Basic Business Statistics Concepts and Applications 11th Edition Test Bank by QuentinAxel

Introduction and Data Collection

CHAPTER 1: INTRODUCTION AND DATA COLLECTION 1. The process of using sample statistics to draw conclusions about true population parameters is called a) statistical inference. b) the scientific method. c) sampling. d) descriptive statistics. ANSWER: a TYPE: MC DIFFICULTY: Easy KEYWORDS: inferential statistics 2. Those methods involving the collection, presentation, and characterization of a set of data in order to properly describe the various features of that set of data are called a) statistical inference. b) the scientific method. c) sampling. d) descriptive statistics. ANSWER: d TYPE: MC DIFFICULTY: Easy KEYWORDS: descriptive statistics 3. The collection and summarization of the socioeconomic and physical characteristics of the employees of a particular firm is an example of a) inferential statistics. b) descriptive statistics. c) a parameter. d) a statistic. ANSWER: b TYPE: MC DIFFICULTY: Easy KEYWORDS: descriptive statistics 4. The estimation of the population average family expenditure on food based on the sample average expenditure of 1,000 families is an example of a) inferential statistics. b) descriptive statistics. c) a parameter. d) a statistic. ANSWER: a TYPE: MC DIFFICULTY: Easy KEYWORDS: inferential statistics

Introduction and Data Collection 5. The universe or "totality of items or things" under consideration is called a) a sample. b) a population. c) a parameter. d) a statistic. ANSWER: b TYPE: MC DIFFICULTY: Easy KEYWORDS: population 6.

The portion of the universe that has been selected for analysis is called a) a sample. b) a frame. c) a parameter. d) a statistic.

ANSWER: a TYPE: MC DIFFICULTY: Easy KEYWORDS: sample 7. A summary measure that is computed to describe a characteristic from only a sample of the population is called a) a parameter. b) a census. c) a statistic. d) the scientific method. ANSWER: c TYPE: MC DIFFICULTY: Easy KEYWORDS: statistic 8. A summary measure that is computed to describe a characteristic of an entire population is called a) a parameter. b) a census. c) a statistic. d) the scientific method. ANSWER: a TYPE: MC DIFFICULTY: Easy KEYWORDS: parameter 9. Which of the following is most likely a population as opposed to a sample? a) respondents to a newspaper survey b) the first 5 students completing an assignment c) every third person to arrive at the bank d) registered voters in a county ANSWER: d

Introduction and Data Collection TYPE: MC DIFFICULTY: Moderate KEYWORDS: population, sample 10. Which of the following is most likely a parameter as opposed to a statistic? a) the average score of the first five students completing an assignment b) the proportion of females registered to vote in a county c) the average height of people randomly selected from a database d) the proportion of trucks stopped yesterday that were cited for bad brakes ANSWER: b TYPE: MC DIFFICULTY: Moderate KEYWORDS: parameter, statistic 11. Which of the following is not an element of descriptive statistical problems? a) an inference made about the population based on the sample b) the population or sample of interest c) tables, graphs, or numerical summary tools d) identification of patterns in the data ANSWER: a TYPE: MC DIFFICULTY: Moderate KEYWORDS: descriptive statistics

12. A study is under way in Yosemite National Forest to determine the adult height of American pine trees. Specifically, the study is attempting to determine what factors aid a tree in reaching heights greater than 60 feet tall. It is estimated that the forest contains 25,000 adult American pines. The study involves collecting heights from 250 randomly selected adult American pine trees and analyzing the results. Identify the population from which the study was sampled. a) the 250 randomly selected adult American pine trees b) the 25,000 adult American pine trees in the forest c) all the adult American pine trees taller than 60 feet d) all American pine trees, of any age, in the forest ANSWER: b TYPE: MC DIFFICULTY: Moderate KEYWORDS: population, sample

13. A study is under way in Yosemite National Forest to determine the adult height of American pine trees. Specifically, the study is attempting to determine what factors aid a tree in reaching heights greater than 60 feet tall. It is estimated that the forest contains 25,000 adult American pines. The study involves collecting heights from 250 randomly selected adult American pine trees and analyzing the results. Identify the variable of interest in the study a) the age of an American pine tree in Yosemite National Forest b) the height of an American pine tree in Yosemite National Forest c) the number of American pine trees in Yosemite National Forest d) the species of trees in Yosemite National Forest ANSWER: b

Introduction and Data Collection TYPE: MC DIFFICULTY: Easy KEYWORDS: data, sampling 14. A study is under way in Yosemite National Forest to determine the adult height of American pine trees. Specifically, the study is attempting to determine what factors aid a tree in reaching heights greater than 60 feet tall. It is estimated that the forest contains 25,000 adult American pines. The study involves collecting heights from 250 randomly selected adult American pine trees and analyzing the results. Identify the sample in the study. a) the 250 randomly selected adult American pine trees b) the 25,000 adult American pine trees in the forest c) all the adult American pine trees taller than 60 feet d) all American pine trees, of any age, in the forest ANSWER: a TYPE: MC DIFFICULTY: Easy KEYWORDS: population, sample 15. Most analysts focus on the cost of tuition as the way to measure the cost of a college education. But incidentals, such as textbook costs, are rarely considered. A researcher at Drummand University wishes to estimate the textbook costs of first-year students at Drummand. To do so, she monitored the textbook cost of 250 first-year students and found that their average textbook cost was $300 per semester. Identify the population of interest to the researcher. a) all Drummand University students b) all college students c) all first-year Drummand University students d) the 250 students that were monitored ANSWER: c TYPE: MC DIFFICULTY: Easy KEYWORDS: population, sample 16. Most analysts focus on the cost of tuition as the way to measure the cost of a college education. But incidentals, such as textbook costs, are rarely considered. A researcher at Drummand University wishes to estimate the textbook costs of first-year students at Drummand. To do so, she monitored the textbook cost of 250 first-year students and found that their average textbook cost was $300 per semester. Identify the variable of interest to the researcher a) the textbook cost of first-year Drummand University students b) the year in school of Drummand University students c) the age of Drummand University students d) the cost of incidental expenses of Drummand University students ANSWER: a TYPE: MC DIFFICULTY: Easy KEYWORDS: data, sampling 17. Most analysts focus on the cost of tuition as the way to measure the cost of a college education. But incidentals, such as textbook costs, are rarely considered. A researcher at Drummand University wishes to estimate the textbook costs of first-year students at Drummand. To do so, she monitored the textbook cost of 250 first-year students and found that their average textbook cost was $300 per semester. Identify the sample in the study.

Introduction and Data Collection

a) all Drummand University students b) all college students c) all first-year Drummand University students d) the 250 students that were monitored ANSWER: d TYPE: MC DIFFICULTY: Easy KEYWORDS: population, sample 18. Researchers suspect that the average number of units earned per semester by college students is rising. A researcher at Calendula College wishes to estimate the number of units earned by students during the spring semester at Calendula. To do so, he randomly selects 100 student transcripts and records the number of units each student earned in the spring term. He found that the average number of semester units completed was 12.96 units per student. Identify the population of interest to the researcher. a) all Calendula College students b) all college students c) all Calendula College students enrolled in the spring d) all college students enrolled in the spring ANSWER: c TYPE: MC DIFFICULTY: Moderate KEYWORDS: population, sample 19. The average number of units earned per semester by college students is suspected to be rising. A researcher at Calendula College wishes to estimate the number of units earned by students during the spring semester at Calendula. To do so, he randomly selects 100 student transcripts and records the number of units each student earned in the spring term. Identify the variable of interest to the researcher. a) the number of students enrolled at Calendula College during the spring term b) the average indebtedness of Calendula College students enrolled in the spring c) the age of Calendula College students enrolled in the spring d) the number of units earned by Calendula College students during the spring term ANSWER: d TYPE: MC DIFFICULTY: Easy KEYWORDS: data, sampling 20. Jared was working on a project to look at global warming and accessed an Internet site where he captured average global surface temperatures from 1866. Which of the four methods of data collection was he using? a) published sources b) experimentation c) surveying d) observation ANSWER: a TYPE: MC DIFFICULTY: Easy

Introduction and Data Collection KEYWORDS: sources of data 21. The British Airways Internet site provides a questionnaire instrument that can be answered electronically. Which of the 4 methods of data collection is involved when people complete the questionnaire? a) published sources b) experimentation c) surveying d) observation ANSWER: c TYPE: MC DIFFICULTY: Easy KEYWORDS: sources of data 22. A marketing research firm, in conducting a comparative taste test, provided three types of peanut butter to a sample of households randomly selected within the state. Which of the 4 methods of data collection is involved when people are asked to compare the three types of peanut butter? a) published sources b) experimentation c) surveying d) observation ANSWER: b TYPE: MC DIFFICULTY: Easy KEYWORDS: sources of data

23. Tim was planning for a meeting with his boss to discuss a raise in his annual salary. In preparation, he wanted to use the Consumer Price Index to determine the percentage increase in his real (inflation-adjusted) salary over the last three years. Which of the 4 methods of data collection was involved when he used the Consumer Price Index? a) published sources b) experimentation c) surveying d) observation ANSWER: a TYPE: MC DIFFICULTY: Easy KEYWORDS: sources of data 24. Which of the 4 methods of data collection is involved when a person counts the number of cars passing designated locations on the Los Angeles freeway system? a) published sources b) experimentation c) surveying d) observation ANSWER: d TYPE: MC DIFFICULTY: Moderate

Introduction and Data Collection

KEYWORDS: sources of data 25. A statistics student found a reference in the campus library that contained the median family incomes for all 50 states. She would report her data as being collected using a) a designed experiment. b) observational data. c) a random sample. d) a published source. ANSWER: d TYPE: MC DIFFICULTY: Easy KEYWORDS: sources of data 26. The personnel director at a large company studied the eating habits of the company’s employees. The director noted whether employees brought their own lunches to work, ate at the company cafeteria, or went out to lunch. The goal of the study was to improve the food service at the company cafeteria. This type of data collection would best be considered as a) an observational study. b) a designed experiment. c) a random sample. d) a quota sample. ANSWER: a TYPE: MC DIFFICULTY: Easy KEYWORDS: sources of data 27. A study attempted to estimate the proportion of Florida residents who were willing to spend more tax dollars on protecting the beaches from environmental disasters. Twenty-five hundred Florida residents were surveyed. What type of data collection procedure was most likely used to collect the data for this study? a) a designed experiment b) a published source c) a random sample d) observational data ANSWER: c TYPE: MC DIFFICULTY: Easy KEYWORDS: sources of data 28. Which of the following is a discrete quantitative variable? a) the Dow Jones Industrial average b) the volume of water released from a dam c) the distance you drove yesterday. d) the number of employees of an insurance company ANSWER: d TYPE: MC DIFFICULTY: Easy KEYWORDS: discrete random variable, types of data

Introduction and Data Collection 29. Which of the following is a continuous quantitative variable? a) the color of a student’s eyes b) the number of employees of an insurance company c) the amount of milk produced by a cow in one 24-hour period d) the number of gallons of milk sold at the local grocery store yesterday ANSWER: c TYPE: MC DIFFICULTY: Easy KEYWORDS: continuous random variable, types of data

30. To monitor campus security, the campus police office is taking a survey of the number of students in a parking lot each 30 minutes of a 24-hour period with the goal of determining when patrols of the lot would serve the most students. If X is the number of students in the lot each period of time, then X is an example of a) a categorical random variable. b) a discrete random variable. c) a continuous random variable. d) a statistic. ANSWER: b TYPE: MC DIFFICULTY: Moderate KEYWORDS: discrete random variable, types of data

31. Researchers are concerned that the weight of the average American school child is increasing

implying, among other things, that children’s clothing should be manufactured and marketed in larger sizes. If X is the weight of school children sampled in a nationwide study, then X is an example of a) a categorical random variable. b) a discrete random variable. c) a continuous random variable. d) a parameter.

ANSWER: c TYPE: MC DIFFICULTY: Moderate KEYWORDS: continuous random variable, types of data 32. The classification of student class designation (freshman, sophomore, junior, senior) is an example of a) a categorical random variable. b) a discrete random variable. c) a continuous random variable. d) a parameter. ANSWER: a TYPE: MC DIFFICULTY: Easy KEYWORDS: categorical random variable, types of data 33. The classification of student major (accounting, economics, management, marketing, other) is an example of

Introduction and Data Collection a) b) c) d)

a categorical random variable. a discrete random variable. a continuous random variable. a parameter.

ANSWER: a TYPE: MC DIFFICULTY: Easy KEYWORDS: categorical random variable, types of data

34. The chancellor of a major university was concerned about alcohol abuse on her campus and wanted to find out the proportion of students at her university who visited campus bars on the weekend before the final exam week. Her assistant took a random sample of 250 students. The total number of students in the sample who visited campus bars on the weekend before the final exam week is an example of a) a categorical random variable. b) a discrete random variable. c) a continuous random variable. d) a parameter. ANSWER: b TYPE: MC DIFFICULTY: Moderate KEYWORDS: discrete random variable, types of data

35. The chancellor of a major university was concerned about alcohol abuse on her campus and wanted to find out the proportion of students at her university who visited campus bars on the weekend before the final exam week. Her assistant took a random sample of 250 students and computed the portion of students in the sample who visited campus bars on the weekend before the final exam. The portion of all students at her university who visited campus bars on the weekend before the final exam week is an example of a) a categorical random variable. b) a discrete random variable. c) a continuous random variable. d) a parameter. ANSWER: d TYPE: MC DIFFICULTY: Moderate KEYWORDS: parameter, types of data

36. The chancellor of a major university was concerned about alcohol abuse on her campus and wanted to find out the proportion of students at her university who visited campus bars on the weekend before the final exam week. Her assistant took a random sample of 250 students. The portion of students in the sample who visited campus bars on the weekend before the final exam week is an example of a) a categorical random variable. b) a discrete random variable. c) a parameter. d) a statistic. ANSWER: d

10 Introduction and Data Collection TYPE: MC DIFFICULTY: Moderate KEYWORDS: statistic, types of data

37. The chancellor of a major university was concerned about alcohol abuse on her campus and wanted to find out the proportion of students at her university who visited campus bars on the weekend before the final exam week. Her assistant took a random sample of 250 students. The portion of students in the sample who visited campus bars on the weekend before the final exam week is an example of a) a categorical random variable. b) a discrete random variable. c) a continuous random variable. d) a parameter. ANSWER: c TYPE: MC DIFFICULTY: Difficult KEYWORDS: continuous random variable, types of data TABLE 1-1 The manager of the customer service division of a major consumer electronics company is interested in determining whether the customers who have purchased a videocassette recorder made by the company over the past 12 months are satisfied with their products. 38. Referring to Table 1-1, the population of interest is a) all the customers who have bought a videocassette recorder made by the company over the past 12 months. b) all the customers who have bought a videocassette recorder made by the company and brought it in for repair over the past 12 months. c) all the customers who have used a videocassette recorder over the past 12 months. d) all the customers who have ever bought a videocassette recorder made by the company. ANSWER: a TYPE: MC DIFFICULTY: Difficult KEYWORDS: population 39. Referring to Table 1-1, the possible responses to the question "How many videocassette recorders made by other manufacturers have you used?" are values from a a) discrete random variable. b) continuous random variable. c) categorical random variable. d) parameter. ANSWER: a TYPE: MC DIFFICULTY: Easy KEYWORDS: discrete random variable, types of data 40. Referring to Table 1-1, the possible responses to the question "Are you happy, indifferent, or unhappy with the performance per dollar spent on the videocassette recorder?" are values from a a) discrete numerical random variable. b) continuous numerical random variable. c) categorical random variable.

Introduction and Data Collection

d) parameter. ANSWER: c TYPE: MC DIFFICULTY: Easy KEYWORDS: categorical random variable, types of data 41. True or False: A population is the totality of items or things under consideration. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: population 42. True or False: A sample is the portion of the universe that is selected for analysis. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: sample 43. True or False: Problems may arise when statistically unsophisticated users who do not understand the assumptions behind the statistical procedures or their limitations are misled by results obtained from computer software. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: statistical package 44. True or False: Managers need an understanding of statistics to be able to present and describe information accurately, draw conclusions about large populations based on small samples, improve processes, and make reliable forecasts. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: reasons for learning statistics 45. True or False: The possible responses to the question “How long have you been living at your current residence?” are values from a continuous variable. ANSWER: True TYPE: TF DIFFICULTY: Moderate KEYWORDS: continuous random variable, types of data 46. True or False: The possible responses to the question “How many times in the past three months have you visited a city park?” are values from a discrete variable. ANSWER: True TYPE: TF DIFFICULTY: Moderate KEYWORDS: discrete random variable, types of data

12 Introduction and Data Collection

47. True or False: A continuous variable may take on any value within its relevant range even though the measurement device may not be precise enough to record it. ANSWER: True TYPE: TF DIFFICULTY: Moderate KEYWORDS: continuous random variable, types of data

48. True or False: Faculty rank (professor to lecturer) is an example of discrete numerical data. ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: categorical random variable, types of data 49. True or False: Student grades (A to F) are an example of continuous numerical data. ANSWER: False TYPE: TF DIFFICULTY: Moderate KEYWORDS: categorical random variables, types of data 50. True or False: The amount of coffee consumed by an individual in a day is an example of a discrete numerical variable. ANSWER: False TYPE: TF DIFFICULTY: Moderate KEYWORDS: continuous random variables, types of data 51. True or False: A statistic is usually used to provide an estimate for a usually unobserved parameter. ANSWER: True TYPE: TF DIFFICULTY: Moderate KEYWORDS: statistic, parameter, inferential statistics 52. True or False: A statistic is usually unobservable while a parameter is usually observable. ANSWER: False TYPE: TF DIFFICULTY: Moderate KEYWORDS: statistic, parameter, inferential statistic 53. True or False: The answer to the question “What is your favorite color?” is an example of an ordinal scaled variable. ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: nominal scale

Introduction and Data Collection

54. True or False: The answer to the question “How do you rate the quality of your business statistics course” is an example of an ordinal scaled variable. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: ordinal scale 55. True or False: The answer to the question “How many hours on average do you spend watching TV every week?” is an example of a ratio scaled variable. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: ratio scale 56. True or False: The answer to the question “What is your sleeping bag temperature rating?” is an example of a ratio scaled variable. ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: interval scale

57. The Human Resources Director of a large corporation wishes to develop an employee benefits package and decides to select 500 employees from a list of all (N = 40,000) workers in order to study their preferences for the various components of a potential package. All the employees in the corporation constitute the . ANSWER: population TYPE: FI DIFFICULTY: Easy KEYWORDS: population

58. The Human Resources Director of a large corporation wishes to develop an employee benefits package and decides to select 500 employees from a list of all (N = 40,000) workers in order to study their preferences for the various components of a potential package. The 500 employees who will participate in this study constitute the . ANSWER: sample TYPE: FI DIFFICULTY: Easy KEYWORDS: sample

59. The Human Resources Director of a large corporation wishes to develop an employee benefits package and decides to select 500 employees from a list of all (N = 40,000) workers in order to study their preferences for the various components of a potential package. The Director will use the data from the sample to compute . ANSWER: statistics TYPE: FI DIFFICULTY: Easy KEYWORDS: statistic

14 Introduction and Data Collection

60. The Human Resources Director of a large corporation wishes to develop an employee benefits package and decides to select 500 employees from a list of all (N = 40,000) workers in order to study their preferences for the various components of a potential package. Information obtained from the sample will be used to draw conclusions about the true population . ANSWER: parameters TYPE: FI DIFFICULTY: Easy KEYWORDS: parameter

61. The Human Resources Director of a large corporation wishes to develop an employee benefits package and decides to select 500 employees from a list of all (N = 40,000) workers in order to study their preferences for the various components of a potential package. In this study, methods involving the collection, presentation, and characterization of the data are called . ANSWER: descriptive statistics/methods TYPE: FI DIFFICULTY: Easy KEYWORDS: descriptive statistics

62. The Human Resources Director of a large corporation wishes to develop an employee benefits package and decides to select 500 employees from a list of all (N = 40,000) workers in order to study their preferences for the various components of a potential package. In this study, methods that result in decisions concerning population characteristics based only on the sample results are called . ANSWER: inferential statistics/methods TYPE: FI DIFFICULTY: Easy KEYWORDS: inferential statistics 63. Mediterranean fruit flies were discovered in California a few years ago and badly damaged the oranges grown in that state. Suppose the manager of a large farm wanted to study the impact of the fruit flies on the orange crops on a daily basis over a 6-week period. On each day a random sample of orange trees were selected from within a random sample of acres. The daily average number of damaged oranges per tree and the proportion of trees having damaged oranges were calculated. The two main measures calculated each day (i.e., average number of damaged oranges per tree and proportion of trees having damaged oranges) are called . ANSWER: statistics TYPE: FI DIFFICULTY: Moderate KEYWORDS: statistic 64. Mediterranean fruit flies were discovered in California a few years ago and badly damaged the oranges grown in that state. Suppose the manager of a large farm wanted to study the impact of the fruit flies on the orange crops on a daily basis over a 6-week period. On each day a random sample of orange trees were selected from within a random sample of acres. The daily average number of damaged oranges per tree and the proportion of trees having damaged oranges were calculated. The two main measures calculated each day (i.e., average number of damaged oranges per tree and proportion of trees having damaged oranges) may be used on a daily basis to estimate the respective true population .

Introduction and Data Collection

ANSWER: parameters TYPE: FI DIFFICULTY: Easy KEYWORDS: parameters 65. Mediterranean fruit flies were discovered in California a few years ago and badly damaged the oranges grown in that state. Suppose the manager of a large farm wanted to study the impact of the fruit flies on the orange crops on a daily basis over a 6-week period. On each day a random sample of orange trees were selected from within a random sample of acres. The daily average number of damaged oranges per tree and the proportion of trees having damaged oranges were calculated. In this study, drawing conclusions on any one day about the true population characteristics based on information obtained from the sample is called . ANSWER: inferential statistics/methods TYPE: FI DIFFICULTY: Moderate KEYWORDS: inferential statistics 66. Mediterranean fruit flies were discovered in California a few years ago and badly damaged the oranges grown in that state. Suppose the manager of a large farm wanted to study the impact of the fruit flies on the orange crops on a daily basis over a 6-week period. On each day a random sample of orange trees were selected from within a random sample of acres. The daily average number of damaged oranges per tree and the proportion of trees having damaged oranges were calculated. In this study, the presentation and characterization of the two main measures calculated each day (i.e., average number of damaged oranges per tree and proportion of trees having damaged oranges) is called . ANSWER: descriptive statistics/methods TYPE: FI DIFFICULTY: Moderate KEYWORDS: descriptive statistics 67. The Quality Assurance Department of a large urban hospital is attempting to monitor and evaluate patient satisfaction with hospital services. Prior to discharge, a random sample of patients is asked to fill out a questionnaire to rate such services as medical care, nursing, therapy, laboratory, food, and cleaning. The Quality Assurance Department prepares weekly reports that are presented at the Board of Directors meetings and extraordinary/atypical ratings are easy to flag. Values computed from the sample results each week are called . ANSWER: statistics TYPE: FI DIFFICULTY: Easy KEYWORDS: statistic 68. The Quality Assurance Department of a large urban hospital is attempting to monitor and evaluate patient satisfaction with hospital services. Prior to discharge, a random sample of patients is asked to fill out a questionnaire to rate such services as medical care, nursing, therapy, laboratory, food, and cleaning. The Quality Assurance Department prepares weekly reports that are presented at the Board of Directors meetings and extraordinary/atypical ratings are easy to flag. True population characteristics estimated from the sample results each week are called . ANSWER:

16 Introduction and Data Collection parameters TYPE: FI DIFFICULTY: Easy KEYWORDS: parameter 69. The Commissioner of Health in New York State wanted to study malpractice litigation in New York. A sample of 31 thousand medical records was drawn from a population of 2.7 million patients who were discharged during the year 1997. The proportion of malpractice claims filed from the sample of 31 thousand patients is a . ANSWER: statistic TYPE: FI DIFFICULTY: Moderate KEYWORDS: statistic 70. The Commissioner of Health in New York State wanted to study malpractice litigation in New York. A sample of 31 thousand medical records was drawn from a population of 2.7 million patients who were discharged during the year 1997. The true proportion of malpractice claims filed from the population of 2.7 million patients is a . ANSWER: parameter TYPE: FI DIFFICULTY: Easy KEYWORDS: parameter 71. The Commissioner of Health in New York State wanted to study malpractice litigation in New York. A sample of 31 thousand medical records was drawn from a population of 2.7 million patients who were discharged during the year 1997. Using the information obtained from the sample to predict population characteristics with respect to malpractice litigation is an example of . ANSWER: inferential statistics TYPE: FI DIFFICULTY: Moderate KEYWORDS: inferential statistics 72. The Commissioner of Health in New York State wanted to study malpractice litigation in New York. A sample of 31 thousand medical records was drawn from a population of 2.7 million patients who were discharged during the year 1997. The collection, presentation, and characterization of the data from patient medical records are examples of . ANSWER: descriptive statistics/methods TYPE: FI DIFFICULTY: Easy KEYWORDS: descriptive statistics 73. An insurance company evaluates many numerical variables about a person before deciding on an appropriate rate for automobile insurance. The number of claims a person has made in the last 3 years is an example of a variable. ANSWER: discrete TYPE: FI DIFFICULTY: Easy KEYWORDS: discrete random variable, types of data

Introduction and Data Collection

74. An insurance company evaluates many numerical variables about a person before deciding on an appropriate rate for automobile insurance. The distance a person drives in a year is an example of a variable. ANSWER: continuous TYPE: FI DIFFICULTY: Easy KEYWORDS: continuous random variable, types of data 75. An insurance company evaluates many numerical variables about a person before deciding on an appropriate rate for automobile insurance. A person's age is an example of a variable. ANSWER: continuous TYPE: FI DIFFICULTY: Easy KEYWORDS: continuous random variable, types of data 76. An insurance company evaluates many numerical variables about a person before deciding on an appropriate rate for automobile insurance. How long a person has been a licensed driver is an example of a variable. ANSWER: continuous TYPE: FI DIFFICULTY: Moderate KEYWORDS: continuous random variable, types of data 77. An insurance company evaluates many numerical variables about a person before deciding on an appropriate rate for automobile insurance. The number of tickets a person has received in the last 3 years is an example of a variable. ANSWER: discrete TYPE: FI DIFFICULTY: Easy KEYWORDS: discrete random variable, types of data 78. In purchasing an automobile, there are a number of variables to consider. The body style of the car (sedan, coupe, wagon, etc.) is an example of a variable. ANSWER: categorical TYPE: FI DIFFICULTY: Easy KEYWORDS: categorical random variable, types of data 79. In purchasing an automobile, there are a number of variables to consider. The classification of the car as a subcompact, compact, standard, or luxury size is an example of a variable. ANSWER: categorical TYPE: FI DIFFICULTY: Easy KEYWORDS: categorical random variable, types of data 80. In purchasing an automobile, there are a number of variables to consider. The color of the car is an example of a variable.

18 Introduction and Data Collection ANSWER: categorical TYPE: FI DIFFICULTY: Easy KEYWORDS: categorical random variable, types of data 81. Most colleges admit students based on their achievements in a number of different areas. Whether a student has taken any advanced placement courses is an example of a variable. ANSWER: categorical TYPE: FI DIFFICULTY: Easy KEYWORDS: categorical random variable, types of data 82. Most colleges admit students based on their achievements in a number of different areas. The grade obtained in senior level English. (A, B, C, D, or F) is an example of a variable. ANSWER: categorical TYPE: FI DIFFICULTY: Moderate KEYWORDS: categorical random variable, types of data 83. Most colleges admit students based on their achievements in a number of different areas. The total SAT score achieved by a student is an example of a numerical variable. ANSWER: discrete TYPE: FI DIFFICULTY: Moderate KEYWORDS: discrete random variable, types of data 84. The Dean of Students conducted a survey on campus. The gender of the student is an example of a variable. ANSWER: categorical TYPE: FI DIFFICULTY: Easy KEYWORDS: categorical random variable, types of data 85. The Dean of Students conducted a survey on campus. Class designation (Freshman, Sophomore, Junior, Senior) is an example of a variable. ANSWER: categorical TYPE: FI DIFFICULTY: Easy KEYWORDS: categorical random variable, types of data 86. The Dean of Students conducted a survey on campus. Major area of study is an example of a variable. ANSWER: categorical TYPE: FI DIFFICULTY: Easy KEYWORDS: categorical random variable, types of data

Introduction and Data Collection

87. The Dean of Students conducted a survey on campus. Average SAT score in mathematics is an example of a numerical variable. ANSWER: continuous TYPE: FI DIFFICULTY: Easy KEYWORDS: continuous random variable, types of data 88. The Dean of Students conducted a survey on campus. Grade point average (GPA) is an example of a numerical variable. ANSWER: continuous TYPE: FI DIFFICULTY: Easy KEYWORDS: continuous random variable, types of data 89. The Dean of Students conducted a survey on campus. Number of credits currently enrolled for is an example of a numerical variable. ANSWER: discrete TYPE: FI DIFFICULTY: Easy KEYWORDS: discrete random variable, types of data 90. The Dean of Students conducted a survey on campus. Number of clubs, groups, teams, and organizations affiliated with on campus is an example of a numerical variable. ANSWER: discrete TYPE: FI DIFFICULTY: Easy KEYWORDS: discrete random variable, types of data 91. A personal computer user survey was conducted. Computer brand primarily used is an example of a variable. ANSWER: categorical TYPE: FI DIFFICULTY: Easy KEYWORDS: categorical random variable, types of data

92. A personal computer user survey was conducted. Number of personal computers owned is an example of a

variable.

ANSWER: discrete TYPE: FI DIFFICULTY: Easy KEYWORDS: discrete random variable, types of data 93. A personal computer user survey was conducted. The number of years using a personal computer is an example of a variable. ANSWER: continuous TYPE: FI DIFFICULTY: Moderate

20 Introduction and Data Collection KEYWORDS: continuous random variable, types of data 94. A personal computer user survey was conducted. Hours of personal computer use per week is an example of a variable ANSWER: continuous TYPE: FI DIFFICULTY: Moderate KEYWORDS: continuous random variable, types of data 95. A personal computer user survey was conducted. Primary word processing package used is an example of a variable ANSWER: categorical TYPE: FI DIFFICULTY: Easy KEYWORDS: categorical random variable, types of data 96. A personal computer user survey was conducted. The number of computer magazine subscriptions is an example of a variable. ANSWER: discrete TYPE: FI DIFFICULTY: Moderate KEYWORDS: discrete random variable, types of data 97. The type of TV one owns is an example of an ordinal scaled variable. ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: nominal scale, ordinal scale

98. The type of TV one owns is an example of a numerical variable. ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: categorical random variable 99. Whether the university is private or public is an example of a nominal scaled variable. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: nominal scale 100.

Whether the university is private or public is an example of a categorical variable.

ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: categorical random variable

Introduction and Data Collection 101.

Marital status is an example of an ordinal scaled variable.

ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: nominal scale, ordinal scale 102.

Marital status is an example of a numerical variable.

ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: categorical random variable

103. The grade level (K-12) of a student is an example of a nominal scaled variable. ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: nominal scale, ordinal scale

104. The grade level (K-12) of a student is an example of a numerical variable. ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: categorical random variable

105. The level of satisfaction (“Very unsatisfied”, “Fairly unsatisfied”, “Fairly satisfied”, and “Very satisfied”) in a class is an example of an ordinal scaled variable.

ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: ordinal scale

106. The level of satisfaction (“Very unsatisfied”, “Fairly unsatisfied”, “Fairly satisfied”, and “Very satisfied”) in a class is an example of a categorical variable.

ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: categorical random variable

107. The quality (“terrible”, “poor”, “fair”, “acceptable”, “very good” and “excellent”) of a day care center is an example of a nominal scaled variable. ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: nominal scale, ordinal scale

22 Introduction and Data Collection

108. The quality (“terrible”, “poor”, “fair”, “acceptable”, “very good” and “excellent”) of a day care center is an example of a numerical variable. ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: categorical random variable 109.

The amount of alcohol consumed by a person per week will be measured on an interval scale.

ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: interval scale, ratio scale 110.

The amount of alcohol consumed by a person per week is an example of a continuous variable.

ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: continuous random variable 111.

The number of defective apples in a single box will be measured on an interval scale.

ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: interval scale, ratio scale 112.

The number of defective apples in a single box is an example of a continuous variable.

ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: discrete random variable, continuous random variable 113. The amount of calories contained in a pack of 12-ounce cheese will be measured on a ratio scale. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: ratio scale 114. The amount of calories contained in a pack of 12-ounce cheese is an example of a discrete variable. ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: discrete random variable, continuous random variable 115.

The amount of time a student spent studying for an exam will be measured on a ratio scale.

Introduction and Data Collection ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: ratio scale 116. The amount of time a student spent studying for an exam is an example of a continuous variable. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: continuous random variable 117. The date when a production line in a factor is out-of-control will be measured with a ratio scale. ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: interval scale, ratio scale

26 Presenting Data in Tables and Charts

CHAPTER 2: PRESENTING DATA IN TABLES AND CHARTS TABLE 2-1 An insurance company evaluates many numerical variables about a person before deciding on an appropriate rate for automobile insurance. A representative from a local insurance agency selected a random sample of insured drivers and recorded, X, the number of claims each made in the last 3 years, with the following results. X f 1 14 2 18 3 12 4 5 5 1 1. Referring to Table 2-1, how many drivers are represented in the sample? a) 5 b) 15 c) 18 d) 50 ANSWER: d TYPE: MC DIFFICULTY: Easy KEYWORDS: frequency distribution 2. Referring to Table 2-1, how many total claims are represented in the sample? a) 15 b) 50 c) 111 d) 250 ANSWER: c TYPE: MC DIFFICULTY: Moderate KEYWORDS: interpretation, frequency distribution 3. A type of vertical bar chart in which the categories are plotted in the descending rank order of the magnitude of their frequencies is called a a) contingency table. b) Pareto diagram. c) dot plot. d) pie chart. ANSWER: b TYPE: MC DIFFICULTY: Easy KEYWORDS: Pareto diagram

27 Presenting Data in Tables and Charts TABLE 2-2 At a meeting of information systems officers for regional offices of a national company, a survey was taken to determine the number of employees the officers supervise in the operation of their departments, where X is the number of employees overseen by each information systems officer. X f_ 1 7 2 5 3 11 4 8 5 9 4. Referring to Table 2-2, how many regional offices are represented in the survey results? a) 5 b) 11 c) 15 d) 40 ANSWER: d TYPE: MC DIFFICULTY: Easy KEYWORDS: interpretation, frequency distribution 5. Referring to Table 2-2, across all of the regional offices, how many total employees were supervised by those surveyed? a) 15 b) 40 c) 127 d) 200 ANSWER: c TYPE: MC DIFFICULTY: Moderate KEYWORDS: interpretation, frequency distribution 6. The width of each bar in a histogram corresponds to the a) differences between the boundaries of the class. b) number of observations in each class. c) midpoint of each class. d) percentage of observations in each class. ANSWER: a TYPE: MC DIFFICULTY: Easy KEYWORDS: frequency distribution TABLE 2-3 Every spring semester, the School of Business coordinates with local business leaders a luncheon for graduating seniors, their families, and friends. Corporate sponsorship pays for the lunches of each of the seniors, but students have to purchase tickets to cover the cost of lunches served to guests they bring with them. The following histogram represents the attendance at the senior luncheon, where X is the number of guests each graduating senior invited to the luncheon and f is the number of graduating seniors in each category.

Presenting Data in Tables and Charts

152

160 140 120 100

80 60 40 20

0 0

2 Guests per Student

7. Referring to the histogram from Table 2-3, how many graduating seniors attended the luncheon? a) 4 b) 152 c) 275 d) 388 ANSWER: c TYPE: MC DIFFICULTY: Difficult EXPLANATION: The number of graduating seniors is the sum of all the frequencies, f. KEYWORDS: interpretation, histogram 8. Referring to the histogram from Table 2-3, if all the tickets purchased were used, how many guests attended the luncheon? a) 4 b) 152 c) 275 d) 388 ANSWER: d TYPE: MC DIFFICULTY: Difficult EXPLANATION: The total number of guests is

 Xf 6

i=1

i i

KEYWORDS: interpretation, histogram 9. A professor of economics at a small Texas university wanted to determine what year in school students were taking his tough economics course. Shown below is a pie chart of the results. What percentage of the class took the course prior to reaching their senior year?

29 Presenting Data in Tables and Charts

Seniors 14%

Freshmen 10%

Juniors 30% Sophomores 46%

a) 14% b) 44% c) 54% d) 86% ANSWER: d TYPE: MC DIFFICULTY: Easy KEYWORDS: interpretation, pie chart 10. When polygons or histograms are constructed, which axis must show the true zero or "origin"? a) The horizontal axis. b) The vertical axis. c) Both the horizontal and vertical axes. d) Neither the horizontal nor the vertical axis. ANSWER: b TYPE: MC DIFFICULTY: Easy KEYWORDS: polygon, histogram 11. When constructing charts, the following is plotted at the class midpoints: a) frequency histograms b) percentage polygons c) cumulative relative frequency ogives d) All of the above. ANSWER: b TYPE: MC DIFFICULTY: Easy KEYWORDS: percentage polygon TABLE 2-4 A survey was conducted to determine how people rated the quality of programming available on television. Respondents were asked to rate the overall quality from 0 (no quality at all) to 100 (extremely good quality). The stem-and-leaf display of the data is shown below. Stem Leaves 3 24 4 03478999 5 0112345

Presenting Data in Tables and Charts

6 7 8 9

12566 01 2

12. Referring to Table 2-4, what percentage of the respondents rated overall television quality with a rating of 80 or above? a) 0 b) 4 c) 96 d) 100 ANSWER: b TYPE: MC DIFFICULTY: Easy KEYWORDS: stem-and-leaf display, interpretation 13. Referring to Table 2-4, what percentage of the respondents rated overall television quality with a rating of 50 or below? a) 11 b) 40 c) 44 d) 56 ANSWER: c TYPE: MC DIFFICULTY: Moderate KEYWORDS: stem-and-leaf display, interpretation 14. Referring to Table 2-4, what percentage of the respondents rated overall television quality with a rating between 50 and 75? a) 11 b) 40 c) 44 d) 56 ANSWER: d TYPE: MC DIFFICULTY: Moderate KEYWORDS: stem-and-leaf display, interpretation TABLE 2-5 The following are the durations in minutes of a sample of long-distance phone calls made within the continental United States reported by one long-distance carrier. Time (in Minutes) 0 but less than 5 5 but less than 10 10 but less than 15 15 but less than 20 20 but less than 25 25 but less than 30 30 or more

Relative Frequency 0.37 0.22 0.15 0.10 0.07 0.07 0.02

31 Presenting Data in Tables and Charts

15. Referring to Table 2-5, what is the width of each class? a) 1 minute b) 5 minutes c) 2% d) 100% ANSWER: b TYPE: MC DIFFICULTY: Easy KEYWORDS: class interval, relative frequency distribution 16. Referring to Table 2-5, if 1,000 calls were randomly sampled, how many calls lasted under 10 minutes? a) 220 b) 370 c) 410 d) 590 ANSWER: d TYPE: MC DIFFICULTY: Moderate KEYWORDS: relative frequency distribution, interpretation 17. Referring to Table 2-5, if 100 calls were randomly sampled, how many calls lasted 15 minutes or longer? a) 10 b) 14 c) 26 d) 74 ANSWER: c TYPE: MC DIFFICULTY: Moderate KEYWORDS: relative frequency distribution, interpretation 18. Referring to Table 2-5, if 10 calls lasted 30 minutes or more, how many calls lasted less than 5 minutes? a) 10 b) 185 c) 295 d) 500 ANSWER: b TYPE: MC DIFFICULTY: Moderate KEYWORDS: relative frequency distribution, interpretation 19. Referring to Table 2-5, what is the cumulative relative frequency for the percentage of calls that lasted under 20 minutes? a) 0.10 b) 0.59 c) 0.76 d) 0.84

Presenting Data in Tables and Charts

ANSWER: d TYPE: MC DIFFICULTY: Easy KEYWORDS: cumulative relative frequency 20. Referring to Table 2-5, what is the cumulative relative frequency for the percentage of calls that lasted 10 minutes or more? a) 0.16 b) 0.24 c) 0.41 d) 0.90 ANSWER: c TYPE: MC DIFFICULTY: Moderate KEYWORDS: cumulative relative frequency

21. Referring to Table 2-5, if 100 calls were randomly sampled,

of them would have lasted

at least 15 minutes but less than 20 minutes a) 0.10 b) 0.16 c) 10 d) 16 ANSWER: c TYPE: MC DIFFICULTY: Easy KEYWORDS: relative frequency distribution, interpretation 22. Referring to Table 2-5, if 100 calls were sampled, 15 minutes. a) 26 b) 74 c) 10 d) None of the above.

of them would have lasted less than

ANSWER: b TYPE: MC DIFFICULTY: Moderate KEYWORDS: relative frequency distribution, interpretation 23. Referring to Table 2-5, if 100 calls were sampled, or more. a) 26 b) 16 c) 74 d) None of the above. ANSWER: b TYPE: MC DIFFICULTY: Moderate KEYWORDS: relative frequency distribution, interpretation

of them would have lasted 20 minutes

33 Presenting Data in Tables and Charts 24. Referring to Table 2-5, if 100 calls were sampled, minutes or at least 30 minutes or more. a) 35 b) 37 c) 39 d) None of the above.

of them would have lasted less than 5

ANSWER: c TYPE: MC DIFFICULTY: Difficult KEYWORDS: relative frequency distribution, interpretation 25. When studying the simultaneous responses to two categorical questions, we should set up a a) contingency table. b) frequency distribution table. c) cumulative percentage distribution table. d) histogram. ANSWER: a TYPE: MC DIFFICULTY: Easy KEYWORDS: contingency table TABLE 2-6 A sample of 200 students at a Big-Ten university was taken after the midterm to ask them whether they went bar hopping the weekend before the midterm or spent the weekend studying, and whether they did well or poorly on the midterm. The following table contains the result.

Studying for Exam Went Bar Hopping

Did Well in Midterm 80 30

Did Poorly in Midterm 20 70

26. Referring to Table 2-6, of those who went bar hopping the weekend before the midterm in the sample, percent of them did well on the midterm. a) 15 b) 27.27 c) 30 d) 55 ANSWER: c TYPE: MC DIFFICULTY: Easy KEYWORDS: contingency table, interpretation 27. Referring to Table 2-6, of those who did well on the midterm in the sample, them went bar hopping the weekend before the midterm. a) 15 b) 27.27 c) 30 d) 50 ANSWER: b

percent of

Presenting Data in Tables and Charts

TYPE: MC DIFFICULTY: Easy KEYWORDS: contingency table, interpretation 28. Referring to Table 2-6, percent of the students in the sample went bar hopping the weekend before the midterm and did well on the midterm. a) 15 b) 27.27 c) 30 d) 50 ANSWER: a TYPE: MC DIFFICULTY: Easy KEYWORDS: contingency table, interpretation 29. Referring to Table 2-6, percent of the students in the sample spent the weekend studying and did well on the midterm. a) 40 b) 50 c) 72.72 d) 80 ANSWER: a TYPE: MC DIFFICULTY: Easy KEYWORDS: contingency table, interpretation 30. Referring to Table 2-6, if the sample is a good representation of the population, we can expect percent of the students in the population to spend the weekend studying and do poorly on the midterm. a) 10 b) 20 c) 45 d) 50 ANSWER: a TYPE: MC DIFFICULTY: Easy KEYWORDS: contingency table, interpretation 31. Referring to Table 2-6, if the sample is a good representation of the population, we can expect percent of those who spent the weekend studying to do poorly on the midterm. a) 10 b) 20 c) 45 d) 50 ANSWER: b TYPE: MC DIFFICULTY: Moderate KEYWORDS: contingency table, interpretation 32. Referring to Table 2-6, if the sample is a good representation of the population, we can expect percent of those who did poorly on the midterm to have spent the weekend studying.

35 Presenting Data in Tables and Charts a) 10 b) 22.22 c) 45 d) 50 ANSWER: b TYPE: MC DIFFICULTY: Moderate KEYWORDS: contingency table, interpretation 33. In a contingency table, the number of rows and columns a) must always be the same. b) must always be 2. c) must add to 100%. d) None of the above. ANSWER: d TYPE: MC DIFFICULTY: Moderate KEYWORDS: contingency table 34. Retailers are always interested in determining why a customer selected their store to make a purchase. A sporting goods retailer conducted a customer survey to determine why its customers shopped at the store. The results are shown in the bar chart below. What proportion of the customers responded that they shopped at the store because of the merchandise or the convenience? 20%

Prices

50%

Merchandise

Convenience

15%

Other

15%

10%

20%

30% Responses

40%

50%

60%

a) 35% b) 50% c) 65% d) 85% ANSWER: c TYPE: MC DIFFICULTY: Easy KEYWORDS: bar chart, interpretation TABLE 2-7 The Stem-and-Leaf display below contains data on the number of months between the date a civil suit is filed and when the case is actually adjudicated for 50 cases heard in superior court.

Presenting Data in Tables and Charts

Stem 1L 1H 2L 2H 3L 3H 4L 4H 5L 5H 6L 6H

Leaves 23444 7899 222234 55678889 001113 5778 0234 5579 1124 66 15 8

Note: 1L means the “low teens” – 10, 11, 12, 13, or 14; 1H means the “high teens” – 15, 16, 17, 18, or 19; 2L means the “low twenties” – 20, 21, 22, 23, or 24, etc.

35. Referring to Table 2-7, locate the first leaf, i.e., the lowest valued leaf with the lowest valued stem. This represents a wait of

months.

ANSWER: 12 TYPE: FI DIFFICULTY: 1 Easy KEYWORDS: stem-and-leaf display, interpretation 36. Referring to Table 2-7, the civil suit with the longest wait between when the suit was filed and when it was adjudicated had a wait of months. ANSWER: 68 TYPE: FI DIFFICULTY: Easy KEYWORDS: stem-and-leaf display, interpretation 37. Referring to Table 2-7, the civil suit with the fourth shortest waiting time between when the suit was filed and when it was adjudicated had a wait of months. ANSWER: 14 TYPE: FI DIFFICULTY: Moderate KEYWORDS: stem-and-leaf display, interpretation 38. Referring to Table 2-7,

percent of the cases were adjudicated within the first 2 years.

ANSWER: 30 TYPE: FI DIFFICULTY: Moderate KEYWORDS: stem-and-leaf display, interpretation 39. Referring to Table 2-7, years. ANSWER: 20 TYPE: FI DIFFICULTY: Moderate

percent of the cases were not adjudicated within the first 4

37 Presenting Data in Tables and Charts KEYWORDS: stem-and-leaf display, interpretation 40. Referring to Table 2-7, if a frequency distribution with equal sized classes was made from this data, and the first class was "10 but less than 20," the frequency of that class would be ANSWER: 9 TYPE: FI DIFFICULTY: Easy KEYWORDS: stem-and-leaf display, interpretation 41. Referring to Table 2-7, if a frequency distribution with equal sized classes was made from this data, and the first class was "10 but less than 20," the relative frequency of the third class would be . ANSWER: 0.20 or 20% or 10/50 TYPE: FI DIFFICULTY: Moderate KEYWORDS: stem-and-leaf display, relative frequency distribution 42. Referring to Table 2-7, if a frequency distribution with equal sized classes was made from this data, and the first class was "10 but less than 20," the cumulative percentage of the second class would be . ANSWER: 46% or 0.46 or 23/50 TYPE: FI DIFFICULTY: Moderate KEYWORDS: stem-and-leaf display, cumulative percentage distribution TABLE 2-8 The Stem-and-Leaf display represents the number of times in a year that a random sample of 100 "lifetime" members of a health club actually visited the facility. Stem Leaves 0 012222233333344566666667789999 1 1111222234444455669999 2 00011223455556889 3 0000446799 4 011345567 5 0077 6 8 7 67 8 3 9 0247 43. Referring to Table 2-8, the person who has the largest leaf associated with the smallest stem visited the facility times. ANSWER: 9 TYPE: FI DIFFICULTY: Moderate KEYWORDS: stem-and-leaf display, interpretation 44. Referring to Table 2-8, the person who visited the health club less than anyone else in the sample visited the facility times.

Presenting Data in Tables and Charts

ANSWER: 0 or no TYPE: FI DIFFICULTY: Easy KEYWORDS: stem-and-leaf display, interpretation 45. Referring to Table 2-8, the person who visited the health club more than anyone else in the sample visited the facility times. ANSWER: 97 TYPE: FI DIFFICULTY: Easy KEYWORDS: stem-and-leaf display, interpretation 46. Referring to Table 2-8, a year.

of the 100 members visited the health club at least 52 times in

ANSWER: 10 TYPE: FI DIFFICULTY: Moderate KEYWORDS: stem-and-leaf display, interpretation 47. Referring to Table 2-8, times in a year.

of the 100 members visited the health club no more than 12

ANSWER: 38 TYPE: FI DIFFICULTY: Moderate KEYWORDS: stem-and-leaf display, interpretation 48. Referring to Table 2-8, if a frequency distribution with equal sized classes was made from this data, and the first class was "0 but less than 10," the frequency of the fifth class would be . ANSWER: 9 TYPE: FI DIFFICULTY: Moderate KEYWORDS: stem-and-leaf display, frequency distribution 49. Referring to Table 2-8, if a frequency distribution with equal sized classes was made from this data, and the first class was "0 but less than 10," the relative frequency of the last class would be . ANSWER: 4% or 0.04 or 4/100 TYPE: FI DIFFICULTY: Moderate KEYWORDS: stem-and-leaf display, relative frequency distribution 50. Referring to Table 2-8, if a frequency distribution with equal sized classes was made from this data, and the first class was "0 but less than 10," the cumulative percentage of the next-to-last class would be . ANSWER: 96% or 0.96 or 96/100

39 Presenting Data in Tables and Charts TYPE: FI DIFFICULTY: Moderate KEYWORDS: stem-and-leaf display, cumulative percentage distribution 51. Referring to Table 2-8, if a frequency distribution with equal sized classes was made from this data, and the first class was "0 but less than 10," the class midpoint of the third class would be . ANSWER: 25 or (20+30)/2 TYPE: FI DIFFICULTY: Moderate KEYWORDS: stem-and-leaf display, class midpoint TABLE 2-9 The frequency distribution below represents the rents of 250 randomly selected federally subsidized apartments in Minneapolis. Rent in $ Frequency 300 but less than 400 113 400 but less than 500 85 500 but less than 600 32 600 but less than 700 16 700 but less than 800 4 52. Referring to Table 2-9,

apartments rented for at least $400 but less than $600.

ANSWER: 117 TYPE: FI DIFFICULTY: Easy KEYWORDS: frequency distribution

53. Referring to Table 2-9,

percent of the apartments rented for no less than $600.

ANSWER: 8% or 20/250 TYPE: FI DIFFICULTY: Easy KEYWORDS: frequency distribution, cumulative percentage distribution 54. Referring to Table 2-9,

percent of the apartments rented for at least $500.

ANSWER: 20.8% or 52/250 TYPE: FI DIFFICULTY: Moderate KEYWORDS: frequency distribution, cumulative percentage distribution 55. Referring to Table 2-9, the class midpoint of the second class is

ANSWER: 450 TYPE: FI DIFFICULTY: Easy KEYWORDS: frequency distribution, class midpoint 56. Referring to Table 2-9, the relative frequency of the second class is

Presenting Data in Tables and Charts

ANSWER: 85/250 or 17/50 or 34% or 0.34 TYPE: FI DIFFICULTY: Easy KEYWORDS: frequency distribution, relative frequency distribution 57. Referring to Table 2-9, the percentage of apartments renting for less than $600 is

ANSWER: 230/250 or 23/25 or 92% or 0.92 TYPE: FI DIFFICULTY: Moderate KEYWORDS: frequency distribution, cumulative percentage distribution TABLE 2-10 The histogram below represents scores achieved by 200 job applicants on a personality profile. 0.30 Rel.Freq.

0.20

0.10

0.00 0

58. Referring to the histogram from Table 2-10, between 10 and 20.

percent of the job applicants scored

ANSWER: 20% TYPE: FI DIFFICULTY: Easy KEYWORDS: histogram, percentage distribution 59. Referring to the histogram from Table 2-10, 50.

percent of the job applicants scored below

ANSWER: 80% TYPE: FI DIFFICULTY: Moderate KEYWORDS: histogram, percentage distribution

60. Referring to the histogram from Table 2-10, the number of job applicants who scored between 30 and below 60 is

ANSWER: 80 TYPE: FI DIFFICULTY: Moderate

41 Presenting Data in Tables and Charts KEYWORDS: histogram 61. Referring to the histogram from Table 2-10, the number of job applicants who scored 50 or above is . ANSWER: 40 TYPE: FI DIFFICULTY: Moderate KEYWORDS: histogram 62. Referring to the histogram from Table 2-10, 90% of the job applicants scored above or equal to . ANSWER: 10 TYPE: FI DIFFICULTY: Moderate KEYWORDS: histogram, cumulative percentage distribution 63. Referring to the histogram from Table 2-10, half of the job applicants scored below

ANSWER: 30 TYPE: FI DIFFICULTY: Moderate KEYWORDS: histogram, cumulative percentage distribution 64. Referring to the histogram from Table 2-10, or at least 50.

percent of the applicants scored below 20

ANSWER: 50% TYPE: FI DIFFICULTY: Moderate KEYWORDS: histogram, cumulative percentage distribution 65. Referring to the histogram from Table 2-10, 20 and below 50.

percent of the applicants scored between

ANSWER: 50% TYPE: FI DIFFICULTY: Moderate KEYWORDS: histogram, cumulative percentage distribution TABLE 2-11 The ordered array below resulted from taking a sample of 25 batches of 500 computer chips and determining how many in each batch were defective. Defects 1 2 4 17 20 21

4 23

5 23

5 25

6 26

7 27

9 27

9 28

12 29

66. Referring to Table 2-11, if a frequency distribution for the defects data is constructed, using "0 but less than 5" as the first class, the frequency of the “20 but less than 25” class would be .

Presenting Data in Tables and Charts

ANSWER: 4 TYPE: FI DIFFICULTY: Easy KEYWORDS: frequency distribution 67. Referring to Table 2-11, if a frequency distribution for the defects data is constructed, using "0 but less than 5" as the first class, the relative frequency of the “15 but less than 20” class would be . ANSWER: 0.08 or 8% or 2/25 TYPE: FI DIFFICULTY: Moderate KEYWORDS: relative frequency distribution 68. Referring to Table 2-11, construct a frequency distribution for the defects data, using "0 but less than 5" as the first class. ANSWER: Defects Frequency 0 but less than 5 4 5 but less than 10 6 10 but less than 15 2 15 but less than 20 2 20 but less than 25 4 25 but less than 30 7 TYPE: PR DIFFICULTY: Easy KEYWORDS: frequency distribution 69. Referring to Table 2-11, construct a relative frequency or percentage distribution for the defects data, using "0 but less than 5" as the first class. ANSWER: Defects Percentage 0 but less than 5 16 5 but less than 10 24 10 but less than 15 8 15 but less than 20 8 20 but less than 25 16 25 but less than 30 28 TYPE: PR DIFFICULTY: Moderate KEYWORDS: relative frequency distribution, percentage distribution 70. Referring to Table 2-11, construct a cumulative percentage distribution for the defects data if the corresponding frequency distribution uses "0 but less than 5" as the first class. ANSWER: Defects CumPct 0 0 5 16 10 40 15 48 20 56 25 72 30 100

43 Presenting Data in Tables and Charts TYPE: PR DIFFICULTY: Moderate KEYWORDS: cumulative percentage distribution 71. Referring to Table 2-11, construct a histogram for the defects data, using "0 but less than 5" as the first class. ANSWER: 7

Frequency 6

6 5 4

4 3 2

2 1 0

15 20 Number o f Defects

TYPE: PR DIFFICULTY: Easy KEYWORDS: histogram, frequency distribution 72. Referring to Table 2-11, construct a cumulative percentage polygon for the defects data if the corresponding frequency distribution uses "0 but less than 5" as the first class.

ANSWER: Cumulative Percentage Polygon 100% 90% 80% 70% 60% 50% 40% 30% 20% 10% 0% 0

Number of Defects

TYPE: PR DIFFICULTY: Moderate

Presenting Data in Tables and Charts

KEYWORDS: cumulative percentage polygon 73. The point halfway between the boundaries of each class interval in a grouped frequency distribution is called the . ANSWER: class midpoint TYPE: FI DIFFICULTY: Easy KEYWORDS: cumulative percentage polygon, frequency distribution 74. A is a vertical bar chart in which the rectangular bars are constructed at the boundaries of each class interval. ANSWER: histogram TYPE: FI DIFFICULTY: Easy KEYWORDS: histogram 75. It is essential that each class grouping or interval in a frequency distribution be

ANSWER: non-overlapping TYPE: FI DIFFICULTY: Easy KEYWORDS: frequency distribution, class interval 76. In order to compare one large batch of numerical data to another, a be developed from the frequency distribution.

distribution must

ANSWER: relative frequency or percentage TYPE: FI DIFFICULTY: Easy KEYWORDS: relative frequency distribution, percentage distribution 77. When comparing two or more large batches of numerical data, the distributions being developed should use the same . ANSWER: class boundaries. TYPE: FI DIFFICULTY: Easy KEYWORDS: class boundaries 78. It is desirable that the width of each class grouping or interval in a frequency distribution be . ANSWER: the same or equal TYPE: FI DIFFICULTY: Easy KEYWORDS: class interval, frequency distribution 79. In constructing a polygon, each class grouping is represented by its consecutively connected to one another. ANSWER: midpoint

and then these are

45 Presenting Data in Tables and Charts TYPE: FI DIFFICULTY: Easy KEYWORDS: polygon, class interval, midpoint 80. A is a summary table in which numerical data are tallied into class intervals or categories. ANSWER: frequency distribution TYPE: FI DIFFICULTY: Easy KEYWORDS: frequency distribution, class interval 81. True or False: In general, grouped frequency distributions should have between 5 and 15 class intervals. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: frequency distribution, number of classes 82. True or False: The sum of relative frequencies in a distribution always equals 1. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: relative frequency 83. True or False: The sum of cumulative frequencies in a distribution always equals 1. ANSWER: False TYPE: TF DIFFICULTY: Moderate KEYWORDS: cumulative frequency distribution 84. True or False: In graphing bivariate categorical data, the side-by-side bar chart is best suited when primary interest is in demonstrating differences in magnitude rather than differences in percentages. ANSWER: True TYPE: TF DIFFICULTY: Moderate KEYWORDS: side-by-side chart 85. True or False: When constructing a frequency distribution, classes should be selected in such a way that they are of equal width. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: frequency distribution 86. True or False: A research analyst was directed to arrange raw data collected on the yield of wheat, ranging from 40 to 93 bushels per acre, in a frequency distribution. He should choose 30 as the class interval width.

Presenting Data in Tables and Charts

ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: frequency distribution, class interval 87. True or False: If the values of the seventh and eighth class in a cumulative frequency distribution are the same, we know that there are no observations in the eighth class. ANSWER: True TYPE: TF DIFFICULTY: Moderate KEYWORDS: cumulative frequency distribution

88. True or False: Research on Human perception concludes that the bar chart is preferred to the pie chart, because the human eye can more accurately judge length comparisons against a fixed scale (as in a bar chart) than angular measures (as in a pie chart). ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: bar chart, pie chart 89. True or False: One of the advantages of a pie chart is that it clearly shows that the total of all the categories of the pie adds to 100%. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: pie chart 90. True or False: The larger the number of observations in a numerical data set, the larger the number of class intervals needed for a grouped frequency distribution. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: class interval, frequency distribution 91. True or False: Determining the class boundaries of a frequency distribution is highly subjective. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: class boundaries, frequency distribution 92. True or False: The original data values cannot be assessed once they are grouped into a frequency distribution table. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: frequency distribution

47 Presenting Data in Tables and Charts 93. True or False: The percentage distribution cannot be constructed from the frequency distribution directly. ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: percentage distribution, frequency distribution 94. True or False: The stem-and-leaf display is often superior to the frequency distribution in that it maintains the original values for further analysis. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: stem-and-leaf display, frequency distribution 95. True or False: The relative frequency is the frequency in each class divided by the total number of observations. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: relative frequency distribution 96. True or False: Ogives are plotted at the midpoints of the class groupings. ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: ogives, midpoint 97. True or False: Percentage polygons are plotted at the boundaries of the class groupings. ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: percentage polygons 98. True or False: The main principle behind the Pareto diagram is the ability to track the "vital few" from the "trivial many." ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: Pareto diagram 99. True or False: A histogram can have gaps between the bars, whereas bar charts cannot have gaps. ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: histogram, bar chart

Presenting Data in Tables and Charts

100. True or False: Histograms are used for numerical data while bar charts are suitable for categorical data. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: histogram, bar chart

101. True or False: A Wal-Mart store in a small town 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 Pareto chart. ANSWER: True TYPE: TF DIFFICULTY: Moderate KEYWORDS: Pareto diagram 102. True or False: Apple Computer, Inc. collected information on the age of their customers. The youngest customer was 12 and the oldest was 72. To study the distribution of the age among its customers, it can use a Pareto diagram. ANSWER: False TYPE: TF DIFFICULTY: Moderate KEYWORDS: Pareto diagram

103. True or False: Apple Computer, Inc. collected information on the age of their customers. The youngest customer was 12 and the oldest was 72. To study the distribution of the age among its customers, it is best to use a pie chart. ANSWER: False TYPE: TF DIFFICULTY: Moderate KEYWORDS: pie chart 104. True or False: Apple Computer, Inc. collected information on the age of their customers. The youngest customer was 12 and the oldest was 72. To study the distribution of the age among its customers, it can use a percentage polygon. ANSWER: True TYPE: TF DIFFICULTY: Moderate KEYWORDS: percentage polygon 105. True or False: Apple Computer, Inc. collected information on the age of their customers. The youngest customer was 12 and the oldest was 72. To study the percentage of their customers who are below a certain age, it can use an ogive. ANSWER: True TYPE: TF DIFFICULTY: Moderate KEYWORDS: ogive

49 Presenting Data in Tables and Charts 106. True or False: If you wish to construct a graph of a relative frequency distribution, you would most likely construct an ogive first. ANSWER: False TYPE: TF DIFFICULTY: Moderate KEYWORDS: ogive 107.

True or False: An ogive is a cumulative percentage polygon.

ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: ogive, cumulative percentage polygon

108. True or False: A side-by-side chart is two histograms plotted side-by-side. ANSWER: False TYPE: TF DIFFICULTY: Moderate KEYWORDS: side-by-side chart

109. True or False: A good choice for the number of class groups to use in constructing frequency distribution is to have at least 5 but no more than 15 class groups. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: number of classes 110. True or False: In general, a frequency distribution should have at least 8 class groups but no more than 20. ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: number of classes 111. True of False: To determine the width of class interval, divide the number of class groups by the range of the data. ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: class interval 112. True or False: The percentage polygon is formed by having the lower boundary of each class represent the data in that class and then connecting the sequence of lower boundaries at their respective class percentages. ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: percentage polygon

Presenting Data in Tables and Charts

113.

True or False: A polygon can be constructed from a bar chart.

ANSWER: False TYPE: TF DIFFICULTY: Moderate KEYWORDS: polygon

114. To evaluate two categorical variables at the same time, a

could be developed.

ANSWER: contingency or cross-classification table or side-by-side bar chart TYPE: FI DIFFICULTY: Easy KEYWORDS: contingency table, cross-classification table 115. Relationships in a contingency table can be examined more fully if the frequencies are converted into . ANSWER: percentages or proportions TYPE: FI DIFFICULTY: Easy KEYWORDS: contingency table TABLE 2-12 The table below contains the opinions of a sample of 200 people broken down by gender about the latest congressional plan to eliminate anti-trust exemptions for professional baseball. Female Male Totals 116.

For 38 12 50

Neutral 54 36 90

Against Totals 12 104 48 96 60 200

Referring to Table 2-12, construct a table of row percentages.

ANSWER: For Neutral Against Female 36.54 51.92 11.54 Male 12.50 37.50 50.00 Totals 25.00 45.00 30.00 TYPE: PR DIFFICULTY: Easy KEYWORDS: row percentages 117.

Totals 100.00 100.00 100.00

Referring to Table 2-12, construct a table of column percentages.

ANSWER: For Neutral Against Totals Female 76.00 60.00 20.00 52.00 Male 24.00 40.00 80.00 48.00 Totals 100.00 100.00 100.00 100.00 TYPE: PR DIFFICULTY: Easy KEYWORDS: column percentages 118.

Referring to Table 2-12, construct a table of total percentages.

51 Presenting Data in Tables and Charts

ANSWER: For Neutral Against Totals Female 19.00 27.00 6.00 52.00 Male 6.00 18.00 24.00 48.00 Totals 25.00 45.00 30.00 100.00 TYPE: PR DIFFICULTY: Easy KEYWORDS: total percentages 119.

Referring to Table 2-12, of those for the plan in the sample,

percent were females.

ANSWER: 76% TYPE: FI DIFFICULTY: Moderate KEYWORDS: contingency table, column percentages 120.

Referring to Table 2-12, of those neutral in the sample,

percent were males.

ANSWER: 40% TYPE: FI DIFFICULTY: Moderate KEYWORDS: contingency table, column percentages 121.

Referring to Table 2-12, of the males in the sample,

percent were for the plan.

ANSWER: 12.50% TYPE: FI DIFFICULTY: Moderate KEYWORDS: contingency table 122.

Referring to Table 2-12, of the females in the sample,

percent were against the plan.

ANSWER: 11.54% TYPE: FI DIFFICULTY: Moderate KEYWORDS: contingency table 123. Referring to Table 2-12, of the females in the sample, against the plan.

percent were either neutral or

ANSWER: 63.46% or (51.92+11.54)% TYPE: FI DIFFICULTY: Moderate KEYWORDS: contingency table 124.

Referring to Table 2-12,

percent of the 200 were females who were against the plan.

ANSWER: 6% TYPE: FI DIFFICULTY: Moderate KEYWORDS: contingency table 125.

Referring to Table 2-12,

percent of the 200 were males who were neutral.

Presenting Data in Tables and Charts

ANSWER: 18% TYPE: FI DIFFICULTY: Moderate KEYWORDS: contingency table 126. Referring to Table 2-12, against the plan.

percent of the 200 were females who were either neutral or

ANSWER: 33% TYPE: FI DIFFICULTY: Difficult KEYWORDS: contingency table 127.

Referring to Table 2-12,

percent of the 200 were males who were not against the plan.

ANSWER: 24% TYPE: FI DIFFICULTY: Difficult KEYWORDS: contingency table 128.

Referring to Table 2-12,

percent of the 200 were not neutral.

ANSWER: 55% TYPE: FI DIFFICULTY: Difficult KEYWORDS: contingency table, row percentages 129.

Referring to Table 2-12,

percent of the 200 were against the plan.

ANSWER: 30% TYPE: FI DIFFICULTY: Moderate KEYWORDS: contingency table, row percentages 130.

Referring to Table 2-12,

percent of the 200 were males.

ANSWER: 48% TYPE: FI DIFFICULTY: Easy KEYWORDS: contingency table, column percentages 131.

Referring to Table 2-12, if the sample is a good representation of the population, we can expect percent of the population will be for the plant.

ANSWER: 25% TYPE: FI DIFFICULTY: Moderate KEYWORDS: contingency table, row percentages 132.

Referring to Table 2-12, if the sample is a good representation of the population, we can expect percent of the population will be males.

ANSWER: 48%

53 Presenting Data in Tables and Charts TYPE: FI DIFFICULTY: Moderate KEYWORDS: contingency table, column percentages 133.

Referring to Table 2-12, if the sample is a good representation of the population, we can expect percent of those for the plan in the population will be males.

ANSWER: 24% TYPE: FI DIFFICULTY: Moderate KEYWORDS: contingency table 134.

Referring to Table 2-12, if the sample is a good representation of the population, we can expect percent of the males in the population will be against the plan.

ANSWER: 50% TYPE: FI DIFFICULTY: Moderate KEYWORDS: contingency table 135.

Referring to Table 2-12, if the sample is a good representation of the population, we can expect percent of the females in the population will not be against the plan.

ANSWER: 88.46% or (36.54+51.92) TYPE: FI DIFFICULTY: Moderate KEYWORDS: contingency table TABLE 2-13 Given below is the stem-and-leaf display representing the amount of detergent used in gallons (with leaves in 10ths of gallons) in a month by 25 drive-through car wash operations in Phoenix. 9 | 147 10 | 02238 11 | 135566777 12 | 223489 13 | 02

136. Referring to Table 2-13, if a frequency distribution for the amount of detergent used is

constructed, using "9.0 but less than 10.0 gallons" as the first class, the frequency of the “11.0 but less than 12.0 gallons” class would be .

ANSWER: 9 TYPE: FI DIFFICULTY: Easy KEYWORDS: frequency distribution

137. Referring to Table 2-13, if a percentage histogram for the detergent data is constructed, using "9.0 but less than 10.0 gallons" as the first class, the percentage of drive-through car wash operations that use “12.0 but less than 13.0 gallons” of detergent would be . ANSWER: 24% TYPE: FI DIFFICULTY: Moderate

Presenting Data in Tables and Charts

KEYWORDS: relative frequency distribution, percentage distribution

138. Referring to Table 2-13, if a percentage histogram for the detergent data is constructed, using "9.0 but less than 10.0 gallons" as the first class, what percentage of drive-through car wash operations use less than 12 gallons of detergent in a month? ANSWER: 68% TYPE: FI DIFFICULTY: Easy KEYWORDS: percentage distribution, cumulative relative frequency

139. Referring to Table 2-13, if a relative frequency or percentage distribution for the detergent data is constructed, using "9.0 but less than 10.0 gallons" as the first class, what percentage of drivethrough car wash operations use at least 10 gallons of detergent in a month? ANSWER: 88% TYPE: FI DIFFICULTY: Easy KEYWORDS: relative frequency distribution, percentage distribution

140. Referring to Table 2-13, if a relative frequency or percentage distribution for the detergent data is constructed, using "9.0 but less than 10.0 gallons" as the first class, what percentage of drivethrough car wash operations use at least 10 gallons but less than 13 gallons of detergent in a month? ANSWER: 80% TYPE: FI DIFFICULTY: Easy KEYWORDS: relative frequency distribution, percentage distribution

141. Referring to Table 2-13, construct a frequency distribution for the detergent data, using "9.0 but less than 10.0 gallons" as the first class. ANSWER: Purchases (gals) Frequency 9.0 but less than 10.0 3 10.0 but less than 11.0 5 11.0 but less than 12.0 9 12.0 but less than 13.0 6 13.0 but less than 14.0 2 TYPE: PR DIFFICULTY: Moderate KEYWORDS: frequency distribution

142. Referring to Table 2-13, construct a relative frequency or percentage distribution for the detergent data, using "9.0 but less than 10.0" as the first class. ANSWER: Gasoline Purchases (gals) 9.0 but less than 10.0 10.0 but less than 11.0 11.0 but less than 12.0 12.0 but less than 13.0

Percentage 12% 20 36 24

55 Presenting Data in Tables and Charts 13.0 but less than 14.0 8 TYPE: PR DIFFICULTY: Moderate KEYWORDS: relative frequency distribution, percentage distribution

143. Referring to Table 2-13, construct a cumulative percentage distribution for the detergent data if the corresponding frequency distribution uses "9.0 but less than 10.0" as the first class. ANSWER: Gasoline Frequency Percentage Purchases (gals) Less Than Less Than 9.0 but less than 10.0 3 12 10.0 but less than 11.0 8 32 11.0 but less than 12.0 17 68 12.0 but less than 13.0 23 92 13.0 but less than 14.0 25 100 TYPE: PR DIFFICULTY: Moderate KEYWORDS: cumulative percentage distribution

144. Referring to Table 2-13, construct a percentage histogram for the detergent data, using "9.0 but less than 10.0" as the first class. ANSWER: % Histogram 40 35 30

25 20 15 10 5 0 9.0 -- 9.9

10.0 -- 10.9 11.0 -- 11.9 12.0 -- 12.9 13.0 -- 13.9 Purchased

TYPE: PR DIFFICULTY: Moderate KEYWORDS: histogram, frequency distribution

145. Referring to Table 2-13, construct a cumulative percentage polygon for the detergent data if the corresponding frequency distribution uses "9.0 but less than 10.0" as the first class.

ANSWER:

Presenting Data in Tables and Charts

Ogive 100 80

60 40 20 0 9

Purchase

TYPE: PR DIFFICULTY: Moderate KEYWORDS: cumulative percentage polygon

146. Referring to Table 2-13, construct a percentage polygon for the detergent data if the corresponding frequency distribution uses "9.0 but less than 10.0" as the first class.

ANSWER:

% Polygon 40 35 30 25 20 15 10 5 0

8.5

9.5

10.5

11.5

12.5

13.5

14.5

Purchase

TYPE: PR DIFFICULTY: Moderate KEYWORDS: percentage distribution TABLE 2-14 The table below contains the number of people who own a portable DVD player in a sample of 600 broken down by gender. Own a Portable DVD Player Yes No 147.

Male 96 224

Female 40 240

Referring to Table 2-14, construct a table of row percentages.

ANSWER: Own Yes

Male 70.59%

Female 29.41%

Total 100.00%

57 Presenting Data in Tables and Charts No Total

48.28% 53.33%

51.72% 46.67%

100.00% 100.00%

TYPE: PR DIFFICULTY: Easy KEYWORDS: row percentages

148. Referring to Table 2-14, construct a table of column percentages. ANSWER: Own Yes No Total

Male 30.00% 70.00% 100.00%

Female 14.29% 85.71% 100.00%

Total 22.67% 77.33% 100.00%

TYPE: PR DIFFICULTY: Easy KEYWORDS: column percentages

149. Referring to Table 2-14, construct a table of total percentages. ANSWER: Own Yes No Total

Male 16.00% 37.33% 53.33%

Female 6.67% 40.00% 46.67%

Total 22.67% 77.33% 100.00%

TYPE: PR DIFFICULTY: Easy KEYWORDS: total percentages

150. Referring to Table 2-14, of those who owned a portable DVD in the sample,

percent

were females. ANSWER: 29.41% TYPE: FI DIFFICULTY: Moderate KEYWORDS: contingency table, row percentages

151. Referring to Table 2-14, of those who did not own a portable DVD in the sample, percent were males. ANSWER: 48.28% TYPE: FI DIFFICULTY: Moderate KEYWORDS: contingency table, row percentages

152. Referring to Table 2-14, of the males in the sample,

percent owned a portable DVD.

ANSWER: 30% TYPE: FI DIFFICULTY: Moderate KEYWORDS: contingency table, column percentages

153. Referring to Table 2-14, of the females in the sample, DVD. ANSWER:

percent did not own a portable

Presenting Data in Tables and Charts

85.71% TYPE: FI DIFFICULTY: Moderate KEYWORDS: contingency table, column percentages

154. Referring to Table 2-14, of the females in the sample,

percent owned a portable

DVD. ANSWER: 14.29% TYPE: FI DIFFICULTY: Moderate KEYWORDS: contingency table, column percentages

155. Referring to Table 2-14,

percent of the 600 were females who owned a portable

DVD. ANSWER: 6.67% TYPE: FI DIFFICULTY: Moderate KEYWORDS: contingency table, total percentage

156. Referring to Table 2-14,

percent of the 600 were males who owned a portable DVD.

ANSWER: 16% TYPE: FI DIFFICULTY: Moderate KEYWORDS: contingency table, total percentage

157. Referring to Table 2-14,

percent of the 600 were females who either owned or did

not own a portable DVD. ANSWER: 46.67% TYPE: FI DIFFICULTY: Moderate KEYWORDS: contingency table, total percentage

158. Referring to Table 2-14,

percent of the 600 were males who did not owned a portable

DVD. ANSWER: 37.33% TYPE: FI DIFFICULTY: Moderate KEYWORDS: contingency table, total percentage

159. Referring to Table 2-14,

percent of the 600 owned a portable DVD.

ANSWER: 22.67% TYPE: FI DIFFICULTY: Moderate KEYWORDS: contingency table, column percentages

160. Referring to Table 2-14, ANSWER:

percent of the 600 did not owned a portable DVD.

59 Presenting Data in Tables and Charts 77.33% TYPE: FI DIFFICULTY: Moderate KEYWORDS: contingency table, column percentages

161. Referring to Table 2-14,

percent of the 600 were females.

ANSWER: 46.67% TYPE: FI DIFFICULTY: Easy KEYWORDS: contingency table, row percentages

162. Referring to Table 2-14, if the sample is a good representation of the population, we can expect percent of the population will own a portable DVD. ANSWER: 22.67% TYPE: FI DIFFICULTY: Moderate KEYWORDS: contingency table, column percentages

163. Referring to Table 2-14, if the sample is a good representation of the population, we can expect percent of the population will be males. ANSWER: 53.33% TYPE: FI DIFFICULTY: Moderate KEYWORDS: contingency table, column percentages

164. Referring to Table 2-14, if the sample is a good representation of the population, we can expect percent of those who own a portable DVD in the population will be males. ANSWER: 70.59% TYPE: FI DIFFICULTY: Moderate KEYWORDS: contingency table, row percentages

165. Referring to Table 2-14, if the sample is a good representation of the population, we can expect percent of the males in the population will own a portable DVD. ANSWER: 30% TYPE: FI DIFFICULTY: Moderate KEYWORDS: contingency table, column percentages

166. Referring to Table 2-14, if the sample is a good representation of the population, we can expect percent of the females in the population will not own a portable DVD. ANSWER: 85.71% TYPE: FI DIFFICULTY: Moderate KEYWORDS: contingency table, column percentages

Numerical Descriptive Measures

CHAPTER 3: NUMERICAL DESCRIPTIVE MEASURES 1. Which of the following statistics is not a measure of central tendency? a) arithmetic mean b) median c) mode d) Q3 ANSWER: d TYPE: MC DIFFICULTY: Easy KEYWORDS: measure of central tendency, arithmetic mean, median, mode, quartiles 2. Which measure of central tendency can be used for both numerical and categorical variables? a) arithmetic mean b) median c) mode d) geometric mean ANSWER: c TYPE: MC DIFFICULTY: Easy KEYWORDS: measure of central tendency, mode, arithmetic mean, median, geometric mean 3. Which of the arithmetic mean, median, mode, and geometric mean are resistant measures of central tendency? a) the arithmetic mean and median only b) the median and mode only c) the mode and geometric mean only d) the arithmetic mean and mode only ANSWER: b TYPE: MC DIFFICULTY: Moderate KEYWORDS: measure of central tendency, resistant to outliers, mean, median, mode 4. In a right-skewed distribution a) the median equals the arithmetic mean. b) the median is less than the arithmetic mean. c) the median is larger than the arithmetic mean. d) None of the above. ANSWER: b TYPE: MC DIFFICULTY: Easy KEYWORDS: shape 5. Which of the following statements about the median is not true? a) It is more affected by extreme values than the arithmetic mean. b) It is a measure of central tendency. c) It is equal to Q2. d) It is equal to the mode in bell-shaped "normal" distributions.

66 Numerical Descriptive Measures

ANSWER: a TYPE: MC DIFFICULTY: Easy KEYWORDS: median, measure of central tendency, resistant to outliers, quartile 6. In a perfectly symmetrical bell-shaped "normal" distribution a) the arithmetic mean equals the median. b) the median equals the mode. c) the arithmetic mean equals the mode. d) All the above. ANSWER: d TYPE: MC DIFFICULTY: Easy KEYWORDS: shape, normal distribution 7. In a perfectly symmetrical distribution a) the range equals the interquartile range. b) the interquartile range equals the arithmetic mean. c) the median equals the arithmetic mean. d) the variance equals the standard deviation. ANSWER: c TYPE: MC DIFFICULTY: Easy KEYWORDS: shape

8. When extreme values are present in a set of data, which of the following descriptive summary measures are most appropriate? a) CV and range b) arithmetic mean and standard deviation c) interquartile range and median d) variance and interquartile range ANSWER: c TYPE: MC DIFFICULTY: Easy KEYWORDS: resistant to outliers, coefficient of variation, range, arithmetic mean, standard deviation, interquartile range, median, variance 9. In general, which of the following descriptive summary measures cannot be easily approximated from a box-and-whisker plot? a) the variance b) the range c) the interquartile range d) the median ANSWER: a TYPE: MC DIFFICULTY: Easy KEYWORDS: box-and-whisker plot, variance, range, interquartile range, median 10. The smaller the spread of scores around the arithmetic mean,

Numerical Descriptive Measures a) b) c) d)

the smaller the interquartile range. the smaller the standard deviation. the smaller the coefficient of variation. All the above.

ANSWER: d TYPE: MC DIFFICULTY: Moderate KEYWORDS: variation, arithmetic mean, interquartile range, standard deviation, coefficient of variation 11. Which descriptive summary measures are considered to be resistant statistics? a) the arithmetic mean and standard deviation b) the interquartile range and range c) the mode and variance d) the median and interquartile range ANSWER: d TYPE: MC DIFFICULTY: Moderate KEYWORDS: resistant to outliers, arithmetic mean, standard deviation, interquartile range, range, mode, variance, median 12. In right-skewed distributions, which of the following is the correct statement? a) The distance from Q1 to Q2 is larger than the distance from Q2 to Q3. b) The distance from Q1 to Q2 is smaller than the distance from Q2 to Q3. c) The arithmetic mean is smaller than the median. d) The mode is larger than the arithmetic mean. ANSWER: b TYPE: MC DIFFICULTY: Easy KEYWORDS: shape, quartiles, arithmetic mean, mode 13. In perfectly symmetrical distributions, which of the following is NOT a correct statement? a) The distance from Q1 to Q2 equals to the distance from Q2 to Q3. b) The distance from the smallest observation to Q1 is the same as the distance from Q3 to the largest observation. c) The distance from the smallest observation to Q2 is the same as the distance from Q2 to the largest observation. d) The distance from Q1 to Q3 is half of the distance from the smallest to the largest observation. ANSWER: d TYPE: MC DIFFICULTY: Easy KEYWORDS: shape, quartiles, five-number summary, quartiles 14. In left-skewed distributions, which of the following is the correct statement? a) The distance from Q1 to Q2 is smaller than the distance from Q2 to Q3. b) The distance from the smallest observation to Q1 is larger than the distance from Q3 to the largest observation.

68 Numerical Descriptive Measures c) The distance from the smallest observation to Q2 is smaller than the distance from Q2 to the largest observation. d) The distance from Q1 to Q3 is twice the distance from the Q1 to Q2. ANSWER: b TYPE: MC DIFFICULTY: Easy KEYWORDS: shape, quartiles, five-number summary, quartiles 15. According to the empirical rule, if the data form a "bell-shaped" normal distribution, percent of the observations will be contained within 2 standard deviations around the arithmetic mean. a) 68.26 b) 88.89 c) 93.75 d) 95.44 ANSWER: d TYPE: MC DIFFICULTY: Moderate KEYWORDS: empirical rule, normal distribution 16. According to the empirical rule, if the data form a "bell-shaped" normal distribution, percent of the observations will be contained within 1 standard deviation around the arithmetic mean. a) 68.26 b) 75.00 c) 88.89 d) 93.75 ANSWER: a TYPE: MC DIFFICULTY: Moderate KEYWORDS: empirical rule, normal distribution 17. According to the empirical rule, if the data form a "bell-shaped" normal distribution, percent of the observations will be contained within 3 standard deviations around the arithmetic mean. a) 68.26 b) 75.00 c) 95.0 d) 99.7 ANSWER: d TYPE: MC DIFFICULTY: Moderate KEYWORDS: empirical rule, normal distribution 18. Which of the following is NOT a measure of central tendency? a) the arithmetic mean b) the geometric mean c) the mode d) the interquartile range

Numerical Descriptive Measures

ANSWER: d TYPE: MC DIFFICULTY: Easy KEYWORDS: measure of central tendency, arithmetic mean, geometric mean, mode, interquartile range 19. Which of the following is NOT sensitive to extreme values? a) the range b) the standard deviation c) the interquartile range d) the coefficient of variation ANSWER: c TYPE: MC DIFFICULTY: Moderate KEYWORDS: resistant to outliers, range, standard deviation, interquartile range, coefficient of variation 20. Which of the following is sensitive to extreme values? a) the median b) the interquartile range c) the arithmetic mean d) the 1st quartile ANSWER: c TYPE: MC DIFFICULTY: Moderate KEYWORDS: resistant to outliers, median, interquartile range, arithmetic mean, quartiles 21. Which of the following is the easiest to compute? a) the arithmetic mean b) the median c) the mode d) the geometric mean ANSWER: c TYPE: MC DIFFICULTY: Easy KEYWORDS: mode, arithmetic mean, median, geometric mean 22. According to the Chebyshev rule, at least 75% of all observations in any data set are contained within a distance of how many standard deviations around the mean? a) 1 b) 2 c) 3 d) 4 ANSWER: b TYPE: MC DIFFICULTY: Easy KEYWORDS: Chebyshev rule

70 Numerical Descriptive Measures 23. According to the Chebyshev rule, at least 93.75% of all observations in any data set are contained within a distance of how many standard deviations around the mean? a) 1 b) 2 c) 3 d) 4 ANSWER: d TYPE: MC DIFFICULTY: Moderate KEYWORDS: Chebyshev rule 24. According to the Chebyshev rule, at least what percentage of the observations in any data set are contained within a distance of 3 standard deviations around the mean? a) 67% b) 75% c) 88.89% d) 99.7% ANSWER: c TYPE: MC DIFFICULTY: Easy KEYWORDS: Chebyshev rule 25. According to the Chebyshev rule, at least what percentage of the observations in any data set are contained within a distance of 2 standard deviations around the mean? a) 67% b) 75% c) 88.89% d) 95% ANSWER: b TYPE: MC DIFFICULTY: Easy KEYWORDS: Chebyshev rule TABLE 3-1 Health care issues are receiving much attention in both academic and political arenas. A sociologist recently conducted a survey of citizens over 60 years of age whose net worth is too high to qualify for Medicaid and have no private health insurance. The ages of 25 uninsured senior citizens were as follows: 60 61 62 63 64 65 66 68 68 69 70 73 73 74 75 76 76 81 81 82 86 87 89 90 92 26. Referring to Table 3-1, calculate the arithmetic mean age of the uninsured senior citizens to the nearest hundredth of a year. ANSWER: 74.04 years TYPE: PR DIFFICULTY: Easy KEYWORDS: arithmetic mean 27. Referring to Table 3-1, identify the median age of the uninsured senior citizens.

Numerical Descriptive Measures

ANSWER: 73 years TYPE: PR DIFFICULTY: Easy KEYWORDS: median 28. Referring to Table 3-1, identify the first quartile of the ages of the uninsured senior citizens. ANSWER: 65.5 years TYPE: PR DIFFICULTY: Moderate KEYWORDS: quartiles 29. Referring to Table 3-1, identify the third quartile of the ages of the uninsured senior citizens. ANSWER: 81.5 years TYPE: PR DIFFICULTY: Moderate KEYWORDS: quartiles 30. Referring to Table 3-1, identify the interquartile range of the ages of the uninsured senior citizens. ANSWER: 16 years TYPE: PR DIFFICULTY: Moderate KEYWORDS: interquartile range 31. Referring to Table 3-1, identify which of the following is the correct statement. a) One fourth of the senior citizens sampled are below 65.5 years of age. b) The middle 50% of the senior citizens sampled are between 65.5 and 73.0 years of age. c) The average age of senior citizens sampled is 73.5 years of age. d) All of the above are correct. ANSWER: a TYPE: MC DIFFICULTY: Moderate KEYWORDS: quartiles, arithmetic mean 32. Referring to Table 3-1, identify which of the following is the correct statement. a) One fourth of the senior citizens sampled are below 64 years of age. b) The middle 50% of the senior citizens sampled are between 65.5 and 73.0 years of age. c) 25% of the senior citizens sampled are older than 81.5 years of age. d) All of the above are correct. ANSWER: c TYPE: MC DIFFICULTY: Moderate KEYWORDS: quartiles 33. Referring to Table 3-1, what type of shape does the distribution of the sample appear to have? ANSWER: Slightly positive or right-skewed. TYPE: PR DIFFICULTY: Moderate

72 Numerical Descriptive Measures KEYWORDS: shape 34. Referring to Table 3-1, calculate the variance of the ages of the uninsured senior citizens correct to the nearest hundredth of a year squared. ANSWER: 94.96 years2 TYPE: PR DIFFICULTY: Moderate KEYWORDS: variance 35. Referring to Table 3-1, calculate the standard deviation of the ages of the uninsured senior citizens correct to the nearest hundredth of a year. ANSWER: 9.74 years TYPE: PR DIFFICULTY: Moderate KEYWORDS: standard deviation 36. Referring to Table 3-1, calculate the coefficient of variation of the ages of the uninsured senior citizens. ANSWER: 13.16% TYPE: PR DIFFICULTY: Moderate KEYWORDS: coefficient of variation 37. True or False: The median of the values 3.4, 4.7, 1.9, 7.6, and 6.5 is 1.9. ANSWER: False TYPE: TF DIFFICULTY: Moderate KEYWORDS: median 38. True or False: The median of the values 3.4, 4.7, 1.9, 7.6, and 6.5 is 4.05. ANSWER: False TYPE: TF DIFFICULTY: Moderate KEYWORDS: median 39. True or False: In a set of numerical data, the value for Q3 can never be smaller than the value for Q1. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: quartiles 40. True or False: In a set of numerical data, the value for Q2 is always halfway between Q1 and Q3. ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: quartiles

Numerical Descriptive Measures

41. True or False: If the distribution of a data set were perfectly symmetrical, the distance from Q1 to the median would always equal the distance from Q3 to the median in a box-and-whisker plot. ANSWER: True TYPE: TF DIFFICULTY: Moderate KEYWORDS: box-and-whisker plot, quartiles, shape 42. True or False: In right-skewed distributions, the distance from Q3 to the largest observation exceeds the distance from the smallest observation to Q1. ANSWER: True TYPE: TF DIFFICULTY: Moderate KEYWORDS: quartiles, shape 43. True or False: In left-skewed distributions, the distance from the smallest observation to Q1 exceeds the distance from Q3 to the largest observation. ANSWER: True TYPE: TF DIFFICULTY: Moderate KEYWORDS: quartiles, shape 44. True or False: A box-and-whisker plot is a graphical representation of a 5-number summary. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: box-and-whisker plot, five-number summary 45. True or False: The 5-number summary consists of the smallest observation, the first quartile, the median, the third quartile, and the largest observation. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: five-number summary 46. True or False: In a box-and-whisker plot, the box portion represents the data between the first and third quartile values. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: box-and-whisker plot 47. True or False: The line drawn within the box of the box-and-whisker plot always represents the arithmetic mean. ANSWER: False TYPE: TF DIFFICULTY: Easy

74 Numerical Descriptive Measures KEYWORDS: box-and-whisker plot, arithmetic mean 48. True or False: The line drawn within the box of the box-and-whisker plot always represents the median. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: box-and-whisker plot, median

49. True or False: In a sample of size 40, the sample mean is 15. In this case, the sum of all observations in the sample is  Xi = 600. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: arithmetic mean 50. True or False: A population with 200 elements has an arithmetic mean of 10. From this information, it can be shown that the population standard deviation is 15. ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: arithmetic mean, standard deviation 51. True or False: In exploratory data analysis, a box-and-whisker plot can be used to illustrate the median, quartiles, and extreme values. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: box-and-whisker plot, five-number summary 52. True or False: An economics professor bases his final grade on homework, two midterm examinations, and a final examination. The homework counts 10% toward the final grade, while each midterm examination counts 25%. The remaining portion consists of the final examination. If a student scored 95% in homework, 70% on the first midterm examination, 96% on the second midterm examination, and 72% on the final, his final average is 79.8%. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: arithmetic mean 53. True or False: The median of a data set with 20 items would be the average of the 10th and the 11th items in the ordered array. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: median, arithmetic mean

Numerical Descriptive Measures

54. True or False: The coefficient of variation measures variability in a data set relative to the size of the arithmetic mean. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: coefficient of variation 55. True or False: The coefficient of variation is expressed as a percentage. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: coefficient of variation 56. True or False: The coefficient of variation is a measure of central tendency in the data. ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: coefficient of variation, measure of variation 57. True or False: The interquartile range is a measure of variation or dispersion in a set of data. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: interquartile range, measure of variation 58. True or False: The interquartile range is a measure of central tendency in a set of data. ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: interquartile range, measure of variation 59. True or False: The geometric mean is a measure of variation or dispersion in a set of data. ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: geometric mean, measure of central tendency 60. True or False: The geometric mean is useful in measuring the rate of change of a variable over time. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: geometric mean 61. True or False: If a set of data is perfectly symmetrical, the arithmetic mean must be identical to the median.

