock will be lower by $0.40 or more? A) 2.27 B) 0.4884 C) 0.0116 D) 0.9884 Answer: C Diff: 3 Keywords: normal, probability, z-value Section: 6-1 The Normal Probability Distribution Outcome: 2 49) Students who have completed a speed reading course have reading speeds that are normally distributed with a mean of 950 words per minute and a standard deviation equal to 220 words per minute. Based on this information, what is the probability of a student reading at more than 1400 words per minute after finishing the course? A) 0.0202 B) 0.5207 C) 0.4798 D) 0.9798 Answer: A Diff: 2 Keywords: normal, probability, z-value Section: 6-1 The Normal Probability Distribution Outcome: 2 50) Students who have completed a speed reading course have reading speeds that are normally distributed with a mean of 950 words per minute and a standard deviation equal to 220 words per minute. If two students were selected at random, what is the probability that they would both read at less than 400 words per minute? A) 0.4938 B) 0.0062 C) 0.00004 D) 0.2438 Answer: C Diff: 3 Keywords: normal, probability, z-value, multiplication Section: 6-1 The Normal Probability Distribution Outcome: 2

51) The manager at a local movie theater has collected data for a long period of time and has concluded that the revenue from concession sales during the first show each evening is normally distributed with a mean equal to $336.25 and a variance equal to 1,456. Based on this information, what are the chances that the revenue on the first show will exceed $800? A) 0.1255 B) Essentially zero C) 0.3745 D) 0.9999 Answer: B Diff: 2 Keywords: normal, z-value, probability Section: 6-1 The Normal Probability Distribution Outcome: 2 52) The manager at a local movie theater has collected data for a long period of time and has concluded that the revenue from concession sales during the first show each evening is normally distributed with a mean equal to $336.25 and a standard deviation equal to $80. Based on this information, what are the chances that the revenue on the first show will be between $300 and $500? A) About 0.3062 B) Approximately 0.6534 C) 0.1736 D) Approximately 0.4798 Answer: B Diff: 2 Keywords: normal, probability, z-value Section: 6-1 The Normal Probability Distribution Outcome: 2 53) The manager of a computer help desk operation has collected enough data to conclude that the distribution of time per call is normally distributed with a mean equal to 8.21 minutes and a standard deviation of 2.14 minutes. Based on this, what is the probability that a call will last longer than 13 minutes? A) About 0.0125 B) Approximately 0.4875 C) About 0.5125 D) About 0.9875 Answer: A Diff: 2 Keywords: normal, z-score, probability Section: 6-1 The Normal Probability Distribution Outcome: 2

54) The manager of a computer help desk operation has collected enough data to conclude that the distribution of time per call is normally distributed with a mean equal to 8.21 minutes and a standard deviation of 2.14 minutes. The manager has decided to have a signal system attached to the phone so that after a certain period of time, a sound will occur on her employees' phone if she exceeds the time limit. The manager wants to set the time limit at a level such that it will sound on only 8 percent of all calls. The time limit should be: A) 10.35 minutes. B) approximately 5.19 minutes. C) about 14.58 minutes. D) about 11.23 minutes. Answer: D Diff: 3 Keywords: normal, probability, z-value Section: 6-1 The Normal Probability Distribution Outcome: 2 55) The manager of a computer help desk operation has collected enough data to conclude that the distribution of time per call is normally distributed with a mean equal to 8.21 minutes and a standard deviation of 2.14 minutes. What is the probability that three randomly monitored calls will each be completed in 4 minutes or less? A) 0.4756 B) Approximately 0.1076 C) About 0.00001 D) Can't be determined without more information. Answer: C Diff: 3 Keywords: normal, probability, z-value Section: 6-1 The Normal Probability Distribution Outcome: 2 56) The makers of Sweet-Things candy sell their candy by the box. Based on company policy, the mean target weight of all boxes is 2.0 pounds. To make sure that they are not putting too much in the boxes, the manager wants no more than 3 percent of all boxes to contain more than 2.10 pounds of candy. In order to do this, with a mean weight of 2 pounds, what must the standard deviation be? Assume that the box weights are normally distributed. A) Approximately 0.05 pounds B) -0.133 pounds C) 1.144 pounds D) None of the above Answer: A Diff: 3 Keywords: normal, probability, z-value, standard deviation Section: 6-1 The Normal Probability Distribution Outcome: 2

57) The makers of Sweet-Things candy sell their candy by the box. Based on company policy, the mean target weight of all boxes is 2.0 pounds. To make sure that they are not putting too much in the boxes, the manager wants no more than 3 percent of all boxes to contain more than 2.10 pounds of candy. In order to do this, what should the mean fill weight be set to if the fill standard deviation is 0.13 pounds? Assume that the box weights are normally distributed. A) Just over 2 pounds B) Approximately 2.33 pounds C) Nearly 1.27 pounds D) Approximately 1.86 pounds Answer: D Diff: 3 Keywords: normal, mean, z-value, probability Section: 6-1 The Normal Probability Distribution Outcome: 2 58) A major U.S. automaker has determined that the city mileage for one of its new SUV models is normally distributed with a mean equal to 15.2 mpg. A report issued by the company indicated that 22 percent of the SUV model vehicles will get more than 17 mpg in the city. Given this information, what is the city mileage standard deviation for this SUV model? A) 0.77 mpg B) Approximately 2.34 mpg C) 1.8 mpg D) Approximately 3.1 mpg Answer: B Diff: 3 Keywords: normal, z-value, standard deviation Section: 6-1 The Normal Probability Distribution Outcome: 1 59) A recent study showed that the length of time required for customers to resolve their computer issues with online support is normally distributed with a mean equal to 0.35 hours with a standard deviation of 0.2 hours. Given this information, what is the probability of resolution will take between 10 and 15 minutes? A) About 0.13 B) Nearly 0.00 C) About 0.31 D) About 0.87 Answer: A Diff: 2 Keywords: normal, z-value, probability Section: 6-1 The Normal Probability Distribution Outcome: 2

60) Suppose that it is believed that investor returns on equity investments at a particular brokerage house are normally distributed with a mean of 9 percent and a standard deviation equal to 3.2 percent. What percent of investors at this brokerage house earned at least 5 percent? A) 89.44 percent B) 10.56 percent C) 39.44 percent D) 100 percent Answer: A Diff: 2 Keywords: normal, z-value, probability Section: 6-1 The Normal Probability Distribution Outcome: 2 61) A major cell phone service provider has determined that the number of minutes that its customers use their phone per month is normally distributed with a mean equal to 445.5 minutes with a standard deviation equal to 177.8 minutes. As a promotion, the company plans to hold a drawing to give away one free vacation to Hawaii for a customer who uses between 400 and 402 minutes during a particular month. Based on the information provided, what proportion of the company's customers would be eligible for the drawing? A) Approximately 0.1026 B) About 0.004 C) Approximately 0.2013 D) About 0.02 Answer: B Diff: 2 Keywords: normal, probability, z-value Section: 6-1 The Normal Probability Distribution Outcome: 2 62) A major cell phone service provider has determined that the number of minutes that its customers use their phone per month is normally distributed with a mean equal to 445.5 minutes with a standard deviation equal to 177.8 minutes. The company is thinking of changing its fee structure so that anyone who uses the phone less than 250 minutes during a given month will pay a reduced monthly fee. Based on the available information, what percentage of current customers would be eligible for the reduced fee? A) About 36.4 percent B) Approximately 52 percent C) About 86.6 percent D) About 13.6 percent Answer: D Diff: 2 Keywords: normal, probability, z-value Section: 6-1 The Normal Probability Distribution Outcome: 2

63) A major cell phone service provider has determined that the number of minutes that its customers use their phone per month is normally distributed with a mean equal to 445.5 minutes with a standard deviation equal to 177.8 minutes. The company is thinking of charging a lower rate for customers who use the phone less than a specified amount. If it wishes to give the rate reduction to no more than 12 percent of its customers, what should the cut-off be? A) About 237 minutes B) About 654 minutes C) About 390 minutes D) About 325 minutes Answer: A Diff: 3 Keywords: normal, probability, z-value Section: 6-1 The Normal Probability Distribution Outcome: 2 64) In a standard normal distribution, the probability that z is greater than 0 is: A) 0.5 B) equal to 1 C) at least 0.5 D) 1.96 Answer: A Diff: 1 Keywords: normal, probability, z-value Section: 6-1 The Normal Probability Distribution Outcome: 2 65) In a standard normal distribution, the probability P(-1.00< z < 1.20) is the same as: A) P(1< z < 1.20) - P(0 < z < 1.00). B) P(1< z < 1.20) - 2*P(0 < z < 1.00). C) 2 ∗ P(1< z < 1.20) - P(0 < z <1.00). D) P(1 < z < 1.20) + 2 ∗ P(0 < z <1.00). Answer: D Diff: 3 Keywords: normal, probability, z-value Section: 6-1 The Normal Probability Distribution Outcome: 2

66) A professor noted that the grades of his students were normally distributed with a mean of 75.07 and a standard deviation of 11.65. If only 10 percent of the students received grades of A, what is the minimum score needed to receive an A? A) 80.00 B) 85.00 C) 90.00 D) 95.00 Answer: C Diff: 3 Keywords: normal, probability, z-value Section: 6-1 The Normal Probability Distribution Outcome: 2 67) A store sells 6 different models of cell phones and have found that they sell an equal number of each model. The probability distribution that would describe this random variable is called: A) uniform distribution. B) Poisson distribution. C) continuous distribution. D) relative frequency distribution. Answer: A Diff: 1 Keywords: uniform distribution Section: 6-2 Other Continuous Probability Distributions Outcome: 1 68) Which of the following probability distributions would most likely be used to describe the time between failures for electronic components? A) Binomial distribution B) Exponential distribution C) Uniform distribution D) Normal distribution Answer: B Diff: 1 Keywords: exponential distribution, continuous Section: 6-2 Other Continuous Probability Distributions Outcome: 5

69) It is assumed that the time customers spend in a record store is uniformly distributed between 3 and 12 minutes. Based on this information, what is the probability that a customer will spend more than 9 minutes in the record store? A) 0.33 B) 0.1111 C) 0.67 D) 0.25 Answer: A Diff: 2 Keywords: uniform, probability Section: 6-2 Other Continuous Probability Distributions Outcome: 4 70) It is assumed that the time customers spend in a record store is uniformly distributed between 3 and 12 minutes. Based on this information, what is the probability that a customer will be exactly 7.50 minutes in the record store? A) 0.1250 B) 0.05 C) Essentially zero D) 0.111 Answer: C Diff: 1 Keywords: uniform, probability Section: 6-2 Other Continuous Probability Distributions Outcome: 4 71) Employees at a large computer company earn sick leave in one-minute increments depending on how many hours per month they work. They can then use the sick leave time any time throughout the year. Any unused time goes into a sick bank account that they or other employees can use in the case of emergencies. The human resources department has determined that the amount of unused sick time for individual employees is uniformly distributed between 0 and 480 minutes. Based on this information, what is the probability that an employee will have less than 20 minutes of unused sick time? A) 0.002 B) 0.966 C) 0.063 D) 0.042 Answer: D Diff: 2 Keywords: uniform, probability Section: 6-2 Other Continuous Probability Distributions Outcome: 4

72) Employees at a large computer company earn sick leave in one-minute increments depending on how many hours per month they work. They can then use the sick leave time any time throughout the year. Any unused time goes into a sick bank account that they or other employees can use in the case of emergencies. The human resources department has determined that the amount of unused sick time for individual employees is uniformly distributed between 0 and 480 minutes. Based on this information, what is the probability that three randomly chosen employees have over 400 unused sick minutes at the end of the year? A) 0.1667 B) 0.0046 C) 0.5001 D) 0.0300 Answer: B Diff: 3 Keywords: uniform, probability, multiplication Section: 6-2 Other Continuous Probability Distributions Outcome: 4 73) Employees at a large computer company earn sick leave in one-minute increments depending on how many hours per month they work. They can then use the sick leave time any time throughout the year. Any unused time goes into a sick bank account that they or other employees can use in the case of emergencies. The human resources department has determined that the amount of unused sick time for individual employees is uniformly distributed between 0 and 480 minutes. The company has decided to give a cash payment to any employee that returns over 400 minutes of sick leave at the end of the year. What percentage of employees could expect a cash payment? A) 16.67 percent B) 0.1667 percent C) Just over 43 percent D) 80 percent Answer: A Diff: 2 Keywords: uniform, probability Section: 6-2 Other Continuous Probability Distributions Outcome: 4

74) Employees at a large computer company earn sick leave in one-minute increments depending on how many hours per month they work. They can then use the sick leave time any time throughout the year. Any unused time goes into a sick bank account that they or other employees can use in the case of emergencies. The human resources department has determined that the amount of unused sick time for individual employees is uniformly distributed between 0 and 480 minutes. The company has decided to give a cash payment to any employee that returns over a specified amount of sick leave minutes. Assuming that the company wishes no more than 5 percent of all employees to get a cash payment, what should the required number of minutes be? A) 24 minutes B) 419 minutes C) 456 minutes D) 470 minutes Answer: C Diff: 2 Keywords: uniform, probability Section: 6-2 Other Continuous Probability Distributions Outcome: 4 75) It is thought that the time between customer arrivals at a fast food business is exponentially distributed with λ equal to 5 customers per hour. Given this information, what is the mean time between arrivals? A) 12 minutes B) 5 minutes C) 5 hours D) 2 minutes Answer: A Diff: 1 Keywords: exponential, expected value Section: 6-2 Other Continuous Probability Distributions Outcome: 5 76) It is assumed that the time failures for an electronic component are exponentially distributed with a mean of 50 hours between consecutive failures. Based on this information, what is the probability that a randomly selected part will fail in less than 10 hours? A) About 0.82 B) About 0.20 C) About 0.33 D) About 0.18 Answer: D Diff: 2 Keywords: exponential, probability Section: 6-2 Other Continuous Probability Distributions Outcome: 5

77) It is assumed that the time failures for an electronic component are exponentially distributed with a mean of 50 hours between consecutive failures. What is the probability that a component will be functioning after 60 hours? A) Approximately 0.30 B) About 0.70 C) About 0.21 D) About 0.49 Answer: A Diff: 2 Keywords: exponential, probability Section: 6-2 Other Continuous Probability Distributions Outcome: 5 78) It is assumed that the time failures for an electronic component are exponentially distributed with a mean of 50 hours between consecutive failures. If one extra component is installed as a backup, what is the probability of at least one of the two components working for at least 60 hours? A) About 0.51 B) About 0.09 C) About 0.06 D) About 0.70 Answer: A Diff: 3 Keywords: exponential, probability Section: 6-2 Other Continuous Probability Distributions Outcome: 5 79) The transportation manager for the State of New Jersey has determined that the time between arrivals at a toll booth on the state's turnpike is exponentially distributed with λ = 4 cars per minute. Based on this, the average time between arrivals is: A) 15 seconds. B) 12 seconds. C) 25 seconds. D) 4 minutes. Answer: A Diff: 2 Keywords: exponential, lambda Section: 6-2 Other Continuous Probability Distributions Outcome: 5

80) The transportation manager for the State of New Jersey has determined that the time between arrivals at a toll booth on the state's turnpike is exponentially distributed with λ = 4 cars per minute. Based on this information, the standard deviation for the time between arrivals is: A) 25 seconds. B) 3.87 seconds. C) 15 seconds. D) 2 minutes. Answer: C Diff: 1 Keywords: exponential, lambda, mean, standard deviation Section: 6-2 Other Continuous Probability Distributions Outcome: 5 81) The transportation manager for the State of New Jersey has determined that the time between arrivals at a toll booth on the state's turnpike is exponentially distributed with λ = 4 cars per minute. Based on this information, what is the probability that the time between any two cars arriving will exceed 11 seconds? A) Approximately 1.0 B) Approximately 0.48 C) About 0.52 D) About 0.75 Answer: B Diff: 2 Keywords: exponential, lambda, probability Section: 6-2 Other Continuous Probability Distributions Outcome: 5 82) The time between calls to an emergency 911-call center is exponentially distributed with a mean time between calls of 645 seconds. Based on this information, what is the probability that the time between the next two calls is between 200 and 400 seconds? A) Approximately 0.47 B) About 0.199 C) About 0.747 D) About 0.801 Answer: B Diff: 3 Keywords: exponential, probability Section: 6-2 Other Continuous Probability Distributions Outcome: 5

83) For a standardized normal distribution, calculate P(z < 1.5). A) 0.9332 B) 0.0668 C) 0.333 D) 0.667 Answer: A Diff: 1 Keywords: normal distribution, probability Section: 6-1 The Normal Probability Distribution Outcome: 1 84) For a standardized normal distribution, calculate P(z ≥ 0.85). A) 0.8033 B) 0.1977 C) 0.2340 D) 0.7660 Answer: B Diff: 1 Keywords: normal distribution, probability Section: 6-1 The Normal Probability Distribution Outcome: 1 85) For a standardized normal distribution, calculate P(-1.28 < z < 1.75). A) 0.3997 B) 0.4599 C) 0.1404 D) 0.8596 Answer: D Diff: 1 Keywords: normal distribution, probability Section: 6-1 The Normal Probability Distribution Outcome: 1 86) For a standardized normal distribution, calculate P(0.00 < z < 2.33). A) 0.7181 B) 0.5099 C) 0.4901 D) 0.2819 Answer: C Diff: 1 Keywords: normal distribution, probability Section: 6-1 The Normal Probability Distribution Outcome: 1

87) For a standardized normal distribution, calculate P(-1.00 < z < 1.00). A) 0.6826 B) 0.6667 C) 0.4572 D) 0.5521 Answer: A Diff: 1 Keywords: normal distribution, probability Section: 6-1 The Normal Probability Distribution Outcome: 1 88) For a standardized normal distribution, calculate P(1.78 < z < 2.34). A) 0.0124 B) 0.0341 C) 0.0412 D) 0.0279 Answer: D Diff: 1 Keywords: normal distribution, probability Section: 6-1 The Normal Probability Distribution Outcome: 1 89) For a standardized normal distribution, determine a value, say z0, so that P(0 < z < z0) = 0.4772. A) 2.00 B) 2.33 C) 1.85 D) 1.66 Answer: A Diff: 1 Keywords: normal distribution, probability Section: 6-1 The Normal Probability Distribution Outcome: 1 90) For a standardized normal distribution, determine a value, say z0, so that P(-z0 ≤ z < 0) = 0.45. A) 1.84 B) 1.645 C) 1.96 D) 1.33 Answer: B Diff: 1 Keywords: normal distribution, probability Section: 6-1 The Normal Probability Distribution Outcome: 1

91) For a standardized normal distribution, determine a value, say z0, so that P(-z0 ≤ z ≤ z0) = 0.95. A) 2.14 B) 1.65 C) 1.96 D) 1.24 Answer: C Diff: 1 Keywords: normal distribution, probability Section: 6-1 The Normal Probability Distribution Outcome: 1 92) For a standardized normal distribution, determine a value, say z0, so that P(z > z0) = 0.025. A) 1.96 B) 1.65 C) 1.24 D) 2.14 Answer: A Diff: 1 Keywords: normal distribution, probability Section: 6-1 The Normal Probability Distribution Outcome: 1 93) For a standardized normal distribution, determine a value, say z0, so that P(z ≤ z0) = 0.01. A) -2.33 B) -1.96 C) 2.33 D) 1.96 Answer: A Diff: 1 Keywords: normal distribution, probability Section: 6-1 The Normal Probability Distribution Outcome: 1 94) Consider a random variable, z, that has a standardized normal distribution. Determine P(0 < z < 1.96). A) 0.1250 B) 0.5250 C) 0.3250 D) 0.4750 Answer: D Diff: 1 Keywords: normal distribution, probability Section: 6-1 The Normal Probability Distribution Outcome: 1

95) Consider a random variable, z, that has a standardized normal distribution. Determine P(z > 1.645). A) 0.05 B) 0.01 C) 0.03 D) 0.45 Answer: A Diff: 1 Keywords: normal distribution, probability Section: 6-1 The Normal Probability Distribution Outcome: 1 96) Consider a random variable, z, that has a standardized normal distribution. Determine P (1.28 < z < 2.33). A) 0.0126 B) 0.3997 C) 0.0904 D) 0.4901 Answer: C Diff: 1 Keywords: normal distribution, probability Section: 6-1 The Normal Probability Distribution Outcome: 1 97) Consider a random variable, z, that has a standardized normal distribution. Determine (-2 ≤ z ≤ 3). A) 0.12414 B) 0.97587 C) 0.47722 D) 0.49865 Answer: B Diff: 1 Keywords: normal distribution, probability Section: 6-1 The Normal Probability Distribution Outcome: 1 98) Consider a random variable, z, that has a standardized normal distribution. Determine P(z > -1). A) 0.8413 B) 0.1251 C) 0.1512 D) 0.2124 Answer: A Diff: 1 Keywords: normal distribution, probability Section: 6-1 The Normal Probability Distribution Outcome: 1

99) A random variable, x, has a normal distribution with μ = 13.6 and σ = 2.90. Determine a value, x0, so that P(x > x0) = 0.05. A) 14.46 B) 15.33 C) 18.37 D) 12.45 Answer: C Diff: 2 Keywords: normal distribution, probability Section: 6-1 The Normal Probability Distribution Outcome: 1 100) A random variable, x, has a normal distribution with μ = 13.6 and σ = 2.90. Determine a value, x0, so that P(x ≤ x0) = 0.975. A) 16.678 B) 19.284 C) 23.360 D) 14.475 Answer: B Diff: 2 Keywords: normal distribution, probability Section: 6-1 The Normal Probability Distribution Outcome: 1 101) A random variable, x, has a normal distribution with μ = 13.6 and σ = 2.90. Determine a value, x0, so that P(μ - x0 ≤ x ≤ μ + x0) = 0.95. A) 7.916 B) 4.535 C) 3.178 D) 9.425 Answer: A Diff: 2 Keywords: normal distribution, probability Section: 6-1 The Normal Probability Distribution Outcome: 1 102) For the normal distribution with parameters μ = 5, σ= 2; calculate P(0 < x < 8). A) 0.8023 B) 0.4152 C) 0.9270 D) 0.8845 Answer: C Diff: 2 Keywords: normal distribution, probability Section: 6-1 The Normal Probability Distribution Outcome: 1

103) For the normal distribution with parameters μ = 5, σ = 4; calculate P(0 < x < 8). A) 0.8841 B) 0.8812 C) 0.4215 D) 0.6678 Answer: D Diff: 2 Keywords: normal distribution, probability Section: 6-1 The Normal Probability Distribution Outcome: 1 104) For the normal distribution with parameters μ = 3, σ = 2; calculate P(0 < x < 8). A) 0.3124 B) 0.9270 C) 0.8123 D) 0.6723 Answer: B Diff: 2 Keywords: normal distribution, probability Section: 6-1 The Normal Probability Distribution Outcome: 1 105) For the normal distribution with parameters μ = 4, σ = 3; calculate P(x > 1). A) 0.8413 B) 0.4562 C) 0.7812 D) 0.4152 Answer: A Diff: 2 Keywords: normal distribution, probability Section: 6-1 The Normal Probability Distribution Outcome: 1 106) For the normal distribution with parameters μ = 0, σ = 3; calculate P(x > 1). A) 0.5812 B) 0.1214 C) 0.3707 D) 0.4412 Answer: C Diff: 2 Keywords: normal distribution, probability Section: 6-1 The Normal Probability Distribution Outcome: 1

107) A randomly selected customer support wait time (from a normal distribution) is calculated to be 1.8 standard deviations below the mean. What is the probability that another randomly selected customer wait time from the distribution will be less than 1.8 standard deviations below the mean? A) 0.0718 B) 0.9281 C) 0.0359 D) 0.0024 Answer: C Diff: 2 Keywords: normal distribution, probability Section: 6-1 The Normal Probability Distribution Outcome: 1 108) A randomly selected customer support wait time (from a normal distribution) is calculated to be 1.8 standard deviations above its mean. What is the probability that another randomly selected customer wait time from the distribution will be less than 1.8 standard deviations from the mean? A) 0.0359 B) 0.9641 C) 0.9281 D) 0.9281 Answer: B Diff: 2 Keywords: normal distribution, probability Section: 6-1 The Normal Probability Distribution Outcome: 1 109) A random variable is normally distributed with a mean of -50 and a standard deviation of 10. If an observation is randomly selected from the distribution, what value will be exceeded 10% of the time? A) -37.20 B) -62.80 C) 62.80 D) 37.20 Answer: A Diff: 2 Keywords: normal distribution, probability Section: 6-1 The Normal Probability Distribution Outcome: 1

110) A random variable is normally distributed with a mean of -50 and a standard deviation of 10. If an observation is randomly selected from the distribution, what value will be exceeded 85% of the time? A) -60.34 B) 60.34 C) -39.66 D) 39.66 Answer: C Diff: 2 Keywords: normal distribution, probability Section: 6-1 The Normal Probability Distribution Outcome: 1 111) A random variable is normally distributed with a mean of 25 and a standard deviation of 5. If an observation is randomly selected from the distribution, determine two values of which the smallest has 25% of the values below it and the largest has 25% of the values above it. A) 18.85 and 27.94 B) 19.31 and 21.12 C) 16.23 and 18.82 D) 21.65 and 28.35 Answer: D Diff: 2 Keywords: normal distribution, probability Section: 6-1 The Normal Probability Distribution Outcome: 1 112) A random variable is normally distributed with a mean of 25 and a standard deviation of 5. If an observation is randomly selected from the distribution, what value will 15% of the observations be below? A) 19.8 B) 16.2 C) 18.7 D) 17.2 Answer: A Diff: 2 Keywords: normal distribution, probability Section: 6-1 The Normal Probability Distribution Outcome: 1 113) If a continuous random variable is said to be exponentially distributed, what would be the easiest way to reduce the standard deviation? Answer: The standard deviation of an exponential distribution is equal to the mean (1/λ) so the easiest way to reduce the standard deviation is to reduce the mean. This would be done by increasing λ. Diff: 1 Keywords: exponential distribution, standard deviation, mean Section: 6-2 Other Continuous Probability Distributions Outcome: 5

114) What is the difference between a normal distribution and the standard normal distribution? Answer: A normal distribution is a bell-shaped distribution defined by two parameters, μ and σ. The values for the mean and standard deviation reflect the population data and may be any values. Thus, a normal distribution may be centered at any value and may have any spread, but it will have the common bell shape. The standard normal distribution is a specific normal distribution with mean = 0 and standard deviation equal to 1.0. The horizontal axis of the standard normal distribution represents z-values. Any normal distribution can be converted to the standard normal by converting the random variable values to z-values. Diff: 2 Keywords: normal distribution, standard normal distribution, mean, standard deviation Section: 6-1 The Normal Probability Distribution Outcome: 1 115) The weight of sacks of potatoes is normally distributed with a mean of 20 pounds and a standard deviation of 2 pounds. The weight of sacks of onions is also normally distributed with a mean of 20 pounds and a standard deviation of 0.50 pounds. Based on this information, which product will yield the highest probability of getting a very heavy sack? Answer: Since both products have the same mean and are both normally distributed, the one with the largest standard deviation will provide the higher probability of a heavy sack. Since potatoes have a standard deviation of 2 pounds compared to 0.50 pounds for onions, you would be more apt to see a very heavy sack of potatoes than onions. Diff: 1 Keywords: normal distribution, mean, standard deviation, probability Section: 6-1 The Normal Probability Distribution Outcome: 2 116) A class takes an exam where the average time to complete the exam is normally distributed with a time of 40 minutes and standard deviation of 9 minutes. If the class lasts 1 hour, what percent of the students will have turned in the exam after 60 minutes? Answer: We are looking for the area under the curve to the left of 60 minutes, which is above the mean so we need the area between 40 and 60 and then 0.5 will be added. z=

= 2.22. Looking up 2.22 in the standard normal table we find 0.4868, which is the area

between 40 and 60. So P(time ≤ 60) = .5 + .4868 = .9868, which means 98.7 percent of class finishes the test before time is up. Diff: 2 Keywords: normal distribution, standard normal distribution, probability Section: 6-1 The Normal Probability Distribution Outcome: 2

117) The fares received by taxi drivers working for the City Taxi line are normally distributed with a mean of $12.50 and a standard deviation of $3.25. Based on this information, what is the probability that a specific fare will exceed $15.00? Answer: Since this is a normal distribution problem, the first step is to convert the normal distribution to a standard normal distribution. We do this by converting the $15.00 value to a z-value using: z=

= 0.7692 ≈ 0.77. Then we go to the standard normal table and locate a z-value =

0.77. The probability corresponding to z = 0.77 is 0.2794. The table in the text always gives the probability between the z-value and the mean. Since we want P(x > $15.00), we need to subtract from 0.50, giving 0.5000 - 0.2794 = 0.2206. Diff: 2 Keywords: normal distribution, standard normal, probability Section: 6-1 The Normal Probability Distribution Outcome: 2 118) The fares received by taxi drivers working for the City Taxi line are normally distributed with a mean of $12.50 and a standard deviation of $3.25. Suppose a driver has four consecutive fares that are less than $6.00. What is the probability of this happening? Answer: Since this is a normal distribution problem, the first step is to convert the normal distribution to a standard normal distribution. We do this by converting the $6.00 value to a z-value using: z=

= -2.00. Then we go to the standard normal table and locate a z-value = -2.00. The

probability corresponding to z = -2.00 is 0.4772. The table in the text always gives the probability between the z-value and the mean. Since we want P(x ≤ $6.00), we need to subtract from 0.50, giving 0.5000 0.4772 = 0.0228. This is the probability of one fare being less than $6.00. To get the probability of four consecutive fares being less than $6.00, we can use the multiplication rule for independent events discussed in Chapter 4. This gives: 0.0228 × 0.0228 × 0.0228 × 0.0228 = 0.0000003. Since this is such a low probability, we would not expect such an event to occur. If it did, then it is likely that the fare distribution has changed. Diff: 2 Keywords: normal distribution, probability Section: 6-1 The Normal Probability Distribution Outcome: 2 119) The money spent by people at an amusement park, after paying to get in the gate, is thought to be uniformly distributed between $5.00 and $25.00. Based on this, what is the probability that someone will spend between $8.00 and $12.00? Answer: The continuous uniform probability distribution has a function: f(x) =

where b is the upper

extreme of the distribution ($25.00) and a is the lower extreme ($5.00). Then f(x) =

= 0.05.

Now, p(8.00 ≤ x ≤ x ≤ 12.00) = f(x)(12.00 - 8.00) = 0.05(4.00) = 0.20. Thus, there is a 0.20 probability that someone will spend between $8.00 and $12.00 after getting into the amusement park. Diff: 2 Keywords: uniform distribution, probability Section: 6-2 Other Continuous Probability Distributions Outcome: 4 6-32 Copyright © 2018 Pearson Education, Inc.

120) In comparing a uniform distribution with a normal distribution where both distributions have the same mean and the same range, explain which distribution will have the larger standard deviation. Answer: A picture of this situation would show a larger area of the uniform distribution farther away from the mean. In the normal distribution, the majority of the distribution is close to the mean and a much smaller proportion is near the edges. As the empirical rule says, about 68 percent should be within one standard deviation of the mean. In the uniform distribution this percentage will be lower (about 58 percent). Further the standard deviation of the normal distribution can be approximated using the 6sigma rule of thumb for the range, meaning that the standard deviation is approximately 1/6 of the range. In the uniform distribution the standard deviation is (range)/3.46, where 3.46 is the square root of 12. All of which gives a larger standard deviation to the uniform distribution. Diff: 3 Keywords: uniform distribution, normal distribution Section: 6-2 Other Continuous Probability Distributions Outcome: 4 121) At the West-Side Drive-Inn, customers arrive at the rate of 10 every 30 minutes. The time between arrivals is exponentially distributed. Given this, what is the mean time between arrivals? Answer: The parameter for the exponential distribution is lambda, λ. This was given as 10 per 30 minutes. Then the mean time between arrivals is

= 0.10. The value 0.10 represents the fraction of

the 30 minutes that occurs between arrivals. Thus, the mean time between arrivals is 0.10(30 minutes) = 3 minutes. On average, customers arrive every 3 minutes. Diff: 2 Keywords: exponential distribution, mean Section: 6-2 Other Continuous Probability Distributions Outcome: 5 122) At the West-Side Drive-Inn, customers arrive at the rate of 10 every 30 minutes. The time between arrivals is exponentially distributed. Based on this information, what is the probability that the time between two customers arriving will exceed 6 minutes? Answer: To solve this problem, we are looking for P(x > 6). It will be helpful to use the complement approach. Therefore, we wish to find 1 - P(x ≤ 6). Since we are dealing with an exponential distribution with arrival rate equal to 10 per 30 minutes, λ = 10 per 30 minutes = 0.33 per minute. Then we use P(x ≤ a) = 1 - e-λa. To do this we can use either the table of exponential values in the back of the text or software such as Excel. We are looking for: 1 - (1 - e-33(6)) = 1 - (1 - e2) = 1 - (1 - 0.1353) = 0.1353 Thus, there is approximately a 0.1353 probability that the time between two customers arriving will exceed 6 minutes. Note that if Excel or other software is used, the answer might be slightly different due to rounding differences. Diff: 2 Keywords: exponential distribution, probability Section: 6-2 Other Continuous Probability Distributions Outcome: 5

Business Statistics, 9e (Groebner/Shannon/Fry) Chapter 7 Introduction to Sampling Distributions 1) The sample mean is a parameter. Answer: FALSE Diff: 1 Keywords: sample, parameter Section: 7-1 Sampling Error—What It Is and Why It Happens Outcome: 1 2) The size of the sampling error that comes from a random sample depends on both the variation in the population and the size of the sample being selected. Answer: TRUE Diff: 1 Keywords: sampling error, random, sample, size Section: 7-1 Sampling Error—What It Is and Why It Happens Outcome: 1 3) Taking a larger sample size will always result in less sampling error but costs more money and takes more time. Answer: FALSE Diff: 1 Keywords: sample, size, sampling error Section: 7-1 Sampling Error—What It Is and Why It Happens Outcome: 1 4) Recently the State Fish and Game planted several thousand tagged fish in a local river. The mean length of these fish, which constitute a population, is 12.6 inches. Yesterday, fishermen caught 100 of these tagged fish. You could expect that the mean length for these fish would be 12.6 inches as well since they come from the population. Answer: FALSE Diff: 2 Keywords: sample, mean, sampling error Section: 7-1 Sampling Error—What It Is and Why It Happens Outcome: 1 5) The actual mean fill volume for all bottles of a soft drink product that were filled on a Tuesday is 11.998 ounces. A sample of 64 bottles was randomly selected and the sample mean fill volume was 12.004 ounces. Based upon this information, the sampling error is .006 ounce. Answer: TRUE Diff: 1 Keywords: mean, sample, sampling error Section: 7-1 Sampling Error—What It Is and Why It Happens Outcome: 1

6) Sampling error is the difference between the sample statistic and the population parameter. Answer: TRUE Diff: 1 Keywords: sampling error, population, sample, parameter Section: 7-1 Sampling Error—What It Is and Why It Happens Outcome: 1 7) The reason that a population mean and the mean of a random sample selected from that population might be different is that the sample mean is found by dividing by n-1 while the population mean is found by dividing by n. Answer: FALSE Diff: 1 Keywords: population, sample, sampling error Section: 7-1 Sampling Error—What It Is and Why It Happens Outcome: 1 8) If the mean age for all students that attend your university is 24.78 years, it would be reasonable to expect that the mean of a sample of students selected from that population would also equal 24.78 years as long at the sampling is done using sound statistical methods. Answer: FALSE Diff: 2 Keywords: population, sample, mean, sampling error Section: 7-1 Sampling Error—What It Is and Why It Happens Outcome: 1 9) If a sample is selected using random sampling methods, the primary reason that the sample mean might be different from the corresponding population mean is that the sample might be biased. Answer: FALSE Diff: 2 Keywords: sample, random, bias, population, mean Section: 7-1 Sampling Error—What It Is and Why It Happens Outcome: 1 10) Suppose it is known that the mean purchase price for all homes sold last year in Blacksburg, Virginia was $203,455. Recently, two studies were done on home sales prices. In the first study, a random sample of 200 homes was selected from the population. In the second study, a random sample of 60 homes was selected. Based on this information, we know that the second study would contain more sampling errors than the first study due to the smaller sample size. Answer: FALSE Diff: 2 Keywords: sample size, population, sample, sampling error Section: 7-1 Sampling Error—What It Is and Why It Happens Outcome: 1

11) A smaller sample might provide less sampling error than a larger sample from a given population. Answer: TRUE Diff: 2 Keywords: sample, sampling error, population Section: 7-1 Sampling Error—What It Is and Why It Happens Outcome: 1 12) A larger sample size reduces the potential for large sampling error. Answer: TRUE Diff: 1 Keywords: sampling error, sample size Section: 7-1 Sampling Error—What It Is and Why It Happens Outcome: 1 13) A local bank has 1,400 checking account customers. Of these, 1,020 also have savings accounts. A sample of 400 checking account customers was selected from the bank of which 302 also had savings accounts. The sampling error in this situation is .0264. Answer: TRUE Diff: 2 Keywords: sampling error, population, sample Section: 7-1 Sampling Error—What It Is and Why It Happens Outcome: 1 14) If a population mean is equal to 200, the sample mean for a random sample selected from the population is about as likely to be higher or lower than 200. Answer: TRUE Diff: 2 Keywords: population, mean, sample Section: 7-1 Sampling Error—What It Is and Why It Happens Outcome: 1 15) An increase in sample size will tend to result in less sampling error. Answer: TRUE Diff: 2 Keywords: sample, size, sampling error Section: 7-1 Sampling Error—What It Is and Why It Happens Outcome: 1 16) A simple random sample is selected in a manner such that each possible sample of a given size has an equal chance of being selected. Answer: TRUE Diff: 2 Keywords: simple random sample Section: 7-1 Sampling Error—What It Is and Why It Happens Outcome: 1

17) Suppose the mean balance of checking accounts at Regions Bank is known to be $4320. A random sample of 10 accounts yields a total of $41,490. This means the sampling error is -$171. Answer: TRUE Diff: 2 Keywords: sampling error Section: 7-1 Sampling Error—What It Is and Why It Happens Outcome: 1 18) Sampling error occurs when the population parameter and the sample statistic are different. Answer: TRUE Diff: 1 Keywords: sampling error, population, parameter, sample, statistic Section: 7-1 Sampling Error—What It Is and Why It Happens Outcome: 1 19) Sampling error can be eliminated if the sampling is done properly. Answer: FALSE Diff: 2 Keywords: sampling error, sample Section: 7-1 Sampling Error—What It Is and Why It Happens Outcome: 1 20) If it is desired that sampling error be reduced, one step that tends to work is to increase the sample size that is selected from the population. Answer: TRUE Diff: 1 Keywords: sampling error, sample size, population Section: 7-1 Sampling Error—What It Is and Why It Happens Outcome: 1 21) It is very unlikely that a nonstatistical sample will ever provide less sampling error than a statistical sample of the same size. Answer: FALSE Diff: 2 Keywords: sampling error, statistical sample, sample size Section: 7-1 Sampling Error—What It Is and Why It Happens Outcome: 1 22) A major automobile manufacturer has developed a new model car that it claims will average 25 mpg on the highway. A random sample of fifty of these cars was tested and they averaged 24 mpg. This means that the claim made by the auto company is incorrect. Answer: FALSE Diff: 2 Keywords: sample, population, sampling error Section: 7-1 Sampling Error—What It Is and Why It Happens Outcome: 1

23) Sampling error can be eliminated if the sample size is large enough. Answer: FALSE Diff: 2 Keywords: sampling error, sample size Section: 7-1 Sampling Error—What It Is and Why It Happens Outcome: 1 24) A sampling distribution is the distribution of the individual values that are included in a sample from a population. Answer: FALSE Diff: 1 Keywords: sampling distribution, sample, population Section: 7-2 Sampling Distribution of the Mean Outcome: 2 25) A sampling distribution for a sample mean shows the distribution of the possible values for the sample mean for a given sample size from a population. Answer: TRUE Diff: 1 Keywords: sampling distribution Section: 7-2 Sampling Distribution of the Mean Outcome: 2 26) Although the concept of sampling distributions is an important concept in statistics, it is very unlikely that a decision maker will actually construct a sampling distribution in any practical business situation. Answer: TRUE Diff: 2 Keywords: sampling distribution Section: 7-2 Sampling Distribution of the Mean Outcome: 2 27) A sampling distribution for is the distribution of all possible sample means that could be computed from the possible samples of a given sample size. Answer: TRUE Diff: 1 Keywords: sampling distribution, sample mean, mean, sample, sample size Section: 7-2 Sampling Distribution of the Mean Outcome: 2 28) The mean of a sampling distribution would be equal to the mean of the population from which the sampling distribution is constructed. Answer: TRUE Diff: 1 Keywords: sampling distribution, mean, population Section: 7-2 Sampling Distribution of the Mean Outcome: 2

29) The population mean of income for adults in a particular community is known to be $28,600. Given this information, the sampling distribution of values will be less than this depending on the size of the sample used in developing the sampling distribution. Answer: FALSE Diff: 2 Keywords: population, mean, sample size, sampling distribution Section: 7-2 Sampling Distribution of the Mean Outcome: 2 30) If a population is normally distributed, then the sampling distribution for the sample mean will always be normally distributed regardless of the sample size. Answer: TRUE Diff: 2 Keywords: population, mean, sample, sampling distribution Section: 7-2 Sampling Distribution of the Mean Outcome: 2 31) The population of soft drink cans filled by a particular machine is known to be normally distributed with a mean equal to 12 ounces and a standard deviation equal to .25 ounce. Given this information, the sampling distribution for a random sample of n = 25 cans will also be normally distributed with a mean equal to 12 ounces and a standard deviation equal to .05 ounce. Answer: TRUE Diff: 2 Keywords: population, mean, standard deviation, sample, sampling distribution Section: 7-2 Sampling Distribution of the Mean Outcome: 2 32) The sampling distribution for is actually the distribution of possible sampling error for samples of a given size selected at random from the population. Answer: TRUE Diff: 2 Keywords: sampling distribution, mean, sampling error, sample, population Section: 7-2 Sampling Distribution of the Mean Outcome: 2 33) A sampling distribution for a sample of n = 4 is normally distributed with a standard deviation equal to 5. Based on this information, the population standard deviation, σ, is equal to 10 mph. Answer: TRUE Diff: 3 Keywords: population, mean, sampling distribution Section: 7-2 Sampling Distribution of the Mean Outcome: 2

34) If a population standard deviation is 100, then the sampling distribution for will have a standard deviation that is less than 100 for all sample sizes greater than 2. Answer: TRUE Diff: 2 Keywords: population, standard deviation, sampling distribution, sample size Section: 7-2 Sampling Distribution of the Mean Outcome: 2 35) The Central Limit Theorem is of most use to decision makers when the population is known to be normally distributed. Answer: FALSE Diff: 1 Keywords: Central Limit Theorem, normal, population Section: 7-2 Sampling Distribution of the Mean Outcome: 3 36) One of the things that the Central Limit Theorem tells us is that about half of the sample means will be greater than the population mean and about half will be less. Answer: TRUE Diff: 2 Keywords: Central Limit Theorem, normal, population, sample, mean Section: 7-2 Sampling Distribution of the Mean Outcome: 3 37) When a population is not normally distributed, the Central Limit Theorem states that a sufficiently large sample will result in the sample mean being normally distributed. Answer: TRUE Diff: 2 Keywords: Central Limit Theorem, population, sampling distribution, spread, sample size Section: 7-2 Sampling Distribution of the Mean Outcome: 3 38) If you are sampling from a very large population, a doubling of the sample size will reduce the standard error of the sampling distribution by one-fourth. Answer: FALSE Diff: 3 Keywords: sample, population, standard error, sampling distribution Section: 7-2 Sampling Distribution of the Mean Outcome: 2

39) The population of incomes in a particular community is thought to be highly right-skewed with a mean equal to $36,789 and a standard deviation equal to $2,490. Based on this, if a sample of size n = 36 is selected, the sampling distribution would have a mean equal to the population mean, but the standard deviation of the sampling distribution will be one-sixth of the population standard deviation. Answer: TRUE Diff: 2 Keywords: Central Limit Theorem, population, sampling distribution standard deviation, mean Section: 7-2 Sampling Distribution of the Mean Outcome: 3 40) The population of incomes in a particular community is thought to be highly right-skewed with a mean equal to $36,789 and a standard deviation equal to $2,490. Based on this, if a sample of size n = 36 is selected, the highest sample mean that we would expect to see would be approximately $38,034. Answer: TRUE Diff: 3 Keywords: Central Limit Theorem, population, sampling distribution standard deviation, mean Section: 7-2 Sampling Distribution of the Mean Outcome: 3 41) If a population is not normally distributed, then the sampling distribution for the mean also cannot be normally distributed. Answer: FALSE Diff: 2 Keywords: Central Limit Theorem, sample size Section: 7-2 Sampling Distribution of the Mean Outcome: 3 42) The Dilmart Company has 8,000 parts in inventory. The mean dollar value of these parts is $10.79 with a standard deviation equal to $3.34. Suppose the inventory manager selected a random sample of n = 64 parts from the inventory and found a sample mean equal to $11.27. The probability of getting a sample mean at least as large as $11.27 is approximately 0.444. Answer: FALSE Diff: 2 Keywords: Central Limit Theorem, sample, population, mean Section: 7-2 Sampling Distribution of the Mean Outcome: 3 43) One of the nation's biggest regional airlines has tracked 4,000 landings and take-offs during the past month. Treating these data as the population of interest, the company found that the average time the planes spent on the ground (called the turn time) was 17.23 minutes with a standard deviation of 3.79 minutes. Further, they determined that the distribution of turn times is normally distributed. Then, the probability that a single turn time selected at random from this population would exceed 20 minutes is approximately 0.2327. Answer: TRUE Diff: 2 Keywords: z-value, probability, mean, standard deviation Section: 7-2 Sampling Distribution of the Mean Outcome: 2 7-8 Copyright © 2018 Pearson Education, Inc.

44) One of the nation's biggest regional airlines has tracked 4,000 landings and take-offs during the past month. Treating these data as the population of interest, the company found that the average time the planes spent on the ground (called the turn time) was 17.23 minutes with a standard deviation of 3.79 minutes. Further, they determined that the distribution of turn times is normally distributed. If a sample of size n = 16 turn times was selected at random from the population, the chances of the mean of this sample exceeding 20 minutes is 0.2327. Answer: FALSE Diff: 2 Keywords: z-value, probability, mean, standard deviation, sample Section: 7-2 Sampling Distribution of the Mean Outcome: 2 45) The Fallbrook Distributing Company has a soft drink bottling plant in Plano, Texas. Based on historical records, if its filling machine is working properly, the mean fill volume per can is 12.0 ounces with a standard deviation equal to 0.13 ounce. Further, the distribution of fill amounts is known to be normally distributed. The State of Texas has a department whose job it is to check on such consumerrelated processes as soft drink filling. The idea is to protect the consumer. The department arrives at the Fallbrook plant once a month on an unscheduled day. When they arrive, they randomly select n = 4 cans and carefully measure the volume in each can. If any of these cans contains less than 11.85 ounces, the plant is shut down until a full inspection of the filling process is performed. Based on this information, the probability that the plant will get shut down if it is operating properly is approximately 0.1251. Answer: TRUE Diff: 3 Keywords: z-value, probability, mean, standard deviation Section: 7-2 Sampling Distribution of the Mean Outcome: 2 46) Regardless of population distribution, the sampling distribution for a random variable X will be approximately normally distributed. Answer: FALSE Diff: 3 Keywords: Central Limit Theorem, sample size Section: 7-2 Sampling Distribution of the Mean Outcome: 3 47) A sample proportion can be assumed normally distributed if np ≥ 5. Answer: FALSE Diff: 2 Keywords: Central Limit Theorem, sampling distribution, mean, proportion Section: 7-3 Sampling Distribution of a Proportion Outcome: 4

48) Suppose it is known that 93 percent of all parts in an inventory of 18,900 parts are in workable order. If a sample of n = 100 parts were selected from the inventory, based on the concept of sampling distributions of proportions, it can be assumed that the sample proportion of workable parts will also be 0.93. Answer: FALSE Diff: 1 Keywords: sampling error, proportion Section: 7-3 Sampling Distribution of a Proportion Outcome: 4 49) The size of the standard error of the sample proportion is dependent on the value of the population proportion and the closer the population proportion is to .50, the larger the standard error for a given sample size will be. Answer: TRUE Diff: 2 Keywords: sample size, standard error, proportion, population Section: 7-3 Sampling Distribution of a Proportion Outcome: 4 50) Assume that n = 18 people are asked a yes/no survey question, and 6 people say "yes" while 12 people say "no." Based on this information the sample proportion can be assumed normally distributed. Answer: TRUE Diff: 2 Keywords: sampling distribution proportion Section: 7-3 Sampling Distribution of a Proportion Outcome: 4 51) In order to assume that the sampling distribution for a proportion is approximately normal, the sampling proportion must be very close to the population proportion. Answer: TRUE Diff: 1 Keywords: sampling distribution, proportion, normal Section: 7-3 Sampling Distribution of a Proportion Outcome: 4 52) In a particular city, the proportion of cars that would fail an air quality emissions test is thought to be 0.13. Given this, the probability that a random sample of n = 200 cars will have a sample proportion between 0.11 and 0.15 is approximately 0.60. Answer: TRUE Diff: 2 Keywords: proportion, probability, sample, z-value Section: 7-3 Sampling Distribution of a Proportion Outcome: 4

53) The quality manager of a fortune cookie company believes that a larger than acceptable proportion of paper fortunes being used are blank. Suppose she takes a sample of 320 fortune cookies from the production line, and 15 of the paper fortunes are blank. The estimate of the proportion is approximately 0.047. Answer: TRUE Diff: 2 Keywords: sample, population, parameter, proportion Section: 7-3 Sampling Distribution of a Proportion Outcome: 4 54) In a campaign speech, a candidate for governor stated that about 63 percent of the people in the state were in favor of spending additional money on higher education. After the speech, a polling agency surveyed a random sample of 400 people and found 234 people who favored more spending on higher education. Based on the candidate's statement, the probability of finding 234 or fewer is approximately 0.97. Answer: FALSE Diff: 2 Keywords: probability, z-value, sample Section: 7-3 Sampling Distribution of a Proportion Outcome: 4 55) The makers of a particular type of candy have stated that 75 percent of their sacks of candy will contain 6 ounces or more of candy. A consumer group that studies such claims recently selected a random sample of 100 sacks of this candy. Of these, 70 sacks actually contained 6 ounces or more. The probability that 70 or fewer sacks would contain 6 ounces or less is approximately 0.1251. Answer: TRUE Diff: 2 Keywords: probability, z-value, sample Section: 7-3 Sampling Distribution of a Proportion Outcome: 4 56) In analyzing the sampling distribution of a proportion, doubling the sample size will cut the standard deviation of the sampling distribution in half. Answer: FALSE Diff: 3 Keywords: sampling distribution, proportion, sample size, standard deviation Section: 7-3 Sampling Distribution of a Proportion Outcome: 4 57) If the standard error for the sampling distribution of a proportion is equal to 0.0327 and if the population proportion, p, is equal to .80, the sample size must be 150. Answer: TRUE Diff: 3 Keywords: proportion, population, sample size Section: 7-3 Sampling Distribution of a Proportion Outcome: 4

58) Regardless of the value of the population proportion, p, (with the obvious exceptions of p = 0 and p = 1) the sampling distribution for the sample proportion, will be approximately normally distributed providing that the sample size is large enough. Answer: FALSE Diff: 3 Keywords: Central Limit Theorem, sample size Section: 7-3 Sampling Distribution of a Proportion Outcome: 4 59) When sampling from a population, the sample mean will: A) typically exceed the population mean. B) likely be different from the population mean. C) always be closer to the population mean as the sample size increases. D) likely be equal to the population mean if proper sampling techniques are employed. Answer: B Diff: 2 Keywords: sample, population, mean, sampling error Section: 7-1 Sampling Error—What It Is and Why It Happens Outcome: 1 60) A measure computed from the entire population is called: A) a statistic. B) a mean. C) a parameter. D) a qualitative value. Answer: C Diff: 1 Keywords: measure, population, parameter Section: 7-1 Sampling Error—What It Is and Why It Happens Outcome: 1 61) The following values represent the population of home mortgage interest rates (in percents) being charged by the banks in a particular city: 3.3 3.9

3.5 3.8

4.5 4.1

4.0 4.2

4.3 3.7

Given this information, what is the most extreme amount of sampling error possible if a random sample of n = 3 banks is surveyed and the mean loan rate is calculated? A) About -0.40 percent B) About 0.13 percent C) About 0.40 percent D) Can't be determined without more information. Answer: C Diff: 3 Keywords: sampling error, mean Section: 7-1 Sampling Error—What It Is and Why It Happens Outcome: 1 7-12 Copyright © 2018 Pearson Education, Inc.

62) The following values represent the population of home mortgage interest rates (in percents) being charged by the banks in a particular city: 3.3 3.9

3.5 3.8

4.5 4.1

4.0 4.2

4.3 3.7

Given this information, what is the smallest amount of sampling error possible if a random sample of n = 3 banks is surveyed and the mean loan rate is calculated? A) About -0.43 percent B) About 0.14 percent C) About 0.43 percent D) Can't be determined without more information. Answer: C Diff: 2 Keywords: sampling error, mean Section: 7-1 Sampling Error—What It Is and Why It Happens Outcome: 1 63) The following values represent the population of home mortgage interest rates (in percents) being charged by the banks in a particular city: 3.3 3.9

3.5 3.8

4.5 4.1

4.0 4.2

4.3 3.7

Given this information, what is the smallest to largest range for the sampling error possible if a random sample of n = 3 banks is surveyed and the mean loan rate is calculated? A) About -0.43 to 0.40 percent B) About -0.40 to 0.43 percent C) About 0.40 to 0.43 percent D) Can't be determined without more information. Answer: A Diff: 2 Keywords: sampling error, mean Section: 7-1 Sampling Error—What It Is and Why It Happens Outcome: 1 64) Suppose the mean of dogs a pet shop grooms each day is know to be 14.2 dogs. If a sample of n = 12 days is chosen and a total of 178 dogs are groomed during those 12 days, then the sampling error is: A) 163.8. B) about 0.63. C) about -0.63. D) -163.8. Answer: B Diff: 2 Keywords: sampling error, mean Section: 7-1 Sampling Error—What It Is and Why It Happens Outcome: 1 7-13 Copyright © 2018 Pearson Education, Inc.

65) A particular subdivision has 20 homes. The number of people living in each of these homes is listed as follows: 2 2 5 2

4 4 4 2

7 5 6 1

3 2 3 4

4 3 4 3

If a random sample of n = 3 homes were selected, what would be the highest possible positive sampling error? A) 6.0 B) 3.0 C) 0.5 D) 2.5 Answer: D Diff: 3 Keywords: sampling error, positive, mean, sample Section: 7-1 Sampling Error—What It Is and Why It Happens Outcome: 1 66) A particular subdivision has 20 homes. The number of people living in each of these homes is listed as follows: 2 2 5 2

4 4 4 2

7 5 6 1

3 2 3 4

4 3 4 3

Which of the following statements is true when comparing a random sample of size three homes selected from the population to a random sample of size 6 homes selected from the population? A) The amount of sampling error that will exist between the sample mean and the population mean will be half for the larger sample. B) The most extreme negative sampling error between and μ is reduced by about 0.167 person. C) We can expect that the larger sample will produce more sampling error due to the potential to make coding errors. D) The sampling error that will result from the smaller sample will be less than what we would see from the larger sample. Answer: B Diff: 3 Keywords: sample, sampling error, sample size Section: 7-1 Sampling Error—What It Is and Why It Happens Outcome: 1

67) Which of the following statements is false? A) Increasing the sample size will always reduce the size of the sampling error when the sample mean is used to estimate the population mean. B) Increasing the sample size will reduce the potential for extreme sampling error. C) Sampling error can occur when differs from μ due to the fact that the sample was not a perfect reflection of the population. D) There is no way to prevent sampling error short of taking a census of the entire population. Answer: A Diff: 2 Keywords: sample size, sampling error, mean, sample, population Section: 7-1 Sampling Error—What It Is and Why It Happens Outcome: 1 68) A particular subdivision has 20 homes. The number of people living in each of these homes is listed as follows: 2 2 5 2

4 4 4 2

7 5 6 1

3 2 3 4

4 3 4 3

If a sample of size n = 5 is selected, the largest possible sample mean is: A) 7 B) 6.5 C) 6 D) 5.4 Answer: D Diff: 2 Keywords: sample mean Section: 7-1 Sampling Error—What It Is and Why It Happens Outcome: 1 69) If the monthly electrical utility bills of all customers for the Far East Power and Light Company are known to be distributed as a normal distribution with mean equal to $87.00 a month and standard deviation of $36.00, which of the following would be the largest individual customer bill that you might expect to find? A) Approximately $811.00 B) About $195.00 C) Nearly $123.00 D) There is no way to determine this without more information. Answer: B Diff: 2 Keywords: z-value, normal distribution Section: 7-2 Sampling Distribution of the Mean Outcome: 2

70) The monthly electrical utility bills of all customers for the Far East Power and Light Company are known to be distributed as a normal distribution with mean equal to $87.00 a month and standard deviation of $36.00. If a statistical sample of n = 100 customers is selected at random, what is the probability that the mean bill for those sampled will exceed $75.00? A) -0.33 B) Approximately 0.63 C) About 1.00 D) 3.33 Answer: C Diff: 2 Keywords: Central Limit Theorem, z-value, mean, standard deviation, normal Section: 7-2 Sampling Distribution of the Mean Outcome: 3 71) The Olsen Agricultural Company has determined that the weight of hay bales is normally distributed with a mean equal to 80 pounds and a standard deviation equal to 8 pounds. Based on this, what is the mean of the sampling distribution for if the sample size is n = 64? A) 80 B) 10 C) Between 72 and 88 D) 8 Answer: A Diff: 1 Keywords: mean, sampling distribution, normal Section: 7-2 Sampling Distribution of the Mean Outcome: 2 72) The Olsen Agricultural Company has determined that the weight of hay bales is normally distributed with a mean equal to 80 pounds and a standard deviation equal to 8 pounds. Based on this, what is the probability that the mean weight of the bales in a sample of n = 64 bales will be between 78 and 82 pounds? A) 0.4772 B) 0.0228 C) 0.6346 D) 0.9544 Answer: D Diff: 2 Keywords: Central Limit Theorem, z-value, mean, standard deviation, sample Section: 7-2 Sampling Distribution of the Mean Outcome: 3

73) The State Department of Weights and Measures is responsible for making sure that commercial weighing and measuring devices, such as scales, are accurate so customers and businesses are not cheated. Periodically, employees of the department go to businesses and test their scales. For example, a dairy bottles milk in 1-gallon containers. Suppose that if the filling process is working correctly, the mean volume of all gallon containers is 1.00 gallon with a standard deviation equal to 0.10 gallon. Based on this information, if the department employee selects a random sample of n = 9 containers, what is the probability that the mean volume for the sample will be greater than 1.01 gallons? A) 0.3821 B) 0.1179 C) 0.6179 D) 0.2358 Answer: A Diff: 2 Keywords: Central Limit Theorem, z-value, mean, standard deviation, sample, probability Section: 7-2 Sampling Distribution of the Mean Outcome: 3 74) The State Department of Weights and Measures is responsible for making sure that commercial weighing and measuring devices, such as scales, are accurate so customers and businesses are not cheated. Periodically, employees of the department go to businesses and test their scales. For example, a dairy bottles milk in 1-gallon containers. Suppose that if the filling process is working correctly, the mean volume of all gallon containers is 1.00 gallon with a standard deviation equal to 0.10 gallon. The department's test process requires that they select a random sample of n = 9 containers. If the sample mean is less than 0.97 gallons, the department will fine the dairy. Based on this information, what is the probability that the dairy will get fined even when its filling process is working correctly? A) 0.90 B) Approximately 0.3159 C) About 0.1841 D) Approximately 0.3821 Answer: C Diff: 2 Keywords: Central Limit Theorem, z-value, mean, standard deviation, sample, type I, alpha, probability Section: 7-2 Sampling Distribution of the Mean Outcome: 3

75) The State Department of Weights and Measures is responsible for making sure that commercial weighing and measuring devices, such as scales, are accurate so customers and businesses are not cheated. Periodically, employees of the department go to businesses and test their scales. For example, a dairy bottles milk in 1-gallon containers. Suppose that if the filling process is working correctly, the mean volume of all gallon containers is 1.00 gallon with a standard deviation equal to 0.10 gallon. The department's test process requires that they select a random sample of n = 9 containers. If the sample mean is less than 0.97 gallon, the department will fine the dairy. Based on this information, suppose that the dairy wants no more than a 0.05 chance of being fined, which of the following options exist if they can't alter the filling standard deviation? A) They can convince the state to decrease the sample size. B) They can change the mean fill level to approximately 1.025 gallons. C) They could lower the mean fill level to a level lower than 1 gallon. D) There is actually nothing that they can do if they can't modify the standard deviation. Answer: B Diff: 3 Keywords: Central Limit Theorem, z-value, mean, standard deviation, sample, type I, alpha, probability Section: 7-2 Sampling Distribution of the Mean Outcome: 3 76) Suppose it is known that the income distribution in a particular region is right-skewed and bi-modal. If bank economists are interested in estimating the mean income, which of the following is true? A) Provided that the sample size is sufficiently large, the sampling distribution for will be approximately normal with a mean equal to the population mean that they wish to estimate. B) The sampling distribution will also be right-skewed for large sample sizes. C) The standard deviation of the sampling distribution for will be proportionally larger than the population standard deviation, depending on the size of the sample. D) The sampling distribution will be left-skewed. Answer: A Diff: 2 Keywords: Central Limit Theorem, z-value, mean, sampling distribution Section: 7-2 Sampling Distribution of the Mean Outcome: 3 77) A company has determined that the mean number of days it takes to collect on its accounts receivable is 36 with a standard deviation of 11 days. The company plans to select a random sample of n = 12 accounts and compute the sample mean. Which of the following statements holds true in this situation? A) There is no way to determine what the mean of the sampling distribution is without knowing the specific shape of the population. B) The sampling distribution will have the same distribution as the population, provided that the population is not normally distributed. C) The sampling error will be larger than if they had sampled n = 64 accounts. D) The sampling distribution may actually be approximately normally distributed depending on what the population distribution is. Answer: D Diff: 2 Keywords: Central Limit Theorem, sampling distribution, normal, population Section: 7-2 Sampling Distribution of the Mean Outcome: 3 7-18 Copyright © 2018 Pearson Education, Inc.

78) A golf course in California has determined that the mean time it takes for a foursome to complete an 18 hole round of golf is 4 hours 35 minutes (275 minutes) with a standard deviation of 14 minutes. The time distribution is also thought to be approximately normal. Every month, the head pro at the course randomly selects a sample of 8 foursomes and monitors the time it takes them to play. Suppose the mean time that was observed for the sample last month was 4 hours 44 minutes (284 minutes). What is the probability of seeing a sample mean this high or higher? A) Approximately 0.4649 B) About 0.9649 C) Approximately 0.0351 D) About 0.9298 Answer: C Diff: 2 Keywords: z-value, probability, mean, Central Limit Theorem Section: 7-2 Sampling Distribution of the Mean Outcome: 3 79) Which of the following statements is not consistent with the Central Limit Theorem? A) The Central Limit Theorem applies without regard to the size of the sample. B) The Central Limit Theorem applies to non-normal distributions. C) The Central Limit Theorem indicates that the sampling distribution will be approximately normal when the sample size is sufficiently large. D) The Central Limit Theorem indicates that the mean of the sampling distribution will be equal to the population mean. Answer: A Diff: 1 Keywords: Central Limit Theorem, sample size Section: 7-2 Sampling Distribution of the Mean Outcome: 3 80) The J.R. Simplot Company produces frozen French fries that are then sold to customers such as McDonald's. The "prime" line of fries has an average length of 6.00 inches with a standard deviation of 0.50 inch. To make sure that Simplot continues to meet the quality standard for "prime" fries, they plan to select a random sample of n = 100 fries each day. The quality analysts will compute the mean length for the sample. They want to establish limits on either side of the 6.00 inch mean so that the chance of the sample mean falling within the limits is 0.99. What should these limits be? A) Approximately ±0.13 inches B) Within the approximate range of 5.87 inches to 6.13 inches C) Within the range of about 4.71 inches to 7.29 inches D) Approximately ±1.29 inches Answer: B Diff: 3 Keywords: z-value, mean, standard deviation, Central Limit Theorem Section: 7-2 Sampling Distribution of the Mean Outcome: 3

81) The J.R. Simplot Company produces frozen French fries that are then sold to customers such as McDonald's. The "prime" line of fries has an average length of 6.00 inches with a standard deviation of 0.50 inch. To make sure that Simplot continues to meet the quality standard for "prime" fries, they plan to select a random sample of n = 100 fries each day. Yesterday, the sample mean was 6.05 inches. What is the probability that the mean would be 6.05 inches or more if they are meeting the quality standards? A) 0.2350 B) 0.3413 C) 0.9413 D) 0.1587 Answer: D Diff: 2 Keywords: z-value, mean, standard deviation, Central Limit Theorem, probability Section: 7-2 Sampling Distribution of the Mean Outcome: 3 82) The St. Joe Company grows pine trees and the average annual increase in tree diameter is 3.1 inches with a standard deviation of 0.5 inch. A random sample of n = 50 trees is collected. What is the probability of the sample mean being less the 2.9 inches? A) 0.4977 B) 0.0023 C) 0.9977 D) 0.9954 Answer: B Diff: 2 Keywords: probability, z-value, mean, standard deviation, Central Limit Theorem Section: 7-2 Sampling Distribution of the Mean Outcome: 3 83) According to the local real estate board, the average number of days that homes stay on the market before selling is 78.4 with a standard deviation equal to 11 days. A prospective seller selected a random sample of 36 homes from the multiple listing service. Above what value for the sample mean should 95 percent of all possible sample means fall? A) About 79.3 days B) About 64 days C) Approximately 75.4 days D) Can't be determined without knowing whether the population is normally distributed. Answer: C Diff: 3 Keywords: probability, z-value, Central Limit Theorem, sample Section: 7-2 Sampling Distribution of the Mean Outcome: 3

84) A population, with an unknown distribution, has a mean of 80 and a standard deviation of 7. For a sample of 49, the probability that the sample mean will be larger than 82 is: A) 0.5228 B) 0.9772 C) 0.4772 D) 0.0228 Answer: D Diff: 2 Keywords: probability, z-value, Central Limit Theorem Section: 7-2 Sampling Distribution of the Mean Outcome: 3 85) Which of the following statements is true with respect to the sampling distribution of a proportion? A) An increase in the sample size will result in a reduction in the size of the standard deviation. B) As long as the sample size is sufficiently large, the sampling distribution will be approximately normal. C) The mean of the sampling distribution will equal the population proportion. D) All of the above are true. Answer: D Diff: 1 Keywords: sampling distribution, proportion, sample size, normal, population Section: 7-3 Sampling Distribution of a Proportion Outcome: 4 86) The Chamber of Commerce in a large Midwestern city has stated that 70 percent of all business owners in the city favor increasing the downtown parking fees. The city council has commissioned a random sample of n = 100 business owners. Of these, 63 said that they favor increasing the parking fees. What is the probability of 63 or fewer favoring the idea if the Chamber's claim is correct? A) Approximately 0.0630 B) About 0.4370 C) Nearly 0.20 D) About 0.9370 Answer: A Diff: 2 Keywords: z-value, proportion, normal, probability Section: 7-3 Sampling Distribution of a Proportion Outcome: 4

87) A claim was recently made on national television that two of every three doctors recommend a particular pain killer. Suppose a random sample of n = 300 doctors revealed that 180 said that they would recommend the painkiller. If the TV claim is correct, what is the probability of 180 or fewer in the sample agreeing? A) 0.4929 B) 0.0049 C) 0.9929 D) 0.0142 Answer: B Diff: 2 Keywords: z-value, proportion, normal, probability Section: 7-3 Sampling Distribution of a Proportion Outcome: 4 88) A major textbook publisher has a contract with a printing company. Part of the contract stipulates that no more than 5 percent of the pages should have any type of printing error. Suppose that the company selects a random sample of 400 pages and finds 33 that have an error. If the printer is meeting the standard, what is the probability that a sample would have 33 or more errors? A) 0.1245 B) 0.4986 C) 0.0014 D) 0.1250 Answer: C Diff: 2 Keywords: z-value, proportion, normal, probability Section: 7-3 Sampling Distribution of a Proportion Outcome: 4 89) A major textbook publisher has a contract with a printing company. Part of the contract stipulates that no more than 5 percent of the pages should have any type of printing error. As a quality control measure, the publisher periodically selects a random sample of n = 100 pages. Then, depending on the proportion of pages with errors, they either say nothing to the printer or they complain that the quality has slipped. Suppose the publisher wants no more than a .10 chance of mistakenly blaming the printer for poor quality, what should the cut-off proportion be? A) About 0.0279 B) Approximately 0.0779 C) About 0.0221 D) About 0.10 Answer: B Diff: 3 Keywords: z-value, proportion, normal, probability, type II, beta Section: 7-3 Sampling Distribution of a Proportion Outcome: 4

90) One of the leading dot-com companies has found that the proportion of customers who come into its Web site that actually makes a purchase is 0.045. The company plans to see whether this rate still holds by selecting a random sample of 200 hits on its Web site. Given that the 0.045 rate still applies, what is the standard deviation of the sampling distribution? A) Approximately 0.0147 B) About 0.0002 C) About 0.0354 D) Can't be determined without knowing the mean. Answer: A Diff: 2 Keywords: z-value, proportion, normal, standard deviation Section: 7-3 Sampling Distribution of a Proportion Outcome: 4 91) In a recent report, it was stated that the proportion of employees who carpool to their work is 0.14 and that the standard deviation of the sampling proportion is 0.0259. However, the report did not indicate what the sample size was. What was the sample size? A) 100 B) 180 C) 460 D) Can't be determined without more information Answer: B Diff: 3 Keywords: proportion, sample size, sampling distribution Section: 7-3 Sampling Distribution of a Proportion Outcome: 4 92) According to an industry report, 26 percent of all households have at least one cell phone. Further, of those that do have a cell phone, the mean monthly bill is $55.90 with a standard deviation equal to $9.60. Recently, a random sample of 400 households was selected. Of these households, 88 indicated that they had cell phones. The mean bill for these 88 households was $57.00. What is the probability of getting 88 or fewer households with cell phones if the numbers provided by the industry report are correct? A) Approximately 0.0344 B) Nearly 0.4656 C) About 0.1345 D) Can't be determined without knowing the standard deviation Answer: A Diff: 2 Keywords: proportion, probability, sampling distribution Section: 7-3 Sampling Distribution of a Proportion Outcome: 4

93) A pharmaceutical company claims that only 5 percent of patients experience nausea when they take a particular drug. In a research study, n = 100 patients were given this drug and 8 experienced nausea. Assuming that the company's claim is true, what is the probability of 8 or more patients experiencing nausea? A) About 0.9162 B) About 0.0300 C) About 0.0838 D) About 0.4162 Answer: C Diff: 2 Keywords: probability, sampling distribution, proportion Section: 7-3 Sampling Distribution of a Proportion Outcome: 4 94) A major shipping company has stated that 96 percent of all parcels are delivered on time. To check this, a random sample of n = 200 parcels were sampled. Of these, 184 arrived on time. If the company's claim is correct, what is the probability of 184 or fewer parcels arriving on time? A) About 0.0019 B) Nearly 0.24 C) Just over 0.98 D) About 0.4981 Answer: A Diff: 2 Keywords: probability, sampling distribution, proportion Section: 7-3 Sampling Distribution of a Proportion Outcome: 4 95) A sample of 25 observations is taken to estimate a population proportion π. The sampling distribution of sample proportion p is: A) not normal since n < 30. B) approximately normal because is always normally distributed. C) approximately normal if np ≥ 5 and n(1 - p) ≥ 5. D) approximately normal if p approaches 0.50. Answer: C Diff: 2 Keywords: sampling distribution, proportion, Central Limit Theorem Section: 7-3 Sampling Distribution of a Proportion Outcome: 4

96) Hillman Management Services manages apartment complexes in Tulsa, Oklahoma. They currently have 30 units available for rent. The monthly rental prices (in dollars) for this population of 30 units are: 455 675 600

690 550 780

450 490 650

495 495 905

550 700 415

780 995 600

800 650 600

395 550 780

500 400 575

405 750 750

What is the range of possible sampling error if a random sample of size n = 6 is selected from the population? A) -194.33 to 225.67 B) -245.23 to 271.86 C) -184.15 to 215.61 D) -172.52 to 234.04 Answer: A Diff: 1 Keywords: sampling error Section: 7-1 Sampling Error—What It Is and Why It Happens Outcome: 1 97) Hillman Management Services manages apartment complexes in Tulsa, Oklahoma. They currently have 30 units available for rent. The monthly rental prices (in dollars) for this population of 30 units are: 455 675 600

690 550 780

450 490 650

495 495 905

550 700 415

780 995 600

800 650 600

395 550 780

500 400 575

405 750 750

What is the range of possible sampling error if a random sample of size n = 10 is selected? A) -174.21 to 191.12 B) -182.59 to 169.91 C) -164.33 to 178.67 D) -162.16 to 171.51 Answer: C Diff: 1 Keywords: sampling error Section: 7-1 Sampling Error—What It Is and Why It Happens Outcome: 1

98) Princess Cruises recently offered a 16-day voyage from Beijing to Bangkok during the time period from May to August. The announced price, excluding airfare, for a room with an ocean view or a balcony was listed as $3,475. Cruise fares usually are quite variable due to discounting by the cruise line and travel agents. A sample of 20 passengers who purchased this cruise paid the following amounts (in dollars): 3,559 3,439

3,005 3,375

3,389 3,349

3,505 3,559

3,605 3,419

3,545 3,569

3,529 3,559

3,709 3,575

3,229 3,449

3,419 3,119

Calculate the sample mean cruise fare. A) 3715.24 B) 3445.30 C) 4581.81 D) 6314.24 Answer: B Diff: 1 Keywords: sampling distribution, mean, sample Section: 7-2 Sampling Distribution of the Mean Outcome: 1 99) Princess Cruises recently offered a 16-day voyage from Beijing to Bangkok during the time period from May to August. The announced price, excluding airfare, for a room with an ocean view or a balcony was listed as $3,475. Cruise fares usually are quite variable due to discounting by the cruise line and travel agents. A sample of 20 passengers who purchased this cruise paid the following amounts (in dollars): 3,559 3,439

3,005 3,375

3,389 3,349

3,505 3,559

3,605 3,419

3,545 3,569

3,529 3,559

3,709 3,575

3,229 3,449

Determine the sampling error for this sample. A) -$29.70 B) -$51.12 C) -$21.71 D) -$31.74 Answer: A Diff: 1 Keywords: sampling error Section: 7-1 Sampling Error—What It Is and Why It Happens Outcome: 1

3,419 3,119

100) A population with a mean of 1,250 and a standard deviation of 400 is known to be highly skewed to the right. If a random sample of 64 items is selected from the population, what is the probability that the sample mean will be less than 1,325? A) 0.8981 B) 0.8141 C) 0.7141 D) 0.9332 Answer: D Diff: 2 Keywords: sampling distribution, mean, sample Section: 7-2 Sampling Distribution of the Mean Outcome: 1 101) Suppose that a population is known to be normally distributed with mean = 2,000 and standard deviation = 230. If a random sample of size n = 8 is selected, calculate the probability that the sample mean will exceed 2,100. A) 0.2141 B) 0.1871 C) 0.0712 D) 0.1093 Answer: D Diff: 2 Keywords: sampling distribution, mean, sample Section: 7-2 Sampling Distribution of the Mean Outcome: 1 102) A normally distributed population has a mean of 500 and a standard deviation of 60. Determine the probability that a random sample of size 16 selected from this population will have a sample mean less than 475. A) 0.3251 B) 0.7124 C) 0.0475 D) 0.0712 Answer: C Diff: 2 Keywords: sampling distribution, mean, sample Section: 7-2 Sampling Distribution of the Mean Outcome: 1

103) A normally distributed population has a mean of 500 and a standard deviation of 60. Determine the probability that a random sample of size 25 selected from the population will have a sample mean greater than or equal to 515. A) 0.1056 B) 0.1761 C) 0.0712 D) 0.0151 Answer: A Diff: 2 Keywords: sampling distribution, mean, sample Section: 7-2 Sampling Distribution of the Mean Outcome: 1 104) Suppose nine items are randomly sampled from a normally distributed population with a mean of 100 and a standard deviation of 20. The nine randomly sampled values are: 125 91

95 102

66 51

116 110

Calculate the probability of getting a sample mean that is smaller than the sample mean for these nine sampled values. A) 0.1411 B) 0.1612 C) 0.1512 D) 0.2266 Answer: D Diff: 2 Keywords: sampling distribution, mean, sample Section: 7-2 Sampling Distribution of the Mean Outcome: 1 105) A random sample of 100 items is selected from a population of size 350. What is the probability that the sample mean will exceed 200 if the population mean is 195 and the population standard deviation equals 20? (Hint: Use the finite correction factor since the sample size is more than 5% of the population size.) A) 0.0415 B) 0.0016 C) 0.0241 D) 0.0171 Answer: B Diff: 2 Keywords: sampling distribution, mean, sample Section: 7-2 Sampling Distribution of the Mean Outcome: 1

106) SeeClear Windows makes windows for use in homes and commercial buildings. The standards for glass thickness call for the glass to average 0.375 inches with a standard deviation equal to 0.050 inch. Suppose a random sample of n = 50 windows yields a sample mean of 0.392 inches. What is the probability if the windows meet the standards? A) 0.0612 B) 0.0082 C) 0.0015 D) 0.0009 Answer: B Diff: 2 Keywords: sampling distribution, mean, sample Section: 7-2 Sampling Distribution of the Mean Outcome: 1 107) Suppose the life of a particular brand of calculator battery is approximately normally distributed with a mean of 75 hours and a standard deviation of 10 hours. What is the probability that a single battery randomly selected from the population will have a life between 70 and 80 hours? A) 0.2412 B) 0.3830 C) 0.1712 D) 0.5121 Answer: B Diff: 3 Keywords: sampling distribution, mean, sample Section: 7-2 Sampling Distribution of the Mean Outcome: 1 108) Suppose the life of a particular brand of calculator battery is approximately normally distributed with a mean of 75 hours and a standard deviation of 10 hours. What is the probability that 16 randomly sampled batteries from the population will have a sample mean life of between 70 and 80 hours? A) 0.9444 B) 0.5121 C) 0.7124 D) 0.1512 Answer: A Diff: 3 Keywords: sampling distribution, mean, sample Section: 7-2 Sampling Distribution of the Mean Outcome: 1

109) Suppose the life of a particular brand of calculator battery is approximately normally distributed with a mean of 75 hours and a standard deviation of 10 hours. If the manufacturer of the battery is able to reduce the standard deviation of battery life from 10 to 9 hours, what would be the probability that 16 batteries randomly sampled from the population will have a sample mean life of between 70 and 80 hours? A) 0.6127 B) 0.8124 C) 0.9736 D) 0.8812 Answer: C Diff: 3 Keywords: sampling distribution, mean, sample Section: 7-2 Sampling Distribution of the Mean Outcome: 1 110) The branch manager for United Savings and Loan in Seaside, Virginia, has worked with her employees in an effort to reduce the waiting time for customers at the bank. Recently, she and the team concluded that average waiting time is now down to 3.5 minutes with a standard deviation equal to 1.0 minute. However, before making a statement at a managers' meeting, this branch manager wanted to double-check that the process was working as thought. To make this check, she randomly sampled 25 customers and recorded the time they had to wait. She discovered that mean wait time for this sample of customers was 4.2 minutes. Based on the team's claims about waiting time, what is the probability that a sample mean for n = 25 people would be as large or larger than 4.2 minutes? A) 0.0214 B) 0.0512 C) 0.0231 D) 0.0011 Answer: D Diff: 3 Keywords: sampling distribution, mean, sample Section: 7-2 Sampling Distribution of the Mean Outcome: 1

111) In an article entitled "Fuel Economy Calculations to Be Altered," James R. Healey indicated that the government planned to change how it calculates fuel economy for new cars and trucks. This is the first modification since 1985. It is expected to lower average mileage for city driving in conventional cars from 10% to 20%. AAA has forecast that the 2008 Ford F-150 would achieve 15.7 mile per gallon (mpg). The 2008 Ford F-150 was tested by AAA members driving the vehicle themselves and was found to have an average of 14.3 mpg. Assume that the mean obtained by AAA members is the true mean for the population of 2008 Ford F-150 trucks and that the population standard deviation is 5 mpg. Suppose 100 AAA members were to test the 2008 F-150. Determine the probability that the average mpg would be at least 15.7. A) 0.0026 B) 0.0121 C) 0.0451 D) 0.0001 Answer: A Diff: 3 Keywords: sampling distribution, mean, sample Section: 7-2 Sampling Distribution of the Mean Outcome: 1 112) In an article entitled "Fuel Economy Calculations to Be Altered," James R. Healey indicated that the government planned to change how it calculates fuel economy for new cars and trucks. This is the first modification since 1985. It is expected to lower average mileage for city driving in conventional cars from 10% to 20%. AAA has forecast that the 2008 Ford F-150 would achieve 15.7 mile per gallon (mpg). The 2008 Ford F-150 was tested by AAA members driving the vehicle themselves and was found to have an average of 14.3 mpg. Assume that the mean obtained by AAA members is the true mean for the population of 2008 Ford F-150 trucks and that the population standard deviation is 5 mpg. The current method of calculating the mpg forecasts that the 2008 F-150 will average 16.8 mpg. Determine the probability that these same 100 AAA members would average more than 16.8 mpg while testing the 2008 F-150. A) 0 B) 0.0155 C) 0.0412 D) None of the above Answer: A Diff: 3 Keywords: sampling distribution, mean, sample Section: 7-2 Sampling Distribution of the Mean Outcome: 1

113) A population has a proportion equal to 0.30. Calculate the following probabilities with n = 100. Find P( ≤ 0.35). A) 0.7244 B) 0.8621 C) 0.7124 D) 0.6126 Answer: B Diff: 2 Keywords: sampling distribution proportion Section: 7-3 Sampling Distribution of a Proportion Outcome: 1 114) A population has a proportion equal to 0.30. Calculate the following probabilities with n = 100. Find P( > 0.40). A) 0.0146 B) 0.0411 C) 0.0521 D) 0.0312 Answer: A Diff: 2 Keywords: sampling distribution proportion Section: 7-3 Sampling Distribution of a Proportion Outcome: 1 115) A population has a proportion equal to 0.30. Calculate the following probabilities with n = 100. Find P(0.25 < ≤ 0.40). A) 0.8121 B) 0.7415 C) 0.8475 D) 0.5612 Answer: C Diff: 2 Keywords: sampling distribution proportion Section: 7-3 Sampling Distribution of a Proportion Outcome: 1 116) A population has a proportion equal to 0.30. Calculate the following probabilities with n = 100. Find P( ≥ 0.27). A) 0.7422 B) 0.8141 C) 0.6125 D) 0.6841 Answer: A Diff: 2 Keywords: sampling distribution proportion Section: 7-3 Sampling Distribution of a Proportion Outcome: 1 7-32 Copyright © 2018 Pearson Education, Inc.

117) If a random sample of 200 items is taken from a population in which the proportion of items having a desired attribute is p = 0.30, what is the probability that the proportion of successes in the sample will be less than or equal to 0.27? A) 0.0841 B) 0.1011 C) 0.1912 D) 0.1762 Answer: D Diff: 2 Keywords: sampling distribution proportion Section: 7-3 Sampling Distribution of a Proportion Outcome: 1 118) The proportion of items in a population that possess a specific attribute is known to be 0.70. If a simple random sample of size n = 100 is selected and the proportion of items in the sample that contain the attribute of interest is 0.65, what is the sampling error? A) -0.03 B) -0.05 C) 0.08 D) 0.01 Answer: B Diff: 2 Keywords: sampling distribution proportion Section: 7-3 Sampling Distribution of a Proportion Outcome: 1 119) Given a population where the proportion of items with a desired attribute is p = 0.25, if a sample of 400 is taken, what is the standard deviation of the sampling distribution of ? A) 0.0512 B) 0.0312 C) 0.0217 D) 0.0412 Answer: C Diff: 2 Keywords: sampling distribution proportion Section: 7-3 Sampling Distribution of a Proportion Outcome: 1

120) Given a population where the proportion of items with a desired attribute is p = 0.25, if a sample of 400 is taken, what is the probability the proportion of successes in the sample will be greater than 0.22? A) 0.9162 B) 0.8812 C) 0.7141 D) 0.8412 Answer: A Diff: 2 Keywords: sampling distribution proportion Section: 7-3 Sampling Distribution of a Proportion Outcome: 1 121) Given a population in which the probability of success is p =0.20, if a sample of 500 items is taken, then calculate the probability the proportion of successes in the sample will be between 0.18 and 0.23. A) 0.7812 B) 0.8221 C) 0.9212 D) 0.6812 Answer: B Diff: 2 Keywords: sampling distribution proportion Section: 7-3 Sampling Distribution of a Proportion Outcome: 1 122) Given a population in which the probability of success is p = 0.20, if a sample of 500 items is taken, then calculate the probability the proportion of successes in the sample will be between 0.18 and 0.23 if the sample size is 200. A) 0.8911 B) 0.7121 C) 0.8712 D) 0.6165 Answer: D Diff: 2 Keywords: sampling distribution proportion Section: 7-3 Sampling Distribution of a Proportion Outcome: 1

123) United Manufacturing and Supply makes sprinkler valves for use in residential sprinkler systems. United supplies these valves to major companies such as Rain Bird and Nelson, who in turn sell sprinkler products to retailers. United recently entered into a contract to supply 40,000 sprinkler valves. The contract called for at least 97% of the valves to be free of defects. Before shipping the valves, United managers tested 200 randomly selected valves and found 190 defect-free valves in the sample. The managers wish to know the probability of finding 190 or fewer defect-free valves if in fact the population of 40,000 valves is 97% defect-free. The probability is: A) 0.0111 B) 0.0612 C) 0.0475 D) 0.0212 Answer: C Diff: 2 Keywords: sampling distribution proportion Section: 7-3 Sampling Distribution of a Proportion Outcome: 1 124) The J R Simplot Company is one of the world's largest privately held agricultural companies, employing over 10,000 people in the United States, Canada, China, Mexico, and Australia. More information can be found at the company's Web site: www.Simplot.com. One of its major products is french fries that are sold primarily on the commercial market to customers such as McDonald's and Burger King. French fries have numerous quality attributes that are important to customers. One of these is called "dark ends," which are the dark-colored ends that can occur when the fries are cooked. Suppose a major customer will accept no more than 0.06 of the fries having dark ends. Recently, the customer called the Simplot Company saying that a recent random sample of 300 fries was tested from a shipment and 27 fries had dark ends. Assuming that the population does meet the 0.06 standard, what is the probability of getting a sample of 300 with 27 or more dark ends? A) 0.0341 B) 0.0162 C) 0.0012 D) 0.0231 Answer: B Diff: 2 Keywords: sampling distribution proportion Section: 7-3 Sampling Distribution of a Proportion Outcome: 1

125) The National Association of Realtors released a survey indicating that a surprising 43% of first-time home buyers purchased their homes with no-money-down loans during 2005. The fear is that house prices will decline and leave homeowners owing more than their homes are worth. PMI Mortgage Insurance estimated that there existed a 50% risk that prices would decline within two years in major metro areas such as San Diego, Boston, Long Island, New York City, Los Angeles, and San Francisco. A survey taken by realtors in the San Francisco area found that 12 out of the 20 first-time home buyers sampled purchased their home with no-money-down loans. Calculate the probability that at least 12 in a sample of 20 first-time buyers would take out no-money-down loans if San Francisco's proportion is the same as the nationwide proportion of no-money-down loans. A) 0.0618 B) 0.0124 C) 0.0512 D) 0.0441 Answer: A Diff: 2 Keywords: sampling distribution proportion Section: 7-3 Sampling Distribution of a Proportion Outcome: 1 126) According to the most recent Labor Department data, 10.5% of engineers (electrical, mechanical, civil, and industrial) were women. Suppose a random sample of 50 engineers is selected. How likely is it that the random sample of 50 engineers will contain 8 or more women in these positions? A) 0.1612 B) 0.0821 C) 0.1020 D) 0.0314 Answer: C Diff: 2 Keywords: sampling distribution proportion Section: 7-3 Sampling Distribution of a Proportion Outcome: 1 127) According to the most recent Labor Department data, 10.5% of engineers (electrical, mechanical, civil, and industrial) were women. Suppose a random sample of 50 engineers is selected. How likely is it that the random sample will contain fewer than 5 women in these positions? A) 0.4522 B) 0.3124 C) 0.5121 D) 0.5512 Answer: A Diff: 2 Keywords: sampling distribution proportion Section: 7-3 Sampling Distribution of a Proportion Outcome: 1

128) Indicate why the Central Limit Theorem is so important in the application of statistical analysis. Answer: The Central Limit Theorem helps decision makers know what the general shape of the sampling distribution is regardless of what the population distribution looks like. This is important because in many cases we won't know the shape of the population distribution. As long as the sample size is sufficiently large, the sampling distribution of will be approximately normally distributed. Diff: 2 Keywords: Central Limit Theorem Section: 7-2 Sampling Distribution of the Mean Outcome: 3 129) Suppose it is known that the ages of all employees working for a very large computer company is normally distributed with a mean of 44.2 and a standard deviation of 5.6 years. Given this information, discuss what the sampling distribution for looks like? Answer: When the population distribution is known to be normally distributed, the sampling distribution of will also be normally distributed. Further, the mean of the sampling distribution will be equal to the population mean, 44.2, and the standard deviation of the sampling distribution will be

Diff: 2 Keywords: sampling distribution, mean, sample Section: 7-2 Sampling Distribution of the Mean Outcome: 2 130) Explain what is meant by the concept of sampling distribution. Answer: A population measure is called a parameter. It is a fixed value as long as the population does not change. A corresponding value computed from a sample is called a statistic. Depending on which sample we select we will get a different statistic. The difference between the statistic and the parameter due to the sample not being a perfect representative of the population is called sampling error. Each sample will generate a different amount of sampling error. For instance, if we are dealing with the mean, and were to take all possible samples of a given size, the distribution of the resulting sample means is called the sampling distribution. It represents the distribution of potential sampling error. The mean of the sampling distribution will equal the population mean and the standard deviation of the sampling distribution will equal the population standard deviation divided by the square root of the sample size. Diff: 2 Keywords: sampling distribution Section: 7-1 Sampling Error—What It Is and Why It Happens Outcome: 1

131) Explain the impact of the size of the sample on the shape of the sampling distribution. Answer: First, if the population distribution is unknown or is not normally distributed, the size of the sample is important in helping us know what the shape of the sampling distribution is. Under these conditions, the Central Limit Theorem tells us that the sampling distribution will be approximately normally distributed provided that the sample size is sufficiently large. Second, the size of the sample is key in determining the spread in the sampling distribution. The larger the sample size, the less variability there is in the sampling distribution. When dealing with means, the standard deviation of the sampling distribution is equal to

Diff: 2 Keywords: sample size, sampling distribution Section: 7-1 Sampling Error—What It Is and Why It Happens Outcome: 1 132) How would you respond to a statement that says that by increasing the sample size, the amount of sampling error will be decreased? Answer: We might respond by saying that in the long run the statement is true. That is, in situations where repeated random samples are selected, larger samples would result in less extreme sampling error on average. However, in a given situation there is no guarantee that a larger sample size will result is less sampling error. By chance, the smaller sample might provide a sample mean that is actually closer to the population mean than would occur from a particular larger sample. Diff: 2 Keywords: sampling error, sample size Section: 7-1 Sampling Error—What It Is and Why It Happens Outcome: 1 133) Suppose a population is normally distributed with a mean 100 and a standard deviation of 15. When a sample of size n = 36 is collected a sampling distribution is created. Explain which is larger: the probability of a value randomly selected from the population being larger than 120, or the probability of a sample mean being larger than 120. Answer: The probability will be larger for the value from the population than for the sample mean distribution because the population has the larger standard deviation. Population: z =

= 20/15 = 1.33, and looking this up in the standard normal table gives us a

probability of 0.4082, so P(x > 120) = .5 - .4082 = 0.0918. Sample mean distribution: z =

= 20/2.5 = 8.00, which is too large to look up in the standard.

Normal table, meaning P( > 120) is so small it's practically 0. Diff: 2 Keywords: sample, population, mean, sampling distribution Section: 7-2 Sampling Distribution of the Mean Outcome: 2

134) The annual income for independent sales representatives in the United States is thought to be highly right-skewed with a mean equal to $144,300 and a standard deviation of $32,450. Given this information, if a sample of 36 independent sales representatives is selected, what is the probability that the mean of the sample will exceed $130,000? Answer: Even though the population distribution is skewed, the Central Limit Theorem tells us that the sampling distribution will be approximately normally distributed for a sample of n = 36. Further, the mean of the sampling distribution should be equal to the population mean (μ = $144,300) and the standard deviation should be equal to = $5,408.33. Then, we convert the

= z=

= $130,000 to a standardized z-value using:

= -2.64.

Now, we can go to the standard normal table for z = -2.64, which gives 0.4959. This is the probability of a sample mean between $130,000 and $144,300. To get the probability we are looking for, we add 0.5000 giving 0.9959. Thus, the probability that the sample mean annual income will exceed $130,000 is 0.9959 or almost a sure thing. Diff: 2 Keywords: probability, mean, sampling distribution, z-value, Central Limit Theorem Section: 7-2 Sampling Distribution of the Mean Outcome: 3 135) The proportion of parts in an inventory that are outdated and no longer useful is thought to be 0.10. To check this, a random sample of n = 100 parts is selected and 14 are found to be outdated. Based upon this information, what is the probability of 14 or more outdated parts? Answer: We are interested in finding P( > 0.14). The sampling distribution for a proportion will be approximately normal as long as both np and n(1 - p) are greater than 5. That applies in this case. The standard deviation for the sampling distribution is given by . Thus, to find the probability, we standardize the sample proportion as follows: z=

= 1.33.

Then we can go to the standard normal table for z = 1.33. We get 0.4082. Subtracting this from 0.5000, we get 0.0918, which is the probability we are looking for. Diff: 2 Keywords: probability, proportion, sampling distribution, sample Section: 7-3 Sampling Distribution of a Proportion Outcome: 4

136) The Good Food chain has a contract to receive eggs from a large egg producer. The eggs come in lots of 4,000 dozen each week. The contract specifies that the rate of broken or defective eggs should not exceed 8 percent. Each time a load comes in, Good Food warehouse employees select a random sample of n = 100 eggs and check to see if they are broken or defective. If Good Food wants no more than a 0.05 chance of rejecting the shipment, what should the cut-off be in terms of proportion of broken or defective eggs so that if the proportion is that value or more, the shipment will be rejected? Answer: Since we are dealing with proportions, and both np and n(1 - p) are greater than 5, the normal distribution can be used to describe the sampling distribution. The standard in the contract calls for a 0.08 defect rate. The cut-off needs to be higher than this so that the chance of exceeding the cut-off is, at most, 0.05. To find what the cut-off should be, we use: =p+z where z is the value from the standard normal table that is associated with 0.45. This zvalue is approximately 1.645. Thus, the cut-off that the company should use is found as: .08 + 1.645

= .1246. Thus, in the sample of 100 if they find more than 12 broken or defective eggs,

the shipment should be rejected. Diff: 3 Keywords: sampling distribution, proportion, probability Section: 7-3 Sampling Distribution of a Proportion Outcome: 4

Business Statistics, 9e (Groebner/Shannon/Fry) Chapter 8 Estimating Single Population Parameters 1) Sampling error is the difference between a statistic computed from a sample and the corresponding parameter computed from the population. Answer: TRUE Diff: 1 Keywords: sampling error, statistic, parameter, sample Section: 8-1 Point and Confidence Interval Estimates for a Population Mean Outcome: 1 2) The higher the level of confidence, the wider the confidence interval must be. Answer: TRUE Diff: 1 Keywords: point estimate, sample Section: 8-1 Point and Confidence Interval Estimates for a Population Mean Outcome: 1 3) The product manager for a large retail store has recently stated that she estimates that the average purchase per visit for the store's customers is between $33.00 and $65.00. The $33.00 and the $65.00 are considered point estimates for the true population mean. Answer: FALSE Diff: 2 Keywords: population, mean, sample, sampling error Section: 8-1 Point and Confidence Interval Estimates for a Population Mean Outcome: 1 4) A point estimate is equally likely to be higher or lower than the population mean if the sampling is done using a statistical sampling procedure. Answer: TRUE Diff: 1 Keywords: point estimate, population, sample Section: 8-1 Point and Confidence Interval Estimates for a Population Mean Outcome: 1 5) A point estimate is an unbiased estimator of the true population value. However, error is associated with this estimate. Answer: TRUE Diff: 2 Keywords: confidence interval, population, point estimate Section: 8-1 Point and Confidence Interval Estimates for a Population Mean Outcome: 1

6) A statement in the newspaper attributed to the leader of a local union stated that the average hourly wage for union members in the region is $13.35. He indicated that this number came from a survey of union members. If an estimate was developed with 95 percent confidence, we can safely conclude that this value is within 95 percent of the true population mean hourly wage. Answer: FALSE Diff: 2 Keywords: confidence interval, population, survey Section: 8-1 Point and Confidence Interval Estimates for a Population Mean Outcome: 2 7) To construct a 99 percent confidence interval where σ is known, the correct critical value is 1.96. Answer: FALSE Diff: 1 Keywords: confidence interval, population Section: 8-1 Point and Confidence Interval Estimates for a Population Mean Outcome: 2 8) A point estimate for a population mean lies exactly halfway between the outer and lower limits of its confidence interval estimate. Answer: TRUE Diff: 1 Keywords: point estimate, confidence interval, population Section: 8-1 Point and Confidence Interval Estimates for a Population Mean Outcome: 1 9) The fact that a point estimate will likely be different from the corresponding population value is due to the fact that point estimates are subject to sampling error. Answer: TRUE Diff: 1 Keywords: sampling error, point estimate, population Section: 8-1 Point and Confidence Interval Estimates for a Population Mean Outcome: 1 10) When using a 95 percent confidence interval for a mean, the area in the upper tail of the distribution that is outside the interval is 5 percent. Answer: FALSE Diff: 2 Keywords: confidence interval Section: 8-1 Point and Confidence Interval Estimates for a Population Mean Outcome: 1 11) If a population is skewed, the point estimate will be pushed to the right or left of the middle of the confidence interval estimate. Answer: FALSE Diff: 1 Keywords: confidence interval, point estimate, population, skewed Section: 8-1 Point and Confidence Interval Estimates for a Population Mean Outcome: 1 8-2 Copyright © 2018 Pearson Education, Inc.

12) A 95 percent confidence interval estimate indicates that there is a 95 percent chance that the true population value will fall within the range defined by the upper and lower limits. Answer: FALSE Diff: 2 Keywords: confidence interval, population Section: 8-1 Point and Confidence Interval Estimates for a Population Mean Outcome: 1 13) Recently, a marketing research company reported that based on a random sample of 300 households the mean number of trips to a major shopping mall per month per household is 4.12 trips. This value is referred to as a parameter and is subject to sampling error. Answer: FALSE Diff: 1 Keywords: parameter, sampling error, mean Section: 8-1 Point and Confidence Interval Estimates for a Population Mean Outcome: 2 14) A report in a consumer magazine indicated that with 90 percent confidence, the mean number of hours that a particular brand light bulb lasts is between 900 and 1,100 hours. Based on this, the sample mean that produced this estimate is 1,000 hours. Answer: TRUE Diff: 2 Keywords: sample, mean, estimate, confidence Section: 8-1 Point and Confidence Interval Estimates for a Population Mean Outcome: 2 15) A random sample of 100 boxes of cereal had a sample mean weight of 396 grams. The standard deviation is known to be 5 grams. The upper end of the confidence interval for the mean is 405.8 grams. Answer: FALSE Diff: 2 Keywords: random sample, mean, standard deviation, confidence interval Section: 8-1 Point and Confidence Interval Estimates for a Population Mean Outcome: 2 16) In a situation where we know the population standard deviation but wish to estimate the population mean using a 90 percent confidence interval, the critical value is z = ±1.645. Answer: TRUE Diff: 2 Keywords: confidence interval, population, mean, standard deviation, critical value Section: 8-1 Point and Confidence Interval Estimates for a Population Mean Outcome: 2

17) When developing a confidence interval estimate, the confidence level is calculated based on the size of the sample and the population standard deviation. Answer: FALSE Diff: 2 Keywords: confidence interval, sample size, population, standard deviation Section: 8-1 Point and Confidence Interval Estimates for a Population Mean Outcome: 2 18) To find a 98 percent confidence interval where the standard deviation is known, the correct z-value to use for the critical value is 2.33. Answer: TRUE Diff: 2 Keywords: confidence interval, critical value Section: 8-1 Point and Confidence Interval Estimates for a Population Mean Outcome: 2 19) In a recent report to the supply-chain manager in a major electronics company, the report writer stated that with 90 percent confidence, the manufacturing lead time for a critical part is between 3.34 hours and 4.14 hours. Based on this, the sample mean that generated the confidence interval was 3.60. Answer: FALSE Diff: 2 Keywords: confidence interval, point estimate Section: 8-1 Point and Confidence Interval Estimates for a Population Mean Outcome: 2 20) In a recent report to the supply-chain manager in a major electronics company, the report writer stated that with 90 percent confidence, the manufacturing lead time for a critical part is between 3.34 hours and 4.14 hours. Based on this information, the margin of error for this estimate is ±.80 hours. Answer: FALSE Diff: 2 Keywords: confidence interval, point estimate, margin of error Section: 8-1 Point and Confidence Interval Estimates for a Population Mean Outcome: 2 21) In developing a confidence interval estimate, the margin of error is directly dependent on the value of the point estimate. Answer: FALSE Diff: 2 Keywords: confidence interval, margin of error, point estimate Section: 8-1 Point and Confidence Interval Estimates for a Population Mean Outcome: 2 22) The margin of error is one-half the width of the confidence interval. Answer: TRUE Diff: 2 Keywords: confidence interval, confidence level, margin of error Section: 8-1 Point and Confidence Interval Estimates for a Population Mean Outcome: 2 8-4 Copyright © 2018 Pearson Education, Inc.