76 Numerical Descriptive Measures ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: shape, arithmetic mean, median 62. True or False: The coefficient of variation is a measure of relative variation. ANSWER: True TYPE: TF DIFFICULTY: Moderate KEYWORDS: coefficient of variation 63. True or False: If the data set is approximately bell-shaped, the empirical rule will more accurately reflect the greater concentration of data close to the mean as compared to the Chebyshev rule. ANSWER: True TYPE: TF DIFFICULTY: Moderate KEYWORDS: empirical rule, Chebyshev rule, normal distribution 64. If the arithmetic mean of a numerical data set exceeds the median, the data are considered to be skewed. ANSWER: positive or right TYPE: FI DIFFICULTY: Easy KEYWORDS: shape, arithmetic mean, median TABLE 3-2 The data below represent the amount of grams of carbohydrates in a serving of breakfast cereal in a sample of 11 different servings. 11 15 23 29 19 22 21 20 15 25 17 65. Referring to Table 3-2, the arithmetic mean carbohydrates in this sample is

grams.

ANSWER: 217/11 = 19.73 TYPE: FI DIFFICULTY: Moderate KEYWORDS: arithmetic mean 66. Referring to Table 3-2, the median carbohydrate amount in the cereal is

grams.

ANSWER: 20 TYPE: FI DIFFICULTY: Moderate KEYWORDS: median 67. Referring to Table 3-2, the first quartile of the carbohydrate amounts is ANSWER: 15 TYPE: FI DIFFICULTY: Moderate KEYWORDS: quartiles

grams.

Numerical Descriptive Measures 68. Referring to Table 3-2, the third quartile of the carbohydrate amounts is

grams.

ANSWER: 23 TYPE: FI DIFFICULTY: Moderate KEYWORDS: quartiles 69. Referring to Table 3-2, the range in the carbohydrate amounts is

grams.

ANSWER: 18 TYPE: FI DIFFICULTY: Easy KEYWORDS: range 70. Referring to Table 3-2, the interquartile range in the carbohydrate amounts is

grams.

ANSWER: 8 TYPE: FI DIFFICULTY: Moderate KEYWORDS: interquartile range 71. Referring to Table 3-2, the variance of the carbohydrate amounts is

(grams squared).

ANSWER: 26.02 TYPE: FI DIFFICULTY: Moderate KEYWORDS: variance 72. Referring to Table 3-2, the standard deviation of the carbohydrate amounts is

grams.

ANSWER: 5.10 TYPE: FI DIFFICULTY: Moderate KEYWORDS: standard deviation 73. Referring to Table 3-2, the coefficient of variation of the carbohydrate amounts is percent. ANSWER: 25.86% TYPE: FI DIFFICULTY: Moderate KEYWORDS: coefficient of variation 74. Referring to Table 3-2, the five-number summary of the carbohydrate amounts consists of , , , , . ANSWER: 11, 15, 20, 23, 29 TYPE: FI DIFFICULTY: Moderate KEYWORDS: five-number summary 75. Referring to Table 3-2, construct a box-and-whisker plot for the carbohydrate amounts.

78 Numerical Descriptive Measures ANSWER: Box-and-whisker Plot

TYPE: PR DIFFICULTY: Moderate KEYWORDS: box-and-whisker plot 76. Referring to Table 3-2, what type of shape does the distribution of the sample appear to have? ANSWER: Slightly positive or right-skewed. TYPE: PR DIFFICULTY: Moderate EXPLANATION: The distribution is considered as slightly right-skewed because the length of the right whisker (6) is slighter longer than the length of the left whisker (4). The distribution could also be considered as slight left-skewed since the mean (19.73) is slightly lower than the median (20). KEYWORDS: shape TABLE 3-3 The stem-and-leaf display below represents the number of vitamin supplements sold by a health food store in a sample of 16 days. Stem Leaves 1H 99 2L 0023 2H 567 3L 034 3H 568 4L 1 Note (1): 1H means the “high teens” – 15, 16, 17, 18, or 19; 2L means the “low twenties” – 20, 21, 22, 23, or 24; 2H means the “high twenties” – 25, 26, 27, 28, or 29, etc. Note (2): For this sample, the sum of the observations is 448, the sum of the squares of the observations is 13,356, and the sum of the squared differences between each observation and the mean is 812. 77. Referring to Table 3-3, the arithmetic mean of the number of vitamin supplements sold in this sample is . ANSWER: 28 TYPE: FI DIFFICULTY: Easy KEYWORDS: arithmetic mean

Numerical Descriptive Measures

78. Referring to Table 3-3, the first quartile of the number of vitamin supplements sold in this sample is . ANSWER: 20 TYPE: FI DIFFICULTY: Moderate KEYWORDS: quartiles 79. Referring to Table 3-3, the third quartile of the number of vitamin supplements sold in this sample is . ANSWER: 35 TYPE: FI DIFFICULTY: Moderate KEYWORDS: quartiles 80. Referring to Table 3-3, the median number of vitamin supplements sold in this sample is . ANSWER: 26.5 TYPE: FI DIFFICULTY: Moderate KEYWORDS: median 81. Referring to Table 3-3, the range of the number of vitamin supplements sold in this sample is . ANSWER: 22 TYPE: FI DIFFICULTY: Easy KEYWORDS: range 82. Referring to Table 3-3, the interquartile range of the number of vitamin supplements sold in this sample is . ANSWER: 15 TYPE: FI DIFFICULTY: Moderate KEYWORDS: interquartile range 83. Referring to Table 3-3, the variance of the number of vitamin supplements sold in this sample is . ANSWER: 54.1 TYPE: FI DIFFICULTY: Easy KEYWORDS: variance 84. Referring to Table 3-3, the standard deviation of the number of vitamin supplements sold in this sample is . ANSWER: 7.4 TYPE: FI DIFFICULTY: Easy

80 Numerical Descriptive Measures KEYWORDS: standard deviation 85. Referring to Table 3-3, the coefficient of variation of the number of vitamin supplements sold in this sample is percent. ANSWER: 26.3% TYPE: FI DIFFICULTY: Moderate KEYWORDS: coefficient of variation 86. Referring to Table 3-3, the five-number summary of the data in this sample consists of , , , , . ANSWER: 19, 20, 26.5, 35, 41 TYPE: FI DIFFICULTY: Moderate KEYWORDS: five-number summary 87. Referring to Table 3-3, construct a box-and-whisker plot for the data in this sample. ANSWER: Box-and-whisker Plot

TYPE: PR DIFFICULTY: Moderate KEYWORDS: box-and-whisker plot 88. Referring to Table 3-3, what type of shape does the distribution of the sample appear to have? ANSWER: Slightly positive or right-skewed. TYPE: PR DIFFICULTY: Moderate KEYWORDS: shape TABLE 3-4 The stem-and-leaf display below represents the number of cargo manifests approved by customs inspectors of the Port of New York in a sample of 35 days. STEM LEAVES 1H 67889 2L 0011122223333444

Numerical Descriptive Measures 2H 3L

5566678899 1122

Note (1): 1H means the “high teens” – 15, 16, 17, 18, or 19; 2L means the “low twenties” – 20, 21, 22, 23, or 24; 2H means the “high twenties” – 25, 26, 27, 28, or 29, etc. Note (2): For this sample, the sum of the observations is 838, the sum of the squares of the observations is 20,684, and the sum of the squared differences between each observation and the mean is 619.89. 89. Referring to Table 3-4, the arithmetic mean of the customs data is

ANSWER: 23.9 TYPE: FI DIFFICULTY: Easy KEYWORDS: arithmetic mean 90. Referring to Table 3-4, the median of the customs data is

ANSWER: 23 TYPE: FI DIFFICULTY: Moderate KEYWORDS: median 91. Referring to Table 3-4, the first quartile of the customs data is

ANSWER: 21 TYPE: FI DIFFICULTY: Moderate KEYWORDS: quartiles 92. Referring to Table 3-4, the third quartile of the customs data is

ANSWER: 27 TYPE: FI DIFFICULTY: Moderate KEYWORDS: quartiles 93. Referring to Table 3-4, the range of the customs data is

ANSWER: 16 TYPE: FI DIFFICULTY: Easy KEYWORDS: range 94. Referring to Table 3-4, the interquartile range of the customs data is

ANSWER: 6 TYPE: FI DIFFICULTY: Moderate KEYWORDS: interquartile range 95. Referring to Table 3-4, the variance of the customs data is ANSWER:

82 Numerical Descriptive Measures 18.2 TYPE: FI DIFFICULTY: Easy KEYWORDS: variance 96. Referring to Table 3-4, the standard deviation of the customs data is

ANSWER: 4.3 TYPE: FI DIFFICULTY: Easy KEYWORDS: standard deviation 97. Referring to Table 3-4, the coefficient of variation of the customs data is

percent.

ANSWER: 17.8% or 18% TYPE: FI DIFFICULTY: Moderate KEYWORDS: coefficient of variation 98. Referring to Table 3-4, the five-number summary for the data in the customs sample consists of , , , , . ANSWER: 16, 21, 23, 27, 32 TYPE: FI DIFFICULTY: Moderate KEYWORDS: five-number summary 99. Referring to Table 3-4, construct a box-and-whisker plot of this sample. ANSWER: Box-and-whisker Plot

TYPE: PR DIFFICULTY: Moderate KEYWORDS: box-and-whisker plot TABLE 3-5 The rate of return of a Fortune 500 company over the past 15 years are: 3.17%, 4.43%, 5.93%, 5.43%, 7.29%, 8.21%, 6.23%, 5.23%, 4.34%, 6.68%, 7.14%, -5.56%, -5.23%, -5.73%, -10.34% 100.

Referring to Table 3-5, compute the arithmetic mean rate of return.

ANWER: 2.48%

Numerical Descriptive Measures

TYPE: PR DIFFICULTY: Easy KEYWORDS: arithmetic mean 101.

Referring to Table 3-5, compute the geometric mean rate of return.

ANWER: 2.31% TYPE: PR DIFFICULTY: moderate KEYWORDS: geometric mean rate of return 102.

Referring to Table 3-5, what is the range of the rate of return?

ANSWER: 18.55% TYPE: PR DIFFICULTY: Easy KEYWORDS: range 103.

Referring to Table 3-5, construct a box-and-whisker plot for the rate of return.

ANSWER: Box-and-whisker Plot

-0.15

-0.1

-0.05

0.05

0.1

TYPE: PR DIFFICULTY: Easy KEYWORDS: box-and-whisker plot 104.

Referring to Table 3-5, what is the shape of the distribution for the rate of return?

ANSWER: Left-skewed TYPE: PR DIFFICULTY: Easy KEYWORDS: shape TABLE 3-6 The rate of return of an Internet Service Provider over a 10 year period are: 10.25%, 12.64%, 8.37%, 9.29%, 6.23%, 42.53%, 29.23%, 15.25%, 21.52%, -2.35%. 105.

Referring to Table 3-6, compute the arithmetic mean rate of return.

ANWER: 15.30% TYPE: PR DIFFICULTY: Easy KEYWORDS: arithmetic mean 106.

Referring to Table 3-6, compute the geometric mean rate of return.

84 Numerical Descriptive Measures

ANSWER: 14.68% TYPE: PR DIFFICULTY: Moderate KEYWORDS: geometric mean rate of return 107.

Referring to Table 3-6, construct a box-and-whisker plot for the rate of return

ANWER: Box-and-whisker Plot

-0.1

0.1

0.2

0.3

0.4

0.5

TYPE: PR DIFFICULTY: Moderate KEYWORDS: box-and-whisker plot 108.

Referring to Table 3-6, what is the shape of the distribution for the rate of return?

ANSWER: Right-skewed TYPE: PR DIFFICULTY: Moderate KEYWORDS: shape

109. The Z scores can be used to identify outliers. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: Z scores

110. The larger the Z score, the farther is the distance from the observation to the median. ANSWER: False TYPE: TF DIFFICULTY: Moderate KEYWORDS: Z scores

111. As a general rule, an observation is considered an extreme value if its Z score is greater than −3.

ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: Z score

112. As a general rule, an observation is considered an extreme value if its Z score is greater than 3. ANSWER:

Numerical Descriptive Measures

True TYPE: TF DIFFICULTY: Easy KEYWORDS: Z score

113. As a general rule, an observation is considered an extreme value if its Z score is less than 3. ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: Z score

114. As a general rule, an observation is considered an extreme value if its Z score is less than −3. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: Z score

115. The Z score of an observation can never be negative. ANSWER: False TYPE: TF DIFFICULTY: Moderate KEYWORDS: Z score

116. The Z score of an observation measures how many standard deviations is the value from the mean. ANSWER: True TYPE: TF DIFFICULTY: Moderate KEYWORDS: Z score

117. The 12-month rate of returns over a nine year period of a particular stock is 0.099, −0.289, 0.089, 0.226, 0.041, 0.161, 0.064, −0.029 and 0.022. The geometric mean rate of return for this stock is . ANSWER: 3.23% or 0.0323 TYPE: FI DIFFICULTY: moderate KEYWORDS: geometric mean rate of return

118. The rate of return for the S&P 500 over a seven year period is −0.029, −0.061, −0.493, −0.286, −0.160, −0.186 and −0.224. The geometric mean rate of return is

ANSWER: −22.03% or −0.2203 TYPE: FI DIFFICULTY: moderate KEYWORDS: geometric mean rate of return

119. The rate of return for Microsoft’s stock over a seven year period is 0.527, 0.145, 0.684, 1.146, 0.564, 0.883 and 0.436. The geometric mean rate of return is

86 Numerical Descriptive Measures

ANSWER: 59.90% or 0.5990 TYPE: FI DIFFICULTY: moderate KEYWORDS: geometric mean rate of return TABLE 3-7 Given below is the frequency distribution of the 12-month rate of return achieved by 30 stocks. 12-month Rate of Return −0.10 but less than 0.0 0.0 but less than 0.1 0.1 but less than 0.2 0.2 but less than 0.3 0.3 but less than 0.4

Number of Funds 3 5 6 10 6

120. Referring to Table 3-7, what is the approximate mean rate of return? ANSWER: 0.1867 TYPE: PR DIFFICULTY: Moderate KEYWORDS: approximate mean

121. Referring to Table 3-7, what is the approximate variance? ANSWER: 0.0162 TYPE: PR DIFFICULTY: Moderate KEYWORDS: approximate variance

122. Referring to Table 3-7, what is the approximate standard deviation? ANSWER: 0.1273 TYPE: PR DIFFICULTY: Moderate KEYWORDS: approximate standard deviation TABLE 3-8 Given below is the frequency distribution of the number of defective tomatoes in 50 different shipments. Defective Items 0 but less than 5 5 but less than 10 10 but less than 15 15 but less than 20 20 but less than 25 25 but less than 30

Frequency 35 7 5 2 1 0

Numerical Descriptive Measures

123. Referring to Table 3-8, what is the approximate mean number of defective tomatoes? ANSWER: 5.2 TYPE: PR DIFFICULTY: Moderate KEYWORDS: approximate mean 124.

Referring to Table 3-8, what is the approximate variance?

ANSWER: 23.6837 TYPE: PR DIFFICULTY: Moderate KEYWORDS: approximate variance 125.

Referring to Table 3-8, what is the approximate standard deviation?

ANSWER: 4.8666 TYPE: PR DIFFICULTY: Moderate KEYWORDS: approximate standard deviation

Basic Probability

CHAPTER 4: BASIC PROBABILITY 1. If two events are collectively exhaustive, what is the probability that one or the other occurs? a) 0 b) 0.50 c) 1.00 d) Cannot be determined from the information given. ANSWER: c TYPE: MC DIFFICULTY: Easy KEYWORDS: collectively exhaustive 2. If two events are collectively exhaustive, what is the probability that both occur at the same time? a) 0 b) 0.50 c) 1.00 d) Cannot be determined from the information given. ANSWER: d TYPE: MC DIFFICULTY: Moderate KEYWORDS: collectively exhaustive, mutually exclusive EXPLANATION: We do not know if they are mutually exclusive. 3. If two events are mutually exclusive, what is the probability that one or the other occurs? a) 0 b) 0.50 c) 1.00 d) Cannot be determined from the information given. ANSWER: d TYPE: MC DIFFICULTY: moderate KEYWORDS: collectively exhaustive, mutually exclusive EXPLANATION: We do not know if they are collectively exhaustive. 4. If two events are mutually exclusive, what is the probability that both occur at the same time? a) 0 b) 0.50 c) 1.00 d) Cannot be determined from the information given. ANSWER: a TYPE: MC DIFFICULTY: Easy KEYWORDS: mutually exclusive 5. If two events are mutually exclusive and collectively exhaustive, what is the probability that both occur?

Basic Probability a) 0 b) 0.50 c) 1.00 d) Cannot be determined from the information given. ANSWER: a TYPE: MC DIFFICULTY: Easy KEYWORDS: collective exhaustive, mutually exclusive 6. If two events are mutually exclusive and collectively exhaustive, what is the probability that one or the other occurs? a) 0 b) 0.50 c) 1.00 d) Cannot be determined from the information given. ANSWER: c TYPE: MC DIFFICULTY: Easy KEYWORDS: collectively exhaustive, mutually exclusive 7. If events A and B are mutually exclusive and collectively exhaustive, what is the probability that event A occurs? a) 0 b) 0.50 c) 1.00 d) Cannot be determined from the information given. ANSWER: d TYPE: MC DIFFICULTY: moderate KEYWORDS: collectively exhaustive, mutually exclusive EXPLANATION: We do not know if they are equally likely events. 8. If two equally likely events A and B are mutually exclusive and collectively exhaustive, what is the probability that event A occurs? a) 0 b) 0.50 c) 1.00 d) Cannot be determined from the information given. ANSWER: b TYPE: MC DIFFICULTY: easy KEYWORDS: collectively exhaustive, mutually exclusive 9. If two equally likely events A and B are mutually exclusive, what is the probability that event A occurs? a) 0 b) 0.50 c) 1.00 d) Cannot be determined from the information given.

Basic Probability

ANSWER: d TYPE: MC DIFFICULTY: moderate KEYWORDS: collectively exhaustive, mutually exclusive EXPLANATION: We do not know if they are collectively exhaustive. 10. If two equally likely events A and B are collectively exhaustive, what is the probability that event A occurs? a) 0 b) 0.50 c) 1.00 d) Cannot be determined from the information given. ANSWER: d TYPE: MC DIFFICULTY: moderate KEYWORDS: collectively exhaustive, mutually exclusive EXPLANATION: We do not know if they are mutually exclusive. 11. Selection of raffle tickets from a large bowl is an example of a) sampling with replacement. b) sampling without replacement. c) subjective probability. d) None of the above. ANSWER: b TYPE: MC DIFFICULTY: Easy KEYWORDS: sampling with replacement, sampling without replacement 12. If two events are independent, what is the probability that they both occur? a) 0 b) 0.50 c) 1.00 d) Cannot be determined from the information given. ANSWER: d TYPE: MC DIFFICULTY: Moderate KEYWORDS: statistical independence

13. If the outcome of event A is not affected by event B, then events A and B are said to be a) b) c) d)

mutually exclusive. statistically independent. collectively exhaustive. None of the above.

ANSWER: b TYPE: MC DIFFICULTY: Easy KEYWORDS: statistical independence

14. If event A and event B cannot occur at the same time, then events A and B are said to be

Basic Probability a) b) c) d)

mutually exclusive. statistically independent. collectively exhaustive. None of the above.

ANSWER: a TYPE: MC DIFFICULTY: Easy KEYWORDS: mutually exclusive

15. If either event A or event B must occur, then events A and B are said to be a) b) c) d)

mutually exclusive. statistically independent. collectively exhaustive. None of the above.

ANSWER: c TYPE: MC DIFFICULTY: Moderate KEYWORDS: collectively exhaustive 16. The collection of all possible events is called a) a simple probability. b) a sample space. c) a joint probability. d) the null set. ANSWER: b TYPE: MC DIFFICULTY: Moderate KEYWORDS: sample space 17. All the events in the sample space that are not part of the specified event are called a) simple events. b) joint events. c) the sample space. d) the complement of the event. ANSWER: d TYPE: MC DIFFICULTY: Moderate KEYWORDS: sample space, complement 18. Simple probability is also called a) marginal probability. b) joint probability. c) conditional probability. d) Bayes' theorem. ANSWER: a TYPE: MC DIFFICULTY: Easy KEYWORDS: marginal probability

Basic Probability

19. When using the general multiplication rule, P(A and B) is equal to a) P(A|B)P(B). b) P(A)P(B). c) P(B)/P(A). d) P(A)/P(B). ANSWER: a TYPE: MC DIFFICULTY: Moderate KEYWORDS: multiplication rule 20. A business venture can result in the following outcomes (with their corresponding chance of occurring in parentheses): Highly Successful (10%), Successful (25%), Break Even (25%), Disappointing (20%), and Highly Disappointing (?). If these are the only outcomes possible for the business venture, what is the chance that the business venture will be considered Highly Disappointing? a) 10% b) 15% c) 20% d) 25% ANSWER: c TYPE: MC DIFFICULTY: Easy KEYWORDS: marginal probability 21. A recent survey of banks revealed the following distribution for the interest rate being charged on a home loan (based on a 30-year mortgage with a 10% down payment). Interest Rate Probability

7.0% 0.12

7.5% 0.23

8.0% 0.24

8.5% 0.35

> 8.5% 0.06

If a bank is selected at random from this distribution, what is the chance that the interest rate charged on a home loan will exceed 8.0%? a) 0.06 b) 0.41 c) 0.59 d) 1.00 ANSWER: b TYPE: MC DIFFICULTY: Easy KEYWORDS: marginal probability, addition rule 22. The employees of a company were surveyed on questions regarding their educational background and marital status. Of the 600 employees, 400 had college degrees, 100 were single, and 60 were single college graduates. The probability that an employee of the company is single or has a college degree is: a) 0.10. b) 0.25. c) 0.667. d) 0.733.

Basic Probability ANSWER: d TYPE: MC DIFFICULTY: Moderate KEYWORDS: addition rule 23. The employees of a company were surveyed on questions regarding their educational background and marital status. Of the 600 employees, 400 had college degrees, 100 were single, and 60 were single college graduates. The probability that an employee of the company is married and has a college degree is: a) 40/600. b) 340/600. c) 400/600. d) 500/600. ANSWER: b TYPE: MC DIFFICULTY: Moderate KEYWORDS: joint probability 24. The employees of a company were surveyed on questions regarding their educational background and marital status. Of the 600 employees, 400 had college degrees, 100 were single, and 60 were single college graduates. The probability that an employee of the company does not have a college degree is: a) 0.10. b) 0.33. c) 0.67. d) 0.75. ANSWER: b TYPE: MC DIFFICULTY: Easy KEYWORDS: complement 25. The probability that house sales will increase in the next 6 months is estimated to be 0.25. The probability that the interest rates on housing loans will go up in the same period is estimated to be 0.74. The probability that house sales or interest rates will go up during the next 6 months is estimated to be 0.89. The probability that both house sales and interest rates will increase during the next 6 months is: a) 0.10. b) 0.185. c) 0.705. d) 0.90. ANSWER: a TYPE: MC DIFFICULTY: Moderate KEYWORDS: joint probability 26. The probability that house sales will increase in the next 6 months is estimated to be 0.25. The probability that the interest rates on housing loans will go up in the same period is estimated to be 0.74. The probability that house sales or interest rates will go up during the next 6 months is estimated to be 0.89. The probability that neither house sales nor interest rates will increase during the next 6 months is:

Basic Probability

a) 0.11. b) 0.195. c) 0.89. d) 0.90. ANSWER: a TYPE: MC DIFFICULTY: Moderate KEYWORDS: joint probability, complement 27. The probability that house sales will increase in the next 6 months is estimated to be 0.25. The probability that the interest rates on housing loans will go up in the same period is estimated to be 0.74. The probability that house sales or interest rates will go up during the next 6 months is estimated to be 0.89. The probability that house sales will increase but interest rates will not during the next 6 months is: a) 0.065. b) 0.15. c) 0.51. d) 0.89. ANSWER: b TYPE: MC DIFFICULTY: Moderate KEYWORDS: joint probability, complement 28. The probability that house sales will increase in the next 6 months is estimated to be 0.25. The probability that the interest rates on housing loans will go up in the same period is estimated to be 0.74. The probability that house sales or interest rates will go up during the next 6 months is estimated to be 0.89. The events of increase in house sales and increase in interest rates in the next 6 months are a) statistically independent. b) mutually exclusive. c) collectively exhaustive. d) None of the above. ANSWER: d TYPE: MC DIFFICULTY: Moderate KEYWORDS: joint probability, statistical independence EXPLANATION: They are not statistically independent. 29. The probability that house sales will increase in the next 6 months is estimated to be 0.25. The probability that the interest rates on housing loans will go up in the same period is estimated to be 0.74. The probability that house sales or interest rates will go up during the next 6 months is estimated to be 0.89. The events of increase in house sales and no increase in house sales in the next 6 months are a) statistically independent. b) mutually exclusive. c) collectively exhaustive. d) (B) and (C) ANSWER: d

Basic Probability TYPE: MC DIFFICULTY: Moderate KEYWORDS: mutually exclusive, collectively exhaustive, complement 30. The probability that a new advertising campaign will increase sales is assessed as being 0.80. The probability that the cost of developing the new ad campaign can be kept within the original budget allocation is 0.40. Assuming that the two events are independent, the probability that the cost is kept within budget and the campaign will increase sales is: a) 0.20. b) 0.32. c) 0.40. d) 0.88. ANSWER: b TYPE: MC DIFFICULTY: Easy KEYWORDS: statistical independence, joint probability, multiplication rule 31. The probability that a new advertising campaign will increase sales is assessed as being 0.80. The probability that the cost of developing the new ad campaign can be kept within the original budget allocation is 0.40. Assuming that the two events are independent, the probability that the cost is kept within budget or the campaign will increase sales is: a) 0.20. b) 0.32. c) 0.68. d) 0.88. ANSWER: d TYPE: MC DIFFICULTY: Easy KEYWORDS: statistical independence, multiplication rule, addition rule 32. The probability that a new advertising campaign will increase sales is assessed as being 0.80. The probability that the cost of developing the new ad campaign can be kept within the original budget allocation is 0.40. Assuming that the two events are independent, the probability that the cost is not kept within budget or the campaign will not increase sales is: a) 0.12. b) 0.32. c) 0.68. d) 0.88. ANSWER: c TYPE: MC DIFFICULTY: Easy KEYWORDS: statistical independence, multiplication rule, addition rule, complement 33. The probability that a new advertising campaign will increase sales is assessed as being 0.80. The probability that the cost of developing the new ad campaign can be kept within the original budget allocation is 0.40. Assuming that the two events are independent, the probability that neither the cost is kept within budget nor the campaign will increase sales is: a) 0.12. b) 0.32. c) 0.68.

Basic Probability

d) 0.88. ANSWER: a TYPE: MC DIFFICULTY: Easy KEYWORDS: statistical independence, multiplication rule, joint probability, complement 34. According to a survey of American households, the probability that the residents own 2 cars if annual household income is over $25,000 is 80%. Of the households surveyed, 60% had incomes over $25,000 and 70% had 2 cars. The probability that the residents of a household own 2 cars and have an income over $25,000 a year is: a) 0.12. b) 0.18. c) 0.22. d) 0.48. ANSWER: d TYPE: MC DIFFICULTY: Moderate KEYWORDS: joint probability, conditional probability 35. According to a survey of American households, the probability that the residents own 2 cars if annual household income is over $25,000 is 80%. Of the households surveyed, 60% had incomes over $25,000 and 70% had 2 cars. The probability that the residents of a household do not own 2 cars and have an income over $25,000 a year is: a) 0.12. b) 0.18. c) 0.22. d) 0.48. ANSWER: a TYPE: MC DIFFICULTY: Difficult KEYWORDS: joint probability, complement, multiplication rule, conditional probability 36. According to a survey of American households, the probability that the residents own 2 cars if annual household income is over $25,000 is 80%. Of the households surveyed, 60% had incomes over $25,000 and 70% had 2 cars. The probability that the residents of a household own 2 cars and have an income less than or equal to $25,000 a year is: a) 0.12. b) 0.18. c) 0.22. d) 0.48. ANSWER: c TYPE: MC DIFFICULTY: Difficult KEYWORDS: joint probability, complement, multiplication rule, conditional probability 37. According to a survey of American households, the probability that the residents own 2 cars if annual household income is over $25,000 is 80%. Of the households surveyed, 60% had incomes

100 Basic Probability over $25,000 and 70% had 2 cars. The probability that annual household income is over $25,000 if the residents of a household own 2 cars is: a) 0.42. b) 0.48. c) 0.50. d) 0.69. ANSWER: d TYPE: MC DIFFICULTY: Difficult KEYWORDS: conditional probability, Bayes’ theorem 38. According to a survey of American households, the probability that the residents own 2 cars if annual household income is over $25,000 is 80%. Of the households surveyed, 60% had incomes over $25,000 and 70% had 2 cars. The probability that annual household income is over $25,000 if the residents of a household do not own 2 cars is: a) 0.12. b) 0.18. c) 0.40. d) 0.70. ANSWER: c TYPE: MC DIFFICULTY: Difficult KEYWORDS: conditional probability, complement, Bayes’ theorem 39. According to a survey of American households, the probability that the residents own 2 cars if annual household income is over $25,000 is 80%. Of the households surveyed, 60% had incomes over $25,000 and 70% had 2 cars. The probability that the residents do not own 2 cars if annual household income is not over $25,000 is: a) 0.12. b) 0.18. c) 0.45. d) 0.70. ANSWER: c TYPE: MC DIFFICULTY: Difficult KEYWORDS: conditional probability, complement 40. A company has 2 machines that produce widgets. An older machine produces 23% defective widgets, while the new machine produces only 8% defective widgets. In addition, the new machine produces 3 times as many widgets as the older machine does. Given that a widget was produced by the new machine, what is the probability it is not defective? a) 0.06 b) 0.50 c) 0.92 d) 0.94 ANSWER: c TYPE: MC DIFFICULTY: Easy KEYWORDS: conditional probability, complement

Basic Probability

101

41. A company has 2 machines that produce widgets. An older machine produces 23% defective widgets, while the new machine produces only 8% defective widgets. In addition, the new machine produces 3 times as many widgets as the older machine does. What is the probability that a randomly chosen widget produced by the company is defective? a) 0.078 b) 0.1175 c) 0.156 d) 0.310 ANSWER: b TYPE: MC DIFFICULTY: Moderate KEYWORDS: marginal probability 42. A company has 2 machines that produce widgets. An older machine produces 23% defective widgets, while the new machine produces only 8% defective widgets. In addition, the new machine produces 3 times as many widgets as the older machine does. Given a randomly chosen widget was tested and found to be defective, what is the probability it was produced by the new machine? a) 0.08 b) 0.15 c) 0.489 d) 0.511 ANSWER: d TYPE: MC DIFFICULTY: Difficult KEYWORDS: conditional probability, Bayes’ theorem TABLE 4-1 Mothers Against Drunk Driving is a very visible group whose main focus is to educate the public about the harm caused by drunk drivers. A study was recently done that emphasized the problem we all face with drinking and driving. Four hundred accidents that occurred on a Saturday night were analyzed. Two items noted were the number of vehicles involved and whether alcohol played a role in the accident. The numbers are shown below:

Did alcohol play a role? Yes No Totals

Number of Vehicles Involved 1 2 3 50 100 20 25 175 30 75 275 50

Totals 170 230 400

43. Referring to Table 4-1, what proportion of accidents involved more than one vehicle? a) 50/400 or 12.5% b) 75/400 or 18.75% c) 275/400 or 68.75% d) 325/400 or 81.25% ANSWER: d

102 Basic Probability TYPE: MC DIFFICULTY: Easy KEYWORDS: contingency table, empirical classical probability, addition rule 44. Referring to Table 4-1, what proportion of accidents involved alcohol and a single vehicle? a) 25/400 or 6.25% b) 50/400 or 12.5% c) 195/400 or 48.75% d) 245/400 or 61.25% ANSWER: b TYPE: MC DIFFICULTY: Easy KEYWORDS: contingency table, empirical classical probability, joint probability 45. Referring to Table 4-1, what proportion of accidents involved alcohol or a single vehicle? a) 25/400 or 6.25% b) 50/400 or 12.5% c) 195/400 or 48.75% d) 245/400 or 61.25% ANSWER: c TYPE: MC DIFFICULTY: Easy KEYWORDS: contingency table, empirical classical probability, addition rule 46. Referring to Table 4-1, given alcohol was involved, what proportion of accidents involved a single vehicle? a) 50/75 or 66.67% b) 50/170 or 29.41% c) 120/170 or 70.59% d) 120/400 or 30% ANSWER: b TYPE: MC DIFFICULTY: Easy KEYWORDS: contingency table, empirical classical probability, marginal probability 47. Referring to Table 4-1, given that multiple vehicles were involved, what proportion of accidents involved alcohol? a) 120/170 or 70.59% b) 120/230 or 52.17% c) 120/325 or 36.92% d) 120/400 or 30% ANSWER: c TYPE: MC DIFFICULTY: Easy KEYWORDS: contingency table, empirical classical probability, conditional probability, addition rule 48. Referring to Table 4-1, given that 3 vehicles were involved, what proportion of accidents involved alcohol?

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103

a) 20/30 or 66.67% b) 20/50 or 40% c) 20/170 or 11.77% d) 20/400 or 5% ANSWER: b TYPE: MC DIFFICULTY: Easy KEYWORDS: contingency table, empirical classical probability, conditional probability 49. Referring to Table 4-1, given that alcohol was not involved, what proportion of the accidents were single vehicle? a) 50/75 or 66.67% b) 25/230 or 10.87% c) 50/170 or 29.41% d) 25/75 or 33.33% ANSWER: b TYPE: MC DIFFICULTY: Easy KEYWORDS: contingency table, empirical classical probability, conditional probability, complement 50. Referring to Table 4-1, given that alcohol was not involved, what proportion of the accidents were multiple vehicle? a) 50/170 or 29.41% b) 120/170 or 70.59% c) 205/230 or 89.13% d) 25/230 or 10.87% ANSWER: c TYPE: MC DIFFICULTY: Easy KEYWORDS: contingency table, empirical classical probability, conditional probability, complement TABLE 4-2 An alcohol awareness task force at a Big-Ten university sampled 200 students after the midterm to ask them whether they went bar hopping the weekend before the midterm or spent the weekend studying, and whether they did well or poorly on the midterm. The following result was obtained. Studying for Exam Went Bar Hopping

Did Well on Midterm 80 30

Did Poorly on Midterm 20 70

51. Referring to Table 4-2, what is the probability that a randomly selected student who went bar hopping will do well on the midterm? a. 30/100 or 30% b. 30/110 or 27.27% c. 30/200 or 15% d. (100/200)*(110/200) or 27.50%

104 Basic Probability

ANSWER: a TYPE: MC DIFFICULTY: Easy KEYWORDS: contingency table, empirical classical probability, conditional probability 52. Referring to Table 4-2, what is the probability that a randomly selected student did well on the midterm or went bar hopping the weekend before the midterm? a) 30/200 or 15% b) (80+30)/200 or 55% c) (30+70)/200 or 50% d) (80+30+70)/200 or 90% ANSWER: d TYPE: MC DIFFICULTY: Easy KEYWORDS: contingency table, empirical classical probability, addition rule 53. Referring to Table 4-2, what is the probability that a randomly selected student did well on the midterm and also went bar hopping the weekend before the midterm? a) 30/200 or 15% b) (80+30)/200 or 55% c) (30+70)/200 or 50% d) (80+30+70)/200 or 90% ANSWER: a TYPE: MC DIFFICULTY: Easy KEYWORDS: contingency table, empirical classical probability, joint probability 54. Referring to Table 4-2, the events "Did Well on Midterm" and "Studying for Exam" are a) statistically dependent. b) mutually exclusive. c) collective exhaustive. d) None of the above. ANSWER: a TYPE: MC DIFFICULTY: Easy KEYWORDS: contingency table, empirical classical probability, statistical independence, joint probability 55. Referring to Table 4-2, the events "Did Well on Midterm" and "Studying for Exam" are a) not statistically dependent. b) not mutually exclusive. c) collective exhaustive. d) None of the above. ANSWER: b TYPE: MC DIFFICULTY: Easy KEYWORDS: contingency table, empirical classical probability, mutually exclusive, joint probability 56. Referring to Table 4-2, the events "Did Well on Midterm" and "Did Poorly on Midterm" are

Basic Probability

105

a) statistically dependent. b) mutually exclusive. c) collective exhaustive. d) All of the above. ANSWER: d TYPE: MC DIFFICULTY: Easy KEYWORDS: contingency table, empirical classical probability, statistical independence, mutually exclusive, collective exhaustive, joint probability

57. True or False: When A and B are mutually exclusive, P(A or B) can be found by adding P(A) and P(B). ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: mutually exclusive, addition rule 58. True or False: The collection of all the possible events is called a sample space. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: sample space

59. True or False: If A and B cannot occur at the same time they are called mutually exclusive. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: mutually exclusive

60. True or False: If either A or B must occur they are called mutually exclusive. ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: mutually exclusive, collective exhaustive

61. True or False: If either A or B must occur they are called collectively exhaustive. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: collective exhaustive

62. True or False: If P(A) = 0.4 and P(B) = 0.6, then A and B must be collectively exhaustive. ANSWER: False TYPE: TF DIFFICULTY: Moderate KEYWORDS: collective exhaustive, mutually exclusive

106 Basic Probability

63. True or False: If P(A) = 0.4 and P(B) = 0.6, then A and B must be mutually exclusive. ANSWER: False TYPE: TF DIFFICULTY: Moderate KEYWORDS: mutually exclusive

64. True or False: If P(A or B) = 1.0, then A and B must be mutually exclusive. ANSWER: False TYPE: TF DIFFICULTY: Moderate KEYWORDS: mutually exclusive, collective exhaustive

65. True or False: If P(A or B) = 1.0, then A and B must be collectively exhaustive. ANSWER: True TYPE: TF DIFFICULTY: Moderate KEYWORDS: collective exhaustive

66. True or False: If P(A and B) = 0, then A and B must be mutually exclusive. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: mutually exclusive

67. True or False: If P(A and B) = 0, then A and B must be collectively exhaustive. ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: collectively exhaustive, mutually exclusive

68. True or False: If P(A and B) = 1, then A and B must be collectively exhaustive. ANSWER: True TYPE: TF DIFFICULTY: Difficult KEYWORDS: collective exhaustive

69. True or False: If P(A and B) = 1, then A and B must be mutually exclusive. ANSWER: False TYPE: TF DIFFICULTY: Difficult KEYWORDS: mutually exclusive, collective exhaustive

70. Suppose A and B are independent events where P(A) = 0.4 and P(B) = 0.5. Then P(A and B) = . ANSWER:

Basic Probability

107

0.2 TYPE: FI DIFFICULTY: Easy KEYWORDS: statistical independence, multiplication rule

71. Suppose A and B are mutually exclusive events where P(A) = 0.4 and P(B) = 0.5. Then P(A and B) =

ANSWER: 0 TYPE: FI DIFFICULTY: Easy KEYWORDS: mutually exclusive, joint probability, multiplication rule

72. Suppose A and B are mutually exclusive events where P(A) = 0.4 and P(B) = 0.5. Then P(A or B) =

ANSWER: 0.9 TYPE: FI DIFFICULTY: Easy KEYWORDS: mutually exclusive, addition rule

73. Suppose A and B are independent events where P(A) = 0.4 and P(B) = 0.5. Then P(A or B) = . ANSWER: 0.7 TYPE: FI DIFFICULTY: Easy KEYWORDS: statistical independence, addition rule

74. Suppose A and B are events where P(A) = 0.4, P(B) = 0.5, and P(A and B) = 0.1. Then P(A or B) =

ANSWER: 0.8 TYPE: FI DIFFICULTY: Easy KEYWORDS: addition rule

75. Suppose A and B are events where P(A) = 0.4, P(B) = 0.5, and P(A and B) = 0.1. Then P(A|B) = . ANSWER: 0.2 TYPE: FI DIFFICULTY: Moderate KEYWORDS: conditional probability

76. Suppose A and B are events where P(A) = 0.4, P(B) = 0.5, and P(A and B) = 0.1. Then P(B|A) = . ANSWER: 0.25 TYPE: FI DIFFICULTY: Moderate KEYWORDS: conditional probability TABLE 4-3

108 Basic Probability

A survey is taken among customers of a fast-food restaurant to determine preference for hamburger or chicken. Of 200 respondents selected, 75 were children and 125 were adults. 120 preferred hamburger and 80 preferred chicken. 55 of the children preferred hamburger. 77. Referring to Table 4-3, the probability that a randomly selected individual is an adult is . ANSWER: 125/200 or 62.5% TYPE: FI DIFFICULTY: Moderate KEYWORDS: empirical classical probability, conditional probability, marginal probability 78. Referring to Table 4-3, the probability that a randomly selected individual is an adult or a child is . ANSWER: 200/200 or 100% TYPE: FI DIFFICULTY: Moderate KEYWORDS: empirical classical probability, addition rule 79. Referring to Table 4-3, the probability that a randomly selected individual is a child and prefers chicken is . ANSWER: 20/200 or 10% TYPE: FI DIFFICULTY: Moderate KEYWORDS: empirical classical probability, joint probability, multiplication rule 80. Referring to Table 4-3, the probability that a randomly selected individual is an adult and prefers chicken is . ANSWER: 60/200 or 30% TYPE: FI DIFFICULTY: Moderate KEYWORDS: empirical classical probability, joint probability, multiplication rule 81. Referring to Table 4-3, the probability that a randomly selected individual is a child or prefers hamburger is . ANSWER: 140/200 or 70% TYPE: FI DIFFICULTY: Moderate KEYWORDS: empirical classical probability, addition rule 82. Referring to Table 4-3, assume we know the person is a child. The probability that this individual prefers hamburger is . ANSWER: 55/75 or 73.33% TYPE: FI DIFFICULTY: Moderate KEYWORDS: empirical classical probability, conditional probability

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109

83. Referring to Table 4-3, assume we know that a person has ordered chicken. The probability that this individual is an adult is . ANSWER: 60/80 or 75% TYPE: FI DIFFICULTY: Moderate KEYWORDS: empirical classical probability, conditional probability, Bayes’ theorem 84. Referring to Table 4-3, assume we know that a person has ordered hamburger. The probability that this individual is a child is . ANSWER: 55/120 or 45.83% TYPE: FI DIFFICULTY: Moderate KEYWORDS: empirical classical probability, conditional probability, Bayes’ theorem TABLE 4-4 Suppose that patrons of a restaurant were asked whether they preferred beer or whether they preferred wine. 70% said that they preferred beer. 60% of the patrons were male. 80% of the males preferred beer. 85. Referring to Table 4-4, the probability a randomly selected patron prefers wine is

ANSWER: 0.30 TYPE: FI DIFFICULTY: Moderate KEYWORDS: addition rule, multiplication rule, marginal probability 86. Referring to Table 4-4, the probability a randomly selected patron is a female is

ANSWER: 0.40 TYPE: FI DIFFICULTY: Moderate KEYWORDS: addition rule, multiplication rule, marginal probability 87. Referring to Table 4-4, the probability a randomly selected patron is a female who prefers wine is . ANSWER: 0.18 TYPE: FI DIFFICULTY: Difficult KEYWORDS: addition rule, multiplication rule, joint probability 88. Referring to Table 4-4, the probability a randomly selected patron is a female who prefers beer is . ANSWER: 0.22 TYPE: FI DIFFICULTY: Difficult KEYWORDS: addition rule, multiplication rule, joint probability 89. Referring to Table 4-4, suppose a randomly selected patron prefers wine. Then the probability the patron is a male is .

110 Basic Probability

ANSWER: 0.40 TYPE: FI DIFFICULTY: Difficult KEYWORDS: addition rule, multiplication rule, conditional probability 90. Referring to Table 4-4, suppose a randomly selected patron prefers beer. Then the probability the patron is a male is . ANSWER: 0.69 TYPE: FI DIFFICULTY: Difficult KEYWORDS: addition rule, multiplication rule, conditional probability 91. Referring to Table 4-4, suppose a randomly selected patron is a female. Then the probability the patron prefers beer is . ANSWER: 0.55 TYPE: FI DIFFICULTY: Difficult KEYWORDS: addition rule, multiplication rule, conditional probability

92. True or False: Referring to Table 4-4, the two events "preferring" beer and "preferring" wine are statistically independent. ANSWER: False TYPE: TF DIFFICULTY: Moderate KEYWORDS: addition rule, multiplication rule, statistical independence

93. True or False: Referring to Table 4-4, the two events "preferring" beer and being a male are statistically independent. ANSWER: False TYPE: TF DIFFICULTY: Moderate KEYWORDS: addition rule, multiplication rule, statistical independence TABLE 4-5 In a meat packaging plant Machine A accounts for 60% of the plant's output, while Machine B accounts for 40% of the plant's output. In total, 4% of the packages are improperly sealed. Also, 3% of the packages are from Machine A and are improperly sealed. 94. Referring to Table 4-5, if a package is selected at random, the probability that it will be properly sealed is . ANSWER: 0.96 TYPE: FI DIFFICULTY: Moderate KEYWORDS: addition rule, marginal probability 95. Referring to Table 4-5, if a package selected at random is improperly sealed, the probability that it came from machine A is .

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111

ANSWER: 0.75 TYPE: FI DIFFICULTY: Difficult KEYWORDS: conditional probability 96. Referring to Table 4-5, if a package selected at random came from Machine A, the probability that it is improperly sealed is . ANSWER: 0.05 TYPE: FI DIFFICULTY: Difficult KEYWORDS: conditional probability 97. Referring to Table 4-5, if a package selected at random came from Machine B, the probability that it is properly sealed is . ANSWER: 0.975 TYPE: FI DIFFICULTY: Difficult KEYWORDS: conditional probability, complement, addition rule 98. Referring to Table 4-5, if a package selected at random came from Machine B, the probability that it is improperly sealed is . ANSWER: 0.025 TYPE: FI DIFFICULTY: Difficult KEYWORDS: conditional probability, complement, addition rule TABLE 4-6 At a Texas college, 60% of the students are from the southern part of the state, 30% are from the northern part of the state, and the remaining 10% are from out-of-state. All students must take and pass an Entry Level Math (ELM) test. 60% of the southerners have passed the ELM, 70% of the northerners have passed the ELM, and 90% of the out-of-staters have passed the ELM. 99. Referring to Table 4-6, the probability that a randomly selected student is someone from northern Texas who has not passed the ELM is . ANSWER: 0.09 TYPE: FI DIFFICULTY: Difficult KEYWORDS: joint probability, multiplication rule 100. Referring to Table 4-6, the probability that a randomly selected student has passed the ELM is . ANSWER: 0.66 TYPE: FI DIFFICULTY: Difficult KEYWORDS: marginal probability, multiplication rule, addition rule

112 Basic Probability 101. Referring to Table 4-6, if a randomly selected student has passed the ELM, the probability the student is from out-of-state is . ANSWER: 0.136 TYPE: FI DIFFICULTY: Difficult KEYWORDS: conditional probability, Bayes’ theorem 102. Referring to Table 4-6, if a randomly selected student has not passed the ELM, the probability the student is from southern Texas is . ANSWER: 0.706 TYPE: FI DIFFICULTY: Difficult KEYWORDS: conditional probability, complement, Bayes’ theorem 103. Referring to Table 4-6, the probability that a randomly selected student is not from southern Texas and has not passed the ELM is . ANSWER: 0.10 TYPE: FI DIFFICULTY: Difficult KEYWORDS: joint probability, complement, multiplication rule, addition 104. Referring to Table 4-6, if a randomly selected student has not passed the ELM, the probability the student is not from northern Texas is . ANSWER: 0.735 TYPE: FI DIFFICULTY: Difficult KEYWORDS: conditional probability, complement, addition rule, multiplication rule, Bayes’ theorem 105. Referring to Table 4-6, if a randomly selected student is not from southern Texas, the probability the student has not passed the ELM is . ANSWER: 0.25 TYPE: FI DIFFICULTY: Difficult KEYWORDS: conditional probability, complement, addition rule, multiplication rule, Bayes’ theorem 106. Referring to Table 4-6, if a randomly selected student is not from out-of-state, the probability the student has passed the ELM is . ANSWER: 0.633 TYPE: FI DIFFICULTY: Difficult KEYWORDS: conditional probability, complement, addition rule, multiplication rule, Bayes’ theorem TABLE 4-7 The next state lottery will have the following payoffs possible with their associated probabilities.

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Payoff Probability $2.00 0.0500 $25.00 0.0100 $100.00 0.0050 $500.00 0.0010 $5,000.00 0.0005 $10,000.00 0.0001 You buy a single ticket. 107. Referring to Table 4-7, the probability that you win any money is

ANSWER: 0.0666 TYPE: FI DIFFICULTY: Easy KEYWORDS: addition rule 108. Referring to Table 4-7, the probability that you win at least $100.00 is

ANSWER: 0.0066 TYPE: FI DIFFICULTY: Easy KEYWORDS: addition rule 109. Referring to Table 4-7, if you have a winning ticket, the probability that you win at least $100.00 is . ANSWER: 0.10 TYPE: FI DIFFICULTY: Moderate KEYWORDS: conditional probability, addition rule 110. A new model car from Ford Motor Company offers a keyless entry system that utilizes a fourletter code. How many different possible combinations are there for the code? ANSWER: 456976 TYPE: PR DIFFICULTY: Easy KEYWORDS: counting rule 111. At the International Pancakes Hut, there are 4 different ways to have an egg cooked, 7 different choices of pancakes, 5 different types of syrups and 8 different beverages. How many different ways are there to order an egg, a pancake with a choice of syrup and a beverage? ANSWER: 1,120 TYPE: PR DIFFICULTY: Easy KEYWORDS: counting rule 112. There are 10 finalists at a national dog show. How many different orders of finishing can there be for all the 10 finalists? ANSWER: 3,628,800 TYPE: PR DIFFICULTY: Easy

114 Basic Probability KEYWORDS: counting rule, permutation 113. Eleven freshmen are to be assigned to eleven empty rooms in a student dormitory. Each room is considered unique so that it matters who is being assigned to which room. How many different ways can those eleven freshmen be allocated? ANSWER: 39,916,800 TYPE: PR DIFFICULTY: Easy KEYWORDS: counting rule, permutation

114. There are only 4 empty rooms available in a student dormitory for eleven new freshmen. Each room is considered unique so that it matters who is being assigned to which room. How many different ways can those 4 empty rooms be filled one student per room? ANSWER: 7,920 TYPE: PR DIFFICULTY: Easy KEYWORDS: counting rule, permutation

115. There are only 4 empty rooms available in a student dormitory for eleven new freshmen. All the rooms are considered as homogenous so that it does not matter who is being assigned to which room. How many different ways can those 4 empty rooms be filled one student per room? ANSWER: 330 TYPE: PR DIFFICULTY: Easy KEYWORDS: counting rule, combination 116. Four freshmen are to be assigned to eleven empty rooms in a student dormitory. All the rooms are considered as homogenous so that it does not matter who is being assigned to which room. How many different ways can those 4 freshmen be assigned? ANSWER: 330 TYPE: PR DIFFICULTY: Easy KEYWORDS: counting rule, combination

117. There are 47 contestants at a national dog show. How many different ways can contestants fill the first place, second place, and third place positions? ANSWER: 97,290 TYPE: PR DIFFICULTY: Easy KEYWORDS: counting rule, permutation 118. Seven passengers are on a waiting list for an overbooked flight. As a result of cancellations, 3 seats become available. How many different ways can those 3 available seats be filled regardless of the order? ANSWER: 35 TYPE: PR DIFFICULTY: Easy

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KEYWORDS: counting rule, combination 119. A high school debate team of 4 is to be chosen from a class of 35. How many possible ways can the team be formed? ANSWER: 52,360 TYPE: PR DIFFICULTY: Easy KEYWORDS: counting rule, combination 120. A debate team of 4 is to be chosen from a class of 35. There are two twin brothers in the class. How many possible ways can the team be formed which will include only one of the twin brothers? ANSWER: 10,912 TYPE: PR DIFFICULTY: Difficult KEYWORDS: counting rule, combination 121. A debate team of 4 is to be chosen from a class of 35. There are two twin brothers in the class. How many possible ways can the team be formed which will not include any of the twin brothers? ANSWER: 40,920 TYPE: PR DIFFICULTY: Moderate KEYWORDS: counting rule, combination 122. A debate team of 4 is to be chosen from a class of 35. There are two twin brothers in the class. How many possible ways can the team be formed which will include both of the twin brothers? ANSWER: 528 TYPE: PR DIFFICULTY: Moderate KEYWORDS: counting rule, combination 123. An exploration team of 2 women and 3 men is to be chosen from a candidate pool of 6 women and 7 men. How many different ways can this team of 5 be formed? ANSWER: 525 TYPE: PR DIFFICULTY: Moderate KEYWORDS: counting rule, combination 124. Twelve students in a Business Statistics class are to be formed into three teams of four. How many different ways can this be done? ANSWER: 5,775 TYPE: PR DIFFICULTY: Difficult

(

)

EXPLANATION: ( 12!/ ( 4!8!) ) ( 8!/ ( 4!4!) ) / 3! KEYWORDS: counting rule, combination

116 Basic Probability

125. The closing price of a company’s stock tomorrow can be lower, higher or the same as today’s closed. Without any prior information that may affect the price of the stock tomorrow, the probability that it will close higher than today’s close is 1/3. This is an example of using which of the following probability approach? a) A priori classical probability b) Empirical classical probability c) Subjective probability d) Conditional probability ANSWER: a TYPE: MC DIFFICULTY: Easy KEYWORDS: a priori classical probability, empirical classical probability, subjective probability

126. The closing price of a company’s stock tomorrow can be lower, higher or the same as today’s closing price. Based on the closing price of the stock collected over the last month, 25% of the days the closing price was higher than previous day’s closing price, 45% was lower than previous day’s and 30% was the same as previous day’s. Based on this information, the probability that tomorrow’s closing price will be higher than today’s is 25%. This is an example of using which of the following probability approach? a) A priori classical probability b) Empirical classical probability c) Subjective probability d) Conditional probability ANSWER: b TYPE: MC DIFFICULTY: Easy KEYWORDS: a priori classical probability, empirical classical probability, subjective probability

127. The closing price of a company’s stock tomorrow can be lower, higher or the same as today’s

closing price. After evaluating all the information available on the company’s fundamentals and the economic environment, an analyst has determined that the probability that tomorrow’s closing price will be higher than today’s is determined to be 25%. This is an example of using which of the following probability approach? a) A priori classical probability b) Empirical classical probability c) Subjective probability d) Conditional probability

ANSWER: c TYPE: MC DIFFICULTY: Easy KEYWORDS: a priori classical probability, empirical classical probability, subjective probability TABLE 4-8 According to the record of the registrar’s office at a state university, 35% of the students are freshman, 25% are sophomore, 16% are junior and the rest are senior. Among the freshmen, sophomores, juniors and seniors, the portion of students who live in the dormitory are, respectively, 80%, 60%, 30% and 20%.

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128. Referring to Table 4-8, what is the probability that a randomly selected student is a freshman who lives in a dormitory? ANSWER: 0.28 TYPE: PR DIFFICULTY: Easy KEYWORDS: joint probability, multiplication rule 129. Referring to Table 4-8, what is the probability that a randomly selected student is a sophomore who does not live in a dormitory? ANSWER: 0.1 TYPE: PR DIFFICULTY: Easy KEYWORDS: joint probability, multiplication rule, complement 130. Referring to Table 4-8, what is the probability that a randomly selected student is a junior who does not live in a dormitory? ANSWER: 0.112 TYPE: PR DIFFICULTY: Easy KEYWORDS: joint probability, multiplication rule, complement 131. Referring to Table 4-8, what is the probability that a randomly selected student is a junior or senior who lives in a dormitory? ANSWER: 0.096 TYPE: PR DIFFICULTY: Moderate KEYWORDS: joint probability, multiplication rule, addition rule 132. Referring to Table 4-8, what percentage of the students live in a dormitory? ANSWER: 0.526 TYPE: PR DIFFICULTY: Easy KEYWORDS: marginal probability, multiplication rule, addition rule 133. Referring to Table 4-8, what percentage of the students do not live in a dormitory? ANSWER: 0.474 TYPE: PR DIFFICULTY: Easy KEYWORDS: marginal probability, multiplication rule, addition rule 134. Referring to Table 4-8, if a randomly selected student lives in the dormitory, what is the probability that the student is a freshman? ANSWER: 0.532 TYPE: PR DIFFICULTY: Moderate KEYWORDS: conditional probability, Bayes’ theorem

118 Basic Probability 135. Referring to Table 4-8, if a randomly selected student lives in the dormitory, what is the probability that the student is not a freshman? ANSWER: 0.468 TYPE: PR DIFFICULTY: Moderate KEYWORDS: conditional probability, Bayes’ theorem, complement, addition rule 136. Referring to Table 4-8, if a randomly selected student does not live in the dormitory, what is the probability that the student is a junior or a senior? ANSWER: 0.641 TYPE: PR DIFFICULTY: Moderate KEYWORDS: conditional probability, Bayes’ theorem, complement, addition rule 137. Referring to Table 4-8, determine whether the class status of a student and whether the student lives in a dormitory are statistically independent. ANSWER: Not statistically independent TYPE: PR DIFFICULTY: Easy KEYWORDS: statistical independence

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CHAPTER 5: SOME IMPORTANT DISCRETE PROBABILITY DISTRIBUTIONS 1. Thirty-six of the staff of 80 teachers at a local intermediate school are certified in CardioPulmonary Resuscitation (CPR). In 180 days of school, about how many days can we expect that the teacher on bus duty will likely be certified in CPR? a) 5 days b) 45 days c) 65 days d) 81 days ANSWER: d TYPE: MC DIFFICULTY: Moderate KEYWORDS: binomial distribution, mean 2. A campus program evenly enrolls undergraduate and graduate students. If a random sample of 4 students is selected from the program to be interviewed about the introduction of a new fast food outlet on the ground floor of the campus building, what is the probability that all 4 students selected are undergraduate students? a) 0.0256 b) 0.0625 c) 0.16 d) 1.00 ANSWER: b TYPE: MC DIFFICULTY: Difficult KEYWORDS: binomial distribution 3. A probability distribution is an equation that a) associates a particular probability of occurrence with each outcome in the sample space. b) measures outcomes and assigns values of X to the simple events. c) assigns a value to the variability in the sample space. d) assigns a value to the center of the sample space. ANSWER: a TYPE: MC DIFFICULTY: Moderate KEYWORDS: probability distribution 4. The connotation "expected value" or "expected gain" from playing roulette at a casino means a) the amount you expect to "gain" on a single play. b) the amount you expect to "gain" in the long run over many plays. c) the amount you need to "break even" over many plays. d) the amount you should expect to gain if you are lucky. ANSWER: b

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TYPE: MC DIFFICULTY: Easy KEYWORDS: expected value

5. Which of the following about the binomial distribution is not a true statement? a) b) c) d)

The probability of success must be constant from trial to trial. Each outcome is independent of the other. Each outcome may be classified as either "success" or "failure." The random variable of interest is continuous.

ANSWER: d TYPE: MC DIFFICULTY: Easy KEYWORDS: binomial distribution, properties 6. In a binomial distribution a) the random variable X is continuous. b) the probability of success p is stable from trial to trial. c) the number of trials n must be at least 30. d) the results of one trial are dependent on the results of the other trials. ANSWER: b TYPE: MC DIFFICULTY: Easy KEYWORDS: binomial distribution, properties

Whenever p = 0.5, the binomial distribution will a) always be symmetric. b) be symmetric only if n is large. c) be right-skewed. d) be left-skewed.

ANSWER: a TYPE: MC DIFFICULTY: Moderate KEYWORDS: binomial distribution, properties

8. Whenever p = 0.1 and n is small, the binomial distribution will be a) b) c) d)

symmetric. right-skewed. left-skewed. None of the above.

ANSWER: b TYPE: MC DIFFICULTY: Moderate KEYWORDS: binomial distribution, properties

9. If n = 10 and p = 0.70, then the mean of the binomial distribution is a) 0.07. b) 1.45. c) 7.00.

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d) 14.29. ANSWER: c TYPE: MC DIFFICULTY: Easy KEYWORDS: binomial distribution, mean

10. If n = 10 and p = 0.70, then the standard deviation of the binomial distribution is a) 0.07. b) 1.45. c) 7.00. d) 14.29. ANSWER: b TYPE: MC DIFFICULTY: Easy KEYWORDS: binomial distribution, standard deviation 11. If the outcomes of a random variable follow a Poisson distribution, then their a) mean equals the standard deviation. b) median equals the standard deviation. c) mean equals the variance. d) median equals the variance. ANSWER: c TYPE: MC DIFFICULTY: Easy KEYWORDS: Poisson distribution, mean, standard deviation, properties 12. What type of probability distribution will the consulting firm most likely employ to analyze the insurance claims in the following problem? An insurance company has called a consulting firm to determine if the company has an unusually high number of false insurance claims. It is known that the industry proportion for false claims is 3%. The consulting firm has decided to randomly and independently sample 100 of the company’s insurance claims. They believe the number of these 100 that are false will yield the information the company desires. a) binomial distribution b) Poisson distribution c) hypergeometric distribution d) None of the above. ANSWER: a TYPE: MC DIFFICULTY: Difficult KEYWORDS: binomial distribution, properties 13. The covariance a) must be between -1 and +1. b) must be positive. c) can be positive or negative.

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d) must be less than +1. ANSWER: c TYPE: MC DIFFICULTY: Easy KEYWORDS: covariance, properties 14. What type of probability distribution will most likely be used to analyze warranty repair needs on new cars in the following problem? The service manager for a new automobile dealership reviewed dealership records of the past 20 sales of new cars to determine the number of warranty repairs he will be called on to perform in the next 90 days. Corporate reports indicate that the probability any one of their new cars needs a warranty repair in the first 90 days is 0.05. The manager assumes that calls for warranty repair are independent of one another and is interested in predicting the number of warranty repairs he will be called on to perform in the next 90 days for this batch of 20 new cars sold. a) binomial distribution b) Poisson distribution c) hypergeometric distribution d) None of the above. ANSWER: a TYPE: MC DIFFICULTY: Difficult KEYWORDS: binomial distribution, properties 15. What type of probability distribution will most likely be used to analyze the number of chocolate chip parts per cookie in the following problem? The quality control manager of Marilyn’s Cookies is inspecting a batch of chocolate chip cookies. When the production process is in control, the average number of chocolate chip parts per cookie is 6.0. The manager is interested in analyzing the probability that any particular cookie being inspected has fewer than 5.0 chip parts. a) binomial distribution b) Poisson distribution c) hypergeometric distribution d) None of the above. ANSWER: b TYPE: MC DIFFICULTY: Moderate KEYWORDS: Poisson distribution, properties 16. What type of probability distribution will most likely be used to analyze the number of cars with defective radios in the following problem? From an inventory of 48 new cars being shipped to local dealerships, corporate reports indicate that 12 have defective radios installed. The sales manager of one dealership wants to predict the probability out of the 8 new cars it just received that, when each is tested, no more than 2 of the cars have defective radios. a) binomial distribution b) Poisson distribution c) hypergeometric distribution

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d) None of the above. ANSWER: c TYPE: MC DIFFICULTY: Moderate KEYWORDS: hypergeometric distribution, properties 17. A stock analyst was provided with a list of 25 stocks. He was expected to pick 3 stocks from the list whose prices are expected to rise by more than 20% after 30 days. In reality, the prices of only 5 stocks would rise by more than 20% after 30 days. If he randomly selected 3 stocks from the list, he would use what type of probability distribution to compute the probability that all of the chosen stocks would appreciate more than 20% after 30 days? a) binomial distribution b) Poisson distribution c) hypergeometric distribution d) None of the above. ANSWER: c TYPE: MC DIFFICULTY: Moderate KEYWORDS: hypergeometric distribution, properties 18. A professor receives, on average, 24.7 e-mails from students the day before the midterm exam. To compute the probability of receiving at least 10 e-mails on such a day, he will use what type of probability distribution? a) binomial distribution b) Poisson distribution c) hypergeometric distribution d) None of the above. ANSWER: b TYPE: MC DIFFICULTY: Moderate KEYWORDS: Poisson distribution, properties 19. A company has 125 personal computers. The probability that any one of them will require repair on a given day is 0.025. To find the probability that exactly 20 of the computers will require repair on a given day, one will use what type of probability distribution? a) binomial distribution b) Poisson distribution c) hypergeometric distribution d) None of the above. ANSWER: a TYPE: MC DIFFICULTY: Moderate KEYWORDS: binomial distribution, properties 20. The portfolio expected return of two investments a) will be higher when the covariance is zero.

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b) will be higher when the covariance is negative. c) will be higher when the covariance is positive. d) does not depend on the covariance. ANSWER: d TYPE: MC DIFFICULTY: Easy KEYWORDS: portfolio, mean

21. A financial analyst is presented with information on the past records of 60 start-up companies and told that in fact only 3 of them have managed to become highly successful. He selected 3 companies from this group as the candidates for success. To analyze his ability to spot the companies that will eventually become highly successful, he will use what type of probability distribution? a) binomial distribution b) Poisson distribution c) hypergeometric distribution d) None of the above. ANSWER: c TYPE: MC DIFFICULTY: Moderate KEYWORDS: hypergeometric distribution, properties 22. On the average, 1.8 customers per minute arrive at any one of the checkout counters of a grocery store. What type of probability distribution can be used to find out the probability that there will be no customer arriving at a checkout counter? a) binomial distribution b) Poisson distribution c) hypergeometric distribution d) None of the above. ANSWER: b TYPE: MC DIFFICULTY: Moderate KEYWORDS: Poisson distribution, properties 23. A multiple-choice test has 30 questions. There are 4 choices for each question. A student who has not studied for the test decides to answer all questions randomly. What type of probability distribution can be used to figure out his chance of getting at least 20 questions right? a) binomial distribution b) Poisson distribution c) hypergeometric distribution d) None of the above. ANSWER: a TYPE: MC DIFFICULTY: Moderate KEYWORDS: binomial distribution, properties

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24. A lab orders 100 rats a week for each of the 52 weeks in the year for experiments that the lab conducts. Suppose the mean cost of rats used in lab experiments turned out to be $13.00 per week. Interpret this value. a) Most of the weeks resulted in rat costs of $13.00. b) The median cost for the distribution of rat costs is $13.00. c) The expected or average cost for all weekly rat purchases is $13.00. d) The rat cost that occurs more often than any other is $13.00. ANSWER: c TYPE: MC DIFFICULTY: Easy KEYWORDS: mean 25. A lab orders 100 rats a week for each of the 52 weeks in the year for experiments that the lab conducts. Prices for 100 rats follow the following distribution: Price: $10.00 $12.50 $15.00 Probability: 0.35 0.40 0.25 How much should the lab budget for next year’s rat orders be, assuming this distribution does not change? a) $520 b) $637 c) $650 d) $780 ANSWER: b TYPE: MC DIFFICULTY: Moderate KEYWORDS: mean 26. The local police department must write, on average, 5 tickets a day to keep department revenues at budgeted levels. Suppose the number of tickets written per day follows a Poisson distribution with a mean of 6.5 tickets per day. Interpret the value of the mean. a) The number of tickets that is written most often is 6.5 tickets per day. b) Half of the days have less than 6.5 tickets written and half of the days have more than 6.5 tickets written. c) If we sampled all days, the arithmetic average or expected number of tickets written would be 6.5 tickets per day. d) The mean has no interpretation since 0.5 ticket can never be written. ANSWER: c TYPE: MC DIFFICULTY: Easy KEYWORDS: mean, Poisson distribution 27. True or False: The Poisson distribution can be used to model a continuous random variable. ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: Poisson distribution, properties

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28. True or False: Another name for the mean of a probability distribution is its expected value. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: mean 29. True or False: The number of customers arriving at a department store in a 5-minute period has a binomial distribution. ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: Poisson distribution, properties 30. True or False: The number of customers arriving at a department store in a 5-minute period has a Poisson distribution. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: Poisson distribution, properties 31. True or False: The number of males selected in a sample of 5 students taken without replacement from a class of 9 females and 18 males has a binomial distribution. ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: hypergeometric distribution, properties 32. True or False: The number of males selected in a sample of 5 students taken without replacement from a class of 9 females and 18 males has a hypergeometric distribution. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: hypergeometric distribution, properties 33. True or False: The diameters of 10 randomly selected bolts have a binomial distribution. ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: binomial distribution, properties

34. True or False: The largest value that a Poisson random variable X can have is n. ANSWER: False TYPE: TF DIFFICULTY: Moderate

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KEYWORDS: Poisson distribution, properties 35. True or False: In a Poisson distribution, the mean and standard deviation are equal. ANSWER: False TYPE: TF DIFFICULTY: Moderate KEYWORDS: Poisson distribution, properties 36. True or False: In a Poisson distribution, the mean and variance are equal. ANSWER: True TYPE: TF DIFFICULTY: Moderate KEYWORDS: Poisson distribution, properties

37. True or False: If p remains constant in a binomial distribution, an increase in n will increase the variance. ANSWER: True TYPE: TF DIFFICULTY: Moderate KEYWORDS: binomial distribution, properties

38. True or False: If p remains constant in a binomial distribution, an increase in n will not change the mean. ANSWER: False TYPE: TF DIFFICULTY: Moderate KEYWORDS: binomial distribution, properties 39. True or False: Suppose that a judge’s decisions follow a binomial distribution and that his verdict is correct 90% of the time. In his next 10 decisions, the probability that he makes fewer than 2 incorrect verdicts is 0.736. ANSWER: True TYPE: TF DIFFICULTY: Moderate KEYWORDS: binomial distribution, probability 40. True or False: Suppose that the number of airplanes arriving at an airport per minute is a Poisson process. The average number of airplanes arriving per minute is 3. The probability that exactly 6 planes arrive in the next minute is 0.0504. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: Poisson distribution, probability

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41. True or False: The covariance between two investments is equal to the sum of the variances of the investments. ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: covariance, variance 42. True or False: If the covariance between two investments is zero, the variance of the sum of the two investments will be equal to the sum of the variances of the investments. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: covariance, variance 43. True or False: The expected return of the sum of two investments will be equal to the sum of the expected returns of the two investments plus twice the covariance between the investments. ANSWER: False TYPE: TF DIFFICULTY: Moderate KEYWORDS: mean, portfolio 44. True or False: The variance of the sum of two investments will be equal to the sum of the variances of the two investments plus twice the covariance between the investments. ANSWER: True TYPE: TF DIFFICULTY: Moderate KEYWORDS: covariance, variance 45. True or False: The variance of the sum of two investments will be equal to the sum of the variances of the two investments when the covariance between the investments is zero. ANSWER: True TYPE: TF DIFFICULTY: Moderate KEYWORDS: covariance, variance 46. True or False: The expected return of a two-asset portfolio is equal to the product of the weight assigned to the first asset and the expected return of the first asset plus the product of the weight assigned to the second asset and the expected return of the second asset. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: mean, portfolio TABLE 5-1

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The probability that a particular type of smoke alarm will function properly and sound an alarm in the presence of smoke is 0.8. You have 2 such alarms in your home and they operate independently. 47. Referring to Table 5-1, the probability that both sound an alarm in the presence of smoke is . ANSWER: 0.64 TYPE: FI DIFFICULTY: Moderate KEYWORDS: binomial distribution 48. Referring to Table 5-1, the probability that neither sound an alarm in the presence of smoke is . ANSWER: 0.04 TYPE: FI DIFFICULTY: Moderate KEYWORDS: binomial distribution 49. Referring to Table 5-1, the probability that at least one sounds an alarm in the presence of smoke is . ANSWER: 0.96 TYPE: FI DIFFICULTY: Difficult KEYWORDS: binomial distribution TABLE 5-2 A certain type of new business succeeds 60% of the time. Suppose that 3 such businesses open (where they do not compete with each other, so it is reasonable to believe that their relative successes would be independent). 50. Referring to Table 5-2, the probability that all 3 businesses succeed is

ANSWER: 0.216 TYPE: FI DIFFICULTY: Moderate KEYWORDS: binomial distribution 51. Referring to Table 5-2, the probability that all 3 businesses fail is

ANSWER: 0.064 TYPE: FI DIFFICULTY: Moderate KEYWORDS: binomial distribution 52. Referring to Table 5-2, the probability that at least 1 business succeeds is ANSWER:

134 Some Important Discrete Probability Distributions

0.936 TYPE: FI DIFFICULTY: Moderate KEYWORDS: binomial distribution 53. Referring to Table 5-2, the probability that exactly 1 business succeeds is

ANSWER: 0.288 TYPE: FI DIFFICULTY: Moderate KEYWORDS: binomial distribution

54. If X has a binomial distribution with n = 4 and p = 0.3, then P(X = 1) =

ANSWER: 0.4116 TYPE: FI DIFFICULTY: Easy KEYWORDS: binomial distribution

55. If X has a binomial distribution with n = 4 and p = 0.3, then P(X > 1) =

ANSWER: 0.3483 TYPE: FI DIFFICULTY: Moderate KEYWORDS: binomial distribution

56. If X has a binomial distribution with n = 5 and p = 0.1, then P(X = 2) =

ANSWER: 0.0729 TYPE: FI DIFFICULTY: Easy KEYWORDS: binomial distribution

57. Suppose that past history shows that 60% of college students prefer Brand C cola. A sample of 5 students is to be selected. The probability that exactly 1 prefers brand C is

ANSWER: 0.0768 TYPE: FI DIFFICULTY: Easy KEYWORDS: binomial distribution

58. Suppose that past history shows that 60% of college students prefer Brand C cola. A sample of 5 students is to be selected. The probability that at least 1 prefers brand C is

ANSWER: 0.9898 TYPE: FI DIFFICULTY: Moderate KEYWORDS: binomial distribution

59. Suppose that past history shows that 60% of college students prefer Brand C cola. A sample of 5 students is to be selected. The probability that exactly 3 prefer brand C is

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ANSWER: 0.3456 TYPE: FI DIFFICULTY: Easy KEYWORDS: binomial distribution

60. Suppose that past history shows that 60% of college students prefer Brand C cola. A sample of 5 students is to be selected. The probability that exactly 4 prefer brand C is

ANSWER: 0.2592 TYPE: FI DIFFICULTY: Easy KEYWORDS: binomial distribution

61. Suppose that past history shows that 60% of college students prefer Brand C cola. A sample of 5 students is to be selected. The probability that at most 2 prefer brand C is

ANSWER: 0.3174 TYPE: FI DIFFICULTY: Moderate KEYWORDS: binomial distribution

62. Suppose that past history shows that 60% of college students prefer Brand C cola. A sample of 5 students is to be selected. The probability that more than 3 prefer brand C is

ANSWER: 0.3370 TYPE: FI DIFFICULTY: Moderate KEYWORDS: binomial distribution

63. Suppose that past history shows that 60% of college students prefer Brand C cola. A sample of 5 students is to be selected. The probability that less than 2 prefer brand C is

ANSWER: 0.0870 TYPE: FI DIFFICULTY: Moderate KEYWORDS: binomial distribution

64. Suppose that past history shows that 60% of college students prefer Brand C cola. A sample of 5 students is to be selected. The average number that you would expect to prefer brand C is . ANSWER: 3 TYPE: FI DIFFICULTY: Easy KEYWORDS: binomial distribution, mean

65. Suppose that past history shows that 60% of college students prefer Brand C cola. A sample of 5 students is to be selected. The variance of the number that prefer brand C is ANSWER:

136 Some Important Discrete Probability Distributions

1.2 TYPE: FI DIFFICULTY: Easy KEYWORDS: binomial distribution, variance TABLE 5-3 The following table contains the probability distribution for X = the number of retransmissions necessary to successfully transmit a 1024K data package through a double satellite media. X 0 1 2 3 P(X) 0.35 0.35 0.25 0.05 66. Referring to Table 5-3, the probability of no retransmissions is

ANSWER: 0.35 TYPE: FI DIFFICULTY: Easy KEYWORDS: probability distribution 67. Referring to Table 5-3, the probability of at least one retransmission is

ANSWER: 0.65 TYPE: FI DIFFICULTY: Easy KEYWORDS: probability distribution 68. Referring to Table 5-3, the mean or expected value for the number of retransmissions is . ANSWER: 1.0 TYPE: FI DIFFICULTY: Easy KEYWORDS: probability distribution, mean 69. Referring to Table 5-3, the variance for the number of retransmissions is

ANSWER: 0.80 TYPE: FI DIFFICULTY: Easy KEYWORDS: probability distribution, variance 70. Referring to Table 5-3, the standard deviation of the number of retransmissions is

ANSWER: 0.894 TYPE: FI DIFFICULTY: Easy KEYWORDS: probability distribution, standard deviation 71. In a game called Taxation and Evasion, a player rolls a pair of dice. If on any turn the sum is 7, 11, or 12, the player gets audited. Otherwise, she avoids taxes. Suppose a player takes 5 turns at rolling the dice. The probability that she does not get audited is .

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ANSWER: 0.2373 TYPE: FI DIFFICULTY: Moderate KEYWORDS: binomial distribution 72. In a game called Taxation and Evasion, a player rolls a pair of dice. If on any turn the sum is 7, 11, or 12, the player gets audited. Otherwise, she avoids taxes. Suppose a player takes 5 turns at rolling the dice. The probability that she gets audited once is . ANSWER: 0.3955 TYPE: FI DIFFICULTY: Moderate KEYWORDS: binomial distribution 73. In a game called Taxation and Evasion, a player rolls a pair of dice. If on any turn the sum is 7, 11, or 12, the player gets audited. Otherwise, she avoids taxes. Suppose a player takes 5 turns at rolling the dice. The probability that she gets audited at least once is . ANSWER: 0.7627 TYPE: FI DIFFICULTY: Moderate KEYWORDS: binomial distribution 74. In a game called Taxation and Evasion, a player rolls a pair of dice. If on any turn the sum is 7, 11, or 12, the player gets audited. Otherwise, she avoids taxes. Suppose a player takes 5 turns at rolling the dice. The probability that she gets audited no more than 2 times is . ANSWER: 0.8965 TYPE: FI DIFFICULTY: Moderate KEYWORDS: binomial distribution 75. In a game called Taxation and Evasion, a player rolls a pair of dice. If on any turn the sum is 7, 11, or 12, the player gets audited. Otherwise, she avoids taxes. Suppose a player takes 5 turns at rolling the dice. The expected number of times she will be audited is . ANSWER: 1.25 TYPE: FI DIFFICULTY: Easy KEYWORDS: binomial distribution, mean 76. In a game called Taxation and Evasion, a player rolls a pair of dice. If on any turn the sum is 7, 11, or 12, the player gets audited. Otherwise, she avoids taxes. Suppose a player takes 5 turns at rolling the dice. The variance of the number of times she will be audited is . ANSWER: 0.9375 TYPE: FI DIFFICULTY: Moderate KEYWORDS: binomial distribution, variance

138 Some Important Discrete Probability Distributions

77. In a game called Taxation and Evasion, a player rolls a pair of dice. If on any turn the sum is 7, 11, or 12, the player gets audited. Otherwise, she avoids taxes. Suppose a player takes 5 turns at rolling the dice. The standard deviation of the number of times she will be audited is . ANSWER: 0.968 TYPE: FI DIFFICULTY: Moderate KEYWORDS: binomial distribution, standard deviation TABLE 5-4 The following table contains the probability distribution for X = the number of traffic accidents reported in a day in Corvallis, Oregon. X 0 1 2 3 4 5 P(X) 0.10 0.20 0.45 0.15 0.05 0.05 78. Referring to Table 5-4, the probability of 3 accidents is

ANSWER: 0.15 TYPE: FI DIFFICULTY: Easy KEYWORDS: probability distribution 79. Referring to Table 5-4, the probability of at least 1 accident is

ANSWER: 0.90 TYPE: FI DIFFICULTY: Easy KEYWORDS: probability distribution 80. Referring to Table 5-4, the mean or expected value of the number of accidents is

ANSWER: 2.0 TYPE: FI DIFFICULTY: Easy KEYWORDS: probability distribution, mean, 81. Referring to Table 5-4, the variance of the number of accidents is

ANSWER: 1.4 TYPE: FI DIFFICULTY: Easy KEYWORDS: probability distribution, variance 82. Referring to Table 5-4, the standard deviation of the number of accidents is ANSWER: 1.18 TYPE: FI DIFFICULTY: Easy

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KEYWORDS: probability distribution, standard deviation 83. The number of power outages at a nuclear power plant has a Poisson distribution with a mean of 6 outages per year. The probability that there will be exactly 3 power outages in a year is . ANSWER: 0.0892 TYPE: FI DIFFICULTY: Easy KEYWORDS: Poisson distribution 84. The number of power outages at a nuclear power plant has a Poisson distribution with a mean of 6 outages per year. The probability that there will be at least 3 power outages in a year is . ANSWER: 0.9380 TYPE: FI DIFFICULTY: Difficult KEYWORDS: Poisson distribution 85. The number of power outages at a nuclear power plant has a Poisson distribution with a mean of 6 outages per year. The probability that there will be at least 1 power outage in a year is . ANSWER: 0.9975 TYPE: FI DIFFICULTY: Moderate KEYWORDS: Poisson distribution 86. The number of power outages at a nuclear power plant has a Poisson distribution with a mean of 6 outages per year. The probability that there will be no more than 1 power outage in a year is . ANSWER: 0.0174 TYPE: FI DIFFICULTY: Moderate KEYWORDS: Poisson distribution

87. The number of power outages at a nuclear power plant has a Poisson distribution with a mean of 6 outages per year. The probability that there will be between 1 and 3 inclusive power outages in a year is . ANSWER: 0.1487 TYPE: FI DIFFICULTY: Difficult KEYWORDS: Poisson distribution 88. The number of power outages at a nuclear power plant has a Poisson distribution with a mean of 6 outages per year. The variance of the number of power outages is .

140 Some Important Discrete Probability Distributions

ANSWER: 6 TYPE: FI DIFFICULTY: Easy KEYWORDS: Poisson distribution 89. The number of 911 calls in Butte, Montana, has a Poisson distribution with a mean of 10 calls a day. The probability of seven 911 calls in a day is . ANSWER: 0.0901 TYPE: FI DIFFICULTY: Easy KEYWORDS: Poisson distribution 90. The number of 911 calls in Butte, Montana, has a Poisson distribution with a mean of 10 calls a day. The probability of seven or eight 911 calls in a day is . ANSWER: 0.2027 TYPE: FI DIFFICULTY: Moderate KEYWORDS: Poisson distribution 91. The number of 911 calls in Butte, Montana, has a Poisson distribution with a mean of 10 calls a day. The probability of 2 or more 911 calls in a day is . ANSWER: 0.9995 TYPE: FI DIFFICULTY: Moderate KEYWORDS: Poisson distribution 92. The number of 911 calls in Butte, Montana, has a Poisson distribution with a mean of 10.0 calls a day. The standard deviation of the number of 911 calls in a day is . ANSWER: 3.16 TYPE: FI DIFFICULTY: Easy KEYWORDS: Poisson distribution 93. An Undergraduate Study Committee of 6 members at a major university is to be formed from a pool of faculty of 18 men and 6 women. If the committee members are chosen randomly, what is the probability that precisely half of the members will be women? ANSWER: 0.1213 TYPE: FI DIFFICULTY: Easy KEYWORDS: hypergeometric distribution

94. An Undergraduate Study Committee of 6 members at a major university is to be formed from a pool of faculty of 18 men and 6 women. If the committee members are chosen randomly, what is the probability that all of the members will be men?

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ANSWER: 0.1379 TYPE: FI DIFFICULTY: Easy KEYWORDS: hypergeometric distribution 95. A debate team of 4 members for a high school will be chosen randomly from a potential group of 15 students. Ten of the 15 students have no prior competition experience while the others have some degree of experience. What is the probability that none of the members chosen for the team have any competition experience? ANSWER: 0.1538 TYPE: FI DIFFICULTY: Easy KEYWORDS: hypergeometric distribution 96. A debate team of 4 members for a high school will be chosen randomly from a potential group of 15 students. Ten of the 15 students have no prior competition experience while the others have some degree of experience. What is the probability that at least 1 of the members chosen for the team have some prior competition experience? ANSWER: 0.8462 TYPE: FI DIFFICULTY: Moderate KEYWORDS: hypergeometric distribution 97. A debate team of 4 members for a high school will be chosen randomly from a potential group of 15 students. Ten of the 15 students have no prior competition experience while the others have some degree of experience. What is the probability that at most 1 of the members chosen for the team have some prior competition experience? ANSWER: 0.5934 TYPE: FI DIFFICULTY: Moderate KEYWORDS: hypergeometric distribution 98. A debate team of 4 members for a high school will be chosen randomly from a potential group of 15 students. Ten of the 15 students have no prior competition experience while the others have some degree of experience. What is the probability that exactly half of the members chosen for the team have some prior competition experience? ANSWER: 0.3297 TYPE: FI DIFFICULTY: Easy KEYWORDS: hypergeometric distribution 99. The Department of Commerce in a particular state has determined that the number of small businesses that declare bankruptcy per month is approximately a Poisson distribution with a mean of 6.4. Find the probability that more than 3 bankruptcies occur next month. ANSWER:

142 Some Important Discrete Probability Distributions

0.881 TYPE: PR DIFFICULTY: Moderate KEYWORDS: Poisson distribution 100.The Department of Commerce in a particular state has determined that the number of small businesses that declare bankruptcy per month is approximately a Poisson distribution with a mean of 6.4. Find the probability that exactly 5 bankruptcies occur next month. ANSWER: 0.149 TYPE: PR DIFFICULTY: Easy KEYWORDS: Poisson distribution 101.The on-line access computer service industry is growing at an extraordinary rate. Current estimates suggest that only 20% of the home-based computers have access to on-line services. This number is expected to grow quickly over the next 5 years. Suppose 25 people with homebased computers were randomly and independently sampled. Find the probability that fewer than 10 of those sampled currently have access to on-line services. ANSWER: 0.983 TYPE: PR DIFFICULTY: Moderate KEYWORDS: binomial distribution

102. The on-line access computer service industry is growing at an extraordinary rate. Current estimates suggest that only 20% of the home-based computers have access to on-line services. This number is expected to grow quickly over the next 5 years. Suppose 25 people with homebased computers were randomly and independently sampled. Find the probability that more than 20 of those sampled currently do not have access to on-line services. ANSWER: 0.421 TYPE: PR DIFFICULTY: Difficult KEYWORDS: binomial distribution 103.A national trend predicts that women will account for half of all business travelers in the next 3 years. To attract these women business travelers, hotels are providing more amenities that women particularly like. A recent survey of American hotels found that 70% offer hairdryers in the bathrooms. Consider a random and independent sample of 20 hotels. Find the probability all of the hotels in the sample offered hairdryers in the bathrooms. ANSWER: 0.0008 TYPE: PR DIFFICULTY: Moderate KEYWORDS: binomial distribution 104.A national trend predicts that women will account for half of all business travelers in the next 3 years. To attract these women business travelers, hotels are providing more amenities that women particularly like. A recent survey of American hotels found that 70% offer hairdryers in the bathrooms. Consider a random and independent sample of 20 hotels. Find the probability that more than 7 but less than 13 of the hotels in the sample offered hairdryers in the bathrooms.

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ANSWER: 0.2264 TYPE: PR DIFFICULTY: Moderate KEYWORDS: binomial distribution

105. A national trend predicts that women will account for half of all business travelers in the next 3 years. To attract these women business travelers, hotels are providing more amenities that women particularly like. A recent survey of American hotels found that 70% offer hairdryers in the bathrooms. Consider a random and independent sample of 20 hotels. Find the probability that at least 9 of the hotels in the sample do not offer hairdryers in the bathrooms. ANSWER: 0.1133 TYPE: PR DIFFICULTY: Difficult KEYWORDS: binomial distribution 106.The local police department must write, on average, 5 tickets a day to keep department revenues at budgeted levels. Suppose the number of tickets written per day follows a Poisson distribution with a mean of 6.4 tickets per day. Find the probability that less than 6 tickets are written on a randomly selected day from this population. ANSWER: 0.384 TYPE: PR DIFFICULTY: Easy KEYWORDS: Poisson distribution 107.The local police department must write, on average, 5 tickets a day to keep department revenues at budgeted levels. Suppose the number of tickets written per day follows a Poisson distribution with a mean of 6.4 tickets per day. Find the probability that exactly 6 tickets are written on a randomly selected day from this population. ANSWER: 0.159 TYPE: PR DIFFICULTY: Easy KEYWORDS: Poisson distribution TABLE 5-5 From an inventory of 48 new cars being shipped to local dealerships, corporate reports indicate that 12 have defective radios installed. 108.Referring to Table 5-5, what is the probability out of the 8 new cars it just received that, when each is tested, no more than 2 of the cars have defective radios? ANSWER: 0.6863 TYPE: PR DIFFICULTY: Moderate KEYWORDS: hypergeometric distribution

144 Some Important Discrete Probability Distributions

109.Referring to Table 5-5, what is the probability out of the 8 new cars it just received that, when each is tested, exactly half of the cars have defective radios? ANSWER: 0.0773 TYPE: PR DIFFICULTY: Easy KEYWORDS: hypergeometric distribution 110.Referring to Table 5-5, what is the probability out of the 8 new cars it just received that, when each is tested, none of the cars have defective radios? ANSWER: 0.08019 TYPE: PR DIFFICULTY: Easy KEYWORDS: hypergeometric distribution 111.Referring to Table 5-5, what is the probability out of the 8 new cars it just received that, when each is tested, at least half of the cars have defective radios? ANSWER: 0.09388 TYPE: PR DIFFICULTY: Moderate KEYWORDS: hypergeometric distribution 112.Referring to Table 5-5, what is the probability out of the 8 new cars it just received that, when each is tested, no more than half of the cars have defective radios? ANSWER: 0.9834 TYPE: PR DIFFICULTY: Moderate KEYWORDS: hypergeometric distribution 113.Referring to Table 5-5, what is the probability out of the 8 new cars it just received that, when each is tested, at most 2 of the cars have defective radios? ANSWER: 0.6863 TYPE: PR DIFFICULTY: Moderate KEYWORDS: hypergeometric distribution 114.Referring to Table 5-5, what is the probability out of the 8 new cars it just received that, when each is tested, exactly two of the cars have non-defective radios? ANSWER: 0.001543 TYPE: PR DIFFICULTY: Difficult KEYWORDS: hypergeometric distribution 115.Referring to Table 5-5, what is the probability out of the 8 new cars it just received that, when each is tested, at most three of the cars have non-defective radios?

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ANSWER: 0.01661 TYPE: PR DIFFICULTY: Difficult KEYWORDS: hypergeometric distribution 116.Referring to Table 5-5, what is the probability out of the 8 new cars it just received that, when each is tested, no more than half of the cars have non-defective radios? ANSWER: 0.09388 TYPE: PR DIFFICULTY: Difficult KEYWORDS: hypergeometric distribution TABLE 5-6 The quality control manager of Marilyn’s Cookies is inspecting a batch of chocolate chip cookies. When the production process is in control, the average number of chocolate chip parts per cookie is 6.0. 117. Referring to Table 5-6, what is the probability that any particular cookie being inspected has 4.0 chip parts. ANSWER: 0.1339 TYPE: PR DIFFICULTY: Easy KEYWORDS: Poisson distribution 118.Referring to Table 5-6, what is the probability that any particular cookie being inspected has fewer than 5.0 chip parts. ANSWER: 0.2851 TYPE: PR DIFFICULTY: Moderate KEYWORDS: Poisson distribution 119.Referring to Table 5-6, what is the probability that any particular cookie being inspected has at least 6.0 chip parts. ANSWER: 0.5543 TYPE: PR DIFFICULTY: Moderate KEYWORDS: Poisson distribution

120. Referring to Table 5-6, what is the probability that any particular cookie being inspected has between 5.0 and 8.0 inclusive chip parts. ANSWER: 0.5622 TYPE: PR DIFFICULTY: Moderate KEYWORDS: Poisson distribution

146 Some Important Discrete Probability Distributions

121.Referring to Table 5-6, what is the probability that any particular cookie being inspected has less than 5.0 or more than 8.0 chip parts. ANSWER: 0.4378 TYPE: PR DIFFICULTY: Difficult KEYWORDS: Poisson distribution TABLE 5-7 There are two houses with almost identical characteristics available for investment in two different neighborhoods with drastically different demographic composition. The anticipated gain in value when the houses are sold in 10 years has the following probability distribution:

Probability .25 .40 .35

Returns Neighborhood A Neighborhood B $30,500 −$22,500 $10,000 $25,000 $40,500 $10,500

122.Referring to Table 5-7, what is the expected value gain for the house in neighborhood A? ANSWER: $ 12,550 TYPE: PR DIFFICULTY: Easy KEYWORDS: mean

123. Referring to Table 5-7, what is the expected value gain for the house in neighborhood B? ANSWER: $ 21,300 TYPE: PR DIFFICULTY: Easy KEYWORDS: mean 124.Referring to Table 5-7, what is the variance of the gain in value for the house in neighborhood A? ANSWER: 583,147,500 TYPE: PR DIFFICULTY: Easy KEYWORDS: variance 125.Referring to Table 5-7, what is the variance of the gain in value for the house in neighborhood B? ANSWER: 67,460,000 TYPE: PR DIFFICULTY: Easy KEYWORDS: variance

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126.Referring to Table 5-7, what is the standard deviation of the value gain for the house in neighborhood A? ANSWER: $24,148.45 TYPE: PR DIFFICULTY: Easy KEYWORDS: standard deviation 127.Referring to Table 5-7, what is the standard deviation of the value gain for the house in neighborhood B? ANSWER: $8,213.40 TYPE: PR DIFFICULTY: Easy KEYWORDS: standard deviation 128. Referring to Table 5-7, what is the covariance of the two houses? ANSWER: −190,040,000 TYPE: PR DIFFICULTY: Easy KEYWORDS: covariance 129. Referring to Table 5-7, what is the expected value gain if you invest in both houses? ANSWER: $33,850 TYPE: PR DIFFICULTY: Moderate KEYWORDS: mean of the sum

130. Referring to Table 5-7, what is the total variance of value gain if you invest in both houses? ANSWER: 270,527,500 TYPE: PR DIFFICULTY: Moderate KEYWORDS: variance of the sum 131.Referring to Table 5-7, what is the total standard deviation of value gain if you invest in both houses? ANSWER: $16,447.72 TYPE: PR DIFFICULTY: Moderate KEYWORDS: standard deviation of the sum 132.Referring to Table 5-7, if you can invest half of your money on the house in neighborhood A and the remaining on the house in neighborhood B, what is the portfolio expected return of your investment?

148 Some Important Discrete Probability Distributions

ANSWER: $16,925 TYPE: PR DIFFICULTY: Moderate KEYWORDS: mean, portfolio

133. Referring to Table 5-7, if you can invest half of your money on the house in neighborhood A and the remaining on the house in neighborhood B, what is the portfolio risk of your investment? ANSWER: $8,223.86 TYPE: PR DIFFICULTY: Moderate KEYWORDS: standard deviation, risk, portfolio 134.Referring to Table 5-7, if you can invest 10% of your money on the house in neighborhood A and the remaining on the house in neighborhood B, what is the portfolio expected return of your investment? ANSWER: $20,425 TYPE: PR DIFFICULTY: Moderate KEYWORDS: mean, portfolio 135.Referring to Table 5-7, if you can invest 10% of your money on the house in neighborhood A and the remaining on the house in neighborhood B, what is the portfolio risk of your investment? ANSWER: $5,125.12 TYPE: PR DIFFICULTY: Moderate KEYWORDS: standard deviation, risk, portfolio 136.Referring to Table 5-7, if you can invest 30% of your money on the house in neighborhood A and the remaining on the house in neighborhood B, what is the portfolio expected return of your investment? ANSWER: $18,675 TYPE: PR DIFFICULTY: Moderate KEYWORDS: mean, portfolio 137.Referring to Table 5-7, if you can invest 30% of your money on the house in neighborhood A and the remaining on the house in neighborhood B, what is the portfolio risk of your investment? ANSWER: $2,392.04 TYPE: PR DIFFICULTY: Moderate KEYWORDS: standard deviation, risk, portfolio 138.Referring to Table 5-7, if you can invest 70% of your money on the house in neighborhood A and the remaining on the house in neighborhood B, what is the portfolio expected return of your investment?

Some Important Discrete Probability Distributions

ANSWER: $15,175 TYPE: PR DIFFICULTY: Moderate KEYWORDS: mean, portfolio

149

150 Some Important Discrete Probability Distributions

139.Referring to Table 5-7, if you can invest 70% of your money on the house in neighborhood A and the remaining on the house in neighborhood B, what is the portfolio risk of your investment? ANSWER: $14,560.11 TYPE: PR DIFFICULTY: Moderate KEYWORDS: standard deviation, risk, portfolio 140.Referring to Table 5-7, if you can invest 90% of your money on the house in neighborhood A and the remaining on the house in neighborhood B, what is the portfolio expected return of your investment? ANSWER: $13,425 TYPE: PR DIFFICULTY: Moderate KEYWORDS: mean, portfolio 141.Referring to Table 5-7, if you can invest 90% of your money on the house in neighborhood A and the remaining on the house in neighborhood B, what is the portfolio risk of your investment? ANSWER: $20,947.96 TYPE: PR DIFFICULTY: Moderate KEYWORDS: standard deviation, risk, portfolio

142. Referring to Table 5-7, if your investment preference is to maximize your expected return while exposing yourself to the minimal amount of risk, will you choose a portfolio that will consist of 10%, 30%, 50%, 70%, or 90% of your money on the house in neighborhood A and the remaining on the house in neighborhood B? ANSWER: 30% TYPE: PR DIFFICULTY: Moderate KEYWORDS: mean, standard deviation, risk, portfolio, coefficient of variation

143. Referring to Table 5-7, if your investment preference is to maximize your expected return and not worry at all about the risk that you have to take, will you choose a portfolio that will consist of 10%, 30%, 50%, 70%, or 90% of your money on the house in neighborhood A and the remaining on the house in neighborhood B? ANSWER: 10% TYPE: PR DIFFICULTY: Moderate KEYWORDS: mean, portfolio

144. Referring to Table 5-7, if your investment preference is to minimize the amount of risk that you have to take and do not care at all about the expected return, will you choose a portfolio that will consist of 10%, 30%, 50%, 70%, or 90% of your money on the house in neighborhood A and the remaining on the house in neighborhood B? ANSWER:

Some Important Discrete Probability Distributions

30% TYPE: PR DIFFICULTY: Moderate KEYWORDS: standard deviation, risk, portfolio

151

The Normal Distribution and Other Continuous Distributions 153

CHAPTER 6: THE NORMAL DISTRIBUTION AND OTHER CONTINUOUS DISTRIBUTIONS 1. In its standardized form, the normal distribution a) has a mean of 0 and a standard deviation of 1. b) has a mean of 1 and a variance of 0. c) has an area equal to 0.5. d) cannot be used to approximate discrete probability distributions. ANSWER: a TYPE: MC DIFFICULTY: Easy KEYWORDS: standardized normal distribution, properties

2. Which of the following about the normal distribution is not true? a) Theoretically, the mean, median, and mode are the same. b) About 2/3 of the observations fall within  1 standard deviation from the mean. c) It is a discrete probability distribution. d) Its parameters are the mean,  , and standard deviation,  . ANSWER: c TYPE: MC DIFFICULTY: Easy KEYWORDS: normal distribution, properties 3. If a particular batch of data is approximately normally distributed, we would find that approximately a) 2 of every 3 observations would fall between  1 standard deviation around the mean. b) 4 of every 5 observations would fall between  1.28 standard deviations around the mean. c) 19 of every 20 observations would fall between  2 standard deviations around the mean. d) All the above. ANSWER: d TYPE: MC DIFFICULTY: Easy KEYWORDS: normal distribution, properties

4. For some positive value of Z, the probability that a standard normal variable is between 0 and Z is 0.3770. The value of Z is a) 0.18. b) 0.81. c) 1.16. d) 1.47. ANSWER: c TYPE: MC DIFFICULTY: Moderate KEYWORDS: standardized normal distribution, value

154 The Normal Distribution and Other Continuous Distributions

5. For some value of Z, the probability that a standard normal variable is below Z is 0.2090. The value of Z is a) – 0.81. b) – 0.31. c) 0.31. d) 1.96. ANSWER: a TYPE: MC DIFFICULTY: Moderate KEYWORDS: standardized normal distribution, value

6. For some positive value of Z, the probability that a standard normal variable is between 0 and Z is 0.3340. The value of Z is a) 0.07. b) 0.37. c) 0.97. d) 1.06. ANSWER: c TYPE: MC DIFFICULTY: Moderate KEYWORDS: standardized normal distribution, value

7. For some positive value of X, the probability that a standard normal variable is between 0 and +2X is 0.1255. The value of X is a) 0.99. b) 0.40. c) 0.32. d) 0.16. ANSWER: d TYPE: MC DIFFICULTY: Difficult KEYWORDS: normal distribution, value

8. For some positive value of X, the probability that a standard normal variable is between 0 and +1.5X is 0.4332. The value of X is a) 0.10. b) 0.50. c) 1.00. d) 1.50. ANSWER: c TYPE: MC DIFFICULTY: Difficult KEYWORDS: normal distribution, value

The Normal Distribution and Other Continuous Distributions 155 9. Given that X is a normally distributed random variable with a mean of 50 and a standard deviation of 2, find the probability that X is between 47 and 54. ANSWER: 0.9104 TYPE: PR DIFFICULTY: Easy KEYWORDS: normal distribution, probability 10. A company that sells annuities must base the annual payout on the probability distribution of the length of life of the participants in the plan. Suppose the probability distribution of the lifetimes of the participants is approximately a normal distribution with a mean of 68 years and a standard deviation of 3.5 years. What proportion of the plan recipients would receive payments beyond age 75? ANSWER: 0.0228 TYPE: PR DIFFICULTY: Easy KEYWORDS: normal distribution, probability 11. A company that sells annuities must base the annual payout on the probability distribution of the length of life of the participants in the plan. Suppose the probability distribution of the lifetimes of the participants is approximately a normal distribution with a mean of 68 years and a standard deviation of 3.5 years. What proportion of the plan recipients die before they reach the standard retirement age of 65? ANSWER: 0.1957 using Excel or 0.1949 using Table E.2 TYPE: PR DIFFICULTY: Moderate KEYWORDS: normal distribution, probability 12. A company that sells annuities must base the annual payout on the probability distribution of the length of life of the participants in the plan. Suppose the probability distribution of the lifetimes of the participants is approximately a normal distribution with a mean of 68 years and a standard deviation of 3.5 years. Find the age at which payments have ceased for approximately 86% of the plan participants. ANSWER: 71.78 years old TYPE: PR DIFFICULTY: Difficult KEYWORDS: normal distribution, value 13. A catalog company that receives the majority of its orders by telephone conducted a study to determine how long customers were willing to wait on hold before ordering a product. The length of time was found to be a random variable best approximated by an exponential distribution with a mean equal to 3 minutes. What proportion of customers having to hold more than 4.5 minutes will hang up before placing an order? a) 0.22313 b) 0.48658 c) 0.51342 d) 0.77687

156 The Normal Distribution and Other Continuous Distributions

ANSWER: a TYPE: MC DIFFICULTY: Moderate KEYWORDS: exponential distribution, probability 14. A catalog company that receives the majority of its orders by telephone conducted a study to determine how long customers were willing to wait on hold before ordering a product. The length of time was found to be a random variable best approximated by an exponential distribution with a mean equal to 3 minutes. What proportion of customers having to hold more than 1.5 minutes will hang up before placing an order? a) 0.86466 b) 0.60653 c) 0.39347 d) 0.13534 ANSWER: b TYPE: MC DIFFICULTY: Moderate KEYWORDS: exponential distribution, probability 15. A catalog company that receives the majority of its orders by telephone conducted a study to determine how long customers were willing to wait on hold before ordering a product. The length of time was found to be a random variable best approximated by an exponential distribution with a mean equal to 3 minutes. Find the waiting time at which only 10% of the customers will continue to hold. a) 2.3 minutes b) 3.3 minutes c) 6.9 minutes d) 13.8 minutes ANSWER: c TYPE: MC DIFFICULTY: Difficult KEYWORDS: exponential distribution, value 16. A catalog company that receives the majority of its orders by telephone conducted a study to determine how long customers were willing to wait on hold before ordering a product. The length of time was found to be a random variable best approximated by an exponential distribution with a mean equal to 2.8 minutes. What proportion of callers is put on hold longer than 2.8 minutes? a) 0.367879 b) 0.50 c) 0.60810 d) 0.632121 ANSWER: a TYPE: MC DIFFICULTY: Moderate KEYWORDS: exponential distribution, probability

The Normal Distribution and Other Continuous Distributions 157

17. A catalog company that receives the majority of its orders by telephone conducted a study to determine how long customers were willing to wait on hold before ordering a product. The length of time was found to be a random variable best approximated by an exponential distribution with a mean equal to 2.8 minutes. What is the probability that a randomly selected caller is placed on hold fewer than 7 minutes? a) 0.0009119 b) 0.082085 c) 0.917915 d) 0.9990881 ANSWER: c TYPE: MC DIFFICULTY: Moderate KEYWORDS: exponential distribution, probability 18. If we know that the length of time it takes a college student to find a parking spot in the library parking lot follows a normal distribution with a mean of 3.5 minutes and a standard deviation of 1 minute, find the probability that a randomly selected college student will find a parking spot in the library parking lot in less than 3 minutes. a) 0.3551 b) 0.3085 c) 0.2674 d) 0.1915 ANSWER: b TYPE: MC DIFFICULTY: Easy KEYWORDS: normal distribution, probability 19. If we know that the length of time it takes a college student to find a parking spot in the library parking lot follows a normal distribution with a mean of 3.5 minutes and a standard deviation of 1 minute, find the probability that a randomly selected college student will take between 2 and 4.5 minutes to find a parking spot in the library parking lot. a) 0.0919 b) 0.2255 c) 0.4938 d) 0.7745 ANSWER: d TYPE: MC DIFFICULTY: Easy KEYWORDS: normal distribution, probability 20. If we know that the length of time it takes a college student to find a parking spot in the library parking lot follows a normal distribution with a mean of 3.5 minutes and a standard deviation of 1 minute, find the point in the distribution in which 75.8% of the college students exceed when trying to find a parking spot in the library parking lot. a) 2.8 minutes b) 3.2 minutes

158 The Normal Distribution and Other Continuous Distributions

c) 3.4 minutes d) 4.2 minutes ANSWER: a TYPE: MC DIFFICULTY: Moderate KEYWORDS: normal distribution, value

21. Let X represent the amount of time it takes a student to park in the library parking lot at the university. If we know that the distribution of parking times can be modeled using an exponential distribution with a mean of 4 minutes, find the probability that it will take a randomly selected student more than 10 minutes to park in the library lot. a) 0.917915 b) 0.670320 c) 0.329680 d) 0.082085 ANSWER: d TYPE: MC DIFFICULTY: Moderate KEYWORDS: exponential distribution, probability

22. Let X represent the amount of time it takes a student to park in the library parking lot at the university. If we know that the distribution of parking times can be modeled using an exponential distribution with a mean of 4 minutes, find the probability that it will take a randomly selected student between 2 and 12 minutes to park in the library lot. a) 0.049787 b) 0.556744 c) 0.606531 d) 0.656318 ANSWER: b TYPE: MC DIFFICULTY: Moderate KEYWORDS: exponential distribution, probability 23. The owner of a fish market determined that the average weight for a catfish is 3.2 pounds with a standard deviation of 0.8 pound. Assuming the weights of catfish are normally distributed, the probability that a randomly selected catfish will weigh more than 4.4 pounds is ? ANSWER: 0.0668 TYPE: FI DIFFICULTY: Easy KEYWORDS: normal distribution, probability 24. The owner of a fish market determined that the average weight for a catfish is 3.2 pounds with a standard deviation of 0.8 pound. Assuming the weights of catfish are normally distributed, the probability that a randomly selected catfish will weigh between 3 and 5 pounds is ? ANSWER:

The Normal Distribution and Other Continuous Distributions 159 0.5865 TYPE: FI DIFFICULTY: Easy KEYWORDS: normal distribution, probability 25. The owner of a fish market determined that the average weight for a catfish is 3.2 pounds with a standard deviation of 0.8 pound. A citation catfish should be one of the top 2% in weight. Assuming the weights of catfish are normally distributed, at what weight (in pounds) should the citation designation be established? a) 1.56 pounds b) 4.84 pounds c) 5.20 pounds d) 7.36 pounds ANSWER: b TYPE: MC DIFFICULTY: Moderate KEYWORDS: normal distribution, value 26. The owner of a fish market determined that the average weight for a catfish is 3.2 pounds with a standard deviation of 0.8 pound. Assuming the weights of catfish are normally distributed, above what weight (in pounds) do 89.80% of the weights occur? ANSWER: 2.184 pounds TYPE: PR DIFFICULTY: Moderate KEYWORDS: normal distribution, value 27. The owner of a fish market determined that the average weight for a catfish is 3.2 pounds with a standard deviation of 0.8 pound. Assuming the weights of catfish are normally distributed, the probability that a randomly selected catfish will weigh less than 2.2 pounds is ? ANSWER: 0.1056 TYPE: FI DIFFICULTY: Easy KEYWORDS: normal distribution, probability 28. The Tampa International Airport (TIA) has been criticized for the waiting times associated with departing flights. While the critics acknowledge that many flights have little or no waiting times, their complaints deal more specifically with the longer waits attributed to some flights. The critics are interested in showing, mathematically, exactly what the problems are. Which type of distribution would best model the waiting times of the departing flights at TIA? a) Uniform distribution b) Binomial distribution c) Normal distribution d) Exponential distribution ANSWER: d TYPE: MC DIFFICULTY: Moderate KEYWORDS: exponential distribution, properties

160 The Normal Distribution and Other Continuous Distributions

29. Scientists in the Amazon are trying to find a cure for a deadly disease that is attacking the rain forests there. One of the variables that the scientists have been measuring involves the diameter of the trunk of the trees that have been affected by the disease. Scientists have calculated that the average diameter of the diseased trees is 42 centimeters. They also know that approximately 95% of the diameters fall between 32 and 52 centimeters and almost all of the diseased trees have diameters between 27 and 57 centimeters. When modeling the diameters of diseased trees, which distribution should the scientists use? a) uniform distribution b) binomial distribution c) normal distribution d) exponential distribution ANSWER: c TYPE: MC DIFFICULTY: Moderate KEYWORDS: normal distribution, properties

30. In the game Wheel of Fortune, which of the following distributions can best be used to compute the probability of winning the special vacation package in a single spin? a) uniform distribution b) binomial distribution c) normal distribution d) exponential distribution ANSWER: a TYPE: MC DIFFICULTY: Moderate KEYWORDS: uniform distribution, properties 31. A food processor packages orange juice in small jars. The weights of the filled jars are approximately normally distributed with a mean of 10.5 ounces and a standard deviation of 0.3 ounce. Find the proportion of all jars packaged by this process that have weights that fall below 10.875 ounces. ANSWER: 0.8944 TYPE: PR DIFFICULTY: Easy KEYWORDS: normal distribution, probability 32. A food processor packages orange juice in small jars. The weights of the filled jars are approximately normally distributed with a mean of 10.5 ounces and a standard deviation of 0.3 ounce. Find the proportion of all jars packaged by this process that have weights that fall above 10.95 ounces. ANSWER: 0.0668 TYPE: PR DIFFICULTY: Easy KEYWORDS: normal distribution, probability

The Normal Distribution and Other Continuous Distributions 161 33. True or False: The probability that a standard normal random variable, Z, falls between – 1.50 and 0.81 is 0.7242. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: standardized normal distribution, probability

34. True or False: The probability that a standard normal random variable, Z, is between 1.50 and 2.10 is the same as the probability Z is between – 2.10 and – 1.50. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: standardized normal distribution, probability

35. True or False: The probability that a standard normal random variable, Z, is below 1.96 is 0.4750. ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: standardized normal distribution, probability

36. True or False: The probability that a standard normal random variable, Z, is between 1.00 and 3.00 is 0.1574. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: standardized normal distribution, probability

37. True or False: The probability that a standard normal random variable, Z, falls between –2.00 and –0.44 is 0.6472. ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: standardized normal distribution, probability

38. True or False: The probability that a standard normal random variable, Z, is less than 50 is approximately 0. ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: standardized normal distribution, probability

162 The Normal Distribution and Other Continuous Distributions

39. True or False: A worker earns $15 per hour at a plant and is told that only 2.5% of all workers make a higher wage. If the wage is assumed to be normally distributed and the standard deviation of wage rates is $5 per hour, the average wage for the plant is $7.50 per hour. ANSWER: False TYPE: TF DIFFICULTY: Moderate KEYWORDS: normal distribution, mean 40. True or False: Theoretically, the mean, median, and the mode are all equal for a normal distribution. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: normal distribution, properties 41. True or False: Any set of normally distributed data can be transformed to its standardized form. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: normal distribution, properties 42. True or False: The "middle spread," that is the middle 50% of the normal distribution, is equal to one standard deviation. ANSWER: False TYPE: TF DIFFICULTY: Moderate KEYWORDS: normal distribution, probability, value 43. True or False: A normal probability plot may be used to assess the assumption of normality for a particular batch of data. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: normal probability plot 44. True or False: If a data batch is approximately normally distributed, its normal probability plot would be S-shaped. ANSWER: False TYPE: TF DIFFICULTY: Moderate KEYWORDS: normal probability plot

45. The probability that a standard normal variable Z is positive is

The Normal Distribution and Other Continuous Distributions 163 ANSWER: 0.50 TYPE: FI DIFFICULTY: Easy KEYWORDS: standardized normal distribution

46. The amount of pyridoxine (in grams) in a multiple vitamin is normally distributed with  = 110 grams and  = 25 grams. What is the probability that a randomly selected vitamin will contain between 100 and 110 grams of pyridoxine? ANSWER: 0.1554 TYPE: PR DIFFICULTY: Easy KEYWORDS: normal distribution, probability

47. The amount of pyridoxine (in grams) in a multiple vitamin is normally distributed with  = 110 grams and  = 25 grams. What is the probability that a randomly selected vitamin will contain between 82 and 100 grams of pyridoxine? ANSWER: 0.2132 TYPE: PR DIFFICULTY: Easy KEYWORDS: normal distribution, probability

48. The amount of pyridoxine (in grams) in a multiple vitamin is normally distributed with  = 110 grams and  = 25 grams. What is the probability that a randomly selected vitamin will contain at least 100 grams of pyridoxine? ANSWER: 0.6554 TYPE: PR DIFFICULTY: Easy KEYWORDS: normal distribution, probability

49. The amount of pyridoxine (in grams) in a multiple vitamin is normally distributed with  = 110 grams and  = 25 grams. What is the probability that a randomly selected vitamin will contain between 100 and 120 grams of pyridoxine? ANSWER: 0.3108 TYPE: PR DIFFICULTY: Moderate KEYWORDS: normal distribution, probability

50. The amount of pyridoxine (in grams) in a multiple vitamin is normally distributed with  = 110 grams and  = 25 grams. What is the probability that a randomly selected vitamin will contain less than 100 grams of pyridoxine? ANSWER: 0.3446 TYPE: PR DIFFICULTY: Easy KEYWORDS: normal distribution, probability

164 The Normal Distribution and Other Continuous Distributions

51. The amount of pyridoxine (in grams) in a multiple vitamin is normally distributed with  = 110 grams and  = 25 grams. What is the probability that a randomly selected vitamin will contain less than 100 grams or more than 120 grams of pyridoxine? ANSWER: 0.6892 TYPE: PR DIFFICULTY: Easy KEYWORDS: normal distribution, probability

52. The amount of pyridoxine (in grams) in a multiple vitamin is normally distributed with  = 110 grams and  = 25 grams. Approximately 83% of the vitamins will have at least how many grams of pyridoxine? ANSWER: 86.15 using Excel or 86.25 using Table E.2 TYPE: PR DIFFICULTY: Moderate KEYWORDS: normal distribution, value 53. The true length of boards cut at a mill with a listed length of 10 feet is normally distributed with a mean of 123 inches and a standard deviation of 1 inch. What proportion of the boards will be between 121 and 124 inches? ANSWER: 0.8186 using Excel or 0.8185 using Table E.2 TYPE: PR DIFFICULTY: Easy KEYWORDS: normal distribution, probability 54. The true length of boards cut at a mill with a listed length of 10 feet is normally distributed with a mean of 123 inches and a standard deviation of 1 inch. What proportion of the boards will be over 125 inches in length? ANSWER: 0.0228 TYPE: PR DIFFICULTY: Easy KEYWORDS: normal distribution, probability 55. The true length of boards cut at a mill with a listed length of 10 feet is normally distributed with a mean of 123 inches and a standard deviation of 1 inch. What proportion of the boards will be less than 124 inches? ANSWER: 0.8413 TYPE: PR DIFFICULTY: Easy KEYWORDS: normal distribution, probability

56. Suppose Z has a standard normal distribution with a mean of 0 and standard deviation of 1. The probability that Z is less than 1.15 is ANSWER: 0.8749

The Normal Distribution and Other Continuous Distributions 165 TYPE: FI DIFFICULTY: Easy KEYWORDS: standardized normal distribution, probability

57. Suppose Z has a standard normal distribution with a mean of 0 and standard deviation of 1. The probability that Z is more than 0.77 is

ANSWER: 0.2207 TYPE: FI DIFFICULTY: Easy KEYWORDS: standardized normal distribution, probability

58. Suppose Z has a standard normal distribution with a mean of 0 and standard deviation of 1. The probability that Z is less than -2.20 is

ANSWER: 0.0139 TYPE: FI DIFFICULTY: Easy KEYWORDS: standardized normal distribution, probability

59. Suppose Z has a standard normal distribution with a mean of 0 and standard deviation of 1. The probability that Z is more than -0.98 is

ANSWER: 0.8365 TYPE: FI DIFFICULTY: Easy KEYWORDS: standardized normal distribution, probability

60. Suppose Z has a standard normal distribution with a mean of 0 and standard deviation of 1. The probability that Z is between -2.33 and 2.33 is

ANSWER: 0.9802 TYPE: FI DIFFICULTY: Easy KEYWORDS: standardized normal distribution, probability

61. Suppose Z has a standard normal distribution with a mean of 0 and standard deviation of 1. The probability that Z is between -2.89 and -1.03 is

ANSWER: 0.1496 TYPE: FI DIFFICULTY: Easy KEYWORDS: standardized normal distribution, probability

62. Suppose Z has a standard normal distribution with a mean of 0 and standard deviation of 1. The probability that Z is between -0.88 and 2.29 is ANSWER: 0.7996 TYPE: FI DIFFICULTY: Easy KEYWORDS: standardized normal distribution, probability

166 The Normal Distribution and Other Continuous Distributions

63. Suppose Z has a standard normal distribution with a mean of 0 and standard deviation of 1. The probability that Z values are larger than

is 0.3483.

ANSWER: 0.39 TYPE: FI DIFFICULTY: Moderate KEYWORDS: standardized normal distribution, probability

64. Suppose Z has a standard normal distribution with a mean of 0 and standard deviation of 1. The probability that Z values are larger than

is 0.6985.

ANSWER: -0.52 TYPE: FI DIFFICULTY: Moderate KEYWORDS: standardized normal distribution, probability

65. Suppose Z has a standard normal distribution with a mean of 0 and standard deviation of 1. So 27% of the possible Z values are smaller than

ANSWER: -0.61 TYPE: FI DIFFICULTY: Moderate KEYWORDS: standardized normal distribution, value

66. Suppose Z has a standard normal distribution with a mean of 0 and standard deviation of 1. So 85% of the possible Z values are smaller than

ANSWER: 1.04 TYPE: FI DIFFICULTY: Moderate KEYWORDS: standardized normal distribution, value

67. Suppose Z has a standard normal distribution with a mean of 0 and standard deviation of 1. So 96% of the possible Z values are between distributed about the mean).

and

(symmetrically

ANSWER: -2.05 and 2.05 or -2.06 and 2.06 TYPE: FI DIFFICULTY: Moderate KEYWORDS: standardized normal distribution, value

68. Suppose Z has a standard normal distribution with a mean of 0 and standard deviation of 1. So 50% of the possible Z values are between distributed about the mean). ANSWER: -0.67 and 0.67 or -0.68 and 0.68 TYPE: FI DIFFICULTY: Moderate KEYWORDS: standardized normal distribution, value

and

(symmetrically

The Normal Distribution and Other Continuous Distributions 167

69. The amount of time necessary for assembly line workers to complete a product is a normal random variable with a mean of 15 minutes and a standard deviation of 2 minutes. The probability is that a product is assembled in less than 12 minutes. ANSWER: 0.0668 TYPE: FI DIFFICULTY: Easy KEYWORDS: normal distribution, probability 70. The amount of time necessary for assembly line workers to complete a product is a normal random variable with a mean of 15 minutes and a standard deviation of 2 minutes. The probability is that a product is assembled in between 14 and 16 minutes. ANSWER: 0.3829 using Excel or 0.3830 using Table E.2 TYPE: FI DIFFICULTY: Easy KEYWORDS: normal distribution, probability 71. The amount of time necessary for assembly line workers to complete a product is a normal random variable with a mean of 15 minutes and a standard deviation of 2 minutes. The probability is that a product is assembled in between 10 and 12 minutes. ANSWER: 0.0606 TYPE: FI DIFFICULTY: Easy KEYWORDS: normal distribution, probability 72. The amount of time necessary for assembly line workers to complete a product is a normal random variable with a mean of 15 minutes and a standard deviation of 2 minutes. The probability is that a product is assembled in between 15 and 21 minutes. ANSWER: 0.49865 TYPE: FI DIFFICULTY: Easy KEYWORDS: normal distribution, probability 73. The amount of time necessary for assembly line workers to complete a product is a normal random variable with a mean of 15 minutes and a standard deviation of 2 minutes. The probability is that a product is assembled in between 16 and 21 minutes. ANSWER: 0.30719 TYPE: FI DIFFICULTY: Easy KEYWORDS: normal distribution, probability 74. The amount of time necessary for assembly line workers to complete a product is a normal random variable with a mean of 15 minutes and a standard deviation of 2 minutes. The probability is that a product is assembled in more than 11 minutes.

168 The Normal Distribution and Other Continuous Distributions

ANSWER: 0.97725 TYPE: FI DIFFICULTY: Easy KEYWORDS: normal distribution, probability 75. The amount of time necessary for assembly line workers to complete a product is a normal random variable with a mean of 15 minutes and a standard deviation of 2 minutes. The probability is that a product is assembled in more than 19 minutes. ANSWER: 0.0228 TYPE: FI DIFFICULTY: Easy KEYWORDS: normal distribution, probability 76. The amount of time necessary for assembly line workers to complete a product is a normal random variable with a mean of 15 minutes and a standard deviation of 2 minutes. The probability is that a product is assembled in less than 20 minutes. ANSWER: 0.9938 TYPE: FI DIFFICULTY: Easy KEYWORDS: normal distribution, probability 77. The amount of time necessary for assembly line workers to complete a product is a normal random variable with a mean of 15 minutes and a standard deviation of 2 minutes. So, 15% of the products require more than minutes for assembly. ANSWER: 17.0729 using Excel or 17.08 using Table E.2 TYPE: FI DIFFICULTY: Moderate KEYWORDS: normal distribution, value 78. The amount of time necessary for assembly line workers to complete a product is a normal random variable with a mean of 15 minutes and a standard deviation of 2 minutes. So, 90% of the products require more than minutes for assembly. ANSWER: 12.44 TYPE: FI DIFFICULTY: Moderate KEYWORDS: normal distribution, value 79. The amount of time necessary for assembly line workers to complete a product is a normal random variable with a mean of 15 minutes and a standard deviation of 2 minutes. So, 60% of the products would be assembled within and minutes (symmetrically distributed about the mean). ANSWER: 13.32 and 16.68 or 13.31 and 16.69 TYPE: FI DIFFICULTY: Difficult KEYWORDS: normal distribution, value

The Normal Distribution and Other Continuous Distributions 169

80. The amount of time necessary for assembly line workers to complete a product is a normal random variable with a mean of 15 minutes and a standard deviation of 2 minutes. So, 17% of the products would be assembled within minutes. ANSWER: 13.1 TYPE: FI DIFFICULTY: Moderate KEYWORDS: normal distribution, value 81. The amount of time necessary for assembly line workers to complete a product is a normal random variable with a mean of 15 minutes and a standard deviation of 2 minutes. So, 70% of the products would be assembled within minutes. ANSWER: 16.0488 using Excel or 16.04 using Table E.2 TYPE: FI DIFFICULTY: Moderate KEYWORDS: normal distribution, value TABLE 6-1 The manager of a surveying company believes that the average number of phone surveys completed per hour by her employees has a normal distribution. She takes a sample of 15 days output from her employees and determines the average number of surveys per hour on these days. The ordered array for this data is: 10.0, 10.1, 10.3, 10.5, 10.7, 11.2, 11.4, 11.5, 11.7, 11.8, 11.8, 12.0, 12.2, 12.2, 12.5. 82. Referring to Table 6-1, the first standard normal quantile is

ANSWER: -1.5341 TYPE: FI DIFFICULTY: Moderate KEYWORDS: standard normal quantile, normal probability plot 83. Referring to Table 6-1, the fourth standard normal quantile is

ANSWER: -0.6745 TYPE: FI DIFFICULTY: Moderate KEYWORDS: standard normal quantile, normal probability plot 84. Referring to Table 6-1, the ninth standard normal quantile is

ANSWER: +0.1573 TYPE: FI DIFFICULTY: Moderate KEYWORDS: standard normal quantile, normal probability plot 85. Referring to Table 6-1, the fourteenth standard normal quantile is ANSWER:

170 The Normal Distribution and Other Continuous Distributions

+1.1503 TYPE: FI DIFFICULTY: Moderate KEYWORDS: standard normal quantile, normal probability plot 86. Referring to Table 6-1, the last standard normal quantile is

ANSWER: +1.5341 TYPE: FI DIFFICULTY: Moderate KEYWORDS: standard normal quantile, normal probability plot 87. Referring to Table 6-1, construct a normal probability plot for the data. ANSWER: 13

Surveys

10 -2

-1.5

-1

-0.5

0.5

1.5

Z -Score

TYPE: PR DIFFICULTY: Moderate KEYWORDS: normal probability plot 88. True or False: Referring to Table 6-1, the data appear reasonably normal but not perfectly normal. ANSWER: True TYPE: TF DIFFICULTY: Difficult KEYWORDS: normal probability plot TABLE 6-2 The city manager of a large city believes that the number of reported accidents on any weekend has a normal distribution. She takes a sample of nine weekends and determines the number of reported accidents during each. The ordered array for this data is: 15, 46, 53, 54, 55, 76, 82, 256, 407. 89. Referring to Table 6-2, the first standard normal quantile is

The Normal Distribution and Other Continuous Distributions 171 ANSWER: -1.28 TYPE: FI DIFFICULTY: Moderate KEYWORDS: standard normal quantile, normal probability plot 90. Referring to Table 6-2, the fifth standard normal quantile is

ANSWER: 0 TYPE: FI DIFFICULTY: Moderate KEYWORDS: standard normal quantile, normal probability plot 91. Referring to Table 6-2, the sixth standard normal quantile is

ANSWER: +0.25 TYPE: FI DIFFICULTY: Moderate KEYWORDS: standard normal quantile, normal probability plot 92. Referring to Table 6-2, the second standard normal quantile is

ANSWER: -0.84 TYPE: FI DIFFICULTY: Moderate KEYWORDS: standard normal quantile, normal probability plot 93. Referring to Table 6-2, the seventh standard normal quantile is ANSWER: +0.52 TYPE: FI DIFFICULTY: Moderate KEYWORDS: standard normal quantile, normal probability plot 94. Referring to Table 6-2, construct a normal probability plot. ANSWER:

172 The Normal Distribution and Other Continuous Distributions

450

Accidents

400 350 300 250 200 150 100 50 0 -1.5

-1

-0.5

0.5

1.5

Z -Scores

TYPE: PR DIFFICULTY: Moderate KEYWORDS: normal probability plot 95. True or False: Referring to Table 6-2, the data appear normal. ANSWER: False TYPE: TF DIFFICULTY: Difficult KEYWORDS: normal probability plot 96. Times spent watching TV every week by first graders follow an exponential distribution with mean 10 hours. The probability that a given first grader spends less than 20 hours watching TV is . ANSWER: 0.8647 TYPE: FI DIFFICULTY: Easy KEYWORDS: exponential distribution, probability 97. Times spent watching TV every week by first graders follow an exponential distribution with mean 10 hours. The probability that a given first grader spends more than 5 hours watching TV is . ANSWER: 0.6065 TYPE: FI DIFFICULTY: Easy KEYWORDS: exponential distribution, probability 98. Times spent watching TV every week by first graders follow an exponential distribution with mean 10 hours. The probability that a given first grader spends between 10 and 15 hours watching TV is . ANSWER: 0.1447

The Normal Distribution and Other Continuous Distributions 173 TYPE: FI DIFFICULTY: Moderate KEYWORDS: exponential distribution, probability 99. Patients arriving at an outpatient clinic follow an exponential distribution with mean 15 minutes. What is the average number of arrivals per minute? ANSWER: 0.06667 TYPE: PR DIFFICULTY: Easy KEYWORDS: exponential distribution, mean 100. Patients arriving at an outpatient clinic follow an exponential distribution with mean 15 minutes. What is the probability that a randomly chosen arrival to be more than 18 minutes? ANSWER: 0.3012 TYPE: PR DIFFICULTY: Easy KEYWORDS: exponential distribution, probability 101. Patients arriving at an outpatient clinic follow an exponential distribution with mean 15 minutes. What is the probability that a randomly chosen arrival to be less than 15 minutes? ANSWER: 0.6321 TYPE: PR DIFFICULTY: Easy KEYWORDS: exponential distribution, probability 102.

Patients arriving at an outpatient clinic follow an exponential distribution at a rate of 15 patients per hour. What is the probability that a randomly chosen arrival to be less than 15 minutes?

ANSWER: 0.9765 TYPE: PR DIFFICULTY: Easy KEYWORDS: exponential distribution, probability 103.

Patients arriving at an outpatient clinic follow an exponential distribution at a rate of 15 patients per hour. What is the probability that a randomly chosen arrival to be more than 5 minutes?

ANSWER: 0.2865 TYPE: PR DIFFICULTY: Easy KEYWORDS: exponential distribution, probability 104.

Patients arriving at an outpatient clinic follow an exponential distribution at a rate of 15 patients per hour. What is the probability that a randomly chosen arrival to be between 5 minutes and 15 minutes?

ANSWER:

174 The Normal Distribution and Other Continuous Distributions

0.2630 TYPE: PR DIFFICULTY: Moderate KEYWORDS: exponential distribution, probability 105.

Patients arriving at an outpatient clinic follow an exponential distribution at a rate of 1 patient per hour. What is the probability that a randomly chosen arrival to be more than 1 hour?

ANSWER: 0.3679 TYPE: PR DIFFICULTY: Easy KEYWORDS: exponential distribution, probability 106.

Patients arriving at an outpatient clinic follow an exponential distribution at a rate of 1 patient per hour. What is the probability that a randomly chosen arrival to be more than 2.5 hours?

ANSWER: 0.0821 TYPE: PR DIFFICULTY: Easy KEYWORDS: exponential distribution, probability 107.

Patients arriving at an outpatient clinic follow an exponential distribution at a rate of 1 patient per hour. What is the probability that a randomly chosen arrival to be less than 20 minutes?

ANSWER: 0.2835 TYPE: PR DIFFICULTY: Easy KEYWORDS: exponential distribution, probability 108.

Patients arriving at an outpatient clinic follow an exponential distribution at a rate of 1.5 patients per hour. What is the probability that a randomly chosen arrival to be less than 10 minutes?

ANSWER: 0.2212 TYPE: PR DIFFICULTY: Easy KEYWORDS: exponential distribution, probability 109.

Patients arriving at an outpatient clinic follow an exponential distribution at a rate of 1.5 patients per hour. What is the probability that a randomly chosen arrival to be between 10 and 15 minutes?

ANSWER: 0.0915 TYPE: PR DIFFICULTY: Moderate KEYWORDS: exponential distribution, probability TABLE 6-3

The Normal Distribution and Other Continuous Distributions 175 The number of column inches of classified advertisements appearing on Mondays in a certain daily newspaper is normally distributed with population mean 320 and population standard deviation 20 inches. 110. Referring to Table 6-3, for a randomly chosen Monday, what is the probability there will be less than 340 column inches of classified advertisement? ANSWER: 0.8413 TYPE: PR DIFFICULTY: Easy KEYWORDS: normal distribution, probability 111. Referring to Table 6-3, for a randomly chosen Monday, what is the probability there will be between 280 and 360 column inches of classified advertisement? ANSWER: 0.9545 using Excel or 0.9544 using Table E.2 TYPE: PR DIFFICULTY: Moderate KEYWORDS: normal distribution, probability 112. Referring to Table 6-3, for a randomly chosen Monday the probability is 0.1 that there will be less than how many column inches of classified advertisements? ANSWER: 294.4 TYPE: PR DIFFICULTY: Moderate KEYWORDS: normal distribution, value 113. Referring to Table 6-3, a single Monday is chosen at random. State in which of the following ranges the number of column inches of classified advertisement is most likely to be: a) 300 --320 b) 310 --330 c) 320 -- 340 d) 330 -- 350 ANSWER: b TYPE: MC Difficulty: Moderate KEYWORDS: normal distribution, probability TABLE 6-4 John has two jobs. For daytime work at a jewelry store he is paid $200 per month, plus a commission. His monthly commission is normally distributed with mean $600 and standard deviation $40. At night he works as a waiter, for which his monthly income is normally distributed with mean $100 and standard deviation $30. John's income levels from these two sources are independent of each other. 114. Referring to Table 6-4, for a given month, what is the probability that John's commission from the jewelry store is less than $640?

176 The Normal Distribution and Other Continuous Distributions

ANSWER: 0.8413 TYPE: PR DIFFICULTY: Easy KEYWORDS: normal distribution, probability 115. Referring to Table 6-4, for a given month, what is the probability that John's income as a waiter is between $70 and $160? ANSWER: 0.8186 using Excel or 0.8185 using Table E.2 TYPE: PR DIFFICULTY: Easy KEYWORDS: normal distribution, probability 116. Referring to Table 6-4, the probability is 0.9 that John's income as a waiter is less than how much in a given month? ANSWER: $138.45 using Excel or $138.40 using Table E.2 TYPE: PR DIFFICULTY: Moderate KEYWORDS: normal distribution, value 117. Referring to Table 6-4, find the mean and standard deviation of John's total income from these two jobs for a given month. ANSWER: $900; $50 TYPE: PR DIFFICULTY: Difficult EXPLANATION: Total mean = $200 + $600 + $100 = $900; Total variance = 402 + 302 = 2,500 KEYWORDS: normal distribution, mean, standard deviation 118. Referring to Table 6-4, for a given month, what is the probability that John's total income from these two jobs is less than $825? ANSWER: 0.0668 TYPE: PR DIFFICULTY: Difficult EXPLANATION: Total mean = $900, Total standard deviation = $50 KEYWORDS: normal distribution, probability 119. Referring to Table 6-4, the probability is 0.2 that John's total income from these two jobs in a given month is less than how much? ANSWER: $857.92 using Excel or $858 using Table E.2 TYPE: PR DIFFICULTY: Difficult KEYWORDS: normal distribution, value TABLE 6-5

The Normal Distribution and Other Continuous Distributions 177 Suppose the time interval between two consecutive defective light bulbs from a production line has a uniform distribution over an interval from 0 to 90 minutes. 120.

Referring to Table 6-5, what is the mean of the time interval?

ANSWER: 45 TYPE: PR DIFFICULTY: Easy KEYWORDS: uniform distribution, mean 121.

Referring to Table 6-5, what is the variance of the time interval?

ANSWER: 675 TYPE: PR DIFFICULTY: Easy KEYWORDS: uniform distribution, variance 122.

Referring to Table 6-5, what is the standard deviation of the time interval?

ANSWER: 25.9808 TYPE: PR DIFFICULTY: Easy KEYWORDS: uniform distribution, standard deviation 123. Referring to Table 6-5, what is the probability that the time interval between two consecutive defective light bulbs will be exactly 10 minutes? ANSWER: 0.0 TYPE: PR DIFFICULTY: Moderate KEYWORDS: uniform distribution, probability 124. Referring to Table 6-5, what is the probability that the time interval between two consecutive defective light bulbs will be less than 10 minutes? ANSWER: 0.1111 TYPE: PR DIFFICULTY: Easy KEYWORDS: uniform distribution, probability 125. Referring to Table 6-5, what is the probability that the time interval between two consecutive defective light bulbs will be between 10 and 20 minutes? ANSWER: 0.1111 TYPE: PR DIFFICULTY: Easy KEYWORDS: uniform distribution, probability 126. Referring to Table 6-5, what is the probability that the time interval between two consecutive defective light bulbs will be between 10 and 35 minutes?

178 The Normal Distribution and Other Continuous Distributions

ANSWER: 0.2778 TYPE: PR DIFFICULTY: Easy KEYWORDS: uniform distribution, probability 127. Referring to Table 6-5, what is the probability that the time interval between two consecutive defective light bulbs will be at least 50 minutes? ANSWER: 0.4444 TYPE: PR DIFFICULTY: Easy KEYWORDS: uniform distribution, probability 128. Referring to Table 6-5, what is the probability that the time interval between two consecutive defective light bulbs will be at least 80 minutes? ANSWER: 0.1111 TYPE: PR DIFFICULTY: Easy KEYWORDS: uniform distribution, probability 129. Referring to Table 6-5, what is the probability that the time interval between two consecutive defective light bulbs will be at least 90 minutes? ANSWER: 0.0 TYPE: PR DIFFICULTY: Moderate KEYWORDS: uniform distribution, probability 130. Referring to Table 6-5, the probability is 50% that the time interval between two consecutive defective light bulbs will fall between which two values that are the same distance from the mean? ANSWER: 22.5 and 67.5 TYPE: PR DIFFICULTY: Difficult KEYWORDS: uniform distribution, value 131. Referring to Table 6-5, the probability is 75% that the time interval between two consecutive defective light bulbs will fall between which two values that are the same distance from the mean? ANSWER: 11.25 and 78.75 TYPE: PR DIFFICULTY: Difficult KEYWORDS: uniform distribution, value

The Normal Distribution and Other Continuous Distributions 179 132. Referring to Table 6-5, the probability is 90% that the time interval between two consecutive defective light bulbs will fall between which two values that are the same distance from the mean? ANSWER: 4.5 and 85.5 TYPE: PR DIFFICULTY: Difficult KEYWORDS: uniform distribution, value TABLE 6-6 The interval between consecutive hits at a web site is assumed to follow an exponential distribution with an average of 40 hits per minute. 133.

Referring to Table 6-6, what is the average time between consecutive hits?

ANSWER: 0.025 minutes TYPE: PR DIFFICULTY: Easy KEYWORDS: exponential distribution, mean 134.Referring to Table 6-6, what is the probability that the next hit at the web site will occur within 10 seconds after just being hit by a visitor? ANSWER: 0.9987 TYPE: PR DIFFICULTY: Easy KEYWORDS: exponential distribution, probability 135.Referring to Table 6-6, what is the probability that the next hit at the web site will occur within no sooner than 5 seconds after just being hit by a visitor? ANSWER: 0.0357 TYPE: PR DIFFICULTY: Moderate KEYWORDS: exponential distribution, probability 136.Referring to Table 6-6, what is the probability that the next hit at the web site will occur between the next 1.2 and 1.5 seconds after just being hit by a visitor? ANSWER: 0.08145 TYPE: PR DIFFICULTY: Moderate KEYWORDS: exponential distribution, probability

137. True or False: One of the reasons that a correction for continuity adjustment is needed when approximating the binomial distribution with a normal distribution is because the normal distribution is used for a discrete random variable while the binomial distribution is used for a continuous random variable.

180 The Normal Distribution and Other Continuous Distributions

ANSWER: False TYPE: TF DIFFICULTY: Moderate KEYWORDS: approximation, normal distribution, binomial distribution, continuity adjustment

138. True or False: One of the reasons that a correction for continuity adjustment is needed when approximating the binomial distribution with a normal distribution is because the probability of getting a specific value of a random variable is zero with the normal distribution. ANSWER: True TYPE: TF DIFFICULTY: Moderate KEYWORDS: approximation, normal distribution, binomial distribution, continuity adjustment

139. True or False: One of the reasons that a correction for continuity adjustment is needed when approximating the binomial distribution with a normal distribution is because a random variable having a binomial distribution can have only a specified value while a random variable having a normal distribution can take on any values within an interval around that specified value. ANSWER: True TYPE: TF DIFFICULTY: Moderate KEYWORDS: approximation, normal distribution, binomial distribution, continuity adjustment

140. True or False: To determine the probability of getting fewer than 3 successes in a binomial distribution, you will find the area under the normal curve for X = 3.5 and below. ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: approximation, normal distribution, binomial distribution, continuity adjustment

141. True or False: To determine the probability of getting more than 3 successes in a binomial distribution, you will find the area under the normal curve for X = 3.5 and above. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: approximation, normal distribution, binomial distribution, continuity adjustment

142. True or False: To determine the probability of getting at least 3 successes in a binomial distribution, you will find the area under the normal curve for X = 2.5 and above. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: approximation, normal distribution, binomial distribution, continuity adjustment

143. True or False: To determine the probability of getting no more than 3 successes in a binomial distribution, you will find the area under the normal curve for X = 2.5 and below.

The Normal Distribution and Other Continuous Distributions 181

ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: approximation, normal distribution, binomial distribution, continuity adjustment

144. True or False: To determine the probability of getting between 3 and 4 successes in a binomial distribution, you will find the area under the normal curve between X = 3.5 and 4.5. ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: approximation, normal distribution, binomial distribution, continuity adjustment

145. True or False: To determine the probability of getting between 2 and 4 successes in a binomial distribution, you will find the area under the normal curve between X = 1.5 and 4.5. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: approximation, normal distribution, binomial distribution, continuity adjustment

146. True or False: As a general rule, one can use the normal distribution to approximate a binomial distribution whenever the sample size is at least 30. ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: approximation, normal distribution, binomial distribution, continuity adjustment

182 The Normal Distribution and Other Continuous Distributions

147. True or False: As a general rule, one can use the normal distribution to approximate a binomial distribution whenever the sample size is at least 15. ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: approximation, normal distribution, binomial distribution, continuity adjustment

148. True or False: As a general rule, one can use the normal distribution to approximate a binomial distribution whenever np is at least 5. ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: approximation, normal distribution, binomial distribution, continuity adjustment

149. True or False: As a general rule, one can use the normal distribution to approximate a binomial distribution whenever n(p-1) is at least 5. ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: approximation, normal distribution, binomial distribution, continuity adjustment

150. True or False: As a general rule, one can use the normal distribution to approximate a binomial distribution whenever n and n(p-1) are at least 5. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: approximation, normal distribution, binomial distribution, continuity adjustment TABLE 6-7 A company has 125 personal computers. The probability that any one of them will require repair on a given day is 0.15.

151. Referring to Table 6-7, which of the following is one of the properties required so that the binomial distribution can be used to compute the probability that no more than 2 computers will require repair on a given day? a) The probability that any one of the computers will require repair on a given day is constant. b) The probability that a computer that will require repair in the morning is the same as that in the afternoon. c) The number of computers that will require repair in the morning is independent of the number of computers that will require repair in the afternoon. d) The probability that two or more computers that will require repair in a given day approaches zero. ANSWER:

The Normal Distribution and Other Continuous Distributions 183 a TYPE: MC DIFFICULTY: Easy KEYWORDS: binomial distribution, properties

152. Referring to Table 6-7, which of the following is one of the properties required so that the binomial distribution can be used to compute the probability that no more than 2 computers will require repair on a given day? a) The probability that a computer that will require repair in the morning is the same as that in the afternoon. b) A randomly selected computer on a given day will either require a repair or will not. c) The number of computers that will require repair in the morning is independent of the number of computers that will require repair in the afternoon. d) The probability that two or more computers that will require repair in a given day approaches zero. ANSWER: b TYPE: MC DIFFICULTY: Easy KEYWORDS: binomial distribution, properties

153. Referring to Table 6-7, which of the following is one of the properties required so that the binomial distribution can be used to compute the probability that no more than 2 computers will require repair on a given day? a) The probability that a computer that will require repair in the morning is the same as that in the afternoon. b) The number of computers that will require repair in the morning is independent of the number of computers that will require repair in the afternoon. c) The probability that any one of the computers that will require repair on a given day will not affect or change the probability that any other computers that will require repair on the same day. d) The probability that two or more computers that will require repair in a given day approaches zero. ANSWER: c TYPE: MC DIFFICULTY: Easy KEYWORDS: binomial distribution, properties 154.Referring to Table 6-7 and assuming that the number of computers that requires repair on a given day follows a binomial distribution, compute the probability that there will be no more than 8 computers that require repair on a given day using a normal approximation. ANSWER: 0.0051 TYPE: PR DIFFICULTY: Easy KEYWORDS: approximation, normal distribution, binomial distribution, continuity adjustment 155.Referring to Table 6-7 and assuming that the number of computers that requires repair on a given day follows a binomial distribution, compute the probability that there will be less than 8 computers that require repair on a given day using a normal approximation.

184 The Normal Distribution and Other Continuous Distributions

ANSWER: 0.0024 TYPE: PR DIFFICULTY: Easy KEYWORDS: approximation, normal distribution, binomial distribution, continuity adjustment 156.Referring to Table 6-7 and assuming that the number of computers that requires repair on a given day follows a binomial distribution, compute the probability that there will be exactly 10 computers that requires repair on a given day using a normal approximation. ANSWER: 0.0091 using Excel or 0.0090 using Table E.2 TYPE: PR DIFFICULTY: Easy KEYWORDS: approximation, normal distribution, binomial distribution, continuity adjustment 157.Referring to Table 6-7 and assuming that the number of computers that requires repair on a given day follows a binomial distribution, compute the probability that there will be at least 25 computers that requires repair on a given day using a normal approximation. ANSWER: 0.0749 TYPE: PR DIFFICULTY: Easy KEYWORDS: approximation, normal distribution, binomial distribution, continuity adjustment 158.Referring to Table 6-7 and assuming that the number of computers that requires repair on a given day follows a binomial distribution, compute the probability that there will be more than 25 computers that requires repair on a given day using a normal approximation. ANSWER: 0.0454 TYPE: PR DIFFICULTY: Easy KEYWORDS: approximation, normal distribution, binomial distribution, continuity adjustment 159.Referring to Table 6-7 and assuming that the number of computers that requires repair on a given day follows a binomial distribution, compute the probability that there will be between 25 and 30 computers that requires repair on a given day using a normal approximation. ANSWER: 0.0733 TYPE: PR DIFFICULTY: Moderate KEYWORDS: approximation, normal distribution, binomial distribution, continuity adjustment 160.Referring to Table 6-7 and assuming that the number of computers that requires repair on a given day follows a binomial distribution, compute the probability that there will be more than 25 but less than 30 computers that requires repair on a given day using a normal approximation. ANSWER: 0.0419 TYPE: PR DIFFICULTY: Moderate KEYWORDS: approximation, normal distribution, binomial distribution, continuity adjustment

The Normal Distribution and Other Continuous Distributions 185

161.Referring to Table 6-7 and assuming that the number of computers that requires repair on a given day follows a binomial distribution, compute the probability that there will be less than 25 or more than 30 computers that requires repair on a given day using a normal approximation. ANSWER: 0.9267 TYPE: PR DIFFICULTY: Difficult KEYWORDS: approximation, normal distribution, binomial distribution, continuity adjustment

Sampling Distributions 187

CHAPTER 7: SAMPLING AND SAMPLING DISTRIBUTIONS 1. Sampling distributions describe the distribution of a) parameters. b) statistics. c) both parameters and statistics. d) neither parameters nor statistics. ANSWER: b TYPE: MC DIFFICULTY: Easy KEYWORDS: statistics, sampling distribution 2. The standard error of the mean a) is never larger than the standard deviation of the population. b) decreases as the sample size increases. c) measures the variability of the mean from sample to sample. d) All of the above. ANSWER: d TYPE: MC DIFFICULTY: Easy KEYWORDS: standard error, mean 3. The Central Limit Theorem is important in statistics because a) for a large n, it says the population is approximately normal. b) for any population, it says the sampling distribution of the sample mean is approximately normal, regardless of the sample size. c) for a large n, it says the sampling distribution of the sample mean is approximately normal, regardless of the shape of the population. d) for any sized sample, it says the sampling distribution of the sample mean is approximately normal. ANSWER: c TYPE: MC DIFFICULTY: Difficult KEYWORDS: central limit theorem

4. If the expected value of a sample statistic is equal to the parameter it is estimating, then we call that sample statistic a) unbiased. b) minimum variance. c) biased. d) random. ANSWER: a TYPE: MC DIFFICULTY: Moderate

188 Sampling Distributions KEYWORDS: unbiased 5. For air travelers, one of the biggest complaints is of the waiting time between when the airplane taxis away from the terminal until the flight takes off. This waiting time is known to have a skewed-right distribution with a mean of 10 minutes and a standard deviation of 8 minutes. Suppose 100 flights have been randomly sampled. Describe the sampling distribution of the mean waiting time between when the airplane taxis away from the terminal until the flight takes off for these 100 flights. a) Distribution is skewed-right with mean = 10 minutes and standard error = 0.8 minutes. b) Distribution is skewed-right with mean = 10 minutes and standard error = 8 minutes. c) Distribution is approximately normal with mean = 10 minutes and standard error = 0.8 minutes. d) Distribution is approximately normal with mean = 10 minutes and standard error = 8 minutes. ANSWER: c TYPE: MC DIFFICULTY: Moderate KEYWORDS: central limit theorem

6. Which of the following statements about the sampling distribution of the sample mean is incorrect? a) The sampling distribution of the sample mean is approximately normal whenever the sample size is sufficiently large (n  30 ). b) The sampling distribution of the sample mean is generated by repeatedly taking samples of size n and computing the sample means. c) The mean of the sampling distribution of the sample mean is equal to  . d) The standard deviation of the sampling distribution of the sample mean is equal to  . ANSWER: d TYPE: MC DIFFICULTY: Easy KEYWORDS: sampling distribution, properties

7. Which of the following is true about the sampling distribution of the sample mean? a) The mean of the sampling distribution is always  . b) The standard deviation of the sampling distribution is always  . c) The shape of the sampling distribution is always approximately normal. d) All of the above are true. ANSWER: a TYPE: MC DIFFICULTY: Moderate KEYWORDS: sampling distribution, properties 8. True or False: The amount of time it takes to complete an examination has a skewed-left distribution with a mean of 65 minutes and a standard deviation of 8 minutes. If 64 students were randomly sampled, the probability that the sample mean of the sampled students exceeds 71 minutes is approximately 0. ANSWER:

Sampling Distributions 189

True TYPE: TF DIFFICULTY: Moderate KEYWORDS: sampling distribution, central limit theorem

9. Suppose the ages of students in Statistics 101 follow a skewed-right distribution with a mean of 23 years and a standard deviation of 3 years. If we randomly sampled 100 students, which of the following statements about the sampling distribution of the sample mean age is incorrect? a) The mean of the sampling distribution is equal to 23 years. b) The standard deviation of the sampling distribution is equal to 3 years. c) The shape of the sampling distribution is approximately normal. d) The standard error of the sampling distribution is equal to 0.3 years. ANSWER: b TYPE: MC DIFFICULTY: Easy KEYWORDS: sampling distribution, central limit theorem 10. Why is the Central Limit Theorem so important to the study of sampling distributions? a) It allows us to disregard the size of the sample selected when the population is not normal. b) It allows us to disregard the shape of the sampling distribution when the size of the population is large. c) It allows us to disregard the size of the population we are sampling from. d) It allows us to disregard the shape of the population when n is large. ANSWER: d TYPE: MC DIFFICULTY: Moderate KEYWORDS: central limit theorem

11. A sample that does not provide a good representation of the population from which it was collected is referred to as a(n)

sample.

ANSWER: biased TYPE: FI DIFFICULTY: Moderate KEYWORDS: unbiased 12. True or False: The Central Limit Theorem is considered powerful in statistics because it works for any population distribution provided the sample size is sufficiently large and the population mean and standard deviation are known. ANSWER: True TYPE: TF DIFFICULTY: Moderate KEYWORDS: central limit theorem

13. Suppose a sample of n = 50 items is drawn from a population of manufactured products and the weight, X, of each item is recorded. Prior experience has shown that the weight has a probability distribution with  = 6 ounces and  = 2.5 ounces. Which of the following is true about the sampling distribution of the sample mean if a sample of size 15 is selected?

190 Sampling Distributions a) b) c) d)

The mean of the sampling distribution is 6 ounces. The standard deviation of the sampling distribution is 2.5 ounces. The shape of the sample distribution is approximately normal. All of the above are correct.

ANSWER: a TYPE: MC DIFFICULTY: Moderate KEYWORDS: sampling distribution, unbiased 14. The average score of all pro golfers for a particular course has a mean of 70 and a standard deviation of 3.0. Suppose 36 golfers played the course today. Find the probability that the average score of the 36 golfers exceeded 71. ANSWER: 0.0228 TYPE: PR DIFFICULTY: Moderate KEYWORDS: sampling distribution, mean, probability, central limit theorem

15. The distribution of the number of loaves of bread sold per day by a large bakery over the past 5 years has a mean of 7,750 and a standard deviation of 145 loaves. Suppose a random sample of n = 40 days has been selected. What is the approximate probability that the average number of loaves sold in the sampled days exceeds 7,895 loaves? ANSWER: Approximately 0 TYPE: PR DIFFICULTY: Moderate KEYWORDS: sampling distribution, mean, probability, central limit theorem 16. Sales prices of baseball cards from the 1960s are known to possess a skewed-right distribution with a mean sale price of $5.25 and a standard deviation of $2.80. Suppose a random sample of 100 cards from the 1960s is selected. Describe the sampling distribution for the sample mean sale price of the selected cards. a) skewed-right with a mean of $5.25 and a standard error of $2.80 b) normal with a mean of $5.25 and a standard error of $0.28 c) skewed-right with a mean of $5.25 and a standard error of $0.28 d) normal with a mean of $5.25 and a standard error of $2.80 ANSWER: b TYPE: MC DIFFICULTY: Easy KEYWORDS: sampling distribution, central limit theorem 17. Major league baseball salaries averaged $1.5 million with a standard deviation of $0.8 million in 1994. Suppose a sample of 100 major league players was taken. Find the approximate probability that the average salary of the 100 players exceeded $1 million. a) approximately 0 b) 0.2357 c) 0.7357 d) approximately 1

Sampling Distributions 191

ANSWER: d TYPE: MC DIFFICULTY: Moderate KEYWORDS: sampling distribution, mean, probability, central limit theorem 18. At a computer manufacturing company, the actual size of computer chips is normally distributed with a mean of 1 centimeter and a standard deviation of 0.1 centimeter. A random sample of 12 computer chips is taken. What is the standard error for the sample mean? a) 0.029 b) 0.050 c) 0.091 d) 0.120 ANSWER: a TYPE: MC DIFFICULTY: Easy KEYWORDS: standard error, mean 19. At a computer manufacturing company, the actual size of computer chips is normally distributed with a mean of 1 centimeter and a standard deviation of 0.1 centimeter. A random sample of 12 computer chips is taken. What is the probability that the sample mean will be between 0.99 and 1.01 centimeters? ANSWER: 0.2710 using Excel or 0.2736 using Table E.2 TYPE: PR DIFFICULTY: Moderate KEYWORDS: sampling distribution, mean, probability 20. At a computer manufacturing company, the actual size of computer chips is normally distributed with a mean of 1 centimeter and a standard deviation of 0.1 centimeter. A random sample of 12 computer chips is taken. What is the probability that the sample mean will be below 0.95 centimeters? ANSWER: 0.0416 using Excel or 0.0418 using Table E.2 TYPE: PR DIFFICULTY: Moderate KEYWORDS: sampling distribution, mean, probability 21. At a computer manufacturing company, the actual size of computer chips is normally distributed with a mean of 1 centimeter and a standard deviation of 0.1 centimeter. A random sample of 12 computer chips is taken. Above what value do 2.5% of the sample means fall? ANSWER: 1.057 TYPE: PR DIFFICULTY: Difficult KEYWORDS: sampling distribution, mean, value 22. The owner of a fish market has an assistant who has determined that the weights of catfish are normally distributed, with mean of 3.2 pounds and standard deviation of 0.8 pound. If a sample of 16 fish is taken, what would the standard error of the mean weight equal? a) 0.003

192 Sampling Distributions b) 0.050 c) 0.200 d) 0.800 ANSWER: c TYPE: MC DIFFICULTY: Easy KEYWORDS: standard error, mean

23. The owner of a fish market has an assistant who has determined that the weights of catfish are normally distributed, with mean of 3.2 pounds and standard deviation of 0.8 pound. If a sample of 25 fish yields a mean of 3.6 pounds, what is the Z-score for this observation? a) 18.750 b) 2.500 c) 1.875 d) 0.750 ANSWER: b TYPE: MC DIFFICULTY: Easy KEYWORDS: sampling distribution, mean

24. The owner of a fish market has an assistant who has determined that the weights of catfish are normally distributed, with mean of 3.2 pounds and standard deviation of 0.8 pound. If a sample of 64 fish yields a mean of 3.4 pounds, what is probability of obtaining a sample mean this large or larger? a) 0.0001 b) 0.0013 c) 0.0228 d) 0.4987 ANSWER: c TYPE: MC DIFFICULTY: Moderate KEYWORDS: sampling distribution, mean, probability 25. The owner of a fish market has an assistant who has determined that the weights of catfish are normally distributed, with mean of 3.2 pounds and standard deviation of 0.8 pound. What percentage of samples of 4 fish will have sample means between 3.0 and 4.0 pounds? a) 84% b) 67% c) 29% d) 16% ANSWER: b TYPE: MC DIFFICULTY: Moderate KEYWORDS: sampling distribution, mean, probability

Sampling Distributions 193

26. The use of the finite population correction factor when sampling without replacement from finite populations will a) increase the standard error of the mean. b) not affect the standard error of the mean. c) reduce the standard error of the mean. d) only affect the proportion, not the mean. ANSWER: c TYPE: MC DIFFICULTY: Easy KEYWORDS: finite population correction 27. For sample size 16, the sampling distribution of the mean will be approximately normally distributed a) regardless of the shape of the population. b) if the shape of the population is symmetrical. c) if the sample standard deviation is known. d) if the sample is normally distributed. ANSWER: b TYPE: MC DIFFICULTY: Moderate KEYWORDS: sampling distribution, mean, central limit theorem 28. The standard error of the mean for a sample of 100 is 30. In order to cut the standard error of the mean to 15, we would a) increase the sample size to 200. b) increase the sample size to 400. c) decrease the sample size to 50. d) decrease the sample to 25. ANSWER: b TYPE: MC DIFFICULTY: Moderate KEYWORDS: standard error, mean 29. Which of the following is true regarding the sampling distribution of the mean for a large sample size? a) It has the same shape, mean, and standard deviation as the population. b) It has a normal distribution with the same mean and standard deviation as the population. c) It has the same shape and mean as the population, but has a smaller standard deviation. d) It has a normal distribution with the same mean as the population but with a smaller standard deviation. ANSWER: d TYPE: MC DIFFICULTY: Moderate KEYWORDS: sampling distribution, mean, central limit theorem 30. For sample sizes greater than 30, the sampling distribution of the mean will be approximately normally distributed

194 Sampling Distributions a) b) c) d)

regardless of the shape of the population. only if the shape of the population is symmetrical. only if the standard deviation of the samples are known. only if the population is normally distributed.

ANSWER: a TYPE: MC DIFFICULTY: Easy KEYWORDS: sampling distribution, mean, central limit theorem 31. For sample size 1, the sampling distribution of the mean will be normally distributed a) regardless of the shape of the population. b) only if the shape of the population is symmetrical. c) only if the population values are positive. d) only if the population is normally distributed. ANSWER: d TYPE: MC DIFFICULTY: Moderate KEYWORDS: sampling distribution, mean, central limit theorem

32. The standard error of the population proportion will become larger a) as population proportion approaches 0. b) as population proportion approaches 0.50. c) as population proportion approaches 1.00. d) as the sample size increases. ANSWER: b TYPE: MC DIFFICULTY: Moderate KEYWORDS: standard error, proportion 33. True or False: As the sample size increases, the standard error of the mean increases. ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: standard error, mean

34. True or False: If the population distribution is symmetric, the sampling distribution of the mean can be approximated by the normal distribution if the samples contain 15 observations. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: sampling distribution, mean, central limit theorem 35. True or False: If the population distribution is unknown, in most cases the sampling distribution of the mean can be approximated by the normal distribution if the samples contain at least 30 observations.

Sampling Distributions 195

ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: sampling distribution, mean, central limit theorem 36. True or False: If the amount of gasoline purchased per car at a large service station has a population mean of $15 and a population standard deviation of $4, then 99.73% of all cars will purchase between $3 and $27. ANSWER: False TYPE: TF DIFFICULTY: Moderate KEYWORDS: sampling distribution, mean, probability 37. True or False: If the amount of gasoline purchased per car at a large service station has a population mean of $15 and a population standard deviation of $4 and a random sample of 4 cars is selected, there is approximately a 68.26% chance that the sample mean will be between $13 and $17. ANSWER: False TYPE: TF DIFFICULTY: Moderate EXPLANATION: The sample is too small for the normal approximation. KEYWORDS: sampling distribution, mean, probability 38. True or False: If the amount of gasoline purchased per car at a large service station has a population mean of $15 and a population standard deviation of $4 and it is assumed that the amount of gasoline purchased per car is symmetric, there is approximately a 68.26% chance that a random sample of 16 cars will have a sample mean between $14 and $16. ANSWER: True TYPE: TF DIFFICULTY: Moderate KEYWORDS: sampling distribution, mean, probability 39. True or False: If the amount of gasoline purchased per car at a large service station has a population mean of $15 and a population standard deviation of $4 and a random sample of 64 cars is selected, there is approximately a 95.44% chance that the sample mean will be between $14 and $16. ANSWER: True TYPE: TF DIFFICULTY: Difficult KEYWORDS: sampling distribution, mean, probability 40. True or False: As the sample size increases, the effect of an extreme value on the sample mean becomes smaller. ANSWER: True TYPE: TF DIFFICULTY: Moderate

196 Sampling Distributions KEYWORDS: sampling distribution, law of large numbers 41. True or False: If the population distribution is skewed, in most cases the sampling distribution of the mean can be approximated by the normal distribution if the samples contain at least 30 observations. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: sampling distribution, mean, central limit theorem 42. True or False: A sampling distribution is a distribution for a statistic. ANSWER: True TYPE: TF DIFFICULTY: Moderate KEYWORDS: sampling distribution

43. True or False: Suppose  = 50 and  2 = 100 for a population. In a sample where n = 100 is randomly taken, 95% of all possible sample means will fall between 48.04 and 51.96. ANSWER: True TYPE: TF DIFFICULTY: Moderate KEYWORDS: sampling distribution, mean, probability

44. True or False: Suppose  = 80 and  2 = 400 for a population. In a sample where n = 100 is randomly taken, 95% of all possible sample means will fall above 76.71. ANSWER: True TYPE: TF DIFFICULTY: Moderate KEYWORDS: sampling distribution, mean, probability

45. True or False: Suppose  = 50 and  2 = 100 for a population. In a sample where n = 100 is randomly taken, 90% of all possible sample means will fall between 49 and 51. ANSWER: False TYPE: TF DIFFICULTY: Moderate KEYWORDS: sampling distribution, mean, probability 46. True or False: The Central Limit Theorem ensures that the sampling distribution of the sample mean approaches normal as the sample size increases. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: central limit theorem

Sampling Distributions 197

47. True or False: The standard error of the mean is also known as the standard deviation of the sampling distribution of the sample mean. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: standard error, mean 48. True or False: A sampling distribution is defined as the probability distribution of possible sample sizes that can be observed from a given population. ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: sampling distribution 49. True or False: As the size of the sample is increased, the standard deviation of the sampling distribution of the sample mean for a normally distributed population will stay the same. ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: standard error, properties 50. True or False: For distributions such as the normal distribution, the arithmetic mean is considered more stable from sample to sample than other measures of central tendency. ANSWER: True TYPE: TF DIFFICULTY: Moderate KEYWORDS: sampling distribution, mean 51. True or False: The fact that the sample means are less variable than the population data can be observed from the standard error of the mean. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: sampling distribution, mean, standard error

52. The amount of pyridoxine (in grams) per multiple vitamin is normally distributed with  = 110 grams and  = 25 grams. A sample of 25 vitamins is to be selected. What is the probability that the sample mean will be between 100 and 120 grams? ANSWER: 0.9545 using Excel or 0.9544 using Table E.2 TYPE: PR DIFFICULTY: Moderate KEYWORDS: sampling distribution, mean, probability

198 Sampling Distributions

53. The amount of pyridoxine (in grams) per multiple vitamin is normally distributed with  = 110 grams and  = 25 grams. A sample of 25 vitamins is to be selected. What is the probability that the sample mean will be less than 100 grams? ANSWER: 0.0228 TYPE: PR DIFFICULTY: Moderate KEYWORDS: sampling distribution, mean, probability

54. The amount of pyridoxine (in grams) per multiple vitamin is normally distributed with  = 110 grams and  = 25 grams. A sample of 25 vitamins is to be selected. What is the probability that the sample mean will be greater than 100 grams? ANSWER: 0.9772 TYPE: PR DIFFICULTY: Moderate KEYWORDS: sampling distribution, mean, probability

55. The amount of pyridoxine (in grams) per multiple vitamin is normally distributed with  = 110 grams and  = 25 grams. A sample of 25 vitamins is to be selected. So, 95% of all sample means will be greater than how many grams? ANSWER: 101.7757 TYPE: PR DIFFICULTY: Difficult KEYWORDS: sampling distribution, mean, value

56. The amount of pyridoxine (in grams) per multiple vitamin is normally distributed with  = 110 grams and  = 25 grams. A sample of 25 vitamins is to be selected. So, the middle 70% of all sample means will fall between what two values? ANSWER: 104.8 and 115.2 TYPE: PR DIFFICULTY: Difficult KEYWORDS: sampling distribution, mean, value 57. The amount of time required for an oil and filter change on an automobile is normally distributed with a mean of 45 minutes and a standard deviation of 10 minutes. A random sample of 16 cars is selected. What would you expect the standard error of the mean to be? ANSWER: 2.5 minutes TYPE: PR DIFFICULTY: Easy KEYWORDS: standard error, mean 58. The amount of time required for an oil and filter change on an automobile is normally distributed with a mean of 45 minutes and a standard deviation of 10 minutes. A random sample of 16 cars is selected. What is the probability that the sample mean is between 45 and 52 minutes? ANSWER: 0.4974

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TYPE: PR DIFFICULTY: Moderate KEYWORDS: sampling distribution, mean, probability 59. The amount of time required for an oil and filter change on an automobile is normally distributed with a mean of 45 minutes and a standard deviation of 10 minutes. A random sample of 16 cars is selected. What is the probability that the sample mean will be between 39 and 48 minutes? ANSWER: 0.8767 TYPE: PR DIFFICULTY: Moderate KEYWORDS: sampling distribution, mean, probability

60. The amount of time required for an oil and filter change on an automobile is normally distributed with a mean of 45 minutes and a standard deviation of 10 minutes. A random sample of 16 cars is selected. So, 95% of all sample means will fall between what two values? ANSWER: 40.1 and 49.9 minutes TYPE: PR DIFFICULTY: Difficult KEYWORDS: sampling distribution, mean, probability 61. The amount of time required for an oil and filter change on an automobile is normally distributed with a mean of 45 minutes and a standard deviation of 10 minutes. A random sample of 16 cars is selected. So, 90% of the sample means will be greater than what value? ANSWER: 41.8 minutes TYPE: PR DIFFICULTY: Difficult KEYWORDS: sampling distribution, mean, probability

62. True or False: The amount of bleach a machine pours into bottles has a mean of 36 oz. with a standard deviation of 0.15 oz. Suppose we take a random sample of 36 bottles filled by this machine. The sampling distribution of the sample mean has a mean of 36 oz. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: sampling distribution, mean, unbiased 63. True or False: The amount of bleach a machine pours into bottles has a mean of 36 oz. with a standard deviation of 0.15 oz. Suppose we take a random sample of 36 bottles filled by this machine. The sampling distribution of the sample mean has a standard error of 0.15. ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: sampling distribution, mean, standard error 64. True or False: The amount of bleach a machine pours into bottles has a mean of 36 oz. with a standard deviation of 0.15 oz. Suppose we take a random sample of 36 bottles filled by this

200 Sampling Distributions machine. The sampling distribution of the sample mean will be approximately normal only if the population sampled is normal. ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: sampling distribution, mean, central limit theorem 65. The amount of bleach a machine pours into bottles has a mean of 36 oz. with a standard deviation of 0.15 oz. Suppose we take a random sample of 36 bottles filled by this machine. The probability that the mean of the sample exceeds 36.01 oz. is . ANSWER: 0.3446 TYPE: FI DIFFICULTY: Moderate KEYWORDS: sampling distribution, mean, probability, central limit theorem 66. The amount of bleach a machine pours into bottles has a mean of 36 oz. with a standard deviation of 0.15 oz. Suppose we take a random sample of 36 bottles filled by this machine. The probability that the mean of the sample is less than 36.03 is . ANSWER: 0.8849 TYPE: FI DIFFICULTY: Moderate KEYWORDS: sampling distribution, mean, probability, central limit theorem 67. The amount of bleach a machine pours into bottles has a mean of 36 oz. with a standard deviation of 0.15 oz. Suppose we take a random sample of 36 bottles filled by this machine. The probability that the mean of the sample is between 35.94 and 36.06 oz. is . ANSWER: 0.9836 TYPE: FI DIFFICULTY: Moderate KEYWORDS: sampling distribution, mean, probability, central limit theorem 68. The amount of bleach a machine pours into bottles has a mean of 36 oz. with a standard deviation of 0.15 oz. Suppose we take a random sample of 36 bottles filled by this machine. The probability that the mean of the sample is between 35.95 and 35.98 oz. is . ANSWER: 0.1891 TYPE: FI DIFFICULTY: Moderate KEYWORDS: sampling distribution, mean, probability, central limit theorem 69. The amount of bleach a machine pours into bottles has a mean of 36 oz. with a standard deviation of 0.15 oz. Suppose we take a random sample of 36 bottles filled by this machine. So, 95% of the sample means based on samples of size 36 will be between and . ANSWER: 35.951 and 36.049 ounces

Sampling Distributions 201

TYPE: FI DIFFICULTY: Difficult KEYWORDS: sampling distribution, mean, value, central limit theorem 70. A manufacturer of power tools claims that the average amount of time required to assemble their top-of-the-line table saw is 80 minutes with a standard deviation of 40 minutes. Suppose a random sample of 64 purchasers of this table saw is taken. The mean of the sampling distribution of the sample mean is minutes. ANSWER: 80 TYPE: FI DIFFICULTY: Easy KEYWORDS: sampling distribution, mean, unbiased 71. A manufacturer of power tools claims that the average amount of time required to assemble their top-of-the-line table saw is 80 minutes with a standard deviation of 40 minutes. Suppose a random sample of 64 purchasers of this table saw is taken. The standard deviation of the sampling distribution of the sample mean is minutes. ANSWER: 5 TYPE: FI DIFFICULTY: Easy KEYWORDS: sampling distribution, mean, standard error 72. A manufacturer of power tools claims that the average amount of time required to assemble their top-of-the-line table saw is 80 minutes with a standard deviation of 40 minutes. Suppose a random sample of 64 purchasers of this table saw is taken. The probability that the sample mean will be less than 82 minutes is . ANSWER: 0.6554 TYPE: FI DIFFICULTY: Moderate KEYWORDS: sampling distribution, mean, probability, central limit theorem 73. A manufacturer of power tools claims that the average amount of time required to assemble their top-of-the-line table saw is 80 minutes with a standard deviation of 40 minutes. Suppose a random sample of 64 purchasers of this table saw is taken. The probability that the sample mean will be between 77 and 89 minutes is . ANSWER: 0.6898 TYPE: FI DIFFICULTY: Moderate KEYWORDS: sampling distribution, mean, probability, central limit theorem 74. A manufacturer of power tools claims that the average amount of time required to assemble their top-of-the-line table saw is 80 minutes with a standard deviation of 40 minutes. Suppose a random sample of 64 purchasers of this table saw is taken. The probability that the sample mean will be greater than 88 minutes is . ANSWER: 0.0548 TYPE: FI DIFFICULTY: Moderate

202 Sampling Distributions KEYWORDS: sampling distribution, mean, probability, central limit theorem 75. A manufacturer of power tools claims that the average amount of time required to assemble their top-of-the-line table saw is 80 minutes with a standard deviation of 40 minutes. Suppose a random sample of 64 purchasers of this table saw is taken. So, 95% of the sample means based on samples of size 64 will be between and . ANSWER: 70.2 and 89.8 minutes TYPE: FI DIFFICULTY: Difficult KEYWORDS: sampling distribution, mean, value, central limit theorem 76. To use the normal distribution to approximate the binomial distribution, we need to be at least 5.

and

ANSWER: np and n(1-p) TYPE: FI DIFFICULTY: Easy KEYWORDS: sampling distribution, proportion, central limit theorem 77. True or False: The sample mean is an unbiased estimate of the population mean. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: mean, unbiased 78. True or False: The sample proportion is an unbiased estimate of the population proportion. ANSWER: True TYPE: TF DIFFICULTY: Difficult KEYWORDS: proportion, unbiased

79. True or False: The mean of the sampling distribution of a sample proportion is the population proportion,  . ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: sampling distribution, proportion

80. True or False: The standard error of the sampling distribution of a sample proportion is p ( 1− p ) where p is the sample proportion. n ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: sampling distribution, proportion

Sampling Distributions 203

81. True or False: The standard deviation of the sampling distribution of a sample proportion is  ( 1−  ) where  is the population proportion. n ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: sampling distribution, proportion 82. True or False: A sample of size 25 provides a sample variance of 400. The standard error, in this case equal to 4, is best described as the estimate of the standard deviation of means calculated from samples of size 25. ANSWER: True TYPE: TF DIFFICULTY: Moderate KEYWORDS: standard error 83. True or False: An unbiased estimator will have a value, on average across samples, equal to the population parameter value. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: unbiased 84. True or False: In inferential statistics, the standard error of the sample mean assesses the uncertainty or error of estimation. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: mean, standard error 85. Assume that house prices in a neighborhood are normally distributed with standard deviation $20,000. A random sample of 16 observations is taken. What is the probability that the sample mean differs from the population mean by more than $5,000? ANSWER: 0.3173 using Excel or 0.3174 using Table E.2 TYPE: PR DIFFICULTY: Moderate KEYWORDS: sampling distribution, mean, probability TABLE 7-1 Times spent studying by students in the week before final exams follow a normal distribution with standard deviation 8 hours. A random sample of 4 students was taken in order to estimate the mean study time for the population of all students.