23) One way to reduce the margin of error in a confidence interval estimate is to lower the level of confidence. Answer: TRUE Diff: 2 Keywords: margin of error, confidence interval Section: 8-1 Point and Confidence Interval Estimates for a Population Mean Outcome: 2 24) The standard deviation for the checking account balances is assumed known to be $357.50. Recently, a bank manager was interested in estimating the mean balance. To do this, she selected a random sample of 81 accounts and found a mean balance of $1,347.20. At a 95 percent confidence level, the lower limit for the confidence interval is $646.50. Answer: FALSE Diff: 2 Keywords: standard deviation, confidence level, lower limit, mean Section: 8-1 Point and Confidence Interval Estimates for a Population Mean Outcome: 2 25) If we are interested in estimating the population mean based on a sample from a population for which we know neither the mean nor the standard deviation, the critical value will be a t value from the tdistribution. Answer: TRUE Diff: 2 Keywords: population, mean, standard deviation, estimate, critical value, t-distribution Section: 8-1 Point and Confidence Interval Estimates for a Population Mean Outcome: 2 26) When calculating a confidence interval, the reason for using the t-distribution rather than the normal distribution for the critical value is that the population standard deviation is unknown. Answer: TRUE Diff: 2 Keywords: population, mean, standard deviation, estimate, critical value, t-distribution Section: 8-1 Point and Confidence Interval Estimates for a Population Mean Outcome: 2 27) Confidence intervals constructed with small samples tend to have greater margins of error than those constructed from larger samples, all else being constant. Answer: TRUE Diff: 1 Keywords: confidence interval, sample, margin of error Section: 8-1 Point and Confidence Interval Estimates for a Population Mean Outcome: 2

28) The bottlers of a new fruit juice daily select a random sample of 12 bottles of the drink to estimate the mean quantity of juice in the bottles filled that day. On one such day, the following results were observed: = 12.03; s = 0.12. Based on this information, the upper limit for a 95 percent confidence interval estimate is approximately 12.106 ounces. Answer: TRUE Diff: 2 Keywords: confidence interval, upper limit, t-statistic Section: 8-1 Point and Confidence Interval Estimates for a Population Mean Outcome: 2 29) The bottlers of a new fruit juice daily select a random sample of 12 bottles of the drink to estimate the mean quantity of juice in the bottles filled that day. On one such day, the following results were observed: = 12.03; s = 0.12. Based on this information, the margin of error associated with a 90 percent confidence interval estimate for the population mean is 1.7959 ounces. Answer: FALSE Diff: 2 Keywords: margin of error, confidence interval, t-statistic Section: 8-1 Point and Confidence Interval Estimates for a Population Mean Outcome: 2 30) A 95 percent confidence interval for a mean can be interpreted that we are 95 percent sure that our answer is correct. Answer: FALSE Diff: 2 Keywords: confidence interval Section: 8-1 Point and Confidence Interval Estimates for a Population Mean Outcome: 2 31) The makers of weight loss product are interested in estimating the mean weight loss for users of their product. To do this, they have selected a random sample of n = 9 people and have provided them with a supply of the product. After six months, the nine people had an average weight loss of 15.3 pounds with a standard deviation equal to 3.5 pounds. The upper limit for the 90 percent confidence interval estimate for the population mean is approximately 17.47 pounds. Answer: TRUE Diff: 2 Keywords: confidence interval, upper limit, t-statistic Section: 8-1 Point and Confidence Interval Estimates for a Population Mean Outcome: 2 32) The t-distribution is used to obtain the critical value in developing a confidence interval when the population distribution is not known or the sample size is small. Answer: TRUE Diff: 1 Keywords: t-distribution, critical value, confidence interval, sample size Section: 8-1 Point and Confidence Interval Estimates for a Population Mean Outcome: 2

33) The impact on the margin of error for a confidence interval for an increase in confidence level and a decrease in sample size is unknown since these changes are contradictory. Answer: FALSE Diff: 2 Keywords: margin of error, confidence level, sample size Section: 8-1 Point and Confidence Interval Estimates for a Population Mean Outcome: 2 34) A 95 percent confidence interval estimate will have a margin of error that is approximately ±47.5 percent of the size of the population mean. Answer: FALSE Diff: 2 Keywords: confidence interval, margin of error Section: 8-1 Point and Confidence Interval Estimates for a Population Mean Outcome: 2 35) A higher confidence level means that the point estimate for the population mean will be closer to the true population mean. Answer: FALSE Diff: 2 Keywords: confidence interval, point estimate Section: 8-1 Point and Confidence Interval Estimates for a Population Mean Outcome: 2 36) If the population is not normally distributed, the t-distribution cannot be used. Answer: FALSE Diff: 3 Keywords: t-distribution, sampling Section: 8-1 Point and Confidence Interval Estimates for a Population Mean Outcome: 2 37) In estimating a population mean, a large sample size is generally preferable to a small sample size because the margin of error is generally smaller. Answer: TRUE Diff: 2 Keywords: sample, estimate, population, margin of error Section: 8-2 Determining the Required Sample Size for Estimating a Population Mean Outcome: 3 38) A pilot sample is one that is used when a decision maker wishes to get an advance idea of what the mean of the population might be. Answer: FALSE Diff: 1 Keywords: pilot sample, population, mean Section: 8-2 Determining the Required Sample Size for Estimating a Population Mean Outcome: 3

39) One factor that plays an important part in determining what the needed sample size is when developing a confidence interval estimate is the level of confidence that you wish to use. Answer: TRUE Diff: 1 Keywords: sample size, confidence interval, confidence level Section: 8-2 Determining the Required Sample Size for Estimating a Population Mean Outcome: 3 40) All of the factors that are needed to determine the required sample size are within the control of the decision maker. Answer: FALSE Diff: 2 Keywords: sample size, factors Section: 8-2 Determining the Required Sample Size for Estimating a Population Mean Outcome: 3 41) A bank manager wishes to estimate the mean waiting time spent by customers at his bank. He knows from previous experience that the standard deviation is about 4.0 minutes. If he desires a 90 percent confidence interval estimate and wishes to have a margin of error of 1 minute, the required sample size will be approximately 143. Answer: FALSE Diff: 3 Keywords: confidence interval, margin of error, sample size Section: 8-2 Determining the Required Sample Size for Estimating a Population Mean Outcome: 3 42) When estimating sample size, a 90 percent confidence level will result in a smaller sample size than a 95 percent confidence level. Answer: TRUE Diff: 2 Keywords: sample size, confidence level Section: 8-2 Determining the Required Sample Size for Estimating a Population Mean Outcome: 3 43) Two confidence interval estimates were developed from the same sample of a population. The wider interval will be the one that has the higher confidence level. Answer: TRUE Diff: 2 Keywords: confidence interval, confidence level, sample Section: 8-2 Determining the Required Sample Size for Estimating a Population Mean Outcome: 3

44) In determining the required sample size in an application involving an estimate for the population mean, if the population standard deviation is known, there is no compelling reason to select a pilot sample. Answer: TRUE Diff: 2 Keywords: sample size, population, standard deviation Section: 8-2 Determining the Required Sample Size for Estimating a Population Mean Outcome: 3 45) The main purpose of a pilot sample in an application involving an estimate for a population mean is to determine what the margin of error will likely be. Answer: FALSE Diff: 2 Keywords: margin of error, pilot sample, estimate Section: 8-2 Determining the Required Sample Size for Estimating a Population Mean Outcome: 3 46) The national sales manager for a textbook publishing company wishes to estimate the mean number of books sold per college. She wishes to have her estimate be within ± 30 copies and wants a 95 percent confidence interval estimate. If a pilot sample of 30 schools gave a sample standard deviation equal to 60 books, the required total sample size is less than the pilot sample already taken. Answer: TRUE Diff: 3 Keywords: pilot sample, sample size, estimate, mean Section: 8-2 Determining the Required Sample Size for Estimating a Population Mean Outcome: 3 47) If a pilot sample of n = 40 items has been used as a first step in determining a required sample size of n = 360, the decision maker can go ahead and use these 40 and take a sample of only 320 more items. Answer: TRUE Diff: 1 Keywords: pilot sample, sample size Section: 8-2 Determining the Required Sample Size for Estimating a Population Mean Outcome: 3 48) When finding sample size, cutting the margin of error in half requires that the sample size be four times larger. Answer: TRUE Diff: 2 Keywords: sample size, margin of error Section: 8-2 Determining the Required Sample Size for Estimating a Population Mean Outcome: 3

49) If a decision maker desires a small margin of error and a high level of confidence, it is certain that the required sample size will be quite large. Answer: FALSE Diff: 2 Keywords: margin of error, confidence level, sample size Section: 8-2 Determining the Required Sample Size for Estimating a Population Mean Outcome: 3 50) In an effort to estimate the mean length of stay for motel guests at a major national motel chain, the decision makers asked for a 95 percent confidence, and a margin of error of ±0.25 days. Further, it was known that the population standard deviation is 0.50 days. Given this, the required sample size to estimate the mean length of stay is about 16 customers. Answer: TRUE Diff: 2 Keywords: sample size, confidence, margin of error Section: 8-2 Determining the Required Sample Size for Estimating a Population Mean Outcome: 3 51) In a sample size determination situation, reducing the margin of error by half will double the required sample size. Answer: FALSE Diff: 2 Keywords: sample size, margin of error Section: 8-2 Determining the Required Sample Size for Estimating a Population Mean Outcome: 3 52) The manager in charge of concessions at an NFL football stadium is interested in estimating the mean dollars that are spent per person attending the games. A pilot sample of n = 50 people has revealed a sample mean and standard deviation of $12.35 and $2.35 respectively. He wishes to estimate the population mean within ± $0.20 of the true mean and wishes to have a confidence level of 95 percent. Given this, he needs to sample an additional 481 people. Answer: TRUE Diff: 2 Keywords: confidence level, sample size, margin of error Section: 8-2 Determining the Required Sample Size for Estimating a Population Mean Outcome: 3 53) A university computer lab manager wishes to estimate the mean time that students stay in the lab per visit. She believes that the population standard deviation would be no larger than 10 minutes. Further, she wishes to have a confidence level of 90 percent and a margin of error of ±2.00 minutes. Given this, the sample size that she uses is approximately 60 students. Answer: FALSE Diff: 2 Keywords: sample size, estimate, confidence level Section: 8-2 Determining the Required Sample Size for Estimating a Population Mean Outcome: 3

54) An analyst for a financial investment firm recently went through the effort to determine the required sample size for estimating the mean number of transactions per year for the clients of his firm. The calculations, which were based on a 95 percent confidence level and a margin of error of ±3, gave a required sample size of 300. Given this information, the value used for the population standard deviation must have been about 26.5 transactions. Answer: TRUE Diff: 3 Keywords: standard deviation, sample size, margin of error Section: 8-2 Determining the Required Sample Size for Estimating a Population Mean Outcome: 3 55) When a decision maker determines the required sample size for estimating a population mean, a change in the confidence level will result in a change in the required sample size, provided that the margin of error is also modified accordingly. Answer: FALSE Diff: 2 Keywords: sample size, confidence level, margin of error Section: 8-2 Determining the Required Sample Size for Estimating a Population Mean Outcome: 3 56) The State Transportation Department wishes to estimate the mean speed of vehicles on a certain stretch of highway. They wish to estimate the mean within ±0.75 mph and they wish to have a confidence level equal to 99 percent. Based on this information only, they can determine that the required sample size is 320 vehicles. Answer: FALSE Diff: 2 Keywords: sample size, confidence level, estimate Section: 8-2 Determining the Required Sample Size for Estimating a Population Mean Outcome: 3 57) After taking a speed-reading course, students are supposed to be able to read faster than they could before taking the course. A pilot sample of n = 25 students showed a mean increase of 300 words per minute with a standard deviation equal to 60 words per minute. To estimate the population mean with 95 percent confidence and a margin of error of ±10 minutes, the required sample size is approximately 139 students. Answer: TRUE Diff: 2 Keywords: sample size, margin of error, standard deviation Section: 8-2 Determining the Required Sample Size for Estimating a Population Mean Outcome: 3

58) An emergency room in a hospital wants to determine the sample size needed for estimating their mean number of visits per day. If they want a 99 percent confidence level the correct critical value to use is 2.33. Answer: FALSE Diff: 2 Keywords: sample size, confidence level Section: 8-2 Determining the Required Sample Size for Estimating a Population Mean Outcome: 3 59) In estimating a population mean, increasing the confidence level will result in a higher margin of error for a given sample size. Answer: TRUE Diff: 2 Keywords: estimate, confidence level, margin of error, sample size Section: 8-2 Determining the Required Sample Size for Estimating a Population Mean Outcome: 3 60) A grocery store manager is interested in estimating the mean weight of apples received in a shipment. If she wishes to have the estimate be within ±.05 pound with 90 percent confidence, the sample size should be 103 apples if she believes that the standard deviation is .08 pound. Answer: FALSE Diff: 3 Keywords: sample size, standard deviation, margin of error Section: 8-2 Determining the Required Sample Size for Estimating a Population Mean Outcome: 3 61) A random sample of 100 people was selected from a population of customers at a local bank. The mean age of these customers was 40. If the population standard deviation is thought to be 5 years, the margin of error for a 95 percent confidence interval estimate is .98 year. Answer: TRUE Diff: 3 Keywords: margin of error, standard deviation, confidence interval Section: 8-2 Determining the Required Sample Size for Estimating a Population Mean Outcome: 3 62) When σ is unknown, one should use an estimate that is considered to be at least as large as the true σ. This will provide a conservative sample size to be on the safe side. Answer: TRUE Diff: 3 Keywords: margin of error, standard deviation Section: 8-2 Determining the Required Sample Size for Estimating a Population Mean Outcome: 3

63) In determining the required sample size when estimating a population proportion, it is necessary to start with some idea of what that proportion is. Answer: TRUE Diff: 2 Keywords: sample size, population proportion, estimate Section: 8-3 Estimating a Population Proportion Outcome: 5 64) The concept of margin of error applies directly when estimating a population mean, but is not appropriate when estimating a population proportion. Answer: FALSE Diff: 1 Keywords: margin of error, population proportion, estimate Section: 8-3 Estimating a Population Proportion Outcome: 5 65) A Parks and Recreation official surveyed 200 people at random who have used one of the city's parks. The survey revealed that 26 resided outside the city limits. If she had to arrive at one single value to estimate the true proportion of park users who are residents of the city it would be 0.13. Answer: TRUE Diff: 1 Keywords: population proportion, estimate Section: 8-3 Estimating a Population Proportion Outcome: 4 66) The one sure thing that can be said about a point estimate is that it will most likely be subject to sampling error and not equal the parameter. Answer: TRUE Diff: 1 Keywords: point estimate, sampling error Section: 8-3 Estimating a Population Proportion Outcome: 4 67) When determining the sample size for a proportion, if you have no previous information available to estimate p, then the best value to use is π = 0.5. Answer: TRUE Diff: 2 Keywords: population proportion, sample size Section: 8-3 Estimating a Population Proportion Outcome: 5 68) In estimating a population proportion, the factors that are needed to determine the required sample size are the confidence level, the margin of error and some idea of what the population proportion is. Answer: TRUE Diff: 2 Keywords: population proportion, sample size, confidence level, margin of error Section: 8-3 Estimating a Population Proportion Outcome: 5 8-13 Copyright © 2018 Pearson Education, Inc.

69) For a given sample size and a given confidence level, the closer p is to 1.0, the greater the margin of error will be. Answer: FALSE Diff: 2 Keywords: sample size, confidence level, margin of error Section: 8-3 Estimating a Population Proportion Outcome: 5 70) The procurement manager for a large company wishes to estimate the proportion of parts from a supplier that are defective. She has selected a random sample of n = 200 incoming parts and has found 11 to be defective. Based on a 95 percent confidence level, the upper and lower limits for the confidence interval estimate are approximately 0.0234 to 0.0866. Answer: TRUE Diff: 2 Keywords: population proportion, confidence interval, estimate Section: 8-3 Estimating a Population Proportion Outcome: 4 71) A local pizza company is interested in estimating the percentage of customers who would take advantage of a coupon offer. To do this, they give the coupon out to a random sample of 100 customers. Of these, 45 actually use the coupon. At the 95 percent confidence level it would be appropriate for the manager to conclude that possibly as many as 50 percent of his customers will redeem the coupon. Answer: TRUE Diff: 3 Keywords: population proportion, confidence interval Section: 8-3 Estimating a Population Proportion Outcome: 4 72) A local pizza company is interested in estimating the percentage of customers who would take advantage of a coupon offer. To do this, they give the coupon out to a random sample of 100 customers. Of these, 45 actually use the coupon. Based on a 95 percent confidence level, the upper and lower confidence interval limits are approximately 0.3525 to 0.5475. Answer: TRUE Diff: 2 Keywords: confidence interval, population proportion Section: 8-3 Estimating a Population Proportion Outcome: 4 73) A publisher is interested in estimating the proportion of textbooks that students resell at the end of the semester. He is interested in making this estimate using a confidence level of 95 percent and a margin of error of ±0.02. Based upon his prior experience, he believes that π is somewhere around 0.60. Given this information, the required sample size is over 2,300 students. Answer: TRUE Diff: 2 Keywords: sample size, population proportion, margin of error Section: 8-3 Estimating a Population Proportion Outcome: 5 8-14 Copyright © 2018 Pearson Education, Inc.

74) When determining sample size for a proportion, using p = 0.5 will produce the smallest possible value for n. Answer: FALSE Diff: 2 Keywords: population proportion, sample size Section: 8-3 Estimating a Population Proportion Outcome: 5 75) The t-distribution is used for the critical value when estimating a population proportion when the standard deviation of the population is not known. Answer: FALSE Diff: 1 Keywords: t-distribution, population proportion Section: 8-3 Estimating a Population Proportion Outcome: 5 76) Recently, a report in a financial journal indicated that the 90 percent confidence interval estimate for the proportion of investors who own one or more mutual funds is between 0.88 and 0.92. Given this information, the sample size that was used in this study was approximately 609 investors. Answer: TRUE Diff: 3 Keywords: sample size, confidence level, population proportion Section: 8-3 Estimating a Population Proportion Outcome: 5 77) A random sample of n = 500 people was surveyed recently to determine an estimate for the proportion of people in the population who had attended at least some college. The estimate concluded that between 0.357 and 0.443 of the population had attended. Given this information, we can determine that the confidence level was approximately 95 percent. Answer: TRUE Diff: 3 Keywords: population proportion, confidence level, sample size Section: 8-3 Estimating a Population Proportion Outcome: 4 78) In determining the sample size requirement for an application involving the estimation of the proportion of department store customers who pay using the store's credit card, the closer the true proportion is to .5, the larger will be the required sample size for a given margin of error and confidence level. Answer: TRUE Diff: 2 Keywords: sample size, population proportion, margin of error, confidence level Section: 8-3 Estimating a Population Proportion Outcome: 5

79) The manager of the local county fair believes that no more than 30 percent of the adults in the county would object to a fee increase to attend the fair if it meant that better entertainment could be secured. To estimate the true proportion, he has selected a random sample of 200 adults. The manager will use a 90 percent confidence level. Assuming his assumption about the 30 percent holds, the margin of error for the estimate will be approximately ±.169. Answer: FALSE Diff: 3 Keywords: population proportion, margin of error, confidence level Section: 8-3 Estimating a Population Proportion Outcome: 5 80) Chicago Connection, a local pizza company, delivers pizzas for free within the market area. The delivery drivers are paid $2.00 per delivery plus they get to keep any tips. To estimate the proportion of deliveries that result in a tip to the driver, a random sample of 64 deliveries was selected. Of these, 48 times a tip was received. Based on this information, and using a 95 percent confidence level, the upper limit for the confidence interval estimate is about .1061. Answer: FALSE Diff: 3 Keywords: confidence interval, upper limit, population proportion Section: 8-3 Estimating a Population Proportion Outcome: 4 81) Chicago Connection, a local pizza company, delivers pizzas for free within the market area. The delivery drivers are paid $2.00 per delivery plus they get to keep any tips. To estimate the proportion of deliveries that result in a tip to the driver, a random sample of 64 deliveries was selected. Of these, 48 times a tip was received. Based on this information, and using a 95 percent confidence level, the margin of error for the estimate is approximately ±.1061. Answer: TRUE Diff: 3 Keywords: margin of error, population proportion, confidence interval Section: 8-3 Estimating a Population Proportion Outcome: 5 82) When determining sample size for a proportion, the farther that p is from 0.5, the smaller the resulting sample size will be. Answer: TRUE Diff: 2 Keywords: sample size, population proportion Section: 8-3 Estimating a Population Proportion Outcome: 4

83) Twenty one (21) customers out of a simple random sample of 50 said that they came to the grocery store within one month after getting a card from the company. Based on the data, the marketing manager thinks that the number of total customers who would come in to the grocery store within one month is 4,200 because 10,000 cards were mailed. Answer: TRUE Diff: 2 Keywords: population proportion, point estimate Section: 8-3 Estimating a Population Proportion Outcome: 4 84) Which of the following statements applies to a point estimate? A) The point estimate is a parameter. B) The point estimate will tend to be accurate if the sample size exceeds 30 for non-normal populations. C) The point estimate is subject to sampling error and will almost always be different from the population value. D) The point estimate is needed to determine the required sample size when estimating the population mean. Answer: C Diff: 2 Keywords: point estimate, sampling error, population Section: 8-1 Point and Confidence Interval Estimates for a Population Mean Outcome: 1 85) Sampling error occurs when: A) a nonstatistical sample is used. B) the statistic computed from the sample is not equal to the parameter for the population. C) a random sample is used rather than a convenience sample. D) a confidence interval is used to estimate a population value rather than a point estimate. Answer: B Diff: 1 Keywords: sampling error, statistic, parameter Section: 8-1 Point and Confidence Interval Estimates for a Population Mean Outcome: 1 86) The general format for a confidence interval is: A) point estimate ± z (standard deviation). B) point estimate ± (critical value)(standard error). C) margin of error ± (confidence coefficient) (standard error). D) point estimate ± (critical value)(standard deviation) Answer: B Diff: 1 Keywords: confidence interval, point estimate, standard error Section: 8-1 Point and Confidence Interval Estimates for a Population Mean Outcome: 2

87) In an application to estimate the mean number of miles that downtown employees commute to work roundtrip each day, the following information is given: n = 20 = 4.33 s = 3.50 If the desired confidence level is 95 percent, the appropriate critical value is: A) z = 1.96 B) t = 2.093 C) t = 2.086 D) .7826 Answer: B Diff: 2 Keywords: t-statistic, confidence level Section: 8-1 Point and Confidence Interval Estimates for a Population Mean Outcome: 2 88) In an application to estimate the mean number of miles that downtown employees commute to work roundtrip each day, the following information is given: n = 20 = 4.33 s = 3.50 The point estimate for the true population mean is: A) 1.638 B) 4.33 ± 1.638 C) 4.33 D) 3.50 Answer: C Diff: 1 Keywords: point estimate, population mean Section: 8-1 Point and Confidence Interval Estimates for a Population Mean Outcome: 2 89) In an application to estimate the mean number of miles that downtown employees commute to work roundtrip each day, the following information is given: n = 20 = 4.33 s = 3.50 Based on this information, the upper limit for a 95 percent confidence interval estimate for the true population mean is: A) about 5.97 miles. B) about 7.83 miles. C) nearly 12.0 miles. D) about 5.86 miles. Answer: A Diff: 3 Keywords: confidence interval, upper limit, t-statistic Section: 8-1 Point and Confidence Interval Estimates for a Population Mean Outcome: 2 8-18 Copyright © 2018 Pearson Education, Inc.

90) In developing a confidence interval estimate for the population mean, which of the following is true? A) The larger the sample standard deviation, the wider will be the interval estimate, all other things being equal. B) If the population standard deviation is unknown, the appropriate critical value should be obtained from the t-distribution. C) The confidence interval developed from a smaller sample size will have a larger margin of error than one obtained using a larger sample size, all other things being equal. D) All of the above are true. Answer: D Diff: 2 Keywords: confidence interval, population mean, t-distribution, sample size Section: 8-1 Point and Confidence Interval Estimates for a Population Mean Outcome: 2 91) Which of the following will increase the width of a confidence interval (assuming that everything else remains constant)? A) Decreasing the confidence level B) Increasing the sample size C) A decrease in the standard deviation D) Decreasing the sample size Answer: D Diff: 2 Keywords: sample size, confidence interval Section: 8-1 Point and Confidence Interval Estimates for a Population Mean Outcome: 2 92) In an effort to estimate the mean dollars spent per visit by customers of a food store, the manager has selected a random sample of 100 cash register receipts. The mean of these was $45.67 with a sample standard deviation equal to $12.30. Assuming that he wants to develop a 90 percent confidence interval estimate, which of the following is the margin of error that will be reported? A) About ±$2.02 B) Nearly $50.20 C) $1.645 D) About $1.43 Answer: A Diff: 3 Keywords: margin of error, confidence interval Section: 8-1 Point and Confidence Interval Estimates for a Population Mean Outcome: 2

93) In an effort to estimate the mean dollars spent per visit by customers of a food store, the manager has selected a random sample of 100 cash register receipts. The mean of these was $45.67 with a sample standard deviation equal to $12.30. Assuming that he wants to develop a 90 percent confidence interval estimate, the upper limit of the confidence interval estimate is: A) about $2.02 B) approximately $65.90 C) about $47.69 D) None of the above Answer: C Diff: 3 Keywords: upper limit, confidence interval, estimate Section: 8-1 Point and Confidence Interval Estimates for a Population Mean Outcome: 2 94) The margin of error is: A) the largest possible sampling error at a specified level of confidence. B) the critical value multiplied by the standard error of the sampling distribution. C) Both A and B D) the difference between the point estimate and the parameter. Answer: C Diff: 2 Keywords: margin of error, sampling error, critical value Section: 8-1 Point and Confidence Interval Estimates for a Population Mean Outcome: 2 95) Which of the following statements is true with respect to the confidence level associated with an estimation application? A) The confidence level is a percentage value between 50 and 100 that corresponds to the percentage of all possible confidence intervals, based on a given sample size, that will contain the true population value. B) The probability that the confidence interval estimate will contain the true population value. C) The degree of accuracy associated with the confidence interval estimate. D) None of the above Answer: A Diff: 2 Keywords: confidence level, population value Section: 8-1 Point and Confidence Interval Estimates for a Population Mean Outcome: 2

96) In a situation where the population standard deviation is known and we wish to estimate the population mean with 90 percent confidence, what is the appropriate critical value to use? A) z = 1.96 B) z = 2.33 C) z = 1.645 D) Can't be determined without knowing the degrees of freedom. Answer: C Diff: 2 Keywords: critical value, confidence interval, population mean Section: 8-1 Point and Confidence Interval Estimates for a Population Mean Outcome: 2 97) In developing a confidence interval estimate for the population mean, the t-distribution is used to obtain the critical value when: A) the sample contains some extreme values that skew the results. B) the population standard deviation is unknown. C) the sampling that is being used is not a statistical sample. D) the confidence level is low. Answer: B Diff: 2 Keywords: t-distribution, confidence interval Section: 8-1 Point and Confidence Interval Estimates for a Population Mean Outcome: 2 98) Which of the following statements is true with respect to the t-distribution? A) The t-distribution is symmetrical. B) The exact shape of the t-distribution depends on the number of degrees of freedom. C) The t-distribution is more spread out than the standard normal distribution. D) All of the above are true. Answer: D Diff: 2 Keywords: t-distribution, degrees of freedom, spread Section: 8-1 Point and Confidence Interval Estimates for a Population Mean Outcome: 2 99) A popular restaurant takes a random sample n = 25 customers and records how long each occupied a table. They found a sample mean of 1.2 hours and a sample standard deviation of 0.3 hour. Find the 95 percent confidence interval for the mean. A) 1.2 ± .118 B) 1.2 ± .124 C) 1.2 ± .588 D) 1.2 ± .609 Answer: B Diff: 2 Keywords: confidence interval, t-distribution Section: 8-1 Point and Confidence Interval Estimates for a Population Mean Outcome: 2

100) If a decision maker wishes to reduce the margin of error associated with a confidence interval estimate for a population mean, she can: A) decrease the sample size. B) increase the confidence level. C) increase the sample size. D) use the t-distribution. Answer: C Diff: 2 Keywords: margin of error, confidence interval, confidence level Section: 8-1 Point and Confidence Interval Estimates for a Population Mean Outcome: 2 101) When small samples are used to estimate a population mean, in cases where the population standard deviation is unknown: A) the t-distribution must be used to obtain the critical value. B) the resulting margin of error for a confidence interval estimate will tend to be fairly small. C) there will be a large amount of sampling error. D) None of the above Answer: A Diff: 2 Keywords: sample size, population mean, t-distribution Section: 8-1 Point and Confidence Interval Estimates for a Population Mean Outcome: 2 102) An educational organization in California is interested in estimating the mean number of minutes per day that children between the age of 6 and 18 spend watching television per day. A previous study showed that the population standard deviation was 21.5 minutes. The organization selected a random sample of n = 200 children between the ages of 6 and 18 and recorded the number of minutes of TV that each person watched on a particular day. The mean time was 191.3 minutes. If the leaders of the organization wish to develop an interval estimate with 98 percent confidence, what critical value should be used? A) z = 1.645 B) t = 2.38 C) Approximately z = 2.33 D) Can't be determined without knowing the margin of error. Answer: C Diff: 2 Keywords: z-value, confidence interval Section: 8-1 Point and Confidence Interval Estimates for a Population Mean Outcome: 2

103) An educational organization in California is interested in estimating the mean number of minutes per day that children between the age of 6 and 18 spend watching television per day. A previous study showed that the population standard deviation was 21.5 minutes. The organization selected a random sample of n = 200 children between the age of 6 and 18 and recorded the number of minutes of TV that each person watched on a particular day. The mean time was 191.3 minutes. If the leaders of the organization wish to develop an interval estimate with 98 percent confidence, what would be the upper and lower limits of the interval estimate? A) Approximately 187.76 minutes - 194.84 minutes B) About 141.21 minutes - 241.40 minutes C) Approximately 188.3 minutes - 194.3 minutes D) None of the above Answer: A Diff: 2 Keywords: confidence interval, upper, lower, limit, z-value Section: 8-1 Point and Confidence Interval Estimates for a Population Mean Outcome: 2 104) An educational organization in California is interested in estimating the mean number of minutes per day that children between the age of 6 and 18 spend watching television per day. A previous study showed that the population standard deviation was 21.5 minutes. The organization selected a random sample of n = 200 children between the age of 6 and 18 and recorded the number of minutes of TV that each person watched on a particular day. The mean time was 191.3 minutes. If the leaders of the organization wish to develop an interval estimate with 95 percent confidence, what will the margin of error be? A) Approximately ±1.52 minutes B) About ±2.98 minutes C) z = 1.96 D) Approximately ±42.14 minutes Answer: B Diff: 2 Keywords: margin of error, confidence interval, z-value Section: 8-1 Point and Confidence Interval Estimates for a Population Mean Outcome: 2

105) The Wisconsin Dairy Association is interested in estimating the mean weekly consumption of milk for adults over the age of 18 in that state. To do this, they have selected a random sample of 300 people from the designated population. The following results were recorded: Given this information, if the leaders wish to estimate the mean milk consumption with 90 percent confidence, what is the approximate margin of error in the estimate? A) z = 1.645 B) ±12.996 ounces C) ±0.456 ounce D) ±0.75 ounce Answer: D Diff: 2 Keywords: margin of error, confidence, z-value Section: 8-1 Point and Confidence Interval Estimates for a Population Mean Outcome: 2 106) The Internal Revenue Service (IRS) is interested in estimating the mean amount of money spent on outside tax service by income tax filers filing as single on their individual form. To do this, they have selected a random sample of n = 16 people from this population and surveyed them to determine the sample mean and sample standard deviation. The following information was observed: Given this information, what is the 95 percent confidence interval for the mean dollars spent on outside tax assistance by taxpayers who file as single? A) Approximately $72.19 - $105.01 B) About $22.97 - $154.23 C) Approximately $80.90 - $96.30 D) About $28.25 - $148.95 Answer: A Diff: 2 Keywords: confidence interval, z-value, estimate Section: 8-1 Point and Confidence Interval Estimates for a Population Mean Outcome: 2

107) A study was recently conducted to estimate the mean cholesterol for adult males over the age of 55 years. The following random sample data were observed: 245 196

304 210

135 188

202 256

300 390

Given this information, what is the point estimate for the population mean? A) About 73.35 B) ±102 C) About 242.6 D) Can't be determined without knowing the confidence level. Answer: C Diff: 2 Keywords: point estimate, mean Section: 8-1 Point and Confidence Interval Estimates for a Population Mean Outcome: 2 108) A study was recently conducted to estimate the mean cholesterol for adult males over the age of 55 years. From a random sample of n = 10 men, the sample mean was found to be 242.6 and the sample standard deviation was 73.33. To find the 95 percent confidence interval estimate for the mean, the correct critical value to use is: A) 1.96 B) 2.2281 C) 2.33 D) 2.2622 Answer: D Diff: 2 Keywords: population mean, confidence interval, t-distribution Section: 8-1 Point and Confidence Interval Estimates for a Population Mean Outcome: 2 109) The following data represent a random sample of bank balances for a population of checking account customers at a large eastern bank. Based on these data, what is the critical value for a 95 percent confidence interval estimate for the true population mean? $2,300 $1,560 $120

$756 $124 $2,760

$325 $356 $998

$1,457 $3,179 $508

$208 $457 $210

$2,345 $789 $789

A) 1.96 B) 2.1009 C) 2.1098 D) None of the above Answer: C Diff: 2 Keywords: confidence interval, population mean Section: 8-1 Point and Confidence Interval Estimates for a Population Mean Outcome: 2 8-25 Copyright © 2018 Pearson Education, Inc.

110) The following data represent a random sample of bank balances for a population of checking account customers at a large eastern bank. Based on these data, what is the 95 percent confidence interval estimate for the true population mean? $2,300 $1,560 $120

$756 $124 $2,760

$325 $356 $998

$1,457 $3,179 $508

$208 $457 $210

$2,345 $789 $789

A) Approximately $1,069 ± $484.41 B) About $839.40 to $1,298.60 C) Approximately $1,069 ± 2.1098 D) None of the above Answer: A Diff: 2 Keywords: confidence interval, population mean Section: 8-1 Point and Confidence Interval Estimates for a Population Mean Outcome: 2 111) When σ is unknown, the margin of error is computed by using: A) normal distribution. B) t-distribution. C) the mean of the sample. D) The margin of error is also unknown. Answer: B Diff: 2 Keywords: margin of error, t-distribution Section: 8-1 Point and Confidence Interval Estimates for a Population Mean Outcome: 2 112) The Hilbert Drug Store owner plans to survey a random sample of his customers with the objective of estimating the mean dollars spent on pharmaceutical products during the past three months. He has assumed that the population standard deviation is known to be $15.50. Given this information, what would be the required sample size to estimate the population mean with 95 percent confidence and a margin of error of ±$2.00? A) 231 B) 163 C) 16 D) 15 Answer: A Diff: 2 Keywords: sample size, confidence, margin of error Section: 8-2 Determining the Required Sample Size for Estimating a Population Mean Outcome: 3

113) A major tire manufacturer wishes to estimate the mean tread life in miles for one of its tires. It wishes to develop a confidence interval estimate that would have a maximum sampling error of 500 miles with 90 percent confidence. A pilot sample of n = 50 tires showed a sample standard deviation equal to 4,000 miles. Based on this information, the required sample size is: A) 124. B) 246. C) 174. D) 196. Answer: C Diff: 2 Keywords: sample size, confidence, margin of error Section: 8-2 Determining the Required Sample Size for Estimating a Population Mean Outcome: 3 114) The purpose of a pilot sample is: A) to provide a better idea of what the population mean will be. B) to help clarify how the sampling process will be performed. C) to provide an idea of what the population standard deviation might be. D) to save time and money instead of having to carry out a full sampling plan. Answer: C Diff: 1 Keywords: pilot sample, standard deviation Section: 8-2 Determining the Required Sample Size for Estimating a Population Mean Outcome: 3 115) Past experience indicates that the variance in the time it takes for a "fast lube" operation to actually complete the lube and oil change for customers is 9.00 minutes. The manager wishes to estimate the mean time with 99 percent confidence and a margin of error of ± 0.50 minutes. Given this, what must the sample size be? A) n = 239 B) n = 2149 C) n = 139 D) n = 1245 Answer: A Diff: 2 Keywords: sample size, mean Section: 8-2 Determining the Required Sample Size for Estimating a Population Mean Outcome: 3

116) A traffic engineer plans to estimate the average number of cars that pass through an intersection each day. Based on previous studies the standard deviation is believed to be 52 cars. She wants to estimate the mean to within ±10 cars with 90 percent confidence. The needed sample size for n is: A) n = 104 days. B) n = 74 days. C) n = 10 days. D) n = 9 days. Answer: B Diff: 2 Keywords: sample size, confidence, mean, margin of error Section: 8-2 Determining the Required Sample Size for Estimating a Population Mean Outcome: 3 117) In order to reduce the cost of a study, a marketing manager wants to reduce the sample size for a study of customer response to a recent advertising campaign. What can she do that would lead to a reduction in sample size? A) Allow a higher margin of error B) Reduce the level of confidence C) Somehow reduce the variation in the population D) All of the above Answer: D Diff: 2 Keywords: sample size, confidence level, margin of error, variation Section: 8-2 Determining the Required Sample Size for Estimating a Population Mean Outcome: 3 118) A large Midwestern university is interested in estimating the mean time that students spend at the student recreation center per week. A previous study indicated that the standard deviation in time is about 40 minutes per week. If the officials wish to estimate the mean time within ± 10 minutes with a 90 percent confidence, what should the sample size be? A) 44 B) 62 C) 302 D) Can't be determined without knowing how many students there are at the university. Answer: A Diff: 2 Keywords: sample size, standard deviation, confidence Section: 8-2 Determining the Required Sample Size for Estimating a Population Mean Outcome: 3

119) A hospital emergency room has collected a sample of n = 40 to estimate the mean number of visits per day. It has found the standard deviation is 32. Using a 90 percent confidence level, what is its margin of error? A) Approximately ±1.5 visits B) About ±9.9 visits C) Approximately ±8.3 visits D) About ±1.3 visits Answer: C Diff: 2 Keywords: margin of error, confidence Section: 8-2 Determining the Required Sample Size for Estimating a Population Mean Outcome: 3 120) A study has indicated that the sample size necessary to estimate the average electricity use by residential customers of a large western utility company is 900 customers. Assuming that the margin of error associated with the estimate will be ±30 watts and the confidence level is stated to be 90 percent, what was the value for the population standard deviation? A) 265 watts B) Approximately 547.1 watts C) About 490 watts D) Can't be determined without knowing the size of the population. Answer: B Diff: 3 Keywords: sample size, standard deviation, margin of error Section: 8-2 Determining the Required Sample Size for Estimating a Population Mean Outcome: 3 121) The U.S. Post Office is interested in estimating the mean weight of packages shipped using the overnight service. They plan to sample 300 packages. A pilot sample taken last year showed that the standard deviation in weight was about 0.15 pound. If they are interested in an estimate that has 95 percent confidence, what margin of error can they expect? A) Approximately 0.017 pound B) About 0.0003 pound C) About 1.96 D) Can't be determined without knowing the population mean. Answer: A Diff: 3 Keywords: margin of error, sample size, standard deviation Section: 8-2 Determining the Required Sample Size for Estimating a Population Mean Outcome: 3

122) A cell phone service provider has selected a random sample of 20 of its customers in an effort to estimate the mean number of minutes used per day. The results of the sample included a sample mean of 34.5 minutes and a sample standard deviation equal to 11.5 minutes. Based on this information, and using a 95 percent confidence level: A) the critical value is z = 1.96 B) the critical value is z = 1.645 C) the critical value is t = 2.093 D) The critical value can't be determined without knowing the margin of error. Answer: C Diff: 2 Keywords: t-statistic, confidence interval, confidence level Section: 8-2 Determining the Required Sample Size for Estimating a Population Mean Outcome: 3 123) A toy store wants to estimate the mean number of purchases online daily. The standard deviation is known. If they want to find a 99 percent confidence interval the critical value to use is: A) 1.645 B) 1.98 C) 2.33 D) 2.575 Answer: D Diff: 2 Keywords: confidence level, critical value Section: 8-2 Determining the Required Sample Size for Estimating a Population Mean Outcome: 3 124) Which of the following will result in a larger margin of error in an application involving the estimation of a population mean? A) Increasing the sample size B) Decreasing the confidence level C) Increasing the sample standard deviation D) All of the above Answer: C Diff: 2 Keywords: margin of error, population mean, standard deviation Section: 8-2 Determining the Required Sample Size for Estimating a Population Mean Outcome: 3

125) A random sample of 121 automobiles traveling on an interstate showed an average speed of 65 mph. From past information, it is known that the standard deviation of the population is 22 mph. The 95 percent confidence interval for μ is determined as (61.08, 68.92). If we are to reduce the sample size to 100 (other factors remain unchanged), the 95 percent confidence interval for μ would: A) become wider. B) become narrower. C) be the same. D) be impossible to determine. Answer: A Diff: 2 Keywords: margin of error, population mean, sample size Section: 8-2 Determining the Required Sample Size for Estimating a Population Mean Outcome: 3 126) The administrator at Sacred Heart Hospital is interested in estimating the proportion of patients who are satisfied with the meals at the hospital. A random sample of 250 patients was selected and the patients were surveyed. Of these, 203 indicated that they were satisfied. Based on this, what is the estimate of the standard error of the sampling distribution? A) 0.8120 B) 0.0247 C) 0.0006 D) Can't be determined without knowing σ. Answer: B Diff: 2 Keywords: standard error, proportion, sampling distribution Section: 8-3 Estimating a Population Proportion Outcome: 4 127) The produce manager for a large retail food chain is interested in estimating the percentage of potatoes that arrive on a shipment with bruises. A random sample of 150 potatoes showed 14 with bruises. Based on this information, what is the margin of error for a 95 percent confidence interval estimate? A) 0.0933 B) 0.0466 C) 0.0006 D) Can't be determined without knowing σ. Answer: B Diff: 2 Keywords: margin of error, proportion Section: 8-3 Estimating a Population Proportion Outcome: 4

128) A random sample of 340 people in Chicago showed that 66 listened to WJKT-1450, a radio station in South Chicago Heights. Based on this sample information, what is the point estimate for the proportion of people in Chicago that listen to WJKT-1450? A) 340 B) About 0.194 C) 1450 D) 66 Answer: B Diff: 1 Keywords: point estimate, proportion Section: 8-3 Estimating a Population Proportion Outcome: 4 129) A random sample of 340 people in Chicago showed that 66 listened to WJKT-1450, a radio station in South Chicago Heights. Based on this information, what is the upper limit for the 95 percent confidence interval estimate for the proportion of people in Chicago that listen to WJKT-1450? A) 1.96 B) Approximately 0.0009 C) About 0.2361 D) About 0.2298 Answer: C Diff: 2 Keywords: proportion, confidence interval, point estimate Section: 8-3 Estimating a Population Proportion Outcome: 4 130) The chamber of commerce in a beach resort town wants to estimate the proportion of visitors who are repeat visitors. From previous experience they believe the portion is in the vicinity of 0.5 and they want to estimate the proportion to within ± 0.03 percentage points with 95 percent confidence. The sample size they should use is: A) n = 1068 B) n = 545 C) n = 33 D) n = 95 Answer: A Diff: 2 Keywords: sample size, proportion, confidence Section: 8-3 Estimating a Population Proportion Outcome: 5

131) The chamber of commerce in a beach resort town wants to estimate the proportion of visitors who are repeat visitors. From previous experience they believe the portion is not larger than 20 percent. They want to estimate the proportion to within ± 0.04 percentage points with 95 percent confidence. The sample size they should use is: A) n = 601 B) n = 97 C) n = 10 D) n = 385 Answer: D Diff: 2 Keywords: sample size, proportion, confidence Section: 8-3 Estimating a Population Proportion Outcome: 5 132) A regional hardware chain is interested in estimating the proportion of their customers who own their own homes. There is some evidence to suggest that the proportion might be around 0.70. Given this, what sample size is required if they wish a 90 percent confidence level with a margin of error of ±.025? A) About 355 B) Approximately 910 C) Almost 1,300 D) 100 Answer: B Diff: 2 Keywords: sample size, confidence level, margin of error, proportion Section: 8-3 Estimating a Population Proportion Outcome: 5 133) Suppose that an internal report submitted to the managers at a bank in Boston showed that with 95 percent confidence, the proportion of the bank's customers who also have accounts at one or more other banks is between .45 and .51. Given this information, what sample size was used to arrive at this estimate? A) About 344 B) Approximately 1,066 C) Just under 700 D) Can't be determined without more information. Answer: B Diff: 3 Keywords: sample size, proportion, confidence, margin of error Section: 8-3 Estimating a Population Proportion Outcome: 5

134) A sample of 250 people resulted in a confidence interval estimate for the proportion of people who believe that the federal government's proposed tax increase is justified is between 0.14 and 0.20. Based on this information, what was the confidence level used in this estimation? A) Approximately 1.59 B) 95 percent C) Approximately 79 percent D) Can't be determined without knowing σ. Answer: C Diff: 3 Keywords: confidence level, proportion, sample size Section: 8-3 Estimating a Population Proportion Outcome: 4 135) The chamber of commerce in a beach resort town wants to estimate the proportion of visitors who are repeat visitors. Suppose that they have estimated that they need a sample size of n=16,577 people to achieve a margin of error of ± .01 percentage points with 99 percent confidence, but this is too large a sample size to be practical. How can they reduce the sample size? A) Use a higher level of confidence B) Use a smaller margin or error C) Use a lower level of confidence D) Conduct a census Answer: C Diff: 2 Keywords: confidence level, proportion, sample size Section: 8-3 Estimating a Population Proportion Outcome: 5 136) Assuming the population of interest is approximately normally distributed, construct a 95% confidence interval estimate for the population mean given the following values: A) (16.73, 20.07) B) (13.22, 23.58) C) (15.86, 20.94) D) (14.20, 22.60) Answer: C Diff: 2 Keywords: point estimate, confidence interval, population Section: 8-1 Point and Confidence Interval Estimates for a Population Mean Outcome: 1

137) Construct a 95% confidence interval estimate for the population mean given the following values: A) (295.54, 304.56) B) (297.42, 302.58) C) (296.52, 303.48) D) (293.18, 306.82) Answer: D Diff: 2 Keywords: point estimate, confidence interval, population Section: 8-1 Point and Confidence Interval Estimates for a Population Mean Outcome: 1 138) Construct a 98% confidence interval estimate for the population mean given the following values: A) (113.41, 126.59) B) (117.46, 122.54) C) (113.67, 126.33) D) (113.13, 126.87) Answer: A Diff: 2 Keywords: point estimate, confidence interval, population Section: 8-1 Point and Confidence Interval Estimates for a Population Mean Outcome: 1 139) Determine the 90% confidence interval estimate for the population mean of a normal distribution given n = 100, σ = 121, and = 1,200. A) (1186.31, 1213.69) B) (1180.10, 1219.90) C) (1182.56, 1217.44) D) (1191.12, 1208.88) Answer: B Diff: 2 Keywords: point estimate, confidence interval, population Section: 8-1 Point and Confidence Interval Estimates for a Population Mean Outcome: 1

140) Allante Pizza delivers pizzas throughout its local market area at no charge to the customer. However, customers often tip the driver. The owner is interested in estimating the mean tip income per delivery. To do this, she has selected a simple random sample of 12 deliveries and has recorded the tips that were received by the drivers. These data are: $2.25 $0.00

$2.50 $2.00

$2.25 $1.50

$2.00 $2.00

$2.00 $3.00

$1.50 $1.50

Based on these sample data, what is the best point estimate to use as an estimate of the true mean tip per delivery? A) 1.875 B) 1.811 C) 1.50 D) 1.312 Answer: A Diff: 2 Keywords: point estimate, confidence interval, population Section: 8-1 Point and Confidence Interval Estimates for a Population Mean Outcome: 1 141) Allante Pizza delivers pizzas throughout its local market area at no charge to the customer. However, customers often tip the driver. The owner is interested in estimating the mean tip income per delivery. To do this, she has selected a simple random sample of 12 deliveries and has recorded the tips that were received by the drivers. These data are: $2.25 $0.00

$2.50 $2.00

$2.25 $1.50

$2.00 $2.00

$2.00 $3.00

$1.50 $1.50

Suppose the owner is interested in developing a 90% confidence interval estimate. Given the fact that the population standard deviation is unknown, what distribution will be used to obtain the critical value? A) s-distribution B) t-distribution C) z-distribution D) k-distribution Answer: B Diff: 1 Keywords: point estimate, confidence interval, population Section: 8-1 Point and Confidence Interval Estimates for a Population Mean Outcome: 1

142) Even before the record gas prices during the summer of 2008, an article written by Will Lester of the Associated Press reported on a poll in which 80% of those surveyed say that Americans who currently own a SUV (sport utility vehicle) should switch to a more fuel-efficient vehicle to ease America's dependency on foreign oil. This study was conducted by the Pew Research Center for the People & the Press. As a follow-up to this report, a consumer group conducted a study of SUV owners to estimate the mean mileage for their vehicles. A simple random sample of 91 SUV owners was selected, and the owners were asked to report their highway mileage. The following results were summarized from the sample data: = 18.2 mpg s = 6.3 mpg Based on these sample data, compute and interpret a 90% confidence interval estimate for the mean highway mileage for SUVs. A) (15.4, 21.0) B) (12.4, 24.0) C) (17.6, 18.8) D) (17.1, 19.3) Answer: D Diff: 2 Keywords: point estimate, confidence interval, population Section: 8-1 Point and Confidence Interval Estimates for a Population Mean Outcome: 1 143) According to USA Today, customers are not settling for automobiles straight off the production lines. As an example, those who purchase a $355,000 Rolls-Royce typically add $25,000 in accessories. One of the affordable automobiles to receive additions is BMW's Mini Cooper. A sample of 179 recent Mini purchasers yielded a sample mean of $5,000 above the $20,200 base sticker price. Suppose the cost of accessories purchased for all Mini Coopers has a standard deviation of $1,500. Calculate a 95% confidence interval for the average cost of accessories on Mini Coopers. A) (4850.33, 5149.67) B) (4878.82, 5121.18) C) (4788.86, 5211.14) D) (4780.25, 5219.75) Answer: D Diff: 2 Keywords: point estimate, confidence interval, population Section: 8-1 Point and Confidence Interval Estimates for a Population Mean Outcome: 1

144) According to USA Today, customers are not settling for automobiles straight off the production lines. As an example, those who purchase a $355,000 Rolls-Royce typically add $25,000 in accessories. One of the affordable automobiles to receive additions is BMW's Mini Cooper. A sample of 179 recent Mini purchasers yielded a sample mean of $5,000 above the $20,200 base sticker price. Suppose the cost of accessories purchased for all Mini Coopers has a standard deviation of $1,500. Determine the margin of error in estimating the average cost of accessories on Mini Coopers. A) 219.75 B) 214.41 C) 231.14 D) 291.11 Answer: A Diff: 2 Keywords: point estimate, confidence interval, population Section: 8-1 Point and Confidence Interval Estimates for a Population Mean Outcome: 1 145) Suppose a study of 196 randomly sampled privately insured adults with incomes over 200% of the current poverty level is to be used to measure out-of-pocket medical expenses for prescription drugs for this income class. The sample data are in the file Drug Expenses. Based on the sample data, construct a 95% confidence interval estimate for the mean annual out-of-pocket expenditures on prescription drugs for this income class. Interpret this interval. A) (162.08, 172.96) B) (163.50, 171.54) C) (164.19, 170.85) D) (161.97, 173.07) Answer: B Diff: 2 Keywords: point estimate, confidence interval, population Section: 8-1 Point and Confidence Interval Estimates for a Population Mean Outcome: 1 146) The file Danish Coffee contains a random sample of 144 Danish coffee drinkers and measures the annual coffee consumption in kilograms for each sampled coffee drinker. A marketing research firm wants to use this information to develop an advertising campaign to increase Danish coffee consumption. Based on the sample's results, what is the best point estimate of average annual coffee consumption for Danish coffee drinkers? A) 6.5368 B) 7.4151 C) 6.1411 D) 7.4127 Answer: A Diff: 2 Keywords: point estimate, confidence interval, population Section: 8-1 Point and Confidence Interval Estimates for a Population Mean Outcome: 1

147) The file Danish Coffee contains a random sample of 144 Danish coffee drinkers and measures the annual coffee consumption in kilograms for each sampled coffee drinker. A marketing research firm wants to use this information to develop an advertising campaign to increase Danish coffee consumption. Develop and interpret a 90% confidence interval estimate for the mean annual coffee consumption of Danish coffee drinkers. A) (6.4257, 6.6479) B) (6.1768, 6.8968) C) (6.3881, 6.6855) D) (6.3366, 6.7370) Answer: C Diff: 2 Keywords: point estimate, confidence interval, population Section: 8-1 Point and Confidence Interval Estimates for a Population Mean Outcome: 1 148) What sample size is needed to estimate a population mean within ±50 of the true mean value using a confidence level of 95%, if the true population variance is known to be 122,500? A) 211 B) 155 C) 214 D) 189 Answer: D Diff: 2 Keywords: point estimate, confidence interval, population Section: 8-1 Point and Confidence Interval Estimates for a Population Mean Outcome: 1 149) An advertising company wishes to estimate the mean household income for all single working professionals who own a foreign automobile. If the advertising company wants a 90% confidence interval estimate with a margin of error of ±$2,500, what sample size is needed if the population standard deviation is known to be $27,500? A) 156 B) 328 C) 251 D) 415 Answer: B Diff: 2 Keywords: sample size, population mean, confidence Section: 8-2 Determining the Required Sample Size for Estimating a Population Mean Outcome: 1

150) The manager of a toy store wants to estimate the population mean of daily online customers by using a 90% confidence interval estimate that has a margin of error of ±50.0. If the population standard deviation is thought to be 300, what is the required sample size? A) 139 B) 100 C) 10 D) 98 Answer: D Diff: 2 Keywords: sample size, population mean, confidence Section: 8-2 Determining the Required Sample Size for Estimating a Population Mean Outcome: 1 151) Suppose a study estimated the population mean for a variable of interest using a 99% confidence interval. If the width of the estimated confidence interval (the difference between the upper limit and the lower limit) is 600 and the sample size used in estimating the mean is 1,000, what is the population standard deviation? A) 26711.14 B) 2451.23 C) 3684.21 D) 5125.11 Answer: C Diff: 2 Keywords: sample size, population mean, confidence Section: 8-2 Determining the Required Sample Size for Estimating a Population Mean Outcome: 1 152) A production process that fills 12-ounce cereal boxes is known to have a population standard deviation of 0.009 ounce. If a consumer protection agency would like to estimate the mean fill, in ounces, for 12-ounce cereal boxes with a confidence level of 92% and a margin of error of 0.001, what size sample must be used? A) 249 B) 351 C) 512 D) 211 Answer: A Diff: 2 Keywords: sample size, population mean, confidence Section: 8-2 Determining the Required Sample Size for Estimating a Population Mean Outcome: 1

153) A public policy research group is conducting a study of health care plans and would like to estimate the average dollars contributed annually to health savings accounts by participating employees. A pilot study conducted a few months earlier indicated that the standard deviation of annual contributions to such plans was $1,225. The research group wants the study's findings to be within $100 of the true mean with a confidence level of 90%. What sample size is required? A) 407 B) 361 C) 512 D) 546 Answer: A Diff: 2 Keywords: sample size, population mean, confidence Section: 8-2 Determining the Required Sample Size for Estimating a Population Mean Outcome: 1 154) Compute the 90% confidence interval estimate for the population proportion, p, based on a sample size of 100 when the sample proportion, is equal to 0.40. A) 0.3880, 0.0412) B) (0.3930, 0.4070) C) (0.3194, 0.4806) D) (0.3886, 0.4114) Answer: C Diff: 2 Keywords: sample size, population mean, confidence Section: 8-2 Determining the Required Sample Size for Estimating a Population Mean Outcome: 1 155) A pilot sample of 75 items was taken, and the number of items with the attribute of interest was found to be 15. How many more items must be sampled to construct a 99% confidence interval estimate for p with a 0.025 margin of error? A) 1512 B) 1612 C) 1698 D) 1623 Answer: D Diff: 2 Keywords: sample size, population mean, confidence Section: 8-2 Determining the Required Sample Size for Estimating a Population Mean Outcome: 1

156) A decision maker is interested in estimating a population proportion. A sample of size n = 150 yields 115 successes. Based on these sample data, construct a 90% confidence interval estimate for the true population proportion. A) (0.714, 0.826) B) (0.717, 0.823) C) (0.737, 0.803) D) (0.750, 0.790) Answer: A Diff: 2 Keywords: population proportion, estimate Section: 8-3 Estimating a Population Proportion Outcome: 1 157) At issue is the proportion of people in a particular county who do not have health care insurance coverage. A simple random sample of 240 people was asked if they have insurance coverage, and 66 replied that they did not have coverage. Based on these sample data, determine the 95% confidence interval estimate for the population proportion. A) (0.239, 0.321) B) (0.259, 0.301) C) (0.224, 0.336) D) (0.268, 0.292) Answer: C Diff: 2 Keywords: population proportion, estimate Section: 8-3 Estimating a Population Proportion Outcome: 1 158) As the automobile accident rate increases, insurers are forced to increase their premium rates. Companies such as Allstate have recently been running a campaign they hope will result in fewer accidents by their policyholders. For each six-month period that a customer goes without an accident, Allstate will reduce the customer's premium rate by a certain percentage. Companies like Allstate have reason to be concerned about driving habits, based on a survey conducted by Drive for Life, a safety group sponsored by Volvo of North America, in which 1,100 drivers were surveyed. Among those surveyed, 74% said that careless or aggressive driving was the biggest threat on the road. One-third of the respondents said that cell phone usage by other drivers was the driving behavior that annoyed them the most. Based on these data, assuming that the sample was a simple random sample, construct and interpret a 95% confidence interval estimate for the true proportion in the population of all drivers who are annoyed by cell phone users. A) (0.313, 0.347) B) (0.306, 0.354) C) (0.302, 0.358) D) (0.316, 0.344) Answer: C Diff: 2 Keywords: population proportion, estimate Section: 8-3 Estimating a Population Proportion Outcome: 1 8-42 Copyright © 2018 Pearson Education, Inc.

159) A survey of 499 women for the American Orthopedic Foot and Ankle Society revealed that 38% wear flats to work. Use this sample information to develop a 99% confidence interval for the population proportion of women who wear flats to work. A) (0.324, 0.436) B) (0.302, 0.458) C) (0.368, 0.392) D) 0.363, 0.397) Answer: A Diff: 2 Keywords: population proportion, estimate Section: 8-3 Estimating a Population Proportion Outcome: 1 160) A survey of 499 women for the American Orthopedic Foot and Ankle Society revealed that 38% wear flats to work. Suppose the society also wishes to estimate the proportion of women who wear athletic shoes to work with a margin of error of 0.01 with 95% confidence. Determine the sample size required. A) 7241 B) 9604 C) 10021 D) 9715 Answer: B Diff: 2 Keywords: sample size, population mean, confidence Section: 8-2 Determining the Required Sample Size for Estimating a Population Mean Outcome: 1 161) Most major airlines allow passengers to carry two pieces of luggage (of a certain maximum size) onto the plane. However, their studies show that the more carry-on baggage passengers have, the longer it takes to unload and load passengers. One regional airline is considering changing its policy to allow only one carry-on per passenger. Before doing so, it decided to collect some data. Specifically, a random sample of 1,000 passengers was selected. The passengers were observed, and the number of bags carried on the plane was noted. Out of the 1,000 passengers, 345 had more than one bag. Based on this sample, develop and interpret a 95% confidence interval estimate for the proportion of the traveling population that would have been impacted had the one-bag limit been in effect. A) (0.3155, 0.3745) B) (0.3220, 0.3680) C) (0.3216, 0.3684) D) (0.3336, 0.3564) Answer: A Diff: 2 Keywords: population proportion, estimate Section: 8-3 Estimating a Population Proportion Outcome: 1

162) Most major airlines allow passengers to carry two pieces of luggage (of a certain maximum size) onto the plane. However, their studies show that the more carry-on baggage passengers have, the longer it takes to unload and load passengers. One regional airline is considering changing its policy to allow only one carry-on per passenger. Before doing so, it decided to collect some data. Specifically, a random sample of 1,000 passengers was selected. The passengers were observed, and the number of bags carried on the plane was noted. Out of the 1,000 passengers, 345 had more than one bag. The domestic version of Boeing's 747 has a capacity for 568 passengers. Determine an interval estimate of the number of passengers that you would expect to carry more than one piece of luggage on the plane. Assume the plane is at its passenger capacity. A) (171.651, 216.214) B) (174.412, 217.218) C) (181.514, 208.313) D) (179.20, 212.716) Answer: D Diff: 2 Keywords: population proportion, estimate Section: 8-3 Estimating a Population Proportion Outcome: 1 163) Most major airlines allow passengers to carry two pieces of luggage (of a certain maximum size) onto the plane. However, their studies show that the more carry-on baggage passengers have, the longer it takes to unload and load passengers. One regional airline is considering changing its policy to allow only one carry-on per passenger. Before doing so, it decided to collect some data. Specifically, a random sample of 1,000 passengers was selected. The passengers were observed, and the number of bags carried on the plane was noted. Out of the 1,000 passengers, 345 had more than one bag. Suppose the airline also noted whether the passenger was male or female. Out of the 1,000 passengers observed, 690 were males. Of this group, 280 had more than one bag. Using this data, obtain and interpret a 95% confidence interval estimate for the proportion of male passengers in the population who would have been affected by the one-bag limit. A) (0.2815, 0.5124) B) (0.3361, 0.4712) C) (0.3692, 0.4424) D) (0.3814, 0.4125) Answer: C Diff: 2 Keywords: population proportion, estimate Section: 8-3 Estimating a Population Proportion Outcome: 1

164) Suppose an airline decides to conduct a survey of its customers to determine their opinion of a proposed one-bag limit. The plan calls for a random sample of customers on different flights to be given a short written survey to complete during the flight. One key question on the survey will be: "Do you approve of limiting the number of carry-on bags to a maximum of one bag?" Airline managers expect that only about 15% will say "yes." Based on this assumption, what size sample should the airline take if it wants to develop a 95% confidence interval estimate for the population proportion who will say "yes" with a margin of error of ±0.02? A) 1151 B) 1341 C) 1512 D) 1225 Answer: D Diff: 3 Keywords: population proportion, estimate Section: 8-3 Estimating a Population Proportion Outcome: 1 165) The sampling distribution for a proportion has a formula for that standard error that involves using p. Yet when a confidence interval is calculated for a proportion, the standard error formula uses the sample proportion. Why do they differ? Answer: The correct way to calculate the standard error for a sampling distribution for a proportion is to use p in the formula. When calculating a confidence interval for a proportion, the problem is that we don't know what p is. The whole purpose of taking the sample and finding the confidence interval is to obtain an estimate of the value of p. The problem is that the standard error formula uses p, so that best that we can do is use the best estimate we have available in place of p. This "best estimate" is the sample proportion. Diff: 2 Keywords: point estimate, proportion, standard error Section: 8-3 Estimating a Population Proportion Outcome: 5 166) In a recent audit report, an accounting firm stated that the mean sale per customer for the client was estimated to be between $14.50 and $28.50. Further, this was based on a random sample of 100 customers and was computed using 95 percent confidence. Provide a correct interpretation of this confidence interval estimate. Answer: A correct interpretation is: "Based on a random sample of 100 customers, with 95 percent confidence, we believe that the true mean sale per customer is in the range $14.50 to $28.50." Diff: 2 Keywords: confidence interval, population mean Section: 8-1 Point and Confidence Interval Estimates for a Population Mean Outcome: 2

167) In discussing a confidence interval estimate for a population mean, is it acceptable to provide an interpretation like the following: "There is a 95 percent chance that μ lies in the range 20 to 40"? Answer: No, this interpretation is not correct. As stated, it implies that μ is a random variable that can take on different values when in fact μ is a parameter and has a fixed value. The 95 percent confidence level indicates that of all possible confidence intervals constructed from a given size sample selected from the population, and using a z critical value associated with 95 percent confidence, 95 percent of the intervals would contain the true population mean. The particular interval that we actually have computed (20 to 40) will either contain μ or it won't. Diff: 2 Keywords: confidence interval, population mean, interpretation Section: 8-1 Point and Confidence Interval Estimates for a Population Mean Outcome: 2 168) In estimating a population mean, under what conditions would the t-distribution be used? Answer: The t-distribution is used to obtain the critical value for a confidence interval estimate for the population mean when the value for the population standard deviation is unknown or the sample size is reasonably small. Technically, the t-distribution can be used when the standard deviation is not known, but since the t-distribution and the z-distribution converge for large samples, it generally does not matter in cases where the sample size is large. It should be noted that the t-distribution is based on the assumption that the population is normally distributed. However, the t-distribution is usable as long as the population is "reasonably" symmetric. Diff: 2 Keywords: population mean, t-distribution, standard deviation, sample size Section: 8-2 Determining the Required Sample Size for Estimating a Population Mean Outcome: 3 169) What are the disadvantages of using a small sample to estimate the population mean? Answer: A small sample impacts the estimation of a population mean in two main ways. First, in developing a confidence interval estimate for μ, we need the standard error of the sampling distribution. The standard error is computed as

. In a given situation, a small value of n will result in larger

standard error. Then, when we develop the confidence interval using ± z

, the width of the interval

is greater than would be the case for a larger sample size. Thus, the margin of error is larger, which is undesirable. Further, in most applications, the population standard deviation, σ, is unknown and we must estimate it using s, the sample standard deviation. In these cases, when the sample size is small, we use the t-distribution to get the critical value for the confidence interval formula of the form: ± t

Since the t-distribution is more spread out than the z-distribution, the width of the interval will be wider when the small sample size is used. Thus, the width is expanded in two ways: larger standard error and larger critical value. Diff: 3 Keywords: sample size, population mean, confidence Section: 8-1 Point and Confidence Interval Estimates for a Population Mean Outcome: 2

170) One of the major oil products companies conducted a study recently to estimate the mean gallons of gasoline purchased by customers per visit to a gasoline station. To do this, a random sample of customers was selected with the following data being recorded that show the gallons of gasoline purchased. 8.7 22 5.7

22.4 14.4 15.7

9.5 35.7 8.9

13.3 19 22.5

18.9 24.9 15.9

Based on these sample data, construct and interpret a 95 percent confidence interval estimate for the population mean. Answer: Since the population standard deviation is unknown, and since the size of the sample is fairly small, the format for the confidence interval estimate is: ±t If we assume that the population distribution is fairly symmetric or normally distributed, the tdistribution can be used. The first thing we need to do is compute both and s from the sample data. First for the sample mean, we get: =

= 17.1667

Next, we compute the sample standard deviation using: s=

= 7.77

Next, for 95 percent confidence and degrees of freedom equal to n-1 = 14, the critical value for the tdistribution is 2.1448. Then the confidence interval estimate is: ±t

= 17.667 ± 2.1448

17.1667 ± 4.3029 Thus, the 95 percent confidence interval estimate is 12.8638 - 21.4696. We interpret this as follows: "Based on a random sample of 15 customers selected from the population of interest, with 95 percent confidence we believe that the true mean purchase per visit to the gas station is between 12.8638 gallons and 21.4696 gallons." Diff: 3 Keywords: confidence interval, population mean Section: 8-1 Point and Confidence Interval Estimates for a Population Mean Outcome: 2

171) If a manager is interested in estimating the mean time customers spend shopping in a store on each visit to the store, she may want to develop a confidence interval estimate. Suppose, she has determined the required sample size and feels that she cannot afford one that large. What options are available? Answer: First, the formula for computing the sample size requirement for estimating a population mean is: n= There are several options that the manager might select to lower the sample size. First, she might lower her desired level of confidence. This gives a smaller critical value and she can get by with a smaller sample size if she does not have to be as confident. Second, she can allow a larger margin of error. This means that the width of the interval will be wider (not as precise). Third, she can do some combination of decreasing the confidence level and increasing the margin of error. Lastly, she may be able to reduce the population standard deviation. However, this option is often out of the decision maker's control. Diff: 2 Keywords: sample size, confidence level, margin of error, standard deviation Section: 8-2 Determining the Required Sample Size for Estimating a Population Mean Outcome: 3 172) A financial analyst is interested in estimating the proportion of publicly traded companies on the New York Stock Exchange that have cash balances that are more than 10 percent of the total assets of the company. A random sample of n = 100 companies shows that 13 had cash balances of more than 10 percent of assets. Based on this information, develop and interpret a 90 percent confidence interval estimate for the population proportion. Answer: We are interested in estimating π, the population proportion. The equation /Thompson_sn3t_WordExportsfor the confidence interval estimate for p is ± z proportion is the point estimate and is computed as: p =

. The sample

= 0.13

The critical value for 90 percent confidence is z = 1.645 from the standard normal table. Thus, the interval estimate is: 0.13 ± 1.645

or 0.13 ± 0.0553. This gives an interval estimate of 0.0747 - 0.1853.

We interpret this as follows: "Based on the random sample of size 100 companies, with 90 percent confidence, we believe that the proportion of all NYSE companies that have cash balances of more than 10 percent of their total assets is between 0.0747 and 0.1853." Diff: 2 Keywords: proportion, confidence interval Section: 8-3 Estimating a Population Proportion Outcome: 4

173) Under what circumstances would you wish to select a pilot sample? Answer: A pilot sample is used when we wish to determine a required sample size to estimate a population mean or a population proportion. The pilot sample is a sample that is smaller than the final sample that will be needed to complete the estimation. It is used to obtain information about the variation in the population. In the case where we wish to estimate a population mean, in most cases we don't know what the population standard deviation is. From a pilot sample, we compute s, the sample standard deviation and use that value in the sample size computation. When we wish to estimate a population proportion, the pilot sample is used to get an idea of what p is. From the pilot sample, we compute p, the sample proportion and use that in the sample size formula. It should be remembered that the pilot sample is selected in the same manner as the final sample will be selected. Thus, the pilot sample can be used as part of the overall sample. Diff: 2 Keywords: pilot sample, sample size, mean, proportion Section: 8-2 Determining the Required Sample Size for Estimating a Population Mean Outcome: 3 174) A human resources manager wishes to estimate the proportion of employees in her large company who have supplemental health insurance. What is the largest size sample she should select if she wants 95 percent confidence and a margin of error of ± 0.01? Answer: She is interested in estimating p, the population proportion. The sample size formula is: n=

. Since we are trying to estimate p, we obviously don't know it. A conservatively large

sample size can be developed by substituting 0.50 for p. This gives: n=

= 9,604

This may be too large for the manager to take. She might take a pilot sample to get a better idea of what p is. For instance, if a pilot sample of 100 employees gives a sample proportion of = 0.10, then the required sample size would be: n=

= 3,458

If this is still too large, she might increase the margin of error or decrease the confidence level. Diff: 2 Keywords: sample size, population proportion, confidence, margin of error Section: 8-3 Estimating a Population Proportion Outcome: 5

Business Statistics, 9e (Groebner/Shannon/Fry) Chapter 9 Introduction to Hypothesis Testing 1) Hypothesis testing and confidence interval estimation are essentially two totally different statistical procedures and share little in common with each other. Answer: FALSE Diff: 1 Keywords: hypothesis, confidence interval Section: 9-1 Hypothesis Tests for Means Outcome: 1 2) In hypothesis testing, the null hypothesis is the hypothesis that the researcher believes is true and wishes to prove. Answer: FALSE Diff: 1 Keywords: hypothesis, null Section: 9-1 Hypothesis Tests for Means Outcome: 1 3) When the decision maker has control over the null and alternative hypotheses, the alternative hypotheses should be the "research" hypothesis. Answer: TRUE Diff: 2 Keywords: hypothesis, alternative, research Section: 9-1 Hypothesis Tests for Means Outcome: 1 4) In testing a hypothesis, statements for the null and alternative hypotheses as well as the selection of the level of significance should precede the collection and examination of the data. Answer: TRUE Diff: 1 Keywords: hypothesis, null, alternative, significance Section: 9-1 Hypothesis Tests for Means Outcome: 1 5) The null and alternate hypotheses must be opposites of each other. Answer: TRUE Diff: 1 Keywords: null, hypothesis, alternate Section: 9-1 Hypothesis Tests for Means Outcome: 1

6) A sample is used to obtain a 95 percent confidence interval for the mean of a population. The confidence interval goes from 15 to 19. If the same sample had been used to test the null hypothesis that the mean of the population is equal to 20 versus the alternative hypothesis that the mean of the population differs from 20, the null hypothesis could be rejected at a level of significance of 0.05. Answer: TRUE Diff: 2 Keywords: hypothesis, null, confidence interval, alternative Section: 9-1 Hypothesis Tests for Means Outcome: 1 7) A conclusion to "not reject" the null hypothesis is the same as the decision to "accept the null hypothesis". Answer: FALSE Diff: 2 Keywords: hypothesis, not reject, accept, null Section: 9-1 Hypothesis Tests for Means Outcome: 1 8) A one-tailed hypothesis for a population mean with a significance level equal to .05 will have a critical value equal to z = .45. Answer: FALSE Diff: 1 Keywords: hypothesis, one-tailed, significance, critical value Section: 9-1 Hypothesis Tests for Means Outcome: 1 9) Whenever possible, in establishing the null and alternative hypotheses, the research hypothesis should be made the alternative hypothesis. Answer: TRUE Diff: 1 Keywords: null, alternative, hypothesis, research Section: 9-1 Hypothesis Tests for Means Outcome: 1 10) If a hypothesis test is conducted for a population mean, a null and alternative hypothesis of the form: H0 : μ = 100 HA : μ ≠ 100 will result in a one-tailed hypothesis test since the sample result can fall in only one tail. Answer: FALSE Diff: 2 Keywords: one-tailed, null, alternative, hypothesis Section: 9-1 Hypothesis Tests for Means Outcome: 1

11) If the sample data lead the decision maker to reject the null hypothesis, the alpha level is the maximum probability of committing a Type II error. Answer: FALSE Diff: 1 Keywords: reject, alpha, type II, probability, hypothesis Section: 9-1 Hypothesis Tests for Means Outcome: 2 12) The following is an appropriate statement of the null and alternate hypotheses for a test of a population mean: H0 : μ = 45 HA : μ < 50 Answer: TRUE Diff: 1 Keywords: hypothesis, null, alternate Section: 9-1 Hypothesis Tests for Means Outcome: 1 13) In a one-tailed hypothesis test, the larger the significance level, the greater the critical value will be. Answer: FALSE Diff: 2 Keywords: hypothesis, significance, critical, one-tailed Section: 9-1 Hypothesis Tests for Means Outcome: 4 14) In a two-tailed hypothesis test the area of both tails in the rejection region is equal to α. Answer: TRUE Diff: 1 Keywords: two-tailed, rejection region, significance Section: 9-1 Hypothesis Tests for Means Outcome: 4 15) A local medical center has advertised that the mean wait for services will be less than 15 minutes. Given this claim, the hypothesis test for the population mean should be a one-tailed test with the rejection region in the lower (left-hand) tail of the sampling distribution. Answer: TRUE Diff: 2 Keywords: one-tailed, population mean, hypothesis, test Section: 9-1 Hypothesis Tests for Means Outcome: 4

16) A local medical center has advertised that the mean wait for services will be less than 15 minutes. In an effort to test whether this claim can be substantiated, a random sample of 100 customers was selected and their wait times were recorded. The mean wait time was 17.0 minutes. Based on this sample result, there is sufficient evidence to reject the medical center's claim. Answer: FALSE Diff: 2 Keywords: hypothesis, reject, mean Section: 9-1 Hypothesis Tests for Means Outcome: 4 17) When using the t-distribution in a hypothesis test, the population does not need to be assumed normally distributed. Answer: FALSE Diff: 2 Keywords: hypothesis, t-distribution Section: 9-1 Hypothesis Tests for Means Outcome: 4 18) In a hypothesis test, the p-value measures the probability that the alternative hypothesis is true. Answer: FALSE Diff: 2 Keywords: hypothesis, p-value, probability, alternative Section: 9-1 Hypothesis Tests for Means Outcome: 4 19) The Adams Shoe Company believes that the mean size for men's shoes is now more than 10 inches. To test this, it has selected a random sample of n = 100 men. Assuming that the test is to be conducted using a .05 level of significance, a p-value of .07 would lead the company to conclude that its belief is correct. Answer: FALSE Diff: 2 Keywords: p-value, significance, hypothesis Section: 9-1 Hypothesis Tests for Means Outcome: 4 20) In conducting a hypothesis test where the conclusion is to reject the null hypothesis, then either a correct decision has been made or else a Type I error. Answer: TRUE Diff: 2 Keywords: type I, null, reject Section: 9-1 Hypothesis Tests for Means Outcome: 2

21) A large tire manufacturing company has claimed that its top line tire will average more than 80,000 miles. If a consumer group wished to test this claim, they would formulate the following null and alternative hypotheses: H0 : μ ≥ 80,000 Hα : μ ≠ 80,000 Answer: FALSE Diff: 2 Keywords: hypothesis, null, alternative Section: 9-1 Hypothesis Tests for Means Outcome: 1 22) A large tire manufacturing company has claimed that its top line tire will average more than 80,000 miles. If a consumer group wished to test this claim, the research hypothesis would be: Ha : μ > 80,000 miles. Answer: TRUE Diff: 2 Keywords: hypothesis, research, alternative Section: 9-1 Hypothesis Tests for Means Outcome: 1 23) A report recently published in a major business periodical stated that the average salary for female managers is less than $50,000. If we were interested in testing this, the following null and alternative hypotheses would be established: H0 : μ ≥ 50,000 Hα : μ < 50,000 Answer: TRUE Diff: 2 Keywords: hypothesis, alternative, null, research Section: 9-1 Hypothesis Tests for Means Outcome: 1 24) When someone has been accused of a crime the null hypothesis is: H0 : innocent. In this case, a Type I error would be convicting an innocent person. Answer: TRUE Diff: 2 Keywords: null, hypothesis, type I Section: 9-1 Hypothesis Tests for Means Outcome: 2 25) Of the two types of statistical errors, the one that decision makers have most control over is Type I error. Answer: TRUE Diff: 1 Keywords: type I, type II, error, control Section: 9-1 Hypothesis Tests for Means Outcome: 2 9-5 Copyright © 2018 Pearson Education, Inc.

26) If a hypothesis test leads to incorrectly rejecting the null hypothesis, a Type II statistical error has been made. Answer: FALSE Diff: 1 Keywords: hypothesis, type II, null Section: 9-1 Hypothesis Tests for Means Outcome: 2 27) A toy store that has a 12% market share launches a marketing campaign. At the end of the campaign period the company conducts a survey in order to assess whether its market share has increased. If this claim is to be tested, the null and alternative hypotheses are: H0 : p = 12% Ha : p > 12% Answer: TRUE Diff: 2 Keywords: hypothesis, null, alternative Section: 9-1 Hypothesis Tests for Means Outcome: 1 28) When a battery company claims that their batteries last longer than 100 hours and a consumer group wants to test this claim, the hypotheses should be: H0 : μ ≤ 100 HA : μ > 100 Answer: TRUE Diff: 2 Keywords: hypothesis, null, alternate Section: 9-1 Hypothesis Tests for Means Outcome: 1 29) A report recently submitted to the managing partner for a market research company stated "the hypothesis test may have resulted in either a Type I or a Type II error. We won't know which one occurred until later." This statement is one that we might correctly make for any hypothesis that we have conducted. Answer: FALSE Diff: 2 Keywords: hypothesis, type I, type II Section: 9-1 Hypothesis Tests for Means Outcome: 2 30) The significance level in a hypothesis test corresponds to the maximum probability that a Type I error will be committed. Answer: TRUE Diff: 1 Keywords: hypothesis, significance, type I Section: 9-1 Hypothesis Tests for Means Outcome: 2 9-6 Copyright © 2018 Pearson Education, Inc.

31) If the probability of a Type I error is set at 0.05, then the probability of a Type II error will be 0.95. Answer: FALSE Diff: 2 Keywords: hypothesis, type I, type II, probability Section: 9-1 Hypothesis Tests for Means Outcome: 2 32) Type II error is failing to reject the null hypothesis when the null is actually false. Answer: TRUE Diff: 1 Keywords: hypothesis, null, type II Section: 9-1 Hypothesis Tests for Means Outcome: 2 33) The critical value in a null hypothesis test is called alpha. Answer: FALSE Diff: 1 Keywords: critical, null, hypothesis, alpha Section: 9-1 Hypothesis Tests for Means Outcome: 4 34) The loan manager for State Bank and Trust has claimed that the mean loan balance on outstanding loans at the bank is over $14,500. To test this at a significance level of 0.05, a random sample of n = 100 loan accounts is selected. Assuming that the population standard deviation is known to be $3,000, the value of that corresponds to the critical value is approximately $14,993.50. Answer: TRUE Diff: 2 Keywords: hypothesis, critical value Section: 9-1 Hypothesis Tests for Means Outcome: 3 35) The loan manager for State Bank and Trust has claimed that the mean loan balance on outstanding loans at the bank is over $14,500. To test this at a significance level of 0.05, a random sample of n = 100 loan accounts is selected. Assuming that the population standard deviation is known to be $3,000, the null and alternative hypotheses to be tested are: H0 : μ ≤ $14,500 HA : μ > $14,500 Answer: TRUE Diff: 2 Keywords: null, alternative, hypothesis Section: 9-1 Hypothesis Tests for Means Outcome: 4

36) The director of the city Park and Recreation Department claims that the mean distance people travel to the city's greenbelt is more than 5.0 miles. Assuming that the population standard deviation is known to be 1.2 miles and the significance level to be used to test the hypothesis is 0.05 when a sample size of n = 64 people are surveyed, the critical value is approximately 4.75 miles. Answer: FALSE Diff: 2 Keywords: hypothesis test, critical value Section: 9-1 Hypothesis Tests for Means Outcome: 3 37) The director of the city Park and Recreation Department claims that the mean distance people travel to the city's greenbelt is more than 5.0 miles. Assume that the population standard deviation is known to be 1.2 miles and the significance level to be used to test the hypothesis is 0.05 when a sample size of n = 64 people are surveyed. Given this information, if the sample mean is 15.90 miles, the null hypothesis should be rejected. Answer: TRUE Diff: 2 Keywords: hypothesis, critical value Section: 9-1 Hypothesis Tests for Means Outcome: 3 38) A two-tailed hypothesis test with α = 0.05 is similar to a 95 percent confidence interval. Answer: TRUE Diff: 2 Keywords: hypothesis, two-tailed, significance Section: 9-1 Hypothesis Tests for Means Outcome: 4 39) The state insurance commissioner believes that the mean automobile insurance claim filed in her state exceeds $1,700. To test this claim, the agency has selected a random sample of 20 claims and found a sample mean equal to $1,733 and a sample standard deviation equal to $400. They plan to conduct the test using a 0.05 significance level. Given this, the appropriate null and alternative hypotheses are H0 : ≤ $1,700 HA :

> $1,700

Answer: FALSE Diff: 2 Keywords: hypothesis, null, alternative Section: 9-1 Hypothesis Tests for Means Outcome: 1

40) The state insurance commissioner believes that the mean automobile insurance claim filed in her state exceeds $1,700. To test this claim, the agency has selected a random sample of 20 claims and found a sample mean equal to $1,733 and a sample standard deviation equal to $400. They plan to conduct the test using a 0.05 significance level. Based on this, the null hypothesis should be rejected if > $1,854.66 approximately. Answer: TRUE Diff: 2 Keywords: hypothesis test, population mean Section: 9-1 Hypothesis Tests for Means Outcome: 1 41) When using the p-value method, the null hypothesis is rejected when the calculated p-value > α. Answer: FALSE Diff: 1 Keywords: null, hypothesis, p-value Section: 9-1 Hypothesis Tests for Means Outcome: 4 42) Generally, it is possible to appropriately test a null and alternative hypotheses using the test statistic approach and reach a different conclusion than would be reached if the p-value approach were used. Answer: FALSE Diff: 1 Keywords: null, alternative, hypothesis, test statistic, p-value Section: 9-1 Hypothesis Tests for Means Outcome: 4 43) A two-tailed hypothesis test is used when the null hypothesis looks like the following: H0 :

= 100.

Answer: FALSE Diff: 1 Keywords: two-tailed, hypothesis, null Section: 9-1 Hypothesis Tests for Means Outcome: 4 44) When using the p-value method for a two-tailed hypothesis, the p-value is found by finding the area in the tail beyond the test statistic, then doubling it. Answer: TRUE Diff: 2 Keywords: hypothesis, two-tailed, p-value Section: 9-1 Hypothesis Tests for Means Outcome: 4

45) Lube-Tech is a major chain whose primary business is performing lube and oil changes for passenger vehicles. The national operations manager has stated in an industry newsletter that the mean number of miles between oil changes for all passenger cars exceeds 4,200 miles. To test this, an industry group has selected a random sample of 100 vehicles that have come into a lube shop and determined the number of miles since the last oil change and lube. The sample mean was 4,278 and the standard deviation was known to be 780 miles. Based on a significance level of 0.10, the critical value for the test is approximately z = 1.28. Answer: TRUE Diff: 2 Keywords: critical value, hypothesis Section: 9-1 Hypothesis Tests for Means Outcome: 4 46) Lube-Tech is a major chain whose primary business is performing lube and oil changes for passenger vehicles. The national operations manager has stated in an industry newsletter that the mean number of miles between oil changes for all passenger cars exceeds 4,200 miles. To test this, an industry group has selected a random sample of 100 vehicles that have come into a lube shop and determined the number of miles since the last oil change and lube. The sample mean was 4,278 and the sample standard deviation was 780 miles. Based on this information, the test statistic is approximately t = 1.000. Answer: TRUE Diff: 2 Keywords: test statistic, t-value, hypothesis Section: 9-1 Hypothesis Tests for Means Outcome: 4 47) Lube-Tech is a major chain whose primary business is performing lube and oil changes for passenger vehicles. The national operations manager has stated in an industry newsletter that the mean number of miles between oil changes for all passenger cars exceeds 4,200 miles. To test this, an industry group has selected a random sample of 100 vehicles that have come into a lube shop and determined the number of miles since the last oil change and lube. The sample mean was 4,278 and the standard deviation was known to be 780 miles. Based on this information, the p-value for the hypothesis test is less than 0.10. Answer: FALSE Diff: 3 Keywords: p-value, hypothesis, test statistic Section: 9-1 Hypothesis Tests for Means Outcome: 4 48) For testing a research hypothesis, the burden of proof that a new product is no better than the original is placed on the new product, and the research hypothesis is formulated as the null hypothesis. Answer: FALSE Diff: 3 Keywords: hypothesis, null, alternative Section: 9-1 Hypothesis Tests for Means Outcome: 1

49) When deciding the null and alternative hypotheses, the rule of thumb is that if the claim contains the equality (e.g., at least, at most, no different from, etc.), the claim becomes the null hypothesis. If the claim does not contain the equality (e.g., less than, more than, different from), the claim is the alternative hypothesis. Answer: TRUE Diff: 2 Keywords: hypothesis, null, alternative Section: 9-1 Hypothesis Tests for Means Outcome: 1 50) The executive director of the United Way believes that more than 24 percent of the employees in the high-tech industry have made voluntary contributions to the United Way. In order to test this statistically, the appropriate null and alternative hypotheses are: H0 : ≤ .24 HA :

> .24

Answer: FALSE Diff: 1 Keywords: proportion, null, alternative, hypothesis Section: 9-2 Hypothesis Tests for Proportions Outcome: none 51) A company that makes and markets a device that is aimed at helping people quit smoking claims that at least 70 percent of the people who have used the product have quit smoking. To test this, a random sample of n = 100 product users was selected. Of these, 65 people were found to have quit smoking. Given these results, the test statistic value is z = -1.0911. Answer: TRUE Diff: 2 Keywords: population, proportion, z-value Section: 9-2 Hypothesis Tests for Proportions Outcome: none 52) A company that makes and markets a device that is aimed at helping people quit smoking claims that at least 70 percent of the people who have used the product have quit smoking. To test this, a random sample of n = 100 product users was selected. The critical value for the hypothesis test using a significance level of 0.05 would be approximately -1.645. Answer: TRUE Diff: 2 Keywords: population, proportion, z test Section: 9-2 Hypothesis Tests for Proportions Outcome: none

53) A cell phone company believes that 90 percent of its customers are satisfied with their service. They survey n = 30 customers. Based on this, it is acceptable to assume the sample distribution is normally distributed. Answer: FALSE Diff: 2 Keywords: population, proportion, sample size, sampling distribution Section: 9-2 Hypothesis Tests for Proportions Outcome: none 54) Aceco has a contract with a supplier to ship parts that contain no more than three percent defects. When a large shipment of parts comes in, Aceco samples n = 150. Based on the results of the sample, they either accept the shipment or reject it. If Aceco wants no more than a 0.10 chance of rejecting a good shipment, the cut-off between accepting and rejecting should be 0.0478 or 4.78 percent of the sample. Answer: TRUE Diff: 3 Keywords: type I, population, proportion, alpha Section: 9-2 Hypothesis Tests for Proportions Outcome: none 55) When testing a hypothesis involving population proportions, an increase in sample size will result in a smaller chance of making a Type I statistical error. Answer: FALSE Diff: 2 Keywords: type I, hypothesis, proportion Section: 9-2 Hypothesis Tests for Proportions Outcome: none 56) When the hypothesized proportion is close to 0.50, the spread in the sampling distribution of greater than when the hypothesized proportion is close to 0.0 or 1.0. Answer: TRUE Diff: 2 Keywords: proportion, spread, sampling distribution Section: 9-2 Hypothesis Tests for Proportions Outcome: none

57) A major package delivery company claims that at least 95 percent of the packages it delivers reach the destination on time. As part of the evidence in a lawsuit against the package company, a random sample of n = 200 packages was selected. A total of 188 of these packages were delivered on time. Using a significance level of 0.05, the critical value for this hypothesis test is approximately 0.90. Answer: FALSE Diff: 2 Keywords: population proportion, hypothesis, critical value Section: 9-2 Hypothesis Tests for Proportions Outcome: none

58) A major package delivery company claims that at least 95 percent of the packages it delivers reach the destination on time. As part of the evidence in a lawsuit against the package company, a random sample of n = 200 packages was selected. A total of 188 of these packages were delivered on time. Using a significance level of .05, the test statistic for this test is approximately z = -0.65. Answer: TRUE Diff: 2 Keywords: population, proportion, hypothesis, test statistic Section: 9-2 Hypothesis Tests for Proportions Outcome: none 59) A cell phone company believes that 90 percent of their customers are satisfied. They survey a sample of n = 100 customers and find that 82 say they are satisfied. In calculating the standard error of the sampling distribution (σp) the proportion to use is 0.82. Answer: FALSE Diff: 2 Keywords: population, proportion, standard error Section: 9-2 Hypothesis Tests for Proportions Outcome: none 60) One claim states the IRS conducts audits for not more than 5 percent of total tax returns each year. In order to test this claim statistically, the appropriate null and alternative hypotheses are: H0 : μ ≤ 0.05 Ha : μ > 0.05 Answer: FALSE Diff: 2 Keywords: proportion, null, alternative, hypothesis Section: 9-2 Hypothesis Tests for Proportions Outcome: none 61) In a hypothesis test, increasing the sample size will generally result in a smaller chance of making a Type I error since sampling error is likely to be reduced. Answer: TRUE Diff: 1 Keywords: hypothesis, sample, type I, error Section: 9-3 Type II Errors Outcome: 5 62) An article in an operations management journal recently stated that a formal hypothesis test rejected the hypothesis that mean employee productivity was less than $45.70 per hour in the wood processing industry. Given this conclusion, it is possible that a Type I statistical error was committed. Answer: TRUE Diff: 2 Keywords: type I, error, hypothesis Section: 9-3 Type II Errors Outcome: 5

63) The chance of making a Type II statistical error increases if the "true" population mean is closer to the hypothesized population mean, all other factors held constant. Answer: TRUE Diff: 2 Keywords: type II, error, hypothesis Section: 9-3 Type II Errors Outcome: 5 64) Choosing an alpha of 0.01 will cause beta to equal 0.99. Answer: FALSE Diff: 1 Keywords: alpha, beta, proportional Section: 9-3 Type II Errors Outcome: 5 65) If a decision maker is concerned that the chance of making a Type II error is too large, one option that will help reduce the risk is to reduce the significance level. Answer: FALSE Diff: 2 Keywords: type II, error, significance Section: 9-3 Type II Errors Outcome: 5 66) Type II errors are typically greater for two-tailed hypothesis tests than for one-tailed tests. Answer: FALSE Diff: 1 Keywords: type II, two-tailed, hypothesis, one-tailed Section: 9-3 Type II Errors Outcome: 5 67) If a decision maker wishes to reduce the chance of making a Type II error, one option is to increase the sample size. Answer: TRUE Diff: 1 Keywords: type II, sample size Section: 9-3 Type II Errors Outcome: 5 68) To calculate beta requires making a "what if" assumption about the true population parameter, where the "what-if" value is one that would cause the null hypothesis to be false. Answer: TRUE Diff: 2 Keywords: beta, null hypothesis Section: 9-3 Type II Errors Outcome: 5

69) The probability of a Type II error decreases as the "true" population value gets farther from the hypothesized population value, given that everything else is held constant. Answer: TRUE Diff: 2 Keywords: type II, hypothesis, population Section: 9-3 Type II Errors Outcome: 5 70) A city newspaper has stated that the average time required to sell a used car advertised in the paper is less than 5 days. Assuming that the population standard deviation is 2.1 days, if the "true" population mean is 4.1 days and a sample size of n = 49 is used with an alpha equal to 0.05, the probability that the hypothesis test will lead to a Type II error is approximately .0869. Answer: TRUE Diff: 3 Keywords: type II, hypothesis, error Section: 9-3 Type II Errors Outcome: 5 71) The director of the city Park and Recreation Department claims that the mean distance people travel to the city's greenbelt is more than 5.0 miles. Assuming that the population standard deviation is known to be 1.2 miles and the significance level to be used to test the hypothesis is 0.05 when a sample size of n = 64 people are surveyed, the probability of a Type II error is approximately .4545 when the "true" population mean is 5.5 miles. Answer: FALSE Diff: 3 Keywords: type II, hypothesis, beta Section: 9-3 Type II Errors Outcome: 5 72) A major airline has stated in an industry report that its mean onground time between domestic flights is less than 18 minutes. To test this, the company plans to sample 36 randomly selected flights and use a significance level of 0.10. Assuming that the population standard deviation is known to be 4.0 minutes, the probability that the null hypothesis will be "accepted" if the true population mean is 16 minutes is approximately 0.955. Answer: TRUE Diff: 3 Keywords: hypothesis, beta, type II Section: 9-3 Type II Errors Outcome: 5

73) A major airline has stated in an industry report that its mean onground time between domestic flights is less than 18 minutes. To test this, the company plans to sample 36 randomly selected flights and use a significance level of .10. Assuming that the population standard deviation is known to be 4.0 minutes, if the true population mean is 16 minutes, the decision maker could end up making either a Type I or a Type II error depending on the sample result. Answer: FALSE Diff: 2 Keywords: hypothesis, beta, alpha, type I, type II Section: 9-3 Type II Errors Outcome: 5 74) When someone is on trial for suspicion of committing a crime, the hypotheses are: H0 : innocent HA : guilty Which of the following is correct? A) Type I error is acquitting a guilty person. B) Type I error is convicting an innocent person. C) Type II error is acquitting an innocent person. D) Type II error is convicting an innocent person. Answer: B Diff: 2 Keywords: hypothesis test, Type I error, Type II error Section: 9-1 Hypothesis Tests for Means Outcome: 2 75) Which of the following statements is true? A) The decision maker controls the probability of making a Type I statistical error. B) Alpha represents the probability of making a Type II error. C) Alpha and beta are directly related such that when one is increased the other will increase also. D) The alternative hypothesis should contain the equality. Answer: A Diff: 2 Keywords: type I, error, probability Section: 9-1 Hypothesis Tests for Means Outcome: 2

76) In a hypothesis test involving a population mean, which of the following would be an acceptable formulation? A) H0 : ≤ $1,700 Ha :

> $1,700

B) H0 : Ha :

> $1,700

≥ $1,700

C) H0 : μ ≤ $1,700 Ha : μ > $1,700 D) None of the above is a correct formulation. Answer: C Diff: 2 Keywords: hypothesis test, mean, null, alternative Section: 9-1 Hypothesis Tests for Means Outcome: 1 77) Which of the following would be an appropriate null hypothesis? A) The mean of a population is equal to 55. B) The mean of a sample is equal to 55. C) The mean of a population is greater than 55. D) The mean of a sample is greater than 55. Answer: A Diff: 1 Keywords: null, hypothesis, mean Section: 9-1 Hypothesis Tests for Means Outcome: 1 78) If we are performing a two-tailed test of whether μ = 100, the probability of detecting a shift of the mean to 105 will be ________ the probability of detecting a shift of the mean to 110. A) less than B) greater than C) equal to D) not comparable to Answer: A Diff: 2 Keywords: probability, two-tailed, hypothesis test Section: 9-1 Hypothesis Tests for Means Outcome: 4

79) If an economist wishes to determine whether there is evidence that average family income in a community near Seattle exceeds $125,000. An appropriate alternative hypothesis is: A) μ = 125,000. B) μ > 125,000. C) μ ≤ 125,000. D) μ ≥ 125,000. Answer: B Diff: 2 Keywords: hypothesis test, one-tailed, null Section: 9-1 Hypothesis Tests for Means Outcome: 4 80) If the p value is less than α in a two-tailed test, A) the null hypothesis should not be rejected. B) the null hypothesis should be rejected. C) a one-tailed test should be used. D) More information is needed to reach a conclusion about the null hypothesis. Answer: B Diff: 2 Keywords: p-value, alpha, hypothesis test Section: 9-1 Hypothesis Tests for Means Outcome: 4 81) A hypothesis test is to be conducted using an alpha = .01 level. This means: A) there is a 1 percent chance that the null hypothesis is true. B) there is a 1 percent chance that the alternative hypothesis is true. C) there is a maximum 1 percent chance that a true null hypothesis will be rejected. D) there is a 99 percent chance that a Type II error has been committed. Answer: C Diff: 2 Keywords: hypothesis test, null, alpha Section: 9-1 Hypothesis Tests for Means Outcome: 4 82) In a two-tailed hypothesis test for a population mean, an increase in the sample size will: A) have no effect on whether the null hypothesis is true or false. B) have no effect on the significance level for the test. C) result in a sampling distribution that has less variability. D) All of the above are true. Answer: D Diff: 2 Keywords: two-tailed, hypothesis test, mean, sample size, significance Section: 9-1 Hypothesis Tests for Means Outcome: 4

83) The reason for using the t-distribution in a hypothesis test about the population mean is: A) the population standard deviation is unknown. B) it results in a lower probability of a Type I error occurring. C) it provides a smaller critical value than the standard normal distribution for a given sample size. D) the population is not normally distributed. Answer: A Diff: 2 Keywords: t-distribution, hypothesis test, mean, standard deviation, sample size Section: 9-1 Hypothesis Tests for Means Outcome: 4 84) A company that makes shampoo wants to test whether the average amount of shampoo per bottle is 16 ounces. The standard deviation is known to be 0.20 ounces. Assuming that the hypothesis test is to be performed using 0.10 level of significance and a random sample of n = 64 bottles, which of the following would be the upper tail critical value? A) 1.28 B) 1.645 C) 1.96 D) 2.575 Answer: B Diff: 2 Keywords: critical value, upper tail, significance level Section: 9-1 Hypothesis Tests for Means Outcome: 3 85) A company that makes shampoo wants to test whether the average amount of shampoo per bottle is 16 ounces. The standard deviation is known to be 0.20 ounces. Assuming that the hypothesis test is to be performed using 0.10 level of significance and a random sample of n = 64 bottles, which of the following would be the correct formulation of the null and alternative hypotheses? A) H0 : = 16 HA :

= 16

B) H0 : μ = 16 HA : μ ≠ 16 C) H0 : μ ≥ 16 HA : μ < 16 D) H0 : HA :

≥ 16

< 16

Answer: B Diff: 1 Keywords: hypothesis test, two-tailed, null, alternative Section: 9-1 Hypothesis Tests for Means Outcome: 4

86) A company that makes shampoo wants to test whether the average amount of shampoo per bottle is 16 ounces. The standard deviation is known to be 0.20 ounces. Assuming that the hypothesis test is to be performed using 0.10 level of significance and a random sample of n = 64 bottles, how large could the sample mean be before they would reject the null hypothesis? A) 16.2 ounces B) 16.049 ounces C) 15.8 ounces D) 16.041 ounces Answer: D Diff: 3 Keywords: mean, hypothesis test, critical value Section: 9-1 Hypothesis Tests for Means Outcome: 4 87) The cost of a college education has increased at a much faster rate than costs in general over the past twenty years. In order to compensate for this, many students work part- or full-time in addition to attending classes. At one university, it is believed that the average hours students work per week exceeds 20. To test this at a significance level of 0.05, a random sample of n = 20 students was selected and the following values were observed: 26 10 40 20 40

15 20 0 32 36

10 30 5 16 10

40 36 10 12 0

Based on these sample data, which of the following statements is true? A) The standard error of the sampling distribution is approximately 3.04. B) The test statistic is approximately t = 0.13. C) The research hypothesis that the mean hours worked exceeds 20 is not supported by these sample data. D) All of the above are true. Answer: D Diff: 3 Keywords: test statistic, hypothesis, mean, standard error Section: 9-1 Hypothesis Tests for Means Outcome: 4

88) The cost of a college education has increased at a much faster rate than costs in general over the past twenty years. In order to compensate for this, many students work part- or full-time in addition to attending classes. At one university, it is believed that the average hours students work per week exceeds 20. To test this at a significance level of 0.05, a random sample of n = 20 students was selected and the following values were observed: 26 10 40 20 40

15 20 0 32 36

10 30 5 16 10

40 36 10 12 0

Based on these sample data, the critical value expressed in hours: A) is approximately equal to 25.26 hours. B) is approximately equal to 25.0 hours. C) cannot be determined without knowing the population standard deviation. D) is approximately 22 hours. Answer: A Diff: 3 Keywords: hypothesis, mean, critical value Section: 9-1 Hypothesis Tests for Means Outcome: 3 89) The R.D. Wilson Company makes a soft drink dispensing machine that allows customers to get soft drinks from the machine in a cup with ice. When the machine is running properly, the average number of fluid ounces in the cup should be 14. Periodically the machines need to be tested to make sure that they have not gone out of adjustment. To do this, six cups are filled by the machine and a technician carefully measures the volume in each cup. In one such test, the following data were observed: 14.25 14.13

13.7 13.99

14.02 14.04

Which of the following would be the correct null hypothesis if the company wishes to test the machine? A) H0 : = 14 ounces B) H0 : μ = 14 ounces C) H0 : μ ≠ 14 ounces D) H0 :

≠ 14 ounces

Answer: B Diff: 2 Keywords: hypothesis test, mean, null Section: 9-1 Hypothesis Tests for Means Outcome: 4

90) The R.D. Wilson Company makes a soft drink dispensing machine that allows customers to get soft drinks from the machine in a cup with ice. When the machine is running properly, the average number of fluid ounces in the cup should be 14. Periodically the machines need to be tested to make sure that they have not gone out of adjustment. To do this, six cups are filled by the machine and a technician carefully measures the volume in each cup. In one such test, the following data were observed: 14.25 14.13

13.7 13.99

14.02 14.04

Based on these sample data, which of the following is true if the significance level is .05? A) No conclusion can be reached about the status of the machine based on a sample size of only six cups. B) The null hypothesis cannot be rejected since the test statistic is approximately t = 0.20, which is not in the rejection region. C) The null hypothesis can be rejected since the sample mean is greater than 14. D) The null can be rejected because the majority of the sample values exceed 14. Answer: B Diff: 3 Keywords: hypothesis test, t-statistic, mean Section: 9-1 Hypothesis Tests for Means Outcome: 4 91) A concern of Major League Baseball is that games last too long. Some executives in the league's headquarters believe that the mean length of games this past year exceeded 3 hours (180 minutes). To test this, the league selected a random sample of 80 games and found the following results: = 193 minutes and s = 16 minutes. Based on these results, if the null hypothesis is tested using an alpha level equal to 0.10, which of the following is true? A) The null hypothesis should be rejected if > 182.31. B) The test statistic is t = 1.2924. C) Based on the sample data, the null hypothesis cannot be rejected. D) It is possible that when the hypothesis test is completed, a Type II statistical error has been made. Answer: A Diff: 2 Keywords: mean, hypothesis test, t-statistic Section: 9-1 Hypothesis Tests for Means Outcome: 4

92) When testing a two-tailed hypothesis using a significance level of 0.05, a sample size of n = 16, and with the population standard deviation unknown, which of the following is true? A) The null hypothesis can be rejected if the sample mean gets too large or too small compared with the hypothesized mean. B) The alpha probability must be split in half and a rejection region must be formed on both sides of the sampling distribution. C) The test statistic will be a t-value. D) All of the above are true. Answer: D Diff: 2 Keywords: two-tailed, hypothesis, t-statistic, null, alpha Section: 9-1 Hypothesis Tests for Means Outcome: 4 93) A major airline is concerned that the waiting time for customers at its ticket counter may be exceeding its target average of 190 seconds. To test this, the company has selected a random sample of 100 customers and times them from when the customer first arrives at the checkout line until he or she is at the counter being served by the ticket agent. The mean time for this sample was 202 seconds with a standard deviation of 28 seconds. Given this information and the desire to conduct the test using an alpha level of 0.02, which of the following statements is true? A) The chance of a Type II error is 1 - 0.02 = 0.98. B) The test to be conducted will be structured as a two-tailed test. C) The test statistic will be approximately t = 4.286, so the null hypothesis should be rejected. D) The sample data indicate that the difference between the sample mean and the hypothesized population mean should be attributed only to sampling error. Answer: C Diff: 2 Keywords: t-statistic, hypothesis Section: 9-1 Hypothesis Tests for Means Outcome: 4 94) A house cleaning service claims that it can clean a four bedroom house in less than 2 hours. A sample of n = 16 houses is taken and the sample mean is found to be 1.97 hours and the sample standard deviation is found to be 0.1 hours. Using a 0.05 level of significance the correct conclusion is: A) reject the null because the test statistic (-1.2) is < the critical value (1.7531). B) do not reject the null because the test statistic (1.2) is > the critical value (-1.7531). C) reject the null because the test statistic (-1.7531) is < the critical value (-1.2). D) do not reject the null because the test statistic (-1.2) is > the critical value (-1.7531). Answer: D Diff: 3 Keywords: T distribution, hypothesis test, critical value Section: 9-1 Hypothesis Tests for Means Outcome: 4

95) The manager of an online shop wants to determine whether the mean length of calling time of its customers is significantly more than 3 minutes. A random sample of 100 customers was taken. The average length of calling time in the sample was 3.1 minutes with a standard deviation of 0.5 minutes. At a 0.05 level of significance, it can be concluded that the mean of the population is: A) significantly greater than 3. B) not significantly greater than 3. C) significantly less than 3. D) not significantly different from 3.10. Answer: A Diff: 2 Keywords: T distribution, hypothesis test, critical value Section: 9-1 Hypothesis Tests for Means Outcome: 4 96) Woof Chow Dog Food Company believes that it has a market share of 25 percent. It surveys n = 100 dog owners and ask whether or not Woof Chow is their regular brand of dog food. The appropriate null and alternate hypotheses are: A) H0 : ρ = .25 Ha : ρ ≠ .25 B) H0 : p = .25 Ha : p ≠ .25 C) H0 : μ = .25 Ha : μ ≠ .25 D) H0 : p ≤ .25 Ha : p > .25 Answer: B Diff: 2 Keywords: population proportion, null, alternative, hypothesis Section: 9-2 Hypothesis Tests for Proportions Outcome: none 97) Woof Chow Dog Food Company believes that it has a market share of 25 percent. It surveys n = 100 dog owners and ask whether or not Woof Chow is their regular brand of dog food, and 23 people say yes. Based upon this information, what is the critical value if the hypothesis is to be tested at the 0.05 level of significance? A) 1.28 B) 1.645 C) 1.96 D) 2.575 Answer: C Diff: 2 Keywords: population proportion, critical value Section: 9-2 Hypothesis Tests for Proportions Outcome: none

98) Woof Chow Dog Food Company believes that it has a market share of 25 percent. It surveys n = 100 dog owners and ask whether or not Woof Chow is their regular brand of dog food, and 23 people say yes. Based upon this information, what is the value of the test statistic? A) -0.462 B) -0.475 C) 0.462 D) 0.475 Answer: A Diff: 2 Keywords: population proportion, test statistic Section: 9-2 Hypothesis Tests for Proportions Outcome: none 99) After completing sales training for a large company, it is expected that the salesperson will generate a sale on at least 15 percent of the calls he or she makes. To make sure that the sales training process is working, a random sample of n = 400 sales calls made by sales representatives who have completed the training have been selected and the null hypothesis is to be tested at 0.05 alpha level. Suppose that a sale is made on 36 of the calls. Based on this information, what is the test statistic for this test? A) Approximately 0.1417 B) About z = -3.35 C) z = -1.645 D) t = -4.567 Answer: B Diff: 2 Keywords: population proportion, test statistic, p bar Section: 9-2 Hypothesis Tests for Proportions Outcome: none 100) After completing sales training for a large company, it is expected that the salesperson will generate a sale on at least 15 percent of the calls he or she makes. To make sure that the sales training process is working, a random sample of n = 400 sales calls made by sales representatives who have completed the training have been selected and the null hypothesis is to be tested at 0.05 alpha level. Suppose that a sale is made on 36 of the calls. Based on these sample data, which of the following is true? A) The null hypothesis should be rejected since the test statistic falls in the lower tail rejection region. B) The null hypothesis is supported since the sample results do not fall in the rejection region. C) There is insufficient evidence to reject the null hypothesis and the sample proportion is different from the hypothesized proportion due to sampling error. D) It is possible that a Type II statistical error has been committed. Answer: A Diff: 2 Keywords: population proportion, hypothesis, lower tail Section: 9-2 Hypothesis Tests for Proportions Outcome: none

101) Mike runs for the president of the student government and is interested to know whether the proportion of the student body in favor of him is significantly more than 50 percent. A random sample of 100 students was taken. Fifty-five of them favored Mike. At a 0.05 level of significance, it can be concluded that the proportion of the students in favor of Mike A) is significantly greater than 50 percent because 55 percent of the sample favored him. B) is not significantly greater than 50 percent. C) is significantly greater than 55 percent. D) is not significantly different from 55 percent. Answer: B Diff: 2 Keywords: population proportion, hypothesis, upper tail Section: 9-2 Hypothesis Tests for Proportions Outcome: none 102) Which of the following is not a required step in finding beta? A) Assuming a true value of the population parameter where the null is false B) Finding the critical value based on the null hypothesis C) Converting the critical value from the standard normal distribution to the units of the data D) Finding the power of the test Answer: D Diff: 2 Keywords: alpha, type II, beta Section: 9-3 Type II Errors Outcome: 5 103) If the Type I error (α) for a given test is to be decreased, then for a fixed sample size n: A) the Type II error (β) will also decrease. B) the Type II error (β) will increase. C) the power of the test will increase. D) a one-tailed test must be utilized. Answer: B Diff: 1 Keywords: type I, alpha, sample size, type II, beta Section: 9-3 Type II Errors Outcome: 5 104) For a given sample size n, if the level of significance (α) is decreased, the power of the test: A) will increase. B) will decrease. C) will remain the same. D) cannot be determined. Answer: B Diff: 2 Keywords: sample size, significance, alpha, power, beta Section: 9-3 Type II Errors Outcome: 5

105) The power of a test is measured by its capability of: A) rejecting a null hypothesis that is true. B) not rejecting a null hypothesis that is true. C) rejecting a null hypothesis that is false. D) not rejecting a null hypothesis that is false. Answer: C Diff: 2 Keywords: power, null, reject Section: 9-3 Type II Errors Outcome: 5 106) Which of the following will be helpful if the decision maker wishes to reduce the chance of making a Type II error? A) Increase the level of significance at which the hypothesis test is conducted. B) Increase the sample size. C) Both A and B will work. D) Neither A nor B will be effective. Answer: C Diff: 2 Keywords: type II, beta, sample size, significance level Section: 9-3 Type II Errors Outcome: 5 107) A consumer group plans to test whether a new passenger car that is advertised to have a mean highway miles per gallon of at least 33 actually meets this level. They plan to test the hypothesis using a significance level of 0.05 and a sample size of n = 100 cars. It is believed that the population standard deviation is 3 mpg. Based upon this information, if the "true" population mean is 32.0 mpg, what is the probability that the test will lead the consumer group to "accept" the claimed mileage for this car? A) About 0.45 B) Approximately 0.0455 C) About 0.9545 D) None of the above Answer: B Diff: 3 Keywords: mean, beta, z-value Section: 9-3 Type II Errors Outcome: 5

108) A consumer group plans to test whether a new passenger car that is advertised to have a mean highway miles per gallon of at least 33 actually meets this level. They plan to test the hypothesis using a significance level of 0.05 and a sample size of n = 100 cars. It is believed that the population standard deviation is 3 mpg. Based upon this information, what is the critical value in terms of miles per gallon that would be needed prior to finding beta? A) 32.5065 B) 33.4935 C) 33.588 D) 32.412 Answer: A Diff: 2 Keywords: mean, critical, z-value Section: 9-3 Type II Errors Outcome: 5 109) Suppose we want to test H0 : μ ≥ 30 versus H1 : μ < 30. Which of the following possible sample results based on a sample of size 36 gives the strongest evidence to reject H0 in favor of H1? A) = 28, s = 6 B) = 27, s = 4 C) = 32, s = 2 D) = 26, s = 9 Answer: B Diff: 3 Keywords: null, alternative, hypothesis Section: 9-3 Type II Errors Outcome: 5 110) A contract calls for the mean diameter of a cylinder to be 1.50 inches. As a quality check, each day a random sample of n = 36 cylinders is selected and the diameters are measured. Assuming that the population standard deviation is thought to be 0.10 inch and that the test will be conducted using an alpha equal to 0.025, what would the probability of a Type II error be? A) Approximately 0.1267 B) About 0.6789 C) 0.975 D) Can't be determined without knowing the "true" population mean. Answer: D Diff: 2 Keywords: type II, beta, hypothesis, mean Section: 9-3 Type II Errors Outcome: 5

111) A company that sells an online course aimed at helping high-school students improve their SAT scores has claimed that SAT scores will improve by more than 90 points on average if students successfully complete the course. To test this, a national school counseling organization plans to select a random sample of n = 100 students who have previously taken the SAT test. These students will take the company's course and then retake the SAT test. Assuming that the population standard deviation for improvement in test scores is thought to be 30 points and the level of significance for the hypothesis test is 0.05, what is the probability that the counseling organization will incorrectly "accept" the null hypothesis when, in fact, the true mean increase is actually 95 points? A) Approximately 0.508 B) About 0.492 C) Approximately 0.008 D) Can't be determined without knowing the sample results. Answer: B Diff: 3 Keywords: type II, beta, mean, difference Section: 9-3 Type II Errors Outcome: 5 112) A company that sells an online course aimed at helping high-school students improve their SAT scores has claimed that SAT scores will improve by more than 90 points on average if students successfully complete the course. To test this, a national school counseling organization plans to select a random sample of n = 100 students who have previously taken the SAT test. These students will take the company's course and then retake the SAT test. Assuming that the population standard deviation for improvement in test scores is thought to be 30 points and the level of significance for the hypothesis test is 0.05, find the critical value in terms of improvement in SAT points, which would be needed prior to finding a beta. A) Reject the null if SAT improvement is > 95 points. B) Reject the null if SAT improvement is < 85.065 points. C) Reject the null if SAT improvement is > 95.88 points. D) Reject the null if SAT improvement is > 94.935 points. Answer: D Diff: 2 Keywords: type II, beta, mean Section: 9-3 Type II Errors Outcome: 5

113) A recent report in which a major pharmaceutical company released the results of testing that had been done on the cholesterol reduction that people could expect if they use the company's new drug indicated that the Type II error probability for a given "true" mean was 0.1250 based on the sample size of n = 64 subjects. Given this, what was the power of the test under these same conditions? The alpha level used in the test was 0.05. A) 0.95 B) 0.875 C) Essentially zero D) Power would be undefined in this case since the hypothesis would be rejected. Answer: B Diff: 2 Keywords: beta, type II, power Section: 9-3 Type II Errors Outcome: 5 114) If the hypothesis test you are conducting is a two-tailed test, which of the following is a possible step that you could take to increase the power of the test? A) Reduce the sample size B) Increase alpha C) Increase beta D) Use the t-distribution Answer: B Diff: 2 Keywords: hypothesis, two-tailed, power, alpha Section: 9-3 Type II Errors Outcome: 5 115) A consumer group plans to test whether a new passenger car that is advertised to have a mean highway miles per gallon of at least 33 actually meets this level. They plan to test the hypothesis using a significance level of 0.05 and a sample size of n = 100 cars. It is believed that the population standard deviation is 3 mpg. Based upon this information, if the "true" population mean is 32.0 mpg, what is the probability that the test will lead the consumer group to reject the claimed mileage for this car? A) About 0.075 B) Approximately 0.95 C) 0.05 D) None of the above Answer: B Diff: 3 Keywords: hypothesis, mean, p-value Section: 9-3 Type II Errors Outcome: 5

116) For the following z-test statistic, compute the p-value assuming that the hypothesis test is a onetailed test: z = 1.34. A) 0.0606 B) 0.0815 C) 0.0124 D) 0.0901 Answer: D Diff: 1 Keywords: hypothesis test, mean Section: 9-1 Hypothesis Tests for Means Outcome: 1 117) For the following z-test statistic, compute the p-value assuming that the hypothesis test is a onetailed test: z = 2.09. A) 0.0172 B) 0.0183 C) 0.0415 D) 0.0611 Answer: B Diff: 1 Keywords: hypothesis test, mean Section: 9-1 Hypothesis Tests for Means Outcome: 1 118) For the following z-test statistic, compute the p-value assuming that the hypothesis test is a onetailed test: z = -1.55. A) 0.0606 B) 0.1512 C) 0.0901 D) 0.0172 Answer: A Diff: 1 Keywords: hypothesis test, mean Section: 9-1 Hypothesis Tests for Means Outcome: 1

119) For the following hypothesis test:

With n = 80, σ = 9, and = 47.1, state the decision rule in terms of the critical value of the test statistic. A) Reject the null hypothesis if the calculated value of the test statistic, z, is greater than the critical value of the test statistic, 1.645. Otherwise, do not reject. B) Reject the null hypothesis if the calculated value of the test statistic, z, is greater than the critical value of the test statistic, 2.05. Otherwise, do not reject. C) Accept the null hypothesis if the calculated value of the test statistic, z, is greater than the critical value of the test statistic, 1.645. Otherwise, do not accept. D) Accept the null hypothesis if the calculated value of the test statistic, z, is greater than the critical value of the test statistic, 2.05. Otherwise, do not accept. Answer: B Diff: 2 Keywords: hypothesis test, mean Section: 9-1 Hypothesis Tests for Means Outcome: 1 120) For the following hypothesis test:

With n = 80, σ = 9, and = 47.1, state the calculated value of the test statistic z. A) 3.151 B) -2.141 C) 2.087 D) -3.121 Answer: C Diff: 2 Keywords: hypothesis test, mean Section: 9-1 Hypothesis Tests for Means Outcome: 1 121) For the following hypothesis test:

With n = 80, σ = 9, and = 47.1, state the appropriate p-value. A) 0.0183 B) 0.0314 C) 0.0512 D) 0.0218 Answer: A Diff: 1 Keywords: hypothesis test, mean Section: 9-1 Hypothesis Tests for Means Outcome: 1 9-32 Copyright © 2018 Pearson Education, Inc.