204 Sampling Distributions 86. Referring to Table 7-1, what is the probability that the sample mean exceeds the population mean by more than 2 hours? ANSWER: 0.3085 TYPE: PR DIFFICULTY: Moderate KEYWORDS: sampling distribution, mean, probability 87. Referring to Table 7-1, what is the probability that the sample mean is more than 3 hours below the population mean? ANSWER: 0.2266 TYPE: PR DIFFICULTY: Moderate KEYWORDS: sampling distribution, mean, probability 88. Referring to Table 7-1, what is the probability that the sample mean differs from the population mean by less than 2 hours? ANSWER: 0.3829 using Excel or 0.3830 using Table E.2 TYPE: PR DIFFICULTY: Moderate KEYWORDS: sampling distribution, mean, probability 89. Referring to Table 7-1, what is the probability that the sample mean differs from the population mean by more than 3 hours? ANSWER: 0.4533 using Excel or 0.4532 using Table E.2 TYPE: PR DIFFICULTY: Moderate KEYWORDS: sampling distribution, mean, probability TABLE 7-2 The mean selling price of new homes in a city over a year was $115,000. The population standard deviation was $25,000. A random sample of 100 new home sales from this city was taken. 90. Referring to Table 7-2, what is the probability that the sample mean selling price was more than $110,000? ANSWER: 0.9772 TYPE: PR DIFFICULTY: Easy KEYWORDS: sampling distribution, mean, probability, central limit theorem 91. Referring to Table 7-2, what is the probability that the sample mean selling price was between $113,000 and $117,000? ANSWER: 0.5763 using Excel or 0.5762 using Table E.2 TYPE: PR DIFFICULTY: Easy

Sampling Distributions 205

KEYWORDS: sampling distribution, mean, probability, central limit theorem 92. Referring to Table 7-2, what is the probability that the sample mean selling price was between $114,000 and $116,000? ANSWER: 0.3108 TYPE: PR DIFFICULTY: Easy KEYWORDS: sampling distribution, mean, probability, central limit theorem 93. Referring to Table 7-2, without doing the calculations, state in which of the following ranges the sample mean selling price is most likely to lie? a) $113,000 -- $115,000 b) $114,000 -- $116,000 c) $115,000 -- $117,000 d) $116,000 -- $118,000 ANSWER: b TYPE: MC DIFFICULTY: Easy KEYWORDS: sampling distribution, mean, probability, central limit theorem TABLE 7-3 The lifetimes of a certain brand of light bulbs are known to be normally distributed with a mean of 1,600 hours and a standard deviation of 400 hours. A random sample of 64 of these light bulbs is taken. 94. Referring to Table 7-3, what is the probability that the sample mean lifetime is more than 1,550 hours? ANSWER: 0.8413 TYPE: PR DIFFICULTY: Easy KEYWORDS: sampling distribution, mean, probability 95. Referring to Table 7-3, the probability is 0.15 that the sample mean lifetime is more than how many hours? ANSWER: 1,651.82 hours using Excel or 1,652 hours using Table E.2 TYPE: PR DIFFICULTY: Moderate KEYWORDS: sampling distribution, mean, probability 96. Referring to Table 7-3, the probability is 0.20 that the sample mean lifetime differs from the population mean lifetime by at least how many hours? ANSWER: 64.08 hours using Excel or 64 hours using Table E.2 TYPE: PR DIFFICULTY: Difficult KEYWORDS: sampling distribution, mean, value

206 Sampling Distributions

TABLE 7-4 According to a survey, only 15% of customers who visited the web site of a major retail store made a purchase. Random samples of size 50 are selected. 97. Referring to Table 7-4, the average of all the sample proportions of customers who will make a purchase after visiting the web site is . ANSWER: 0.15 or 15% TYPE: FI DIFFICULTY: Moderate KEYWORDS: sampling distribution, proportion, mean 98. Referring to Table 7-4, the standard deviation of all the sample proportions of customers who will make a purchase after visiting the web site is . ANSWER: 0.05050 TYPE: FI DIFFICULTY: Moderate KEYWORDS: sampling distribution, proportion, standard error 99. True of False: Referring to Table 7-4, the requirements for using a normal distribution to approximate a binomial distribution is fulfilled. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: sampling distribution, proportion, central limit theorem 100. Referring to Table 7-4, what proportion of the samples will have between 20% and 30% of customers who will make a purchase after visiting the web site? ANSWER: 0.1596 TYPE: PR DIFFICULTY: Easy KEYWORDS: sampling distribution, proportion, probability 101. Referring to Table 7-4, what proportion of the samples will have less than 15% of customers who will make a purchase after visiting the web site? ANSWER: 0.5 TYPE: PR DIFFICULTY: Easy KEYWORDS: sampling distribution, proportion, probability 102. Referring to Table 7-4, what is the probability that a random sample of 50 will have at least 30% of customers who will make a purchase after visiting the web site? ANSWER: 0.0015

Sampling Distributions 207

TYPE: PR DIFFICULTY: Easy KEYWORDS: sampling distribution, proportion, probability 103. Referring to Table 7-4, 90% of the samples will have less than what percentage of customers who will make a purchase after visiting the web site? ANSWER: 21.47% TYPE: PR DIFFICULTY: Difficult KEYWORDS: sampling distribution, proportion, value 104. Referring to Table 7-4, 90% of the samples will have more than what percentage of customers who will make a purchase after visiting the web site? ANSWER: 8.528% using Excel or 8.536 using Table E.2 TYPE: PR DIFFICULTY: Difficult KEYWORDS: sampling distribution, proportion, value 105.A study at a college in the west coast reveals that, historically, 45% of their students are minority students. The expected percentage of minority students in their next batch of freshmen is . ANSWER: 45% TYPE: FI DIFFCULTY: Moderate KEYWORDS: sampling distribution, proportion, mean 106.A study at a college in the west coast reveals that, historically, 45% of their students are minority students. If random samples of size 75 are selected, the standard error of the proportions of students in the samples who are minority students is . ANSWER: 0.05745 TYPE: FI DIFFCULTY: Moderate KEYWORDS: sampling distribution, proportion, standard error 107.A study at a college in the west coast reveals that, historically, 45% of their students are that minority students. If a random sample of size 75 is selected, the probability is between 30% and 50% of the students in the sample will be minority students. ANSWER: 0.8034 using Excel or 0.8033 using Table E.2 TYPE: FI DIFFCULTY: Easy KEYWORDS: sampling distribution, proportion, probability 108.A study at a college in the west coast reveals that, historically, 45% of their students are minority students. If a random sample of size 75 is selected, the probability is that more than half of the students in the sample will be minority students. ANSWER:

208 Sampling Distributions 0.1920 using Excel or 0.1922 using Table E.2 TYPE: FI DIFFCULTY: Easy KEYWORDS: sampling distribution, proportion, probability 109.A study at a college in the west coast reveals that, historically, 45% of their students are minority students. If random samples of size 75 are selected, 80% of the samples will have less than % of minority students. ANSWER: 49.83 TYPE: FI DIFFCULTY: Difficult KEYWORDS: sampling distribution, proportion, value 110.A study at a college in the west coast reveals that, historically, 45% of their students are minority students. If random samples of size 75 are selected, 95% of the samples will have more than % of minority students. ANSWER: 35.55 TYPE: FI DIFFCULTY: Difficult KEYWORDS: sampling distribution, proportion, value 111. Which of the following is NOT a reason for the need for sampling? a) It is usually too costly to study the whole population. b) It is usually too time consuming to look at the whole population. c) It is sometimes destructive to observe the entire population. d) It is always more informative by investigating a sample than the entire population. ANSWER: d TYPE: MC DIFFICULTY: Moderate KEYWORDS: reasons for sampling 112. Which of the following is NOT a reason for drawing a sample? a) A sample is less time consuming than a census. b) A sample is less costly to administer than a census. c) A sample is usually not a good representation of the target population. d) A sample is less cumbersome and more practical to administer. ANSWER: c TYPE: MC DIFFICULTY: Easy KEYWORDS: reasons for sampling 113. Which of the following sampling methods is a probability sample? a) Chunk b) Quota sample c) Stratified sample d) Judgment sample ANSWER: c

Sampling Distributions 209 TYPE: MC DIFFICULTY: Easy KEYWORDS: probability sample 114.A sample of 300 subscribers to a particular magazine is selected from a population frame of 9,000 subscribers. If, upon examining the data, it is determined that no subscriber had been selected in the sample more than once, a) the sample could not have been random. b) the sample may have been selected without replacement or with replacement. c) the sample had to have been selected with replacement. d) the sample had to have been selected without replacement. ANSWER: b TYPE: MC DIFFICULTY: Moderate KEYWORDS: sampling method 115.For a population frame containing N = 1,007 individuals, what code number should you assign to the first person on the list in order to use a table of random numbers? a) 0 b) 1 c) 01 d) 0001 ANSWER: d TYPE: MC DIFFICULTY: Easy KEYWORDS: random number 116.Which of the following types of samples can you use if you want to make valid statistical inferences from a sample to a population? a) A judgment sample b) A quota sample c) A chunk d) A probability sample ANSWER: d TYPE: MC DIFFICULTY: Easy KEYWORDS: probability sample, sampling method 117.The evening host of a dinner dance reached into a bowl, mixed all the tickets around, and selected the ticket to award the grand door prize. What sampling method was used? a) Simple random sample b) Systematic sample c) Stratified sample d) Cluster sample ANSWER: a TYPE: MC DIFFICULTY: Easy KEYWORDS: simple random sample, probability sample, sampling method

210 Sampling Distributions 118.The Dean of Students mailed a survey to a total of 400 students. The sample included 100 students randomly selected from each of the freshman, sophomore, junior, and senior classes on campus last term. What sampling method was used? a) Simple random sample b) Systematic sample c) Stratified sample d) Cluster sample ANSWER: c TYPE: MC DIFFICULTY: Easy KEYWORDS: stratified sample, probability sample, sampling method 119.A telemarketer set the company’s computerized dialing system to contact every 25th person listed in the local telephone directory. What sampling method was used? a) Simple random sample b) Systematic sample c) Stratified sample d) Cluster sample ANSWER: b TYPE: MC DIFFICULTY: Easy KEYWORDS: systematic sample, probability sample, sampling method 120.Since a is not a randomly selected probability sample, there is no way to know how well it represents the overall population. a) Simple random sample b) Quota sample c) Stratified sample d) Cluster sample ANSWER: b TYPE: MC DIFFICULTY: Easy KEYWORDS: quota sample, nonprobability sample, sampling method 121.A population frame for a survey contains a listing of 72,345 names. Using a table of random numbers, how many digits will the code numbers for each member of your population contain? a) 3 b) 4 c) 5 d) 6 ANSWER: c TYPE: MC DIFFICULTY: Easy KEYWORDS: random number 122.A population frame for a survey contains a listing of 6,179 names. Using a table of random numbers, which of the following code numbers will appear on your list? a) 06 b) 0694 c) 6946

Sampling Distributions 211 d) 61790 ANSWER: b TYPE: MC DIFFICULTY: Easy KEYWORDS: random number 123.Which of the following can be reduced by proper interviewer training? a) Sampling error b) Measurement error c) Both of the above d) None of the above ANSWER: b TYPE: MC DIFFICULTY: Difficult KEYWORDS: measurement error, survey worthiness TABLE 7-5 The manager of the customer service division of a major consumer electronics company is interested in determining whether the customers who have purchased a videocassette recorder made by the company over the past 12 months are satisfied with their products.

124. Referring to Table 7-5, the manager decides to ask a sample of customers, who have bought a videocassette recorder made by the company and filed a complaint over the past year, to fill in a survey about whether they are satisfied with the product. This method will most likely suffer from a) non-response error. b) measurement error. c) coverage error. d) non-probability sampling. ANSWER: c TYPE: MC DIFFICULTY: Difficult KEYWORDS: coverage error, survey worthiness

125. Referring to Table 7-5, if there are 4 different brands of videocassette recorders made by the company, the best sampling strategy would be to use a a) a simple random sample. b) a stratified sample. c) a cluster sample. d) a systematic sample. ANSWER: b TYPE: MC DIFFICULTY: Difficult KEYWORDS: stratified sample, probability sample, sampling method

126. Referring to Table 7-5, which of the following questions in the survey will NOT likely induce a measurement error? a) How many times have you illegally copied copyrighted sporting events?

212 Sampling Distributions b) What is your exact annual income? c) How many times have you brought the videocassette recorder back for service? d) How many times have you failed to set the time on the videocassette recorder? ANSWER: c TYPE: MC DIFFICULTY: Difficult KEYWORDS: measurement error, survey worthiness

127. Referring to Table 7-5, if a customer survey questionnaire is included in all the videocassette recorders made and sold by the company over the past 12 months, this method of collecting data will most like suffer from a) nonresponse error. b) measurement error. c) coverage error. d) nonprobability sampling. ANSWER: a TYPE: MC DIFFICULTY: Difficult KEYWORDS: nonresponse error, survey worthiness 128.True or False: As a population becomes large, it is usually better to obtain statistical information from the entire population. ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: population, sample, reasons for samplings 129.True or False: If a simple random sample is chosen with replacement, each individual has the same chance of selection on every draw. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: simple random sample, probability sample, sampling method 130.True or False: When dealing with human surveys, we are usually interested in sampling with replacement. ANSWER: False TYPE: TF DIFFICULTY: Moderate KEYWORDS: sampling with replacement, sampling method, survey worthiness 131.True or False: The only reliable way a researcher can make statistical inferences from a sample to a population is to use nonprobability sampling methods. ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: nonprobability, probability sample, sampling method

Sampling Distributions 213 132.True or False: A sample is always a good representation of the target population. ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: sample, population, sampling method 133.True or False: There can be only one sample drawn from a population. ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: sample, sampling method 134.True or False: Using different frames to generate data can lead to totally different conclusions. ANSWER: True TYPE:TF DIFFICULTY: Easy KEYWORDS: frame, sampling method 135.True or False: Sampling error can be completely eliminated by taking larger sample sizes. ANSWER: False TYPE: TF DIFFICULTY: Difficult KEYWORDS: sampling error 136.True or False: Sampling error can be reduced by taking larger sample sizes. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: sampling error 137.True or False: Chunk sample is a type of probability sample. ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: chunk sample 138.True or False: Items or individuals in a judgment sample are chosen with regard to their probability of occurrence. ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: judgment sample, nonprobability sample 139.True or False: When participants are allowed to self-select into the sample, you have a nonprobability sample. ANSWER:

214 Sampling Distributions True TYPE: TF DIFFICULTY: Easy KEYWORDS: nonprobability sample 140. True or False: Systematic samples are less efficient than stratified sample. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: systematic sample, stratified sample 141.True or False: The professor of a business statistics class wanted to find out the average amount of time per week her students spent studying for the class. Among the 50 students in her class, 20% were freshmen, 50% were sophomores and 30% were juniors. She decided to draw 2 students randomly from the freshmen, 5 randomly from the sophomores and 3 randomly from the juniors. This is an example of a systematic sample. ANSWER: False TYPE: TF DIFFICULTY: Moderate KEYWORDS: stratified sample 142.

results from the exclusion of certain groups of subjects from a population frame.

ANSWER: Coverage error TYPE: FI DIFFICULTY: Difficult KEYWORDS: coverage error, survey worthiness 143. Coverage error results in a

ANSWER: selection bias TYPE: FI DIFFICULTY: Difficult KEYWORDS: selection bias, survey worthiness 144.

results from the failure to collect data on all subjects in the sample.

ANSWER: Nonresponse error or bias TYPE: FI DIFFICULTY: Moderate KEYWORDS: nonresponse error, survey worthiness 145. The sampling process begins by locating appropriate data sources called

ANSWER: frames TYPE: FI DIFFICULTY: Easy KEYWORDS: frames, sampling method

146. True or False: If you randomly select a student from the first row of a business statistics class and then every other 5 students thereafter until you get a sample of 20 students, this is an example of a chunk sample.

Sampling Distributions 215 ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: chunk sample

147. True or False: You stand at the main entrance to a departmental store and pick the first 20 customers that enter the store after it has opened its door for business on a single day. This is an example of a systematic sample. ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: systematic sample

148. True or False: An electronic appliance chain gathered customer opinions on their services using the customer feedback forms that are attached to the product registration forms. This is an example of a convenience sample. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: convenience sample

149. True or False: To gather opinions on the efficacy of U.S. foreign policies, a sample of 50 faculty members is selected from the pool of university professors who have taught political science at the graduate level. This is an example of a judgment sample. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: judgment sample

150. True or False: A sample is selected by including everybody who sits in the first row of a business statistics class. This is an example of a cluster sample. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: cluster sample

151. True or False: The question “How many times have you abused your spouse in the last 6 months?” will most likely result in nonresponse error.

ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: nonresponse error, survey worthiness

152. True or False: The question “Is your household income last year somewhere in between $25,000 and $35,000?” will most likely result in coverage error. ANSWER: False

216 Sampling Distributions TYPE: TF DIFFICULTY: Easy KEYWORDS: coverage error, survey worthiness

153. True or False: The only way one can eliminate sampling error is to take the whole population as the sample. ANSWER: True TYPE: TF DIFFICULTY: Moderte KEYWORDS: sampling error, survey worthiness

154. True or False: Coverage error can become an ethical issue if a particular group is intentionally excluded from the frame. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: ethical issue, coverage error, survey worthiness

155. True or False: Measurement error will become an ethical issue when the findings are presented without reference to sample size and margin of error. ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: ethical issue, measurement error, sampling error, survey worthiness 156. Which of the following sampling methods will more likely be susceptible to ethical violation? a) Simple random sample b) Cluster sample c) Convenience sample d) Stratified sample ANSWER: c TYPE: MC DIFFICULTY: Easy KEYWORDS: ethical issues, sampling method 157.True or False: The professor of a business statistics class wanted to find out the average amount of time per week her students spent studying for the class. She divided the students into the left, right and center groups according to the location they sat in the class that day. One of these 3 groups was randomly selected and everyone in the group was asked the average amount of time per week he/she spent studying for the class. This is an example of a cluster sample. ANSWER: True TYPE: TF DIFFICULTY: Moderate KEYWORDS: cluster sample 158.True or False: The professor of a business statistics class wanted to find out the average amount of time per week her students spent studying for the class. She divided the fifty students on her roster into ten groups starting from the first student on the roster. The first student was randomly selected from the first group. Then every tenth student was selected from the remaining students. This is an example of a cluster sample.

Sampling Distributions 217

ANSWER: False TYPE: TF DIFFICULTY: Moderate KEYWORDS: systematic sample 159.True or False: Selection bias occurs more frequently in systematic samples than in simple random samples. ANSWER: True TYPE: TF DIFFICULTY: easy KEYWORDS: simple random sample, systematic sample

160. True or False: The question: “Have you used any form of illicit drugs over the past 2 months?” will most likely result in measurement error if the question is answered. ANSWER: True TYPE: TF DIFFICULTY: Moderate KEYWORDS: measurement error, survey worthiness 161.True or False: The question: “How much did you make last year rounded to the nearest hundreds of dollars?” will most likely result in measurement error. ANSWER: True TYPE: TF DIFFICULTY: Moderate KEYWORDS: measurement error, survey worthiness TABLE 7-6 According to an article, 19% of the entire U.S. population have high-speed access to the Internet. Random samples of size 200 are selected from the U.S. population.

162. Referring to Table 7-6, the population mean of all the sample proportions is

ANSWER: 19% or 0.19 TYPE: FI DIFFICULTY: Easy KEYWORDS: mean, sampling distribution, proportion, central limit theorem

163. Referring to Table 7-6, the standard error of all the sample proportions is

ANSWER: 0.0277 TYPE: FI DIFFICULTY: Easy KEYWORDS: standard error, sampling distribution, proportion, central limit theorem

164. Referring to Table 7-6, among all the random samples of size 200, between 14% and 24% who have high-speed access to the Internet.

% will have

218 Sampling Distributions ANSWER: 92.85 using Excel or 92.82 using Table E.2 TYPE: FI DIFFICULTY: Easy KEYWORDS: sampling distribution, proportion, central limit theorem

165. Referring to Table 7-6, among all the random samples of size 200,

% will have

between 9% and 29% who have high-speed access to the Internet. ANSWER: 99.97 TYPE: FI DIFFICULTY: Easy KEYWORDS: sampling distribution, proportion, central limit theorem

166. Referring to Table 7-6, among all the random samples of size 200,

% will have more

than 30% who have high-speed access to the Internet. ANSWER: 0.0000 or virtually zero TYPE: FI DIFFICULTY: Easy KEYWORDS: sampling distribution, proportion, central limit theorem

167. Referring to Table 7-6, among all the random samples of size 200,

% will have less

than 20% who have high-speed access to the Internet. ANSWER: 64.08 using Excel or 64.06 using Table E.2 TYPE: FI DIFFICULTY: Easy KEYWORDS: sampling distribution, proportion, central limit theorem 168. Referring to Table 7-6, among all the random samples of size 200, 90 % will have less than % who have high-speed access to the Internet. ANSWER: 22.56 using Excel or 22.55 using Table E.2 TYPE: FI DIFFICULTY: Moderate KEYWORDS: sampling distribution, proportion, central limit theorem

169. Referring to Table 7-6, among all the random samples of size 200, 90 % will have more than % who have high-speed access to the Internet. ANSWER: 15.45 TYPE: FI DIFFICULTY: Moderate KEYWORDS: sampling distribution, proportion, central limit theorem TABLE 7-7 Online customer service is a key element to successful online retailing. According to a marketing survey, 37.5% of online customers take advantage of the online customer service. Random samples of 200 customers are selected.

170. Referring to Table 7-7, the population mean of all possible sample proportions is

Sampling Distributions 219

ANSWER: 0.375 or 37.5% TYPE: FI DIFFICULTY: Easy KEYWORDS: mean, sampling distribution, proportion, central limit theorem

171. Referring to Table 7-7, the standard error of all possible sample proportions is

ANSWER: 0.0342 TYPE: FI DIFFICULTY: Easy KEYWORDS: standard error, sampling distribution, proportion, central limit theorem

172. Referring to Table 7-7,

% of the samples are likely to have between 35% and 40% who take advantage of online customer service.

ANSWER: 53.48 using Excel or 53.46 using Table E.2 TYPE: FI DIFFICULTY: Easy KEYWORDS: sampling distribution, proportion, central limit theorem

173. Referring to Table 7-7,

% of the samples are likely to have less than 37.5% who take advantage of online customer service.

ANSWER: 50 TYPE: FI DIFFICULTY: Easy KEYWORDS: sampling distribution, proportion, central limit theorem

174. Referring to Table 7-7, 90% of the samples proportions symmetrically around the population proportion will have between online customer service.

% and

% of the customers who take advantage of

ANSWER: 31.87 and 43.13 TYPE: FI DIFFICULTY: Moderate KEYWORDS: sampling distribution, proportion, central limit theorem

175. Referring to Table 7-7, 95% of the samples proportions symmetrically around the population proportion will have between online customer service.

% and

% of the customers who take advantage of

ANSWER: 30.79 and 44.21 TYPE: FI DIFFICULTY: Moderate KEYWORDS: sampling distribution, proportion, central limit theorem

224 Confidence Interval Estimation

CHAPTER 8: CONFIDENCE INTERVAL ESTIMATION 1. The width of a confidence interval estimate for a proportion will be a) narrower for 99% confidence than for 95% confidence. b) wider for a sample size of 100 than for a sample size of 50. c) narrower for 90% confidence than for 95% confidence. d) narrower when the sample proportion is 0.50 than when the sample proportion is 0.20. ANSWER: c TYPE: MC DIFFICULTY: Moderate KEYWORDS: confidence interval, proportion, properties, width

2. When determining the sample size for a proportion for a given level of confidence and sampling error, the closer to 0.50 that p is estimated to be, the sample size required a) is smaller b) is larger c) is not affected d) can be smaller, larger or unaffected

ANSWER: b TYPE: MC DIFFICULTY: Moderate KEYWORDS: confidence interval, proportion, properties 3. A 99% confidence interval estimate can be interpreted to mean that a) if all possible samples are taken and confidence interval estimates are developed, 99% of them would include the true population mean somewhere within their interval. b) we have 99% confidence that we have selected a sample whose interval does include the population mean. c) Both of the above. d) None of the above. ANSWER: c TYPE: MC DIFFICULTY: Moderate KEYWORDS: confidence interval, interpretation

4. If you were constructing a 99% confidence interval of the population mean based on a sample of n=25 where the standard deviation of the sample s = 0.05, the critical value of t will be a) 2.7969 b) 2.7874 c) 2.4922 d) 2.4851 ANSWER: a TYPE: MC DIFFICULTY: Easy KEYWORDS: critical value, t distribution

5. Which of the following is not true about the Student’s t distribution? a) It has more area in the tails and less in the center than does the normal distribution.

225 Confidence Interval Estimation b) It is used to construct confidence intervals for the population mean when the population standard deviation is known. c) It is bell shaped and symmetrical. d) As the number of degrees of freedom increases, the t distribution approaches the normal distribution. ANSWER: b TYPE: MC DIFFICULTY: Easy KEYWORDS: t distribution, properties

6. True or False: The t distribution is used to construct confidence intervals for the population mean when the population standard deviation is unknown. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: confidence interval, mean, standard deviation unknown

7. The t distribution a) b) c) d)

assumes the population is normally distributed. approaches the normal distribution as the sample size increases. has more area in the tails than does the normal distribution. All of the above.

ANSWER: d TYPE: MC DIFFICULTY: Easy KEYWORDS: t distribution, properties 8. It is desired to estimate the average total compensation of CEOs in the Service industry. Data were randomly collected from 18 CEOs and the 97% confidence interval was calculated to be ($2,181,260, $5,836,180). Which of the following interpretations is correct? a) 97% of the sampled total compensation values fell between $2,181,260 and $5,836,180. b) We are 97% confident that the mean of the sampled CEOs falls in the interval $2,181,260 to $5,836,180. c) In the population of Service industry CEOs, 97% of them will have total compensations that fall in the interval $2,181,260 to $5,836,180. d) We are 97% confident that the average total compensation of all CEOs in the Service industry falls in the interval $2,181,260 to $5,836,180. ANSWER: d TYPE: MC DIFFICULTY: Difficult KEYWORDS: confidence interval, interpretation

9. It is desired to estimate the average total compensation of CEOs in the Service industry. Data were randomly collected from 18 CEOs and the 97% confidence interval was calculated to be ($2,181,260, $5,836,180). Based on the interval above, do you believe the average total compensation of CEOs in the Service industry is more than $3,000,000? a) Yes, and I am 97% confident of it. b) Yes, and I am 78% confident of it. c) I am 97% confident that the average compensation is $3,000,000.

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d) I cannot conclude that the average exceeds $3,000,000 at the 97% confidence level. ANSWER: d TYPE: MC DIFFICULTY: Difficult KEYWORDS: confidence interval, interpretation 10. A confidence interval was used to estimate the proportion of statistics students that are females. A random sample of 72 statistics students generated the following 90% confidence interval: (0.438, 0.642). Based on the interval above, is the population proportion of females equal to 0.60? a) No, and we are 90% sure of it. b) No. The proportion is 54.17%. c) Maybe. 0.60 is a believable value of the population proportion based on the information above. d) Yes, and we are 90% sure of it. ANSWER: c TYPE: MC DIFFICULTY: Moderate KEYWORDS: confidence interval, proportion, testing

11. A confidence interval was used to estimate the proportion of statistics students that are female. A random sample of 72 statistics students generated the following 90% confidence interval: (0.438, 0.642). Using the information above, what total size sample would be necessary if we wanted to estimate the true proportion to within 0.08 using 95% confidence? a) 105 b) 150 c) 420 d) 597 ANSWER: b TYPE: MC DIFFICULTY: Moderate KEYWORDS: proportion, sample size determination

12. When determining the sample size necessary for estimating the true population mean, which factor is not considered when sampling with replacement? a) The population size. b) The population standard deviation. c) The level of confidence desired in the estimate. d) The allowable or tolerable sampling error. ANSWER: a TYPE: MC DIFFICULTY: Easy KEYWORDS: mean, sample size determination

13. Suppose a 95% confidence interval for  turns out to be (1,000, 2,100). Give a definition of

what it means to be “95% confident” in an inference. a) In repeated sampling, the population parameter would fall in the given interval 95% of the time.

227 Confidence Interval Estimation b) In repeated sampling, 95% of the intervals constructed would contain the population mean. c) 95% of the observations in the entire population fall in the given interval. d) 95% of the observations in the sample fall in the given interval. ANSWER: b TYPE: MC DIFFICULTY: Moderate KEYWORDS: confidence interval, interpretation

14. Suppose a 95% confidence interval for  turns out to be (1,000, 2,100). To make more useful inferences from the data, it is desired to reduce the width of the confidence interval. Which of the following will result in a reduced interval width? a) Increase the sample size. b) Decrease the confidence level. c) Both increase the sample size and decrease the confidence level. d) Both increase the confidence level and decrease the sample size. ANSWER: c TYPE: MC DIFFICULTY: Moderate KEYWORDS: confidence interval, properties, width

15. Suppose a 95% confidence interval for  has been constructed. If it is decided to take a larger sample and to decrease the confidence level of the interval, then the resulting interval width would . (Assume that the sample statistics gathered would not change very much for the new sample.) a) be larger than the current interval width b) be narrower than the current interval width c) be the same as the current interval width d) be unknown until actual sample sizes and reliability levels were determined ANSWER: b TYPE: MC DIFFICULTY: Moderate KEYWORDS: confidence interval, properties, width

16. In the construction of confidence intervals, if all other quantities are unchanged, an increase in the sample size will lead to a a) narrower b) wider c) less significant d) biased

interval.

ANSWER: a TYPE: MC DIFFICULTY: Easy KEYWORDS: confidence interval, properties, width

17. A major department store chain is interested in estimating the average amount its credit card customers spent on their first visit to the chain’s new store in the mall. Fifteen credit card accounts were randomly sampled and analyzed with the following results: X = $50.50 and s 2 = 400 . Assuming the distribution of the amount spent on their first visit is approximately

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normal, what is the shape of the sampling distribution of the sample mean that will be used to create the desired confidence interval for  ? a) Approximately normal with a mean of $50.50 b) A standard normal distribution c) A t distribution with 15 degrees of freedom d) A t distribution with 14 degrees of freedom ANSWER: d TYPE: MC DIFFICULTY: Easy KEYWORDS: confidence interval, mean, t distribution

18. A major department store chain is interested in estimating the average amount its credit card

customers spent on their first visit to the chain’s new store in the mall. Fifteen credit card accounts were randomly sampled and analyzed with the following results: X = $50.50 and

s 2 = 400 . Construct a 95% confidence interval for the average amount its credit card customers

spent on their first visit to the chain’s new store in the mall assuming that the amount spent follows a normal distribution. a) $50.50  $9.09 b) $50.50  $10.12 c) $50.50  $11.00 d) $50.50  $11.08 ANSWER: d TYPE: MC DIFFICULTY: Easy KEYWORDS: confidence interval, mean, t distribution

19. Private colleges and universities rely on money contributed by individuals and corporations for their operating expenses. Much of this money is put into a fund called an endowment, and the college spends only the interest earned by the fund. A recent survey of 8 private colleges in the United States revealed the following endowments (in millions of dollars): 60.2, 47.0, 235.1, 490.0, 122.6, 177.5, 95.4, and 220.0. What value will be used as the point estimate for the mean endowment of all private colleges in the United States? a) $1,447.8 b) $180.975 c) $143.042 d) $8 ANSWER: b TYPE: MC DIFFICULTY: Easy KEYWORDS: point estimate, mean

20. Private colleges and universities rely on money contributed by individuals and corporations for their operating expenses. Much of this money is put into a fund called an endowment, and the college spends only the interest earned by the fund. A recent survey of 8 private colleges in the United States revealed the following endowments (in millions of dollars): 60.2, 47.0, 235.1, 490.0, 122.6, 177.5, 95.4, and 220.0. Summary statistics yield X = 180.975 and s = 143.042 . Calculate a 95% confidence interval for the mean endowment of all the private colleges in the United States assuming a normal distribution for the endowments. a) $180.975  $94.066

229 Confidence Interval Estimation

b) $180.975  $99.123 c) $180.975  $116.621 d) $180.975  $119.586 ANSWER: d TYPE: MC DIFFICULTY: Moderate KEYWORDS: confidence interval, mean, t distribution

21. A university system enrolling hundreds of thousands of students is considering a change in the way students pay for their education. Currently, the students pay $55 per credit hour. The university system administrators are contemplating charging each student a set fee of $750 per quarter, regardless of how many credit hours each takes. To see if this proposal would be economically feasible, the administrators would like to know how many credit hours, on the average, each student takes per quarter. A random sample of 250 students yields a mean of 14.1 credit hours per quarter and a standard deviation of 2.3 credit hours per quarter. Suppose the administration wanted to estimate the mean to within 0.1 hours at 95% reliability and assumed that the sample standard deviation provided a good estimate for the population standard deviation. How large a total sample would they need to take? ANSWER: n = 2033 TYPE: PR DIFFICULTY: Easy KEYWORDS: mean, sample size determination

22. As an aid to the establishment of personnel requirements, the director of a hospital wishes to estimate the mean number of people who are admitted to the emergency room during a 24-hour period. The director randomly selects 64 different 24-hour periods and determines the number of admissions for each. For this sample, X = 19.8 and s2 = 25. Which of the following assumptions is necessary in order for a confidence interval to be valid? a) The population sampled from has an approximate normal distribution. b) The population sampled from has an approximate t distribution. c) The mean of the sample equals the mean of the population. d) None of these assumptions are necessary. ANSWER: d TYPE: MC DIFFICULTY: Moderate KEYWORDS: confidence interval, mean, t distribution

23. As an aid to the establishment of personnel requirements, the director of a hospital wishes to estimate the mean number of people who are admitted to the emergency room during a 24-hour period. The director randomly selects 64 different 24-hour periods and determines the number of admissions for each. For this sample, X = 19.8 and s2 = 25. Estimate the mean number of admissions per 24-hour period with a 95% confidence interval. ANSWER: 19.8  1.249 TYPE: PR DIFFICULTY: Moderate KEYWORDS: confidence interval, mean, t distribution

24. As an aid to the establishment of personnel requirements, the director of a hospital wishes to estimate the mean number of people who are admitted to the emergency room during a 24-hour

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period. The director randomly selects 64 different 24-hour periods and determines the number of admissions for each. For this sample, X = 19.8 and s2 = 25. If the director wishes to estimate the mean number of admissions per 24-hour period to within 1 admission with 99% reliability, what size sample should she choose? ANSWER: n = 166 TYPE: PR DIFFICULTY: Moderate KEYWORDS: mean, sample size determination 25. A university dean is interested in determining the proportion of students who receive some sort of financial aid. Rather than examine the records for all students, the dean randomly selects 200 students and finds that 118 of them are receiving financial aid. Use a 90% confidence interval to estimate the true proportion of students who receive financial aid. ANSWER: 0.59  0.057 or 0.533    0.647 TYPE: PR DIFFICULTY: Moderate KEYWORDS: confidence interval, proportion

26. A university dean is interested in determining the proportion of students who receive some sort of financial aid. Rather than examine the records for all students, the dean randomly selects 200 students and finds that 118 of them are receiving financial aid. The 95% confidence interval for p is 0.59  0.07. Interpret this interval. a) We are 95% confident that the true proportion of all students receiving financial aid is between 0.52 and 0.66. b) 95% of the students get between 52% and 66% of their tuition paid for by financial aid. c) We are 95% confident that between 52% and 66% of the sampled students receive some sort of financial aid. d) We are 95% confident that 59% of the students are on some sort of financial aid. ANSWER: a TYPE: MC DIFFICULTY: Moderate KEYWORDS: confidence interval, proportion, interpretation

231 Confidence Interval Estimation 27. A university dean is interested in determining the proportion of students who receive some sort of financial aid. Rather than examine the records for all students, the dean randomly selects 200 students and finds that 118 of them are receiving financial aid. If the dean wanted to estimate the proportion of all students receiving financial aid to within 3% with 99% reliability, how many students would need to be sampled? a) n = 1,844 b) n = 1,784 c) n = 1,503 d) n = 1,435 ANSWER: b TYPE: MC DIFFICULTY: Moderate KEYWORDS: proportion, sample size determination 28. An economist is interested in studying the incomes of consumers in a particular region. The population standard deviation is known to be $1,000. A random sample of 50 individuals resulted in an average income of $15,000. What is the upper end point in a 99% confidence interval for the average income? a) $15,052 b) $15,141 c) $15,330 d) $15,364 ANSWER: d TYPE: MC DIFFICULTY: Easy KEYWORDS: confidence interval, mean, standardized normal distribution 29. An economist is interested in studying the incomes of consumers in a particular region. The population standard deviation is known to be $1,000. A random sample of 50 individuals resulted in an average income of $15,000. What is the width of the 90% confidence interval? a) $232.60 b) $364.30 c) $465.23 d) $728.60 ANSWER: c TYPE: MC DIFFICULTY: Moderate KEYWORDS: width, confidence interval, mean, standardized normal distribution

30. An economist is interested in studying the incomes of consumers in a particular region. The population standard deviation is known to be $1,000. A random sample of 50 individuals resulted in an average income of $15,000. What total sample size would the economist need to use for a 95% confidence interval if the width of the interval should not be more than $100? a) n = 1537 b) n = 385 c) n = 40 d) n = 20 ANSWER: a

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TYPE: MC DIFFICULTY: Easy KEYWORDS: mean, sample size determination 31. The head librarian at the Library of Congress has asked her assistant for an interval estimate of the mean number of books checked out each day. The assistant provides the following interval estimate: from 740 to 920 books per day. What is an efficient, unbiased point estimate of the number of books checked out each day at the Library of Congress? a) 740 b) 830 c) 920 d) 1,660 ANSWER: b TYPE: MC DIFFICULTY: Easy KEYWORDS: point estimate, mean 32. The head librarian at the Library of Congress has asked her assistant for an interval estimate of the mean number of books checked out each day. The assistant provides the following interval estimate: from 740 to 920 books per day. If the head librarian knows that the population standard deviation is 150 books checked out per day, approximately how large a sample did her assistant use to determine the interval estimate? a) 2 b) 3 c) 12 d) It cannot be determined from the information given. ANSWER: d TYPE: MC DIFFICULTY: Difficult KEYWORDS: mean, sample size determination 33. The head librarian at the Library of Congress has asked her assistant for an interval estimate of the mean number of books checked out each day. The assistant provides the following interval estimate: from 740 to 920 books per day. If the head librarian knows that the population standard deviation is 150 books checked out per day, and she asked her assistant for a 95% confidence interval, approximately how large a sample did her assistant use to determine the interval estimate? a) 125 b) 13 c) 11 d) 4 ANSWER: c TYPE: MC DIFFICULTY: Difficult KEYWORDS: mean, sample size determination 34. The head librarian at the Library of Congress has asked her assistant for an interval estimate of the mean number of books checked out each day. The assistant provides the following interval estimate: from 740 to 920 books per day. If the head librarian knows that the population standard deviation is 150 books checked out per day, and she asked her assistant to use 25 days of data to construct the interval estimate, what confidence level can she attach to the interval estimate? a) 99.7%

233 Confidence Interval Estimation b) 99.0% c) 98.0% d) 95.4% ANSWER: a TYPE: MC DIFFICULTY: Difficult KEYWORDS: mean, sample size determination 35. True or False: A race car driver tested his car for time from 0 to 60 mph, and in 20 tests obtained an average of 4.85 seconds with a standard deviation of 1.47 seconds. A 95% confidence interval for the 0 to 60 time is 4.52 seconds to 5.18 seconds. ANSWER: False TYPE: TF DIFFICULTY: Moderate KEYWORDS: confidence interval, mean, t distribution

36. True or False: Given a sample mean of 2.1 and a sample standard deviation of 0.7 from a sample of 10 data points, a 90% confidence interval will have a width of 2.36. ANSWER: False TYPE: TF DIFFICULTY: Moderate KEYWORDS: confidence interval, mean, t distribution

37. True or False: Given a sample mean of 2.1 and a population standard deviation of 0.7 from a sample of 10 data points, a 90% confidence interval will have a width of 2.36. ANSWER: False TYPE: TF DIFFICULTY: Moderate KEYWORDS: confidence interval, mean, standardized normal distribution 38. True or False: A sample size of 5 provides a sample mean of 9.6. If the population variance is known to be 5 and the population distribution is assumed to be normal, the lower limit for a 92% confidence interval is 7.85. ANSWER: True TYPE: TF DIFFICULTY: Moderate KEYWORDS: confidence interval, mean, standardized normal distribution

39. True or False: A random sample of 50 provides a sample mean of 31 with a standard deviation of s=14. The upper bound of a 90% confidence interval estimate of the population mean is 34.32. ANSWER: True TYPE: TF DIFFICULTY: Moderate KEYWORDS: confidence interval, mean, t distribution

40. True or False: In forming a 90% confidence interval for a population mean from a sample size of 22, the number of degrees of freedom from the t distribution equals 22.

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ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: confidence interval, mean, t distribution 41. True or False: Other things being equal, as the confidence level for a confidence interval increases, the width of the interval increases. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: confidence interval, properties

42. True or False: The t distribution allows the calculation of confidence intervals for means when the actual standard deviation is not known. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: confidence interval, mean, t distribution

43. True or False: The t distribution allows the calculation of confidence intervals for means for small samples when the population variance is not known, regardless of the shape of the distribution in the population. ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: confidence interval, mean, t distribution

44. True or False: For a t distribution with 12 degrees of freedom, the area between – 2.6810 and 2.1788 is 0.980. ANSWER: False TYPE: TF DIFFICULTY: Moderate KEYWORDS: t distribution

45. True or False: A sample of 100 fuses from a very large shipment is found to have 10 that are defective. The 95% confidence interval would indicate that, for this shipment, the proportion of defective fuses is between 0 and 0.28. ANSWER: False TYPE: TF DIFFICULTY: Moderate KEYWORDS: confidence interval, proportion 46. True or False: The sample mean is a point estimate of the population mean. ANSWER: True TYPE: TF DIFFICULTY: Easy

235 Confidence Interval Estimation KEYWORDS: point estimate, mean 47. True or False: The confidence interval estimate of the population mean is constructed around the sample mean. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: confidence interval, mean 48. True or False: The confidence interval estimate of the population proportion is constructed around the sample proportion. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: confidence interval, proportion 49. True or False: A point estimate consists of a single sample statistic that is used to estimate the true population parameter. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: point estimate 50. True or False: The confidence interval obtained will always correctly estimate the population parameter. ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: confidence interval, interpretation 51. True or False: Other things being equal, the confidence interval for the mean will be wider for 95% confidence than for 90% confidence. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: confidence interval, properties, width

52. True or False: The t distribution is used to develop a confidence interval estimate of the population mean when the population standard deviation is unknown. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: confidence interval, mean, t distribution

53. True or False: The t distribution is used to develop a confidence interval estimate of the population proportion when the population standard deviation is unknown. ANSWER:

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False TYPE: TF DIFFICULTY: Easy KEYWORDS: confidence interval, proportion, t distribution 54. True or False: The standardized normal distribution is used to develop a confidence interval estimate of the population proportion regardless of whether the population standard deviation is known. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: confidence interval, proportion, standardized normal distribution

55. True or False: The standardized normal distribution is used to develop a confidence interval estimate of the population proportion when the sample size is sufficiently large. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: confidence interval, proportion, standardized normal distribution

56. True or False: The t distribution approaches the standardized normal distribution when the number of degrees of freedom increases. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: t distribution, degrees of freedom, properties, standardized normal distribution

57. True or False: The t distribution is used to develop a confidence interval estimate of the population mean when the population standard deviation is unknown. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: t distribution, standardized normal distribution 58. True or False: In estimating the population mean with the population standard deviation unknown, if the sample size is 12, there will be 6 degrees of freedom. ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: confidence interval, mean, t distribution 59. True or False: The difference between the sample mean and the population mean is called the sampling error. ANSWER: False TYPE: TF DIFFICULTY: Difficult KEYWORDS: sampling error

237 Confidence Interval Estimation 60. True or False: The difference between the sample proportion and the population proportion is called the sampling error. ANSWER: False TYPE: TF DIFFICULTY: Moderate KEYWORDS: sampling error 61. True or False: The difference between the sample size and the population size is called the sampling error. ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: sampling error 62. True or False: The difference between the upper limit of a confidence interval and the point estimate used in constructing the confidence interval is called the sampling error. ANSWER: True TYPE: TF DIFFICULTY: Moderate KEYWORDS: sampling error 63. True or False: The difference between the lower limit of a confidence interval and the point estimate used in constructing the confidence interval is called the sampling error. ANSWER: True TYPE: TF DIFFICULTY: Moderate KEYWORDS: sampling error 64. True or False: Sampling error equals half the width of a confidence interval. ANSWER: True TYPE: TF DIFFICULTY: Moderate KEYWORDS: sampling error 65. True or False: The width of a confidence interval equals twice the sampling error. ANSWER: True TYPE: TF DIFFICULTY: Moderate KEYWORDS: sampling error 66. True or False: The sampling error can either be positive or negative. ANSWER: True TYPE: TF DIFFICULTY: Difficult KEYWORDS: sampling error 67. True or False: A population parameter is used to estimate a confidence interval.

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ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: point estimate, confidence interval

68. True or False: For a given data set, the confidence interval will be wider for 95% confidence than for 90% confidence. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: confidence interval, properties, width TABLE 8-1 A random sample of 100 stores from a large chain of 1,000 garden supply stores was selected to determine the average number of lawnmowers sold at an end-of-season clearance sale. The sample results indicated an average of 6 and a standard deviation of 2 lawnmowers sold. A 95% confidence interval (5.623 to 6.377) was established based on these results. 69. True or False: Referring to Table 8-1, if the population had consisted of 10,000 stores, the confidence interval estimate of the mean would have been wider in range. ANSWER: False TYPE: TF DIFFICULTY: Moderate KEYWORDS: confidence interval, mean, properties, width 70. True or False: Referring to Table 8-1, of all possible samples of 100 stores drawn from the population of 1,000 stores, 95% of the sample means will fall between 5.623 and 6.377 lawnmowers. ANSWER: False TYPE: TF DIFFICULTY: Moderate KEYWORDS: confidence interval, mean, interpretation 71. True or False: Referring to Table 8-1, of all possible samples of 100 stores taken from the population of 1,000 stores, 95% of the confidence intervals developed will contain the true population mean within the interval. ANSWER: True TYPE: TF DIFFICULTY: Moderate KEYWORDS: confidence interval, mean, interpretation 72. True or False: Referring to Table 8-1, there are 10 possible samples of 100 stores that can be selected out of the population of 1,000 stores. ANSWER: False TYPE: TF DIFFICULTY: Moderate KEYWORDS: confidence interval, mean, interpretation

239 Confidence Interval Estimation

73. True or False: Referring to Table 8-1, 95% of the stores have sold between 5.623 and 6.377 lawnmowers. ANSWER: False TYPE: TF DIFFICULTY: Moderate KEYWORDS: confidence interval, mean, interpretation 74. True or False: Referring to Table 8-1, we do not know for sure whether the true population mean is between 5.623 and 6.377 lawnmowers sold. ANSWER: True TYPE: TF DIFFICULTY: Moderate KEYWORDS: confidence interval, mean, interpretation TABLE 8-2 The managers of a company are worried about the morale of their employees. In order to determine if a problem in this area exists, they decide to evaluate the attitudes of their employees with a standardized test. They select the Fortunato test of job satisfaction, which has a known standard deviation of 24 points. 75. Referring to Table 8-2, they should sample employees if they want to estimate the mean score of the employees within 5 points with 90% confidence. ANSWER: 63 TYPE: FI DIFFICULTY: Easy KEYWORDS: mean, sample size determination 76. Referring to Table 8-2, due to financial limitations, the managers decide to take a sample of 45 employees. This yields a mean score of 88.0 points. A 90% confidence interval would go from to . ANSWER: 82.12 to 93.88 TYPE: FI DIFFICULTY: Moderate KEYWORDS: confidence interval, mean, standardized normal distribution 77. True or False: Referring to Table 8-2, this confidence interval is only valid if the scores on the Fortunato test are normally distributed. ANSWER: False TYPE: TF DIFFICULTY: Difficult EXPLANATION: With a sample size of 45, this confidence interval will still be valid if the scores are not normally distributed due to the central limit theorem. KEYWORDS: confidence interval, mean, standardized normal distribution, central limit theorem TABLE 8-3

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A quality control engineer is interested in the mean length of sheet insulation being cut automatically by machine. The desired length of the insulation is 12 feet. It is known that the standard deviation in the cutting length is 0.15 feet. A sample of 70 cut sheets yields a mean length of 12.14 feet. This sample will be used to obtain a 99% confidence interval for the mean length cut by machine. 78. Referring to Table 8-3, the critical value to use in obtaining the confidence interval is

ANSWER: 2.58 TYPE: FI DIFFICULTY: Easy KEYWORDS: confidence interval, mean, standardized normal distribution 79. Referring to Table 8-3, the confidence interval goes from

ANSWER: 12.09 to 12.19 TYPE: FI DIFFICULTY: Moderate KEYWORDS: confidence interval, mean, standardized normal distribution 80. True or False: Referring to Table 8-3, the confidence interval indicates that the machine is not working properly. ANSWER: True TYPE: TF DIFFICULTY: Moderate KEYWORDS: confidence interval, mean, standardized normal distribution, interpretation 81. True or False: Referring to Table 8-3, the confidence interval is valid only if the lengths cut are normally distributed. ANSWER: False TYPE: TF DIFFICULTY: Moderate EXPLANATION: With a sample size of 70, this confidence interval will still be valid if the lengths cut are not normally distributed due to the central limit theorem. KEYWORDS: confidence interval, mean, standardized normal distribution, central limit theorem 82. Referring to Table 8-3, suppose the engineer had decided to estimate the mean length to within 0.03 with 99% confidence. Then the sample size would be . ANSWER: 165.8724 rounds up to 166 TYPE: FI DIFFICULTY: Moderate KEYWORDS: mean, sample size determination TABLE 8-4 To become an actuary, it is necessary to pass a series of 10 exams, including the most important one, an exam in probability and statistics. An insurance company wants to estimate the mean score on this exam for actuarial students who have enrolled in a special study program. They take a sample of 8 actuarial students in this program and determine that their scores are: 2, 5, 8, 8, 7, 6, 5, and 7. This sample will be used to calculate a 90% confidence interval for the mean score for actuarial students in the special study program.

241 Confidence Interval Estimation 83. Referring to Table 8-4, the mean of the sample is the sample is .

, while the standard deviation of

ANSWER: 6.0; 2.0 TYPE: FI DIFFICULTY: Moderate KEYWORDS: confidence interval 84. Referring to Table 8-4, the confidence interval will be based on

degrees of freedom.

ANSWER: 7 TYPE: FI DIFFICULTY: Easy KEYWORDS: confidence interval, mean, t distribution 85. Referring to Table 8-4, the critical value used in constructing a 90% confidence interval is . ANSWER: 1.8946 TYPE: FI DIFFICULTY: Easy KEYWORDS: confidence interval, mean, t distribution 86. Referring to Table 8-4, a 90% confidence interval for the mean score of actuarial students in the special program is from to . ANSWER: 4.66 to 7.34 TYPE: FI DIFFICULTY: Moderate KEYWORDS: confidence interval, mean, t distribution 87. True or False: Referring to Table 8-4, for the confidence interval to be valid, it is necessary that test scores of students in the special study program on the actuarial exam be normally distributed. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: confidence interval, mean, t distribution 88. True or False: Referring to Table 8-4, it is possible that the confidence interval obtained will not contain the mean score for all actuarial students in the special class. ANSWER: True TYPE: TF DIFFICULTY: Moderate KEYWORDS: confidence interval, mean, interpretation 89. True or False: Referring to Table 8-4, if we use the same sample information to obtain a 95% confidence interval, the resulting interval would be narrower than the one obtained here with 90% confidence. ANSWER: False TYPE: TF DIFFICULTY: Moderate

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KEYWORDS: confidence interval, mean, properties, width

90. Suppose a department store wants to estimate the average age of the customers of its contemporary apparel department, correct to within 2 years, with level of confidence equal to 95%. Management believes that the standard deviation is 8 years. The sample size they should take is . ANSWER: 62 TYPE: FI DIFFICULTY: Easy KEYWORDS: mean, sample size determination TABLE 8-5 The actual voltages of power packs labeled as 12 volts are as follows: 11.77, 11.90, 11.64, 11.84, 12.13, 11.99, and 11.77.

91. Referring to Table 8-5, a confidence interval for this sample would be based on the t distribution with

degrees of freedom.

ANSWER: 6 TYPE: FI DIFFICULTY: Easy KEYWORDS: confidence interval, mean, t distribution 92. Referring to Table 8-5, the critical value for a 99% confidence interval for this sample is . ANSWER: 3.7074 TYPE: FI DIFFICULTY: Easy KEYWORDS: confidence interval, mean, t distribution 93. Referring to Table 8-5, a 99% confidence interval for the mean voltage of the power packs is from to . ANSWER: 11.6367 to 12.0891 TYPE: FI DIFFICULTY: Moderate KEYWORDS: confidence interval, mean, t distribution 94. True or False: Referring to Table 8-5, a 95% confidence interval for the mean voltage of the power pack is wider than a 99% confidence interval. ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: confidence interval, mean, properties, width 95. True or False: Referring to Table 8-5, a 99% confidence interval will contain 99% of the voltages for all such power packs. ANSWER: False

243 Confidence Interval Estimation TYPE: TF DIFFICULTY: Moderate KEYWORDS: confidence interval, mean, interpretation 96. True or False: Referring to Table 8-5, a confidence interval estimate of the population mean would only be valid if the distribution of voltages is normal. ANSWER: True TYPE: TF DIFFICULTY: Moderate KEYWORDS: confidence interval, mean, t distribution 97. True or False: Referring to Table 8-5, a 90% confidence interval calculated from the same data would be narrower than a 99% confidence interval. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: confidence interval, mean, properties, width 98. True or False: Referring to Table 8-5, it is possible that the 99% confidence interval calculated from the data will not contain the mean voltage for the sample. ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: confidence interval, mean, interpretation 99. True or False: Referring to Table 8-5, it is possible that the 99% confidence interval calculated from the data will not contain the mean voltage for the entire population. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: confidence interval, mean, interpretation TABLE 8-6 A sample of salary offers (in thousands of dollars) given to management majors is: 28, 31, 26, 32, 27, 28, 27, 30, 31, and 29. Using this data to obtain a 95% confidence interval resulted in an interval from 27.5 to 30.3. 100. True or False: Referring to Table 8-6, 95% of the time, the sample mean salary offer to management majors will fall between 27.5 and 30.3. ANSWER: False TYPE: TF DIFFICULTY: Moderate KEYWORDS: confidence interval, mean, interpretation 101.

True or False: Referring to Table 8-6, 95% of the salary offers are between 27.5 and 30.3.

ANSWER: False TYPE: TF DIFFICULTY: Moderate

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KEYWORDS: confidence interval, mean, interpretation 102.

True or False: Referring to Table 8-6, it is possible that the mean of the population is between 27.5 and 30.3.

ANSWER: True TYPE: TF DIFFICULTY: Moderate KEYWORDS: confidence interval, mean, interpretation 103. True or False: Referring to Table 8-6, it is possible that the mean of the population is not between 27.5 and 30.3. ANSWER: True TYPE: TF DIFFICULTY: Moderate KEYWORDS: confidence interval, mean, interpretation 104. True or False: Referring to Table 8-6, 95% of the sample means will fall between 27.5 and 30.3. ANSWER: False TYPE: TF DIFFICULTY: Moderate KEYWORDS: confidence interval, mean, interpretation

105. True or False: Referring to Table 8-6, 95% of all confidence intervals constructed similarly to this one with a sample size of 10 will contain the mean of the population. ANSWER: True TYPE: TF DIFFICULTY: Moderate KEYWORDS: confidence interval, mean, interpretation 106. True or False: Referring to Table 8-6, a 99% confidence interval for the mean of the population from the same sample would be wider than 27.5 to 30.3. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: confidence interval, mean, properties, width 107. True or False: Referring to Table 8-6, the confidence interval obtained is valid only if the distribution of the population of salary offers is normal. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: confidence interval, mean, t distribution 108. A prison official wants to estimate the proportion of cases of recidivism. Examining the records of 250 convicts, the official determines that there are 65 cases of recidivism. A confidence interval will be obtained for the proportion of cases of recidivism. Part of this

245 Confidence Interval Estimation calculation includes the estimated standard error of the sample proportion. For this sample, the estimated standard error is . ANSWER: 0.028 TYPE: FI DIFFICULTY: Easy KEYWORDS: confidence interval, proportion 109. A prison official wants to estimate the proportion of cases of recidivism. Examining the records of 250 convicts, the official determines that there are 65 cases of recidivism. A 99% confidence interval for the proportion of cases of recidivism would go from to . ANSWER: 0.189 to 0.331 TYPE: FI DIFFICULTY: Moderate KEYWORDS: confidence interval, proportion TABLE 8-7 After an extensive advertising campaign, the manager of a company wants to estimate the proportion of potential customers that recognize a new product. She samples 120 potential consumers and finds that 54 recognize this product. She uses this sample information to obtain a 95% confidence interval that goes from 0.36 to 0.54. 110. True or False: Referring to Table 8-7, the parameter of interest to the manager is the proportion of potential customers in this sample that recognize the new product. ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: confidence interval, proportion, parameter 111.

True or False: Referring to Table 8-7, the parameter of interest is 54/120 = 0.45.

ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: confidence interval, proportion, parameter

112. True or False: Referring to Table 8-7, this interval requires the use of the t distribution to obtain the confidence coefficient. ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: confidence interval, proportion, t distribution 113. True or False: Referring to Table 8-7, this interval requires the assumption that the distribution of the number of people recognizing the product has a normal distribution. ANSWER: False

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TYPE: TF DIFFICULTY: Moderate KEYWORDS: confidence interval, proportion 114. True or False: Referring to Table 8-7, 95% of the time, the proportion of people that recognize the product will fall between 0.36 and 0.54. ANSWER: False TYPE: TF DIFFICULTY: Moderate KEYWORDS: confidence interval, proportion, interpretation 115. True or False: Referring to Table 8-7, 95% of the time, the sample proportion of people that recognize the product will fall between 0.36 and 0.54. ANSWER: False TYPE: TF DIFFICULTY: Moderate KEYWORDS: confidence interval, proportion, interpretation 116. True or False: Referring to Table 8-7, 95% of the people will recognize the product between 36% and 54% of the time. ANSWER: False TYPE: TF DIFFICULTY: Difficult KEYWORDS: confidence interval, proportion, interpretation 117. True or False: Referring to Table 8-7, it is possible that the true proportion of people that recognize the product is between 0.36 and 0.54. ANSWER: True TYPE: TF DIFFICULTY: Moderate KEYWORDS: confidence interval, proportion, interpretation 118. True or False: Referring to Table 8-7, it is possible that the true proportion of people that recognize the product is not between 0.36 and 0.54. ANSWER: True TYPE: TF DIFFICULTY: Moderate KEYWORDS: confidence interval, proportion, interpretation 119. The head of a computer science department is interested in estimating the proportion of students entering the department who will choose the new computer engineering option. Suppose there is no information about the proportion of students who might choose the option. What size sample should the department head take if she wants to be 95% confident that the estimate is within 0.10 of the true proportion? ANSWER: 97 TYPE: PR DIFFICULTY: Moderate KEYWORDS: proportion, sample size determination

247 Confidence Interval Estimation 120. The head of a computer science department is interested in estimating the proportion of students entering the department who will choose the new computer engineering option. A preliminary sample indicates that the proportion will be around 0.25. Therefore, what size sample should the department head take if she wants to be 95% confident that the estimate is within 0.10 of the true proportion? ANSWER: 73 TYPE: PR DIFFICULTY: Easy KEYWORDS: proportion, sample size determination TABLE 8-8 A hotel chain wants to estimate the average number of rooms rented daily in each month. The population of rooms rented daily is assumed to be normally distributed for each month with a standard deviation of 24 rooms. 121. Referring to Table 8-8, during January, a sample of 16 days has a sample mean of 48 rooms. This information is used to calculate an interval estimate for the population mean to be from 40 to 56 rooms. What is the level of confidence of this interval? ANSWER: 81.76% TYPE: PR DIFFICULTY: Difficult KEYWORDS: confidence interval, mean 122. Referring to Table 8-8, during February, a sample of 25 days has a sample mean of 37 rooms. Use this information to calculate a 92% confidence interval for the population mean. ANSWER: 28.60 to 45.40 TYPE: PR DIFFICULTY: Easy KEYWORDS: confidence interval, mean 123. The county clerk wants to estimate the proportion of retired voters who will need special election facilities. The clerk wants to find a 95% confidence interval for the population proportion which extends at most 0.07 to either side of the sample proportion. How large a sample must be taken to assure these conditions are met? ANSWER: 196 TYPE: PR DIFFICULTY: Easy KEYWORDS: proportion, sample size determination 124. The county clerk wants to estimate the proportion of retired voters who will need special election facilities. Suppose a sample of 400 retired voters was taken. If 150 need special election facilities, calculate an 80% confidence interval for the population proportion. ANSWER: 0.344 to 0.406 TYPE: PR DIFFICULTY: Easy KEYWORDS: confidence interval, proportion TABLE 8-9

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A wealthy real estate investor wants to decide whether it is a good investment to build a high-end shopping complex in a suburban county in Chicago. His main concern is the total market value of the 3,605 houses in the suburban county. He commissioned a statistical consulting group to take a sample of 200 houses and obtained a sample average market price of $225,000 and a sample standard deviation of $38,700. The consulting group also found out that the average differences between market prices and appraised prices was $125,000 with a standard deviation of $3,400. Also the proportion of houses in the sample that are appraised for higher than the market prices is 0.24. 125. Referring to Table 8-9, if he wants a 95% confidence on estimating the true population average market price of the houses in the suburban county to be within $10,000, how large a sample will he need? ANSWER: 57 TYPE: PR DIFFICULTY: Moderate EXPLANATION: This is a sample size with finite population correction factor. KEYWORDS: mean, sample size determination, finite population correction 126. Referring to Table 8-9, what will be the 90% confidence interval for the average market price of the houses in the suburban county constructed by the consulting group? ANSWER: $220,604.42 to $229,395.58 TYPE: PR DIFFICULTY: Moderate EXPLANATION: This is a confidence interval estimate for the mean with unknown standard deviation and finite population correction factor. KEYWORDS: confidence interval, mean, finite population correction 127. Referring to Table 8-9, what will be the 90% confidence interval for the total market price of the houses in the suburban county constructed by the consulting group? ANSWER: $795,278,945.08 to $826,971,054.92 TYPE: PR DIFFICULTY: Moderate EXPLANATION: This confidence interval is computed with the t critical value of 1.652546747 obtained in Excel. KEYWORDS: confidence interval, total amount, finite population correction 128. Referring to Table 8-9, what will be the 90% confidence interval for the total difference between the market prices and appraised prices of the houses in the suburban county constructed by the consulting group? ANSWER: $449,232,840.14 to $452,017,159.86 TYPE: PR DIFFICULTY: Moderate EXPLANATION: This confidence interval is computed with the t critical value of 1.652546747 obtained in Excel. KEYWORDS: confidence interval, total difference, finite population correction 129. Referring to Table 8-9, what will be the 90% confidence interval for the population proportion of houses that will be appraised for higher than the market prices? ANSWER:

249 Confidence Interval Estimation 0.1917 to 0.2883 TYPE: PR DIFFICULTY: Moderate EXPLANATION: This is a confidence interval estimate for the proportion with finite population correction factor. KEYWORDS: confidence interval, proportion, finite population correction 130. A quality control engineer is interested in estimating the proportion of defective items coming off a production line. In a sample of 300 items, 27 are defective. A 90% confidence interval for the proportion of defectives from this production line would go from to . ANSWER: 0.063 to 0.117 TYPE: FI DIFFICULTY: Moderate KEYWORDS: confidence interval, proportion TABLE 8-10 The president of a university would like to estimate the proportion of the student population who owns a personal computer. In a sample of 500 students, 417 own a personal computer. 131. True or False: Referring to Table 8-10, the parameter of interest is the average number of students in the population who own a personal computer. ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: confidence interval, proportion, parameter 132. True or False: Referring to Table 8-10, the parameter of interest is the proportion of student population who own a personal computer. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: confidence interval, proportion, parameter 133.

Referring to Table 8-10, the critical value for a 99% confidence interval for this sample is .

ANSWER: 2.5758 TYPE: FI DIFFICULTY: Easy KEYWORDS: confidence interval, proportion 134. Referring to Table 8-10, a 99% confidence interval for the proportion of student population who own a personal computer is from to . ANSWER: 0.7911 to 0.8769 TYPE: FI DIFFICULTY: Moderate KEYWORDS: confidence interval, proportion

Confidence Interval Estimation

250

135. Referring to Table 8-10, the sampling error of a 99% confidence interval for the proportion of student population who own a personal computer is . ANSWER: 0.04286 TYPE: FI DIFFICULTY: Moderate KEYWORDS: confidence interval, proportion, sampling error 136. True or False: Referring to Table 8-10, a 95% confidence interval for the proportion of student population who own a personal computer is narrower than a 99% confidence interval. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: confidence interval, proportion, properties, width 137. True or False: Referring to Table 8-10, a 99% confidence interval will contain 99% of the student population who own a personal computer. ANSWER: False TYPE: TF DIFFICULTY: Moderate KEYWORDS: confidence interval, proportion, interpretation 138. True or False: Referring to Table 8-10, a confidence interval estimate of the population proportion would only be valid if the distribution of the number of students who own a personal computer is normal. ANSWER: False TYPE: TF DIFFICULTY: Moderate KEYWORDS: confidence interval, proportion 139. True or False: Referring to Table 8-10, a 90% confidence interval calculated from the same data would be narrower than a 99% confidence interval. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: confidence interval, proportion, properties, width 140. True or False: Referring to Table 8-10, it is possible that the 99% confidence interval calculated from the data will not contain the sample proportion of students who own a personal computer. ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: confidence interval, proportion, interpretation 141. True or False: Referring to Table 8-10, it is possible that the 99% confidence interval calculated from the data will not contain the proportion of student population who own a personal computer.

251 Confidence Interval Estimation ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: confidence interval, proportion, interpretation

142. True or False: Referring to Table 8-10, we are 99% confident that the average numbers of student population who own a personal computer is between 0.7911 and 0.8769. ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: confidence interval, proportion, interpretation

143. True or False: Referring to Table 8-10, we are 99% confident that between 79.11% and 87.69% of the student population own a personal computer. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: confidence interval, proportion, interpretation TABLE 8-11 The superintendent of a unified school district of a small town wants to make sure that no more than 5% of the students skip more than 10 days of school in a year. A random sample of 145 students from a population of 800 showed that 12 students skipped more than 10 days of school last year. 144. Referring to Table 8-11, what is the critical value for the 95% one-sided confidence interval for the proportion of students who skipped more than 10 days of school last year? ANSWER: 1.6449 TYPE: PR DIFFICULTY: Easy KEYWORDS: confidence interval, proportion, one-sided, critical value 145. Referring to Table 8-11, what is the upper bound of the 95% one-sided confidence interval for the proportion of students who skipped more than 10 days of school last year? ANSWER: 0.1168 TYPE: PR DIFFICULTY: Easy KEYWORDS: confidence interval, proportion, one-sided 146. True or False: Referring to Table 8-11, the superintendent can conclude with 95% level of confidence that no more than 5% of the students in the unified school district skipped more than 10 days of school last year. ANSWER: False TYPE: TF DIFFICULTY: Moderate KEYWORDS: confidence interval, proportion, one-sided, interpretation TABLE 8-12

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The president of a university is concerned that illicit drug use on campus is higher than the 5% acceptable level. A random sample of 250 students from a population of 2000 revealed that 7 of them had used illicit drug during the last 12 months. 147. Referring to Table 8-12, what is the critical value for the 90% one-sided confidence interval for the proportion of students who had used illicit drug during the last 12 months? ANSWER: 1.2816 TYPE: PR DIFFICULTY: Easy KEYWORDS: confidence interval, proportion, one-sided, critical value 148. Referring to Table 8-12, what is the upper bound of the 90% one-sided confidence interval for the proportion of students who had used illicit drug during the last 12 months? ANSWER: 0.0405 TYPE: PR DIFFICULTY: Easy KEYWORDS: confidence interval, proportion, one-sided

149. True or False: Referring to Table 8-12, the president can be 90% confident that no more than 5% of the students at the university had used illicit drug during the last 12 months. ANSWER: True TYPE: TF DIFFICULTY: Moderate KEYWORDS: confidence interval, proportion, one-sided, interpretation

150. True or False: Referring to Table 8-12, using the 90% one-sided confidence interval, the president can be 95% confident that no more than 5% of the students at the university had used illicit drug during the last 12 months. ANSWER: False TYPE: TF DIFFICULTY: Difficult EXPLANATION: Since the upper limit of the 95% one-sided confidence interval will be larger than 0.0405, the president cannot be 95% confident that no more than 5% of the students at the university had used illicit drug during the last 12 months. KEYWORDS: confidence interval, proportion, one-sided, interpretation

151. True or False: Referring to Table 8-12, using the 90% one-sided confidence interval, the president can be 85% confident that no more than 5% of the students at the university had used illicit drug during the last 12 months. ANSWER: True TYPE: TF DIFFICULTY: Difficult EXPLANATION: Since the upper limit of the 85% one-sided confidence interval will be smaller than 0.0405, the president can be 85% confident that no more than 5% of the students at the university had used illicit drug during the last 12 months. KEYWORDS: confidence interval, proportion, one-sided, interpretation TABLE 8-13

253 Confidence Interval Estimation

A university wanted to find out the percentage of students who felt comfortable reporting cheating by their fellow students. A surveyed of 2,800 students was conducted and the students were asked if they felt comfortable reporting cheating by their fellow students. The results were 1,344 answered “Yes” and 1,456 answered “no”. 152. True or False: Referring to Table 8-13, the parameter of interest is the total number of students in the population who feel comfortable reporting cheating by their fellow students. ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: confidence interval, proportion, parameter 153. True or False: Referring to Table 8-13, the parameter of interest is the proportion of student population who feel comfortable reporting cheating by their fellow students. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: confidence interval, proportion, parameter 154.

Referring to Table 8-13, the critical value for a 99% confidence interval for this sample is .

ANSWER: 2.5758 TYPE: FI DIFFICULTY: Easy KEYWORDS: confidence interval, proportion 155. Referring to Table 8-13, a 99% confidence interval for the proportion of student population who feel comfortable reporting cheating by their fellow students is from to . ANSWER: 0.4557 to 0.5043 TYPE: FI DIFFICULTY: Moderate KEYWORDS: confidence interval, proportion 156. Referring to Table 8-13, the sampling error of a 99% confidence interval for the proportion of student population who feel comfortable reporting cheating by their fellow students is . ANSWER: 0.0243 TYPE: FI DIFFICULTY: Moderate KEYWORDS: confidence interval, proportion, sampling error 157. True or False: Referring to Table 8-13, a 95% confidence interval for the proportion of student population who feel comfortable reporting cheating by their fellow students is narrower than a 99% confidence interval. ANSWER: True

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TYPE: TF DIFFICULTY: Easy KEYWORDS: confidence interval, proportion, properties, width 158. True or False: Referring to Table 8-13, a 99% confidence interval will contain 99% of the student population who feel comfortable reporting cheating by their fellow students. ANSWER: False TYPE: TF DIFFICULTY: Moderate KEYWORDS: confidence interval, proportion, interpretation 159. True or False: Referring to Table 8-13, a confidence interval estimate of the population proportion would only be valid if the distribution of the number of students who feel comfortable reporting cheating by their fellow students is normal. ANSWER: False TYPE: TF DIFFICULTY: Moderate KEYWORDS: confidence interval, proportion 160. True or False: Referring to Table 8-13, a 90% confidence interval calculated from the same data would be narrower than a 99% confidence interval. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: confidence interval, proportion, properties, width 161. True or False: Referring to Table 8-13, it is possible that the 99% confidence interval calculated from the data will not contain the sample proportion of students who feel comfortable reporting cheating by their fellow students. ANSWER: False TYPE: TF DIFFICULTY: Moderate EXPLANATION: The confidence interval always contain the sample proportion. KEYWORDS: confidence interval, proportion, interpretation 162. True or False: Referring to Table 8-13, it is possible that the 99% confidence interval calculated from the data will not contain the proportion of student population who feel comfortable reporting cheating by their fellow students. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: confidence interval, proportion, interpretation

163. True or False: Referring to Table 8-13, we are 99% confident that the total numbers of student population who feel comfortable reporting cheating by their fellow students is between 0.4557 and 0.5043. ANSWER: False TYPE: TF DIFFICULTY: Easy

255 Confidence Interval Estimation KEYWORDS: confidence interval, proportion, interpretation

164. True or False: Referring to Table 8-13, we are 99% confident that between 45.57% and 50.43% of the student population feel comfortable reporting cheating by their fellow students. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: confidence interval, proportion, interpretation TABLE 8-14 The president of a university is concerned that the percentage of students who have cheated on an exam is higher than the 1% acceptable level. A confidential random sample of 1000 students from a population of 7000 revealed that 6 of them said that they had cheated on an exam during the last semester. 165. Referring to Table 8-14, what is the critical value for the 90% one-sided confidence interval for the proportion of students who had cheated on an exam during the last 12 months? ANSWER: 1.2816 TYPE: PR DIFFICULTY: Easy KEYWORDS: confidence interval, proportion, one-sided, critical value 166. Referring to Table 8-14, what is the upper bound of the 90% one-sided confidence interval for the proportion of students who had cheated on an exam during the last 12 months? ANSWER: 0.0089 TYPE: PR DIFFICULTY: Easy KEYWORDS: confidence interval, proportion, one-sided

Confidence Interval Estimation

256

167. True or False: Referring to Table 8-14, the president can be 90% confident that no more than 1% of the students at the university had cheated on an exam during the last semester. ANSWER: True TYPE: TF DIFFICULTY: Moderate KEYWORDS: confidence interval, proportion, one-sided, interpretation

168. True or False: Referring to Table 8-14, using on the 90% one-sided confidence interval, the president can be 95% confident that no more than 1% of the students at the university had cheated on an exam during the last semester. ANSWER: False TYPE: TF DIFFICULTY: Difficult EXPLANATION: Since the upper limit of the 95% one-sided confidence interval will be larger than 0.0089, the president cannot be 95% confident that no more than 1% of the students at the university had cheated on an exam during the last semester. KEYWORDS: confidence interval, proportion, one-sided, interpretation

169. True or False: Referring to Table 8-14, using on the 90% one-sided confidence interval, the superintendent can be 85% confident that no more than 1% of the students at the university had cheated on an exam during the last semester. ANSWER: True TYPE: TF DIFFICULTY: Difficult EXPLANATION: Since the upper limit of the 85% one-sided confidence interval will be smaller than 0.0089, the president can be 85% confident that no more than 1% of the students at the university had cheated on an exam during the last semester. KEYWORDS: confidence interval, proportion, one-sided, interpretation

Fundamentals of Hypothesis Testing: One-Sample Tests

CHAPTER 9: FUNDAMENTALS OF HYPOTHESIS TESTING: ONE-SAMPLE TESTS 1. 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) Only (a) and (c) are true. ANSWER: a TYPE: MC DIFFICULTY: Easy KEYWORDS: form of hypothesis 2. Which of the following would be an appropriate null hypothesis? a) The population proportion is less than 0.65. b) The sample proportion is less than 0.65. c) The population proportion is no less than 0.65. d) The sample proportion is no less than 0.65. ANSWER: c TYPE: MC DIFFICULTY: Easy KEYWORDS: form of hypothesis 3. Which of the following would be an appropriate alternative 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: c TYPE: MC DIFFICULTY: Easy KEYWORDS: form of hypothesis 4. Which of the following would be an appropriate alternative hypothesis? a) The population proportion is less than 0.65. b) The sample proportion is less than 0.65. c) The population proportion is no less than 0.65. d) The sample proportion is no less than 0.65. ANSWER: a TYPE: MC DIFFICULTY: Easy KEYWORDS: form of hypothesis 5. A Type II error is committed when a) we reject a null hypothesis that is true. b) we don't reject a null hypothesis that is true. c) we reject a null hypothesis that is false.

Fundamentals of Hypothesis Testing: One-Sample Tests d) we don't reject a null hypothesis that is false. ANSWER: d TYPE: MC DIFFICULTY: Easy KEYWORDS: type II error 6. A Type I error is committed when a) we reject a null hypothesis that is true. b) we don't reject a null hypothesis that is true. c) we reject a null hypothesis that is false. d) we don't reject a null hypothesis that is false. ANSWER: a TYPE: MC DIFFICULTY: Easy KEYWORDS: type I error 7. 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 TYPE: MC DIFFICULTY: Easy KEYWORDS: power 8.

If we are performing a two-tailed test of whether  = 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 TYPE: MC DIFFICULTY: Moderate KEYWORDS: power 9. True or False: For a given level of significance, if the sample size is increased, the power of the test will increase. ANSWER: True TYPE: TF DIFFICULTY: Moderate KEYWORDS: power, level of significance, sample size 10. True or False: For a given level of significance, if the sample size is increased, the probability of committing a Type I error will increase. ANSWER: False

Fundamentals of Hypothesis Testing: One-Sample Tests

TYPE: TF DIFFICULTY: Moderate KEYWORDS: level of significance, sample size, type I error 11. True or False: For a given level of significance, if the sample size is increased, the probability of committing a Type II error will increase. ANSWER: False TYPE: TF DIFFICULTY: Moderate KEYWORDS: level of significance, sample size, type II error 12. True or False: For a given sample size, the probability of committing a Type II error will increase when the probability of committing a Type I error is reduced. ANSWER: True TYPE: TF DIFFICULTY: Moderate KEYWORDS: sample size, type I error, type II error 13. If an economist wishes to determine whether there is evidence that average family income in a community exceeds $25,000 a) either a one-tailed or two-tailed test could be used with equivalent results. b) a one-tailed test should be utilized. c) a two-tailed test should be utilized. d) None of the above. ANSWER: b TYPE: MC DIFFICULTY: Easy KEYWORDS: one-tailed test 14. If an economist wishes to determine whether there is evidence that average family income in a community equals $25,000 a) either a one-tailed or two-tailed test could be used with equivalent results. b) a one-tailed test should be utilized. c) a two-tailed test should be utilized. d) None of the above. ANSWER: c TYPE: MC DIFFICULTY: Easy KEYWORDS: two-tailed test 15. If the p-value is less than  in a two-tailed test,

a) b) c) d)

the null hypothesis should not be rejected. the null hypothesis should be rejected. a one-tailed test should be used. no conclusion should be reached.

ANSWER: b TYPE: MC DIFFICULTY: Easy KEYWORDS: p-value, level of significance

Fundamentals of Hypothesis Testing: One-Sample Tests 16. If a test of hypothesis has a Type I error probability (  ) of 0.01, we mean

a) b) c) d)

if the null hypothesis is true, we don't reject it 1% of the time. if the null hypothesis is true, we reject it 1% of the time. if the null hypothesis is false, we don't reject it 1% of the time. if the null hypothesis is false, we reject it 1% of the time.

ANSWER: b TYPE: MC DIFFICULTY: Moderate KEYWORDS: type I error, level of significance 17. If the Type I error ( ) for a given test is to be decreased, then for a fixed sample size n a) the Type II error (  ) will also decrease. b) the Type II error (  ) will increase.

c) the power of the test will increase. d) a one-tailed test must be utilized. ANSWER: b TYPE: MC DIFFICULTY: Moderate KEYWORDS: type I error, type II error, sample size 18. For a given sample size n, if the level of significance ( ) is decreased, the power of the test

a) b) c) d)

will increase. will decrease. will remain the same. cannot be determined.

ANSWER: b TYPE: MC DIFFICULTY: Moderate KEYWORDS: level of significance, power, sample size 19. For a given level of significance ( ), if the sample size n is increased, the probability of a Type

II error (  ) a) will decrease. b) will increase. c) will remain the same. d) cannot be determined.

ANSWER: a TYPE: MC DIFFICULTY: Moderate KEYWORDS: level of significance, beta-risk, sample size 20. If a researcher rejects a true null hypothesis, she has made a

error.

ANSWER: Type I TYPE: FI DIFFICULTY: Easy KEYWORDS: type I error 21. If a researcher accepts a true null hypothesis, she has made a

decision.

Fundamentals of Hypothesis Testing: One-Sample Tests

ANSWER: correct TYPE: FI DIFFICULTY: Easy KEYWORDS: decision 22. If a researcher rejects a false null hypothesis, she has made a

decision.

ANSWER: correct TYPE: FI DIFFICULTY: Easy KEYWORDS: decision 23. If a researcher accepts a false null hypothesis, she has made a

error.

ANSWER: Type II TYPE: FI DIFFICULTY: Easy KEYWORDS: type II error 24. It is possible to directly compare the results of a confidence interval estimate to the results obtained by testing a null hypothesis if a) a two-tailed test for  is used. b) a one-tailed test for  is used. c) Both of the previous statements are true. d) None of the previous statements is true. ANSWER: a TYPE: MC DIFFICULTY: Moderate KEYWORDS: confidence interval, two-tailed test 25. The power of a statistical test is a) the probability of not rejecting H0 when it is false. b) the probability of rejecting H0 when it is true. c) the probability of not rejecting H0 when it is true. d) the probability of rejecting H0 when it is false. ANSWER: d TYPE: MC DIFFICULTY: Moderate KEYWORDS: power 26. The symbol for the power of a statistical test is a)  . b) 1 –  . c)  . d) 1 –  . ANSWER: d TYPE: MC DIFFICULTY: Easy KEYWORDS: power

Fundamentals of Hypothesis Testing: One-Sample Tests 27. Suppose we wish to test H0:   47 versus H1:  > 47. What will result if we conclude that the

mean is greater than 47 when its true value is really 52? a) We have made a Type I error. b) We have made a Type II error. c) We have made a correct decision d) None of the above are correct. ANSWER: c TYPE: MC DIFFICULTY: Moderate KEYWORDS: one-tailed test, conclusion 28. How many Kleenex should the Kimberly Clark Corporation package of tissues contain?

Researchers determined that 60 tissues is the average number of tissues used during a cold. Suppose a random sample of 100 Kleenex users yielded the following data on the number of tissues used during a cold: X = 52, s = 22. Give the null and alternative hypotheses to determine if the number of tissues used during a cold is less than 60. a) H0 :   60 and H1 :   60. b) H0 :   60 and H1 :   60. c) H0 : X  60 and H1 : X  60. d) H0 : X = 52 and H1 : X  52. ANSWER: b TYPE: MC DIFFICULTY: Moderate KEYWORDS: one-tailed test, form of hypothesis, form of hypothesis, mean, t test 29. How many Kleenex should the Kimberly Clark Corporation package of tissues contain?

b) t = ( 52 −60 ) / ( 22 /100 )

c) t = ( 52 −60 ) / ( 22 /1002 ) d) t = ( 52 −60 ) / ( 22 /10 ) ANSWER: d TYPE: MC DIFFICULTY: Easy KEYWORDS: one-tailed test, mean, t test 30. How many Kleenex should the Kimberly Clark Corporation package of tissues contain?

Researchers determined that 60 tissues is the average number of tissues used during a cold. Suppose a random sample of 100 Kleenex users yielded the following data on the number of tissues used during a cold: X = 52, s = 22. Suppose the alternative we wanted to test was H1 :   60 . State the correct rejection region for  = 0.05. a) Reject H0 if t > 1.6604. b) Reject H0 if t < – 1.6604. c) Reject H0 if t > 1.9842 or Z < – 1.9842. d) Reject H0 if t < – 1.9842.

Fundamentals of Hypothesis Testing: One-Sample Tests

ANSWER: b TYPE: MC DIFFICULTY: Moderate KEYWORDS: one-tailed test, mean, t test, rejection region 31. How many Kleenex should the Kimberly Clark Corporation package of tissues contain?

Researchers determined that 60 tissues is the average number of tissues used during a cold. Suppose a random sample of 100 Kleenex users yielded the following data on the number of tissues used during a cold: X = 52, s = 22. Suppose the test statistic does fall in the rejection region at  = 0.05. Which of the following conclusion is correct? a) At  = 0.05, there is not sufficient evidence to conclude that the average number of tissues used during a cold is 60 tissues. b) At  = 0.05, there is sufficient evidence to conclude that the average number of tissues used during a cold is 60 tissues. c) At  = 0.05, there is not sufficient evidence to conclude that the average number of tissues used during a cold is not 60 tissues. d) At  = 0.10, there is sufficient evidence to conclude that the average number of tissues used during a cold is not 60 tissues. ANSWER: d TYPE: MC DIFFICULTY: Moderate KEYWORDS: one-tailed test, mean, t test, conclusion 33. We have created a 95% confidence interval for  with the result (10, 15). What decision will we

make if we test H0 :  = 16 versus H1 :   16 at  = 0.05? a) Reject H0 in favor of H1. b) Accept H0 in favor of H1. c) Fail to reject H0 in favor of H1. d) We cannot tell what our decision will be from the information given.

ANSWER: a TYPE: MC DIFFICULTY: Easy KEYWORDS: two-tailed test, confidence interval, mean, t test, decision

Fundamentals of Hypothesis Testing: One-Sample Tests 34. We have created a 95% confidence interval for  with the result (10, 15). What decision will we

make if we test H0 :  = 16 versus H1 :   16 at  = 0.10? a) Reject H0 in favor of H1. b) Accept H0 in favor of H1. c) Fail to reject H0 in favor of H1. d) We cannot tell what our decision will be from the information given.

ANSWER: a TYPE: MC DIFFICULTY: Moderate EXPLANATION: The 90% confidence interval is narrower than (10, 15), which still does not contain 16. KEYWORDS: two-tailed test, confidence interval, mean, t test, decision 35. We have created a 95% confidence interval for  with the result (10, 15). What decision will we

make if we test H0 :  = 16 versus H1 :   16 at  = 0.025? a) Reject H0 in favor of H1. b) Accept H0 in favor of H1. c) Fail to reject H0 in favor of H1. d) We cannot tell what our decision will be from the information given.

ANSWER: d TYPE: MC DIFFICULTY: Moderate EXPLANATION: The 97.5% confidence interval is wider than (10, 15), which could have contained 16 or not have contained 16. KEYWORDS: two-tailed test, confidence interval, mean, t test, decision 36. Suppose we want to test H0 :   30 versus H1 :   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) X = 28, s = 6 b) X = 27, s = 4 c) X = 32, s = 2 d) X = 26, s = 9 ANSWER: b TYPE: MC DIFFICULTY: Moderate KEYWORDS: one-tailed test, rejection region 37. Which of the following statements is not true about the level of significance in a hypothesis test?

a) The larger the level of significance, the more likely you are to reject the null hypothesis. b) The level of significance is the maximum risk we are willing to accept in making a Type I error. c) The significance level is also called the  level. d) The significance level is another name for Type II error. ANSWER: d TYPE: MC DIFFICULTY: Moderate KEYWORDS: level of significance

Fundamentals of Hypothesis Testing: One-Sample Tests

38. If, as a result of a hypothesis test, we reject the null hypothesis when it is false, then we have committed a) a Type II error. b) a Type I error. c) no error. d) an acceptance error. ANSWER: c TYPE: MC DIFFICULTY: Moderate KEYWORDS: decision, type I error, type II error 39. The value that separates a rejection region from a non-rejection region is called the

ANSWER: critical value TYPE: FI DIFFICULTY: Easy KEYWORDS: critical value, rejection region 40. A

is a numerical quantity computed from the data of a sample and is used in reaching a decision on whether or not to reject the null hypothesis. a) significance level b) critical value c) test statistic d) parameter

ANSWER: c TYPE: MC DIFFICULTY: Easy KEYWORDS: test statistic 41. The owner of a local nightclub has recently surveyed a random sample of n = 250 customers of

H0 : X  30 versus H1 : X  30 .

H0 : X  30 versus H1 : X  30 .

ANSWER: b TYPE: MC DIFFICULTY: Easy KEYWORDS: one-tailed test, form of hypothesis, form of hypothesis, mean 42. The owner of a local nightclub has recently surveyed a random sample of n = 250 customers of

the club. She would now like to determine whether or not the mean age of her customers is over 30. If so, she plans to alter the entertainment to appeal to an older crowd. If not, no entertainment changes will be made. If she wants to be 99% confident in her decision, what rejection region should she use? a) Reject H0 if t < – 2.34. b) Reject H0 if t < – 2.55. c) Reject H0 if t > 2.34.

Fundamentals of Hypothesis Testing: One-Sample Tests d)

Reject H0 if t > 2.58.