122) For the following hypothesis test:

With n = 80, σ = 9, and = 47.1, state the conclusion. A) Because the computed value of z = 3.012 is greater than 2.05, accept the null hypothesis and conclude the mean is greater than 45. Also because the p-value is less than 0.02 B) Because the computed value of z = 3.012 is greater than 2.05, reject the null hypothesis and conclude the mean is greater than 45. Also because the p-value is less than 0.02 C) Because the computed value of z = 2.087 is greater than 2.05, accept the null hypothesis and conclude the mean is greater than 45. Also because the p-value is less than 0.02 D) Because the computed value of z = 2.087 is greater than 2.05, reject the null hypothesis and conclude the mean is greater than 45. Also because the p-value is less than 0.02 Answer: D Diff: 2 Keywords: hypothesis test, mean Section: 9-1 Hypothesis Tests for Means Outcome: 1 123) For the following hypothesis test:

With n = 15, s = 7.5, and = 62.2, state the decision rule in terms of the critical value of the test statistic A) This is a two-tailed test of the population mean with σ unknown. Therefore, the decision rule is: reject the null hypothesis if the calculated value of the test statistic, t, is less than -2.1448 or greater than 2.1448. Otherwise, do not reject. B) This is a two-tailed test of the population mean with σ unknown. Therefore, the decision rule is: accept the null hypothesis if the calculated value of the test statistic, t, is less than -2.1448 or greater than 2.1448. Otherwise, do not accept. C) This is a two-tailed test of the population mean with σ unknown. Therefore, the decision rule is: reject the null hypothesis if the calculated value of the test statistic, t, is less than -1.1828 or greater than 1.1828. Otherwise, do not reject. D) This is a two-tailed test of the population mean with σ unknown. Therefore, the decision rule is: reject the null hypothesis if the calculated value of the test statistic, t, is less than -1.1828 or greater than 1.1828. Otherwise, do not accept. Answer: A Diff: 2 Keywords: hypothesis test, mean Section: 9-1 Hypothesis Tests for Means Outcome: 1

124) For the following hypothesis test:

With n = 15, s = 7.5, and = 62.2, state the calculated value of the test statistic t. A) 1.014 B) 0.012 C) 0.878 D) 1.312 Answer: C Diff: 2 Keywords: hypothesis test, mean Section: 9-1 Hypothesis Tests for Means Outcome: 1 125) For the following hypothesis test:

With n = 15, s = 7.5, and = 62.2, state the conclusion. A) Because the computed value of t = 0.878 is not less than -2.1448 and not greater than 2.1448, do not reject the null hypothesis. B) Because the computed value of t = 1.312 is not less than -2.1448 and not greater than 2.1448, do not reject the null hypothesis. C) Because the computed value of t = 0.878 is not less than -2.1448 and not greater than 2.1448, reject the null hypothesis D) Because the computed value of t = 1.312 is not less than -2.1448 and not greater than 2.1448, reject the null hypothesis Answer: A Diff: 2 Keywords: hypothesis test, mean Section: 9-1 Hypothesis Tests for Means Outcome: 1

126) For the following hypothesis:

With n = 20, = 71.2, s = 6.9, and α = 0.1, state the decision rule in terms of the critical value of the test statistic. A) This is a one-tailed test of the population mean with σ unknown. Therefore, the decision rule is: accept the null hypothesis if the calculated value of the test statistic, t, is greater than 1.3277. Otherwise, reject. B) This is a one-tailed test of the population mean with σ unknown. Therefore, the decision rule is: accept the null hypothesis if the calculated value of the test statistic, t, is greater than 2.1727. Otherwise, reject. C) This is a one-tailed test of the population mean with σ unknown. Therefore, the decision rule is: reject the null hypothesis if the calculated value of the test statistic, t, is greater than 1.3277. Otherwise, do not reject. D) This is a one-tailed test of the population mean with σ unknown. Therefore, the decision rule is: reject the null hypothesis if the calculated value of the test statistic, t, is greater than 2.1727. Otherwise, do not reject. Answer: C Diff: 2 Keywords: hypothesis test, mean Section: 9-1 Hypothesis Tests for Means Outcome: 1 127) For the following hypothesis:

With n = 20, = 71.2, s = 6.9, and α = 0.1, state the calculated value of the test statistic t. A) 1.58 B) 0.78 C) 1.14 D) 0.41 Answer: B Diff: 2 Keywords: hypothesis test, mean Section: 9-1 Hypothesis Tests for Means Outcome: 1

128) For the following hypothesis:

With n = 20, = 71.2, s = 6.9, and α = 0.1, state the conclusion. A) Because the computed value of t = 0.78 is not greater than 2.1727, reject the null hypothesis. B) Because the computed value of t = 0.78 is not greater than 2.1727, do not reject the null hypothesis. C) Because the computed value of t = 0.78 is not greater than 1.3277, reject the null hypothesis. D) Because the computed value of t = 0.78 is not greater than 1.3277, do not reject the null hypothesis. Answer: D Diff: 2 Keywords: hypothesis test, mean Section: 9-1 Hypothesis Tests for Means Outcome: 1 129) The National Club Association does periodic studies on issues important to its membership. The 2012 Executive Summary of the Club Managers Association of America reported that the average country club initiation fee was $31,912. Suppose a random sample taken in 2009 of 12 country clubs produced the following initiation fees: $29,121 $26,205

$31,472 $33,299

$28,054 $25,602

$31,005 $33,726

$36,295 $39,731

$32,771 $27,816

Based on the sample information, can you conclude at the α = 0.05 level of significance that the average 2009 country club initiation fees are lower than the 2008 average? Conduct your test at the level of significance. A) Because t = -0.4324 is not less than t critical = -1.4512, do not reject Ho. The 2009 average country club initiation fee is not less than the 2008 average. B) Because t = -0.4324 is not less than t critical = -1.4512, reject Ho. The 2009 average country club initiation fee is less than the 2008 average. C) Because t = -0.5394 is not less than t critical = -1.7959, do not reject Ho. The 2009 average country club initiation fee is not less than the 2008 average. D) Because t = -0.5394 is not less than t critical = -1.7959, reject Ho. The 2009 average country club initiation fee is less than the 2008 average. Answer: C Diff: 2 Keywords: hypothesis test, mean Section: 9-1 Hypothesis Tests for Means Outcome: 1

130) The director of a state agency believes that the average starting salary for clerical employees in the state is less than $30,000 per year. To test her hypothesis, she has collected a simple random sample of 100 starting clerical salaries from across the state and found that the sample mean is $29,750. State the appropriate null and alternative hypotheses. A) H0 : μ ≥ 30,000 HA : μ < 30,000 B) H0 : μ ≥ 29,750 HA : μ < 29,750 C) H0 : μ ≤ 30,000 HA : μ > 30,000 D) H0 : μ ≤ 29,750 HA : μ > 29,750 Answer: A Diff: 2 Keywords: hypothesis test, mean Section: 9-1 Hypothesis Tests for Means Outcome: 1 131) The director of a state agency believes that the average starting salary for clerical employees in the state is less than $30,000 per year. To test her hypothesis, she has collected a simple random sample of 100 starting clerical salaries from across the state and found that the sample mean is $29,750. Assuming the population standard deviation is known to be $2,500 and the significance level for the test is to be 0.05, what is the critical value (stated in dollars)? A) For alpha = .05 and a one tailed, lower tail test, the critical value is z = -1.645. Solving for the critical xbar: -1.645 = (x-bar - 30,000)/250, x-bar = $29,588.75 B) For alpha = .05 and a one tailed, lower tail test, the critical value is z = -1.96. Solving for the critical xbar: -1.96 = (x-bar - 30,000)/250, x-bar = $34,211.14 C) For alpha = .05 and a one tailed, lower tail test, the critical value is z = -1.645. Solving for the critical xbar: -1.645 = (x-bar - 30,000)/250, x-bar = $34,211.14 D) For alpha = .05 and a one tailed, lower tail test, the critical value is z = -1.96. Solving for the critical xbar: -1.96 = (x-bar - 30,000)/250, x-bar = $30,411.25 Answer: A Diff: 2 Keywords: hypothesis test, mean Section: 9-1 Hypothesis Tests for Means Outcome: 1

132) A mail-order business prides itself in its ability to fill customers' orders in six calendar days or less on the average. Periodically, the operations manager selects a random sample of customer orders and determines the number of days required to fill the orders. Based on this sample information, he decides if the desired standard is not being met. He will assume that the average number of days to fill customers' orders is six or less unless the data suggest strongly otherwise. Establish the appropriate null and alternative hypotheses. A) H0 : μ ≥ 6 days Ha : μ < 6 days B) H0 : μ ≤ 6 days Ha : μ > 6 days C) H0 : μ > 6 days Ha : μ ≤ 6 days D) H0 : μ < 6 days Ha : μ ≥ 6 days Answer: B Diff: 2 Keywords: hypothesis test, mean Section: 9-1 Hypothesis Tests for Means Outcome: 1 133) A mail-order business prides itself in its ability to fill customers' orders in six calendar days or less on the average. Periodically, the operations manager selects a random sample of customer orders and determines the number of days required to fill the orders. Based on this sample information, he decides if the desired standard is not being met. He will assume that the average number of days to fill customers' orders is six or less unless the data suggest strongly otherwise. On one occasion where a sample of 40 customers was selected, the average number of days was 6.65, with a sample standard deviation of 1.5 days. Can the operations manager conclude that his mail-order business is achieving its goal? Use a significance level of 0.025 to answer this question. A) Since 2.2216 < 2.4511, reject H0 and conclude that the mail-order business is not achieving its goal B) Since 2.7406 > 2.023, reject H0 and conclude that the mail-order business is not achieving its goal. C) Since 2.4421 > 2.023, reject H0 and conclude that the mail-order business is not achieving its goal. D) Since 2.2346 < 2.5113, reject H0 and conclude that the mail-order business is not achieving its goal. Answer: B Diff: 2 Keywords: hypothesis test, mean Section: 9-1 Hypothesis Tests for Means Outcome: 1

134) A mail-order business prides itself in its ability to fill customers' orders in six calendar days or less on the average. Periodically, the operations manager selects a random sample of customer orders and determines the number of days required to fill the orders. Based on this sample information, he decides if the desired standard is not being met. He will assume that the average number of days to fill customers' orders is six or less unless the data suggest strongly otherwise. On one occasion where a sample of 40 customers was selected, the average number of days was 6.65, with a sample standard deviation of 1.5 days. Can the operations manager conclude that his mail-order business is achieving its goal? Use a significance level of 0.025 to answer this question. Conduct the test using this p-value. A) Since 0.024 > 0.0041, reject the null hypothesis. B) Since 0.046 > 0.0025, reject the null hypothesis. C) Since 0.0046 < 0.025, reject the null hypothesis. D) Since 0.0024 < 0.041, reject the null hypothesis. Answer: C Diff: 2 Keywords: hypothesis test, mean Section: 9-1 Hypothesis Tests for Means Outcome: 1 135) The makers of Mini-Oats Cereal have an automated packaging machine that can be set at any targeted fill level between 12 and 32 ounces. Every box of cereal is not expected to contain exactly the targeted weight, but the average of all boxes filled should. At the end of every shift (eight hours), 16 boxes are selected at random and the mean and standard deviation of the sample are computed. Based on these sample results, the production control manager determines whether the filling machine needs to be readjusted or whether it remains all right to operate. Use α = 0.05. Establish the appropriate null and alternative hypotheses to be tested for boxes that are supposed to have an average of 24 ounces. A) H0 : μ = 32 ounces Ha : μ ≠ 32 ounces B) H0 : μ = 16 ounces Ha : μ ≠ 16 ounces C) H0 : μ = 22 ounces Ha : μ ≠ 22 ounces D) H0 : μ = 24 ounces Ha : μ ≠ 24 ounces Answer: D Diff: 2 Keywords: hypothesis test, mean Section: 9-1 Hypothesis Tests for Means Outcome: 1

136) The makers of Mini-Oats Cereal have an automated packaging machine that can be set at any targeted fill level between 12 and 32 ounces. Every box of cereal is not expected to contain exactly the targeted weight, but the average of all boxes filled should. At the end of every shift (eight hours), 16 boxes are selected at random and the mean and standard deviation of the sample are computed. Based on these sample results, the production control manager determines whether the filling machine needs to be readjusted or whether it remains all right to operate. Use α= 0.05. At the end of a particular shift during which the machine was filling 24-ounce boxes of Mini-Oats, the sample mean of 16 boxes was 24.32 ounces, with a standard deviation of 0.70 ounce. Assist the production control manager in determining if the machine is achieving its targeted average using test statistic and critical value t. A) Since -1.2445 < 1.013 < 1.2445, do not reject H0 and conclude that the filling machine remains all right to operate. B) Since -1.2445 < 1.013 < 1.2445, reject H0 and conclude that the filling machine needs to be moderated. C) Since -2.1315 < 1.83 < 2.1315, do not reject H0 and conclude that the filling machine remains all right to operate. D) Since -2.1315 < 1.83 < 2.1315, reject H0 and conclude that the filling machine needs to be moderated. Answer: C Diff: 2 Keywords: hypothesis test, mean Section: 9-1 Hypothesis Tests for Means Outcome: 1 137) The makers of Mini-Oats Cereal have an automated packaging machine that can be set at any targeted fill level between 12 and 32 ounces. Every box of cereal is not expected to contain exactly the targeted weight, but the average of all boxes filled should. At the end of every shift (eight hours), 16 boxes are selected at random and the mean and standard deviation of the sample are computed. Based on these sample results, the production control manager determines whether the filling machine needs to be readjusted or whether it remains all right to operate. At the end of a particular shift during which the machine was filling 24-ounce boxes of Mini-Oats, the sample mean of 16 boxes was 24.32 ounces, with a standard deviation of 0.70 ounce. Assist the production control manager in determining if the machine is achieving its targeted average using test statistic and critical value t. Conduct the test using a p-value. A) p-value = 0.0872 > 0.025; therefore do not reject H0 B) p-value = 0.0422 > 0.005; therefore do not reject H0 C) p-value = 0.0314 < 0.105; therefore reject H0 D) p-value = 0.0121< 0. 0805; therefore reject H0 Answer: A Diff: 2 Keywords: hypothesis test, mean Section: 9-1 Hypothesis Tests for Means Outcome: 1

138) At a recent meeting, the manager of a national call center for a major Internet bank made the statement that the average past-due amount for customers who have been called previously about their bills is now no larger than $20.00. Other bank managers at the meeting suggested that this statement may be in error and that it might be worthwhile to conduct a test to see if there is statistical support for the call center manager's statement. The file called Bank Call Center contains data for a random sample of 67 customers from the call center population. Assuming that the population standard deviation for past due amounts is known to be $60.00, what should be concluded based on the sample data? Test using α = 0.10. A) Because p-value = 0.4121 > alpha = 0.10, we do not reject the null hypothesis. The sample data do not provide sufficient evidence to reject the call center manager's statement that the mean past due amount is $20.00 or less. B) Because p-value = 0.4121 > alpha = 0.10, we reject the null hypothesis. The sample data provide sufficient evidence to reject the call center manager's statement that the mean past due amount is $20.00 or less. C) Because p-value = 0.2546 > alpha = 0.10, we do not reject the null hypothesis. The sample data do not provide sufficient evidence to reject the call center manager's statement that the mean past due amount is $20.00 or less. D) Because p-value = 0.2546 > alpha = 0.10, we reject the null hypothesis. The sample data provide sufficient evidence to reject the call center manager's statement that the mean past due amount is $20.00 or less. Answer: C Diff: 2 Keywords: hypothesis test, mean Section: 9-1 Hypothesis Tests for Means Outcome: 1

139) The U.S. Bureau of Labor Statistics (www.bls.gov) released its Consumer Expenditures report in October 2008. Among its findings is that average annual household spending on food at home was $3,624. Suppose a random sample of 137 households in Detroit was taken to determine whether the average annual expenditure on food at home was less for consumer units in Detroit than in the nation as a whole. The sample results are in the file Detroit Eats. Based on the sample results, can it be concluded at the α = 0.02 level of significance that average consumer-unit spending for food at home in Detroit is less than the national average? A) Because t = -13.2314 is less than the critical t value of -1.4126, do not reject H0. The annual average consumer unit spending for food at home in Detroit is not less than the 2006 national consumer unit average B) Because t = -13.2314 is less than the critical t value of -1.4126, reject H0. The annual average consumer unit spending for food at home in Detroit is less than the 2006 national consumer unit average C) Because t = -15.7648 is less than the critical t value of -2.0736, do not reject H0. The annual average consumer unit spending for food at home in Detroit is not less than the 2006 national consumer unit average. D) Because t = -15.7648 is less than the critical t value of -2.0736, reject H0. The annual average consumer unit spending for food at home in Detroit is less than the 2006 national consumer unit average. Answer: D Diff: 2 Keywords: hypothesis test, mean Section: 9-1 Hypothesis Tests for Means Outcome: 1

140) The Center on Budget and Policy Priorities (www.cbpp.org) reported that average out-of-pocket medical expenses for prescription drugs for privately insured adults with incomes over 200% of the poverty level was $173 in 2002. Suppose an investigation was conducted in 2012 to determine whether the increased availability of generic drugs, Internet prescription drug purchases, and cost controls have reduced out-of-pocket drug expenses. The investigation randomly sampled 196 privately insured adults with incomes over 200% of the poverty level, and the respondents' 2012 out-of-pocket medical expenses for prescription drugs were recorded. These data are in the file Drug Expenses. Based on the sample data, can it be concluded that 2012 out-of-pocket prescription drug expenses are lower than the 2002 average reported by the Center on Budget and Policy Priorities? Use a level of significance of 0.01 to conduct the hypothesis test. A) Because t = -2.69 is less than -2.3456, do not reject H0 Conclude that 2012 average out-of-pocket prescription drug expenses are not lower than the 2002 average. B) Because t = -2.69 is less than -2.3456, reject H0 Conclude that 2012 average out-of-pocket prescription drug expenses are lower than the 2002 average. C) Because t = -1.69 is less than -0.8712, reject H0 Conclude that 2012 average out-of-pocket prescription drug expenses are lower than the 2002 average. D) Because t = -1.69 is less than -0.8712, do not reject H0 Conclude that 2012 average out-of-pocket prescription drug expenses are not lower than the 2002 average. Answer: B Diff: 2 Keywords: hypothesis test, mean Section: 9-1 Hypothesis Tests for Means Outcome: 1 141) Hono Golf is a manufacturer of golf products in Taiwan and China. One of the golf accessories it produces at its plant in Tainan Hsing, Taiwan, is plastic golf tees. The injector molder produces golf tees that are designed to have an average height of 66 mm. To determine if this specification is met, random samples are taken from the production floor. One sample is contained in the file labeled THeight. Determine if the process is not producing the tees to specification. Use a significance level of 0.01. A) Since t = 2.1953 < 2.8073 do not reject H0. There is not sufficient evidence to conclude that the average height of the plastic tees is different from 66 mm. B) Since t = 2.1953 < 2.8073 reject H0. There is sufficient evidence to conclude that the average height of the plastic tees is different from 66 mm. C) Since t = 1.2814 < 1.9211 do not reject H0. There is not sufficient evidence to conclude that the average height of the plastic tees is different from 66 mm. D) Since t = 1.2814 < 1.9211 reject H0. There is sufficient evidence to conclude that the average height of the plastic tees is different from 66 mm. Answer: A Diff: 2 Keywords: hypothesis test, mean Section: 9-1 Hypothesis Tests for Means Outcome: 1

142) Hono Golf is a manufacturer of golf products in Taiwan and China. One of the golf accessories it produces at its plant in Tainan Hsing, Taiwan, is plastic golf tees. The injector molder produces golf tees that are designed to have an average height of 66 mm. To determine if this specification is met, random samples are taken from the production floor. One sample is contained in the file labeled THeight. If the hypothesis test determines the specification is not being met, the production process will be shut down while causes and remedies are determined. At times this occurs even though the process is functioning to specification. What type of statistical error would this be? A) The null hypothesis, the specification not being met, was not rejected when in fact it was not being met, this is a Type II error. B) The null hypothesis, the specification not being met, was not rejected when in fact it was not being met, this is a Type I error. C) The null hypothesis, the specification is being met, was rejected when in fact it was being met, this is a Type II error. D) The null hypothesis, the specification is being met, was rejected when in fact it was being met, this is a Type I error. Answer: D Diff: 2 Keywords: hypothesis test, mean Section: 9-1 Hypothesis Tests for Means Outcome: 1 143) Given the following null and alternative

Test the hypothesis using α = 0.01 assuming that a sample of n = 200 yielded x = 105 items with the desired attribute. A) Since -2.17 > -2.33, the null hypothesis is not rejected. B) Since -1.86 > -1.02, the null hypothesis is not rejected. C) Since -2.17 > -2.33, the null hypothesis is rejected. D) Since -1.86 > -1.02, the null hypothesis is rejected. Answer: A Diff: 2 Keywords: hypothesis test, proportion Section: 9-2 Hypothesis Tests for Proportions Outcome: 1

144) For the following hypothesis test:

With n= 64 and p= 0.42, state the decision rule in terms of the critical value of the test statistic A) The decision rule is: reject the null hypothesis if the calculated value of the test statistic, z, is greater than 2.013 or less than -2.013. Otherwise, do not reject. B) The decision rule is: reject the null hypothesis if the calculated value of the test statistic, z, is less than 2.013 or greater than -2.013. Otherwise, do not reject. C) The decision rule is: reject the null hypothesis if the calculated value of the test statistic, z, is greater than 2.575 or less than -2.575. Otherwise, do not reject. D) The decision rule is: reject the null hypothesis if the calculated value of the test statistic, z, is less than 2.575 or greater than -2.575. Otherwise, do not reject. Answer: C Diff: 2 Keywords: hypothesis test, proportion Section: 9-2 Hypothesis Tests for Proportions Outcome: 1 145) For the following hypothesis test:

With n = 64 and p = 0.42, state the calculated value of the test statistic A) t = 0.4122 B) t = 1.7291 C) z = 0.3266 D) z = 1.2412 Answer: C Diff: 1 Keywords: hypothesis test, proportion Section: 9-2 Hypothesis Tests for Proportions Outcome: 1

146) For the following hypothesis test:

With n= 0.42 and p = 0.42, state the conclusion A) Because the calculated value of the test statistic, t=0.4122, is neither greater than 2.013 nor less than 2.013, do not reject the null hypothesis and conclude that the population proportion is not different from 0.40. B) Because the calculated value of the test statistic, t=1.7291, is neither greater than 2.013 nor less than 2.013, do not reject the null hypothesis and conclude that the population proportion is not different from 0.40. C) Because the calculated value of the test statistic, z = 1.2412, is neither greater than 2.575 nor less than 2.575, do not reject the null hypothesis and conclude that the population proportion is not different from 0.40. D) Because the calculated value of the test statistic, z = 0.3266, is neither greater than 2.575 nor less than 2.575, do not reject the null hypothesis and conclude that the population proportion is not different from 0.40. Answer: D Diff: 2 Keywords: hypothesis test, proportion Section: 9-2 Hypothesis Tests for Proportions Outcome: 1 147) For the following hypothesis test

With n = 100 and p = 0.66, state the decision rule in terms of the critical value of the test statistic A) The decision rule is: reject the null hypothesis if the calculated value of the test statistic, z, is less than the critical value of the test statistic z = -1.96. Otherwise, do not reject. B) The decision rule is: reject the null hypothesis if the calculated value of the test statistic, z, is less than the critical value of the test statistic z = -1.645. Otherwise, do not reject. C) The decision rule is: reject the null hypothesis if the calculated value of the test statistic, z, is greater than the critical value of the test statistic z = 1.96. Otherwise, do not reject. D) The decision rule is: reject the null hypothesis if the calculated value of the test statistic, z, is greater than the critical value of the test statistic z = 1.645. Otherwise, do not reject. Answer: A Diff: 2 Keywords: hypothesis test, proportion Section: 9-2 Hypothesis Tests for Proportions Outcome: 1

148) For the following hypothesis test

With n = 100 and p = 0.66, state the calculated value of the test statistic. A) 2.7299 B) -2.0785 C) 1.4919 D) -0.3421 Answer: B Diff: 2 Keywords: hypothesis test, proportion Section: 9-2 Hypothesis Tests for Proportions Outcome: 1 149) For the following hypothesis test

With n = 100 and p = 0.66, state the conclusion. A) Because the computed value of z = -2.0785 is less than the critical value of z = -1.96, reject the null hypothesis and conclude that the population proportion is less than 0.75. B) Because the computed value of z = -0.3412 is less than the critical value of z = -1.645, reject the null hypothesis and conclude that the population proportion is less than 0.75. C) Because the computed value of z = 1.4919 is greater than the critical value of z = -1.96, accept the null hypothesis and conclude that the population proportion is greater than 0.75. D) Because the computed value of z = -0.3412 is greater than the critical value of z = -1.645, accept the null hypothesis and conclude that the population proportion is greater than 0.75. Answer: A Diff: 2 Keywords: hypothesis test, proportion Section: 9-2 Hypothesis Tests for Proportions Outcome: 1

150) Suppose a recent random sample of employees nationwide that have a 401(k) retirement plan found that 18% of them had borrowed against it in the last year. A random sample of 100 employees from a local company who have a 401(k) retirement plan found that 14 had borrowed from their plan. Based on the sample results, is it possible to conclude, at the α = 0.025 level of significance, that the local company had a lower proportion of borrowers from its 401(k) retirement plan than the 18% reported nationwide? A) The z-critical value for this lower tailed test is z = -1.96. Because -1.5430 is greater than the z-critical value we do not reject the null hypothesis and conclude that the proportion of employees at the local company who borrowed from their 401(k) retirement plan is not less than the national average. B) The z-critical value for this lower tailed test is z = -1.96. Because -1.0412 is greater than the z-critical value we do not reject the null hypothesis and conclude that the proportion of employees at the local company who borrowed from their 401(k) retirement plan is not less than the national average. C) The z-critical value for this lower tailed test is z = 1.96. Because 1.5430 is less than the z-critical value we do not reject the null hypothesis and conclude that the proportion of employees at the local company who borrowed from their 401(k) retirement plan is not less than the national average. D) The z-critical value for this lower tailed test is z = 1.96. Because 1.0412 is less than the z-critical value we do not reject the null hypothesis and conclude that the proportion of employees at the local company who borrowed from their 401(k) retirement plan is not less than the national average. Answer: B Diff: 2 Keywords: hypothesis test, proportion Section: 9-2 Hypothesis Tests for Proportions Outcome: 1 151) An issue that faces individuals investing for retirement is allocating assets among different investment choices. Suppose a study conducted 10 years ago showed that 65% of investors preferred stocks to real estate as an investment. In a recent random sample of 900 investors, 540 preferred real estate to stocks. Is this new data sufficient to allow you to conclude that the proportion of investors preferring stocks to real estate has declined from 10 years ago? Conduct your analysis at the α = 0.02 level of significance. A) Because z = -1.915 is not less than -2.055, do not reject H0. A higher proportion of investors prefer stocks today than 10 years ago. B) Because z = -1.915 is not less than -2.055, do not reject H0. A lower proportion of investors prefer stocks today than 10 years ago. C) Because z = -3.145 is less than -2.055, reject H0. A lower proportion of investors prefer stocks today than 10 years ago. D) Because z = -3.145 is less than -2.055, reject H0. A higher proportion of investors prefer stocks today than 10 years ago. Answer: C Diff: 2 Keywords: hypothesis test, proportion Section: 9-2 Hypothesis Tests for Proportions Outcome: 1

152) A major issue facing many states is whether to legalize casino gambling. Suppose the governor of one state believes that more than 55% of the state's registered voters would favor some form of legal casino gambling. However, before backing a proposal to allow such gambling, the governor has instructed his aides to conduct a statistical test on the issue. To do this, the aides have hired a consulting firm to survey a simple random sample of 300 voters in the state. Of these 300 voters, 175 actually favored legalized gambling. State the appropriate null and alternative hypotheses. A) H0 : p = 0.58 Ha : p ≠ 0.58 B) H0 : p ≤ 0.55 Ha : p > 0.55 C) H0 : p = 0.55 Ha : p ≠ 0.55 D) H0 : p ≤ 0.58 Ha : p > 0.58 Answer: B Diff: 1 Keywords: hypothesis test, proportion Section: 9-2 Hypothesis Tests for Proportions Outcome: 1 153) A major issue facing many states is whether to legalize casino gambling. Suppose the governor of one state believes that more than 55% of the state's registered voters would favor some form of legal casino gambling. However, before backing a proposal to allow such gambling, the governor has instructed his aides to conduct a statistical test on the issue. To do this, the aides have hired a consulting firm to survey a simple random sample of 300 voters in the state. Of these 300 voters, 175 actually favored legalized gambling. Assuming that a significance level of 0.05 is used, what conclusion should the governor reach based on these sample data? A) Since z = 1.1594 < 1.645, do not reject the null hypothesis. The sample data do not provide sufficient evidence to conclude that more than 55 percent of the population favor legalized gambling. B) Since z = 2.1316 > 1.645, reject the null hypothesis. The sample data provide sufficient evidence to conclude that more than 55 percent of the population favor legalized gambling. C) Since z = 1.1594 < 1.645, do not reject the null hypothesis. The sample data do not provide sufficient evidence to conclude that more than 58 percent of the population favor legalized gambling. D) Since z = 2.1316 > 1.645, reject the null hypothesis. The sample data provide sufficient evidence to conclude that more than 58 percent of the population favor legalized gambling. Answer: A Diff: 2 Keywords: hypothesis test, proportion Section: 9-2 Hypothesis Tests for Proportions Outcome: 1

154) A recent article in The Wall Street Journal entitled "As Identity Theft Moves Online, Crime Rings Mimic Big Business" states that 39% of the consumer scam complaints by American consumers are about identity theft. Suppose a random sample of 90 complaints is obtained. Of these complaints, 40 were regarding identity theft. Based on these sample data, what conclusion should be reached about the statement made in The Wall Street Journal? (Test using α= 0.10.) A) Since z = 1.947 > 1.645, we reject the null hypothesis. There is sufficient evidence to conclude that the 0.39 rate quoted in the WSJ article is wrong. B) Since z = 2.033 > 1.96, we reject the null hypothesis. There is sufficient evidence to conclude that the 0.39 rate quoted in the WSJ article is wrong. C) Since z = 1.341 < 1.645, we do not reject the null hypothesis. There is insufficient evidence to conclude that the 0.39 rate quoted in the WSJ article is wrong. D) Since z = 0.97 < 1.645, we do not reject the null hypothesis. There is insufficient evidence to conclude that the 0.39 rate quoted in the WSJ article is wrong. Answer: D Diff: 2 Keywords: hypothesis test, proportion Section: 9-2 Hypothesis Tests for Proportions Outcome: 1 155) Because of the complex nature of the U.S. income tax system, many people have questions for the Internal Revenue Service (IRS). Yet, an article published by the Detroit Free Press entitled "Assistance: IRS Help Centers Give the Wrong Information" discusses the propensity of IRS staff employees to give incorrect tax information to tax-payers who call with questions. Then IRS Inspector General Pamela Gardiner told a Senate subcommittee that "the IRS employees at 400 taxpayer assistance centers nationwide encountered 8.5 million taxpayers face-to-face last year. The problem: When inspector general auditors posing as taxpayers asked them to answer tax questions, the answers were right 69% of the time." Suppose an independent commission was formed to test whether the 0.69 accuracy rate is correct or whether it is actually higher or lower. The commission has randomly selected n = 180 tax returns that were completed by IRS assistance employees and found that 105 of the returns were accurately completed. State the appropriate null and alternative hypotheses. A) H0 : p = 0.69 Ha : p ≠ 0.69 B) H0 : p = 0.58 Ha : p ≠ 0.58 C) H0 : p > 0.69 Ha : p ≤ 0.69 D) H0 : p > 0.58 Ha : p ≤ 0.58 Answer: A Diff: 1 Keywords: hypothesis test, proportion Section: 9-2 Hypothesis Tests for Proportions Outcome: 1

156) Because of the complex nature of the U.S. income tax system, many people have questions for the Internal Revenue Service (IRS). Yet, an article published by the Detroit Free Press entitled "Assistance: IRS Help Centers Give the Wrong Information" discusses the propensity of IRS staff employees to give incorrect tax information to tax-payers who call with questions. Then IRS Inspector General Pamela Gardiner told a Senate subcommittee that "the IRS employees at 400 taxpayer assistance centers nationwide encountered 8.5 million taxpayers face-to-face last year. The problem: When inspector general auditors posing as taxpayers asked them to answer tax questions, the answers were right 69% of the time." Suppose an independent commission was formed to test whether the 0.69 accuracy rate is correct or whether it is actually higher or lower. The commission has randomly selected n = 180 tax returns that were completed by IRS assistance employees and found that 105 of the returns were accurately completed. Using an α= 0.05 level, based on the sample data, what conclusion should be reached about the IRS rate of correct tax returns? A) The z-critical values from the standard normal table for a two-tailed test with alpha = 0.05 are and z = -1.96. Since z = -0.96 > -1.96, we do not reject the null hypothesis. Thus, based on these sample data, we believe that the accuracy rate is actually higher than the 0.69 rate quoted in the Detroit Free Press article B) The z-critical values from the standard normal table for a two-tailed test with alpha = 0.05 are and z = -1.96. Since z = -0.96 > -1.96, we do not reject the null hypothesis. Thus, based on these sample data, we believe that the accuracy rate is actually higher than the 0.58 rate quoted in the Detroit Free Press article C) The z-critical values from the standard normal table for a two-tailed test with alpha = 0.05 are and z = -1.96. Since z= -3.19 < -1.96, we reject the null hypothesis. Thus, based on these sample data, we believe that the accuracy rate is actually lower than the 0.69 rate quoted in the Detroit Free Press article. D) The z-critical values from the standard normal table for a two-tailed test with alpha = 0.05 are and z = -1.96. Since z = -3.19 < -1.96, we reject the null hypothesis. Thus, based on these sample data, we believe that the accuracy rate is actually lower than the 0.58 rate quoted in the Detroit Free Press article. Answer: C Diff: 2 Keywords: hypothesis test, proportion Section: 9-2 Hypothesis Tests for Proportions Outcome: 1

157) You are given the following null and alternative hypotheses:

If the true population mean is 1.25, determine the value of beta. Assume the population standard deviation is known to be 0.50 and the sample size is 60. A) 0.40 B) 0.04 C) 0.51 D) 0.80 Answer: D Diff: 3 Keywords: type II, error, hypothesis Section: 9-3 Type II Errors Outcome: 1 158) You are given the following null and alternative hypotheses:

If the true population mean is 1.25, calculate the power of the test. Assume the population standard deviation is known to be 0.50 and the sample size is 60. A) 0.49 B) 0.20 C) 0.96 D) 0.60 Answer: B Diff: 3 Keywords: type II, error, hypothesis Section: 9-3 Type II Errors Outcome: 1 159) You are given the following null and alternative hypotheses:

If the true population mean is 4,345, determine the value of beta. Assume the population standard deviation is known to be 200 and the sample size is 100. A) 0.9192 B) 0.8233 C) 0.6124 D) 0.0314 Answer: A Diff: 3 Keywords: type II, error, hypothesis Section: 9-3 Type II Errors Outcome: 1 9-52 Copyright © 2018 Pearson Education, Inc.

160) You are given the following null and alternative hypotheses:

If the true population mean is 4,345, calculate the power of the test. Assume the population standard deviation is known to be 200 and the sample size is 100. A) 0.1766 B) 0.3876 C) 0.0808 D) 0.9686 Answer: C Diff: 3 Keywords: type II, error, hypothesis Section: 9-3 Type II Errors Outcome: 1 161) You are given the following null and alternative hypotheses:

Calculate the probability of committing a Type II error when the population mean is 505, the sample size is 64, and the population standard deviation is known to be 36 A) 0.1562 B) 0.5997 C) 0.3426 D) 0.8888 Answer: D Diff: 3 Keywords: type II, error, hypothesis Section: 9-3 Type II Errors Outcome: 1 162) According to data from the Environmental Protection Agency, the average daily water consumption for a household of four people in the United States is approximately at least 243 gallons. (Source: http://www.catskillcenter.org/programs/csp/H20/Lesson3/house3.htm) Suppose a state agency plans to test this claim using an alpha level equal to 0.05 and a random sample of 100 households with four people. State the appropriate null and alternative hypotheses. A) H0 : μ > 243 Ha : μ ≤ 243 B) H0 : μ < 243 Ha : μ ≥ 243 C) H0 : μ ≤ 243 Ha : μ > 243 D) H0 : μ ≥ 243 Ha : μ < 243 Answer: D Diff: 1 Keywords: hypothesis test, mean Section: 9-1 Hypothesis Tests for Means Outcome: 1 9-53 Copyright © 2018 Pearson Education, Inc.

163) According to data from the Environmental Protection Agency, the average daily water consumption for a household of four people in the United States is approximately at least 243 gallons. (Source: http://www.catskillcenter.org/programs/csp/H20/Lesson3/house3.htm) Suppose a state agency plans to test this claim using an alpha level equal to 0.05 and a random sample of 100 households with four people. Calculate the probability of committing a Type II error if the true population mean is 230 gallons. Assume that the population standard deviation is known to be 40 gallons. A) 0.0331 B) 0.0712 C) 0.0537 D) 0.1412 Answer: C Diff: 3 Keywords: type II error Section: 9-3 Type II Errors Outcome: 1 164) Swift is the holding company for Swift Transportation Co., Inc., a truckload carrier headquartered in Phoenix, Arizona. Swift operates the largest truckload fleet in the United States. Before Swift switched to its current computer-based billing system, the average payment time from customers was approximately 40 days. Suppose before purchasing the present billing system, it performed a test by examining a random sample of 24 invoices to see if the system would reduce the average billing time. The sample indicates that the average payment time is 38.7 days. The company that created the billing system indicates that the system would reduce the average billing time to less than 40 days. Conduct a hypothesis test to determine if the new computer-based billing system would reduce the average billing time to less than 40 days. Assume the standard deviation is known to be 6 days. Use a significance level of 0.025. A) Since z = -0.423 > -1.96, we will not reject H0, there is not sufficient evidence to conclude that the new computer-based billing system would reduce the average billing time to less than 40 days. B) Since z = -1.0614 > -1.96, we will not reject H0, there is not sufficient evidence to conclude that the new computer-based billing system would reduce the average billing time to less than 40 days. C) z = -1.0231 > -1.96, we will reject H0, there is sufficient evidence to conclude that the new computerbased billing system would reduce the average billing time to less than 40 days. D) z = 0.341 > -1.96, we will reject H0, there is sufficient evidence to conclude that the new computerbased billing system would reduce the average billing time to less than 40 days. Answer: B Diff: 2 Keywords: hypothesis test, mean Section: 9-1 Hypothesis Tests for Means Outcome: 1

165) Waiters at Finegold's Restaurant and Lounge earn most of their income from tips. Each waiter is required to "tip-out" a portion of tips to the table bussers and hostesses. The manager has based the "tipout" rate on the assumption that the mean tip is at least 15% of the customer bill. To make sure that this is the correct assumption, he has decided to conduct a test by randomly sampling 60 bills and recording the actual tips. State the appropriate null and alternative hypotheses. A) H0 : μ ≥ 15 Ha : μ < 15 B) H0 : μ ≤ 15 Ha : μ > 15 C) H0 : μ ≥ 9 Ha : μ < 9 D) H0 : μ ≤ 9 Ha : μ > 9 Answer: A Diff: 1 Keywords: hypothesis test, mean Section: 9-1 Hypothesis Tests for Means Outcome: 1 166) Waiters at Finegold's Restaurant and Lounge earn most of their income from tips. Each waiter is required to "tip-out" a portion of tips to the table bussers and hostesses. The manager has based the "tipout" rate on the assumption that the mean tip is at least 15% of the customer bill. To make sure that this is the correct assumption, he has decided to conduct a test by randomly sampling 60 bills and recording the actual tips. Calculate the probability of a Type II error if the true mean is 14%. Assume that the population standard deviation is known to be 2% and that a significance level equal to 0.01 will be used to conduct the hypothesis test. A) 0.0041 B) 0.1251 C) 0.0606 D) 0.4123 Answer: C Diff: 3 Keywords: type II error Section: 9-3 Type II Errors Outcome: 1

167) Nationwide Mutual Insurance, based in Columbus, Ohio, is one of the largest diversified insurance and financial services organizations in the world, with more than $157 billion in assets. Nationwide ranked 108th on the Fortune 100 list in 2008. The company provides a full range of insurance and financial services. In a recent news release Nationwide reported the results of a new survey of 1,097 identity theft victims. The survey shows victims spend an average of 81 hours trying to resolve their cases. If the true average time spent was 81 hours, determine the probability that a test of hypothesis designed to test that the average was less than 85 hours would reject the research hypothesis. Use α= 0.05 and a standard deviation of 50. A) 0.0123 B) 0.5182 C) 0.1241 D) 0.1587 Answer: D Diff: 3 Keywords: type II error Section: 9-3 Type II Errors Outcome: 1 168) According to CNN business partner Careerbuilder.com, the average starting salary for accounting graduates in 2008 was at least $47,413. Suppose that the American Society for Certified Public Accountants planned to test this claim by randomly sampling 200 accountants who graduated in 2008. State the appropriate null and alternative hypotheses. A) H0 : μ ≥ $47,413 HA : μ < $47,413 B) H0 : μ < $47,413 HA : μ ≥ $47,413 C) H0 : μ ≤ $47,413 HA : μ > $47,413 D) H0 : μ > $47,413 HA : μ ≤ $47,413 Answer: A Diff: 1 Keywords: hypothesis test, mean Section: 9-1 Hypothesis Tests for Means Outcome: 1 169) According to CNN business partner Careerbuilder.com, the average starting salary for accounting graduates in 2008 was at least $47,413. Suppose that the American Society for Certified Public Accountants planned to test this claim by randomly sampling 200 accountants who graduated in 2008. Compute the power of the hypothesis test to reject the null hypothesis if the true average starting salary is only $47,000. Assume that the population standard deviation is known to be $4,600 and the test is to be conducted using an alpha level equal to 0.01. A) 0.0872 B) 0.1323 C) 0.8554 D) 0.9812 Answer: C Diff: 3 Keywords: type II error Section: 9-3 Type II Errors Outcome: 1 9-56 Copyright © 2018 Pearson Education, Inc.

170) What is meant by the terms Type I and Type II statistical error? Answer: When a null hypothesis is tested using sample information, it is expected that sampling error will exist. It is possible that the sampling error will lead to an error in the conclusion that is reached with respect to the null hypothesis. For example, if the null hypothesis is true, extreme sampling error will push the test statistic into the rejection region. Rejecting a true null hypothesis is called a Type I error. Also, if the null hypothesis is false, the sample data may be such that we do not reject the null hypothesis. "Accepting" a false null hypothesis is referred to as a Type II error. Diff: 1 Keywords: type I, type II, error Section: 9-3 Type II Errors Outcome: 5 171) The produce manager for a large retail grocery store believes that an average head of lettuce weighs more than 1.7 pounds. If she were to test this statistically, what would the null and alternative hypotheses be and what is the research hypothesis? Answer: In this case, the manager wants to show that the mean weight of a head of lettuce exceeds 1.7 pounds, this will be the research hypothesis. The research hypothesis is the claim that we wish to "prove." Thus, we would set up the following null and alternative hypotheses: H0 : μ ≤ 1.7 H1 : μ >1.7 Note that the null and alternative hypotheses are stated in terms of μ, the population mean, and that the null hypothesis contains the equality. The research hypothesis is the alternative hypothesis. If the sample data lead us to reject the null, we will have evidence to suggest that the mean weight of lettuce exceeds 1.7 pounds. Diff: 2 Keywords: hypothesis, hypotheses, null, alternative Section: 9-1 Hypothesis Tests for Means Outcome: 1

172) One of the factors that a company will use in determining whether it will locate a new facility in a community is the status of the real estate market. The managers believe that an important measure of the real estate market is the average length of time that homes stay on the market before selling. They believe that if the mean time on the market is less than 45 days, the real estate market is favorable. To test this in a particular area, a random sample of n = 100 homes that sold during the past six months was selected. The mean for this sample was 40 days. It is believed that the population standard deviation is 15 days. If the test is conducted using a 0.05 level of significance, what conclusion should be reached? Answer: This situation involves a test of the population mean. The research hypothesis is that the mean time on the market is less than 45 days. Thus, we can form the null and alternative hypothesis as follows: H0 : μ ≥ 45 days Ha : μ < 45 days This test is a one-tailed test with the rejection region in the lower (left-hand) tail of the distribution. The significance level is 0.05 meaning that the area in the rejection region is 0.05. Since we assume that the population standard deviation is known, we can compute the critical value as follows: μ-z 45 - 1.645

= 42.5325 days

Note, z = 1.645 is from the standard normal distribution for a one-tail area of 0.05 (0.45 between the critical value and the hypothesized mean). Then the decision rule is: If < 42.5325 days, reject the null hypothesis. Otherwise, do not reject. Since = 40 days, which is less than 42.5325 days, we reject the null hypothesis. Thus, based on the sample data, at a 0.05 level of significance, we conclude that the mean time that homes stay on the market is less than 45 days. Diff: 2 Keywords: mean, hypothesis test, null, alternative Section: 9-1 Hypothesis Tests for Means Outcome: 4

173) A cell phone manufacturer claims that its phone will last for more than 8 hours of continuous talk time when the battery is fully charged. To test this claim a sample of n = 18 phones were tested. The results showed a sample mean of 8.2 hours and a sample standard deviation of 0.4 hour. Conduct the hypothesis test using a 0.5 level of significance and determine whether or not the company's claim is supported. Answer: Since the claim "more than 8 hours" is a strict inequality, this needs to be the alternate hypothesis so that that "equal to 8 hours" is included in the null: H0 : μ ≤ 8 HA : μ > 8 So this will be a one-tailed test with the rejection region in the upper tail. In addition, the t-distribution must be used since a sample standard deviation is all we have available. There are 18 - 1 = 17 degrees of freedom. So the critical value for α = 0.05 and 17 d.f. is t = 1.7396 and the decision rule is: Reject the null if t > 1.7396 The formula for the test statistic is:

t= Then the test statistic is:

= 2.12

Since the test statistic t = 2.12 is greater than the critical value (1.7396) we should reject the null hypothesis and conclude that the mean is significantly larger than 8 hours, which means the manufacturer's claim is supported. Diff: 2 Keywords: mean difference, hypothesis, t-test Section: 9-1 Hypothesis Tests for Means Outcome: 4

174) If a real estate market is strong there will be a close relationship between the asking price for homes and the selling price. Suppose that one analyst believes that the mean difference between asking price and selling price for homes in a particular market area is less than $2,000. To test this using an alpha level equal to 0.05, a random sample of n = 15 homes that have sold recently was selected. The sample differences between asking price and selling price are in the following table: $2,053 $1,396 $1,038

$1,693 $2,473 $2,755

$1,854 $1,931 $2,084

$1,747 $2,303 $1,664

$869 $1,502 $2,104

Based on these sample data, what is the critical value expressed in dollars? Answer: The analyst claims that the mean difference will be less than $2,000. Given this research hypothesis, we can form the following null and alternative hypotheses to test statistically: Ho : μ ≥ $2,000 Ha : μ < $2,000 The test is a one-tail, lower-tail test with 0.05 area in the rejection region. Since we do not know the population standard deviation and the sample size is relatively small, if we assume that the population is normally distributed, the t-distribution can be used. The critical t-value for degrees of freedom 15-1 = 14 and 0.05 in one tail is -1.7613. We find the critical value as follows: μ–t

= 2000 - 1.7613

We compute the sample standard deviation, s, from the data as follows: s=

= $506.59.

Then the critical value is 2000 - 1.7613

= $1,769.62

Thus, if < $1,769.62 reject the null hypothesis; otherwise do not reject. We find as follows: =

= $1,831.067.

Since $1,831.067 ≥ $1,769.62, we do not reject the null hypothesis. Diff: 3 Keywords: mean difference, hypothesis test, t-test, critical value Section: 9-1 Hypothesis Tests for Means Outcome: 4

175) Explain what is meant by a p-value. Answer: A p-value is the probability of getting a sample mean that is as extreme or more extreme than the one observed from a population with the hypothesized parameter. For instance, if the sample mean that we find in our sample is "substantially" different from the hypothesized population mean, a small pvalue will be computed. If the calculated p-value is less than alpha (α), the null hypothesis should be rejected. Diff: 2 Keywords: p-value, probability Section: 9-1 Hypothesis Tests for Means Outcome: 4 176) A company makes a device that can be fitted to automobile engines to improve the mileage. The company claims that if the device is installed, owners will observe a mean increase of more than 3.0 mpg. Assuming that the population standard deviation of increase is known to be 0.75 mpg, and a sample of size 64 cars is selected with an = 3.25 mpg, use the p-value approach to test the null hypothesis using a significance level of 0.05. Answer: The research hypothesis is based on the claim made by the company that mean mileage will increase by more than 3.0 mpg. Thus, we set up the following null and alternative hypotheses: H0 : μ ≤ 3.0 mpg Ha : μ > 3.0 mpg. The next step in determining the p-value is to compute the test statistic. Since we assume that the population standard deviation is known, the test statistic will be a z-value computed as follows:

= 2.67

We then go to the standard normal table to find the probability associated with z = 2.68. This value is 0.4962. The p-value is found by taking 0.5000 - 0.4962 = 0.0038. Since 0.0038 < α = 0.05, we reject the null hypothesis. The difference between 3.25 mpg and 3.0 is too much to attribute to sampling error alone. This conclusion means that the manufacturer's claim is supported by the sample data. Diff: 2 Keywords: p-value, hypothesis, significance level, mean difference Section: 9-1 Hypothesis Tests for Means Outcome: 4

177) A company makes a device that can be fitted to automobile engines to improve the mileage. The company claims that if the device is installed, owners will observe a mean increase of more than 3.0 mpg. Assuming that the population standard deviation of increase is known to be 0.75 mpg, and a sample of size 64 cars is selected, what is the probability of "accepting" the null hypothesis if the true population mean is 3.10 mpg increase? Assume that the test will be performed using a 0.05 level of significance. Answer: The objective here is to compute beta, the probability of a Type II error. We begin by formulating the null and alternative hypotheses: H0 : μ ≤ 3.0 mpg Ha : μ > 3.0 mpg Next, we need to determine the critical value as follows: μ+z

= 3.0 + 1.645

= 3.1542.

The next step in solving for beta is to determine the probability of finding a sample mean of 3.1542 or less given a true population mean of 3.10 mpg. We do this by finding the test statistic z-value associated with 3.1542 as follows:

= .5781 ≈ .58.

From the standard normal table, the probability for a z-value equal to 0.58 is 0.2190. To find beta, we add 0.5000 to 0.2190, giving 0.7190. Thus, if the true mean is 3.10 mpg, there is a 71.9 percent chance that the hypothesis test will not detect that and, instead, will lead to not rejecting the null hypothesis that the mean is 3.0 or less. Diff: 3 Keywords: mean difference, hypothesis, beta, type II Section: 9-1 Hypothesis Tests for Means Outcome: 4 178) Explain why an increase in sample size will reduce the probability of a Type II error but will not impact the probability of a Type I error. Answer: The probability of a Type I error is set by the decision maker and is based on his/her willingness to reject a true null hypothesis. Thus, alpha is independent of the size of the sample. However, the Type II error probability, beta, is affected by the size of the sample. The reason for this is that the standard error for the sampling distribution is found using

. Thus, if n is increased, the standard error is reduced.

Then, if the standard error is reduced, the "true" population parameter is relatively farther from the hypothesized population value making it easier for the sample data to distinguish between the hypothesized value and the "true" value. Thus, an increase in sample size will reduce beta. Diff: 2 Keywords: sample size, type II, type I, alpha, beta Section: 9-3 Type II Errors Outcome: 5

179) A small city is considering breaking away from the county school system and starting its own city school system. City leaders believe that more than 60 percent of residents support the idea. A poll of n = 215 residents is taken and 134 people say they support starting a city school district. Using a 0.10 level of significance, conduct a hypothesis test to determine whether this poll supports the belief of city leaders. Answer: Since "more than 60 percent" is a strict inequality, this must go in the alternate hypothesis, so the hypotheses are: H0 : p ≤ 0.6 HA : p > 0.6 Since this a proportion, the normal distribution is used, and the test statistic for α = 0.10 is z = 1.28 and the decision rule is: Reject the null if Z > 1.28. The test statistic formula is z=

And the sample proportion is p = 134/215 = 0.6233. So the test statistic is z =

= 0.6976

This does not fall in the rejection region since 0.6976 < 1.28, so we do not reject the null. This indicates the proportion of residents who support a city school district is not significantly greater than 0.6. Diff: 2 Keywords: null, alternative, hypothesis, proportion Section: 9-2 Hypothesis Tests for Proportions Outcome: none

180) The Gordon Beverage Company bottles soft drinks using an automatic filling machine. When the process is running properly, the mean fill is 12 ounces per can. The machine has a known standard deviation of 0.20 ounces. Each day, the company selects a random sample of 36 cans and measures the volume in each can. They then test to determine whether the filling process is working properly. The test is conducted using a 0.05 significance level. What is the critical value in ounces? Answer: The company will take action if it determines that the fill process is putting too little or too much in the cans on average. Therefore, the research hypothesis is that the population mean is not equal to 12 ounces. Given this, the null and alternative hypotheses are: H0 : μ = 12 ounces Ha : μ ≠ 12 ounces. We note that the hypothesis test is two-tailed since we can reject it if the sample mean gets too large or too small. Thus, there are actually two critical values, one on each side of the distribution. We find the critical values as follows: μ±z Because this is a two-tailed test, we must split the alpha into two parts of 0.025 each. The z-value from the standard normal distribution for 0.5000 - 0.025 = 0.475 is 1.96. Given this, the critical values are determined as: 12.0 ± 1.96

or 12.0 ± .0653

Thus, the decision rule is: If 11.9347 ≤ ≤ 12.0653 do not reject the null hypothesis, otherwise reject. Diff: 2 Keywords: critical value, mean, z-value Section: 9-1 Hypothesis Tests for Means Outcome: 3

181) The Gordon Beverage Company bottles soft drinks using an automatic filling machine. When the process is running properly, the mean fill is 12 ounces per can. The machine has a known standard deviation of 0.20 ounces. Each day, the company selects a random sample of 36 cans and measures the volume in each can. They then test to determine whether the filling process is working properly. The test is conducted using a 0.05 significance level. Using the test statistic approach, what conclusion should the company reach if the sample mean is 12.02 ounces? What type of statistical error may have been committed? Answer: The company will take action if it determines that the fill process is putting too little or too much in the cans on average. Therefore, the research hypothesis is that the population mean is not equal to 12 ounces. Given this, the null and alternative hypotheses are: H0 : μ = 12 ounces Ha : μ ≠ 12 ounces. We note that the hypothesis test is two-tailed since we can reject if the sample mean gets too large or too small. The critical z-value from the standard normal table for a two-tailed test with alpha = 0.05 is 1.96, the decision rule is reject the null if z < -1.96 or z > 1.96. The test statistic is computed as:

= .60

Since 0.60 < 1.96, we do not reject the null hypothesis based on these sample data. So we can conclude that the filling process does not need adjusting. Thus, the statistical error that could have been committed is a Type II error since the null hypothesis was not rejected. If a Type II error was committed this would mean that the filling machine actually needs adjusting, but we failed to detect it. Diff: 2 Keywords: test statistic, z-value Section: 9-1 Hypothesis Tests for Means Outcome: 4

182) A national car rental chain believes that more than 80 percent of its customers are satisfied with the check-in process that the company is using. To test this, a random sample of n = 200 customers are surveyed. These sample results show 168 that say they were satisfied. If the test is to be conducted using a .05 level of significance, what is the critical value? Answer: This is a hypothesis about a single population proportion. The appropriate null and alternative hypotheses are: H0 : p ≤ .80 Ha : p > .80 The critical value for this one-tailed test is found as: p + z The z-value from the standard normal table for a 0.05 level of significance and one-tailed test is 1.645. Then the critical value is: .80 + 1.645 we reject the null hypothesis, since =

= .8465. Thus, if the sample proportion exceeds 0.8465, =

= .84

Since 0.84 is just slightly less than 0.8465, we do not reject the null hypothesis. Diff: 2 Keywords: population proportion, hypothesis test, critical value Section: 9-2 Hypothesis Tests for Proportions Outcome: none 183) A contract with a parts supplier calls for no more than .04 defects in the large shipment of parts. To test whether the shipment meets the contract, the receiving company has selected a random sample of n = 100 parts and found 6 defects. If the hypothesis test is to be conducted using a significance level equal to 0.05, what is the test statistic and what conclusion should the company reach based on the sample data? Answer: This is a hypothesis about a single population proportion. The appropriate null and alternative hypotheses are: H0 : p ≤ .04 Ha : p > .04 Note that the research hypothesis is that the shipment contains more than the suggested proportion of defects. The test statistic when testing about proportions is a z-value computed as: z=

For this situation, the calculated z-value is z =

= 1.02.

We compare this value to the z-value from the standard normal distribution table for 0.45 which is z = 1.645. The decision rule is: If z calculated > 1.645, reject the null hypothesis, otherwise do not reject. Since 1.02 < 1.645, we cannot reject the null hypothesis based on the sample data. Thus, there is insufficient evidence to conclude that the proportion of defects in the shipment exceeds 0.04. Diff: 2 Keywords: test statistic, hypothesis test, z-value Section: 9-1 Hypothesis Tests for Means Outcome: 4

Business Statistics, 9e (Groebner/Shannon/Fry) Chapter 10 Estimation and Hypothesis Testing for Two Population Parameters 1) The Cranston Hardware Company is interested in estimating the difference in the mean purchase for men customers versus women customers. It wishes to estimate this difference using a 95 percent confidence level. If the sample size is n = 10 from each population, the samples are independent, and sample standard deviations are used, and the variances are assumed equal, then the critical value will be t = 2.1009. Answer: TRUE Diff: 2 Keywords: confidence interval, mean difference, independent, sample Section: 10-1 Estimation for Two Population Means Using Independent Samples Outcome: 1 2) To find a confidence interval for the difference between the means of independent samples, when the variances are unknown but assumed equal, the sample sizes of the two groups must be the same. Answer: FALSE Diff: 2 Keywords: confidence interval, mean difference, independent Section: 10-1 Estimation for Two Population Means Using Independent Samples Outcome: 1 3) The Cranston Hardware Company is interested in estimating the difference in the mean purchase for men customers versus women customers. It wishes to estimate this difference using a 95 percent confidence level. Assume that the variances are equal and the populations normally distributed. The following data represent independent samples from each population: Men 16.49 33.34 20.18 26.39 28.03 26.02 16.08 32.27 21.66 32.32 29.79 33.37

Women 17.21 17.46 15.65 10.67 14.07 19.61 15.90 11.17 24.66 13.35 20.87 22.57

Based on these data, the upper limit of the interval estimate is approximately $13.82. Answer: TRUE Diff: 3 Keywords: confidence interval, mean difference, independent Section: 10-1 Estimation for Two Population Means Using Independent Samples Outcome: 1 10-1 Copyright © 2018 Pearson Education, Inc.

4) The Cranston Hardware Company is interested in estimating the difference in the mean purchase for men customers versus women customers. It wishes to estimate this difference using a 95 percent confidence level. Assume that the variances are equal and the populations normally distributed. The following data represent independent samples from each population: Men 16.49 33.34 20.18 26.39 28.03 26.02 16.08 32.27 21.66 32.32 29.79 33.37

Women 17.21 17.46 15.65 10.67 14.07 19.61 15.90 11.17 24.66 13.35 20.87 22.57

Based on these data, the company can conclude that there is a statistical difference between men and women with regard to mean spending at the hardware store with men tending to spend more on average than women. Answer: TRUE Diff: 3 Keywords: confidence interval, mean difference independent Section: 10-1 Estimation for Two Population Means Using Independent Samples Outcome: 1 5) In estimating the difference between two population means, if a 95 percent confidence interval includes zero, then we can conclude that there is a 95 percent chance that the difference between the two population means is zero. Answer: FALSE Diff: 2 Keywords: confidence interval, mean difference Section: 10-1 Estimation for Two Population Means Using Independent Samples Outcome: 1 6) Marketing managers for a toy store located in two separate cities is interested in estimating the difference in the mean daily sales for the two cities. They want to calculate a 90 percent confidence interval and will select a sample of 10 days in each store for the study. If the marketing managers assume that the population standard deviations are known, the critical value for the confidence interval is z = 1.645. Answer: TRUE Diff: 2 Keywords: confidence interval, mean difference, independent Section: 10-1 Estimation for Two Population Means Using Independent Samples Outcome: 1 10-2 Copyright © 2018 Pearson Education, Inc.

7) The NCAA is interested in estimating the difference in mean number of daily training hours for men and women athletes on college campuses. They want 95 percent confidence and will select a sample of 10 men and 10 women for the study. The sample results are: Men n1 = 10 students

Women n2 = 10 students

1 = 2.7 hours s1 = .30 hours

2 = 2.4 hours s2 = .40 hours

Based on these data, the point estimate is .30 hours. Answer: TRUE Diff: 1 Keywords: mean difference, confidence interval, independent Section: 10-1 Estimation for Two Population Means Using Independent Samples Outcome: 1 8) The NCAA is interested in estimating the difference in mean number of daily training hours for men and women athletes on college campuses. It wants 95 percent confidence and will select a sample of 10 men and 10 women for the study. The variances are assumed equal and the populations normally distributed. The sample results are: Men n1 = 10 students

Women n2 = 10 students

1 = 2.7 hours s1 = .30 hours

2 = 2.4 hours s2 = .40 hours

Based on these sample data, the critical value for developing the confidence interval is z = 1.96. Answer: FALSE Diff: 1 Keywords: mean difference, confidence interval, independent Section: 10-1 Estimation for Two Population Means Using Independent Samples Outcome: 1

9) The NCAA is interested in estimating the difference in mean number of daily training hours for men and women athletes on college campuses. It wants 95 percent confidence and will select a sample of 10 men and 10 women for the study. The variances are assumed equal and the populations normally distributed. The sample results are: Men n1 = 10 students

Women n2 = 10 students

1 = 2.7 hours s1 = .30 hours

2 = 2.4 hours s2 = .40 hours

Based on these sample data, the estimate for the standard deviation of the sampling distribution is found by taking the square root of the sum of the two sample variances. Answer: FALSE Diff: 2 Keywords: mean difference, confidence interval, independent, standard deviation Section: 10-1 Estimation for Two Population Means Using Independent Samples Outcome: 1 10) The NCAA is interested in estimating the difference in mean number of daily training hours for men and women athletes on college campuses. It wants 95 percent confidence and will select a sample of 10 men and 10 women for the study. The variances are assumed equal and the populations normally distributed. The sample results are: Men n1 = 10 students

Women n2 = 10 students

1 = 2.7 hours s1 = .30 hours

2 = 2.4 hours s2 = .40 hours

Based on these data, the lower limit for the difference between population means is 0.15 hours. Answer: FALSE Diff: 3 Keywords: mean difference, confidence interval, independent, limit Section: 10-1 Estimation for Two Population Means Using Independent Samples Outcome: 1 11) In estimating the difference between two population means based on small, independent samples from the two populations, two important assumptions are that the populations each be normally distributed and the populations have equal variances. Answer: TRUE Diff: 1 Keywords: mean difference, confidence interval, independent, assumption Section: 10-1 Estimation for Two Population Means Using Independent Samples Outcome: 1

12) Box and whisker plots are often useful for determining whether two populations have distributions that might be skewed. Answer: TRUE Diff: 1 Keywords: mean difference, confidence interval, independent, box, whisker, normal Section: 10-1 Estimation for Two Population Means Using Independent Samples Outcome: 1 13) In estimating a confidence interval for the difference between two means, when the samples are independent and the standard deviations are unknown, it can be acceptable for there to be small violations of the assumptions of normality and equal variances, especially when the sample sizes are equal. Answer: TRUE Diff: 2 Keywords: mean difference, confidence interval, independent, equal variances Section: 10-1 Estimation for Two Population Means Using Independent Samples Outcome: 1 14) Recently the managers for a large retail department store stated that a study has revealed that female shoppers spend on average 23.5 minutes longer in the store per visit than do male shoppers. Based on this information, the managers can be confident that female shoppers, as a population, do spend longer times in the store than do males shoppers, as a population. Answer: FALSE Diff: 2 Keywords: mean difference, confidence interval, independent Section: 10-1 Estimation for Two Population Means Using Independent Samples Outcome: 1 15) Increasing the size of the samples in a study to estimate the difference between two population means will increase the level of confidence that a decision maker can have regarding the interval estimate. Answer: FALSE Diff: 2 Keywords: mean difference, confidence interval, independent, sample, size Section: 10-1 Estimation for Two Population Means Using Independent Samples Outcome: 1 16) All other things held constant, decreasing the level of confidence for a confidence interval estimate for the difference between two population means will result in a smaller margin of error. Answer: TRUE Diff: 2 Keywords: mean difference, confidence interval, independent Section: 10-1 Estimation for Two Population Means Using Independent Samples Outcome: 1

17) The t-distribution is still applicable even when there are small violations of the assumptions for the case when the variances for two populations are unknown. This is particularly true when the sample sizes are approximately equal. Answer: TRUE Diff: 2 Keywords: independent, sample, t-distribution Section: 10-1 Estimation for Two Population Means Using Independent Samples Outcome: 1 18) In order to make the test for the difference between two population means valid, the sample size in each independent sample must be the same. Answer: FALSE Diff: 1 Keywords: mean difference, independent, sample, size Section: 10-2 Hypothesis Tests for Two Population Means Using Independent Samples Outcome: none 19) When performing a hypothesis test for the difference between the means of two independent populations, where the standard deviations are known, the variances must be assumed equal. Answer: FALSE Diff: 2 Keywords: mean difference, independent, known standard deviations Section: 10-2 Hypothesis Tests for Two Population Means Using Independent Samples Outcome: none 20) Two samples are said to be independent if they are collected at different points in time. Answer: FALSE Diff: 1 Keywords: samples, independent Section: 10-2 Hypothesis Tests for Two Population Means Using Independent Samples Outcome: none 21) The t-distribution can be used to test hypotheses about the difference between two population means given the following two assumptions: - each population is normally distributed, and - the two populations have equal variances. Answer: TRUE Diff: 2 Keywords: mean difference, t-distribution, normal, variance Section: 10-2 Hypothesis Tests for Two Population Means Using Independent Samples Outcome: none

22) When performing a hypothesis test for the difference between the means of two independent populations where the standard deviations are known, it is necessary to use the pooled standard deviation in calculating the test statistic. Answer: FALSE Diff: 2 Keywords: hypothesis, mean, independent, known standard deviations Section: 10-2 Hypothesis Tests for Two Population Means Using Independent Samples Outcome: none 23) There is interest at the American Savings and Loan as to whether there is a difference between average daily balances in checking accounts that are joint accounts (two or more members per account) versus single accounts (one member per account). To test this, a random sample of checking accounts was selected with the following results: Single Accounts n = 20 s = $256 = $1,123

Joint Accounts n = 30 s = $300 = $1,245

Based upon these data, assuming that the populations are normally distributed with equal variances, the test statistic for testing whether the two populations have equal means is approximately -1.49. Answer: TRUE Diff: 2 Keywords: mean difference, equal variance, test statistic Section: 10-2 Hypothesis Tests for Two Population Means Using Independent Samples Outcome: none 24) There is interest at the American Savings and Loan as to whether there is a difference between average daily balances in checking accounts that are joint accounts (two or more members per account) versus single accounts (one member per account). To test this, a random sample of checking accounts was selected with the following results: Single Accounts n = 20 s = $256 = $1,123

Joint Accounts n = 30 s = $300 = $1,245

Based upon these data, the critical value from the t-distribution for testing the difference between the two population means using a significance level of 0.05 is t = 1.6772. Answer: FALSE Diff: 2 Keywords: mean difference, equal variance, significance, critical value Section: 10-2 Hypothesis Tests for Two Population Means Using Independent Samples Outcome: none

25) To find the pooled standard deviation involves taking a weighted average of the two sample variances, then finding its square root. Answer: TRUE Diff: 2 Keywords: mean difference, pooled, standard deviation Section: 10-2 Hypothesis Tests for Two Population Means Using Independent Samples Outcome: none 26) The Sergio Lumber Company manufactures plywood. One step in the process is the one where the veneer is dried by passing through a huge dryer (similar to an oven) where much of the moisture in the veneer is extracted. At the end of this step, samples of veneer are tested for moisture content. It is believed that pine veneer will be less moist on average than will fir veneer. The following data were reported recently where the values represent the percent of moisture in the wood: Pine 13.1 10.2 14.7 17.8 17.6 19.2 7.4 13.3 17.3 10.7

Fir 19.8 13.5 20.3 20.5 21.4 23.4 14.6 18.8

The null and alternative hypotheses to be tested are H0 : μp ≥ μf Ha : μ p < μ f Answer: TRUE Diff: 2 Keywords: independent, null, alternative Section: 10-2 Hypothesis Tests for Two Population Means Using Independent Samples Outcome: none 27) The Sergio Lumber Company manufactures plywood. One step in the process is the one where the veneer is dried by passing through a huge dryer (similar to an oven) where much of the moisture in the veneer is extracted. At the end of this step, samples of veneer are tested for moisture content. It is believed that pine veneer will be less moist on average than will fir veneer. The hypothesis test that will be conducted using an alpha = 0.05 level will be a two-tailed test. Answer: FALSE Diff: 1 Keywords: independent, hypothesis, two tailed Section: 10-2 Hypothesis Tests for Two Population Means Using Independent Samples Outcome: none

28) The Sergio Lumber Company manufactures plywood. One step in the process is the one where the veneer is dried by passing through a huge dryer (similar to an oven) where much of the moisture in the veneer is extracted. At the end of this step, samples of veneer are tested for moisture content. It is believed that pine veneer will be less moist on average than will fir veneer. The following data were reported recently where the values represent the percent moisture in the wood: Pine 13.1 10.2 14.7 17.8 17.6 19.2 7.4 13.3 17.3 10.7

Fir 19.8 13.5 20.3 20.5 21.4 23.4 14.6 18.8

Based on these data, the critical t value from the t-distribution will be 1.7459 if the significance level is set at 0.05 and variances are presumed equal. Answer: TRUE Diff: 2 Keywords: independent, critical value, t-distribution Section: 10-2 Hypothesis Tests for Two Population Means Using Independent Samples Outcome: none 29) In order to test the difference in populations means, samples were collected for two independent populations where the variances are assumed equal and the population normally distributed. The following data resulted: Population 1

Population 2

= 112 s = 14 n = 25

= 107 s = 17 n = 28

The value of the pooled standard deviation is 15.66. Answer: TRUE Diff: 2 Keywords: independent, pooled standard deviation Section: 10-2 Hypothesis Tests for Two Population Means Using Independent Samples Outcome: none

30) If the sample data lead us to suspect that the variances of the two populations are not equal, the t-test statistic and the degrees of freedom must be adjusted accordingly. Answer: TRUE Diff: 2 Keywords: variances, critical value, t-distribution Section: 10-2 Hypothesis Tests for Two Population Means Using Independent Samples Outcome: none 31) If you are interested in estimating the difference between the means of two samples that have been paired, the point estimate for this difference is the mean value of the paired differences. Answer: TRUE Diff: 1 Keywords: mean difference, confidence interval, paired, point estimate Section: 10-3 Interval Estimation and Hypothesis Tests for Paired Samples Outcome: 2 32) In comparing two populations using paired differences, after the difference is found for each pair, the method for testing whether the mean difference is equal to 0 becomes the same as was used for a onesample hypothesis test with unknown standard deviation. Answer: TRUE Diff: 2 Keywords: mean difference, paired, hypothesis test Section: 10-3 Interval Estimation and Hypothesis Tests for Paired Samples Outcome: 2 33) The point estimate in a paired difference estimation example is an estimation of the population differences lying halfway between the interval limits of a confidence interval. Answer: TRUE Diff: 2 Keywords: mean difference, confidence interval, paired Section: 10-3 Interval Estimation and Hypothesis Tests for Paired Samples Outcome: 2 34) When conducting a hypothesis test to determine whether or not two groups differ, using paired samples rather than independent samples has the advantage of controlling for sources of variation that might distort the conclusions of the study. Answer: TRUE Diff: 2 Keywords: paired, sample, mean Section: 10-3 Interval Estimation and Hypothesis Tests for Paired Samples Outcome: 2

35) Two placement exams are available that students can take to determine which math class they should begin with in their freshman year. It is believed that there is no difference in the mean scores that would be received for the two tests. To test this using a 0.05 level of significance, a randomly selected group of students took both tests and had their scores recorded. The following data were obtained: Student 1 2 3 4 5 6 7

Test A 78 86 74 72 75 68 77

Test B 82 74 79 93 80 82 99

Based on these data, the test statistic is approximately t = -1.892. Answer: TRUE Diff: 3 Keywords: mean difference, paired, test statistic Section: 10-3 Interval Estimation and Hypothesis Tests for Paired Samples Outcome: 2 36) To test whether Model A and Model B cars have the same MPG, the first step is to select two independent random samples of drivers and assign one of them to drive Model A and the other Model B. Answer: FALSE Diff: 3 Keywords: paired, sampling Section: 10-3 Interval Estimation and Hypothesis Tests for Paired Samples Outcome: 2 37) The test statistic that is used when testing hypotheses about the difference between two population proportions is the t-value from the t-distribution. Answer: FALSE Diff: 1 Keywords: mean difference, proportion, hypothesis Section: 10-4 Estimation and Hypothesis Tests for Two Population Proportions Outcome: 3

38) An accounting firm has been hired by a large computer company to determine whether the proportion of accounts receivables with errors in one division (Division 1) exceeds that of the second division (Division 2). The managers believe that such a difference may exist because of the lax standards employed by the first division. The null and alternative hypotheses that will be tested are: H0 : μ1 ≥ μ2 Ha : μ1 < μ2 Answer: FALSE Diff: 1 Keywords: mean difference, proportion, hypothesis, null, alternative Section: 10-4 Estimation and Hypothesis Tests for Two Population Proportions Outcome: 3 39) An accounting firm has been hired by a large computer company to determine whether the proportion of accounts receivables with errors in one division (Division 1) exceeds that of the second division (Division 2). The managers believe that such a difference may exist because of the lax standards employed by the first division. To conduct the test, the accounting firm has selected random samples of accounts from each division with the following results.

Sample Size

Division 1 n1 = 100

Division 2 n2 = 100

Errors found

x1 = 13

x2 = 8

Based on this information and using a significance level equal to 0.05, the critical value from the standard normal table is z = 1.645. Answer: TRUE Diff: 2 Keywords: mean difference, proportion, hypothesis, critical value Section: 10-4 Estimation and Hypothesis Tests for Two Population Proportions Outcome: 3

40) An accounting firm has been hired by a large computer company to determine whether the proportion of accounts receivables with errors in one division (Division 1) exceeds that of the second division (Division 2). The managers believe that such a difference may exist because of the lax standards employed by the first division. To conduct the test, the accounting firm has selected random samples of accounts from each division with the following results.

Sample Size

Division 1 n1 = 100

Division 2 n2 = 100

Errors found

x1 = 13

x2 = 8

Based on this information, and using a significance level equal to 0.05, the pooled estimator for the overall proportion is = .1050. Answer: TRUE Diff: 2 Keywords: mean difference, proportion, hypothesis, pooled Section: 10-4 Estimation and Hypothesis Tests for Two Population Proportions Outcome: 3 41) An accounting firm has been hired by a large computer company to determine whether the proportion of accounts receivables with errors in one division (Division 1) exceeds that of the second division (Division 2). The managers believe that such a difference may exist because of the lax standards employed by the first division. To conduct the test, the accounting firm has selected random samples of accounts from each division with the following results.

Sample Size

Division 1 n1 = 100

Division 2 n2 = 100

Errors found

x1 = 13

x2 = 8

Based on this information and using a significance level equal to 0.05, the test statistic for the hypothesis test is approximately 1.153 and, therefore, the null hypothesis is not rejected. Answer: TRUE Diff: 2 Keywords: mean difference, proportion, hypothesis, test statistic Section: 10-4 Estimation and Hypothesis Tests for Two Population Proportions Outcome: 3

42) A direct retailer that sells clothing on the Internet has two distribution centers and wants to determine if there is a difference between the proportion of customer order shipments that contain errors (wrong color, wrong size, etc.). It takes a sample of orders from each distribution center and obtain the following results:

Number of orders Number of errors

Distribution Center 1 120 4

Distribution Center 2 145 6

Based on these data it can proceed with assuming the normal distribution for each of the proportion sampling distributions. Answer: FALSE Diff: 2 Keywords: proportion, hypothesis, normality assumption Section: 10-4 Estimation and Hypothesis Tests for Two Population Proportions Outcome: 3 43) A major manufacturer of home electronics is interested in determining whether customers have a preference between two new speaker designs for their home entertainment centers. To test this, the design department manager has selected a random sample of customers and shown them the first design. A second sample of customers is shown design 2. The manager then asks each customer whether they prefer the new design they were shown over the one they currently own. The following results were observed:

Sample size

Design 1 n1 = 150

Design 2 n2 = 80

Number preferring new

x1 = 65

x2 = 58

Based on these data and a significance level equal to 0.05, the appropriate null and alternative hypotheses are: H0 : 1 ≥ 2 Ha : 1 < 2 Answer: TRUE Diff: 2 Keywords: mean difference, proportion, hypothesis, null, alternative Section: 10-4 Estimation and Hypothesis Tests for Two Population Proportions Outcome: 3

44) A major manufacturer of home electronics is interested in determining whether customers have a preference between two new speaker designs for their home entertainment centers. To test this, the design department manager has selected a random sample of customers and shown them the first design. A second sample of customers is shown design 2. The manager then asks each customer whether they prefer the new design they were shown over the one they currently own. The following results were observed:

Sample size

Design 1 n1 = 150

Design 2 n2 = 80

Number preferring new

x1 = 65

x2 = 58

Based on these data and a significance level equal to 0.05, the test statistic is approximately -4.22 and thus the null hypothesis should be rejected. Answer: TRUE Diff: 3 Keywords: mean difference, proportion, hypothesis, test statistic Section: 10-4 Estimation and Hypothesis Tests for Two Population Proportions Outcome: 3 45) An Internet service provider is interested in estimating the proportion of homes in a particular community that have computers but do not already have Internet access. To do this, the company has selected a random sample of n = 200 homes and made calls. A total of 188 homes responded to the survey question with 38 saying that they had a computer with no Internet access. The 95 percent confidence interval estimate for the true population proportion is approximately 0.1447 - 0.2595. Answer: TRUE Diff: 2 Keywords: population, proportion, confidence interval Section: 10-4 Estimation and Hypothesis Tests for Two Population Proportions Outcome: 3 46) A direct retailer that sells clothing on the Internet has two distribution centers and wants to determine if there is a difference between the proportion of customer order shipments that contain errors (wrong color, wrong size, etc.). It calculates a 95 percent confidence interval on the difference in the sample proportions to be -0.012 to 0.037. Based on this, it can conclude that the distribution centers differ significantly for the proportion of orders with errors. Answer: FALSE Diff: 2 Keywords: mean difference, proportion, confidence interval Section: 10-4 Estimation and Hypothesis Tests for Two Population Proportions Outcome: 3

47) If a manager wishes to develop a confidence interval estimate for estimating the difference between two population means, an increase in confidence level will result in: A) a decrease in the size of the critical value. B) a wider confidence interval. C) a need for an increased sample size. D) all of the above. Answer: B Diff: 1 Keywords: mean difference, population, confidence interval Section: 10-1 Estimation for Two Population Means Using Independent Samples Outcome: 1 48) If the population variances are assumed to be known in an application where a manager wishes to estimate the difference between two population means, the 90 percent confidence interval estimate can be developed using which of the following critical values? A) z = 1.645 B) z = 1.96 C) t value that depends on the sample sizes from the two populations D) z = 2.575 Answer: B Diff: 1 Keywords: mean difference, confidence interval, population, variance, critical value Section: 10-1 Estimation for Two Population Means Using Independent Samples Outcome: 1

49) A company in Maryland has developed a device that can be attached to car engines, which it believes will increase the miles per gallon that cars will get. The owners are interested in estimating the difference between mean mpg for cars using the device versus those that are not using the device. The following data represent the mpg for independent random samples of cars from each population. The variances are assumed equal and the populations normally distributed. With Device 22.6 23.4 28.4 29.0 29.3 20.0

Without Device 26.9 24.4 20.8 20.8 20.2 26.0 28.1 25.6

Given this data, what is the critical value if the owners wish to have a 90 percent confidence interval estimate? A) t = 2.015 B) t = 1.7823 C) z = 1.645 D) z = 1.96 Answer: B Diff: 2 Keywords: mean difference, confidence interval, critical value Section: 10-1 Estimation for Two Population Means Using Independent Samples Outcome: 1

50) A company in Maryland has developed a device that can be attached to car engines, which it believes will increase the miles per gallon that cars will get. The owners are interested in estimating the difference between mean mpg for cars using the device versus those that are not using the device. The following data represent the mpg for random independent samples of cars from each population. The variances are assumed equal and the populations normally distributed. With Device 22.6 23.4 28.4 29.0 29.3 20.0

Without Device 26.9 24.4 20.8 20.8 20.2 26.0 28.1 25.6

Given this data, what is the upper limit for a 95 percent confidence interval estimate for the difference in mean mpg? A) Approximately 3.88 mpg B) About 5.44 mpg C) Just under 25.0 D) None of the above Answer: B Diff: 3 Keywords: mean difference, confidence interval, population Section: 10-1 Estimation for Two Population Means Using Independent Samples Outcome: 1 51) When estimating a confidence interval for the difference between 2 means using the method where sample variances are pooled, which of the following assumptions is not needed? A) The populations are normally distributed. B) The populations have equal variances. C) The samples are independent. D) The sample sizes are equal. Answer: D Diff: 2 Keywords: mean difference, confidence interval, independent, variances unknown Section: 10-1 Estimation for Two Population Means Using Independent Samples Outcome: 1

52) A major retail clothing store is interested in estimating the difference in mean monthly purchases by customers who use the store's in-house credit card versus using a Visa, Mastercard, or one of the other major credit cards. To do this, it has randomly selected a sample of customers who have made one or more purchases with each of the types of credit cards. The following represents the results of the sampling:

Sample Size: Mean Monthly Purchases: Standard Deviation:

In-House Credit Card 86 $45.67 $10.90

National Credit Card 113 $39.87 $12.47

Based on these sample data, what is the lower limit for the 95 percent confidence interval estimate for the difference between population means? A) About $5.28 B) Approximately $4.85 C) Approximately $2.54 D) Approximately $3.41 Answer: C Diff: 2 Keywords: mean difference, confidence interval, population, lower limit Section: 10-1 Estimation for Two Population Means Using Independent Samples Outcome: 1 53) A major retail clothing store is interested in estimating the difference in mean monthly purchases by customers who use the store's in-house credit card versus using a Visa, Mastercard, or one of the other major credit cards. To do this, it has randomly selected a sample of customers who have made one or more purchases with each of the types of credit cards. The following represents the results of the sampling:

Sample Size: Mean Monthly Purchases: Standard Deviation:

In-House Credit Card 86 $45.67 $10.90

National Credit Card 113 $39.87 $12.47

Given this information, which of the following statements is true? A) If either of the sample sizes is increased, the resulting confidence interval will have a smaller margin of error. B) If the confidence level were changed from 95 percent to 90 percent, the margin of error in the estimate would be reduced. C) Neither A nor B are true. D) Both A and B are true. Answer: D Diff: 2 Keywords: mean difference, confidence interval, population, margin of error Section: 10-1 Estimation for Two Population Means Using Independent Samples Outcome: 1

54) The management of a department store is interested to estimate the difference in the amount of money spent by female and male shoppers. You are given the following information.

Sample size Sample mean Population standard deviation

Female Shoppers 64 $140 $10

Male Shoppers 49 $125 $8

A 95 percent confidence interval estimate for the difference between the average purchases of the customers using the two different credit cards is: A) 49 to 64 B) 11.68 to 18.32 C) 125 to 140 D) 8 to 10 Answer: B Diff: 2 Keywords: mean difference, confidence interval, population Section: 10-1 Estimation for Two Population Means Using Independent Samples Outcome: 1 55) A major retail clothing store is interested in estimating the difference in mean monthly purchases by customers who use the store's in-house credit card versus using a Visa, Mastercard, or one of the other major credit cards. To do this, it has randomly selected a sample of customers who have made one or more purchases with each of the types of credit cards. The following represents the results of the sampling:

Sample Size: Mean Monthly Purchases: Standard Deviation:

In-House Credit Card 86 $45.67 $10.90

National Credit Card 113 $39.87 $12.47

Suppose that the managers wished to test whether there is a statistical difference in the mean monthly purchases by customers using the two types of credit cards, using a significance level of .05, what is the value of the test statistic assuming the standard deviations are known? A) t = 3.49 B) z = 11.91 C) z = 2.86 D) z = 3.49 Answer: D Diff: 2 Keywords: mean difference, known standard deviations, test statistic Section: 10-2 Hypothesis Tests for Two Population Means Using Independent Samples Outcome: none

56) If we are testing for the difference between the means of two independent populations with samples of n1 = 20 and n2 = 20, the number of degrees of freedom is equal to: A) 39 B) 38 C) 19 D) 18 Answer: B Diff: 1 Keywords: mean difference, independent, degrees of freedom Section: 10-2 Hypothesis Tests for Two Population Means Using Independent Samples Outcome: none 57) Given the following information, calculate the degrees of freedom that should be used in the pooledstandard deviation t-test.

A) df = 41 B) df = 39 C) df = 16 D) df = 25 Answer: B Diff: 2 Keywords: degrees of freedom, pooled, standard deviation, t-test Section: 10-2 Hypothesis Tests for Two Population Means Using Independent Samples Outcome: none 58) A recent study posed the question about whether Japanese managers are more motivated than American managers. A randomly selected sample of each was administered the Sarnoff Survey of Attitudes Toward Life (SSATL), which measures motivation for upward mobility. The SSATL scores are summarized below.

Sample Size Mean SSATL Score Population Std. Dev

American 211 65.75 11.07

Japanese 100 79.83 6.41

Judging from the way the data were collected, which test would likely be most appropriate? A) Related samples t-test for mean difference B) Pooled-variance t-test for the difference in means C) Independent samples Z-test for the difference in means D) Related samples Z-test for mean difference Answer: C Diff: 1 Keywords: independent, samples, z-test, mean difference Section: 10-2 Hypothesis Tests for Two Population Means Using Independent Samples Outcome: none

59) A recent study posed the question about whether Japanese managers are more motivated than American managers. A randomly selected sample of each was administered the Sarnoff Survey of Attitudes Toward Life (SSATL), which measures motivation for upward mobility. The SSATL scores are summarized below.

Sample Size Mean SSATL Score Population Std. Dev

American 211 65.75 11.07

Japanese 100 79.83 6.41

Which of the following is the correct the null and alternative hypotheses to determine if the average SSATL score of Japanese managers differs from the average SSATL score of American managers? A) H0 : μA - μJ ≥ 0 versus H1 : μA - μJ < 0 B) H0 : μA - μJ ≤ 0 versus H1 : μA - μJ > 0 C) H0 : μA - μJ = 0 versus H1 : μA - μJ ≠ 0 D) H0 : A - J = 0 versus H1 : A - J ≠ 0 Answer: C Diff: 1 Keywords: null, alternative, hypothesis, mean difference Section: 10-2 Hypothesis Tests for Two Population Means Using Independent Samples Outcome: none 60) A commuter has two different routes available to drive to work. She wants to test whether route A is faster than route B. The best hypotheses are: A) H0 : μA - μβ ≥ 0 HA : μA - μβ < 0 B) H0 : μA - μβ ≤ 0 HA : μA - μβ > 0 C) H0 : μA - μβ = 0 HA : μA - μβ ≠ 0 D) H0 : μA - μβ < 0 HA : μA - μβ ≥ 0 Answer: A Diff: 2 Keywords: independent, sample, hypotheses Section: 10-2 Hypothesis Tests for Two Population Means Using Independent Samples Outcome: none

61) In conducting a hypothesis test for the difference between two population means where the standard deviations are known and the null hypothesis is: H0 : μA - μβ ≥ 0 What is the p-value assuming that the test statistic has been found to be z = 2.52? A) 0.0059 B) 0.9882 C) 0.0118 D) 0.4941 Answer: A Diff: 2 Keywords: mean difference, independent, sample, p-value Section: 10-2 Hypothesis Tests for Two Population Means Using Independent Samples Outcome: none 62) Under what conditions can the t-distribution be correctly employed to test the difference between two population means? A) When the samples from the two populations are small and the population variances are unknown B) When the two populations of interest are assumed to be normally distributed C) When the population variances are assumed to be equal D) All of the above Answer: D Diff: 2 Keywords: variance, small sample, normal, t-distribution Section: 10-2 Hypothesis Tests for Two Population Means Using Independent Samples Outcome: none 63) A hypothesis test for the difference between two means is considered a two-tailed test when: A) the population variances are equal. B) the null hypothesis states that the population means are equal. C) the alpha level is 0.10 or higher. D) the standard deviations are unknown. Answer: B Diff: 1 Keywords: mean difference, hypothesis, two-tailed, null Section: 10-2 Hypothesis Tests for Two Population Means Using Independent Samples Outcome: none

64) There have been complaints recently from homeowners in the north end claiming that their homes have been assessed at values that are too high compared with other parts of town. They say that the mean increase from last year to this year has been higher in their part of town than elsewhere. To test this, the assessor's office staff plans to select a random sample of north end properties (group 1) and a random sample of properties from other areas within the city (group 2) and perform a hypothesis test. Based on the information provided, the research (or alternate) hypothesis is: A) μ1 = μ2 B) μ1 ≠ μ2 C) μ1 > μ2 D) μ1 < μ2 Answer: C Diff: 2 Keywords: independent, mean, research, alternate, hypothesis Section: 10-2 Hypothesis Tests for Two Population Means Using Independent Samples Outcome: none 65) There have been complaints recently from homeowners in the north end claiming that their homes have been assessed at values that are too high compare with other parts of town. They say that the mean increase from last year to this year has been higher in their part of town than elsewhere. To test this, the assessor's office staff plans to select a random sample of north end properties (group 1) and a random sample of properties from other areas within the city (group 2) and perform a hypothesis test. The following sample information is available:

Sample Size Sample Mean Increase Sample St. Deviations

North End 20 $4,010 $1,800

Other 10 $3,845 $1,750

Assuming that the null hypothesis will be tested using an alpha level equal to 0.05, what is the value of the test statistic? A) z = 1.578 B) t = 1.7011 C) t = 0.2388 D) t = 0.3944 Answer: C Diff: 2 Keywords: test statistic, hypothesis, null, independent, mean Section: 10-2 Hypothesis Tests for Two Population Means Using Independent Samples Outcome: none

66) There have been complaints recently from homeowners in the north end claiming that their homes have been assessed at values that are too high compare with other parts of town. They say that the mean increase from last year to this year has been higher in their part of town than elsewhere. To test this, the assessor's office staff plans to select a random sample of north end properties (group 1) and a random sample of properties from other areas within the city (group 2) and perform a hypothesis test. The following sample information is available:

Sample Size Sample Mean Increase Sample St. Deviations

North End 20 $4,010 $1,800

Other 10 $3,845 $1,750

Assuming that the null hypothesis will be tested using an alpha level equal to 0.05, what is the critical value? A) z = 1.578 B) t = 1.7011 C) t = 0.2388 D) t = 2.0484 Answer: B Diff: 2 Keywords: critical value, hypothesis, mean, independent Section: 10-2 Hypothesis Tests for Two Population Means Using Independent Samples Outcome: none 67) Assume that you are testing the difference in the means of two independent populations at the 0.05 level of significance. The null hypothesis is: H0 : μA - μβ ≥ 0 and you have found the test statistic is What should you conclude? A) The mean of pop. A is greater than the mean of pop. B because p-value < α. B) The mean of pop. A is greater than the mean of pop. B because p-value > α. C) There is no significant difference in the two means because p-value > α. D) The mean of pop. B is greater than the mean of pop. A because p-value < α. Answer: D Diff: 2 Keywords: hypothesis, one tailed, mean, independent, p-value Section: 10-2 Hypothesis Tests for Two Population Means Using Independent Samples Outcome: none

68) The U.S. Golf Association provides a number of services for its members. One of these is the evaluation of golf equipment to make sure that the equipment satisfies the rules of golf. For example, they regularly test the golf balls made by the various companies that sell balls in the United States. Recently they undertook a study of two brands of golf balls with the objective to see whether there is a difference in the mean distance that the two golf ball brands will fly off the tee. To conduct the test, the U.S.G.A. uses a robot named "Iron Byron," which swings the club at the same speed and with the same swing pattern each time it is used. The following data reflect sample data for a random sample of balls of each brand. Brand A: Brand B:

234 240

236 236

230 241

227 236

234 239

233 243

228 230

229 239

230 243

238 240

Given this information, what is the test statistic for testing whether the two population means are equal? A) t = 1.115 B) t = 1.96 C) t = -4.04 D) t = -2.58 Answer: C Diff: 3 Keywords: test statistic, mean, independent Section: 10-2 Hypothesis Tests for Two Population Means Using Independent Samples Outcome: none 69) Most companies that make golf balls and golf clubs use a one-armed robot named "Iron Byron" to test their balls for length and accuracy, but because of swing variations by real golfers, these test robots don't always indicate how the clubs will perform in actual use. One company in the golfing industry is interested in testing its new driver to see how it compares with the best-selling driver. To do this, it has selected a group of golfers of differing abilities and ages. Its plan is to have each player use each of the two clubs and hit five balls. It will record the average length of the drives with each club for each player. Given this description of the planned test, which of the following statements is true? A) The test won't be meaningful if only five balls are hit by each player with each club. B) The samples in this case are called paired samples since the same players are hitting both golf clubs. C) The test will be invalid unless different players are used to hit each club so that the samples will be independent. D) The samples are independent because each player is independent of the other players. Answer: B Diff: 1 Keywords: mean difference, paired Section: 10-3 Interval Estimation and Hypothesis Tests for Paired Samples Outcome: 2

70) Most companies that make golf balls and golf clubs use a one-armed robot named "Iron Byron" to test their balls for length and accuracy, but because of swing variations by real golfers, these test robots don't always indicate how the clubs will perform in actual use. One company in the golfing industry is interested in testing its new driver to see if has greater length off the tee than the best-selling driver. To do this, it has selected a group of golfers of differing abilities and ages. Its plan is to have each player use each of the two clubs and hit five balls. It will record the average length of the drives with each club for each player. The resulting data for a sample of 10 players is: Player 1 2 3 4 5 6 7 8 9 10

New Club 236.4 202.5 245.6 257.4 223.5 205.3 266.7 240 278.9 211.4

Leading Club 237.2 200.4 240.8 259.3 218.9 200.6 258.9 236.5 280.5 206.5

Based on these sample data, what is the point estimate for the difference between the mean distance for the new driver versus the leading driver? A) 2.81 B) 1.55 C) -3.45 D) 233.4 Answer: A Diff: 2 Keywords: mean difference, point estimate, paired Section: 10-3 Interval Estimation and Hypothesis Tests for Paired Samples Outcome: 2 71) Suppose that a group of 10 people join a weight loss program for 3 months. Each person's weight is recorded at the beginning and at the end of the 3-month program. To test whether the weight loss program is effective, the data should be treated as: A) independent samples using the normal distribution. B) paired samples using the t-distribution. C) independent samples using the t-distribution. D) independent proportions. Answer: B Diff: 1 Keywords: mean difference, population, independent, paired Section: 10-3 Interval Estimation and Hypothesis Tests for Paired Samples Outcome: 2

72) The t-test for the mean difference between 2 related populations assumes that the respective: A) sample sizes are equal. B) sample variances are equal. C) populations are approximately normal or sample sizes are large. D) All of the above Answer: C Diff: 2 Keywords: mean difference, sample size, normal Section: 10-3 Interval Estimation and Hypothesis Tests for Paired Samples Outcome: 2 73) If we are testing for the difference between the means of two paired populations with samples of n1 = 20 and n2 = 20, the number of degrees of freedom is equal to: A) 39 B) 38 C) 19 D) 18 Answer: C Diff: 1 Keywords: mean difference, population, paired, degrees of freedom Section: 10-3 Interval Estimation and Hypothesis Tests for Paired Samples Outcome: 2 74) In testing for differences between the means of two paired populations, an appropriate null hypothesis would be: A) H0 : μD = 2 B) H0 : μD = 0 C) H0 : μD < 0 D) H0 : μD > 0 Answer: B Diff: 1 Keywords: paired, mean difference, null, hypothesis Section: 10-3 Interval Estimation and Hypothesis Tests for Paired Samples Outcome: 2

75) Most companies that make golf balls and golf clubs use a one-armed robot named "Iron Byron" to test their balls for length and accuracy, but because of swing variations by real golfers, these test robots don't always indicate how the clubs will perform in actual use. One company in the golfing industry is interested in testing its new driver to see if it has greater length off the tee than the best-selling driver. To do this, it has selected a group of golfers of differing abilities and ages. Its plan is to have each player use each of the two clubs and hit five balls. It will record the average length of the drives with each club for each player. The resulting data for a sample of 10 players are: Player 1 2 3 4 5 6 7 8 9 10

New Club 236.4 202.5 245.6 257.4 223.5 205.3 266.7 240 278.9 211.4

Leading Club 237.2 200.4 240.8 259.3 218.9 200.6 258.9 236.5 280.5 206.5

What is an appropriate null hypothesis to be tested? A) H0 : μ1 = μ2 B) H0 : μ1 ≥ μ2 C) H0 : μd ≤ 0 D) H0 : μD = 0 Answer: C Diff: 2 Keywords: null, hypothesis, mean, paired Section: 10-3 Interval Estimation and Hypothesis Tests for Paired Samples Outcome: 2

76) Most companies that make golf balls and golf clubs use a one-armed robot named "Iron Mike" to test their balls for length and accuracy, but because of swing variations by real golfers, these test robots don't always indicate how the clubs will perform in actual use. One company in the golfing industry is interested in testing its new driver to see if it has greater length off the tee than the best-selling driver. To do this, it has selected a group of golfers of differing abilities and ages. Its plan is to have each player use each of the two clubs and hit five balls. It will record the average length of the drives with each club for each player. The resulting data for a sample of 10 players are: Player 1 2 3 4 5 6 7 8 9 10

New Club 236.4 202.5 245.6 257.4 223.5 205.3 266.7 240 278.9 211.4

Leading Club 237.2 200.4 240.8 259.3 218.9 200.6 258.9 236.5 280.5 206.5

What is the critical value for the appropriate hypothesis test if the test is conducted using a 0.05 level of significance? A) z = 1.645 B) t = 1.7341 C) t = 1.8331 D) t = 2.2622 Answer: C Diff: 2 Keywords: critical value, hypothesis, t-test, means, paired Section: 10-3 Interval Estimation and Hypothesis Tests for Paired Samples Outcome: 2

77) Assume that 20 people participated in a speed reading class for 6 months. Each person's speed of reading words per minute both before and after the program is determined and recorded. The following summarizes the results for the 20 people: Mean weight lost = 9 pounds Sample standard deviation of weight lost = 4.6 pounds Assume that the hypothesis test will be conducted to determine whether or not the weight loss program is effective using a 0.05 level of significance. What is the value of the test statistic? A) t = 18.634 B) t = 1.96 C) z = 1.96 D) z = 6.19 Answer: A Diff: 2 Keywords: test statistic, means, hypothesis Section: 10-3 Interval Estimation and Hypothesis Tests for Paired Samples Outcome: 2

78) To increase productivity, workers went through a training program. The management wanted to know the effectiveness of the program. A sample of seven workers was taken and their daily production rates before and after the training are shown below. Worker 1 2 3 4 5 6 7

Before 18 23 25 22 20 21 19

After 22 25 27 25 24 19 20

Based on the data, the training program is: A) effective. B) ineffective. C) neither effective nor ineffective. D) None of the above Answer: A Diff: 3 Keywords: test statistic, paired difference Section: 10-3 Interval Estimation and Hypothesis Tests for Paired Samples Outcome: 2

79) An advertising company has developed a new ad for one of the national car manufacturing companies. The ad agency is interested in testing whether the proportion of favorable responses to the ad is the same between male adults versus female adults. It plans on conducting the test using an alpha level equal to 0.05. A sample of 100 adults of each gender will be used in the study. Each person will be asked to view the ad and indicate whether they find the ad to be "pleasing" or not. Given this, what is the appropriate null hypothesis? A) H0 : μ1 = μ2 B) H0 : p1 ≠ p2 C) H0 : 1 = 2 D) H0 : p1 = p2 Answer: D Diff: 1 Keywords: null, hypothesis, proportion Section: 10-4 Estimation and Hypothesis Tests for Two Population Proportions Outcome: 3 80) An advertising company has developed a new ad for one of the national car manufacturing companies. The ad agency is interested in testing whether the proportion of favorable response to the ad is the same between male adults versus female adults. It plans on conducting the test using an alpha level equal to 0.05. A sample of 100 adults of each gender will be used in the study. Each person will be asked to view the ad and indicate whether they find the ad to be "pleasing" or not. Given this information, what is the critical value? A) z = 1.645 B) t = 1.96 C) z = 1.96 D) Can't be determined without knowing the results of the sample. Answer: C Diff: 2 Keywords: proportion, critical value, z-test Section: 10-4 Estimation and Hypothesis Tests for Two Population Proportions Outcome: 3 81) An advertising company has developed a new ad for one of the national car manufacturing companies. The ad agency is interested in testing whether the proportion of favorable response to the ad is the same between male adults versus female adults. It plans on conducting the test using an alpha level equal to 0.05. A sample of 100 adults of each gender will be used in the study. Each person will be asked to view the ad and indicate whether they find the ad to be "pleasing" or not. The samples resulted in 57 males that liked the ad and 47 females that liked the ad. Based on this information, what is the value of the test statistic? A) z = 1.645 B) z = 1.42 C) t = 2.234 D) z = 1.024 Answer: B Diff: 2 Keywords: proportion, test statistic, z-value Section: 10-4 Estimation and Hypothesis Tests for Two Population Proportions Outcome: 3 10-32 Copyright © 2018 Pearson Education, Inc.