ANSWER: c TYPE: MC DIFFICULTY: Moderate KEYWORDS: one-tailed test, mean, Z test, t test, rejection region 43. The owner of a local nightclub has recently surveyed a random sample of n = 250 customers of

the club. She would now like to determine whether or not the mean age of her customers is over 30. If so, she plans to alter the entertainment to appeal to an older crowd. If not, no entertainment changes will be made. Suppose she found that the sample mean was 30.45 years and the sample standard deviation was 5 years. If she wants to be 99% confident in her decision, what decision should she make? a) Reject H0. b) Accept H0. c) Fail to reject H0. d) We cannot tell what her decision should be from the information given. ANSWER: c TYPE: MC DIFFICULTY: Moderate KEYWORDS: one-tailed test, mean, t test, decision 44. The owner of a local nightclub has recently surveyed a random sample of n = 250 customers of

the club. She would now like to determine whether or not the mean age of her customers is over 30. If so, she plans to alter the entertainment to appeal to an older crowd. If not, no entertainment changes will be made. Suppose she found that the sample mean was 30.45 years and the sample standard deviation was 5 years. If she wants to be 99% confident in her decision, what conclusion can she make? a) There is not sufficient evidence that the mean age of her customers is over 30. b) There is sufficient evidence that the mean age of her customers is over 30. c) There is not sufficient evidence that the mean age of her customers is not over 30. d) There is sufficient evidence that the mean age of her customers is not over 30. ANSWER: a TYPE: MC DIFFICULTY: Moderate KEYWORDS: one-tailed test, mean, t test, conclusion 45. The owner of a local nightclub has recently surveyed a random sample of n = 250 customers of

Fundamentals of Hypothesis Testing: One-Sample Tests

46. A survey claims that 9 out of 10 doctors recommend aspirin for their patients with headaches. To test this claim against the alternative that the actual proportion of doctors who recommend aspirin is less than 0.90, a random sample of 100 doctors results in 83 who indicate that they recommend aspirin. The value of the test statistic in this problem is approximately equal to: a) – 4.12 b) – 2.33 c) – 1.86 d) – 0.07 ANSWER: b TYPE: MC DIFFICULTY: Easy KEYWORDS: one-tailed test, proportion, test statistic 47. A survey claims that 9 out of 10 doctors recommend aspirin for their patients with headaches. To

test this claim against the alternative that the actual proportion of doctors who recommend aspirin is less than 0.90, a random sample of 100 doctors was selected. Suppose that the test statistic is – 2.20. Can we conclude that H0 should be rejected at the (a)  = 0.10, (b)  = 0.05, and (c)  = 0.01 level of Type I error? a) (a) yes; (b) yes; (c) yes b) (a) no; (b) no; (c) no c) (a) no; (b) no; (c) yes d) (a) yes; (b) yes; (c) no ANSWER: d TYPE: MC DIFFICULTY: Moderate KEYWORDS: one-tailed test, proportion, decision 48. A survey claims that 9 out of 10 doctors recommend aspirin for their patients with headaches. To

test this claim against the alternative that the actual proportion of doctors who recommend aspirin is less than 0.90, a random sample of 100 doctors was selected. Suppose you reject the null hypothesis. What conclusion can you draw? a) There is not sufficient evidence that the proportion of doctors who recommend aspirin is not less than 0.90. b) There is sufficient evidence that the proportion of doctors who recommend aspirin is not less than 0.90. c) There is not sufficient evidence that the proportion of doctors who recommend aspirin is less than 0.90. d) There is sufficient evidence that the proportion of doctors who recommend aspirin is less than 0.90. ANSWER: d TYPE: MC DIFFICULTY: Moderate KEYWORDS: one-tailed test, proportion, conclusion 49. A major videocassette rental chain is considering opening a new store in an area that currently does not have any such stores. The chain will open if there is evidence that more than 5,000 of the 20,000 households in the area are equipped with videocassette recorders (VCRs). It conducts a telephone poll of 300 randomly selected households in the area and finds that 96 have VCRs. State the test of interest to the rental chain. a) H 0 :   0.32 versus H1 :   0.32

Fundamentals of Hypothesis Testing: One-Sample Tests b) H 0 :   0.25 versus H1 :   0.25

H 0 :   5,000 versus H1 :   5,000 d) H0 :   5,000 versus H1 :   5,000 c)

ANSWER: b TYPE: MC DIFFICULTY: Easy KEYWORDS: one-tailed test, proportion, form of hypothesis, form of hypothesis 50. A major videocassette rental chain is considering opening a new store in an area that currently does not have any such stores. The chain will open if there is evidence that more than 5,000 of the 20,000 households in the area are equipped with videocassette recorders (VCRs). It conducts a telephone poll of 300 randomly selected households in the area and finds that 96 have VCRs. The value of the test statistic in this problem is approximately equal to: a) 2.80 b) 2.60 c) 1.94 d) 1.30 ANSWER: a TYPE: MC DIFFICULTY: Easy KEYWORDS: one-tailed test, proportion, Z test, test statistic 51. A major videocassette rental chain is considering opening a new store in an area that currently

does not have any such stores. The chain will open if there is evidence that more than 5,000 of the 20,000 households in the area are equipped with videocassette recorders (VCRs). It conducts a telephone poll of 300 randomly selected households in the area and finds that 96 have VCRs. The p-value associated with the test statistic in this problem is approximately equal to: a) 0.0100 b) 0.0051 c) 0.0026 d) 0.0013 ANSWER: c TYPE: MC DIFFICULTY: Easy KEYWORDS: one-tailed test, proportion, Z test, p-value 52. A major videocassette rental chain is considering opening a new store in an area that currently does not have any such stores. The chain will open if there is evidence that more than 5,000 of the 20,000 households in the area are equipped with videocassette recorders (VCRs). It conducts a telephone poll of 300 randomly selected households in the area and finds that 96 have VCRs. The decision on the hypothesis test using a 3% level of significance is: a) to reject H0 in favor of H1. b) to accept H0 in favor of H1. c) to fail to reject H0 in favor of H1. d) we cannot tell what the decision should be from the information given. ANSWER: a TYPE: MC DIFFICULTY: Easy KEYWORDS: one-tailed test, proportion, Z test, decision

Fundamentals of Hypothesis Testing: One-Sample Tests

53. A major videocassette rental chain is considering opening a new store in an area that currently does not have any such stores. The chain will open if there is evidence that more than 5,000 of the 20,000 households in the area are equipped with videocassette recorders (VCRs). It conducts a telephone poll of 300 randomly selected households in the area and finds that 96 have VCRs. The rental chain's conclusion from the hypothesis test using a 3% level of significance is: a) to open a new store. b) not to open a new store. c) to delay opening a new store until additional evidence is collected. d) we cannot tell what the decision should be from the information given. ANSWER: a TYPE: MC DIFFICULTY: Easy KEYWORDS: one-tailed test, proportion, Z test, conclusion 54. An entrepreneur is considering the purchase of a coin-operated laundry. The current owner

claims that over the past 5 years, the average daily revenue was $675 with a standard deviation of $75. A sample of 30 days reveals a daily average revenue of $625. If you were to test the null hypothesis that the daily average revenue was $675, which test would you use? a) Z-test of a population mean b) Z-test of a population proportion c) t-test of a population mean d) t-test of a population proportion ANSWER: a TYPE: MC DIFFICULTY: Moderate KEYWORDS: two-tailed test, mean, Z test 55. An entrepreneur is considering the purchase of a coin-operated laundry. The current owner

claims that over the past 5 years, the average daily revenue was $675 with a standard deviation of $75. A sample of 30 days reveals a daily average revenue of $625. If you were to test the null hypothesis that the daily average revenue was $675 and decide not to reject the null hypothesis, what can you conclude? a) There is not enough evidence to conclude that the daily average revenue was $675. b) There is not enough evidence to conclude that the daily average revenue was not $675. c) There is enough evidence to conclude that the daily average revenue was $675. d) There is enough evidence to conclude that the daily average revenue was not $675. ANSWER: b TYPE: MC DIFFICULTY: Moderate KEYWORDS: two-tailed test, mean, Z test, conclusion 56. A manager of the credit department for an oil company would like to determine whether the average monthly balance of credit card holders is equal to $75. An auditor selects a random sample of 100 accounts and finds that the average owed is $83.40 with a sample standard deviation of $23.65. If you wanted to test whether the auditor should conclude that there is evidence that the average balance is different from $75, which test would you use? a) Z-test of a population mean b) Z-test of a population proportion c) t-test of a population mean d) t-test of a population proportion

Fundamentals of Hypothesis Testing: One-Sample Tests ANSWER: c TYPE: MC DIFFICULTY: Easy KEYWORDS: two-tailed test, mean, t test 57. A manager of the credit department for an oil company would like to determine whether the

average monthly balance of credit card holders is equal to $75. An auditor selects a random sample of 100 accounts and finds that the average owed is $83.40 with a sample standard deviation of $23.65. If you were to conduct a test to determine whether the average balance is different from $75 and decided to reject the null hypothesis, what conclusion could you draw? a) There is not evidence that the average balance is $75. b) There is not evidence that the average balance is not $75. c) There is evidence that the average balance is $75. d) There is evidence that the average balance is not $75. ANSWER: d TYPE: MC DIFFICULTY: moderate KEYWORDS: two-tailed test, mean, t test, conclusion 58. The marketing manager for an automobile manufacturer is interested in determining the proportion of new compact-car owners who would have purchased a passenger-side inflatable air bag if it had been available for an additional cost of $300. The manager believes from previous information that the proportion is 0.30. Suppose that a survey of 200 new compact-car owners is selected and 79 indicate that they would have purchased the inflatable air bags. If you were to conduct a test to determine whether there is evidence that the proportion is different from 0.30, which test would you use? a) Z-test of a population mean b) Z-test of a population proportion c) t-test of population mean d) t-test of a population proportion ANSWER: b TYPE: MC DIFFICULTY: Easy KEYWORDS: two-tailed test, proportion 59. The marketing manager for an automobile manufacturer is interested in determining the proportion of new compact-car owners who would have purchased a passenger-side inflatable air bag if it had been available for an additional cost of $300. The manager believes from previous information that the proportion is 0.30. Suppose that a survey of 200 new compact-car owners is selected and 79 indicate that they would have purchased the inflatable air bags. If you were to conduct a test to determine whether there is evidence that the proportion is different from 0.30 and decided not to reject the null hypothesis, what conclusion could you draw? a) There is sufficient evidence that the proportion is 0.30. b) There is not sufficient evidence that the proportion is 0.30. c) There is sufficient evidence that the proportion is 0.30. d) There is not sufficient evidence that the proportion is not 0.30. ANSWER: d TYPE: MC DIFFICULTY: Easy KEYWORDS: two-tailed test, proportion, conclusion

Fundamentals of Hypothesis Testing: One-Sample Tests

TABLE 9-1 Microsoft Excel was used on a set of data involving the number of parasites found on 46 Monarch butterflies captured in Pismo Beach State Park. A biologist wants to know if the mean number of parasites per butterfly is over 20. She will make her decision using a test with a level of significance of 0.10. The following information was extracted from the Microsoft Excel output for the sample of 46 Monarch butterflies: n = 46; Arithmetic Mean = 28.00; Standard Deviation = 25.92; Standard Error = 3.82; Null Hypothesis: H0 :   20.000 ;  = 0.10; df = 45; T Test Statistic = 2.09; One-Tailed Test Upper Critical Value = 1.3006; p-value = 0.021; Decision = Reject. 60. Referring to Table 9-1, the parameter the biologist is interested in is: a) the mean number of butterflies in Pismo Beach State Park. b) the mean number of parasites on these 46 butterflies. c) the mean number of parasites on Monarch butterflies in Pismo Beach State Park. d) the proportion of butterflies with parasites. ANSWER: c TYPE: MC DIFFICULTY: Easy KEYWORDS: mean, t test, parameter 61. Referring to Table 9-1, state the alternative hypothesis for this study. ANSWER:

H1 :   20.000 TYPE: PR DIFFICULTY: Easy KEYWORDS: one-tailed test, mean, t test, form of hypothesis 62. Referring to Table 9-1, what critical value should the biologist use to determine the rejection region? a) 1.6794 b) 1.3011 c) 1.3006 d) 0.6800 ANSWER: c TYPE: MC DIFFICULTY: Easy KEYWORDS: one-tailed test, mean, t test, critical value 63. True or False: Referring to Table 9-1, the null hypothesis would be rejected. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: one-tailed test, mean, t test, decision 64. True or False: Referring to Table 9-1, the null hypothesis would be rejected if a 4% probability of committing a Type I error is allowed. ANSWER: True

Fundamentals of Hypothesis Testing: One-Sample Tests TYPE: TF DIFFICULTY: Easy KEYWORDS: one-tailed test, mean, t test, decision 65. True or False: Referring to Table 9-1, the null hypothesis would be rejected if a 1% probability of committing a Type I error is allowed. ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: one-tailed test, mean, t test, decision 66. Referring to Table 9-1, the lowest level of significance at which the null hypothesis can be

rejected is

ANSWER: 0.021 TYPE: FI DIFFICULTY: Moderate KEYWORDS: one-tailed test, mean, t test, p-value 67. True or False: Referring to Table 9-1, the evidence proves beyond a doubt that the mean number

of parasites on butterflies in Pismo Beach State Park is over 20. ANSWER: False TYPE: TF DIFFICULTY: Moderate KEYWORDS: one-tailed test, mean, t test, conclusion 68. True of False: Referring to Table 9-1, the biologist can conclude that there is sufficient evidence

to show that the average number of parasites per butterfly is over 20 using a level of significance of 0.10. ANSWER: True TYPE: TF DIFFICULTY: Moderate KEYWORDS: one-tailed test, mean, t test, conclusion 69. True or False: Referring to Table 9-1, the biologist can conclude that there is sufficient evidence

to show that the average number of parasites per butterfly is over 20 with no more than a 5% probability of incorrectly rejecting the true null hypothesis. ANSWER: True TYPE: TF DIFFICULTY: Moderate KEYWORDS: one-tailed test, mean, t test, decision 70. True or False: Referring to Table 9-1, the biologist can conclude that there is sufficient evidence

to show that the average number of parasites per butterfly is over 20 with no more than a 1% probability of incorrectly rejecting the true null hypothesis. ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: one-tailed test, mean, t test, decision

Fundamentals of Hypothesis Testing: One-Sample Tests

71. True or False: Referring to Table 9-1, the value of  is 0.90.

ANSWER: False TYPE: TF DIFFICULTY: Moderate KEYWORDS: one-tailed test, mean, t test, beta-risk 72. True or False: Referring to Table 9-1, if these data were used to perform a two-tailed test, the p-

value would be 0.042. ANSWER: True TYPE: TF DIFFICULTY: Moderate KEYWORDS: one-tailed test, mean, t test, p-value 73. Referring to Table 9-1, the power of the test is if the mean number of parasites per butterfly on Monarch butterflies in Pismo Beach State Park is 18 using a 0.1 level of significance and assuming that the population standard deviation is 25.92. ANSWER: 0.0355 TYPE: FI DIFFICULTY: Difficult EXPLANATION: The power is computed for the Z test because the population standard deviation is given. KEYWORDS: one-tailed test, mean, Z test, power 74. Referring to Table 9-1, the probability of committing a Type II error is if the mean number of parasites per butterfly on Monarch butterflies in Pismo Beach State Park is 18 using a 0.1 level of significance and assuming that the population standard deviation is 25.92. ANSWER: 0.9645 TYPE: FI DIFFICULTY: Moderate KEYWORDS: one-tailed test, mean, Z test, type II error 75. Referring to Table 9-1, the power of the test is if the mean number of parasites per butterfly on Monarch butterflies in Pismo Beach State Park is 24 using a 0.1 level of significance and assuming that the population standard deviation is 25.92. ANSWER: 0.4071 TYPE: FI DIFFICULTY: Moderate KEYWORDS: one-tailed test, mean, Z test, power 76. Referring to Table 9-1, the probability of committing a Type II error is if the mean number of parasites per butterfly on Monarch butterflies in Pismo Beach State Park is 24 using a 0.1 level of significance and assuming that the population standard deviation is 25.92. ANSWER: 0.5929 TYPE: FI DIFFICULTY: Moderate KEYWORDS: one-tailed test, mean, Z test, type II error

Fundamentals of Hypothesis Testing: One-Sample Tests 77. Referring to Table 9-1, the power of the test is if the mean number of parasites per butterfly on Monarch butterflies in Pismo Beach State Park is 30 using a 0.1 level of significance and assuming that the population standard deviation is 25.92. ANSWER: 0.9091 TYPE: FI DIFFICULTY: Moderate KEYWORDS: one-tailed test, mean, Z test, power 78. Referring to Table 9-1, the probability of committing a Type II error is if the mean number of parasites per butterfly on Monarch butterflies in Pismo Beach State Park is 30 using a 0.1 level of significance and assuming that the population standard deviation is 25.92. ANSWER: 0.0909 TYPE: FI DIFFICULTY: Moderate KEYWORDS: one-tailed test, mean, Z test, type II error 79. Referring to Table 9-1, the power of the test is if the mean number of parasites per butterfly on Monarch butterflies in Pismo Beach State Park is 18 using a 0.05 level of significance and assuming that the population standard deviation is 25.92. ANSWER: 0.0151 TYPE: FI DIFFICULTY: Moderate KEYWORDS: one-tailed test, mean, Z test, power 80. Referring to Table 9-1, the probability of committing a Type II error is if the mean number of parasites per butterfly on Monarch butterflies in Pismo Beach State Park is 18 using a 0.05 level of significance and assuming that the population standard deviation is 25.92. ANSWER: 0.9849 TYPE: FI DIFFICULTY: Moderate KEYWORDS: one-tailed test, mean, Z test, type II error 81. Referring to Table 9-1, the power of the test is if the mean number of parasites per butterfly on Monarch butterflies in Pismo Beach State Park is 24 using a 0.05 level of significance and assuming that the population standard deviation is 25.92. ANSWER: 0.2749 TYPE: FI DIFFICULTY: Moderate KEYWORDS: one-tailed test, mean, Z test, power 82. Referring to Table 9-1, the probability of committing a Type II error is if the mean number of parasites per butterfly on Monarch butterflies in Pismo Beach State Park is 24 using a 0.05 level of significance and assuming that the population standard deviation is 25.92. ANSWER: 0.7251 TYPE: FI DIFFICULTY: Moderate KEYWORDS: one-tailed test, mean, Z test, type II error

Fundamentals of Hypothesis Testing: One-Sample Tests

83. Referring to Table 9-1, the power of the test is if the mean number of parasites per butterfly on Monarch butterflies in Pismo Beach State Park is 30 using a 0.05 level of significance and assuming that the population standard deviation is 25.92. ANSWER: 0.8344 TYPE: FI DIFFICULTY: Moderate KEYWORDS: one-tailed test, mean, Z test, power 84. Referring to Table 9-1, the probability of committing a Type II error is if the mean number of parasites per butterfly on Monarch butterflies in Pismo Beach State Park is 30 using a 0.05 level of significance and assuming that the population standard deviation is 25.92. ANSWER: 0.1656 TYPE: FI DIFFICULTY: Moderate KEYWORDS: one-tailed test, mean, Z test, type II error 85. True or False: Suppose, in testing a hypothesis about a proportion, the p-value is computed to be

0.043. The null hypothesis should be rejected if the chosen level of significance is 0.05. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: mean, t test, p-value, level of significance, decision 86. True or False: Suppose, in testing a hypothesis about a proportion, the p-value is computed to be

0.034. The null hypothesis should be rejected if the chosen level of significance is 0.01. ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: p-value, level of significance, decision 87. True or False: Suppose, in testing a hypothesis about a proportion, the Z test statistic is computed

to be 2.04. The null hypothesis should be rejected if the chosen level of significance is 0.01 and a two-tailed test is used. ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: proportion, Z test, test statistic, critical value, decision 88. True or False: 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 TYPE: TF DIFFICULTY: Easy KEYWORDS: ethical issues 89. True or False: The test statistic measures how close the computed sample statistic has come to the hypothesized population parameter.

Fundamentals of Hypothesis Testing: One-Sample Tests

ANSWER: True TYPE: TF DIFFICULTY: Moderate KEYWORDS: test statistic 90. True or False: The statement of the null hypothesis always contains an equality. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: form of null hypoehsis 91. True or False: The larger is the p-value, the more likely one is to reject the null hypothesis.

ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: p-value 92. True or False: The smaller is the p-value, the stronger is the evidence against the null hypothesis.

ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: p-value 93. True or False: A sample is used to obtain a 95% 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 TYPE: TF DIFFICULTY: Easy KEYWORDS: confidence interval, two-tailed test, decision 94. True or False: A sample is used to obtain a 95% 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 18 versus the alternative hypothesis that the mean of the population differs from 18, the null hypothesis could be rejected at a level of significance of 0.05. ANSWER: False TYPE: TF DIFFICULTY: Moderate KEYWORDS: confidence interval, two-tailed test, decision 95. True or False: A sample is used to obtain a 95% 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 0.10.

Fundamentals of Hypothesis Testing: One-Sample Tests

ANSWER: True TYPE: TF DIFFICULTY: Moderate KEYWORDS: confidence interval, two-tailed test, decision 96. True or False: A sample is used to obtain a 95% 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 0.02. ANSWER: False TYPE: TF DIFFICULTY: Moderate EXPLANATION: We are not sure if 20 will be in the wider confidence interval. KEYWORDS: confidence interval, two-tailed test, decision 97. True or False: A sample is used to obtain a 95% 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 accepted at a level of significance of 0.02. ANSWER: False TYPE: TF DIFFICULTY: Moderate EXPLANATION: We are not sure if 20 will be in the wider confidence interval. KEYWORDS: confidence interval, two-tailed test, decision TABLE 9-2 A student claims that he can correctly identify whether a person is a business major or an agriculture major by the way the person dresses. Suppose in actuality that if someone is a business major, he can correctly identify that person as a business major 87% of the time. When a person is an agriculture major, the student will incorrectly identify that person as a business major 16% of the time. Presented with one person and asked to identify the major of this person (who is either a business or agriculture major), he considers this to be a hypothesis test with the null hypothesis being that the person is a business major and the alternative that the person is an agriculture major. 98. Referring to Table 9-2, what would be a Type I error? a) Saying that the person is a business major when in fact the person is a business major. b) Saying that the person is a business major when in fact the person is an agriculture major. c) Saying that the person is an agriculture major when in fact the person is a business major. d) Saying that the person is an agriculture major when in fact the person is an agriculture major. ANSWER: c TYPE: MC DIFFICULTY: Difficult KEYWORDS: form of hypothesis, form of hypothesis

Fundamentals of Hypothesis Testing: One-Sample Tests 99. Referring to Table 9-2, what would be a Type II error? a) Saying that the person is a business major when in fact the person is a business major. b) Saying that the person is a business major when in fact the person is an agriculture major. c) Saying that the person is an agriculture major when in fact the person is a business major. d) Saying that the person is an agriculture major when in fact the person is an agriculture major. ANSWER: b TYPE: MC DIFFICULTY: Difficult KEYWORDS: type II error 100.

Referring to Table 9-2, what is the “actual level of significance” of the test? a) 0.13 b) 0.16 c) 0.84 d) 0.87

ANSWER: a TYPE: MC DIFFICULTY: Difficult KEYWORDS: level of significance 101.

Referring to Table 9-2, what is the “actual confidence coefficient”? a) 0.13 b) 0.16 c) 0.84 d) 0.87

ANSWER: d TYPE: MC DIFFICULTY: Difficult KEYWORDS: confidence coefficient 102. Referring to Table 9-2, what is the value of  ?

a) 0.13 b) 0.16 c) 0.84 d) 0.87 ANSWER: a TYPE: MC DIFFICULTY: Difficult KEYWORDS: level of significance 103. Referring to Table 9-2, what is the value of  ?

a) 0.13 b) 0.16 c) 0.84 d) 0.87 ANSWER:

Fundamentals of Hypothesis Testing: One-Sample Tests b TYPE: MC DIFFICULTY: Difficult KEYWORDS: beta-risk 104.

Referring to Table 9-2, what is the power of the test? a) 0.13 b) 0.16 c) 0.84 d) 0.87

ANSWER: c TYPE: MC DIFFICULTY: Difficult KEYWORDS: power TABLE 9-3 An appliance manufacturer claims to have developed a compact microwave oven that consumes an average of no more than 250 W. From previous studies, it is believed that power consumption for microwave ovens is normally distributed with a standard deviation of 15 W. A consumer group has decided to try to discover if the claim appears true. They take a sample of 20 microwave ovens and find that they consume an average of 257.3 W. 105.

Referring to Table 9-3, the population of interest is a) the power consumption in the 20 microwave ovens. b) the power consumption in all such microwave ovens. c) the mean power consumption in the 20 microwave ovens. d) the mean power consumption in all such microwave ovens.

ANSWER: b TYPE: MC DIFFICULTY: Moderate KEYWORDS: one-tailed test, mean, Z test 106.

Referring to Table 9-3, the parameter of interest is a) the mean power consumption of the 20 microwave ovens. b) the mean power consumption of all such microwave ovens. c) 250 d) 257.3

ANSWER: b TYPE: MC DIFFICULTY: Easy KEYWORDS: one-tailed test, mean, Z test, parameter 107. Referring to Table 9-3, the appropriate hypotheses to determine if the manufacturer's claim appears reasonable are: a) H 0 :  = 250 versus H1 :   250 b) H 0 :   250 versus H1 :   250 c) H 0 :   250 versus H1 :   250 d) H 0 :   257.3 versus H1 :   257.3

Fundamentals of Hypothesis Testing: One-Sample Tests ANSWER: c TYPE: MC DIFFICULTY: Moderate KEYWORDS: one-tailed test, mean, Z test, form of hypothesis, form of hypothesis 108.

Referring to Table 9-3, for a test with a level of significance of 0.05, the critical value would be .

ANSWER: Z = 1.645 TYPE: FI DIFFICULTY: Moderate KEYWORDS: one-tailed test, mean, Z test, critical value 109. Referring to Table 9-3, the value of the test statistic is

ANSWER: 2.18 TYPE: FI DIFFICULTY: Easy KEYWORDS: one-tailed test, mean, Z test, test statistic 110. Referring to Table 9-3, the p-value of the test is

ANSWER: 0.0148 using Excel or 0.0146 using Table E.2 TYPE: FI DIFFICULTY: Moderate KEYWORDS: one-tailed test, mean, Z test, p-value 111. True or False: Referring to Table 9-3, for this test to be valid, it is necessary that the power consumption for microwave ovens has a normal distribution. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: one-tailed test, mean, Z test, assumptions 112. True or False: Referring to Table 9-3, the null hypothesis will be rejected at 5% level of significance. ANSWER: True TYPE: TF DIFFICULTY: Moderate KEYWORDS: one-tailed test, mean, Z test, decision 113. True or False: Referring to Table 9-3, the null hypothesis will be rejected at 1% level of significance. ANSWER: False TYPE: TF DIFFICULTY: Moderate KEYWORDS: one-tailed test, mean, Z test, decision 114. True or False: Referring to Table 9-3, the consumer group can conclude that there is enough evidence to prove that the manufacturer’s claim is not true when allowing for a 5% probability of committing a type I error.

Fundamentals of Hypothesis Testing: One-Sample Tests

ANSWER: True TYPE: TF DIFFICULTY: Difficult KEYWORDS: one-tailed test, mean, Z test, conclusion 115. Referring to Table 9-3, what is the power of the test if the average power consumption of all such microwave ovens is in fact 257.3 W using a 0.05 level of significance? ANSWER: 0.7025 TYPE: PR DIFFICULTY: Difficult KEYWORDS: one-tailed test, mean, Z test, power 116. Referring to Table 9-3, what is the probability of making a Type II error if the average power consumption of all such microwave ovens is in fact 257.3 W using a 0.05 level of significance? ANSWER: 0.2975 TYPE: PR DIFFICULTY: Difficult KEYWORDS: one-tailed test, mean, Z test, type II error 117. Referring to Table 9-3, what is the power of the test if the average power consumption of all such microwave ovens is in fact 248 W using a 0.05 level of significance? ANSWER: 0.0125 TYPE: PR DIFFICULTY: Difficult KEYWORDS: one-tailed test, mean, Z test, power 118. Referring to Table 9-3, what is the probability of making a Type II error if the average power consumption of all such microwave ovens is in fact 248 W using a 0.05 level of significance? ANSWER: 0.9875 TYPE: PR DIFFICULTY: Difficult KEYWORDS: one-tailed test, mean, Z test, type II error 119. Referring to Table 9-3, what is the power of the test if the average power consumption of all such microwave ovens is in fact 257.3 W using a 0.10 level of significance? ANSWER: 0.8146 TYPE: PR DIFFICULTY: Difficult KEYWORDS: one-tailed test, mean, Z test, power 120. Referring to Table 9-3, what is the probability of making a Type II error if the average power consumption of all such microwave ovens is in fact 257.3 W using a 0.10 level of significance? ANSWER: 0.1854 TYPE: PR DIFFICULTY: Difficult KEYWORDS: one-tailed test, mean, Z test, type II error

Fundamentals of Hypothesis Testing: One-Sample Tests 121. Referring to Table 9-3, what is the power of the test if the average power consumption of all such microwave ovens is in fact 248 W using a 0.10 level of significance? ANSWER: 0.0302 TYPE: PR DIFFICULTY: Difficult KEYWORDS: one-tailed test, mean, Z test, power 122. Referring to Table 9-3, what is the probability of making a Type II error if the average power consumption of all such microwave ovens is in fact 248 W using a 0.10 level of significance? ANSWER: 0.9698 TYPE: PR DIFFICULTY: Difficult KEYWORDS: one-tailed test, mean, Z test, type II error TABLE 9-4 A drug company is considering marketing a new local anesthetic. The effective time of the anesthetic the drug company is currently producing has a normal distribution with an average of 7.4 minutes with a standard deviation of 1.2 minutes. The chemistry of the new anesthetic is such that the effective time should be normal with the same standard deviation, but the mean effective time may be lower. If it is lower, the drug company will market the new anesthetic; otherwise, they will continue to produce the older one. A sample of size 36 results in a sample mean of 7.1. A hypothesis test will be done to help make the decision. 123. Referring to Table 9-4, the appropriate hypotheses are: a) H 0 :  = 7.4 versus H1 :   7.4 b) H 0 :   7.4 versus H 1 :   7.4 c) H 0 :   7.4 versus H1 :   7.4 d) H 0 :   7.4 versus H1 :   7.4 ANSWER: c TYPE: MC DIFFICULTY: Easy KEYWORDS: one-tailed test, mean, Z test, form of hypothesis, form of hypothesis 124. Referring to Table 9-4, for a test with a level of significance of 0.10, the critical value would be . ANSWER: -1.28 TYPE: FI DIFFICULTY: Easy KEYWORDS: one-tailed test, mean, Z test, critical value 125. Referring to Table 9-4, the value of the test statistic is

ANSWER: -1.50 TYPE: FI DIFFICULTY: Easy KEYWORDS: one-tailed test, mean, Z test, test statistic

Fundamentals of Hypothesis Testing: One-Sample Tests

126. Referring to Table 9-4, the p-value of the test is

ANSWER: 0.0668 TYPE: FI DIFFICULTY: Moderate KEYWORDS: one-tailed test, mean, Z test, p-value 127. True or False: Referring to Table 9-4, the null hypothesis will be rejected with a level of

significance of 0.10. ANSWER: True TYPE: TF DIFFICULTY: Moderate KEYWORDS: one-tailed test, mean, Z test, decision 128. True or False: Referring to Table 9-4, if the level of significance had been chosen as 0.05, the

null hypothesis would be rejected. ANSWER: False TYPE: TF DIFFICULTY: Moderate KEYWORDS: one-tailed test, mean, Z test, decision 129. True or False: Referring to Table 9-4, if the level of significance had been chosen as 0.05, the

company would market the new anesthetic. ANSWER: False TYPE: TF DIFFICULTY: Moderate KEYWORDS: one-tailed test, mean, Z test, conclusion 130. Referring to Table 9-4, what is the power of the test if the average effective time of the anesthetic

is 7.0 using a 0.05 level of significance? ANSWER: 0.6388 TYPE: PR DIFFICULTY: Difficult KEYWORDS: one-tailed test, mean, Z test, power 131. Referring to Table 9-4, what is the probability of making a Type II error if the average effective time of the anesthetic is 7.0 using a 0.05 level of significance? ANSWER: 0.3612 TYPE: PR DIFFICULTY: Difficult KEYWORDS: one-tailed test, mean, Z test, type II error 132. Referring to Table 9-4, what is the power of the test if the average effective time of the anesthetic is 7.5 using a 0.05 level of significance? ANSWER: 0.0160 TYPE: PR DIFFICULTY: Difficult

Fundamentals of Hypothesis Testing: One-Sample Tests KEYWORDS: one-tailed test, mean, Z test, power 133. Referring to Table 9-4, what is the probability of making a Type II error if the average effective time of the anesthetic is 7.5 using a 0.05 level of significance? ANSWER: 0.9840 TYPE: PR DIFFICULTY: Difficult KEYWORDS: one-tailed test, mean, Z test, type II error 134. Referring to Table 9-4, what is the power of the test if the average effective time of the anesthetic is 7.0 using a 0.10 level of significance? ANSWER: 0.7638 TYPE: PR DIFFICULTY: Difficult KEYWORDS: one-tailed test, mean, Z test, power 135. Referring to Table 9-4, what is the probability of making a Type II error if the average effective time of the anesthetic is 7.0 using a 0.10 level of significance? ANSWER: 0.2362 TYPE: PR DIFFICULTY: Difficult KEYWORDS: one-tailed test, mean, Z test, type II error 136. Referring to Table 9-4, what is the power of the test if the average effective time of the anesthetic is 7.5 using a 0.10 level of significance? ANSWER: 0.0374 TYPE: PR DIFFICULTY: Difficult KEYWORDS: one-tailed test, mean, Z test, power 137. Referring to Table 9-4, what is the probability of making a Type II error if the average effective time of the anesthetic is 7.5 using a 0.10 level of significance? ANSWER: 0.9626 TYPE: PR DIFFICULTY: Difficult KEYWORDS: one-tailed test, mean, Z test, type II error TABLE 9-5 A bank tests the null hypothesis that the mean age of the bank's mortgage holders is less than or equal to 45, versus an alternative that the mean age is greater than 45. They take a sample and calculate a p-value of 0.0202. 138. True or False: Referring to Table 9-5, the null hypothesis would be rejected at a significance

level of  = 0.05.

ANSWER: True TYPE: TF DIFFICULTY: Easy

Fundamentals of Hypothesis Testing: One-Sample Tests

KEYWORDS: one-tailed test, mean, decision 139. True or False: Referring to Table 9-5, the null hypothesis would be rejected at a significance

level of  = 0.01.

ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: one-tailed test, mean, decision 140. True or False: Referring to Table 9-5, the bank can conclude that the average age is greater than

45 at a significance level of  = 0.01.

ANSWER: False TYPE: TF DIFFICULTY: Moderate KEYWORDS: one-tailed test, mean, conclude 141. Referring to Table 9-5, if the same sample was used to test the opposite one-tailed test, what

would be that test's p-value? a) 0.0202 b) 0.0404 c) 0.9596 d) 0.9798 ANSWER: d TYPE: MC DIFFICULTY: Moderate KEYWORDS: one-tailed test, mean, p-value TABLE 9-6 The quality control engineer for a furniture manufacturer is interested in the mean amount of force necessary to produce cracks in stressed oak furniture. She performs a two-tailed test of the null hypothesis that the mean for the stressed oak furniture is 650. The calculated value of the Z test statistic is a positive number that leads to a p-value of 0.080 for the test. 142. True or False: Referring to Table 9-6, if the test is performed with a level of significance of 0.10, the null hypothesis would be rejected. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: one-tailed test, mean, decision 143. True or False: Referring to Table 9-6, if the test is performed with a level of significance of 0.10, the engineer can conclude that the mean amount of force necessary to produce cracks in stressed oak furniture is 650. ANSWER: False TYPE: TF DIFFICULTY: Moderate KEYWORDS: one-tailed test, mean, conclusion

Fundamentals of Hypothesis Testing: One-Sample Tests 144. True or False: Referring to Table 9-6, if the test is performed with a level of significance of 0.05, the null hypothesis would be rejected. ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: one-tailed test, mean, decision 145. True or False: Referring to Table 9-6, if the test is performed with a level of significance of 0.05, the engineer can conclude that the mean amount of force necessary to produce cracks in stressed oak furniture is 650. ANSWER: False TYPE: TF DIFFICULTY: Moderate EXPLANATION: the engineer can conclude that there is insufficient evidence to show that the mean amount of force needed is not 650 but cannot conclude that there is evidence to show that the force needed is 650. KEYWORDS: one-tailed test, mean, conclusion 146. True or False: Referring to Table 9-6, suppose the engineer had decided that the alternative hypothesis to test was that the mean was greater than 650. Then if the test is performed with a level of significance of 0.10, the null hypothesis would be rejected. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: one-tailed test, mean, decision 147. Referring to Table 9-6, suppose the engineer had decided that the alternative hypothesis to test

was that the mean was greater than 650. What would be the p-value of this one-tailed test? a) 0.040 b) 0.160 c) 0.840 d) 0.960 ANSWER: a TYPE: MC DIFFICULTY: Moderate KEYWORDS: one-tailed test, mean, p-value 148. Referring to Table 9-6, suppose the engineer had decided that the alternative hypothesis to test

was that the mean was less than 650. What would be the p-value of this one-tailed test? a) 0.040 b) 0.160 c) 0.840 d) 0.960 ANSWER: d TYPE: MC DIFFICULTY: Moderate KEYWORDS: one-tailed test, mean, p-value

Fundamentals of Hypothesis Testing: One-Sample Tests

149. True or False: Referring to Table 9-6, suppose the engineer had decided that the alternative hypothesis to test was that the mean was less than 650. Then if the test is performed with a level of significance of 0.10, the null hypothesis would be rejected. ANSWER: False TYPE: TF DIFFICULTY: Moderate KEYWORDS: one-tailed test, mean, decision TABLE 9-7 A major home improvement store conducted its biggest brand recognition campaign in the company’s history. A series of new television advertisements featuring well-known entertainers and sports figures were launched. A key metric for the success of television advertisements is the proportion of viewers who “like the ads a lot”. A study of 1,189 adults who viewed the ads reported that 230 indicated that they “like the ads a lot.” The percentage of a typical television advertisement receiving the “like the ads a lot” score is believed to be 22%. Company officials wanted to know if there is evidence that the series of television advertisements are less successful than the typical ad (i.e. if there is evidence that the population proportion of “like the ads a lot” for the company’s ads is less than 0.22) at a 0.01 level of significance. 150. Referring to Table 9-7, the parameter the company officials is interested in is: a) the mean number of viewers who “like the ads a lot”. b) the total number of viewers who “like the ads a lot”. c) the mean number of company officials who “like the ads a lot”. d) the proportion of viewers who “like the ads a lot”. ANSWER: d TYPE: MC DIFFICULTY: Easy KEYWORDS: proportion, t test, parameter 151. Referring to Table 9-7, state the null hypothesis for this study. ANSWER:

H 0 :   0.22 TYPE: PR DIFFICULTY: Easy KEYWORDS: one-tailed test, proportion, t test, form of hypothesis 152. Referring to Table 9-7, state the alternative hypothesis for this study. ANSWER:

H1 :   0.22 TYPE: PR DIFFICULTY: Easy KEYWORDS: one-tailed test, proportion, t test, form of hypothesis 153. Referring to Table 9-7, what critical value should the company officials use to determine the rejection region? ANSWER: −2.3263 TYPE: PR DIFFICULTY: Easy KEYWORDS: one-tailed test, proportion, t test, critical value

Fundamentals of Hypothesis Testing: One-Sample Tests 154. Referring to Table 9-7, the null hypothesis will be rejected if the test statistics is a) greater than 2.3263 b) less than 2.3263 c) greater than −2.3263 d) less than −2.3263 ANSWER: d TYPE: MC DIFFICULTY: Easy KEYWORDS: one-tailed test, proportion, t test, rejection region 155. True or False: Referring to Table 9-7, the null hypothesis would be rejected. ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: one-tailed test, proportion, t test, decision 156. Referring to Table 9-7, the lowest level of significance at which the null hypothesis can be rejected is . ANSWER: 0.0135 TYPE: FI DIFFICULTY: Moderate KEYWORDS: one-tailed test, proportion, t test, p-value 157. Referring to Table 9-7, the largest level of significance at which the null hypothesis will not be rejected is . ANSWER: 0.0135 TYPE: FI DIFFICULTY: Moderate KEYWORDS: one-tailed test, proportion, t test, p-value 158. True of False: Referring to Table 9-7, the company officials can conclude that there is sufficient

evidence to show that the series of television advertisements are less successful than the typical ad using a level of significance of 0.01. ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: one-tailed test, proportion, t test, conclusion 159. True or False: Referring to Table 9-7, the company officials can conclude that there is sufficient

evidence to show that the series of television advertisements are less successful than the typical ad using a level of significance of 0.05. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: one-tailed test, proportion, t test, decision 160. True or False: Referring to Table 9-7, the value of  is 0.90.

Fundamentals of Hypothesis Testing: One-Sample Tests

ANSWER: False TYPE: TF DIFFICULTY: Moderate KEYWORDS: one-tailed test, mean, t test, beta-risk 161. Referring to Table 9-7, what will be the p-value if these data were used to perform a two-tailed

test? ANSWER: 0.027 TYPE: TF DIFFICULTY: Moderate KEYWORDS: one-tailed test, proportion, t test, p-value

Two-Sample Tests

CHAPTER 10: TWO-SAMPLE TESTS 1. The t test for the difference between the means of 2 independent populations assumes that the respective a) sample sizes are equal. b) sample variances are equal. c) populations are approximately normal. d) All of the above. ANSWER: c TYPE: MC DIFFICULTY: Moderate KEYWORDS: pooled variance, t test, assumption

2. The t test for the mean difference between 2 related populations assumes that the a) population sizes are equal. b) sample variances are equal. c) population of differences is approximately normal or sample sizes are large enough. d) All of the above. ANSWER: c TYPE: MC DIFFICULTY: Moderate KEYWORDS: t test, mean difference, assumption

3. If we are testing for the difference between the means of 2 related 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 TYPE: MC DIFFICULTY: Easy KEYWORDS: t test, mean difference, degrees of freedom

4. If we are testing for the difference between the means of 2 independent populations presumes equal variances 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 TYPE: MC DIFFICULTY: Easy KEYWORDS: pooled variance, t test, degrees of freedom

44 Two-Sample Tests 5. In what type of test is the variable of interest the difference between the values of the observations rather than the observations themselves? a) A test for the equality of variances from 2 independent populations. b) A test for the difference between the means of 2 related populations. c) A test for the difference between the means of 2 independent populations. d) All of the above. ANSWER: b TYPE: MC DIFFICULTY: Moderate KEYWORDS: t test, mean difference 6. In testing for the differences between the means of 2 independent populations where the variances in each population are unknown but assumed equal, the degrees of freedom are a) n – 1. b) n1 + n2 – 1. c) n1 + n2 – 2. d) n – 2. ANSWER: c TYPE: MC DIFFICULTY: Easy KEYWORDS: pooled variance, t test, degrees of freedom 7. In testing for differences between the means of 2 related populations where the variance of the differences is unknown, the degrees of freedom are a) n – 1. b) n1 + n2 – 1. c) n1 + n2 – 2. d) n – 2. ANSWER: a TYPE: MC DIFFICULTY: Easy KEYWORDS: t test, mean difference, degrees of freedom 8. In testing for differences between the means of two related populations, the null hypothesis is a) H0 : D = 2 .

b) H0 : D = 0 . c) H0 : D  0 . d) H0 : D  0 . ANSWER: b TYPE: MC DIFFICULTY: Easy KEYWORDS: t test, mean difference, form of hypothesis 9. In testing for differences between the means of two independent populations, the null hypothesis is: a) H0 : 1 − 2 = 2.

Two-Sample Tests

b) H0 : 1 – 2 = 0. c) H0 : 1 – 2 > 0. d) H0 : 1 – 2 < 2. ANSWER: b TYPE: MC DIFFICULTY: Easy KEYWORDS: pooled variance, t test, form of hypothesis

10. When testing for the difference between 2 population variances with sample sizes of n1 = 8 and n2 = 10, the number of degrees of freedom are a) 8 and 10. b) 7 and 9. c) 18. d) 16. ANSWER: b TYPE: MC DIFFICULTY: Easy KEYWORDS: F test, difference between two variances, degrees of freedom 11. The statistical distribution used for testing the difference between two population variances is the distribution. a) t b) standardized normal c) binomial d) F ANSWER: d TYPE: MC DIFFICULTY: Easy KEYWORDS: F test, difference between two variances, degrees of freedom 12. The test for the equality of two population variances is based on a) the difference between the 2 sample variances. b) the ratio of the 2 sample variances. c) the difference between the 2 population variances. d) the difference between the sample variances divided by the difference between the sample means. ANSWER: b TYPE: MC DIFFICULTY: Easy KEYWORDS: F test, difference between two variances

13. True or False: The F test used for testing the difference in two population variances is always a one-tailed test. ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: F test, difference between two variances, rejection region

46 Two-Sample Tests 14. In testing for the differences between the means of two related populations, the hypothesis is the hypothesis of "no differences." ANSWER: null TYPE: FI DIFFICULTY: Easy KEYWORDS: t test, mean difference, form of hypothesis 15. In testing for the differences between the means of two related populations, we assume that the differences follow a distribution. ANSWER: normal TYPE: FI DIFFICULTY: Easy KEYWORDS: t test, mean difference, assumption 16. In testing for the differences between the means of two independent populations, we assume that the 2 populations each follow a distribution. ANSWER: normal TYPE: FI DIFFICULTY: Easy KEYWORDS: pooled variance, t test, assumption

17. Given the following information, calculate the degrees of freedom that should be used in the pooled-variance t test. s12 = 4 s 2 2= 6 n1 = 16 n2 = 25 a) df = 41 b) df = 39 c) df = 16 d) df = 25 ANSWER: b TYPE: MC DIFFICULTY: Easy KEYWORDS: pooled variance, t test, degrees of freedom

18. Given the following information, calculate s p2, the pooled sample variance that should be used in the pooled-variance t test. s12 = 4 s 2 2= 6 n1 = 16 n2 = 25 a) sp2 = 6.00 b) sp 2 = 5.00 c) sp2 = 5.23 d) sp 2 = 4.00 ANSWER: c TYPE: MC DIFFICULTY: Easy KEYWORDS: pooled variance, t test TABLE 10-1

Two-Sample Tests

Are Japanese managers more motivated than American managers? A randomly selected group of each were 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

19. Referring to Table 10-1, judging from the way the data were collected, which test would likely be most appropriate to employ? a) Paired t test b) Pooled-variance t test for the difference between two means c) Independent samples Z test for the difference between two means d) Related samples Z test for the mean difference ANSWER: c TYPE: MC DIFFICULTY: Easy KEYWORDS: Z test, difference between two means 20. Referring to Table 10-1, give 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 :  A –  J  0 versus H1 :  A –  J  0 b) H0 :  A –  J  0 versus H1 :  A –  J  0 c) H0 :  A –  J = 0 versus H1 :  A –  J  0 d) H0 : X A – XJ = 0 versus H1 : X A – XJ  0 ANSWER: c TYPE: MC DIFFICULTY: Easy KEYWORDS: Z test, mean differences, form of hypothesis 21. Referring to Table 10-1, assuming the independent samples procedure was used, calculate the value of the test statistic.

65.75 – 79.83 9.82 9.82 a) + 211 100 65.75 – 79.83 Z= 11.07 6.41 b) + 211 100 65.75 – 79.83 Z= c) 9.822 9.822 + 211 100 Z=

48 Two-Sample Tests

65.75 – 79.83

11.072 6.412 + 211 100

ANSWER: d TYPE: MC DIFFICULTY: Easy KEYWORDS: Z test, mean differences, test statistic

22. Referring to Table 10-1, suppose that the test statistic is Z = 2.45. Find the p-value if we assume that the alternative hypothesis was a two-tailed test ( H1 :  A –  J  0 ). a) 0.0071 b) 0.0142 c) 0.4929 d) 0.9858 ANSWER: b TYPE: MC DIFFICULTY: Moderate KEYWORDS: Z test, mean differences, p-value TABLE 10-2 A researcher randomly sampled 30 graduates of an MBA program and recorded data concerning their starting salaries. Of primary interest to the researcher was the effect of gender on starting salaries. Analysis of the mean salaries of the females and males in the sample is given below. Hypothesized Difference Level of Significance Population 1 Sample Sample Size Sample Mean Sample Standard Deviation Population 2 Sample Sample Size Sample Mean Sample Standard Deviation Difference in Sample Means t-Test Statistic Lower-Tail Test Lower Critical Value p-Value

0 0.05 18 48266.7 13577.63 12 55000 11741.29 -6733.3 -1.40193 -1.70113 0.085962

23. Referring to Table 10-2, the researcher was attempting to show statistically that the female MBA graduates have a significantly lower mean starting salary than the male MBA graduates. According to the test run, which of the following is an appropriate alternative hypothesis? a) H1 : females  males

H1 : females  males c) H1 : females  males d) H1 : females = males b)

Two-Sample Tests

ANSWER: b TYPE: MC DIFFICULTY: Moderate KEYWORDS: pooled variance, t test, form of hypothesis 24. Referring to Table 10-2, the researcher was attempting to show statistically that the female MBA graduates have a significantly lower mean starting salary than the male MBA graduates. From the analysis in Table 10-2, the correct test statistic is: a) 0.0860 b) – 1.4019 c) – 1.7011 d) – 6,733.33 ANSWER: b TYPE: MC DIFFICULTY: Moderate KEYWORDS: pooled variance, t test, test statistic 25. Referring to Table 10-2, the researcher was attempting to show statistically that the female MBA graduates have a significantly lower mean starting salary than the male MBA graduates. The proper conclusion for this test is: a) At the  = 0.10 level, there is sufficient evidence to indicate a difference in the mean starting salaries of male and female MBA graduates. b) At the  = 0.10 level, there is sufficient evidence to indicate that females have a lower mean starting salary than male MBA graduates. c) At the  = 0.10 level, there is sufficient evidence to indicate that females have a higher mean starting salary than male MBA graduates. d) At the  = 0.10 level, there is insufficient evidence to indicate any difference in the mean starting salaries of male and female MBA graduates. ANSWER: b TYPE: MC DIFFICULTY: Moderate KEYWORDS: pooled variance, t test, conclusion 26. Referring to Table 10-2, the researcher was attempting to show statistically that the female MBA graduates have a significantly lower mean starting salary than the male MBA graduates. What assumptions were necessary to conduct this hypothesis test? a) Both populations of salaries (male and female) must have approximate normal distributions. b) The population variances are approximately equal. c) The samples were randomly and independently selected. d) All of the above assumptions were necessary. ANSWER: d TYPE: MC DIFFICULTY: Moderate KEYWORDS: pooled variance, t test, assumption

27. Referring to Table 10-2, what is the 99% confidence interval estimate for the difference between two means?

50 Two-Sample Tests ANSWER: −$20,004.90 to $6,538.30 TYPE: PR DIFFICULTY: Moderate KEYWORDS: confidence interval, difference between the means 28. Referring to Table 10-2, what is the 95% confidence interval estimate for the difference between two means? ANSWER: −$16571.55 to $3,104.95 TYPE: PR DIFFICULTY: Moderate KEYWORDS: confidence interval, difference between the means 29. Referring to Table 10-2, what is the 90% confidence interval estimate for the difference between two means? ANSWER: −$14,903.61 to $1,437.01 TYPE: PR DIFFICULTY: Moderate KEYWORDS: confidence interval, difference between the means TABLE 10-3 The use of preservatives by food processors has become a controversial issue. Suppose 2 preservatives are extensively tested and determined safe for use in meats. A processor wants to compare the preservatives for their effects on retarding spoilage. Suppose 15 cuts of fresh meat are treated with preservative A and 15 are treated with preservative B, and the number of hours until spoilage begins is recorded for each of the 30 cuts of meat. The results are summarized in the table below. Preservative A Preservative B X A = 106.4 hours X B = 96.54 hours S A = 10.3 hours S B = 13.4 hours 30. Referring to Table 10-3, state the test statistic for determining if the population variances differ for preservatives A and B. a) F = – 3.10 b) F = 0.5908 c) F = 0.7687 d) F = 0.8250 ANSWER: b TYPE: MC DIFFICULTY: Moderate KEYWORDS: F test, difference between two variances, test statistic 31. Referring to Table 10-3, what assumptions are necessary for a comparison of the population variances to be valid? a) Both sampled populations are normally distributed. b) Both samples are random and independent. c) Neither (a) nor (b) is necessary. d) Both (a) and (b) are necessary. ANSWER:

Two-Sample Tests

d TYPE: MC DIFFICULTY: Easy KEYWORDS: F test, difference between two variances, assumption TABLE 10-4 A real estate company is interested in testing whether, on average, families in Gotham have been living in their current homes for less time than families in Metropolis have. Assume that the two population variances are equal. A random sample of 100 families from Gotham and a random sample of 150 families in Metropolis yield the following data on length of residence in current homes. Gotham: X G = 35 months, sG 2 = 900 Metropolis: s 2 = 1050 M = 50 months,

X M

32. Referring to Table 10-4, which of the following represents the relevant hypotheses tested by the real estate company? a) H0 :  G – M  0 versus H1 :  G –  M  0 b) H0 : G –  M  0 versus H1 :  G –  M  0 c) H0 :  G – M = 0 versus H1 :  G –  M  0 d) H0 : XG – X M  0 versus H1 : XG – XM  0 ANSWER: a TYPE: MC DIFFICULTY: Easy KEYWORDS: pooled variance, t test, form of hypothesis 33. Referring to Table 10-4, what is the estimated standard error of the difference between the 2 sample means? a) 4.06 b) 5.61 c) 8.01 d) 16.00 ANSWER: a TYPE: MC DIFFICULTY: Moderate KEYWORDS: pooled variance, t test, standard error 34. Referring to Table 10-4, what is an unbiased point estimate for the mean of the sampling distribution of the difference between the 2 sample means? a) – 22 b) – 10 c) – 15 d) 0 ANSWER: c TYPE: MC DIFFICULTY: Easy KEYWORDS: pooled variance, t test, unbiased, point estimate 35. Referring to Table 10-4, what is(are) the critical value(s) of the relevant hypothesis test if the level of significance is 0.05? a) t  Z = – 1.645

52 Two-Sample Tests

b) t  Z =  1.96 c) t  Z = – 1.96 d) t  Z = – 2.080 ANSWER: a TYPE: MC DIFFICULTY: Easy KEYWORDS: pooled variance, t test, critical value 36. Referring to Table 10-4, what is(are) the critical value(s) of the relevant hypothesis test if the level of significance is 0.01? a) t  Z = – 1.96 b) t  Z =  1.96 c) t  Z = – 2.080 d) t  Z = – 2.33 ANSWER: d TYPE: MC DIFFICULTY: Easy KEYWORDS: pooled variance, t test, critical value 37. Referring to Table 10-4, what is the standardized value of the estimate of the mean of the sampling distribution of the difference between sample means? a) – 8.75 b) – 3.69 c) – 2.33 d) – 1.96 ANSWER: b TYPE: MC DIFFICULTY: Moderate KEYWORDS: pooled variance, t test, test statistic

38. Referring to Table 10-4, suppose  = 0.10. Which of the following represents the result of the relevant hypothesis test? a) The alternative hypothesis is rejected. b) The null hypothesis is rejected. c) The null hypothesis is not rejected. d) Insufficient information exists on which to make a decision. ANSWER: b TYPE: MC DIFFICULTY: Easy KEYWORDS: pooled variance, t test, decision

39. Referring to Table 10-4, suppose  = 0.05. Which of the following represents the result of the relevant hypothesis test? a) The alternative hypothesis is rejected. b) The null hypothesis is rejected. c) The null hypothesis is not rejected. d) Insufficient information exists on which to make a decision. ANSWER:

Two-Sample Tests

b TYPE: MC DIFFICULTY: Easy KEYWORDS: pooled variance, t test, decision

40. Referring to Table 10-4, suppose  = 0.01. Which of the following represents the result of the relevant hypothesis test? a) The alternative hypothesis is rejected. b) The null hypothesis is rejected. c) The null hypothesis is not rejected. d) Insufficient information exists on which to make a decision. ANSWER: b TYPE: MC DIFFICULTY: Easy KEYWORDS: pooled variance, t test, decision

41. Referring to Table 10-4, suppose  = 0.1. Which of the following represents the correct conclusion? a) There is not enough evidence that, on average, families in Gotham have been living in their current homes for less time than families in Metropolis have. b) There is enough evidence that, on average, families in Gotham have been living in their current homes for less time than families in Metropolis have. c) There is not enough evidence that, on average, families in Gotham have been living in their current homes for no less time than families in Metropolis have. d) There is enough evidence that, on average, families in Gotham have been living in their current homes for no less time than families in Metropolis have. ANSWER: b TYPE: MC DIFFICULTY: Easy KEYWORDS: pooled variance, t test, conclusion

42. Referring to Table 10-4, suppose  = 0.05. Which of the following represents the correct conclusion? a) There is not enough evidence that, on average, families in Gotham have been living in their current homes for less time than families in Metropolis have. b) There is enough evidence that, on average, families in Gotham have been living in their current homes for less time than families in Metropolis have. c) There is not enough evidence that, on average, families in Gotham have been living in their current homes for no less time than families in Metropolis have. d) There is enough evidence that, on average, families in Gotham have been living in their current homes for no less time than families in Metropolis have. ANSWER: b TYPE: MC DIFFICULTY: Easy KEYWORDS: pooled variance, t test, conclusion

43. Referring to Table 10-4, suppose  = 0.01. Which of the following represents the correct conclusion? a) There is not enough evidence that, on average, families in Gotham have been living in their current homes for less time than families in Metropolis have. b) There is enough evidence that, on average, families in Gotham have been living in their current homes for less time than families in Metropolis have.

54 Two-Sample Tests c) There is not enough evidence that, on average, families in Gotham have been living in their current homes for no less time than families in Metropolis have. d) There is enough evidence that, on average, families in Gotham have been living in their current homes for no less time than families in Metropolis have. ANSWER: b TYPE: MC DIFFICULTY: Easy KEYWORDS: pooled variance, t test, conclusion 44. Referring to Table 10-4, what is the 99% confidence interval estimate for the difference in the two means? ANSWER: −25.54 to −4.46 months TYPE: PR DIFFICULTY: Moderate KEYWORDS: confidence interval, difference between two means 45. Referring to Table 10-4, what is the 95% confidence interval estimate for the difference in the two means? ANSWER: −23.00 to −7.00 months TYPE: PR DIFFICULTY: Moderate KEYWORDS: confidence interval, difference between two means TABLE 10-5 To test the effectiveness of a business school preparation course, 8 students took a general business test before and after the course. The results are given below. Exam Score Exam Score Student Before Course (1) After Course (2) 1 530 670 2 690 770 3 910 1,000 4 700 710 5 450 550 6 820 870 7 820 770 8 630 610 46. Referring to Table 10-5, the number of degrees of freedom is a) 14. b) 13. c) 8. d) 7. ANSWER: d TYPE: MC DIFFICULTY: Easy KEYWORDS: t test, mean difference, degrees of freedom

Two-Sample Tests 47. Referring to Table 10-5, the value of the sample mean difference is if the difference scores reflect the results of the exam after the course minus the results of the exam before the course. a) 0 b) 50 c) 68 d) 400 ANSWER: b TYPE: MC DIFFICULTY: Easy KEYWORDS: t test, mean difference, test statistic 48. Referring to Table 10-5, the value of the standard error of the difference scores is a) 65.027 b) 60.828 c) 22.991 d) 14.696 ANSWER: c TYPE: MC DIFFICULTY: Easy KEYWORDS: t test, mean difference, standard error

49. Referring to Table 10-5, what is the critical value for testing at the 5% level of significance whether the business school preparation course is effective in improving exam scores? a) 2.365 b) 2.145 c) 1.761 d) 1.895 ANSWER: d TYPE: MC DIFFICULTY: Easy KEYWORDS: t test, mean difference, critical value 50. Referring to Table 10-5, at the 0.05 level of significance, the decision for this hypothesis test would be: a) reject the null hypothesis. b) do not reject the null hypothesis. c) reject the alternative hypothesis. d) It cannot be determined from the information given. ANSWER: a TYPE: MC DIFFICULTY: Easy KEYWORDS: t test, mean difference, decision 51. Referring to Table 10-5, at the 0.05 level of significance, the conclusion for this hypothesis test would be: a) the business school preparation course does improve exam score. b) the business school preparation course does not improve exam score. c) the business school preparation course has no impact on exam score. d) It cannot be drawn from the information given.

56 Two-Sample Tests

ANSWER: a TYPE: MC DIFFICULTY: Easy KEYWORDS: t test, mean difference, conclusion 52. True or False: Referring to Table 10-5, one must assume that the population of difference scores is normally distributed. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: t test, mean difference, assumption 53. Referring to Table 10-5, the calculated value of the test statistic is

ANSWER: 2.175 TYPE: FI DIFFICULTY: Easy KEYWORDS: t test, mean difference, test statistic

54. Referring to Table 10-5, the p-value of the test statistic is

ANSWER: 0.0331 TYPE: FI DIFFICULTY: Moderate EXPLANATION: p-value obtained from Excel KEYWORDS: t test, mean difference, p-value 55. True or False: Referring to Table 10-5, in examining the differences between related samples we are essentially sampling from an underlying population of difference "scores." ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: t test, mean difference, sampling distribution 56. True or False: The sample size in each independent sample must be the same if we are to test for differences between the means of 2 independent populations. ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: pooled variance, t test, sample size 57. True or False: When we test for differences between the means of 2 independent populations, we can only use a two-tailed test. ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: pooled variance, t test, rejection region

Two-Sample Tests

58. True or False: When testing for differences between the means of 2 related populations, we can use either a one-tailed or two-tailed test. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: t test, mean difference, rejection region 59. True or False: Repeated measurements from the same individuals is an example of data collected from 2 related populations. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: t test, mean difference 60. True or False: The test for the equality of 2 population variances assumes that each of the 2 populations is normally distributed. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: F test, difference between two variances, assumption 61. True or False: For all two-sample tests, the sample sizes must be equal in the 2 groups. ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: sample size 62. True or False: When the sample sizes are equal, the pooled variance of the 2 groups is the average of the 2 sample variances. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: pooled variance, t test, sample size

63. True or False: The F distribution is symmetric. ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: F distribution, properties

64. True or False: The F distribution can only have positive values. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: F distribution, properties

58 Two-Sample Tests

65. True or False: All F tests are one-tailed tests. ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: F distribution, properties

66. True of False: When performing a two-tailed test, the lower-tailed critical value of the F distribution with n1 −1 degrees of freedom in the numerator and n2 −1 degrees of freedom in the denominator is exactly equivalent to the reciprocal of the upper-tailed critical value of the F distribution with n2 −1 degrees of freedom in the numerator and n1 −1 degrees of freedom in the denominator. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: F distribution, properties

67. Given the upper-tailed critical value of an F test with 3 degrees of freedom in the numerator and 8 degrees of freedom in the denominator being 4.07, the lower-tailed critical value of an F test with 8 degrees of freedom in the numerator and 3 degrees of freedom in the denominator for the same level of significance will be . ANSWER: 0.2457 TYPE: FI DIFFICULTY: Easy KEYWORDS: F distribution, critical value, properties 68. True or False: A researcher is curious about the effect of sleep on students’ test performances. He chooses 60 students and gives each 2 tests: one given after 2 hours’ sleep and one after 8 hours’ sleep. The test the researcher should use would be a related samples test. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: t test, mean difference

69. When testing H0 : 1 – 2 = 0 versus H1 : 1 –  2  0 , the observed value of the Z-score was found to be – 2.13. The p-value for this test would be a) 0.0166. b) 0.0332. c) 0.9668. d) 0.9834. ANSWER: b TYPE: MC DIFFICULTY: Easy KEYWORDS: Z test, difference between two means, p-value

70. When testing H0 : 1 − 2  0 versus H1 : 1 − 2  0 , the observed value of the Z-score was found to be – 2.13. The p-value for this test would be a) 0.0166.

Two-Sample Tests

b) 0.0332. c) 0.9668. d) 0.9834. ANSWER: d TYPE: MC DIFFICULTY: Moderate KEYWORDS: Z test, difference between two means, p-value

71. When testing H0 : 1 − 2  0 versus H1 : 1 − 2  0 , the observed value of the Z-score was found to be – 2.13. The p-value for this test would be a) 0.0166. b) 0.0332. c) 0.9668. d) 0.9834. ANSWER: a TYPE: MC DIFFICULTY: Easy KEYWORDS: Z test, difference between two means, p-value

72. True or False: A statistics professor wanted to test whether the grades on a statistics test were the same for upper and lower classmen. The professor took a random sample of size 10 from each, conducted a test and found out that the variances were equal. For this situation, the professor should use a t test with related samples. ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: pooled variance, t test

73. True or False: A statistics professor wanted to test whether the grades on a statistics test were the same for upper and lower classmen. The professor took a random sample of size 10 from each, conducted a test and found out that the variances were equal. For this situation, the professor should use a t test with independent samples. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: pooled variance, t test

74. True or False: A Marine drill instructor recorded the time in which each of 11 recruits completed an obstacle course both before and after basic training. To test whether any improvement occurred, the instructor would use a t-distribution with 11 degrees of freedom. ANSWER: False TYPE: TF DIFFICULTY: Moderate KEYWORDS: t test, mean difference, degrees of freedom

75. True or False: A Marine drill instructor recorded the time in which each of 11 recruits completed an obstacle course both before and after basic training. To test whether any improvement occurred, the instructor would use a t-distribution with 10 degrees of freedom.

60 Two-Sample Tests

ANSWER: True TYPE: TF DIFFICULTY: Moderate KEYWORDS: t test, mean difference, degrees of freedom TABLE 10-6 Two samples each of size 25 are taken from independent populations assumed to be normally distributed with equal variances. The first sample has a mean of 35.5 and standard deviation of 3.0 while the second sample has a mean of 33.0 and standard deviation of 4.0. 76. Referring to Table 10-6, the pooled (i.e., combined) variance is

ANSWER: 12.5 TYPE: FI DIFFICULTY: Easy KEYWORDS: pooled variance, t test

77. Referring to Table 10-6, the computed t statistic is

ANSWER: 2.50 TYPE: FI DIFFICULTY: Easy KEYWORDS: pooled variance, t test, test statistic 78. Referring to Table 10-6, there are

degrees of freedom for this test.

ANSWER: 48 TYPE: FI DIFFICULTY: Easy KEYWORDS: pooled variance, t test, degrees of freedom

79. Referring to Table 10-6, the critical values for a two-tailed test of the null hypothesis of no difference in the population means at the  = 0.05 level of significance are . ANSWER:  2.0106 TYPE: FI DIFFICULTY: Easy KEYWORDS: pooled variance, t test, critical value

80. Referring to Table 10-6, a two-tailed test of the null hypothesis of no difference would (be rejected/not be rejected) at the  = 0.05 level of significance. ANSWER: be rejected TYPE: FI DIFFICULTY: Easy KEYWORDS: pooled variance, t test, decision

81. Referring to Table 10-6, the p-value for a two-tailed test whose computed t statistic is 2.50 is between ANSWER:

and

Two-Sample Tests

0.01 and 0.02 TYPE: FI DIFFICULTY: Moderate KEYWORDS: pooled variance, t test, p-value

82. Referring to Table 10-6, if we were interested in testing against the one-tailed alternative that 1   2 at the  = 0.01 level of significance, the null hypothesis would . ANSWER: be rejected TYPE: FI DIFFICULTY: Easy KEYWORDS: pooled variance, t test, decision

83. Referring to Table 10-6, the p-value for a one-tailed test whose computed statistic is 2.50 (in the hypothesized direction) is between

ANSWER: 0.005 and 0.01 TYPE: FI DIFFICULTY: Moderate KEYWORDS: pooled variance, t test, p-value

84. Referring to Table 10-6, what is the 99% confidence interval estimate for the difference in the two means? ANSWER: −0.1822 to 5.1822 TYPE: PR DIFFICULTY: Moderate KEYWORDS: confidence interval, difference between two means

85. Referring to Table 10-6, what is the 95% confidence interval estimate for the difference in the two means? ANSWER: 0.4894 to 4.5106 TYPE: PR DIFFICULTY: Moderate KEYWORDS: confidence interval, difference between two means

86. Referring to Table 10-6, what is the 90% confidence interval estimate for the difference in the two means? ANSWER: 0.8228 to 4.1772 TYPE: PR DIFFICULTY: Moderate KEYWORDS: confidence interval, difference between two means TABLE 10-7 To investigate the efficacy of a diet, a random sample of 16 male patients is drawn from a population of adult males using the diet. The weight of each individual in the sample is taken at the start of the diet and at a medical follow-up 4 weeks later. Assuming that the population of differences in weight before versus after the diet follow a normal distribution, the t-test for related samples can be used to determine if there was a significant decrease in the mean weight during this period. Suppose the mean decrease in weights over all 16 subjects in the study is 3.0 pounds with the standard deviation of differences computed as 6.0 pounds.

62 Two-Sample Tests

87. Referring to Table 10-7, the t test should be

-tailed.

ANSWER: one TYPE: FI DIFFICULTY: Easy KEYWORDS: t test, mean difference, rejection region

88. Referring to Table 10-7, the computed t statistic is

ANSWER: 2.00 TYPE: FI DIFFICULTY: Easy KEYWORDS: t test, mean difference, test statistic

89. Referring to Table 10-7, there are

degrees of freedom for this test.

ANSWER: 15 TYPE: FI DIFFICULTY: Easy KEYWORDS: t test, mean difference, degrees of freedom

90. Referring to Table 10-7, the critical value for a one-tailed test of the null hypothesis of no difference at the  = 0.05 level of significance is . ANSWER: 1.7531 TYPE: FI DIFFICULTY: Easy KEYWORDS: t test, mean difference, critical value

91. Referring to Table 10-7, a one-tailed test of the null hypothesis of no difference would (be rejected/not be rejected) at the  = 0.05 level of significance. ANSWER: be rejected TYPE: FI DIFFICULTY: Easy KEYWORDS: t test, mean difference, decision

92. Referring to Table 10-7, the p-value for a one-tailed test whose computed t statistic is 2.00 is between

ANSWER: 0.025 and 0.05 TYPE: FI DIFFICULTY: Moderate KEYWORDS: t test, mean difference, p-value

93. Referring to Table 10-7, if we were interested in testing against the two-tailed alternative that D is not equal to zero at the  = 0.05 level of significance, the null hypothesis would (be rejected/not be rejected). ANSWER: not be rejected TYPE: FI DIFFICULTY: Easy

Two-Sample Tests

KEYWORDS: t test, mean difference, decision

94. Referring to Table 10-7, the p-value for a two-tailed test whose computed statistic is 2.00 is between

ANSWER: 0.05 and 0.10 TYPE: FI DIFFICULTY: Moderate KEYWORDS: t test, mean difference, p-value

95. Referring to Table 10-7, what is the 95% confidence interval estimate for the mean difference in weight before and after the diet? ANSWER: −0.20 to 6.20 TYPE: PR DIFFICULTY: Moderate KEYWORDS: confidence interval, mean difference

96. Referring to Table 10-7, what is the 99% confidence interval estimate for the mean difference in weight before and after the diet? ANSWER: − to 7.42 TYPE: PR DIFFICULTY: Moderate KEYWORDS: confidence interval, mean difference

97. Referring to Table 10-7, what is the 90% confidence interval estimate for the mean difference in weight before and after the diet? ANSWER: 0 to 5.63 TYPE: PR DIFFICULTY: Moderate KEYWORDS: confidence interval, mean difference TABLE 10-8 A buyer for a manufacturing plant suspects that his primary supplier of raw materials is overcharging. In order to determine if his suspicion is correct, he contacts a second supplier and asks for the prices on various identical materials. He wants to compare these prices with those of his primary supplier. The data collected is presented in the table below, with some summary statistics presented (all of these might not be necessary to answer the questions which follow). The buyer believes that the differences are normally distributed and will use this sample to perform an appropriate test at a level of significance of 0.01. Primary Secondary Material Supplier Supplier Difference 1 $55 $45 $10 2 $48 $47 $1 3 $31 $32 – $1 4 $83 $77 $6 5 $37 $37 $0 6 $55 $54 $1 Sum: $309 $292 $17 Sum of Squares: $17,573 $15,472 $139

64 Two-Sample Tests

98. Referring to Table 10-8, the hypotheses that the buyer should test are a null hypothesis that versus an alternative hypothesis that

ANSWER:

H0 : D  0, H1 : D  0 TYPE: FI DIFFICULTY: Moderate KEYWORDS: t test, mean difference, form of hypothesis

99. Referring to Table 10-8, the test to perform is a a) pooled-variance t test for differences between two means. b) separate-variance t test for differences between two means. c) Z test for the difference between two means. d) paired t-test for the mean difference. ANSWER: d TYPE: MC DIFFICULTY: Easy KEYWORDS: t test, mean difference

100. Referring to Table 10-8, the decision rule is to reject the null hypothesis if

ANSWER: t > 3.3649 TYPE: FI DIFFICULTY: Easy KEYWORDS: t test, mean difference, decision

101. Referring to Table 10-8, the calculated value of the test statistic is

ANSWER: 1.628 TYPE: FI DIFFICULTY: Easy KEYWORDS: t test, mean difference, test statistic

102. Referring to Table 10-8, the p-value of the test is between

and

ANSWER: 0.05 and 0.1 TYPE: FI DIFFICULTY: Moderate KEYWORDS: t test, mean difference, p-value

103. True or False: Referring to Table 10-8, the null hypothesis should be rejected. ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: t test, mean difference, decision

104. Referring to Table 10-8, the buyer should decide that the primary supplier is a) overcharging because there is strong evidence that this is the case. b) overcharging because there is insufficient evidence to prove otherwise. c) not overcharging because there is insufficient evidence to prove otherwise.

Two-Sample Tests

d) not overcharging because there is strong evidence to prove otherwise. ANSWER: c TYPE: MC DIFFICULTY: Moderate KEYWORDS: t test, mean difference, conclusion

105. Referring to Table 10-8, if the buyer had decided to perform a two-tailed test, the p-value would have been between

and

ANSWER: 0.01 and 0.2 TYPE: FI DIFFICULTY: Moderate KEYWORDS: t test, mean difference, p-value

106. Referring to Table 10-8, what is the 99% confidence interval estimate for the mean difference in prices? ANSWER: −$4.18 to $9.85 TYPE: PR DIFFICULTY: Moderate KEYWORDS: confidence interval, mean difference

107. Referring to Table 10-8, what is the 95% confidence interval estimate for the mean difference in prices? ANSWER: −$1.64 to $7.31 TYPE: PR DIFFICULTY: Moderate KEYWORDS: confidence interval, mean difference

108. Referring to Table 10-8, what is the 90% confidence interval estimate for the mean difference in prices? ANSWER: −$0.67 to $6.34 TYPE: PR DIFFICULTY: Moderate KEYWORDS: confidence interval, mean difference 109. If we wish to determine whether there is evidence that the proportion of successes is higher in group 1 than in group 2, the appropriate test to use is a) the Z test for the difference between two proportions. b) the F test for the difference between two variances. c) the pooled-variance t test for the difference between two proportions. d) the F test for the difference between two proportions. ANSWER: a TYPE: MC DIFFICULTY: Easy KEYWORDS: Z test for the difference between two proportions

66 Two-Sample Tests 110. Moving companies are required by the government to publish a Carrier Performance Report each year. One of the descriptive statistics they must include is the annual percentage of shipments on which a $50 or greater claim for loss or damage was filed. Suppose two companies, Econo-Move and On-the-Move, each decide to estimate this figure by sampling their records, and they report the data shown in the following table. Econo-Move

On-the-Move

Total shipments sampled

900

750

Number of shipments with a claim  $50

162

The owner of On-the-Move is hoping to use these data to show that the company is superior to Econo-Move with regard to the percentage of claims filed. Which test would be used to properly analyze the data in this experiment? a) Z test for the difference between two means b) F test for the difference between two variances c) Separate variance t test for the difference between two means d) Z test for the difference between two proportions ANSWER: d TYPE: MC DIFFICULTY: Moderate KEYWORDS: Z test, difference between two proportions

111. The Wall Street Journal recently ran an article indicating differences in perception of sexual harassment on the job between men and women. The article claimed that women perceived the problem to be much more prevalent than did men. One question asked of both men and women was: “Do you think sexual harassment is a major problem in the American workplace?” Some 24% of the men compared to 62% of the women responded “Yes.” Assuming W designates women’s responses and M designates men’s, what hypothesis should The Wall Street Journal test in order to show that its claim is true? a) H0: W −  M  0 versus H1: W −  M < 0

b) H0: W −  M  0 versus H1: W −  M > 0 c) H0: W −  M = 0 versus H1: W −  M  0 d) H0: W −  M  0 versus H1: W −  M > 0 ANSWER: b TYPE: MC DIFFICULTY: Easy KEYWORDS: Z test, difference between two proportions, form of hypothesis

112. The Wall Street Journal recently ran an article indicating differences in perception of sexual harassment on the job between men and women. The article claimed that women perceived the problem to be much more prevalent than did men. One question asked to both men and women was: “Do you think sexual harassment is a major problem in the American workplace?” Some 24% of the men compared to 62% of the women responded “Yes.” Suppose that 150 women and 200 men were interviewed. For a 0.01 level of significance, what is the critical value for the rejection region? a) 7.173 b) 7.106

Two-Sample Tests

c) 6.635 d) 2.33 ANSWER: d TYPE: MC DIFFICULTY: Moderate KEYWORDS: Z test, difference between two proportions, critical value

113. The Wall Street Journal recently ran an article indicating differences in perception of sexual harassment on the job between men and women. The article claimed that women perceived the problem to be much more prevalent than did men. One question asked to both men and women was: “Do you think sexual harassment is a major problem in the American workplace?” Some 24% of the men compared to 62% of the women responded “Yes.” Suppose that 150 women and 200 men were interviewed. What is the value of the test statistic? a) 7.173 b) 7.106 c) 6.635 d) 2.33 ANSWER: a TYPE: MC DIFFICULTY: Moderate KEYWORDS: Z test, difference between two proportions, test statistic

114. The Wall Street Journal recently ran an article indicating differences in perception of sexual harassment on the job between men and women. The article claimed that women perceived the problem to be much more prevalent than did men. One question asked to both men and women was: “Do you think sexual harassment is a major problem in the American workplace?” Some 24% of the men compared to 62% of the women responded “Yes.” Suppose that 150 women and 200 men were interviewed. Construct a 99% confidence interval estimate of the difference between the proportion of women and men who think sexual harassment is a major problem in the American workplace. ANSWER: 0.25 to 0.51 TYPE: PR DIFFICULTY: Moderate KEYWORDS: confidence interval, difference between two proportions

115. The Wall Street Journal recently ran an article indicating differences in perception of sexual harassment on the job between men and women. The article claimed that women perceived the problem to be much more prevalent than did men. One question asked to both men and women was: “Do you think sexual harassment is a major problem in the American workplace?” Some 24% of the men compared to 62% of the women responded “Yes.” Suppose that 150 women and 200 men were interviewed. Construct a 95% confidence interval estimate of the difference between the proportion of women and men who think sexual harassment is a major problem in the American workplace. ANSWER: 0.28 to 0.48 TYPE: PR DIFFICULTY: Moderate KEYWORDS: confidence interval, difference between two proportions

68 Two-Sample Tests

116. The Wall Street Journal recently ran an article indicating differences in perception of sexual harassment on the job between men and women. The article claimed that women perceived the problem to be much more prevalent than did men. One question asked to both men and women was: “Do you think sexual harassment is a major problem in the American workplace?” Some 24% of the men compared to 62% of the women responded “Yes.” Suppose that 150 women and 200 men were interviewed. Construct a 90% confidence interval estimate of the difference between the proportion of women and men who think sexual harassment is a major problem in the American workplace. ANSWER: 0.30 to 0.46 TYPE: PR DIFFICULTY: Moderate KEYWORDS: confidence interval, difference between two proportions

117. The Wall Street Journal recently ran an article indicating differences in perception of sexual harassment on the job between men and women. The article claimed that women perceived the problem to be much more prevalent than did men. One question asked to both men and women was: “Do you think sexual harassment is a major problem in the American workplace?” Some 24% of the men compared to 62% of the women responded “Yes.” Suppose that 150 women and 200 men were interviewed. What conclusion should be reached? a) Using a 0.01 level of significance, there is sufficient evidence to conclude that women perceive the problem of sexual harassment on the job as much more prevalent than do men. b) There is insufficient evidence to conclude with at least 99% confidence that women perceive the problem of sexual harassment on the job as much more prevalent than do men. c) There is no evidence of a significant difference between the men and women in their perception. d) More information is needed to draw any conclusions from the data set. ANSWER: a TYPE: MC DIFFICULTY: Moderate KEYWORDS: Z test, difference between two proportions, decision, conclusion

118. A powerful women’s group has claimed that men and women differ in attitudes about sexual discrimination. A group of 50 men (group 1) and 40 women (group 2) were asked if they thought sexual discrimination is a problem in the United States. Of those sampled, 11 of the men and 19 of the women did believe that sexual discrimination is a problem. Assuming W designates women’s responses and M designates men’s, which of the following are the appropriate null and alternative hypotheses to test the group’s claim? a) H0: W −  M  0 versus H1: W −  M < 0

b) H0: W −  M  0 versus H1: W −  M > 0 c) H0: W −  M = 0 versus H1: W −  M  0 d) H0: W −  M  0 versus H1: W −  M = 0 ANSWER: c TYPE: MC DIFFICULTY: Moderate KEYWORDS: Z test, difference between two proportions, form of hypothesis

Two-Sample Tests

119. A powerful women’s group has claimed that men and women differ in attitudes about sexual discrimination. A group of 50 men (group 1) and 40 women (group 2) were asked if they thought sexual discrimination is a problem in the United States. Of those sampled, 11 of the men and 19 of the women did believe that sexual discrimination is a problem. Find the value of the test statistic. a) Z = – 2.55 b) Z = – 0.85 c) Z = – 1.05 d) Z = – 1.20 ANSWER: a TYPE: MC DIFFICULTY: Moderate KEYWORDS: Z test, difference between two proportions, test statistic

120. A powerful women’s group has claimed that men and women differ in attitudes about sexual discrimination. A group of 50 men (group 1) and 40 women (group 2) were asked if they thought sexual discrimination is a problem in the United States. Of those sampled, 11 of the men and 19 of the women did believe that sexual discrimination is a problem. If the p-value turns out to be 0.35 (which is not the real value in this data set), then a) at  = 0.05, we should fail to reject H0 b) at  = 0.04, we should reject H0 c) at  = 0.03, we should reject H0 d) None of the above would be correct statements. ANSWER: b TYPE: MC DIFFICULTY: Moderate KEYWORDS: Z test, difference between two proportions, decision, conclusion

121. A powerful women’s group has claimed that men and women differ in attitudes about sexual discrimination. A group of 50 men (group 1) and 40 women (group 2) were asked if they thought sexual discrimination is a problem in the United States. Of those sampled, 11 of the men and 19 of the women did believe that sexual discrimination is a problem. Construct a 99% confidence interval estimate of the difference between the proportion of men and women who believe that sexual discrimination is a problem. ANSWER: −0.508 to −0.002 or 0.002 to 0.508 TYPE: PR DIFFICULTY: Moderate KEYWORDS: confidence interval, difference between two proportions

122. A powerful women’s group has claimed that men and women differ in attitudes about sexual discrimination. A group of 50 men (group 1) and 40 women (group 2) were asked if they thought sexual discrimination is a problem in the United States. Of those sampled, 11 of the men and 19 of the women did believe that sexual discrimination is a problem. Construct a 95% confidence interval estimate of the difference between the proportion of men and women who believe that sexual discrimination is a problem. ANSWER: −0.448 to −0.062 or 0.062 to 0.448 TYPE: PR DIFFICULTY: Moderate

70 Two-Sample Tests KEYWORDS: confidence interval, difference between two proportions

123. A powerful women’s group has claimed that men and women differ in attitudes about sexual discrimination. A group of 50 men (group 1) and 40 women (group 2) were asked if they thought sexual discrimination is a problem in the United States. Of those sampled, 11 of the men and 19 of the women did believe that sexual discrimination is a problem. Construct a 90% confidence interval estimate of the difference between the proportion of men and women who believe that sexual discrimination is a problem. ANSWER: −0.417 to −0.093 or 0.093 to 0.417 TYPE: PR DIFFICULTY: Moderate KEYWORDS: confidence interval, difference between two proportions TABLE 10-9 A few years ago, Pepsi invited consumers to take the “Pepsi Challenge.” Consumers were asked to decide which of two sodas, Coke or Pepsi, they preferred in a blind taste test. Pepsi was interesting in determining what factors played a role in people’s taste preferences. One of the factors studied was the gender of the consumer. Below are the results of analyses comparing the taste preferences of men and women with the proportions depicting preference for Pepsi. Males: n = 109, pM = 0.422018 Females: n = 52, pF = 0.25 pM – pF = 0.172018 Z = 2.11825

124. Referring to Table 10-9, to determine if a difference exists in the taste preferences of men and women, give the correct alternative hypothesis that Pepsi would test. a) H1: M –  F  0 b) H1: M –  F  0 c) H1:  M −  F  0 d) H1:  M −  F = 0 ANSWER: c TYPE: MC DIFFICULTY: Easy KEYWORDS: Z test, difference between two proportions, form of hypothesis

125. Referring to Table 10-9, suppose Pepsi wanted to test to determine if the males preferred Pepsi more than the females. Using the test statistic given, compute the appropriate p-value for the test. a) 0.0171 b) 0.0340 c) 0.2119 d) 0.4681 ANSWER: a TYPE: MC DIFFICULTY: Moderate KEYWORDS: Z test, difference between two proportions, p-value

126. Referring to Table 10-9, suppose Pepsi wanted to test to determine if the males preferred Pepsi less than the females. Using the test statistic given, compute the appropriate p-value for the test. a) 0.0170

Two-Sample Tests

b) 0.0340 c) 0.9660 d) 0.9830 ANSWER: d TYPE: MC DIFFICULTY: Moderate KEYWORDS: Z test, difference between two proportions, p-value

127. Referring to Table 10-9, suppose that the two-tailed p-value was really 0.0734. State the proper conclusion. a) At  = 0.05, there is sufficient evidence to indicate the proportion of males preferring Pepsi differs from the proportion of females preferring Pepsi. b) At  = 0.10, there is sufficient evidence to indicate the proportion of males preferring Pepsi differs from the proportion of females preferring Pepsi. c) At  = 0.05, there is sufficient evidence to indicate the proportion of males preferring Pepsi equals the proportion of females preferring Pepsi. d) At  = 0.08, there is insufficient evidence to indicate the proportion of males preferring Pepsi differs from the proportion of females preferring Pepsi. ANSWER: b TYPE: MC DIFFICULTY: Moderate KEYWORDS: Z test, difference between two proportions, decision, conclusion 128. Referring to Table 10-9, construct a 90% confidence interval estimate of the difference between the proportion of males and females who prefer Pepsi. ANSWER: 0.05 to 0.30 TYPE: PR DIFFICULTY: Moderate KEYWORDS: confidence interval, difference between two proportions 129. Referring to Table 10-9, construct a 95% confidence interval estimate of the difference between the proportion of males and females who prefer Pepsi. ANSWER: 0.02 to 0.32 TYPE: PR DIFFICULTY: Moderate KEYWORDS: confidence interval, difference between two proportions

130. Referring to Table 10-9, construct a 99% confidence interval estimate of the difference between the proportion of males and females who prefer Pepsi. ANSWER: −0.03 to 0.37 TYPE: PR DIFFICULTY: Moderate KEYWORDS: confidence interval, difference between two proportions

72 Two-Sample Tests TABLE 10-10 The following EXCEL output contains the results of a test to determine if the proportions of satisfied guests at two resorts are the same or different. Hypothesized Difference Level of Significance

0 0.05 Group 1

Number of Successes Sample Size Group 2 Number of Successes Sample Size Group 1 Proportion Group 2 Proportion Difference in Two Proportions Average Proportion Test Statistic Two-Tailed Test Lower Critical Value Upper Critical Value p-Value

163 227 154 262 0.718061674 0.58778626 0.130275414 0.648261759 3.00875353 −1.959961082 1.959961082 0.002623357

131. Referring to Table 10-10, allowing for 0.75% probability of committing a Type I error, what are the decision and conclusion on testing whether there is any difference in the proportions of satisfied guests in the two resorts? a) Do not reject the null hypothesis; there is enough evidence to conclude that there is significant difference in the proportions of satisfied guests at the two resorts. b) Do not reject the null hypothesis; there is not enough evidence to conclude that there is significant difference in the proportions of satisfied guests at the two resorts. c) Reject the null hypothesis; there is enough evidence to conclude that there is significant difference in the proportions of satisfied guests at the two resorts. d) Reject the null hypothesis; there is not enough evidence to conclude that there is significant difference in the proportions of satisfied guests at the two resorts. ANSWER: c TYPE: MC DIFFICULTY: Easy KEYWORDS: Z test, difference between two proportions, decision, conclusion

132. Referring to Table 10-10, if you want to test the claim that "Resort 1 (Group 1) has a higher proportion of satisfied guests compared to Resort 2 (Group 2)", the p-value of the test will be a) 0.00262 b) 0.00262/2 c) 2*(0.00262) d) 1-(0.00262/2) ANSWER: b TYPE: MC DIFFICULTY: Moderate KEYWORDS: Z test, difference between two proportions, p-value

Two-Sample Tests

133. Referring to Table 10-10, if you want to test the claim that "Resort 1 (Group 1) has a lower proportion of satisfied guests compared to Resort 2 (Group 2)", you will use a) a t-test for the difference between two proportions. b) a z-test for the difference between two proportions. c) an F test for the difference between two proportions. d) a  2 test for the difference between two proportions. ANSWER: b TYPE: MC DIFFICULTY: Moderate KEYWORDS: Z test, difference between two proportions 134. Referring to Table 10-10, construct a 99% confidence interval estimate of the difference in the population proportion of satisfied guests between the two resorts. ANSWER: 0.02 to 0.24 TYPE: PR DIFFICULTY: Moderate KEYWORDS: confidence interval, difference between two proportions 135. Referring to Table 10-10, construct a 95% confidence interval estimate of the difference in the population proportion of satisfied guests between the two resorts. ANSWER: 0.05 to 0.21 TYPE: PR DIFFICULTY: Moderate KEYWORDS: confidence interval, difference between two proportions 136. Referring to Table 10-10, construct a 90% confidence interval estimate of the difference in the population proportion of satisfied guests between the two resorts. ANSWER: 0.06 to 0.20 TYPE: PR DIFFICULTY: Moderate KEYWORDS: confidence interval, difference between two proportions TABLE 10-11 A corporation randomly selects 150 salespeople and finds that 66% who have never taken a selfimprovement course would like such a course. The firm did a similar study 10 years ago in which 60% of a random sample of 160 salespeople wanted a self-improvement course. The groups are assumed to be independent random samples. Let 1 and  2 represent the true proportion of workers who would like to attend a self-improvement course in the recent study and the past study, respectively.

137. Referring to Table 10-11, if the firm wanted to test whether this proportion has changed from the previous study, which represents the relevant hypotheses? a) H0: 1 −  2 = 0 versus H1: 1 −  2  0 b) H0: 1 −  2  0 versus H1: 1 −  2 = 0

c) H0: 1 −  2  0 versus H1: 1 −  2 > 0 d) H0: 1 −  2  0 versus H1: 1 −  2 < 0

74 Two-Sample Tests

ANSWER: a TYPE: MC DIFFICULTY: Easy KEYWORDS: Z test, difference between two proportions, form of hypothesis

138. Referring to Table 10-11, if the firm wanted to test whether a greater proportion of workers would currently like to attend a self-improvement course than in the past, which represents the relevant hypotheses? a) H0: 1 −  2 = 0 versus H1: 1 −  2  0

b) H0: 1 −  2  0 versus H1: 1 −  2 = 0 c) H0: 1 −  2  0 versus H1: 1 −  2 > 0 d) H0: 1 −  2  0 versus H1: 1 −  2 < 0

ANSWER: c TYPE: MC DIFFICULTY: Easy KEYWORDS: Z test, difference between two proportions, form of hypothesis

139. Referring to Table 10-11, what is the unbiased point estimate for the difference between the two population proportions? a) 0.06 b) 0.10 c) 0.15 d) 0.22 ANSWER: a TYPE: MC DIFFICULTY: Easy KEYWORDS: point estimate

140. Referring to Table 10-11, what is/are the critical value(s) when performing a Z test on whether population proportions are different if  = 0.05? a)  1.645 b)  1.96 c) −1.96 d)  2.08 ANSWER: b TYPE: MC DIFFICULTY: Easy KEYWORDS: Z test, difference between two proportions, critical value

141. Referring to Table 10-11, what is/are the critical value(s) when testing whether population proportions are different if  = 0.10? a)  1.645 b)  1.96 c) -1.96 d)  2.08 ANSWER:

Two-Sample Tests

a TYPE: MC DIFFICULTY: Easy KEYWORDS: Z test, difference between two proportions, critical value

142. Referring to Table 10-11, what is/are the critical value(s) when testing whether the current population proportion is higher than before if  = 0.05? a)  1.645 b) + 1.645 c)  1.96 d) + 1.96 ANSWER: b TYPE: MC DIFFICULTY: Easy KEYWORDS: Z test for the difference between two proportions, critical value

143. Referring to Table 10-11, what is the estimated standard error of the difference between the two sample proportions? a) 0.629 b) 0.500 c) 0.055 d) 0 ANSWER: c TYPE: MC DIFFICULTY: Moderate KEYWORDS: Z test, difference between two proportions, standard error

144. Referring to Table 10-11, what is the value of the test statistic to use in evaluating the alternative hypothesis that there is a difference in the two population proportions? a) 4.335 b) 1.96 c) 1.093 d) 0 ANSWER: c TYPE: MC DIFFICULTY: Moderate KEYWORDS: Z test, difference between two proportions, test statistic

145. Referring to Table 10-11, the company tests to determine at the 0.05 level whether the population proportion has changed from the previous study. Which of the following is most correct? a) Reject the null hypothesis and conclude that the proportion of employees who are interested in a self-improvement course has changed over the intervening 10 years. b) Do not reject the null hypothesis and conclude that the proportion of employees who are interested in a self-improvement course has not changed over the intervening 10 years. c) Reject the null hypothesis and conclude that the proportion of employees who are interested in a self-improvement course has increased over the intervening 10 years. d) Do not reject the null hypothesis and conclude that the proportion of employees who are interested in a self-improvement course has increased over the intervening 10 years. ANSWER:

76 Two-Sample Tests b TYPE: MC DIFFICULTY: Moderate KEYWORDS: Z test, difference between two proportions, decision, conclusion

146. Referring to Table 10-11, construct a 99% confidence interval estimate of the difference in proportion of workers who would like to attend a self-improvement course in the recent study and the past study. ANSWER: −0.08 to 0.20 TYPE: PR DIFFICULTY: Moderate KEYWORDS: confidence interval, difference between two proportions

147. Referring to Table 10-11, construct a 95% confidence interval estimate of the difference in proportion of workers who would like to attend a self-improvement course in the recent study and the past study. ANSWER: −0.05 to 0.17 TYPE: PR DIFFICULTY: Moderate KEYWORDS: confidence interval, difference between two proportions

148. Referring to Table 10-11, construct a 90% confidence interval estimate of the difference in proportion of workers who would like to attend a self-improvement course in the recent study and the past study. ANSWER: −0.03 to 0.15 TYPE: PR DIFFICULTY: Moderate KEYWORDS: confidence interval, difference between two proportions

149. True or False: In testing the difference between two proportions using the normal distribution, we may use a two-tailed Z test. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: Z test, difference between two proportions

150. If we wish to determine whether there is evidence that the proportion of successes is higher in Group 1 than in Group 2, and the test statistic for Z = +2.07 where the difference is defined as Group 1’s proportion minus Group 2’s proportion, the p-value is equal to . ANSWER: 0.0192 TYPE: FI DIFFICULTY: Moderate KEYWORDS: Z test, difference between two proportions, p-value

151. If we wish to determine whether there is evidence that the proportion of successes is higher in Group 1 than in Group 2, and the test statistic for Z = −2.07 where the difference is defined as Group 1’s proportion minus Group 2’s proportion, the p-value is equal to .

ANSWER:

Two-Sample Tests

0.9808 TYPE: FI DIFFICULTY: Moderate KEYWORDS: Z test, difference between two proportions, p-value TABLE 10-12 The dean of a college is interested in the proportion of graduates from his college who have a job offer on graduation day. He is particularly interested in seeing if there is a difference in this proportion for accounting and economics majors. In a random sample of 100 of each type of major at graduation, he found that 65 accounting majors and 52 economics majors had job offers. If the accounting majors are designated as “Group 1” and the economics majors are designated as “Group 2,” perform the appropriate hypothesis test using a level of significance of 0.05.