82) Suppose a survey is taken of two groups of people where each person is asked a yes/no question and the proportion of people who answer yes is calculated for each group. Which of the following is true about a hypothesis test of the difference in the two proportions? A) Normality can be assumed if the sample size for each population is at least 30. B) The t-distribution should be used if the standard deviations are unknown. C) The standard deviation must be assumed equal. D) Normality can be assumed if, in each group, at least 5 people say yes, and at 5 people say no. Answer: D Diff: 2 Keywords: proportion, difference, assumptions Section: 10-4 Estimation and Hypothesis Tests for Two Population Proportions Outcome: 3 83) There are a number of highly touted search engines for finding things of interest on the Internet. Recently a consumer rating system ranked two search engines ahead of the others. Now, a computer user's magazine wishes to make the final determination regarding which one is actually better at finding particular information. To do this, each search engine was used in an attempt to locate specific information using specified keywords. Both search engines were subjected to 100 queries. Search engine 1 successfully located the information 88 times and search engine 2 located the information 80 times. Using a significance level equal to 0.05, what is the null hypothesis to be tested? A) H0 : μ1 = μ2 B) H0 : p1 ≠ p2 C) H0 : 1 = 2 D) H0 : p1 = p2 Answer: B Diff: 1 Keywords: proportion, hypothesis, null Section: 10-4 Estimation and Hypothesis Tests for Two Population Proportions Outcome: 3 84) There are a number of highly touted search engines for finding things of interest on the Internet. Recently, a consumer rating system ranked two search engines ahead of the others. Now, a computer user's magazine wishes to make the final determination regarding which one is actually better at finding particular information. To do this, each search engine was used in an attempt to locate specific information using specified keywords. Both search engines were subjected to 100 queries. Search engine 1 successfully located the information 88 times and search engine 2 located the information 80 times. Using a significance level equal to 0.05, what is the critical value for the hypothesis test? A) z = ±196 B) z = 1.645 C) t = ±2.0456 D) z = ±2.575 Answer: A Diff: 2 Keywords: proportion, critical value, z-value Section: 10-4 Estimation and Hypothesis Tests for Two Population Proportions Outcome: 3 10-33 Copyright © 2018 Pearson Education, Inc.

85) There are a number of highly touted search engines for finding things of interest on the Internet. Recently, a consumer rating system ranked two search engines ahead of the others. Now, a computer user's magazine wishes to make the final determination regarding which one is actually better at finding particular information. To do this, each search engine was used in an attempt to locate specific information using specified keywords. Both search engines were subjected to 100 queries. Search engine 1 successfully located the information 88 times and search engine 2 located the information 80 times. Using a significance level equal to 0.05, which of the following is true? A) Based the sample data, the null hypothesis of equal population proportions is rejected since the test statistic exceeds the critical value. B) Based on the sample data, the null hypothesis should not be rejected since the test statistic of z = 2.04 exceeds the critical value of z = 1.96. C) According to the test results, the hypothesis should be rejected since the test statistic value, z = 1.54, falls in the rejection region. D) Based on the sample data, there is not sufficient evidence to conclude that a difference exists between the proportion of search hits since the test statistic, z = 1.54, does not fall in the rejection region. Answer: D Diff: 2 Keywords: proportion, test statistic, z-value Section: 10-4 Estimation and Hypothesis Tests for Two Population Proportions Outcome: 3 86) Suppose that two population proportions are being compared to test whether there is any difference between them. Assume that the test statistic has been calculated to be z = 2.21. Find the p-value for this situation. A) p-value = 0.0136 B) p-value = 0.4864 C) p-value = 0.0272 D) p-value = 0.9728 Answer: C Diff: 2 Keywords: proportion, p-value, hypothesis Section: 10-4 Estimation and Hypothesis Tests for Two Population Proportions Outcome: 3

87) A study was recently conducted at a major university to determine whether there is a difference in the proportion of business school graduates who go on to graduate school within five years after graduation and the proportion of non-business school graduates who attend graduate school. A random sample of 400 business school graduates showed that 75 had gone to graduate school while in a random sample of 500 non-business graduates, 137 had gone on to graduate school. Based on these sample data, and testing at the 0.10 level of significance, what is the value of the test statistic? A) Approximately z = 1.645 B) About z = -3.04 C) Approximately z = 3.45 D) About z = 1.96 Answer: B Diff: 2 Keywords: proportion, test statistic, z-value Section: 10-4 Estimation and Hypothesis Tests for Two Population Proportions Outcome: 3 88) A study was recently conducted at a major university to estimate the difference in the proportion of business school graduates who go on to graduate school within five years after graduation and the proportion of non-business school graduates who attend graduate school. A random sample of 400 business school graduates showed that 75 had gone to graduate school while in a random sample of 500 non-business graduates, 137 had gone on to graduate school. Based on a 95 percent confidence level, what is the upper limit of the confidence interval estimate? A) 0.2340 B) 0.1034 C) -0.031 D) -0.018 Answer: C Diff: 2 Keywords: proportion, confidence interval Section: 10-4 Estimation and Hypothesis Tests for Two Population Proportions Outcome: 3

89) The following information is based on independent random samples taken from two normally distributed populations having equal variances: n1 = 15

n2 = 13

1 = 50 s1 = 5

2 = 53 s2 = 6

Based on the sample information, determine the 90% confidence interval estimate for the difference between the two population means. A) -6.54 ≤ (µ1 - µ2) ≤ 0.54 B) -4.25 ≤ (µ1 - µ2) ≤ 1.25 C) -5.98 ≤ (µ1 - µ2) ≤ 1.88 D) -2.25 ≤ (µ1 - µ2) ≤ 5.25 Answer: A Diff: 2 Keywords: population, mean, sample Section: 10-1 Estimation for Two Population Means Using Independent Samples Outcome: 1 90) The following information is based on independent random samples taken from two normally distributed populations having equal variances: n1 = 24

n2 = 28

1 = 130 s1 = 19

2 = 125 s2 = 17.5

Based on the sample information, determine the 95% confidence interval estimate for the difference between the two population means. A) -6.23 < (µ1 - µ2) < 14.23 B) -4.81 < (µ1 - µ2) < 16.81 C) -5.17 < (µ1 - µ2) < 15.17 D) -3.25 < (µ1 - µ2) < 17.25 Answer: C Diff: 2 Keywords: population, mean, sample Section: 10-1 Estimation for Two Population Means Using Independent Samples Outcome: 1

91) Construct a 95% confidence interval estimate for the difference between two population means based on the following information: Population 1

Population 2

1 = 355 σ1 = 50

2 = 320 σ2 = 40

n1 = 50

n2 = 80

A) 25.41 ≤ (μ1 - μ2) ≤ 44.59 B) 35.41 ≤ (μ1 - μ2) ≤ 40.59 C) 15.741 ≤ (μ1 - μ2) ≤ 54.16 D) 22.13 ≤ (μ1 - μ2) ≤ 47.87 Answer: D Diff: 2 Keywords: population, mean, sample Section: 10-1 Estimation for Two Population Means Using Independent Samples Outcome: 1 92) A credit card company operates two customer service centers: one in Boise and one in Richmond. Callers to the service centers dial a single number, and a computer program routs callers to the center having the fewest calls waiting. As part of a customer service review program, the credit card center would like to determine whether the average length of a call (not including hold time) is different between the two centers. The managers of the customer service centers are willing to assume that the populations of interest are normally distributed with equal variances. Suppose a random sample of phone calls to the two centers is selected and the following results are reported:

Sample Size Sample Mean(seconds) Sample St. Dev. (seconds)

Boise 120 195 35.10

Richmond 135 216 37.80

Using the sample results, develop a 90% confidence interval estimate for the difference between the two population means. A) -29.3124 ≤ (µ1 - µ2) ≤ -18.6876 B) -24.2412 ≤ (µ1 - µ2) ≤ -17.7588 C) -26.2941 ≤ (µ1 - µ2) ≤ -11.8059 D) -28.5709 ≤ (µ1 - µ2) ≤ -13.4291 Answer: D Diff: 2 Keywords: population, mean, sample Section: 10-1 Estimation for Two Population Means Using Independent Samples Outcome: 1

93) A pet food producer manufactures and then fills 25-pound bags of dog food on two different production lines located in separate cities. In an effort to determine whether differences exist between the average fill rates for the two lines, a random sample of 19 bags from line 1 and a random sample of 23 bags from line 2 were recently selected. Each bag's weight was measured and the following summary measures from the samples are reported:

Sample Size, n Sample Mean, Sample Standard Deviation, s

Production Line 1 19

Production Line2 23

24.96 0.07

25.01 0.08

Management believes that the fill rates of the two lines are normally distributed with equal variances. Calculate the point estimate for the difference between the population means of the two lines. A) 0.040 B) 0.034 C) -0.050 D) -0.042 Answer: C Diff: 2 Keywords: population, mean, sample Section: 10-1 Estimation for Two Population Means Using Independent Samples Outcome: 1 94) A pet food producer manufactures and then fills 25-pound bags of dog food on two different production lines located in separate cities. In an effort to determine whether differences exist between the average fill rates for the two lines, a random sample of 19 bags from line 1 and a random sample of 23 bags from line 2 were recently selected. Each bag's weight was measured and the following summary measures from the samples are reported:

Sample Size, n Sample Mean, Sample Standard Deviation, s

Production Line 1 19

Production Line2 23

24.96 0.07

25.01 0.08

Management believes that the fill rates of the two lines are normally distributed with equal variances. Develop a 95% confidence interval estimate of the true mean difference between the two lines. A) -0.1412 ≤ (µ1 - µ2) ≤ -0.0912 B) -0.0974 ≤ (µ1 - µ2) ≤ -0.0026 C) -0.0231 ≤ (µ1 - µ2) ≤ -0.0069 D) -0.0812 ≤ (µ1 - µ2) ≤ -0.0188 Answer: B Diff: 2 Keywords: population, mean, sample Section: 10-1 Estimation for Two Population Means Using Independent Samples Outcome: 1 10-38 Copyright © 2018 Pearson Education, Inc.

95) A decision maker wishes to test the following null and alternative hypotheses using an alpha level equal to 0.05: H0 : μ1 - μ2 = 0 HA : μ1 - μ2 ≠ 0 The population standard deviations are assumed to be known. After collecting the sample data, the test statistic is computed to be z = 1.78 Using the test statistic approach, what conclusion should be reached about the null hypothesis? A) Because z = 2.40 > 1.96, we reject the null hypothesis. B) Because z = 2.03 > 1.96, we reject the null hypothesis. C) Because z = 1.78 < 1.96, we do not reject the null hypothesis. D) Because z = 1.45 < 1.645, we do not reject the null hypothesis. Answer: C Diff: 2 Keywords: hypothesis test, population mean Section: 10-2 Hypothesis Tests for Two Population Means Using Independent Samples Outcome: 1 96) A decision maker wishes to test the following null and alternative hypotheses using an alpha level equal to 0.05: H0 : μ1 - μ2 = 0 HA : μ1 - μ2 ≠ 0 The population standard deviations are assumed to be known. After collecting the sample data, the test statistic is computed to be z = 1.78 Using the p-value approach, what decision should be reached about the null hypothesis? A) Since p-value = 0.0018 < α/2 = 0.025, reject the null hypothesis B) Since p-value = 0.0415 > α/2 = 0.025, do not reject the null hypothesis. C) Since p-value = 0.0033 < α/2 = 0.025, reject the null hypothesis. D) Since p-value = 0.0375 > α/2 = 0.025, do not reject the null hypothesis. Answer: D Diff: 2 Keywords: hypothesis test, population mean Section: 10-2 Hypothesis Tests for Two Population Means Using Independent Samples Outcome: 1

97) Given the following null and alternative hypotheses H0 : μ1 ≥ μ2 HA : μ1 < μ2 Together with the following sample information Sample 1 n1 = 14

Sample 2 n2 = 18

1 = 565 s1 = 28.9

2 = 578 s2 = 26.3

Assuming that the populations are normally distributed with equal variances, test at the 0.10 level of significance whether you would reject the null hypothesis based on the sample information. Use the test statistic approach. A) Because the calculated value of t = -1.415 is less than the critical value of t=-1.3104, reject the null hypothesis. Based on these sample data, at the α = 0.10 level of significance there is sufficient evidence to conclude that the mean for population 1 is less than the mean for population 2. B) Because the calculated value of t = -1.329 is less than the critical value of t=-1.3104, reject the null hypothesis. Based on these sample data, at the α = 0.10 level of significance there is sufficient evidence to conclude that the mean for population 1 is less than the mean for population 2. C) Because the calculated value of t = -0.429 is not less than the critical value of t=-1.3104, do not reject the null hypothesis. Based on these sample data, at the α = 0.10 level of significance there is not sufficient evidence to conclude that the mean for population 1 is less than the mean for population 2. D) Because the calculated value of t = -0.021 is not less than the critical value of t=-1.3104, do not reject the null hypothesis. Based on these sample data, at the α = 0.10 level of significance there is not sufficient evidence to conclude that the mean for population 1 is less than the mean for population 2. Answer: B Diff: 2 Keywords: hypothesis test, population mean Section: 10-2 Hypothesis Tests for Two Population Means Using Independent Samples Outcome: 1

98) Given the following null and alternative hypotheses H0 : μ1 ≥ μ2 HA : μ1 < μ2 Together with the following sample information Sample 1 n1 = 14

Sample 2 n2 = 18

1 = 565 s1 = 28.9

2 = 578 s2 = 26.3

Assuming that the populations are normally distributed with equal variances, test at the 0.05 level of significance whether you would reject the null hypothesis based on the sample information. Use the test statistic approach. A) Because the calculated value of t = -2.145 is less than the critical value of t = -1.6973, reject the null hypothesis. Based on these sample data, at the α = 0.05 level of significance there is sufficient evidence to conclude that the mean for population 1 is less than the mean for population 2. B) Because the calculated value of t = -1.814 is less than the critical value of t = -1.6973, reject the null hypothesis. Based on these sample data, at the α = 0.05 level of significance there is sufficient evidence to conclude that the mean for population 1 is less than the mean for population 2. C) Because the calculated value of t = -1.329 is not less than the critical value of t = -1.6973, do not reject the null hypothesis. Based on these sample data, at the α = 0.05 level of significance there is not sufficient evidence to conclude that the mean for population 1 is less than the mean for population 2. D) Because the calculated value of t = -1.415 is not less than the critical value of t = -1.6973, do not reject the null hypothesis. Based on these sample data, at the α = 0.05 level of significance there is not sufficient evidence to conclude that the mean for population 1 is less than the mean for population 2. Answer: C Diff: 2 Keywords: hypothesis test, population mean Section: 10-2 Hypothesis Tests for Two Population Means Using Independent Samples Outcome: 1

99) Given the following null and alternative hypotheses H0 : μ1 - μ2 = 0 HA : μ1 - μ2 ≠ 0 Together with the following sample information Sample 1 n1 = 125

Sample 2 n2 = 120

s1 = 31

s2 = 38

1 = 130

2 = 105

Develop the appropriate decision rule, assuming a significance level of 0.05 is to be used. A) If t > 1.9698 or t < -1.9698 reject H0, otherwise do not reject H0 B) If t > 1.5412 or t < -1.5412 reject H0, otherwise do not reject H0 C) If t > 1.8157 or t < -1.8157 reject H0, otherwise do not reject H0 D) If t > 2.0124 or t < -2.0124 reject H0, otherwise do not reject H0 Answer: A Diff: 2 Keywords: hypothesis test, population mean Section: 10-2 Hypothesis Tests for Two Population Means Using Independent Samples Outcome: 1 100) Given the following null and alternative hypotheses H0 : μ1 - μ2 = 0 HA : μ1 - μ2 ≠ 0 Together with the following sample information Sample 1 n1 = 125

Sample 2 n2 = 120

s1 = 31

s2 = 38

1 = 130

2 = 105

Test the null hypothesis and indicate whether the sample information leads you to reject or fail to reject the null hypothesis, assuming a significance level of 0.05 is to be used. Use the test statistic approach. A) Since 0.812 < 1.9698 reject H0 B) Since 1.041 < 1.9698 reject H0 C) Since 5.652 > 1.9698 reject H0 D) Since 4.418 > 1.9698 reject H0 Answer: C Diff: 2 Keywords: hypothesis test, population mean Section: 10-2 Hypothesis Tests for Two Population Means Using Independent Samples Outcome: 1 10-42 Copyright © 2018 Pearson Education, Inc.

101) Descent, Inc., produces a variety of climbing and mountaineering equipment. One of its products is a traditional three-strand climbing rope. An important characteristic of any climbing rope is its tensile strength. Descent produces the three-strand rope on two separate production lines: one in Bozeman and the other in Challis. The Bozeman line has recently installed new production equipment. Descent regularly tests the tensile strength of its ropes by randomly selecting ropes from production and subjecting them to various tests. The most recent random sample of ropes, taken after the new equipment was installed at the Bozeman plant, revealed the following: Bozeman

Challis

1 = 7,200 lb s1 = 425

2 = 7,087 lb s2 = 415

n1 = 25

n2 = 20

Descent's production managers are willing to assume that the population of tensile strengths for each plant is approximately normally distributed with equal variances. Based on the sample results, can Descent's managers conclude that there is a difference between the mean tensile strengths of ropes produced in Bozeman and Challis? Conduct the appropriate hypothesis test at the 0.05 level of significance. A) Because the calculated value of t = 0.896 is neither less than the lower tail critical value of t = -2.0167, nor greater than the upper tail critical value of t = 2.0167, do not reject the null hypothesis. Based on these sample data, at the α = 0.05 level of significance there is not sufficient evidence to conclude that the average tensile strength of ropes produced at the two plants is different. B) Because the calculated value of t = 0.451 is neither less than the lower tail critical value of t = -2.0167, nor greater than the upper tail critical value of t = 2.0167, do not reject the null hypothesis. Based on these sample data, at the α = 0.05 level of significance there is not sufficient evidence to conclude that the average tensile strength of ropes produced at the two plants is different. C) Because the calculated value of t = -2.8126 is less than the lower tail critical value of t = -2.0167, reject the null hypothesis. Based on these sample data, at the α = 0.05 level of significance there is sufficient evidence to conclude that the average tensile strength of ropes produced at the two plants is different. D) Because the calculated value of t = 2.8126 is greater than the lower tail critical value of t = -2.0167, reject the null hypothesis. Based on these sample data, at the α = 0.05 level of significance there is sufficient evidence to conclude that the average tensile strength of ropes produced at the two plants is different. Answer: A Diff: 2 Keywords: hypothesis test, population mean Section: 10-2 Hypothesis Tests for Two Population Means Using Independent Samples Outcome: 1

102) The management of the Seaside Golf Club regularly monitors the golfers on its course for speed of play. Suppose a random sample of golfers was taken in 2011 and another random sample of golfers was selected in 2006. The results of the two samples are as follows: 2011

2012

1 = 225 s1 = 20.25

2 = 219 s2 = 21.70

n1 = 36

n2 = 31

Based on the sample results, can the management of the Seaside Golf Club conclude that average speed of play was different in 2012 than in 2011? Conduct the appropriate hypothesis test at the 0.10 level of significance. Assume that the management of the club is willing to accept the assumption that the populations of playing times for each year are approximately normally distributed with equal variances. A) Because the calculated value of t = -2.03 is less than the lower tail critical value of t = - 1.6686, reject the null hypothesis. Based on these sample data, at the α = 0.10 level of significance there is sufficient evidence to conclude that the average speed of play is different in 2012 than in 2011. B) Because the calculated value of t = 1.84 is greater than the upper tail critical value of t = 1.6686, reject the null hypothesis. Based on these sample data, at the α = 0.10 level of significance there is sufficient evidence to conclude that the average speed of play is different in 2012 than in 2011. C) Because the calculated value of t = 0.89 is neither less than the lower tail critical value of t = - 1.6686, nor greater than the upper tail critical value of t = 1.6686, do not reject the null hypothesis. Based on these sample data, at the α = 0.10 level of significance there is not sufficient evidence to conclude that the average speed of play is different in 2012 than in 2011. D) Because the calculated value of t = 1.17 is neither less than the lower tail critical value of t = - 1.6686, nor greater than the upper tail critical value of t = 1.6686, do not reject the null hypothesis. Based on these sample data, at the α = 0.10 level of significance there is not sufficient evidence to conclude that the average speed of play is different in 2012 than in 2011. Answer: D Diff: 2 Keywords: hypothesis test, population mean Section: 10-2 Hypothesis Tests for Two Population Means Using Independent Samples Outcome: 1

103) The following paired samples have been obtained from normally distributed populations. Construct a 90% confidence interval estimate for the mean paired difference between the two population means. Sample # 1 2 3 4 5 6 7

Population 1 3,693 3,679 3,921 4,106 3,808 4,394 3,878

Population 2 4,635 4,262 4,293 4,197 4,536 4,494 4,094

A) -571.92 ≤ µ ≤ -172.51 B) -487.41 ≤ µ ≤ -283.89 C) -812.21 ≤ µ ≤ -72.61 D) -674.41 ≤ µ ≤ -191.87 Answer: D Diff: 2 Keywords: population, mean, sample Section: 10-1 Estimation for Two Population Means Using Independent Samples Outcome: 1 104) You are given the following results of a paired-difference test: = -4.6 sd = 0.25 n = 16 Construct a 99% confidence interval estimate for the paired difference in mean values. A) -2.912 -------- -2.718 B) -4.784 -------- -4.416 C) -5.241 -------- -4.971 D) -3.141 -------- -2.812 Answer: B Diff: 2 Keywords: population, mean, sample Section: 10-1 Estimation for Two Population Means Using Independent Samples Outcome: 1

105) You are given the following results of a paired-difference test: = -4.6 sd = 0.25 n = 16 Construct a 90% confidence interval estimate for the paired difference in mean values. A) -2.86 ------- -2.53 B) -5.72 ------- -5.18 C) -3.81 ------- -3.71 D) -4.71 ------- -4.49 Answer: D Diff: 2 Keywords: population, mean, sample Section: 10-1 Estimation for Two Population Means Using Independent Samples Outcome: 1 106) A paired sample study has been conducted to determine whether two populations have equal means. Twenty paired samples were obtained with the following sample results: Based on these sample data and a significance level of 0.05, what conclusion should be made about the population means? A) Because t = 5.06 > 2.0930, reject the null hypothesis. B) Because t = 3.41 > 2.0930, reject the null hypothesis. C) Because t = 1.82 < 2.0930, do not reject the null hypothesis. D) Because t = 2.02 < 2.0930, do not reject the null hypothesis. Answer: A Diff: 2 Keywords: hypothesis test, population mean Section: 10-2 Hypothesis Tests for Two Population Means Using Independent Samples Outcome: 1

107) The following samples are observations taken from the same elements at two different times: Unit 1 2 3 4 5 6

Sample 1 15.1 12.3 14.9 17.5 18.1 18.4

Sample 2 4.8 5.7 6.2 9.4 2.3 4.7

Assume that the populations are normally distributed and construct a 90% confidence interval for the difference in the means of the distribution at the times in which the samples were taken. A) (7.6232, 13.4434) B) (5.2825, 15.8127) C) (6.8212, 14.7821) D) (7.4122, 14.6801) Answer: A Diff: 2 Keywords: interval estimation, hypothesis test, paired samples Section: 10-3 Interval Estimation and Hypothesis Tests for Paired Samples Outcome: 1 108) The following samples are observations taken from the same elements at two different times: Unit 1 2 3 4 5 6

Sample 1 15.1 12.3 14.9 17.5 18.1 18.4

Sample 2 4.8 5.7 6.2 9.4 2.3 4.7

Perform a test of hypothesis to determine if the difference in the means of the distribution at the first time period is 10 units larger than at the second time period. Use a level of significance equal to 0.10. A) Because t = 1.98 > 2.0150, the null hypothesis must be rejected. B) Because t = 1.67 > 2.0150, the null hypothesis must be rejected C) Because t = 1.02 < 2.0150, the null hypothesis cannot be rejected. D) Because t = 0.37 < 2.0150, the null hypothesis cannot be rejected. Answer: D Diff: 2 Keywords: interval estimation, hypothesis test, paired samples Section: 10-3 Interval Estimation and Hypothesis Tests for Paired Samples Outcome: 1

109) In an article entitled "Childhood Pastimes Are Increasingly Moving Indoors," Dennis Cauchon asserts that there have been huge declines in spontaneous outdoor activities such as bike riding, swimming, and touch football. In the article, he cites separate studies by the national Sporting Goods Association and American Sports Data that indicate bike riding alone is down 31% from 1995 to 2004. According to the surveys, 68% of 7- to 11-year-olds rode a bike at least six times in 1995 and only 47% did in 2004. Assume the sample sizes were 1,500 and 2,000, respectively. Calculate a 95% confidence interval to estimate the proportion of 7- to 11-year-olds who rode their bike at least six times in 2004. A) (0.4481, 0.4919) B) (0.4324, 0.4676) C) (0.4021, 0.5179) D) (0.4712, 0.4888). Answer: A Diff: 3 Keywords: population, mean, sample Section: 10-1 Estimation for Two Population Means Using Independent Samples Outcome: 1 110) Suppose as part of a national study of economic competitiveness a marketing research firm randomly sampled 200 adults between the ages of 27 and 35 living in metropolitan Seattle and 180 adults between the ages of 27 and 35 living in metropolitan Minneapolis. Each adult selected in the sample was asked, among other things, whether they had a college degree. From the Seattle sample 66 adults answered yes and from the Minneapolis sample 63 adults answered yes when asked if they had a college degree. Based on the sample data, can we conclude that there is a difference between the population proportions of adults between the ages of 27 and 35 in the two cities with college degrees? Use a level of significance of 0.10 to conduct the appropriate hypothesis test. A) Since the test statistic, 1.8214, is greater than the critical value of 1.645, reject the null hypothesis and conclude that there is a higher proportion of Seattle adults that have a college degree B) Since the test statistic, 2.0112, is greater than the critical value of 1.645, reject the null hypothesis and conclude that there is a higher proportion of Seattle adults that have a college degree. C) Since the test statistic, 0.7001, is not greater than the critical value of 1.645, do not reject the null hypothesis and conclude that there is not a higher proportion of Seattle adults that have a college degree. D) Since the test statistic, 0.8921, is not greater than the critical value of 1.645, do not reject the null hypothesis and conclude that there is not a higher proportion of Seattle adults that have a college degree. Answer: C Diff: 2 Keywords: hypothesis test, population mean Section: 10-2 Hypothesis Tests for Two Population Means Using Independent Samples Outcome: 1

111) The American College Health Association produced the National College Health Assessment (Andy Gardiner, "Surfacing from Depression," February 6, 2006). The assessment indicates that the percentage of U.S. college students who report having been diagnosed with depression has risen from 2000. The assessment surveyed 47,202 students at 74 campuses. It discovered that 10.3% and 14.9% of students indicated that they had been diagnosed with depression in 2000 and 2004, respectively. Assume that half of the students surveyed were surveyed in 2004. Conduct a hypothesis test to determine if there has been more than a 0.04 increase in the proportion of students who indicated they have been diagnosed with depression. Use a significance level of 0.05 and a p-value approach to this test. A) Since p-value = 0.065 > 0.05, do not reject H0. There is not sufficient evidence to conclude that there has been more than a 0.04 increase in the proportion of students that indicate they have been diagnosed with depression. B) Since p-value = 0.025 < 0.05, reject H0. There is sufficient evidence to conclude that there has been more than a 0.04 increase in the proportion of students that indicate they have been diagnosed with depression. C) Since p-value = 0.072 < 0.05, reject H0. There is sufficient evidence to conclude that there has been more than a 0.04 increase in the proportion of students that indicate they have been diagnosed with depression. D) Since p-value = 0.071 > 0.05, do not reject H0. There is not sufficient evidence to conclude that there has been more than a 0.04 increase in the proportion of students that indicate they have been diagnosed with depression. Answer: B Diff: 2 Keywords: hypothesis test, population mean Section: 10-2 Hypothesis Tests for Two Population Means Using Independent Samples Outcome: 1 112) The American College Health Association produced the National College Health Assessment (Andy Gardiner, "Surfacing from Depression," February 6, 2006). The assessment indicates that the percentage of U.S. college students who report having been diagnosed with depression has risen from 2000. The assessment surveyed 47,202 students at 74 campuses. It discovered that 10.3% and 14.9% of students indicated that they had been diagnosed with depression in 2000 and 2004, respectively. Assume that half of the students surveyed were surveyed in 2004. Indicate the margin of error for estimating p1 - p2 with 1 - 2. A) 0.04156 B) 0.00121 C) 0.03418 D) 0.00597 Answer: D Diff: 2 Keywords: hypothesis test, population mean Section: 10-2 Hypothesis Tests for Two Population Means Using Independent Samples Outcome: 1

113) A major U.S. oil company has developed two blends of gasoline. Managers are interested in estimating the difference in mean gasoline mileage that will be obtained from using the two blends. As part of their study, they have decided to run a test using the Chevrolet Impala automobile with automatic transmissions. They selected a random sample of 100 Impalas using Blend 1 and another 100 Impalas using Blend 2. Each car was first emptied of all the gasoline in its tank and then filled with the designated blend of the new gasoline. The car was then driven 200 miles on a specified route involving both city and highway roads. The cars were then filled and the actual miles per gallon were recorded. The following summary data were recorded:

Sample Size Sample Mean Sample St. Dev.

Blend 1 100 23.4 mpg 4.0 mpg

Blend 2 100 25.7 mpg 4.2 mpg

Based on these sample data, compute and interpret the 95 percent confidence interval estimate for the difference in mean mpg for the two blends. Answer: The general format for a confidence interval estimate is: Point estimate ± critical value (standard error). In this case, the point estimate is 1 - 2 = 23.4 - 25.7 = -2.3 mpg. This value forms the center of the interval estimate. To complete the confidence interval, we need to compute the standard error. Since the sample sizes in this example are large, we compute the standard error using:

sc =

= 0.58

The critical value for a 95 percent confidence interval based on large samples is z = 1.96 from the standard normal table. Therefore, the 95 percent confidence interval estimate for the difference in population means is:

1- 2±z (23.4 - 25.7) ± 1.96

. Substituting we get which gives -2.3 ± 1.96(.58) or -2.3 ± 1.1368.

Thus we get: -3.4368 mpg to -1.1632 mpg. We interpret this to mean that based on the sample information, with 95 percent confidence, Impalas using Blend 1 will get between 1.1632 and 3.4368 mpg less than Impalas using Blend 2. Diff: 2 Keywords: confidence interval, mean difference Section: 10-1 Estimation for Two Population Means Using Independent Samples Outcome: 1

114) A major U.S. oil company has developed two blends of gasoline. Managers are interested in determining whether a difference in mean gasoline mileage will be obtained from using the two blends. As part of their study, they have decided to run a test using the Chevrolet Impala automobile with automatic transmissions. They selected a random sample of 100 Impalas using Blend 1 and another 100 Impalas using Blend 2. Each car was first emptied of all the gasoline in its tank and then filled with the designated blend of the new gasoline. The car was then driven 200 miles on a specified route involving both city and highway roads. The cars were then filled and the actual miles per gallon were recorded. The following summary data were recorded:

Sample Size Sample Mean Sample St. Dev.

Blend 1 100 23.4 mpg 4.0 mpg

Blend 2 100 25.7 mpg 4.2 mpg

Based on the sample data, using a 0.05 level of significance, what conclusion should the company reach about whether the population mean mpg is the same or different for the two blends? Use the test statistic approach to test the null hypothesis. Answer: The hypothesis test involves a test for the difference between population means and is based on large samples from each population. The appropriate null and alternative hypotheses are: H0 : μ1 = μ2 Ha : μ1 ≠ μ2 This will be a two-tailed test since we are interested in testing whether there is a difference between the two population means with respect to miles per gallon and neither blend of gasoline is predicted to be superior to the other. The critical value for a two-tailed test with a significance level of 0.05 is found in the Standard Normal table to be z = ±1.96. The test statistic in this case is computed using: z=

= -3.9655

Since z = -3.9655 < -1.96, we reject the null hypothesis and conclude that the population means are not equal. Based on the samples, we infer that Blend 2 will provide higher mean mileage. Diff: 3 Keywords: mean difference, test statistic, null hypothesis Section: 10-2 Hypothesis Tests for Two Population Means Using Independent Samples Outcome: none

115) A major U.S. oil company has developed two blends of gasoline. Managers are interested in determining whether a difference in mean gasoline mileage will be obtained from using the two blends. As part of their study, they have decided to run a test using the Chevrolet Impala automobile with automatic transmissions. They selected a random sample of 100 Impalas using Blend 1 and another 100 Impalas using Blend 2. Each car was first emptied of all the gasoline in its tank and then filled with the designated blend of the new gasoline. The car was then driven 200 miles on a specified route involving both city and highway roads. The cars were then filled and the actual miles per gallon were recorded. The following summary data were recorded:

Sample Size Sample Mean Sample St. Dev.

Blend 1 100 23.4 mpg 4.0 mpg

Blend 2 100 25.7 mpg 4.2 mpg

Based on the sample data, using a 0.05 level of significance, what conclusion should the company reach about whether the population mean mpg is the same or different for the two blends? Use the p-value approach to test the null hypothesis. Answer: The hypothesis test involves a test for the difference between population means and is based on large samples from each population. The appropriate null and alternative hypotheses The appropriate null and alternative hypotheses are: H0 : μ1 = μ2 Ha: μ1 ≠ μ2 This will be a two-tailed test since we are interested in testing whether there is a difference between the two population means with respect to miles per gallon and neither blend of gasoline is predicted to be superior to the other. To obtain the p-value, we first compute the test statistic. The test statistic in this case is computed using: z=

= -3.9655.

From the standard normal table, the probability associated with a z = -3.9655 value is essentially 0.5000. This is the area between z = 3.9655 and the mean of the standard normal distribution. We want the area in the tail of the distribution. Thus the p-value for this test is 0.5000 - 0.5000 ≈ 0.0000. This value can be compared with α/2 = 0.025. Since 0.0000 < 0.025, we reject the null hypothesis and conclude that the population means are not equal. Diff: 3 Keywords: p-value, hypothesis, null, mean difference Section: 10-2 Hypothesis Tests for Two Population Means Using Independent Samples Outcome: none

116) A PC company uses two suppliers for rechargeable batteries for its notebook computers. Two factors are important quality features of the batteries: mean use time and variation. It is desirable that the mean use time be high and the variability be low. Recently, the PC maker conducted a test on batteries from the two suppliers. In the test, 9 randomly selected batteries from Supplier 1 were tested and 12 randomly selected batteries from Supplier 2 were tested. The following results were observed: Supplier 1 n1 = 9

Supplier 2 n2 = 12

1 = 67.25 min. S1= 11.2 min

2 = 72.4 min. S2 = 9.9 min

Based on these sample results, can the PC maker conclude that a difference exists between the two batteries with respect to the population mean use time? Test using a 0.10 level of significance. Answer: This test calls for us to test whether the population means are equal. Since the sample sizes are small and the population variances are unknown, the t-distribution should be employed if we assume that the populations are normally distributed and the populations have equal variances. We can test the following null and alternative hypotheses: H0 : μ1 = μ2 Ha: μ1 ≠ μ2 The critical value for a t-test for a two-tailed test is found from the t-distribution with degrees of freedom equal to n1 + n2 - 2 = 9 + 12 - 2 = 19. Given a 0.10 significance level, the critical value from the tdistribution with degrees of freedom equal to 19 is equal to ±1.7291. The test statistic is computed using: t=

where Sp is the pooled standard deviation and is computed as:

sp =

= 10.4671

Then the test statistic is: t=

= -1.1158

Since t = -1.1158 > -1.7291, we do not reject the null hypothesis. Thus, based on the sample data, we do not have evidence to indicate that the two batteries differ in mean use time. Diff: 2 Keywords: t-test, mean difference, independent Section: 10-2 Hypothesis Tests for Two Population Means Using Independent Samples Outcome: none

117) A real estate agent believes that home with swimming pools take longer to sell than home without swimming pools. A random sample of each type of recently sold homes was taken where the number of days on the market is recorded. Results are:

Sample mean Sample Standard deviation Sample size

Homes with pool 77 days 8.4 days 19

Homes w/o pool 65 days 7.2 days 23

Assuming that the populations are normally distributed and the variances are equal, conduct the appropriate hypothesis test to determine if the real estate agent is correct. Use the 0.05 level of significance. Answer: The agent believes that homes with pools have a higher average number of days on the market. Let μ1 =the mean days for homes with pools and μ2 = the mean days for homes without pools. Then the hypotheses are: H0 : μ1 - μ2 ≤ 0 Ha : μ1 - u2 > 0 Since we have sample standard deviations we need to use the t-distribution with (19+23-2) = 40 degrees of freedom. This is a one-tailed test with the rejection region of size 0.05 in the upper tail. So the critical value is t = 1.6839 and we will reject the null if the test statistic exceeds 1.6839. The test statistic is computed using: t=

where Sp is the pooled standard deviation and is computed as:

sp =

= 7.763

Then the test statistic is: t=

= 4.986

Since 4.986 > 1.6839 we reject the null hypothesis and can conclude that the alternate hypothesis is correct, which says that homes with pools have a significantly higher number of average days on the market. So the real estate agent is correct. Diff: 3 Keywords: t-test, mean difference, independent, pooled Section: 10-2 Hypothesis Tests for Two Population Means Using Independent Samples Outcome: none

118) The National Football League (NFL) is interested in testing to see whether there is a difference in the proportion of male fans that prefer instant replay to review officials' calls and the proportion of female fans that prefer instant replay. It is believed that males tend to favor the practice to a higher degree than do females. To test this, random samples of male fans and female fans were selected and the following results were obtained:

Sample size Number Preferring

Male Fans 300 234

Female Fans 100 57

Using a significance level equal to 0.05, what conclusion should be reached based on the sample data? Answer: The parameter of interest in this situation is the difference in population proportions. The research hypothesis states that the proportion of males favoring instant replay exceeds the proportion of females who favor it. Thus, we can formulate the following null and alternative hypotheses: H0 : p1 ≤ p2 Hα: p1 > p2 This means that we are dealing with a one-tailed hypothesis test. The critical value for a one-tailed test for the difference between population proportions with a 0.05 significance level is z = 1.645 from the standard normal distribution. The test statistic is computed using: z=

where = 1=

= =

= .7275 and = .78 and 2 =

= .57

Then, the test statistic is: z=

= 4.0846

Since 4.0846 > 1.645, we reject the null hypothesis and conclude that based on the sample data, the male fans do favor instant replay by a higher proportion than do female fans. Diff: 2 Keywords: proportion, hypothesis, null, alternative Section: 10-4 Estimation and Hypothesis Tests for Two Population Proportions Outcome: 3

119) A maker of toothpaste is interested in testing whether the proportion of adults (over age 18) who use its toothpaste and have no cavities within a six-month period is any different from the proportion of children (18 and under) who use the toothpaste and have no cavities within a six-month period. To test this, it has selected a sample of adults and a sample of children randomly from the population of those customers who use their toothpaste. The following results were observed.

Sample size Number with 0 cavities

Adults 100 83

Children 200 165

Based on these sample data and using a significance level of 0.05, what conclusion should be reached? Use the p-value approach to conduct the test. Answer: The parameter of interest in this situation is the difference in population proportions. Since we are interested in whether there is a difference in the proportion of adults who get cavities versus children, we can formulate the following null and alternative hypotheses: H0 : p1 = p2 Hα: p1 ≠ p2 This means that we are dealing with a two-tailed hypothesis test. The critical value for a two-tailed test for the difference between population proportions with a 0.05 significance level is z=1.96 from the standard normal distribution. The test statistic is computed using:

where =

= .83 and 2 =

= .8267 and

= .8250

Then, the test statistic is: z=

= .1079

To find the p-value, we go to the standard normal table for z = 0.11 getting 0.0438. Subtracting 0.0438 from 0.5000 gives 0.4562, which is the p-value. This value should be compared to alpha/2 = 0.025. Since 0.4562 > 0.025, we do not reject the null hypothesis and conclude that based on the sample data, there is no basis for concluding that the proportion of adults with no cavities is any different from the proportion of children with no cavities. Diff: 3 Keywords: proportion, p-value, hypothesis Section: 10-4 Estimation and Hypothesis Tests for Two Population Proportions Outcome: 3

120) A market research firm has come up with two coupon designs that could be inserted into the envelopes that go out with the monthly statement to credit card customers. The coupons offer the customer an opportunity to become a member of a travel club. The research firm is interested in estimating the difference in proportion of customers who will join the club after receiving one or the other of the two coupons. To obtain this estimate, the research firm has sent out coupon design 1 to a random sample of 100 customers. A second random sample of 100 customers received coupon 2. For the first coupon, 11 customers joined the travel club, while 15 customers who received coupon 2 joined. Based upon this sample information, develop and interpret the desired 95 percent confidence interval estimate. Answer: The parameter of interest in this situation is the difference in population proportions. The general format for any confidence interval estimate is: Point estimate ± critical value (standard error). We are given the following sample information: 1=

= .11 and 2 =

= .15

In this case, the point estimate will be the difference between the two sample proportions which is: 1- 2 = .11 - .15 = -.04. The standard error for the difference in proportions is: =

= 0.0475

The critical value for an estimate for the difference in population proportions is a z-value from the standard normal table. In this case, given a 95 percent confidence level, the critical value is z=1.96. The confidence interval estimate is found as: 1- 2±z Substituting values, we get: -.04 ± 1.96 -.04 ± 0.0931. Thus, the 95 percent confidence interval is: -0.133 to +0.0531. This indicates that we estimate the difference in proportions of redeemed coupons to be between -0.1331 and +0.0531. Thus, the range of possibilities for the difference in population proportions ranges from over 0.13 more for coupon 2 versus coupon 1 all the way to over 0.05 more for coupon 1 versus coupon 2. Therefore, since the interval crosses zero, we are unable to conclude that a difference exists between the two coupons with respect to the proportions that will be redeemed. Diff: 2 Keywords: proportion, confidence interval, difference Section: 10-4 Estimation and Hypothesis Tests for Two Population Proportions Outcome: 3 10-57 Copyright © 2018 Pearson Education, Inc.

121) When estimating the difference between two population means, when should the normal distribution be used and when should the t-distribution be used? Answer: If the two standard deviations are known, then the normal distribution should be used. In most real situations the standard deviations are not know and sample standard deviations must be used. In this case the t-distribution should be used. However, as the number of degrees of freedom increases, the t-distribution approaches the normal distribution. So if the sample sizes are large enough the normal distribution can be used because there will be very little difference between the two. A sample size of n > 30 is often used a rule of thumb for a sufficiently large sample. Diff: 2 Keywords: population mean, mean difference, normal distribution, t-distribution Section: 10-1 Estimation for Two Population Means Using Independent Samples Outcome: 1

Business Statistics, 9e (Groebner/Shannon/Fry) Chapter 11 Hypothesis Tests and Estimation for Population Variances 1) The test statistic that is used when testing a null hypothesis for a population variance is the standard normal z-value. Answer: FALSE Diff: 1 Keywords: variance, test statistic, chi-square Section: 11-1 Hypothesis Tests and Estimation for a Single Population Variance Outcome: 1 2) A one-tailed hypothesis test for a population variance always has the rejection region in the upper tail. Answer: TRUE Diff: 1 Keywords: hypothesis, variance, chi-square Section: 11-1 Hypothesis Tests and Estimation for a Single Population Variance Outcome: 1 3) A contract calls for the strength of a steel rod to stand up to pressure of 200 lbs per square inch on average. The contract also requires that the variability in strength for individual steel rods be no more than 5 pounds per square inch. If a random sample of n = 15 rods is selected and the sample standard deviation is 6.7 pounds, the test statistic is approximately χ2 = 25.138. Answer: TRUE Diff: 2 Keywords: test statistic, chi-square, variance Section: 11-1 Hypothesis Tests and Estimation for a Single Population Variance Outcome: 1 4) A machine that is used to fill soda pop cans with pop has an adjustable mean fill setting, but the standard deviation is not supposed to exceed 0.18 ounces. To make sure that this is the case, the managers at the beverage company each day select a random sample of n = 6 cans and measure the fill volume carefully. In one such case, the following data (ounces per can) were observed. 12.29

11.88

12.03

12.22

11.76

11.98

Based on these sample data, the test statistic is approximately χ2 = 5.01. Answer: FALSE Diff: 3 Keywords: test statistic, chi-square, variance Section: 11-1 Hypothesis Tests and Estimation for a Single Population Variance Outcome: 1

5) If we are interested in performing a one-tailed, upper-tail hypothesis test about a population variance where the level of significance is .01 and the sample size is n = 25, the critical chi-square value to be used is 42.9798. Answer: TRUE Diff: 2 Keywords: chi-square, critical value Section: 11-1 Hypothesis Tests and Estimation for a Single Population Variance Outcome: 1 6) The variance in the diameter of a bolt should not exceed 0.500 mm. A random sample of n = 12 bolts showed a sample variance of 0.505 mm. The test statistic is χ2 = 11.11. Answer: TRUE Diff: 2 Keywords: chi square, test statistic, hypothesis Section: 11-1 Hypothesis Tests and Estimation for a Single Population Variance Outcome: 1 7) Assume a sample of size n = 12 has been collected. To perform a hypothesis test of a population variance using a 0.05 level of significance, where the null hypothesis is: H0 : σ2 = 25 The upper tail critical value is 21.92. Answer: TRUE Diff: 2 Keywords: chi-square, critical value, two-tailed Section: 11-1 Hypothesis Tests and Estimation for a Single Population Variance Outcome: 1 8) One of the most important aspects of quality improvement is the idea of reducing the variability in a product or service. For instance, a major bank has worked to reduce the variability in the service time at the drive-through. The managers believe that the standard deviation in service time should not exceed 30 seconds. To test whether this goal is being achieved, a random sample of n = 25 cars is selected each week and the service time for each car is measured. Last week, the mean time was 345 seconds with a standard deviation equal to 38 seconds. Given this information, if the significance level is 0.10, the critical value from the chi-square table is about 34.3. Answer: FALSE Diff: 2 Keywords: chi-square, critical value, one tailed Section: 11-1 Hypothesis Tests and Estimation for a Single Population Variance Outcome: 1 9) For a given significance level, increasing the sample size will make the chi-square distribution more skewed. Answer: FALSE Diff: 1 Keywords: sample size, critical value, hypothesis Section: 11-1 Hypothesis Tests and Estimation for a Single Population Variance Outcome: 1 11-2 Copyright © 2018 Pearson Education, Inc.

10) When using a chi-square test for the variance of one population, we are assuming that the population is normally distributed. Answer: TRUE Diff: 2 Keywords: standard deviation, variance, chi-square, hypothesis Section: 11-1 Hypothesis Tests and Estimation for a Single Population Variance Outcome: 1 11) The central location and shape of the chi-square distribution depend only on the population variance. Answer: FALSE Diff: 1 Keywords: chi-square, degrees of freedom Section: 11-1 Hypothesis Tests and Estimation for a Single Population Variance Outcome: 1 The central location and shape of the chi-square distribution depend only on the degrees of freedom, n - 1 12) The F-distribution is used to test whether two sample variances are equal. Answer: TRUE Diff: 1 Keywords: F-distribution, variances, hypothesis Section: 11-2 Hypothesis Tests for Two Population Variances Outcome: 3 13) In a hypothesis test involving two population variances, if the null hypothesis states that the two variances are strictly equal, then the test statistic is a chi-square statistic. Answer: FALSE Diff: 1 Keywords: Null hypothesis, chi-square hypothesis, 2 population variances Section: 11-2 Hypothesis Tests for Two Population Variances Outcome: 3 14) In a two-tailed hypothesis test for the difference between two population variances, the test statistic is an F-ratio formed by putting the larger sample variance in numerator. Answer: TRUE Diff: 2 Keywords: hypothesis test, test statistic, F-ratio, variance Section: 11-2 Hypothesis Tests for Two Population Variances Outcome: 3 15) In a two-tailed hypothesis test for the difference between two population variances, if s1 = 3 and s2 = 5, then the test statistic is F = 1.6667. Answer: FALSE Diff: 2 Keywords: two-tailed, hypothesis, test statistic, F-test Section: 11-2 Hypothesis Tests for Two Population Variances Outcome: 3

16) In a two-tailed hypothesis test for the difference between two population variances, if s1 = 3 and s2 = 5, then the test statistic is F = 2.7778. Answer: TRUE Diff: 2 Keywords: test statistic, F-test, two-tailed, hypothesis Section: 11-2 Hypothesis Tests for Two Population Variances Outcome: 3 17) A potato chip manufacturer has two packaging lines and wants to determine if the variances differ between the two lines. They take a sample of n= 15 bags from each line and find the following: The value of the test statistic is F = 1.5 Answer: FALSE Diff: 2 Keywords: test statistic, variances, F-value Section: 11-2 Hypothesis Tests for Two Population Variances Outcome: 3 18) In a two-tailed test for the equality of two variances, the critical value is determined by going to the Fdistribution table with an upper-tail area equal to alpha divided by two. Answer: TRUE Diff: 2 Keywords: two-tailed, hypothesis test, alpha, critical value Section: 11-2 Hypothesis Tests for Two Population Variances Outcome: 3 19) The managers for a fruit juice facility claim that the standard deviation for the ounces per bottle on the new automated line is not the same as the older manual line. Given this, the correct null and alternative hypotheses for performing the statistical test are: H0 : σ1 = σ2 Ha : σ1 ≠ σ2 Answer: TRUE Diff: 2 Keywords: hypothesis test, null, alternative, F-test Section: 11-2 Hypothesis Tests for Two Population Variances Outcome: 3 20) In a hypothesis test for the equality of two variances, the lower-tail critical value does not need to be found as long as the larger sample variance is placed in the denominator of the test statistic. Answer: FALSE Diff: 2 Keywords: variance, hypothesis test, F-test, critical value, test statistic Section: 11-2 Hypothesis Tests for Two Population Variances Outcome: 3

21) In a test for determining whether two population variances are the same or different, the larger the sample sizes from the two populations, the lower will be the chance of making a Type I statistical error. Answer: FALSE Diff: 1 Keywords: Type I, alpha, sample size, F-test Section: 11-2 Hypothesis Tests for Two Population Variances Outcome: 3 22) The null hypothesis that two population variances are equal will tend to be rejected if the ratio of the sample variances from each population is substantially larger than 1.0. Answer: TRUE Diff: 1 Keywords: hypothesis, null, variance, F-test Section: 11-2 Hypothesis Tests for Two Population Variances Outcome: 3 23) A potato chip manufacturer has two packaging lines and wants to determine if the variances differ between the two lines. They take samples of n1 = 10 bags from line 1 and n2 = 8 bags from line 2. To perform the hypothesis test at the 0.05 level of significance, the critical value is F = 3.68. Answer: FALSE Diff: 1 Keywords: variance, critical value, F-test Section: 11-2 Hypothesis Tests for Two Population Variances Outcome: 3 24) If a one-tailed F-test is employed when testing a null hypothesis about two population variances, the test statistic is an F-value formed by taking the ratio of the two sample variances so that the sample variance predicted to be larger is placed in the numerator. Answer: TRUE Diff: 2 Keywords: F-value, test statistic, variance, numerator Section: 11-2 Hypothesis Tests for Two Population Variances Outcome: 3 25) A two-tailed test for two population variances could have a null hypothesis like the following: H0 :

Answer: TRUE Diff: 1 Keywords: two-tailed, hypothesis, null Section: 11-2 Hypothesis Tests for Two Population Variances Outcome: 3

26) One of the major automobile makers has developed two new engines. At question is whether the two engines have the same variability with respect to miles per gallon. To test this, the following information is available: Engine 1 n1 = 7

Engine 2 n2 = 9

1 = 28.7 s1 = 3.4

2 = 33.4 s2 = 4.12

Based on this situation and the information provided, the appropriate null and alternative hypotheses are: H0 :

≠

Ha :

Answer: FALSE Diff: 1 Keywords: F-test, two-tailed, null, alternative, hypothesis Section: 11-2 Hypothesis Tests for Two Population Variances Outcome: 3 27) One of the major automobile makers has developed two new engines. At question is whether the two engines have the same variability with respect to miles per gallon. To test this using a significance level equal to 0.10, the following information is available: Engine 1 n1 = 7

Engine 2 n2 = 9

1 = 28.7 s1 = 3.4

2 = 33.4 s2 = 4.12

Based on this situation and the information provided, the critical value is F = 4.147 . Answer: TRUE Diff: 2 Keywords: critical value, F-test, hypothesis, two-tailed Section: 11-2 Hypothesis Tests for Two Population Variances Outcome: 3

28) One of the major automobile makers has developed two new engines. At question is whether the two engines have the same variability with respect to miles per gallon. To test this, the following information is available: Engine 1 n1 = 7

Engine 2 n2 = 9

1 = 28.7 s1 = 3.4

2 = 33.4 s2 = 4.12

Based on this situation and the information provided, the test statistic is 1.2118. Answer: FALSE Diff: 2 Keywords: hypothesis, F-test, test statistic Section: 11-2 Hypothesis Tests for Two Population Variances Outcome: 3 29) One of the major automobile makers has developed two new engines. At question is whether the two engines have the same variability with respect to miles per gallon. To test this using a 0.10 level of significance, the following information is available: Engine 1 n1 = 7

Engine 2 n2 = 9

1 = 28.7 s1 = 3.4

2 = 33.4 s2 = 4.12

Based on this situation and the information provided, the null hypothesis cannot be rejected and it is possible that the two engines produce the same variation in mpg. Answer: TRUE Diff: 2 Keywords: F-test, variance, hypothesis Section: 11-2 Hypothesis Tests for Two Population Variances Outcome: 3 30) In a two-tailed hypothesis test involving two population variances, if the null hypothesis is true then the F-test statistic should be approximately equal to 1.0. Answer: TRUE Diff: 2 Keywords: two-tailed, hypothesis, F-test statistic Section: 11-2 Hypothesis Tests for Two Population Variances Outcome: 3

31) The logic behind the F-test for testing whether two populations have equal variances is to determine whether sample variances computed from random samples selected from the two populations differ due to sampling error, or whether the difference is more than can be attributed to sampling error alone, in which case, we conclude that the populations have different variances. Answer: TRUE Diff: 2 Keywords: F-test, variance, sampling error Section: 11-2 Hypothesis Tests for Two Population Variances Outcome: 3 32) One of the key quality characteristics in many service environments is that the variation in service time be reasonably small. Recently, a major amusement park company initiated a new line system at one of its parks. It then wished to compare this new system with the old system in place at a comparable park in another state. At issue is whether the standard deviation in waiting time is less under the new line system than under the old line system. The following information was collected: System 1 (old) n1 = 6 customers

System 2 (new) n2 = 8 customers

1 = 15.6 minutes s1 = 4.0 minutes

2 = 17.8 minutes s2 = 3.2 minutes

Assuming that it wishes to conduct the test using a 0.05 level of significance, the correct null and alternative hypotheses would be: H0 :

≠

Ha :

Answer: FALSE Diff: 2 Keywords: hypothesis test, null, alternative, F-test Section: 11-2 Hypothesis Tests for Two Population Variances Outcome: 3

33) One of the key quality characteristics in many service environments is that the variation in service time be reasonably small. Recently, a major amusement park company initiated a new line system at one of its parks. It then wished to compare this new system with the old system in place at a comparable park in another state. At issue is whether the standard deviation in waiting time is less under the new line system than under the old line system. The following information was collected: System 1 (old) n1 = 6 customers

System 2 (new) n2 = 8 customers

1 = 15.6 minutes s1 = 4.0 minutes

2 = 17.8 minutes s2 = 3.2 minutes

Assuming that it wishes to conduct the test using a 0.05 level of significance, the test statistic will be . Answer: TRUE Diff: 2 Keywords: hypothesis test, null, alternative, F-test Section: 11-2 Hypothesis Tests for Two Population Variances Outcome: 3 34) One of the key quality characteristics in many service environments is that the variation in service time be reasonably small. Recently, a major amusement park company initiated a new line system at one of its parks. It then wished to compare this new system with the old system in place at a comparable park in another state. At issue is whether the standard deviation in waiting time is less under the new line system than under the old line system. The following information was collected: System 1 (old) n1 = 6 customers

System 2 (new) n2 = 8 customers

1 = 15.6 minutes s1 = 4.0 minutes

2 = 17.8 minutes s2 = 3.2 minutes

Assuming that it wishes to conduct the test using a 0.05 level of significance, the null hypothesis should be rejected since the test statistic exceeds the F-critical value from the F-distribution table. Answer: FALSE Diff: 2 Keywords: hypothesis, F-test, p-value, variance Section: 11-2 Hypothesis Tests for Two Population Variances Outcome: 3

35) A frozen food company that makes burritos currently has employees making burritos by hand. It is considering purchasing equipment to automate the process and wants to determine if the automated process would result in lower variability of burrito weights. To conduct a hypothesis test using a 0.05 level of significance, the proper format for the null and alternative hypotheses is (where the current byhand process is process 1 and the automated process is process 2). H0 :

≤

Ha :

Answer: TRUE Diff: 2 Keywords: null, alternative, hypothesis, F-test Section: 11-2 Hypothesis Tests for Two Population Variances Outcome: 3 36) A frozen food company that makes burritos currently has employees making burritos by hand. It is considering purchasing equipment to automate the process and wants to determine if the automated process would result in lower variability of burrito weights. It takes a random sample from each process as shown below. Process 1 (by hand) n = 10 s = 0.2 ounces

Process 2 (automated) n=8 s = 0.17 ounces

To conduct a hypothesis test using a 0.05 level of significance, the critical value is 3.347. Answer: FALSE Diff: 2 Keywords: F-test, one-tailed, critical value Section: 11-2 Hypothesis Tests for Two Population Variances Outcome: 3

37) A frozen food company that makes burritos currently has employees making burritos by hand. It is considering purchasing equipment to automate the process and wants to determine if the automated process would result in lower variability of burrito weights. It takes a random sample from each process as shown below. Process 1 (by hand) n = 10 s = 0.22 ounces

Process 2 (automated) n=8 s = 0.16 ounces

In conducting the hypothesis test, the test statistic is F = 1.375. Answer: FALSE Diff: 2 Keywords: F-distribution, test statistic Section: 11-2 Hypothesis Tests for Two Population Variances Outcome: 3 38) The F-distribution can only have negative values. Answer: FALSE Diff: 1 Keywords: F-distribution, positive Section: 11-2 Hypothesis Tests for Two Population Variances Outcome: 3 39) Similar to the chi-square distribution, all F-tests are one-tailed tests. Answer: FALSE Diff: 1 Keywords: F-distribution, F-test, one-tailed Section: 11-2 Hypothesis Tests for Two Population Variances Outcome: 3 40) A first step in testing whether two populations have the same mean value using the t-distribution is to use the chi-square distribution to test whether the populations have equal variances. Answer: FALSE Diff: 2 Keywords: hypothesis test, chi-square distribution, F-distribution, variance Section: 11-2 Hypothesis Tests for Two Population Variances Outcome: 3

41) There is interest at the American Savings and Loan as to whether there is a difference between average daily balances in checking accounts that are joint accounts (two or more members per account) versus single accounts (one member per account). To test this, a random sample of checking accounts was selected with the following results: Single Accounts n = 20 s = $256

Joint Accounts n = 30 s = $300

= $1,123

= $1,245

Based upon these data, the test statistic for testing whether the two populations have equal variances is F = 1.3733. Answer: TRUE Diff: 2 Keywords: test statistic, hypothesis, variance Section: 11-2 Hypothesis Tests for Two Population Variances Outcome: 3 42) There is interest at the American Savings and Loan as to whether there is a difference between average daily balances in checking accounts that are joint accounts (two or more members per account) versus single accounts (one member per account). To test this, a random sample of checking accounts was selected with the following results: Single Accounts n = 20 s = $256

Joint Accounts n = 30 s = $300

= $1,123

= $1,245

Based upon these data, if tested using a significance level equal to 0.10, the assumption of equal population variances should be rejected. Answer: FALSE Diff: 2 Keywords: F-test, critical value, test statistic, variance Section: 11-2 Hypothesis Tests for Two Population Variances Outcome: 3 43) The F test statistic for testing whether the variances of two populations are the same is always positive. Answer: TRUE Diff: 2 Keywords: F-test, test statistic, variance Section: 11-2 Hypothesis Tests for Two Population Variances Outcome: 3

44) An analyst plans to test whether the standard deviation for the time it takes bank tellers to provide service to customers exceeds the standard of 1.5 minutes. The correct null and alternative hypotheses for this test are: A) H0 : σ > 1.5 HA : σ ≥ 1.5 B) H0 : σ ≤ 1.5 HA : σ > 1.5 C) H0 : σ2 ≤ 2.25 HA : σ2 > 2.25 D) H0 : σ2 > 2.25 HA : σ2 ≤ 2.25 Answer: C Diff: 1 Keywords: null, alternative, hypothesis, chi-square, population variance Section: 11-1 Hypothesis Tests and Estimation for a Single Population Variance Outcome: 1 45) When a hypothesis test is to be conducted regarding a population variance, the test statistic will be: A) a t-value from the t-distribution. B) an x2 value from the chi-square distribution. C) a z-value from the standard normal distribution. D) a binomial distribution p value. Answer: B Diff: 1 Keywords: hypothesis, population, variance, test statistic, chi-square Section: 11-1 Hypothesis Tests and Estimation for a Single Population Variance Outcome: 1 46) If a hypothesis test for a single population variance is to be conducted using a significance level of 0.10, a sample size of n = 16, and the test is a one-tailed upper-tail test, the critical value is: A) z = 1.28 B) t = 1.345 C) x2 = 22.3071 D) x2 = 24.9958 Answer: C Diff: 2 Keywords: hypothesis test, critical value, chi-square Section: 11-1 Hypothesis Tests and Estimation for a Single Population Variance Outcome: 1

47) When conducting a one-tailed hypothesis test of a population variance using a sample size of n = 24 and a 0.10 level of significance, the critical value is: A) 32.0069 B) 35.1725 C) 33.1962 D) 36.4150 Answer: A Diff: 2 Keywords: hypothesis, variance, critical value Section: 11-1 Hypothesis Tests and Estimation for a Single Population Variance Outcome: 1 48) A potato chip manufacturer has found that in the past the standard deviation of bag weight has been 0.2 ounce. They want to test whether the standard deviation has changed. The null hypothesis is: A) H0 : σ2 = 0.2 B) H0 : σ = 0.2 C) H0 : σ = 0.04 D) H0 : σ2 = 0.04 Answer: D Diff: 2 Keywords: test statistic, chi-square, null, variance, two-tailed Section: 11-1 Hypothesis Tests and Estimation for a Single Population Variance Outcome: 1 49) A fast food restaurant that sells burritos is concerned about the variability in the amount of filling that different employees place in the burritos. To achieve product consistency it needs this variability to be no more than 1.7 ounces. A sample of n = 18 burritos showed a sample variance of 2.89 ounces. Using a 0.10 level of significance, what can you conclude? A) The standards are being met since (test statistic) < (critical value). B) The standards are not being met since (test statistic) > (critical value). C) The standards are being met since (test statistic) > (critical value). D) The standards are not being met since (test statistic) < (critical value). Answer: B Diff: 2 Keywords: hypothesis test, variance, chi-square Section: 11-1 Hypothesis Tests and Estimation for a Single Population Variance Outcome: 1

50) If a hypothesis test for a single population variance is to be conducted, which of the following statements is true? A) The null hypothesis must be stated in terms of the population variance. B) The chi-square distribution is used. C) If the sample size is increased, the critical value is also increased for a given level of statistical significance. D) All of the above are true. Answer: D Diff: 2 Keywords: hypothesis, population, variance, chi-square, critical value, sample size Section: 11-1 Hypothesis Tests and Estimation for a Single Population Variance Outcome: 1 51) A manufacturer of industrial plywood has a contract to supply a boat maker with a large amount of plywood. One of the specifications calls for the standard deviation in thickness to not exceed .03 inch. A sample of n = 30 sheets was sampled randomly from a recent production run. The mean thickness was right at the 3/4 inch target thickness and the sample standard deviation was .05 inch. Testing at the 0.05 level of significance, which of the following is true? A) The test statistic is approximately 80.56. B) The critical value is approximately χ2 = 43.773. C) The test statistic is approximately 48.333. D) Based on the sample data, there is no evidence to suggest that the plywood is not meeting the specifications. Answer: A Diff: 2 Keywords: chi-square, test statistic, standard deviation Section: 11-1 Hypothesis Tests and Estimation for a Single Population Variance Outcome: 1 52) To test the following hypotheses at the 0.05 level of significance, using a sample size of n = 15. H0 : σ2 = 0.05 HA : σ2 ≠ 0.05 What is the upper tail critical value? A) 23.685 B) 24.996 C) 27.488 D) 26.119 Answer: D Diff: 2 Keywords: standard deviation, hypothesis test, critical value, two-tailed Section: 11-1 Hypothesis Tests and Estimation for a Single Population Variance Outcome: 1

53) A consulting report that was recently submitted to a company indicated that a hypothesis test for a single population variance was conducted. The report indicated that the test statistic was 34.79, the hypothesized variance was 345 and the sample variance 600. However, the report did not indicate what the sample size was. What was it? A) n = 100 B) Approximately n = 18 C) Approximately 21 D) Can't be determined without knowing what alpha is. Answer: C Diff: 3 Keywords: sample size, test statistic, variance, chi-square Section: 11-1 Hypothesis Tests and Estimation for a Single Population Variance Outcome: 1 54) If the variance of the contents of cans of orange juice is significantly more than 0.003, the manager has to order to stop the filling machine. A sample of 26 cans of orange juice showed a standard deviation of 0.06 ounce. Based on the sample and at the 0.05 level of significance, the filling machine should be A) stopped. B) kept going. C) upgraded. D) downgraded. Answer: B Diff: 2 Keywords: test statistic, variance, chi-square Section: 11-1 Hypothesis Tests and Estimation for a Single Population Variance Outcome: 1 55) Which of the following is the appropriate null hypothesis when testing whether two population variances are equal? A) H0 :

B) H0 :

≠

C) H0 :

D) H0 :

≥

Answer: A Diff: 1 Keywords: hypothesis, null, F-test, variance Section: 11-2 Hypothesis Tests for Two Population Variances Outcome: 3

56) Which distribution is used in testing the hypotheses about the equality of two population variances? A) z-distribution B) F-distribution C) x2 distribution D) t-distribution Answer: B Diff: 1 Keywords: distribution, f-distribution, F-test, population, variance Section: 11-2 Hypothesis Tests for Two Population Variances Outcome: 3 57) It is believed that the SAT scores for students entering two state universities may have different standard deviations. Specifically, it is believed that the standard deviation at University A is greater than the standard deviation at University B. If a statistical test is to be conducted, which of the following would be the proper way to formulate the null hypothesis? A) H0 :

B) H0 :

≤

C) H0 : σA ≤ σB D) H0 :

Answer: B Diff: 2 Keywords: population, variance, F-test, null, hypothesis Section: 11-2 Hypothesis Tests for Two Population Variances Outcome: 3

58) It is believed that the SAT scores for students entering two state universities may have different standard deviations. Specifically, it is believed that the standard deviation at University A is greater than the standard deviation at University B. To test this using an alpha = 0.05 level, a sample of 14 student SAT scores from University A was selected and a sample of 8 SAT scores from University B was selected. The following sample results were observed: University A

University B

= 1104 s = 134

= 1254 s = 108

Based on this information, what is the critical value that will be used to test the hypothesis? A) = 3.55 B) = 2.832 C) z = 1.645 D) = 3.237 Answer: A Diff: 2 Keywords: critical value, F-test, standard deviation, hypothesis, one-tailed Section: 11-2 Hypothesis Tests for Two Population Variances Outcome: 3 59) It is believed that the SAT scores for students entering two state universities may have different standard deviations. Specifically, it is believed that the standard deviation at University A is greater than the standard deviation at University B. To test this using an alpha = 0.05 level, a sample of 14 student SAT scores from University A was selected and a sample of 8 SAT scores from University B was selected. The following sample results were observed: University A

University B

= 1104 s = 134

= 1254 s = 108

Based on this information, what is the value of the test statistic? A) 1.2407 B) 0.6496 C) 1.5394 D) None of the above. Answer: C Diff: 2 Keywords: test statistic, F-statistic, hypothesis, standard deviation Section: 11-2 Hypothesis Tests for Two Population Variances Outcome: 3

60) The Russet Potato Company has been working on the development of a new potato seed that is hoped to be an improvement over the existing seed that is being used. Specifically, the company hopes that the new seed will result in less variability in individual potato length than the existing seed without reducing the mean length. To test whether this is the case, a sample of each seed is used to grow potatoes to maturity. The following information is given: Old Seed Number of Seeds = 11 Average length = 6.25 inches Standard Deviation = 1.0 inches

New Seed Number of Seeds = 16 Average length = 5.95 inches Standard Deviation = 0.80 inches

The correct null hypothesis for testing whether the variability of the new seed is less than the old seed is: A) H0 :

≤

B) H0 :

≥

C) H0 : σO ≥ σN D) H0 :

Answer: A Diff: 2 Keywords: null, hypothesis, F-test Section: 11-2 Hypothesis Tests for Two Population Variances Outcome: 3 61) The Russet Potato Company has been working on the development of a new potato seed that is hoped to be an improvement over the existing seed that is being used. Specifically, the company hopes that the new seed will result in less variability in individual potato length than the existing seed without reducing the mean length. To test whether this is the case, a sample of each seed is used to grow potatoes to maturity. The following information is given: Old Seed Number of Seeds = 11 Average length = 6.25 inches Standard Deviation = 1.0 inches

New Seed Number of Seeds = 16 Average length = 5.95 inches Standard Deviation = 0.80 inches

The on these data, if the hypothesis test is conducted using a 0.05 level of significance, the calculated test statistic is: A) = 1.25 B) = 0.80 C) = 0.64 D) = 1.56 Answer: D Diff: 2 Keywords: test statistic, F-test, hypothesis, variance Section: 11-2 Hypothesis Tests for Two Population Variances Outcome: 3 11-19 Copyright © 2018 Pearson Education, Inc.

62) In performing a one-tailed test for the difference between two population variances, which of the following statements is true? A) The level of alpha needs to be doubled before finding the F-critical value in the table. B) The sample variance that is predicted to be larger in the alternative hypothesis goes in the numerator when forming the F-test statistic. C) You always place the larger of the two sample variances in the numerator. D) The alternative hypothesis must contain the equality. Answer: B Diff: 2 Keywords: hypothesis, one-tailed, variance, F-statistic, numerator Section: 11-2 Hypothesis Tests for Two Population Variances Outcome: 3 63) Two toy companies are being compared with respect to the time it takes them to build a popular wooden train set. The study is interested in determining whether there is a difference in the variability between the two toy companies production times. They wish to conduct the hypothesis test using an alpha = 0.05. If random samples of 20 train sets are selected from each toy company, what is the appropriate F critical value? A) 2.526 B) 2.938 C) 2.168 D) 2.124 Answer: A Diff: 2 Keywords: critical value, F-test, hypothesis, variance Section: 11-2 Hypothesis Tests for Two Population Variances Outcome: 3 64) A small business owner has two fast food restaurants. The owner wants to determine if there is any difference in the variability of service times at the drive-thru window of each restaurant. A sample of size n = 9 is taken from each restaurant's drive-thru window. To perform a hypothesis test using the 0.05 level of significance the critical value is: A) 3.438 B) 3.197 C) 4.026 D) 4.433 Answer: D Diff: 2 Keywords: F-test, two-tailed, variances Section: 11-2 Hypothesis Tests for Two Population Variances Outcome: 3

65) The U.S. Golf Association provides a number of services for its members. One of these is the evaluation of golf equipment to make sure that the equipment satisfies the rules of golf. For example, they regularly test the golf balls made by the various companies that sell balls in the United States. Recently, they undertook a study of two brands of golf balls with the objective to see whether there is a difference in the mean distance that the two golf ball brands will fly off the tee. To conduct the test, the U.S.G.A. uses a robot named "Iron Byron," which swings the club at the same speed and with the same swing pattern each time it is used. The following data reflect sample data for a random sample of balls of each brand. Brand A: Brand B:

234 240

236 236

230 241

227 236

234 239

233 243

228 230

229 239

230 243

238 240

Given this information, what is the test statistic for testing whether the two population variances are equal? A) Approximately F = 1.145 B) t = 1.96 C) t = -4.04 D) None of the above Answer: A Diff: 3 Keywords: test statistic, F-test, variance Section: 11-2 Hypothesis Tests for Two Population Variances Outcome: 3

66) The Department of Weights and Measures in a southern state has the responsibility for making sure that all commercial weighing and measuring devices are working properly. For example, when a gasoline pump indicates that 1 gallon has been pumped, it is expected that 1 gallon of gasoline will actually have been pumped. The problem is that there is variation in the filling process. The state's standards call for the mean amount of gasoline to be 1.0 gallon with a standard deviation not to exceed 0.010 gallons. Recently, the department came to a gasoline station and filled 10 cans until the pump read 1.0 gallon. It then measured precisely the amount of gasoline in each can. The following data were recorded: 0.991 1.052

0.962 0.934

1.007 0.993

1.038 1.033

1.036 0.967

Based on these data, what should the Department of Weights and Measures conclude if it wishes to test whether the standard deviation exceeds 0.010 gallons or not, using a 0.05 level of significance? Answer: The variation requirement specifies that the standard deviation cannot exceed 0.01 gallon. We need to use a test for a single population variance. The appropriate null and alternative hypotheses are: H0 : σ2 ≤ .0001 Ha : σ2 > .0001 Note that we have converted from standard deviation to variances since there is no test for standard deviation directly. In this situation, the test statistic is a chi-square value computed as follows: x2 =

= 135.49

This test statistic can be compared to the chi-square value from the table with 10 - 1 = 9 degrees of freedom and a one-tail area of 0.05. That value is 16.9190. Since x2 = 135.49 > 16.9190, the null hypothesis is rejected. This means that the variation standard is being exceeded based on these data. The gasoline station will need to get the pump checked out and the variation of fill around the mean reduced. Diff: 3 Keywords: chi-square, hypothesis, variation Section: 11-1 Hypothesis Tests and Estimation for a Single Population Variance Outcome: 1

67) A PC company uses two suppliers for rechargeable batteries for its notebook computers. Two factors are important quality features of the batteries: mean use time and variation. It is desirable that the mean use time be high and the variability be low. Recently, the PC maker conducted a test on batteries from the two suppliers. In the test, 9 randomly selected batteries from Supplier 1 were tested and 12 randomly selected batteries from Supplier 2 were tested. The following results were observed: Supplier 1 n1 = 9

Supplier 2 n2 = 12

1 = 67.25 min s1 = 11.2 min

2 = 72.4 min s2 = 9.9 min

Based on these sample results, can the PC maker conclude that a difference exists between the two batteries with respect to the population standard deviations? Test using a 0.10 level of significance. Answer: Ideally, we would like to test whether the two population standard deviations are equal. However, there is no test available for doing this test directly. Instead, we must convert to variances and perform the F-test. The following null and alternative hypotheses would be appropriate in this situation: H0 :

Ha :

≠

This will be a two-tailed test since we are testing to see whether a difference exists and are not predicting which battery will have more or less variability. In a two-tailed test of variances, the test statistic is an Fvalue that is formed by the ratio of the two sample variances and by placing the larger sample variance in the numerator:

= 1.2799.

We then compare this value to a critical value from the F-distribution table. Since this is a two-tailed test with alpha = 0.10, we use the table with 0.05 in the upper-tail of the F-distribution. Two sets of degrees of freedom are used with the F-distribution. Across the top of the F-table we look for degrees of freedom corresponding to n-1 where n is the sample size associated with the sample variance that was placed in the numerator of the F-test statistic. In our case, n = 9, so the degrees of freedom is 9 - 1 = 8 for the numerator. The degrees of freedom down the side in the F-table is n-1 where n corresponds to the sample size for the sample variance in the denominator of the F-test statistic. In our case, that would be n = 12. Thus, the denominator degrees of freedom are 12 - 1 = 11. Then the critical F from the 0.05 table with 8 and 11 degrees of freedom is 2.948. The decision rule is: If F calculated > 2.948 reject H0, otherwise do not reject. Since our calculated F-test statistic is F = 1.2799 < 2.948, we do not reject the null hypothesis. Thus, based on the sample information, we have no basis for believing that there is a difference in the two batteries with respect to variability in use time. Diff: 2 Keywords: F-test, hypothesis, standard deviation Section: 11-2 Hypothesis Tests for Two Population Variances Outcome: 3 11-23 Copyright © 2018 Pearson Education, Inc.

68) A PC company uses two suppliers for rechargeable batteries for its notebook computers. Two factors are important quality features of the batteries: mean use time and variation. It is desirable that the mean use time be high and the variability be low. Recently, the PC maker conducted a test on batteries from the two suppliers. In the test, 9 randomly selected batteries from Supplier 1 were tested and 12 randomly selected batteries from Supplier 2 were tested. The following results were observed: Supplier 1 n1 = 9

Supplier 2 n2 = 12

1 = 67.25 min s1 = 11.2 min

2 = 72.4 min s2 = 9.9 min

Based on these sample results, can the PC maker conclude that a difference exists between the two batteries with respect to the population mean use time? Test using a 0.10 level of significance. Answer: This test calls for us to test whether the population means are equal. Since the sample sizes are small and the population variances are unknown, the t-distribution should be employed if we assume that the populations are normally distributed and the populations have equal variances. We can test for the equal variance assumptions as follows. Ideally, we would like to test whether the two population standard deviations are equal. However, there is no test available for doing this test directly. Instead, we must convert to variances and perform the Ftest. The following null and alternative hypotheses would be appropriate in this situation: H0 :

H0 :

≠

= 1.2799.

Since our calculated F-test statistic is F = 1.2799 < 2.948, we do not reject the null hypothesis. Thus, based on the sample information, we have no basis for believing that there is a difference in the two batteries with respect to variability in use time. Given this conclusion we can now test the following null and alternative hypotheses: H0 : μ1 = μ2 Ha : μ1 ≠ μ2 The critical value for a t-test for a two-tailed test is found from the t-distribution with degrees of freedom equal to n1 + n2 - 2=9+12-2=19. Given a 0.10 significance level, the critical value from the t-distribution with degrees of freedom equal to 19 is equal to ±1.7291. The test statistic is computed using:

where sp is the pooled standard deviation and is computed as:

sp =

= 10.4671

Then the test statistic is:

= -1.1158

Since t = -1.1158 > t - 1.7291, we do not reject the null hypothesis. Thus, based on the sample data, we do not have evidence to indicate that the two batteries differ in mean use time. Diff: 2 Keywords: t-test, t-distribution, F-test, hypothesis Section: 11-2 Hypothesis Tests for Two Population Variances Outcome: 3

69) There are two major companies that provide SAT test tutoring for high school students. At issue is whether Company 1, which has been in business for the longer time, provides better results than Company 2, the newer company. Specifically of interest is whether the mean increase in SAT scores for students who have already taken the SAT test one time is higher for Company 1 than for Company 2. Two random samples of students are selected. The following data reflect the number of points higher (or lower) that the students scored on the SAT test after taking the tutoring. Prior to conducting the test, which compares the means, we should determine if the assumption of equal variances is supported. Conduct the appropriate hypothesis test to determine if the assumption of equal variances is supported using a 0.10 level of significance. Company 1 Company 2 98 65 60 11 -14 30 80 52 55 27 71 16 37 27 41 47 Answer: The following null and alternative hypotheses would be appropriate in this situation: H0 :

Ha :

≠

This will be a two-tailed test since we are testing to see whether a difference exists and are not predicting which company will have more or less variability. We first need to compute the sample variances for each sample using: s2 = The following variances are computed from the sample data: = 1145.43 and

= 345.7

In a two-tailed test of variances, the test statistic is an F-value that is formed by the ratio of the two sample variances and by placing the larger sample variance in the numerator:

= 3.3134

We then compare this value to a critical value from the F-distribution table. Since this is a two-tailed test with alpha = .10, we use the table with 0.05 in the upper-tail of the F-distribution.

Two sets of degrees of freedom are used with the F-distribution. Across the top of the F-table we look for degrees of freedom corresponding to n-1 where n is the sample size associated with the sample variance that was placed in the numerator of the F-test statistic. In our case, n = 8, so the degrees of freedom is 8 - 1 = 7 for the numerator. The degrees of freedom down the side in the F table is n-1 where n corresponds to the sample size for the sample variance in the denominator of the F-test statistic. In our case that would be n = 8. Thus, the denominator degrees of freedom are 8 - 1 = 7. Then the critical F from the .05 table with 7 and 7 degrees of freedom is 3.787. The decision rule is: If F calculated > 3.787, reject H0; otherwise do not reject. Since our calculated F-test statistic is F = 3.3134 < 3.787, we do not reject the null hypothesis. Thus, based on the sample information, we have no basis for believing that there is a difference in the two companies with respect to population variance. Diff: 3 Keywords: F-test, hypothesis, test statistic, two-tailed Section: 11-2 Hypothesis Tests for Two Population Variances Outcome: 3

Business Statistics, 9e (Groebner/Shannon/Fry) Chapter 12 Analysis of Variance 1) The term one-way analysis of variance refers to the fact that in conducting the test, there is only one way to set up the null and alternative hypotheses. Answer: FALSE Diff: 1 Keywords: ANOVA, hypothesis, one-way, analysis of variance Section: 12-1 One-Way Analysis of Variance Outcome: 1 2) In a one-way analysis of variance design, there is a single factor of interest but there may be multiple levels of the factor. Answer: TRUE Diff: 1 Keywords: ANOVA, analysis of variance, one-way, factor, level Section: 12-1 One-Way Analysis of Variance Outcome: 1 3) In conducing one-way analysis of variance, the population distributions are assumed normally distributed. Answer: TRUE Diff: 1 Keywords: ANOVA, analysis of variance, one-way Section: 12-1 One-Way Analysis of Variance Outcome: 1 4) Recently, a company tested three different machine types to see if there was a difference in the mean thickness of products produced by the three. A random sample of ten products was selected from the output from each machine. Given this information, the proper design to test whether the means are equal is a one-way ANOVA balanced design. Answer: TRUE Diff: 2 Keywords: ANOVA, analysis of variance, one-way, balanced Section: 12-1 One-Way Analysis of Variance Outcome: 1 5) The one-way ANOVA test involves assuming that the population variances are equal. Answer: TRUE Diff: 1 Keywords: ANOVA, analysis of variance, one-way Section: 12-1 One-Way Analysis of Variance Outcome: 2

6) In a completely randomized analysis of variance design, the observations from each factor are selected in an independent and random fashion. Answer: TRUE Diff: 2 Keywords: ANOVA, analysis of variance, randomized, factor, random Section: 12-1 One-Way Analysis of Variance Outcome: 2 7) In conducting one-way analysis of variance, the sample size for each group must be equal. Answer: FALSE Diff: 2 Keywords: ANOVA, analysis of variance, one-way Section: 12-1 One-Way Analysis of Variance Outcome: 1 8) In a one-way analysis of variance design, the total variation in the data across the various factor levels can be partitioned into two parts, the within sample variation and the between sample variation. Answer: TRUE Diff: 2 Keywords: ANOVA, analysis of variance, factor, within, between, variation Section: 12-1 One-Way Analysis of Variance Outcome: 1 9) The within sample variation is the dispersion that exists because the sample means for the various factor levels are not all equal. Answer: FALSE Diff: 1 Keywords: ANOVA, analysis of variance, within, factor, between Section: 12-1 One-Way Analysis of Variance Outcome: 1 10) In a recent one-way ANOVA test, SSW was equal to 15,900 and the SSB was equal to 3,100. Therefore, SST is equal to 12,800. Answer: FALSE Diff: 2 Keywords: ANOVA, analysis of variance. Sum of squares, total, within, between Section: 12-1 One-Way Analysis of Variance Outcome: 1 11) Under the basic logic of one-way analysis of variance, if the within variation is large relative to the between variation, it is an indication that the population means are likely to be different. Answer: FALSE Diff: 2 Keywords: ANOVA, analysis of variance, within, between, population, mean Section: 12-1 One-Way Analysis of Variance Outcome: 1

12) In a one-way analysis of variance test, the following null and alternative hypotheses are appropriate: H0 : μ1 = μ2 = μ3 Hα : μ1 ≠ μ2 ≠ μ3 Answer: FALSE Diff: 2 Keywords: ANOVA, analysis of variance, null, alternative, hypothesis Section: 12-1 One-Way Analysis of Variance Outcome: 2 13) In one-way analysis of variance, the within-sample variation is not affected by whether the null hypothesis is true or not. Answer: TRUE Diff: 2 Keywords: ANOVA, analysis of variance, one-way, Section: 12-1 One-Way Analysis of Variance Outcome: 2 14) In conducting a one-way analysis of variance, if the null hypothesis is true then the variance between groups (MSB) should be approximately equal to the variance within groups (MSW). Answer: TRUE Diff: 2 Keywords: ANOVA, analysis of variance, F-test Section: 12-1 One-Way Analysis of Variance Outcome: 2 15) A study was recently conducted to see whether the mean starting salaries for graduates of engineering, business, healthcare, and computer information systems majors differ. A random sample of 8 graduates was selected from each major. If the test is to be conducted using an alpha = 0.05 level, the critical value will be F = 3.838. Answer: FALSE Diff: 2 Keywords: ANOVA, analysis of variance, critical value Section: 12-1 One-Way Analysis of Variance Outcome: 2

16) A study was recently conducted to see whether the mean starting salaries for graduates of engineering, business, healthcare, and computer information systems majors differ. A random sample of 8 graduates was selected from each major. The following shows the results of the ANOVA computations. However, the degrees of freedom column has been omitted. The correct number of degrees of freedom for the within variation is 28. ANOVA: Single Factor

Answer: TRUE Diff: 2 Keywords: ANOVA, analysis of variance, degrees of freedom Section: 12-1 One-Way Analysis of Variance Outcome: 2 17) A marketing study was recently conducted to see whether the mean monthly sales for a department store located in four cities differ. A random sample of 6 months were selected from each store. Based upon this information, the appropriate null hypothesis to be tested is H0 : μ1 = μ2 = μ3 = μ4 Answer: TRUE Diff: 2 Keywords: null, hypothesis, ANOVA Section: 12-1 One-Way Analysis of Variance Outcome: 2

18) A study was recently conducted to see whether the mean starting salaries for graduates of engineering, business, healthcare, and computer information systems majors differ. A random sample of 8 graduates was selected from each major. The following chart shows some of the results of the ANOVA computations; however, some of the output is missing. If it had been included, the calculated test statistic would be F = 8.33. ANOVA: Single Factor

Answer: TRUE Diff: 2 Keywords: ANOVA, analysis of variance, F-statistic, degrees of freedom Section: 12-1 One-Way Analysis of Variance Outcome: 2

19) A study was recently conducted to see whether the mean starting salaries for graduates of engineering, business, healthcare, and computer information systems majors differ. A random sample of 8 graduates was selected from each major. The following chart shows some of the results of the ANOVA computations; however, some of the output is missing. Given what is available, the proper conclusion to reach based on the sample data is that the population means could be equal using a 0.05 level of significance. ANOVA: Single Factor

Answer: FALSE Diff: 2 Keywords: ANOVA, analysis of variance, F-statistic, degrees of freedom, F-critical Section: 12-1 One-Way Analysis of Variance Outcome: 2 20) A company’s customer service department wants to compare the average service times at four different service locations. They want to conduct a hypothesis test to determine if all four means are the same or not. If n = 20 observations are taken at each of the four locations, then the degrees of freedom are D1 = 3 and D2 = 19. Answer: FALSE Diff: 1 Keywords: ANOVA, analysis of variance, degrees of freedom, one-way Section: 12-1 One-Way Analysis of Variance Outcome: 2

21) A national car rental agency is interested in determining whether the mean days that customers rent cars is the same between three of its major cities. The following data reflect the number of days people rented a car for a sample of people in each of three cities. Given this information, the correct null and alternative hypotheses are: H0 : 1 = 2 = 3 Ha : not all j are equal. Boston 5 3 7 1 2

Dallas 7 7 11 5 7 3

Seattle 7 5 8 11

Answer: FALSE Diff: 1 Keywords: ANOVA, analysis of variance, null, alternative, hypothesis Section: 12-1 One-Way Analysis of Variance Outcome: 2 22) A national car rental agency is interested in determining whether the mean days that customers rent cars is the same between three of its major cities. The following data reflect the number of days people rented a car for a sample of people in each of three cities. Assuming that a one-way analysis of variance is to be performed, the total sum of squares is computed to be approximately 120.9. Boston 5 3 7 1 2

Dallas 7 7 11 5 7 3

Seattle 7 5 8 11

Answer: TRUE Diff: 2 Keywords: ANOVA, analysis of variance, sum of squares Section: 12-1 One-Way Analysis of Variance Outcome: 2

23) A national car rental agency is interested in determining whether the mean days that customers rent cars is the same between three of its major cities. The following data reflect the number of days people rented a car for a sample of people in each of three cities. Assuming that a one-way analysis of variance is to be performed, the value of the test statistic is approximately F = 3.4. Boston 5 3 7 1 2

Dallas 7 7 11 5 7 3

Seattle 7 5 8 11

Answer: TRUE Diff: 3 Keywords: ANOVA, analysis of variance, F-statistic, test statistic Section: 12-1 One-Way Analysis of Variance Outcome: 2 24) A company’s customer service department wants to compare the average service times at three different service locations. It wants to conduct a hypothesis test to determine if all three means are the same or not, at the 0.05 level of significance. If n =10 observations are taken at each of the three restaurants, the critical value is F = 3.354. Answer: TRUE Diff: 2 Keywords: ANOVA, analysis of variance, critical value, one-way Section: 12-1 One-Way Analysis of Variance Outcome: 2

25) As a step in establishing its rates, an automobile insurance company is interested in determining whether there is a difference in the mean highway speeds that drivers of different age groups drive. To help answer this question, it has selected a random sample of drivers in three age categories: under 21, 21-50, and over 50. The engineers then recorded the drivers' speeds at a designated point on a highway in the state. The subjects were unaware that their speed was being recorded. The following one-way ANOVA output was generated from the sample data. Based upon this output, if the significance level is 0.05, the engineers should conclude that the mean speeds may all be equal since the p-value is less than alpha. ANOVA: Single Factor

Answer: FALSE Diff: 2 Keywords: ANOVA, analysis of variance, p-value, alpha Section: 12-1 One-Way Analysis of Variance Outcome: 2 26) A fixed effects analysis of variance differs from a random-effects analysis of variance in the way in which the sums of squares are computed. Answer: FALSE Diff: 2 Keywords: ANOVA, analysis of variance, fixed, random, sum of squares Section: 12-1 One-Way Analysis of Variance Outcome: 2 27) The Tukey-Kramer method for multiple comparison is used after the analysis of variance F-test has lead us to reject the null hypothesis that all population means are equal. Answer: TRUE Diff: 2 Keywords: ANOVA, analysis of variance, Tukey-Kramer, F-test Section: 12-1 One-Way Analysis of Variance Outcome: 3

28) If the null hypothesis that all population means are equal is rejected by the analysis of variance F-test, the alternative hypothesis that all population means differ is concluded to be true. Answer: FALSE Diff: 2 Keywords: ANOVA, analysis of variance, null, alternative Section: 12-1 One-Way Analysis of Variance Outcome: 2 29) As a step in establishing its rates, an automobile insurance company is interested in determining whether there is a difference in the mean highway speeds that drivers of different age groups drive. To help answer this question, it has selected a random sample of drivers in three age categories: under 21, 21-50, and over 50. The engineers then recorded the drivers' speeds at a designated point on a highway in the state. The subjects were unaware that their speed was being recorded. The following one-way ANOVA output was generated from the sample data. Based upon this output, it is possible that a Type II statistical error has been committed if the null hypothesis is tested at the alpha equal 0.05 level. ANOVA: Single Factor

Answer: FALSE Diff: 2 Keywords: ANOVA, analysis of variance, type II, beta, null Section: 12-1 One-Way Analysis of Variance Outcome: 2 30) The Tukey-Kramer method for multiple comparisons can only be used when the analysis of variance design is balanced. Answer: FALSE Diff: 2 Keywords: ANOVA, analysis of variance, Tukey-Kramer, balanced Section: 12-1 One-Way Analysis of Variance Outcome: 3

31) The experiment-wide error rate will be higher than the 0.05 significance level if the multiple comparison tests for the mean difference between any two populations use the 0.05 level. Answer: TRUE Diff: 3 Keywords: ANOVA, analysis of variance, Tukey-Kramer Section: 12-1 One-Way Analysis of Variance Outcome: 3 32) Analysis of variance can only be done for fixed effects. Answer: FALSE Diff: 3 Keywords: ANOVA, fixed effects, randomized effects Section: 12-1 One-Way Analysis of Variance Outcome: 2 33) A randomized complete block analysis of variance allows the analyst to control for sources of variation that might adversely affect the analysis by using the concept of paired samples. Answer: TRUE Diff: 2 Keywords: ANOVA, analysis of variance, randomized block, paired Section: 12-2 Randomized Complete Block Analysis of Variance Outcome: 4 34) Three brands of running shoes are each tested by 10 different runners. The amount of wear on the sole of the shoes is then measured. The objective is to determine if there is any difference among the three brands of shoes based on how long the soles last. This means that there is one factor with 10 levels and 3 blocks. Answer: FALSE Diff: 2 Keywords: ANOVA, analysis of variance, factor, treatment, randomized block Section: 12-2 Randomized Complete Block Analysis of Variance Outcome: 4 35) Three brands of running shoes are each tested by 10 different runners. The amount of wear on the sole of the shoes is then measured. The objective is to determine if there is any difference among the three brands of shoes based on how long the soles last. The null hypothesis is: H0 : μ1 = μ2 = μ3. Answer: TRUE Diff: 2 Keywords: ANOVA, analysis of variance, randomized block Section: 12-2 Randomized Complete Block Analysis of Variance Outcome: 4

36) A company has established an experiment with its production process in which three temperature settings are used and five elapsed times are used for each setting. The company then produces one product under each and measures the resulting strength of the product. This experimental design is called a randomized complete block design. Answer: TRUE Diff: 1 Keywords: ANOVA, analysis of variance, randomized block, complete block Section: 12-2 Randomized Complete Block Analysis of Variance Outcome: 4 37) A company has established an experiment with its production process in which three temperature settings are used and five elapsed times are used for each setting. The company then produces one product under each and measures the resulting strength of the product. The managers are mainly interested in determining whether the mean strength is the same at all temperature settings, but they know that controlling for process time is important. The following data were observed from the experiment:

Time 1 Time 2 Time 3 Time 4 Time 5

Temperature 1 Temperature 2 Temperature 3 140 150 170 160 160 180 150 150 160 170 140 180 150 150 170

Based on these data and experimental design, the primary null hypothesis to be tested is H0 : μ1 = μ2 = μ3. Answer: TRUE Diff: 2 Keywords: ANOVA, analysis of variance, randomized block, null, hypothesis Section: 12-2 Randomized Complete Block Analysis of Variance Outcome: 4 38) Three brands of running shoes are each tested by 10 different runners. The amount of wear on the sole of the shoes is then measured. The objective is to determine if there is any difference among the three brands of shoes based on how long the soles last. The degrees of freedom for testing whether the brands of shoes differ are D1 = 9 and D2 = 2. Answer: FALSE Diff: 2 Keywords: ANOVA, analysis of variance, blocking, degrees of freedom Section: 12-2 Randomized Complete Block Analysis of Variance Outcome: 4

39) Three brands of running shoes are each tested by 10 different runners. The amount of wear on the sole of the shoes is then measured. The objective is to determine if there is any difference among the three brands of shoes based on how long the soles last. The degrees of freedom for testing whether there is any blocking effect D1 = 9 and D2 = 18. Answer: TRUE Diff: 2 Keywords: ANOVA, analysis of variance, degrees of freedom, F-test, blocking Section: 12-2 Randomized Complete Block Analysis of Variance Outcome: 4 40) A company has established an experiment with its production process in which three temperature settings are used and five elapsed times are used for each setting. The company then produces one product under each and measures the resulting strength of the product. The managers are mainly interested in determining whether the mean strength is the same at all temperature settings, but they know that controlling for process time is important. The following data were observed from the experiment:

Time 1 Time 2 Time 3 Time 4 Time 5

Temperature 1 Temperature 2 Temperature 3 140 150 170 160 160 180 150 150 160 170 140 180 150 150 170

Based on these data and experimental design, for a significance level of 0.05, the managers should conclude that they were justified in blocking on the basis of processing time. Answer: FALSE Diff: 3 Keywords: ANOVA, analysis of variance, blocking Section: 12-2 Randomized Complete Block Analysis of Variance Outcome: 4 41) An advertising company is interested in determining if there is a difference in the mean sales that will be generated for a soft drink company based on which shelf the soft drinks are located. There are four possible shelf levels. The ad company wants to control for store size. The following data reflect the sales for one week at each combination of shelf level and store size.

Small Store Medium Store Large Store

Level 1 300 400 600

Level 2 200 180 300

Level 3 500 600 900

Level 4 200 200 400

Based on the experimental design, the number of treatments is two. Answer: FALSE Diff: 2 Keywords: ANOVA, analysis of variance, blocking, treatments Section: 12-2 Randomized Complete Block Analysis of Variance Outcome: 4 12-13 Copyright © 2018 Pearson Education, Inc.

42) An advertising company is interested in determining if there is a difference in the mean sales that will be generated for a soft drink company based on which shelf the soft drinks are located. There are four possible shelf levels. The ad company wants to control for store size. The following data reflect the sales for one week at each combination of shelf level and store size.

Small Store Medium Store Large Store

Level 1 300 400 600

Level 2 200 180 300

Level 3 500 600 900

Level 4 200 200 400

Based on the experimental design, the calculated F-test statistic value for testing whether blocking on store size was effective is approximately 16.3. Answer: TRUE Diff: 3 Keywords: ANOVA, analysis of variance, blocking, F-test, test statistic Section: 12-2 Randomized Complete Block Analysis of Variance Outcome: 4 43) An advertising company is interested in determining if there is a difference in the mean sales that will be generated for a soft drink company based on which shelf the soft drinks are located. There are four possible shelf levels. The ad company wants to control for store size. The following data reflect the sales for one week at each combination of shelf level and store size.

Small Store Medium Store Large Store

Level 1 300 400 600

Level 2 200 180 300

Level 3 500 600 900

Level 4 200 200 400

Based on the experimental design, the managers should conclude that they were justified in blocking on store size if they test using a 0.05 level of significance. Answer: TRUE Diff: 3 Keywords: ANOVA, analysis of variance, blocking Section: 12-2 Randomized Complete Block Analysis of Variance Outcome: 4

44) Given the partially completed ANOVA table below, the test statistic for determining if there is any blocking effect is F = 4.38. Source Between blocks Between samples Within samples Total

SS 387 422 265 1374

df 9 3 27 39

Answer: TRUE Diff: 2 Keywords: ANOVA, analysis of variance, blocking, test statistic Section: 12-2 Randomized Complete Block Analysis of Variance Outcome: 4 45) Based on the partially completed ANOVA table below, we know that 3 samples are being compared using 9 blocks. Source Between blocks Between samples Within samples Total

SS 387 422 265 1374

df 9 3 27 39

Answer: FALSE Diff: 2 Keywords: ANOVA, analysis of variance, degrees of freedom, blocking Section: 12-2 Randomized Complete Block Analysis of Variance Outcome: 4 46) To test for the stopping distances of four brake systems, 10 of the same make and model of car are selected randomly and then are assigned randomly to each of four brake systems. This is a randomized complete block design. Answer: FALSE Diff: 2 Keywords: analysis of variance, blocking, single factor Section: 12-2 Randomized Complete Block Analysis of Variance Outcome: 4 47) In order to analyze any potential interactions between factors in an analysis of variance study, it is necessary to have at least two measurements at each level of each factor. Answer: TRUE Diff: 1 Keywords: ANOVA, analysis of variance, factor, replication Section: 12-3 Two-Factor Analysis of Variance With Replication Outcome: 5

48) The number of cells in a two-factor analysis of variance design is equal to the number of levels of factor A plus the number of levels of factor B. Answer: FALSE Diff: 1 Keywords: ANOVA, analysis of variance, two-factor, cell, level Section: 12-3 Two-Factor Analysis of Variance With Replication Outcome: 5 49) Six food critics each visited and rated four different restaurants. Each critic visited each restaurant on three separate occasions and recorded a score for each visit. The correct method for analyzing this data is two-factor ANOVA. Answer: TRUE Diff: 1 Keywords: ANOVA, analysis of variance, cells, two-factor Section: 12-3 Two-Factor Analysis of Variance With Replication Outcome: 5 50) Six food critics each visited and rated four different restaurants. Each critic visited each restaurant on three separate occasions and recorded a score for each visit. The number of degrees of freedom in the interaction row of the ANOVA table is 15. Answer: TRUE Diff: 1 Keywords: ANOVA, analysis of variance, two-way, degrees of freedom Section: 12-3 Two-Factor Analysis of Variance With Replication Outcome: 5 51) Six food critics each visited and rated four different restaurants. Each critic visited each restaurant on three separate occasions and recorded a score for each visit. The total number of degrees of freedom is 71. Answer: TRUE Diff: 1 Keywords: ANOVA, analysis of variance, replication, degrees of freedom Section: 12-3 Two-Factor Analysis of Variance With Replication Outcome: 5 52) In a two-factor ANOVA, the minimum number of replications required in any cell is two, but all cells must have the same number of replications. Answer: TRUE Diff: 1 Keywords: ANOVA, analysis of variance, two-factor, replication, cell Section: 12-3 Two-Factor Analysis of Variance With Replication Outcome: 5

53) In a two-factor ANOVA, the total sum of squares can be partitioned into four parts; the variation due to factor A, the variation due to factor B, the variation due to blocking, and the error variation. Answer: FALSE Diff: 2 Keywords: ANOVA, analysis of variance, sum of squares Section: 12-3 Two-Factor Analysis of Variance With Replication Outcome: 5 54) In a two-factor ANOVA design with replications, there are three hypotheses to be tested; test for factor A, test for factor B, and test for interaction between factors A and B. Answer: TRUE Diff: 2 Keywords: ANOVA, analysis of variance, hypothesis, interaction, two-factor Section: 12-3 Two-Factor Analysis of Variance With Replication Outcome: 5 55) In a two-factor ANOVA design, the variances of the populations are assumed to be equal unless there is interaction present. Answer: FALSE Diff: 2 Keywords: ANOVA, analysis of variance, two-factor, variance Section: 12-3 Two-Factor Analysis of Variance With Replication Outcome: 5 56) A study recently conducted by a marketing firm analyzed three different advertising designs and four different income levels of potential customers. At each combination of factor A and factor B, 5 customers are observed and the number of products produced are recorded. The total number of degrees of freedom associated with this two-factor ANOVA design is 59. Answer: TRUE Diff: 2 Keywords: ANOVA, analysis of variance Section: 12-3 Two-Factor Analysis of Variance With Replication Outcome: 5 57) A study recently conducted by a marketing firm analyzed three different advertising designs and four different income levels of potential customers. At each combination of factor A and factor B, 5 customers are observed and the number of products produced are recorded. The degrees of freedom for the MSE in this two-factor ANOVA design is 48. Answer: TRUE Diff: 2 Keywords: ANOVA, analysis of variance, degrees of freedom Section: 12-3 Two-Factor Analysis of Variance With Replication Outcome: 5

58) Six food critics each visited and rated four different restaurants. Each critic visited each restaurant on three separate occasions and recorded a score for each visit. The critical value for testing whether there is any difference among the four restaurants, using the 0.05 level of significance is approximately F = 2.8. Answer: TRUE Diff: 2 Keywords: ANOVA, analysis of variance, critical value, F-test Section: 12-3 Two-Factor Analysis of Variance With Replication Outcome: 5 59) Interaction is the term that is used in a two-factor ANOVA design when the two factors have different means. Answer: FALSE Diff: 2 Keywords: ANOVA, analysis of variance, interaction, two-factor Section: 12-3 Two-Factor Analysis of Variance With Replication Outcome: 5 60) In a two-factor ANOVA study, if the two factors do not interact, then neither factor A nor factor B can be considered statistically significant. Answer: FALSE Diff: 2 Keywords: ANOVA, analysis of variance, two-factor, interact Section: 12-3 Two-Factor Analysis of Variance With Replication Outcome: 5 61) A study recently conducted by a marketing firm analyzed three different advertising designs (factor A) and four different income levels (factor B) of potential customers. At each combination of factor A and factor B, 5 customers are observed and the number of products produced is recorded. Interaction between the two factors would exist if low income customers have higher mean buying when design 1 is used, but higher income customers have higher mean buying when designs 2 and 3 are used. Answer: TRUE Diff: 2 Keywords: ANOVA, analysis of variance, two-factor, mean Section: 12-3 Two-Factor Analysis of Variance With Replication Outcome: 5 62) The general idea is that interaction between two factors means that the effect due to one of the factors is not uniform across all levels of the other factor. Answer: TRUE Diff: 2 Keywords: two factor, effect, factor Section: 12-3 Two-Factor Analysis of Variance With Replication Outcome: 5

63) Six food critics each visited and rated four different restaurants. Each critic visited each restaurant on three separate occasions and recorded a score for each visit. Assume that results show that there is an interaction. This would mean that, for example, which restaurant is rated the highest depends on which critic does the rating. Answer: TRUE Diff: 2 Keywords: ANOVA, analysis of variance, two-factor, interaction Section: 12-3 Two-Factor Analysis of Variance With Replication Outcome: 5 64) In a two-factor ANOVA with replication in which all hypotheses are to be tested using an alpha = .05 level, if the p-value for interaction is .03467, the decision maker should conclude that no interaction is present. Answer: FALSE Diff: 2 Keywords: ANOVA, analysis of variance, replication, interaction, p-value Section: 12-3 Two-Factor Analysis of Variance With Replication Outcome: 5 65) In a two-factor ANOVA design with replication, the null hypothesis for testing whether interaction exists is that no interaction exists. The alternative hypothesis is that interaction does exist. Answer: TRUE Diff: 1 Keywords: ANOVA, analysis of variance, two-factor, null, hypothesis, interaction, alternative Section: 12-3 Two-Factor Analysis of Variance With Replication Outcome: 5 66) In a two-factor ANOVA design with replication, if the null hypothesis pertaining to interaction between factors A and B is rejected, then it is recommended that the hypothesis tests for factor A and factor B individually should not be conducted because the conclusions might be misleading. Answer: TRUE Diff: 2 Keywords: ANOVA, analysis of variance, two-factor, null, hypothesis Section: 12-3 Two-Factor Analysis of Variance With Replication Outcome: 5 67) Which of the following is an assumption for the one-way analysis of variance experimental design? A) All populations are normally distributed. B) The populations have equal variances. C) The observations are independent. D) All of the above Answer: D Diff: 1 Keywords: ANOVA, analysis of variance, independent, variance, normal Section: 12-1 One-Way Analysis of Variance Outcome: 1

68) A hotel chain has four hotels in Oregon. The general manager is interested in determining whether the mean length of stay is the same or different for the four hotels. She selects a random sample of n = 20 guests at each hotel and determines the number of nights they stayed. Assuming that she plans to test this using an alpha level equal to 0.05, which of the following is the correct critical value? A) F = 3.04 B) F = 2.76 C) t = 1.9917 D) F = 2.56 Answer: B Diff: 2 Keywords: mean, random, alpha, F-value, F-critical Section: 12-1 One-Way Analysis of Variance Outcome: 2 69) A hotel chain has four hotels in Oregon. The general manager is interested in determining whether the mean length of stay is the same or different for the four hotels. She selects a random sample of n = 20 guests at each hotel and determines the number of nights they stayed. Assuming that she plans to test this using an alpha level equal to 0.05, which of the following is the appropriate alternative hypothesis? A) H0 : μ1 = μ2 = μ3 = μ4 B) H0 : μ1 ≠ μ2 ≠ μ3 ≠ μ4 C) Not all population means are equal. D) σ1 = σ2 = σ3 = σ4 Answer: C Diff: 2 Keywords: mean, null, alternative, hypothesis, alpha Section: 12-1 One-Way Analysis of Variance Outcome: 2 70) The State Transportation Department is thinking of changing its speed limit signs. It is considering two new options in addition to the existing sign design. At question is whether the three sign designs will produce the same mean speed. To test this, the department has conducted a limited test in which a stretch of roadway was selected. With the original signs up, a random sample of 30 cars was selected and the speeds were measured. Then, on different days, the two new designs were installed, 30 cars each day were sampled, and their speeds were recorded. Suppose that the following summary statistics were computed based on the data:

Based on these sample results and significance level equal to 0.05, what is the critical value for this hypothesis test? A) F = approximately 3.15 B) F = approximately 4.90 C) F = approximately 29.47 D) F = approximately 2.70 Answer: A Diff: 2 Keywords: ANOVA, analysis of variance, critical value, F-critical Section: 12-1 One-Way Analysis of Variance Outcome: 2 12-20 Copyright © 2018 Pearson Education, Inc.

71) The State Transportation Department is thinking of changing its speed limit signs. It is considering two new options in addition to the existing sign design. At question is whether the three sign designs will produce the same mean speed. To test this, the department has conducted a limited test in which a stretch of roadway was selected. With the original signs up, a random sample of 30 cars was selected and the speeds were measured. Then, on different days, the two new designs were installed, 30 cars each day were sampled, and their speeds were recorded. Suppose that the following summary statistics were computed based on the data:

The appropriate test to conduct to determine if the population means are equal is: A) Hartley's F-max test. B) one-way analysis of variance. C) three sample t-test. D) randomized complete block analysis of variance. Answer: B Diff: 2 Keywords: ANOVA, analysis of variance, population, mean Section: 12-1 One-Way Analysis of Variance Outcome: 2 72) The State Transportation Department is thinking of changing its speed limit signs. It is considering two new options in addition to the existing sign design. At question is whether the three sign designs will produce the same mean speed. To test this, the department has conducted a limited test in which a stretch of roadway was selected. With the original signs up, a random sample of 30 cars was selected and the speeds were measured. Then, on different days, the two new designs were installed, 30 cars each day were sampled, and their speeds were recorded. Suppose that the following summary statistics were computed based on the data:

Based on these sample results and significance level equal to 0.05, the sum of squares between is: A) approximately 3,586. B) approximately 2,430. C) approximately 1,215. D) None of the above Answer: B Diff: 3 Keywords: ANOVA, analysis of variance, sum of squares Section: 12-1 One-Way Analysis of Variance Outcome: 2

73) In conducting a one-way analysis of variance where the test statistic is less than the critical value, which of the following is correct? A) Conclude that the means are not all the same and that that the Tukey-Kramer procedure should be conducted. B) Conclude that the means are not all the same and that that the Tukey-Kramer procedure is not needed. C) Conclude that all means are the same and that the Tukey-Kramer procedure should be conducted. D) Conclude that all means are the same and there is no need to conduct the Tukey-Kramer procedure. Answer: D Diff: 2 Keywords: ANOVA, analysis of variance, one-way, Tukey-Kramer Section: 12-1 One-Way Analysis of Variance Outcome: 3 74) The State Transportation Department is thinking of changing its speed limit signs. It is considering two new options in addition to the existing sign design. At question is whether the three sign designs will produce the same mean speed. To test this, the department has conducted a limited test in which a stretch of roadway was selected. With the original signs up, a random sample of 30 cars was selected and the speeds were measured. Then, on different days, the two new designs were installed, 30 cars each day were sampled, and their speeds were recorded. Suppose that the following summary statistics were computed based on the data:

Based on these sample results and a significance level equal to 0.05, assuming that the null hypothesis of equal means has been rejected, the Tukey-Kramer critical range is: A) 1.96. B) approximately 4.0. C) Can't be determined without more information D) None of the above Answer: B Diff: 3 Keywords: ANOVA, analysis of variance, Tukey-Kramer, mean, null Section: 12-1 One-Way Analysis of Variance Outcome: 3 75) In a one-way analysis of variance test in which the levels of the factor being analyzed are randomly selected from a large set of possible factors, the design is referred to as: A) a fixed-effects design. B) a random-effects design. C) an undetermined results design. D) a balanced design. Answer: B Diff: 2 Keywords: ANOVA, analysis of variance, one-way, random, random-effects Section: 12-1 One-Way Analysis of Variance Outcome: 2

76) An Internet service provider is interested in testing to see if there is a difference in the mean weekly connect time for users who come into the service through a dial-up line, DSL, or cable Internet. To test this, the ISP has selected random samples from each category of user and recorded the connect time during a week period. The following data were collected: Dial Up 19.2 17.7 17.2 18.9 26.9 22.6 31.2

DSL 40.6 40 41.5 30.5 46.8

Cable 39.5 42.3 47 45.4 41.1 43.2 39.9 41.9 49.3

Which of the following would be the correct alternative hypotheses for the test to be conducted? A) H0 : μ1 = μ2 = μ3 B) H0 : μ1 ≠ μ2 ≠μ3 C) Not all population means are equal. D) σ1 = σ2 = σ3 = σ4 Answer: C Diff: 2 Keywords: ANOVA, analysis of variance, alternative, hypothesis Section: 12-1 One-Way Analysis of Variance Outcome: 2

77) An Internet service provider is interested in testing to see if there is a difference in the mean weekly connect time for users who come into the service through a dial-up line, DSL, or cable Internet. To test this, the ISP has selected random samples from each category of user and recorded the connect time during a week period. The following data were collected: Dial Up 19.2 17.7 17.2 18.9 26.9 22.6 31.2

DSL 40.6 40 41.5 30.5 46.8

Cable 39.5 42.3 47 45.4 41.1 43.2 39.9 41.9 49.3

Assuming that the test is to be conducted at a 0.01 level of significance, what would the critical value be for this test? A) F = 1.93 B) F = 3.555 C) t = 2.8784 D) F = 6.013 Answer: D Diff: 2 Keywords: ANOVA, analysis of variance, critical value Section: 12-1 One-Way Analysis of Variance Outcome: 2 78) Assume you are conducting a one-way analysis of variance using a 0.05 level of significance and have found that the p-value = 0.06. Which of the follow is correct regarding what you can conclude? A) Do not reject the null hypothesis; the means are all the same. B) Reject the null hypothesis; the means are not all the same. C) Do not reject the null hypothesis; the means are not all the same. D) Reject the null hypothesis; the means are all the same. Answer: A Diff: 2 Keywords: ANOVA, analysis of variance, p-value Section: 12-1 One-Way Analysis of Variance Outcome: 2

79) An Internet service provider is interested in testing to see if there is a difference in the mean weekly connect time for users who come into the service through a dial-up line, DSL, or cable Internet. To test this, the ISP has selected random samples from each category of user and recorded the connect time during a week period. The following data were collected: Dial Up 19.2 17.7 17.2 18.9 26.9 22.6 31.2

DSL 40.6 40 41.5 30.5 46.8

Cable 39.5 42.3 47 45.4 41.1 43.2 39.9 41.9 49.3

Based upon these data and a significance level of 0.05, which of the following statements is true? A) The F-critical value for the test is 3.555 B) The test statistic is approximately 43.9 C) The null hypothesis should be rejected and conclude that the mean connect times for the three user categories are not all equal. D) All of the above are true. Answer: D Diff: 3 Keywords: ANOVA, analysis of variance, test statistic, F-critical, null, hypothesis Section: 12-1 One-Way Analysis of Variance Outcome: 2

80) A fast food chain operation is interested in determining whether the mean per customer purchase differs by day of the week. To test this, it has selected random samples of customers for each day of the week. The analysts then ran a one-way analysis of variance generating the following output: ANOVA: Single Factor

Based upon this output, which of the following statements is true if the test is conducted at the 0.05 level of significance? A) There is no basis for concluding that mean sales is different for the different days of the week. B) Based on the p-value, the null hypothesis should be rejected since the p-value exceeds the alpha level. C) The experiment is conducted as an unbalanced design. D) Based on the critical value, the null should be rejected. Answer: A Diff: 2 Keywords: ANOVA, analysis of variance, F-critical, F-test, null, hypothesis Section: 12-1 One-Way Analysis of Variance Outcome: 2 81) In order for a one-way analysis of variance to be considered a balanced design, which of the following must hold? A) The population variances must be equal. B) The sample sizes selected from each population must be equal. C) The study must have the same number of rows as it does columns. D) All of the above are true. Answer: B Diff: 2 Keywords: ANOVA, analysis of variance, balanced, sample size Section: 12-1 One-Way Analysis of Variance Outcome: 2

82) In a one-way design, which of the following is true? A) The populations must have equal means. B) The sample sizes must be equal. C) The mean squares between will be larger than the mean squares within if the null hypothesis is rejected. D) The sample sizes must all differ. Answer: C Diff: 2 Keywords: ANOVA, analysis of variance, one-way, mean squares Section: 12-1 One-Way Analysis of Variance Outcome: 2 83) In a one-way ANOVA, which of the following is true? A) The degrees of freedom associated with the between sum of squares is equal to one less than the number of populations. B) The critical value will be an F-value from the F distribution. C) If the null hypothesis is rejected, it may still be possible that two or more of the population means are equal. D) All of the above Answer: D Diff: 2 Keywords: ANOVA, analysis of variance, one-way, degrees of freedom, F-value, hypothesis Section: 12-1 One-Way Analysis of Variance Outcome: 2 84) To test the mileage efficiency of three new car models, random samples of various sample sizes were selected from each of the three cars and the mpg data obtained are shown below. Model A 37 33 36 38

Model B 43 39 35 38 40

Model C 28 32 33

Based on the sample date, one can conclude that A) all three car models have the same mean mpg. B) at least two car models have different mpgs. C) Model C has a higher mpg than Model A. D) None of the above Answer: B Diff: 3 Keywords: ANOVA, one-way, hypothesis Section: 12-1 One-Way Analysis of Variance Outcome: 2

85) Which of the following describes a treatment in a randomized complete block analysis of variance? A) A treatment is a combination of one level of each factor. B) A treatment is a level associated with each factor of the experiment. C) A treatment is another term for the data that are collected in the experiment. D) A treatment is considered to be the analysis that is performed on the sample data. Answer: A Diff: 2 Keywords: ANOVA, analysis of variance randomized block, factor, treatment Section: 12-2 Randomized Complete Block Analysis of Variance Outcome: 4 86) In a randomized complete block design analysis of variance, how many factors are there to be analyzed? A) Depends on the sample size in each treatment B) One factor, but multiple levels C) Two factors D) Can't be determined without additional information Answer: C Diff: 1 Keywords: ANOVA, analysis of variance, randomized, block, factor Section: 12-2 Randomized Complete Block Analysis of Variance Outcome: 4 87) In a randomized complete block design analysis of variance, which of the following correctly describes the number of degrees of freedom associated with the between sum of squares? A) One less than the number of populations involved B) One less than the number of blocks C) One less than the combined sample size in the experiment D) One less than the total number of observations Answer: A Diff: 2 Keywords: ANOVA, analysis of variance, randomized, block, degrees of freedom, sum of squares, between Section: 12-2 Randomized Complete Block Analysis of Variance Outcome: 4

88) A test is conducted to compare three different income tax software packages to determine whether there is any difference in the average time it takes to prepare income tax returns using the three different software packages. Ten different person's income tax returns are done by each of the three software packages and the time is recorded for each. Which of the following is true? A) The total degrees of freedom is 30. B) The between blocks degrees of freedom is 2. C) The between samples degrees of freedom is 2. D) The three software packages are the blocks. Answer: C Diff: 2 Keywords: ANOVA, analysis of variance, block, degrees of freedom Section: 12-2 Randomized Complete Block Analysis of Variance Outcome: 4 89) A test is conducted to compare three different income tax software packages to determine whether there is any difference in the average time it takes to prepare income tax returns using the three different software packages. Ten different person's income tax returns are done by each of the three software packages and the time is recorded for each. Assuming that results show that blocking was effective, this means that: A) there are significant differences in the average times needed by the 3 different software packages. B) there are significant differences in the average times needed for the 10 different person's tax returns. C) the analysis should be redone using a one-way analysis of variance. D) the randomized complete block was the wrong method to use. Answer: B Diff: 2 Keywords: ANOVA, analysis of variance, randomized, complete block Section: 12-2 Randomized Complete Block Analysis of Variance Outcome: 4

90) Recently, a department store chain was interested in determining if there was a difference in the mean number of customers who enter the three stores in Seattle. The analysts set up a study in which the number of people entering the stores was counted depending on whether the day of the week was Saturday, Sunday, or a weekday. The following data were collected:

Saturday Sunday Weekday

Store A 176 145 108

Store B 300 290 150

Store C 56 40 40

Given this format, which of the following is true? A) The day of the week would be considered the blocking factor in the study. B) There are six treatments. C) This is a balanced design since the number of rows and columns is equal. D) All of the above are true. Answer: A Diff: 2 Keywords: ANOVA, analysis of variance, block, factor Section: 12-2 Randomized Complete Block Analysis of Variance Outcome: 4 91) Recently, a department store chain was interested in determining if there was a difference in the mean number of customers who enter the three stores in Seattle. The analysts set up a study in which the number of people entering the stores was counted depending on whether the day of the week was Saturday, Sunday, or a weekday. The following data were collected:

Saturday Sunday Weekday

Store A 176 145 108

Store B 300 290 150

Store C 56 40 40

Given this format, what is the null hypothesis for testing whether blocking is effective? A) H0 : μA = μB = μC B) H0 : μSat = μSun = μWeek C) Not all means are equal. D) H0 : σ1 = σ2 = σ3 Answer: B Diff: 2 Keywords: ANOVA, analysis of variance, block, null, hypothesis Section: 12-2 Randomized Complete Block Analysis of Variance Outcome: 4

92) Recently, a department store chain was interested in determining if there was a difference in the mean number of customers who enter the three stores in Seattle. The analysts set up a study in which the number of people entering the stores was counted depending on whether the day of the week was Saturday, Sunday, or a weekday. The following data were collected:

Saturday Sunday Weekday

Store A 176 145 108

Store B 300 290 150

Store C 56 40 40

Given this format and testing using an alpha level equal to 0.05, which of the following statements is true? A) The total degrees of freedom is 9. B) The between blocks degrees of freedom equals 8. C) The between samples degrees of freedom equals 3. D) The within sample degrees of freedom equals 4. Answer: D Diff: 2 Keywords: ANOVA, analysis of variance, block, degrees of freedom Section: 12-2 Randomized Complete Block Analysis of Variance Outcome: 4 93) A test is conducted to compare three different income tax software packages to determine whether there is any difference in the average time it takes to prepare income tax returns using the three different software packages. Ten different person's income tax returns are done by each of the three software packages and the time is recorded for each. Given this format and testing using an alpha level equal to 0.05, the critical value associated with the primary hypothesis test is: A) 3.555 B) 2.456 C) 19.385 D) 4.256 Answer: A Diff: 2 Keywords: ANOVA, analysis of variance, critical value, F-test, block Section: 12-2 Randomized Complete Block Analysis of Variance Outcome: 4

94) A major consumer group recently undertook a study to determine whether automobile customers would rate the quality of cars differently whether they were manufactured in the U.S., Europe, or Japan. To conduct this test, a sample of 20 individuals was asked to look at mid-range model cars made in each of the three countries. The individuals in the sample were then asked to provide a rating for each car on a scale of 1 to 1000. The following computer output resulted, and the tests were conducted using a significance level equal to 0.05. ANOVA: Two-Factor Without Replication

Based upon the data, which of the following statements is true? A) Blocking was effective. B) Blocking was not effective. C) The primary null hypothesis should not be rejected. D) The averages for the 20 people are not all the same. Answer: B 12-32 Copyright © 2018 Pearson Education, Inc.

Diff: 2 Keywords: ANOVA, analysis of variance, block, hypothesis Section: 12-2 Randomized Complete Block Analysis of Variance Outcome: 4 95) A golf ball manufacturer has three dimple patterns it is interested in analyzing to see whether one results in longer driving distances. However, it also wishes to control for the cover material the ball is made from since it believes that the material might affect driving distance. Four materials can be used. The following data represent the results of tests in which each combination of dimple pattern and cover material were used and the length of the ball hit by a robot has been recorded. The test will be conducted using an alpha = 0.05 level.

Material A Material B Material C Material D

Pattern 1 257 250 230 266

Pattern 2 248 247 260 256

Pattern 3 260 255 240 280

Given these data, which of the following statements is true? A) There is no basis for concluding that mean driving distance is different for the different dimple patterns. B) There is no basis for concluding that mean driving distance is different for the different cover materials. C) Both A and B are true. D) Neither A nor B is true. Answer: C Diff: 3 Keywords: ANOVA, analysis of variance, factor, block, hypothesis Section: 12-2 Randomized Complete Block Analysis of Variance Outcome: 4

96) A golf ball manufacturer has three dimple patterns it is interested in analyzing to see whether one results in longer driving distances. However, it also wishes to control for the material the ball is made from since it believes that the material might affect driving distance. Four materials can be used. The following data represent the results of tests in which each combination of dimple pattern and cover material were used and the length of the ball hit by a robot has been recorded. The test will be conducted using an alpha = 0.05 level.

Material A Material B Material C Material D

Pattern 1 257 250 230 266

Pattern 2 248 247 260 256

Pattern 3 260 255 240 280

Given these data, what is the value of Fisher's Least Significant Difference critical value? A) Approximately 19.06 B) 2.4469 C) About 7.65 D) None of the above Answer: A Diff: 3 Keywords: ANOVA, analysis of variance, block, Fisher, Fisher's Least Significant Difference, critical value Section: 12-2 Randomized Complete Block Analysis of Variance Outcome: 4 97) A large orchard owner in the state of Washington is interested in determining whether the mean number of bushels of peaches per acre is the same or different depending on the type of tree that is used. He also thinks that production may be affected by the type of fertilizer that is used. To test, he has set up a test in which a one-acre plot of peach trees with a combination each of 5 varieties and 3 fertilizer types are studied. In this case, the number of treatments is: A) 5 B) 3 C) 15 D) Can't be determined without knowing how many trees are planted on each acre. Answer: C Diff: 2 Keywords: ANOVA, analysis of variance, treatment, block Section: 12-2 Randomized Complete Block Analysis of Variance Outcome: 4

98) A large orchard owner in the state of Washington is interested in determining whether the mean number of bushels of peaches per acre is the same or different depending on the type of tree that is used. He also thinks that production may be affected by the type of fertilizer that is used. To test, he has set up a test in which a one-acre plot of peach trees with a combination each of 5 varieties and 3 fertilizer types are studied. The following data reflect the number of bushels of peaches on each acre plot.

Fertilizer A Fertilizer B Fertilizer C

Tree Type 1 300 150 300

Tree Type 2 400 200 300

Tree Type 3 200 100 400

Tree Type 4 500 150 200

Tree Type 5 400 200 50

Assuming that the hypothesis tests will be conducted using an alpha equal 0.05 level, which of the following is true? A) The total sum of squares is approximately 4,570,900. B) The grower was justified in controlling for the fertilizer type since the test shows that blocking was effective. C) Based on the data, the grower can conclude that there is a difference in mean production of peaches across the different types of tree. D) A, B and C are all true. Answer: B Diff: 3 Keywords: ANOVA, analysis of variance, block Section: 12-2 Randomized Complete Block Analysis of Variance Outcome: 4 99) A large orchard owner in the state of Washington is interested in determining whether the mean number of bushels of peaches per acre is the same or different depending on the type of tree that is used. He also thinks that production may be affected by the type of fertilizer that is used. To test, he has set up a test in which a one-acre plot of peach trees with a combination each of 5 varieties and 3 fertilizer types are studied. The following data reflect the number of bushels of peaches on each acre plot.

Fertilizer A Fertilizer B Fertilizer C

Tree Type 1 Tree Type 2 Tree Type 3 Tree Type 4 Tree Type 5 300 400 200 500 400 150 200 100 150 200 300 300 400 200 500

Assuming that the hypothesis tests will be conducted using an alpha equal 0.05 level, what is the value of the Fisher's LSD critical value for doing the multiple comparisons? A) Approximately 16.78 B) About 11.30 C) Approximately 186.7 D) Need to know the number of trees planted on each acre. Answer: C Diff: 3 Keywords: ANOVA, analysis of variance, LSD, Fisher, block Section: 12-2 Randomized Complete Block Analysis of Variance Outcome: 4 12-35 Copyright © 2018 Pearson Education, Inc.

100) A car company is interested in testing to see whether the mean miles that a car engine will last without changing oil is the same or different depending on which brand of oil is used. The engineers also wish to control for the type of transmission (manual or automatic) that is used. To conduct this test, the car company obtains enough engines so that all four oil brands can be tested in a design that involves no replication. Based on this information, how many engines will be needed to conduct the test? A) 4 B) 8 C) One for each oil type D) One for each factor in the study Answer: B Diff: 1 Keywords: ANOVA, analysis of variance, block, factor Section: 12-2 Randomized Complete Block Analysis of Variance Outcome: 4

101) A test is conducted to compare three different income tax software packages to determine whether there is any difference in the average time it takes to prepare income tax returns using the three different software packages. Ten different persons' income tax returns are done by each of the three software packages and the time is recorded for each. The computer results are shown below. SUMMARY

Count 1 2 3 4 5 6 7 8 9 10

Software A Software B Software C ANOVA Source of Variation Rows Columns Error Total

Sum

Average

Variance

3 3 3 3 3 3 3 3 3 3

9 30 12 6.5 25 7 10 18 33.5 4.5

3 10 4 2.166667 8.333333 2.333333 3.333333 6 11.16667 1.5

1 1 0 0.583333 2.333333 1.083333 0.333333 1 0.583333 0.25

47.5

4.75

12.95833

47.5

4.75

10.79167

60.5

6.05

13.46944

36.65648

130.227

1.6E-14

2.456281

11.26667

5.633333

20.01316

2.66E-05

3.554557

5.066667

0.281481

346.2417

329.9083

P-value

F crit

Based on these results and using a 0.05 level of significance which is correct regarding blocking? A) Blocking was not effective because p-value = 2.66 is greater than 0.05. B) Blocking was effective because p-value = 2.66E - 5 is less than 0.05. C) Blocking was not effective because p-value = 1.6 is greater than 0.05. D) Blocking was effective because p-value = 1.6E-14 is less than 0.05. Answer: D Diff: 2 Keywords: ANOVA, analysis of variance randomized block, p-value Section: 12-2 Randomized Complete Block Analysis of Variance Outcome: 4

102) A test is conducted to compare three different income tax software packages to determine whether there is any difference in the average time it takes to prepare income tax returns using the three different software packages. Ten different persons' income tax returns are done by each of the three software packages and the time is recorded for each. The computer results are shown below. SUMMARY

Count 1 2 3 4 5 6 7 8 9 10

Software A Software B Software C ANOVA Source of Variation Rows Columns Error Total

Sum

Average

Variance

3 3 3 3 3 3 3 3 3 3

9 30 12 6.5 25 7 10 18 33.5 4.5

3 10 4 2.166667 8.333333 2.333333 3.333333 6 11.16667 1.5

1 1 0 0.583333 2.333333 1.083333 0.333333 1 0.583333 0.25

47.5

4.75

12.95833

47.5

4.75

10.79167

60.5

6.05

13.46944

36.65648

130.227

1.6E-14

2.456281

11.26667

5.633333

20.01316

2.66E-05

3.554557

5.066667

0.281481

346.2417

329.9083

P-value

F crit

Based on these results and using a 0.05 level of significance which is correct regarding the primary hypothesis? A) The three software packages are not all the same because p-value = 1.6E-14 is less than 0.05. B) The three software packages are all the same because p-value = 1.6 is greater than 0.05. C) The three software packages are not all the same because p-value = 2.66E-5 is less than 0.05. D) The three software packages are all the same because p-value = 2.66 is greater than 0.05. Answer: C Diff: 2 Keywords: ANOVA, analysis of variance, block, effect, p-value Section: 12-2 Randomized Complete Block Analysis of Variance Outcome: 4

103) A test is conducted to compare three different income tax software packages to determine whether there is any difference in the average time it takes to prepare income tax returns using the three different software packages. Ten different persons' income tax returns are done by each of the three software packages and the time is recorded for each. The computer results are shown below. SUMMARY

Count 1 2 3 4 5 6 7 8 9 10

Software A Software B Software C ANOVA Source of Variation Rows Columns Error Total

Sum

Average

Variance

3 3 3 3 3 3 3 3 3 3

9 30 12 6.5 25 7 10 18 33.5 4.5

3 10 4 2.166667 8.333333 2.333333 3.333333 6 11.16667 1.5

1 1 0 0.583333 2.333333 1.083333 0.333333 1 0.583333 0.25

47.5

4.75

12.95833

47.5

4.75

10.79167

60.5

6.05

13.46944

36.65648

130.227

1.6E-14

2.456281

11.26667

5.633333

20.01316

2.66E-05

3.554557

5.066667

0.281481

346.2417

329.9083

P-value

F crit

Assuming that the hypothesis tests are conducted using a significance level equal to 0.05, the Fisher's LSD value for multiple comparisons is: A) approximately 0.4985. B) about 0.91. C) approximately 1.91. D) about 0.5387. Answer: A Diff: 2 Keywords: ANOVA, analysis of variance, Fisher, LSD Section: 12-2 Randomized Complete Block Analysis of Variance Outcome: 4

104) A car company is interested in testing to see whether the mean miles that a car engine will last without changing oil is the same or different depending on which brand of oil is used. The engineers also wish to control for the type of transmission (manual or automatic) that is used. To conduct this test, the car company obtains enough engines so that all four oil brands can be tested in a design that involves no replication. The following data reflect the miles the engine lasted until problems were encountered. Data are in thousands of miles.

Manual Automatic

Oil 1 58 83

Oil 2 40.8 65.9

Oil 3 60.8 90.5

Oil 4 40 68.6

Assuming that the hypothesis tests are conducted using a significance level equal to 0.05, which of the following statements is true? A) Based on the data, Oil 1 and Oil 3 give statistically more miles on average than do the other two oils. B) The type of transmission does seem to have an impact on the mean miles that an engine will last. C) The F-critical value for testing whether blocking is effective is 10.128. D) All of the above are true. Answer: D Diff: 3 Keywords: ANOVA, analysis of variance, F-critical, block, hypothesis Section: 12-2 Randomized Complete Block Analysis of Variance Outcome: 4 105) Which of the following is NOT one of the assumptions required by the randomized block design? A) The populations are normally distributed. B) The populations have equal variances. C) The observations within samples are dependent. D) The data measurement must be interval or ratio level. Answer: C Diff: 1 Keywords: ANOVA, analysis of variance, randomized block design Section: 12-2 Randomized Complete Block Analysis of Variance Outcome: 4 106) The correct null hypothesis to determine whether interaction exists in a two-factor ANOVA is A) H0: Factors A and B interact to affect the mean response B) H0: Factors A and B do not interact to affect the mean response C) HA: Factors A and B do not interact to affect the mean response D) None of the above Answer: B Diff: 1 Keywords: ANOVA, analysis of variance, two-factor, interaction Section: 12-3 Two-Factor Analysis of Variance With Replication Outcome: 5

107) Which of the following is the minimum number of required replications per cell for a two-factor ANOVA design if you plan to test for interactive effects between factors A and B? A) 3 B) 1 C) 2 D) 5 Answer: C Diff: 1 Keywords: ANOVA, analysis of variance, interaction, factor Section: 12-3 Two-Factor Analysis of Variance With Replication Outcome: 5 108) Considering the following printout from a two-factor ANOVA design, how many levels of factor A (Sample) were there in this study?

A) 4 B) 3 C) 2 D) 6 Answer: A Diff: 2 Keywords: ANOVA, analysis of variance, level, factor Section: 12-3 Two-Factor Analysis of Variance With Replication Outcome: 5

109) Considering the following printout for a two-factor ANOVA design, which of the following is a proper conclusion to reach?

A) There is no significant interaction between the two factors. B) The levels of factor A (Sample) have significantly different means. C) The levels of factor B (Columns) have significantly different means. D) The total number of observations is 47. Answer: A Diff: 2 Keywords: ANOVA, analysis of variance, interaction, factor Section: 12-3 Two-Factor Analysis of Variance With Replication Outcome: 5 110) Considering the following printout for a two-factor ANOVA study, which of the following is the number of replications used?

A) 2 B) 5 C) 4 D) Can't be determined without more information. Answer: C Diff: 2 Keywords: ANOVA, analysis of variance, replication Section: 12-3 Two-Factor Analysis of Variance With Replication Outcome: 5 12-42 Copyright © 2018 Pearson Education, Inc.

111) A national car rental company recently conducted a study recently in which cars with automatic and standard transmissions (factor A-Sample) were rented to male and female customers (factor B-Columns). Three customers in each category were randomly selected and the miles driven per day was recorded as follows:

Based on the design of this study, how many degrees of freedom will be associated with the mean square for factor A? A) 1 B) 2 C) 3 D) 8 Answer: A Diff: 2 Keywords: ANOVA, analysis of variance, degrees of freedom, mean square Section: 12-3 Two-Factor Analysis of Variance With Replication Outcome: 5 112) A national car rental company recently conducted a study recently in which cars with automatic and standard transmissions (factor A-Sample) were rented to male and female customers (factor B-Columns). Three customers in each category were randomly selected and the miles driven per day was recorded as follows:

Based on these sample data, and alpha = .05, which of the following statements is true? A) The means for factor A are significantly different. B) There is no significant interaction between factors A and B. C) The means for factor B are significantly different. D) All of the above statements are true. Answer: B Diff: 3 Keywords: ANOVA, analysis of variance, interaction, factor Section: 12-3 Two-Factor Analysis of Variance With Replication Outcome: 5 12-43 Copyright © 2018 Pearson Education, Inc.

113) A two-factor analysis of variance is conducted to test the effect the price and advertising have on sales of a particular brand of bottled water. Each week a combination of particular levels of price and advertising are used and the sales level is recorded. The computer results are shown below. ANOVA Source of Variation Sample (advertising) Columns (price) Interaction Within

SS 99.73324 1150.432 1577.526 341.835

Total

3169.526

df 1 2 2 18

MS F p-value F-crit 99.73324 5.251652 0.034201 4.413873 575.2161 30.28914 1.74E-06 3.554557 788.7629 41.53387 1.8E-07 3.554557 18.99083

How many replications were used in this study? A) 2 B) 3 C) 4 D) 5 Answer: C Diff: 2 Keywords: ANOVA, analysis of variance, two-way, degrees of freedom Section: 12-3 Two-Factor Analysis of Variance With Replication Outcome: 5 114) A two-factor analysis of variance is conducted to test the effect the price and advertising have on sales of a particular brand of bottled water. Each week a combination of particular levels of price and advertising are used and the sales level is recorded. The computer results are shown below. ANOVA Source of Variation Sample (advertising) Columns (price) Interaction Within

SS 99.73324 1150.432 1577.526 341.835

Total

3169.526

df 1 2 2 18

MS F p-value F-crit 99.73324 5.251652 0.034201 4.413873 575.2161 30.28914 1.74E-06 3.554557 788.7629 41.53387 1.8E-07 3.554557 18.99083

Based on the results above, which of the following is correct? A) 1 level of advertising and 2 levels of price were used. B) 3 levels of adverting and 2 levels of price were used. C) 2 levels of advertising and 3 levels of price were used. D) There were a total of 6 different treatments. Answer: C Diff: 2 Keywords: ANOVA, analysis of variance, hypothesis, null Section: 12-3 Two-Factor Analysis of Variance With Replication Outcome: 5 12-44 Copyright © 2018 Pearson Education, Inc.

115) A two-factor analysis of variance is conducted to test the effect that price and advertising have on sales of a particular brand of bottled water. Each week a combination of particular levels of price and advertising are used and the sales amount is recorded. The computer results are shown below. ANOVA Source of Variation Sample (advertising) Columns (price) Interaction Within

SS 99.73324 1150.432 1577.526 341.835

Total

3169.526

df 1 2 2 18

MS F p-value F-crit 99.73324 5.251652 0.034201 4.413873 575.2161 30.28914 1.74E-06 3.554557 788.7629 41.53387 1.8E-07 3.554557 18.99083

Based on the results above and a 0.05 level of significance, which of the following is correct? A) There is no interaction between price and advertising, so results for individual factors may be misleading. B) There is interaction between price and advertising, so the above results for individual factors may be misleading. C) There is no interaction between price and advertising, and both factors significantly affect sales. D) There is interaction between price and advertising, so the above results conclusively show that both factors affect price. Answer: B Diff: 2 Keywords: ANOVA, analysis of variance, interaction, factor Section: 12-3 Two-Factor Analysis of Variance With Replication Outcome: 5

116) A start-up cell phone applications company is interested in determining whether house-hold incomes are different for subscribers to three different service providers. A random sample of 25 subscribers to each of the three service providers was taken, and the annual household income for each subscriber was recorded. The partially completed ANOVA table for the analysis is shown here:

Complete the ANOVA table by filling in the missing sums of squares, the degrees of freedom for each source, the mean square, and the calculated F-test statistic. A) Between Groups df = 2, Within Groups df = 74, SSW = 6,322,592,933, MSB = 1,474,542,579, MSW = 87,813,791, F = 16.79 B) Between Groups df = 2, Within Groups df = 72, SSW = 6,322,592,933, MSB = 1,474,542,579, MSW = 87,813,791, F = 16.79. C) Between Groups df = 2, Within Groups df = 72, SSW = 6,322,592,933, MSB = 1,474,542,579, MSW = 87,813,791, F = 9.73 D) Between Groups df = 2, Within Groups df = 72, SSW = 6,322,592,933, MSB = 87,813,791, MSW = 87,813,791, F = 16.79 Answer: B Diff: 2 Keywords: ANOVA, analysis of variance, one-way Section: 12-1 One-Way Analysis of Variance Outcome: 1

117) A start-up cell phone applications company is interested in determining whether house-hold incomes are different for subscribers to three different service providers. A random sample of 25 subscribers to each of the three service providers was taken, and the annual household income for each subscriber was recorded. The partially completed ANOVA table for the analysis is shown here:

Based on the sample results, can the start-up firm conclude that there is a difference in household incomes for subscribers to the three service providers? You may assume normal distributions and equal variances. Conduct your test at the alpha= 0.10 level of significance. Be sure to state a critical F-statistic, a decision rule, and a conclusion. A) H0 : µ1 = µ2 = µ3 HA: Not all populations have the same mean F = MSB/MSW = 1,474,542,579/87,813,791 = 16.79 Because the F test statistic = 16.79 > Fα = 2.3778, we do reject the null hypothesis based on these sample data. B) H0 : µ1 = µ2 = µ3 HA : Not all populations have the same mean F = MSB/MSW = 87,813,791 /1,474,542,579= 0.060 Because the F test statistic = 0.060 < Fα = 2.3778, we do not reject the null hypothesis based on these sample data. C) H0 : µ1 = µ2 = µ3 HA : Not all populations have the same mean F = SSW/MSW = 6,322,592,933/87,813,791 = 72 Because the F test statistic = 72 > Fα = 2.3778, we do reject the null hypothesis based on these sample data. D) H0 : µ1 = µ2 = µ3 HA : Not all populations have the same mean F = SSW/MSW = 6,322,592,933/1,474,542,579= 4.28 Because the F test statistic = 4.28 > Fα = 2.3778, we do reject the null hypothesis based on these sample data Answer: A Diff: 2 Keywords: ANOVA, analysis of variance, one-way Section: 12-1 One-Way Analysis of Variance Outcome: 1

118) A manager is interested in testing whether three populations of interest have equal population means. Simple random samples of size 10 were selected from each population. The following ANOVA table and related statistics were computed:

Conduct the appropriate test of the null hypothesis assuming that the populations have equal variances and the populations are normally distributed. Use a 0.05 level of significance. A) Using the F test approach, because F = 3.354 < critical F = 9.84, we reject the null hypothesis and conclude that the population means are not all equal. B) Using the F test approach, because F = 3.354 < critical F = 9.84, we do not reject the null hypothesis and conclude that the population means are all equal. C) Using the F test approach, because F = 9.84 > critical F = 3.35, we reject the null hypothesis and conclude that the population means are not all equal. D) Using the F test approach, because F = 9.84 > critical F = 3.35, we do not reject the null hypothesis and conclude that the population means are all equal. Answer: C Diff: 2 Keywords: ANOVA, analysis of variance, randomized block Section: 12-2 Randomized Complete Block Analysis of Variance Outcome: 1

119) Respond to the following questions using this partially completed one-way ANOVA table: Source of Variation Between Samples Within Samples Total

SS 1,745 ______ 6,504

F-ratio

240 246

How many different populations are being considered in this analysis? A) 5 B) 8 C) 7 D) 6 Answer: C Diff: 2 Keywords: ANOVA, analysis of variance, one-way Section: 12-1 One-Way Analysis of Variance Outcome: 1 120) Respond to the following questions using this partially completed one-way ANOVA table: Source of Variation Between Samples Within Samples Total

SS 1,745 ______ 6,504

F-ratio

240 246

Fill in the ANOVA table with the missing values. A) Between Samples df = 6, MSB = 290.833, F-ratio = 14.667, SSW = 4,759, MSW = 19.829 B) Between Samples df = 6, MSB = 290.833, F-ratio = 7.948, SSW = 4,759, MSW = 19.829 C) Between Samples df = 5, MSB = 290.833, F-ratio = 14.667, SSW = 4,759, MSW = 19.829 D) Between Samples df = 5, MSB = 290.833, F-ratio = 7.948, SSW = 4,759, MSW = 19.829 Answer: A Diff: 2 Keywords: ANOVA, analysis of variance, one-way Section: 12-1 One-Way Analysis of Variance Outcome: 1

121) Respond to the following questions using this partially completed one-way ANOVA table: Source of Variation Between Samples Within Samples Total

SS 1,745 ______ 6,504

F-ratio

240 246

Based on the analysis of variance F-test, what conclusion should be reached regarding the null hypothesis? Test using a significance level of 0.01. A) Since 7.948 > 2.8778 accept H0 and conclude that all population means are the same. B) Since 14.667 > 2.8778 accept H0 and conclude that all population means are the same. C) Since 7.948 > 2.8778 reject H0 and conclude that at least two populations means are different. D) Since 14.667 > 2.8778 reject H0 and conclude that at least two populations means are different. Answer: D Diff: 2 Keywords: ANOVA, analysis of variance, one-way Section: 12-1 One-Way Analysis of Variance Outcome: 1 122) Respond to the following questions using this partially completed one-way ANOVA table: Source of Variation Between Samples Within Samples Total

SS 405 888

df 3 ____ 31

F-ratio

How many different populations are being considered in this analysis? A) 2 B) 4 C) 6 D) 5 Answer: B Diff: 2 Keywords: ANOVA, analysis of variance, one-way Section: 12-1 One-Way Analysis of Variance Outcome: 1

123) Respond to the following questions using this partially completed one-way ANOVA table: Source of Variation Between Samples Within Samples Total

SS 405 888

df 3 ____ 31

F-ratio

Fill in the ANOVA table with the missing values. A) SSB = 483, MSB = 161, F-ratio = 11.1309, Within Samples df = 28, MSW = 14.464 B) SSB = 483, MSB = 161, F- ratio = 8.1629, Within Samples df = 28, MSW = 14.464 C) SSB = 483, MSB = 161, F-ratio = 8.1629, Within Samples df = 25, MSW = 14.464 D) SSB = 504, MSB = 161, F-ratio = 8.1629, Within Samples df = 28, MSW = 14.464 Answer: A Diff: 2 Keywords: ANOVA, analysis of variance, one-way Section: 12-1 One-Way Analysis of Variance Outcome: 1 124) Respond to the following questions using this partially completed one-way ANOVA table: Source of Variation Between Samples Within Samples Total

SS 405 888

df 3 ____ 31

F-ratio

Based on the analysis of variance F-test, what conclusion should be reached regarding the null hypothesis? Test using alpha = 0.05. A) Since 11.1309 > 2.9467 accept H0 and conclude that all population means are the same. B) Since 2.9467 > 11.1309 accept H0 and conclude that all population means are the same. C) Since 11.1309 > 2.9467 reject H0 and conclude that at least two populations means are different. D) Since 2.9467 > 11.1309 reject H0 and conclude that at least two populations means are different. Answer: C Diff: 2 Keywords: ANOVA, analysis of variance, one-way Section: 12-1 One-Way Analysis of Variance Outcome: 1

125) Given the following sample data Item 1 2 3 4 5 6 7

Group 1 20.9 27.2 26.6 22.1 25.3 30.1 23.8

Group 2 28.2 26.2 21.6 29.7 30.3 25.9

Group 3 17.8 15.9 18.4 20.2 14.1

Group 4 21.2 23.9 19.5 17.4

Based on the computations for the within- and between-sample variation, develop the ANOVA table and test the appropriate null hypothesis using alpha= 0.05. Use the p-value approach. A) Since p-value = 0.0678 > 0.05 reject H0 and conclude that at least two population means are different. B) Since p-value = 0.000136 < 0.05 reject H0 and conclude that at least two population means are different. C) Since p-value = 0.0678 > 0.05 accept H0 and conclude that all population means are the same. D) Since p-value = 0.000136 < 0.05 accept H0 and conclude that all population means are the same. Answer: B Diff: 2 Keywords: ANOVA, analysis of variance, randomized block Section: 12-2 Randomized Complete Block Analysis of Variance Outcome: 1 126) In conjunction with the housing foreclosure crisis of 2009, many economists expressed increasing concern about the level of credit card debt and efforts of banks to raise interest rates on these cards. The banks claimed the increases were justified. A Senate subcommittee decided to determine if the average credit card balance depends on the type of credit card used. Under consideration are Visa, MasterCard, Discover, and American Express. The sample sizes to be used for each level are 25, 25, 26, and 23, respectively. State the number of degrees of freedom available for determining the between-samples variation. A) 6 B) 5 C) 2 D) 3 Answer: D Diff: 1 Keywords: ANOVA, analysis of variance, one-way Section: 12-1 One-Way Analysis of Variance Outcome: 1

127) In conjunction with the housing foreclosure crisis of 2009, many economists expressed increasing concern about the level of credit card debt and efforts of banks to raise interest rates on these cards. The banks claimed the increases were justified. A Senate subcommittee decided to determine if the average credit card balance depends on the type of credit card used. Under consideration are Visa, MasterCard, Discover, and American Express. The sample sizes to be used for each level are 25, 25, 26, and 23, respectively. State the number of degrees of freedom available for determining the within-samples variation. A) 93 B) 95 C) 97 D) 98 Answer: B Diff: 1 Keywords: ANOVA, analysis of variance, one-way Section: 12-1 One-Way Analysis of Variance Outcome: 1 128) In conjunction with the housing foreclosure crisis of 2009, many economists expressed increasing concern about the level of credit card debt and efforts of banks to raise interest rates on these cards. The banks claimed the increases were justified. A Senate subcommittee decided to determine if the average credit card balance depends on the type of credit card used. Under consideration are Visa, MasterCard, Discover, and American Express. The sample sizes to be used for each level are 25, 25, 26, and 23, respectively. State the number of degrees of freedom available for determining the total variation. A) 93 B) 95 C) 98 D) 97 Answer: C Diff: 1 Keywords: ANOVA, analysis of variance, one-way Section: 12-1 One-Way Analysis of Variance Outcome: 1

129) Damage to homes caused by burst piping can be expensive to repair. By the time the leak is discovered, hundreds of gallons of water may have already flooded the home. Automatic shutoff valves can prevent extensive water damage from plumbing failures. The valves contain sensors that cut off water flow in the event of a leak, thereby preventing flooding. One important characteristic is the time (in milliseconds) required for the sensor to detect the water leak. Sample data obtained for four different shutoff valves are contained in the file entitled Waterflow. Produce the relevant ANOVA table and conduct a hypothesis test to determine if the mean detection time differs among the four shutoff valve models. Use a significance level of 0.05. A) The ANOVA produces a p-value of 0.033 < alpha = 0.05. Therefore, the null hypothesis is not rejected. There is not sufficient evidence to indicate that the mean detection time differs among the four shutoff valve models B) The ANOVA produces a p-value of 0.033 < alpha = 0.05. Therefore, the null hypothesis is rejected. There is sufficient evidence to indicate that the mean detection time differs among the four shutoff valve models C) The ANOVA produces a p-value of 0.000 < alpha = 0.05. Therefore, the null hypothesis is not rejected. There is not sufficient evidence to indicate that the mean detection time differs among the four shutoff valve models D) The ANOVA produces a p-value of 0.000 < alpha = 0.05. Therefore, the null hypothesis is rejected. There is sufficient evidence to indicate that the mean detection time differs among the four shutoff valve models Answer: D Diff: 2 Keywords: ANOVA, analysis of variance, one-way Section: 12-1 One-Way Analysis of Variance Outcome: 1

130) Damage to homes caused by burst piping can be expensive to repair. By the time the leak is discovered, hundreds of gallons of water may have already flooded the home. Automatic shutoff valves can prevent extensive water damage from plumbing failures. The valves contain sensors that cut off water flow in the event of a leak, thereby preventing flooding. One important characteristic is the time (in milliseconds) required for the sensor to detect the water leak. Sample data obtained for four different shutoff valves are contained in the file entitled Waterflow. Use the Tukey-Kramer multiple comparison technique to discover any differences in the average detection time. Use a significance level of 0.05. A) The confidence intervals indicate that there is not sufficient evidence to conclude that the average detection time for valve 1, 2, and 3 differ. There is, however, enough evidence to indicate that the average detection time for valve 4 is larger than the other three means. B) The confidence intervals indicate that there is not sufficient evidence to conclude that the average detection time for valve 1, 2, and 4 differ. There is, however, enough evidence to indicate that the average detection time for valve 3 is larger than the other three means. C) The confidence intervals indicate that there is not sufficient evidence to conclude that the average detection time for valve 2 and 4 differ. There is, however, enough evidence to indicate that the average detection time for valve 1 and 3 are larger than the other two means. D) All mean detection times are equal. Answer: B Diff: 2 Keywords: ANOVA, analysis of variance, one-way Section: 12-1 One-Way Analysis of Variance Outcome: 1 131) The Lottaburger restaurant chain in central New Mexico is conducting an analysis of its restaurants, which take pride in serving burgers and fries to go faster than the competition. As a part of its analysis, Lottaburger wants to determine if its speed of service is different across its four outlets. Orders at Lottaburger restaurants are tracked electronically, and the chain is able to determine the speed with which every order is filled. The chain decided to randomly sample 20 orders from each of the four restaurants it operates. The speed of service for each randomly sampled order was noted and is contained in the file Lottaburger. At the alpha = 0.05 level of service, can Lottaburger conclude that the speed of service is different across the four restaurants in the chain? A) Since F = 18.418 > Fα=0.05 = 2.725, reject the null hypothesis. Based on these sample data we can conclude that the average service time is different across the four restaurants in the chain. B) Since F = 22.666 > Fα=0.05 = 2.725, reject the null hypothesis. Based on these sample data we can conclude that the average service time is different across the four restaurants in the chain. C) Since F = 22.666 > Fα=0.05 = 2.725, do not reject the null hypothesis. Based on these sample data there is not sufficient evidence to conclude that the average service time is different across the four restaurants in the chain. D) Since F = 18.418 > Fα=0.05 = 2.725, do not reject the null hypothesis. Based on these sample data there is not sufficient evidence to conclude that the average service time is different across the four restaurants in the chain. Answer: B Diff: 2 Keywords: ANOVA, analysis of variance, one-way Section: 12-1 One-Way Analysis of Variance Outcome: 1 12-55 Copyright © 2018 Pearson Education, Inc.

132) A study was conducted to determine if differences in new textbook prices exist between on-campus bookstores, off-campus bookstores, and Internet bookstores. To control for differences in textbook prices that might exist across disciplines, the study randomly selected 12 textbooks and recorded the price of each of the 12 books at each of the three retailers. You may assume normality and equal-variance assumptions have been met. The partially completed ANOVA table based on the study's findings is shown here: ANOVA Source of Variation Textbooks Retailer Error Total

SS 16,624 2.4

17,477.6

Complete the ANOVA table by filling in the missing sums of squares, the degrees of freedom for each source, the mean square, and the calculated F-test statistic for each possible hypothesis test. A) Textbooks df = 11, MSBL = 1,511.3, F (Textbooks) = 40.05, Retailer df = 2, MSB = 1.2, , SSW = 851.2, Error df = 22, MSW = 38.7, Total df = 35 B) Textbooks df = 11, MSBL = 1,511.3, F (Textbooks) = 39.05, Retailer df = 2, MSB = 1.2, , SSW = 851.2, Error df = 22, MSW = 38.7, Total df = 35 C) Textbooks df = 12, MSBL = 1,511.3, F (Textbooks) = 39.05, Retailer df = 2, MSB = 1.2, , SSW = 851.2, Error df = 22, MSW = 38.7, Total df = 36 D) Textbooks df = 11, MSBL = 1,511.3, F (Textbooks) = 39.05, Retailer df = 2, MSB = 1.2, , SSW = 831.2, Error df = 22, MSW = 38.7, Total df = 36 Answer: B Diff: 2 Keywords: ANOVA, analysis of variance, one-way Section: 12-1 One-Way Analysis of Variance Outcome: 1

133) A study was conducted to determine if differences in new textbook prices exist between on-campus bookstores, off-campus bookstores, and Internet bookstores. To control for differences in textbook prices that might exist across disciplines, the study randomly selected 12 textbooks and recorded the price of each of the 12 books at each of the three retailers. You may assume normality and equal-variance assumptions have been met. The partially completed ANOVA table based on the study's findings is shown here: ANOVA Source of Variation Textbooks Retailer Error Total

SS 16,624 2.4

17,477.6

Based on the study's findings, was it correct to block for differences in textbooks? Conduct the appropriate test at the alpha = 0.10 level of significance. A) Since F = 39.05 > Fα=0.10 = 1.88, reject the null hypothesis. This means that based on these sample data we can conclude that blocking is effective. B) Since F = 39.05 > Fα=0.10 = 1.88, do not reject the null hypothesis. This means that based on these sample data we can conclude that blocking is not effective. C) Since F = 40.05 > Fα=0.10 = 1.88, reject the null hypothesis. This means that based on these sample data we can conclude that blocking is effective. D) Since F = 40.05 > Fα=0.10 = 1.88, do not reject the null hypothesis. This means that based on these sample data we can conclude that blocking is not effective Answer: A Diff: 2 Keywords: ANOVA, analysis of variance, one-way Section: 12-1 One-Way Analysis of Variance Outcome: 1

134) A study was conducted to determine if differences in new textbook prices exist between on-campus bookstores, off-campus bookstores, and Internet bookstores. To control for differences in textbook prices that might exist across disciplines, the study randomly selected 12 textbooks and recorded the price of each of the 12 books at each of the three retailers. You may assume normality and equal-variance assumptions have been met. The partially completed ANOVA table based on the study's findings is shown here: ANOVA Source of Variation Textbooks Retailer Error Total

SS 16,624 2.4

17,477.6

Based on the study's findings, can it be concluded that there is a difference in the average price of textbooks across the three retail outlets? Conduct the appropriate hypothesis test at the alpha = 0.10 level of significance. A) F = 0.0411 < Fα=0.10 = 2.56, reject the null hypothesis. Thus, based on these sample data we can conclude that there is a difference in textbook prices at the three different types of retail outlets. B) F = 0.0411 < Fα=0.10 = 2.56, do not reject the null hypothesis. Thus, based on these sample data we cannot conclude that there is a difference in textbook prices at the three different types of retail outlets. C) F = 0.031 < Fα=0.10 = 2.56, reject the null hypothesis. Thus, based on these sample data we can conclude that there is a difference in textbook prices at the three different types of retail outlets. D) F = 0.031 < Fα=0.10 = 2.56, do not reject the null hypothesis. Thus, based on these sample data we cannot conclude that there is a difference in textbook prices at the three different types of retail outlets. Answer: D Diff: 2 Keywords: ANOVA, analysis of variance, one-way Section: 12-1 One-Way Analysis of Variance Outcome: 1

135) Consider the following: Summary 1 2 3 4 5 6 7 8 Sample 1 Sample 2 Sample 3 Sample 4

Count 4 4 4 4 4 4 4 4 8 8 8 8

ANOVA Source of Variation Rows Columns Error Total

SS 199,899 11,884 5,317 217,100

Sum 443 275 1,030 300 603 435 1,190 460 1,120 1,236 1,400 980

df 7 3 21 31

Average 110.8 68.8 257.5 75.0 150.8 108.8 297.5 115.0 140.0 154.5 175.0 122.5

MS 28557.0 3961.3 253.2

F 112.8 15.7

Variance 468.9 72.9 1891.7 433.3 468.9 72.9 1891.7 433.3 7142.9 8866.6 9000.0 4307.1

p-value 0.0000 0.0000

How many blocks were used in this study? A) 10 B) 9 C) 7 D) 8 Answer: D Diff: 1 Keywords: ANOVA, analysis of variance, randomized block Section: 12-2 Randomized Complete Block Analysis of Variance Outcome: 1

F-crit 2.488 3.073

136) Consider the following: Summary 1 2 3 4 5 6 7 8 Sample 1 Sample 2 Sample 3 Sample 4

Count 4 4 4 4 4 4 4 4 8 8 8 8

ANOVA Source of Variation Rows Columns Error Total

SS 199,899 11,884 5,317 217,100

Sum 443 275 1,030 300 603 435 1,190 460 1,120 1,236 1,400 980

df 7 3 21 31

Average 110.8 68.8 257.5 75.0 150.8 108.8 297.5 115.0 140.0 154.5 175.0 122.5

MS 28557.0 3961.3 253.2

F 112.8 15.7

Variance 468.9 72.9 1891.7 433.3 468.9 72.9 1891.7 433.3 7142.9 8866.6 9000.0 4307.1

p-value 0.0000 0.0000

How many populations are involved in this test? A) 4 B) 2 C) 5 D) 3 Answer: A Diff: 1 Keywords: ANOVA, analysis of variance, randomized block Section: 12-2 Randomized Complete Block Analysis of Variance Outcome: 1

F-crit 2.488 3.073

137) Consider the following: Summary 1 2 3 4 5 6 7 8 Sample 1 Sample 2 Sample 3 Sample 4

Count 4 4 4 4 4 4 4 4 8 8 8 8

ANOVA Source of Variation Rows Columns Error Total

SS 199,899 11,884 5,317 217,100

Sum 443 275 1,030 300 603 435 1,190 460 1,120 1,236 1,400 980

df 7 3 21 31

Average 110.8 68.8 257.5 75.0 150.8 108.8 297.5 115.0 140.0 154.5 175.0 122.5

MS 28557.0 3961.3 253.2

F 112.8 15.7

Variance 468.9 72.9 1891.7 433.3 468.9 72.9 1891.7 433.3 7142.9 8866.6 9000.0 4307.1

p-value 0.0000 0.0000

F-crit 2.488 3.073

Test to determine whether blocking is effective using an alpha level equal to 0.05 A) Because F = 14.81 > critical F = 2.5, we do not reject the null hypothesis and conclude that blocking is not effective. B) Because F = 14.81 > critical F = 2.5, we reject the null hypothesis and conclude that blocking is effective. C) Because F = 112.79 > critical F = 2.5, we do not reject the null hypothesis and conclude that blocking is not effective. D) Because F = 112.79 > critical F = 2.5, we reject the null hypothesis and conclude that blocking is effective. Answer: D Diff: 2 Keywords: ANOVA, analysis of variance, randomized block Section: 12-2 Randomized Complete Block Analysis of Variance Outcome: 1

138) Consider the following: Summary 1 2 3 4 5 6 7 8 Sample 1 Sample 2 Sample 3 Sample 4

Count 4 4 4 4 4 4 4 4 8 8 8 8

ANOVA Source of Variation Rows Columns Error Total

SS 199,899 11,884 5,317 217,100

Sum 443 275 1,030 300 603 435 1,190 460 1,120 1,236 1,400 980

df 7 3 21 31

Average 110.8 68.8 257.5 75.0 150.8 108.8 297.5 115.0 140.0 154.5 175.0 122.5

MS 28557.0 3961.3 253.2

F 112.8 15.7

Variance 468.9 72.9 1891.7 433.3 468.9 72.9 1891.7 433.3 7142.9 8866.6 9000.0 4307.1

p-value 0.0000 0.0000

F-crit 2.488 3.073

Test the main hypothesis of interest using α = 0.05 A) Because F = 15.65 > critical F = 3.0, we reject the null hypothesis and conclude that the four populations do not have the same mean. B) Because F = 15.65 > critical F = 3.0, we do not reject the null hypothesis and conclude that the four populations have the same mean. C) Because F = 125.82 > critical F = 3.0, we reject the null hypothesis and conclude that the four populations do not have the same mean. D) Because F = 125.82 > critical F = 3.0, we do not reject the null hypothesis and conclude that the four populations have the same mean. Answer: A Diff: 2 Keywords: ANOVA, analysis of variance, randomized block Section: 12-2 Randomized Complete Block Analysis of Variance Outcome: 1

139) The following sample data were recently collected in the course of conducting a randomized block analysis of variance. Based on these sample data, what conclusions should be reached about blocking effectiveness and about the means of the three populations involved? Test using a significance level equal to 0.05. Block 1 2 3 4 5 6

Sample 1 30 50 60 40 80 20

Sample 2 40 70 40 40 70 10

Sample 3 40 50 70 30 90 10

A) Because F = 0.4195 < critical F = 4.103, we do not reject the null hypothesis and conclude that the three populations may have the same mean value. B) Because F = 0.4195 < critical F = 4.103, we reject the null hypothesis and conclude that the three populations do not have the same mean value. C) Because F = 0.1515 < critical F = 4.103, we do not reject the null hypothesis and conclude that the three populations may have the same mean value. D) Because F = 0.1515 < critical F = 4.103, we reject the null hypothesis and conclude that the three populations do not have the same mean value. Answer: C Diff: 2 Keywords: ANOVA, analysis of variance, randomized block Section: 12-2 Randomized Complete Block Analysis of Variance Outcome: 1

140) Frasier and Company manufactures four different products that it ships to customers throughout the United States. Delivery times are not a driving factor in the decision as to which type of carrier to use (rail, plane, or truck) to deliver the product. However, breakage cost is very expensive, and Frasier would like to select a mode of delivery that reduces the amount of product breakage. To help it reach a decision, the managers have decided to examine the dollar amount of breakage incurred by the three alternative modes of transportation under consideration. Because each product's fragility is different, the executives conducting the study wish to control for differences due to type of product. The company randomly assigns each product to each carrier and monitors the dollar breakage that occurs over the course of 100 shipments. The dollar breakage per shipment (to the nearest dollar) is as follows:

Product 1 Product 2 Product 3 Product 4

Rail $7,960 $8,399 $9,429 $6,022

Plane $8,053 $7,764 $9,196 $5,821

Truck $8,818 $9,432 $9,260 $5,676

Was Frasier and Company correct in its decision to block for type of product? Conduct the appropriate hypothesis test using a level of significance of 0.01. A) Because F = 32.12 > Fα=0.01 = 9.78, reject the null hypothesis. Thus, based on these sample data we conclude that blocking is effective. B) Because F = 28.14 > Fα=0.01 = 7.63, reject the null hypothesis. Thus, based on these sample data we conclude that blocking is effective. C) Because F = 32.12 > Fα=0.01 = 9.78, do not reject the null hypothesis. Thus, based on these sample data we conclude that blocking is not effective. D) Because F = 28.14 > Fα=0.01 = 7.63, do not reject the null hypothesis. Thus, based on these sample data we conclude that blocking is not effective. Answer: A Diff: 2 Keywords: ANOVA, analysis of variance, randomized block Section: 12-2 Randomized Complete Block Analysis of Variance Outcome: 1

141) Applebee's International, Inc., is a U.S. company that develops, franchises, and operates the Applebee's Neighborhood Grill and Bar restaurant chain. It is the largest chain of casual dining restaurants in the country, with over 1,500 restaurants across the United States. The headquarters is located in Overland Park, Kansas. The company is interested in determining if mean weekly revenue differs among three restaurants in a particular city. The file entitled Applebees contains revenue data for a sample of weeks for each of the three locations. Test to determine if blocking the week on which the testing was done was necessary. Use a significance level of 0.05. A) The p-value = 0.078 > α = 0.05. This indicates that inserting the week on which the testing was done was necessary. B) The p-value = 0.078 > α = 0.05. This indicates that inserting the week on which the testing was done was not necessary. C) The p-value = 0.000 < α = 0.05. This indicates that inserting the week on which the testing was done was necessary. D) The p-value = 0.000 < α = 0.05. This indicates that inserting the week on which the testing was done was not necessary. Answer: C Diff: 2 Keywords: ANOVA, analysis of variance, randomized block Section: 12-2 Randomized Complete Block Analysis of Variance Outcome: 1 142) Applebee's International, Inc., is a U.S. company that develops, franchises, and operates the Applebee's Neighborhood Grill and Bar restaurant chain. It is the largest chain of casual dining restaurants in the country, with over 1,500 restaurants across the United States. The headquarters is located in Overland Park, Kansas. The company is interested in determining if mean weekly revenue differs among three restaurants in a particular city. The file entitled Applebees contains revenue data for a sample of weeks for each of the three locations. Based on the data gathered by Applebee's, can it be concluded that there is a difference in the average revenue among the three restaurants? A) The p-value = 0.004 < alpha = 0.05. This indicates that we should not reject the null hypothesis and conclude that there is not a difference in the average revenue among the three restaurants. B) The p-value = 0.004 < alpha = 0.05. This indicates that we should reject the null hypothesis and conclude that there exists a difference in the average revenue among the three restaurants. C) The p-value = 0.084 > alpha = 0.05. This indicates that we should not reject the null hypothesis and conclude that there is not a difference in the average revenue among the three restaurants. D) The p-value = 0.084 > alpha = 0.05. This indicates that we should reject the null hypothesis and conclude that there exists a difference in the average revenue among the three restaurants. Answer: B Diff: 2 Keywords: ANOVA, analysis of variance, one-way Section: 12-1 One-Way Analysis of Variance Outcome: 1

143) Applebee's International, Inc., is a U.S. company that develops, franchises, and operates the Applebee's Neighborhood Grill and Bar restaurant chain. It is the largest chain of casual dining restaurants in the country, with over 1,500 restaurants across the United States. The headquarters is located in Overland Park, Kansas. The company is interested in determining if mean weekly revenue differs among three restaurants in a particular city. The file entitled Applebees contains revenue data for a sample of weeks for each of the three locations. If you did conclude that there was a difference in the average revenue, use Fisher's LSD approach to determine which restaurant has the lowest mean sales. A) There is no difference between the average revenues. B) Restaurant 1 has the highest average revenue while there is no evidence of a difference between Restaurant 2's and 3's average revenues. C) Restaurant 3 has the highest average revenue while there is no evidence of a difference between Restaurant 1's and 2's average revenues. D) Restaurant 2 has the highest average revenue while there is no evidence of a difference between Restaurant 1's and 3's average revenues. Answer: C Diff: 2 Keywords: ANOVA, analysis of variance, one-way Section: 12-1 One-Way Analysis of Variance Outcome: 1 144) In a local community there are three grocery chain stores. The three have been carrying out a spirited advertising campaign in which each claims to have the lowest prices. A local news station recently sent a reporter to the three stores to check prices on several items. She found that for certain items each store had the lowest price. This survey didn't really answer the question for consumers. Thus, the station set up a test in which 20 shoppers were given different lists of grocery items and were sent to each of the three chain stores. The sales receipts from each of the three stores are recorded in the data file Groceries. Based on a significance level of 0.05 and these sample data, test to determine whether blocking was necessary in this example. Use a test-statistic approach. A) Since 952.6155 > 1.8673 do not reject H0 and conclude that there is an indication that blocking was not effective. B) Since 952.6155 > 1.8673 reject H0 and conclude that there is an indication that blocking was effective. C) Since 102.2912 > 1.8673 do not reject H0 and conclude that there is an indication that blocking was not effective. D) Since 102.2912 > 1.8673 reject H0 and conclude that there is an indication that blocking was effective. Answer: B Diff: 2 Keywords: ANOVA, analysis of variance, randomized block Section: 12-2 Randomized Complete Block Analysis of Variance Outcome: 1

145) In a local community there are three grocery chain stores. The three have been carrying out a spirited advertising campaign in which each claims to have the lowest prices. A local news station recently sent a reporter to the three stores to check prices on several items. She found that for certain items each store had the lowest price. This survey didn't really answer the question for consumers. Thus, the station set up a test in which 20 shoppers were given different lists of grocery items and were sent to each of the three chain stores. The sales receipts from each of the three stores are recorded in the data file Groceries. Based on these sample data, can you conclude the three grocery stores have different sample means? Test using a significance level of 0.05. Use a p-value approach. A) Since the p-value of 2.68E - 10 < 0.05 reject H0 and conclude that at least two means are different. B) Since the p-value of 2.68E-10 < 0.05 do not reject H0 and conclude that means are all the same. C) Since the p-value of 0.027 < 0.05 reject H0 and conclude that at least two means are different. D) Since the p-value of 0.027 < 0.05 do not reject H0 and conclude that means are all the same. Answer: A Diff: 2 Keywords: ANOVA, analysis of variance, randomized block Section: 12-2 Randomized Complete Block Analysis of Variance Outcome: 1 146) In a local community there are three grocery chain stores. The three have been carrying out a spirited advertising campaign in which each claims to have the lowest prices. A local news station recently sent a reporter to the three stores to check prices on several items. She found that for certain items each store had the lowest price. This survey didn't really answer the question for consumers. Thus, the station set up a test in which 20 shoppers were given different lists of grocery items and were sent to each of the three chain stores. The sales receipts from each of the three stores are recorded in the data file Groceries. Based on the sample data, which store has the highest average prices? Use Fisher's LSD test if appropriate. A) Store 1 B) Store 2 C) Store 3 D) There is no difference between the average prices. Answer: B Diff: 3 Keywords: ANOVA, analysis of variance, randomized block Section: 12-2 Randomized Complete Block Analysis of Variance Outcome: 1

147) When the world's largest retailer, Walmart, decided to enter the grocery marketplace in a big way with its "Super Stores," it changed the retail grocery landscape in a major way. The other major chains such as Albertsons have struggled to stay competitive. In addition, regional discounters such as WINCO in the western United States have made it difficult for the traditional grocery chains. Recently, a study was conducted in which a "market basket" of products was selected at random from those items offered in three stores in Boise, Idaho: Walmart, Winco, and Albertsons. At issue was whether the mean prices at the three stores are equal or whether there is a difference in prices. The sample data are in the data file called Food Price Comparisons. Using an alpha level equal to 0.05, test to determine whether the three stores have equal population mean prices. If you conclude that there are differences in the mean prices, perform the appropriate posttest to determine which stores have different means. A) There is no difference between the three mean prices. B) Based on the sample data, we conclude that Winco is significantly different (higher) than Albertsons and Walmart in terms of average prices. However, we can make no conclusion about Albertsons and Walmart. C) Based on the sample data, we conclude that Walmart is significantly different (higher) than Albertsons and Winco in terms of average prices. However, we can make no conclusion about Albertsons and Winco. D) Based on the sample data, we conclude that Albertsons is significantly different (higher) than Walmart and Winco in terms of average prices. However, we can make no conclusion about Walmart and Winco. Answer: D Diff: 3 Keywords: ANOVA, analysis of variance, randomized block Section: 12-2 Randomized Complete Block Analysis of Variance Outcome: 1 148) Examine the following two-factor analysis of variance table: Source Factor A Factor B AB Interaction Error Total

SS 162.79

df 4

F-Ratio

28.12 262.31 ________ 1,298.74

12 ___ 84

Complete the analysis of variance table. A) MSA = 40.928, F Factor A =3.35, SSB = 85.35, Factor B df = 3, F Factor B = 2.316, MSAB = 21.859, F Factor AB = 1.8, SSE = 789.29, SSE df = 66, MSE = 12.143 B) MSA = 40.928, F Factor A = 3.35, SSB = 85.35, Factor B df = 4, F Factor B = 2.316, MSAB = 21.859, F Factor AB = 2.1 SSE = 789.29, SSE df = 66, MSE = 12.143 C) MSA = 40.698, F Factor A = 3.35, SSB = 84.35, Factor B df = 5, F Factor B = 2.316, MSAB = 21.859, F Factor AB = 2.1, SSE = 789.29, SSE df = 65, MSE = 12.143 D) MSA = 40.698, F Factor A = 3.35, SSB = 84.35, Factor B df = 3, F Factor B = 2.316, MSAB = 21.859, F Factor AB = 1.8, SSE = 789.29, SSE df = 65, MSE = 12.143 Answer: D Diff: 2 Keywords: ANOVA, analysis of variance, two-factor, replication, cell Section: 12-3 Two-Factor Analysis of Variance With Replication Outcome: 1 12-68 Copyright © 2018 Pearson Education, Inc.

149) Examine the following two-factor analysis of variance table: Source Factor A Factor B AB Interaction Error Total

SS 162.79

df 4

262.31 ________ 1,298.74

12 ___ 84

F-Ratio

28.12

Determine if interaction exists between factor A and factor B. Use alpha = 0.05. A) Fail to reject H0. Conclude that there is not sufficient evidence to indicate interaction exists between Factor A and Factor B B) Reject H0. Conclude that there is sufficient evidence to indicate interaction exists between Factor A and Factor B C) Fail to reject H0. Conclude that there is sufficient evidence to indicate interaction exists between Factor A and Factor B D) Reject H0. Conclude that there is not sufficient evidence to indicate interaction exists between Factor A and Factor B Answer: A Diff: 2 Keywords: ANOVA, analysis of variance, two-factor, replication, cell Section: 12-3 Two-Factor Analysis of Variance With Replication Outcome: 1 150) Examine the following two-factor analysis of variance table: Source Factor A Factor B AB Interaction Error Total

SS 162.79

df 4

262.31 ________ 1,298.74

12 ___ 84

F-Ratio

28.12

Determine if the levels of factor A have equal means. Use a significance level of 0.05. A) Fail to reject H0. Conclude that there is sufficient evidence to indicate that at least two levels of Factor A have different mean responses. B) Fail to reject H0. Conclude that there is not sufficient evidence to indicate that at least two levels of Factor A have different mean responses. C) Reject H0. Conclude that there is sufficient evidence to indicate that at least two levels of Factor A have different mean responses. D) Reject H0. Conclude that there is not sufficient evidence to indicate that at least two levels of Factor A have different mean responses. Answer: C Diff: 2 Keywords: ANOVA, analysis of variance, two-factor, replication, cell Section: 12-3 Two-Factor Analysis of Variance With Replication Outcome: 1 12-69 Copyright © 2018 Pearson Education, Inc.

151) Examine the following two-factor analysis of variance table: Source Factor A Factor B AB Interaction Error Total

SS 162.79

df 4

262.31 ________ 1,298.74

12 ___ 84

F-Ratio

28.12

Does the ANOVA table indicate that the levels of factor B have equal means? Use a significance level of 0.05. A) Fail to reject H0. Conclude that there is not sufficient evidence to indicate that at least two levels of Factor B have different mean responses. B) Reject H0. Conclude that there is sufficient evidence to indicate that at least two levels of Factor B have different mean responses. C) Fail to reject H0. Conclude that there is sufficient evidence to indicate that at least two levels of Factor B have different mean responses. D) Reject H0. Conclude that there is not sufficient evidence to indicate that at least two levels of Factor B have different mean responses. Answer: A Diff: 2 Keywords: ANOVA, analysis of variance, two-factor, replication, cell Section: 12-3 Two-Factor Analysis of Variance With Replication Outcome: 1 152) What is meant by the term balanced design in an analysis of variance application? Answer: A balanced design is one in which the sample sizes from each population are equal. Diff: 1 Keywords: ANOVA, analysis of variance, balanced Section: 12-1 One-Way Analysis of Variance Outcome: 1 153) What are the assumptions for a one-way analysis of variance design? Answer: There are four assumption for the one-way ANOVA. These are: 1. The populations are normally distributed. 2. The populations have equal variances. 3. The observations are independent. 4. The data are interval or ratio level. Diff: 1 Keywords: ANOVA, analysis of variance, assumption Section: 12-1 One-Way Analysis of Variance Outcome: 1

154) Explain what is meant by partitioning the sum of squares in a one-way analysis of variance application. Answer: In a one-way analysis of variance, sample data are collected from each of the populations. The total variation in all the sample data is measured by the total sum-of-squares, which is computed using: SST =

. This total can be divided into two parts: the between variation, which is the

variation that exists between the sample means of each population, and the within variation, which is the variation that exists within each sample. The between variation is measured by the sum-of-squares: SSB =

. If the population means are not equal, the SSB value will tend to be larger since we

would expect more difference in the sample means. The within variation is measured by the sum-ofsquares using SSW = SST - SSB. Thus the total sum-of-square is composed of two parts as follows: SST = SSB + SSW. Diff: 2 Keywords: ANOVA, analysis of variance, sum of squares, partitioning Section: 12-1 One-Way Analysis of Variance Outcome: 2

155) A study has been conducted to determine whether the mean spending for recreational activities during the month of August differs for residents of three cities. Random samples of 30 people were selected from each city and their spending on recreation was recorded during August. The following output was generated using Excel: ANOVA: Single Factor SUMMARY Groups City 1 City 2 City 3

Count 30 30 30

Sum 7897.179 10322.1 6045.102

ANOVA Source of Variation Between Groups Within Groups

SS 306701.8 224803.5

Total

531505.4

df 2 87

Average 236.2393 344.0701 201.5034

Variance 3334.11 2201.818 2215.919

MS 153350.9 2583.949

F 59.3475

P-value 5.54E-17

F crit 3.101292

Based on the information provided, should we conclude that the three populations (cities) have equal mean spending during August? Test at the 0.05 level of significance. Answer: The null and alternative hypotheses to be tested are: H0 : μ1 = μ2 = μ3 Ha : not all μj are equal. Since three populations are involved and the samples that have been selected are independent, if we assume the following: 1. The populations are normally distributed. 2. The populations have equal variances. 3. The observations are independent. Then the one-way analysis of variance test can be used to test the null hypothesis. The test statistic for this test is provided in the Excel output to be F = 59.3475. This value is compared to the F-critical value for 2 and 87 degrees of freedom, which is 3.10. Since F = 59.3475 > 3.10, we reject the null hypothesis and conclude that not all means are equal. We could also use the p-value approach to conduct the test. Since the p-value is shown to be virtually zero, which is less than alpha = 0.05, we would reject the null hypothesis. Diff: 2 Keywords: ANOVA, analysis of variance, one-way, null, hypothesis, F-test Section: 12-1 One-Way Analysis of Variance Outcome: 2

156) A study has been conducted to determine whether the mean spending for recreational activities during the month of August differs for residents of three cities. Random samples of 30 people were selected from each city and their spending on recreation was recorded during August. The following output was generated using Excel: ANOVA: Single Factor SUMMARY Groups City 1 City 2 City 3

Count 30 30 30

Sum 7897.179 10322.1 6045.102

ANOVA Source of Variation Between Groups Within Groups

SS 306701.8 224803.5

Total

531505.4

df 2 87

Average 236.2393 344.0701 201.5034

Variance 3334.11 2201.818 2215.919

MS 153350.9 2583.949

F 59.3475

P-value 5.54E-17

F crit 3.101292

The Excel output shows that the null hypothesis of equal means should be rejected. Given this, perform the appropriate method for determining which population means are different. Conduct the test using an alpha = 0.05 level. Answer: The appropriate method for determining which populations have different means is the TukeyKramer method of multiple comparisons. Using this method is superior to conducting multiple t-tests for all pairs of means since we can control the experiment-wide error rate to 0.05 by using Tukey-Kramer. We begin by computing the Tukey-Kramer critical range as follows: Critical Range = q1-α

From the Excel output we can get MSW = 2583.95. The q-value is extrapolated from the Studentized range table with k = 3 and N - k = 87 degrees of freedom, which is approximately 3.38. Since the sample sizes were equal, we can find one critical range for testing all pairs of means. The critical range is: Critical Range = 3.38

= 31.369.

Now, we compare the absolute difference between each pair of sample means to this critical value. In any case where the absolute difference exceeds the critical range, we conclude that those population means are different. Thus, we get the following: =

= 80.83 > 31.369

= 61.74 > 31.369

= 142.57 > 31.369

Thus, based on the Tukey-Kramer test, we conclude that all three population means are different. Diff: 2 Keywords: ANOVA, analysis of variance, one-way, Tukey-Kramer, critical Section: 12-1 One-Way Analysis of Variance Outcome: 3 12-73 Copyright © 2018 Pearson Education, Inc.

157) Under what circumstances would you use a randomized complete block analysis of variance design instead of a one-way analysis of variance? Answer: Analysis of variance is a statistical tool that is used in cases where we are interested in testing whether three or more populations have the same mean. One-way analysis of variance is used when the samples from the populations are independent and we are not interested in controlling for a second factor that might adversely influence the analysis. With one-way, we test only whether the population means are all equal. However, there are cases in which we will be interested in controlling for a second factor that could have an adverse influence on our results. In these cases, we will want to use the concept of paired samples. The second factor (called the blocking factor) can have multiple levels. In the randomized complete block design (without replication), one observation is obtained for each combination of the two-factors. There are actually two hypotheses tests of interest: the primary test to see whether the populations of interest have different means, and the secondary test to determine whether we were justified in blocking. If the hypothesis test for blocking shows that we're justified in controlling for the second factor, then we look to the primary hypothesis test. If blocking is not determined to be effective, this means that we could (should) have selected independent samples. There is a cost in terms of lost degrees of freedom and it becomes more likely that we will not reject the primary hypothesis. It may be necessary to re-do the experiment as a one-way design if blocking is deemed not effective and the primary null hypothesis is not rejected. Diff: 2 Keywords: ANOVA, analysis of variance, one-way, randomized, complete block Section: 12-2 Randomized Complete Block Analysis of Variance Outcome: 4

158) A real estate broker is interested in determining whether there is a difference in the mean number of days a home stays on the market before selling based on which area of the city it is located in. However, she is also concerned that the price of the house may be an issue in determining how long it takes to sell a house, so she wants to control for this. To carry out the test, she plans to randomly select one house from each part of the city in each price range. The following data show the number of days for the sample of houses selected.

under $70,000 $70,000 < $90,000 $90,000 < $120,000 $120,000 < $180,000 $180,000 and over

East 42 40 50 30 56

West 60 70 80 40 33

North 29 40 60 56 40

South 50 37 30 40 20

Using a significance level equal to 0.05, determine whether the broker was justified in controlling for house prices. Be sure to indicate what type of statistical test should be used and why. Answer: The primary issue is whether or not the mean days that it takes to sell a house is related to the area of town in which the home is located. However, the broker believes that selling price may influence the time a house stays on the market, so she has controlled for that. The correct design under these circumstances is a randomized complete block analysis of variance design in which the area of town represents the levels of the primary factor, location. The levels of house price are the blocks. This question asks us to determine whether blocking was effective. This test is conducted before testing the primary hypothesis. Thus, we need to determine whether the mean days on the market is the same or different for the five different price levels. The null hypothesis to be tested is: H0 : μ1 = μ2 = μ3 = μ4 Ha : not all μj are equal. The total variation in the data can be partitioned into three parts: between area variation (SSB), between price (block) variation (SSBl), and within variation (SSW). The total variation is measured by the total sum-of-squares (SST), which is computed as:

SST= The SSB is computed as: SSB= The SSbl is computed as: SSbl=

We can then create an analysis of variance table as follows: ANOVA Source of Variation Rows Columns Error

SS 701.3 1142.95 2420.3

Total

4264.55

df 4 3 12

MS 175.325 380.9833 201.6917

F 0.869272 1.888939

p-value 0.510085 0.185295

F-crit 3.25916 3.4903

To test the hypothesis regarding the effectiveness of blocking we calculate a test statistic as: F=

= .869

This value is compared to the F-critical value for 4 and 12 degrees of freedom and an alpha = 0.05 level which is 3.259. Since 0.869 < 3.259 we do not reject the null hypothesis. Thus, based on the sample data, there is no basis for having blocked on house price. Thus, blocking was not effective. Diff: 3 Keywords: ANOVA, analysis of variance, randomized complete block Section: 12-2 Randomized Complete Block Analysis of Variance Outcome: 4 159) Assume you are conducting a two-way ANOVA and have found a significant interaction between the two factors. Explain what conclusions you can make from the results and what if any further steps should be taken. Answer: When there is a significant interaction between the factors it means the effect of each factor differs based on the level of the other factor. For example, when factor A is low, increasing factor B may cause the response variable to increase. But when factor A is high, factor B may have the opposite effect where the response variable decreases when factor B is increased. For this reason the main effects of each individual factor cannot necessarily be interpreted from their results in this particular ANOVA table. A follow up study should be conducted where a one-way ANOVA is conducted for a fixed level of one factor while the other factor is varied to investigate the effect of a particular factor at a fixed level of the other. Diff: 3 Keywords: ANOVA, analysis of variance, two-way, interaction Section: 12-3 Two-Factor Analysis of Variance With Replication Outcome: 5

160) Suppose that you have just performed an experiment using the randomized complete block analysis of variance design and the output from Excel is shown as follows. The primary null hypothesis to be tested involves whether the means of the three groups are equal. ANOVA: Two-Factor Without Replication Summary Block 1 Block 2 Block 3 Block 4

Count 3 3 3 3

Sum

Average Variance 150 468.9 70 72.9 203.3333 633.3333 233.3333 3333.333

450 210 610 700

Group 1 Group 2 Group 3

4 4 4

560 810 600

ANOVA Source of Variation Rows Columns Error

SS 46158.33 9016.667 4116.667

Total

5929.67

3 2 6

140.0 202.5 150

5066.667 8425 3266.667

MS 15386.11 4508.333 686.1111

F 22.4251 6.57085

p-value 0.001163 0.030797

F-crit 4.757055 5.143249

Using a significance level of 0.05, what conclusions should be made about the three groups? Which groups have different means? Discuss. Answer: The test has been conducted using a randomized complete block analysis of variance design without replication. The first step is to test to see whether blocking was effective. The appropriate null and alternative hypotheses are: H0 : μ1 = μ2 = μ3 = μ4 Ha : not all μj are equal. The analysis of variance table produced by Excel contains the information necessary to test this. The Ftest statistic is formed by: F=

= 22.425

The p-value for this is shown to be .0011, which is less than alpha = 0.05, so we reject the null hypothesis and conclude that blocking is effective. Next, we test the primary hypothesis about the three group means. The null and alternative hypotheses are: H0 : μ1 = μ2 = μ3 Ha : not all μj are equal.

The F-test statistic is computed using:

= 6.57

The p-value for this test statistic value is 0.0307, which is less than alpha = 0.05. Thus, based on the sample data, we reject the null hypothesis and conclude that the means for Groups 1-3 are not all equal. Now we can employ the Fisher's Least Significant Difference test to determine which groups have different means. The LSD value is computed using: LSD = ta/2 where n = number of blocks, in this case 4. The t-value comes from the upper end of the t-distribution table for α/2 = .025 and (k-1)(n-1) = 6 degrees of freedom. The t-value is 2.4469. Thus the LSD value is: LSD = 2.4469

= 45.3209

We now form all possible pairwise comparisons for the absolution difference of the sample means for the three groups. Any absolute difference between sample means that exceeds LSD = 45.3209 will provide grounds for saying that those two population means are different. The following pairwise comparisons exist: = = =

= 62.05 since 62.05 > 45.3209 = 10 since 10 < 45.3209 = 52.5 since 52.5 > 45.3209

Thus, based on the sample data, we conclude that group 1 and group 2 do not have the same mean and that group 2 and group 3 do not have the same mean. However, we are not able to conclude that group 1 and group 3 have different means. Diff: 3 Keywords: ANOVA, analysis of variance, randomized, complete block hypothesis Section: 12-2 Randomized Complete Block Analysis of Variance Outcome: 4 161) Explain what is meant by interaction between two factors in a two-way analysis of variance study. Answer: If there is no interaction between two factors (A and B), then the levels of one factor will have uniformly higher average values across all levels of the second factor. However, if interaction exists between the factors, we would see the means for some levels of one factor would be higher when associated with certain levels of the second factor and lower for other levels of the second factor. In general, interaction occurs if the differences in the averages of the response variable for the various levels of one factor, say Factor A, are not the same for each level of the other factor, say Factor B. The general idea is that interaction between two factors means that the effect due to one of them is not uniform across all levels of the other factor. Diff: 3 Keywords: ANOVA, analysis of variance, interaction, factor, two-way Section: 12-3 Two-Factor Analysis of Variance With Replication Outcome: 5 12-78 Copyright © 2018 Pearson Education, Inc.

162) A personal computer assembly company is interested in studying the time it takes to assemble a computer under different conditions. Specifically, two factors are to be controlled: Factor A (Sample) is the number of individual work stations in the line and Factor B (Columns) is whether the assembly was done on day shift, swing shift, or graveyard shift. The following data were obtained in the study where the response variable is the time required to assemble the PC from start to final test.

Carry out the appropriate statistical tests using an alpha equal .05 level. Answer: The experimental design is a two-factor ANOVA design with replications. The design is balanced since each cell has the same number of replications. Excel or Minitab can be used to perform the computations. The following results are provided: ANOVA Source of Variation Sample Columns Interaction Within

SS 5.895556 0.937222 1.702778 15.3475

Total

23.88306

df 2 2 4 27

MS F p-value F-crit 2.947778 5.185861 0.012419 3.354131 0.468611 0.824401 0.449235 3.354131 0.425694 0.7489 0.567377 2.727766 0.568426

The first test to be conducted is to test whether the number of workstations and the shift interact. The following null and alternative hypothesis are tested: H0 : Factors A and B do not interact to affect the mean response HA : Factors A and B do interact. The output shows a calculated F-ratio of .7489 for the interaction portion. The F-critical for alpha = .05 and 4 and 27 degrees of freedom is 2.728. Since .7489 < 2.728, we do not reject the null hypothesis. Thus, we conclude based on these sample data that the number of workstations and shift do not interact. We then proceed to test whether the means differ by number of workstations. The null and alternative hypotheses are: H0 : μA1 = μA2 = μA3 Ha : Not all menas are equal. 12-79 Copyright © 2018 Pearson Education, Inc.

The Excel printout refers to factor A (Work Stations) as the Sample. The calculated F = 5.1858 and this is compared to the critical F = 3.354. Since 5.1858 > 3.354, we reject the null hypothesis and conclude that the mean time to complete assembly of the PCs is not equal across all number of workstations. To check whether differences exist across the three shifts, we test the following null and alternative hypotheses: H0 : μB1 = μB2 = μB3 HA : Not all means are equal. Since the p-value for this test is seen to be .449 > alpha = .05, we do not reject the null hypothesis. Thus, the mean time to assemble PCs does not differ by shift. Diff: 3 Keywords: ANOVA, analysis of variance, two-factor, replication Section: 12-3 Two-Factor Analysis of Variance With Replication Outcome: 5

Business Statistics, 9e (Groebner/Shannon/Fry) Chapter 13 Goodness-of-Fit Tests and Contingency Analysis 1) A goodness-of-fit test can be used to determine whether a set of sample data comes from a specific hypothesized population distribution. Answer: TRUE Diff: 1 Keywords: fit, goodness-of-fit, distribution, population Section: 13-1 Introduction to Goodness-of-fit Tests Outcome: 1 2) If the test statistic for a chi-square goodness-of-fit test is larger than the critical value, the null hypothesis should be rejected. Answer: TRUE Diff: 1 Keywords: fit, goodness-of-fit, test statistic, critical value, null Section: 13-1 Introduction to Goodness-of-fit Tests Outcome: 1 3) The logic behind the chi-square goodness-of-fit test is based on determining how far the actual observed frequencies are from the expected frequencies. Answer: TRUE Diff: 2 Keywords: fit, goodness-of-fit, chi-square Section: 13-1 Introduction to Goodness-of-fit Tests Outcome: 1 4) The goodness-of-fit test is always a one-tail test with the rejection region in the lower tail. Answer: FALSE Diff: 2 Keywords: fit, goodness-of-fit, chi-square, rejection region Section: 13-1 Introduction to Goodness-of-fit Tests Outcome: 1 5) When the expected cell frequencies are smaller than 5, the cells should be combined in a meaningful way such that the expected cell frequencies do exceed 5. Answer: TRUE Diff: 2 Keywords: fit, goodness-of-fit, cell, chi-square, frequency Section: 13-1 Introduction to Goodness-of-fit Tests Outcome: 1

6) A sample size of at least 30 is sufficient in most cases provided that none of the expected frequencies are less than 5. Answer: TRUE Diff: 2 Keywords: fit, goodness-of-fit, null, group, test statistic Section: 13-1 Introduction to Goodness-of-fit Tests Outcome: 1 7) In a goodness-of-fit test, when the null hypothesis is true, the expected value for the chi-square test statistic is < zero. Answer: FALSE Diff: 2 Keywords: fit, goodness-of-fit, test statistic Section: 13-1 Introduction to Goodness-of-fit Tests Outcome: 1 8) The Conrad Real Estate Company recently conducted a statistical test to determine whether the number of days that homes are on the market prior to selling is normally distributed with a mean equal to 50 days and a standard deviation equal to 10 days. The sample of 200 homes was divided into 8 groups to form a grouped data frequency distribution. The degrees of freedom for the test will be 7. Answer: TRUE Diff: 2 Keywords: fit, goodness-of-fit, normal, degrees of freedom Section: 13-1 Introduction to Goodness-of-fit Tests Outcome: 1 9) The Conrad Real Estate Company recently conducted a statistical test to determine whether the number of days that homes are on the market prior to selling is normally distributed with a mean equal to 50 days and a standard deviation equal to 10 days. The sample of 200 homes was divided into 8 groups to form a grouped data frequency distribution. If a chi-square goodness-of-fit test is to be conducted using an alpha = .05, the critical value is 14.0671. Answer: TRUE Diff: 2 Keywords: fit, goodness-of-fit, chi-square, critical value, degrees of freedom Section: 13-1 Introduction to Goodness-of-fit Tests Outcome: 1

10) A business with 5 copy machines keeps track of how many copy machines need service on a given day. It believes this is binomially distributed with a probability of p = 0.2 of each machine needing service on any given day. It has collected the following based on a random sample of 100 days. X 0 1 2 3 4 or 5 Total

Frequency 28 38 22 7 5 100

Given this information, assuming that all expected values are sufficiently large to use the classes as shown above, the critical value for testing the hypothesis will be based on 5 degrees of freedom. Answer: FALSE Diff: 2 Keywords: fit, goodness-of-fit, degrees of freedom Section: 13-1 Introduction to Goodness-of-fit Tests Outcome: 1 11) A business with 5 copy machines keeps track of how many copy machines need service on a given day. It believes this is binomially distributed with a probability of p = 0.2 of each machine needing service on any given day. It has collected the following based on a random sample of 100 days. X 0 1 2 3 4 or 5 Total

Frequency 28 38 22 7 5 100

Given this information the expected number of days on which exactly 1 machine breaks down is 40.96. Answer: TRUE Diff: 3 Keywords: fit, goodness-of-fit, chi-square, expected value Section: 13-1 Introduction to Goodness-of-fit Tests Outcome: 1

12) A business with 5 copy machines keeps track of how many copy machines need service on a given day. It believes this is binomially distributed with a probability of p = 0.2 of each machine needing service on any given day. It has collected the following based on a random sample of 100 days. X 0 1 2 3 4 or 5 Total

Frequency 28 38 22 7 5 100

Given this information, assuming that all expected values are sufficiently large to use the classes as shown above, the critical value based on a 0.05 level of significance is 9.4877. Answer: TRUE Diff: 2 Keywords: fit, goodness-of-fit, critical value Section: 13-1 Introduction to Goodness-of-fit Tests Outcome: 1 13) It is believed that the number of drivers who are ticketed for speeding on a particular stretch of highway is a Poisson distribution with a mean of 3.5 per hour. A random sample of 100 hours is selected with the following results: X 0 1 2 3 4 5 6 7 8 9

Frequency 5 10 20 18 20 15 4 6 1 2 100

Given this information, and without regard to whether there is a need to combine cells due to expected cell frequencies, the critical value for testing whether the distribution is Poisson with a mean of 3.5 per hour at an alpha level of .05 is x2 = 15.5073. Answer: FALSE Diff: 2 Keywords: fit, goodness-of-fit, chi-square, critical value Section: 13-1 Introduction to Goodness-of-fit Tests Outcome: 1

14) It is believed that the number of drivers who are ticketed for speeding on a particular stretch of highway is a Poisson distribution with a mean of 3.5 per hour. A random sample of 100 hours is selected with the following results: X 0 1 2 3 4 5 6 7 8 9

Frequency 5 10 20 18 20 15 4 6 1 2 100

Given this information, it can be seen that the cells will need to be combined since the actual number of occurrences at some levels of x is less than 5. Answer: FALSE Diff: 2 Keywords: fit, goodness-of-fit, chi-square, cell Section: 13-1 Introduction to Goodness-of-fit Tests Outcome: 1 15) If the sample size is large, the standard normal distribution can be used in place of the chi-square in a goodness-of-fit test for testing whether the population is normally distributed. Answer: FALSE Diff: 2 Keywords: fit, goodness-of-fit, normal, chi-square, standard normal Section: 13-1 Introduction to Goodness-of-fit Tests Outcome: 1 16) By combining cells we guard against having an inflated test statistic that could have led us to incorrectly accept the null hypothesis. Answer: FALSE Diff: 3 Keywords: fit, goodness-of-fit, null, test statistic Section: 13-1 Introduction to Goodness-of-fit Tests Outcome: 1 17) If any of the observed frequencies are smaller than 5, then categories should be combined until all observed frequencies are at least 5. Answer: FALSE Diff: 2 Keywords: fit, goodness-of-fit, frequency Section: 13-1 Introduction to Goodness-of-fit Tests Outcome: 1 13-5 Copyright © 2018 Pearson Education, Inc.

18) A lube and oil change business believes that the number of cars that arrive for service is the same each day of the week. If the business is open six days a week (Monday - Saturday) and a random sample of n = 200 customers is selected, the expected number that will arrive on Monday is about 33.33. Answer: TRUE Diff: 1 Keywords: fit, goodness-of-fit, expected value, random sample Section: 13-1 Introduction to Goodness-of-fit Tests Outcome: 1 19) A lube and oil change business believes that the number of cars that arrive for service is the same each day of the week. If the business is open six days a week (Monday - Saturday) and a random sample of n = 200 customers is selected, the sum of the expected frequencies over the six days cannot be determined without seeing the actual sample data. Answer: FALSE Diff: 2 Keywords: fit, goodness-of-fit, expected frequency Section: 13-1 Introduction to Goodness-of-fit Tests Outcome: 1 20) A lube and oil change business believes that the number of cars that arrive for service is the same each day of the week. If the business is open six days a week (Monday - Saturday) and a random sample of n = 200 customers is selected, the critical value for testing the hypothesis using a goodness-of-fit test is x2 = 9.2363 if the alpha level for the test is set at .10. Answer: TRUE Diff: 2 Keywords: fit, goodness-of-fit, chi-square, critical value Section: 13-1 Introduction to Goodness-of-fit Tests Outcome: 1 21) A goodness-of-fit test can decide whether a set of data comes from a specific hypothesized distribution. Answer: TRUE Diff: 2 Keywords: fit, goodness-of-fit, distribution Section: 13-1 Introduction to Goodness-of-fit Tests Outcome: 1 22) If the calculated chi-square statistic for specific degrees of freedom is a large value, this supports evidence that the fit of the actual data to the hypothesized distribution is not good, and HA should be rejected. Answer: FALSE Diff: 2 Keywords: fit, goodness-of-fit, chi-square, null Section: 13-1 Introduction to Goodness-of-fit Tests Outcome: 1

23) The goodness-of-fit test is essentially determining if the test statistic is significantly larger than zero. Answer: FALSE Diff: 2 Keywords: fit, goodness-of-fit, test statistic Section: 13-1 Introduction to Goodness-of-fit Tests Outcome: 1 24) By combining cells we guard against having an inflated test statistic that could have led us to incorrectly reject the H0. Answer: TRUE Diff: 2 Keywords: fit, goodness-of-fit, test statistic, null Section: 13-1 Introduction to Goodness-of-fit Tests Outcome: 1 25) If one or more parameters are left unspecified in a goodness-of-fit test, they must be estimated from the sample data and one degree of freedom is lost for each parameter that must be estimated. Answer: TRUE Diff: 2 Keywords: fit, goodness-of-fit, parameter, degrees of freedom Section: 13-1 Introduction to Goodness-of-fit Tests Outcome: 1 26) The sampling distribution for a goodness-of-fit test is the Poisson distribution. Answer: FALSE Diff: 2 Keywords: goodness-of-fit, sampling distribution Section: 13-1 Introduction to Goodness-of-fit Tests Outcome: 1 27) Contingency analysis helps to make decisions when multiple proportions are involved. Answer: TRUE Diff: 1 Keywords: contingency analysis, proportions Section: 13-2 Introduction to Contingency Analysis Outcome: 2 28) Contingency analysis is used only for numerical data. Answer: FALSE Diff: 1 Keywords: contingency analysis, interval, ratio, measurement Section: 13-2 Introduction to Contingency Analysis Outcome: 2

29) Managers use contingency analysis to determine whether two categorical variables are independent of each other. Answer: TRUE Diff: 1 Keywords: contingency analysis, categorical, independent Section: 13-2 Introduction to Contingency Analysis Outcome: 2 30) A survey was recently conducted in which males and females were asked whether they owned a laptop personal computer. The following data were observed:

Have Laptop No Laptop

Males 120 50

Females 70 60

Given this information, the sample size in the survey was 300 people. Answer: TRUE Diff: 1 Keywords: contingency analysis, sample size Section: 13-2 Introduction to Contingency Analysis Outcome: 2 31) A survey was recently conducted in which males and females were asked whether they owned a laptop personal computer. The following data were observed:

Have Laptop No Laptop

Males 120 50

Females 70 60

Given this information, if having a laptop is independent of gender, the expected number of males with laptops in this survey is 150. Answer: FALSE Diff: 2 Keywords: contingency analysis, expected value Section: 13-2 Introduction to Contingency Analysis Outcome: 2

32) A survey was recently conducted in which males and females were asked whether they owned a laptop personal computer. The following data were observed:

Have Laptop No Laptop

Males 120 50

Females 70 60

Given this information, if an alpha level of .05 is used, the critical value for testing whether the two variables are independent is x2 = 3.8415. Answer: TRUE Diff: 2 Keywords: contingency analysis, critical value, degrees of freedom Section: 13-2 Introduction to Contingency Analysis Outcome: 2 33) A survey was recently conducted in which males and females were asked whether they owned a laptop personal computer. The following data were observed:

Have Laptop No Laptop

Males 120 50

Females 70 60

Given this information, if an alpha level of .05 is used, the sum of the expected cell frequencies will be equal to the sum of the observed cell frequencies. Answer: TRUE Diff: 2 Keywords: contingency analysis, expected, observed, cell frequencies Section: 13-2 Introduction to Contingency Analysis Outcome: 2 34) A survey was recently conducted in which males and females were asked whether they owned a laptop personal computer. The following data were observed:

Have Laptop No Laptop

Males 120 50

Females 70 60

Given this information, if an alpha level of .05 is used, the test statistic for determining whether having a laptop is independent of gender is approximately 14.23. Answer: FALSE Diff: 2 Keywords: contingency analysis, test statistic Section: 13-2 Introduction to Contingency Analysis Outcome: 2

35) When the variables of interest are both categorical and the decision maker is interested in determining whether a relationship exists between the two, a statistical technique known as contingency analysis is useful. Answer: TRUE Diff: 1 Keywords: contingency analysis, categorical Section: 13-2 Introduction to Contingency Analysis Outcome: 2 36) In conducting a test of independence for a contingency table that has 4 rows and 3 columns, the number of degrees of freedom is 11. Answer: FALSE Diff: 2 Keywords: contingency analysis, degrees of freedom, contingency, table Section: 13-2 Introduction to Contingency Analysis Outcome: 2 37) A study was recently conducted in which people were asked to indicate which new medium was their preferred choice for national news. The following data were observed:

under 21 21-40 41 and over

radio 30 20 30

television newspaper 50 5 25 30 30 50

Given this data, if we wish to test whether the preferred news source is independent of age, the expected frequency in the cell, radio—under 21 cell is 30. Answer: FALSE Diff: 2 Keywords: contingency analysis, expected frequency Section: 13-2 Introduction to Contingency Analysis Outcome: 2

38) A cell phone company wants to determine if the use of text messaging is independent of age. The follow data has been collected from a random sample of their customers.

Under 21 21-39 40 and over

Regularly use text Do not regularly messaging use text messaging 82 38 57 34 6 83

Using the data above, in order to test for the independence of age and the use of text messaging, the expected value for the "under 21 and regularly use text messaging" cell is 82. Answer: FALSE Diff: 2 Keywords: contingency analysis, hypothesis, expected number Section: 13-2 Introduction to Contingency Analysis Outcome: 2 39) A study was recently conducted in which people were asked to indicate which news medium was their preferred choice for national news. The following data were observed:

under 21 21-40 41 and over

radio 30 20 30

television newspaper 50 5 25 30 30 50

Given this data, if we wish to test whether the preferred news source is independent of age with an alpha equal to .05, the critical value will be a chi-square value with 9 degrees of freedom. Answer: FALSE Diff: 2 Keywords: contingency analysis, chi-square, degrees of freedom Section: 13-2 Introduction to Contingency Analysis Outcome: 2 40) A study was recently conducted in which people were asked to indicate which news medium was their preferred choice for national news. The following data were observed:

under 21 21-40 41 and over

radio 30 20 30

television newspaper 50 5 25 30 30 50

Given this data, if we wish to test whether the preferred news source is independent of age, the cell with the largest expected cell frequency is also the cell with the largest observed frequency. Answer: FALSE Diff: 3 Keywords: contingency analysis, expected, observed, frequency Section: 13-2 Introduction to Contingency Analysis Outcome: 2 13-11 Copyright © 2018 Pearson Education, Inc.

41) A study was recently conducted in which people were asked to indicate which news medium was their preferred choice for national news. The following data were observed:

under 21 21-40 41 and over

radio 30 20 30

television newspaper 50 5 25 30 30 50

Given this data, if we wish to test whether the preferred news source is independent of age, for an alpha = .05 level, the critical value from the chi-square table is based on 8 degrees of freedom. Answer: FALSE Diff: 2 Keywords: contingency analysis, chi-square, degrees of freedom, critical value Section: 13-2 Introduction to Contingency Analysis Outcome: 2 42) A study was recently conducted in which people were asked to indicate which news medium was their preferred choice for national news. The following data were observed:

under 21 21-40 41 and over

radio 30 20 30

television newspaper 50 5 25 30 30 50

Given this data, if we wish to test whether the preferred news source is independent of age, for an alpha = .05 level, the critical value from the chi-square table is 9.4877. Answer: TRUE Diff: 2 Keywords: contingency analysis, critical value, chi-square Section: 13-2 Introduction to Contingency Analysis Outcome: 2

43) A study was recently conducted in which people were asked to indicate which news medium was their preferred choice for national news. The following data were observed:

under 21 21-40 41 and over

radio 30 20 30

television newspaper 50 5 25 30 30 50

Given this data, if we wish to test whether the preferred news source is independent of age, for an alpha = .05 level, the test statistic is computed to be approximately 40.70. Answer: TRUE Diff: 3 Keywords: contingency analysis, test statistic, chi-square Section: 13-2 Introduction to Contingency Analysis Outcome: 2 44) A cell phone company wants to determine if the use of text messaging is independent of age. The following data has been collected from a random sample of customers.

Under 21 21-39 40 and over

Regularly use text Do not regularly messaging use text messaging 82 38 57 34 6 8

Using this data, if we wish to test whether the preferred news source is independent of age using a 0.05 level of significance, the critical value is 5.9915. Answer: TRUE Diff: 2 Keywords: contingency analysis, chi-square, null, critical value Section: 13-2 Introduction to Contingency Analysis Outcome: 2 45) In order to apply the chi-square contingency methodology for quantitative variables, we must first break the quantitative variable down into discrete categories. Answer: TRUE Diff: 2 Keywords: contingency analysis, variables, chi-square Section: 13-2 Introduction to Contingency Analysis Outcome: 2

46) A study was recently done in the United States in which car owners were asked to indicate whether their most recent car purchase was a U.S. car, a German car, or a Japanese car. The people in the survey were divided by geographic region in the United States. The following data were recorded.

East Coast Central West Coast

US 200 250 80

Japanese 200 100 300

German 50 20 40

Given this situation, the sample size used in this study was nine. Answer: FALSE Diff: 1 Keywords: contingency analysis, sample size Section: 13-2 Introduction to Contingency Analysis Outcome: 2 47) A study was recently done in the United States in which car owners were asked to indicate whether their most recent car purchase was a U.S. car, a German car, or a Japanese car. The people in the survey were divided by geographic region in the United States. The following data were recorded.

East Coast Central West Coast

US 200 250 80

Japanese 200 100 300

German 50 20 40

Given this situation, the null hypothesis to be tested is that the car origin is dependent on the geographical location of the buyer. Answer: FALSE Diff: 2 Keywords: contingency analysis, null, hypothesis Section: 13-2 Introduction to Contingency Analysis Outcome: 2

48) A study was recently done in the United States in which car owners were asked to indicate whether their most recent car purchase was a U.S. car, a German car, or a Japanese car. The people in the survey were divided by geographic region in the United States. The following data were recorded.

East Coast Central West Coast

US 200 250 80

Japanese 200 100 300

German 50 20 40

Given this situation, to test whether the car origin is independent of the geographical location of the buyer, the sum of the expected cell frequencies will equal 1,240. Answer: TRUE Diff: 2 Keywords: contingency analysis, cell, cell frequencies, sample size Section: 13-2 Introduction to Contingency Analysis Outcome: 2 49) A study was recently done in the United States in which car owners were asked to indicate whether their most recent car purchase was a U.S. car, a German car, or a Japanese car. The people in the survey were divided by geographic region in the United States. The following data were recorded.

East Coast Central West Coast

US 200 250 80

Japanese 200 100 300

German 50 20 40

Given this situation, to test whether the car origin is independent of the geographical location of the buyer, the critical value for alpha = .10 is 14.6837. Answer: FALSE Diff: 2 Keywords: contingency analysis, critical value, chi-square Section: 13-2 Introduction to Contingency Analysis Outcome: 2

50) A study was recently done in the United States in which car owners were asked to indicate whether their most recent car purchase was a U.S. car, a German car, or a Japanese car. The people in the survey were divided by geographic region in the United States. The following data were recorded.

East Coast Central West Coast

US 200 250 80

Japanese 200 100 300

German 50 20 40

Given this situation, to test whether the car origin is independent of the geographical location of the buyer, the expected number of people in the sample who bought a German made car and who lived on the East Coast is just under 40 people. Answer: TRUE Diff: 2 Keywords: contingency analysis, expected value Section: 13-2 Introduction to Contingency Analysis Outcome: 2 51) A cell phone company wants to determine if the use of text messaging is independent of age. The following data has been collected from a random sample of customers.

Under 21 21-39 40 and over

Regularly use text Do not regularly messaging use text messaging 82 38 57 34 6 83

To conduct a test of independence, the difference expected value for the "40 and over and regularly use text messaging" cell is just over 43 people. Answer: TRUE Diff: 2 Keywords: contingency analysis, chi-square, expected frequency Section: 13-2 Introduction to Contingency Analysis Outcome: 2 52) Contingency analysis can be used when the level of data measurement is nominal or ordinal. Answer: TRUE Diff: 1 Keywords: contingency analysis, nominal, ordinal, data Section: 13-2 Introduction to Contingency Analysis Outcome: 2 53) To employ contingency analysis, we set up a 2-dimensional table called a contingency table. Answer: TRUE Diff: 1 Keywords: contingency analysis, table Section: 13-2 Introduction to Contingency Analysis Outcome: 2 13-16 Copyright © 2018 Pearson Education, Inc.