152. Referring to Table 10-12, the hypotheses the dean should use are: a) H0: 1 −  2 = 0 versus H1: 1 −  2  0 b) H0: 1 −  2  0 versus H1: 1 −  2 = 0 c) H0: 1 −  2  0 versus H1: 1 −  2 > 0 d) H0: 1 −  2  0 versus H1: 1 −  2 < 0 ANSWER: a TYPE: MC DIFFICULTY: Easy KEYWORDS: Z test, difference between two proportions, form of hypothesis

153. Referring to Table 10-12, the null hypothesis will be rejected if the test statistic is ANSWER: Z > 1.96 or < -1.96 TYPE: FI DIFFICULTY: Easy KEYWORDS: Z test, difference between two proportions, critical value

154. Referring to Table 10-12, the value of the test statistic is

ANSWER: Z = 1.866 TYPE: FI DIFFICULTY: Moderate KEYWORDS: Z test, difference between two proportions, test statistic

155. Referring to Table 10-12, the p-value of the test is

ANSWER: 0.0621 TYPE: FI DIFFICULTY: Moderate KEYWORDS: Z test, difference between two proportions, p-value

156. True or False: Referring to Table 10-12, the null hypothesis should be rejected. ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: Z test, difference between two proportions, decision

78 Two-Sample Tests

157. True or False: Referring to Table 10-12, the same decision would be made with this test if the level of significance had been 0.01 rather than 0.05. ANSWER: True TYPE: TF DIFFICULTY: Moderate KEYWORDS: Z test, difference between two proportions, decision

158. True or False: Referring to Table 10-12, the same decision would be made with this test if the level of significance had been 0.10 rather than 0.05. ANSWER: False TYPE: TF DIFFICULTY: Moderate KEYWORDS: Z test, difference between two proportions, decision 159. Referring to Table 10-12, construct a 99% confidence interval estimate of the difference in proportion between accounting majors and economic majors who have a job offer on graduation day. ANSWER: −0.05 to 0.31 TYPE: PR DIFFICULTY: Moderate KEYWORDS: confidence interval, difference between two proportions

160. Referring to Table 10-12, construct a 95% confidence interval estimate of the difference in proportion between accounting majors and economic majors who have a job offer on graduation day. ANSWER: −0.01 to 0.27 TYPE: PR DIFFICULTY: Moderate KEYWORDS: confidence interval, difference between two proportions

161. Referring to Table 10-12, construct a 90% confidence interval estimate of the difference in proportion between accounting majors and economic majors who have a job offer on graduation day. ANSWER: 0.02 to 0.24 TYPE: PR DIFFICULTY: Moderate KEYWORDS: confidence interval, difference between two proportions TABLE 10-13 A quality control engineer is in charge of the manufacture of computer disks. Two different processes can be used to manufacture the disks. He suspects that the Kohler method produces a greater proportion of defects than the Russell method. He samples 150 of the Kohler and 200 of the Russell disks and finds that 27 and 18 of them, respectively, are defective. If Kohler is designated as “Group 1” and Russell is designated as “Group 2,” perform the appropriate test at a level of significance of 0.01.

162. Referring to Table 10-13, the hypotheses that should be tested are:

Two-Sample Tests

a) H0: 1 −  2 = 0 versus H1: 1 −  2  0 b) H0: 1 −  2  0 versus H1: 1 −  2 = 0 c) H0: 1 −  2  0 versus H1: 1 −  2 > 0 d) H0: 1 −  2  0 versus H1: 1 −  2 < 0 ANSWER: c TYPE: MC DIFFICULTY: Easy KEYWORDS: Z test, difference between two proportions, form of hypothesis

163. Referring to Table 10-13, the null hypothesis will be rejected if the test statistic is

ANSWER: Z > 2.33 TYPE: FI DIFFICULTY: Easy KEYWORDS: Z test, difference between two proportions, critical value

164. Referring to Table 10-13, the value of the test statistic is

ANSWER: 2.49 TYPE: FI DIFFICULTY: Moderate KEYWORDS: Z test, difference between two proportions, test statistic

165. Referring to Table 10-13, the p-value of the test is

ANSWER: 0.0064 TYPE: FI DIFFICULTY: Moderate KEYWORDS: Z test, difference between two proportions, p-value

166. True or False: Referring to Table 10-13, the null hypothesis should be rejected. ANSWER: True TYPE: TF DIFFICULTY: Moderate KEYWORDS: Z test, difference between two proportions, decision

167. True or False: Referring to Table 10-13, the same decision would be made with this test if the level of significance had been 0.05 rather than 0.01. ANSWER: True TYPE: TF DIFFICULTY: Moderate KEYWORDS: Z test, difference between two proportions, decision

168. True or False: Referring to Table 10-13, the same decision would be made if this had been a two-tailed test at a level of significance of 0.01. ANSWER: False TYPE: TF DIFFICULTY: Moderate

80 Two-Sample Tests KEYWORDS: Z test, difference between two proportions, decision 169. Referring to Table 10-13, construct a 90% confidence interval estimate of the difference in proportion between the Kohler and Russell disks that are defective. ANSWER: 0.03 to 0.15 TYPE: PR DIFFICULTY: Moderate KEYWORDS: confidence interval, difference between two proportions 170. Referring to Table 10-13, construct a 95% confidence interval estimate of the difference in proportion between the Kohler and Russell disks that are defective. ANSWER: 0.02 to 0.16 TYPE: PR DIFFICULTY: Moderate KEYWORDS: confidence interval, difference between two proportions 171. Referring to Table 10-13, construct a 99% confidence interval estimate of the difference in proportion between the Kohler and Russell disks that are defective. ANSWER: -0.01 to 0.19 TYPE: PR DIFFICULTY: Moderate KEYWORDS: confidence interval, difference between two proportions

86 Analysis of Variance

Two-Sample Tests

CHAPTER 11: ANALYSIS OF VARIANCE 1. In a one-way ANOVA, if the computed F statistic exceeds the critical F value we may a) reject H0 since there is evidence all the means differ. b) reject H0 since there is evidence of a treatment effect. c) not reject H0 since there is no evidence of a difference. d) not reject H0 because a mistake has been made. ANSWER: b TYPE: MC DIFFICULTY: Easy KEYWORDS: one-way analysis of variance, decision 2. Which of the following components in an ANOVA table are not additive? a) Sum of squares. b) Degrees of freedom. c) Mean squares. d) It is not possible to tell. ANSWER: c TYPE: MC DIFFICULTY: Easy KEYWORDS: one-way analysis of variance, sum of squares, properties 3. Why would you use the Tukey-Kramer procedure? a) To test for normality. b) To test for homogeneity of variance. c) To test independence of errors. d) To test for differences in pairwise means. ANSWER: d TYPE: MC DIFFICULTY: Easy KEYWORDS: Tukey-Kramer procedure 4. A completely randomized design a) has only one factor with several treatment groups. b) can have more than one factor, each with several treatment groups. c) has one factor and one block. d) has one factor and one block and multiple values. ANSWER: a TYPE: MC DIFFICULTY: Moderate KEYWORDS: completely randomized design

5. The F test statistic in a one-way ANOVA is a) b) c) d)

MSW/MSA. SSW/SSA. MSA/MSW. SSA/SSW.

Analysis of Variance

ANSWER: c TYPE: MC DIFFICULTY: Easy KEYWORDS: one-way analysis of variance, F test for factor

6. The degrees of freedom for the F test in a one-way ANOVA are a) (n – c) and (c – 1). b) (c – 1) and (n – c). c) (c – n) and (n – 1). d) (n – 1) and (c – n). ANSWER: b TYPE: MC DIFFICULTY: Easy KEYWORDS: one-way analysis of variance, degrees of freedom 7. In a one-way ANOVA, the null hypothesis is always a) there is no treatment effect. b) there is some treatment effect. c) all the population means are different. d) some of the population means are different. ANSWER: a TYPE: MC DIFFICULTY: Easy KEYWORDS: one-way analysis of variance, form of hypothesis 8. In a one-way ANOVA a) an interaction term is present. b) an interaction effect can be tested. c) there is no interaction term. d) the interaction term has (c – 1)(n – 1) degrees of freedom. ANSWER: c TYPE: MC DIFFICULTY: Moderate KEYWORDS: one-way analysis of variance, properties, interaction 9. Interaction in an experimental design can be tested in a) a completely randomized model. b) a randomized block model. c) a two-factor model. d) all ANOVA models. ANSWER: c TYPE: MC DIFFICULTY: Easy KEYWORDS: two-factor analysis of variance, interaction, properties 10. In a two-way ANOVA the degrees of freedom for the interaction term is a) (r – 1)(c – 1). b) rc(n – 1).

Analysis of Variance

c) (r – 1). d) rcn + 1. ANSWER: a TYPE: MC DIFFICULTY: Easy KEYWORDS: two-factor analysis of variance, interaction, degrees of freedom 11. In a two-way ANOVA the degrees of freedom for the "error" term is a) (r – 1)(c – 1). b) rc(n’ – 1). c) (r – 1). d) rcn' + 1. ANSWER: b TYPE: MC DIFFICULTY: Easy KEYWORDS: two-factor analysis of variance, degrees of freedom TABLE 11-1 Psychologists have found that people are generally reluctant to transmit bad news to their peers. This phenomenon has been termed the “MUM effect.” To investigate the cause of the MUM effect, 40 undergraduates at Duke University participated in an experiment. Each subject was asked to administer an IQ test to another student and then provide the test taker with his or her percentile score. Unknown to the subject, the test taker was a bogus student who was working with the researchers. The experimenters manipulated two factors: subject visibility and success of test taker, each at two levels. Subject visibility was either visible or not visible to the test taker. Success of the test taker was either top 20% or bottom 20%. Ten subjects were randomly assigned to each of the 2 x 2 = 4 experimental conditions, then the time (in seconds) between the end of the test and the delivery of the percentile score from the subject to the test taker was measured. (This variable is called the latency to feedback.) The data were subjected to appropriate analyses with the following results. Source df Subject visibility 1 Test taker success 1 Interaction 1 Error 36 Total 39

SS 1380.24 1325.16 3385.80 11,664.00 17,755.20

MS 1380.24 1325.16 3385.80 324.00

F 4.26 4.09 10.45

PR > F 0.043 0.050 0.002

12. Referring to Table 11-1, what type of experimental design was employed in this study? a) Completely randomized design with 4 treatments b) Randomized block design with four treatments and 10 blocks c) 2 x 2 factorial design with 10 replications d) None of the above ANSWER: c TYPE: MC DIFFICULTY: Moderate KEYWORDS: two-factor analysis of variance, two-factor factorial design 13. Referring to Table 11-1, at the 0.01 level, what conclusions can you draw from the analyses?

Analysis of Variance a) At the 0.01 level, subject visibility and test taker success are significant predictors of latency feedback. b) At the 0.01 level, the model is not useful for predicting latency to feedback. c) At the 0.01 level, there is evidence to indicate that subject visibility and test taker success interact. d) At the 0.01 level, there is no evidence of interaction between subject visibility and test taker success. ANSWER: c TYPE: MC DIFFICULTY: Moderate KEYWORDS: two-factor analysis of variance, F test for interaction, decision, conclusion, interaction 14. Referring to Table 11-1, in the context of this study, interpret the statement: “Subject visibility and test taker success interact.” a) The difference between the mean feedback time for visible and nonvisible subjects depends on the success of the test taker. b) The difference between the mean feedback time for test takers scoring in the top 20% and bottom 20% depends on the visibility of the subject. c) The relationship between feedback time and subject visibility depends on the success of the test taker. d) All of the above are correct interpretations. ANSWER: d TYPE: MC DIFFICULTY: Difficult KEYWORDS: two-factor analysis of variance, interaction, conclusion 15. An airline wants to select a computer software package for its reservation system. Four software packages (1, 2, 3, and 4) are commercially available. The airline will choose the package that bumps as few passengers, on the average, as possible during a month. An experiment is set up in which each package is used to make reservations for 5 randomly selected weeks. (A total of 20 weeks was included in the experiment.) The number of passengers bumped each week is given below. How should the data be analyzed? Package 1: 12, 14, 9, 11, 16 Package 2: 2, 4, 7, 3, 1 Package 3: 10, 9, 6, 10, 12 Package 4: 7, 6, 6, 15, 12

a) b) c) d)

F test for differences in variances. One-way ANOVA F test. t test for the differences in means. t test for the mean difference.

ANSWER: b TYPE: MC DIFFICULTY: Easy KEYWORDS: one-way analysis of variance, F test for factor TABLE 11-2 An airline wants to select a computer software package for its reservation system. Four software packages (1, 2, 3, and 4) are commercially available. The airline will choose the package that bumps

Analysis of Variance

as few passengers, on the average, as possible during a month. An experiment is set up in which each package is used to make reservations for 5 randomly selected weeks. (A total of 20 weeks was included in the experiment.) The number of passengers bumped each week is obtained, which gives rise to the following Excel output: ANOVA Source of Variation Between Groups Within Groups

SS 212.4 136.4

Total

348.8

MS 3

F 8.304985

P-value 0.001474

F crit 3.238867

8.525

16. Referring to Table 11-2, the within groups degrees of freedom is a) 3 b) 4 c) 16 d) 19 ANSWER: c TYPE: MC DIFFICULTY: Easy KEYWORDS: one-way analysis of variance, degrees of freedom 17. Referring to Table 11-2, the total degrees of freedom is a) 3 b) 4 c) 16 d) 19 ANSWER: d TYPE: MC DIFFICULTY: Easy KEYWORDS: one-way analysis of variance, degrees of freedom

18. Referring to Table 11-2, the among-group (between-group) mean squares is a) 8.525 b) 70.8 c) 212.4 d) 637.2 ANSWER: b TYPE: MC DIFFICULTY: Easy KEYWORDS: one-way analysis of variance, mean squares 19. Referring to Table 11-2, at a significance level of 1%, a) there is insufficient evidence to conclude that the average numbers of customers bumped by the 4 packages are not all the same. b) there is insufficient evidence to conclude that the average numbers of customers bumped by the 4 packages are all the same. c) there is sufficient evidence to conclude that the average numbers of customers bumped by the 4 packages are not all the same.

Analysis of Variance d) there is sufficient evidence to conclude that the average numbers of customers bumped by the 4 packages are all the same. ANSWER: c TYPE: MC DIFFICULTY: Easy KEYWORDS: one-way analysis of variance, F test for factor, decision, conclusion TABLE 11-3 A realtor wants to compare the average sales-to-appraisal ratios of residential properties sold in four neighborhoods (A, B, C, and D). Four properties are randomly selected from each neighborhood and the ratios recorded for each, as shown below. A: 1.2, 1.1, 0.9, 0.4 C: 1.0, 1.5, 1.1, 1.3 B: 2.5, 2.1, 1.9, 1.6 D: 0.8, 1.3, 1.1, 0.7 Interpret the results of the analysis summarized in the following table: Source Neighborhoods Error Total

SS 3.1819

MS 1.0606

F 10.76

PR > F 0.001

12 4.3644

20. Referring to Table 11-3, the among group degrees of freedom is a) 3 b) 4 c) 12 d) 16 ANSWER: a TYPE: MC DIFFICULTY: Easy KEYWORDS: one-way analysis of variance, degrees of freedom 21. Referring to Table 11-3, the within group sum of squares is a) 1.0606 b) 1.1825 c) 3.1819 d) 4.3644 ANSWER: b TYPE: MC DIFFICULTY: Easy KEYWORDS: one-way analysis of variance, sum of squares 22. Referring to Table 11-3, the within group mean squares is a) 0.10 b) 0.29 c) 1.06 d) 1.18 ANSWER: a

Analysis of Variance

TYPE: MC DIFFICULTY: Easy KEYWORDS: one-way analysis of variance, mean squares 23. Referring to Table 11-3, a) at the 0.05 level of significance, the mean ratios for the 4 neighborhoods are not all the same. b) at the 0.01 level of significance, the mean ratios for the 4 neighborhoods are all the same. c) at the 0.10 level of significance, the mean ratios for the 4 neighborhoods are not significantly different. d) at the 0.05 level of significance, the mean ratios for the 4 neighborhoods are not significantly different from 0. ANSWER: a TYPE: MC DIFFICULTY: Easy KEYWORDS: one-way analysis of variance, F test for factor, decision, conclusion 24. Referring to Table 11-3, the null hypothesis for Levene’s test for homogeneity of variances is a) H0 :  A = B = C = D b) H0 : M A = M B = M C = M D c) H0 :  2 = 2 =  2 =  2 A

d) H0 :  A =  B =  C =  D ANSWER: c TYPE: MC DIFFICULTY: Easy KEYWORDS: Levene’s test, form of hypothesis 25. Referring to Table 11-3, the value of the test statistic for Levene’s test for homogeneity of variances is a) 0.25 b) 0.37 c) 4.36 d) 10.76 ANSWER: a TYPE: MC DIFFICULTY: Moderate KEYWORDS: Levene’s test, test statistic 26. Referring to Table 11-3, the numerator and denominator degrees of freedom for Levene’s test for homogeneity of variances at a 5% level of significance are, respective, a) 3, 12 b) 12, 3 c) 3, 15 d) 15, 3 ANSWER: a TYPE: MC DIFFICULTY: Moderate KEYWORDS: Levene’s test, degrees of freedom

Analysis of Variance 27. Referring to Table 11-3, the critical value of Levene’s test for homogeneity of variances at a 5% level of significance is a) 0.64 b) 2.48 c) 3.29 d) 3.49 ANSWER: d TYPE: MC DIFFICULTY: Moderate KEYWORDS: Levene’s test, critical value

28. Referring to Table 11-3, the p-value of the test statistic for Levene’s test for homogeneity of variances is a) 0.25 b) 0.64 c) 0.86 d) 3.49 ANSWER: c TYPE: MC DIFFICULTY: Moderate KEYWORDS: Levene’s test, p-value 29. Referring to Table 11-3, what should be the decision for the Levene’s test for homogeneity of variances at a 5% level of significance? a) Reject the null hypothesis because the p-value is smaller than the level of significance. b) Reject the null hypothesis because the p-value is larger than the level of significance. c) Do not eject the null hypothesis because the p-value is smaller than the level of significance. d) Do not reject the null hypothesis because the p-value is larger than the level of significance. ANSWER: d TYPE: MC DIFFICULTY: Easy KEYWORDS: Levene’s test, decision 30. Referring to Table 11-3, what should be the conclusion for the Levene’s test for homogeneity of variances at a 5% level of significance? a) There is not sufficient evidence that the variances are all the same. b) There is sufficient evidence that the variances are all the same. c) There is not sufficient evidence that the variances are not all the same. d) There is sufficient evidence that the variances are not all the same. ANSWER: c TYPE: MC DIFFICULTY: Easy KEYWORDS: Levene’s test, conclusion 31. A campus researcher wanted to investigate the factors that affect visitor travel time in a complex, multilevel building on campus. Specifically, he wanted to determine whether different building

Analysis of Variance

signs (building maps versus wall signage) affect the total amount of time visitors require to reach their destination and whether that time depends on whether the starting location is inside or outside the building. Three subjects were assigned to each of the combinations of signs and starting locations, and travel time in seconds from beginning to destination was recorded. How should the data be analyzed? Starting Room Interior Exterior Wall Signs 141, 119, 238 224, 339, 139 Map 85, 94, 126 226, 129, 130 a) Completely randomized design b) Randomized block design c) 2 x 2 factorial design d) Levene’s test ANSWER: c TYPE: MC DIFFICULTY: Moderate KEYWORDS: two-factor factorial design TABLE 11-4 A campus researcher wanted to investigate the factors that affect visitor travel time in a complex, multilevel building on campus. Specifically, he wanted to determine whether different building signs (building maps versus wall signage) affect the total amount of time visitors require to reach their destination and whether that time depends on whether the starting location is inside or outside the building. Three subjects were assigned to each of the combinations of signs and starting locations, and travel time in seconds from beginning to destination was recorded. An Excel output of the appropriate analysis is given below: ANOVA Source of Variation Signs Starting Location Interaction Within

SS 14008.33 12288 48 35305.33

Total

61649.67

MS 14008.33 48 4413.167

P-value 0.11267 2.784395 0.13374 0.919506

F crit 5.317645 5.317645 5.317645

32. Referring to Table 11-4, the degrees of freedom for the different building signs (factor A) is a) 1 b) 2 c) 3 d) 8 ANSWER: a TYPE: MC DIFFICULTY: Easy KEYWORDS: two-factor analysis of variance, degrees of freedom 33. Referring to Table 11-4, the within (error) degrees of freedom is a) 1

Analysis of Variance b) 4 c) 8 d) 11 ANSWER: c TYPE: MC DIFFICULTY: Easy KEYWORDS: two-factor analysis of variance, degrees of freedom 34. Referring to Table 11-4, the mean squares for starting location (factor B) is a) 48 b) 4,413.17 c) 12,288 d) 14,008.3 ANSWER: c TYPE: MC DIFFICULTY: Easy KEYWORDS: two-factor analysis of variance, mean squares

35. Referring to Table 11-4, the F test statistic for testing the main effect of types of signs is a) 0.0109 b) 2.7844 c) 3.1742 d) 5.3176 ANSWER: c TYPE: MC DIFFICULTY: Easy KEYWORDS: two-factor analysis of variance, F test for factor

36. Referring to Table 11-4, the F test statistic for testing the interaction effect between the types of signs and the starting location is a) 0.0109 b) 2.7844 c) 3.1742 d) 5.3176 ANSWER: a TYPE: MC DIFFICULTY: Easy KEYWORDS: two-factor analysis of variance, F test for interaction, interaction 37. Referring to Table 11-4, at 1% level of significance, a) there is insufficient evidence to conclude that the difference between the average traveling time for the different starting locations depends on the types of signs. b) there is insufficient evidence to conclude that the difference between the average traveling time for the different types of signs depends on the starting locations. c) there is insufficient evidence to conclude that the relationship between traveling time and the types of signs depends on the starting locations. d) All of the above. ANSWER: d

Analysis of Variance

TYPE: MC DIFFICULTY: Easy KEYWORDS: two-factor analysis of variance, F test for factor, F test for interaction, decision, conclusion 38. Referring to Table 11-4, at 10% level of significance, a) there is sufficient evidence to conclude that the difference between the average traveling time for the different starting locations depends on the types of signs. b) there is insufficient evidence to conclude that the difference between the average traveling time for the different types of signs depends on the starting locations. c) there is sufficient evidence to conclude that the difference between the average traveling time for the different starting locations does not depend on the types of signs. d) None of the above. ANSWER: b TYPE: MC DIFFICULTY: Easy KEYWORDS: two-factor analysis of variance, F test for factor, decision, conclusion TABLE 11-5 A physician and president of a Tampa Health Maintenance Organization (HMO) are attempting to show the benefits of managed health care to an insurance company. The physician believes that certain types of doctors are more cost-effective than others. One theory is that Primary Specialty is an important factor in measuring the cost-effectiveness of physicians. To investigate this, the president obtained independent random samples of 20 HMO physicians from each of 4 primary specialties General Practice (GP), Internal Medicine (IM), Pediatrics (PED), and Family Physicians (FP) - and recorded the total charges per member per month for each. A second factor which the president believes influences total charges per member per month is whether the doctor is a foreign or USA medical school graduate. The president theorizes that foreign graduates will have higher mean charges than USA graduates. To investigate this, the president also collected data on 20 foreign medical school graduates in each of the 4 primary specialty types described above. So information on charges for 40 doctors (20 foreign and 20 USA medical school graduates) was obtained for each of the 4 specialties. The results for the ANOVA are summarized in the following table. Source Specialty Med school Interaction Error Total

df 3 1 3 152 159

SS 22,855 105 890 18,950 42,800

MS 7,618 105 297

F 60.94 0.84 2.38

PR > F 0.0001 0.6744 0.1348

39. Referring to Table 11-5, what was the total number of doctors included in the study? a) 20 b) 40 c) 159 d) 160 ANSWER: d TYPE: MC DIFFICULTY: Easy KEYWORDS: two-factor analysis of variance, properties

Analysis of Variance

40. Referring to Table 11-5, what degrees of freedom should be used to determine the critical value of the F ratio against which to test for interaction between the two factors? a) numerator df = 1, denominator df = 159 b) numerator df = 3, denominator df = 159 c) numerator df = 1, denominator df = 152 d) numerator df = 3, denominator df = 152 ANSWER: d TYPE: MC DIFFICULTY: Easy KEYWORDS: two-factor analysis of variance, F test for interaction, degrees of freedom 41. Referring to Table 11-5, interpret the test for interaction. a) There is insufficient evidence to say at the 0.10 level of significance that the difference between the mean charges for foreign and USA graduates depends on primary specialty. b) There is sufficient evidence to say at the 0.10 level of significance that the difference between the mean charges for foreign and USA graduates depends on primary specialty. c) There is sufficient evidence at the 0.10 level of significance of a difference between the mean charges for foreign and USA medical graduates. d) There is sufficient evidence to say at the 0.10 level of significance that mean charges depend on both primary specialty and medical school. ANSWER: a TYPE: MC DIFFICULTY: Difficult KEYWORDS: two-factor analysis of variance, interaction, interpretation

42. Referring to Table 11-5, what degrees of freedom should be used to determine the critical value of the F ratio against which to test for differences in the mean charges for doctors among the four primary specialty areas? a) numerator df = 1, denominator df = 159 b) numerator df = 3, denominator df = 159 c) numerator df = 1, denominator df = 152 d) numerator df = 3, denominator df = 152 ANSWER: d TYPE: MC DIFFICULTY: Easy KEYWORDS: two-factor analysis of variance, F test for factor, degrees of freedom

43. Referring to Table 11-5, what degrees of freedom should be used to determine the critical value of the F ratio against which to test for differences between the mean charges of foreign and USA medical school graduates? a) numerator df = 1, denominator df = 159 b) numerator df = 3, denominator df = 159 c) numerator df = 1, denominator df = 152 d) numerator df = 3, denominator df = 152 ANSWER: c TYPE: MC DIFFICULTY: Easy KEYWORDS: two-factor analysis of variance, F test for factor, degrees of freedom

Analysis of Variance

44. Referring to Table 11-5, is there evidence of a difference between the mean charges of foreign and USA medical school graduates? a) Yes, the test for the main effect for primary specialty is significant at  = 0.10. b) No, the test for the main effect for medical school is not significant at  = 0.10. c) No, the test for the interaction is not significant at  = 0.10. d) Maybe, but we need information on the  -estimates to fully answer the question. ANSWER: b TYPE: MC DIFFICULTY: Easy KEYWORDS: two-factor analysis of variance, F test for factor, decision, conclusion 45. Referring to Table 11-5, what assumption(s) need(s) to be made in order to conduct the test for differences between the mean charges of foreign and USA medical school graduates? a) There is no significant interaction effect between the area of primary specialty and the medical school on the doctors’ mean charges. b) The charges in each group of doctors sampled are drawn from normally distributed populations. c) The charges in each group of doctors sampled are drawn from populations with equal variances. d) All of the above are necessary assumptions. ANSWER: d TYPE: MC DIFFICULTY: Moderate KEYWORDS: two-factor analysis of variance, assumptions 46. True or False: The analysis of variance (ANOVA) tests hypotheses about the population variance. ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: one-way analysis of variance

47. True or False: The F test in a completely randomized model is just an expansion of the t test for independent samples. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: completely randomized design, F test for factor

48. True or False: When the F test is used for ANOVA, the rejection region is always in the right tail. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: F test for factor, rejection region

Analysis of Variance 49. True or False: A completely randomized design with 4 groups would have 6 possible pairwise comparisons. ANSWER: True TYPE: TF DIFFICULTY: Moderate KEYWORDS: completely randomized design, properties 50. True or False: If you are comparing the average sales among 3 different brands you are dealing with a three-way ANOVA design. ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: one-way analysis of variance, properties 51. True or False: The MSE must always be positive. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: mean squares, properties 52. True or False: In a two-way ANOVA, it is easier to interpret main effects when the interaction component is not significant. ANSWER: True TYPE: TF DIFFICULTY: Moderate KEYWORDS: two-factor analysis of variance, interpretation 53. True or False: In a one-factor ANOVA analysis, the among sum of squares and within sum of squares must add up to the total sum of squares. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: one-way analysis of variance, sum of squares, properties 54. True or False: In a two-factor ANOVA analysis, the sum of squares due to both factors, the interaction sum of squares and the within sum of squares must add up to the total sum of squares. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: two-factor analysis of variance, sum of squares, properties TABLE 11-6 As part of an evaluation program, a sporting goods retailer wanted to compare the downhill coasting speeds of 4 brands of bicycles. She took 3 of each brand and determined their maximum downhill speeds. The results are presented in miles per hour in the table below. Trial Barth Tornado Reiser Shaw 1 43 37 41 43

Analysis of Variance 2 3

46 43

38 39

45 42

100

45 46

55. Referring to Table 11-6, the sporting goods retailer decided to perform an ANOVA F test. The amount of total variation or SST is

ANSWER: 102.67 TYPE: FI DIFFICULTY: Moderate KEYWORDS: one-way analysis of variance, sum of squares, interpretation 56. Referring to Table 11-6, the among group variation or SSA is

ANSWER: 81.33 TYPE: FI DIFFICULTY: Moderate KEYWORDS: one-way analysis of variance, sum of squares 57. Referring to Table 11-6, the within group variation or SSW is

ANSWER: 21.33 TYPE: FI DIFFICULTY: Moderate KEYWORDS: one-way analysis of variance, sum of squares 58. Referring to Table 11-6, the value of MSA is

, while MSW is

ANSWER: 27.11; 2.67 TYPE: FI DIFFICULTY: Moderate KEYWORDS: one-way analysis of variance, mean squares 59. Referring to Table 11-6, the null hypothesis that the average downhill coasting speeds of the 4 brands of bicycles are equal will be rejected at a level of significance of 0.05 if the value of the test statistic is greater than . ANSWER: 4.07 TYPE: FI DIFFICULTY: Easy KEYWORDS: one-way analysis of variance, F test for factor, decision 60. Referring to Table 11-6, in testing the null hypothesis that the average downhill coasting speeds of the 4 brands of bicycles are equal, the value of the test statistic is . ANSWER: 10.17 TYPE: FI DIFFICULTY: Moderate KEYWORDS: one-way analysis of variance, F test for factor, test statistic 61. Referring to Table 11-6, construct the ANOVA table from the sample data. ANSWER: Analysis of Variance

101 Analysis of Variance

Source Bicycle Brands Error Total

df SS MS F 3 81.33 27.11 10.17 8 21.33 2.67 11 102.67 * or p < 0.005, tabular value TYPE: PR DIFFICULTY: Moderate KEYWORDS: one-way analysis of variance, properties

p 0.004*

62. True or False: Referring to Table 11-6, the null hypothesis should be rejected at a 5% level of significance. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: one-way analysis of variance, decision 63. True or False: Referring to Table 11-6, the decision made implies that all 4 means are significantly different. ANSWER: False TYPE: TF DIFFICULTY: Moderate KEYWORDS: one-way analysis of variance, conclusion, interpretation 64. True or False: Referring to Table 11-6, the test is valid only if the population of speeds has the same variance for the 4 brands. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: one-way analysis of variance, assumption 65. True or False: Referring to Table 11-6, the test is less sensitive to the assumption that the population of speeds has the same variance for the 4 brands if the sample sizes of the 4 brands are equal. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: one-way analysis of variance, assumption 66. True or False: Referring to Table 11-6, the test is valid only if the population of speeds is normally distributed. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: one-way analysis of variance, assumption 67. True or False: Referring to Table 11-6, the test is robust to the violation of the assumption that the population of speeds is normally distributed. ANSWER:

Analysis of Variance

102

True TYPE: TF DIFFICULTY: Easy KEYWORDS: one-way analysis of variance, assumption 68. Referring to Table 11-6, the sporting goods retailer decided to compare the 4 treatment means by using the Tukey-Kramer procedure with an overall level of significance of 0.05. There are pairwise comparisons that can be made. ANSWER: 6 TYPE: FI DIFFICULTY: Moderate KEYWORDS: one-way analysis of variance, Tukey-Kramer procedure

69. Referring to Table 11-6, using an overall level of significance of 0.05, the critical value of the Studentized range Q used in calculating the critical range for the Tukey-Kramer procedure is . ANSWER: 4.53 TYPE: FI DIFFICULTY: Moderate KEYWORDS: one-way analysis of variance, Tukey-Kramer procedure, critical value 70. Referring to Table 11-6, using an overall level of significance of 0.05, the critical range for the Tukey-Kramer procedure is . ANSWER: 4.27 TYPE: FI DIFFICULTY: Easy KEYWORDS: one-way analysis of variance,, Tukey-Kramer procedure, critical value 71. True or False: Referring to Table 11-6, based on the Tukey-Kramer procedure with an overall level of significance of 0.05, the retailer would decide that there is a significant difference between all pairs of mean speeds. ANSWER: False TYPE: TF DIFFICULTY: Moderate KEYWORDS: one-way analysis of variance, Tukey-Kramer procedure, decision, conclusion 72. True or False: Referring to Table 11-6, based on the Tukey-Kramer procedure with an overall level of significance of 0.05, the retailer would decide that there is no significant difference between any pair of mean speeds. ANSWER: False TYPE: TF DIFFICULTY: Moderate KEYWORDS: one-way analysis of variance, Tukey-Kramer procedure, decision, conclusion 73. True or False: Referring to Table 11-6, based on the Tukey-Kramer procedure with an overall level of significance of 0.05, the retailer would decide that the mean speed for the Tornado brand is significantly different from each of the mean speeds for other brands. ANSWER: True

103 Analysis of Variance TYPE: TF DIFFICULTY: Moderate KEYWORDS: one-way analysis of variance, Tukey-Kramer procedure, decision, conclusion 74. True or False: Referring to Table 11-6, based on the Tukey-Kramer procedure with an overall level of significance of 0.05, the retailer would decide that the 3 means other than the mean for Tornado are not significantly different from each other. ANSWER: True TYPE: TF DIFFICULTY: Moderate KEYWORDS: one-way analysis of variance, Tukey-Kramer procedure, decision, conclusion 75. Referring to Table 11-6, the null hypothesis for Levene’s test for homogeneity of variances is a) H0 :  A = B = C = D b) H0 : M A = M B = M C = M D c) H0 :  2 = 2 =  2 =  2 A

d) H0 :  A =  B =  C =  D ANSWER: c TYPE: MC DIFFICULTY: Easy KEYWORDS: Levene’s test, form of hypothesis 76. Referring to Table 11-6, what is the value of the test statistic for Levene’s test for homogeneity of variances? ANSWER: 0.1333 TYPE: PR DIFFICULTY: Moderate KEYWORDS: Levene’s test, test statistic 77. Referring to Table 11-6, what are the numerator and denominator degrees of freedom for Levene’s test for homogeneity of variances respectively? ANSWER: 3, 8 TYPE: PR DIFFICULTY: Moderate KEYWORDS: Levene’s test, degrees of freedom 78. Referring to Table 11-6, what is the critical value of Levene’s test for homogeneity of variances at a 5% level of significance? ANSWER: 4.07 TYPE: PR DIFFICULTY: Moderate KEYWORDS: Levene’s test, critical value

79. Referring to Table 11-6, what is the p-value of the test statistic for Levene’s test for homogeneity of variances? ANSWER: 0.94

Analysis of Variance

104

TYPE: PR DIFFICULTY: Moderate KEYWORDS: Levene’s test, p-value 80. Referring to Table 11-6, what should be the decision for the Levene’s test for homogeneity of variances at a 5% level of significance? a) Reject the null hypothesis because the p-value is smaller than the level of significance. b) Reject the null hypothesis because the p-value is larger than the level of significance. c) Do not eject the null hypothesis because the p-value is smaller than the level of significance. d) Do not reject the null hypothesis because the p-value is larger than the level of significance. ANSWER: d TYPE: MC DIFFICULTY: Easy KEYWORDS: Levene’s test, decision 81. Referring to Table 11-6, what should be the conclusion for the Levene’s test for homogeneity of variances at a 5% level of significance? a) There is not sufficient evidence that the variances are all the same. b) There is sufficient evidence that the variances are all the same. c) There is not sufficient evidence that the variances are not all the same. d) There is sufficient evidence that the variances are not all the same. ANSWER: c TYPE: MC DIFFICULTY: Easy KEYWORDS: Levene’s test, conclusion TABLE 11-7 An agronomist wants to compare the crop yield of 3 varieties of chickpea seeds. She plants 15 fields, 5 with each variety. She then measures the crop yield in bushels per acre. Treating this as a completely randomized design, the results are presented in the table that follows. Trial Smith Walsh Trevor 1 11.1 19.0 14.6 2 13.5 18.0 15.7 3 15.3 19.8 16.8 4 14.6 19.6 16.7 5 9.8 16.6 15.2

82. Referring to Table 11-7, the agronomist decided to perform an ANOVA F test. The amount of total variation or SST is

ANSWER: 114.82 TYPE: FI DIFFICULTY: Moderate KEYWORDS: one-way analysis of variance, sum of squares 83. Referring to Table 11-7, the among-group variation or SSA is ANSWER: 82.39

105 Analysis of Variance TYPE: FI DIFFICULTY: Moderate KEYWORDS: one-way analysis of variance, sum of squares 84. Referring to Table 11-7, the within-group variation or SSW is

ANSWER: 32.43 TYPE: FI DIFFICULTY: Moderate KEYWORDS: one-way analysis of variance, sum of squares 85. Referring to Table 11-7, the value of MSA is

, while MSW is

ANSWER: 41.19; 2.70 TYPE: FI DIFFICULTY: Moderate KEYWORDS: one-way analysis of variance, mean squares 86. Referring to Table 11-7, the null hypothesis will be rejected at a level of significance of 0.01 if the value of the test statistic is greater than . ANSWER: 6.93 TYPE: FI DIFFICULTY: Easy KEYWORDS: one-way analysis of variance, F test for factor, critical value 87. Referring to Table 11-7, the value of the test statistic is

ANSWER: 15.24 TYPE: FI DIFFICULTY: Moderate KEYWORDS: one-way analysis of variance, F test for factor, test statistic 88. Referring to Table 11-7, construct the ANOVA table from the sample data. ANSWER: Analysis of Variance Source df SS MS F p Seed Varieties 2 82.39 41.19 15.24 0.000508* Error 12 32.43 2.70 Total 14 114.82 * or p < 0.005, tabular value TYPE: PR DIFFICULTY: Moderate KEYWORDS: one-way analysis of variance, F test for factor, properties 89. Referring to Table 11-7, state the null hypothesis that can be tested. ANSWER: H0: Smith = Walsh = Trevor TYPE: PR DIFFICULTY: Easy KEYWORDS: one-way analysis of variance, F test for factor, form of hypothesis 90. True or False: Referring to Table 11-7, the null hypothesis should be rejected at 0.005 level of significance.

Analysis of Variance

106

ANSWER: True TYPE: TF DIFFICULTY: Moderate KEYWORDS: one-way analysis of variance, F test for factor, decision 91. True or False: Referring to Table 11-7, the decision made at 0.005 level of significance implies that all 3 means are significantly different. ANSWER: False TYPE: TF DIFFICULTY: Moderate KEYWORDS: one-way analysis of variance, F test for factor, conclusion 92. True or False: Referring to Table 11-7, the test is valid only if the population of crop yields has the same variance for the 3 varieties. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: one-way analysis of variance, F test for factor, assumption 93. True or False: Referring to Table 11-7, the test is valid only if the population of crop yields is normally distributed for the 3 varieties. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: one-way analysis of variance, F test for factor, assumption 94. Referring to Table 11-7, the agronomist decided to compare the 3 treatment means by using the Tukey-Kramer procedure with an overall level of significance of 0.01. There are pairwise comparisons that can be made. ANSWER: 3 TYPE: FI DIFFICULTY: Moderate KEYWORDS: one-way analysis of variance, Tukey-Kramer procedure, properties

95. Referring to Table 11-7, using an overall level of significance of 0.01, the critical value of the Studentized range Q used in calculating the critical range for the Tukey-Kramer procedure is . ANSWER: 5.04 TYPE: FI DIFFICULTY: Easy KEYWORDS: one-way analysis of variance, Tukey-Kramer procedure, critical value 96. Referring to Table 11-7, using an overall level of significance of 0.01, the critical range for the Tukey-Kramer procedure is . ANSWER: 3.70 TYPE: FI DIFFICULTY: Moderate

107 Analysis of Variance KEYWORDS: one-way analysis of variance, Tukey-Kramer procedure, critical value 97. True or False: Referring to Table 11-7, based on the Tukey-Kramer procedure with an overall level of significance of 0.01, the agronomist would decide that there is a significant difference between the crop yield of Smith and Walsh seeds. ANSWER: True TYPE: TF DIFFICULTY: Moderate KEYWORDS: one-way analysis of variance, Tukey-Kramer procedure, decision, conclusion 98. True or False: Referring to Table 11-7, based on the Tukey-Kramer procedure with an overall level of significance of 0.01, the agronomist would decide that there is a significant difference between the crop yield of Smith and Trevor seeds. ANSWER: False TYPE: TF DIFFICULTY: Moderate KEYWORDS: one-way analysis of variance, Tukey-Kramer procedure, decision, conclusion 99. True or False: Referring to Table 11-7, based on the Tukey-Kramer procedure with an overall level of significance of 0.01, the agronomist would decide that there is a significant difference between the crop yield of Walsh and Trevor seeds. ANSWER: False TYPE: TF DIFFICULTY: Moderate KEYWORDS: one-way analysis of variance, Tukey-Kramer procedure, decision, conclusion TABLE 11-8 A hotel chain has identically sized resorts in 5 locations. The data that follow resulted from analyzing the hotel occupancies on randomly selected days in the 5 locations. ROW Caymen Pennkamp California Mayaguez Maui 1 28 40 21 37 22 2 33 35 21 47 19 3 41 33 27 45 25 Analysis of Variance Source df SS Location 4 963.6 Error 10 210.0 Total

F 11.47

p 0.001

100. True or False: Referring to Table 11-8, if a level of significance of 0.05 is chosen, the null hypothesis should be rejected. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: one-way analysis of variance, F test for factor, decision

Analysis of Variance

108

101. True or False: Referring to Table 11-8, if a level of significance of 0.05 is chosen, the decision made indicates that all 5 locations have different mean occupancy rates. ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: one-way analysis of variance, F test for factor, conclusion 102. True or False: Referring to Table 11-8, if a level of significance of 0.05 is chosen, the decision made indicates that at least 2 of the 5 locations have different mean occupancy rates. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: one-way analysis of variance, F test for factor, conclusion 103. Referring to Table 11-8, the among-group variation or SSA is

ANSWER: 963.6 TYPE: FI DIFFICULTY: Easy KEYWORDS: one-way analysis of variance, sum of squares 104. Referring to Table 11-8, the within-group variation or SSW is

ANSWER: 210.0 TYPE: FI DIFFICULTY: Easy KEYWORDS: one-way analysis of variance, sum of squares 105. Referring to Table 11-8, the total variation or SST is

ANSWER: 1,173.6 TYPE: FI DIFFICULTY: Easy KEYWORDS: one-way analysis of variance, sum of squares 106. Referring to Table 11-8, the value of MSA is

while MSW is

ANSWER: 240.9; 21.0 TYPE: FI DIFFICULTY: Easy KEYWORDS: one-way analysis of variance, mean squares 107. Referring to Table 11-8, the numerator and denominator degrees of freedom of the test ratio are and , respectively. ANSWER: 4 and 10 TYPE: FI DIFFICULTY: Easy KEYWORDS: one-way analysis of variance, degrees of freedom 108. True or False: Referring to Table 11-8, the total mean squares is 261.90.

109 Analysis of Variance ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: one-way analysis of variance, sum of squares 109. Referring to Table 11-8, the null hypothesis for Levene’s test for homogeneity of variances is a) H0 :  A = B = C = D b) H0 : M A = M B = M C = M D c) H0 :  2 = 2 =  2 =  2 A

d) H0 :  A =  B =  C =  D ANSWER: c TYPE: MC DIFFICULTY: Easy KEYWORDS: Levene’s test, form of hypothesis 110. Referring to Table 11-8, what is the value of the test statistic for Levene’s test for homogeneity of variances? ANSWER: 0.28 TYPE: PR DIFFICULTY: Moderate KEYWORDS: Levene’s test, test statistic 111. Referring to Table 11-8, what are the numerator and denominator degrees of freedom for Levene’s test for homogeneity of variances respectively? ANSWER: 4, 10 TYPE: PR DIFFICULTY: Moderate KEYWORDS: Levene’s test, degrees of freedom 112. Referring to Table 11-8, what is the critical value of Levene’s test for homogeneity of variances at a 5% level of significance? ANSWER: 3.48 TYPE: PR DIFFICULTY: Moderate KEYWORDS: Levene’s test, critical value

113. Referring to Table 11-8, what is the p-value of the test statistic for Levene’s test for homogeneity of variances? ANSWER: 0.88 TYPE: PR DIFFICULTY: Moderate KEYWORDS: Levene’s test, p-value 114. Referring to Table 11-8, what should be the decision for the Levene’s test for homogeneity of variances at a 5% level of significance? a) Reject the null hypothesis because the p-value is smaller than the level of significance. b) Reject the null hypothesis because the p-value is larger than the level of significance.

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110

c) Do not eject the null hypothesis because the p-value is smaller than the level of significance. d) Do not reject the null hypothesis because the p-value is larger than the level of significance. ANSWER: d TYPE: MC DIFFICULTY: Easy KEYWORDS: Levene’s test, decision 115. Referring to Table 11-8, what should be the conclusion for the Levene’s test for homogeneity of variances at a 5% level of significance? a) There is not sufficient evidence that the variances are all the same. b) There is sufficient evidence that the variances are all the same. c) There is not sufficient evidence that the variances are not all the same. d) There is sufficient evidence that the variances are not all the same. ANSWER: c TYPE: MC DIFFICULTY: Easy KEYWORDS: Levene’s test, conclusion TABLE 11-9 The marketing manager of a company producing a new cereal aimed for children wants to examine the effect of the color and shape of the box's logo on the approval rating of the cereal. He combined 4 colors and 3 shapes to produce a total of 12 designs. Each logo was presented to 2 different groups (a total of 24 groups) and the approval rating for each was recorded and is shown below. The manager analyzed these data using the  = 0.05 level of significance for all inferences. SHAPES Circle Square Diamond

Red 54 44 34 36 46 48

Analysis of Variance Source df SS Colors 3 2711.17 Shapes 2 579.00 Interaction 6 150.33 Error 12 150.00 Total 23 3590.50

COLORS Green Blue 67 36 61 44 56 36 58 30 60 34 60 38

MS 903.72 289.50 25.06 12.50

F 72.30 23.16 2.00

Yellow 45 41 21 25 31 33

p 0.000 0.000 0.144

116. Referring to Table 11-9, the mean square for the factor color is ANSWER: 903.72 TYPE: FI DIFFICULTY: Easy KEYWORDS: two-factor analysis of variance, mean squares

111 Analysis of Variance

117. Referring to Table 11-9, the mean square for the factor shape is

ANSWER: 289.50 TYPE: FI DIFFICULTY: Easy KEYWORDS: two-factor analysis of variance, mean squares 118. Referring to Table 11-9, the mean square for the interaction of color and shape is

ANSWER: 25.06 TYPE: FI DIFFICULTY: Easy KEYWORDS: two-factor analysis of variance, mean squares 119. Referring to Table 11-9, the mean square for error is

ANSWER: 12.50 TYPE: FI DIFFICULTY: Easy KEYWORDS: two-factor analysis of variance, mean squares 120. Referring to Table 11-9, the critical value of the test for significant differences between colors is . ANSWER: 3.49 TYPE: FI DIFFICULTY: Easy KEYWORDS: two-factor analysis of variance, F test for factor, critical value 121. Referring to Table 11-9, the value of the statistic used to test for significant differences between colors is . ANSWER: 72.30 TYPE: FI DIFFICULTY: Easy KEYWORDS: two-factor analysis of variance, F test for factor, test statistic 122. True or False: Referring to Table 11-9, based on the results of the hypothesis test, it appears that there is a significant effect on the approval rating associated with the color of the logo. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: two-factor analysis of variance, F test for factor, decision, conclusion 123. Referring to Table 11-9, the critical value in the test for significant differences between shapes is . ANSWER: 3.89 TYPE: FI DIFFICULTY: Easy KEYWORDS: two-factor analysis of variance, F test for factor, critical value

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124. Referring to Table 11-9, the value of the statistic used to test for significant differences between shapes is . ANSWER: 23.16 TYPE: FI DIFFICULTY: Easy KEYWORDS: two-factor analysis of variance, F test for factor, test statistic 125. True or False: Referring to Table 11-9, based on the results of the hypothesis test, it appears that there is a significant effect associated with the shape of the logo. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: two-factor analysis of variance, F test for factor, decision, conclusion 126. Referring to Table 11-9, the critical value in the test for a significant interaction is

ANSWER: 3.00 TYPE: FI DIFFICULTY: Easy KEYWORDS: two-factor analysis of variance, F test for interaction, critical value 127. Referring to Table 11-9, the value of the statistic used to test for an interaction is

ANSWER: 2.00 TYPE: FI DIFFICULTY: Easy KEYWORDS: two-factor analysis of variance, F test for interaction, test statistic 128. True or False: Referring to Table 11-9, based on the results of the hypothesis test, it appears that there is a significant interaction. ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: two-factor analysis of variance, F test for interaction, decision, conclusion TABLE 11-10 An agronomist wants to compare the crop yield of 3 varieties of chickpea seeds. She plants all 3 varieties of the seeds on each of 5 different patches of fields. She then measures the crop yield in bushels per acre. Treating this as a randomized block design, the results are presented in the table that follows. Fields Smith Walsh Trevor 1 11.1 19.0 14.6 2 13.5 18.0 15.7 3 15.3 19.8 16.8 4 14.6 19.6 16.7 5 9.8 16.6 15.2

129. Referring to Table 11-10, the agronomist decided to perform a randomized block F test for the difference in the means. The amount of total variation or SST is

113 Analysis of Variance ANSWER: 114.82 TYPE: FI DIFFICULTY: Moderate KEYWORDS: randomized block design, sum of squares 130. Referring to Table 11-10, the among-group variation or SSA is

ANSWER: 82.39 TYPE: FI DIFFICULTY: Moderate KEYWORDS: randomized block design, sum of squares 131. Referring to Table 11-10, the among-block variation or SSBL is

ANSWER: 24.46 TYPE: FI DIFFICULTY: Moderate KEYWORDS: randomized block design, sum of squares 132. Referring to Table 11-10, the value of MSA is

, while MSBL is

ANSWER: 41.19; 6.11 TYPE: FI DIFFICULTY: Moderate KEYWORDS: randomized block design, mean squares

133. Referring to Table 11-10, the null hypothesis for the randomized block F test for the difference in the means is a) H0 : Field 1 = Field 2 = Field 3 = Field 4 = Field 5 b) c) d)

H0 : Smith = Walsh = Trevor H0 : MField 1 = MField 2 = MField 3 = MField 4 = MField 5 H0 : MSmith = MWalsh = MTrevor

ANSWER: b TYPE: MC DIFFICULTY: Easy KEYWORDS: randomized block design, F test for factor, form of hypothesis

134. Referring to Table 11-10, what are the degrees of freedom of the randomized block F test for the difference in the means at a level of significance of 0.01? ANSWER: 2 numerator and 8 denominator degrees of freedom TYPE: PR DIFFICULTY: Easy KEYWORDS: randomized block design, F test for factor, degrees of freedom

135. Referring to Table 11-10, what is the critical value of the randomized block F test for the difference in the means at a level of significance of 0.01? ANSWER: 8.65 TYPE: PR DIFFICULTY: Easy

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KEYWORDS: randomized block design, F test for factor, critical value

136. Referring to Table 11-10, what is the value of the test statistic for the randomized block F test for the difference in the means? ANSWER: 41.32 TYPE: PR DIFFICULTY: Moderate KEYWORDS: randomized block design, F test for factor, test statistic

137. Referring to Table 11-10, what is the p-value of the test statistic for the randomized block F test for the difference in the means? ANSWER: 6.07E-05 TYPE: PR DIFFICULTY: Moderate KEYWORDS: randomized block design, F test for factor, p-value

138. True or False: Referring to Table 11-10, the null hypothesis for the randomized block F test for the difference in the means should be rejected at a 0.01 level of significance. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: randomized block design, F test for factor, decision

139. True or False: Referring to Table 11-10, the decision made at a 0.01 level of significance on the randomized block F test for the difference in means implies that all 3 means are significantly different. ANSWER: False TYPE: TF DIFFICULTY: Moderate KEYWORDS: randomized block design, F test for factor, conclusion

140. True or False: Referring to Table 11-10, the randomized block F test is valid only if the population of crop yields has the same variance for the 3 varieties. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: randomized block design, F test for factor, assumption

141. True or False: Referring to Table 11-10, the randomized block F test is valid only if the population of crop yields is normally distributed for the 3 varieties. ANSWER: True TYPE: TF DIFFICULTY: Easy v KEYWORDS: randomized block design, F test for factor, assumption

142. True or False: Referring to Table 11-10, the randomized block F test is valid only if there is no interaction between the variety of seeds and the patches of fields.

115 Analysis of Variance

ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: randomized block design, F test for factor, assumption 143. Referring to Table 11-10, the agronomist decided to compare the 3 treatment means by using the Tukey multiple comparison procedure with an overall level of significance of 0.01. How many pair-wise comparisons can be made? ANSWER: 3 TYPE: PR DIFFICULTY: Moderate KEYWORDS: randomized block design, Tukey procedure, properties

144. Referring to Table 11-10, using an overall level of significance of 0.01, what is the critical value of the Studentized range Q used in calculating the critical range for the Tukey multiple comparison procedure? ANSWER: 5.63 TYPE: PR DIFFICULTY: Easy KEYWORDS: randomized block design, Tukey procedure, critical value 145. Referring to Table 11-10, using an overall level of significance of 0.01, what is the critical range for the Tukey multiple comparison procedure? ANSWER: 2.51 TYPE: PR DIFFICULTY: Moderate KEYWORDS: randomized block design, Tukey procudure, critical value 146. True or False: Referring to Table 11-10, based on the Tukey multiple comparison procedure with an overall level of significance of 0.01, the agronomist would decide that there is a significant difference between the crop yield of Smith and Walsh seeds. ANSWER: True TYPE: TF DIFFICULTY: Moderate KEYWORDS: randomized block design, Tukey procedure, decision, conclusion 147. True or False: Referring to Table 11-10, based on the Tukey-Kramer procedure with an overall level of significance of 0.01, the agronomist would decide that there is a significant difference between the crop yield of Smith and Trevor seeds. ANSWER: True TYPE: TF DIFFICULTY: Moderate KEYWORDS: randomized block design, Tukey procedure, decision, conclusion 148. True or False: Referring to Table 11-10, based on the Tukey multiple comparison procedure with an overall level of significance of 0.01, the agronomist would decide that there is a significant difference between the crop yield of Walsh and Trevor seeds.

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ANSWER: True TYPE: TF DIFFICULTY: Moderate KEYWORDS: randomized block design, Tukey procedure, decision, conclusion 149. Referring to Table 11-10, what is the null hypothesis for testing the block effects? a) H0 : Field 1 = Field 2 = Field 3 = Field 4 = Field 5 b) c) d)

H0 : Smith = Walsh = Trevor H0 : M Field 1 = MField 2 = MField 3 = MField 4 = MField 5 H0 : MSmith = MWalsh = MTrevor

ANSWER: a TYPE: MC DIFFICULTY: Easy KEYWORDS: randomized block design, F test for block effects, form of hypothesis

150. Referring to Table 11-10, what are the degrees of freedom of the F test statistic for testing the block effects? ANSWER: 4 numerator and 8 denominator degrees of freedom TYPE: PR DIFFICULTY: Easy KEYWORDS: randomized block design, F test for block effects, degrees of freedom

151. Referring to Table 11-10, what is the value of the F test statistic for testing the block effects? ANSWER: 6.13 TYPE: PR DIFFICULTY: Moderate KEYWORDS: randomized block design, F test for block effects, test statistic 152. Referring to Table 11-10, what is the critical value for testing the block effects at a 0.01 level of significance? ANSWER: 7.01 TYPE: PR DIFFICULTY: Moderate KEYWORDS: randomized block design, F test for block effects, critical value

153. Referring to Table 11-10, what is the p-value of the F test statistic for testing the block effects? ANSWER: 0.015 or between 0.01 and 0.025 TYPE: PR DIFFICULTY: Moderate KEYWORDS: randomized block design, F test for block effects, p-value

154. True or False: Referring to Table 11-10, the null hypothesis for the F test for the block effects should be rejected at a 0.01 level of significance. ANSWER: False

117 Analysis of Variance TYPE: TF DIFFICULTY: Easy KEYWORDS: randomized block design, F test for block effects, decision

155. True or False: Referring to Table 11-10, the decision made at a 0.01 level of significance on the F test for the block effects implies that the blocking has been advantageous in reducing the experiment error. ANSWER: False TYPE: TF DIFFICULTY: Moderate KEYWORDS: randomized block design, F test for block effects, conclusion 156. Referring to Table 11-10, what is the estimated relative efficiency? ANSWER: 2.47 TYPE: PR DIFFICULTY: Moderate KEYWORDS: randomized block design, F test for block effects, relative efficiency 157. True or False: Referring to Table 11-10, the relative efficiency means that 2.47 times as many observations in each variety group would be needed in a one-way ANOVA design as compared to the randomized block design in order to obtain the same precision for comparison of the variety means. ANSWER: True TYPE: TF DIFFICULTY: Moderate KEYWORDS: randomized block design, F test for block effects, relative efficiency, interpretation

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CHAPTER 12: CHI-SQUARE TESTS AND NONPARAMETRIC TESTS 1. 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 TYPE: MC DIFFICULTY: Easy KEYWORDS: Chi-square test of independence, degrees of freedom

2. If we use the  2 method of analysis to test for the differences among 4 proportions, the degrees of freedom are equal to: a) 3 b) 4 c) 5 d) 1 ANSWER: a TYPE: MC DIFFICULTY: Easy KEYWORDS: Chi-square test for difference in proportions, degrees of freedom 3. If we wish to determine whether there is evidence that the proportion of successes is the same in group 1 as in group 2, the appropriate test to use is a) the Z test. b) the  2 test. c) Both a) and b). d) Neither a) nor b). ANSWER: c TYPE: MC DIFFICULTY: Easy KEYWORDS: Chi-square test for difference in proportions

4. In testing a hypothesis using the  2 test, the theoretical frequencies are based on the a) b) c) d)

null hypothesis. alternative hypothesis. normal distribution. None of the above.

ANSWER: a TYPE: MC DIFFICULTY: Easy

124 Chi-Square Tests and Nonparametric Tests KEYWORDS: Chi-square test for difference in proportions, Chi-square test of independence, properties TABLE 12-1 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

5. Referring to Table 12-1, which test would be used to properly analyze the data in this experiment? a)  2 test for independence

b)  2 test for differences among more than two proportions c) Wilcoxon rank sum test for independent populations d) McNemar test for the difference between two proportions ANSWER: a TYPE: MC DIFFICULTY: Moderate KEYWORDS: Chi-square test of independence, 6. Referring to Table 12-1, the calculated test statistic is a) -0.9991 b) -0.1368 c) 48.1849 d) 72.8063 ANSWER: c TYPE: MC DIFFICULTY: Easy KEYWORDS: Chi-square test of independence, test statistic 7. Referring to Table 12-1, at 5% level of significance, the critical value of the test statistic is a) 3.8415 b) 5.9914 c) 9.4877 d) 13.2767 ANSWER: a TYPE: MC DIFFICULTY: Easy KEYWORDS: Chi-square test of independence, critical value 8. Referring to Table 12-1, at 5% level of significance, there is sufficient evidence to conclude that

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a) b) c) d)

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use of seat belts in motor vehicles is related to ethnic status in San Diego County. use of seat belts in motor vehicles depends on ethnic status in San Diego County. use of seat belts in motor vehicles is associated with ethnic status in San Diego County. All of the above.

ANSWER: d TYPE: MC DIFFICULTY: Moderate KEYWORDS: Chi-square test of independence, decision, conclusion TABLE 12-2 Many companies use well-known celebrities as spokespersons 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

9. Referring to Table 12-2, which test would be used to properly analyze the data in this experiment? a)  2 test for independence

b)  2 test for differences among more than two proportions c) Wilcoxon rank sum test for independent populations d) McNemar test for the difference between two proportions ANSWER: a TYPE: MC DIFFICULTY: Moderate KEYWORDS: Chi-square test of independence 10. Referring to Table 12-2, the calculated test statistic is a) -0.1006 b) 0.00 c) 5.9418 d) 6.1194 ANSWER: c TYPE: MC DIFFICULTY: Easy KEYWORDS: Chi-square test of independence, test statistic 11. Referring to Table 12-2, at 5% level of significance, the critical value of the test statistic is a) 3.8415 b) 5.9914 c) 9.4877 d) 13.2767

126 Chi-Square Tests and Nonparametric Tests ANSWER: a TYPE: MC DIFFICULTY: Easy KEYWORDS: Chi-square test of independence, critical value 12. Referring to Table 12-2, the degrees of freedom of the test statistic are a) 1 b) 2 c) 4 d) 299 ANSWER: a TYPE: MC DIFFICULTY: Easy KEYWORDS: Chi-square test of independence, degrees of freedom 13. Referring to Table 12-2, at 5% level of significance, the conclusion is that a) brand awareness of female TV viewers and the gender of the spokesperson are independent. b) brand awareness of female TV viewers and the gender of the spokesperson are not independent. c) brand awareness of female TV viewers and the gender of the spokesperson are related. d) Both (b) and (c). ANSWER: d TYPE: MC DIFFICULTY: Moderate KEYWORDS: Chi-square test of independence, decision, conclusion TABLE 12-3 A computer used by a 24-hour banking service is supposed to randomly assign each transaction to one of 5 memory locations. A check at the end of a day’s transactions gave the counts shown in the table to each of the 5 memory locations, along with the number of reported errors. Memory Location:

Number of Transactions:

100

102

Number of Reported Errors

The bank manager wanted to test whether the proportion of errors in transactions assigned to each of the 5 memory locations differ. 14. Referring to Table 12-3, which test would be used to properly analyze the data in this experiment? a)  2 test of independence

b)  2 test for difference among more than two proportions c) McNemar test for the difference between two proportions d) McNemar test for the difference among more than two proportions

Chi-Square Tests and Nonparametric Tests

ANSWER: b TYPE: MC DIFFICULTY: Moderate KEYWORDS: Chi-square test for difference in proportions 15. Referring to Table 12-3, the degrees of freedom of the test statistic is a) 4 b) 8 c) 10 d) 448 ANSWER: a TYPE: MC DIFFICULTY: Easy KEYWORDS: Chi-square test for difference in proportions, degrees of freedom 16. Referring to Table 12-3, the critical value of the test statistic at 1% level of significance is a) 7.7794 b) 13.2767 c) 20.0902 d) 23.2093 ANSWER: b TYPE: MC DIFFICULTY: Easy KEYWORDS: Chi-square test for difference in proportions, critical value 17. Referring to Table 12-3, the calculated value of the test statistic is a) -0.1777 b) -0.0185 c) 1.4999 d) 1.5190 ANSWER: c TYPE: MC DIFFICULTY: Moderate KEYWORDS: Chi-square test for difference in proportions, test statistic 18. Referring to Table 12-3, at 1% level of significance a) there is sufficient evidence to conclude that the proportions of errors in transactions assigned to each of the 5 memory locations are all different. b) there is insufficient evidence to conclude that the proportions of errors in transactions assigned to each of the 5 memory locations are all different. c) there is sufficient evidence to conclude that the proportion of errors in transactions assigned to each of the 5 memory locations are not all the same. d) there is insufficient evidence to conclude that the proportion of errors in transactions assigned to each of the 5 memory locations are not all the same. ANSWER: d

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128 Chi-Square Tests and Nonparametric Tests TYPE: MC DIFFICULTY: Moderate KEYWORDS: Chi-square test for difference in proportions, decision, conclusion TABLE 12-4 One criterion used to evaluate employees in the assembly section of a large factory is the number of defective pieces per 1,000 parts produced. The quality control department wants to find out whether there is a relationship between years of experience and defect rate. Since the job is repetitious, after the initial training period any improvement due to a learning effect might be offset by a loss of motivation. A defect rate is calculated for each worker in a yearly evaluation. The results for 100 workers are given in the table below. Years Since Training Period < 1 Year 1 – 4 Years 5 – 9 Years High 6 9 9 Defect Rate: Average 9 19 23 Low 7 8 10

19. Referring to Table 12-4, which test would be used to properly analyze the data in this experiment to determine whether there is a relationship between defect rate and years of experience? a)  2 test for independence

20. Referring to Table 12-4, find the rejection region necessary for testing at the 0.05 level of significance whether there is a relationship between defect rate and years of experience. a) Reject H0 if  2 > 16.919

b) Reject H0 if  2 > 15.507 c) Reject H0 if  2 > 11.143 d) Reject H0 if  2 > 9.488 ANSWER: d TYPE: MC DIFFICULTY: Easy KEYWORDS: Chi-square test of independence, critical value

21. Referring to Table 12-4, what is the expected number of employees with less than 1 year of training time and a high defect rate? a) 4.17 b) 4.60 c) 5.28 d) 9.17 ANSWER:

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c TYPE: MC DIFFICULTY: Moderate KEYWORDS: Chi-square test of independence, contingency table, properties

22. Referring to Table 12-4, what is the expected number of employees with 1 to 4 years of training time and a high defect rate? a) 12.00 b) 8.64 c) 6.67 d) 6.00 ANSWER: b TYPE: MC DIFFICULTY: Moderate KEYWORDS: Chi-square test of independence, contingency table, properties

23. Referring to Table 12-4, of the cell for 1 to 4 years of training time and a high defect rate, what is 2 the contribution to the overall  statistic for the independence test? a) 0.36 b) 0.1296 c) 0.015 d) 0.0144 ANSWER: c TYPE: MC DIFFICULTY: Moderate KEYWORDS: Chi-square test of independence, contingency table, properties

24. Referring to Table 12-4, a test was conducted to determine if a relationship exists between defect rate and years of experience. Which of the following p-values would indicate that defect rate and years of experience are dependent? Assume you are testing at  = 0.05. a) 0.045 b) 0.055 c) 0.074 d) 0.080 ANSWER: a TYPE: MC DIFFICULTY: Easy KEYWORDS: Chi-square test of independence, p-value, decision TABLE 12-5 A corporation randomly selects 150 salespeople and finds that 66% who have never taken a selfimprovement course would like such a course. The firm did a similar study 10 years ago in which 60% of a random sample of 160 salespeople wanted a self-improvement course. The groups are assumed to be independent random samples. Let 1 and  2 represent the true proportion of workers who would like to attend a self-improvement course in the recent study and the past study, respectively.

130 Chi-Square Tests and Nonparametric Tests

25. Referring to Table 12-5, if the firm wanted to test whether this proportion has changed from the previous study, which represents the relevant hypotheses? a) H0: 1 −  2 = 0 versus H1: 1 −  2  0 b) H0: 1 −  2  0 versus H1: 1 −  2 = 0 c) H0: 1 −  2  0 versus H1: 1 −  2 > 0 d) H0: 1 −  2  0 versus H1: 1 −  2 < 0 ANSWER: a TYPE: MC DIFFICULTY: Easy KEYWORDS: Chi-square test for difference in proportions

26. Referring to Table 12-5, what is the critical value when performing a chi-square test on whether population proportions are different if  = 0.05? a)  1.645 b)  1.96 c) 3.842 d) 2.706 ANSWER: c TYPE: MC DIFFICULTY: Easy KEYWORDS: Chi-square test for difference in proportions, critical value

27. Referring to Table 12-5, what is the critical value when testing whether population proportions are different if  = 0.10? a)  1.645 b)  1.96 c) 3.842 d) 2.706 ANSWER: d TYPE: MC DIFFICULTY: Easy KEYWORDS: Chi-square test for difference in proportions, critical value

28. Referring to Table 12-5, what is the value of the test statistic to use in evaluating the alternative hypothesis that there is a difference in the two population proportions using  = 0.10? a) 4.335 b) 2.706 c) 1.194 d) 0.274 ANSWER: c TYPE: MC DIFFICULTY: Moderate KEYWORDS: Chi-square test for difference in proportions, test statistic

29. Referring to Table 12-5, the company tests to determine at the 0.05 level whether the population proportion has changed from the previous study. Which of the following is most correct?

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a) Reject the null hypothesis and conclude that the proportion of employees who are interested in a self-improvement course has changed over the intervening 10 years. b) Do not reject the null hypothesis and conclude that the proportion of employees who are interested in a self-improvement course has not changed over the intervening 10 years. c) Reject the null hypothesis and conclude that the proportion of employees who are interested in a self-improvement course has increased over the intervening 10 years. d) Do not reject the null hypothesis and conclude that the proportion of employees who are interested in a self-improvement course has increased over the intervening 10 years. ANSWER: b TYPE: MC DIFFICULTY: Moderate KEYWORDS: Chi-square test for difference in proportions, decision, conclusion

30. True or False: In testing the difference between two proportions using the normal distribution, we may use either a one-tailed Chi-square test or two-tailed Z test. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: Z test for difference in two proportions, Chi-square test for difference in proportions 31. True or False: The squared difference between the observed and theoretical frequencies should be large if there is no significant difference between the proportions. ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: Chi-square test for difference in proportions, properties 32. True or False: A test for the difference between two proportions can be performed using the chisquare distribution. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: Chi-square test for difference in proportions 33. True or False: A test for whether one proportion is higher than the other can be performed using the chi-square distribution. ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: Chi-square test for difference in proportions

34. True or False: When using the  2 tests for independence, one should be aware that expected frequencies that are too small will lead to too big a type I error. ANSWER:

132 Chi-Square Tests and Nonparametric Tests True TYPE: TF DIFFICULTY: Difficult KEYWORDS: Chi-square test of independence, properties, assumption 35. True or False: If we use the chi-square method of analysis to test for the difference between proportions, we must assume that there are at least 5 observed frequencies in each cell of the contingency table. ANSWER: False TYPE: TF DIFFICULTY: Moderate KEYWORDS: Chi-square test of independence, properties, assumption TABLE 12-6 The dean of a college is interested in the proportion of graduates from his college who have a job offer on graduation day. He is particularly interested in seeing if there is a difference in this proportion for accounting and economics majors. In a random sample of 100 of each type of major at graduation, he found that 65 accounting majors and 52 economics majors had job offers. If the accounting majors are designated as “Group 1” and the economics majors are designated as “Group 2,” perform the appropriate hypothesis test using a level of significance of 0.05.

36. Referring to Table 12-6, the hypotheses the dean should use are: a) H0: 1 −  2 = 0 versus H1: 1 −  2  0 b) H0: 1 −  2  0 versus H1: 1 −  2 = 0 c) H0: 1 −  2  0 versus H1: 1 −  2 > 0 d) H0: 1 −  2  0 versus H1: 1 −  2 < 0 ANSWER: a TYPE: MC DIFFICULTY: Easy KEYWORDS: Chi-square test for difference in proportions, form of hypothesis

37. Referring to Table 12-6, the null hypothesis will be rejected if the test statistic is ANSWER:  2 > 3.841 TYPE: FI DIFFICULTY: Easy KEYWORDS: Chi-square test for difference in proportions, critical value

38. Referring to Table 12-6, the value of the test statistic is

ANSWER:  2 = 3.4806 TYPE: FI DIFFICULTY: Moderate KEYWORDS: Chi-square test for difference in proportions, test statistic

39. Referring to Table 12-6, the p-value of the test is ANSWER: 0.0621

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133

TYPE: FI DIFFICULTY: Moderate KEYWORDS: Chi-square test for difference in proportions, p-value

40. True or False: Referring to Table 12-6, the null hypothesis should be rejected. ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: Chi-square test for difference in proportions, decision

41. True or False: Referring to Table 12-6, the same decision would be made with this test if the level of significance had been 0.01 rather than 0.05. ANSWER: True TYPE: TF DIFFICULTY: Moderate KEYWORDS: Chi-square test for difference in proportions, decision

42. True or False: Referring to Table 12-6, the same decision would be made with this test if the level of significance had been 0.10 rather than 0.05. ANSWER: False TYPE: TF DIFFICULTY: Moderate KEYWORDS: Chi-square test for difference in proportions, decision TABLE 12-7 The director of transportation of a large company is interested in the usage of her van pool. She considers her routes to be divided into local and non-local. She is particularly interested in learning if there is a difference in the proportion of males and females who use the local routes. She takes a sample of a day's riders and finds the following:

Local Non-Local Total

Male 27 33 60

Female 44 25 69

Total 71 58 129

She will use this information to perform a chi-square hypothesis test using a level of significance of 0.05.

43. Referring to Table 12-7, the test will involve

degree(s) of freedom.

ANSWER: 1 TYPE: FI DIFFICULTY: Easy KEYWORDS: Chi-square test for difference in proportions, degrees of freedom

44. Referring to Table 12-7, the overall or average proportion of local riders is

134 Chi-Square Tests and Nonparametric Tests ANSWER: 0.550 TYPE: FI DIFFICULTY: Easy KEYWORDS: Chi-square test for difference in proportions, properties

45. Referring to Table 12-7, the expected cell frequency in the Male/Local cell is

ANSWER: 33.02 TYPE: FI DIFFICULTY: Moderate KEYWORDS: Chi-square test for difference in proportions, contingency table, properties

46. Referring to Table 12-7, the expected cell frequency in the Female/Non-Local cell is . ANSWER: 31.02 TYPE: FI DIFFICULTY: Moderate KEYWORDS: Chi-square test for difference in proportions, contingency table, properties

47. Referring to Table 12-7, the critical value of the test is

ANSWER: 3.841 TYPE: FI DIFFICULTY: Easy KEYWORDS: Chi-square test for difference in proportions, critical value

48. Referring to Table 12-7, the value of the test statistic is

ANSWER: 4.568 TYPE: FI DIFFICULTY: Moderate KEYWORDS: Chi-square test for difference in proportions, test statistic

49. True or False: Referring to Table 12-7, the null hypothesis will be rejected. ANSWER: True TYPE: TF DIFFICULTY: Moderate KEYWORDS: Chi-square test for difference in proportions, decision

50. True or False: Referring to Table 12-7, the decision made suggests that there is a difference between the proportion of males and females who ride local versus non-local routes. ANSWER: True TYPE: TF DIFFICULTY: Median KEYWORDS: Chi-square test for difference in proportions, conclusion

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51. Referring to Table 12-7, the director now wants to know if the proportion of users who are male and the proportion of users who use the local routes are the same. Which test should she use? a)  2 -test for difference in proportions b) Z-test for difference in proportions c) McNemar test for difference in proportions d) Wilcoxon rank sum test ANSWER: c TYPE: MC DIFFICULTY: Moderate KEYWORDS: McNemar test, assumption 52. Referring to Table 12-7, the director now wants to know if the proportion of users who are male and the proportion of users who use the local routes are the same. What is the value of the test statistic using a 5% level of significance? ANSWER: 0.1429 or -0.1429 TYPE: PR DIFFICULTY: Moderate KEYWORD: McNemar test, test statistic

53. Referring to Table 12-7, the director now wants to know if the proportion of users who are male and the proportion of users who use the local routes are the same. What is the p-value of the test statistic using a 5% level of significance? ANSWER: 0.8864 TYPE: PR DIFFICULTY: Moderate KEYWORD: McNemar test, p-vlaue

54. True or False: Referring to Table 12-7, the director now wants to know if the proportion of users who are male and the proportion of users who use the local routes are the same. She should reject the null hypothesis using a 5% level of significance. ANSWER: False TYPE: TF DIFFICULTY: Moderate KEYWORD: McNemar test, decision 55. Referring to Table 12-7, the director now wants to know if the proportion of users who are male and the proportion of users who use the local routes are the same. What should be her conclusion? a) There is sufficient evidence that the proportion of users who are male is the same as the proportion of users who use the local routes. b) There is insufficient evidence that the proportion of users who are male is the same as the proportion of users who use the local routes. c) There is sufficient evidence that the proportion of users who are male is not the same as the proportion of users who use the local routes. d) There is insufficient evidence that the proportion of users who are male is not the same as the proportion of users who use the local routes.

136 Chi-Square Tests and Nonparametric Tests

ANSWER: d TYPE: TF DIFFICULTY: Moderate KEYWORD: McNemar test, conclusion 56. Referring to Table 12-7, the director now wants to know if the proportion of users who are female and the proportion of users who use the local routes are the same. Which test should she use? a)  2 -test for difference in proportions b) Z-test for difference in proportions c) McNemar test for difference in proportions d) Wilcoxon rank sum test ANSWER: c TYPE: MC DIFFICULTY: Moderate KEYWORDS: McNemar test, assumption

57. Referring to Table 12-7, the director now wants to know if the proportion of users who are female and the proportion of users who use the local routes are the same. What is the value of the test statistic using a 5% level of significance? ANSWER: 0.0385 or −0.0385 TYPE: PR DIFFICULTY: Moderate KEYWORD: McNemar test, test statistic

58. Referring to Table 12-7, the director now wants to know if the proportion of users who are female and the proportion of users who use the local routes are the same. What is the p-value of the test statistic using a 5% level of significance? ANSWER: 0.9693 TYPE: PR DIFFICULTY: Moderate KEYWORD: McNemar test, p-value

59. True or False: Referring to Table 12-7, the director now wants to know if the proportion of users who are female and the proportion of users who use the local routes are the same. She should reject the null hypothesis using a 5% level of significance. ANSWER: False TYPE: TF DIFFICULTY: Moderate KEYWORD: McNemar test, decision

60. Referring to Table 12-7, the director now wants to know if the proportion of users who are female and the proportion of users who use the local routes are the same. What should be her conclusion? a) There is sufficient evidence that the proportion of users who are female is the same as the proportion of users who use the local routes.

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b) There is insufficient evidence that the proportion of users who are female is the same as the proportion of users who use the local routes. c) There is sufficient evidence that the proportion of users who are female is not the same as the proportion of users who use the local routes. d) There is insufficient evidence that the proportion of users who are female is not the same as the proportion of users who use the local routes. ANSWER: d TYPE: TF DIFFICULTY: Moderate KEYWORD: McNemar test, conclusion TABLE 12-8 Four surgical procedures currently are used to install pacemakers. If the patient does not need to return for follow-up surgery, the operation is called a "clear" operation. A heart center wants to compare the proportion of clear operations for the 4 procedures, and collects the following numbers of patients from their own records:

Clear Return Total

A 27 11 38

Procedure B C 41 21 15 9 56 30

D 7 11 18

Total 96 46 142

They will use this information to test for a difference among the proportion of clear operations using a chi-square test with a level of significance of 0.05.

61. Referring to Table 12-8, the test will involve

degrees of freedom.

ANSWER: 3 TYPE: FI DIFFICULTY: Easy KEYWORDS: Chi-square test for difference in proportions, degrees of freedom

62. Referring to Table 12-8, the overall or average proportion of clear operations is ANSWER: 0.676 TYPE: FI DIFFICULTY: Easy KEYWORDS: Chi-square test for difference in proportions, properties

63. Referring to Table 12-8, the expected cell frequency for the Procedure A/Clear cell is . ANSWER: 25.69 TYPE: FI DIFFICULTY: Moderate KEYWORDS: Chi-square test for difference in proportions, contingency table, properties

138 Chi-Square Tests and Nonparametric Tests

64. Referring to Table 12-8, the expected cell frequency for the Procedure D/Return cell is . ANSWER: 5.83 TYPE: FI DIFFICULTY: Moderate KEYWORDS: Chi-square test for difference in proportions, contingency table, properties

65. Referring to Table 12-8, the critical value of the test is

ANSWER: 7.815 TYPE: FI DIFFICULTY: Easy KEYWORDS: Chi-square test for difference in proportions, critical value

66. Referring to Table 12-8, the value of the test statistic is

ANSWER: 7.867 TYPE: FI DIFFICULTY: Moderate KEYWORDS: Chi-square test for difference in proportions, test statistic

67. True or False: Referring to Table 12-8, the null hypothesis will be rejected. ANSWER: True TYPE: TF DIFFICULTY: Moderate KEYWORDS: Chi-square test for difference in proportions, decision

68. True or False: Referring to Table 12-8, the decision made suggests that the 4 procedures all have different proportions of clear operations. ANSWER: False TYPE: TF DIFFICULTY: Moderate KEYWORDS: Chi-square test for difference in proportions, conclusion

69. True or False: Referring to Table 12-8, the decision made suggests that the 4 procedures do not all have the same proportion of clear operations. ANSWER: True TYPE: TF DIFFICULTY: Moderate KEYWORDS: Chi-square test for difference in proportions, conclusion

70. Referring to Table 12-8, what is the value of the critical range for the Marascuilo procedure to test for the difference in proportions between procedure A and procedure B using a 0.05 level of significance? ANSWER: 0.0560

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TYPE: PR DIFFICULTY: Difficult KEYWORDS: Chi-square test for difference in proportions, Marascuilo procedure, critical value

71. True or False: Referring to Table 12-8, there is sufficient evidence to conclude that the proportions between procedure A and procedure B are different at a 0.05 level of significance. ANSWER: False TYPE: TF DIFFICULTY: Moderate KEYWORDS: Chi-square test for difference in proportions, Marascuilo procedure, decision, conclusion

72. Referring to Table 12-8, what is the value of the critical range for the Marascuilo procedure to test for the difference in proportions between procedure A and procedure C using a 0.05 level of significance? ANSWER: 0.0570 TYPE: PR DIFFICULTY: Difficult KEYWORDS: Chi-square test for difference in proportions, Marascuilo procedure, critical value

73. True or False: Referring to Table 12-8, there is sufficient evidence to conclude that the proportions between procedure A and procedure C are different at a 0.05 level of significance. ANSWER: False TYPE: TF DIFFICULTY: Moderate KEYWORDS: Chi-square test for difference in proportions, Marascuilo procedure, decision, conclusion

74. Referring to Table 12-8, what is the value of the critical range for the Marascuilo procedure to test for the difference in proportions between procedure A and procedure D using a 0.05 level of significance? ANSWER: 0.0589 TYPE: PR DIFFICULTY: Difficult KEYWORDS: Chi-square test for difference in proportions, Marascuilo procedure, critical value

75. True or False: Referring to Table 12-8, there is sufficient evidence to conclude that the proportions between procedure A and procedure D are different at a 0.05 level of significance. ANSWER: True TYPE: TF DIFFICULTY: Moderate KEYWORDS: Chi-square test for difference in proportions, Marascuilo procedure, decision, conclusion

76. Referring to Table 12-8, what is the value of the critical range for the Marascuilo procedure to test for the difference in proportions between procedure B and procedure C using a 0.05 level of significance? ANSWER:

140 Chi-Square Tests and Nonparametric Tests 0.0563 TYPE: PR DIFFICULTY: Difficult KEYWORDS: Chi-square test for difference in proportions, Marascuilo procedure, critical value

77. True or False: Referring to Table 12-8, there is sufficient evidence to conclude that the proportions between procedure B and procedure C are different at a 0.05 level of significance. ANSWER: False TYPE: TF DIFFICULTY: Moderate KEYWORDS: Chi-square test for difference in proportions, Marascuilo procedure, decision, conclusion

78. Referring to Table 12-8, what is the value of the critical range for the Marascuilo procedure to test for the difference in proportions between procedure B and procedure D using a 0.05 level of significance? ANSWER: 0.0582 TYPE: PR DIFFICULTY: Difficult KEYWORDS: Chi-square test for difference in proportions, Marascuilo procedure, critical value

79. True or False: Referring to Table 12-8, there is sufficient evidence to conclude that the proportions between procedure B and procedure D are different at a 0.05 level of significance. ANSWER: True TYPE: TF DIFFICULTY: Moderate KEYWORDS: Chi-square test for difference in proportions, Marascuilo procedure, decision, conclusion

80. Referring to Table 12-8, what is the value of the critical range for the Marascuilo procedure to test for the difference in proportions between procedure C and procedure D using a 0.05 level of significance? ANSWER: 0.0591 TYPE: PR DIFFICULTY: Difficult KEYWORDS: Chi-square test for difference in proportions, Marascuilo procedure, critical value

81. True or False: Referring to Table 12-8, there is sufficient evidence to conclude that the proportions between procedure C and procedure D are different at a 0.05 level of significance. ANSWER: True TYPE: TF DIFFICULTY: Moderate KEYWORDS: Chi-square test for difference in proportions, Marascuilo procedure, decision, conclusion

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TABLE 12-9 The director of admissions at a state college is interested in seeing if admissions status (admitted, waiting list, denied admission) at his college is independent of the type of community in which an applicant resides. He takes a sample of recent admissions decisions and forms the following table:

Urban Rural Suburban Total

Admitted 45 33 34 112

Wait List 21 13 12 46

Denied 17 24 39 80

Total 83 70 85 238

He will use this table to do a chi-square test of independence with a level of significance of 0.01.

82. Referring to Table 12-9, the test will involve

degrees of freedom.

ANSWER: 4 TYPE: FI DIFFICULTY: Easy KEYWORDS: Chi-square test of independence, degrees of freedom

83. Referring to Table 12-9, the critical value of the test is

ANSWER: 13.277 TYPE: FI DIFFICULTY: Easy KEYWORDS: Chi-square test of independence, critical value

84. Referring to Table 12-9, the expected cell frequency for the Admitted/Urban cell is ANSWER: 39.06 TYPE: FI DIFFICULTY: Moderate KEYWORDS: Chi-square test of independence, contingency table, properties

85. Referring to Table 12-9, the value of the test statistic is

ANSWER: 12.624 TYPE: FI DIFFICULTY: Moderate KEYWORDS: Chi-square test of independence, test statistic

86. True or False: Referring to Table 12-9, the null hypothesis will be rejected. ANSWER: False TYPE: TF DIFFICULTY: Moderate KEYWORDS: Chi-square test of independence, decision

87. True or False: Referring to Table 12-9, the p-value of this test is greater than 0.01.

142 Chi-Square Tests and Nonparametric Tests

ANSWER: True TYPE: TF DIFFICULTY: Moderate KEYWORDS: Chi-square test of independence, p-value

88. True or False: Referring to Table 12-9, the decision made suggests that admissions status at the college is independent of the type of community in which an applicant resides. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: Chi-square test of independence, conclusion

89. True or False: Referring to Table 12-9, the same decision would be made with this test if the level of significance had been 0.005. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: Chi-square test of independence, decision

90. True or False: Referring to Table 12-9, the same decision would be made with this test if the level of significance had been 0.05. ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: Chi-square test of independence, decision

91. True or False: Referring to Table 12-9, the null hypothesis claims that "there is no association between admission status at the college and the type of community in which an applicant resides." ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: Chi-square test of independence, form of hypothesis, conclusion

92. True or False: Referring to Table 12-9, the alternative hypothesis claims that "there is some connection between admission status at the college and the type of community in which an applicant resides." ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: Chi-square test of independence, form of hypothesis, conclusion 93. True or False: The chi-square test of independence requires that the number of expected frequency in each cell to be at least 5.

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ANSWERS: False TYPE: TF DIFFICULTY: Easy KEYWORDS: Chi-square test of independence, assumption 94. True or False: The chi-square test of independence requires that the number of expected frequency in each cell to be at least 1. ANSWERS: True TYPE: TF DIFFICULTY: Easy KEYWORDS: Chi-square test of independence, assumption TABLE 12-10 Parents complain that children read too few storybooks and watch too much television nowadays. A survey of 1,000 children reveals the following information on average time spent watching TV and average time spent reading storybooks

Average time spent watching TV Less than 2 hours More than 2 hours

Average time spent reading story books Less than 1 hour Between 1and 2 hours

More than 2 hours

90 655

130 8

85 32

95. Referring to Table 12-10, how many children in the survey spent less than 2 hours watching TV and more than 2 hours reading story books on average? a) 8 b) 130 c) 175 d) 687 ANSWER: b TYPE: MC DIFFICULTY: Easy KEYWORDS: contingency table, properties

96. Referring to Table 12-10, how many children in the survey spent less than 2 hours watching TV and no more than 2 hours reading storybooks on average? a) 8 b) 130 c) 175 d) 687 ANSWER: c TYPE: MC DIFFICULTY: Easy KEYWORDS: contingency table, properties

144 Chi-Square Tests and Nonparametric Tests

97. Referring to Table 12-10, if the null hypothesis of no connection between time spent watching TV and time spent reading story books is true, how many children watching less than 2 hours of TV and reading no more than 2 hours of story books on average can we expect? a) 35.69 b) 227.23 c) 262.91 d) 969.75 ANSWER: c TYPE: MC DIFFICULTY: Difficult KEYWORDS: Chi-square test of independence, contingency table, properties

98. Referring to Table 12-10, if the null hypothesis of no connection between time spent watching TV and time spent reading story books is true, how many children watching less than 2 hours of TV and reading more than 2 hours of story books on average can we expect? a) 42.09 b) 155.25 c) 262.92 d) 987.75 ANSWER: a TYPE: MC DIFFICULTY: Moderate KEYWORDS: Chi-square test of independence, contingency table, properties

99. Referring to Table 12-10, to test whether there is any relationship between average time spent watching TV and average time spent reading story books, the value of the measured test statistic is a) -12.59 b) 1.61 c) 481.49 d) 1,368.06 ANSWER: c TYPE: MC DIFFICULTY: Moderate KEYWORDS: Chi-square test of independence, test statistic

100. Referring to Table 12-10, suppose we want to constrain the probability of committing a Type I error to 5% when testing whether there is any relationship between average time spent watching TV and average time spent reading story books, the critical value will be a) 5.991 b) 7.378 c) 12.592 d) 14.449 ANSWER: a TYPE: MC DIFFICULTY: Easy KEYWORDS: Chi-square test of independence, critical value

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101. Referring to Table 12-10, we want to test whether there is any relationship between average time spent watching TV and average time spent reading storybooks. Suppose the value of the test statistic was 164 (which is not the correct answer) and the critical value was 19.00 (which is not the correct answer), then we could conclude that a) There is connection between time spent reading storybooks and time spent watching TV. b) There is no connection between time spent reading storybooks and time spent watching TV. c) More time spent reading storybooks leads to less time spent watching TV. d) More time spent watching TV leads to less time spent reading storybooks. ANSWER: a TYPE: MC DIFFICULTY: Moderate KEYWORDS: Chi-square test of independence, decision, conclusion TABLE 12-11 Recent studies have found that American children are more obese than in the past. The amount of time children spent watching television has received much of the blame. A survey of 100 ten-yearolds revealed the following with regards to weights and average number of hours a day spent watching television. We are interested in testing whether the average number of hours spent watching TV and weights are independent at 1% level of significance. Weights More than 10 lbs. overweight Within 10 lbs. of normal weight More than 10 lbs. underweight Total

0-3 1 20 10 31

TV Hours 3-6 9 15 5 29

6+ 20 15 5 40

Total 30 50 20 100

102. Referring to Table 12-11, if there is no connection between weights and average number of hours spent watching TV, we should expect how many children to be spending 3-6 hours on average watching TV and are more than 10 lbs. underweight? a) 5 b) 5.8 c) 6.2 d) 8 ANSWER: b TYPE: MC DIFFICULTY: Easy KEYWORDS: Chi-square test of independence, contingency table, properties

103. Referring to Table 12-11, if there is no connection between weights and average number of hours spent watching TV, we should expect how many children to be spending no more than 6 hours on average watching TV and are more than 10 lbs. underweight? a) 5.8 b) 6.2 c) 8 d) 12

146 Chi-Square Tests and Nonparametric Tests ANSWER: d TYPE: MC DIFFICULTY: Moderate KEYWORDS: Chi-square test of independence, contingency table, properties

104. Referring to Table 12-11, how many children in the survey spend more than 6 hours watching TV and are more than 10 lbs. overweight? a) 1 b) 9 c) 20 d) 40 ANSWER: c TYPE: MC DIFFICULTY: Easy KEYWORDS: Chi-square test of independence, contingency table, properties

105. Referring to Table 12-11, how many children in the survey spend no more than 6 hours watching TV and are more than 10 lbs. underweight? a) 5 b) 10 c) 15 d) 20 ANSWER: c TYPE: MC DIFFICULTY: Easy KEYWORDS: Chi-square test of independence, contingency table, properties

106. Referring to Table 12-11, the value of the test statistic is a) 8.532 b) 15.483 c) 18.889 d) 69.744 ANSWER: c TYPE: MC DIFFICULTY: moderate KEYWORDS: Chi-square test of independence, test statistic

107. Referring to Table 12-11, the critical value of the test will be a) 6.635 b) 13.277 c) 14.860 d) 21.666 ANSWER: b TYPE: MC DIFFICULTY: Easy KEYWORDS: Chi-square test of independence, critical value

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108. Referring to Table 12-11, suppose the value of the test statistic was 30.00 (which is not the correct value) and the critical value at 1% level of significance was 10.00 (which is not the correct value), which of the following conclusions would be correct? a) We will accept the null and conclude that average number of hours spent watching TV and weights are independent. b) We will reject the null and conclude that average number of hours spent watching TV and weights are independent. c) We will accept the null and conclude that average number of hours spent watching TV and weights are not independent. d) We will reject the null and conclude that average number of hours spent watching TV and weights are not independent. ANSWER: d TYPE: MC DIFFICULTY: Moderate KEYWORDS: Chi-square test of independence, decision, conclusion

109. Referring to Table 12-11, which of the following statements is correct? a) b) c) d)

We can accept the null for any level of significance greater than 0.005 We can reject the null for any level of significance greater than 0.005. We can accept the null for any level of significance smaller than 0.005 We can reject the null for any level of significance smaller than 0.005.

ANSWER: b TYPE: MC DIFFICULTY: Difficult KEYWORDS: Chi-square test of independence, decision

110. Referring to Table 12-11, the degrees of freedom of the test statistic are a) b) c) d)

1 2 4 9

ANSWER: c TYPE: MC DIFFICULTY: Easy KEYWORDS: Chi-square test of independence, degrees of freedom

111. True or False: Referring to Table 12-11, the test is always a one-tailed test. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: Chi-square test of independence, properties Table 12-12 According to an article in Marketing News, fewer checks are being written at grocery store checkout stands than in the past. To determine whether there is a difference in the proportion of shoppers who pay by check among three consecutive years at a 0.05 level of significance, the results of a survey of 500 shoppers in three consecutive years are obtained and presented below.

148 Chi-Square Tests and Nonparametric Tests

Check Written Yes No

Year 1 225 275

Year Year 2 175 325

Year 3 125 375

112. Referring to Table 12-12, what is the expected number of shoppers who pay by check in year 1 if there is no difference in the proportion of shoppers who pay by check among the three years? ANSER: 175 TYPE: PR DIFFICULTY: Easy KEYWORDS: Chi-square test for difference in proportions, properties

113. Referring to Table 12-12, what is the expected number of shoppers who do not pay by check in year 3 if there is no difference in the proportion of shoppers who pay by check among the three years? ANSER: 325 TYPE: PR DIFFICULTY: Easy KEYWORDS: Chi-square test for difference in proportions, properties

114. Referring to Table 12-12, what is the form of the null hypothesis? a) H0 : 1 =  2 = 3 b) H0 : 1 =  2  3 c) H 0 : 1   2 = 3 d) H 0 : 1   2  3 ANSWER: a TYPE: MC DIFFICULTY: Easy KEYWORDS: Chi-square test for difference in proportions, forms of hypothesis

115. Referring to Table 12-12, what is the form of the alternative hypothesis? a) H1 : 1   2  3 b) H1 : 1   2 = 3 c) H1 : 1 =  2  3 d) H1 : not all  j are the same ANSWER: d TYPE: MC DIFFICULTY: Easy KEYWORDS: Chi-square test for difference in proportions, forms of hypothesis

116. True or False: Referring to Table 12-12, the assumptions needed to perform the test are satisfied. ANSWER: True

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TYPE: TF DIFFICULTY: Moderate KEYWORDS: Chi-square test for difference in proportions, assumption

117. Referring to Table 12-12, what are the degrees of freedom of the test statistic? ANSWER: 2 TYPE: PR DIFFICULTY: Easy KEYWORDS: Chi-square test for difference in proportions, degrees of freedom

118. Referring to Table 12-12, what is the value of the test statistic? ANSWER: 43.96 TYPE: PR DIFFICULTY: Moderate KEYWORDS: Chi-square test for difference in proportions, test statistic

119. Referring to Table 12-12, what is the critical value? ANSWER: 5.99 TYPE: PR DIFFICULTY: Easy KEYWORDS: Chi-square test for difference in proportions, critical value

120. Referring to Table 12-12, what is the p-value of the test statistic? ANSWER: 2.9E-10 or smaller than 0.005 TYPE: PR DIFFICULTY: Moderate KEYWORDS: Chi-square test for difference in proportions, p-value

121. True or False: Referring to Table 12-12, the null hypothesis cannot be rejected. ANSWER: False TYPE: TF DIFFICULTY: Moderate KEYWORDS: Chi-square test for difference in proportions, decision

122. Referring to Table 12-12, what is the correct conclusion? a) b) c) d)

There is enough evidence that the proportions are all different in the 3 years. There is not enough evidence that the proportions are all different in the 3 years. There is enough evidence that at least two of the proportions are not equal. There is not enough evidence that at least two of the proportions are not equal.

ANSWER: c TYPE: MC DIFFICULTY: Moderate KEYWORDS: Chi-square test for difference in proportions, conclusion

123. Referring to Table 12-12, what is the value of the critical range for the Marascuilo procedure to test for the difference in proportions between year 1 and year 2 using a 0.05 level of significance? ANSWER:

150 Chi-Square Tests and Nonparametric Tests 0.0754 TYPE: PR DIFFICULTY: Difficult KEYWORDS: Chi-square test for difference in proportions, Marascuilo procedure, critical value

124. Referring to Table 12-12, what is the value of the critical range for the Marascuilo procedure to test for the difference in proportions between year 1 and year 3 using a 0.05 level of significance? ANSWER: 0.0722 TYPE: PR DIFFICULTY: Difficult KEYWORDS: Chi-square test for difference in proportions, Marascuilo procedure, critical range

125. Referring to Table 12-12, what is the value of the critical range for the Marascuilo procedure to test for the difference in proportions between year 2 and year 3 using a 0.05 level of significance? ANSWER: 0.0705 TYPE: PR DIFFICULTY: Difficult KEYWORDS: Chi-square test for difference in proportions, Marascuilo procedure, critical value

126. True or False: Referring to Table 12-12, there is sufficient evidence to conclude that the proportions between year 1 and year 2 are different at a 0.05 level of significance. ANSWER: True TYPE: TF DIFFICULTY: Moderate KEYWORDS: Chi-square test for difference in proportions, Marascuilo procedure, decision, conclusion

127. True or False: Referring to Table 12-12, there is not sufficient evidence to conclude that the proportions between year 1 and year 3 are different at a 0.05 level of significance. ANSWER: False TYPE: TF DIFFICULTY: Moderate KEYWORDS: Chi-square test for difference in proportions, Marascuilo procedure, decision, conclusion

128. True or False: Referring to Table 12-12, there is sufficient evidence to conclude that the proportions between year 2 and year 3 are different at a 0.05 level of significance. ANSWER: True TYPE: TF DIFFICULTY: Moderate KEYWORDS: Chi-square test for difference in proportions, Marascuilo procedure, decision conclusion TABLE 12-13 A filling machine at a local soft drinks company is calibrated to fill the cans at an average amount of 12 fluid ounces and a standard deviation of 0.5 ounces. The company wants to test whether the

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standard deviation of the amount filled by the machine is indeed 0.5 ounces. A random sample of 15 cans filled by the machine reveals a standard deviation of 0.67 ounces.

129. Referring to Table 12-13, the parameter of interest in the test is

ANSWER: Population standard deviation or population variance TYPE: FI DIFFICULTY: Easy KEYWORDS: chi-square test, two-tailed test, variance, parameter

130. Referring to Table 12-13, which is the appropriate test to use? a)  2 -test of independence b) McNemar test for the difference between two proportions c) Wilcoxon rank sum test

d)  2 -test of population variance ANSWER: d TYPE: MC DIFFICULTY: Moderate KEYWORDS: chi-square test, two-tailed test, variance

131. True or False: Referring to Table 12-13, in order to perform the test, we need to assume that the amount filled by the machine has a normal distribution. ANSWER: True TYPE: TF DIFFICULTY: Moderate KEYWORDS: two-tailed test, variance, assumption

132. Referring to Table 12-13, what type of test should be performed? a) b) c) d)

Lower-tailed test Upper-tailed test Two-tailed test None of the above

ANSWER: c TYPE: MC DIFFICULTY: Easy KEYWORDS: chi-square test, two-tailed test, variance, form of hypothesis

133. Referring to Table 12-13, what are the lower and upper critical values of the test when allowing for 5% probability of committing a type I error? ANSWER: 5.6287 and 26.1189 TYPE: PR DIFFICULTY: Easy KEYWORDS: chi-square test, two-tailed test, variance, critical value

134. Referring to Table 12-13, what is the value of the test statistic? ANSWER: 25.1384

152 Chi-Square Tests and Nonparametric Tests TYPE: PR DIFFICULTY: Easy KEYWORDS: chi-square test, two-tailed test, variance, test statistic

135. True or False: Referring to Table 12-13, the decision is to reject the null hypothesis when using a 5% level of significance. ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: chi-square test, two-tailed test, variance, decision

136. True or False: Referring to Table 12-13, there is sufficient evidence to conclude that the standard deviation of the amount filled by the machine is not exactly 0.5 ounces when using a 5% level of significance. ANSWER: False TYPE: TF DIFFICULTY: Moderate KEYWORDS: chi-square test, two-tailed test, variance, conclusion

137. True or False: Referring to Table 12-13, the decision is to reject the null hypothesis when using a 10% level of significance. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: chi-square test, two-tailed test, variance, decision

138. True or False: Referring to Table 12-13, there is sufficient evidence to conclude that the standard deviation of the amount filled by the machine is not exactly 0.5 ounces when using a 10% level of significance. ANSWER: True TYPE: TF DIFFICULTY: Moderate KEYWORDS: chi-square test, two-tailed test, variance, conclusion

139. True or False: Referring to Table 12-13, the p-value of the test is somewhere in between 5% and 10%. ANSWER: True TYPE: TF DIFFICULTY: Moderate

KEYWORDS: chi-square test, two-tailed test, variance, p-value 140. In testing for differences between the median of two independent populations, the null hypothesis is a) H0 : M D = 0 .

b) H0 : M D  0 . c) H0 : M1 − M2 = 0 . d) H0 : M1 − M2  0 .

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ANSWER: c TYPE: MC DIFFICULTY: Easy KEYWORDS: Wilcoxon rank sum test, form of hypothesis 141. In testing for whether the median difference of two related populations is zero, the null hypothesis is a) H0 : M D = 0 .

b) H0 : M D  0 . c) H0 : M1 − M2 = 0 . d) H0 : M1 − M2  0 . ANSWER: a TYPE: MC DIFFICULTY: Easy KEYWORDS: Wilcoxon signs test, form of hypothesis 142.

To use the Wilcoxon Rank Sum Test as a test for location, we must assume that a) the obtained data are either ranks or numerical measurements that will be converted to combined ranks. b) both samples are randomly and independently drawn from their respective populations. c) both underlying populations from which the samples were drawn are equivalent in shape and dispersion. d) All the above.

ANSWER: d TYPE: MC DIFFICULTY: Moderate KEYWORDS: Wilcoxon rank sum test, assumption 143. Which of the following is a “robust” test procedure against the violation of distribution assumptions? a)  2 -test of independence. b)  2 -test of a variance c)  2 -test for the differences among more than two proportions d) Wilcoxon rank sum test for difference in medians ANSWER: d TYPE: MC DIFFICULTY: Easy KEYWORDS: robust test 144. A local real estate appraiser analyzed the sales prices of homes in 2 neighborhoods to the corresponding appraised values of the homes. The goal of the analysis was to compare the distribution of sale-to-appraised ratios from homes in the 2 neighborhoods. Random and independent samples were selected from the 2 neighborhoods from last year’s homes sales, 8 from each of the 2 neighborhoods. Identify the nonparametric method that would be used to analyze the data.

154 Chi-Square Tests and Nonparametric Tests

a) The McNemar test for the difference between two proportions, using the test statistic T1 b) The McNemar test for the difference between two proportions, using the test statistic Z c) The Wilcoxon Rank Sum Test, using the test statistic T1 d) The Wilcoxon Rank Sum Test, using the test statistic Z ANSWER: c TYPE: MC DIFFICULTY: Moderate KEYWORDS: Wilcoxon rank sum test, test statistic 145. True or False: The procedure for the Wilcoxon rank sum test requires that we rank each group separately rather than together. ANSWER: False TYPE: TF DIFFICULTY: Moderate KEYWORDS: Wilcoxon rank sum test 146. True or False: A researcher is curious about the effect of sleep on students’ test performances. He chooses 60 students and gives each 2 tests: one given after 2 hours’ sleep and one after 8 hours’ sleep. The test the researcher should use would be a related samples test. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: t test for mean difference TABLE 12-14 A perfume manufacturer is trying to choose between 2 magazine advertising layouts. An expensive layout would include a small package of the perfume. A cheaper layout would include a "scratchand-sniff" sample of the product. The manufacturer would use the more expensive layout only if there is evidence that it would lead to a higher approval rate. The manufacturer presents the more expensive layout to 4 groups and determines the approval rating for each group. He presents the "scratch-and-sniff" layout to 5 groups and again determines the approval rating of the perfume for each group. The data are given below. Use this to test the appropriate hypotheses with the Wilcoxon Rank Sum Test with a level of significance of 0.05. Package 52 68 43 48

Scratch 37 43 53 39 47

147. Referring to Table 12-14, the hypotheses that should be used are: a) H0 : 1 =  2 versus H1 : 1   2 b) H0 : 1  2 versus H1 : 1   2 c) H0 : M1 = M2 versus H1 : M1  M 2 d) H0 : M1  M2 versus H1 : M1  M2

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ANSWER: d TYPE: MC DIFFICULTY: Easy KEYWORDS: Wilcoxon rank sum test, form of hypothesis

148. Referring to Table 12-14, the rank given to the last observation in the "scratch-and-sniff" group is . ANSWER: 5 TYPE: FI DIFFICULTY: Easy KEYWORDS: Wilcoxon rank sum test, test statistic

149. Referring to Table 12-14, the rank given to the second observation in the "scratch-and-sniff" group is

ANSWER: 3.5 TYPE: FI DIFFICULTY: Easy KEYWORDS: Wilcoxon rank sum test, test statistic

150. Referring to Table 12-14, the calculated value of the test statistic is

ANSWER: 25.5 TYPE: FI DIFFICULTY: Easy KEYWORDS: Wilcoxon rank sum test, test statistic

151. Referring to Table 12-14, the critical value of the test is

ANSWER: 28 TYPE: FI DIFFICULTY: Easy KEYWORDS: Wilcoxon rank sum test, critical value

152. True or False: Referring to Table 12-14, the null hypothesis should be rejected. ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: Wilcoxon rank sum test, decision

153. Referring to Table 12-14, the perfume manufacturer will a) use the "scratch-and-sniff" layout because there is insufficient evidence to do otherwise. b) use the package layout because there is insufficient evidence to do otherwise. c) use the "scratch-and-sniff" layout because there is sufficient evidence to conclude that this is the best course of action. d) use the package layout because there is sufficient evidence to conclude that this is the best course of action.

156 Chi-Square Tests and Nonparametric Tests ANSWER: a TYPE: MC DIFFICULTY: Moderate KEYWORDS: Wilcoxon rank sum test, conclusion

154. True or False: The McNemar test is used to determine whether there is evidence of a difference between the proportions of two related samples. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: McNemar test, assumption

155. True or False: To test whether one proportion is higher than the other in two related sample, you can use the Marascuilo procedure. ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: Marascuilo procedure, McNemar test, assumption

156. True or False: To test whether one proportion is higher than the other in two related sample, you can use the McNemar test. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: McNemar test, assumption

157. True or False: The McNemar test is approximately chi-square distributed. ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: McNemar test, properties

158. True or False The McNemar test is approximately distributed as a chi-square random variable. ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: McNemar test, properties

159. True or False The McNemar test is approximately distributed as a standardized normal random variable. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: McNemar test, properties TABLE 12-15

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The director of the MBA program of a state university wanted to know if a one week orientation would change the proportion among potential incoming students who would perceive the program as being good. Given below is the result from 215 students’ view of the program before and after the orientation. Before the Orientation Good Not Good Total

After the Orientation Good Not Good 93 37 71 14 164 51

Total 130 85 215

160. Referring to Table 12-15, which test should she use? a)  2 -test for difference in proportions b) Z-test for difference in proportions c) McNemar test for difference in proportions d) Wilcoxon rank sum test ANSWER: c TYPE: MC DIFFICULTY: Moderate KEYWORDS: McNemar test, assumption

161. Referring to Table 12-15, what is the value of the test statistic using a 1% level of significance? ANSWER: 3.37 or −3.37 TYPE: PR DIFFICULTY: Moderate KEYWORD: McNemar test, test statistic

162. Referring to Table 12-15, what is the p-value of the test statistic using a 5% level of significance? ANSWER: 0.0008 TYPE: PR DIFFICULTY: Moderate KEYWORD: McNemar test, p-value

163. True or False: Referring to Table 12-15, the director should reject the null hypothesis using a 1% level of significance. ANSWER: True TYPE: TF DIFFICULTY: Moderate KEYWORD: McNemar test, decision

164. True or False: Referring to Table 12-15, what should be the director’s conclusion? a) There is sufficient evidence that the proportion of potential incoming students who perceive the program as being good is the same before and after the orientation. b) There is insufficient evidence that the proportion of potential incoming students who perceive the program as being good is the same before and after the orientation.

158 Chi-Square Tests and Nonparametric Tests c) There is sufficient evidence that the proportion of potential incoming students who perceive the program as being good is not the same before and after the orientation. d) There is insufficient evidence that the proportion of potential incoming students who perceive the program as being good is not the same before and after the orientation. ANSWER: c TYPE: TF DIFFICULTY: Moderate KEYWORD: McNemar test, conclusion

165. If the assumptions of the one-way ANOVA F test are not met, which of the following test could be used? a) McNemar test b) Kruskal-Wallis test c) Friedman rank test d) Marascuilo procedure ANSWER: b TYPE: MC DIFFICULTY: Easy KEYWORD: Kruskal-Wallis rank test, nonparametric test, assumption 166. The Kruskal-Wallis test is an extension of which of the following for two independent samples? a) Pooled-variance t test b) Paired-sample t test c) Wilcoxon rank sum test d) McNemar test ANSWER: c TYPE: MC DIFFICULTY: Easy KEYWORD: Kruskal-Wallis rank test, nonparametric test, Wilcoxon rank sum test 167. When the normality assumption is not met in a randomized block design, which of the following tests should be used? a) Wilcoxon rank sum test b) McNemar test c) Kruskal-Wallis test d) None of the above ANSWER: d TYPE: MC DIFFICULTY: Easy KEYWORD: Kruskal-Wallis rank test, nonparametric test, Wilcoxon rank sum test, assumption 168. True or False: If the sample sizes in each group is larger than 5, the Kruskal-Wallis rank test statistic can be approximated by a standardized normal distribution. ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORD: Kruskal-Wallis rank test, nonparametric test, properties

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169. True or False: If the sample sizes in each group is larger than 5, the Kruskal-Wallis rank test statistic can be approximated by a chi-square distribution. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORD: Kruskal-Wallis rank test, nonparametric test, properties 170. True or False: When the parametric assumption on the distribution is met, a parametric test is usually more powerful than a nonparametric test. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORD: nonparametric test 171. Suppose there is interest in comparing the median response time for three independent groups learning a specific task. The appropriate nonparametric procedure is a) Wilcoxon Rank Sums Test. b) McNemar test. c) Kruskal-Wallis Rank Test. d) None of the above. ANSWER: c TYPE: MC DIFFICULTY: Easy KEYWORDS: Kruskal-Wallis rank test, nonparametric test

172. The Kruskal-Wallis Rank Test for differences in more than two medians is a nonparametric alternative to a) ANOVA F test for completely randomized experiments. b) Student's t test for related samples. c) Student's t test for independent samples. d) Wilcoxon's Rank Sum Test. ANSWER: a TYPE: MC DIFFICULTY: Easy KEYWORDS: Kruskal-Wallis rank test, nonparametric test

173. The Journal of Business Venturing reported on the activities of entrepreneurs during the organization creation process. As part of a designed study, a total of 71 entrepreneurs were interviewed and divided into 3 groups: those that were successful in founding a new firm (n1 = 34), those still actively trying to establish a firm (n2 = 21), and those who tried to start a new firm but eventually gave up (n3 = 16). The total number of activities undertaken (e.g., developed a business plan, sought funding, looked for facilities) by each group over a specified time period during organization creation was measured. The objective is to compare the mean number of activities of the 3 groups of entrepreneurs. Because of concerns over necessary assumption of the parametric analysis, it was decided to use a nonparametric analysis. Identify the nonparametric method that would be used to analyze the data..

160 Chi-Square Tests and Nonparametric Tests a) Wilcoxon Rank Sums Test b) McNemar test for the difference between two proportions c) Kruskal-Wallis Rank Test d) One-way ANOVA F test ANSWER: c TYPE: MC DIFFICULTY: Easy KEYWORDS: Kruskal-Wallis rank test, nonparametric test TABLE 12-16 As part of an evaluation program, a sporting goods retailer wanted to compare the downhill coasting speeds of 4 brands of bicycles. She took 3 of each brand and determined their maximum downhill speeds. The results are presented in miles per hour in the table below. Trial Barth Tornado Reiser Shaw 1 43 37 41 43 2 46 38 45 45 3 43 39 42 46

174. Referring to Table 12-16, the sporting goods retailer decided to perform a Kruskal-Wallis test. The null hypothesis of the test is

ANSWER: H0: M1 = M2 = M3 = M4 TYPE: FI DIFFICULTY: Easy KEYWORDS: Kruskal-Wallis rank test, nonparametric test, form of hypothesis

175. Referring to Table 12-16, the alternative hypothesis of the Kruskal-Wallis test is that . ANSWER: not all the medians are equal. TYPE: FI DIFFICULTY: Easy KEYWORDS: Kruskal-Wallis rank test, nonparametric test, form of hypothesis

176. Referring to Table 12-16, to use the chi-square critical value for the Kruskal-Wallis test, the sample size in each group has to be greater than

ANSWER: 5 TYPE: FI DIFFICULTY: Easy KEYWORDS: Kruskal-wallis rank test, nonparametric test, properties

177. Referring to Table 12-16, the decision rule for a level of significance of 0.05 using the KruskalWallis test is to reject the null hypothesis if the test statistic H is sample sizes are large enough to use a chi-square approximation. ANSWER: greater than 7.815 TYPE: FI DIFFICULTY: Easy

assuming that the

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KEYWORDS: Kruskal-Wallis rank test, nonparametric test, critical value, decision

178. Referring to Table 12-16, the calculation of the Kruskal-Wallis test statistic H involves ranking the observations. Construct a table containing these ranks. ANSWER: Barth Tornado Reiser Shaw 7.0 1.0 4.0 7.0 11.5 2.0 9.5 9.5 7.0 3.0 5.0 11.5 TYPE: PR DIFFICULTY: Moderate KEYWORDS: Kruskal-Wallis rank test, nonparametric test, test statistic

179. Referring to Table 12-16, the calculation of the Kruskal-Wallis test statistic H involves obtaining the total of the ranks for each sample. These totals are , and .

ANSWER: 25.5, 6.0, 18.5, and 28.0 TYPE: FI DIFFICULTY: Moderate KEYWORDS: Kruskal-Wallis rank test, nonparametric test, test statistic

180. Referring to Table 12-16, the calculated value of the Kruskal-Wallis test statistic H is . ANSWER: 7.47 TYPE: FI DIFFICULTY: Moderate KEYWORDS: Kruskal-Wallis rank test, nonparametric test, test statistic TABLE 12-17 An agronomist wants to compare the crop yield of 3 varieties of chickpea seeds. She plants all 3 varieties of the seeds on each of 5 different patches of fields. She then measures the crop yield in bushels per acre. Treating this as a randomized block design, the results are presented in the table that follows. Fields Smith Walsh Trevor 1 11.1 19.0 14.6 2 13.5 18.0 15.7 3 15.3 19.8 16.8 4 14.6 19.6 16.7 5 9.8 16.6 15.2 Below is the Minitab output of the Friedman rank test: Friedman Test: Yield versus Varieties, Fields Friedman test for Yield by Variety blocked by Fields S = 10.00

DF = 2

P = 0.007 Est

Sum of

162 Chi-Square Tests and Nonparametric Tests Variety Smith Trevor Walsh

N 5 5 5

Median 13.500 15.667 18.533

Grand median

15.900

Ranks 5.0 10.0 15.0

181. Referring to Table 12-17, the null hypothesis for the Friedman rank test is a) H0 : Field 1 = Field 2 = Field 3 = Field 4 = Field 5 b) c) d)

H0 : Smith = Walsh = Trevor H0 : MField 1 = MField 2 = MField 3 = MField 4 = MField 5 H0 : MSmith = MWalsh = MTrevor

ANSWER: d TYPE: MC DIFFICULTY: Easy KEYWORDS: Friedman rank test, nonparametric test, form of hypothesis

182. Referring to Table 12-17, what are the degrees of freedom of the Friedman rank test for the difference in the medians at a level of significance of 0.01? ANSWER: 2 TYPE: PR DIFFICULTY: Easy KEYWORDS: Friedman rank test, nonparametric test, degrees of freedom

183. Referring to Table 12-17, what is the critical value of the Friedman rank test for the difference in the medians at a level of significance of 0.01? ANSWER: 9.210 TYPE: PR DIFFICULTY: Easy KEYWORDS: Friedman rank test, nonparametric test, critical value

184. Referring to Table 12-17, what is the value of the test statistic for the Friedman rank test for the difference in the medians? ANSWER: 10 TYPE: PR DIFFICULTY: Moderate KEYWORDS: Friedman rank test, nonparametric test, test statistic

185. True or False: Referring to Table 12-17, the null hypothesis for the Friedman rank test for the difference in the means should be rejected at a 0.01 level of significance. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: Friedman rank test, nonparametric test, decision

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186. True or False: Referring to Table 12-17, the decision made at a 0.01 level of significance on the Friedman rank test for the difference in medians implies that all 3 medians are significantly different. ANSWER: False TYPE: TF DIFFICULTY: Moderate KEYWORDS: Friedman rank test, nonparametric test, conclusion

187. True or False: Referring to Table 12-17, the decision made at a 0.01 level of significance on the Friedman rank test for the difference in medians implies that the 3 medians are not all the same. ANSWER: True TYPE: TF DIFFICULTY: Moderate KEYWORDS: Friedman rank test, nonparametric test, conclusion

188. True or False: Referring to Table 12-17, the Friedman rank test is valid only if there is no interaction between the 5 blocks and the 3 treatment levels. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: Friedman rank test, nonparametric test, assumptions

189. True or False: Referring to Table 12-17, the Friedman rank test is valid only if the 5 blocks are independent so that the yields in one block have no influence on the yields in any other block. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: Friedman rank test, nonparametric test, assumptions

175 Simple Linear Regression

CHAPTER 13: SIMPLE LINEAR REGRESSION 1. The Y-intercept (b0) represents the a) predicted value of Y when X = 0. b) change in estimated average Y per unit change in X. c) predicted value of Y. d) variation around the sample regression line. ANSWER: a TYPE: MC DIFFICULTY: Easy KEYWORDS: intercept, interpretation

2. The Y-intercept (b0) represents the a) estimated average Y when X = 0. b) change in estimated average Y per unit change in X. c) predicted value of Y. d) variation around the sample regression line. ANSWER: a TYPE: MC DIFFICULTY: Easy KEYWORDS: intercept, interpretation

3. The slope (b1) represents a) predicted value of Y when X = 0. b) the estimated average change in Y per unit change in X. c) the predicted value of Y. d) variation around the line of regression. ANSWER: b TYPE: MC DIFFICULTY: Easy KEYWORDS: slope, interpretation 4. The least squares method minimizes which of the following? a) SSR b) SSE c) SST d) All of the above ANSWER: b TYPE: MC DIFFICULTY: Difficult KEYWORDS: least squares, properties TABLE 13-1 A large national bank charges local companies for using their services. A bank official reported the results of a regression analysis designed to predict the bank’s charges (Y) -- measured in dollars per

174

Simple Linear Regression month -- for services rendered to local companies. One independent variable used to predict service charge to a company is the company’s sales revenue (X) -- measured in millions of dollars. Data for 21 companies who use the bank’s services were used to fit the model:

E(Y ) = 0 + 1X

The results of the simple linear regression are provided below.

Yµ = −2, 700 + 20 X , SYX = 65, two-tailed p value = 0.034 (for testing  )

5. Referring to Table 13-1, interpret the estimate of 0 , the Y-intercept of the line. a) b) c) d)

All companies will be charged at least $2,700 by the bank. There is no practical interpretation since a sales revenue of $0 is a nonsensical value. About 95% of the observed service charges fall within $2,700 of the least squares line. For every $1 million increase in sales revenue, we expect a service charge to decrease $2,700.

ANSWER: b TYPE: MC DIFFICULTY: Moderate KEYWORDS: intercept, interpretation

6. Referring to Table 13-1, interpret the estimate of  , the standard deviation of the random error term (standard error of the estimate) in the model. a) About 95% of the observed service charges fall within $65 of the least squares line. b) About 95% of the observed service charges equal their corresponding predicted values. c) About 95% of the observed service charges fall within $130 of the least squares line. d) For every $1 million increase in sales revenue, we expect a service charge to increase $65. ANSWER: c TYPE: MC DIFFICULTY: Difficult KEYWORDS: standard error of the estimate, interpretation

7. Referring to Table 13-1, interpret the p-value for testing whether 1 exceeds 0. a) There is sufficient evidence (at the  = 0.05) to conclude that sales revenue (X) is a useful linear predictor of service charge (Y).

b) There is insufficient evidence (at the  = 0.10) to conclude that sales revenue (X) is a useful linear predictor of service charge (Y). c) Sales revenue (X) is a poor predictor of service charge (Y). d) For every $1 million increase in sales revenue, we expect a service charge to increase $0.034. ANSWER: a TYPE: MC DIFFICULTY: Moderate KEYWORDS: p-value, interpretation

8. Referring to Table 13-1, a 95% confidence interval for 1 is (15, 30). Interpret the interval. a) We are 95% confident that the mean service charge will fall between $15 and $30 per month.

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b) We are 95% confident that the sales revenue (X) will increase between $15 and $30 million for every $1 increase in service charge (Y).

c) We are 95% confident that average service charge (Y) will increase between $15 and $30 for every $1 million increase in sales revenue (X). d) At the  = 0.05 level, there is no evidence of a linear relationship between service charge (Y) and sales revenue (X). ANSWER: c TYPE: MC DIFFICULTY: Moderate KEYWORDS: confidence interval, interpretation TABLE 13-2 A candy bar manufacturer is interested in trying to estimate how sales are influenced by the price of their product. To do this, the company randomly chooses 6 small cities and offers the candy bar at different prices. Using candy bar sales as the dependent variable, the company will conduct a simple linear regression on the data below: City Price ($) Sales River Falls 1.30 100 Hudson 1.60 90 Ellsworth 1.80 90 Prescott 2.00 40 Rock Elm 2.40 38 Stillwater 2.90 32 9. Referring to Table 13-2, what is the estimated slope parameter for the candy bar price and sales data? a) 161.386 b) 0.784 c) – 3.810 d) – 48.193 ANSWER: d TYPE: MC DIFFICULTY: Moderate KEYWORDS: slope, estimation 10. Referring to Table 13-2, what is the estimated average change in the sales of the candy bar if price goes up by $1.00? a) 161.386 b) 0.784 c) – 3.810 d) – 48.193 ANSWER: d TYPE: MC DIFFICULTY: Moderate KEYWORDS: slope, interpretation 11. Referring to Table 13-2, what is the coefficient of correlation for these data? a) – 0.8854 b) – 0.7839

177 Simple Linear Regression c) 0.7839 d) 0.8854 ANSWER: a TYPE: MC DIFFICULTY: Moderate KEYWORDS: coefficient of correlation, estimation 12. Referring to Table 13-2, what is the percentage of the total variation in candy bar sales explained by the regression model? a) 100% b) 88.54% c) 78.39% d) 48.19% ANSWER: c TYPE: MC DIFFICULTY: Moderate KEYWORDS: coefficient of determination, interpretation 13. Referring to Table 13-2, what percentage of the total variation in candy bar sales is explained by prices? a) 100% b) 88.54% c) 78.39% d) 48.19% ANSWER: c TYPE: MC DIFFICULTY: Moderate KEYWORDS: coefficient of determination, interpretation

14. Referring to Table 13-2, what is the standard error of the estimate, SYX, for the data? a) 0.784 b) 0.885 c) 12.650 d) 16.299 ANSWER: d TYPE: MC DIFFICULTY: Moderate KEYWORDS: standard error of the estimate, estimation

15. Referring to Table 13-2, what is the standard error of the regression slope estimate, sb ? 1

a) 0.784 b) 0.885 c) 12.650 d) 16.299 ANSWER: c TYPE: MC DIFFICULTY: Moderate KEYWORDS: standard error, slope

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16. Referring to Table 13-2, what is (X – X )2 for these data? a) 0 b) 1.66 c) 2.54 d) 25.66 ANSWER: b TYPE: MC DIFFICULTY: Moderate KEYWORDS: sum of squares

17. Referring to Table 13-2, to test that the regression coefficient, 1, is not equal to 0, what would be the critical values? Use  = 0.05. a)  2.5706 b)  2.7765 c)  3.1634 d)  3.4954 ANSWER: b TYPE: MC DIFFICULTY: Difficult KEYWORDS: t test on slope, critical value

18. Referring to Table 13-2, to test whether a change in price will have any impact on average sales, what would be the critical values? Use  = 0.05. a)  2.5706 b)  2.7765 c)  3.1634 d)  3.4954 ANSWER: b TYPE: MC DIFFICULTY: Difficult EXPLANATION: This is a test for 1 is not equal to 0. KEYWORDS: t test on slope, critical value 19. Referring to Table 13-2, if the price of the candy bar is set at $2, the estimated average sales will be a) 30 b) 65 c) 90 d) 100 ANSWER: b TYPE: MC DIFFICULTY: Moderate KEYWORDS: estimation of mean values 20. Referring to Table 13-2, if the price of the candy bar is set at $2, the predicted sales will be a) 30 b) 65

179 Simple Linear Regression c) 90 d) 100 ANSWER: b TYPE: MC DIFFICULTY: Moderate KEYWORDS: prediction of individual values

21. True of False: The Chancellor of a university has commissioned a team to collect data on

students’ GPAs and the amount of time they spend bar hopping every week (measured in minutes). He wants to know if imposing much tougher regulations on all campus bars to make it more difficult for students to spend time in any campus bar will have a significant impact on general students' GPAs. His team should use a t test on the slope of the population regression.

ANSWER: True TYPE: TF DIFFICULTY: Moderate KEYWORDS: t test on slope 22. The residual represents the discrepancy between the observed dependent variable and its value. ANSWER: predicted or estimated average TYPE: FI DIFFICULTY: Moderate KEYWORDS: estimation of mean values, prediction of individual values TABLE 13-3 The director of cooperative education at a state college wants to examine the effect of cooperative education job experience on marketability in the work place. She takes a random sample of 4 students. For these 4, she finds out how many times each had a cooperative education job and how many job offers they received upon graduation. These data are presented in the table below. Student 1 2 3 4

CoopJobs 1 2 1 0

JobOffer 4 6 3 1

23. Referring to Table 13-3, set up a scatter diagram. ANSWER:

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Scatter Diagram 6 5 4 3 2 1 0 0

Coop Jobs

TYPE: PR DIFFICULTY: Moderate KEYWORDS: scatter plot 24. Referring to Table 13-3, the least squares estimate of the slope is

ANSWER: 2.50 TYPE: FI DIFFICULTY: 2 Moderate KEYWORDS: slope, estimation, least squares

25. Referring to Table 13-3, the least squares estimate of the Y-intercept is

ANSWER: 1.00 TYPE: FI DIFFICULTY: Moderate KEYWORDS: intercept, estimation, least squares 26. Referring to Table 13-3, the prediction for the number of job offers for a person with 2 coop jobs is . ANSWER: 6 TYPE: FI DIFFICULTY: Moderate KEYWORDS: prediction of individual values 27. Referring to Table 13-3, the total sum of squares (SST) is

ANSWER: 13.0 TYPE: FI DIFFICULTY: Moderate KEYWORDS: sum of squares 28. Referring to Table 13-3, the regression sum of squares (SSR) is

181 Simple Linear Regression ANSWER: 12.5 TYPE: FI DIFFICULTY: Moderate KEYWORDS: sum of squares 29. Referring to Table 13-3, the error or residual sum of squares (SSE) is

ANSWER: 0.50 TYPE: FI DIFFICULTY: Moderate KEYWORDS: sum of squares 30. Referring to Table 13-3, the coefficient of determination is

ANSWER: 0.962 TYPE: FI DIFFICULTY: Moderate KEYWORDS: coefficient of determination 31. Referring to Table 13-3, the standard error of estimate is

ANSWER: 0.50 TYPE: FI DIFFICULTY: Moderate KEYWORDS: standard error of estimate 32. Referring to Table 13-3, the coefficient of correlation is

ANSWER: 0.981 TYPE: FI DIFFICULTY: Moderate KEYWORDS: coefficient of correlation

33. Referring to Table 13-3, suppose the director of cooperative education wants to obtain a 95% confidence-interval estimate for the mean number of job offers received by people who have had exactly one cooperative education job. The t critical value she would use is . ANSWER: 4.3027 TYPE: FI DIFFICULTY: Easy KEYWORDS: confidence interval, estimation of mean values, critical value 34. Referring to Table 13-3, suppose the director of cooperative education wants to obtain a 95% confidence interval estimate for the mean number of job offers received by people who have had exactly one cooperative education job. The confidence interval is from to . ANSWER: 2.424 to 4.576 TYPE: FI DIFFICULTY: Moderate KEYWORDS: confidence interval, estimation of mean values, critical value 35. Referring to Table 13-3, suppose the director of cooperative education wants to obtain a 95% prediction interval estimate for the number of job offers received by people who have had exactly one cooperative education job. The prediction interval is from to .

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ANSWER: 1.0947 to 5.9053 TYPE: FI DIFFICULTY: Moderate KEYWORDS: prediction interval, prediction of individual values 36. True or False: Referring to Table 13-3, suppose the director of cooperative education wants to obtain two 95% confidence interval estimates. One is for the mean number of job offers received by people who have had exactly one cooperative education job and one for people who have had two. The confidence interval for people who have had one cooperative education job would be the wider of the two intervals. ANSWER: False TYPE: TF DIFFICULTY: Moderate KEYWORDS: confidence interval, estimation of mean values

37. Referring to Table 13-3, suppose the director of cooperative education wants to obtain a 95% prediction interval for the number of job offers received by a person who has had exactly two cooperative education jobs. The t critical value she would use is . ANSWER: 4.3027 TYPE: FI DIFFICULTY: Moderate KEYWORDS: prediction interval, prediction of individual values, critical value 38. Referring to Table 13-3, suppose the director of cooperative education wants to obtain a 95% prediction interval for the number of job offers received by a person who has had exactly two cooperative education jobs. The prediction interval is from to . ANSWER: 3.154 to 8.846 TYPE: FI DIFFICULTY: Moderate KEYWORDS: prediction interval, prediction of individual values

39. True or False: Referring to Table 13-3, suppose the director of cooperative education wants to obtain both a 95% confidence interval estimate and a 95% prediction interval for X = 2. The confidence interval estimate would be the wider of the two intervals. ANSWER: False TYPE: TF DIFFICULTY: Moderate KEYWORDS: confidence interval, prediction interval, estimation of mean values, prediction of individual values, properties

40. Referring to Table 13-3, the director of cooperative education wanted to test the hypothesis that the true slope was equal to 0. The denominator of the test statistic is sb . The value of sb in this 1

sample is

ANSWER: 0.3536 TYPE: FI DIFFICULTY: Moderate KEYWORDS: t test on slope, standard error, slope

183 Simple Linear Regression

41. Referring to Table 13-3, the director of cooperative education wanted to test the hypothesis that the true slope was equal to 0. The value of the test statistic is . ANSWER: 7.07 TYPE: FI DIFFICULTY: Moderate KEYWORDS: t test on slope, test statistic, slope 42. Referring to Table 13-3, the director of cooperative education wanted to test the hypothesis that the true slope was equal to 3.0. The value of the test statistic is . ANSWER: -1.4142 TYPE: FI DIFFICULTY: Moderate KEYWORDS: t test on slope, test statistic, slope 43. Referring to Table 13-3, the director of cooperative education wanted to test the hypothesis that the true slope was equal to 0. For a test with a level of significance of 0.05, the null hypothesis should be rejected if the value of the test statistic is . ANSWER: > 4.3027 or < -4.3027 TYPE: FI DIFFICULTY: Easy KEYWORDS: t test on slope, critical value, slope, decision 44. Referring to Table 13-3, the director of cooperative education wanted to test the hypothesis that the true slope was equal to 3.0. For a test with a level of significance of 0.05, the null hypothesis should be rejected if the value of the test statistic is . ANSWER: > 4.3027 or < -4.3027 TYPE: FI DIFFICULTY: Easy KEYWORDS: t test on slope, critical value, slope, decision

45. Referring to Table 13-3, the director of cooperative education wanted to test the hypothesis that the true slope was equal to 0. The p-value of the test is between

and

ANSWER: 0.01 and 0.02 using the Table, or 0.0194 using Excel TYPE: FI DIFFICULTY: Moderate KEYWORDS: t test on slope, p-value, slope

46. Referring to Table 13-3, the director of cooperative education wanted to test the hypothesis that the true slope was equal to 3.0. The p-value of the test is between ANSWER: 0.2 and 0.5 using the Table, or 0.2929 using Excel TYPE: FI DIFFICULTY: Difficult EXPLANATION: The t-test statistic is t =

( b1 − 1 ) = ( 2.5 − 3) = −1.4142

KEYWORDS: t test on slope, p-value, slope

Sb1

0.3536

and

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TABLE 13-4 The managers of a brokerage firm are interested in finding out if the number of new clients a broker brings into the firm affects the sales generated by the broker. They sample 12 brokers and determine the number of new clients they have enrolled in the last year and their sales amounts in thousands of dollars. These data are presented in the table that follows. Broker Clients Sales 1 27 52 2 11 37 3 42 64 4 33 55 5 15 29 6 15 34 7 25 58 8 36 59 9 28 44 10 30 48 11 17 31 12 22 38 47. Referring to Table 13-4, set up a scatter diagram. ANSWER: Scatter Diagram 70 60 50 40 30 20 10 0 0

Clients

TYPE: PR DIFFICULTY: Moderate KEYWORDS: scatter plot 48. Referring to Table 13-4, the least squares estimate of the slope is ANSWER: 1.12 TYPE: FI DIFFICULTY: Moderate

185 Simple Linear Regression KEYWORDS: slope, estimation, least squares

49. Referring to Table 13-4, the least squares estimate of the Y-intercept is

ANSWER: 17.7 TYPE: FI DIFFICULTY: Moderate KEYWORDS: intercept, estimation, least squares 50. Referring to Table 13-4, the prediction for the amount of sales (in $1,000s) for a person who brings 25 new clients into the firm is . ANSWER: 45.66 TYPE: FI DIFFICULTY: Moderate KEYWORDS: prediction of individual values 51. Referring to Table 13-4, the total sum of squares (SST) is

ANSWER: 1,564.2 TYPE: FI DIFFICULTY: Moderate KEYWORDS: sum of squares 52. Referring to Table 13-4, the regression sum of squares (SSR) is

ANSWER: 1,227.4 TYPE: FI DIFFICULTY: Moderate KEYWORDS: sum of squares 53. Referring to Table 13-4, the error or residual sum of squares (SSE) is

ANSWER: 336.9 TYPE: FI DIFFICULTY: Moderate KEYWORDS: sum of squares 54. Referring to Table 13-4, the coefficient of determination is

ANSWER: 0.785 TYPE: FI DIFFICULTY: Moderate KEYWORDS: coefficient of determination 55. Referring to Table 13-4, % of the total variation in sales generated can be explained by the number of new clients brought in. ANSWER: 78.5 TYPE: FI DIFFICULTY: Moderate KEYWORDS: coefficient of determination, interpretation 56. Referring to Table 13-4, the standard error of the estimated slope coefficient is

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186

ANSWER: 0.185 TYPE: FI DIFFICULTY: Moderate KEYWORDS: slope, standard error 57. Referring to Table 13-4, the standard error of estimate is

ANSWER: 5.804 TYPE: FI DIFFICULTY: Moderate KEYWORDS: standard error of estimate 58. Referring to Table 13-4, the coefficient of correlation is

ANSWER: 0.886 TYPE: FI DIFFICULTY: Moderate KEYWORDS: coefficient of correlation

59. Referring to Table 13-4, suppose the managers of the brokerage firm want to obtain a 99% confidence interval estimate for the mean sales made by brokers who have brought into the firm 24 new clients. The t critical value they would use is . ANSWER: 3.1693 TYPE: FI DIFFICULTY: Easy KEYWORDS: estimation of mean values, confidence interval, critical value 60. Referring to Table 13-4, suppose the managers of the brokerage firm want to obtain a 99% confidence interval estimate for the mean sales made by brokers who have brought into the firm 24 new clients. The confidence interval is from to . ANSWER: 39.19 to 49.89 TYPE: FI DIFFICULTY: Moderate KEYWORDS: estimation of mean values, confidence interval

61. Referring to Table 13-4, suppose the managers of the brokerage firm want to obtain a 99% prediction interval for the sales made by a broker who has brought into the firm 18 new clients. The t critical value they would use is . ANSWER: 3.1693 TYPE: FI DIFFICULTY: Moderate KEYWORDS: prediction of individual values, prediction interval, critical value 62. Referring to Table 13-4, suppose the managers of the brokerage firm want to obtain a 99% prediction interval for the sales made by a broker who has brought into the firm 18 new clients. The prediction interval is from to . ANSWER: 18.23 to 57.42

187 Simple Linear Regression TYPE: FI DIFFICULTY: Moderate KEYWORDS: prediction of individual values, prediction interval

63. Referring to Table 13-4, suppose the managers of the brokerage firm want to obtain both a 99% confidence interval estimate and a 99% prediction interval for X = 24. The confidence interval estimate would be the (wider or narrower) of the two intervals. ANSWER: narrower TYPE: FI DIFFICULTY: Moderate KEYWORDS: prediction of individual values, prediction interval, estimation of mean values, confidence interval, properties

64. Referring to Table 13-4, the managers of the brokerage firm wanted to test the hypothesis that the true slope was equal to 0. The denominator of the test statistic is sb . The value of sb in this 1

sample is

ANSWER: 0.1853 TYPE: FI DIFFICULTY: Moderate KEYWORDS: t test on slope, slope, standard error 65. Referring to Table 13-4, the managers of the brokerage firm wanted to test the hypothesis that the true slope was equal to 0. The value of the test statistic is . ANSWER: 6.04 TYPE: FI DIFFICULTY: Moderate KEYWORDS: t test on slope, slope, test statistic 66. Referring to Table 13-4, the managers of the brokerage firm wanted to test the hypothesis that the number of new clients brought in did not affect the amount of sales generated. The value of the test statistic is . ANSWER: 6.04 TYPE: FI DIFFICULTY: Moderate KEYWORDS: t test on slope, slope, test statistic 67. Referring to Table 13-4, the managers of the brokerage firm wanted to test the hypothesis that the true slope was equal to 0. For a test with a level of significance of 0.01, the null hypothesis should be rejected if the value of the test statistic is . ANSWER: > 3.1693 or < -3.1693 TYPE: FI DIFFICULTY: Moderate KEYWORDS: t test on slope, slope, critical value, decision

68. Referring to Table 13-4, the managers of the brokerage firm wanted to test the hypothesis that the true slope was equal to 0. The p-value of the test is ANSWER: < 0.01 using the Table, or 0.000126 using Excel

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TYPE: FI DIFFICULTY: Moderate KEYWORDS: t test on slope, slope, p-value 69. Referring to Table 13-4, the managers of the brokerage firm wanted to test the hypothesis that the true slope was equal to 0. At a level of significance of 0.01, the null hypothesis should be (accepted or rejected). ANSWER: rejected TYPE: FI DIFFICULTY: Moderate KEYWORDS: t test on slope, slope, decision 70. Referring to Table 13-4, the managers of the brokerage firm wanted to test the hypothesis that the true slope was equal to 0. At a level of significance of 0.01, the decision that should be made implies that (there is or there is no) linear dependent relation between the independent and dependent variables. ANSWER: there is TYPE: FI DIFFICULTY: Moderate KEYWORDS: t test on slope, slope, conclusion 71. Referring to Table 13-4, the managers of the brokerage firm wanted to test the hypothesis that the number of new clients brought in had a positive impact on the amount of sales generated. The value of the test statistic is . ANSWER: 6.04 TYPE: FI DIFFICULTY: Moderate KEYWORDS: t test on slope, slope, test statistic 72. Referring to Table 13-4, the managers of the brokerage firm wanted to test the hypothesis that the number of new clients brought in had a positive impact on the amount of sales generated. For a test with a level of significance of 0.01, the null hypothesis should be rejected if the value of the test statistic is . ANSWER: > 2.7638 TYPE: FI DIFFICULTY: Moderate KEYWORDS: t test on slope, slope, critical value, decision

73. Referring to Table 13-4, the managers of the brokerage firm wanted to test the hypothesis that the number of new clients brought in had a positive impact on the amount of sales generated. The pvalue of the test is . ANSWER: < 0.005 using the Table, or 0.000126/2 using Excel TYPE: FI DIFFICULTY: Moderate KEYWORDS: t test on slope, slope, p-value 74. Referring to Table 13-4, the managers of the brokerage firm wanted to test the hypothesis that the number of new clients brought in had a positive impact on the amount of sales generated. At a level of significance of 0.01, the null hypothesis should be (accepted or rejected).

189 Simple Linear Regression ANSWER: rejected TYPE: FI DIFFICULTY: Moderate KEYWORDS: t test on slope, slope, decision 75. Referring to Table 13-4, the managers of the brokerage firm wanted to test the hypothesis that the number of new clients brought in had a positive impact on the amount of sales generated. At a level of significance of 0.01, the decision that should be made implies that the number of new clients brought in (had or did not have) a positive impact on the amount of sales generated. ANSWER: had TYPE: FI DIFFICULTY: Moderate KEYWORDS: t test on slope, slope, conclusion TABLE 13-5 The managing partner of an advertising agency believes that his company's sales are related to the industry sales. He uses Microsoft Excel’s Data Analysis tool to analyze the last 4 years of quarterly data (i.e., n = 16) with the following results: Regression Statistics Multiple R 0.802 R Square 0.643 Adjusted R Square 0.618 Standard Error SYX Observations

0.9224 16

ANOVA df Regression 1 Error 14 Total 15 Predictor Coef Intercept 3.962 Industry 0.040451

SS 21.497 11.912 33.409 StdError 1.440 0.008048

Durbin-Watson Statistic

MS 21.497 0.851 t Stat 2.75 5.03

F 25.27

Sig.F 0.000

P-value 0.016 0.000

1.59

76. Referring to Table 13-5, the value of the quantity that the least squares regression line minimizes is . ANSWER: 11.912 TYPE: FI DIFFICULTY: Moderate KEYWORDS: least squares, properties, sum of squares

77. Referring to Table 13-5, the estimates of the Y-intercept and slope are respectively. ANSWER:

and

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190

3.962 and 0.040451 TYPE: FI DIFFICULTY: Easy KEYWORDS: intercept, slope, estimation

78. Referring to Table 13-5, the prediction for a quarter in which X = 120 is Y =

ANSWER: 8.816 TYPE: FI DIFFICULTY: Moderate KEYWORDS: prediction of individual values 79. Referring to Table 13-5, the standard error of the estimate is

ANSWER: 0.9224 TYPE: FI DIFFICULTY: Easy KEYWORDS: standard error of estimate 80. Referring to Table 13-5, the coefficient of determination is

ANSWER: 0.643 TYPE: FI DIFFICULTY: Easy KEYWORDS: coefficient of determination 81. Referring to Table 13-5, the standard error of the estimated slope coefficient is

ANSWER: 0.008 TYPE: FI DIFFICULTY: Easy KEYWORDS: slope, standard error 82. Referring to Table 13-5, the correlation coefficient is

ANSWER: 0.802 TYPE: FI DIFFICULTY: Easy KEYWORDS: coefficient of correlation

83. Referring to Table 13-5, the partner wants to test for autocorrelation using the Durbin-Watson statistic. Using a level of significance of 0.05, the critical values of the test are dL = and dU = . ANSWER: 1.10; 1.37 TYPE: FI DIFFICULTY: Easy KEYWORDS: Durbin-Watson statistic, critical value, autocorrelation 84. Referring to Table 13-5, the partner wants to test for autocorrelation using the Durbin-Watson statistic. Using a level of significance of 0.05, the decision he should make is: a) there is evidence of autocorrelation. b) the test is unable to make a definite conclusion. c) there is no evidence of autocorrelation.

191 Simple Linear Regression d) there is not enough information to perform the test. ANSWER: c TYPE: MC DIFFICULTY: Moderate KEYWORDS: Durbin-Watson statistic, decision, autocorrelation 85. If the Durbin-Watson statistic has a value close to 0, which assumption is violated? a) Normality of the errors. b) Independence of errors. c) Homoscedasticity. d) None of the above. ANSWER: b TYPE: MC Difficulty: Easy KEYWORDS: Durbin-Watson statistic, autocorrelation, assumption 86. If the Durbin-Watson statistic has a value close to 4, which assumption is violated? a) Normality of the errors. b) Independence of errors. c) Homoscedasticity. d) None of the above. ANSWER: b TYPE: MC Difficulty: Easy KEYWORDS: Durbin-Watson statistic, assumption, autocorrelation 87. The standard error of the estimate is a measure of a) total variation of the Y variable. b) the variation around the sample regression line. c) explained variation. d) the variation of the X variable. ANSWER: b TYPE: MC DIFFICULTY: Easy KEYWORDS: standard error of estimate, interpretation

88. The coefficient of determination (r2) tells us a) that the coefficient of correlation (r) is larger than 1. b) whether r has any significance. c) that we should not partition the total variation. d) the proportion of total variation that is explained. ANSWER: d TYPE: MC DIFFICULTY: Easy KEYWORDS: coefficient of determination, interpretation TABLE 13-6

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The following EXCEL tables are obtained when "Score received on an exam (measured in percentage points)" (Y) is regressed on "percentage attendance" (X) for 22 students in a Statistics for Business and Economics course. Regression Statistics Multiple R 0.142620229 R Square 0.02034053 Adjusted R Square -0.028642444 Standard Error 20.25979924 Observations 22

Intercept Attendance

Coefficients Standard Error 39.39027309 37.24347659 0.340583573 0.52852452

T Stat 1.057642216 0.644404489

P-value 0.302826622 0.526635689

89. Referring to Table 13-6, which of the following statements is true? a) -2.86% of the total variability in score received can be explained by percentage attendance. b) -2.86% of the total variability in percentage attendance can be explained by score received. c) 2% of the total variability in score received can be explained by percentage attendance. d) 2% of the total variability in percentage attendance can be explained by score received. ANSWER: c TYPE: MC DIFFICULTY: Easy KEYWORDS: coefficient of determination, interpretation 90. Referring to Table 13-6, which of the following statements is true? a) If attendance increases by 0.341%, the estimated average score received will increase by 1 percentage point. b) If attendance increases by 1%, the estimated average score received will increase by 39.39 percentage points. c) If attendance increases by 1%, the estimated average score received will increase by 0.341 percentage points. d) If the score received increases by 39.39%, the estimated average attendance will go up by 1%. ANSWER: c TYPE: MC DIFFICULTY: Easy KEYWORDS: slope, interpretation 91. True or False: The Regression Sum of Squares (SSR) can never be greater than the Total Sum of Squares (SST). ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: sum of squares, properties 92. True or False: The coefficient of determination represents the ratio of SSR to SST.

193 Simple Linear Regression ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: coefficient of determination, properties 93. True or False: Regression analysis is used for prediction, while correlation analysis is used to measure the strength of the association between two numerical variables. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: coefficient of correlation, slope, interpretation

94. True or False: The value of r is always positive. ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: coefficient of correlation, properties 95. In performing a regression analysis involving two numerical variables, we are assuming a) the variances of X and Y are equal. b) the variation around the line of regression is the same for each X value. c) that X and Y are independent. d) All of the above. ANSWER: b TYPE: MC DIFFICULTY: Moderate KEYWORDS: assumption 96. Which of the following assumptions concerning the probability distribution of the random error term is stated incorrectly? a) The distribution is normal. b) The mean of the distribution is 0. c) The variance of the distribution increases as X increases. d) The errors are independent. ANSWER: c TYPE: MC DIFFICULTY: Moderate KEYWORDS: assumption

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97. Based on the residual plot below, you will conclude that there might be a violation of which of the following assumptions.

Footage Residual Plot 6000

Residuals

4000 2000 0 0

1,000 2,000 3,000 4,000 5,000 6,000

-2000 -4000 Footage

a) b) c) d)

Linearity of the relationship Normality of errors Homoscedasticity Independence of errors

ANSWER: c TYPE: MC DIFFICULTY: Moderate KEYWORDS: assumption, homoscedasticity 98. True or False: Data that exhibit an autocorrelation effect violate the regression assumption of independence. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: assumption, autocorrelation

99. True or False: The Durbin-Watson D statistic is used to check the assumption of normality. ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: Durbin-Watson statistic, assumption, autocorrelation

100. If the residuals in a regression analysis of time ordered data are not correlated, the value of the Durbin-Watson D statistic should be near

ANSWER: 2 TYPE: FI DIFFICULTY: Moderate KEYWORDS: Durbin-Watson statistic, autocorrelation, properties 101. The residuals represent

195 Simple Linear Regression

a) the difference between the actual Y values and the mean of Y. b) the difference between the actual Y values and the predicted Y values. c) the square root of the slope. d) the predicted value of Y for the average X value. ANSWER: b TYPE: MC DIFFICULTY: Easy KEYWORDS: residual, properties 102. If the plot of the residuals is fan shaped, which assumption is violated? a) Normality. b) Homoscedasticity. c) Independence of errors. d) No assumptions are violated, the graph should resemble a fan. ANSWER: b TYPE: MC DIFFICULTY: Moderate KEYWORDS: residual plot, homoscedasticity 103. What do we mean when we say that a simple linear regression model is “statistically” useful? a) All the statistics computed from the sample make sense. b) The model is an excellent predictor of Y. c) The model is “practically” useful for predicting Y. d) The model is a better predictor of Y than the sample mean, Y . ANSWER: d TYPE: MC DIFFICULTY: Difficult KEYWORDS: slope, interpretation

104. If the correlation coefficient (r) = 1.00, then a) the Y-intercept (b0) must equal 0. b) the explained variation equals the unexplained variation. c) there is no unexplained variation. d) there is no explained variation. ANSWER: c TYPE: MC DIFFICULTY: Moderate KEYWORDS: coefficient of correlation, properties, interpretation

105. If the correlation coefficient (r) = 1.00, then a) b) c) d)

all the data points must fall exactly on a straight line with a slope that equals 1.00. all the data points must fall exactly on a straight line with a negative slope. all the data points must fall exactly on a straight line with a positive slope. all the data points must fall exactly on a horizontal straight line with a zero slope.

ANSWER: c TYPE: MC DIFFICULTY: - Moderate KEYWORDS: coefficient of correlation, properties

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106. Assuming a linear relationship between X and Y, if the coefficient of correlation (r) equals – 0.30, a) there is no correlation. b) the slope (b1) is negative. c) variable X is larger than variable Y. d) the variance of X is negative. ANSWER: b TYPE: MC DIFFICULTY: Moderate KEYWORDS: coefficient of correlation, properties, interpretation 107. Testing for the existence of correlation is equivalent to a) testing for the existence of the slope (  ).

b) testing for the existence of the Y-intercept (  ). c) the confidence interval estimate for predicting Y. d) None of the above.

ANSWER: a TYPE: MC DIFFICULTY: Moderate KEYWORDS: coefficient of correlation, t test for correlation coefficient, t test on slope 108. The strength of the linear relationship between two numerical variables may be measured by the a) scatter diagram. b) coefficient of correlation. c) slope. d) Y-intercept. ANSWER: b TYPE: MC DIFFICULTY: Easy KEYWORDS: coefficient of correlation, interpretation

109. In a simple linear regression problem, r and b1 a) b) c) d)

may have opposite signs. must have the same sign. must have opposite signs. are equal.

ANSWER: b TYPE: MC DIFFICULTY: Easy KEYWORDS: coefficient of correlation, slope, properties

110. The sample correlation coefficient between X and Y is 0.375. It has been found out that the pvalue is 0.256 when testing H0 :  = 0 against the two-sided alternative H1 :   0 . To test H0 :  = 0 against the one-sided alternative H1 :   0 at a significance level of 0.2, the p-value is

197 Simple Linear Regression

a) 0.256 / 2 b) ( 0.256 ) 2 c) 1− 0.256 d) 1− 0.256 / 2 ANSWER: d TYPE: MC DIFFICULTY: Difficult KEYWORDS: coefficient of correlation, t test for correlation coefficient, p-value

111. The sample correlation coefficient between X and Y is 0.375. It has been found out that the pvalue is 0.256 when testing H0 :  = 0 against the two-sided alternative H1 :   0 . To test H0 :  = 0 against the one-sided alternative H1 :   0 at a significance level of 0.2, the p-value is

a) 0.256 / 2 b) (0.256) 2 c) 1− 0.256 d) 1− 0.256 / 2 ANSWER: a TYPE: MC DIFFICULTY: Difficult KEYWORDS: coefficient of correlation, t test for correlation coefficient, p-value

112. The sample correlation coefficient between X and Y is 0.375. It has been found out that the pvalue is 0.256 when testing H0 :  = 0 against the one-sided alternative H1 :   0 . To test H0 :  = 0 against the two-sided alternative H1 :   0 at a significance level of 0.2, the pvalue is

a) 0.256 / 2 b) 0.256 * 2 c) 1− 0.256 d) 1− 0.256 / 2 ANSWER: b TYPE: MC DIFFICULTY: Difficult KEYWORDS: coefficient of correlation, t test for correlation coefficient, p-value

113. The sample correlation coefficient between X and Y is 0.375. It has been found out that the pvalue is 0.744 when testing H0 :  = 0 against the one-sided alternative H1 :   0 . To test H0 :  = 0 against the two-sided alternative H1 :   0 at a significance level of 0.2, the pvalue is

a) 0.744 / 2 b) 0.744 * 2 c) 1− 0.744 d) ( 1− 0.744 ) * 2

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ANSWER: d TYPE: MC DIFFICULTY: Difficult KEYWORDS: coefficient of correlation, t test for correlation coefficient, p-value

114. If you wanted to find out if alcohol consumption (measured in fluid oz.) and grade point average on a 4-point scale are linearly related, you would perform a a.  2 test for the difference in two proportions.

 2 test for independence.

a Z test for the difference in two proportions. d. a t test for a correlation coefficient. ANSWER: d TYPE: MC DIFFICULTY: Difficult KEYWORDS: coefficient of correlation, t test for correlation coefficient

115. True or False: When r = – 1, it indicates a perfect relationship between X and Y. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: coefficient of correlation, properties TABLE 13-7 An investment specialist claims that if one holds a portfolio that moves in the opposite direction to the market index like the S&P 500, then it is possible to reduce the variability of the portfolio's return. In other words, one can create a portfolio with positive returns but less exposure to risk. A sample of 26 years of S&P 500 index and a portfolio consisting of stocks of private prisons, which are believed to be negatively related to the S&P 500 index, is collected. A regression analysis was performed by regressing the returns of the prison stocks portfolio (Y) on the returns of S&P 500 index (X) to prove that the prison stocks portfolio is negatively related to the S&P 500 index at a 5% level of significance. The results are given in the following EXCEL output.

Intercept S&P

Coefficients Standard Error 4.866004258 0.35743609 -0.502513506 0.071597152

T Stat 13.61363441 -7.01862425

P-value 8.7932E-13 2.94942E-07

116. Referring to Table 13-7, to test whether the prison stocks portfolio is negatively related to the S&P 500 index, the appropriate null and alternative hypotheses are, respectively, a) H 0 :   0 vs. H1 :   0 b) H 0 :   0 vs. H1 :   0

H0 : r  0 vs. H1 : r  0 d) H 0 : r  0 vs. H1 : r  0 c)

ANSWER: a TYPE: MC DIFFICULTY: Easy KEYWORDS: coefficient of correlation, t test for correlation coefficient, form of hypothesis

199 Simple Linear Regression

117. Referring to Table 13-7, to test whether the prison stocks portfolio is negatively related to the S&P 500 index, the measured value of the test statistic is a) -7.019 b) -0.503 c) 0.072 d) 0.357 ANSWER: a TYPE: MC DIFFICULTY: Easy KEYWORDS: coefficient of correlation, t test for correlation coefficient, test statistic

118. Referring to Table 13-7, to test whether the prison stocks portfolio is negatively related to the S&P 500 index, the p-value of the associated test statistic is a) 2.94942E-07 b) 2.94942E-07 / 2 c)

( 2.94942E-07 ) 2

d) 8.7932E-13 ANSWER: b TYPE: MC DIFFICULTY: Easy KEYWORDS: coefficient of correlation, t test for correlation coefficient, p-value 119. Referring to Table 13-7, which of the following will be a correct conclusion? a) We cannot reject the null hypothesis and, therefore, conclude that there is sufficient evidence to show that the prisons stock portfolio and S&P 500 index are negatively related. b) We can reject the null hypothesis and, therefore, conclude that there is sufficient evidence to show that the prisons stock portfolio and S&P 500 index are negatively related. c) We cannot reject the null hypothesis and, therefore, conclude that there is not sufficient evidence to show that the prisons stock portfolio and S&P 500 index are negatively related. d) We can reject the null hypothesis and conclude that there is not sufficient evidence to show that the prisons stock portfolio and S&P 500 index are negatively related. ANSWER: b TYPE: MC DIFFICULTY: Easy KEYWORDS: coefficient of correlation, t test for correlation coefficient, decision, conclusion TABLE 13-8 It is believed that GPA (grade point average, based on a four point scale) should have a positive linear relationship with ACT scores. Given below is the Excel output from regressing GPA on ACT scores using a data set of 8 randomly chosen students from a Big-Ten university. Regressing GPA on ACT Regression Statistics

Simple Linear Regression

Multiple R R Square Adjusted R Square Standard Error Observations

200

0.7598 0.5774 0.5069 0.2691 8

ANOVA df Regression Residual Total

Intercept ACT

SS 1 6 7

0.5940 0.4347 1.0287

Coefficients Standard Error 0.5681 0.9284 0.1021 0.0356

MS 0.5940 0.0724

F Significance F 8.1986 0.0286

t Stat P-value 0.6119 0.5630 2.8633 0.0286

Lower 95% Upper 95% -1.7036 2.8398 0.0148 0.1895

120. Referring to Table 13-8, the interpretation of the coefficient of determination in this regression is a) 57.74% of the total variation of ACT scores can be explained by GPA. b) ACT scores account for 57.74% of the total fluctuation in GPA. c) GPA accounts for 57.74% of the variability of ACT scores. d) None of the above. ANSWER: b TYPE: MC DIFFICULTY: Easy KEYWORDS: coefficient of determination, interpretation 121. Referring to Table 13-8, the value of the measured test statistic to test whether there is any linear relationship between GPA and ACT is a) 0.0356 b) 0.1021 c) 0.7598 d) 2.8633 ANSWER: d TYPE: MC DIFFICULTY: Easy KEYWORDS: coefficient of correlation, t test for correlation coefficient, test statistic 122. Referring to Table 13-8, what is the predicted average value of GPA when ACT = 20? a. 2.61 b. 2.66 c. 2.80 d. 3.12 ANSWER: a TYPE: MC DIFFICULTY: Easy KEYWORDS: prediction of individual values

201 Simple Linear Regression 123. Referring to Table 13-8, what are the decision and conclusion on testing whether there is any linear relationship at 1% level of significance between GPA and ACT scores? a) Do not reject the null hypothesis; hence there is not sufficient evidence to show that ACT scores and GPA are linearly related. b) Reject the null hypothesis; hence there is not sufficient evidence to show that ACT scores and GPA are linearly related. c) Do not reject the null hypothesis; hence there is sufficient evidence to show that ACT scores and GPA are linearly related. d) Reject the null hypothesis; hence there is sufficient evidence to show that ACT scores and GPA are linearly related. ANSWER: a TYPE: MC DIFFICULTY: Easy KEYWORDS: coefficient of correlation, t test for correlation coefficient, decision, conclusion

124. Referring to Table 13-8, the value of the measured (observed) test statistic of the F-test for H0 :  = 0 vs. H1 :   0 a) may be negative. b) is always positive. c) is always negative. d) has the same sign as the corresponding t test statistic. ANSWER: b TYPE: MC DIFFICULTY: Easy KEYWORDS: F test on slope, test statistic TABLE 13-9 It is believed that, the average numbers of hours spent studying per day (HOURS) during undergraduate education should have a positive linear relationship with the starting salary (SALARY, measured in thousands of dollars per month) after graduation. Given below is the Excel output from regressing starting salary on number of hours spent studying per day for a sample of 51 students. NOTE: Some of the numbers in the output are purposely erased. Regression Statistics Multiple R R Square Adjusted R Square Standard Error Observations

0.8857 0.7845 0.7801 1.3704 51

ANOVA df Regression Residual Total

SS 335.0472

427.0798

Coefficients

Standard Error

MS 335.0473 1.8782

F Significance F 178.3859

t Stat

P-value

Lower 95%

Upper 95%

Simple Linear Regression

Intercept Hours

-1.8940 0.9795

0.4018 0.0733

-4.7134 13.3561

2.051E-05 5.944E-18

202

-2.7015 0.8321

125. Referring to Table 13-9, the estimated average change in salary (in thousands of dollars) as a result of spending an extra hour per day studying is a. -1.8940 b. 0.7845 c. 0.9795 d. 335.0473 ANSWER: c TYPE: MC DIFFICULTY: Moderate KEYWORDS: slope, interpretation

126. Referring to Table 13-9, the value of the measured t-test statistic to test whether average SALARY depends linearly on HOURS is a) -4.7134 b) -1.8940 c) 0.9795 d) 13.3561 ANSWER: d TYPE: MC DIFFICULTY: Moderate KEYWORDS: t test on slope, slope, test statistic

127. Referring to Table 13-9, the p-value of the measured F-test statistic to test whether HOURS affects SALARY is a) (5.944E-18)/2 b) 5.944E-18 c) (2.051E-05)/2 d) 2.051E-05 ANSWER: b TYPE: MC DIFFICULTY: Moderate KEYWORDS: F test on slope, slope, test statistic 128. Referring to Table 13-9, the degrees of freedom for testing whether HOURS affects SALARY are a) 1, 49 b) 1, 50 c) 49, 1 d) 50, 1 ANSWER: a TYPE: MC DIFFICULTY: Moderate KEYWORDS: F test on slope, slope, degrees of freedom 129. Referring to Table 13-9, the error sum of squares (SSE) of the above regression is a) 1.878215

-1.0865 1.1269

203 Simple Linear Regression b) 92.0325465 c) 335.047257 d) 427.079804 ANSWER: b TYPE: MC DIFFICULTY: Moderate KEYWORDS: sum of squares 130. Referring to Table 13-9, the 90% confidence interval for the average change in SALARY (in thousands of dollars) as a result of spending an extra hour per day studying is a) wider than [-2.70159, -1.08654]. b) narrower than [-2.70159, -1.08654]. c) wider than [0.8321927, 1.12697]. d) narrower than [0.8321927, 1.12697]. ANSWER: d TYPE: MC DIFFICULTY: Moderate KEYWORDS: slope, interpretation, confidence interval

131. Referring to Table 13-9, to test the claim that average SALARY depends positively on HOURS against the null hypothesis that average SALARY does not depend linearly on HOURS, the p-value of the test statistic is a) (5.944E-18)/2 b) 5.944E-18 c) (2.051E-05)/2 d) 2.051E-05 ANSWER: a TYPE: MC DIFFICULTY: Moderate KEYWORDS: t test on slope, slope, p-value 132. True or False: A zero population correlation coefficient between a pair of random variables means that there is no linear relationship between the random variables. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: coefficient of correlation, interpretation

133. True or False: You give a pre-employment examination to your applicants. The test is scored from 1 to 100. You have data on their sales at the end of one year measured in dollars. You want to know if there is any linear relationship between pre-employment examination score and sales. An appropriate test to use is the t test on the population correlation coefficient. ANSWER: True TYPE: MC DIFFICULTY: Moderate KEYWORDS: coefficient of correlation, t test for correlation coefficient

134. The width of the prediction interval for the predicted value of Y is dependent on

Simple Linear Regression

204

a) the standard error of the estimate. b) the value of X for which the prediction is being made. c) the sample size. d) All of the above. ANSWER: d TYPE: MC DIFFICULTY: Moderate KEYWORDS: prediction of individual values, prediction interval, properties

135. True or False: The confidence interval for the mean of Y is always narrower than the prediction interval for an individual response Y given the same data set, X value, and confidence level. ANSWER: True TYPE: TF DIFFICULTY: Moderate KEYWORDS: estimation of mean values, confidence interval, prediction of individual values, prediction interval, properties Table 13-10 The management of a chain electronic store would like to develop a model for predicting the weekly sales (in thousand of dollars) for individual stores based on the number of customers who made purchases. A random sample of 12 stores yields the following results:

Customers Sales (Thousands of Dollars) 907 11.20 926 11.05 713 8.21 741 9.21 780 9.42 898 10.08 510 6.73 529 7.02 460 6.12 872 9.52 650 7.53 603 7.25 136. Referring to Table 13-10, generate the scatter plot.

205 Simple Linear Regression ANSWER: Scatter Diagram

Salses (Thousands of Dollars)

12 10 8 6 4 2 0 0

200

400

600

800

1000

Customers

TYPE: PR DIFFICULTY: Moderate KEYWORDS: scatter plot 137. Referring to Table 13-10, what are the values of the estimated intercept and slope? ANSWER: 1.4464, 0.0100 TYPE: PR DIFFICULTY: Moderate KEYWORDS: intercept, slope, estimation 138. Referring to Table 13-10, what is the value of the coefficient of determination? ANSWER: 0.9453 TYPE: PR DIFFICULTY: Moderate KEYWORDS: coefficient of determination 139. Referring to Table 13-10, what is the value of the coefficient of correlation? ANSWER: 0.9723 TYPE: PR DIFFICULTY: Moderate KEYWORDS: coefficient of correlation 140. Referring to Table 13-10, what is the value of the standard error of the estimate? ANSWER: 0.4191 TYPE: PR DIFFICULTY: Moderate KEYWORDS: standard error of estimate 141. Referring to Table 13-10, which is the correct null hypothesis for testing whether the number of customers who make purchase affects weekly sales? a) H0 : 0 = 0 b) H0 : 1 = 0 c) H0 :  = 0

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206

d) H0 :  = 0 ANSWER: b TYPE: MC DIFFICULTY: Easy KEYWORDS: t test on slope, slope, form of hypothesis

142. Referring to Table 13-10, what is the value of the t test statistic when testing whether the number of customers who make purchase affects weekly sales? ANSWER: 13.1464 TYPE: PR DIFFICULTY: Moderate KEYWORDS: t test on slope, slope, test statistic

143. Referring to Table 13-10, what are the degrees of freedom of the t test statistic when testing whether the number of customers who make purchase affects weekly sales? ANSWER: 10 TYPE: PR DIFFICULTY: Easy KEYWORDS: t test on slope, slope, degrees of freedom

144. Referring to Table 13-10, what is the p-value of the t test statistic when testing whether the number of customers who make purchase affects weekly sales? ANSWER: 1.23323E-07 or smaller than 0.01 TYPE: PR DIFFICULTY: Moderate KEYWORDS: t test on slope, slope, p-value 145. True or False: Referring to Table 13-10, the null hypothesis for testing whether the number of customers who make purchase effects weekly sales cannot be rejected if 1% probability of committing a type I error is desired. ANSWER: False TYPE: TF DIFFICULTY: Moderate KEYWORDS: t test on slope, slope, decision 146. True or False: Referring to Table 13-10, the average weekly sales will increase by an estimated $0.01 for each additional purchasing customer. ANSWER: False TYPE: TF DIFFICULTY: Moderate KEYWORDS: t test on slope, slope, interpretation 147. True or False: Referring to Table 13-10, the average weekly sales will increase by an estimated $10 for each additional purchasing customer. ANSWER: True

207 Simple Linear Regression TYPE: TF DIFFICULTY: Moderate KEYWORDS: t test on slope, slope, interpretation 148. True or False: Referring to Table 13-10, 93.98% of the total variation in weekly sales can be explained by the variation in the number of customers who make purchases. ANSWER: False TYPE: TF DIFFICULTY: Moderate KEYWORDS: coefficient of determination, interpretation

149. Referring to Table 13-10, what are the degrees of freedom of the F test statistic when testing whether the number of customers who make purchases is a good predictor for weekly sales? ANSWER: 1 in the numerator and 10 in the denominator TYPE: PR DIFFICULTY: Easy KEYWORDS: F test on slope, slope, test statistic

150. Referring to Table 13-10, what is the value of the F test statistic when testing whether the number of customers who make purchases is a good predictor for weekly sales? ANSWER: 172.8265 TYPE: PR DIFFICULTY: Moderate KEYWORDS: F test on slope, slope, test statistic

151. Referring to Table 13-10, what is the p-value of the F test statistic when testing whether the number of customers who make purchases is a good predictor for weekly sales? ANSWER: 1.23323E-07 or smaller than 0.01 TYPE: PR DIFFICULTY: Moderate KEYWORDS: F test on slope, slope, p-value

152. True or False: Referring to Table 13-10, the p-value of the t test and F test should be the same when testing whether the number of customers who make purchases is a good predictor for weekly sales. ANSWER: True TYPE: TF DIFFICULTY: Moderate KEYWORDS: t test on slope, F test on slope, p-value

153. True or False: Referring to Table 13-10, the value of the t test statistic and F test statistic should be the same when testing whether the number of customers who make purchases is a good predictor for weekly sales. ANSWER: False TYPE: TF DIFFICULTY: Moderate KEYWORDS: t test on slope, F test on slope, test statistic

Simple Linear Regression

208

154. True or False: Referring to Table 13-10, the value of the F test statistic equals the square of the t test statistic when testing whether the number of customers who make purchases is a good predictor for weekly sales. ANSWER: True TYPE: TF DIFFICULTY: Moderate KEYWORDS: t test on slope, F test on slope, test statistic 155. Referring to Table 13-10, generate the residual plot. ANSWER: Customers Residual Plot 0.8 0.6

Residuals

0.4 0.2 0 -0.2 -0.4 -0.6 -0.8 0

200

400

600

800

1000

Customers

TYPE: PR DIFFICULTY: Moderate KEYWORDS: residual plot 156.Referring to Table 13-10, the residual plot indicates possible violation of which assumptions? a) Linearity of the relationship b) Homoscedasticity c) Autocorrelation d) Normality ANSWER: b TYPE: MC DIFFICULTY: Moderate KEYWORDS: homoscedasticity, residual plot, assumption 157.True or False: Referring to Table 13-10, it is inappropriate to compute the Durbin-Watson statistic and test for autocorrelation in this case. ANSWER: True TYPE: TF DIFFICULTY: Moderate KEYWORDS: Durbin-Watson statistic, autocorrelation 158.Referring to Table 13-10, construct a 95% confidence interval for the change in average weekly sales when the number of customers who make purchases increases by one.

209 Simple Linear Regression ANSWER: 0.0083 to 0.0117 TYPE: PR DIFFICULTY: Difficult KEYWORDS: slope, confidence interval, interpretation 159.Referring to Table 13-10, construct a 95% confidence interval for the average weekly sales when the number of customers who make purchases is 600. ANSWER: 7.1194 to 7.7864 thousands of dollars TYPE: PR DIFFICULTY: Difficult KEYWORDS: estimate of mean values, confidence interval 160.Referring to Table 13-10, construct a 95% prediction interval for the weekly sales of a store that has 600 purchasing customers. ANSWER: 6.4614 to 8.4444 thousands of dollars TYPE: PR DIFFICULTY: Difficult KEYWORDS: prediction of individual values, prediction interval TABLE 13-11 A company that has the distribution rights to home video sales of previously released movies would like to use the box office gross (in millions of dollars) to estimate the number of units (in thousands of units) that it can expect to sell. Following is the output from a simple linear regression along with the residual plot and normal probability plot obtained from a data set of 30 different movie titles: Regression Statistics Multiple R 0.8531 R Square 0.7278 0.7180 Adjusted R Square 47.8668 Standard Error Observations 30 ANOVA df Regression Residual Total

Intercept Gross

1 28 29

171499.78 64154.42 235654.20

171499.78 2291.23

Coefficients

Standard Error 11.8318 0.5008

76.5351 4.3331

F 74.8505

Significance F 2.1259E-09

t Stat

P-value

Lower 95%

6.4686 8.6516

5.24E-07 2.13E-09

52.2987 3.3072

Upper 95% 100.7716 5.3590

Simple Linear Regression

210

Gross Residual Plot 150

Residuals

100 50 0 -50 -100 0

Gross

Normal Probability Plot

Residuals

120 100 80 60 40 20 0 -20-2.5

-2

-1.5

-1

-0.5

0.5

1.5

2.5

-40 -60 -80 -100

Z Value

161.Referring to table 13-11, which of the following is the correct interpretation for the slope coefficient? a) For each increase of 1 million dollars in box office gross, expected home video units sold are estimated to increase by 4.3331 units. b) For each increase of 1 million dollars in box office gross, expected home video units sold are estimated to increase by 4.3331 thousand units. c) For each increase of 1 dollar in box office gross, expected home video units sold are estimated to increase by 4.3331 units. d) For each increase of 1 dollar in box office gross, expected home video units sold are estimated to increase by 4.3331 thousand units. ANSWER: b TYPE: MC DIFFICULTY: Easy KEYWORDS: slope, interpretation 162.Referring to Table 13-11, predict the video unit sales for a movie that had a box office gross of $30 millions. ANSWER: 206.53 thousand units TYPE: PR DIFFICULTY: Easy KEYWORDS: prediction of individual values

211 Simple Linear Regression 163.Referring to Table 13-11, which of the following is the correct interpretation for the coefficient of determination? a) 71.8% of the variation in the video unit sales can be explained by the variation in the box office gross. b) 72.8% of the variation in the video unit sales can be explained by the variation in the box office gross. c) 71.8% of the variation in the box office gross can be explained by the variation in the video unit sales. d) 72.8% of the variation in the box office gross can be explained by the variation in the video unit sales. ANSWER: b TYPE: MC DIFFICULTY: Easy KEYWORDS: interpretation, coefficient of determination 164. Referring to Table 13-11, what is the standard error of estimate? ANSWER: 47.8668 thousand units TYPE: PR DIFFICULTY: Moderate KEYWORDS: standard error of estimate 165. Referring to Table 13-11, what is the standard deviation around the regression line? ANSWER: 47.8668 thousand units TYPE: PR DIFFICULTY: Difficult KEYWORDS: interpretation, standard error of estimate 166. Referring to Table 13-11, which of the following assumptions appears to have been violated? a) Normality of error b) Homoscedasticity c) Independence of errors d) None of the above ANSWER: d TYPE: MC DIFFICULTY: Moderate KEYWORDS: interpretation, assumption, residual plot, homoscedasticity 167.True or False: Referring to Table 13-11, the normality of error assumption appears to have been violated. ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: interpretation, assumption 168.True or False: Referring to Table 13-11, the homoscedasticity of error assumption appears to have been violated. ANSWER:

Simple Linear Regression

212

False TYPE: TF DIFFICULTY: Easy KEYWORDS: interpretation, assumption, homoscedasticity 169.True or False: Referring to Table 13-11, there appears to be autocorrelation in the residuals. ANSWER: False TYPE: TF DIFFICULTY: Moderate KEYWORDS: interpretation, autocorrelation 170.True or False: Referring to Table 13-11, the Durbin-Watson statistic is inappropriate for this data set. ANSWER: True TYPE: TF DIFFICULTY: Moderate KEYWORDS: Durbin-Watson statistic, assumption 171.True or False: Referring to Table 13-11, the null hypothesis for testing whether there is a linear relationship between box office gross and home video unit sales is “There is no linear relationship between box office gross and home video unit sales”. ANSWER: True TYPE: TF DIFFICULTY: Moderate KEYWORDS: slope, t test for slope, form of hypothesis 172.Referring to Table 13-11, which of the following is the correct null hypothesis for testing whether there is a linear relationship between box office gross and home video unit sales? a) H0 : b1 = 0 b) H0 : b1  0 c) H0 : 1 = 0 d) H0 : 1  0 ANSWER: c TYPE: MC DIFFICULTY: Easy KEYWORDS: slope, t test for slope, form of hypothesis 173.Referring to Table 13-11, which of the following is the correct alternative hypothesis for testing whether there is a linear relationship between box office gross and home video unit sales? a) H1 : b1 = 0 b) H1 : b1  0 c) H1 : 1 = 0 d) H1 : 1  0 ANSWER: d TYPE: MC DIFFICULTY: Easy

213 Simple Linear Regression KEYWORDS: slope, t test for slope, form of hypothesis 174.Referring to Table 13-11, what is the value of the test statistic for testing whether there is a linear relationship between box office gross and home video unit sales? ANSWER: 8.65 TYPE: PR DIFFICULTY: Easy KEYWORDS: test statistic, slope, t test for slope 175.Referring to Table 13-11, what is the critical value for testing whether there is a linear relationship between box office gross and home video unit sales at a 5% level of significance? ANSWER: 2.0484 TYPE: PR DIFFICULTY: Easy KEYWORDS: critical value, slope, t test for slope

176. Referring to Table 13-11, what is the p-value for testing whether there is a linear relationship between box office gross and home video unit sales at a 5% level of significance? ANSWER: virtually zero or 2.13E-09 TYPE: PR DIFFICULTY: Easy KEYWORDS: p-value , slope, t test for slope

177. True or False: Referring to Table 13-11, the null hypothesis that there is no linear relationship between box office gross and home video unit sales should be reject at a 5% level of significance. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: decision, slope, t test for slope

178. True or False: Referring to Table 13-11, there is sufficient evidence that box office gross and home video unit sales are linearly related at a 5% level of significance. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: conclusion, slope, t test for slope 179.Referring to Table 13-11, what are, respectively, the lower and upper limits of the 95% confidence interval estimate for population slope? ANSWER: 3.3072 and 5.3590 TYPE: PR DIFFICULTY: Easy KEYWORDS: slope, confidence interval 180.Referring to Table 13-11, what are, respectively, the lower and upper limits of the 95% confidence interval estimate for the average change in video unit sales as a result of a one million dollars increase in box office?

Simple Linear Regression

ANSWER: 3.3072 and 5.3590 thousand units TYPE: PR DIFFICULTY: Difficult KEYWORDS: slope, confidence interval

214

221 Multiple Regression Models

CHAPTER 14: INTRODUCTION TO MULTIPLE REGRESSION 1. In a multiple regression problem involving two independent variables, if b1 is computed to be +2.0, it means that a) the relationship between X1 and Y is significant. b) the estimated average of Y increases by 2 units for each increase of 1 unit of X1, holding X2 constant. c) the estimated average of Y increases by 2 units for each increase of 1 unit of X1, without regard to X2. d) the estimated average of Y is 2 when X1 equals zero. ANSWER: b TYPE: MC DIFFICULTY: Easy KEYWORDS: slope, interpretation

2. The coefficient of multiple determination r2Y.12 a) measures the variation around the predicted regression equation. b) measures the proportion of variation in Y that is explained by X1 and X2. c) measures the proportion of variation in Y that is explained by X1 holding X2 constant. d) will have the same sign as b1. ANSWER: b TYPE: MC DIFFICULTY: Easy KEYWORDS: coefficient of multiple determination, interpretation 3. In a multiple regression model, the value of the coefficient of multiple determination a) has to fall between -1 and +1. b) has to fall between 0 and +1. c) has to fall between -1 and 0. d) can fall between any pair of real numbers. ANSWER: b TYPE: MC DIFFICULTY: Easy KEYWORDS: coefficient of multiple determination, properties

4. In a multiple regression model, which of the following is correct regarding the value of the adjusted r 2 ? a) b) c) d)

It can be negative. It has to be positive. It has to be larger than the coefficient of multiple determination. It can be larger than 1.