54) A contingency table and a cross-tabulation table are two separate things and should not be used for the same purpose. Answer: FALSE Diff: 1 Keywords: contingency analysis, cross-tabulation Section: 13-2 Introduction to Contingency Analysis Outcome: 2 55) In a contingency analysis, we expect the actual frequencies in each cell to approximately match the corresponding expected cell frequencies when H0 is true. Answer: TRUE Diff: 2 Keywords: contingency analysis, observed, expected, frequency, null Section: 13-2 Introduction to Contingency Analysis Outcome: 2 56) In a chi-square contingency test, the number of degrees of freedom is equal to the number of cells minus 1. Answer: FALSE Diff: 2 Keywords: contingency analysis, degrees of freedom, cell Section: 13-2 Introduction to Contingency Analysis Outcome: 2 57) In a chi-square contingency analysis application, the expected cell frequencies will be equal in all cells if the null hypothesis is true. Answer: FALSE Diff: 2 Keywords: contingency analysis, expected frequency, null, hypothesis Section: 13-2 Introduction to Contingency Analysis Outcome: 2 58) Unlike the case of goodness-of-fit testing, with contingency analysis there is no restriction on the minimum size for an expected cell frequency. Answer: FALSE Diff: 2 Keywords: contingency analysis, goodness-of-fit, expected frequency Section: 13-2 Introduction to Contingency Analysis Outcome: 2 59) In a contingency analysis the expected values are based on the assumption that the two variables are independent of each other. Answer: TRUE Diff: 2 Keywords: contingency analysis, chi-square, expected values Section: 13-2 Introduction to Contingency Analysis Outcome: 2 13-17 Copyright © 2018 Pearson Education, Inc.

60) If a contingency analysis test is performed with a 4 × 6 design, and if alpha = .05, the critical value from the chi-square distribution is 24.9958 Answer: TRUE Diff: 2 Keywords: contingency analysis, critical value, chi-square Section: 13-2 Introduction to Contingency Analysis Outcome: 2 61) If a contingency analysis test performed with a 4 × 6 design results in a test statistic value of 18.72, and if alpha = .05, the null hypothesis that the row and column variable are independent should be rejected. Answer: FALSE Diff: 2 Keywords: contingency analysis, chi-square, test statistic, critical value, null Section: 13-2 Introduction to Contingency Analysis Outcome: 2 62) If the null hypothesis is not rejected, you do not need to worry when the expected cell frequencies drop below 5.0 Answer: TRUE Diff: 3 Keywords: contingency analysis, chi-square, null Section: 13-2 Introduction to Contingency Analysis Outcome: 2 63) The degrees of freedom for the chi-square goodness-of-fit test are equal to ________, where k is the number of categories. A) k + 1 B) k - 1 C) k + 2 D) k - 2 Answer: B Diff: 1 Keywords: chi-square, goodness of fit, degrees of freedom Section: 13-1 Introduction to Goodness-of-fit Tests Outcome: 1 64) Which of the following statements is true in the context of a chi-square goodness-of-fit test? A) The degrees of freedom for determining the critical value will be the number of categories minus 1. B) The critical value will come from the standard normal table if the sample size exceeds 30. C) The null hypothesis will be rejected for a small value of the test statistic. D) A very large test statistic will result in the null not being rejected. Answer: A Diff: 2 Keywords: chi-square, goodness of fit, degrees of freedom Section: 13-1 Introduction to Goodness-of-fit Tests Outcome: 1

65) A walk-in medical clinic believes that arrivals are uniformly distributed over weekdays (Monday through Friday). It has collected the following data based on a random sample of 100 days.

Mon Tue Wed Thu Fri Total

Frequency 25 22 19 18 16 100

Based on this information how many degrees for freedom are involved in this goodness of fit test? A) 99 B) 100 C) 4 D) 5 Answer: C Diff: 1 Keywords: chi-square, goodness of fit, degrees of freedom Section: 13-1 Introduction to Goodness-of-fit Tests Outcome: 1 66) A walk-in medical clinic believes that arrivals are uniformly distributed over weekdays (Monday through Friday). It has collected the following data based on a random sample of 100 days.

Mon Tue Wed Thu Fri Total

Frequency 25 22 19 18 16 100

Assuming that a goodness-of-fit test is to be conducted using a 0.10 level of significance, the critical value is: A) 9.4877 B) 11.0705 C) 7.7794 D) 9.2363 Answer: C Diff: 2 Keywords: chi-square, goodness-of-fit, critical value Section: 13-1 Introduction to Goodness-of-fit Tests Outcome: 1

67) A walk-in medical clinic believes that arrivals are uniformly distributed over weekdays (Monday through Friday). It has collected the following data based on a random sample of 100 days.

Mon Tue Wed Thu Fri Total

Frequency 25 22 19 18 16 100

To conduct a goodness-of-fit test, what is the expected value for Friday? A) 20 B) 25 C) 16 D) 100 Answer: A Diff: 1 Keywords: chi-square, goodness-of-fit, expected frequency Section: 13-1 Introduction to Goodness-of-fit Tests Outcome: 1 68) A walk-in medical clinic believes that arrivals are uniformly distributed over weekdays (Monday through Friday). It has collected the following data based on a random sample of 100 days.

Mon Tue Wed Thu Fri Total

Frequency 25 22 19 18 16 100

What is the value of the test statistic needed to conduct a goodness-of-fit test? A) 8.75 B) 7.7794 C) 2.46 D) 2.50 Answer: D Diff: 2 Keywords: chi-square, goodness-of-fit, test statistic Section: 13-1 Introduction to Goodness-of-fit Tests Outcome: 1

69) A walk-in medical clinic believes that arrivals are uniformly distributed over weekdays (Monday through Friday). It has collected the following data based on a random sample of 100 days.

Mon Tue Wed Thu Fri Total

Frequency 25 22 19 18 16 100

Based on these data, conduct a goodness-of-fit test using a 0.10 level of significance. Which conclusion is correct? A) Arrivals are not uniformly distributed over the weekday because (test statistic) > (critical value). B) Arrivals are uniformly distributed over the weekday because (test statistic) > (critical value). C) Arrivals are not uniformly distributed over the weekday because (test statistic) < (critical value). D) Arrives are uniformly distributed over the weekday because (test statistic) < (critical value). Answer: D Diff: 3 Keywords: chi-square, goodness-of-fit, critical value, test statistic Section: 13-1 Introduction to Goodness-of-fit Tests Outcome: 1 70) In a chi-square goodness-of-fit test, by combining cells we guard against having an inflated test statistic that could have caused us to: A) incorrectly reject the H0. B) incorrectly accept the H0. C) incorrectly reject the H1. D) incorrectly accept the H1. Answer: A Diff: 2 Keywords: chi-square, goodness-of-fit, null, cells Section: 13-1 Introduction to Goodness-of-fit Tests Outcome: 1 71) In a goodness-of-fit test about a population distribution, if one or more parameters are left unspecified in H0, they must be estimated from the sample data. This will reduce the degrees of freedom by ________ for each estimated parameter. A) 1 B) 2 C) 3 D) None of the above Answer: A Diff: 2 Keywords: chi-square, goodness-of-fit, degrees of freedom, parameter Section: 13-1 Introduction to Goodness-of-fit Tests Outcome: 1 13-21 Copyright © 2018 Pearson Education, Inc.

72) If a sample with n = 30 subjects distributed over 5 categories was selected, a chi-square test for goodness-of-fit will be used. How many degrees of freedom will be used in determining the chi-square test statistic? A) 3 B) 4 C) 5 D) 25 Answer: B Diff: 2 Keywords: chi-square, goodness-of-fit, degrees of freedom Section: 13-1 Introduction to Goodness-of-fit Tests Outcome: 1 73) Consider a goodness-of-fit test with a computed value of chi-square = 27.385 and a critical value = 13.388, the appropriate conclusion would be to: A) reject H0. B) fail to reject H0. C) reject HA. D) take a larger sample. Answer: A Diff: 2 Keywords: chi-square, goodness-of-fit, null, critical value, test statistic Section: 13-1 Introduction to Goodness-of-fit Tests Outcome: 1 74) A researcher is using a chi-square test to determine whether there are any preferences among 4 brands of orange juice. With alpha = 0.05 and n = 30, the critical region for the hypothesis test would have a boundary of: A) 7.81 B) 8.71 C) 8.17 D) 42.25 Answer: A Diff: 2 Keywords: chi-square, goodness-of-fit, critical value Section: 13-1 Introduction to Goodness-of-fit Tests Outcome: 1

75) A chi-square test for goodness-of-fit is used to test whether or not there are any preferences among 3 brands of peas. If the study uses a sample of n = 60 subjects, then the expected frequency for each category would be: A) 20 B) 30 C) 60 D) 33 Answer: A Diff: 2 Keywords: chi-square, goodness-of-fit, expected frequency Section: 13-1 Introduction to Goodness-of-fit Tests Outcome: 1 76) We are interested in determining whether the opinions of the individuals on gun control (as to Yes, No, and No Opinion) are uniformly distributed. A sample of 150 was taken and the following data were obtained. Do you support gun control Yes No No Opinion

Number of Responses 40 60 50

The conclusion of the test with alpha = 0.05 is that the views of people on gun control are: A) uniformly distributed. B) not uniformly distributed. C) inconclusive. D) None of the above Answer: A Diff: 2 Keywords: chi-square, goodness-of-fit, expected frequency Section: 13-1 Introduction to Goodness-of-fit Tests Outcome: 1 77) To use contingency analysis for numerical data, which of the following is true? A) Contingency analysis cannot be used for numerical data. B) Numerical data must be broken up into specific categories. C) Contingency analysis can be used for numerical data only if both variables are numerical. D) Contingency analysis can be used for numerical data only if it is interval data. Answer: B Diff: 2 Keywords: contingency analysis, nominal, ordinal, data Section: 13-2 Introduction to Contingency Analysis Outcome: 2

78) What does the term observed cell frequencies refer to? A) The frequencies found in the population being examined B) The frequencies found in the sample being examined C) The frequencies computed from H0 D) The frequencies computed from H1 Answer: B Diff: 2 Keywords: contingency analysis, observed, cell frequency Section: 13-2 Introduction to Contingency Analysis Outcome: 2 79) What does the term expected cell frequencies refer to? A) The frequencies found in the population being examined B) The frequencies found in the sample being examined C) The frequencies computed from H0 D) the frequencies computed from H1 Answer: C Diff: 2 Keywords: contingency analysis, expected, cell frequency Section: 13-2 Introduction to Contingency Analysis Outcome: 2 80) We expect the actual frequencies in each cell to approximately match the corresponding expected cell frequencies when: A) H0 is false. B) H0 is true. C) H0 is falsely accepted. D) the variables are related to each other. Answer: B Diff: 1 Keywords: contingency analysis, null, hypothesis, observed, actual, expected, frequency Section: 13-2 Introduction to Contingency Analysis Outcome: 2 81) In a contingency analysis, the greater the difference between the actual and the expected frequencies, the more likely: A) H0 should be rejected. B) H0 should be accepted. C) we cannot determine H0. D) the smaller the test statistic will be. Answer: A Diff: 1 Keywords: contingency analysis, actual, expected, observed, null Section: 13-2 Introduction to Contingency Analysis Outcome: 2

82) In a chi-square contingency analysis, when expected cell frequencies drop below 5, the calculated chisquare value tends to be inflated and may inflate the true probability of ________ beyond the stated significance level. A) committing a Type I error B) committing a Type II error C) Both A and B D) All of the above Answer: A Diff: 2 Keywords: contingency analysis, expected, frequency, type I, alpha Section: 13-2 Introduction to Contingency Analysis Outcome: 2 83) In performing chi-square contingency analysis, to overcome a small expected cell frequency problem, we: A) combine the categories of the row and/or column variables. B) increase the sample size. C) Both A and B D) None of the above Answer: C Diff: 2 Keywords: contingency analysis, combine, sample size, expected, frequency Section: 13-2 Introduction to Contingency Analysis Outcome: 2 84) How can the degrees of freedom be found in a contingency table with cross-classified data? A) When df are equal to rows minus columns B) When df are equal to rows multiplied by columns C) When df are equal to rows minus 1 multiplied by columns minus 1 D) Total number of cell minus 1 Answer: C Diff: 2 Keywords: contingency analysis, degrees of freedom, table, cross-classified Section: 13-2 Introduction to Contingency Analysis Outcome: 2

85) A cell phone company wants to determine if the use of text messaging is independent of age. The following data has been collected from a random sample of customers.

Under 21 21-39 40 and over

Regularly use text Do not regularly messaging use text messaging 82 38 57 34 6 83

Based on the data above what is the expected value for the "under 21 and regularly use text messaging" cell? A) 82 B) 50 C) 120 D) 58 Answer: D Diff: 2 Keywords: contingency analysis, expected frequency Section: 13-2 Introduction to Contingency Analysis Outcome: 2 86) A cell phone company wants to determine if the use of text messaging is independent of age. The following data has been collected from a random sample of customers.

Under 21 21-39 40 and over

Regularly use text Do not regularly messaging use text messaging 82 38 57 34 6 83

To conduct a contingency analysis, the number of degrees of freedom is: A) 6 B) 5 C) 3 D) 2 Answer: D Diff: 2 Keywords: contingency analysis, critical value, chi-square Section: 13-2 Introduction to Contingency Analysis Outcome: 2

87) A cell phone company wants to determine if the use of text messaging is independent of age. The following data has been collected from a random sample of customers.

Under 21 21-39 40 and over

Regularly use text Do not regularly messaging use text messaging 82 38 57 34 6 83

To conduct a contingency analysis using a 0.01 level of significance, the value of the critical value is: A) 15.0863 B) 5.9915 C) 9.2104 D) 11.0705 Answer: C Diff: 2 Keywords: contingency analysis, chi-square, critical value Section: 13-2 Introduction to Contingency Analysis Outcome: 2 88) A cell phone company wants to determine if the use of text messaging is independent of age. The following data has been collected from a random sample of customers.

Under 21 21-39 40 and over

Regularly use text Do not regularly messaging use text messaging 82 38 57 34 6 83

To conduct a contingency analysis, the value of the test statistic is: A) 9.2104 B) 88.3 C) 275.02 D) 14.6 Answer: B Diff: 2 Keywords: contingency analysis, test statistic, chi-square Section: 13-2 Introduction to Contingency Analysis Outcome: 2

89) For a chi-square test involving a contingency table, suppose H0 is rejected. We conclude that the two variables are: A) curvilinear. B) linear. C) related. D) not related. Answer: C Diff: 2 Keywords: contingency analysis, chi-square, null, alternative, hypothesis Section: 13-2 Introduction to Contingency Analysis Outcome: 2 90) When testing for independence in a contingency table with 3 rows and 4 columns, there are ________ degrees of freedom. A) 5 B) 6 C) 7 D) 12 Answer: B Diff: 1 Keywords: contingency analysis, degrees of freedom Section: 13-2 Introduction to Contingency Analysis Outcome: 2 91) In testing a hypothesis that two categorical variables are independent using the x2 test, the expected cell frequencies are based on assuming: A) the null hypothesis. B) the alternative hypothesis. C) the normal distribution. D) the variable are related. Answer: A Diff: 1 Keywords: contingency analysis, chi-square, expected, frequency, null Section: 13-2 Introduction to Contingency Analysis Outcome: 2

92) A study published in the American Journal of Public Health was conducted to determine whether the use of seat belts in motor vehicles depends on ethnic status in San Diego County. A sample of 792 children treated for injuries sustained from motor vehicle accidents was obtained, and each child was classified according to (1) ethnic status (Hispanic or non-Hispanic) and (2) seat belt usage (worn or not worn) during the accident. The number of children in each category is given in the table below.

Seat belts worn Seat belts not worn

Hispanic 31 283

Non-Hispanic 148 330

Referring to these data, which test would be used to properly analyze the data in this experiment? A) x2 test for independence in a two-way contingency table B) x2 test for equal proportions in a one-way table C) ANOVA F-test for interaction in a 2 × 2 factorial design D) x2 goodness-of-fit test Answer: A Diff: 2 Keywords: contingency analysis, chi-square, independence Section: 13-2 Introduction to Contingency Analysis Outcome: 2 93) A study published in the American Journal of Public Health was conducted to determine whether the use of seat belts in motor vehicles depends on ethnic status in San Diego County. A sample of 792 children treated for injuries sustained from motor vehicle accidents was obtained, and each child was classified according to (1) ethnic status (Hispanic or non-Hispanic) and (2) seat belt usage (worn or not worn) during the accident. The number of children in each category is given in the table below.

Seat belts worn Seat belts not worn

Hispanic 31 283

Non-Hispanic 148 330

Referring to these data, the calculated test statistic is: A) approximately -0.9991 B) nearly -0.1368 C) about 48.1849 D) approximately 72.8063 Answer: C Diff: 3 Keywords: contingency analysis, test statistic, chi-square Section: 13-2 Introduction to Contingency Analysis Outcome: 2

94) A study published in the American Journal of Public Health was conducted to determine whether the use of seat belts in motor vehicles depends on ethnic status in San Diego County. A sample of 792 children treated for injuries sustained from motor vehicle accidents was obtained, and each child was classified according to (1) ethnic status (Hispanic or non-Hispanic) and (2) seat belt usage (worn or not worn) during the accident. The number of children in each category is given in the table below.

Seat belts worn Seat belts not worn

Hispanic 31 283

Non-Hispanic 148 330

Referring to these data, which of the following conclusions should be reached if the appropriate hypothesis is conducted using an alpha = .05 level? A) The mean value for Hispanics is the same as for Non-Hispanics. B) There is no relationship between whether someone is Hispanic and whether they wear a seat belt. C) The use of seat belts and whether a person is Hispanic or not is statistically related. D) None of the above Answer: C Diff: 3 Keywords: contingency analysis, chi-square, hypothesis Section: 13-2 Introduction to Contingency Analysis Outcome: 2 95) Many companies use well-known celebrities as spokespeople in their TV advertisements. A study was conducted to determine whether brand awareness of female TV viewers and the gender of the spokesperson are independent. Each in a sample of 300 female TV viewers was asked to identify a product advertised by a celebrity spokesperson. The gender of the spokesperson and whether or not the viewer could identify the product was recorded. The numbers in each category are given below.

Identified product Could not identify

Male Celebrity 41 109

Female Celebrity 61 89

Referring to these sample data, which test would be used to properly analyze the data in this experiment? A) x2 test for independence in a two-way contingency table B) x2 test for equal proportions in a one-way table C) ANOVA F-test for main treatment effect D) x2 goodness-of-fit test Answer: A Diff: 2 Keywords: contingency analysis, chi-square, table Section: 13-2 Introduction to Contingency Analysis Outcome: 2

96) Many companies use well-known celebrities as spokespeople in their TV advertisements. A study was conducted to determine whether brand awareness of female TV viewers and the gender of the spokesperson are independent. Each in a sample of 300 female TV viewers was asked to identify a product advertised by a celebrity spokesperson. The gender of the spokesperson and whether or not the viewer could identify the product was recorded. The numbers in each category are given below.

Identified product Could not identify

Male Celebrity 41 109

Female Celebrity 61 89

Referring to these sample data, if the appropriate hypothesis test is to be conducted using a .05 level of significance, which of the following is correct critical value? A) 9.4877 B) 3.8415 C) 1.96 D) 7.8147 Answer: B Diff: 2 Keywords: contingency analysis, chi-square, critical value Section: 13-2 Introduction to Contingency Analysis Outcome: 2 97) Many companies use well-known celebrities as spokespeople in their TV advertisements. A study was conducted to determine whether brand awareness of female TV viewers and the gender of the spokesperson are independent. Each in a sample of 300 female TV viewers was asked to identify a product advertised by a celebrity spokesperson. The gender of the spokesperson and whether or not the viewer could identify the product was recorded. The numbers in each category are given below.

Identified product Could not identify

Male Celebrity 41 109

Female Celebrity 61 89

Referring to these sample data, which of the following values is the correct value of the test statistic? A) Approximately 9.48 B) Nearly 23.0 C) About 3.84 D) Approximately 5.94 Answer: D Diff: 3 Keywords: contingency analysis, chi-square, test statistic Section: 13-2 Introduction to Contingency Analysis Outcome: 2

98) Many companies use well-known celebrities as spokespeople in their TV advertisements. A study was conducted to determine whether brand awareness of female TV viewers and the gender of the spokesperson are independent. Each in a sample of 300 female TV viewers was asked to identify a product advertised by a celebrity spokesperson. The gender of the spokesperson and whether or not the viewer could identify the product was recorded. The numbers in each category are given below.

Identified product Could not identify

Male Celebrity 41 109

Female Celebrity 61 89

Referring to these sample data, if the appropriate null hypothesis is tested using a significance level equal to .05, which of the following conclusions should be reached? A) There is a relationship between gender of the celebrity and product identification. B) There is no relationship between gender of the celebrity and product identification. C) The mean number of products identified for males is different than the mean number for females. D) Females have higher brand awareness than males. Answer: A Diff: 3 Keywords: contingency analysis, chi-square, hypothesis, null Section: 13-2 Introduction to Contingency Analysis Outcome: 2 99) The degrees of freedom for a contingency table with 15 rows and 11 columns is: A) 165 B) 117 C) 140 D) 60 Answer: C Diff: 1 Keywords: contingency analysis, chi-square Section: 13-2 Introduction to Contingency Analysis Outcome: 2

100) We want to test whether type of car owned (domestic or foreign) is independent of gender. A contingence table is obtained from a sample of 990 people as

At alpha = 0.05 level, we conclude that: A) x2= 3.34 and type of car owned is independent of gender. B) x2 = 3.34 and type of car owned is dependent of gender. C) x2 = 3.84 and type of car owned is independent of gender. D) x2 = 3.84 and type of car owned is dependent of gender. Answer: A Diff: 3 Keywords: contingency analysis, chi-square, null, critical value Section: 13-2 Introduction to Contingency Analysis Outcome: 2 101) The billing department of a national cable service company is conducting a study of how customers pay their monthly cable bills. The cable company accepts payment in one of four ways: in person at a local office, by mail, by credit card, or by electronic funds transfer from a bank account. The cable company randomly sampled 400 customers to determine if there is a relationship between the customer's age and the payment method used. The following sample results were obtained:

Based on the sample data, can the cable company conclude that there is a relationship between the age of the customer and the payment method used? Conduct the appropriate test at the alpha= 0.01 level of significance. A) Because x2 = 42.2412 > 21.666, do not reject the null hypothesis. Based on the sample data conclude that age and type of payment are independent. B) Because x2 = 42.2412 > 21.666, reject the null hypothesis. Based on the sample data conclude that age and type of payment are not independent C) Because x2 = 50.3115 > 21.666, do not reject the null hypothesis. Based on the sample data conclude that age and type of payment are independent. D) Because x2 = 50.3115 > 21.666, reject the null hypothesis. Based on the sample data conclude that age and type of payment are not independent. Answer: D Diff: 2 Keywords: contingency analysis Section: 13-2 Introduction to Contingency Analysis Outcome: 1 13-33 Copyright © 2018 Pearson Education, Inc.

102) Explain the basic logic behind the chi-square goodness-of-fit test. Answer: The logic of the chi-square goodness-of-fit test is as follows: If the sample data selected randomly from a hypothesized population distribution have a distribution that "closely" matches the hypothesized distribution, then the hypothesis should be "accepted." If the sample data do not closely fit the hypothesized distribution, then we would reject the hypothesis. The decision is based on the chisquare test statistic, which compares the actual frequency of sample occurrences as prescribed levels of the variable against the expected frequencies at those same levels assuming the hypothesized distribution is correct. The test statistics is computed as follows: x2 = where: oi = Observed cell frequency for category i ei = Expected cell frequency for category i k = Number of categories. If the test statistic is larger than the chi-square critical value for a specified level of significance, the null hypothesis is rejected. Diff: 2 Keywords: chi-square, goodness-of-fit Section: 13-1 Introduction to Goodness-of-fit Tests Outcome: 1

103) The Bradfield Container Company makes "cardboard" boxes for commercial use (i.e., pizza boxes). One of the big issues for the company is the set-up time required to change over from one order to the next. At one particular machine, the set-up time is thought to be uniformly distributed between 10 and 21 minutes. To test whether this is true or not, a random sample of 180 set-ups on this machine was selected with set-up time rounded to the nearest two-minute intervals. The following results occurred: Set-up Time 10-11 minutes 12-13 minutes 14-15 minutes 16-17 minutes 18-19 minutes 20-21 minutes

Frequency 13 23 40 44 40 20

a. What are the appropriate null and alternative hypothesis to be tested? b. Based on the null and alternative hypotheses stated in part a, determine the expected frequencies for each set-up time category. c. Assuming that we wish to conduct the hypothesis test at the .05 level, what is the critical value that should be used? d. Compute the test statistic and carry out the hypothesis test. Answer: a. The null and alternative hypotheses are: H0 : set-up times on this machine are uniformly distributed between 10 and 21 minutes HA : set-up times are not uniformly distributed between 10 and 21 minutes. b. Because the hypothesized distribution is a uniform distribution, the expected number of values at each set-up time level is found by dividing number of levels by the sample size: Expected = 180/6 = 30 Set-up Time 10-11 minutes 12-13 minutes 14-15 minutes 16-17 minutes 18-19 minutes 20-21 minutes

Frequency 13 23 40 44 40 20

Expected 30 30 30 30 30 30

c. The critical value will be a chi-square value with degrees of freedom equal to the number of levels minus one. The degrees of freedom is 6 - 1 = 5. From the chi-square table with alpha = .05 and 5 degrees of freedom, the critical value is x2= 11.0705. Thus, if the test statistic > 11.0705, reject the null hypothesis. d. The test statistic is found using: x2 =

The computations are shown as follows: Set-up Time

Frequency

Expected

10-11 minutes

12-13 minutes

14-15 minutes

(O-E)2 (13-30)2=289 (23-30)2 = 49 (40-30)2 =100

16-17 minutes

18-19 minutes

(44-30)2=196 (40-30)2=100

(20-30)2=100

20-21 minutes

(O-E)2/E 289/30=9.63 49/30=1.63 100/30 = 3.33 196/30 = 6.53 100/30 = 3.33 Total =27.78

100/30 = 3.33

Since x2 = 27.78 > 11.07, we reject the null hypothesis and conclude that the distribution of set-up times is not uniform. Diff: 3 Keywords: null, alternative, hypothesis, test statistic, chi-square, critical value Section: 13-1 Introduction to Goodness-of-fit Tests Outcome: 1 104) Explain why, in performing a goodness-of-fit test, it is sometimes necessary to combine categories. Answer: Because of the way in which the chi-square test statistic is computed by squaring the difference between the observed and expected frequencies, when the expected frequencies are small (less than 5), the calculated test statistic can become artificially large and therefore may lead to an increased chance of committing a Type I statistical error. That is, a true null hypothesis may be rejected at a higher rate than indicated by the selected significance level. By combining categories, the small expected frequencies are grouped to become larger than five and thus the issue of inflated Type I error probability dissolves. Note: An alternative to combining categories is to increase the sample size. Large sample sizes result in greater expected cell frequencies in all categories. Diff: 2 Keywords: goodness-of-fit, combine, categories, cell Section: 13-1 Introduction to Goodness-of-fit Tests Outcome: 1

105) Recently a survey was conducted involving customers of a fitness center in Dallas, Texas. Participants were asked to indicate how often they use the club by checking one of the following categories: 0-1 time per week; 2-3 times per week; 4-5 times per week; more than 5 times. The following data show how males and females responded to this question. 0-1 41 109

Males Females

2-3 61 89

4-5 50 60

over 5 20 45

One of the purposes of the survey was to determine whether there is a relationship between the gender of the customer and the number of visits made each week. a. State the appropriate null and alternative hypothesis. b. What test procedure is appropriate to use to conduct this test? c. Conduct the hypothesis test using an alpha = .05 level. Answer: a. H0 : number of visits is independent of gender HA : number of visits is related to gender b. Because the data are observed frequencies in various discrete categories, the appropriate test to use is chi-square contingency analysis. This involves determining the expected frequencies assuming the null hypothesis is true and then comparing these expected frequencies, cell by cell, to the observed frequencies. If these closely match, then the null hypothesis should not be rejected. However, if there is a big difference between the expected and observed cell frequencies, we should reject the null hypothesis. c. The test statistic for performing a chi-square contingency analysis is computed as follows: x2 =

with d.f. = (r - 1)(c - 1).

The first step needed is to compute the expected cell frequencies. This is done under the assumption that the null hypothesis is true and that the proportion of customers in each use level is the same regardless of gender. The expected frequencies can be computed using: Expected Frequency = For example, for the cell corresponding to males who use the center 0-1 times per week, we get: Expected Frequency =

= 54.3158.

The following shows the expected cell frequencies for each cell:

Males Females

0-1 54.31579 96.68421

2-3 54.34179 95.68421

4-5 39.83158 70.16842

over 5 23.53684 41.46316

Next for each cell we compute:

. For example in the cell for males and use between 0 and 1, we get:

3.26 Below we show the computation for each cell:

Males Females

0-1 3.264433 1.853077

2-3 0.822572 0.466939

4-5 2.59585 1.473552

over 5 0.531475 0.301696

The chi-square test statistic is computed by summing these values giving x2 =

11.309

The critical value for the contingency analysis test with (2 - 1) × (4 - 1) = 3 degrees of freedom and alpha equal .05 is found in the chi-square table to be 7.8147. The decision rule is: If x2 > 7.8147, reject the null hypothesis Otherwise, do not reject. Since x2 = 11.309 > 7.8147, reject the null hypothesis and conclude that use rate is related to gender of the customer. Diff: 3 Keywords: null, alternative, hypothesis, contingency table, crosstabulation Section: 13-2 Introduction to Contingency Analysis Outcome: 2

Business Statistics, 9e (Groebner/Shannon/Fry) Chapter 14 Introduction to Linear Regression and Correlation Analysis 1) The scatter plot is a two dimensional graph that is used to graphically represent the relationship between two variables. Answer: TRUE Diff: 1 Keywords: scatter plot, relationship Section: 14-1 Correlation Outcome: 1 2) The difference between a scatter plot and a scatter diagram is that the scatter plot has the independent variable on the x-axis while the independent variable is on the Y-axis in a scatter diagram. Answer: FALSE Diff: 1 Keywords: scatter plot, scatter diagram, independent Section: 14-1 Correlation Outcome: 1 3) A dependent variable is the variable that we wish to predict or explain in a regression model. Answer: TRUE Diff: 1 Keywords: dependent, predict, regression Section: 14-1 Correlation Outcome: 1 4) When constructing a scatter plot, the dependent variable is placed on the vertical axis and the independent variable is placed on the horizontal axis. Answer: TRUE Diff: 1 Keywords: scatter plot, dependent, independent, axis Section: 14-1 Correlation Outcome: 1 5) Both a scatter plot and the correlation coefficient can distinguish between a curvilinear and a linear relationship. Answer: FALSE Diff: 1 Keywords: scatter plot, linear, relationship, curvilinear, correlation Section: 14-1 Correlation Outcome: 1 6) In developing a scatter plot, the decision maker has the option of connecting the points or not. Answer: FALSE Diff: 1 Keywords: scatter plot, points, connect Section: 14-1 Correlation Outcome: 1 14-1 Copyright © 2018 Pearson Education, Inc.

7) A scatter plot is useful for identifying a linear relationship between the independent and dependent variable, but it is not particularly useful if the relationship is curvilinear. Answer: FALSE Diff: 2 Keywords: scatter plot, relationship, independent, dependent Section: 14-1 Correlation Outcome: 1 8) If two variables are related in a positive linear manner, the scatter plot will show points on the x,y space that are generally moving from the lower left to the upper right. Answer: TRUE Diff: 2 Keywords: scatter plot, positive, linear Section: 14-1 Correlation Outcome: 1 9) When a correlation is found between a pair of variables, this always means that there is a direct cause and effect relationship between the variables. Answer: FALSE Diff: 2 Keywords: correlation, cause and effect relation Section: 14-1 Correlation Outcome: 1 10) A research study has stated that the income earned by working individuals is correlated at a .64 value with the years of experience of the individual. Given this, the points of the scatter plot would fall on an inclined straight line that has a slope equal to .64. Answer: FALSE Diff: 2 Keywords: scatter plot, correlated, slope Section: 14-1 Correlation Outcome: 1 11) A study was recently conducted by a Minor League Baseball team to determine whether there is a correlation between money spent at games on concessions and the game inning. In this study, the independent variable would be the game inning. Answer: TRUE Diff: 2 Keywords: correlation, dependent variable Section: 14-1 Correlation Outcome: 1 12) A correlation of -0.9 indicates a weak linear relationship between the variables. Answer: FALSE Diff: 2 Keywords: correlation, Pearson, relationship Section: 14-1 Correlation Outcome: 1 14-2 Copyright © 2018 Pearson Education, Inc.

13) Two variables have a correlation coefficient that is very close to zero. This means that there is no relationship between the two variables. Answer: FALSE Diff: 2 Keywords: correlation, zero, relationship Section: 14-1 Correlation Outcome: 1 14) A perfect correlation between two variables will always produce a correlation coefficient of +1.0 Answer: FALSE Diff: 2 Keywords: correlation, perfect Section: 14-1 Correlation Outcome: 1 15) If the correlation coefficient for two variables is computed to be a -0.70, the scatter plot will show the data to be downward sloping from left to right. Answer: TRUE Diff: 2 Keywords: correlation, scatter plot Section: 14-1 Correlation Outcome: 1 16) A correlation coefficient computed from a sample of data values selected from a population is called a statistic and is subject to sampling error. Answer: TRUE Diff: 2 Keywords: correlation, statistic, sampling error Section: 14-1 Correlation Outcome: 1 17) A bank is interested in determining whether its customers' checking balances are linearly related to their savings balances. A sample of n = 20 customers was selected and the correlation was calculated to be +0.40. If the bank is interested in testing to see whether there is a significant linear relationship between the two variables using a significance level of .05, the correct null and alternative hypotheses to test are: H0 : r = 0.0 Ha : r ≠ 0.0 Answer: FALSE Diff: 2 Keywords: correlation, null, alternative, hypothesis, parameter Section: 14-1 Correlation Outcome: 1

18) A cellular phone service provider believes that there is negative correlation between the minutes used by its customers and the age of the customer. To test this, the following would be the appropriate null and alternative hypotheses: H0 : ρ = 0.0 Ha : ρ ≠ 0.0 Answer: FALSE Diff: 2 Keywords: correlation, null, alternative, hypothesis Section: 14-1 Correlation Outcome: 1 19) In conducting a hypothesis test for a correlation, the correct probability distribution to use is the F distribution. Answer: FALSE Diff: 2 Keywords: correlation, hypothesis Section: 14-1 Correlation Outcome: 1 20) A bank is interested in determining whether its customers' checking balances are linearly related to their savings balances. A sample of n = 20 customers was selected and the correlation was calculated to be +0.40. If the bank is interested in testing to see whether there is a significant linear relationship between the two variables using a significance level of 0.05, the value of the test statistic is approximately t = 1.8516. Answer: TRUE Diff: 2 Keywords: correlation, test statistic Section: 14-1 Correlation Outcome: 2

21) You are given the following sample data for two variables: Y 10 8 12 15 16 10 10 8 12

X 100 110 90 200 150 100 80 90 150

The sample correlation coefficient for these data is approximately r = 0.755. Answer: FALSE Diff: 3 Keywords: correlation, coefficient Section: 14-1 Correlation Outcome: 2 22) You are given the following sample data for two variables: Y 10 8 12 15 16 10 10 8 12

X 100 110 90 200 150 100 80 90 150

Based upon these sample data, and testing at the 0.05 level of significance, the critical value for testing whether the population correlation coefficient is equal to zero is t = 2.2622. Answer: FALSE Diff: 2 Keywords: correlation, critical value Section: 14-1 Correlation Outcome: 2 23) Given a sample of size n = 15 and a sample correlation of r = 0.7, the value of the test statistic for conducting a hypothesis test of the correlation is t = 3.53. Answer: TRUE Diff: 2 Keywords: correlation, test statistic Section: 14-1 Correlation Outcome: 2 14-5 Copyright © 2018 Pearson Education, Inc.

24) State University recently randomly sampled ten students and analyzed grade point average (GPA) and number of hours worked off-campus per week. The following data were observed: GPA 3.14 2.75 3.68 3.22 2.45 2.80 3.00 2.23 3.14 2.90

HOURS 25 30 11 18 22 40 15 29 10 0

In this study the independent variable is the number of hours worked off campus per week. Answer: TRUE Diff: 1 Keywords: independent, variable Section: 14-1 Correlation Outcome: 2 25) A university study indicated that a correlation of -0.7 was found between the number of hours spent playing video games and a student’s GPA. This means that the more time spent playing video games, the lower the overall GPA. Answer: TRUE Diff: 2 Keywords: correlation, negative Section: 14-1 Correlation Outcome: 1

26) State University recently randomly sampled ten students and analyzed grade point average (GPA) and number of hours worked off-campus per week. The following data were observed: GPA 3.14 2.75 3.68 3.22 2.45 2.80 3.00 2.23 3.14 2.90

HOURS 25 30 11 18 22 40 15 29 10 0

If the university wished to test the claim that the correlation between hours worked and GPA is negative, the following null and alternative hypotheses would be appropriate: H0 : ρ < 0.0 Ha : ρ ≥ 0.0 Answer: FALSE Diff: 2 Keywords: correlation, hypothesis, null, alternative Section: 14-1 Correlation Outcome: 1 27) State University recently randomly sampled ten students and analyzed grade point average (GPA) and number of hours worked off-campus per week. The following data were observed: GPA 3.14 2.75 3.68 3.22 2.45 2.80 3.00 2.23 3.14 2.90

HOURS 25 30 11 18 22 40 15 29 10 0

The correlation between these two variables is approximately -.461 Answer: TRUE Diff: 3 Keywords: correlation coefficient Section: 14-1 Correlation Outcome: 2

28) State University recently randomly sampled ten students and analyzed grade point average (GPA) and number of hours worked off-campus per week. The following data were observed: GPA 3.14 2.75 3.68 3.22 2.45 2.80 3.00 2.23 3.14 2.90

HOURS 25 30 11 18 22 40 15 29 10 0

The test statistic for testing whether the two variables are significantly correlated is approximately z = 1.56. Answer: FALSE Diff: 2 Keywords: correlation, test statistic Section: 14-1 Correlation Outcome: 2 29) Given a sample correlation r = -0.5 and a sample size of n = 30, the test statistic for testing whether the two variables are significantly correlated is approximately t = -3.055. Answer: TRUE Diff: 2 Keywords: correlation, test statistic Section: 14-1 Correlation Outcome: 2 30) If two variables are highly correlated, it not only means that they are linearly related, it also means that a change in one variable will cause a change in the other variable. Answer: FALSE Diff: 2 Keywords: correlation, cause, causation Section: 14-1 Correlation Outcome: 2 31) If two variables are spuriously correlated, it means that the correlation coefficient between them is near zero. Answer: FALSE Diff: 1 Keywords: correlation, spurious Section: 14-1 Correlation Outcome: 2

32) The fact that teachers' salaries in Wisconsin are correlated with egg prices in Texas means that the two variables are spuriously correlated since a change in one variable could not cause the change in the other. Answer: TRUE Diff: 2 Keywords: correlation, spurious Section: 14-1 Correlation Outcome: 2 33) In conducting a hypothesis test on the correlation between a pair of variables, we assume that each variable is normally and independently distributed. Answer: FALSE Diff: 2 Keywords: correlation, bivariate normal assumption Section: 14-1 Correlation Outcome: 2 34) If two variables are uncorrelated, the sample correlation coefficient will be r = 0.00. Answer: FALSE Diff: 2 Keywords: correlation, sample Section: 14-1 Correlation Outcome: 2 35) When the correlation coefficient for the two variables was -0.23, it implies that the two variables are not correlated because the correlation coefficient cannot be negative. Answer: FALSE Diff: 2 Keywords: correlation, sample Section: 14-1 Correlation Outcome: 2 36) When a pair of variables has a positive correlation, the slope in the regression equation will always be positive. Answer: TRUE Diff: 1 Keywords: regression, independent, dependent, slope, correlation Section: 14-2 Simple Linear Regression Analysis Outcome: 3 37) In a simple regression model, the slope coefficient represents the average change in the independent variable for a one-unit change in the dependent variable. Answer: FALSE Diff: 2 Keywords: regression, slope, dependent, independent Section: 14-2 Simple Linear Regression Analysis Outcome: 3

38) In developing a simple linear regression model it is assumed that the distribution of error terms will be normally distributed for all levels of x. Answer: TRUE Diff: 2 Keywords: regression, error, normal, distribution Section: 14-2 Simple Linear Regression Analysis Outcome: 3 39) If a set of data contains no values of x that are equal to zero, then the regression coefficient, b0, has no particular meaning. Answer: TRUE Diff: 2 Keywords: regression, coefficient, intercept Section: 14-2 Simple Linear Regression Analysis Outcome: 3 40) If the correlation between two variables is known to be statistically significant at the 0.05 level, then the regression slope coefficient will also be significant at the 0.05 level. Answer: TRUE Diff: 2 Keywords: regression, slope, correlation, significant Section: 14-2 Simple Linear Regression Analysis Outcome: 3 41) The sign on the intercept coefficient in a simple regression model will always be the same as the sign on the correlation coefficient. Answer: FALSE Diff: 2 Keywords: regression, intercept, correlation Section: 14-2 Simple Linear Regression Analysis Outcome: 3 42) The following regression model has been computed based on a sample of twenty observations: = 34.2 + 19.3x. Given this model, the predicted value for y when x = 40 is 806.2. Answer: TRUE Diff: 1 Keywords: regression, predicted value Section: 14-2 Simple Linear Regression Analysis Outcome: 3 43) Given a regression equation of = 16 + 2.3x we would expect that an increase in x of 2.0 would lead to an average increase of y of 4.6. Answer: TRUE Diff: 2 Keywords: regression, slope, predicted value Section: 14-2 Simple Linear Regression Analysis Outcome: 3 14-10 Copyright © 2018 Pearson Education, Inc.

44) The following regression model has been computed based on a sample of fifty observations of parttime workers in which the individual weekly income was the dependent variable and hours worked is the independent variable : = 43.2 + 25.2x. The first observations in the sample for y and x were 350 and 15, respectively. Given this, the residual value for the first observation is approximately 71.2. Answer: TRUE Diff: 2 Keywords: regression, residual Section: 14-2 Simple Linear Regression Analysis Outcome: 3 45) In a study of 30 customers' utility bills in which the monthly bill was the dependent variable and the number of square feet in the house is the independent variable, the resulting regression model is = 23.40 + 0.4x. Given this, the sample correlation coefficient is known to be positive. Answer: TRUE Diff: 2 Keywords: regression, slope, correlation Section: 14-2 Simple Linear Regression Analysis Outcome: 3 46) In a study of 30 customers' utility bills in which the monthly bill was the dependent variable and the number of square feet in the house is the independent variable, the resulting regression model is = 23.40 + 0.04x. Given this model, for a customer with a 2,000 square foot house and a monthly utility bill equal to $100.00, the residual from the regression model is approximately -$3.40. Answer: TRUE Diff: 2 Keywords: regression, residual Section: 14-2 Simple Linear Regression Analysis Outcome: 3 47) In a study of 30 customers' utility bills in which the monthly bill was the dependent variable and the number of square feet in the house is the independent variable, the resulting regression model is = 23.40 + 0.4x. Based on this model, the expected utility bill for a customer with a home with 2,300 square feet is approximately $92.00. Answer: FALSE Diff: 2 Keywords: regression, expected, predicted Section: 14-2 Simple Linear Regression Analysis Outcome: 3 48) Given a sample of data for use in simple linear regression, the values for the slope and the intercept are chosen to minimize the sum of squared errors. Answer: TRUE Diff: 2 Keywords: regression, equation, slope, intercept Section: 14-2 Simple Linear Regression Analysis Outcome: 3

49) You are given the following sample data for two variables: Y 10 8 12 15 16 10 10 8 12

X 100 110 90 200 150 100 80 90 150

The regression model based on these sample data explains approximately 75 percent of the variation in the dependent variable. Answer: FALSE Diff: 3 Keywords: regression, variation, correlation Section: 14-2 Simple Linear Regression Analysis Outcome: 3 50) If it is known that a simple linear regression model explains 56 percent of the variation in the dependent variable and that the slope on the regression equation is negative, then we also know that the correlation between x and y is approximately -0.75. Answer: TRUE Diff: 2 Keywords: regression, correlation, slope Section: 14-2 Simple Linear Regression Analysis Outcome: 3 51) The sum of the residuals in a least squares regression model will be zero only when the correlation between the x and y variables is statistically significant. Answer: FALSE Diff: 2 Keywords: regression, residuals, correlation Section: 14-2 Simple Linear Regression Analysis Outcome: 3 52) If the sample value of the intercept turns out to be an illogical value, this is acceptable as long as x = 0 is not within the range of the data. Answer: TRUE Diff: 2 Keywords: regression, intercept Section: 14-2 Simple Linear Regression Analysis Outcome: 3

53) A study was recently done in which the following regression output was generated using Excel. SUMMARY OUTPUT

Given this, we know that approximately 57 percent of the variation in the y variable is explained by the x variable. Answer: TRUE Diff: 2 Keywords: regression analysis, R-square, coefficient of determination Section: 14-2 Simple Linear Regression Analysis Outcome: 3

54) A study was recently done in which the following regression output was generated using Excel. SUMMARY OUTPUT

Given this output, we would reject the null hypothesis that the population regression slope coefficient is equal to zero at the alpha = 0.05 level. Answer: TRUE Diff: 2 Keywords: regression, slope, null, hypothesis Section: 14-2 Simple Linear Regression Analysis Outcome: 3 55) The standard error of the estimate for a simple linear regression model measures the variation in the slope coefficient from sample to sample. Answer: FALSE Diff: 2 Keywords: regression, standard error, slope Section: 14-2 Simple Linear Regression Analysis Outcome: 3 56) If the correlation of x and y is -0.65, then coefficient of determination is -0.4225. Answer: FALSE Diff: 2 Keywords: regression, variation, correlation, coefficient of determination Section: 14-2 Simple Linear Regression Analysis Outcome: 3

57) If the correlation between the dependent variable and the independent variable is negative, the standard error of the regression slope coefficient in a simple linear regression model will also be negative. Answer: FALSE Diff: 2 Keywords: regression, correlation, standard error, independent, dependent Section: 14-2 Simple Linear Regression Analysis Outcome: 3 58) In a simple regression model, if the regression model is deemed to be statistically significant, it means that the regression slope coefficient is significantly greater than zero. Answer: FALSE Diff: 2 Keywords: regression, slope, significant Section: 14-2 Simple Linear Regression Analysis Outcome: 4 59) Assume that we have found a regression equation of = 3.6 - 2.4x, and that the coefficient of determination is 0.72, then the correlation of x and y must be about 0.849. Answer: FALSE Diff: 2 Keywords: regression, equation, correlation, coefficient of determination Section: 14-2 Simple Linear Regression Analysis Outcome: 3 60) State University recently randomly sampled seven students and analyzed grade point average (GPA) and number of hours worked off-campus per week. The following data were observed: GPA 3.14 2.75 3.68 3.22 2.45 2.80 3.00 2.23 3.14 2.90

HOURS 25 30 11 18 22 40 15 29 10 0

A regression model with HOURS as the independent variable has an R-square equal to approximately .46. Answer: FALSE Diff: 3 Keywords: regression, R-square, correlation Section: 14-2 Simple Linear Regression Analysis Outcome: 3

61) State University recently randomly sampled seven students and analyzed grade point average (GPA) and number of hours worked off-campus per week. The following data were observed: GPA 3.14 2.75 3.68 3.22 2.45 2.80 3.00 2.23 3.14 2.90

HOURS 25 30 11 18 22 40 15 29 10 0

In testing the significance of the regression slope coefficient for the independent variable, HOURS, the calculated test statistic is approximately t = -1.47. Answer: TRUE Diff: 3 Keywords: regression, slope, test statistic Section: 14-2 Simple Linear Regression Analysis Outcome: 4 62) If a simple least squares regression model is developed based on a sample where the two variables are known to be positively correlated, the sign on the regression coefficient will be positive also. Answer: TRUE Diff: 1 Keywords: regression, regression, slope, correlation Section: 14-2 Simple Linear Regression Analysis Outcome: 4 63) If a simple least squares regression model is developed based on a sample where the two variables are known to be positively correlated, the sum of the residuals will be positive. Answer: FALSE Diff: 1 Keywords: regression, correlation, residuals Section: 14-2 Simple Linear Regression Analysis Outcome: 4 64) The values of the regression coefficients are found such the sum of the residuals is minimized. Answer: FALSE Diff: 1 Keywords: regression, residuals Section: 14-2 Simple Linear Regression Analysis Outcome: 4

65) If the correlation between two variables in a simple linear regression model is calculated to be .73, the R-square value is .85. Answer: FALSE Diff: 2 Keywords: regression, R-square, correlation Section: 14-2 Simple Linear Regression Analysis Outcome: 4 66) In a simple linear regression analysis, if the test statistic for testing the significance of the regression slope coefficient is 3.6, the F ratio from the analysis of variance table is known to be 12.96 Answer: TRUE Diff: 2 Keywords: regression, test statistic, ratio Section: 14-2 Simple Linear Regression Analysis Outcome: 4 67) In simple linear regression, the t-test for the slope and the F-test are both conducting the same hypothesis test. Answer: TRUE Diff: 2 Keywords: regression, slope, t-test, F-test Section: 14-2 Simple Linear Regression Analysis Outcome: 4 68) A positive population slope of 12 (β1 = 12) means that a 1-unit increase in x causes an average 12-unit increase in y. Answer: FALSE Diff: 3 Keywords: regression, slope, correlation Section: 14-2 Simple Linear Regression Analysis Outcome: 4

69) A study was recently performed by the Internal Revenue Service to determine how much tip income waiters and waitresses should make based on the size of the bill at each table. A random sample of bills and resulting tips were collected and the following regression results were observed: SUMMARY OUTPUT

Given this output, the point estimate for the average tip per dollar amount of the bill is approximately $0.21. Answer: TRUE Diff: 2 Keywords: regression, slope Section: 14-3 Uses for Regression Analysis Outcome: 6

70) A study was recently performed by the Internal Revenue Service to determine how much tip income waiters and waitresses should make based on the size of the bill at each table. A random sample of bills and resulting tips were collected and the following regression results were observed: SUMMARY OUTPUT

Given this output, the upper limit for the 95 percent confidence interval estimate for the true regression slope coefficient is approximately 0.28. Answer: TRUE Diff: 3 Keywords: regression, slope, confidence interval Section: 14-3 Uses for Regression Analysis Outcome: 6 71) When regression analysis is used for descriptive purposes, two of the main items of interest are whether the sign on the regression slope coefficient is positive or negative and whether the regression slope coefficient is significantly different from zero. Answer: TRUE Diff: 2 Keywords: regression, slope Section: 14-3 Uses for Regression Analysis Outcome: 6 72) When calculating prediction intervals for predicted values of y based on a given x, all 95 percent prediction intervals will be of equal width. Answer: FALSE Diff: 2 Keywords: regression, prediction interval Section: 14-3 Uses for Regression Analysis Outcome: 5

73) If the R-squared value for a regression model is high, the regression model will necessarily provide accurate forecasts of the y variable. Answer: FALSE Diff: 1 Keywords: regression, R-squared Section: 14-3 Uses for Regression Analysis Outcome: 6 74) When the intercept in a regression equation is deemed not significantly different from 0, then in making predictions for y, 0.0 should be used as the value of the intercept rather than the estimated intercept value. Answer: FALSE Diff: 2 Keywords: regression, intercept, prediction Section: 14-3 Uses for Regression Analysis Outcome: 5 75) The prediction interval developed from a simple linear regression model will be at its narrowest point when the value of x used to predict y is equal to the mean value of x. Answer: TRUE Diff: 2 Keywords: regression, prediction interval, mean Section: 14-3 Uses for Regression Analysis Outcome: 5

76) A manufacturing company is interested in predicting the number of defects that will be produced each hour on the assembly line. The managers believe that there is a relationship between the defect rate and the production rate per hour. The managers believe that they can use production rate to predict the number of defects. The following data were collected for 10 randomly selected hours. Defects 20 30 10 20 30 25 30 20 10 40

Production Rate Per Hour 400 450 350 375 400 400 450 300 300 300

Given these sample data, the simple linear regression model for predicting the number of defects is approximately = 5.67 + 0.048x. Answer: TRUE Diff: 3 Keywords: regression, model Section: 14-3 Uses for Regression Analysis Outcome: 6 77) Given the following regression equation, the predicted value for y when x = 0.5 is about 4.57

78) A regression model that is deemed to have a regression slope coefficient that could be equal to zero should not be used for prediction since there is no established linear relationship between the x and y variable. Answer: TRUE Diff: 2 Keywords: regression, slope, hypothesis Section: 14-3 Uses for Regression Analysis Outcome: 5 79) A high coefficient of determination (R2) implies that the regression model will be a good predictor for future values of the dependent variable given the value of the independent variable. Answer: FALSE Diff: 2 Keywords: regression, prediction, R-squared Section: 14-3 Uses for Regression Analysis Outcome: 5 80) In analyzing the relationship between two variables, a scatter plot cannot be used to detect which of the following? A) A positive linear relationship B) A curvilinear relationship C) A negative linear relationship D) A causal relationship Answer: D Diff: 2 Keywords: scatter plot, relationship Section: 14-1 Correlation Outcome: 1 81) If the population correlation between two variables is determined to be -0.70, which of the following is known to be true? A) There is a positive linear relationship between the two variables. B) There is a fairly strong negative linear relationship between the two variables. C) An increase in one of the variables will cause the other variable to decline by 70 percent. D) The scatter diagram for the two variables will be upward sloping from left to right. Answer: B Diff: 2 Keywords: correlation, negative, relationship Section: 14-1 Correlation Outcome: 1

82) If a pair of variables have a strong curvilinear relationship, which of the following is true? A) The correlation coefficient will be able to indicate that curvature is present. B) A scatter plot will not be needed to indicate that curvature is present. C) The correlation coefficient will not be able to indicate the relationship is curved. D) The correlation coefficient will be equal to zero. Answer: C Diff: 2 Keywords: correlation, curvilinear relationship Section: 14-1 Correlation Outcome: 1 83) If a sample of n = 30 people is selected and the sample correlation between two variables is r = 0.468, what is the test statistic value for testing whether the true population correlation coefficient is equal to zero? A) About t = 2.80 B) About t = -.3.01 C) t = 2.0484 D) Can't be determined without knowing the level of significance for the test. Answer: A Diff: 2 Keywords: sample, correlation, test statistic Section: 14-1 Correlation Outcome: 1 84) Which of the following statements is correct? A) A scatter plot showing two variables with a positive linear relationship will have all points on a straight line. B) The stronger the linear relationship between two variables, the closer the correlation coefficient will be to 1.0. C) Two variables that are uncorrelated with one another may still be related in a nonlinear manner. D) All of the above are correct. Answer: C Diff: 2 Keywords: correlation, related, nonlinear Section: 14-1 Correlation Outcome: 1

85) Recently, an automobile insurance company performed a study of a random sample of 15 of its customers to determine if there is a positive relationship between the number of miles driven and the age of the driver. The sample correlation coefficient is r = .38. Given this information, which of the following is appropriate critical value for testing the null hypothesis at an alpha = .05 level? A) t = 2.6104 B) t = 1.7613 C) t = 1.7531 D) t = 1.7709 Answer: D Diff: 2 Keywords: correlation, critical value, hypothesis Section: 14-1 Correlation Outcome: 1 86) Recently, an automobile insurance company performed a study of a random sample of 15 of its customers to determine if there is a positive relationship between the number of miles driven and the age of the driver. The sample correlation coefficient is r = .38. Given this information, and assuming that the test is to be performed at the .05 level of significance, which of the following is the correct test statistic? A) t = 1.4812 B) t = 1.7709 C) z = 2.114 D) t = 1.74 Answer: A Diff: 2 Keywords: correlation, test statistic Section: 14-1 Correlation Outcome: 1 87) Assume that a medical research study found a correlation of -0.73 between consumption of vitamin A and the cancer rate of a particular type of cancer. This could be interpreted to mean: A) the more vitamin A consumed, the lower a person's chances are of getting this type of cancer. B) the less vitamin A consumed, the lower a person's chances are of getting this type of cancer. C) the more vitamin A consumed, the higher a person's chances are of getting this type of cancer. D) vitamin A causes this type of cancer. Answer: A Diff: 2 Keywords: correlation, Section: 14-1 Correlation Outcome: 1

88) The term that is given when two variables are correlated but there is no apparent connection between them is: A) spontaneous correlation. B) random correlation. C) spurious correlation. D) linear correlation. Answer: C Diff: 1 Keywords: correlation, spurious Section: 14-1 Correlation Outcome: 1 89) A recent study of 15 shoppers showed that the correlation between the time spent in the store and the dollars spent was 0.235. Using a significance level equal to 0.05, which of the following is the appropriate null hypothesis to test whether the population correlation is zero? A) H0 : r = 0.0 B) H0 : ρ = 0.0 C) H0 : μ = 0.0 D) H0 : ρ ≠ 0 Answer: B Diff: 2 Keywords: correlation, hypothesis, null Section: 14-1 Correlation Outcome: 1 90) A recent study of 15 shoppers showed that the correlation between the time spent in the store and the dollars spent was 0.235. Using a significance level equal to 0.05, which of the following is the test statistic for testing whether the true population correlation is equal to zero? A) t = 0.245 B) t = 1.76 C) t = 2.1604 D) t = 0.872 Answer: D Diff: 2 Keywords: correlation, test statistic Section: 14-1 Correlation Outcome: 2

91) A recent study of 15 shoppers showed that the correlation between the time spent in the store and the dollars spent was 0.235. Using a significance level equal to 0.05, which of the following is true? A) The null hypothesis that the population mean is equal to zero should be rejected and we should conclude that the true correlation is not equal to zero. B) Based on the sample data there is not enough evidence to conclude that the true correlation is different from zero. C) The sample correlation coefficient could be zero since the test statistic does not fall in the rejection region. D) The null hypothesis should be rejected because the test statistic exceeds the critical value. Answer: B Diff: 2 Keywords: correlation, test statistic, critical value Section: 14-1 Correlation Outcome: 2 92) A study was done in which the high daily temperature and the number of traffic accidents within the city were recorded. These sample data are shown as follows: High Temperature 91 56 75 68 50 39 98

Traffic Accidents 7 4 9 11 3 5 8

Given this data the sample correlation is: A) -0.57 B) 0.64 C) 1.54 D) 0.57 Answer: D Diff: 3 Keywords: correlation, sample Section: 14-1 Correlation Outcome: 2

93) Which of the following is NOT an assumption for the simple linear regression model? A) The individual error terms are statistically independent. B) The distribution of the error terms will be skewed left or right depending on the shape of the dependent variable. C) The error terms have equal variances for all values of the independent variable. D) The mean of the dependent variable value for all levels of x can be connected by a straight line. Answer: B Diff: 2 Keywords: regression, assumption, skewed Section: 14-2 Simple Linear Regression Analysis Outcome: 3 94) Which of the following statements is true with respect to a simple linear regression model? A) The percent of variation in the dependent variable that is explained by the regression model is equal to the square of the correlation coefficient between the x and y variables. B) If the correlation coefficient between the x and y variables is negative, the sign on the regression slope coefficient will also be negative. C) If the correlation between the dependent and the independent variable is determined to be significant, the regression model for y given x will also be significant. D) All of the above are true. Answer: D Diff: 2 Keywords: regression, correlation, variation, slope Section: 14-2 Simple Linear Regression Analysis Outcome: 3

95) Use the following regression results to answer the question below.

Which of the following is true? A) The correlation between x and y must be approximately 0.8851. B) The correlation between x and y must be approximately -0.8851. C) The correlation between x and y must be approximately 0.7835. D) The correlation between x and y must be approximately -0.7835. Answer: B Diff: 2 Keywords: regression, R-squared, correlation, slope Section: 14-2 Simple Linear Regression Analysis Outcome: 3

96) Use the following regression results to answer the question below.

Which of the following is true? A) x explains about 88.5 percent of the variation in y. B) y explains about 88.5 percent of the variation in x. C) x explains about 78.4 percent of the variation in y. D) y explains about 78.4 percent of the variation in x. Answer: C Diff: 2 Keywords: regression, coefficient of determination Section: 14-2 Simple Linear Regression Analysis Outcome: 3

97) Use the following regression results to answer the question below.

How many observations were involved in this regression? A) 7 B) 8 C) 9 D) 10 Answer: B Diff: 1 Keywords: regression, sample size Section: 14-2 Simple Linear Regression Analysis Outcome: 3

98) Use the following regression results to answer the question below.

In conducting a hypothesis test of the slope using a 0.05 level of significance, which of the following is correct? A) The slope differs significantly from 0 because p-value = 0.205 is greater than 0.05 B) The slope does not differ significantly from 0 because p-value = 0.205 is greater than 0.05 C) The slope differs significantly from 0 because p-value = 0.003 is less than 0.05 D) The slope does not differ significantly from 0 because p-value = 0.003 is less than 0.05 Answer: C Diff: 2 Keywords: regression, hypothesis test of slope, p-value Section: 14-2 Simple Linear Regression Analysis Outcome: 3

99) The following regression output is available. Notice that some of the values are missing.

Given this information, what percent of the variation in the y variable is explained by the independent variable? A) About 75 percent B) Approximately 57 percent C) Can't be determined without having the actual data available. D) About 25 percent Answer: B Diff: 2 Keywords: regression, R-squared, variation Section: 14-2 Simple Linear Regression Analysis Outcome: 3

100) The following regression output is available. Notice that some of the values are missing.

Given this information, what was the sample size used in the study? A) 8 B) 18 C) 9 D) 16 Answer: C Diff: 1 Keywords: regression, degrees of freedom, sample size Section: 14-2 Simple Linear Regression Analysis Outcome: 3

101) The following regression output is available. Notice that some of the values are missing.

Given this information, what is the test statistic for testing whether the regression slope coefficient is equal to zero? A) Approximately t = 3.04 B) About t = 2.19 C) About t = 9.24 D) About t = 2.39 Answer: A Diff: 2 Keywords: regression, standard error, coefficient, test statistic Section: 14-2 Simple Linear Regression Analysis Outcome: 3

102) The following regression output is available. Notice that some of the values are missing.

Given this information, what is the standard error of the estimate for the regression model? A) About 36.18 B) Approximately 6.02 C) About 1.98 D) 3.91 Answer: C Diff: 3 Keywords: regression, standard error Section: 14-2 Simple Linear Regression Analysis Outcome: 3 103) Which of the following statements is true with respect to a simple linear regression model? A) The regression slope coefficient is the square of the correlation coefficient. B) The percentage of variation in the dependent variable that is explained by the independent variable can be determined by squaring the correlation coefficient. C) It is possible that the correlation between a y and x variable might be statistically significant, but the regression slope coefficient could be determined to be zero since they measure different things. D) The standard error of the estimate is equal to the standard error of the slope. Answer: B Diff: 2 Keywords: regression, R-squared, correlation Section: 14-2 Simple Linear Regression Analysis Outcome: 3

104) In a regression analysis situation, the standard error of the slope is: A) a measure of the variation in the regression slope from sample to sample. B) equal to the square root of the standard error of the estimate. C) a measure of the amount of change in y that will occur for a one-unit change in x. D) All of the above Answer: A Diff: 2 Keywords: regression, standard error, slope Section: 14-2 Simple Linear Regression Analysis Outcome: 3 105) A recent study by a major financial investment company was interested in determining whether the annual percentage change in stock price for companies is linearly related to the annual percent change in profits for the company. The following data was determined for 7 randomly selected companies: % Change Stock Price 8.4 9.5 13.6 -3.2 7 18.4 -2.1

% Change in Profit 4.2 5.6 11.2 4.5 12.2 12 -13.4

Based upon this sample information, what portion of variation in stock price percentage change is explained by the percent change in yearly profit? A) Approximately 70 percent B) Nearly 19 percent C) About 49 percent D) None of the above Answer: C Diff: 3 Keywords: regression, R-squared, variation Section: 14-2 Simple Linear Regression Analysis Outcome: 3

106) A recent study by a major financial investment company was interested in determining whether the annual percentage change in stock price for companies is linearly related to the annual percent change in profits for the company. The following data was determined for 7 randomly selected companies: % Change Stock Price 8.4 9.5 13.6 -3.2 7 18.4 -2.1

% Change in Profit 4.2 5.6 11.2 4.5 12.2 12 -13.4

Based upon this sample information, which of the following is the regression equation? A)

= 4.19 + .61x

= 15.04 + 4.25x

= 1.19 - 3.00x

D) = 20.19 + .005x Answer: A Diff: 3 Keywords: regression, equation, slope, intercept Section: 14-2 Simple Linear Regression Analysis Outcome: 3 107) Assuming that a regression has been conducted for a group of small companies where x = the number of employees at the company, y = annual revenue of the company (recorded in thousands of dollars), and the largest company included in the study had 82 employees. The resulting regression equation is = 59.2 + 83.4x. Which of the following is true? A) For each additional employee, revenue on average will increase by $83.4 B) A company with 2100 employees could be predicted to have average revenue of about $175 million. C) For each additional employee, revenue on average will increase by $59.2 thousand. D) This model should not be used to make predictions for companies with more than 82 employees. Answer: D Diff: 2 Keywords: regression, range of data, slope Section: 14-2 Simple Linear Regression Analysis Outcome: 3

108) Consider the following partially completed computer printout for a regression analysis where the dependent variable is the price of a personal computer and the independent variable is the size of the hard drive.

Based on the information provided, what percentage of the variation in the price of the personal computers is accounted for by the regression model using hard drive capacity as the independent variable? A) About 82 percent B) About 67 percent C) 217.75 D) About 66 percent Answer: B Diff: 2 Keywords: regression, R-squared Section: 14-2 Simple Linear Regression Analysis Outcome: 3

109) Consider the following partially completed computer printout for a regression analysis where the dependent variable is the price of a personal computer and the independent variable is the size of the hard drive.

Based on the information provided, what is the estimate for the standard error of the estimate for the regression model? A) Approximately 690.50 B) About 4,026 C) Just under 376.23 D) 476,800 Answer: A Diff: 3 Keywords: regression, standard error Section: 14-2 Simple Linear Regression Analysis Outcome: 3

110) Consider the following partially completed computer printout for a regression analysis where the dependent variable is the price of a personal computer and the independent variable is the size of the hard drive.

Based on the information provided, what is the F statistic? A) About 8 .33 B) Just over 2.35 C) About 4.76 D) About 69.5 Answer: D Diff: 2 Keywords: regression, F-statistic, F-test Section: 14-2 Simple Linear Regression Analysis Outcome: 3

111) Consider the following partially completed computer printout for a regression analysis where the dependent variable is the price of a personal computer and the independent variable is the size of the hard drive.

Based on the information provided, which of the following statements is true if alpha = .05? A) The slope is not significantly different from 0 because p-value = 0.84 is greater than 0.05 B) The slope is significantly different from 0 because p-value = 9.95 is greater than 0.05 C) The slope is not significantly different from 0 because p-value = 9.95 is greater than 0.05 D) The slope is significantly different from 0 because p-value = 9.95En - 10 is less than 0.05 Answer: D Diff: 2 Keywords: regression, correlation, slope, variation Section: 14-2 Simple Linear Regression Analysis Outcome: 3 112) Which of the following statements is true in simple linear regression? A) The standard error of the estimate is equal to the standard error of the slope. B) The total degrees of freedom are (n-2). C) The coefficient of determination is equal to the correlation of x and y. D) The p-value of the F test will equal the p-value of the t-test of the slope. Answer: D Diff: 2 Keywords: regression, slope, correlation Section: 14-2 Simple Linear Regression Analysis Outcome: 3

113) A health insurance company conducted a linear regression analysis between the number of claims per customer (Y) and customer age (X) that resulted in the following equation analyzing customer risk: = 2 + 3X The above equation implies that an A) increase of one year in age is correlated with an increase of 5 claims. B) increase of one year in age is correlated with an increase of 1 claim. C) increase one year in age is correlated with an increase of 50. D) increase of one year in age is correlated with a decrease of 5 claims. Answer: A Diff: 2 Keywords: regression, slope, correlation Section: 14-2 Simple Linear Regression Analysis Outcome: 3 114) Given the data below, one ran the simple regression analysis of Y on X. Y 4 3 4 6 8

X 2 1 4 3 5

The relationship between Y and X is A) significant at the alpha = 1 percent level. B) significant at the alpha = 5 percent level. C) significant at the alpha = 10 percent level. D) not significant at the alpha = 10 percent level. Answer: D Diff: 3 Keywords: regression, slope, correlation Section: 14-2 Simple Linear Regression Analysis Outcome: 3 115) When using regression analysis for descriptive purposes, which of the following is of importance? A) The size of the regression slope coefficient B) The sign of the regression slope coefficient C) The standard error of the regression slope coefficient D) All of the above Answer: D Diff: 2 Keywords: regression, slope, slope coefficient, standard error Section: 14-3 Uses for Regression Analysis Outcome: 6

116) The National Football League has performed a study in which the total yards gained by teams in games was used as an independent variable to explain the variation in total points scored by teams during games. The points scored ranged from 0 to 57 and the yards gained ranged from 187 to 569. The following regression model was determined: = 12.3 + .12x Given this model, which of the following statements is true? A) The average points scored for teams who gain zero yards during a game is -12.3 points. B) The average yards gained will increase by .12 for every additional point scored. C) The average change in points scored for each increase of one yard will be 0.12 D) The average number of points scored per game is 12.3 Answer: C Diff: 2 Keywords: regression, slope Section: 14-3 Uses for Regression Analysis Outcome: 6

117) The following regression output was generated based on a sample of utility customers. The dependent variable was the dollar amount of the monthly bill and the independent variable was the size of the house in square feet.

Based on this regression output, which of the following statements is not true? A) The number of square feet in the house explains only about 2 percent of the variation in the monthly power bill. B) At an alpha level equal to 0.05, there is no basis for rejecting the hypothesis that the slope coefficient is equal to zero. C) The average increase in the monthly power bill is about 66.4 for each additional square foot of space in the house. D) The correlation of the monthly power bill with the square footage of the house is 0.149 Answer: C Diff: 2 Keywords: regression, slope, R-squared, variation, hypothesis Section: 14-3 Uses for Regression Analysis Outcome: 6

118) The following regression output was generated based on a sample of utility customers. The dependent variable was the dollar amount of the monthly bill and the independent variable was the size of the house in square feet.

Based on this regression output, what is the 95 percent confidence interval estimate for the population regression slope coefficient? A) Approximately -0.0003 ----- +0.0103 B) About -0.0082 ----- +0.0188 C) Approximately -32.76 ----- +32.79 D) None of the above Answer: B Diff: 2 Keywords: regression, confidence interval, slope coefficient Section: 14-3 Uses for Regression Analysis Outcome: 6 119) Which of the following statements is true? A) If the confidence interval estimate for the regression slope coefficient crosses over zero, the average change in y for a one-unit change in x may be zero. B) If the regression slope coefficient is very close to zero, this means that the relationship between the x and y variables is statistically insignificant. C) A statistically significant regression slope coefficient indicates that for a one-unit change in y there will be a positive change in x by an amount equal to the regression slope coefficient. D) It is acceptable to make predictions for y using x values that are outside the range of the data that was used in the regression. Answer: A Diff: 2 Keywords: regression, confidence interval Section: 14-3 Uses for Regression Analysis Outcome: 6 14-45 Copyright © 2018 Pearson Education, Inc.

120) It is believed that number of people who attend a Mardi Gras parade each year depends on the temperature that day. A regression has been conducted on a sample of years where the temperature ranged from 28 to 64 degrees and the number of people attending ranged from 8400 to 14,600. The regression equation was found to be = 2378 + 191x. Which of the following is true? A) The average change in parade attendance is an additional 2378 people per one-degree increase in temperature. B) The average change in parade attendance is an additional 191 people per one-degree increase in temperature. C) If the temperature is 75 degrees, we can expect that 16,703 people will attend. D) If the temperature is 0 degrees this year, then we should expect 2378 people to attend. Answer: B Diff: 2 Keywords: regression, predicted value, slope Section: 14-3 Uses for Regression Analysis Outcome: 5 121) Assume that you have calculated a prediction of = 110 where the specific value for x is equal to the average value of x. Also assume that n = 201 and that the standard error of the estimate is sε = 4.5. Find the approximate 95 percent prediction interval. A) About 101 ----- 119 B) About 109.4 ----- 110.6 C) About 105.5 ----- 104.5 D) About 98.4 ----- 121.6 Answer: A Diff: 2 Keywords: regression, prediction interval Section: 14-3 Uses for Regression Analysis Outcome: 5 122) Which of the following statements is true? A) The interval estimate for predicting a particular value of y given a specific x will be narrower than the interval estimate for the average value of y given a particular x. B) The higher the r-square value, the wider will be the prediction interval based on a simple linear regression model. C) The prediction interval generated from a simple linear regression model will be narrowest when the value of x used to generate the predicted y value is close to the mean value of x. D) The prediction interval generated from a simple linear regression model will be widest when the value of x used to generate the predicted y value is close to the mean value of x. Answer: C Diff: 2 Keywords: regression, prediction interval Section: 14-3 Uses for Regression Analysis Outcome: 5

123) A study was recently performed by the Internal Revenue Service to determine how much tip income waiters and waitresses should make based on the size of the bill at each table. A random sample of bills and resulting tips were collected. These data are shown as follows: Total Bill $126 $58 $86 $20 $59 $120 $14 $17 $26 $74

Tip $19 $11 $20 $3 $14 $30 $2 $4 $2 $16

Based upon these data, what is the approximate predicted value for tips if the total bill is $100? A) $15.55 B) $20.61 C) $26.03 D) $12.88 Answer: B Diff: 3 Keywords: regression, predicted Section: 14-3 Uses for Regression Analysis Outcome: 5 124) In analyzing the residuals to determine whether the simple regression analysis satisfies the regression assumptions, which of the following is true? A) The histogram of the residuals should be approximately bell-shaped. B) The scatter plot of the residuals against the dependent variable should illustrate that the variation in residuals is the same over all levels of . C) Neither A nor B are true. D) Both A and B are true. Answer: D Diff: 2 Keywords: regression, residuals, scatter plot Section: 14-3 Uses for Regression Analysis Outcome: 6

125) Residual analysis is conducted to check whether regression assumptions are met. Which of the following is not an assumption made in simple linear regression? A) Errors are independent of each other. B) Errors are normally distributed. C) Errors are linearly related to x. D) Errors have constant variance. Answer: C Diff: 2 Keywords: regression, residuals, assumptions Section: 14-3 Uses for Regression Analysis Outcome: 5 126) If you were going to develop a scatter plot for the purpose of determining whether one of the assumptions of the regression model is being satisfied, which of the following is true? A) The plot should illustrate a bell-shaped distribution to show that the residuals are normally distributed. B) The horizontal axis should show the fitted values for the dependent variable. C) The plot should illustrate a cone shaped look. D) The points should fall in a straight line. Answer: B Diff: 2 Keywords: regression, scatter plot Section: 14-3 Uses for Regression Analysis Outcome: 5 127) Which of the following statements is true? A) When using a simple linear regression analysis model for prediction purposes, the potential error in the forecast will be less when the value of x used to forecast y is closer to . B) The accuracy of the regression forecast is improved if the standard error for the regression slope coefficient is reduced. C) The use of regression analysis as a means of predicting the value for a dependent variable is not impacted by sampling error since the regression model uses all sample data to arrive at the regression model. D) None of the above Answer: A Diff: 2 Keywords: regression, error Section: 14-3 Uses for Regression Analysis Outcome: 5

128) Which of the following is a correct interpretation for the regression slope coefficient? A) For a one-unit change in y, we can expect the value of the independent variable to change by b1 units on average. B) For each unit change in x, the dependent variable will change by b1 units. C) The average change in y of a one-unit change in x will be b1 units. D) The average change in x of a one-unit change in y will be b1 units. Answer: C Diff: 2 Keywords: regression, slope, coefficient Section: 14-3 Uses for Regression Analysis Outcome: 6 129) If a residual plot exhibits a curved pattern in the residuals, this means that: A) the errors are not normally distributed. B) there must be a curvilinear relation between x and y. C) there is no significant relation between x and y. D) there is a problem with constant variance. Answer: B Diff: 2 Keywords: regression, residual plot, assumptions Section: 14-3 Uses for Regression Analysis Outcome: 6 130) An industry study was recently conducted in which the sample correlation between units sold and marketing expenses was 0.57. The sample size for the study included 15 companies. Based on the sample results, test to determine whether there is a significant positive correlation between these two variables. Use an alpha = 0.05 A) Because t = 2.50 > 1.7709, do not reject the null hypothesis. There is not sufficient evidence to conclude there is a positive linear relationship between sales units and marketing expense for companies in this industry. B) Because t = 2.50 > 1.7709, reject the null hypothesis. There is sufficient evidence to conclude there is a positive linear relationship between sales units and marketing expense for companies in this industry. C) Because t = 3.13 > 1.7709, do not reject the null hypothesis. There is not sufficient evidence to conclude there is a positive linear relationship between sales units and marketing expense for companies in this industry. D) Because t = 3.13 > 1.7709, reject the null hypothesis. There is sufficient evidence to conclude there is a positive linear relationship between sales units and marketing expense for companies in this industry. Answer: B Diff: 2 Keywords: correlation, scatter plot Section: 14-1 Correlation Outcome: 1

131) The following data for the dependent variable, y, and the independent variable, x, have been collected using simple random sampling: x 10 14 16 12 20 18 16 14 16 18

y 120 130 170 150 200 180 190 150 160 200

Compute the correlation coefficient. A) 0.52 B) 0.71 C) 0.62 D) 0.89 Answer: D Diff: 2 Keywords: correlation, scatter plot Section: 14-1 Correlation Outcome: 1

132) A random sample of two variables, x and y, produced the following observations: x 19 13 17 9 12 25 20 17

y 7 9 8 11 9 6 7 8

Compute the correlation coefficient for these sample data. A) -0.9707 B) -0.2141 C) 0.5133 D) 0.8612 Answer: A Diff: 2 Keywords: correlation, scatter plot Section: 14-1 Correlation Outcome: 1

133) A random sample of two variables, x and y, produced the following observations: x 19 13 17 9 12 25 20 17

y 7 9 8 11 9 6 7 8

Test to determine whether the population correlation coefficient is negative. Use a significance level of 0.05 for the hypothesis test. A) Because t = -4.152 < -1.9432, reject the null hypothesis. Because the null hypothesis is rejected, the sample data does support the hypothesis that there is a negative linear relationship between x and y. B) Because t = -4.152 < -1.9432, do not reject the null hypothesis. Because the null hypothesis is not rejected, the sample data support the hypothesis that there is a negative linear relationship between x and y. C) Because t = -9.895 < -1.9432, reject the null hypothesis. Because the null hypothesis is rejected, the sample data does support the hypothesis that there is a negative linear relationship between x and y. D) Because t = -9.895 < -1.9432, do not reject the null hypothesis. Because the null hypothesis is not rejected, the sample data support the hypothesis that there is a negative linear relationship between x and y. Answer: C Diff: 2 Keywords: regression analysis, linear Section: 14-2 Simple Linear Regression Analysis Outcome: 1 134) Explain what the correlation coefficient measures and some detail of the key issues associated with it. Be sure to also discuss the concept of spurious correlation. Answer: Correlation is the term used to indicate the extent to which two variables are linearly related. The correlation coefficient is a measure of the strength of the linear relationship. The range on the correlation coefficient is ±1.0, and the closer the correlation is to either +1.00 or -1.00, the stronger the linear relationship. A correlation close to zero implies a weak linear relationship. However, there may be a curvilinear relationship. We should always develop a scatter plot of the two variables to determine whether a nonlinear relationship exists. Correlation does not imply cause and effect. Two variables that have absolutely nothing to do with each other could be correlated, but a change in one variable does not cause a change in the other. When two unrelated variables are statistically correlated, the term that is used is spurious correlation. The population correlation coefficient is ρ (pronounced rho) and the sample correlation coefficient is r. The sample correlation coefficient is subject to sampling error and will likely be higher or lower than the true population correlation coefficient. Diff: 2 Keywords: correlation, spurious Section: 14-1 Correlation Outcome: 1 14-52 Copyright © 2018 Pearson Education, Inc.

135) Explain why it is important to construct scatter plots prior to conducting regression analysis. Answer: Simple linear regression, which includes the word linear in its name, assumes there is linear relationship between x and y. If there is a curvilinear relation of some type between x and y this will not be detected by the correlation of the regression results. It may be that x and y have a very strong relationship, but if it's curved, the strength of relation will be underestimated. It is even possible for the correlation to equal 0 when there is a strong curvilinear relation. So a scatter plot should be constructed prior to doing regression analysis to check for curvature in the relationship. If this is not done, the curvature should actually show up in the residual plot, but it would have been better to check on the front end rather than waste time on fitting a linear model to curved data. Diff: 2 Keywords: scatter plot, linear, curvilinear relation Section: 14-1 Correlation Outcome: 1

136) A national job placement company is interested in developing a model that might be used to explain the variation in starting salaries for college graduates based on the college GPA. The following data were collected through a random sample of the clients with which this company has been associated. GPA 3.20 3.40 2.90 3.60 2.80 2.50 3.00 3.60 2.90 3.50

Starting Salary $35,000 $29,500 $30,000 $36,400 $31,500 $29,000 $33,200 $37,600 $32,000 $36,000

Based on this sample information, determine the least squares regression model, determine what percent of the variation in starting salaries is explained by GPA, and test to determine whether the regression model is statistically significant at the 0.05 level of significance. Also, develop a scatter plot of the data and locate the regression line on the scatter plot. Answer: In this situation, the dependent variable, y, is the starting salary for college graduates. The independent variable, x, is the graduate's college GPA. A random sample of n = 10 people was selected and the data were recorded. The least squares regression model seeks to fit a straight line to the data that "best" fits the data. The least squares criterion states that the regression line will minimize the sum of the squared residuals. The residual is the difference between the fitted y value, as determined by the regression line, and the actual y value. The sample regression model will take the form, = b0 + b1x. The values b0 and b1 are called the regression coefficients. They are the y intercept and the slope respectively. The following equations are used to arrive at b0 and b1: b1 = b0 = - b1 By using these least squares equations, we will arrive at the "optimal" values for regression coefficients. The following regression model is determined: = 13,524.6 + 6,208.72x The scatter plot for the data with the regression line fitted to the data is shown as follows:

As seen in the scatter plot, the regression line intercepts the y-axis at about 13,524 and the slope of the regression line is about 6,208. To determine the percentage of variation in starting salary that is explained by the regression model with GPA as the independent variable, we must compute the coefficient of determination or R2. The following equation is used to compute the coefficient of determination: R2 = The quantity SSR stands for sum of squares regression and SST is the total sum of squares. These are computed as follows: SSR =

and SST =

For these sample data we get: SSR=

= 49,495,938

SST =

= 86,256,000

and

Thus the coefficient of determination is: R2=

= .5738

Therefore, knowing GPA explains just over 57 percent of the variation in starting salaries for the sample data.

To determine whether the regression model is statistically significant, we need to test the following null and alternative hypotheses: H0 : β1 = 0.0 Ha : β1 ≠ 0.0 We can test this hypothesis in two ways: using a t-test approach or using the related F-test approach. We will use the F-test approach. The following table provides the required values: ANOVA df 1 8 9

Regression Residual Total

SS 49495938 45950008

MS 49495938 4595008

F Significance F 10.77168 0.01115406

The calculated F test statistic is found by taking the ratio of the mean square regression over the mean square error: F=

= 10.77168

We then compare this calculated F to a critical F for an alpha = 0.05 and degrees of freedom equal to 1 and 8, which is 5.318. Since F = 10.77 > 5.318, we reject the null hypothesis and conclude that the regression slope coefficient is not zero. This means that GPA is a significant variable for explaining the variation in the dependent variable, starting salaries. Diff: 3 Keywords: regression, R-squared, scatter plot, hypothesis Section: 14-2 Simple Linear Regression Analysis Outcome: 3

137) The Public Utility Commission in a southern state is interested in describing the relationship between household monthly utility bills and the size of the house. A recent study of 30 randomly selected household resulted in the following regression results:

Based on the information provided, indicate what, if any, conclusions can be reached about the relationship between utility bill and the size of the house in square feet. Answer: The output from a simple linear regression analysis provides information necessary to assess what, if any, linear relationship might exist between utility bill and square feet in the home. The first item to examine is the R-square value, which measures the percent of variation in the dependent variable that is explained by the independent variable. It is seen that R-square = 0.0224. Thus, for these sample data, only slightly more than 2 percent of the variation in utility bill is explained by the square feet. Next, we could test the following null hypothesis: H0 : β1= 0.0 Ha : β1≠ 0.0 We can look to the calculated F value (F = 0.6424) and to the t value (t = 0.8015) for the regression slope coefficient. Both values are quite small and would fall well short of the associated critical value at any reasonable level of alpha, so we would not reject the null hypothesis. This means that the true population slope coefficient could be zero, implying that there is not a significant linear relationship between the two variables. We might also construct a 95 percent confidence interval estimate for the true population regression slope coefficient using: b1 ± t sb1 The t value from the t-distribution with 95 percent confidence and n - 2 = 28 degrees of freedom is 2.0484. Then from the regression output, we get: .0052 ± 2.0484(.00656) .0052 ± .0134

The resulting confidence interval estimate is: -0.0082 ----- +0.0186 Thus, based on the sample information, with 95 percent confidence, we believe that for a one foot increase in square feet, the utility bill will change by an amount between -$0.0082 and +$0.0186. Since this interval crosses zero, we are unable to conclude that the number of square feet has either a positive or negative impact on utility bill. Given that we have not been able to establish a linear relationship between the two variables, we might determine that a curvilinear relationship exists. To see, we can develop a scatter plot and check visually for such a relationship. This plot is given as follows:

Based on the scatter plot, there does not appear to be any relationship, curvilinear or linear, between the two variables. Diff: 2 Keywords: regression, scatter plot, hypothesis Section: 14-2 Simple Linear Regression Analysis Outcome: 3

138) What factors are of importance to an analyst when linear regression analysis is used for descriptive purposes? Answer: The primary objective when using regression analysis for descriptive purposes is to measure the relationship between the dependent and independent variables. There are several statistics that are useful for this purpose. The first is the correlation coefficient. The closer this measure is to positive 1.0 or minus 1.0, the stronger is the linear relationship. The R-square value measures the percentage of variation in the dependent variable that is explained by the independent variable. Values close to 1.0 (or 100 percent) indicate that a strong linear relationship exists. However, the primary statistics of interest are the regression coefficients. The intercept value (b0) can be interpreted to be the average value of y when x = 0.0. Note: This interpretation is valid only when the data used to develop the regression equation can legitimately have values of the x variable equal to zero. If not, then the intercept has no particular interpretation. The regression slope coefficient (b1) is usually of prime importance in a descriptive analysis. We interpret it to mean the average change in y for a one-unit change in x. We are particularly interested in the sign on the slope coefficient and the actual value of the coefficient too. We remember that the value of b1 is a point estimate and, thus, subject to sampling error. Therefore, we would likely develop a confidence interval estimate for the true population regression slope coefficient. If this interval does not cross zero, we say that there is a significant linear relationship. Diff: 2 Keywords: regression, slope Section: 14-2 Simple Linear Regression Analysis Outcome: 3 139) Explain what is meant by the term least squares regression model. Answer: When we look at a scatter plot, we see that there are an infinite number of possible lines that could pass through the data points. For each of these lines we could measure the "fit" by comparing the value of y from the regression line ( ) to the actual value of y. These differences are called the residuals or errors. Some of the residuals are positive (y was greater than ) and some are negative (y was less than ). If we add these residuals, the plus and minus tend to cancel out. To overcome this, we can square each residual. Then for each possible line we would get the following: SSE = This sum will be larger for some possible lines through the data, smaller for others. The line of best fit will be the one which provides the smallest sum of squared errors and is called the least squares regression line. Differential calculus could be used to determine the minimum point for the SSE function. The regression equations that actually do minimize the SSE, derived using calculus are: b1 = b0 = - b1 The regression line that minimizes the SSE will pass through the mean of the x variable and the mean of the y variable. Diff: 2 Keywords: regression, least squares Section: 14-2 Simple Linear Regression Analysis Outcome: 3 14-59 Copyright © 2018 Pearson Education, Inc.

Business Statistics, 9e (Groebner/Shannon/Fry) Chapter 15 Multiple Regression Analysis and Model Building 1) In multiple regression analysis, the residual is the absolute difference between the actual value of y and the predicted value of y. Answer: FALSE Diff: 1 Keywords: regression, multiple, residual, actual, predicted Section: 15-1 Introduction to Multiple Regression Analysis Outcome: 1 2) In a multiple regression model, R-square can be computed by squaring the highest correlation coefficient between the dependent variable and any independent variable. Answer: FALSE Diff: 1 Keywords: regression, multiple, R-square, correlation Section: 15-1 Introduction to Multiple Regression Analysis Outcome: 3 3) The multiple coefficient of determination is the average of all the squared correlations of the independent variables. Answer: FALSE Diff: 1 Keywords: regression, multiple, coefficient of determination Section: 15-1 Introduction to Multiple Regression Analysis Outcome: 3 4) In multiple regression analysis, the model will be developed with one dependent variable and two or more independent variables. Answer: TRUE Diff: 1 Keywords: regression, multiple, dependent, independent Section: 15-1 Introduction to Multiple Regression Analysis Outcome: 1 5) The multiple coefficient of determination measures the percentage of variation in the dependent variable that is explained by the independent variables in the model. Answer: TRUE Diff: 1 Keywords: regression, multiple, coefficient of determination Section: 15-1 Introduction to Multiple Regression Analysis Outcome: 3

6) In a multiple regression model, each regression slope coefficient measures the average change in the dependent variable for a one-unit change in the independent variable, all other variables held constant. Answer: TRUE Diff: 2 Keywords: regression, multiple, slope Section: 15-1 Introduction to Multiple Regression Analysis Outcome: 1 7) Residuals are calculated by e = y - . Answer: TRUE Diff: 2 Keywords: regression, residuals Section: 15-1 Introduction to Multiple Regression Analysis Outcome: 1 8) The three components of the regression model-building process are model specification, model fitting, and model diagnosis. Answer: TRUE Diff: 2 Keywords: regression, multiple, specification, fitting, diagnosis Section: 15-1 Introduction to Multiple Regression Analysis Outcome: 1 9) Where there are two independent variables in a multiple regression, the regression equation forms a plane. Answer: TRUE Diff: 2 Keywords: regression, regression plane Section: 15-1 Introduction to Multiple Regression Analysis Outcome: 1 10) It is generally suggested that the sample size in developing a multiple regression model should be at least four times the number of independent variables. Answer: TRUE Diff: 2 Keywords: regression, multiple, sample size Section: 15-1 Introduction to Multiple Regression Analysis Outcome: 1 11) In conducting multiple regression analysis, t-tests are one way to test for significance between x and y variables prior to conducting the F-test. Answer: TRUE Diff: 2 Keywords: regression, multiple, t-tests, F-test Section: 15-1 Introduction to Multiple Regression Analysis Outcome: 1

12) The standard error of the estimate is a term that is used for the standard deviation of the residuals in a multiple regression model. Answer: TRUE Diff: 1 Keywords: regression, multiple, standard error, standard deviation, residuals Section: 15-1 Introduction to Multiple Regression Analysis Outcome: 1 13) In a multiple regression model, it is assumed that the errors or residuals are normally distributed. Answer: TRUE Diff: 1 Keywords: regression, multiple, residuals, normal, errors Section: 15-1 Introduction to Multiple Regression Analysis Outcome: 1 14) In a multiple regression model, the regression coefficients are calculated such that the quantity, , is minimized. Answer: FALSE Diff: 2 Keywords: regression, multiple, coefficients Section: 15-1 Introduction to Multiple Regression Analysis Outcome: 1 15) In simple linear regression analysis, the regression model forms a straight line in two-dimensional space through the x,y data points, while a multiple regression model forms a plane through multidimensional space. Answer: TRUE Diff: 2 Keywords: regression, multiple, linear, line, plane Section: 15-1 Introduction to Multiple Regression Analysis Outcome: 1 16) A model is a representation of an actual system. Answer: TRUE Diff: 1 Keywords: model, system Section: 15-1 Introduction to Multiple Regression Analysis Outcome: 1 17) Multicollinearity occurs when one or more independent variables is highly correlated with the dependent variable. Answer: FALSE Diff: 2 Keywords: regression, multicollinearity Section: 15-1 Introduction to Multiple Regression Analysis Outcome: 5

18) The correlation matrix is an effective means of determining whether any of the independent variables has a curvilinear relationship with the dependent variable. Answer: FALSE Diff: 2 Keywords: regression, multiple, correlation matrix, multicollinearity Section: 15-1 Introduction to Multiple Regression Analysis Outcome: 2 19) In the model diagnosis step in regression modeling, we are interested in whether the overall regression model and individual variables are significant in addition to satisfying regression analysis assumptions. Answer: TRUE Diff: 2 Keywords: regression, multiple, sign, size, slope Section: 15-1 Introduction to Multiple Regression Analysis Outcome: 5 20) A correlation matrix shows the correlation between each independent variable and the dependent variable but gives no information about the potential for multicollinearity problems. Answer: FALSE Diff: 2 Keywords: regression, multiple, correlation, matrix, multicollinearity Section: 15-1 Introduction to Multiple Regression Analysis Outcome: 2 21) If a decision maker has several potential independent variables to select from in building a regression model, the variable that, by itself, will always be the most effective in explaining the variation in the dependent variable will be the variable that has a correlation closest to positive 1.00. Answer: FALSE Diff: 2 Keywords: regression, multiple, correlation Section: 15-1 Introduction to Multiple Regression Analysis Outcome: 2 22) The adjusted R2 value is a particularly important measure when the number of independent variables is large relative to the sample size. Answer: TRUE Diff: 2 Keywords: regression, multiple, adjusted R-square Section: 15-1 Introduction to Multiple Regression Analysis Outcome: 4

23) A major car magazine has recently collected data on 30 leading cars in the U.S. market. It is interested in building a multiple regression model to explain the variation in highway miles. The following correlation matrix has been computed from the data collected: mileage, highway mileage, highway mileage, city Curb Weight cylinders Horsepower

1 0.857550598 -0.739110566 -0.694837149 -0.549172956

mileage, city

1 -0.70765104 -0.866135056 -0.684199197

Curb Weight

1 0.596475711 0.293202385

cylinders

Horsepower

1 0.840347219

Based on this information, if we test using a 0.05 level of significance, the critical value for testing whether any of the independent variables are significantly correlated with the dependent variable is t = 2.0484. Answer: TRUE Diff: 2 Keywords: regression, multiple, critical value Section: 15-1 Introduction to Multiple Regression Analysis Outcome: 4 24) A major car magazine has recently collected data on 30 leading cars in the U.S. market. It is interested in building a multiple regression model to explain the variation in highway miles. The following correlation matrix has been computed from the data collected: mileage, highway mileage, highway mileage, city Curb Weight cylinders Horsepower

1 0.857550598 -0.739110566 -0.694837149 -0.549172956

mileage, city

1 -0.70765104 -0.866135056 -0.684199197

Curb Weight

1 0.596475711 0.293202385

cylinders

1 0.840347219

Horsepower

If only one independent variable (ignoring city mileage) is to be used in explaining the dependent variable in a regression model, the percentage of variation that will be explained will be nearly 74 percent. Answer: FALSE Diff: 2 Keywords: regression, multiple, variation Section: 15-1 Introduction to Multiple Regression Analysis Outcome: 4

25) A major car magazine has recently collected data on 30 leading cars in the U.S. market. It is interested in building a multiple regression model to explain the variation in highway miles. The following correlation matrix has been computed from the data collected: mileage, highway mileage, highway mileage, city Curb Weight cylinders Horsepower

1 0.857550598 -0.739110566 -0.694837149 -0.549172956

mileage, city

1 -0.70765104 -0.866135056 -0.684199197

Curb Weight

1 0.596475711 0.293202385

cylinders

1 0.840347219

Horsepower

If the independent variables, curb weight, cylinders, and horsepower are used together in a multiple regression model, there may be a potential problem with multicollinearity since horsepower and cylinders are highly correlated. Answer: TRUE Diff: 2 Keywords: regression, multiple, multicollinearity Section: 15-1 Introduction to Multiple Regression Analysis Outcome: 5 26) In a multiple regression model where three independent variables are included in the model, the percentage of explained variation will be equal to the square of the sum of the correlations between the independent variables and the dependent variable. Answer: FALSE Diff: 2 Keywords: regression, multiple, R-square Section: 15-1 Introduction to Multiple Regression Analysis Outcome: 3

27) A major car magazine has recently collected data on 30 leading cars in the U.S. market. It is interested in building a multiple regression model to explain the variation in highway miles. The following correlation matrix has been computed from the data collected: mileage, highway mileage, highway mileage, city Curb Weight cylinders Horsepower

1 0.857550598 -0.739110566 -0.694837149 -0.549172956

mileage, city

1 -0.70765104 -0.866135056 -0.684199197

Curb Weight

1 0.596475711 0.293202385

cylinders

1 0.840347219

Horsepower

The analysts also produced the following multiple regression output using curb weight, cylinders, and horsepower as the three independent variables. Note that a number of the output fields are missing, but can be determined from the information provided.

Based on the information provided, the three independent variables explain approximately 67 percent of the variation in the highway mileage among these 30 cars. Answer: TRUE Diff: 2 Keywords: regression, multiple, R-square Section: 15-1 Introduction to Multiple Regression Analysis Outcome: 4 28) When an independent variable, that has a positive correlation with the dependent variable, receives a negative slope in a multiple regression, this is probably caused by multicollinearity. Answer: TRUE Diff: 2 Keywords: regression, multiple, multicollinearity Section: 15-1 Introduction to Multiple Regression Analysis Outcome: 5

29) In a multiple regression model, the adjusted R-square value measures the explained variation in the dependent variable after taking into account the relationship between the sample size and the number of independent variables in the model. Answer: TRUE Diff: 2 Keywords: regression, multiple, adjusted, R-square, sample size Section: 15-1 Introduction to Multiple Regression Analysis Outcome: 4 30) Based on the correlations below: Y X1 X2

0.8 0.7

we could say that x1 accounts for 64 percent of the variation in y and x2 accounts for 49 percent of the variation in y. So if both xs are included in a multiple regression model, then the resulting R-square = 1.13. Answer: FALSE Diff: 2 Keywords: regression, multiple, correlation, R-square Section: 15-1 Introduction to Multiple Regression Analysis Outcome: 4 31) If the R-square for a multiple regression model with two independent variables is .64, the correlation between the two independent variables will be .80 Answer: FALSE Diff: 2 Keywords: regression, multiple, R-square, correlation Section: 15-1 Introduction to Multiple Regression Analysis Outcome: 3 32) The variance inflation factor is an indication of how much multicollinearity there is in the regression model. Answer: TRUE Diff: 2 Keywords: regression, multiple, multicollinearity, inflation, variance Section: 15-1 Introduction to Multiple Regression Analysis Outcome: 5

33) A study has recently been conducted by a major computer magazine publisher in which the objective was to develop a multiple regression model to explain the variation in price of personal computers. Three independent variables were used. The following computer printout shows the final output. However, several values are omitted from the printout.

Given this information, the regression model explains just under 70 percent of the variation in the price of personal computers. Answer: TRUE Diff: 3 Keywords: regression, multiple, R-square Section: 15-1 Introduction to Multiple Regression Analysis Outcome: 4 34) In a multiple regression analysis with three independent variables the alternative hypothesis for conducting the test of the overall model is: HA : More than one βi ≠ 0. Answer: FALSE Diff: 2 Keywords: regression, multiple, F-test, null hypothesis Section: 15-1 Introduction to Multiple Regression Analysis Outcome: 4

35) A study has recently been conducted by a major computer magazine publisher in which the objective was to develop a multiple regression model to explain the variation in price of personal computers. Three independent variables were used. The following computer printout shows the final output. However, several values are omitted from the printout.

Given this information, the calculated test statistic for the regression slope coefficient on the variable RAM, is approximately 1.54. Answer: TRUE Diff: 2 Keywords: regression, multiple, test statistic Section: 15-1 Introduction to Multiple Regression Analysis Outcome: 4

36) A study has recently been conducted by a major computer magazine publisher in which the objective was to develop a multiple regression model to explain the variation in price of personal computers. Three independent variables were used. The following computer printout shows the final output. However, several values are omitted from the printout.

Given this information, using an alpha = .05 level, you can conclude that the overall regression model is statistically significant. Answer: TRUE Diff: 2 Keywords: regression, multiple, F-test Section: 15-1 Introduction to Multiple Regression Analysis Outcome: 4 37) The variance inflation factor (VIF) provides a measure for each independent variable of how much multicollinearity is associated with that particular independent variable. Answer: TRUE Diff: 2 Keywords: regression, multiple, variance inflation factor Section: 15-1 Introduction to Multiple Regression Analysis Outcome: 5

38) A major car magazine has recently collected data on 30 leading cars in the U.S. market. It is interested in building a multiple regression model to explain the variation in highway miles. The following correlation matrix has been computed from the data collected: mileage, highway mileage, highway mileage, city Curb Weight cylinders Horsepower

1 0.857550598 -0.739110566 -0.694837149 -0.549172956

mileage, city

1 -0.70765104 -0.866135056 -0.684199197

Curb Weight

1 0.596475711 0.293202385

cylinders

1 0.840347219

Horsepower

The analysts also produced the following multiple regression output using curb weight, cylinders, and horsepower as the three independent variables. Note, a number of the output fields are missing, but can be determined from the information provided.

Based on this information, the standard error of the estimate for the regression model is approximately 6.46 miles per gallon. Answer: FALSE Diff: 2 Keywords: regression, multiple, standard error Section: 15-1 Introduction to Multiple Regression Analysis Outcome: 4

39) A major car magazine has recently collected data on 30 leading cars in the U.S. market. It is interested in building a multiple regression model to explain the variation in highway miles. The following correlation matrix has been computed from the data collected: mileage, highway mileage, highway mileage, city Curb Weight cylinders Horsepower

1 0.857550598 -0.739110566 -0.694837149 -0.549172956

mileage, city

1 -0.70765104 -0.866135056 -0.684199197

Curb Weight

1 0.596475711 0.293202385

cylinders

1 0.840347219

Horsepower

If the analysts are interested in testing whether the overall regression model is statistically significant, the appropriate null and alternative hypotheses are: H0 : β1 = β2 = β3 Ha : β1 ≠ β2 ≠ β3 Answer: FALSE Diff: 2 Keywords: regression, multiple, null, alternative, hypothesis Section: 15-1 Introduction to Multiple Regression Analysis Outcome: 4

40) A major car magazine has recently collected data on 30 leading cars in the U.S. market. It is interested in building a multiple regression model to explain the variation in highway miles. The following correlation matrix has been computed from the data collected: mileage, highway mileage, highway mileage, city Curb Weight cylinders Horsepower

1 0.857550598 -0.739110566 -0.694837149 -0.549172956

mileage, city

1 -0.70765104 -0.866135056 -0.684199197

Curb Weight

1 0.596475711 0.293202385

cylinders

1 0.840347219

Horsepower

Based on the above information, the test statistic for testing whether the overall model is statistically significant is approximately F = 17.4 Answer: TRUE Diff: 3 Keywords: regression, multiple, test statistic, F-test Section: 15-1 Introduction to Multiple Regression Analysis Outcome: 4

41) A major car magazine has recently collected data on 30 leading cars in the U.S. market. It is interested in building a multiple regression model to explain the variation in highway miles. The following correltion matrix has been computed from the data collected: mileage, highway mileage, highway mileage, city Curb Weight cylinders Horsepower

1 0.857550598 -0.739110566 -0.694837149 -0.549172956

mileage, city

1 -0.70765104 -0.866135056 -0.684199197

Curb Weight

1 0.596475711 0.293202385

cylinders

1 0.840347219

Horsepower

Based on the information provided, holding the other variables constant, increasing horsepower by one unit results in an average decrease in highway mileage by 0.016 miles per gallon. Answer: TRUE Diff: 2 Keywords: regression, multiple, slope, coefficient Section: 15-1 Introduction to Multiple Regression Analysis Outcome: 4

42) A major car magazine has recently collected data on 30 leading cars in the U.S. market. It is interested in building a multiple regression model to explain the variation in highway miles. The following correlation matrix has been computed from the data collected: mileage, highway mileage, highway mileage, city Curb Weight cylinders Horsepower

1 0.857550598 -0.739110566 -0.694837149 -0.549172956

mileage, city

1 -0.70765104 -0.866135056 -0.684199197

Curb Weight

1 0.596475711 0.293202385

cylinders

1 0.840347219

Horsepower

Based on the information provided, using a 0.05 level of statistical significance, both curb weight and horsepower are statistically significant variables in explaining the variation in the dependent variable when they are included in the model along with cylinders. Answer: FALSE Diff: 2 Keywords: regression, multiple, slope, t-test Section: 15-1 Introduction to Multiple Regression Analysis Outcome: 4

43) A major car magazine has recently collected data on 30 leading cars in the U.S. market. It is interested in building a multiple regression model to explain the variation in highway miles. The following correlation matrix has been computed from the data collected: mileage, highway mileage, highway mileage, city Curb Weight cylinders Horsepower

1 0.857550598 -0.739110566 -0.694837149 -0.549172956

mileage, city

1 -0.70765104 -0.866135056 -0.684199197

Curb Weight

1 0.596475711 0.293202385

cylinders

1 0.840347219

Horsepower

Based on the information provided, the 95 percent confidence interval estimate for regression slope coefficient for horsepower is approximately - 0.041 to 0.009 and since this interval crosses zero, we are unable to conclude that the regression slope coefficient for this variable is different from zero. Answer: TRUE Diff: 2 Keywords: regression, multiple, confidence interval, slope Section: 15-1 Introduction to Multiple Regression Analysis Outcome: 5

44) Models can be specified as linear or nonlinear. Answer: TRUE Diff: 2 Keywords: regression, multiple, model specification Section: 15-1 Introduction to Multiple Regression Analysis Outcome: 1 45) A dummy variable is a dependent variable whose value is set at either zero or one. Answer: FALSE Diff: 1 Keywords: dummy, regression Section: 15-2 Using Qualitative Independent Variables Outcome: 6 46) The method used in regression analysis for incorporating a categorical variable into the model is by organizing the categorical variable into one or more dummy variables. Answer: TRUE Diff: 1 Keywords: categorical, dummy Section: 15-2 Using Qualitative Independent Variables Outcome: 6 47) One of the variables that are being considered for inclusion in a multiple regression model is marital status of the customer. There are four possible responses listed for this variable. Based on this, three dummy variables will need to be created and incorporated into the regression model. Answer: TRUE Diff: 2 Keywords: regression, multiple, dummy Section: 15-2 Using Qualitative Independent Variables Outcome: 6 48) On a survey there is a question that asks whether someone lives in a house, apartment, or condominium. These three responses could be coded in a dummy variable using value 0, 1, and 2. Answer: FALSE Diff: 2 Keywords: regression, multiple, dummy Section: 15-2 Using Qualitative Independent Variables Outcome: 6 49) Consider the following regression equation: = 356 + 180x1 - 2.5x2. The x1 variable is a quantitative variable and the x2 variable is a dummy with values 1 and 0. Given this, we can interpret the slope coefficient on variable x2 as follows: Holding x1 constant, if the value of x2 is changed from 0 to 1, the average value of y will decrease by 2.5 units. Answer: TRUE Diff: 2 Keywords: regression, multiple, dummy, slope, coefficient Section: 15-2 Using Qualitative Independent Variables Outcome: 6 15-18 Copyright © 2018 Pearson Education, Inc.

50) A study has recently been conducted by a major computer magazine publisher in which the objective was to develop a multiple regression model to explain the variation in price of personal computers. Three quantitative independent variables were used along with one qualitative variable. The qualitative variable was coded 1 if the computer included a monitor, 0 otherwise. The following computer printout shows the final output.

Based on this information, it is possible to determine that 4 of the PCs in the data set did not have a monitor included. Answer: FALSE Diff: 2 Keywords: regression, multiple, dummy Section: 15-2 Using Qualitative Independent Variables Outcome: 6

51) A study has recently been conducted by a major computer magazine publisher in which the objective was to develop a multiple regression model to explain the variation in price of personal computers. Three quantitative independent variables were used along with one qualitative variable. The qualitative variable was coded 1 if the computer included a monitor, 0 otherwise. The following computer printout shows the final output.

Based on this information, given the other variables in the model, whether or not a monitor is included has a significant impact on the price of the personal computer. Answer: FALSE Diff: 2 Keywords: regression, multiple, dummy, t-test Section: 15-2 Using Qualitative Independent Variables Outcome: 6 52) A regression equation that predicts the price of homes in thousands of dollars is ŷ = 24.6 + 0.055x1 -

3.6x2, where x2 is a dummy variable that represents whether the house is on a busy street or not. Here x2 = 1 means the house is on a busy street and x2 = 0 means it is not. From this we can conclude that on average homes that are on busy streets are worth $3600 more than homes that are not on busy streets. Answer: FALSE Diff: 2 Keywords: regression, multiple, slope, dummy Section: 15-2 Using Qualitative Independent Variables Outcome: 6

53) In a study of individuals’ television viewing hours per week, the predictors are defined as income, job in terms of hours per week, number of people living in the household and stress level. The stress level is the only categorical variable with self-reported levels of stress as 1 = none to 5 = extreme. The stress level can be used as a number in the regression equation. Answer: TRUE Diff: 2 Keywords: regression, multiple, dummy Section: 15-2 Using Qualitative Independent Variables Outcome: 6 54) In a study of employees at a local company, the human resource manager wants to develop a multiple regression model to explain the difference in employee wage rates. She is thinking of including a variable, degree status, in which the following categories exist: no degree, H.S. degree, junior college degree, bachelor degree, graduate degree. Two other variables are being considered; Age and Years With the Company. Given this, the appropriate number of variables in the model will be six. Answer: TRUE Diff: 2 Keywords: regression, multiple, independent, dummy Section: 15-2 Using Qualitative Independent Variables Outcome: 6 55) You should not include more than one categorical variable in a multiple regression because the use of two or more will cause misleading results. Answer: FALSE Diff: 2 Keywords: regression, multiple, dummy Section: 15-2 Using Qualitative Independent Variables Outcome: 6 56) If given a choice in collecting data on age for use as an independent variable in a regression model, a decision maker would generally prefer to record the actual age rather than an age category so as to avoid using dummy variables. Answer: TRUE Diff: 2 Keywords: regression, multiple, dummy Section: 15-2 Using Qualitative Independent Variables Outcome: 6 57) To describe variable credit status that has three levels: Excellent, Good, and Poor, we need to use two different dummy variables. Answer: TRUE Diff: 2 Keywords: regression, multiple, dummy Section: 15-2 Using Qualitative Independent Variables Outcome: 6

58) A regression model of the form: = B0 + B1x1 + B2

+ B3

+ ε is called a 3rd order polynomial

model. Answer: TRUE Diff: 1 Keywords: regression, multiple, polynomial Section: 15-3 Working With Nonlinear Relationships Outcome: 7 59) A complete polynomial model contains terms of all orders less than or equal to the pth order. Answer: TRUE Diff: 2 Keywords: regression, multiple, polynomial, complete Section: 15-3 Working With Nonlinear Relationships Outcome: 7 60) Consider the following regression model:

= B0 + B1x1 + B1

+ ε. If B2 > 0, then the parabola will

open downward and if B2 < 0, then the parabola will open downward. Answer: FALSE Diff: 2 Keywords: regression, multiple, parabola Section: 15-3 Working With Nonlinear Relationships Outcome: 7 61) If one independent variable affects the relationship between a second independent variable and the dependent variable, it is said that there is interaction between the two independent variables. Answer: TRUE Diff: 2 Keywords: regression, multiple, interaction, independent Section: 15-3 Working With Nonlinear Relationships Outcome: 7

62) Consider the following scatter plot:

Given the apparent relationship between the x and y variable, a possible curvilinear regression model to consider would be a second-order polynomial model. Answer: TRUE Diff: 2 Keywords: regression, multiple, curvilinear, polynomial Section: 15-3 Working With Nonlinear Relationships Outcome: 7 63) When a regression equation includes a term such as x1x2 where two independent variables are multiplied, this is an interaction term. Answer: FALSE Diff: 2 Keywords: regression, multiple, R-square, polynomial Section: 15-3 Working With Nonlinear Relationships Outcome: 7

64) The following output is for a second-order polynomial regression model where the independent variables are x and x2 (x^2 in output). Some of the output has been omitted.

Considering the above information, the model explains approximately 56.7 percent of the variation in the y variable. Answer: FALSE Diff: 2 Keywords: regression, multiple, R-square, polynomial Section: 15-3 Working With Nonlinear Relationships Outcome: 7 65) If a polynomial model has a larger R-square than a linear model for the same set of data, this is one indication that the polynomial model fits the data better than the linear model. Answer: TRUE Diff: 2 Keywords: regression, multiple, R-square Section: 15-3 Working With Nonlinear Relationships Outcome: 7

66) The following output is for a second-order polynomial regression model where the independent variables are x and x2 (x^2 in output). Some of the output has been omitted.

Considering the above information, both independent variables in the model are considered statistically significant at the alpha = 0.05 level. Answer: TRUE Diff: 2 Keywords: regression, multiple, t-test, test statistic Section: 15-3 Working With Nonlinear Relationships Outcome: 7

67) The following output is for a second-order polynomial regression model where the independent variables are x and x2 (x^2 in output). Some of the output has been omitted.

Considering the above information, it is clear that the second-order polynomial model will be a more effective regression model for explaining the variation in the y variable than would a linear regression model involving only one independent variable, x. Answer: FALSE Diff: 2 Keywords: regression, multiple, polynomial Section: 15-3 Working With Nonlinear Relationships Outcome: 7 68) A multiple regression model of the form = B0 + B1x + B2x2 + ε is called a second-degree polynomial model. Answer: TRUE Diff: 2 Keywords: regression, multiple, polynomial Section: 15-3 Working With Nonlinear Relationships Outcome: 7 69) In a second-order polynomial regression model, the regression coefficient, B2, will be positive if the parabola opens downward and negative when the parabola opens upward. Answer: FALSE Diff: 2 Keywords: regression, multiple, parabola, polynomial Section: 15-3 Working With Nonlinear Relationships Outcome: 7

70) In curvilinear regression modeling, a composite model is one that contains either the basic terms or the interactive terms but not both. Answer: FALSE Diff: 2 Keywords: regression, multiple, composite, curvilinear Section: 15-3 Working With Nonlinear Relationships Outcome: 7 71) A multiple regression model of the form = B0 + B1x + B2x2 + B3x3 + ε is called an expanded secondorder polynomial since it contains all the terms up to x3 in the model at one time. Answer: FALSE Diff: 2 Keywords: regression, multiple, polynomial, expanded Section: 15-3 Working With Nonlinear Relationships Outcome: 7 72) Interaction terms and polynomial terms should not be included in the same multiple regression model. Answer: FALSE Diff: 2 Keywords: regression, multiple, polynomial, interaction Section: 15-3 Working With Nonlinear Relationships Outcome: 7 73) In regression model: y = β0 + β1x1 + β2

+ ε , if β2 < 0, then the value of y is expected to increase with

x until x reaches a certain point after which the value of y is expected to decrease. Answer: TRUE Diff: 3 Keywords: regression, multiple, second order polynomial model Section: 15-3 Working With Nonlinear Relationships Outcome: 7 74) Stepwise regression is the approach that is always taken when developing a regression model to fit a curvilinear relationship between the dependent and potential independent variables. Answer: FALSE Diff: 1 Keywords: regression, multiple, curvilinear, stepwise Section: 15-4 Stepwise Regression Outcome: 8

75) A decision maker is considering constructing a multiple regression model with two independent variables. The correlation between x1 and y is 0.70, and the correlation between variable x2 and y is 0.50. Based on this, the regression model containing both independent variables will explain 74 percent of the variation in the dependent variable. Answer: FALSE Diff: 2 Keywords: regression, multiple, correlation Section: 15-4 Stepwise Regression Outcome: 8 76) In a forward selection stepwise regression process, the first variable to be selected will be the variable that can, by itself, do the most to explain the variation in the dependent variable. This will be the variable that provided the highest possible R-square value by itself. Answer: TRUE Diff: 2 Keywords: regression, multiple, stepwise, R-square, forward Section: 15-4 Stepwise Regression Outcome: 8 77) In a forward selection stepwise regression process, the second variable to be selected from the list of potential independent variables is always the one that has the second highest correlation with the dependent variable. Answer: FALSE Diff: 2 Keywords: regression, multiple, stepwise, forward, correlation Section: 15-4 Stepwise Regression Outcome: 8 78) Backward elimination is the reverse of the forward stepwise selection procedure. The resulting model can be different than the forward model. Answer: TRUE Diff: 2 Keywords: regression, multiple, stepwise Section: 15-4 Stepwise Regression Outcome: 8

79) In a forward stepwise regression process, it is actually possible for the R-square value to decline if variables are added to the regression model that do not help to explain the variation in the dependent variable. Answer: FALSE Diff: 2 Keywords: regression, multiple, stepwise, forward, R-square Section: 15-4 Stepwise Regression Outcome: 8

80) Standard stepwise regression combines attributes of both forward selection and backward elimination. Answer: TRUE Diff: 2 Keywords: regression, multiple, stepwise Section: 15-4 Stepwise Regression Outcome: 8 81) It is possible for the standard error of the estimate to actually increase if variables are added to the model that do not aid in explaining the variation in the dependent variable. Answer: TRUE Diff: 2 Keywords: regression, multiple, stepwise, standard error Section: 15-4 Stepwise Regression Outcome: 8 82) In the best subsets approach to regression analysis, if we start with 4 independent variables, a total of 33 different regression models will actually be computed for possible selection as the best model to use. Answer: FALSE Diff: 2 Keywords: regression, multiple, stepwise Section: 15-4 Stepwise Regression Outcome: 8 83) When the best subsets approach is used in a regression application, one method for determining which of the many possible models to select for potential use is called the Cp statistic. Answer: TRUE Diff: 2 Keywords: regression, multiple, Cp, subset, stepwise Section: 15-4 Stepwise Regression Outcome: 8 84) One reason for examining the adjusted R-square value in a multiple regression analysis is that the Rsquare value will increase just by adding additional independent variables to the model, whereas the adjusted R-square accounts for the relationship between the number of independent variables and the sample size and may actually decline if inappropriate independent variables are included in the model. Answer: TRUE Diff: 2 Keywords: regression, multiple, R-square, adjusted Section: 15-4 Stepwise Regression Outcome: 8

85) The forward selection method and the backward elimination method will always lead to choosing the same final regression model. Answer: FALSE Diff: 2 Keywords: regression, multiple, forward selection, backward elimination Section: 15-4 Stepwise Regression Outcome: 8 86) Standard stepwise regression is a good way of identifying potential multicollinearity problems since we are able to see the impact on the model at each step that occurs when a new variable is added to the model. For instance, if bringing in a new variable causes the sign to change on a previously entered variable, we have evidence of multicollinearity. Answer: TRUE Diff: 2 Keywords: regression, multiple, stepwise, multicollinearity Section: 15-4 Stepwise Regression Outcome: 8 87) If a stepwise regression approach is used to enter, one at a time, four variables into a regression model, the resulting regression equation may differ from the regression equation that occurs when all four of the variables are entered at one step. Answer: FALSE Diff: 2 Keywords: regression, multiple, stepwise Section: 15-4 Stepwise Regression Outcome: 8 88) The best subsets method will involve trying fewer different regression models than stepwise regression. Answer: FALSE Diff: 2 Keywords: regression, multiple, stepwise, best subsets Section: 15-4 Stepwise Regression Outcome: 8 89) When we say that we wish to determine the aptness of a regression model, we are actually saying that we wish to check to see whether the resulting model meets the basic assumptions of regression analysis. Answer: TRUE Diff: 2 Keywords: regression, multiple, assumptions Section: 15-5 Determining the Aptness of the Model Outcome: 9

90) A useful method for determining whether a linear function is the appropriate function to describe the relationship between the x and y variable is a residual plot in which the residuals are plotted on the vertical axis and the independent variable is on the horizontal axis. Answer: TRUE Diff: 2 Keywords: regression, multiple, linear, residual Section: 15-5 Determining the Aptness of the Model Outcome: 9 91) If one or more of the regression assumptions has been violated this means that the current regression model is not the best one for this data set, and another model should be sought. Answer: TRUE Diff: 2 Keywords: regression, residuals Section: 15-5 Determining the Aptness of the Model Outcome: 9 92) The following residual plot is an output of a regression model.

Based on this residual plot, there is evidence to suggest that the underlying relationship between the y variable and the x variable is nonlinear. Answer: TRUE Diff: 2 Keywords: regression, residual, nonlinear, plot Section: 15-5 Determining the Aptness of the Model Outcome: 9 15-31 Copyright © 2018 Pearson Education, Inc.

93) The following residual plot was constructed based on a simple linear regression model.

Based on this plot, there appears to be no basis for concluding that a curvilinear model may be more appropriate than a linear model to explain the variation in the y variable. Answer: FALSE Diff: 2 Keywords: regression, residual, plot, linear, curvilinear Section: 15-5 Determining the Aptness of the Model Outcome: 9 94) To check out whether the regression assumption involving normality of the error terms is valid, it is appropriate to construct a normal probability plot. If this plot forms a straight line from the lower lefthand corner diagonally up to the upper right-hand corner, the error terms may be assumed to be normally distributed. Answer: TRUE Diff: 2 Keywords: regression, normal, probability, plot, error Section: 15-5 Determining the Aptness of the Model Outcome: 9

95) Which of the following is not an assumption of the multiple regression model? A) The mean of the residuals is equal to the variance at all combinations of levels of the independent variables. B) The regression error terms are normally distributed. C) The model error terms are independent. D) The residuals have a constant variance for all combinations of values for the independent variables. Answer: A Diff: 2 Keywords: regression, multiple, assumption, mean, variance Section: 15-1 Introduction to Multiple Regression Analysis Outcome: 1 96) In a multiple regression model, which of the following is true? A) The coefficient of determination will be equal to the square of the highest correlation in the correlation matrix. B) Adding variables that have a low correlation with the dependent variable will cause the R-square value to decline. C) The sum of the residuals computed for the least squares regression equation will be zero. D) The adjusted R-square might be higher or lower than the value of the R-square. Answer: C Diff: 2 Keywords: regression, multiple, residuals Section: 15-1 Introduction to Multiple Regression Analysis Outcome: 1 97) The editors of a national automotive magazine recently studied 30 different automobiles sold in the United States with the intent of seeing whether they could develop a multiple regression model to explain the variation in highway mileage per gallon. A number of different independent variables were collected. The following correlation matrix was developed:

If only one variable were to be brought into the model, which variable should it be if the goal is to explain the highest possible percentage of variation in the dependent variable? A) 0 to 60 mph B) Horsepower C) Curb weight D) Displacement Answer: C Diff: 2 Keywords: regression, multiple, correlation Section: 15-1 Introduction to Multiple Regression Analysis Outcome: 2 15-33 Copyright © 2018 Pearson Education, Inc.

98) The editors of a national automotive magazine recently studied 30 different automobiles sold in the United States with the intent of seeing whether they could develop a multiple regression model to explain the variation in highway miles per gallon. A number of different independent variables were collected. The following regression output (with some values missing) was recently presented to the editors by the magazine's analysts:

Based on this output and your understanding of multiple regression analysis, what percentage of variation in the dependent variable is explained by the regression model? A) Approximately 82 percent B) Over 90 percent C) About 37 percent D) None of the above Answer: A Diff: 2 Keywords: regression, multiple, R-square, variation Section: 15-1 Introduction to Multiple Regression Analysis Outcome: 2

99) The editors of a national automotive magazine recently studied 30 different automobiles sold in the United States with the intent of seeing whether they could develop a multiple regression model to explain the variation in highway miles per gallon. A number of different independent variables were collected. The following regression output (with some values missing) was recently presented to the editors by the magazine's analysts:

Based on this output and your understanding of multiple regression analysis, what is the adjusted Rsquare value for this model? A) About 0.82 B) Approximately 0.90 C) Just under 0.48 D) None of the above Answer: D Diff: 2 Keywords: regression, multiple, adjusted R-square Section: 15-1 Introduction to Multiple Regression Analysis Outcome: 4

100) The editors of a national automotive magazine recently studied 30 different automobiles sold in the United States with the intent of seeing whether they could develop a multiple regression model to explain the variation in highway miles per gallon. A number of different independent variables were collected. The following regression output (with some values missing) was recently presented to the editors by the magazine's analysts:

Based on this output and your understanding of multiple regression analysis, how many degrees of freedom are associated with the Residual in the ANOVA table? A) 19 B) 22 C) 7 D) 29 Answer: B Diff: 2 Keywords: regression, multiple, degrees of freedom, residual Section: 15-1 Introduction to Multiple Regression Analysis Outcome: 4

101) The editors of a national automotive magazine recently studied 30 different automobiles sold in the United States with the intent of seeing whether they could develop a multiple regression model to explain the variation in highway miles per gallon. A number of different independent variables were collected. The following regression output (with some values missing) was recently presented to the editors by the magazine's analysts:

Based on this output and your understanding of multiple regression analysis, what is the value of the standard error of the estimate for this model? A) Approximately 2.02 B) About 5.97 C) Approximately 14.05 D) Nearly 8.0 Answer: A Diff: 3 Keywords: regression, multiple, standard error Section: 15-1 Introduction to Multiple Regression Analysis Outcome: 4

102) The editors of a national automotive magazine recently studied 30 different automobiles sold in the United States with the intent of seeing whether they could develop a multiple regression model to explain the variation in highway miles per gallon. A number of different independent variables were collected. The following regression output (with some values missing) was recently presented to the editors by the magazine's analysts:

Based on this output and your understanding of multiple regression analysis, what is the critical value for testing the significance of the overall regression model at a 0.05 level of statistical significance? A) About F = 5.92 B) Nearly F = 3.80 C) Approximately F = 2.50 D) None of the above Answer: C Diff: 2 Keywords: regression, multiple, critical value, F-test Section: 15-1 Introduction to Multiple Regression Analysis Outcome: 4

103) The editors of a national automotive magazine recently studied 30 different automobiles sold in the United States with the intent of seeing whether they could develop a multiple regression model to explain the variation in highway miles per gallon. A number of different independent variables were collected. The following regression output (with some values missing) was recently presented to the editors by the magazine's analysts:

Based on this output and your understanding of multiple regression analysis, which of the following statements is true? A) The overall multiple regression model explains a significant portion of the variation in highway mileage when tested at a significance level of 0.05. B) Only the two independent variables are statistically significant in the presence of the others when a significance level of 0.05 is used to test. C) The standard error of the estimate is a negative value due to the multicollinearity problems in the model. D) None of the above is true. Answer: A Diff: 3 Keywords: regression, multiple Section: 15-1 Introduction to Multiple Regression Analysis Outcome: 4

104) The editors of a national automotive magazine recently studied 30 different automobiles sold in the United States with the intent of seeing whether they could develop a multiple regression model to explain the variation in highway miles per gallon. A number of different independent variables were collected. The following regression output (with some values missing) was recently presented to the editors by the magazine's analysts:

Based on this output and your understanding of multiple regression analysis, which of the independent variables is not considered statistically significant if the test is conducted at the 0.05 level of statistical significance? A) All the variables in the model are statistically significant. B) None of the variables in the model is statistically significant. C) Torque and price as tested D) Cylinders, torque, and 0 to 60 Answer: D Diff: 3 Keywords: regression, multiple, slope, significant Section: 15-1 Introduction to Multiple Regression Analysis Outcome: 4

105) A multiple regression is shown for a data set of yachts where the dependent variable is the price in thousands of dollars.

Based on this output, which of the independent variables appear to be significantly helping to predict the price of a yacht, using a 0.10 level of significance? A) Age B) Age and length C) Rooms and Nav. Equip. D) Length and Nav. Equip. Answer: B Diff: 2 Keywords: regression, multiple, t-tests, slope Section: 15-1 Introduction to Multiple Regression Analysis Outcome: 4 106) Which of the following is not an indication of potential multicollinearity problems? A) The sign on the standard error of the estimate is positive. B) A sign on a regression slope coefficient is negative when the sign on the correlation coefficient was positive. C) The standard error of the estimate increases when a variable enters the model in the presence of other independent variables. D) An independent variable goes from being statistically significant to being insignificant when a new variable is added to the model. Answer: A Diff: 2 Keywords: regression, multiple, multicollinearity Section: 15-1 Introduction to Multiple Regression Analysis Outcome: 5

107) Under what circumstances does the variance inflation factor signal that multicollinearity may be a problem? A) When the value of VIF exceeds the size of the sample from which the regression model was developed B) When the VIF value is approximately 1.0 C) When the VIF is greater than or equal to 5 D) When the VIF is a negative value Answer: C Diff: 2 Keywords: regression, multiple, multicollinearity, inflation, variance Section: 15-1 Introduction to Multiple Regression Analysis Outcome: 5 108) Which of the following regression output values is used in computing the variance inflation factors? A) The standard error of the estimate B) The regression intercept value C) The F critical value from the F distribution for the appropriate number of degrees of freedom and the appropriate level of significance D) The R-squared value Answer: D Diff: 2 Keywords: regression, multiple, inflation, R-square Section: 15-1 Introduction to Multiple Regression Analysis Outcome: 5 109) Which of the following statements is true? A) If the confidence interval estimate for the regression slope coefficient, based on the sample information, crosses over zero, the true population regression slope coefficient could be zero. B) R-square will tend to be smaller than the adjusted R-squared values when insignificant independent variables are included in the model. C) The y-intercept will usually be negative in a multiple regression model when the regression slope coefficients are predominately positive. D) None of the above Answer: A Diff: 2 Keywords: regression, multiple, confidence interval, slope Section: 15-1 Introduction to Multiple Regression Analysis Outcome: 5

110) A multiple regression is shown below for a data set of yachts where the dependent variable is the price of the boat in thousands of dollars.

Given this information, what percentage of variation in the dependent variable is explained by the regression model? A) Approximately 68 percent B) About 83 percent C) About 37 percent D) About 60 percent Answer: A Diff: 2 Keywords: regression, multiple, R-square Section: 15-1 Introduction to Multiple Regression Analysis Outcome: 5

111) A multiple regression is shown for a data set of yachts where the dependent variable is the price in thousands of dollars.

Given this information, which of the following is true regarding the slope coefficient for Age, where Age represents how many years old the yacht is? A) On average the price of the yacht falls by $1.778 per year. B) On average the yacht is 1.778 years older per $1000 price change. C) On average the price of the yacht falls by $1778 per year. D) On average the price of the yacht increases by $1778 per year. Answer: C Diff: 2 Keywords: regression, multiple, slope coefficient Section: 15-1 Introduction to Multiple Regression Analysis Outcome: 5

112) A multiple regression is shown for a data set of yachts where the dependent variable is the price in thousands of dollars.

Given this information, what is the null hypothesis for testing the overall model? A) H0 : β0 = β1 = β2 = β3 = β4 = 0 B) H0 : β1 = β2 = β3 = β4 = 0 C) H0 : β1 = 0 D) H0 : β0 = 0 Answer: B Diff: 2 Keywords: regression, multiple, F-test, null hypothesis Section: 15-1 Introduction to Multiple Regression Analysis Outcome: 5

113) A multiple regression is shown for a data set of yachts where the dependent variable is the price in thousands of dollars.

Given this information, which is correct regarding the test of the overall model using the 0.10 level of significance? A) The overall model does not have significant ability to predict the price of a yacht because p-value = .163 is greater than 0.10 B) The overall model has significant ability to predict the price of a yacht because p-value = 0.163 is greater than 0.10 C) The overall model does not have significant ability to predict the price of a yacht because p-value = .001 is less than 0.10 D) The overall model has significant ability to predict the price of a yacht because p-value = .001 is less than 0.10 Answer: D Diff: 2 Keywords: regression, multiple, F-test, p-value Section: 15-1 Introduction to Multiple Regression Analysis Outcome: 4 114) In a multiple regression analysis involving 15 independent variables and 200 observations, SST = 800 and SSE = 240. The adjusted coefficient of determination is A) 0.15 B) 0.50 C) 0.66 D) 0.70 Answer: C Diff: 2 Keywords: regression, multiple, adjusted R-square Section: 15-1 Introduction to Multiple Regression Analysis Outcome: 4

115) If a decision maker wishes to develop a regression model in which the University Class Standing is a categorical variable with 5 possible levels of response, then he will need to include how many dummy variables? A) 5 B) 4 C) 1 D) 3 Answer: B Diff: 2 Keywords: regression, multiple, dummy Section: 15-2 Using Qualitative Independent Variables Outcome: 6 116) Which of the following statements is true? A) Dummy variables are used to incorporate categorical variables into a regression model. B) You should use one fewer dummy variables than are categories for the qualitative variable in question. C) It is appropriate to compute a correlation coefficient for the relationship between a dependent variable and a dummy variable. D) All of the above are true. Answer: D Diff: 2 Keywords: regression, multiple, dummy, categorical, correlation Section: 15-2 Using Qualitative Independent Variables Outcome: 6

117) The editors of a national automotive magazine recently studied 30 different automobiles sold in the United States with the intent of seeing whether they could develop a multiple regression model to explain the variation in highway miles per gallon. A number of different independent variables were collected. Included in these were two variables described as follows: Car Type (categorical) 1 = Four door sedan 2 = Luxury car 3 = Compact truck 4 = Full size truck 5 = Sports car

Whether Car has ABS Brakes (categorical) 1 = All wheel ABS 2 = Rear wheel ABS 3 = No ABS

If these two variables are to be included in a regression model, how many additional variables will be needed? A) The model explains nearly 63 percent of the variation in the dependent variable. B) If tested at the 0.05 significance level, the overall model would be considered statistically significant. C) The variable X1 has a slope coefficient that is significantly different from zero if tested at the 0.05 level of significance. D) All of the above are true. Answer: A Diff: 2 Keywords: regression, multiple, dummy, R-square, slope Section: 15-2 Using Qualitative Independent Variables Outcome: 6

118) The following multiple regression output was generated from a study in which two independent variables are included. The first independent variable (X1) is a quantitative variable measured on a continuous scale. The second variable (X2) is qualitative coded 0 if Yes, 1 if No.

Based on this information, which of the following statements is true? A) The model explains nearly 63 percent of the variation in the dependent variable B) If tested at the 0.05 significance level, the overall model would be considered statistically significant. C) The variable X1 has a slope coefficient that is significantly different from zero if tested at the 0.05 level of significance. D) All of the above are true. Answer: D Diff: 2 Keywords: regression, multiple, dummy, R-square, slope Section: 15-2 Using Qualitative Independent Variables Outcome: 6

119) A regression equation that predicts the price of homes in thousands of dollars is t = 24.6 + 0.055x1 3.6x2, where x2 is a dummy variable that represents whether the house in on a busy street or not. Here x2 = 1 means the house is on a busy street and x2 = 0 means it is not. Based on this information, which of the following statements is true? A) On average, homes that are on busy streets are worth $3600 less than homes that are not on busy streets. B) On average, homes that are on busy streets are worth $3.6 less than homes that are not on busy streets. C) On average, homes that are on busy streets are worth $3600 more than homes that are not on busy streets. D) On average, homes that are on busy streets are worth $3.6 more than homes that are not on busy streets. Answer: A Diff: 2 Keywords: regression, multiple, dummy Section: 15-2 Using Qualitative Independent Variables Outcome: 6

120) A study has recently been conducted by a major computer magazine publisher in which the objective was to develop a multiple regression model to explain the variation in price of personal computers. Three quantitative independent variables were used along with one qualitative variable. The qualitative variable was coded 1 if the computer included a monitor, 0 otherwise. The following computer printout shows the final output.

Based on this information, and with a 0.05 level of significance, which of the following conclusions can be justified? A) Knowing whether the computer comes with a monitor or not is a significant factor in explaining the variation in price of the computer. B) There are substantial multicollinearity effects in this regression model. C) The only significant variable in the model at the .05 level of significance is Hard Drive Capacity. D) Removing the Monitor dummy variable would reduce the standard error of the estimate considerably. Answer: C Diff: 2 Keywords: regression, multiple, slope, t-value Section: 15-2 Using Qualitative Independent Variables Outcome: 6

121) Golf handicaps are used to allow players of differing abilities to play against one another in a fair match. Recently a sample of golfers was selected in an effort to develop a model for explaining the difference in handicaps. One independent variable of interest is the number of rounds played per year. Another is whether or not the player is using an "original" name brand club or a copy. In recent years, a number of smaller golf club manufacturers have attempted to copy major golf club designs and sell "copies" of original clubs such as the Big Bertha by Calloway. To incorporate the type of club used, which of the following methods could be used? A) Create a dummy variable called "Club Used" and code it "O" for original and "C" for copy. B) Create a dummy variable called "Club Used" and code it 1 for copy and 0 for original. C) Create a dummy variable called Club Used" and code it 1 for original and 0 for copy. D) Either B or C would work. Answer: D Diff: 2 Keywords: regression, multiple, dummy Section: 15-2 Using Qualitative Independent Variables Outcome: 6

122) Golf handicaps are used to allow players of differing abilities to play against one another in a fair match. Recently a sample of golfers was selected in an effort to develop a model for explaining the difference in handicaps. One independent variable of interest is the number of rounds played per year. Another is whether or not the player is using an "original" name brand club or a copy. In recent years, a number of smaller golf club manufacturers have attempted to copy major golf club designs and sell "copies" of original clubs such as the Big Bertha by Calloway. The resulting regression analysis containing both Rounds Played and a Dummy variable for Club Used is shown as follows:

Given this information, which of the following statements is not correct? A) The overall regression model is insignificant at the alpha = .05 level. B) The Club Dummy variable is statistically significant in the model meaning that knowing that a player used an original club or copy is of value in knowing the player's handicap. C) The two independent variables do not explain a statistically significant portion of the variation in golf handicap. D) All of the above are not correct. Answer: B Diff: 2 Keywords: regression, multiple, dummy Section: 15-2 Using Qualitative Independent Variables Outcome: 6

123) A decision maker is considering including two additional variables into a regression model that has as the dependent variable, Total Sales. The first additional variable is the region of the country (North, South, East, or West) in which the company is located. The second variable is the type of business (Manufacturing, Financial, Information Services, or Other). Given this, how many additional variables will be incorporated into the model? A) 2 B) 6 C) 8 D) 9 Answer: B Diff: 2 Keywords: regression, multiple, dummy Section: 15-2 Using Qualitative Independent Variables Outcome: 6

124) A multiple regression was conducted to predict the price of yachts in thousands of dollars. A dummy variable was included to indicate whether or not the yacht has a flying bridge, where 0 means "no" and 1 means "yes."

Which of the following statements is correct using the 0.10 level of significance? A) Having a flying bridge significantly increases the price of a yacht by an average of $17.7, given the other variables present. B) Having a flying bridge significantly increases the price of a yacht by an average of $17,708, given the other variables present. C) We can tell that 17 out of 20 yachts have a flying bridge. D) Whether or not the yacht has a flying bridge does not significantly affect the price of a yacht, given the other variables present. Answer: D Diff: 2 Keywords: regression, multiple, dummy, qualitative, t-test, p-value Section: 15-2 Using Qualitative Independent Variables Outcome: 6

125) In a multiple regression, the dependent variable is house value (in '000$) and one of the independent variables is a dummy variable, which is defined as 1 if a house has a garage and 0 if not. The coefficient of the dummy variable is found to be 5.4 but the t-test reveals that it is not significant at the 0.05 level. Which of the following is true? A) A garage increases the house value by $5,400. B) A garage increases the house value by $5,400, holding all other independent variables constant. C) The house value remains the same with or without a garage. D) We need to include other dummy variables. Answer: C Diff: 3 Keywords: regression, multiple, dummy, qualitative, t-test Section: 15-2 Using Qualitative Independent Variables Outcome: 6 126) A forecasting model of the following form was developed: y = B0 + B1xj + B2

+ B3

+ε

Which of the following best describes the form of this model? A) Quadratic model B) 3rd degree polynomial model C) 3rd level regression model D) Tri-slope regression model Answer: B Diff: 2 Keywords: regression, multiple, polynomial, curvilinear Section: 15-3 Working With Nonlinear Relationships Outcome: 7 127) Which of the following would best describe the situation that a second-degree polynomial regression equation would be used to model? A) An exponential growth trend B) A cosine function C) A parabola D) It depends on the number of independent variables. Answer: C Diff: 2 Keywords: regression, multiple, parabola, polynomial Section: 15-3 Working With Nonlinear Relationships Outcome: 7

128) Assume that a time-series plot takes the form of that shown in the following graph:

Given this plot, which of the following models would likely give the best fit? A) = b0b1 + b1t B) = b0 + b1t + b1t2 + b1t3 C) = b0 + b1t + b1t2 D) = b0 + b1t + b1t2 + b3t3 + b4t4 Answer: B Diff: 2 Keywords: regression, multiple, polynomial Section: 15-3 Working With Nonlinear Relationships Outcome: 7

129) The following regression output is from a multiple regression model:

The variables t, t2, and t3 represent the t, t-squared, and t-cubed respectively where t is the indicator of time from periods t = 1 to t = 20. Which of the following best describes the type of forecasting model that has been developed? A) A complete third-order polynomial model B) A tri-variate smoothed regression model C) A nonlinear trend model D) A qualitative regression model Answer: A Diff: 2 Keywords: regression, multiple, polynomial, complete Section: 15-3 Working With Nonlinear Relationships Outcome: 7 130) Interaction exists in a multiple regression model when: A) one independent variable affects the relationship between another independent variable and the dependent variable. B) multicollinearity is present in a regression model. C) the regression model is overall insignificant. D) a polynomial model used. Answer: A Diff: 2 Keywords: regression, multiple, interaction Section: 15-3 Working With Nonlinear Relationships Outcome: 7

131) Second-order polynomial models: A) always curve upward. B) always curve downward. C) can curve upward or downward depending on the data. D) measure interaction between variables. Answer: C Diff: 2 Keywords: regression, multiple, polynomial Section: 15-3 Working With Nonlinear Relationships Outcome: 7 132) The following model: y = β0 + β1x1 + β2x2 + β3x1x2 + ε A) is a linear model with interaction. B) is a second order polynomial model. C) is a composite model. D) is a convex model. Answer: A Diff: 2 Keywords: regression, multiple, polynomial, parabola Section: 15-3 Working With Nonlinear Relationships Outcome: 7 133) Which of the following is not considered to be a stepwise regression technique? A) Forward selection regression B) Optimal variable entry and removal regression C) Backward elimination D) Standard stepwise regression Answer: B Diff: 1 Keywords: regression, multiple, stepwise, forward, backward, standard Section: 15-4 Stepwise Regression Outcome: 8 134) Standard stepwise regression A) is the same as forward selection. B) involves trying more regressions that the best subsets method. C) always finds the best regression model. D) combines attributes of both forward selection and backward elimination. Answer: D Diff: 2 Keywords: regression, multiple, stepwise Section: 15-4 Stepwise Regression Outcome: 8

135) A decision maker has five potential independent variables with which to build a regression model to explain the variation in the dependent variable. At step 1, variable x3 enters the regression model. Which of the following indicates which of the four remaining independent variables will be next to enter the model? A) The variable that has the next highest correlation with the dependent variable B) The variable that will provide the next largest value for the slope coefficient C) The variable with the highest coefficient of partial determination D) Can't be determined without seeing the correlation matrix. Answer: C Diff: 2 Keywords: regression, multiple, coefficient, partial determination Section: 15-4 Stepwise Regression Outcome: 8 136) Which of the following is the difference between forward selection and standard stepwise regression? A) In the standard stepwise regression, variables that were added at earlier steps can be removed at later steps, which is not the case with forward selection. B) The standard stepwise approach will generally produce a regression model with a higher R-square value than the forward selection approach. C) Forward selection begins by selecting the variable with the highest correlation with the dependent variable and then proceeds to select subsequent variables in order of their F-to-enter value, while standard stepwise selects the variables in the order specified by the decision maker and then removes them from the model as needed. D) There are no appreciable differences between the two methods, just different names for the same technique. Answer: A Diff: 2 Keywords: regression, multiple, stepwise, forward Section: 15-4 Stepwise Regression Outcome: 8 137) Which of the following is an advantage of using stepwise regression compared to just entering all the independent variables at one time? A) Stepwise will generally produce a model with a larger R-square value. B) The standard error of the estimate for a model constructed with stepwise regression will be larger than the one generated when all variables are entered at the same time. C) The stepwise regression allows the decision maker to observe the effects of multicollinearity more easily than when all the variables are entered at one time. D) There are no advantages of using stepwise regression over entering all variables at one time. Answer: C Diff: 2 Keywords: regression, multiple, stepwise, multicollinearity Section: 15-4 Stepwise Regression Outcome: 8

138) The editors of a national automotive magazine recently studied 30 different automobiles sold in the United States with the intent of seeing whether they could develop a multiple regression model to explain the variation in highway miles per gallon. A number of different independent variables were collected. The following regression output is the result of using a forward selection stepwise regression approach.

Based on the regression output, which of the following statements is true? A) There is a multicollinearity problem since the standard error of the estimate actually increased when the second variable, "Price as Tested," entered the model. B) The R-square value increased when the second variable entered the model. C) Neither variable in the model is statistically significant at the alpha = 0.05 level. D) The reason that only two variables entered the model is due to the small sample size used in this study. Answer: B Diff: 2 Keywords: regression, multiple, R-square, stepwise Section: 15-4 Stepwise Regression Outcome: 8

139) The editors of a national automotive magazine recently studied 30 different automobiles sold in the United States with the intent of seeing whether they could develop a multiple regression model to explain the variation in highway miles per gallon. A number of different independent variables were collected. The following regression output is the result of using a forward selection stepwise regression approach.

Which of the following might explain why no other independent variables entered the model? A) No other variable had a correlation with the dependent variable that was close to 1.0. B) None of the remaining variables had a positive correlation with y. C) The remaining variables must be nearly perfectly correlated with the two variables already in the model. D) Given the two variables already in the model, none of the others could add significantly to the percentage of variation in the y variable that would be explained by the model. Answer: D Diff: 2 Keywords: regression, multiple, stepwise Section: 15-4 Stepwise Regression Outcome: 8

140) To determine the aptness of the model, which of the following would most likely be performed? A) Check to see whether the residuals have a constant variance B) Determine whether the residuals are normally distributed C) Check to determine whether the regression model meets the assumption of linearity D) All of the above Answer: D Diff: 2 Keywords: regression, multiple, residuals, linearity, normal Section: 15-5 Determining the Aptness of the Model Outcome: 9 141) The assumption that the errors or residuals are independent is best checked by: A) looking at a normal probability plot of the residuals. B) looking a scatter plot of each x versus y. C) looking at a residual plot versus x and checking for curvature. D) looking at a plot of the residuals versus time and checking for trends. Answer: A Diff: 2 Keywords: regression, multiple, residuals, linearity Section: 15-5 Determining the Aptness of the Model Outcome: 9 142) If the residuals have a constant variance, which of the following should be evident? A) The residuals should have a variance equal to zero for all levels of the independent variable. B) The plot of the residuals against each x variable should show that the spread in the residuals is about the same at all levels of each of the independent variables. C) The average residual should be about zero and the residual standard deviation should be approximately 1. D) None of the above Answer: B Diff: 2 Keywords: regression, multiple, residuals, variance Section: 15-5 Determining the Aptness of the Model Outcome: 9

143) Consider the following residual plot.

Given this plot, what conclusion should be reached? A) There appears to be no basis for concluding that the relationship between the x and y variable is not linear. B) The assumption of constant variance seems to be supported by this plot. C) Both A and B are true. D) Neither A nor B are true. Answer: C Diff: 2 Keywords: regression, multiple, plot, residual, linearity, variance Section: 15-5 Determining the Aptness of the Model Outcome: 9 144) A standardized residual is: A) equal to the sum of the residuals divided by n-1. B) the ratio of each residual divided by an estimate for the standard deviation of the residuals. C) a value that is normally distributed with a mean equal to zero and a standard deviation equal to one. D) None of the above Answer: B Diff: 2 Keywords: regression, multiple, residual, standardized Section: 15-5 Determining the Aptness of the Model Outcome: 9

145) Based on the residual plot below, which of the following is correct?

The above residual plot shows: A) linearity and nonconstant variance. B) nonlinearity and constant variance. C) linearity and constant variance. D) nonlinearity and nonconstant variance. Answer: D Diff: 2 Keywords: regression, multiple, residual, normal Section: 15-5 Determining the Aptness of the Model Outcome: 9

146) Explain the difference between forward stepwise regression (standard stepwise), forward selection, and all possible subsets regression approaches. Answer: For lack of a better term, "normal" regression analysis is the regression approach that includes all the independent variables at one time in the regression model. The regression equation is calculated such that the sum of squared errors (SSE) is minimized. However, there are other approaches for constructing the regression models. The forward selection is one in which we start with no x variables in the model. The first x variable added is the one most highly correlated with the dependent variable. Then, the next variable is the one that can do the most to explain the yet unexplained variation in the dependent variable given that the first variable is in the model. The process continues until either all x variables are included or until none of the remaining variables can meet the entering criteria. The standard stepwise, also called forward stepwise regression model, enters variables in the same manner as the forward selection approach. The difference between the two is that with the standard stepwise approach, a variable that was entered on a previous step can actually be removed if its contribution is diminished after including other independent variables. This approach has an entry criteria and an exit criteria that the software checks at each step to determine which variables to add and which variables to remove. The best subsets regression constructs the regression models for all combinations of regression models starting with all the models with one x variable, then all the possible models with two independent variables and so forth. Criteria such as highest R-square, lowest standard error of the estimate, and the Cp are used to determine which of the possible models is preferred. Diff: 2 Keywords: regression, multiple, stepwise, forward, subsets, standard Section: 15-4 Stepwise Regression Outcome: 8

147) The following regression output is the result of a multiple regression application in which we are interested in explaining the variation in retail price of personal computers based on three independent variables, CPU speed, RAM, and hard drive capacity. However, some of the regression output has been omitted.

Given this information and your knowledge of multiple regression, what percentage of variation in the dependent variable is explained by the three independent variables in the model? Answer: R-squared is the percentage of variation in the dependent variable that is explained by the independent variables in the model. The formula for R-square is: R2 = where SSR = sum of squares regression and SST = total sum of squares. These values are provided in the ANOVA section of the regression output above. Thus, we get: R2 =

= .6961

We could also find the R-square value by squaring the multiple correlation. This is given as: R2 = (multiple correlation coefficient)2 = (8343)2 R2 = .6961 Thus, the three independent variables explain approximately 69.6 percent of the variation in the dependent variable. Diff: 2 Keywords: regression, multiple, variation, R-square Section: 15-1 Introduction to Multiple Regression Analysis Outcome: 3

148) The following regression output is the result of a multiple regression application in which we are interested in explaining the variation in retail price of personal computers based on three independent variables, CPU speed, RAM, and hard drive capacity. However, some of the regression output has been omitted.

Given this information and your knowledge of multiple regression, what is the adjusted R-square value? Answer: R-squared is the percentage of variation in the dependent variable that is explained by the independent variables in the model. The formula for R-square is: R2 = where SSR = sum of squares regression and SST = total sum of squares. These values are provided in the ANOVA section of the regression output above. Thus, we get: = .6961 R2 = The adjusted R-square measures the percentage of explained variation after adjusting for the relationship between sample size and the number of independent variables in the model. This is an important statistic since R-square will always increase when more independent variables are added to the model, even if those variables are of no significance in explaining the dependent variable. Adjusted R-square takes this into account. The formula for adjusted R-square is: = 1 - (1 - R2)(

)

where n = 36 and k = 3 Therefore, we get: = 1 - (1 - .6961)(

) = .6676

The adjusted R-square value is .6676

Diff: 2 Keywords: regression, multiple, adjusted, R-square Section: 15-1 Introduction to Multiple Regression Analysis Outcome: 4

149) The following regression output is the result of a multiple regression application in which we are interested in explaining the variation in retail price of personal computers based on three independent variables, CPU speed, RAM, and hard drive capacity. However, some of the regression output has been omitted.

Given this information and your knowledge of multiple regression, what is the value for the standard error of the estimate? Answer: The standard error of the estimate is a measure of the deviation of the fitted y values around the actual y values. The formula for the standard error is: SEE = However, the MSE is not given directly in the above output. We can compute it using the following formula: MSE= where SSE = Sum of Squares Error or Sum of Squares Residual and n = 36 and k = 3. The Sum of Squares Residual is not given either, but can be computed as: SSE = SST - SSR and TSS and SSR are given in the ANOVA section of the regression output. Thus, we get: SSE = 49,327,250 - 34,335,283 = 14,991,967 Therefore, we get: MSE = (

) = 468,499

Then, the standard error of the estimate is: = = 684.47 SSE = Diff: 3 Keywords: regression, multiple, standard, error Section: 15-1 Introduction to Multiple Regression Analysis Outcome: 4 15-70 Copyright © 2018 Pearson Education, Inc.

150) The following multiple regression was conducted to attempt to predict the price of yachts based on the independent variables shown.

Given this information and your knowledge of multiple regression, conduct the appropriate test to determine whether the overall regression model is statistically significant at the 0.05 level of significance using the critical value method. Answer: The null and alternative hypotheses to be tested are: H0 : B1 = B2 = B3 = B4 = 0.0 Hα : not all Bj are equal to zero The test statistic for testing the overall model is the F value from the ANOVA table, which is F = 14.38, which is associated with the F-distribution with 4 and 15 degrees of freedom. So the critical value is F = 3.056 and the decision rule is: Reject the null if F > 3.056. Since 14.38 is much larger than 3.056, we clearly reject the null hypothesis and conclude that the overall model has at least some significant ability to predict y, which is to say that at least one of the four independent variables is significantly related to y. Diff: 2 Keywords: regression, multiple, F-test, model, null, alternative, hypothesis Section: 15-1 Introduction to Multiple Regression Analysis Outcome: 4

151) The following multiple regression was conducted to attempt to predict the price of yachts based on the independent variables shown.

Given this information and your knowledge of multiple regression, determine which, if any, of the four independent variables are statistically significant in explaining the variation in the dependent variable. Use a 0.05 level of significance and use the p-value method. Answer: For each of the four independent variables the hypotheses are: H0 : Bj = 0.0 Hα : Bj ≠ 0.0 For example, for Age we have: H0 : B1 = 0.0 Hα : B1 ≠ 0.0 Where the null assumes that the slope coefficient for Age is zero. The test statistic for Age is t = -2.59 and the corresponding p-value is 0.02. Using the p-value method we reject the null if p < α, so we reject the null and conclude that Age's slope coefficient is significantly different from 0, which means that Age is helping to predict Price given that the other 3 independent variables are in the model. For Length and for Nav. Equip. the same conclusion is reached because each of them has a p-value that is smaller than 0.05. However, for Flying Bridge p-value = 0.27 is greater than 0.05, so we cannot reject the null hypothesis. So even though Flying Bridge has the largest slope coefficient, we are concluding that this slope of 17.7 is not significantly different from zero. In the presence of the other three independent variables, Flying Bridge is not significantly contributing to predicting y and we should consider removing it from the model.

In summary, the first three independent variables (everything except Flying Bridge) are significant in this model. Diff: 2 Keywords: regression, multiple, slope, t-test, hypothesis, null, alternative Section: 15-1 Introduction to Multiple Regression Analysis Outcome: 4

152) The following regression output is the result of a multiple regression application in which we are interested in explaining the variation in retail price of personal computers based on three independent variables, CPU speed, RAM, hard drive capacity, and-monitor included (1=Yes, 0=No).

Given this output, what is the variable, Monitor, called? Also, given the other variables in the model, is Monitor significant in explaining the variation in the dependent variable? Test using a .05 level of alpha. Answer: The variable Monitor included is a qualitative variable since the actual value for this variable is either yes or no depending on whether the listed price includes a monitor or not. In order to include this variable, only one dummy is needed and it has been coded 1 = yes and 0 = no. Given this, the regression result shows that the coefficient for the dummy variable, Monitor, is 19.004. To test this we establish the following null and alternative hypotheses and test statistic: H0 : B1 = 0.0 Ha : B1 ≠ 0.0 The test statistic is: t=

= .0799

We compare this value to the table t value for 0.05 level of statistical significance and n-k-1 = 31 degrees of freedom. The critical value from the table is 2.0395. Since 0.0799 < 2.0395, we do not reject the null hypothesis. Thus, in the presence of the other variables, knowing whether the computer has a monitor included or not is of no statistical significance in explaining the variation in the dependent variable. Diff: 2 Keywords: regression, multiple, dummy, hypothesis, null, alternative Section: 15-2 Using Qualitative Independent Variables Outcome: 6

Business Statistics, 9e (Groebner/Shannon/Fry) Chapter 16 Analyzing and Forecasting Time-Series Data 1) Model specification is the process of determining how well a forecasting model fits the past data. Answer: FALSE Diff: 1 Keywords: forecasting, model, specification Section: 16-1 Introduction to Forecasting and Time-Series Data Outcome: 1 2) If a manager is planning for an expansion of the factory, a forecast model with a long-term planning horizon would probably be used. Answer: TRUE Diff: 1 Keywords: forecast, horizon Section: 16-1 Introduction to Forecasting and Time-Series Data Outcome: 1 3) If the historical data on which the model is being built consist of weekly data, the forecasting period would also be weekly. Answer: TRUE Diff: 1 Keywords: forecast, data Section: 16-1 Introduction to Forecasting and Time-Series Data Outcome: 1 4) The forecasting interval is the unit of time for which forecasts are made. Answer: FALSE Diff: 2 Keywords: forecast, interval, period Section: 16-1 Introduction to Forecasting and Time-Series Data Outcome: 1 5) A stockbroker at a large brokerage firm recently analyzed the combined annual profits for all firms in the airline industry. One time-series component that may have been present in these annual data was a seasonal component. Answer: FALSE Diff: 2 Keywords: forecast, seasonal Section: 16-1 Introduction to Forecasting and Time-Series Data Outcome: 1

6) In order for a time series to exhibit a seasonal component, the data must be measured in periods as short or shorter than quarterly. Answer: TRUE Diff: 2 Keywords: time series, forecast, seasonal Section: 16-1 Introduction to Forecasting and Time-Series Data Outcome: 1 7) The time-series component that implies a long-term upward or downward pattern is called the trend component. Answer: TRUE Diff: 1 Keywords: time series, forecast, trend Section: 16-1 Introduction to Forecasting and Time-Series Data Outcome: 1 8) An annual time series cannot exhibit a seasonal component. Answer: TRUE Diff: 2 Keywords: time series, seasonal Section: 16-1 Introduction to Forecasting and Time-Series Data Outcome: 1 9) In a recent meeting, a manager indicated that sales tend to be higher during October, November, and December and lower in the spring. In making this statement, she is indicating that sales for the company are cyclical. Answer: FALSE Diff: 2 Keywords: time series, forecast, cyclical Section: 16-1 Introduction to Forecasting and Time-Series Data Outcome: 1 10) While virtually all time series exhibit a random component, not all time series exhibit other components. Answer: TRUE Diff: 1 Keywords: time series, forecast, random Section: 16-1 Introduction to Forecasting and Time-Series Data Outcome: 1

11) The Gilbert Company chief financial officer has been tracking annual sales for each of the company's three divisions for the past 10 years. At a recent meeting, he pointed to the annual data and indicated that it clearly showed the seasonality associated with its business. Given the data, this statement may have been very appropriate. Answer: FALSE Diff: 2 Keywords: time series, forecast, seasonality Section: 16-1 Introduction to Forecasting and Time-Series Data Outcome: 1 12) Harrison Hollow, an upscale eatery in Atlanta, tracks its sales on a daily basis. Recently, the manager stated that sales over the past three weeks have been very cyclical. Given the data she has, this statement is not a reasonable one to make. Answer: TRUE Diff: 2 Keywords: time series, forecast, cycle, cyclical Section: 16-1 Introduction to Forecasting and Time-Series Data Outcome: 1 13) Some stocks are referred to as cyclical stock because they tend to be in favor for several years and then out of favor for several years. This is a correct use of the term cyclical. Answer: TRUE Diff: 2 Keywords: time series, forecast, cyclical Section: 16-1 Introduction to Forecasting and Time-Series Data Outcome: 1 14) If a forecast period is one month, then we will provide a new forecast each month. Answer: TRUE Diff: 1 Keywords: time series, forecast, cycle, cyclical Section: 16-1 Introduction to Forecasting and Time-Series Data Outcome: 1 15) Stock analysts have recently stated in a meeting on Wall Street that over the past 50 years there have been periods of high market prices followed by periods of lower prices but over time prices have moved upwards. Given their statement, stock prices most likely exhibit only trend and cyclical components. Answer: FALSE Diff: 2 Keywords: time series, forecast, trend, cycle, cyclical Section: 16-1 Introduction to Forecasting and Time-Series Data Outcome: 1

16) To compare one value measured at one point in time with other values measured at different points in time, index numbers must be used. Answer: TRUE Diff: 1 Keywords: time series, forecast, index Section: 16-1 Introduction to Forecasting and Time-Series Data Outcome: 2 17) Two common unweighted indexes are the Paasche Index and the Laspeyres Index. Answer: FALSE Diff: 1 Keywords: time series, forecast, index Section: 16-1 Introduction to Forecasting and Time-Series Data Outcome: 2 18) One of the basic tools for creating a trend-based forecasting model is regression analysis. Answer: TRUE Diff: 1 Keywords: time series, forecast, trend, regression Section: 16-2 Trend-Based Forecasting Techniques Outcome: 4 19) A time-series graph shows that monthly income data have decreased gradually over the past 5 years. Given this, if a linear trend model is used to forecast future monthly income, the sign on the regression slope coefficient will be negative. Answer: TRUE Diff: 1 Keywords: time series, forecast, trend, slope Section: 16-2 Trend-Based Forecasting Techniques Outcome: 4 20) You are given the following linear trend model: Ft = 345.60 - 200.5(t). The forecast for period 15 is approximately -2,662. Answer: TRUE Diff: 1 Keywords: time series, forecast, trend Section: 16-2 Trend-Based Forecasting Techniques Outcome: 4 21) From an annual time series of a company's employee income, the linear trend model Ft = 165 - 54(t) has been developed. This means that on average income has been increasing by 165 per year. Answer: FALSE Diff: 1 Keywords: time series, forecast, trend, positive Section: 16-2 Trend-Based Forecasting Techniques Outcome: 4

22) You are given the following linear trend model: Ft = 345.60 - 200.5(t). This model implies that in year 1, the dependent variable had a value of 145.1. Answer: FALSE Diff: 2 Keywords: time series, forecast, trend Section: 16-2 Trend-Based Forecasting Techniques Outcome: 4 23) Renton Industries makes replacement parts for the automobile industry. As part of the company's capacity planning, it needs a long-range total demand forecast. The following information was generated based on 10 years of historical data on total number of parts sold each year.

Based on this information, the percent of variation in the number of parts sold that is explained by the linear trend model is approximately 90.9. Answer: TRUE Diff: 2 Keywords: time series, forecast, trend, variation, R-square Section: 16-2 Trend-Based Forecasting Techniques Outcome: 4

24) Renton Industries makes replacement parts for the automobile industry. As part of the company's capacity planning, it needs a long-range total demand forecast. The following information was generated based on 10 years of historical data on total number of parts sold each year.

Based on this information we can conclude that the linear trend model explains a significant proportion of the variation in the number of parts sold, because the p-value is much smaller than any reasonable α that we might use. Answer: TRUE Diff: 2 Keywords: time series, forecast, trend, p-value, alpha, variation Section: 16-2 Trend-Based Forecasting Techniques Outcome: 4

25) Renton Industries makes replacement parts for the automobile industry. As part of the company's capacity planning, it needs a long-range total demand forecast. The following information was generated based on 10 years of historical data on total number of parts sold each year.

Based on this information, we can conclude that sales on average have been growing by more than 48 thousand annually. Answer: TRUE Diff: 2 Keywords: time series, forecast, trend Section: 16-2 Trend-Based Forecasting Techniques Outcome: 4 26) In using simple linear regression to find the linear trend in an annual time series from 2000 to 2015, the values 2000, 2001, etc. are used as the values of the independent variable t when the regression is conducted. Answer: FALSE Diff: 2 Keywords: time series, forecast, trend Section: 16-2 Trend-Based Forecasting Techniques Outcome: 4 27) In comparing two or more forecasting models, the MAD value is useful in determining how successful the models were in fitting historical data. Answer: TRUE Diff: 1 Keywords: time series, forecast, trend, MAD Section: 16-2 Trend-Based Forecasting Techniques Outcome: 4

28) In measuring forecast errors, the MAD and the square root of the MSE will provide similar (but not identical) values, in that both provide a measure of the "typical" amount of error in forecasts. Answer: TRUE Diff: 2 Keywords: time series, forecast, trend, MAD, MSE Section: 16-2 Trend-Based Forecasting Techniques Outcome: 4 29) Gibson, Inc. is a holding company that owns several businesses. One such business is a truck sales company. To help in managing this operation, managers at Gibson have collected sales data for the past 20 years showing the number of trucks sold each year. They have then developed the linear trend forecasting model shown as follows:

Based on this information, it appears that the time series has a strong positive linear trend component. Answer: TRUE Diff: 2 Keywords: time series, forecast, trend, linear Section: 16-2 Trend-Based Forecasting Techniques Outcome: 4

30) Gibson, Inc. is a holding company that owns several businesses. One such business is a truck sales company. To help in managing this operation, managers at Gibson have collected sales data for the past 20 years showing the number of trucks sold each year. They have then developed the linear trend forecasting model shown as follows:

Based on this information, the fitted value for year 1 is about 99. Answer: TRUE Diff: 2 Keywords: time series, forecast, trend Section: 16-2 Trend-Based Forecasting Techniques Outcome: 4 31) Forecast bias measures the average amount of error per forecast, so a positive value means that forecasts tended to be too low. Answer: TRUE Diff: 2 Keywords: time series, forecast, trend, bias Section: 16-2 Trend-Based Forecasting Techniques Outcome: 4 32) One of the disadvantages of a regression-based linear trend forecasting model is that the forecast errors are computed for time periods that were used in developing the forecasting model and thus do not truly measure the forecasting ability of the model. Answer: TRUE Diff: 2 Keywords: time series, forecast, trend, disadvantage Section: 16-2 Trend-Based Forecasting Techniques Outcome: 4

33) The reason for using split samples in developing a forecasting model is to eliminate the potential for bias in the resulting model. Answer: FALSE Diff: 2 Keywords: time series, forecast, trend, split samples, bias Section: 16-2 Trend-Based Forecasting Techniques Outcome: 4 34) It is possible to use linear regression analysis to develop a forecasting model for nonlinear data if we can effectively transform the data. Answer: TRUE Diff: 2 Keywords: time series, forecast, trend, transform Section: 16-2 Trend-Based Forecasting Techniques Outcome: 4 35) A scatter plot of time series data with time on the vertical axis is an effective means of assessing whether the series is linear or nonlinear. Answer: FALSE Diff: 2 Keywords: time series, forecast, trend, plot Section: 16-2 Trend-Based Forecasting Techniques Outcome: 4 36) A seasonal index is a statistic that is computed from time-series data to indicate the effect of the seasonality in the time-series data. Answer: TRUE Diff: 1 Keywords: time series, forecast, trend, seasonal, index Section: 16-2 Trend-Based Forecasting Techniques Outcome: 4

37) The Baker's Candy Company has been in business for three years. The quarterly sales data for the company are shown as follows: Quarter 1 2 3 4 5 6 7 8 9 10 11 12

Sales 3000 3500 4000 2900 3100 3400 4100 2800 3000 3600 4000 2800

Based on this information, the data reflect both a linear trend and seasonal components. Answer: FALSE Diff: 2 Keywords: time series, forecast, trend, season Section: 16-2 Trend-Based Forecasting Techniques Outcome: 4 38) In a time series with monthly sales data, a spring quarter seasonal index of 1.21 can be interpreted to mean that sales tend to be 21 percent higher in the spring quarter when compared to the other quarters. Answer: TRUE Diff: 2 Keywords: time series, forecast, seasonal, index Section: 16-2 Trend-Based Forecasting Techniques Outcome: 4

39) The Baker's Candy Company has been in business for three years. The quarterly sales data for the company are shown as follows: Quarter 1 2 3 4 5 6 7 8 9 10 11 12

Sales 3000 3500 4000 2900 3100 3400 4100 2800 3000 3600 4000 2800

As a first step in computing a seasonal index, the four-period moving average corresponding to the midpoint between periods 2 and 3 is 3,350. Answer: TRUE Diff: 2 Keywords: time series, forecast, trend, moving average, seasonal Section: 16-2 Trend-Based Forecasting Techniques Outcome: 4

40) The Baker's Candy Company has been in business for three years. The quarterly sales data for the company are shown as follows: Quarter 1 2 3 4 5 6 7 8 9 10 11 12

Sales 3000 3500 4000 2900 3100 3400 4100 2800 3000 3600 4000 2800

Based on these data, the seasonal index for quarters 3, 7, and 11 is approximately 1.61 Answer: FALSE Diff: 3 Keywords: time series, forecast, trend, seasonal index Section: 16-2 Trend-Based Forecasting Techniques Outcome: 4 41) Recently, a manager for a major retailer computed the following seasonal indexes: Fall Qtr 1.21

Winter Qtr 0.85

Spring Qtr 0.91

Summer Qtr

Note that the index for Summer Qtr is missing. However, it can be determined that the index for that period is approximately 1.03 Answer: TRUE Diff: 2 Keywords: time series, forecast, trend, seasonal index Section: 16-2 Trend-Based Forecasting Techniques Outcome: 4 42) If a time series involves monthly data there will be a total of 12 seasonal indexes. Answer: TRUE Diff: 1 Keywords: time series, forecast, trend, seasonal index Section: 16-2 Trend-Based Forecasting Techniques Outcome: 4

43) A seasonally unadjusted forecast is one that is made from seasonal data without any adjustment for the seasonal component in the time series. Answer: TRUE Diff: 1 Keywords: time series, forecast, trend, seasonal Section: 16-2 Trend-Based Forecasting Techniques Outcome: 4 44) Recently, a manager for a major retailer computed the following seasonal indexes: Fall Qtr 1.61

Winter Qtr 0.75

Spring Qtr 0.91

Summer Qtr 0.73

The manager then developed the following least squares trend model based on the past five years of quarterly data: = 200 + 11.5t. Based on this, the seasonally adjusted forecast for quarter 25, which is the winter quarter, is 489.11 Answer: FALSE Diff: 2 Keywords: time series, forecast, trend, seasonal index Section: 16-2 Trend-Based Forecasting Techniques Outcome: 4 45) When using the multiplicative time-series model to determine the seasonal indexes, the first step is to isolate the seasonal and random components from the cyclical and trend components. Answer: TRUE Diff: 2 Keywords: time series, forecast, trend, seasonal, random, cyclical Section: 16-2 Trend-Based Forecasting Techniques Outcome: 4 46) Herb Criner, an analyst for the Folgerty Company, recently gave a report in which he stated that the annual sales forecast based on 20 years of annual sales data was done using a seasonally adjusted, trendbased forecasting technique. Given the information presented here, this statement has the potential to be credible. Answer: FALSE Diff: 2 Keywords: time series, forecast, trend, seasonal Section: 16-2 Trend-Based Forecasting Techniques Outcome: 4 47) To deseasonalize a time series, assuming a multiplicative model, the observed values are divided by the appropriate seasonal index. Answer: TRUE Diff: 2 Keywords: time series, forecast, deseasonalizing Section: 16-2 Trend-Based Forecasting Techniques Outcome: 4 16-14 Copyright © 2018 Pearson Education, Inc.

48) If the forecast errors are autocorrelated, this is a good indication that the model has been specified correctly. Answer: FALSE Diff: 2 Keywords: time series, forecast, autocorrelated, error Section: 16-2 Trend-Based Forecasting Techniques Outcome: 4 49) If the Durbin-Watson test leads you to reject the null hypothesis, then you are concluding that the forecast errors are positively autocorrelated. Answer: TRUE Diff: 2 Keywords: Durbin-Watson, null, hypothesis, autocorrelation Section: 16-2 Trend-Based Forecasting Techniques Outcome: 4 50) It is possible to conduct a statistical test for autocorrelation using the Durbin-Watson test and not be able to make a definitive conclusion about whether there is autocorrelation or not based on the data. Answer: TRUE Diff: 2 Keywords: Durbin-Watson, autocorrelation Section: 16-2 Trend-Based Forecasting Techniques Outcome: 4 51) Large values of the Durbin-Watson d statistic indicate that positive autocorrelation among the forecast errors exists. Answer: FALSE Diff: 2 Keywords: Durbin-Watson, autocorrelation, error Section: 16-2 Trend-Based Forecasting Techniques Outcome: 4 52) If you suspect that a nonlinear trend exists in your data, one way to deal with it in a trend-based forecasting application is to transform the independent variable, for example by squaring the time measure or maybe taking the square-root of the time measure. Answer: TRUE Diff: 2 Keywords: time series, forecast, trend, transform, nonlinear Section: 16-2 Trend-Based Forecasting Techniques Outcome: 4 53) The purpose of deseasonalizing a time series is that a strong seasonal pattern may make it difficult to see a trend in the time series. Answer: TRUE Diff: 1 Keywords: time series, trend, deseasonalizing Section: 16-2 Trend-Based Forecasting Techniques Outcome: 4 16-15 Copyright © 2018 Pearson Education, Inc.

54) If the Durbin-Watson d statistic has a value close to 2, there is reason to believe that there is no autocorrelation between the forecast errors. Answer: TRUE Diff: 2 Keywords: Durbin-Watson, autocorrelation, error Section: 16-2 Trend-Based Forecasting Techniques Outcome: 4 55) If a forecasting model produces forecast errors (residuals) that are negatively correlated, then we expect a negative residual to be followed by another negative residual to be followed by another negative residual and so forth. Answer: FALSE Diff: 2 Keywords: time series, forecast, residual, correlation Section: 16-2 Trend-Based Forecasting Techniques Outcome: 4 56) The reason for testing for the presence of autocorrelation in a regression-based trend forecasting model is that one assumption of the regression analysis is that the residuals are not correlated. Answer: TRUE Diff: 2 Keywords: time series, forecast, trend, autocorrelation, residuals Section: 16-2 Trend-Based Forecasting Techniques Outcome: 4 57) If the observed value in a time series for period 3 is yt = 128, and the seasonal index that applies to period 3 is 1.20, then the deseasonalized value for period 3 is 153.6 Answer: FALSE Diff: 2 Keywords: time series, forecast, seasonal, deseasonalize Section: 16-2 Trend-Based Forecasting Techniques Outcome: 4 58) The Durbin-Watson test for autocorrelation can be reliably applied to any sample sizes. Answer: FALSE Diff: 2 Keywords: Durbin-Watson, autocorrelation, error Section: 16-2 Trend-Based Forecasting Techniques Outcome: 4

59) An advantage of exponential smoothing techniques over a regression-based trend model is that the exponential smoothing model allows us to weigh each observation equally, thereby giving a fairer method of developing a forecast. Answer: FALSE Diff: 1 Keywords: time series, forecast, trend, exponential, smoothing Section: 16-3 Forecasting Using Smoothing Methods Outcome: 5 60) In a single exponential smoothing model, finding the forecast value for each period requires having the actual and forecasted values from the proceeding period. The first period should use an estimated average value based on previous estimates. Answer: FALSE Diff: 1 Keywords: time series, forecast, exponential, smoothing Section: 16-3 Forecasting Using Smoothing Methods Outcome: 5 61) In a single exponential smoothing model, one smoothing constant is used to weigh the historical data, and the model is of primary value when the data do not exhibit trend or seasonal components. Answer: TRUE Diff: 2 Keywords: time series, forecast, exponential, smoothing, trend, seasonal Section: 16-3 Forecasting Using Smoothing Methods Outcome: 5 62) In a single exponential smoothing model, a large value for the smoothing constant will result in greater smoothing of the data than will a smoothing constant close to zero. Answer: FALSE Diff: 2 Keywords: time series, forecast, exponential, smoothing Section: 16-3 Forecasting Using Smoothing Methods Outcome: 5 63) If a time series contains substantial irregular movement, the smoothing constant for a single exponential smoothing model that is close to 1.0 will result in forecasts that are not as smoothed out as those that would occur if a smaller smoothing constant was used. Answer: TRUE Diff: 2 Keywords: time series, forecast, exponential, smoothing, constant Section: 16-3 Forecasting Using Smoothing Methods Outcome: 5

64) In establishing a single exponential smoothing forecasting model, a starting point for the forecast value for period 1 is required. One method for arriving at this starting point is to use the first data point as the forecast for that period. If we do that, then the first data point should be ignored when computing measures of forecast error. Answer: TRUE Diff: 2 Keywords: time series, forecast, exponential, smoothing, constant Section: 16-3 Forecasting Using Smoothing Methods Outcome: 5 65) Because simple exponential smoothing models require a starting point for the first period forecast that will be arbitrary, it is important to have as much data as possible to dampen out the effect of the starting point. Answer: TRUE Diff: 2 Keywords: time series, forecast, exponential, smoothing, constant Section: 16-3 Forecasting Using Smoothing Methods Outcome: 5 66) The owners of Hal's Cookie Company have collected sales data for the past 8 months. These data are shown as follows: Month 1 2 3 4 5 6 7 8

Sales 100 130 90 120 100 80 120 90

Using a smoothing constant equal to 0.20 and starting forecast in period 1 of 100, the forecast value for period 9 is approximately 104.2. Answer: FALSE Diff: 2 Keywords: time series, forecast, exponential, smoothing Section: 16-3 Forecasting Using Smoothing Methods Outcome: 5

67) The owners of Hal's Cookie Company have collected sales data for the past 8 months. These data are shown as follows: Month 1 2 3 4 5 6 7 8

Sales 100 130 90 120 100 80 120 90

Using a smoothing constant of 0.4, the forecast value for period 3 is 112. Answer: TRUE Diff: 2 Keywords: time series, forecast, exponential, smoothing Section: 16-3 Forecasting Using Smoothing Methods Outcome: 5 68) Double exponential smoothing is used instead of single exponential smoothing when extra smooth forecasts are desired. Answer: FALSE Diff: 2 Keywords: time series, forecast, double, exponential smoothing Section: 16-3 Forecasting Using Smoothing Methods Outcome: 5 69) The owners of Hal's Cookie Company have collected sales data for the past 8 months. These data are shown as follows: Month 1 2 3 4 5 6 7 8

Sales 100 130 90 120 100 80 120 90

Using a starting forecast in period 1 of 100, the forecast bias over periods 2-8 is negative when a single exponential smoothing model is used with a smoothing constant of 0.20 Answer: FALSE Diff: 3 Keywords: time series, forecast, exponential, smoothing, bias Section: 16-3 Forecasting Using Smoothing Methods Outcome: 5 16-19 Copyright © 2018 Pearson Education, Inc.

70) If a smoothing model is applied for a stable time series, a smoothing constant of 0.20 would be an appropriate value. Answer: TRUE Diff: 2 Keywords: time series, forecast, exponential, smoothing, bias Section: 16-3 Forecasting Using Smoothing Methods Outcome: 5 71) The Morgan Company is interested in developing a forecast for next month's sales. It has collected sales data for the past 12 months. Month 1 2 3 4 5 6 7 8 9 10 11 12

Sales 100 120 90 150 170 150 180 205 220 195 230 240

After analyzing these data, if the company wishes to use exponential smoothing, it should employ a single smoothing model since there is evidence of a linear trend in the data. Answer: FALSE Diff: 2 Keywords: time series, forecast, exponential, smoothing, trend Section: 16-3 Forecasting Using Smoothing Methods Outcome: 5 72) In a double smoothing model, the second smoothing constant is introduced to account for the trend in the data if one exists. Answer: TRUE Diff: 1 Keywords: time series, forecast, exponential, smoothing, double, constant Section: 16-3 Forecasting Using Smoothing Methods Outcome: 5

73) If a time-series plot indicates that the data do not appear to exhibit a trend, then a double exponential smoothing model would likely be the most appropriate to use rather than simple exponential smoothing model. Answer: FALSE Diff: 2 Keywords: time series, forecast, exponential, smoothing, double, trend Section: 16-3 Forecasting Using Smoothing Methods Outcome: 5 74) If you suspect that your time-series data has a strong downward trend, you should set the beta smoothing constant at value fairly close to negative 1.0 Answer: FALSE Diff: 2 Keywords: time series, forecast, exponential, smoothing, beta, trend Section: 16-3 Forecasting Using Smoothing Methods Outcome: 5 75) In a double smoothing model, large values for the two smoothing constants will result in greater smoothing of the time series. Answer: FALSE Diff: 2 Keywords: time series, forecast, exponential, smoothing, double Section: 16-3 Forecasting Using Smoothing Methods Outcome: 5

76) The Morgan Company is interested in developing a forecast for next month's sales. It has collected sales data for the past 12 months. Month 1 2 3 4 5 6 7 8 9 10 11 12

Sales 100 120 90 150 170 150 180 205 220 195 230 240

After analyzing these data, if the company wishes to use double exponential smoothing with alpha = 0.20 and beta = 0.20, the starting values for the constant process and the trend process can be derived from a linear trend regression model by using the intercept and slope coefficient respectively. Answer: TRUE Diff: 2 Keywords: time series, forecast, exponential, smoothing, double Section: 16-3 Forecasting Using Smoothing Methods Outcome: 5 77) Prior to conducting double exponential smoothing a simple linear regression is conducted and the trend equation is = 42 + 38.3t, so the smoothed constant process value should be C0 = 38.3 and the smoothed trend value should be T0 = 42. Answer: FALSE Diff: 2 Keywords: time series, forecast, exponential, smoothing, double Section: 16-3 Forecasting Using Smoothing Methods Outcome: 5

78) The Morgan Company is interested in developing a forecast for next month's sales. It has collected sales data for the past 12 months. Month 1 2 3 4 5 6 7 8 9 10 11 12

Sales 100 120 90 150 170 150 180 205 220 195 230 240

Assuming that the company plans to use double exponential smoothing with starting values for the smoothed constant process value and smoothed trend value of 98.97 and 13.16 respectively, the MAD value for periods 2-12 is greater when alpha = 0.20 and beta = 0.20 than when alpha = 0.10 and beta = 0.10. Answer: TRUE Diff: 3 Keywords: time series, forecast, exponential, smoothing, double, alpha, beta Section: 16-3 Forecasting Using Smoothing Methods Outcome: 5 79) The process of selecting the forecasting technique to use in a particular application is called: A) model diagnosis. B) model specification. C) validity checking. D) model fitting. Answer: B Diff: 1 Keywords: time series, forecast, specification, model Section: 16-1 Introduction to Forecasting and Time-Series Data Outcome: 1 80) In a recent meeting, the marketing manager for a large hardware company stated that he needed to have a forecast prepared for the next three months. The three-month period is called: A) the forecasting horizon. B) the forecasting period. C) the planning time. D) a business cycle. Answer: A Diff: 2 Keywords: time series, forecast, horizon Section: 16-1 Introduction to Forecasting and Time-Series Data Outcome: 1 16-23 Copyright © 2018 Pearson Education, Inc.

81) In a recent meeting, the sales manager for a drug company stated that she needed to have a forecast prepared for each week for the next six weeks. The week in this case is the: A) forecasting horizon. B) forecasting period. C) planning time. D) forecasting interval. Answer: B Diff: 2 Keywords: time series, forecast, period Section: 16-1 Introduction to Forecasting and Time-Series Data Outcome: 1 82) The Cresswell Company updates its annual sales forecast every month as new sales data becomes available. The one-month update is called: A) the planning horizon. B) the forecast period. C) the forecast interval. D) the forecasting horizon. Answer: C Diff: 2 Keywords: time series, forecast, interval Section: 16-1 Introduction to Forecasting and Time-Series Data Outcome: 1 83) A time-series plot that exhibits a general increase in value from the early periods to the latest time periods: A) most likely is exhibiting a cyclical component. B) has a trend component present in the data. C) will probably start to decline in a few periods. D) has a seasonal component present in the data Answer: B Diff: 1 Keywords: time series, forecast, trend Section: 16-1 Introduction to Forecasting and Time-Series Data Outcome: 1 84) A company has a monthly time series that regularly shows sales being higher in the summer months. This is an example of which component? A) Trend B) Seasonal C) Cyclical D) Random Answer: B Diff: 2 Keywords: time series, forecast, seasonal, random Section: 16-1 Introduction to Forecasting and Time-Series Data Outcome: 1 16-24 Copyright © 2018 Pearson Education, Inc.

85) Which of the following time-series components can be identified when a company has 12 weeks of data beginning January and extending through March? A) Seasonal component B) Random component C) Trend component D) Both B and C Answer: D Diff: 2 Keywords: time series, forecast, random, trend Section: 16-1 Introduction to Forecasting and Time-Series Data Outcome: 1 86) Which of the following is true about index numbers? Index numbers are: A) used to measure the trend component. B) used to measure the seasonal component. C) used to measure the cyclical component. D) used to make a relative comparison of different time periods. Answer: D Diff: 2 Keywords: time series, index numbers Section: 16-1 Introduction to Forecasting and Time-Series Data Outcome: 2 87) Which of the following time-series components is almost always present in sales data for electronics companies? A) Random B) Trend C) Seasonal D) All of the above Answer: A Diff: 1 Keywords: time series, forecast, random Section: 16-1 Introduction to Forecasting and Time-Series Data Outcome: 1 88) Assume that the year 2000 is used as the index base period and that sales were 12 million in the year 2000. If sales were 18 million in the year 2006, the simple index number for the year 2006 is: A) 1.5 B) 6 million C) 150 D) 0.666 Answer: C Diff: 2 Keywords: time series, simple index number Section: 16-1 Introduction to Forecasting and Time-Series Data Outcome: 2

89) A company has recorded annual sales for the past 14 years and found the following linear trend model: = 5.23 + 144.60t. This means that: A) on average sales are increasing by of 5.23 per year. B) on average sales are increasing by of 144.6 per year. C) there is a seasonal component in the data. D) sales in year 1 were 5.23. Answer: B Diff: 2 Keywords: time series, linear trend Section: 16-2 Trend-Based Forecasting Techniques Outcome: 4 90) If a company has 12 years of annual sales data and is in need of a three-year forecast, which of the following forecasting techniques might be useful? A) Forecast generalization B) Seasonality processing C) Regression-based trend forecasting D) Random component analysis Answer: C Diff: 1 Keywords: time series, forecast, regression Section: 16-2 Trend-Based Forecasting Techniques Outcome: 4 91) A company has recorded 12 months of sales data for the past year and has found the linear trend equation is = 286 + 64.9t. Based on this information, which of the following is the forecast for period 13? A) 3718 B) 842.4 C) 350.8 D) 1129.7 Answer: D Diff: 2 Keywords: time series, forecast, regression Section: 16-2 Trend-Based Forecasting Techniques Outcome: 4

92) A company has recorded the 12 months of sales data for the past year. It wishes to use the regression approach to develop a trend-based approach for forecasting future months. The following data are provided: Month 1 2 3 4 5 6 7 8 9 10 11 12

Sales 200 260 210 340 450 230 300 500 360 600 460 700

Based on this information, which of the following is forecast for month 15? A) Approximately 692 B) About 596 C) Just over 825 D) None of the above Answer: A Diff: 3 Keywords: time series, forecast, regression Section: 16-2 Trend-Based Forecasting Techniques Outcome: 4 93) Which of the following statistical measures is used to help decision makers assess the potential for their model to provide usable forecasts? A) The mean absolute deviation B) The mean square error C) Both A and B D) Neither A nor B Answer: C Diff: 1 Keywords: time series, forecast, MAD, MSE, mean, squared, absolute, error, deviation Section: 16-2 Trend-Based Forecasting Techniques Outcome: 4

94) A company has developed a linear trend model to forecast monthly sales. The following data show the actual sales and the "fitted" sales for months 1-12. Month Sales 1 200 2 260 3 210 4 340 5 450 6 230 7 300 8 500 9 360 10 600 11 460 12 700

Predicted Sales 185 221 258 294 330 366 402 438 475 511 547 583

Based on these data, what is the mean square error for the linear trend model? A) Approximately 7,968 B) About 99,615 C) Nearly 18,000 D) None of the above Answer: A Diff: 2 Keywords: time series, forecast, mean squared error, MSE Section: 16-2 Trend-Based Forecasting Techniques Outcome: 4

95) A company has developed a linear trend model to forecast monthly sales. The following data show the actual sales and the "fitted" sales for months 1-12. Month Sales 1 200 2 260 3 210 4 340 5 450 6 230 7 300 8 500 9 360 10 600 11 460 12 700

Predicted Sales 185 221 258 294 330 366 402 438 475 511 547 583

Based on these data, what is the value for the mean absolute deviation for months 1-12? A) Approximately 975 B) About 81 C) Nearly 150 D) None of the above Answer: B Diff: 2 Keywords: time series, forecast, mean absolute deviation, MAD Section: 16-2 Trend-Based Forecasting Techniques Outcome: 4 96) Which of the following is true regarding the MSE and MAD in forecasting? A) The MAD will tend to be a larger value than the MSE. B) The MSE is much more useful than the MAD. C) The square root of the MSE provides a value that will be similar to the MAD. D) The MAD will be equal to the MSE. Answer: C Diff: 2 Keywords: time series, forecast, MSE, MAD Section: 16-2 Trend-Based Forecasting Techniques Outcome: 4

97) The following output is provided for a linear trend regression-based forecasting model based on 12 months of data:

Suppose that the actual sales for months 13-15 are: 720, 680, 800. Given this, which of the following is the forecast bias value for months 13-15? A) Approximately 77.7 B) Approximately 233.2 C) About -56.8 D) Just under 80 Answer: A Diff: 3 Keywords: time series, forecast, bias Section: 16-2 Trend-Based Forecasting Techniques Outcome: 4 98) Which of the following is true about forecast bias? A) It is the mean absolute deviation. B) It is equal to the square root of the MSE. C) When it is negative it means that the forecasts tended to be high. D) When it is positive it means that the forecasts tended to be high. Answer: C Diff: 2 Keywords: time series, forecast, bias Section: 16-2 Trend-Based Forecasting Techniques Outcome: 4

99) If you suspect that your trend forecasting model may have autocorrelated forecast errors, which the following should you compute? A) The MAD B) The standard error of the estimate C) The Durbin-Watson test statistic D) The correlation coefficient Answer: C Diff: 1 Keywords: time series, forecast, trend, Durbin-Watson, autocorrelation Section: 16-2 Trend-Based Forecasting Techniques Outcome: 4 100) Which of the following statements about the Durbin-Watson d statistic is true? A) It can assume any value between 0 and 4. B) It may assume any value between -4 and + 4. C) It may assume any positive value. D) It has a maximum value of 2. Answer: A Diff: 1 Keywords: time series, forecast, Durbin-Watson Section: 16-2 Trend-Based Forecasting Techniques Outcome: 4 101) Suppose an economist has developed a model for forecasting annual consumption, yt, as function of total labor income, x1t, and total property income, x2t based on 20 years on annual data. The following regression model has been developed: t= 7.81 + 0.91x1t + 0.57x2t with the standard error = 1.29 and the Durbin-Watson d statistic = 2.09. Using an alpha = .05, which of the following is the correct critical value for testing whether the residuals are autocorrelated? A) 1.20 and 1.41 B) 1.10 and 1.54 C) 1.08 and 1.53 D) Can't be determined without seeing the residuals. Answer: B Diff: 2 Keywords: time series, forecast, Durbin-Watson, autocorrelation Section: 16-2 Trend-Based Forecasting Techniques Outcome: 4

102) Suppose an economist has developed a model for forecasting annual consumption, yt, as function of total labor income, x1t , and total property income, x2t based on 20 years on annual data. The following regression model has been developed: t = 7.81 + 0.91x1t + 0.57x2t with the standard error = 1.29 and the Durbin-Watson d statistic = 2.09. Using an alpha = .05, which of the following conclusions should be reached? A) Conclude that no positive autocorrelation exists B) Conclude that significant positive autocorrelation exists C) Uncertain whether positive or negative autocorrelation exists D) No significant autocorrelation exists. Answer: A Diff: 3 Keywords: time series, forecast, autocorrelation, Durbin-Watson Section: 16-2 Trend-Based Forecasting Techniques Outcome: 4 103) After a linear forecasting model is found for a time series, if the Durbin-Watson statistic is less than dL this means that: A) no positive autocorrelation exists, the linear model is adequate. B) positive autocorrelation exists, a nonlinear model should be tried. C) no positive autocorrelation exists, a nonlinear model should be tried. D) positive autocorrelation exists, the linear model is adequate. Answer: B Diff: 2 Keywords: time series, forecast, residuals, Durbin-Watson Section: 16-2 Trend-Based Forecasting Techniques Outcome: 4 104) If time-series data exhibit a seasonal pattern, which of the following approaches could be used to compute season indexes? A) The exponential smoothing technique B) The multiplicative model C) The Pearson product-moment approach D) None of the above Answer: B Diff: 1 Keywords: time series, forecast, season, multiplicative Section: 16-2 Trend-Based Forecasting Techniques Outcome: 4

105) The Boxer Company has been in business since 1998. The following sales data are recorded by quarter for the years 2010-2012. Quarter Winter 10 Spring 10 Summer 10 Fall 10 Winter 11 Spring 11 Summer 11 Fall 11 Winter 12 Spring 12 Summer 12 Fall 12

Sales 50 70 100 60 60 70 120 80 70 90 140 100

Which of the following time-series components are present in these data? A) Trend component B) Seasonal component C) Random component D) All of the above Answer: D Diff: 2 Keywords: time series, forecast, trend, seasonal, random Section: 16-2 Trend-Based Forecasting Techniques Outcome: 4

106) The Boxer Company has been in business since 1998. The following sales data are recorded by quarter for the years 2010-2012. Quarter Winter 10 Spring 10 Summer 10 Fall 10 Winter 11 Spring 11 Summer 11 Fall 11 Winter 12 Spring 12 Summer 12 Fall 12

Sales 50 70 100 60 60 70 120 80 70 90 140 100

The managers at the company wish to determine the seasonal indexes for each quarter during the year. The first step in the process is to remove the seasonal and random components. To do this, they will begin by computing a four period moving average. What is the four-period moving average based on Winter 99 - Fall 99? A) 280 B) 70 C) 60 D) 55 Answer: B Diff: 1 Keywords: time series, forecast, moving average Section: 16-2 Trend-Based Forecasting Techniques Outcome: 4

107) The Boxer Company has been in business since 1998. The following sales data are recorded by quarter for the years 2010-2012. Quarter Winter 10 Spring 10 Summer 10 Fall 10 Winter 11 Spring 11 Summer 11 Fall 11 Winter 12 Spring 12 Summer 12 Fall 12

Sales 50 70 100 60 60 70 120 80 70 90 140 100

108) The Boxer Company has been in business since 1998. The following sales data are recorded by quarter for the years 2010-2012. Quarter Winter 10 Spring 10 Summer 10 Fall 10 Winter 11 Spring 11 Summer 11 Fall 11 Winter 12 Spring 12 Summer 12 Fall 12

Sales 50 70 100 60 60 70 120 80 70 90 140 100

The managers at the company wish to determine the seasonal indexes for each quarter during the year. The first step in the process is to remove the seasonal and random components. To do this, they will begin by computing a four-period moving average. They then compute the centered moving average. What is the next step in applying the multiplicative model? A) Compute the grand mean B) Divide the centered moving average by the forecasted value C) Compute the ratio-to-moving average value D) Normalize the data Answer: C Diff: 2 Keywords: time series, forecast, moving average, seasonal index Section: 16-2 Trend-Based Forecasting Techniques Outcome: 4 109) Which of the following is true about deseasonalizing data? A) The sum of the seasonal indexes should be 1.0. B) The time series is deseasonalized by multiplying each value by its seasonal index. C) Deseasonalizing a time series reduces autocorrelation. D) Deseasonalizing a time series can make it easier to see a trend. Answer: D Diff: 2 Keywords: time series, seasonal index, deseasonalizing Section: 16-2 Trend-Based Forecasting Techniques Outcome: 4

110) The Boxer Company has been in business since 1998. The following sales data are recorded by quarter for the years 2010-2012. Quarter Winter 10 Spring 10 Summer 10 Fall 10 Winter 11 Spring 11 Summer 11 Fall 11 Winter 12 Spring 12 Summer 12 Fall 12

Sales 50 70 100 60 60 70 120 80 70 90 140 100

The managers at the company wish to determine the seasonal indexes for each quarter during the year. The index for the summer quarter is: A) approximately 1.39 B) less than 1.1 C) about 087 D) approximately 159 Answer: A Diff: 3 Keywords: time series, forecast, seasonal index Section: 16-2 Trend-Based Forecasting Techniques Outcome: 4 111) A company has developed a linear trend regression model based on 16 quarters of data. The independent variable is the measure of time (t = 1 thru 16 where quarter 1 is winter quarter, 2 is spring, etc.). The company has also developed seasonal indexes for each quarter as follows: Winter 1.20

Spring 1.00

Summer 0.70

Fall 1.10

The linear trend forecast equation is: = 120 + 56t. Given this information, what is the seasonally unadjusted forecast for period 19? A) 1,064 B) 1,184 C) 828.80 D) 986.7 Answer: B Diff: 2 Keywords: time series, forecast, trend, regression Section: 16-2 Trend-Based Forecasting Techniques Outcome: 4 16-37 Copyright © 2018 Pearson Education, Inc.

112) A company has developed a linear trend regression model based on 16 quarters of data. The independent variable is the measure of time (t = 1 thru 16 where quarter 1 is winter quarter, 2 is spring, etc.). The company has also developed seasonal indexes for each quarter as follows: Winter 1.20

Spring 1.00

Summer 0.70

Fall 1.10

The linear trend forecast equation is: = 120 + 56t. Given this information, what is the seasonally adjusted forecast for period 19? A) 1,064 B) 1,184 C) 828.80 D) 986.7 Answer: C Diff: 2 Keywords: time series, forecast, trend, seasonal index Section: 16-2 Trend-Based Forecasting Techniques Outcome: 4 113) A company has developed a linear trend regression model based on 16 quarters of data. The independent variable is the measure of time (t = 1 thru 16 where quarter 1 is winter quarter, 2 is spring, etc.). The company has also developed seasonal indexes for each quarter as follows: Winter 1.20

Spring 1.00

Summer 0.70

Fall 1.10

The linear trend forecast equation is: = 120 + 56t. Given this information, which of the four quarters beginning next winter (period t = 17) will have the highest seasonally adjusted forecast? A) Winter B) Spring C) Summer D) Fall Answer: D Diff: 2 Keywords: time series, forecast, seasonal index, regression Section: 16-2 Trend-Based Forecasting Techniques Outcome: 4

114) The most useful tool to identify the trend component of a time series is A) moving averages. B) exponential smoothing. C) regression analysis. D) the MAD approach. Answer: C Diff: 2 Keywords: time series, forecast, seasonal index, regression Section: 16-2 Trend-Based Forecasting Techniques Outcome: 4 115) Which of the following forecasting methods allows the decision maker to weigh the past time series differently to make the model more sensitive to more recent data? A) Linear trend regression model B) Moving average model C) Exponential smoothing D) Deseasonalizing the time series Answer: C Diff: 1 Keywords: time series, forecast, exponential smoothing Section: 16-3 Forecasting Using Smoothing Methods Outcome: 5 116) In a simple exponential smoothing model, which of the following statements is true? A) The forecast values are determined by computing a moving average of the previous 4 data periods if we are working with quarterly data. B) The larger the smoothing constant, the more smoothing that takes place in the model. C) If the data contain a lot of random or irregular ups and downs, then a larger smoothing constant should be used in an attempt to model these fluctuations. D) More smoothing of the data will take place if a smoothing constant value close to zero is used. Answer: D Diff: 2 Keywords: time series, forecast, exponential smoothing, alpha Section: 16-3 Forecasting Using Smoothing Methods Outcome: 5 117) Which of the following approaches might be used to determine which value for the smoothing constant to use in a single exponential smoothing model? A) Select the value that produces the smallest MAD value B) Select the value that maximizes the MSE C) Use the value that gives the largest MAD value D) Use the value that smoothes the data the most Answer: A Diff: 2 Keywords: time series, forecast, exponential smoothing, MAD Section: 16-3 Forecasting Using Smoothing Methods Outcome: 5

118) Under which of the following conditions would you suggest that a double exponential smoothing model should be used instead of a single exponential smoothing model? A) When the time-series data exhibit a trend B) When there are two main time-series variables involved in the forecast application C) When the data exhibit only random variation D) When the data set is extremely uneven Answer: A Diff: 2 Keywords: time series, forecast, exponential smoothing, double Section: 16-3 Forecasting Using Smoothing Methods Outcome: 5 119) The Zocor Company is interested in forecasting period 13 sales for a product. It has 12 months of historical data. The following shows the data and the forecasted values for periods 1-12 using a single exponential smoothing model with a smoothing constant value equal to 0.20.

Month Sales 1 20 2 40 3 30 4 20 5 50 6 30 7 50 8 10 9 60 10 50 11 30 12 50

Forecast for Period t 20.0 20.0 24.0 25.2 24.2 29.3 29.5 33.6 28.9 35.1 38.1 36.5

Which of the following would be the forecast for period 13? A) 40.3 B) 39.2 C) 34.7 D) 36.5 Answer: B Diff: 2 Keywords: time series, forecast, exponential smoothing Section: 16-3 Forecasting Using Smoothing Methods Outcome: 5

120) The Zocor Company is interested in forecasting period 13 sales for a product. It has 12 months of historical data. The following shows the data and the forecasted valued for periods 1-12 using a single exponential smoothing model with a smoothing constant value equal to 0.20

Month Sales 1 20 2 40 3 30 4 20 5 50 6 30 7 50 8 10 9 60 10 50 11 30 12 50

Forecast for Period t 20.0 20.0 24.0 25.2 24.2 29.3 29.5 33.6 28.9 35.1 38.1 36.5

What is the value of the MAD for periods 2-12? A) About 19.6 B) Nearly 6 C) Approximately 15.4 D) Can't be determined without more information Answer: C Diff: 2 Keywords: time series, forecast, exponential smoothing, MAD, mean absolute deviation Section: 16-3 Forecasting Using Smoothing Methods Outcome: 5

121) A time series is shown below. Perform single exponential smoothing for this data set using α = 0.2 Period 1 2 3 4 5 6

Y 28 32 31 35 36 40

What is the value of the forecast for period 6? A) 40 B) 36 C) 31.5 D) 33.2 Answer: C Diff: 2 Keywords: time series, forecast, exponential smoothing Section: 16-3 Forecasting Using Smoothing Methods Outcome: 5

122) The Zocor Company is interested in forecasting period 13 sales for a product. It has 12 months of historical data. The following shows the data and the forecasted values for periods 1-12 using a single exponential smoothing model with a smoothing constant value equal to 0.20

Month Sales 1 20 2 40 3 30 4 20 5 50 6 30 7 50 8 10 9 60 10 50 11 30 12 50

Forecast for Period t 20.0 20.0 24.0 25.2 24.2 29.3 29.5 33.6 28.9 35.1 38.1 36.5

What is the forecast bias value for the model over periods 2-12? A) About 8.7 B) Approximately -11.9 C) About ± 8.9 D) None of the above Answer: A Diff: 2 Keywords: time series, forecast, exponential smoothing, bias Section: 16-3 Forecasting Using Smoothing Methods Outcome: 5

123) The Wilson Company is interested in forecasting demand for its XG-667 product for quarter 13 based on 12 quarters of data. The following shows the data and the double exponential smoothing model results for periods 1-12 using alpha = 0.20 and beta = 0.40.

Quarter 1 2 3 4 5 6 7 8 9 10 11 12

Demand 800 890 790 900 880 960 900 920 1040 1090 1200 1140

C 781.6 830.4 854.1 890.0 915.5 949.1 964.8 976.2 1004.8 1040.5 1095.1 1135.1

T 33.8 39.8 33.4 34.4 30.8 31.9 25.4 19.8 23.3 28.3 38.8 39.3

Forecast Time t 815.4 870.2 887.5 924.4 946.4 981.0 990.3 996.0 1028.2 1068.8 1133.8

Based on this information, which of the following is the forecast for period 13? A) About 1,345 B) Just under 1,300 C) Approximately 1,174 D) Nearly 1,225 Answer: C Diff: 2 Keywords: time series, forecast, exponential smoothing, double Section: 16-3 Forecasting Using Smoothing Methods Outcome: 5

124) The Wilson Company is interested in forecasting demand for its XG-667 product for quarter 13 based on 12 quarters of data. The following shows the data and the double exponential smoothing model results for periods 1-12 using alpha = 0.20 and beta = 0.40.

Quarter 1 2 3 4 5 6 7 8 9 10 11 12

Demand 800 890 790 900 880 960 900 920 1040 1090 1200 1140

C 781.6 830.4 854.1 890.0 915.5 949.1 964.8 976.2 1004.8 1040.5 1095.1 1135.1

T 33.8 39.8 33.4 34.4 30.8 31.9 25.4 19.8 23.3 28.3 38.8 39.3

Forecast Time t 815.4 870.2 887.5 924.4 946.4 981.0 990.3 996.0 1028.2 1068.8 1133.8

Based on this information, what is the MAD value for quarters 2-12? A) About 56.3 B) Approximately -16.7 C) Approximately 71.2 D) About 38.5 Answer: A Diff: 2 Keywords: time series, forecast, exponential smoothing, double, MAD, mean absolute deviation Section: 16-3 Forecasting Using Smoothing Methods Outcome: 5

125) The Wilson Company is interested in forecasting demand for its XG-667 product for quarter 13 based on 12 quarters of data. The following shows the data and the double exponential smoothing model results for periods 1-12 using alpha = 0.20 and beta = 0.40

Quarter 1 2 3 4 5 6 7 8 9 10 11 12

Demand 800 890 790 900 880 960 900 920 1040 1090 1200 1140

C 781.6 830.4 854.1 890.0 915.5 949.1 964.8 976.2 1004.8 1040.5 1095.1 1135.1

T 33.8 39.8 33.4 34.4 30.8 31.9 25.4 19.8 23.3 28.3 38.8 39.3

Forecast Time t 815.4 870.2 887.5 924.4 946.4 981.0 990.3 996.0 1028.2 1068.8 1133.8

Based on this information, what is the difference between the forecast for period 13 using smoothing constants of alpha = 0.20 and beta = 0.40 and smoothing constants of alpha = 0.10 and beta = 0.30? (Assume that the starting values for period 0 are C = 745 and T = 32.) A) About 108 units B) Approximately 9 units C) Just under 32 units D) About 85 units Answer: B Diff: 3 Keywords: time series, forecast, exponential smoothing, double Section: 16-3 Forecasting Using Smoothing Methods Outcome: 5

126) The Wilson Company is interested in forecasting demand for its XG-667 product for quarter 13 based on 12 quarters of data. The following shows the data and the double exponential smoothing model results for periods 1-12 using alpha = 0.20 and beta = 0.40.

Quarter 1 2 3 4 5 6 7 8 9 10 11 12

Demand 800 890 790 900 880 960 900 920 1040 1090 1200 1140

C 781.6 830.4 854.1 890.0 915.5 949.1 964.8 976.2 1004.8 1040.5 1095.1 1135.1

T 33.8 39.8 33.4 34.4 30.8 31.9 25.4 19.8 23.3 28.3 38.8 39.3

Forecast Time t 815.4 870.2 887.5 924.4 946.4 981.0 990.3 996.0 1028.2 1068.8 1133.8

Based on this information, which of the following statements is true? A) If beta is changed to 0.20 and alpha remains at 0.20, the MAD for periods 2-12 is reduced. B) If beta is changed to 0.20 and alpha remains at 0.20, the forecast for period 13 is slightly reduced. C) The forecast bias for periods 2-12 is approximately 6.18. D) All of the above are true. Answer: D Diff: 3 Keywords: time series, forecast, exponential smoothing, double, bias. Beta, alpha Section: 16-3 Forecasting Using Smoothing Methods Outcome: 5 127) Given the following time series data: Time Period (t) 1 2

Timer Series Value (Y t) 18 22

and using smoothing constants of alpha = 0.3, the exponential smoothing forecast for time period 3 is A) 18 B) 19.2 C) 20 D) 40 Answer: B Diff: 2 Keywords: time series, forecast, exponential smoothing Section: 16-3 Forecasting Using Smoothing Methods Outcome: 5

128) In analyzing a forecast model, what is the main advantage of using the MAD instead of the MSE? Answer: The MAD stands for mean absolute deviation. Its advantage is that the resulting value is in the same units as the time-series data. That is, if the time-series is in dollars, the MAD value is in dollars. The MSE stands for mean square error and is computed by averaging the squared errors. The resulting value is in squared units rather than the original units. Diff: 2 Keywords: time series, forecast, MAD, MSE, mean absolute deviation, mean squared error Section: 16-2 Trend-Based Forecasting Techniques Outcome: 4 129) A company's annual sales are shown below in thousands of dollars for a period of 10 years. Year 1 2 3 4 5 6 7 8 9 10

Sales 138 150 142 151 161 188 191 224 254 322

Plot the time series; find the linear regression model, and also the forecast value and error for each of the years. Also discuss whether you think a linear model appears to be appropriate. Answer: The time series plot shown below appears to somewhat nonlinear as sales appears to be increasing at an increasing rate.

Linear regression can be used to develop a trend-based forecasting model. The model will take the following form: = b0 + b1t The intercept and slope are calculated using the least squares equations:

b1 =

and b0 =

- b1

The results of the calculations are: = 94.07 + 17.82t. The forecast and error for each year, based on this model are shown below.

Based both on the time-series plot and the pattern in the error, it appears that a nonlinear model would probably fit this model better. The pattern in the errors refers to the fact that they are negative for the center values and positive for values on each end of the series, which means they are not randomly positive or negative. Diff: 2 Keywords: time series, forecast, plot, trend, errors Section: 16-2 Trend-Based Forecasting Techniques Outcome: 1

130) The All American Toy Company has a very seasonal sales pattern. Sales are high during the fall quarter, drop off substantially in the winter quarter and are more typical during spring and summer quarters. The following historical data exist for the past 16 quarters. Quarter Winter-09 Spring-09 Summer-09 Fall-09 Winter 10 Spring 10 Summer 10 Fall 10 Winter 11 Spring 11 Summer 11 Fall 11 Winter 12 Spring 12 Summer 12 Fall 12

Sales 3000 5000 5500 7000 3400 6000 6200 8100 3600 6700 6400 9000 4000 7000 7300 9900

Based on these data, develop a seasonally adjusted forecast for the four quarters of 2013 using a linear trend regression model. Answer: A time-series plot illustrates that there is a pronounced seasonal component as well as a trend component. The first step in developing a seasonally adjusted forecast for the four quarters of 2013 is to calculate the seasonal indexes for the four quarters. We can do this using the multiplicative model: yt = Tt × St × Ct × It. To compute the seasonal indexes, we go through several steps. First, for quarterly data, we compute a four-period moving average shown in Table 1:

Table 1

Quarter 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16

Sales 3000 5000 5500 7000 3400 6000 6200 8100 3600 6700 6400 9000 4000 7000 7300 9900

4 Period MA Centered MA Ration-to-MA 5125 5225 5475 5650 5925 5975 6150 6200 6425 6525 6600 6825 7050

5175 5350 5562.5 5787.5 5950 6062.5 6175 6312.5 6475 6562.5 6712.5 6937.5

1.06 1.31 0.61 1.04 1.04 1.34 0.58 1.06 0.99 1.37 0.60 1.01

Deseasonalized Data 5027.93 4823.15 5339.81 5223.88 5698.32 5787.78 6019.42 6044.78 6033.52 6463.02 6213.59 6716.42 6703.91 6752.41 7087.38 7388.06

Now, we need to compute the centered moving averages by averaging each successive pair of moving averages as shown in Table 1. The centered moving averages provide an estimate for the Tt × Ct portion of the multiplicative model. To isolate the seasonal and irregular portion, we compute the ratio-to-moving-average by dividing each y value by the centered moving average value as shown in Table 1. Note: We now have three ratio to moving average values for each quarter. The next step is to average these ratio-to-moving-averages for each quarter as follows:

average =

Summer 1.03

Fall 1.34

Winter 0.60

Spring 1.04

The averages represent the seasonal indexes for each quarter. For example, the index for Fall quarter is 1.34, indicating that fall sales are 134 percent of the "normal" quarter sales. These indexes can be used to adjust the forecast made for each quarter. Next, we need to deseasonalize the data by dividing each sales value by its appropriate seasonal index. The results are shown in Table 1 above. Linear regression can be used to develop a trend-based forecasting model. The results of the calculations are: = 4779.3 + 153.3t. To generate the forecast for periods 17-20, we substitute t = 17, t = 18, t = 19, and t = 20 into the forecast equation as follows: For Winter Quarter 2013 (t = 17)

= 4779.3 + 153.3(17) = 7385.4

For Spring Quarter 2013 (t = 18)

= 4779.3 + 153.3(18) = 7538.70

For Summer Quarter 2013 (t = 19)

= 4779.3 + 153.3(19) = 7692.0

For Fall Quarter 2013 (t = 20)

These forecasts are referred to as seasonally unadjusted forecasts. The final step is to adjust these for the seasonal component by multiplying each forecast by the associated seasonal index. Winter Quarter 2002: 7385.4 × 0.60 = 4431.24 Spring Quarter 2002: 7538.7 × 1.04 = 7840.25 Summer Quarter 2002: 7692.0 × 1.03 = 7922.76 Fall Quarter 2002: 7845.3 × 1.34 = 10512.70 These are the seasonally adjusted forecasts for the 2013 year. Diff: 3 Keywords: time series, forecast, seasonal indices, regression Section: 16-2 Trend-Based Forecasting Techniques Outcome: 4

131) The Norton Industrial Company has 12 quarters of historical sales data and is interested in forecasting sales for quarter 13. The following data are available: Quarter 1 2 3 4 5 6 7 8 9 10 11 12

Sales 500 400 700 300 400 300 500 300 600 400 500 200

Based on this data, use simple exponential smoothing with a smoothing constant value equal to 0.10 and a starting forecast value equal to the sales in quarter 1 to forecast sales for period 13. Answer: The simple exponential smoothing equation is: Ft+1 = F1 + α(yt -Ft) Given a starting forecast value for period 1 equal to sales in period 1 we get the following: Quarter 1 2 3 4 5 6 7 8 9 10 11 12

Sales 500 400 700 300 400 300 500 300 600 400 500 200

alpha =

.01

F(t) 500.0 500.0 490.0 511.0 489.9 480.9 462.8 466.5 449.9 464.9 458.4 462.6

F(t + 1) 500.0 490.0 511.0 489.9 480.9 462.8 466.5 449.9 464.9 458.4 462.6 436.3

Thus, the forecast for period 13 is 436.3 Diff: 3 Keywords: time series, forecast, exponential smoothing Section: 16-3 Forecasting Using Smoothing Methods Outcome: 5

132) The Norton Industrial Company has 12 quarters of historical sales data and is interested in forecasting sales for quarter 13. The following data are available: Quarter 1 2 3 4 5 6 7 8 9 10 11 12

Sales 500 400 700 300 400 300 500 300 600 400 500 200

Based on this data, using simple exponential smoothing with a smoothing constant value equal to 0.10 and a starting forecast value equal to the sales in quarter 1 to forecast sales for period 13, what is the MAD value for periods 2-12? Answer: The simple exponential smoothing equation is: Ft + 1 = F1 + α(yt -Ft) Given a starting forecast value for period 1 equal to sales in period 1 we get the following: Quarter 1 2 3 4 5 6 7 8 9 10 11 12

Sales 500 400 700 300 400 300 500 300 600 400 500 200

alpha =

0.1

F(t) 500.0 500.0 490.0 511.0 489.9 480.9 462.8 466.5 449.9 464.9 458.4 462.6

F(t + 1) 500.0 490.0 511.0 489.9 480.9 462.8 466.5 449.9 464.9 458.4 462.6 436.3

Thus, the forecast for period 13 is 436.3. The MAD is the mean absolute deviation that is computed as follows: MAD = The results are shown as follows:

Quarter 1 2 3 4 5 6 7 8 9 10 11 12

Sales 500 400 700 300 400 300 500 300 600 400 500 200

alpha =

0.1

F(t) 500.0 500.0 490.0 511.0 489.9 480.9 462.8 466.5 449.9 464.9 458.4 462.6

F(t + 1) 500.0 490.0 511.0 489.9 480.9 462.8 466.5 449.9 464.9 458.4 462.6 436.3

Error

Absolute Error

-100.00 210.00 -211.00 -89.90 -189.91 37.18 -166.54 150.12 -64.90 41.59 -262.56

100.00 210.00 211.00 89.90 180.91 37.18 166.54 150.12 64.90 41.59 262.56

MAD =

137.70

Diff: 3 Keywords: time series, forecast, exponential smoothing, MAD, mean absolute deviation Section: 16-3 Forecasting Using Smoothing Methods Outcome: 5

133) The All American Toy Company has records for quarterly sales for the past 4 years. These data are shown as follows: Quarter 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16

Sales 3000 5000 5500 7000 3400 6000 6200 8100 3600 6700 6400 9000 4000 7000 7300 9900

Given these data, develop a forecast for quarter 17 using a double exponential smoothing model with alpha = 0.20, beta = 0.30 and starting values for C and T based on the linear trend regression model for periods 1-16. Answer: The first step is to develop the linear trend model to obtain the starting values for C and T, which will be needed to get the double exponential smoothing model started. The model will take the following form: = b0 + b1t The intercept and slope are calculated using the least squares equations:

b1 =

and b0 =

- b1

The results of the calculations are: = 4117.5 + 236.9t Thus, the C0 = 4117.5 and T0 = 236.9

The double exponential smoothing equations are: C1 = αy1 + (1 - α)(Ct-1 + Tt-1) T1 = β(Ct - Ct-1) + (1 - β) Tt-1 Ft-1 - Ct + Ft The following shows the results of applying these equations along with the starting values for C and T. Quarter 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 alpha = beta =

Sales 3000 5000 5500 7000 3400 6000 6200 8100 3600 6700 6400 9000 4000 7000 7300 9900

C 4083.5 4391.3 4774.1 5423.9 5318.2 5638.6 5956.7 6605.7 6314.5 6538.8 6667.8 7275.0 6864.2 6963.8 7105.6 7750.7

T 155.6 201.3 255.7 373.9 230.1 527.2 275.4 387.5 183.9 196.0 175.9 305.3 90.5 93.2 701.8 269.0

F(t) 4239.2 4592.6 5029.8 5797.8 5548.3 5895.8 6232.1 6993.2 6498.4 6734.8 6843.7 7580.3 6954.7 7057.0 7213.3 8019.7

0.2 0.3

Therefore, the forecast for period 17 is 8,019.7 The following plot shows the forecasts and the actual sales data for periods 2-16. This plot illustrates how the double smoothing has picked up the trend in the data and has smoothed out the random fluctuations in the sales data.

Diff: 3 Keywords: time series, forecast, double, exponential smoothing Section: 16-3 Forecasting Using Smoothing Methods Outcome: 5

Business Statistics, 9e (Groebner/Shannon/Fry) Chapter 17 Introduction to Nonparametric Statistics 1) The Wilcoxon signed rank test is used for testing hypotheses about a population median if the data are at nominal level. Answer: FALSE Diff: 1 Keywords: nonparametric, Wilcoxon, rank Section: 17-1 The Wilcoxon Signed Rank Test for One Population Median Outcome: 1 2) The Wilcoxon signed rank test is used to test hypotheses about the population median. Answer: TRUE Diff: 1 Keywords: nonparametric, rank, Wilcoxon, median Section: 17-1 The Wilcoxon Signed Rank Test for One Population Median Outcome: 1 3) The basic logic of the Wilcoxon signed rank test is that if about half the data values fall above the hypothesized median, and about half fall below, the null hypothesis should not be rejected. Answer: TRUE Diff: 2 Keywords: nonparametric, rank, Wilcoxon, median Section: 17-1 The Wilcoxon Signed Rank Test for One Population Median Outcome: 1 4) The t-test is not appropriate for ordinal data or if the populations examined are not believed to be approximately normally distributed and other methods such as the Wilcoxon test can be used. Answer: TRUE Diff: 2 Keywords: nonparametric, Wilcoxon, sample size Section: 17-1 The Wilcoxon Signed Rank Test for One Population Median Outcome: 1 5) If a decision maker believes that the population is normally distributed and the data are known to be ratio level, then the either the t-test or the Wilcoxon signed rank test can be used to test null hypotheses about a single population mean. Answer: FALSE Diff: 2 Keywords: nonparametric, rank, Wilcoxon, t-test, ratio, normal Section: 17-1 The Wilcoxon Signed Rank Test for One Population Median Outcome: 1

6) The following null and alternative hypotheses are appropriate when using a Wilcoxon signed rank test. H0 : Population Median equals 14.5 HA : Population Median is not equal to 14.5 Answer: TRUE Diff: 2 Keywords: nonparametric, rank, Wilcoxon, hypothesis, null, alternative Section: 17-1 The Wilcoxon Signed Rank Test for One Population Median Outcome: 1 7) All Wilcoxon signed rank tests are two-tailed tests since we are testing whether the population median is the exact center of the population distribution. Answer: FALSE Diff: 1 Keywords: nonparametric, rank, Wilcoxon, median, two-tailed Section: 17-1 The Wilcoxon Signed Rank Test for One Population Median Outcome: 1 8) Recently, Major League Baseball officials stated that the median cost for a family of four to attend a baseball game including, parking, tickets, food, and drinks did not exceed $125.00. As long as the data are considered interval or ratio level, the t-test can be used to test MLB's claim. Answer: FALSE Diff: 2 Keywords: nonparametric, rank, Wilcoxon, t-test, interval, ratio Section: 17-1 The Wilcoxon Signed Rank Test for One Population Median Outcome: 1 9) The Wilcoxon signed rank test can be a one or two-tailed test similar to hypothesis tests for population means. Answer: TRUE Diff: 2 Keywords: nonparametric, rank, Wilcoxon, rejection region Section: 17-1 The Wilcoxon Signed Rank Test for One Population Median Outcome: 1

10) Recently, Major League Baseball officials stated that the median cost for a family of four to attend a baseball game including, parking, tickets, food, and drinks did not exceed $125.00. The following sample data were collected: Dollars Spent 142.00 99.00 134.00 175.00 100.00 225.00 80.00 Assuming that the test of the owners' claim is going to be conducted using an alpha = .05 level, the critical value is t = 1.9432. Answer: FALSE Diff: 2 Keywords: nonparametric, rank, Wilcoxon, critical value Section: 17-1 The Wilcoxon Signed Rank Test for One Population Median Outcome: 1 11) Recently, Major League Baseball officials stated that the median cost for a family of four to attend a baseball game including, parking, tickets, food, and drinks did not exceed $125.00. The following sample data were collected: Dollars Spent 142.00 99.00 134.00 175.00 100.00 225.00 80.00 Assuming that the test of the owners' claim is going to be conducted using an alpha = .05 level, the null hypothesis that the median cost does not exceed $125 should not be rejected. Answer: TRUE Diff: 3 Keywords: nonparametric, rank, Wilcoxon, hypothesis Section: 17-1 The Wilcoxon Signed Rank Test for One Population Median Outcome: 1

12) Managers for a company that produces a weight loss product claim that the median weight lost over six weeks for people who use this product will be at least 20 pounds. The following data were collected from a sample of nine people who used the product. Pounds Lost 6.00 22.00 14.00 5.00 30.00 12.00 7.00 21.00 25.00 In order to test the manager's claim, they will need to assume that the population is normally distributed. Answer: FALSE Diff: 2 Keywords: nonparametric, rank, Wilcoxon, normal Section: 17-1 The Wilcoxon Signed Rank Test for One Population Median Outcome: 1 13) Managers for a company that produces a weight loss product claim that the median weight lost over six weeks for people who use this product will be at least 20 pounds. The following data were collected from a sample of nine people who used the product. Pounds Lost 6.00 22.00 14.00 5.00 30.00 12.00 7.00 21.00 25.00 Based on these data, the test statistic is W = 12. Answer: TRUE Diff: 3 Keywords: nonparametric, rank, Wilcoxon, test statistic Section: 17-1 The Wilcoxon Signed Rank Test for One Population Median Outcome: 1

14) In conducting the Wilcoxon signed rank test, after collecting the sample data the next step is to find the sample median and subtract this value from each data value to obtain the deviations. Answer: FALSE Diff: 2 Keywords: nonparametric, rank, Wilcoxon, deviations Section: 17-1 The Wilcoxon Signed Rank Test for One Population Median Outcome: 1 15) In the Wilcoxon signed rank test for testing about a single population median, if the sample size is large (n > 20), the test statistic can be approximated by the standard normal distribution. Answer: TRUE Diff: 2 Keywords: nonparametric, rank, Wilcoxon, normal, test statistic Section: 17-1 The Wilcoxon Signed Rank Test for One Population Median Outcome: 1 16) In the Wilcoxon signed rank test using either small or large sample sizes, any value that equals the hypothesized median is discarded from the analysis. Answer: TRUE Diff: 2 Keywords: nonparametric, rank, Wilcoxon, median Section: 17-1 The Wilcoxon Signed Rank Test for One Population Median Outcome: 1 17) In a large sample test about a single population median, it is appropriate to employ the standard normal distribution so long as the population is also normally distributed. Answer: FALSE Diff: 2 Keywords: nonparametric, rank, Wilcoxon, normal distribution Section: 17-1 The Wilcoxon Signed Rank Test for One Population Median Outcome: 1 18) The procedure of the Wilcoxon signed rank test is the same for either small or large sample sizes. Answer: FALSE Diff: 1 Keywords: nonparametric, rank, Wilcoxon, median Section: 17-1 The Wilcoxon Signed Rank Test for One Population Median Outcome: 1 19) The Mann-Whitney U test is a nonparametric test that is used to test whether two related populations have the same median. Answer: FALSE Diff: 2 Keywords: nonparametric, rank, Mann-Whitney, median Section: 17-2 Nonparametric Tests for Two Population Medians Outcome: 2

20) The Mann-Whitney U test can be used to test whether two independent populations have the same median so long as the data are measured on at least an ordinal scale. Answer: TRUE Diff: 2 Keywords: nonparametric, rank, Mann-Whitney, ordinal, median Section: 17-2 Nonparametric Tests for Two Population Medians Outcome: 2 21) In employing the Mann-Whitney U test, the sample data from the two populations are first combined and the ranks of the data are determined, but we keep track of which population each ranked item came from. Answer: TRUE Diff: 2 Keywords: nonparametric, rank, Mann-Whitney Section: 17-2 Nonparametric Tests for Two Population Medians Outcome: 2 22) The logic behind the Mann-Whitney U test is that if the hypothesis is true that the populations have equal central locations, then the sum of the ranks from each population will be approximately equal. Answer: TRUE Diff: 2 Keywords: nonparametric, rank, Mann-Whitney Section: 17-2 Nonparametric Tests for Two Population Medians Outcome: 2 23) One of the assumptions of the Mann-Whitney U test is that the population distributions are the same for shape and spread. Answer: TRUE Diff: 2 Keywords: nonparametric, rank, Mann-Whitney, assumptions Section: 17-2 Nonparametric Tests for Two Population Medians Outcome: 2 24) In a Mann-Whitney U test, the test statistic will be equal to the sum of the ranks from sample one, or sample two, whichever is larger. Answer: FALSE Diff: 2 Keywords: nonparametric, rank, Mann-Whitney, test statistic Section: 17-2 Nonparametric Tests for Two Population Medians Outcome: 2 25) In a Mann-Whitney U test, if the sample sizes are large then the test statistic can be approximated by the student's t-distribution. Answer: FALSE Diff: 1 Keywords: nonparametric, rank, Mann-Whitney, sample size, t-distribution Section: 17-2 Nonparametric Tests for Two Population Medians Outcome: 2 17-6 Copyright © 2018 Pearson Education, Inc.

26) The Mann-Whitney U test is always a one-tailed test. Answer: FALSE Diff: 2 Keywords: nonparametric, rank, Mann-Whitney, rejection region Section: 17-2 Nonparametric Tests for Two Population Medians Outcome: 2 27) A claim was recently made that stated that the median income for male and female graduates is the same for those graduating with an MBA degree. In order to test this claim using the Mann-Whitney U test, it is not necessary to select an equal number of males and females. Answer: TRUE Diff: 2 Keywords: nonparametric, rank, Mann-Whitney, sample size Section: 17-2 Nonparametric Tests for Two Population Medians Outcome: 2 28) The critical value for a one-tailed Mann-Whitney U test with sample sizes of n1 = 10 and n2 = 10 is 23 for a 0.05 level of significance. Answer: FALSE Diff: 2 Keywords: nonparametric, rank, Mann-Whitney, critical value Section: 17-2 Nonparametric Tests for Two Population Medians Outcome: 2 29) A claim was recently made that stated that the median income for male and female graduates is the same for those graduating with a degree in operations management. The following sample data were collected: Males $42,000 $36,000 $40,000 $32,000 $50,000 $47,000 $47,000

Females $39,000 $39,000 $41,000 $42,000 $44,000 $38,000 $51,000

In employing the Mann-Whitney U test, the sum of the ranks for the males is 53.5 Answer: TRUE Diff: 3 Keywords: Mann-Whitney U test Section: 17-2 Nonparametric Tests for Two Population Medians Outcome: 2

30) A claim was recently made that stated that the median income for male and female graduates is the same for those graduating with a degree in operations management. The following sample data were collected: Males $42,000 $36,000 $40,000 $32,000 $50,000 $47,000 $47,000

Females $39,000 $39,000 $41,000 $42,000 $44,000 $38,000 $51,000

In employing the Mann-Whitney U test, the U test statistic for the males is 43.5 Answer: FALSE Diff: 3 Keywords: nonparametric, rank, Mann-Whitney, U test statistic Section: 17-2 Nonparametric Tests for Two Population Medians Outcome: 2 31) A claim was recently made that stated that the median income for male and female graduates is the same for those graduating with a degree in operations management. The following sample data were collected: Males $42,000 $36,000 $40,000 $32,000 $50,000 $47,000 $47,000

Females $39,000 $39,000 $41,000 $42,000 $44,000 $38,000 $51,000

In employing the Mann-Whitney U test, the U test statistic is 25.5 Answer: TRUE Diff: 3 Keywords: nonparametric, rank, Mann-Whitney, U test statistic Section: 17-2 Nonparametric Tests for Two Population Medians Outcome: 2

32) Recently, a study was done to determine whether the median speed on a section of highway is the same for cars versus trucks. A sample of 12 cars (n1 = 12) and 15 trucks (n2 = 15) was collected. If the Mann-Whitney U test is to be performed using an alpha = .05, the test critical U value is 49. Answer: TRUE Diff: 2 Keywords: nonparametric, rank, Mann-Whitney, critical value Section: 17-2 Nonparametric Tests for Two Population Medians Outcome: 2 33) Recently, a study was done to determine whether the median speed on a section of highway is the same for cars versus trucks. A sample of 12 cars (n1 = 12) and 15 trucks (n2 = 15) was collected. If the Mann-Whitney U test is to be performed using an alpha = .05 and if the U test statistic is calculated to be 68, the null hypothesis should be rejected. Answer: TRUE Diff: 2 Keywords: nonparametric, rank, Mann-Whitney, null, hypothesis Section: 17-2 Nonparametric Tests for Two Population Medians Outcome: 2 34) In conducting a Mann-Whitney U test when the sample size is greater than 20, the U test statistic can be assumed normally distributed. Answer: TRUE Diff: 2 Keywords: nonparametric, rank, Mann-Whitney, z-value, U test statistic Section: 17-2 Nonparametric Tests for Two Population Medians Outcome: 2 35) A large sample Mann-Whitney U test should be used when the sample sizes exceed 20. Answer: TRUE Diff: 1 Keywords: nonparametric, rank, Mann-Whitney, sample size Section: 17-2 Nonparametric Tests for Two Population Medians Outcome: 2 36) If a decision maker wishes to test to determine whether the medians are equal for two populations the Mann-Whitney U test is appropriate for either independent or dependent sampling situations. Answer: FALSE Diff: 1 Keywords: nonparametric, rank, Mann-Whitney, independent, dependent, sample Section: 17-2 Nonparametric Tests for Two Population Medians Outcome: 2

37) The Wilcoxon Matched-Pairs Signed rank test is an alternative to the paired sample t-test when we are unwilling to assume that the populations are normally distributed. Answer: TRUE Diff: 1 Keywords: nonparametric, rank, Wilcoxon, matched pairs Section: 17-2 Nonparametric Tests for Two Population Medians Outcome: 3 38) In order to determine whether the median distance for the X-Special golf ball exceeds the median distance for the best-selling golf ball, six golfers were selected and asked to hit each ball with their driver. The distance was recorded. The following data were observed. Golfer 1 2 3 4 5 6

X-Special 240 267 255 234 250 285

Best Seller 233 270 240 230 260 270

Based on these, the appropriate test is the Kruskal-Wallis one-way analysis of variance. Answer: FALSE Diff: 1 Keywords: nonparametric, rank, Kruskal-Wallis, ANOVA, analysis of variance Section: 17-2 Nonparametric Tests for Two Population Medians Outcome: 3 39) In order to determine whether the median distance for the X-Special golf ball exceeds the median distance for the best-selling golf ball, six golfers were selected and asked to hit each ball with their driver. The distance was recorded. The following data were observed. Golfer 1 2 3 4 5 6

X-Special 240 267 255 234 250 285

Best Seller 233 270 240 230 260 270

The appropriate null and alternative hypotheses are: H0 : 1 ≤ 2 H0 : 1 > 2 Answer: TRUE Diff: 2 Keywords: nonparametric, rank, null, alternative, hypothesis Section: 17-2 Nonparametric Tests for Two Population Medians Outcome: 3 17-10 Copyright © 2018 Pearson Education, Inc.

40) One of the assumptions associated with the Wilcoxon Matched-Pairs Signed Rank test is that the distribution of the population differences is symmetric about their median. Answer: TRUE Diff: 2 Keywords: nonparametric, rank, Wilcoxon, matched pairs, signed rank Section: 17-2 Nonparametric Tests for Two Population Medians Outcome: 3 41) In order to determine whether the median distance for the X-Special golf ball exceeds the median distance for the best-selling golf ball, six golfers were selected and asked to hit each ball with their driver. The distance was recorded. The following data were observed. Golfer 1 2 3 4 5 6

X-Special 240 267 255 234 250 285

Best Seller 233 270 240 230 260 270

Based on these data, and testing at an alpha = 0.025 level, the critical value for the Wilcoxon Matched Pairs Signed Rank test is 2. Answer: FALSE Diff: 2 Keywords: nonparametric, rank, Wilcoxon, matched pairs, signed rank Section: 17-2 Nonparametric Tests for Two Population Medians Outcome: 3 42) In conducting the Wilcoxon Matched-Pairs Signed Rank test, the difference between each pair of values must be found prior to conducting any ranking. Answer: TRUE Diff: 2 Keywords: nonparametric, rank, Wilcoxon, matched pairs, signed rank Section: 17-2 Nonparametric Tests for Two Population Medians Outcome: 3 43) When testing whether two paired populations have equal medians and the sample sizes are large, it is appropriate to convert the Wilcoxon Matched-Pairs Signed Rank test to a paired sample t-test. Answer: FALSE Diff: 1 Keywords: nonparametric, rank, Wilcoxon, matched pairs, signed rank, t-test Section: 17-2 Nonparametric Tests for Two Population Medians Outcome: 3

44) The distribution of T-values in the Wilcoxon Matched-Pairs Signed Rank test is approximately normal if the sample size (number of matched pairs) exceeds 25. Answer: TRUE Diff: 2 Keywords: nonparametric, Wilcoxon, matched pairs, signed rank Section: 17-2 Nonparametric Tests for Two Population Medians Outcome: 3 45) If a decision maker wishes to test whether four independent populations have the same center and is unwilling to make the assumption that the populations are normally distributed with equal variances, she can use the Kruskal-Wallis test. Answer: TRUE Diff: 1 Keywords: nonparametric, Kruskal-Wallis Section: 17-3 Kruskal-Wallis One-Way Analysis of Variance Outcome: 4 46) The Kruskal-Wallis test is used to test whether the centers of 3 or more populations are equal so long as that is the only possible difference between the population distributions. Answer: TRUE Diff: 1 Keywords: nonparametric, Kruskal-Wallis Section: 17-3 Kruskal-Wallis One-Way Analysis of Variance Outcome: 4 47) In using the Kruskal-Wallis test the sample sizes for each population must be equal. Answer: FALSE Diff: 2 Keywords: nonparametric, Kruskal-Wallis Section: 17-3 Kruskal-Wallis One-Way Analysis of Variance Outcome: 4 48) The critical value for a Kruskal Wallis test comes from the chi-square distribution. Answer: TRUE Diff: 2 Keywords: nonparametric, Kruskal-Wallis, F-distribution Section: 17-3 Kruskal-Wallis One-Way Analysis of Variance Outcome: 4

49) A recent study was conducted to determine if any of three suppliers of electronic components has a different median delivery time on special orders. To test this, five orders were given to each supplier and the delivery days were recorded. These data are shown as follows: Supplier 1 Supplier 2 Supplier 3 15 11 15 19 7 9 13 19 5 10 10 12 20 12 10 If a Kruskal-Wallis test is to be performed, the critical value for a test conducted using an alpha = .05 level is χ2 = 11.0705 Answer: FALSE Diff: 2 Keywords: nonparametric, Kruskal-Wallis, chi-squared Section: 17-3 Kruskal-Wallis One-Way Analysis of Variance Outcome: 4 50) A recent study was conducted to determine if any of three suppliers of electronic components has a different median delivery time on special orders. To test this, five orders were given to each supplier and the delivery days were recorded. These data are shown as follows: Supplier 1 Supplier 2 Supplier 3 15 11 15 19 7 9 13 19 5 10 10 12 20 12 10 If a Kruskal-Wallis test is to be performed, the number of degrees of freedom for determining the critical value is 2. Answer: TRUE Diff: 2 Keywords: nonparametric, Kruskal-Wallis, degrees of freedom Section: 17-3 Kruskal-Wallis One-Way Analysis of Variance Outcome: 4

51) A recent study was conducted to determine if any of three suppliers of electronic components has a different median delivery time on special orders. To test this, five orders were given to each supplier and the delivery days were recorded. These data are shown as follows: Supplier 1 Supplier 2 Supplier 3 15 11 15 19 7 9 13 19 5 10 10 12 20 12 10 If a Kruskal-Wallis test is to be performed, the sum of the rankings for Supplier 1 is 45. Answer: FALSE Diff: 2 Keywords: nonparametric, Kruskal-Wallis, rankings Section: 17-3 Kruskal-Wallis One-Way Analysis of Variance Outcome: 4 52) Assume that a Kruskal-Wallis test is being conducted to determine whether or not the medians of three populations are equal. The sum of rankings and the sample size for each group are below. Group 1 R1 = 60

Group 2 R2 = 36

Group 3 R3 = 24

n1 = 6

n2 = 5

n3 = 4

The value of the test statistic is H = 0.68 Answer: FALSE Diff: 2 Keywords: nonparametric, Kruskal-Wallis, test statistic Section: 17-3 Kruskal-Wallis One-Way Analysis of Variance Outcome: 4

53) The makers of furnace filters recently conducted a test to determine whether the median number of particulates that would pass through their four leading filters was the same. A random sample of 6 of each type of filter was used with the following data being recorded: Filter 1 40 25 70 47 55 88

Filter 2 100 110 89 67 77 102

Filter 3 165 90 40 200 103 110

Filter 4 55 20 90 105 90 120

The Kruskal-Wallis test can be used in this case since it requires that the sample sizes be equal. Answer: FALSE Diff: 1 Keywords: nonparametric, Kruskal-Wallis, sample size Section: 17-3 Kruskal-Wallis One-Way Analysis of Variance Outcome: 4 54) The makers of furnace filters recently conducted a test to determine whether the median number of particulates that would pass through their four leading filters was the same. A random sample of 6 of each type of filter was used with the following data being recorded: Filter 1 40 25 70 47 55 88

Filter 2 100 110 89 67 77 102

Filter 3 165 90 40 200 103 110

Filter 4 55 20 90 105 90 120

If the managers are unwilling to assume that the populations are normally distributed, the appropriate test in this case would be the Mann-Whitney U test. Answer: FALSE Diff: 1 Keywords: nonparametric, Kruskal-Wallis, Mann-Whitney Section: 17-3 Kruskal-Wallis One-Way Analysis of Variance Outcome: 4

55) The makers of furnace filters recently conducted a test to determine whether the median number of particulates that would pass through their four leading filters was the same. A random sample of 6 of each type of filter was used with the following data being recorded: Filter 1 40 25 70 47 55 88

Filter 2 100 110 89 67 77 102

Filter 3 165 90 40 200 103 110

Filter 4 55 20 90 105 90 120

If the Kruskal-Wallis test is used, the critical value for an alpha = .05 is 7.814 Answer: TRUE Diff: 2 Keywords: nonparametric, Kruskal-Wallis, critical value Section: 17-3 Kruskal-Wallis One-Way Analysis of Variance Outcome: 4 56) The makers of furnace filters recently conducted a test to determine whether the median number of particulates that would pass through their four leading filters was the same. A random sample of 6 of each type of filter was used with the following data being recorded: Filter 1 40 25 70 47 55 88

Filter 2 100 110 89 67 77 102

Filter 3 165 90 40 200 103 110

Filter 4 55 20 90 105 90 120

If the Kruskal-Wallis test is used, the test statistic is approximately H = 7.814 Answer: FALSE Diff: 3 Keywords: nonparametric, Kruskal-Wallis, test statistic Section: 17-3 Kruskal-Wallis One-Way Analysis of Variance Outcome: 4

57) The makers of furnace filters recently conducted a test to determine whether the median number of particulates that would pass through their four leading filters was the same. A random sample of 6 of each type of filter was used with the following data being recorded: Filter 1 40 25 70 47 55 88

Filter 2 100 110 89 67 77 102

Filter 3 165 90 40 200 103 110

Filter 4 55 20 90 105 90 120

If the Kruskal-Wallis test is used with an alpha = .01, the null hypothesis should be rejected and the managers should conclude that the four filters do not allow an equal median number of particulates. Answer: FALSE Diff: 3 Keywords: nonparametric, Kruskal-Wallis, hypothesis, null Section: 17-3 Kruskal-Wallis One-Way Analysis of Variance Outcome: 4 58) Kruskal-Wallis One-Way Analysis of Variance is the nonparametric counterpart to the one-way ANOVA procedure in which the assumptions of normally distributed populations with equal variances are satisfied. Answer: TRUE Diff: 2 Keywords: nonparametric, Kruskal-Wallis, ANOVA Section: 17-3 Kruskal-Wallis One-Way Analysis of Variance Outcome: 4 59) If you are interested in testing whether the median of a population is equal to a specific value, an appropriate test to use is: A) the Mann-Whitney U test. B) the t-test. C) the Wilcoxon signed rank test. D) the Wilcoxon Matched-Pairs Signed Rank test. Answer: C Diff: 1 Keywords: nonparametric, rank, Wilcoxon, median Section: 17-1 The Wilcoxon Signed Rank Test for One Population Median Outcome: 1

60) Nonparametric statistical tests are used when: A) the sample sizes are small. B) we are unwilling to make the assumptions of parametric tests. C) the standard normal distribution cannot be computed. D) the population parameters are unknown. Answer: B Diff: 1 Keywords: nonparametric, rank, assumptions Section: 17-1 The Wilcoxon Signed Rank Test for One Population Median Outcome: 1 61) Which of the following is not a step involved in the Wilcoxon signed rank test? A) Find the deviations from the hypothesized median B) Rank the deviations C) Convert the deviations to absolute values D) Find the deviations from the sample median Answer: D Diff: 2 Keywords: nonparametric, rank, Wilcoxon, signed rank test Section: 17-1 The Wilcoxon Signed Rank Test for One Population Median Outcome: 1 62) The Wilcoxon signed rank test is used to test which of the following type of hypotheses? A) Tests about a single population mean B) Tests involving three or more population medians C) Tests about a single population median D) Tests about two or more population proportions Answer: C Diff: 1 Keywords: nonparametric, rank, Wilcoxon, median Section: 17-1 The Wilcoxon Signed Rank Test for One Population Median Outcome: 1

63) The General Electric service department believes that the median time for a service call should be 30 or fewer minutes. To test this, the following random sample of service times was collected: Time 33 27 40 34 22 19 40 73 26 Given that the managers do not wish to make the assumption that the population is normally distributed, the appropriate statistical test for testing about service times is: A) the t-test. B) the Kruskal-Wallis test. C) the Wilcoxon signed rank sum test. D) the F-test. Answer: C Diff: 2 Keywords: nonparametric, rank, Wilcoxon, signed rank, median Section: 17-1 The Wilcoxon Signed Rank Test for One Population Median Outcome: 1

64) The General Electric service department believes that the median time for a service call should be 30 or fewer minutes. To test this, the following random sample of service times was collected: Time 33 27 40 34 22 19 40 73 26 Given that the managers do not wish to make the assumption that the population is normally distributed, the critical value for the test about median service times, using a .05 level of significance, is: A) 5 B) 40 C) 8 D) 37 Answer: D Diff: 2 Keywords: nonparametric, rank, median, critical value, Wilcoxon Section: 17-1 The Wilcoxon Signed Rank Test for One Population Median Outcome: 1

65) The General Electric service department believes that the median time for a service call should be 30 or fewer minutes. To test this, the following random sample of service times was collected: Time 33 27 40 34 22 19 40 73 26 Given that the managers do not wish to make the assumption that the population is normally distributed, the test statistic for the Wilcoxon signed rank sum test is: A) W = 43.0 B) W = 27.0 C) W = 18.0 D) None of the above Answer: B Diff: 3 Keywords: nonparametric, rank, Wilcoxon, signed rank Section: 17-1 The Wilcoxon Signed Rank Test for One Population Median Outcome: 1 66) In the finding the critical value for the Wilcoxon signed rank test, what does "n" represent? A) The number of observations in the sample B) The number of pairs C) The number of nonzero deviations D) The number of positive ranks Answer: C Diff: 2 Keywords: nonparametric, rank, Wilcoxon, signed rank, critical value Section: 17-1 The Wilcoxon Signed Rank Test for One Population Median Outcome: 1 67) Which of the following tests would not be an example of nonparametric method? A) Wilcoxon signed-rank test B) Mann-Whitney U-Test C) Kruskal-Wallis One-Way Analysis of Variance D) χ2 test Answer: D Diff: 2 Keywords: nonparametric Section: 17-1 The Wilcoxon Signed Rank Test for One Population Median Outcome: 1 17-21 Copyright © 2018 Pearson Education, Inc.

68) A marketing firm is interested to know whether the median age of college students is 21 years. A sample of 80 college students is taken. Thirty of the students were under 21, 45 of the students were over 21, and 10 were 21 years old. The conclusion is that A) the median age of college students is significantly different from 21. B) the median age of college students is not significantly different from 21. C) the median age of college students is significantly older than from 21. D) the median age of college students is significantly younger than from 21. Answer: B Diff: 3 Keywords: nonparametric, rank, Wilcoxon, signed rank, critical value Section: 17-1 The Wilcoxon Signed Rank Test for One Population Median Outcome: 1 69) When the Mann-Whitney U test is performed, which of the following is true? A) We assume that the populations are normally distributed. B) We are interested in testing whether the medians from two populations are equal. C) The data are nominal level. D) The samples are independent. Answer: D Diff: 2 Keywords: nonparametric, rank, Mann-Whitney Section: 17-2 Nonparametric Tests for Two Population Medians Outcome: 2 70) Which of the following is not an assumption of the Mann-Whitney U test? A) The sample sizes are equal. B) The samples are independent. C) The value measured is continuous. D) The population distributions are the same for shape and spread. Answer: A Diff: 2 Keywords: nonparametric, rank, Mann-Whitney, assumptions Section: 17-2 Nonparametric Tests for Two Population Medians Outcome: 2

71) Consider the situation in which a study was recently conducted to determine whether the median price of houses is the same in Seattle and Phoenix. The following data were collected. Seattle 233,000 567,800 145,600 234,000 356,000 203,000

Phoenix 309,000 422,000 209,000 187,000 165,000 189,000

Given these data, if a Mann-Whitney U test is to be used, the sum of the ranks for Seattle is: A) 43 B) 35 C) 25.5 D) 40 Answer: A Diff: 2 Keywords: nonparametric, rank, Mann-Whitney Section: 17-2 Nonparametric Tests for Two Population Medians Outcome: 2 72) Consider the situation in which a study was recently conducted to determine whether the median price of houses is the same in Seattle and Phoenix. The following data were collected. Seattle 233,000 567,800 145,600 234,000 356,000 203,000

Phoenix 309,000 422,000 209,000 187,000 165,000 189,000

Given these data, if a Mann-Whitney U test is to be used, the U statistic for Phoenix is: A) 14 B) 22 C) 35 D) 27 Answer: B Diff: 3 Keywords: nonparametric, rank, Mann-Whitney Section: 17-2 Nonparametric Tests for Two Population Medians Outcome: 2

73) Consider the situation in which a study was recently conducted to determine whether the median price of houses is the same in Seattle and Phoenix. The following data were collected. Seattle 233,000 567,800 145,600 234,000 356,000 203,000

Phoenix 309,000 422,000 209,000 187,000 165,000 189,000

Given these data, if a Mann-Whitney U test is to be used, the U statistic for Seattle is: A) 45 B) 35 C) 22 D) 14 Answer: D Diff: 3 Keywords: nonparametric, rank, Mann-Whitney Section: 17-2 Nonparametric Tests for Two Population Medians Outcome: 2 74) Consider the situation in which a study was recently conducted to determine whether the median price of houses is the same in Seattle and Phoenix. The following data were collected. Seattle 233,000 567,800 145,600 234,000 356,000 203,000

Phoenix 309,000 422,000 209,000 187,000 165,000 189,000

Given these data, if a Mann-Whitney U test is to be used, the test statistic is: A) 22 B) 14 C) approximately 1.96 D) 34 Answer: B Diff: 3 Keywords: nonparametric, rank, Mann-Whitney, test statistic Section: 17-2 Nonparametric Tests for Two Population Medians Outcome: 2

75) Under what circumstances should the standard normal distribution be used when employing the Mann-Whitney U test? A) When the sample sizes are equal from the two populations B) When the sample sizes are greater than 20 C) When the populations are normally distributed D) You would never use the standard normal distribution. Answer: B Diff: 2 Keywords: nonparametric, rank, Mann-Whitney, sample size, normal Section: 17-2 Nonparametric Tests for Two Population Medians Outcome: 2 76) Consider the situation in which a human resources manager wishes to determine whether the median number of days of sick leave per year is the same for female employees as for male employees. The following data represent random samples of males and females: Males 5 10 0 9 2 7 5 10 3 1

Females 14 5 13 7 8 0 7 10 6 5

If the manager is unwilling to assume that the populations are normally distributed, which of the following would be the appropriate null hypothesis to be tested? A) H0 : μ = 0 B) H0 :

C) H0 : μ1 = μ2 D) H0 : 1 = 2 Answer: D Diff: 2 Keywords: nonparametric, rank, null, hypothesis, Mann-Whitney Section: 17-2 Nonparametric Tests for Two Population Medians Outcome: 2

77) Assume you are conducting a two-tailed Mann-Whitney U test for a small sample and have found that U1 = 58 and U2 = 86. What is the value of the test statistic? A) 58 B) 86 C) 72 D) 144 Answer: A Diff: 2 Keywords: nonparametric, rank, Mann-Whitney, test statistic Section: 17-2 Nonparametric Tests for Two Population Medians Outcome: 2 78) Consider the situation in which a human resources manager wishes to determine whether the median number of days of sick leave per year is greater for female employees than for male employees. The following data represent random samples of males and females: Males 5 10 0 9 2 7 5 10 3 1

Females 14 5 13 7 8 0 7 10 6 5

If the manager is unwilling to assume that the populations are normally distributed, which of the following is the correct conclusion to reach if the test is conducted using a .05 level of significance? A) Reject the null hypothesis B) Conclude that females do have a higher median than males C) Do not reject the null hypothesis D) Conclude that males have a higher median than females Answer: C Diff: 3 Keywords: nonparametric, rank, Mann-Whitney, hypothesis Section: 17-2 Nonparametric Tests for Two Population Medians Outcome: 2

79) In a large sample Mann-Whitney U test in which the sample size from the first population is 30 and the sample size from the second population is 40, which of the following is the expected U value if the null hypothesis of equal median values is true? A) 1,200 B) 70 C) 35 D) 600 Answer: D Diff: 2 Keywords: nonparametric, rank, Mann-Whitney, null, hypothesis Section: 17-2 Nonparametric Tests for Two Population Medians Outcome: 2 80) Recently, a legislative committee commissioned a study of incomes in a western state. At issue was whether the ratings of the legislature's performance differed between rural citizens and city residents. A random sample of 25 city residents and 35 rural residents was asked to rate the performance of the legislature on a scale of 1 to 100. The analysts believe that the population distribution of ratings would be highly skewed so they decided to use the Mann-Whiney U test to test whether there is a difference in median ratings by the two groups. Given this information, which of the following is the correct critical value if the test is to be conducted at the .10 level of significance? A) z = 1.96 B) t = 2.0357 C) U = 437.5 D) z = 1.645 Answer: D Diff: 2 Keywords: nonparametric, rank, Mann-Whitney, critical value Section: 17-2 Nonparametric Tests for Two Population Medians Outcome: 2 81) Assume that you are conducting a small sample Mann-Whitney U test where n1 = 14 and n2 = 16 and that U1 = 98. Assuming that U1 has been found correctly, what is the value of U2? A) 112 B) 126 C) 224 D) Insufficient information to determine U2. Answer: B Diff: 2 Keywords: nonparametric, rank, Mann-Whitney, test statistic Section: 17-2 Nonparametric Tests for Two Population Medians Outcome: 2

82) If we wish to test whether two related populations have equal medians, an appropriate nonparametric test to use is: A) the Mann-Whitney U test. B) the Kruskal-Wallis test. C) the Wilcoxon signed rank test. D) the Wilcoxon matched-pairs signed rank test. Answer: D Diff: 1 Keywords: nonparametric, rank, Wilcoxon, matched pairs, signed rank Section: 17-2 Nonparametric Tests for Two Population Medians Outcome: 3 83) If a two-tailed Wilcoxon Matched-Pairs Signed Rank test is conducted for a sample of n = 8 and an alpha level equal to .05, the critical value is: A) 4 B) 1.96 C) 30 D) 2 Answer: A Diff: 2 Keywords: nonparametric, rank, Wilcoxon, matched, pairs, signed, critical value Section: 17-2 Nonparametric Tests for Two Population Medians Outcome: 3 84) The Mann-Whitney U test assumes that the 2 samples are: A) equal in size. B) independent and random. C) matched or paired. D) normally distributed. Answer: B Diff: 2 Keywords: nonparametric, rank, Mann-Whitney Section: 17-2 Nonparametric Tests for Two Population Medians Outcome: 2 85) The Wilcoxon matched-pairs signed rank test assumes that the two samples are: A) equal in size. B) independent and random. C) paired. D) Both A and C Answer: D Diff: 1 Keywords: nonparametric, rank, Wilcoxon, matched, pairs, signed Section: 17-2 Nonparametric Tests for Two Population Medians Outcome: 3

86) When employing a small sample Mann-Whitney U test for a two-tailed test, which of the following is true? A) The sample sizes need to be equal. B) Select as the test statistic the smaller of the two U values. C) Select either of the U values to be the test statistic. D) The alpha level should be doubled. Answer: B Diff: 2 Keywords: nonparametric, rank, Mann-Whitney, test statistic Section: 17-2 Nonparametric Tests for Two Population Medians Outcome: 2 87) If a Mann-Whitney U test was performed and U1 = 50 and U2 = 40, if the sample from population 1 was 10, the sample size from population 2 was: A) 10 B) 15 C) 9 D) Can't be determined without more information. Answer: C Diff: 2 Keywords: nonparametric, rank, Mann-Whitney, sample size Section: 17-2 Nonparametric Tests for Two Population Medians Outcome: 2 88) In conducting a Kruskal-Wallis one-way analysis of variance, the test statistic is assumed to have approximately which distribution when the null hypothesis is true? A) A t-distribution B) An F-distribution C) A normal distribution D) A chi-square distribution Answer: D Diff: 2 Keywords: nonparametric, rank, Kruskal-Wallis, ANOVA, test statistic Section: 17-3 Kruskal-Wallis One-Way Analysis of Variance Outcome: 4 89) Which of the following is not an assumption of the Kruskal-Wallis one-way analysis of variance? A) Variables have a continuous distribution. B) Samples are independent. C) Sample sizes are equal for all populations. D) Population distributions are identical except for possible differences in center. Answer: C Diff: 2 Keywords: nonparametric, rank, Kruskal-Wallis, ANOVA, analysis of variance, sample Section: 17-3 Kruskal-Wallis One-Way Analysis of Variance Outcome: 4

90) The Kruskal-Wallis test is usually limited to comparing sample values from ________ or more populations. A) 2 B) 3 C) 4 D) 5 Answer: B Diff: 1 Keywords: nonparametric, rank, Kruskal-Wallis, ANOVA, analysis of variance Section: 17-3 Kruskal-Wallis One-Way Analysis of Variance Outcome: 4 91) In a Kruskal-Wallis test when ties occur, each observation is given the ________ for which it is tied. A) highest rank B) lowest rank C) mean rank D) median rank Answer: C Diff: 2 Keywords: nonparametric, rank, Kruskal-Wallis, ANOVA, analysis of variance, mean Section: 17-3 Kruskal-Wallis One-Way Analysis of Variance Outcome: 4 92) Assume that 4 populations are to be compared using a Kruskal-Wallis one-way analysis of variance. What is the critical value using a 0.05 level of significance? A) 5.9915 B) 6.2514 C) 7.8147 D) 9.4877 Answer: C Diff: 2 Keywords: nonparametric, rank, Kruskal-Wallis, ANOVA, analysis of variance, null, hypothesis Section: 17-3 Kruskal-Wallis One-Way Analysis of Variance Outcome: 4

93) Assume that a Kruskal-Wallis test is being conducted to determine whether or not the medians of three populations are equal. The sum of rankings and the sample size for each group are below. Group 1 R1 = 60

Group 2 R2 = 36

Group 3 R3 = 24

n1 = 6

n2 = 5

n3 = 4

What is the value of the test statistic? A) 7.8147 B) 2.16 C) 48.68 D) 12.59 Answer: B Diff: 2 Keywords: nonparametric, rank, Kruskal-Wallis, ANOVA, analysis of variance, test statistic Section: 17-3 Kruskal-Wallis One-Way Analysis of Variance Outcome: 4 94) Assume that a Kruskal-Wallis test is being conducted to determine whether or not the medians of three populations are equal. The sum of rankings and the sample size for each group are below. Group 1 R1 = 60

Group 2 R2 = 36

Group 3 R3 = 24

n1 = 6

n2 = 5

n3 = 4

What is the critical value for this test using a 0.10 level of significance? A) 6.2514 B) 5.9915 C) 7.8147 D) 4.6052 Answer: D Diff: 2 Keywords: nonparametric, rank, Kruskal-Wallis, ANOVA, analysis of variance, critical value Section: 17-3 Kruskal-Wallis One-Way Analysis of Variance Outcome: 4 95) If we are interested in testing to determine whether the center of three or more populations is equal when the data in the samples are ordinal, what is the appropriate test to conduct? A) A t-test B) An ANOVA C) A Kruskal-Wallis D) A Wilcoxon Matched-Pairs Sign Rank test Answer: C Diff: 1 Keywords: nonparametric, rank, Kruskal-Wallis, ANOVA, analysis of variance Section: 17-3 Kruskal-Wallis One-Way Analysis of Variance Outcome: 4 17-31 Copyright © 2018 Pearson Education, Inc.

96) A survey was recently conducted in which random samples of car owners of Chrysler, GM, and Ford cars were surveyed to determine their satisfaction. Each owner was asked to rate overall satisfaction on a scale of 1 (poor) to 1000 (excellent). The following data were recorded: Chrysler 650 700 500 800 900 750

GM 400 800 500 400 600 900

Ford 700 750 650 800 900 700

If the analysts are not willing to assume that the population ratings are normally distributed and will use the Kruskal-Wallis test to determine if the three companies have different median ratings, which company has the smallest sum of ranks? A) Chrysler B) GM C) Ford D) All three are equal. Answer: B Diff: 3 Keywords: nonparametric, rank, Kruskal-Wallis, ANOVA, analysis of variance Section: 17-3 Kruskal-Wallis One-Way Analysis of Variance Outcome: 4

97) A survey was recently conducted in which random samples of car owners of Chrysler, GM, and Ford cars were surveyed to determine their satisfaction. Each owner was asked to rate overall satisfaction on a scale of 1 (poor) to 1000 (excellent). The following data were recorded: Chrysler 650 700 500 800 900 750

GM 400 800 500 400 600 900

Ford 700 750 650 800 900 700

98) A survey was recently conducted in which random samples of car owners of Chrysler, GM, and Ford cars were surveyed to determine their satisfaction. Each owner was asked to rate overall satisfaction on a scale of 1 (poor) to 1000 (excellent). The following data were recorded: Chrysler 650 700 500 800 900 750

GM 400 800 500 400 600 900

Ford 700 750 650 800 900 700

If the analysts are not willing to assume that the population ratings are normally distributed and will use the Kruskal-Wallis test to determine if the three companies have different median ratings, what is correct test statistic for these data? A) H = 1.965 B) t = 1.96 C) H = 3.34 D) H = .65 Answer: A Diff: 3 Keywords: nonparametric, rank, Kruskal-Wallis, ANOVA, analysis of variance, test statistic, tied rankings Section: 17-3 Kruskal-Wallis One-Way Analysis of Variance Outcome: 4

99) A survey was recently conducted in which random samples of car owners of Chrysler, GM, and Ford cars were surveyed to determine their satisfaction. Each owner was asked to rate overall satisfaction on a scale of 1 (poor) to 1000 (excellent). The following data were recorded: Chrysler 650 700 500 800 900 750

GM 400 800 500 400 600 900

Ford 700 750 650 800 900 700

If the analysts are not willing to assume that the population ratings are normally distributed and will use the Kruskal-Wallis test to determine if the three companies have different median ratings, what is the correct conclusion if the test is to be conducted using an alpha = .05 level? A) H0 should be rejected and we conclude that there is no significant difference between the 3 companies. B) H0 should not be rejected and we conclude that there is no significant difference between the 3 companies. C) H0 should be rejected and we conclude that there is a significant difference between the 3 companies. D) H0 should not be rejected and we conclude that there is a significant difference between the 3 companies. Answer: B Diff: 3 Keywords: nonparametric, rank, Kruskal-Wallis, ANOVA, analysis of variance, null, hypothesis Section: 17-3 Kruskal-Wallis One-Way Analysis of Variance Outcome: 4 100) Under what conditions should a decision maker use a nonparametric statistical procedure? Answer: Generally, nonparametric tests are used when you are unwilling to make the assumptions that are required for the corresponding parametric test. For example, in testing a null hypothesis about a population mean when the population standard deviation is unknown, we might like to use the t-test. However, the t-test requires that the population be normally distributed. Nonparametric tests are also used when the level of data measurement is lower. For instance parametric tests require that the data level be at least interval. However, nonparametric tests are typically based with ranked data and can be used with ordinal data. Diff: 1 Keywords: nonparametric Section: 17-1 The Wilcoxon Signed Rank Test for One Population Median Outcome: 1

101) A real estate broker claims that the median days that one of his listings stays on the market is 45 or less days. To test this, he has collected the following random sample of properties sold showing the days they were on the market prior to selling: Days 50 30 70 20 30 40 60 80 The broker is unwilling to assume that the population data are normally distributed. a. What is the correct null and alternative hypothesis to be tested? b. What statistical test would you recommend be used to test this hypothesis? c. Conduct the test and indicate what conclusion should be reached if we test at an alpha = .05 level? Answer: a. H0 : ≤ 45 H0 :

> 45

b. Because we are unwilling to assume that the population is normally distributed and because we wish to test about the median, an appropriate test to use is the Wilcoxon signed rank test. c. The first step in conducting the Wilcoxon signed rank test is to compute the differences between each value and the hypothesized median. These are shown as follows: Days 50 30 70 20 30 40 60 80

d 5 -15 25 -25 -15 -5 15 35

Next, we convert the differences to absolute values. Days 50 30 70 20 30 40 60 80

d 5 -15 25 -25 -15 -5 15 35

asb(d) 5 15 25 25 15 5 15 35

Now, we rank each of these absolute differences from smallest to largest. Note: If there are ties, we supply the mean rank for the tied items. Days 50 30 70 20 30 40 60 80

d 5 -15 25 -25 -15 -5 15 35

asb(d) 5 15 25 25 15 5 15 35

Rank 1.5 4 6.5 6.5 4 1.5 4 8

Next, keeping track of the original sign on the differences, we place the ranks in + or - columns as shown: Days d asb(d) Rank + Rank - Rank 50 5 5 1.5 1.5 30 -15 15 4 4 70 25 25 7.5 6.5 20 -25 25 7.5 6.5 30 -15 15 4 4 40 -5 5 1.5 1.5 60 15 15 4 4 80 35 35 8 8 Totals = 20 16 The test statistic is the sum of + Ranks. This is 21. For an alpha level = .05, the critical value for a sample size of n = 8 is 6 and 30. Since 20 < 30, we do not reject the null hypothesis. Diff: 3 Keywords: nonparametric, null, alternative, hypothesis, rank, Wilcoxon Section: 17-1 The Wilcoxon Signed Rank Test for One Population Median Outcome: 1

102) A paint manufacturer is interested in determining whether there is a difference in the median time it takes for two different brands of paint to dry once they have been applied to a wall surface. To test this, the company has selected a random sample of 5 walls and applied brand 1 and another 5 walls and applied brand 2. The following data reflect the actual drying time in hours: Brand 1 3.4 3.2 2.8 3.7 2.5

Brand 2 2.0 3.0 3.0 2.2 2.1

a. If you are unwilling to make the assumptions necessary to use a t-test, what test would you recommend in this situation? b. What would be the appropriate null and alternative hypothesis? c. Using an alpha level equal to .05, conduct the hypothesis test and state the conclusion. Answer: a. Since the samples are independent, an appropriate test would be the Mann-Whitney U test. b. H0 : 1 = 2 H0 : 1 ≠ 2 c. The first step is to establish the ranks of the data by combining all sample data from the two populations together. Then show the ranks broken out by sample. Note: Use the mean rank when data values are tied. Total the ranks for each sample.

Next, use equations 17.2 and 17.3 to find the U statistics for the two samples: u1 = n1n2 +

- 36 = 4

u1 = n1n2 +

- 19 = 21.

We select the smaller U value to be the test statistic. Since our samples sizes are small, we use the MannWhitney U table for samples n < 9. For sample sizes of 5 and 5 with U equals 4, the p-value is .048. Since .048 is > alpha/2 = .025, we do not reject the null hypothesis. Thus, based upon the sample information, we are unable to state that a difference exists between the two brands of paint with respect to median drying time. Diff: 3 Keywords: nonparametric, null, alternative, hypothesis, rank, Mann-Whitney Section: 17-2 Nonparametric Tests for Two Population Medians Outcome: 2 17-38 Copyright © 2018 Pearson Education, Inc.

103) A restaurant has three separate dining areas: the patio, the alcove, and the main hall. At question is whether the median dollar amount per customer is the same or different between these three restaurant locations. To test this, the manager has randomly selected samples from each location. These data are shown as follows: Patio Alcove Main Hall $22.45 $20.20 $40.00 $35.70 $19.50 $18.50 $17.90 $30.00 $19.60 $25.50 $18.25 $34.50 $19.50 $22.50 $32.00 $14.50 $14.50 $28.70 $30.20 $17.80 $25.00 $36.50 $21.30 $15.75 a. If the manager is unwilling to make the assumption that the bill amounts are normally distributed at all three locations, what statistical technique would you suggest to test whether the median bill amounts are the same or different? b. State the appropriate null and alternative hypotheses. c. Using an alpha level equal to .05, test the null hypothesis and state your conclusion. Answer: a. An appropriate statistical tool is the Kruskal-Wallis one way analysis of variance. This test does not require the normality assumptions but fits this case since the samples are independent and there are three or more groups to be tested. b. H0 : 1 = 2 = 3 HA: not all medians are equal c. The following summary values would be computed: Sum of Ranks-Patio = 108 Sum of Ranks-Alcove = 75 Sum of Ranks-Main = 117 The calculated test statistic is H = 2.445 The critical value from the chi-square distribution with 2 degrees of freedom and alpha = .05 is 5.99 Then, since H = 2.445 < 5.99, do not reject the null hypothesis. Thus, based on these sample data, we are unable to conclude that a difference exists in median bill amount between the three seating areas in the restaurant. Diff: 2 Keywords: nonparametric, rank, Kruskal-Wallis, ANOVA, analysis of variance, null, alternative, hypothesis Section: 17-3 Kruskal-Wallis One-Way Analysis of Variance Outcome: 4

Business Statistics, 9e (Groebner/Shannon/Fry) Chapter 18 Introducing Business Analytics 1) We define business analytics as the transformation of large and complex data sets into information that is useful for making decisions. Answer: TRUE Diff: 1 Keywords: business analytics, transformation, decisions Section: 18-1 Introducing Business Analytics Outcome: 1 2) Business intelligence is the process of collecting data from inside or outside an organization that are thought to relate to or influence the performance of the organization. Answer: TRUE Diff: 1 Keywords: business intelligence, collecting data, performance Section: 18-1 Introducing Business Analytics Outcome: 1 3) Business intelligence uses the same methods as business analytics. Answer: FALSE Diff: 1 Keywords: business intelligence, business analytics, methods Section: 18-1 Introducing Business Analytics Outcome: 1 4) It is a good idea to include potentially irrelevant data so that the business analytics organization can identify meaningful data and reach proper conclusions. Answer: TRUE Diff: 2 Keywords: irrelevant data, business analytics, meaningful data Section: 18-1 Introducing Business Analytics Outcome: 1 5) Business analytics can provide answers to unknown questions. Answer: FALSE Diff: 2 Keywords: business analytics, unknown questions Section: 18-1 Introducing Business Analytics Outcome: 1 6) Business analytics methods can examine data using deeper methods of analysis in order to draw conclusions about relationships. Answer: TRUE Diff: 2 Keywords: business analytics, methods of analysis, conclusions about relationships Section: 18-1 Introducing Business Analytics Outcome: 1 18-1 Copyright © 2018 Pearson Education, Inc.

7) Usually the early output from the business analytics effort generates an “answer”. Answer: FALSE Diff: 2 Keywords: business analytics, output, answer Section: 18-1 Introducing Business Analytics Outcome: 1 8) The process of business analytics starts with questions posed by decision makers, followed by more detailed questions until an “answer” is hopefully found. Answer: TRUE Diff: 2 Keywords: business analytics, questions, decision makers, answer Section: 18-1 Introducing Business Analytics Outcome: 1 9) Business analytics offers powerful ways of digging through data to examine relationships to a far greater extent than was possible in the past. Answer: TRUE Diff: 2 Keywords: business analytics, relationships, decision makers Section: 18-1 Introducing Business Analytics Outcome: 1 10) Descriptive analytics and statistical analysis are two categories of business analytics. Answer: FALSE Diff: 2 Keywords: descriptive analytics, statistical analysis, business analytics Section: 18-1 Introducing Business Analytics Outcome: 2 11) Descriptive analytics combined with analysis through sophisticated statistical software can provide analysts with both visual and numerical tools to examine relationships in data. Answer: TRUE Diff: 1 Keywords: descriptive analytics, statistical software, numerical tools Section: 18-1 Introducing Business Analytics Outcome: 2 12) An important outgrowth of descriptive analytics has been the expanded use of dashboards, which provide historical summaries of key performance indicators for an organization. Answer: FALSE Diff: 2 Keywords: descriptive analytics, dashboards, key performance indicators Section: 18-1 Introducing Business Analytics Outcome: 2

13) The dashboards can provide powerful real time displays used to access multiple levels below highlevel metrics. Answer: TRUE Diff: 2 Keywords: dashboards, real time displays, high-level metrics Section: 18-1 Introducing Business Analytics Outcome: 2 14) Predictive analytics can identify patterns and relationships in the data that allow decision makers to forecast a behavior or action between independent indicator variables and a dependent variable of interest. Answer: TRUE Diff: 2 Keywords: predictive analytics, patterns, forecast Section: 18-1 Introducing Business Analytics Outcome: 2 15) Predictive analytics is restricted to business forecasting. Answer: FALSE Diff: 2 Keywords: predictive analytics, forecasting Section: 18-1 Introducing Business Analytics Outcome: 2 16) Forecasting methods are applied to data to answer questions about will happen. Answer: FALSE Diff: 2 Keywords: predictive analytics, patterns, forecast Section: 18-1 Introducing Business Analytics Outcome: 2 17) Thanks to predictive analytics, a retail grocer has a better idea of who his customers are, in terms of behavior and spending power, which can be aligned with promotional campaigns. Answer: TRUE Diff: 2 Keywords: predictive analytics, spending power, promotional campaigns Section: 18-1 Introducing Business Analytics Outcome: 2 18) High-level key performance indicator (KPIs) measure the success of an organization’s subunits (departments, centers, etc.) while low-level KPIs measure the overall performance of the organization. Answer: FALSE Diff: 2 Keywords: key performance indicator (KPI), performance Section: 18-1 Introducing Business Analytics Outcome: 2

19) Power BI Desktop is one of several Power BI platforms developed by Microsoft that enable users to create, share, and use information embedded in the data available to their organization. Answer: TRUE Diff: 2 Keywords: Power BI Desktop, BI platforms, embedded data Section: 18-2 Data Visualization Using Microsoft Power BI Desktop Outcome: 3 20) An important component of Power BI’s capability includes data visualization. Answer: TRUE Diff: 1 Keywords: Power BI, data visualization Section: 18-2 Data Visualization Using Microsoft Power BI Desktop Outcome: 3 21) Dashboards display real-time graphic and numerical displays where most commonly, the most important information is the largest visual and most often displayed on the lower corner. Answer: FALSE Diff: 2 Keywords: dashboard, real-time displays, graphic and numerical displays Section: 18-2 Data Visualization Using Microsoft Power BI Desktop Outcome: 3 22) A single graph, chart, or table is referred to as a tile in Power BI. Answer: TRUE Diff: 1 Section: 18-2 Data Visualization Using Microsoft Power BI Desktop Outcome: 3 23) Power BI can import a wide range of data including Excel work-books and CSV files, and a range of databases including Access, SQL, and Sybase. Answer: TRUE Diff: 2 Keywords: quality, process, control, specification Section: 18-2 Data Visualization Using Microsoft Power BI Desktop Outcome: 3 24) One of the disadvantages of business analytics software such as Microsoft Power BI is its inability to pull information from multiple data tables. Answer: FALSE Diff: 2 Keywords: business analytics software, multiple data tables, Microsoft Power BI, Section: 18-2 Data Visualization Using Microsoft Power BI Desktop Outcome: 3

25) Which of the following is not a component of Power BI? A) Reports B) Visualizations C) Automatic Analysis D) Dashboards Answer: C Diff: 2 Keywords: Power BI, visualizations, reports, dashboards Section: 18-2 Data Visualization Using Microsoft Power BI Desktop Outcome: 2 26) Visualizations in Power BI can include the following: A) Maps B) Tables C) Charts D) All of the above Answer: D Diff: 1 Keywords: Power BI, visualizations, maps, tables, charts Section: 18-2 Data Visualization Using Microsoft Power BI Desktop Outcome: 2 27) Fill in the blank: A Power BI report includes ________. A) single or multiple pages B) visual items that relate to each other C) A & B D) no visual items Answer: C Diff: 2 Keywords: Power BI, report, single or multiple pages, visual items Section: 18-2 Data Visualization Using Microsoft Power BI Desktop Outcome: 2 28) Which of the following are types of data that can be imported into Power BI? A) a single chart B) html files C) CSV files D) databases including Access, SQL, and Sybase Answer: B Diff: 2 Keywords: Power BI, chart, html, CSV, Access, SQL, Sybase databases Section: 18-2 Data Visualization Using Microsoft Power BI Desktop Outcome: 3

29) Microsoft Power BI has the capability to pull information from multiple data tables if those tables are related. Which of the following defines how those tables are related to each other? A) Power BI checks to see if the file names are the same. B) Power BI checks to see if the rows and columns are the same size for the tables. C) Power BI autodetects relationships by checking variable names (Fields) to see if there is a match. D) Power BI checks to see if the author of the files are the same. Answer: C Diff: 3 Keywords: Power BI, data tables, variable names (Fields) Section: 18-2 Data Visualization Using Microsoft Power BI Desktop Outcome: 3 30) For multiple tables that are related, Power BI can draw from two or more of them at the same time to create A) visualization tiles. B) reports. C) dashboards. D) All of the above Answer: D Diff: 2 Keywords: Power BI, visualization tiles, reports, dashboards Section: 18-2 Data Visualization Using Microsoft Power BI Desktop Outcome: 4 31) Using Power BI, you can create a new variable from existing variables as long as A) the new variable is created from variables that exist either in the same data table or data tables that are related with no restriction on how many variables can be combined. B) the new variable is created from new information. C) the new variable is created from existing variables from unrelated data tables. D) the new variable is created from existing variables either in the same data table or data tables that are related with the restriction that no more than two variables can be combined. Answer: A Diff: 2 Keywords: Power BI, variable, related or unrelated data tables Section: 18-2 Data Visualization Using Microsoft Power BI Desktop Outcome: 5 32) Measures are a feature of Power BI that allows you to A) visualize created variables. B) perform calculations such as sums and averages on your data when you interact with the visuals that have been created. C) related variables from multiple data tables. D) All of the above Answer: B Diff: 2 Keywords: Power BI, variable, visualization, measures Section: 18-2 Data Visualization Using Microsoft Power BI Desktop Outcome: 5 18-6 Copyright © 2018 Pearson Education, Inc.

33) Power BI allows you to create a report that offers multiple perspectives on the data analyzed. One thing that Power BI cannot do is A) visualize the data. B) perform calculations such as sums and averages on your data. C) allow more than one report to be opened at a time. D) All of the above Answer: C Diff: 2 Keywords: Power BI, variable, visualization, measures Section: 18-2 Data Visualization Using Microsoft Power BI Desktop Outcome: 5 34) Subsequent pages of the Power BI report will allow the managers to examine multiple territory groups in more detail. To do this, we can make use of A) the Power BI filters. B) the Slicer tool. C) neither A) or B) D) either A) or B) Answer: D Diff: 3 Keywords: Power BI report, pages of BI report, territory groups, BI filters, Slicer tool Section: 18-2 Data Visualization Using Microsoft Power BI Desktop Outcome: 5 35) Once you save your work as a Power BI Desktop file, it is considered to be a A) file. B) desktop. C) report. D) tile. Answer: C Diff: 2 Keywords: Power BI Desktop, report, tile Section: 18-2 Data Visualization Using Microsoft Power BI Desktop Outcome: 5 36) The Dashboard shown below in the Power BI Desktop is displaying A) a report. B) tiles. C) files. D) A) and B). Answer: D Diff: 2 Keywords: Power BI Desktop, report, tile Section: 18-2 Data Visualization Using Microsoft Power BI Desktop Outcome: 5

37) The Dashboard shown below in the Power BI Desktop is displaying A) files. B) a desktop. C) a report. D) tiles. Answer: D Diff: 2 Keywords: Power BI Desktop, report, tile Section: 18-2 Data Visualization Using Microsoft Power BI Desktop Outcome: 5 38) Power BI has five key components. List the components along with a brief description of each. Answer: The Power BI key components are: 1. Visualizations 2. Data Sets 3. Reports 4. Dashboards 5. Tiles 1.

2. 3. 4.

Visualizations - as the name implies, a visualization is a pictorial display of the data using charts, graphs, tables, and other visual tools. Visualizations vary in complexity—from a single value or table of values to something more complex like a map depicting total order quantity by sales territory. Data sets - a data set can be as simple as a single Excel file or a collection of files from a variety of sources. Reports - a report is a set of visual items that are related to each other. A report can consist of a single page or multiple pages. Dashboards - dashboards are real-time graphical and numerical displays of key performance indicators. A dashboard is one or more pages of important information and measures that are commonly shared with others in the organization. Tile - a single graph, chart, or table is referred to as a tile in Power BI.

Diff: 2 Keywords: Power BI, visualizations, data sets, reports, dashboards, tile Section: 18-2 Data Visualization Using Microsoft Power BI Desktop Outcome: 3 18-8 Copyright © 2018 Pearson Education, Inc.

39) There is no specific way to create a dashboard in Power BI but there are some commonly accepted practices. List three of them. Answer: The three commonly accepted practices for creating a dashboard in Power Bi are: 1. The most important information is displayed as the biggest visual. 2. The most important information is in the upper corner. 3. Appropriate charts and graphs convey the information effectively. Diff: 2 Keywords: Power BI, dashboard, visual display Section: 18-2 Data Visualization Using Microsoft Power BI Desktop Outcome: 3 40) Discuss the differences between business analytics and business intelligence. Answer: We have defined business analytics as the application of statistical tools, combined with powerful software, to transform large and complex data sets into information that is useful for making decisions. But before business analytics can be put into practice, an organization must have data to analyze. The concepts associated with data collection and different types and levels of data form the basis of what has become known as business intelligence. Business intelligence within an organization uses data acquisition efforts. Thus, although the field of business intelligence is critically important to the field of business analytics, the two are not identical. Diff: 2 Keywords: business intelligence, business analytics, methodology, data collection Section: 18-1 Introducing Business Analytics Outcome: 1 41) Although a key component of business intelligence is gathering data that are relevant to the decision makers in your organization, it is considered to be a good idea to include extra potentially irrelevant data. Why? Answer: It is a good idea to include potentially irrelevant data so that business analytics can be used to identify the meaningful data and to provide effective ways of communicating the information in the data in order that the decision makers can reach the proper conclusions. Diff: 2 Keywords: business intelligence, business analytics, irrelevant data, data collection Section: 18-1 Introducing Business Analytics Outcome: 1 42) Describe the difference between descriptive analytics and predictive analytics. Answer: Business analytics is typically divided into two categories: descriptive analytics and predictive analytics. Descriptive analytics uses a variety of graphical tools, such as bar charts, histograms, pie charts, scatter plots, and trend charts as tools to help us visualize data. Descriptive analytics added a discussion of numerical measures of the center and spread in the data using such measures as the mean, median, and standard deviation. Using descriptive analytics along with powerful software such as Microsoft Power BI Desktop, analysts can combine the visual and numerical tools to examine relationships in the data that lay far below the surface level. Diff: 2 Keywords: business analytics, descriptive analytics, predictive analytics 18-9 Copyright © 2018 Pearson Education, Inc.

Section: 18-1 Introducing Business Analytics Outcome: 2 43) Business analytics is typically divided into two categories. Identify and describe each of the categories in addition to a possible third category. Answer: Business analytics is typically divided into two categories: descriptive analytics and predictive analytics. Some business analytics experts add a third category they refer to as prescriptive analytics, which involves the use of optimization tools and simulation models to answer the question. Descriptive analytics uses a variety of graphical tools, such as bar charts, histograms, pie charts, scatter plots, and trend charts as tools to help us visualize data. Descriptive analytics added a discussion of numerical measures of the center and spread in the data using such measures as the mean, median, and standard deviation. Using descriptive analytics along with powerful software such as Microsoft Power BI Desktop, analysts can combine the visual and numerical tools to examine relationships in the data that lay far below the surface level. Diff: 3 Keywords: business analytics, descriptive analytics, predictive analytics Section: 18-1 Introducing Business Analytics Outcome: 2 44) Predictive analytics is a growing field and can be of potential value to organizations in what type of situations? Identify at least two. Answer: Predictive analytics can be of potential value to organizations where the outcomes of repetitive decisions may have significant financial impact, where there are lots of available data, and where predictive models can be developed to automate these repetitive decisions. Diff: 2 Keywords: predictive analytics, predictive models Section: 18-1 Introducing Business Analytics Outcome: 2 45) A Key Performance Indicator (KPI) is a measure of how well an organization is achieving its most important business objectives. Identify at least three. Answer: Answers will vary. KPIs include primarily financial metrics such as profit, revenue, comparison between actual and projected revenue, cost, sales by regions Diff: 2 Keywords: Key Performance Indicator (KPI), measure, business objectives Section: 18-1 Introducing Business Analytics Outcome: 2 46) Describe in two specific examples how predictive analytics are used to understand a current business situation and then use this knowledge to help determine a future result. Answer: For example, through an analysis of large quantities of customer purchases, a grocery store chain may determine that a customer who buys beer also tends to buy chips and other snack products. The range of applications is very wide, including such areas as health care, crime prevention, tax collections, and public welfare fraud detection. In health care, predictive analytics is used to tie symptoms and treatments to long-term patient outcomes. Insurance companies use predictive analytics to 18-10 Copyright © 2018 Pearson Education, Inc.

help identify fraudulent claims before payments are paid and to recover previously paid claims that are deemed fraudulent after the fact. The IRS uses predictive analytics to identify tax returns that are either underreporting income or overstating deductions. Diff: 2 Keywords: predictive analytics Section: 18-1 Introduction to Statistical Process Control Charts Outcome: 3 47) A 2010 white paper published by Oracle called “Predictive analytics: Bringing the tools to the data” presents a nice way of organizing the predictive analytics tools. A good predictive analyst uses fairly advanced mathematics and statistics along with special software. Identify a minimum of three predictive analytics functions or techniques identified through this white paper. Answer: Functions include classification, regression, anomaly detection, attribute importance, association rules, clustering, and feature extraction Diff: 3 Keywords: predictive analytics, functions, classification, regression, anomaly detection, attribute importance, association rules, clustering, and feature extraction Section: 18-1 Introduction to Statistical Process Control Charts Outcome: 3 48) The figure below displays the process of creating a new variable to be used in analyzing data in the Power BI desktop. Describe the specific measure being created, identifying each part of the formula: ProfitMargin = ProductData[ListPrice] - ProductData[StandardCost] Answer: ProfitMargin is the new calculated variable which subtracts the Standard Cost from the List Price. ListPrice is from the ProductData data table and Standard Cost is from the ProductData data table. Diff: 3 Keywords: Power BI desktop, measure, calculated variable Section: 18-2 Data Visualization Using Microsoft Power BI Desktop Outcome: 5

49) To create a new variable in Power BI, the managers can use the DAX (Data Analysis Expressions) programming language. Describe how DAX is similar to formulas in Excel and what the term RELATED means.

Answer: The DAX methodology is very similar to how equations are created in Excel and includes constants, operators, and functions that allow users to create new variables from the data they already have available. The term RELATED is used to access information from the two related tables, related through a common variable or product ID. Diff: 3 Keywords: Power BI desktop, measure, calculated variable Section: 18-2 Data Visualization Using Microsoft Power BI Desktop Outcome: 5 50) We can create new measures in the Power BI desktop from original variables in our data set or from variables that we created earlier. Let’s say we need to create a variable in the SalesData table called Order Cost which represents the OrderQty variable from the SalesData table multipled by the StandardCost variable in the related ProductData table. The DAX com-mand to do this is: Answer: Order Cost = SalesData [OrderQty] * RELATED(ProductData[StandardCost] Diff: 3 Keywords: Power BI desktop, measure, calculated variable Section: 18-2 Data Visualization Using Microsoft Power BI Desktop Outcome: 5

Business Statistics, 9e (Groebner/Shannon/Fry) Chapter 19 Introduction to Decision Analysis 1) Not all decisions require the use of decision analysis. Answer: TRUE Diff: 1 Keywords: decisions, decision analysis Section: 19-1 Decision-Making Environments and Decision Criteria Outcome: 1 2) Certainty represents a situation where the results of selecting each alternative are known before the decision is made. Answer: TRUE Diff: 1 Keywords: certainty, decision, alternative Section: 19-1 Decision-Making Environments and Decision Criteria Outcome: 1 3) In business, you will encounter a wide variety of decision situations. The two primary decision environments are certainty and uncertainty. Answer: TRUE Diff: 1 Keywords: certainty, uncertainty, decision environments. Section: 19-1 Decision-Making Environments and Decision Criteria Outcome: 1 4) In a certainty environment, the best decision is not necessarily associated with the best outcome. Answer: FALSE Diff: 2 Keywords: certainty environment, best decision, best outcome. Section: 19-1 Decision-Making Environments and Decision Criteria Outcome: 1 5) By “best decision,” we mean the alternative course of action, using all available information that best satisfies the decision criterion. Answer: TRUE Diff: 1 Keywords: best decision, decision criterion, alternative course of action. Section: 19-1 Decision-Making Environments and Decision Criteria Outcome: 1 6) Uncertainty is a decision environment in which the decision maker does not know what outcome will occur when an alternative is selected. Answer: TRUE Diff: 2 Keywords: decision environment, outcome, alternative. Section: 19-1 Decision-Making Environments and Decision Criteria Outcome: 1 19-1 Copyright © 2018 Pearson Education, Inc.

7) In reality, the typical business decision-making environment is one of certainty rather than uncertainty. Answer: FALSE Diff: 1 Keywords: decision-making environment, certainty, uncertainty. Section: 19-1 Decision-Making Environments and Decision Criteria Outcome: 1 8) The established basis for making a decision gives us the criteria for analyzing a decision and making a “best” choice. Answer: TRUE Diff: 1 Keywords: decision, criteria, “best choice”. Section: 19-1 Decision-Making Environments and Decision Criteria Outcome: 2 9) The possible outcomes in a decision situation over which the decision maker has no control is referred to as the choices. Answer: FALSE Diff: 2 Keywords: outcomes, decision situation, choices, states of nature. Section: 19-1 Decision-Making Environments and Decision Criteria Outcome: 2 10) The outcome that is associated with any combination of a particular state of nature and an alternative is called a payoff. Answer: TRUE Diff: 1 Keywords: outcome, state of nature, alternative, payoff. Section: 19-1 Decision-Making Environments and Decision Criteria Outcome: 2 11) The payoffs associated with all possible combinations of alternatives and states of nature constitute a profit or loss table. Answer: FALSE Diff: 2 Keywords: payoffs, combinations of alternatives, states of nature, payoff table. Section: 19-1 Decision-Making Environments and Decision Criteria Outcome: 2 12) Nonprobabilistic criteria are used when either the probabilities associated with the possible payoffs are unknown or the decision maker lacks confidence and/or information with which to assess probabilities for the various payoffs. Answer: TRUE Diff: 2 Keywords: nonprobabilistic criteria, probabilities, possible payoffs. Section: 19-1 Decision-Making Environments and Decision Criteria Outcome: 2 19-2 Copyright © 2018 Pearson Education, Inc.

13) Probabilistic criteria incorporates the decision maker’s probability of each state of nature occurring. Answer: TRUE Diff: 2 Keywords: probabilistic criteria, probabilities, state of nature. Section: 19-1 Decision-Making Environments and Decision Criteria Outcome: 2 14) A decision criterion that considers the results of selecting the “wrong” alternative is referred to as the maximax regret criterion. Answer: FALSE Diff: 2 Keywords: decision criteria, alternative, maximax regret criterion. Section: 19-1 Decision-Making Environments and Decision Criteria Outcome: 2 15) A conservative decision criterion for dealing with uncertainty without using probability is referred to as the maximin criterion. Answer: TRUE Diff: 2 Keywords: conservative decision criterion, uncertainty, maximin criterion, maximax criterion. Section: 19-1 Decision-Making Environments and Decision Criteria Outcome: 2 16) An optimistic decision criterion for dealing with uncertainty without using probability is the maximax criterion. Answer: TRUE Diff: 2 Keywords: optimistic decision criterion, uncertainty, maximax criterion. Section: 19-1 Decision-Making Environments and Decision Criteria Outcome: 2 17) A loss of convenience is the difference between the actual payoff that occurs for a decision and the optimal payoff for the same state of nature. Answer: FALSE Diff: 2 Keywords: opportunity loss, actual payoff, optimal payoff, state of nature. Section: 19-1 Decision-Making Environments and Decision Criteria Outcome: 2 18) A disadvantage of the maximax and maximin criteria is that they use only one value from the payoff table to make a decision. Answer: TRUE Diff: 2 Keywords: maximax criteria, maximin criteria, payoff table, decision criteria. Section: 19-1 Decision-Making Environments and Decision Criteria Outcome: 2

19) All three decision criteria (maximax, maximin, minimax regret) lead to the same decisions, describing a specific decision-making philosophy. Answer: FALSE Diff: 2 Keywords: maximax criteria, maximin criteria, minimax regret, decision criteria. Section: 19-1 Decision-Making Environments and Decision Criteria Outcome: 2 20) Some decision criteria take into account the probabilities associated with each outcome. One of the options, the expected-value criterion, refers to the long-run average outcome for a given alternative. Answer: FALSE Diff: 2 Keywords: decision criteria, probabilities, expected-value criteria, long-run average outcome. Section: 19-1 Decision-Making Environments and Decision Criteria Outcome: 3 21) When we apply the expected-value criterion, the best decision is to select the alternative with the highest average payoff or the lowest average loss. Answer: TRUE Diff: 3 Keywords: expected-value criterion, average payoff, average loss Section: 19-1 Decision-Making Environments and Decision Criteria Outcome: 4 22) The disadvantage of the expected-value criterion is that it does not take into account the decision maker’s attitude toward profit. Answer: FALSE Diff: 3 Keywords: expected-value criterion, profit, risk Section: 19-1 Decision-Making Environments and Decision Criteria Outcome: 4 23) In an uncertain environment, the best decision might not result in the best outcome. Answer: TRUE Diff: 2 Keywords: uncertain environment, best outcome Section: 19-2 Cost of Uncertainty Outcome: 5 24) A decision tree is a diagram that illustrates the correct ordering of actions and events in a decision analysis problem. Each act or event is represented by a branch on the decision tree. Answer: TRUE Diff: 1 Keywords: decision tree, decision analysis, branch on decision tree Section: 19-3 Decision Tree Analysis Outcome: 6

25) A decision tree provides a “road map” for organizing events, starting with the most recent events and choices first. Answer: FALSE Diff: 1 Keywords: decision tree, road map if events and choices Section: 19-3 Decision Tree Analysis Outcome: 6 26) Decision-tree analysis provides a way to take into account future decisions when making the most current decision. Answer: TRUE Diff: 2 Keywords: decision-tree analysis, future and current decisions Section: 19-3 Decision Tree Analysis Outcome: 6 27) A cash flow is defined as usually any dollar change (positive or negative) in the decision maker’s asset position. Answer: TRUE Diff: 2 Keywords: cash flow, decision maker, asset position Section: 19-3 Decision Tree Analysis Outcome: 6 28) Sensitivity analysis measures how sensitive a decision is to the number of possible outcomes. Answer: FALSE Diff: 2 Keywords: cash flow, decision maker, asset position Section: 19-3 Decision Tree Analysis Outcome: 6 29) Which one of the factors below would not affect the complexity of a decision? A) The number of alternatives available to the decision maker. B) The number of possible outcomes associated with each alternative. C) The general level of uncertainty associated with the decision. D) The probability of having a positive outcome. Answer: D Diff: 2 Keywords: complexity, decision, alternative, possible outcomes, positive outcome, uncertainty, probability Section: 19-1 Decision-Making Environments and Decision Criteria Outcome: 1 30) Decision makers must continually remind themselves that A) there is a difference between a good decision and a good outcome. B) not all decisions require the use of decision analysis. C) the complexity of the decision situation usually determines the usefulness of decision analysis. D) All of the above. 19-5 Copyright © 2018 Pearson Education, Inc.

Answer: D Diff: 2 Keywords: decision makers, good decision, good outcome, decision analysis Section: 19-1 Decision-Making Environments and Decision Criteria Outcome: 1 31) The expected-value criterion does not include which of the following steps: A) Compute the expected value for each decision alternative. B) Assign probabilities to the possible outcomes associated with each alternative. C) Define only the positive outcomes. D) Define the decision alternatives. Answer: C Diff: 2 Keywords: expected-value criterion, expected value, decision alternative, possible outcomes, positive outcomes Section: 19-1 Decision-Making Environments and Decision Criteria Outcome: 2 32) Which of the following are not examples of non-probabilistic decision criteria? A) Maximin decision criteria. B) Minimum outcome criteria. C) Maximax decision criteria. D) Minimax regret decision criteria. Answer: B Diff: 3 Keywords: non-probabilistic decision criteria, maximin decision criteria, maximax decision criteria, minimax regret decision criteria, outcome Section: 19-1 Decision-Making Environments and Decision Criteria Outcome: 2 33) In constructing a payoff table to compute the cost of uncertainty, which of the following defines the correct order of steps taken? A) Add the probabilities, compute the expected value for each decision, and determine the best decision using expected-value decision criterion. B) Compute the expected value for each decision alternative, calculate the probabilities, and determine the best decision using expected-value decision criterion. C) Add the probabilities and determine the best decision using expected-value decision criterion. D) Add the probabilities, compute the expected value for each decision, and determine the best decision using highest values. Answer: A Diff: 3 Keywords: cost of uncertainty, payoff table, probabilities, expected values, expected-value decision criterion. Section: 19-2 Cost of Uncertainty Outcome: 5

34) Select the best answer. A decision maker must identify the outcomes for each decision alternative and A) keeps the choices organized by constructing a table. B) determine the biggest money making alternative. C) assess probabilities associated with each outcome. D) all of the above. Answer: C Diff: 2 Keywords: outcomes, decision alternative, probabilities Section: 19-3 Decision-Tree Analysis Outcome: 6 35) Explain the primary difference between the business decision-making environments of certainty versus uncertainty. Answer: The certainty environment is predicated on the fact that we know the outcome for each alternative course of action. In the uncertainty environment, the decision maker does not know what outcome will occur when an alternative is selected. Diff: 2 Keywords: decision-making environments, environment of certainty, environment of uncertainty. Section: 19-1 Decision-Making Environments and Decision Criteria Outcome: 1 36) Explain the goal of decision analysis and identify three factors that affect the complexity of a decision. Answer: The goal of decision analysis is to focus on making good decisions, which in the long run should result in an increased number of good outcomes. Three factors that affect the complexity of a decision: the number of alternatives available to the decision maker, the number of possible outcomes associated with each alternative, and the general level of uncertainty associated with the decision. Diff: 3 Keywords: decision analysis, factors, complexity of decision Section: 19-1 Decision-Making Environments and Decision Criteria Outcome: 1 37) Explain as clearly as possible what the “best decision” is for a business problem and what the best decision will be associated with in a certainty environment. Answer: By “best decision,” we mean the alternative course of action, using all available information that best satisfies the decision criterion. In a certainty environment, the best decision will always be associated with the best outcome. Diff: 2 Keywords: decision analysis, factors, complexity of decision Section: 19-1 Decision-Making Environments and Decision Criteria Outcome: 1

38) Jamie Anderson is analyzing the following payoff table. She is trying to choose how many salads to make in advance each day before she knows the actual demand.

Table (Kaplan Financial) Her choice is between 40, 50, 60 and 70 salads. The actual demand can also vary between 40, 50, 60 and 70 with the probabilities as shown in the table - e.g. P(demand = 40) is 0.1. • The table then shows the profit or loss - for example, if he chooses to make 70 but demand is only 50, then he will make a loss of $60. Determine the maximax selection from this payoff table. • •

Answer: The maximax rule involves selecting the alternative that maximises the maximum payoff available. This approach would be suitable for an optimist, or 'risk-seeking' investor, who seeks to achieve the best results if the best happens. Looking at the payoff table, the highest maximum possible pay-off is $140. This happens if we make 70 salads and demand is also 70. Jamie should therefore decide to supply 70 salads every day. Diff: 3 Keywords: payoff table, maximax rule, maximum payoff Section: 19-1 Decision-Making Environments and Decision Criteria Outcome: 2 39) Jamie Anderson is analyzing the following payoff table. She is trying to choose how many salads to make in advance each day before she knows the actual demand.

Table (Kaplan Financial) • •

Her choice is between 40, 50, 60 and 70 salads. The actual demand can also vary between 40, 50, 60 and 70 with the probabilities as shown in the table - e.g. P(demand = 40) is 0.1. • The table then shows the profit or loss - for example, if he chooses to make 70 but demand is only 50, then he will make a loss of $60. Determine the maximin selection from this payoff table. Answer: The maximin rule involves selecting the alternative that maximizes the minimum pay-off achievable. The investor would look at the worst possible outcome at each supply level, then selects the highest one of these. The decision maker therefore chooses the outcome which is guaranteed to minimize his losses. In the process, he loses out on the opportunity of making big profits. This approach would be appropriate for a pessimist who seeks to achieve the best results if the worst happens. Looking at the payoff table, If we decide to supply 40 salads, the minimum pay-off is $80. If we decide to supply 50 salads, the minimum pay-off is $0. If we decide to supply 60 salads, the minimum pay-off is ($80). 19-8 Copyright © 2018 Pearson Education, Inc.

If we decide to supply 70 salads, the minimum pay-off is ($160). The highest minimum payoff arises from supplying 40 salads. This ensures that the worst possible scenario still results in a gain of at least $80. Diff: 3 Keywords: payoff table, maximin rule, maximum payoff Section: 19-1 Decision-Making Environments and Decision Criteria Outcome: 2 40) Explain the difference between the two main categories of decision criteria. Answer: The two main categories of decision criteria are nonprobabilistic and probabilistic. Nonprobabilistic criteria are used when either the probabilities associated with the possible payoffs are unknown or the decision maker lacks confidence and/or information with which to assess probabilities for the various payoffs. Probabilistic criteria incorporate the decision maker’s assessment of the probability of each state of nature occurring. Diff: 2 Keywords: decision criteria, nonprobabilistic, probabilistic, payoffs, probabilities, state of nature Section: 19-1 Decision-Making Environments and Decision Criteria Outcome: 2 41) What is the disadvantage of the maximax and maximin criteria? What is another alternative approach? Answer: A disadvantage of the maximax and maximin criteria is that they use only one value from the payoff table to make a decision. In analyzing the decision situation, it may be useful to determine how much damage making the wrong choice would cause. If someone decided to use the minimax regret criterion, they would need to know the value of the opportunity loss. Diff: 2 Keywords: decision criteria, nonprobabilistic, probabilistic, payoffs, probabilities, state of nature Section: 19-1 Decision-Making Environments and Decision Criteria Outcome: 3 42) Identify the expected value formula and include a description of the calculation and each of the variables. k

Answer: E ( x) =  xi P( xi ) i =1

The variables in the expected value formula are: i = 1 to k indicates the number of outcomes xi is alternative values for each outcome P(xi) is the probabilities associated with each possible outcome The formula adds the results of the value of each outcome times its probability. Diff: 3 Keywords: expected value, outcomes, alternatives, probabilities Section: 19-1 Decision-Making Environments and Decision Criteria Outcome: 3

43) The following table defines two outcomes for two specified alternatives and their associated probabilities. Calculate the expected value based on these numbers and its associated interpretation. x $25 -$2

P(x) 0.40 0.60

Answer: The expected value is computed using the formula: k

E ( x) =  xi P( xi ) i =1

Resulting in E(x) = ($25)(0.40) + (-$2)(0.60) = $10 + $1.2 = $11.20 This means that over the long run, we would have an expected $11.20 gain in units if we elected option 1. Diff: 2 Keywords: outcomes, alternatives, probabilities payoffs, expected value Section: 19-1 Decision-Making Environments and Decision Criteria Outcome: 4 44) Identify the 4 steps of the expected-value criterion. Answer: The 4 steps of the expected-value criterion: Step 1: Define the decision alternatives. Step 2: Define the possible outcomes (states of nature) associated with each alternative. Step 3: Assign probabilities to the possible outcomes associated with each alternative. Step 4: Compute the expected value for each decision alternative. Diff: 3 Keywords: expected-value criterion, decision alternatives, states of nature, possible outcomes, probabilities, expected value Section: 19-1 Decision-Making Environments and Decision Criteria Outcome: 4 45) Identify the 8 steps of computing the cost of uncertainty. Answer: The 4 steps of the expected-value criterion: Step 1: Determine the possible outcomes (states of nature). Step 2: Assess the probability for each possible outcome. Step 3: Assign appropriate revenue and cost values and establish the payoff function. Step 4: Construct the payoff table. Step 5: Add the probabilities to the payoff table and compute the expected value for each production decision. Step 6: Determine the best decision using the expected-value decision criterion. Step 7: Compute the expected value under certainty. Step 8: Compute the cost of certainty. Diff: 3 Keywords: cost of uncertainty, expected-value criterion, states of nature, probability for each outcome, payoff function, payoff table, expected value under uncertainty Section: 19-2 Cost of Uncertainty Outcome: 5

46) Identify the 4 steps to constructing a decision tree. Answer: The 4 steps to constructing a decision tree: Step 1: Grow the decision tree - organize the decisions and events in chronological order. Step 2: Assign probabilities to the event outcomes on the tree. Step 3: Assign the cash flows to the tree. Step 4: Fold back the decision tree and compute the expected values for each decision. Diff: 3 Keywords: decision tree, event outcomes, probabilities, cash flows, expected values. Section: 19-3 Decision-Tree Analysis Outcome: 6 47) Describe the process of growing a decision tree and why it is such a valuable method of making a business decision. Answer: We develop the decision tree by organizing the decisions and events in chronological order. When finished, the decision tree should show all the decisions and events. The value of decision-tree analysis is that a tree helps you visually structure the decision problem and systematically analyze the decision alternatives. The decision tree should reflect the decision problem as accurately as possible before any further analysis is performed using the tree. Decision-tree analysis also provides a way for you to take into account future decisions when making the most current decision. Diff: 2 Keywords: decision tree, business decisions, decision alternatives Section: 19-3 Decision-Tree Analysis Outcome: 6 48) Define sensitivity analysis and how it related to cash flow. Answer: Sensitivity analysis determines how sensitive the decision to the probabilities being assessed. Sensitivity analysis can also investigate how much a cash-flow value would have to change before the decision would change. Diff: 3 Keywords: sensitivity analysis, cash flow, probabilities cash-flow value Section: 19-3 Decision-Tree Analysis Outcome: 6 49) Define the process of “folding back the tree” in decision tree analysis. Answer: To fold back the decision tree, we begin with the end values at the right of the tree and work our way back to the initial decision at the far left. To do this, we must determine the expected value of each decision branch. Diff: 2 Keywords: decision tree analysis, expected value, decision branch Section: 19-3 Decision-Tree Analysis Outcome: 6 50) Decision tree analysis could be missing branches on the tree. Define one important consideration that could be left out. Answer: Unfortunately, some important branches of the tree, such as ethical considerations, may be missing since they are not easily converted into financial terms. Diff: 2 Keywords: decision tree analysis, expected value, decision branch 19-11 Copyright © 2018 Pearson Education, Inc.

Section: 19-3 Decision-Tree Analysis Outcome: 6

Business Statistics, 9e (Groebner/Shannon/Fry) Chapter 20 Introduction to Quality and Statistical Process Control 1) We expect virtually all the data in a stable process to fall within 2 standard deviations of the mean. Answer: FALSE Diff: 1 Keywords: quality, standard deviation, stable, process Section: 18-1 Introduction to Statistical Process Control Charts Outcome: 1 2) Increased variation in processes and products can be detected as finer gauge measurement devices are used. Answer: TRUE Diff: 1 Keywords: variation, processes, measurement devices Section: 18-1 Introduction to Statistical Process Control Charts Outcome: 1 3) Special cause variation is variation in the output of a process that is naturally occurring and expected and that may be the result of random causes. Answer: FALSE Diff: 2 Keywords: quality, special, random, variation Section: 18-1 Introduction to Statistical Process Control Charts Outcome: 1 4) Common cause variation is variation in the output of a process that is unexpected and has an assignable cause. Answer: FALSE Diff: 2 Keywords: quality, common, variation, process, assignable Section: 18-1 Introduction to Statistical Process Control Charts Outcome: 1 5) In process improvement efforts, the goal is to first remove the common cause variation and then to reduce the special cause variation in a system. Answer: FALSE Diff: 3 Keywords: quality, process, common, special, cause Section: 18-1 Introduction to Statistical Process Control Charts Outcome: 1 6) The six most common sources of variation are people, machines, materials, methods, measurement, and environment. Answer: TRUE Diff: 2 Keywords: sources of variation 20-1 Copyright © 2018 Pearson Education, Inc.

Section: 18-1 Introduction to Statistical Process Control Charts Outcome: 1 7) Process control charts are used to provide signals to indicate when the output of a process is out of control. Answer: TRUE Diff: 1 Keywords: quality, process, control Section: 18-1 Introduction to Statistical Process Control Charts Outcome: 1 8) If a process control chart has only one point outside the upper or lower control limits, there is insufficient evidence to conclude that the process was out of control at the time that the measurement was taken. Answer: FALSE Diff: 1 Keywords: quality, process control Section: 18-1 Introduction to Statistical Process Control Charts Outcome: 1 9) The issue with a process always involves a quantitative variable. Answer: FALSE Diff: 1 Keywords: process, quantitative, variation, process Section: 18-1 Introduction to Statistical Process Control Charts Outcome: 1 10) Total process variation is made up of the sum of common cause variation and special cause variation. Answer: TRUE Diff: 2 Keywords: quality, constant-cause, system Section: 18-1 Introduction to Statistical Process Control Charts Outcome: 1 11) A stable process is one that has had all its variation removed through quality improvement efforts on the part of the organization. Answer: FALSE Diff: 2 Keywords: quality, stable, process, variation Section: 18-1 Introduction to Statistical Process Control Charts Outcome: 1 12) A stable process is typically defined as one in which all output is operating within ± 3 standard deviations of the process center. Answer: TRUE Diff: 2 Keywords: quality, stable, process, standard deviation Section: 18-1 Introduction to Statistical Process Control Charts 20-2 Copyright © 2018 Pearson Education, Inc.

Outcome: 1 13) A p-chart would potentially be used to monitor the diameters of bolts made by a bolt manufacturing plant. Answer: FALSE Diff: 2 Keywords: quality, p-chart, control chart Section: 18-1 Introduction to Statistical Process Control Charts Outcome: 2 14) Both p-charts and c-charts are designed for use when the data we are working with are referred to as attribute data. Answer: TRUE Diff: 2 Keywords: quality, p-chart, c-chart, attribute Section: 18-1 Introduction to Statistical Process Control Charts Outcome: 2 15) The frequency distribution of most processes' statistics will begin to resemble the shape of the normal distribution as the values are collected and grouped into classes. Answer: TRUE Diff: 2 Keywords: variation, normal distribution Section: 18-1 Introduction to Statistical Process Control Charts Outcome: 2 16) It is entirely possible for the R-chart to show that a process is in statistical control and the -chart to show that the same process is out of control. Answer: TRUE Diff: 2 Keywords: quality, R-chart, x-bar chart, process, control Section: 18-1 Introduction to Statistical Process Control Charts Outcome: 2 17) The control limits in the x-bar chart are set so that 95 percent of the values will fall inside the control limits when there is only common cause variation. Answer: FALSE Diff: 2 Keywords: quality, x-bar chart, control limits Section: 18-1 Introduction to Statistical Process Control Charts Outcome: 2 18) One of the roles of managers who are overseeing the statistical process control analysis is to set the control limits at the desired levels prior to collecting data from the process. Answer: FALSE Diff: 2 Keywords: quality, control, limits, process Section: 18-1 Introduction to Statistical Process Control Charts 20-3 Copyright © 2018 Pearson Education, Inc.

Outcome: 2 19) Control charts monitor a process as it currently operates, not necessarily how you would like it to operate. Answer: TRUE Diff: 2 Keywords: control charts, process, control, Section: 18-1 Introduction to Statistical Process Control Charts Outcome: 3 20) A process control chart can be used to determine whether the process average has shifted up or down, but is not useful for determining whether the process is just drifting in an upward or downward direction. Answer: TRUE Diff: 2 Keywords: quality, process, control chart, drift, shift Section: 18-1 Introduction to Statistical Process Control Charts Outcome: 3 21) The statistical process control (SPC) chart is one of the most important tools for identifying important issues to improve quality. Answer: TRUE Diff: 1 Keywords: quality, statistical, process control Section: 18-1 Introduction to Statistical Process Control Charts Outcome: 3 22) Because variations are unavoidable in a system, the output of the system is always unpredictable. Answer: FALSE Diff: 3 Keywords: quality, variation, process, control Section: 18-1 Introduction to Statistical Process Control Charts Outcome: 3 23) Which of the following process changes cannot be detected with a process control chart? A) The process average is trending up or down from normal. B) The process average has shifted up or down from normal. C) The process is behaving in such a manner that the existing variation is not random in nature. D) All of the above. Answer: D Diff: 2 Keywords: process control chart, variation, random Section: 18-1 Introduction to Statistical Process Control Charts Outcome: 1

24) Which of the following is not among the most common sources of variation? A) People B) Materials C) Methods D) Quotas Answer: D Diff: 2 Keywords: quality, variation, common cause Section: 18-1 Introduction to Statistical Process Control Charts Outcome: 1 25) The x-bar chart is based on the principles of which distribution? A) t-distribution B) Chi square distribution C) F distribution D) Normal distribution Answer: D Diff: 2 Keywords: quality, x-bar chart, normal distribution Section: 18-1 Introduction to Statistical Process Control Charts Outcome: 1 26) Which of the following is the most complete group of commonly used process control charts other than the x-bar chart? A) R-chart, c-chart B) R-chart, p-chart, c-chart C) p-chart, R-chart, n-chart D) R-chart, p-chart Answer: B Diff: 2 Keywords: quality, control charts Section: 18-1 Introduction to Statistical Process Control Charts Outcome: 2 27) Which of the following statements is correct? A) A process can be in statistical control, yet it can be producing defects in abundance. B) At least three points outside the upper or lower control limits on a control chart are required before the process is deemed to be out of control. C) If a process is out of control, then the variation that is present is limited to common cause variation. D) When special cause variation is present, the process can be expected to be in control. Answer: A Diff: 3 Keywords: quality, statistical control Section: 18-1 Introduction to Statistical Process Control Charts Outcome: 1

28) The main process change that can be detected with a process control chart is: A) the process average has shifted up or down from normal. B) the process average is trending up or down from normal. C) the process is behaving in such a manner that the existing variation is not random in nature. D) All of the above Answer: D Diff: 2 Keywords: quality, control chart Section: 18-1 Introduction to Statistical Process Control Charts Outcome: 1 29) The control limits in a control chart can be interpreted to mean: A) any value falling outside the limits is a defect and the product should be discarded. B) the range of virtually all special cause variation. C) any value falling within the limits means the product is high quality. D) the range of virtually all common cause variation. Answer: D Diff: 2 Keywords: quality, control chart, control limits Section: 18-1 Introduction to Statistical Process Control Charts Outcome: 1 30) A p-chart is useful for: A) analyzing whether a process for a measurable variable is in or out of control. B) analyzing processes which yield attribute data. C) determining what the most likely cause of defects is. D) All of the above Answer: B Diff: 2 Keywords: quality, p-chart, attribute, control chart Section: 18-1 Introduction to Statistical Process Control Charts Outcome: 2 31) A plywood manufacturer is interested in monitoring the thickness of the plywood. Which of the following would be most useful for doing this? A) p-charts B) c-charts C) -charts D) Histograms Answer: C Diff: 2 Keywords: quality, control chart, x-bar chart Section: 18-1 Introduction to Statistical Process Control Charts Outcome: 2

32) Which of the following in not an out of control signal for an x-bar chart? A) One or more points outside the control limits B) Seven or more consecutive points that all fall on the same side of the center line C) Six or more consecutive points moving in the same direction (an upward or downward trend) D) Fourteen points in a row, alternating up and down Answer: B Diff: 2 Keywords: quality, out of control signals Section: 18-1 Introduction to Statistical Process Control Charts Outcome: 1 33) Recently a shipping company took 30 samples, each of size n = 100, of packages that it was responsible for delivering. Out of the 3,000 total packages, 300 were delivered late. In setting up an appropriate process control chart, what would be the correct 3-sigma upper control limit value? A) 0.03 B) 0.13 C) 0.19 D) 0.07 Answer: C Diff: 3 Keywords: quality, p-chart, control limit Section: 18-1 Introduction to Statistical Process Control Charts Outcome: 2 34) Each evening, a nationwide retail chain randomly calls 100 of the customers who came to their store that day to ask whether they were satisfied with the service they had received. The customers respond yes or no. Suppose the company has found over time that 8 percent of the customers are not satisfied ("no" answers), what is the 3-sigma upper and lower control limits for the appropriate control chart? A) About .053 and .107 B) 0 to about .16 C) -.0.14 to about .16 D) About -1.96 to 1.96 Answer: B Diff: 3 Keywords: quality, control limit, p-chart Section: 18-1 Introduction to Statistical Process Control Charts Outcome: 2

35) Each evening, a nationwide retail chain randomly calls 100 of the customers who came to their store that day to ask whether they were satisfied with the service they had received. The customers respond yes or no. Suppose the company has found over time that 8 percent of the customers are not satisfied ("no" answers). If they have established a process control chart, what conclusion should be reached if the percentage of customers surveyed tonight that say no is 14 percent? A) This result indicates that a special cause situation exists. B) Although this point is above the upper control limit, there is no cause for alarm if this is the first time. C) While this value is higher than "normal," it is still within the range of common cause variation and no action is needed. D) This is outside the control limits and action should be taken Answer: C Diff: 3 Keywords: quality, p-chart, common cause Section: 18-1 Introduction to Statistical Process Control Charts Outcome: 2 36) A company that fills soft drinks into bottles wishes to establish an -chart to monitor the average fill level in the bottles. To do this, the company has taken a series of samples of size n = 4 bottles. The overall average fill is 12.03 ounces. The average range for the subgroups has been .06 ounces. Based on this information, what is the upper limit of the 3-sigma control limit? A) .729 B) .0437 C) 12.09 D) 12.074 Answer: D Diff: 3 Keywords: quality, x-bar, control limit Section: 18-1 Introduction to Statistical Process Control Charts Outcome: 1 37) A company that fills soft drinks into bottles has established an -chart and an R-chart to monitor the average fill level in the bottles. To do this, the company has taken a series of samples of size n = 4 bottles. The overall average fill is 12.03 ounces. The average range for the subgroups has been .06 ounces. Suppose, after developing the control chart, a subgroup of size 4 yields a sample mean of 12.09 ounces and a range of .08, which of the following statements is true? A) The process is in control on both the -chart and the R-chart. B) The process is out of control on the R-chart but in control on the -chart. C) The process is out of control on the -chart but in control on the R-chart. D) The process is out of control on both the -chart and the R-chart. Answer: C Diff: 3 Keywords: quality, control chart, x-bar, r, control Section: 18-1 Introduction to Statistical Process Control Charts Outcome: 1

38) Identify the four primary signals that indicate a change and that, if observed, will cause us to reject the null hypothesis. State the probability of a Type I error for this definition. Answer: There are four primary signals that indicate a change and that, if observed, will cause us to reject the null hypothesis. 1. One or more points outside the upper or lower control limits 2. Nine or more points in a row above or below the centerline 3. Six or more consecutive points moving in the same direction (increasing or decreasing) 4. Fourteen points in a row, alternating up and down These signals were derived such that the probability of a Type I error is less than 0.01. Diff: 3 Keywords: null hypothesis, signals, variation, control limits, centerline Section: 18-1 Introduction to Statistical Process Control Charts Outcome: 1 39) Statistical Process Control charts are used to detect whether a process remains in control or whether it has gone out of control. Explain how the SPC signals work. Answer: The control limits on an SPC chart are typically set at the 3-sigma level. This means that if the process remains unchanged, approximately 99.7 percent of the observations will fall within the control limits. Thus, under normal circumstances it would be most unusual for a measurement to fall outside the upper or lower control limits if nothing has changed. The signals that something has changed are: 1. one or more points outside the control limits 2. nine or more points in a row above or below the center line 3. six or more points moving in the same direction (up or down). These limits are all set such that the probability of one of these events happening if the process is actually unchanged (in control) is very small. Therefore, if such an event exists, we should conclude that something has changed and the process is out of control. These rules guard against the Type I error— saying the process is out of control when it isn't. Diff: 2 Keywords: quality, SPC, statistical process control Section: 18-1 Introduction to Statistical Process Control Charts Outcome: 1

40) Referring to the SPC chart signals that a process is out of control, what type of problem does each signal indicate? List the signals. Answer: The signals that something has changed are: 1. one or more points outside the control limits 2. nine or more points in a row above or below the center line 3. six or more points moving in the same direction (up or down) 4. 14 points in a row, alternating up and down. The first signal means that something has caused the process to produce a higher (or lower) measure than would be normally expected. This could be a problem with variability or a jump one way or the other in the process center. Only one such point above or below the control limits is needed to reach the conclusion that a special cause situation exists. The second signal implies that the process center is drifting up or down. For example, this type of situation could occur in a process that is sensitive to temperature. If for some reason temperature gradually is increasing, the output measure will gradually change with the temperature change. The third signal indicates that the process average has shifted (higher or lower) due to some special circumstance. For instance, a control setting may have been changed causing the output measures to increase or decrease consistently. The fourth signal indicates something is causing the process to change between measurements. Diff: 2 Keywords: quality, SPC, control chart, signal Section: 18-1 Introduction to Statistical Process Control Charts Outcome: 1

41) A tire company is interested in monitoring the process that produced tread thickness on its tires. Every hour 4 tires are selected from production and the tread thickness is measured. Data for the past 25 days is shown as follows:

a. What type of control chart would you recommend be used in this case? b. Compute the upper and lower control limits for these data. Answer: a. The appropriate statistical process control chart to use in this situation is an -chart and an R-chart. The reason for this is that the variable of interest is a measurable variable (tread thickness in inches) and the subgroup or sample size is n = 4. b. The control limits are calculated using: UCL = + A2 and LCL = - A2 The first step is to compute the mean and range for each subgroup. This is shown as follows:

Now, we find the grand mean by: =

= .508

Thus, for the data in the sample, the overall mean value is .508 inches. This will form the centerline on the process control chart. Next, we find the average range in the 25 samples using: =

= .23642

Next, we go to the table of control factors to find the A2 value for subgroup size of n = 4. This value is .729. So the control limits are: UCL = + A2 = .508 + .729(.236) = .68 and LCL = - A2 = .508 - .729(.236) = .336 Thus, the control limits for the chart are LCL = .336 and UCL = .68 The R chart is computed by first finding the D3 and D4 factors from the table of control chart factors for subgroup size n = 4. These are: D3 = 0 and D4 = 2.282.

The R-chart control limits are computed as follows: LCL = D3 = 0(.236) = 0.00 UCL = D4 = 2.282(.236) = .5386 Diff: 3 Keywords: quality, control chart, SPC, x-bar chart, r-chart Section: 18-1 Introduction to Statistical Process Control Charts Outcome: 2 42) Explain the relationship between control limits and specification limits. Answer: Control limits are computed from data collected from the actual process. These control limits define the range of inherent variation in the process as it actually exists. The key is they are computed from the data. Managers do not get to determine what these limits are. They have nothing directly to do with the product specifications. On the other hand, specification limits are the range within which the product of service is considered to be acceptable quality. The specification limits are set by the customer, by management, or by engineers based on what defines a quality product. These limits have no direct relationship to how the process may actually be performing other than they represent the range within which we hope all product or service measurements fall. Thus, it is very possible for a process to be deemed to be in control yet be producing a product that is not within specifications. Diff: 2 Keywords: quality, SPC, control limit, specification limit Section: 18-1 Introduction to Statistical Process Control Charts Outcome: 3

43) A major airline is interested in monitoring customer satisfaction with its baggage handling process. To do so, each day the airline randomly selects 100 customers and surveys them to determine if they are satisfied or not with the service provided. After 20 samples, a total of 260 unsatisfied customers were surveyed. a. If the airline wishes to use a control chart, which chart would you recommend and why? b. Determine the 3-sigma control limits for the appropriate control chart. Answer: a. The appropriate control chart in this case would be a p-chart since the variable of interest is an attribute variable (satisfied or unsatisfied). b. For the p-chart, the first step is to calculate the process average. This is computed using: =

= .13

Next, the standard deviation for a proportion is computed using: s=

= .0336

The control limits are: LCL = - 3 = .13 - 3(.0336) = .0292 and UCL = + 3 = .13 + 3(.0336) = .2308 Thus, as long as the level of dissatisfaction remains within .0292 and .2308 each day, the process should be considered in control. Does that mean the airline should be pleased with the level of dissatisfaction? No, not necessarily, but since the process is in control they should now go to work to improve it. Diff: 2 Keywords: quality, SPC, control limit, specification limit Section: 18-1 Introduction to Statistical Process Control Charts Outcome: 2