ANSWER: a TYPE: MC DIFFICULTY: Moderate

220

Multiple Regression Models KEYWORDS: adjusted r-square, properties TABLE 14-1 A manager of a product sales group believes the number of sales made by an employee (Y) depends on how many years that employee has been with the company (X1) and how he/she scored on a business aptitude test (X2). A random sample of 8 employees provides the following: Employee Y X1 X2 1 100 10 7 2 90 3 10 3 80 8 9 4 70 5 4 5 60 5 8 6 50 7 5 7 40 1 4 8 30 1 1

5. Referring to Table 14-1, for these data, what is the value for the regression constant, b0? a) 0.998 b) 3.103 c) 4.698 d) 21.293 ANSWER: d TYPE: MC DIFFICULTY: Moderate KEYWORDS: intercept, estimation

6. Referring to Table 14-1, for these data, what is the estimated coefficient for the variable representing years an employee has been with the company, b1? a) 0.998 b) 3.103 c) 4.698 d) 21.293 ANSWER: b TYPE: MC DIFFICULTY: Moderate KEYWORDS: slope, estimation

7. Referring to Table 14-1, for these data, what is the estimated coefficient for the variable representing scores on the aptitude test, b2? a) 0.998 b) 3.103 c) 4.698 d) 21.293 ANSWER: c TYPE: MC DIFFICULTY: Moderate KEYWORDS: slope, estimation 8. Referring to Table 14-1, if an employee who had been with the company 5 years scored a 9 on the aptitude test, what would his estimated expected sales be?

Multiple Regression Models

222

a) 79.09 b) 60.88 c) 55.62 d) 17.98 ANSWER: a TYPE: MC DIFFICULTY: Moderate KEYWORDS: estimation of mean values TABLE 14-2 A professor of industrial relations believes that an individual’s wage rate at a factory (Y) depends on his performance rating (X1) and the number of economics courses the employee successfully completed in college (X2). The professor randomly selects 6 workers and collects the following information: Employee Y ($) X1 X2 1 10 3 0 2 12 1 5 3 15 8 1 4 17 5 8 5 20 7 12 6 25 10 9

9. Referring to Table 14-2, for these data, what is the value for the regression constant, b0? a) 0.616 b) 1.054 c) 6.932 d) 9.103 ANSWER: c TYPE: MC DIFFICULTY: Moderate KEYWORDS: intercept, estimation

10. Referring to Table 14-2, for these data, what is the estimated coefficient for performance rating, b1? a) 0.616 b) 1.054 c) 6.932 d) 9.103 ANSWER: b TYPE: MC DIFFICULTY: Moderate KEYWORDS: slope, estimation

11. Referring to Table 14-2, for these data, what is the estimated coefficient for the number of economics courses taken, b2? a) 0.616 b) 1.054 c) 6.932 d) 9.103

223 Multiple Regression Models

ANSWER: a TYPE: MC DIFFICULTY: Moderate KEYWORDS: slope, estimation 12. Referring to Table 14-2, suppose an employee had never taken an economics course and managed to score a 5 on his performance rating. What is his estimated expected wage rate? a) 10.90 b) 12.20 c) 17.23 d) 25.11 ANSWER: b TYPE: MC DIFFICULTY: Moderate KEYWORDS: estimation of mean values 13. Referring to Table 14-2, an employee who took 12 economics courses scores 10 on the performance rating. What is her estimated expected wage rate? a) 10.90 b) 12.20 c) 24.87 d) 25.70 ANSWER: c TYPE: MC DIFFICULTY: Moderate KEYWORDS: estimation of mean values 14. The variation attributable to factors other than the relationship between the independent variables and the explained variable in a regression analysis is represented by a) regression sum of squares. b) error sum of squares. c) total sum of squares. d) regression mean squares. ANSWER: b TYPE: MC DIFFICULTY: Moderate KEYWORDS: sum of squares TABLE 14-3 An economist is interested to see how consumption for an economy (in $ billions) is influenced by gross domestic product ($ billions) and aggregate price (consumer price index). The Microsoft Excel output of this regression is partially reproduced below. SUMMARY OUTPUT Regression Statistics Multiple R 0.991 R Square 0.982 Adjusted R Square 0.976

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Standard Error Observations

224

0.299 10

ANOVA Regression Residual Total Intercept GDP Price

df 2 7 9 Coeff – 0.0861 0.7654 – 0.0006

SS 33.4163 0.6277 34.0440

MS 16.7082 0.0897

StdError 0.5674 0.0574 0.0028

t Stat – 0.152 13.340 – 0.219

F 186.325

Signif F 0.0001

P-value 0.8837 0.0001 0.8330

15. Referring to Table 14-3, when the economist used a simple linear regression model with consumption as the dependent variable and GDP as the independent variable, he obtained an r2 value of 0.971. What additional percentage of the total variation of consumption has been explained by including aggregate prices in the multiple regression? a) 98.2 b) 11.1 c) 2.8 d) 1.1 ANSWER: d TYPE: MC DIFFICULTY: Moderate KEYWORDS: coefficient of multiple determination

16. Referring to Table 14-3, the p-value for GDP is a) 0.05 b) 0.01 c) 0.001 d) None of the above. ANSWER: d TYPE: MC DIFFICULTY: Easy KEYWORDS: t test on slope, p-value

17. Referring to Table 14-3, the p-value for the aggregated price index is a) 0.05 b) 0.01 c) 0.001 d) None of the above. ANSWER: d TYPE: MC DIFFICULTY: Easy KEYWORDS: t test on slope, p-value

18. Referring to Table 14-3, the p-value for the regression model as a whole is a) 0.05 b) 0.01

225 Multiple Regression Models c) 0.001 d) None of the above. ANSWER: d TYPE: MC DIFFICULTY: Easy KEYWORDS: F test for the entire regression, p-value 19. Referring to Table 14-3, what is the predicted consumption level for an economy with GDP equal to $4 billion and an aggregate price index of 150? a) $1.39 billion b) $2.89 billion c) $4.75 billion d) $9.45 billion ANSWER: b TYPE: MC DIFFICULTY: Moderate KEYWORDS: prediction of individual values 20. Referring to Table 14-3, what is the estimated average consumption level for an economy with GDP equal to $4 billion and an aggregate price index of 150? a) $1.39 billion b) $2.89 billion c) $4.75 billion d) $9.45 billion ANSWER: b TYPE: MC DIFFICULTY: Moderate KEYWORDS: estimation of mean values 21. Referring to Table 14-3, what is the estimated average consumption level for an economy with GDP equal to $2 billion and an aggregate price index of 90? a) $1.39 billion b) $2.89 billion c) $4.75 billion d) $9.45 billion ANSWER: a TYPE: MC DIFFICULTY: Moderate KEYWORDS: estimation of mean values 22. Referring to Table 14-3, one economy in the sample had an aggregate consumption level of $3 billion, a GDP of $3.5 billion, and an aggregate price level of 125. What is the residual for this data point? a) $2.52 billion b) $0.48 billion c) – $1.33 billion d) – $2.52 billion ANSWER: b

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TYPE: MC DIFFICULTY: Moderate KEYWORDS: residual 23. Referring to Table 14-3, one economy in the sample had an aggregate consumption level of $4 billion, a GDP of $6 billion, and an aggregate price level of 200. What is the residual for this data point? a) $4.39 billion b) $0.39 billion c) – $0.39 billion d) – $1.33 billion ANSWER: c TYPE: MC DIFFICULTY: Moderate KEYWORDS: residual

24. Referring to Table 14-3, to test for the significance of the coefficient on aggregate price index, the value of the relevant t-statistic is a) 2.365 b) 0.143 c) – 0.219 d) – 1.960 ANSWER: c TYPE: MC DIFFICULTY: Easy KEYWORDS: t test on slope, test statistic

25. Referring to Table 14-3, to test for the significance of the coefficient on aggregate price index, the p-value is a) 0.0001 b) 0.8330 c) 0.8837 d) 0.9999 ANSWER: b TYPE: MC DIFFICULTY: Easy KEYWORDS: t test on slope, p-value

26. Referring to Table 14-3, to test for the significance of the coefficient on gross domestic product, the p-value is a) 0.0001 b) 0.8330 c) 0.8837 d) 0.9999 ANSWER: a TYPE: MC DIFFICULTY: Easy KEYWORDS: t test on slope, p-value

227 Multiple Regression Models

27. Referring to Table 14-3, to test whether aggregate price index has a negative impact on consumption, the p-value is a) 0.0001 b) 0.4165 c) 0.8330 d) 0.8837

ANSWER: b TYPE: MC DIFFICULTY: Moderate KEYWORDS: t test on slope, p-value

28. Referring to Table 14-3, to test whether aggregate price index has a positive impact on consumption, the p-value is a) 0.0001 b) 0.4165 c) 0.5835 d) 0.8330 ANSWER: c TYPE: MC DIFFICULTY: Moderate KEYWORDS: t test on slope, p-value

29. Referring to Table 14-3, to test whether gross domestic product has a positive impact on consumption, the p-value is a) 0.00005 b) 0.0001 c) 0.9999 d) 0.99995 ANSWER: a TYPE: MC DIFFICULTY: Easy KEYWORDS: t test on slope, p-value TABLE 14-4 A real estate builder wishes to determine how house size (House) is influenced by family income (Income), family size (Size), and education of the head of household (School). House size is measured in hundreds of square feet, income is measured in thousands of dollars, and education is in years. The builder randomly selected 50 families and ran the multiple regression. Microsoft Excel output is provided below: SUMMARY OUTPUT Regression Statistics Multiple R 0.865 R Square 0.748 Adjusted R Square 0.726 Standard Error 5.195 Observations 50 ANOVA

Multiple Regression Models

df Regression Residual Total

Intercept Income Size School

Coeff – 1.6335 0.4485 4.2615 – 0.6517

SS 3605.7736 1214.2264 4820.0000 StdError 5.8078 0.1137 0.8062 0.4319

MS 901.4434 26.9828

t Stat – 0.281 3.9545 5.286 – 1.509

P-value 0.7798 0.0003 0.0001 0.1383

228

Signif F 0.0001

30. Referring to Table 14-4, what fraction of the variability in house size is explained by income, size of family, and education? a) 27.0% b) 33.4% c) 74.8% d) 86.5% ANSWER: c TYPE: MC DIFFICULTY: Easy KEYWORDS: coefficient of multiple determination, interpretation 31. Referring to Table 14-4, which of the independent variables in the model are significant at the 2% level? a) Income, Size, School b) Income, Size c) Size, School d) Income, School ANSWER: b TYPE: MC DIFFICULTY: Easy KEYWORDS: t test on slope, critical value, p-value, decision, conclusion

32. Referring to Table 14-4, when the builder used a simple linear regression model with house size (House) as the dependent variable and education (School) as the independent variable, he obtained an r2 value of 23.0%. What additional percentage of the total variation in house size has been explained by including family size and income in the multiple regression? a) 2.8% b) 51.8% c) 72.6% d) 74.8% ANSWER: b TYPE: MC DIFFICULTY: Moderate KEYWORDS: coefficient of multiple determination 33. Referring to Table 14-4, which of the following values for the level of significance is the smallest for which all explanatory variables are significant individually? a) 0.01 b) 0.025

229 Multiple Regression Models c) 0.05 d) 0.15 ANSWER: d TYPE: MC DIFFICULTY: Moderate KEYWORDS: p-value, interpretation 34. Referring to Table 14-4, which of the following values for the level of significance is the smallest for which at least two explanatory variables are significant individually? a) 0.01 b) 0.025 c) 0.05 d) 0.15 ANSWER: a TYPE: MC DIFFICULTY: Moderate KEYWORDS: p-value, interpretation 35. Referring to Table 14-4, which of the following values for the level of significance is the smallest for which the regression model as a whole is significant? a) 0.00005 b) 0.001 c) 0.01 d) 0.05 ANSWER: b TYPE: MC DIFFICULTY: Moderate KEYWORDS: p-value, interpretation 36. Referring to Table 14-4, what is the predicted house size (in hundreds of square feet) for an individual earning an annual income of $40,000, having a family size of 4, and going to school a total of 13 years? a) 11.43 b) 15.15 c) 24.88 d) 53.87 ANSWER: c TYPE: MC DIFFICULTY: Moderate KEYWORDS: prediction of individual values 37. Referring to Table 14-4, what minimum annual income would an individual with a family size of 4 and 16 years of education need to attain a predicted 10,000 square foot home (House = 100)? a) $44.14 thousand b) $56.75 thousand c) $178.33 thousand d) $211.85 thousand ANSWER: d

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TYPE: MC DIFFICULTY: Moderate KEYWORDS: prediction of individual values 38. Referring to Table 14-4, what minimum annual income would an individual with a family size of 9 and 10 years of education need to attain a predicted 5,000 square foot home (House = 50)? a) $44.14 thousand b) $56.75 thousand c) $178.33 thousand d) $211.85 thousand ANSWER: a TYPE: MC DIFFICULTY: Moderate KEYWORDS: prediction of individual values 39. Referring to Table 14-4, one individual in the sample had an annual income of $100,000, a family size of 10, and an education of 16 years. This individual owned a home with an area of 7,000 square feet (House = 70.00). What is the residual (in hundreds of square feet) for this data point? a) 7.40 b) 2.52 c) – 2.52 d) – 5.40 ANSWER: d TYPE: MC DIFFICULTY: Moderate KEYWORDS: residual 40. Referring to Table 14-4, one individual in the sample had an annual income of $10,000, a family size of 1, and an education of 8 years. This individual owned a home with an area of 1,000 square feet (House = 10.00). What is the residual (in hundreds of square feet) for this data point? a) 8.10 b) 5.40 c) – 5.40 d) – 8.10 ANSWER: a TYPE: MC DIFFICULTY: Moderate KEYWORDS: residual

41. Referring to Table 14-4, suppose the builder wants to test whether the coefficient on Income is significantly different from 0. What is the value of the relevant t-statistic? a) 5.286 b) 5.195 c) 3.945 d) – 1.509 ANSWER: c TYPE: MC DIFFICULTY: Easy KEYWORDS: t test on slope, test statistic

231 Multiple Regression Models 42. Referring to Table 14-4, at the 0.01 level of significance, what conclusion should the builder draw regarding the inclusion of Income in the regression model? a) Income is significant in explaining house size and should be included in the model because its p-value is less than 0.01. b) Income is significant in explaining house size and should be included in the model because its p-value is more than 0.01. c) Income is not significant in explaining house size and should not be included in the model because its p-value is less than 0.01. d) Income is not significant in explaining house size and should not be included in the model because its p-value is more than 0.01. ANSWER: a TYPE: MC DIFFICULTY: Moderate KEYWORDS: t test on slope, decision, conclusion

43. Referring to Table 14-4, suppose the builder wants to test whether the coefficient on School is significantly different from 0. What is the value of the relevant t-statistic? a) 5.286 b) 5.195 c) 3.945 d) – 1.509 ANSWER: d TYPE: MC DIFFICULTY: Easy KEYWORDS: t test on slope, test statistic

44. Referring to Table 14-4, what is the value of the calculated F test statistic that is missing from the output for testing whether the whole regression model is significant? a) 0.0001 b) 0.0299 c) 0.726 d) 33.408 ANSWER: d TYPE: MC DIFFICULTY: Moderate KEYWORDS: F test on the entire regression, test statistic

45. Referring to Table 14-4, the observed value of the F-statistic is missing from the printout. What are the degrees of freedom for this F-statistic? a) 46 for the numerator, 3 for the denominator b) 3 for the numerator, 49 for the denominator c) 46 for the numerator, 49 for the denominator d) 3 for the numerator, 46 for the denominator ANSWER: d TYPE: MC DIFFICULTY: Easy KEYWORDS: F test on the entire regression, degrees of freedom

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232

46. Referring to Table 14-4, at the 0.01 level of significance, what conclusion should the builder draw regarding the inclusion of School in the regression model? a) School is significant in explaining house size and should be included in the model because its p-value is less than 0.01. b) School is significant in explaining house size and should be included in the model because its p-value is more than 0.01. c) School is not significant in explaining house size and should not be included in the model because its p-value is less than 0.01. d) School is not significant in explaining house size and should not be included in the model because its p-value is more than 0.01. ANSWER: d TYPE: MC DIFFICULTY: Moderate KEYWORDS: t test on slope, critical value, p-value, decision, conclusion 47. Referring to Table 14-4, what are the regression degrees of freedom that are missing from the output? a) 3 b) 46 c) 49 d) 50 ANSWER: a TYPE: MC DIFFICULTY: Easy KEYWORDS: degrees of freedom 48. Referring to Table 14-4, what are the residual degrees of freedom that are missing from the output? a) 3 b) 46 c) 49 d) 50 ANSWER: b TYPE: MC DIFFICULTY: Easy KEYWORDS: degrees of freedom TABLE 14-5 A microeconomist wants to determine how corporate sales are influenced by capital and wage spending by companies. She proceeds to randomly select 26 large corporations and record information in millions of dollars. The Microsoft Excel output below shows results of this multiple regression. SUMMARY OUTPUT Regression Statistics Multiple R 0.830 R Square 0.689 Adjusted R Square 0.662

233 Multiple Regression Models Standard Error Observations

17501.643 26

ANOVA Regression Residual Total

df 2 23 25

SS 15579777040 7045072780 22624849820

Intercept Capital Wages

Coeff 15800.0000 0.1245 7.0762

StdError 6038.2999 0.2045 1.4729

MS 7789888520 306307512 t Stat 2.617 0.609 4.804

F 25.432

Signif F 0.0001

P-value 0.0154 0.5485 0.0001

49. Referring to Table 14-5, what fraction of the variability in sales is explained by spending on capital and wages? a) 27.0% b) 50.9% c) 68.9% d) 83.0% ANSWER: c TYPE: MC DIFFICULTY: Easy KEYWORDS: coefficient of multiple determination, interpretation 50. Referring to Table 14-5, which of the independent variables in the model are significant at the 5% level? a) Capital, Wages b) Capital c) Wages d) None of the above ANSWER: c TYPE: MC DIFFICULTY: Easy KEYWORDS: t test on slope, critical value, p-value, decision, conclusion

51. Referring to Table 14-5, when the microeconomist used a simple linear regression model with sales as the dependent variable and wages as the independent variable, he obtained an r2 value of 0.601. What additional percentage of the total variation of sales has been explained by including capital spending in the multiple regression? a) 60.1% b) 31.1% c) 22.9% d) 8.8% ANSWER: d TYPE: MC DIFFICULTY: Moderate KEYWORDS: coefficient of multiple determination

52. Referring to Table 14-5, what is the p-value for Wages?

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234

a) 0.01 b) 0.05 c) 0.0001 d) None of the above ANSWER: c TYPE: MC DIFFICULTY: Easy KEYWORDS: t test on slope, p-value

53. Referring to Table 14-5, what is the p-value for testing whether Wages have a positive impact on corporate sales? a) 0.01 b) 0.05 c) 0.0001 d) 0.00005 ANSWER: d TYPE: MC DIFFICULTY: Easy KEYWORDS: t test on slope, p-value

54. Referring to Table 14-5, what is the p-value for testing whether Wages have a negative impact on corporate sales? a) 0.05 b) 0.0001 c) 0.00005 d) 0.99995 ANSWER: d TYPE: MC DIFFICULTY: Moderate KEYWORDS: t test on slope, p-value

55. Referring to Table 14-5, what is the p-value for Capital? a) 0.01 b) 0.025 c) 0.05 d) None of the above ANSWER: d TYPE: MC DIFFICULTY: Easy KEYWORDS: t test on slope, p-value

56. Referring to Table 14-5, what is the p-value for testing whether Capital has a positive influence on corporate sales? a) 0.025 b) 0.05 c) 0.2743 d) 0.5485 ANSWER:

235 Multiple Regression Models c TYPE: MC DIFFICULTY: Easy KEYWORDS: t test on slope, p-value

57. Referring to Table 14-5, what is the p-value for testing whether Capital has a negative influence on corporate sales? a) 0.05 b) 0.2743 c) 0.5485 d) 0.7258 ANSWER: d TYPE: MC DIFFICULTY: Moderate KEYWORDS: t test on slope, p-value

58. Referring to Table 14-5, which of the following values for  is the smallest for which the regression model as a whole is significant? a) 0.00005 b) 0.001 c) 0.01 d) 0.05 ANSWER: b TYPE: MC DIFFICULTY: Moderate KEYWORDS: F test on the entire regression, p-value, interpretation 59. Referring to Table 14-5, what are the predicted sales (in millions of dollars) for a company spending $100 million on capital and $100 million on wages? a) 15,800.00 b) 16,520.07 c) 17,277.49 d) 20,455.98 ANSWER: b TYPE: MC DIFFICULTY: Moderate KEYWORDS: prediction of individual values 60. Referring to Table 14-5, what are the predicted sales (in millions of dollars) for a company spending $500 million on capital and $200 million on wages? a) 15,800.00 b) 16,520.07 c) 17,277.49 d) 20,455.98 ANSWER: c TYPE: MC DIFFICULTY: Moderate KEYWORDS: prediction of individual values

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236

61. Referring to Table 14-5, one company in the sample had sales of $20 billion (Sales = 20,000). This company spent $300 million on capital and $700 million on wages. What is the residual (in millions of dollars) for this data point? a) 874.55 b) 622.87 c) –790.69 d) –983.56 ANSWER: c TYPE: MC DIFFICULTY: Moderate KEYWORDS: residual 62. Referring to Table 14-5, one company in the sample had sales of $21.439 billion (Sales = 21,439). This company spent $300 million on capital and $700 million on wages. What is the residual (in millions of dollars) for this data point? a) 790.69 b) 648.31 c) – 648.31 d) – 790.69 ANSWER: b TYPE: MC DIFFICULTY: Moderate KEYWORDS: residual

63. Referring to Table 14-5, suppose the microeconomist wants to test whether the coefficient on Capital is significantly different from 0. What is the value of the relevant t-statistic? a) 0.609 b) 2.617 c) 4.804 d) 25.432 ANSWER: a TYPE: MC DIFFICULTY: Easy KEYWORDS: t test on slope, test statistic 64. Referring to Table 14-5, at the 0.01 level of significance, what conclusion should the microeconomist draw regarding the inclusion of Capital in the regression model? a) Capital is significant in explaining corporate sales and should be included in the model because its p-value is less than 0.01. b) Capital is significant in explaining corporate sales and should be included in the model because its p-value is more than 0.01. c) Capital is not significant in explaining corporate sales and should not be included in the model because its p-value is less than 0.01. d) Capital is not significant in explaining corporate sales and should not be included in the model because its p-value is more than 0.01. ANSWER: d TYPE: MC DIFFICULTY: Moderate KEYWORDS: t test on the slope, critical value, p-value, decision, conclusion

237 Multiple Regression Models

65. Referring to Table 14-5, the observed value of the F-statistic is given on the printout as 25.432. What are the degrees of freedom for this F-statistic? a) 25 for the numerator, 2 for the denominator b) 2 for the numerator, 23 for the denominator c) 23 for the numerator, 25 for the denominator d) 2 for the numerator, 25 for the denominator ANSWER: b TYPE: MC DIFFICULTY: Easy KEYWORDS: F test on the entire regression, degrees of freedom TABLE 14-6 One of the most common questions of prospective house buyers pertains to the average cost of heating in dollars (Y). To provide its customers with information on that matter, a large real estate firm used the following 4 variables to predict heating costs: the daily minimum outside temperature in degrees of Fahrenheit ( X1 ), the amount of insulation in inches ( X 2 ), the number of windows in the house ( X 3 ), and the age of the furnace in years ( X 4 ). Given below are the EXCEL outputs of two regression models. Model 1 Regression Statistics R Square 0.8080 Adjusted R Square 0.7568 Observations 20 ANOVA df Regression 4 Residual 15 Total 19

Intercept X1 (Temperature) X2 (Insulation) X3 (Windows) X4 (Furnace Age)

SS MS 169503.4241 42375.86 40262.3259 2684.155 209765.75

Coefficients Standard Error 421.4277 77.8614 -4.5098 0.8129 -14.9029 5.0508 0.2151 4.8675 6.3780 4.1026

F Significance F 15.7874 2.96869E-05

t Stat P-value Lower 90.0% Upper 90.0% 5.4125 7.2E-05 284.9327 557.9227 -5.5476 5.58E-05 -5.9349 -3.0847 -2.9505 0.0099 -23.7573 -6.0485 0.0442 0.9653 -8.3181 8.7484 1.5546 0.1408 -0.8140 13.5702

Model 2 Regression Statistics R Square 0.7768 Adjusted R Square 0.7506 Observations 20 ANOVA df Regression Residual

SS MS F 2 162958.2277 81479.11 29.5923 17 46807.5222 2753.384

Significance F 2.9036E-06

Multiple Regression Models

Total

Intercept X1 (Temperature) X2 (Insulation)

Coefficients 489.3227 -5.1103 -14.7195

238

209765.75 Standard Error t Stat P-value Lower 95% Upper 95% 43.9826 11.1253 3.17E-09 396.5273 582.1180 0.6951 -7.3515 1.13E-06 -6.5769 -3.6437 4.8864 -3.0123 0.0078 -25.0290 -4.4099

66. Referring to Table 14-6, the estimated value of the partial regression parameter 1 in Model 1 means that a) all else equal, an estimated expected $1 increase in average heating costs is associated with a decrease in the daily minimum outside temperature by 4.51 degrees. b) all else equal, a 1 degree increase in the daily minimum outside temperature results in a decrease in average heating costs by $4.51. c) all else equal, a 1 degree increase in the daily minimum outside temperature results in an estimated expected decrease in average heating costs by $4.51. d) all else equal, a 1% increase in the daily minimum outside temperature results in an estimated expected decrease in average heating costs by 4.51%. ANSWER: c TYPE: MC DIFFICULTY: Easy KEYWORDS: slope, interpretation

67. Referring to Table 14-6, what can we say about Model 1? a) The model explains 77.7% of the sample variability of average heating costs; after correcting for the degrees of freedom, the model explains 75.1% of the sample variability of average heating costs. b) The model explains 75.1% of the sample variability of average heating costs; after correcting for the degrees of freedom, the model explains 77.7% of the sample variability of average heating costs. c) The model explains 80.8% of the sample variability of average heating costs; after correcting for the degrees of freedom, the model explains 75.7% of the sample variability of average heating costs. d) The model explains 75.7% of the sample variability of average heating costs; after correcting for the degrees of freedom, the model explains 80.8% of the sample variability of average heating costs. ANSWER: c TYPE: MC DIFFICULTY: Moderate KEYWORDS: adjusted r-square

68. Referring to Table 14-6, what is your decision and conclusion for the test H0 : 2 = 0 vs. H1 : 2  0 at the  = 0.01 level of significance using Model 1? a) Do not reject H0 and conclude that the amount of insulation has a linear effect on average heating costs.

b) Reject H0 and conclude that the amount of insulation does not have a linear effect on average heating costs.

c) Reject H0 and conclude that the amount of insulation has a negative linear effect on average heating costs.

239 Multiple Regression Models

d) Do not reject H0 and conclude that the amount of insulation has a negative linear effect on average heating costs. ANSWER: c TYPE: MC DIFFICULTY: Easy KEYWORDS: t test on slope, critical value, p-value, decision, conclusion

69. Referring to Table 14-6, what is the 90% confidence interval for the expected change in average heating costs as a result of a 1 degree Fahrenheit change in the daily minimum outside temperature using Model 1? a) [−6.58, −3.65] b) [−6.24, −2.78] c) [−5.94, −3.08] d) [−2.37, 15.12] ANSWER: c TYPE: MC DIFFICULTY: Moderate KEYWORDS: confidence interval, slope, interpretation

70. Referring to Table 14-6 and allowing for a 1% probability of committing a type I error, what is the decision and conclusion for the test H 0 : 1 =  2 = 3 = 4 = 0 vs. H1 : At least one  j  0, j = 1, 2,L, 4 using Model 1? a) Do not reject H0 and conclude that the 4 independent variables have significant individual linear effects on average heating costs. b) Reject H0 and conclude that the 4 independent variables taken as a group have significant linear effects on average heating costs. c) Do not reject H0 and conclude that the 4 independent variables taken as a group do not have significant linear effects on average heating costs. d) Reject H0 and conclude that the 4 independent variables taken as a group do not have significant linear effects on average heating costs. ANSWER: b TYPE: MC DIFFICULTY: Easy KEYWORDS: F test on the entire regression, critical value, p-value, decision, conclusion

71. Referring to Table 14-6, what is the value of the partial F test statistic for H0 : 3 = 4 = 0 vs. H1 : At least one j  0, j = 3, 4 ? a) 0.820 b) 1.219 c) 1.382 d) 15.787 ANSWER: b TYPE: MC DIFFICULTY: Difficult KEYWORDS: partial F test, test statistic

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72. Referring to Table 14-6, what are the degrees of freedom of the partial F test for H0 : 3 = 4 = 0 vs. H1 : At least one j  0, j = 3, 4 ? a) b) c) d)

2 numerator degrees of freedom and 15 denominator degrees of freedom 15 numerator degrees of freedom and 2 denominator degrees of freedom 2 numerator degrees of freedom and 17 denominator degrees of freedom 17 numerator degrees of freedom and 2 denominator degrees of freedom

ANSWER: a TYPE: MC DIFFICULTY: Moderate KEYWORDS: partial F test, degrees of freedom 73. True or False: The interpretation of the slope is different in a multiple linear regression model as compared to a simple linear regression model. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: slope, interpretation

74. True or False: The coefficient of multiple determination r2 Y.12 measures the proportion of variation in Y that is explained by X1 and X2. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: coefficient of multiple determination 75. True or False: When an additional explanatory variable is introduced into a multiple regression model, the coefficient of multiple determination will never decrease. ANSWER: True TYPE: TF DIFFICULTY: Moderate KEYWORDS: coefficient of multiple determination, properties

76. True or False: When an additional explanatory variable is introduced into a multiple regression model, the adjusted r 2 can never decrease. ANSWER: False TYPE: TF DIFFICULTY: Moderate KEYWORDS: adjusted r-square, properties 77. True or False: When an explanatory variable is dropped from a multiple regression model, the coefficient of multiple determination can increase. ANSWER: False TYPE: TF DIFFICULTY: Moderate KEYWORDS: coefficient of multiple determination, properties

241 Multiple Regression Models

78. True or False: When an explanatory variable is dropped from a multiple regression model, the adjusted r 2 can increase. ANSWER: True TYPE: TF DIFFICULTY: Moderate KEYWORDS: adjusted r-square, properties 79. True or False: The slopes in a multiple regression model are called net regression coefficients. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: slope

80. True or False: In calculating the standard error of the estimate, SYX =

MSE , there are n – p –

1 degrees of freedom, where n is the sample size and p represents the number of independent variables in the model. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: standard error of estimate 81. True or False: The total sum of squares (SST) in a regression model will never exceed the regression sum of squares (SSR). ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: sum of squares, properties 82. True or False: The coefficient of multiple determination measures the fraction of the total variation in the dependent variable that is explained by the regression plane. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: coefficient of multiple determination, interpretation 83. True or False: The coefficient of multiple determination is calculated by taking the ratio of the regression sum of squares over the total sum of squares (SSR/SST) and subtracting that value from 1. ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: coefficient of multiple determination, properties 84. True or False: In a particular model, the sum of the squared residuals was 847. If the model had 5 independent variables, and the data set contained 40 points, the value of the standard error of the estimate is 24.911.

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242

ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: sum of squares, standard error of estimate, properties 85. True or False: A multiple regression is called “multiple” because it has several data points. ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: properties, interpretation 86. True or False: A multiple regression is called “multiple” because it has several explanatory variables. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: properties, interpretation 87. True or False: If we have taken into account all relevant explanatory factors, the residuals from a multiple regression should be random. ANSWER: True TYPE: TF DIFFICULTY: Moderate KEYWORDS: residual, properties 88. True or False: Multiple regression is the process of using several independent variables to predict a number of dependent variables. ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: properties

89. True or False: You have just run a regression in which the value of coefficient of multiple determination is 0.57. To determine if this indicates that the independent variables explain a significant portion of the variation in the dependent variable, you would perform an F-test. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: F test on the entire regression

90. True or False: From the coefficient of multiple determination, we cannot detect the strength of the relationship between Y and any individual independent variable. ANSWER: True TYPE: TF DIFFICULTY: Moderate KEYWORDS: coefficient of multiple determination, interpretation

243 Multiple Regression Models

91. True or False: Consider a regression in which b2 = – 1.5 and the standard error of this coefficient equals 0.3. To determine whether X2 is a significant explanatory variable, you would compute an observed t-value of – 5.0. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: t test on slope, test statistic 92. True or False: A regression had the following results: SST = 82.55, SSE = 29.85. It can be said that 73.4% of the variation in the dependent variable is explained by the independent variables in the regression. ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: coefficient multiple determination, sum of squares, properties 93. True or False: A regression had the following results: SST = 82.55, SSE = 29.85. It can be said that 63.84% of the variation in the dependent variable is explained by the independent variables in the regression. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: coefficient multiple determination, sum of squares, properties 94. True or False: A regression had the following results: SST = 102.55, SSE = 82.04. It can be said that 90.0% of the variation in the dependent variable is explained by the independent variables in the regression. ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: coefficient multiple determination, sum of squares, properties 95. True or False: A regression had the following results: SST = 102.55, SSE = 82.04. It can be said that 20.0% of the variation in the dependent variable is explained by the independent variables in the regression. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: coefficient multiple determination, sum of squares, properties TABLE 14-7 The department head of the accounting department wanted to see if she could predict the GPA of students using the number of course units (credits) and total SAT scores of each. She takes a sample of students and generates the following Microsoft Excel output: SUMMARY OUTPUT Regression Statistics

Multiple Regression Models

Multiple R R Square Adjusted R Square Standard Error Observations

244

0.916 0.839 0.732 0.24685 6

ANOVA df 2 3 5

Regression Residual Total

Coeff Intercept 4.593897 Units – 0.247270 SAT Total 0.001443

SS 0.95219 0.18281 1.13500 StdError 1.13374542 0. 06268485 0.00101241

MS 0.47610 0.06094 t Stat 4.052 – 3.945 1.425

F 7.813

Signif F 0.0646

P-value 0.0271 0.0290 0.2494

96. Referring to Table 14-7, the estimate of the unit change in the mean of Y per unit change in X1, holding X2 constant, is

ANSWER: -0.247 TYPE: FI DIFFICULTY: Easy KEYWORDS: slope, interpretation

97. Referring to Table 14-7, the net regression coefficient of X2 is

ANSWER: 0.001443 TYPE: FI DIFFICULTY: Easy KEYWORDS: slope, interpretation 98. Referring to Table 14-7, the predicted GPA for a student carrying 15 course units and who has a total SAT of 1,100 is . ANSWER: 2.472 TYPE: FI DIFFICULTY: Moderate KEYWORDS: prediction of individual values

99. Referring to Table 14-7, the value of the coefficient of multiple determination, r2 Y.12, is ANSWER: 0.839 TYPE: FI DIFFICULTY: Easy KEYWORDS: coefficient of multiple determination

100. Referring to Table 14-7, the value of the adjusted coefficient of multiple determination, r2adj, is ANSWER: 0.732

245 Multiple Regression Models TYPE: FI DIFFICULTY: Easy KEYWORDS: adjusted r-square

101. Referring to Table 14-7, the department head wants to test H0: 1 = 2 = 0 . The appropriate alternative hypothesis is

ANSWER: at least one  i  0 . TYPE: FI DIFFICULTY: Easy KEYWORDS: F test on the entire regression, form of hypothesis

102. Referring to Table 14-7, the department head wants to test H0: 1 = 2 = 0 . The critical value of the F test for a level of significance of 0.05 is

ANSWER: 9.55 TYPE: FI DIFFICULTY: Easy KEYWORDS: F test on the entire regression, critical value

103. Referring to Table 14-7, the department head wants to test H0: 1 = 2 = 0 . The value of the F-test statistic is

ANSWER: 7.813 TYPE: FI DIFFICULTY: Easy KEYWORDS: F test on the entire regression, test statistic

104. Referring to Table 14-7, the department head wants to test H0: 1 = 2 = 0 . The p-value of the test is

ANSWER: 0.0646 TYPE: FI DIFFICULTY: Easy KEYWORDS: F test on the entire regression, p-value

105. True or False: Referring to Table 14-7, the department head wants to test H0: 1 = 2 = 0 . At a level of significance of 0.05, the null hypothesis is rejected. ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: F test on the entire regression, decision

106. Referring to Table 14-7, the department head wants to use a t test to test for the significance of the coefficient of X1. For a level of significance of 0.05, the critical values of the test are . ANSWER:  3.1824 TYPE: FI DIFFICULTY: Easy KEYWORDS: t test on slope, critical value

Multiple Regression Models

246

107. Referring to Table 14-7, the department head wants to use a t test to test for the significance of the coefficient of X1. The value of the test statistic is

ANSWER: -3.945 TYPE: FI DIFFICULTY: Easy KEYWORDS: t test on slope, test statistic

108. Referring to Table 14-7, the department head wants to use a t test to test for the significance of the coefficient of X1. The p-value of the test is

ANSWER: 0.0290 TYPE: FI DIFFICULTY: Easy KEYWORDS: t test on slope, p-value

109. True or False: Referring to Table 14-7, the department head wants to use a t test to test for the significance of the coefficient of X1. At a level of significance of 0.05, the department head would decide that 1  0 . ANSWER: True TYPE: TF DIFFICULTY: Moderate KEYWORDS: t test on slope, decision, conclusion

110. Referring to Table 14-7, the department head decided to obtain a 95% confidence interval for 1 . The confidence interval is from to . ANSWER: -0.4468 to -0.0478 TYPE: FI DIFFICULTY: Moderate KEYWORDS: slope, confidence interval TABLE 14-8 A financial analyst wanted to examine the relationship between salary (in $1,000) and 4 variables: age (X1 = Age), experience in the field (X2 = Exper), number of degrees (X3 = Degrees), and number of previous jobs in the field (X4 = Prevjobs). He took a sample of 20 employees and obtained the following Microsoft Excel output: SUMMARY OUTPUT Regression Statistics Multiple R 0.992 R Square 0.984 Adjusted R Square 0.979 Standard Error 2.26743 Observations 20 ANOVA Regression Residual

df 4 15

SS 4609.83164 77.11836

MS 1152.45791 5.14122

F 224.160

Signif F 0.0001

247 Multiple Regression Models Total

4686.95000

Intercept Age Exper Degrees Prevjobs

Coeff – 9.611198 1.327695 – 0.106705 7.311332 – 0.504168

StdError t Stat 2.77988638 – 3.457 0.11491930 11.553 0.14265559 – 0.748 0.80324187 9.102 0.44771573 – 1.126

P-value 0.0035 0.0001 0.4660 0.0001 0.2778

111. Referring to Table 14-8, the estimate of the unit change in the mean of Y per unit change in X4, taking into account the effects of the other 3 variables, is

ANSWER: -0.504 TYPE: FI DIFFICULTY: Easy KEYWORDS: slope, interpretation

112. Referring to Table 14-8, the net regression coefficient of X2 is

ANSWER: -0.107 TYPE: FI DIFFICULTY: Easy KEYWORDS: slope, interpretation 113. Referring to Table 14-8, the predicted salary for a 35-year-old person with 10 years of experience, 3 degrees, and 1 previous job is . ANSWER: 57.2 TYPE: FI DIFFICULTY: Moderate KEYWORDS: prediction of individual values

114. Referring to Table 14-8, the value of the coefficient of multiple determination, r2 Y.1234, is . ANSWER: 0.984 TYPE: FI DIFFICULTY: Easy KEYWORDS: coefficient of multiple determination

115. Referring to Table 14-8, the value of the adjusted coefficient of multiple determination, adj r2, is

ANSWER: 0.979 TYPE: FI DIFFICULTY: Easy KEYWORDS: adjusted r-square

116. Referring to Table 14-8, the analyst wants to use an F-test to test H0 : 1 = 2 = 3 =  4 = 0 . The appropriate alternative hypothesis is ANSWER: at least one  i  0 .

Multiple Regression Models

248

TYPE: FI DIFFICULTY: Easy KEYWORDS: F test on the entire regression, form of hypothesis

117. Referring to Table 14-8, the critical value of an F test on the entire regression for a level of significance of 0.01 is

ANSWER: 4.89 TYPE: FI DIFFICULTY: Easy KEYWORDS: F test on the entire regression, critical value

118. Referring to Table 14-8, the value of the F-statistic for testing the significance of the entire regression is

ANSWER: 224.160 TYPE: FI DIFFICULTY: Easy KEYWORDS: F test on the entire regression, test statistic

119. Referring to Table 14-8, the p-value of the F test for the significance of the entire regression is . ANSWER: 0.0001 TYPE: FI DIFFICULTY: Easy KEYWORDS: F test on the entire regression, p-value

120. True or False: Referring to Table 14-8, the F test for the significance of the entire regression performed at a level of significance of 0.01 leads to a rejection of the null hypothesis. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: F test on the entire regression, decision

121. Referring to Table 14-8, the analyst wants to use a t test to test for the significance of the coefficient of X3. For a level of significance of 0.01, the critical values of the test are ANSWER:  2.9467 TYPE: FI DIFFICULTY: Easy KEYWORDS: t test on the slope, critical value

122. Referring to Table 14-8, the analyst wants to use a t test to test for the significance of the coefficient of X3. The value of the test statistic is

ANSWER: 9.102 TYPE: FI DIFFICULTY: Easy KEYWORDS: t test on the slope, test statistic

123. Referring to Table 14-8, the analyst wants to use a t test to test for the significance of the coefficient of X3. The p-value of the test is

249 Multiple Regression Models

ANSWER: 0.0001 TYPE: FI DIFFICULTY: Easy KEYWORDS: t test on the slope, p-value

124. True or False: Referring to Table 14-8, the analyst wants to use a t test to test for the significance of the coefficient of X3. At a level of significance of 0.01, the department head would decide that  3  0 . ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: t test on the slope, decision, conclusion

125. Referring to Table 14-8, the analyst decided to obtain a 99% confidence interval for  3. The confidence interval is from

ANSWER: 4.944 to 9.678 TYPE: FI DIFFICULTY: Moderate KEYWORDS: slope, confidence interval TABLE 14-9 You decide to predict gasoline prices in different cities and towns in the United States for your term project. Your dependent variable is price of gasoline per gallon and your explanatory variables are per capita income, the number of firms that manufacture automobile parts in and around the city, the number of new business starts in the last year, population density of the city, percentage of local taxes on gasoline, and the number of people using public transportation. You collected data of 32 cities and obtained a regression sum of squares SSR= 122.8821. Your computed value of standard error of the estimate is 1.9549. 126. Referring to Table 14-9, what is the value of the coefficient of multiple determination? a) 0.2225 b) 0.4576 c) 0.5626 d) 0.6472 ANSWER: c TYPE: MC DIFFICULTY: Easy KEYWORDS: coefficient of multiple determination

127. Referring to Table 14-9, the value of adjusted r 2 is a) 0.4576 b) 0.5626 c) 0.6472 d) 95.5414 ANSWER: a TYPE: MC DIFFICULTY: Moderate

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250

KEYWORDS: adjusted r-square

128. Referring to Table 14-9, if variables that measure the number of new business starts in the last year and population density of the city were removed from the multiple regression model, which of the following would be true? a) The adjusted r 2 will definitely increase. b) The adjusted r 2 can not increase. c) The coefficient of multiple determination will not increase. d) The coefficient of multiple determination will definitely increase. ANSWER: c TYPE: MC DIFFICULTY: Moderate KEYWORDS: coefficient of multiple determination, adjusted r-square, properties TABLE 14-10 You worked as an intern at We Always Win Car Insurance Company last summer. You notice that individual car insurance premiums depend very much on the age of the individual, the number of traffic tickets received by the individual, and the population density of the city in which the individual lives. You performed a regression analysis in EXCEL and obtained the following information: Regression Analysis Regression Statistics Multiple R 0.63 R Square 0.40 Adjusted R Square 0.23 Standard Error 50.00 Observations 15.00 ANOVA df Regression Residual Total

Intercept AGE TICKETS DENSITY

SS 3 11

MS 5994.24

F 2.40

Significance F 0.12

P-value

Lower 99.0%

Upper 99.0%

0.03 0.36 0.07 0.64

-27.47 -3.51 -11.86 -23.19

275.07 1.87 54.37 16.91

27496.82 45479.54

Coefficients Standard Error 123.80 48.71 -0.82 0.87 21.25 10.66 -3.14 6.46

t Stat 2.54 -0.95 1.99 -0.49

129. Referring to Table 14-10, the proportion of the total variability in insurance premiums that can be explained by AGE, TICKETS, and DENSITY is ANSWER: 0.40 TYPE: FI DIFFICULTY: Easy KEYWORDS: coefficient of multiple determination, interpretation

251 Multiple Regression Models

130. Referring to Table 14-10, the adjusted r 2 is

ANSWER: 0.23 TYPE: FI DIFFICULTY: Easy KEYWORDS: adjusted r-square 131. Referring to Table 14-10, the standard error of the estimate is

ANSWER: 50.0 TYPE: FI DIFFICULTY: Easy KEYWORDS: standard error of estimate

132. Referring to Table 14-10, the estimated average change in insurance premiums for every 10 additional tickets received is

ANSWER: 212.5 TYPE: FI DIFFICULTY: Difficult KEYWORDS: slope, interpretation

133. Referring to Table 14-10, the 99% confidence interval for the change in average insurance premiums of a person who has become 1 year older (i.e., the slope coefficient for AGE) is . ANSWER:

-0.82  2.69

TYPE: FI DIFFICULTY: Moderate KEYWORDS: slope, interpretation, confidence interval 134. Referring to Table 14-10, the total degrees of freedom that are missing in the ANOVA table should be . ANSWER: 14 TYPE: FI DIFFICULTY: Easy KEYWORDS: degrees of freedom 135. Referring to Table 14-10, the regression sum of squares that is missing in the ANOVA table should be . ANSWER: 17,982.72 TYPE: FI DIFFICULTY: Easy KEYWORDS: sum of squares 136. Referring to Table 14-10, the residual mean squares (MSE) that are missing in the ANOVA table should be . ANSWER: 2,499.71

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252

TYPE: FI DIFFICULTY: Easy KEYWORDS: mean squares 137. Referring to Table 14-10, to test the significance of the multiple regression model, what is the form of the null hypothesis? a) H0 : 0 b) H0 : 1

H0 : 1 = 2 = 3 d) H0 : 0 = 1 = 2 = 3 c)

ANSWER: c TYPE: MC DIFFICULTY: Easy KEYWORDS: F test on the entire regression, form of hypothesis 138. Referring to Table 14-10, to test the significance of the multiple regression model, the value of the test statistic is . ANSWER: 2.40 TYPE: FI DIFFICULTY: Easy KEYWORDS: F test on the entire regression, test statistic

139. Referring to Table 14-10, to test the significance of the multiple regression model, the p-value of the test statistic in the sample is

ANSWER: 0.12 TYPE: FI DIFFICULY: Easy KEYWORDS: F test on the entire regression, p-value 140. Referring to Table 14-10, to test the significance of the multiple regression model, what are the degrees of freedom? ANSWER: 3 for the numerator, 11 for the denominator TYPE: PR DIFFICULY: Easy KEYWORDS: F test on the entire regression, degrees of freedom 141. True or False: Referring to Table 14-10, to test the significance of the multiple regression model, the null hypothesis should be rejected while allowing for 1% probability of committing a type I error. ANSWER: False TYPE: TF DIFFICULY: Easy KEYWORDS: F test on the entire regression, decision 142. True or False: Referring to Table 14-10, the multiple regression model is significant at a 10% level of significance. ANSWER:

253 Multiple Regression Models False TYPE: TF DIFFICULY: Easy KEYWORDS: F test on the entire regression, decision, conclusion 143. If a categorical independent variable contains 2 categories, then will be needed to uniquely represent these categories. a) 1 b) 2 c) 3 d) 4

dummy variable(s)

ANSWER: a TYPE: MC DIFFICULTY: Easy KEYWORDS: dummy variable, properties 144. If a categorical independent variable contains 4 categories, then will be needed to uniquely represent these categories. a) 1 b) 2 c) 3 d) 4

dummy variable(s)

ANSWER: c TYPE: MC DIFFICULTY: Easy KEYWORDS: dummy variable, properties 145. A dummy variable is used as an independent variable in a regression model when a) the variable involved is numerical. b) the variable involved is categorical. c) a curvilinear relationship is suspected. d) when 2 independent variables interact. ANSWER: b TYPE: MC DIFFICULTY: Easy KEYWORDS: dummy variable, properties 146. An interaction term in a multiple regression model may be used when a) the coefficient of determination is small. b) there is a curvilinear relationship between the dependent and independent variables. c) neither one of 2 independent variables contribute significantly to the regression model. d) the relationship between X1 and Y changes for differing values of X2. ANSWER: d TYPE: MC DIFFICULTY: Moderate KEYWORDS: interaction, properties 147. To explain personal consumption (CONS) measured in dollars, data is collected for INC: CRDTLIM:

personal income in dollars $1 plus the credit limit in dollars available to the individual

Multiple Regression Models

APR: ADVT: SEX:

254

average annualized percentage interest rate for borrowing for the individual per person advertising expenditure in dollars by manufacturers in the city where the individual lives gender of the individual; 1 if female, 0 if male

A regression analysis was performed with CONS as the dependent variable and ln(CRDTLIM), ln(APR), ln(ADVT), and SEX as the independent variables. The estimated model was $y = 2.28 - 0.29 ln(CRDTLIM) + 5.77 ln(APR) + 2.35 ln(ADVT) + 0.39 SEX

What is the correct interpretation for the estimated coefficient for SEX? a) Holding everything else fixed, personal consumption for females is estimated to be $0.39 higher than males on the average. b) Holding everything else fixed, personal consumption for males is estimated to be $0.39 higher than females on the average. c) Holding everything else fixed, personal consumption for females is estimated to be 0.39% higher than males on the average. d) Holding everything else fixed, personal consumption for males is estimated to be 0.39% higher than females on the average. ANSWER: a TYPE: MC DIFFICULTY: Easy KEYWORDS: dummy variable, slope, interpretation TABLE 14-11 A weight-loss clinic wants to use regression analysis to build a model for weight-loss of a client (measured in pounds). Two variables thought to affect weight-loss are client’s length of time on the weight loss program and time of session. These variables are described below: Y = Weight-loss (in pounds) X1 = Length of time in weight-loss program (in months) X2 = 1 if morning session, 0 if not X3 = 1 if afternoon session, 0 if not (Base level = evening session) Data for 12 clients on a weight-loss program at the clinic were collected and used to fit the interaction model: Y = 0 + 1X1 + 2 X2 +  3X 3 + 4 X1 X2 + 5 X1X3 +  Partial output from Microsoft Excel follows: Regression Statistics Multiple R 0.73514 R Square 0.540438 Adjusted R Square 0.157469 Standard Error 12.4147 Observations 12 ANOVA F = 5.41118 Intercept Length (X1) Morn Ses (X2)

Significance F = 0.040201 Coeff 0.089744 6.22538 2.217272

StdError 14.127 2.43473 22.1416

t Stat 0.0060 2.54956 0.100141

P-value 0.9951 0.0479 0.9235

255 Multiple Regression Models Aft Ses (X3) Length*Morn Ses Length*Aft Ses

11.8233 0.77058 – 0.54147

3.1545 3.562 3.35988

3.558901 0.216334 – 0.161158

0.0165 0.8359 0.8773

148. Referring to Table 14-11, what is the experimental unit for this analysis? a) b) c) d)

A clinic A client on a weight-loss program A month A morning, afternoon, or evening session

ANSWER: b TYPE: MC DIFFICULTY: Moderate KEYWORDS: model building

149. Referring to Table 14-11, what null hypothesis would you test to determine whether the slope of the linear relationship between weight-loss (Y) and time in the program (X1) varies according to time of session? a) H0 : 1 = 2 = 3 = 4 = 5 = 0 b) H0 : 2 = 3 = 4 = 5 = 0 c) H0 : 4 = 5 = 0 d) H0 : 2 = 3 = 0 ANSWER: c TYPE: MC DIFFICULTY: Moderate KEYWORDS: interaction, form of hypothesis

150. Referring to Table 14-11, in terms of the  's in the model, give the average change in weightloss (Y) for every 1 month increase in time in the program (X1) when attending the evening session. a) 1 + 4 b) 1 + 5 c) 1 d) 4 + 5 ANSWER: c TYPE: MC DIFFICULTY: Moderate KEYWORDS: interaction, slope, interpretation

151. Referring to Table 14-11, in terms of the  's in the model, give the average change in weightloss (Y) for every 1 month increase in time in the program (X1) when attending the morning session. a) 1 + 4 b) 1 + 5 c) 1 d) 4 + 5 ANSWER: a

Multiple Regression Models

TYPE: MC DIFFICULTY: Moderate KEYWORDS: interaction, slope, interpretation

256

257 Multiple Regression Models

152. Referring to Table 14-11, in terms of the  's in the model, give the average change in weightloss (Y) for every 1 month increase in time in the program (X1) when attending the afternoon session. a) 1 + 4 b) 1 + 5 c) 1 d) 4 + 5 ANSWER: b TYPE: MC DIFFICULTY: Moderate KEYWORDS: interaction, slope, interpretation

153. True or False: Referring to Table 14-11, the overall model for predicting weight-loss (Y) is statistically significant at the 0.05 level. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: interaction, F test on the entire regression, decision, conclusion

154. Referring to Table 14-11, which of the following statements is supported by the analysis shown?

a) There is sufficient evidence (at  = 0.05) of curvature in the relationship between weight-loss (Y) and months in program(X1).

b) There is sufficient evidence (at  = 0.05) to indicate that the relationship between weight-loss (Y) and months in program (X1) depends on session time. c) There is sufficient evidence (at  = 0.10) to indicate that the session time (morning, afternoon, evening) affects weight-loss (Y). d) There is insufficient evidence (at  = 0.10) to indicate that the relationship between weight-loss (Y) and months in program(X1) depends on session time. ANSWER: d TYPE: MC DIFFICULTY: Difficult KEYWORDS: interaction, t test on slope, decision, conclusion

155. In a multiple regression model, the adjusted r 2 a) b) c) d)

cannot be negative. can sometimes be negative. can sometimes be greater than +1. has to fall between 0 and +1.

ANSWER: b TYPE: MC DIFFICULTY: Difficult KEYWORDS: adjusted r-square, properties TABLE 14-12 As a project for his business statistics class, a student examined the factors that determined parking meter rates throughout the campus area. Data were collected for the price per hour of parking, blocks

Multiple Regression Models

258

to the quadrangle, and one of the three jurisdictions: on campus, in downtown and off campus, or outside of downtown and off campus. The population regression model hypothesized is Yi =  + 1 x1i + 2 x2i + 3 x3i +i where Y is the meter price x1 is the number of blocks to the quad x2 is a dummy variable that takes the value 1 if the meter is located in downtown and off campus and the value 0 otherwise x3 is a dummy variable that takes the value 1 if the meter is located outside of downtown and off campus, and the value 0 otherwise The following Excel results are obtained. Regression Statistics Multiple R 0.9659 R Square 0.9331 Adjusted R Square 0.9294 Standard Error 0.0327 Observations 58 ANOVA Df Regression 3 Residual 54 Total 57

Intercept X1 X2 X3

SS 0.8094 0.0580 0.8675

Coefficients Standard Error 0.5118 0.0136 -0.0045 0.0034 -0.2392 0.0123 -0.0002 0.0123

MS 0.2698 0.0010

t Stat 37.4675 -1.3276 -19.3942 -0.0214

F Significance F 251.1995 1.0964E-31

P-value 2.4904 0.1898 5.3581E-26 0.9829

156. Referring to Table 14-12, what is the correct interpretation for the estimated coefficient for x2 ? a) All else equal, the estimated average difference in costs between parking on campus, and

parking outside of downtown and off campus is −$0.24 per hour. b) All else equal, the estimated average difference in costs between parking in downtown and off campus, and parking on campus is −$0.24 per hour. c) All else equal, the estimated average difference in costs between parking in downtown and off campus, and parking outside of downtown and off campus is −$0.24 per hour. d) All else equal, the estimated average difference in costs between parking in downtown and off campus, and parking either outside of downtown and off campus or on campus is −$0.24 per hour. ANSWER: b TYPE: MC DIFFICULTY: Moderate KEYWORDS: dummy variable, slope, interpretation

157. Referring to Table 14-12, predict the meter rate per hour if one parks outside of downtown and off campus 3 blocks from the quad. a) $−0.0139 b) $0.2589

259 Multiple Regression Models c) $0.2604 d) $0.4981 ANSWER: d TYPE: MC DIFFICULTY: Easy KEYWORDS: prediction of individual values

158. Referring to Table 14-12, if one is already outside of downtown and off campus but decides to park 3 more blocks from the quad, the estimated average parking meter rate will a) decrease by 0.0045. b) decrease by 0.0135 . c) decrease by 0.0139. d) decrease by 0.4979. ANSWER b TYPE: MC DIFFICULTY: Moderate KEYWORDS: dummy variable, slope, interpretation

159. True or False: An interaction term in a multiple regression model may be used when the relationship between X1 and Y changes for differing values of X2. ANSWER: True TYPE: TF DIFFICULTY: Moderate KEYWORDS: interaction, properties 160. True or False: When a dummy variable is included in a multiple regression model, the interpretation of the estimated slope coefficient does not make any sense anymore. ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: dummy variable, slope, interpretation 161. True or False: In trying to obtain a model to estimate grades on a statistics test, a professor wanted to include, among other factors, whether the person had taken the course previously. To do this, the professor included a dummy variable in her regression that was equal to 1 if the person had previously taken the course, and 0 otherwise. The interpretation of the coefficient associated with this dummy variable would be the average amount the repeat students tended to be above or below non-repeaters, with all other factors the same. ANSWER: True TYPE: TF DIFFICULTY: Moderate KEYWORDS: dummy variable, slope, interpretation TABLE 14-13 An econometrician is interested in evaluating the relation of demand for building materials to mortgage rates in Los Angeles and San Francisco. He believes that the appropriate model is Y = 10 + 5X1 + 8X2

Multiple Regression Models

where

260

X1 = mortgage rate in % X2 = 1 if SF, 0 if LA Y = demand in $100 per capita

162. Referring to Table 14-13, holding constant the effect of city, each additional increase of 1% in the mortgage rate would lead to an estimated increase of

in the mean demand.

ANSWER: $500 per capita TYPE: FI DIFFICULTY: Moderate KEYWORDS: slope, interpretation

163. Referring to Table 14-13, the effect of living in San Francisco rather than Los Angeles is to increase the mean demand by an estimated

ANSWER: $800 per capita TYPE: FI DIFFICULTY: Moderate KEYWORDS: dummy variable, slope, interpretation

164. Referring to Table 14-13, the predicted demand in Los Angeles when the mortgage rate is 8% is

ANSWER: $5,000 per capita TYPE: FI DIFFICULTY: Moderate KEYWORDS: prediction of individual values, dummy variable

165. Referring to Table 14-13, the predicted demand in San Francisco when the mortgage rate is 10% is

ANSWER: $6,800 per capita TYPE: FI DIFFICULTY: Moderate KEYWORDS: prediction of individual values, dummy variable

166. Referring to Table 14-13, the fitted model for predicting demand in Los Angeles is a) 10 + 5X1 b) 10 + 13X1 c) 15 + 8X2 d) 18 + 5X2 ANSWER: a TYPE: MC DIFFICULTY: Moderate KEYWORDS: prediction of individual values, dummy variable

167. Referring to Table 14-13, the fitted model for predicting demand in San Francisco is . a) 10 + 5X1 b) 10 + 13X1

261 Multiple Regression Models

c) 15 + 8X2 d) 18 + 5X1 ANSWER: d TYPE: MC DIFFICULTY: Moderate KEYWORDS: prediction of individual values, dummy variable TABLE 14-14 An automotive engineer would like to be able to predict automobile mileages. She believes that the two most important characteristics that affect mileage are horsepower and the number of cylinders (4 or 6) of a car. She believes that the appropriate model is Y = 40 – 0.05X1 + 20X2 – 0.1X1X2 where X1 = horsepower X2 = 1 if 4 cylinders, 0 if 6 cylinders Y = mileage.

168. Referring to Table 14-14, the predicted mileage for a 300 horsepower, 6-cylinder car is . ANSWER: 25 TYPE: FI DIFFICULTY: Moderate KEYWORDS: prediction of individual values, dummy variable

169. Referring to Table 14-14, the predicted mileage for a 200 horsepower, 4-cylinder car is . ANSWER: 30 TYPE: FI DIFFICULTY: Moderate KEYWORDS: prediction of individual values, dummy variable

170. Referring to Table 14-14, the fitted model for predicting mileages for 6-cylinder cars is .

a) 40 – 0.05X1 b) 40 – 0.10X1 c) 60 – 0.10X1 d) 60 – 0.15X1 ANSWER: a TYPE: MC DIFFICULTY: Moderate KEYWORDS: prediction of individual values, dummy variable

171. Referring to Table 14-14, the fitted model for predicting mileages for 4-cylinder cars is .

a) 40 – 0.05X1 b) 40 – 0.10X1 c) 60 – 0.10X1 d) 60 – 0.15X1

Multiple Regression Models

262

ANSWER: d TYPE: MC DIFFICULTY: Moderate KEYWORDS: prediction of individual values, dummy variable TABLE 14-15 The superintendent of a school district wanted to predict the percentage of students passing a sixthgrade proficiency test. She obtained the data on percentage of students passing the proficiency test (% Passing), daily average of the percentage of students attending class (% Attendance), average teacher salary in dollars (Salaries), and instructional spending per pupil in dollars (Spending) of 47 schools in the state. Following is the multiple regression output with Y = % Passing as the dependent variable, X1 =:% Attendance, X 2 = Salaries and X 3 = Spending: Regression Statistics Multiple R 0.7930 R Square 0.6288 0.6029 Adjusted R Square Standard 10.4570 Error Observations 47 ANOVA df Regression Residual Total

3 43 46 Coefficients

Intercept % Attendance Salary Spending

-753.4225 8.5014 6.85E-07 0.0060

SS 7965.08 4702.02 12667.11

MS 2655.03 109.35

F 24.2802

Significance F 2.3853E-09

Standard Error 101.1149 1.0771 0.0006 0.0046

t Stat

P-value

Lower 95%

-7.4511 7.8929 0.0011 1.2879

2.88E-09 6.73E-10 0.9991 0.2047

-957.3401 6.3292 -0.0013 -0.0034

Upper 95% -549.5050 10.6735 0.0013 0.0153

172. Referring to Table 14-15, which of the following is a correct statement? a) The average percentage of students passing the proficiency test is estimated to go up by 8.50% when daily average of percentage of students attending class increases by 1%. b) The daily average of the percentage of students attending class is expected to go up by an estimated 8.50% when the percentage of students passing the proficiency test increases by 1%. c) The average percentage of students passing the proficiency test is estimated to go up by 8.50% when daily average of the percentage of students attending class increases by 1% holding constant the effects of all the remaining independent variables. d) The daily average of the percentage of students attending class is expected to go up by an estimated 8.50% when the percentage of students passing the proficiency test increases by 1% holding constant the effects of all the remaining independent variables.

263 Multiple Regression Models ANSWER: c TYPE: MC DIFFICULTY: Moderate KEYWORDS: interpretation, slope

173. Referring to Table 14-15, which of the following is a correct statement? a) 62.88% of the total variation in the percentage of students passing the proficiency test can be explained by daily average of the percentage of students attending class, average teacher salary, and instructional spending per pupil. b) 62.88% of the total variation in the percentage of students passing the proficiency test can be explained by daily average of the percentage of students attending class, average teacher salary, and instructional spending per pupil after adjusting for the number of predictors and sample size. c) 62.88% of the total variation in the percentage of students passing the proficiency test can be explained by daily average of the percentage of students attending class holding constant the effect of average teacher salary, and instructional spending per pupil. d) 62.88% of the total variation in the percentage of students passing the proficiency test can be explained by daily average of the percentage of students attending class after adjusting for the effect of average teacher salary, and instructional spending per pupil. ANSWER: a TYPE: MC DIFFICULTY: Moderate KEYWORDS: interpretation, coefficient of multiple determination

174. Referring to Table 14-15, which of the following is a correct statement? a) 60.29% of the total variation in the percentage of students passing the proficiency test can be explained by daily average of the percentage of students attending class, average teacher salary, and instructional spending per pupil. b) 60.29% of the total variation in the percentage of students passing the proficiency test can be explained by daily average of the percentage of students attending class, average teacher salary, and instructional spending per pupil after adjusting for the number of predictors and sample size. c) 60.29% of the total variation in the percentage of students passing the proficiency test can be explained by daily average of the percentage of students attending class holding constant the effect of average teacher salary, and instructional spending per pupil. d) 60.29% of the total variation in the percentage of students passing the proficiency test can be explained by daily average of the percentage of students attending class after adjusting for the effect of average teacher salary, and instructional spending per pupil. ANSWER: b TYPE: MC DIFFICULTY: Moderate KEYWORDS: interpretation, adjusted r-square

175. Referring to Table 14-15, what is the standard error of estimate? ANSWER: 10.46 (%) TYPE: PR DIFFICULTY: Easy KEYWORDS: standard error of estimate

Multiple Regression Models

264

176. Referring to Table 14-15, predict the percentage of students passing the proficiency test for a school which has a daily average of 95% of students attending class, an average teacher salary of 40,000 dollars, and an instructional spending per pupil of 2000 dollars. ANSWER: 66.20 (%) TYPE: PR DIFFICULTY: Easy KEYWORDS: prediction of individual values

177. Referring to Table 14-15, estimate the average percentage of students passing the proficiency test for all the schools that have a daily average of 95% of students attending class, an average teacher salary of 40,000 dollars, and an instructional spending per pupil of 2000 dollars. ANSWER: 66.20 (%) TYPE: PR DIFFICULTY: Easy KEYWORDS: estimation of mean values

178. Referring to Table 14-15, which of the following is the correct null hypothesis to test whether instructional spending per pupil has any effect on percentage of students passing the proficiency test? a) H0 : 0 = 0 b) H0 : 1 = 0 c) H0 : 2 = 0 d) H0 : 3 = 0 ANSWER: d TYPE: MC DIFFICULTY: Easy KEYWORDS: t test on slope, form of hypothesis

179. Referring to Table 14-15, which of the following is the correct alternative hypothesis to test whether instructional spending per pupil has any effect on percentage of students passing the proficiency test? a) H1 : 0  0 b) H1 : 1  0 c) H1 : 2  0 d) H1 : 3  0 ANSWER: d TYPE: MC DIFFICULTY: Easy KEYWORDS: t test on slope, form of hypothesis

180. Referring to Table 14-15, what is the value of the test statistics when testing whether instructional spending per pupil has any effect on percentage of students passing the proficiency test? ANSWER: 1.2879 TYPE: PR DIFFICULTY: Easy KEYWORDS: t test on slope, test statistic

265 Multiple Regression Models

181. Referring to Table 14-15, what is the p-value of the test statistics when testing whether instructional spending per pupil has any effect on percentage of students passing the proficiency test? ANSWER: 0.2047 TYPE: PR DIFFICULTY: Easy KEYWORDS: t test on slope, p-value

182. True or False: Referring to Table 14-15, the null hypothesis should be rejected at a 5% level of significance when testing whether instructional spending per pupil has any effect on percentage of students passing the proficiency test. ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: t test on slope, decision

183. True or False: Referring to Table 14-15, there is sufficient evidence that instructional spending per pupil has an effect on percentage of students passing the proficiency test while holding constant the effect of all the other independent variables at a 5% level of significance. ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: t test on slope, conclusion

184. Referring to Table 14-15, which of the following is the correct null hypothesis to test whether daily average of the percentage of students attending class has any effect on percentage of students passing the proficiency test? a) H0 : 0 = 0 b) H0 : 1 = 0 c) H0 : 2 = 0 d) H0 : 3 = 0 ANSWER: b TYPE: MC DIFFICULTY: Easy KEYWORDS: t test on slope, form of hypothesis

185. Referring to Table 14-15, which of the following is the correct alternative hypothesis to test whether daily average of the percentage of students attending class has any effect on percentage of students passing the proficiency test? a) H1 : 0  0 b) H1 : 1  0 c) H1 : 2  0 d) H1 : 3  0 ANSWER:

Multiple Regression Models

266

b TYPE: MC DIFFICULTY: Easy KEYWORDS: t test on slope, form of hypothesis

186. Referring to Table 14-15, what is the value of the test statistics when testing whether daily average of the percentage of students attending class has any effect on percentage of students passing the proficiency test? ANSWER: 7.8929 TYPE: PR DIFFICULTY: Easy KEYWORDS: t test on slope, test statistic

187. Referring to Table 14-15, what is the p-value of the test statistics when testing whether daily average of the percentage of students attending class has any effect on percentage of students passing the proficiency test? ANSWER: 6.73E−10 or virtually zero TYPE: PR DIFFICULTY: Easy KEYWORDS: t test on slope, p-value

188. True or False: Referring to Table 14-15, the null hypothesis should be rejected at a 5% level of significance when testing whether daily average of the percentage of students attending class has any effect on percentage of students passing the proficiency test. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: t test on slope, decision

189. True or False: Referring to Table 14-15, there is sufficient evidence that daily average of the percentage of students attending class has an effect on percentage of students passing the proficiency test while holding constant the effect of all the other independent variables at a 5% level of significance. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: t test on slope, conclusion

190. Referring to Table 14-15, which of the following is the correct null hypothesis to determine whether there is a significant relationship between percentage of students passing the proficiency test and the entire set of explanatory variables? a) H0 : 0 = 1 = 2 = 3 = 0 b) H0 : 1 = 2 = 3 = 0

H0 : 0 = 1 = 2 = 3  0 d) H0 : 1 = 2 = 3  0 c)

ANSWER: b

267 Multiple Regression Models TYPE: MC DIFFICULTY: Easy KEYWORDS: F test on the entire regression, form of hypothesis

191. Referring to Table 14-15, which of the following is the correct alternative hypothesis to determine whether there is a significant relationship between percentage of students passing the proficiency test and the entire set of explanatory variables? a) H1 : 0 = 1 = 2 = 3  0 b) H1 : 1 = 2 = 3  0

H1 : At least one of  j  0 for j = 0, 1, 2, 3 d) H1 : At least one of  j  0 for j = 1, 2, 3 c)

ANSWER: d TYPE: MC DIFFICULTY: Easy KEYWORDS: F test on the entire regression, form of hypothesis

192. Referring to Table 14-15, the null hypothesis H0 : 1 = 2 = 3 = 0 implies that percentage of students passing the proficiency test is not affected by any of the explanatory variables. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: F test on the entire regression, interpretation, form of hypothesis

193. Referring to Table 14-15, the null hypothesis H0 : 1 = 2 = 3 = 0 implies that percentage of students passing the proficiency test is not affected by some of the explanatory variables. ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: F test on the entire regression, interpretation, form of hypothesis

194. Referring to Table 14-15, the null hypothesis H0 : 1 = 2 = 3 = 0 implies that percentage of students passing the proficiency test is not related to any of the explanatory variables. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: F test on the entire regression, interpretation, form of hypothesis

195. Referring to Table 14-15, the null hypothesis H0 : 1 = 2 = 3 = 0 implies that percentage of students passing the proficiency test is not related to one of the explanatory variables. ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: F test on the entire regression, interpretation, form of hypothesis

Multiple Regression Models

268

196. Referring to Table 14-15, the alternative hypothesis H1 : At least one of  j  0 for j = 1, 2, 3 implies that percentage of students passing the proficiency test is related to all of the explanatory variables. ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: F test on the entire regression, interpretation, form of hypothesis

197. Referring to Table 14-15, the alternative hypothesis H1 : At least one of  j  0 for j = 1, 2, 3 implies that percentage of students passing the proficiency test is related to at least one of the explanatory variables. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: F test on the entire regression, interpretation, form of hypothesis

198. Referring to Table 14-15, the alternative hypothesis H1 : At least one of  j  0 for j = 1, 2, 3 implies that percentage of students passing the proficiency test is affected by all of the explanatory variables. ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: F test on the entire regression, interpretation, form of hypothesis

199. Referring to Table 14-15, the alternative hypothesis H1 : At least one of  j  0 for j = 1, 2, 3 implies that percentage of students passing the proficiency test is affected by at least one of the explanatory variables. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: F test on the entire regression, interpretation, form of hypothesis

200. Referring to Table 14-15, what are the numerator and denominator degrees of freedom, respectively, for the test statistic to determine whether there is a significant relationship between percentage of students passing the proficiency test and the entire set of explanatory variables? ANSWER: 3 and 43 TYPE: PR DIFFICULTY: Easy KEYWORDS: F test on the entire regression, degrees of freedom

201. Referring to Table 14-15, what is the value of the test statistic to determine whether there is a significant relationship between percentage of students passing the proficiency test and the entire set of explanatory variables? ANSWER: 24.2802

269 Multiple Regression Models TYPE: PR DIFFICULTY: Easy KEYWORDS: F test on the entire regression, test statistic

202. Referring to Table 14-15, what is the p-value of the test statistic to determine whether there is a significant relationship between percentage of students passing the proficiency test and the entire set of explanatory variables? ANSWER: 2.3853E−09 or virtually zero TYPE: PR DIFFICULTY: Easy KEYWORDS: F test on the entire regression, p-value

203. True or False: Referring to Table 14-15, the null hypothesis should be rejected at a 5% level of significance when testing whether there is a significant relationship between percentage of students passing the proficiency test and the entire set of explanatory variables. ANSWER: True TYPE: PR DIFFICULTY: Easy KEYWORDS: F test on the entire regression, decision

204. True or False: Referring to Table 14-15, there is sufficient evidence that at least one of the explanatory variables is related to the percentage of students passing the proficiency test. ANSWER: True TYPE: PR DIFFICULTY: Easy KEYWORDS: F test on the entire regression, conclusion

205. True or False: Referring to Table 14-15, there is sufficient evidence that the percentage of students passing the proficiency test depends on at least one of the explanatory variables. ANSWER: True TYPE: PR DIFFICULTY: Easy KEYWORDS: F test on the entire regression, conclusion

206. True or False: Referring to Table 14-15, there is sufficient evidence that all of the explanatory variables is related to the percentage of students passing the proficiency test. ANSWER: False TYPE: PR DIFFICULTY: Moderate KEYWORDS: F test on the entire regression, conclusion

207. True or False: Referring to Table 14-15, there is sufficient evidence that the percentage of students passing the proficiency test depends on all of the explanatory variables. ANSWER: False TYPE: PR DIFFICULTY: Moderate KEYWORDS: F test on the entire regression, conclusion

Multiple Regression Models

270

208. Referring to Table 14-15, what are the lower and upper limits of the 95% confidence interval estimate for the effect of a one dollar increase in instructional spending per pupil on average percentage of students passing the proficiency test? ANSWER: −0.0034 (%) to 0.0153 (%) TYPE: PR DIFFICULTY: Moderate KEYWORDS: F test on the entire regression, confidence interval, slope, interpretation

209. Referring to Table 14-15, what are the lower and upper limits of the 95% confidence interval estimate for the effect of a one dollar increase in average teacher salary on average percentage of students passing the proficiency test? ANSWER: −0.0013 (%) to 0.0013 (%) TYPE: PR DIFFICULTY: Moderate KEYWORDS: F test on the entire regression, confidence interval, slope, interpretation

210. True or False: Referring to Table 14-15, we can conclude that average teacher salary has no impact on average percentage of students passing the proficiency test at a 5% level of significance using the 95% confidence interval estimate for 2 . ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: F test on the entire regression, confidence interval, t test on slope, decision

211. True or False: Referring to Table 14-15, we can conclude that instructional spending per pupil has no impact on average percentage of students passing the proficiency test at a 5% level of significance using the 95% confidence interval estimate for 3 . ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: F test on the entire regression, confidence interval, t test on slope, decision

212. True or False: Referring to Table 14-15, we can conclude that average teacher salary has no impact on average percentage of students passing the proficiency test at a 1% level of significance based solely on the 95% confidence interval estimate for 2 . ANSWER: True TYPE: TF DIFFICULTY: Moderate KEYWORDS: F test on the entire regression, confidence interval, t test on slope, decision

213. True or False: Referring to Table 14-15, we can conclude that instructional spending per pupil has no impact on average percentage of students passing the proficiency test at a 1% level of significance based solely on the 95% confidence interval estimate for 3 . ANSWER: True

271 Multiple Regression Models TYPE: TF DIFFICULTY: Moderate KEYWORDS: F test on the entire regression, confidence interval, t test on slope, decision

214. True or False: Referring to Table 14-15, we can conclude that average teacher salary has no impact on average percentage of students passing the proficiency test at a 10% level of significance based solely on the 95% confidence interval estimate for 2 . ANSWER: False TYPE: TF DIFFICULTY: Moderate KEYWORDS: F test on the entire regression, confidence interval, t test on slope, decision

215. True or False: Referring to Table 14-15, we can conclude that instructional spending per pupil has no impact on average percentage of students passing the proficiency test at a 10% level of significance based solely on the 95% confidence interval estimate for 3 . ANSWER: False TYPE: TF DIFFICULTY: Moderate KEYWORDS: F test on the entire regression, confidence interval, t test on slope, decision

Time-Series Analysis and Index Numbers

CHAPTER 16: TIME-SERIES ANALYSIS AND INDEX NUMBERS 1. The effect of an unpredictable, rare event will be contained in the a) trend b) cyclical c) irregular d) seasonal

component.

ANSWER: c TYPE: MC DIFFICULTY: Easy KEYWORDS: component factors, properties 2. The overall upward or downward pattern of the data in an annual time series will be contained in the component. a) trend b) cyclical c) irregular d) seasonal ANSWER: a TYPE: MC DIFFICULTY: Easy KEYWORDS: component factors, properties 3. The fairly regular fluctuations that occur within each year would be contained in the component. a) trend b) cyclical c) irregular d) seasonal ANSWER: d TYPE: MC DIFFICULTY: Easy KEYWORDS: component factors, properties 4. The annual multiplicative time-series model does not possess a) a trend b) a cyclical c) an irregular d) a seasonal ANSWER: d TYPE: MC DIFFICULTY: Easy KEYWORDS: component factors, properties

component.

48 Time-Series Analysis and Index Numbers 5. Based on the following scatter plot, which of the time-series components is not present in this quarterly time series? 350

Stock Returns

300 250 200 150 100 50 0 0

Quarters

a. b. c. d.

Trend Seasonal Cyclical Irregular

ANSWER: c TYPE: MC DIFFICULTY: Easy KEYWORDS: component factors, properties 6. The method of moving averages is used a) to plot a series. b) to exponentiate a series. c) to smooth a series. d) in regression analysis. ANSWER: c TYPE: MC DIFFICULTY: Easy KEYWORDS: moving averages 7. When using the exponentially weighted moving average for purposes of forecasting rather than smoothing, a) the previous smoothed value becomes the forecast. b) the current smoothed value becomes the forecast. c) the next smoothed value becomes the forecast. d) None of the above. ANSWER: b TYPE: MC DIFFICULTY: Moderate KEYWORDS: exponential smoothing, properties 8. In selecting an appropriate forecasting model, the following approaches are suggested: a) Perform a residual analysis. b) Measure the size of the forecasting error. c) Use the principle of parsimony.

Time-Series Analysis and Index Numbers

d) All of the above. ANSWER: d TYPE: MC DIFFICULTY: Moderate KEYWORDS: model selection 9. To assess the adequacy of a forecasting model, one measure that is often used is a) quadratic trend analysis. b) the MAD. c) exponential smoothing. d) moving averages. ANSWER: b TYPE: MC DIFFICULTY: Easy KEYWORDS: model selection 10. Which of the following methods should not be used for short-term forecasts into the future? a) Exponential smoothing b) Moving averages c) Linear trend model d) Autoregressive modeling ANSWER: b TYPE: MC DIFFICULTY: Moderate KEYWORDS: moving averages, properties 11. A model that can be used to make predictions about long-term future values of a time series is a) linear trend. b) quadratic trend. c) exponential trend. d) All of the above. ANSWER: d TYPE: MC DIFFICULTY: Easy KEYWORDS: least-squares trend fitting, properties 12. You need to decide whether you should invest in a particular stock. You would like to invest if the price is likely to rise in the long run. You have data on the daily average price of this stock over the past 12 months. Your best action is to a) compute moving averages b) perform exponential smoothing c) estimate a least square trend model d) compute the MAD statistic ANSWER: c TYPE: MC DIFFICULTY: Moderate KEYWORDS: least-squares trend fitting, properties

50 Time-Series Analysis and Index Numbers 13. When a time series appears to be increasing at an increasing rate, such that percentage difference from observation to observation is constant, the appropriate model to fit is the a. linear trend. b. quadratic trend. c. exponential trend. d. None of the above. ANSWER: c TYPE: MC DIFFICULTY: Easy KEYWORDS: least-squares trend fitting, model selection 14. The method of least squares is used on time-series data for a) eliminating irregular movements. b) deseasonalizing the data. c) obtaining the trend equation. d) exponentially smoothing a series. ANSWER: c TYPE: MC DIFFICULTY: Moderate KEYWORDS: least-squares trend fitting, properties

15. Which of the following statements about the method of exponential smoothing is not true? a) b) c) d)

It gives greater weight to more recent data. It can be used for forecasting. It uses all earlier observations in each smoothing calculation. It gives greater weight to the earlier observations in the series.

ANSWER: d TYPE: MC DIFFICULTY: Moderate KEYWORDS: exponential smoothing, properties

16. Which of the following is not an advantage of exponential smoothing? a) b) c) d)

It enables us to perform one-period ahead forecasting. It enables us to perform more than one-period ahead forecasting. It enables us to smooth out seasonal components. It enables us to smooth out cyclical components.

ANSWER: b TYPE: MC DIFFICULTY: Moderate KEYWORDS: exponential smoothing, properties

17. Which of the following statements about moving averages is not true? a) b) c) d)

It can be used to smooth a series. It gives equal weight to all values in the computation. It is simpler than the method of exponential smoothing. It gives greater weight to more recent data.

ANSWER: d TYPE: MC DIFFICULTY: Moderate

Time-Series Analysis and Index Numbers

KEYWORDS: moving averages, properties

18. After estimating a trend model for annual time-series data, you obtain the following residual plot against time: 1 0.8

Residuals

0.6 0.4 0.2 0 -0.2 0

-0.4 -0.6 -0.8 Tim e (Year)

The problem with your model is that: a) The cyclical component has not been accounted for. b) The seasonal component has not been accounted for. c) The trend component has not been accounted for. d) The irregular component has not been accounted for. ANSWER: a TYPE: MC DIFFICULTY: Moderate KEYWORDS: component factors TABLE 16-1 The number of cases of chardonnay wine sold by a Paso Robles winery in an 8-year period follows. Year Cases of Wine 1991 270 1992 356 1993 398 1994 456 1995 438 1996 478 1997 460 1998 480

19. Referring to Table 16-1, set up a scatter diagram (i.e., a time-series plot) with year on the horizontal X-axis. ANSWER:

52 Time-Series Analysis and Index Numbers

500 Number of Cases of Wine 450 400 350 300 250 200 150 100 50 0 1991

1992

1993

1994

Year

1996

1995

1997

1998

TYPE: PR DIFFICULTY: Easy KEYWORDS: scatter plot 20. Referring to Table 16-1, does there appear to be a relationship between year and the number of cases of wine sold? a) No, there appears to be no relationship between the year and the number of cases of wine sold by the vintner. b) Yes, there appears to be a slight negative linear relationship between the year and the number of cases of wine sold by the vintner. c) Yes, there appears to be a slight positive relationship between the year and the number of cases of wine sold by the vintner. d) Yes, there appears to be a negative nonlinear relationship between the year and the number of cases of wine sold by the vintner. ANSWER: c TYPE: MC DIFFICULTY: Easy KEYWORDS: scatter plot, component factor

21. After estimating a trend model for annual time-series data, you obtain the following residual plot against time: 1 0.8 0.6 0.4 0.2 0 -0.2 0

-0.4 -0.6 -0.8 -1 T ime ( Y ear )

The problem with your model is that a) the cyclical component has not been accounted for. b) the seasonal component has not been accounted for.

Time-Series Analysis and Index Numbers

c) the trend component has not been accounted for. d) the irregular component has not been accounted for. ANSWER: c TYPE: MC DIFFICULTY: Moderate KEYWORDS: component factor 22. The cyclical component of a time series a) represents periodic fluctuations which reoccur within 1 year. b) represents periodic fluctuations which usually occur in 2 or more years. c) is obtained by adding up the seasonal indexes. d) is obtained by adjusting for calendar variation. ANSWER: b TYPE: MC DIFFICULTY: Easy KEYWORDS: component factor 23. Which of the following terms describes the overall long-term tendency of a time series? a) Trend b) Cyclical component c) Irregular component d) Seasonal component ANSWER: a TYPE: MC DIFFICULTY: Easy KEYWORDS: component factor 24. Which of the following terms describes the up and down movements of a time series that vary both in length and intensity? a) Trend b) Cyclical component c) Irregular component d) Seasonal component ANSWER: b TYPE: MC DIFFICULTY: Easy KEYWORDS: component factor 25. The following is the list of MAD statistics for each of the models you have estimated from time-series data: Model MAD Linear Trend 1.38 Quadratic Trend 1.22 Exponential Trend 1.39 AR(2) 0.71 Based on the MAD criterion, the most appropriate model is a) linear trend. b) quadratic trend. c) exponential trend.

54 Time-Series Analysis and Index Numbers d) AR(2). ANSWER: d TYPE: MC DIFFICULTY: Easy KEYWORDS: model selection TABLE 16-2 The monthly advertising expenditures of a department store chain (in $1,000,000s) were collected over the last decade. The last 14 months of this time series follows: Month Expenditures ($) 1 1.4 2 1.8 3 1.6 4 1.5 5 1.8 6 1.7 7 1.9 8 2.2 9 1.9 10 1.9 11 2.1 12 2.4 13 2.8 14 3.1

26. Referring to Table 16-2, set up a scatter diagram (i.e., time-series plot) with months on the horizontal X-axis. ANSWER: 3.5

Advertising Expenditures

3 2.5 2 1.5 1 0.5 0 0

6 8 Number of M onths

TYPE: PR DIFFICULTY: Easy KEYWORDS: scatter plot 27. True or False: Referring to Table 16-2, advertising expenditures appear to be increasing in a linear rather than curvilinear manner over time.

Time-Series Analysis and Index Numbers

ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: least-squares trend fitting TABLE 16-3 The following table contains the number of complaints received in a department store for the first 6 months of last year. Month Complaints January 36 February 45 March 81 April 90 May 108 June 144 28. Referring to Table 16-3, if a three-term moving average is used to smooth this series, what would be the second calculated term? a) 36 b) 40.5 c) 54 d) 72 ANSWER: d TYPE: MC DIFFICULTY: Easy KEYWORDS: moving averages 29. Referring to Table 16-3, if a three-term moving average is used to smooth this series, what would be the last calculated term? a) 72 b) 93 c) 114 d) 126 ANSWER: c TYPE: MC DIFFICULTY: Easy KEYWORDS: moving averages 30. Referring to Table 16-3, if a three-term moving average is used to smooth this series, how many terms would it have? a) 2 b) 3 c) 4 d) 5 ANSWER: c TYPE: MC DIFFICULTY: Easy KEYWORDS: moving averages, properties

56 Time-Series Analysis and Index Numbers 31. Referring to Table 16-3, if this series is smoothed using exponential smoothing with a smoothing constant of 1/3, how many terms would it have? a) 3 b) 4 c) 5 d) 6 ANSWER: d TYPE: MC DIFFICULTY: Easy KEYWORDS: exponential smoothing, properties 32. Referring to Table 16-3, if this series is smoothed using exponential smoothing with a smoothing constant of 1/3, what would be the first term? a) 36 b) 39 c) 42 d) 45 ANSWER: a TYPE: MC DIFFICULTY: Moderate KEYWORDS: exponential smoothing 33. Referring to Table 16-3, if this series is smoothed using exponential smoothing with a smoothing constant of 1/3, what would be the second term? a) 39 b) 42 c) 45 d) 53 ANSWER: a TYPE: MC DIFFICULTY: Moderate KEYWORDS: exponential smoothing 34. Referring to Table 16-3, if this series is smoothed using exponential smoothing with a smoothing constant of 1/3, what would be the third term? a) 53 b) 65.33 c) 68 d) 81 ANSWER: a TYPE: MC DIFFICULTY: Moderate KEYWORDS: exponential smoothing 35. Referring to Table 16-3, suppose the last two smoothed values are 81 and 96 (Note: they are not). What would you forecast as the value of the time series for July? a) 81 b) 86 c) 91 d) 96

Time-Series Analysis and Index Numbers

ANSWER: d TYPE: MC DIFFICULTY: Moderate KEYWORDS: exponential smoothing, forecasting 36. Referring to Table 16-3, suppose the last two smoothed values are 81 and 96 (Note: they are not). What would you forecast as the value of the time series for September? a) 81 b) 86 c) 91 d) 96 ANSWER: d TYPE: MC DIFFICULTY: Moderate KEYWORDS: exponential smoothing, forecasting 37. If you want to recover the trend using exponential smoothing, you will choose a weight (W) that falls in the range a)  0, 0.2

 0.2, 0.4 c)  0.6, 0.8 d)  0.8,1.0 b)

ANSWER: a TYPE: MC DIFFICULTY: Easy KEYWORDS: exponential smoothing, forecasting, properties TABLE 16-4 Given below are EXCEL outputs for various estimated autoregressive models for Coca-Cola's real operating revenues (in billions of dollars) from 1975 to 1998. From the data, we also know that the real operating revenues for 1996, 1997, and 1998 are 11.7909, 11.7757 and 11.5537, respectively. AR(1) Model: Intercept XLag1

Coefficients Standard Error t Stat P-value 0.1802077 0.39797154 0.452815546 0.655325119 1.011222533 0.049685158 20.35260757 2.64373E-15

AR(2) Model: Intercept X Lag 1 X Lag 2

Coefficients Standard Error t Stat P-value 0.30047473 0.4407641 0.681713257 0.503646149 1.17322186 0.234737881 4.998008229 7.98541E-05 -0.183028189 0.250716669 -0.730020026 0.474283347

AR(3) Model: Intercept

Coefficients Standard Error t Stat P-value 0.313043288 0.514437257 0.608515972 0.550890271

58 Time-Series Analysis and Index Numbers XLag1 XLag2 XLag3

1.173719587 -0.069378567 -0.122123515

0.246490594 4.761721601 0.000180926 0.373086508 -0.185958391 0.854678245 0.282031297 -0.433014053 0.670448392

38. Referring to Table 16-4 and using a 5% level of significance, what is the appropriate AR model for Coca-Cola's real operating revenue? a) AR(1) b) AR(2) c) AR(3) d) Any of the above ANSWER: a TYPE: MC DIFFICULTY: Moderate KEYWORDS: autoregressive model, t test on slope 39. Referring to Table 16-4, if one decides to use AR(3), what will the predicted real operating revenue for Coca-Cola be in 2001? a) $11.59 billion b) $11.68 billion c) $11.84 billion d) $12.47 billion ANSWER: c TYPE: MC DIFFICULTY: Easy KEYWORDS: autoregressive model, forecasting TABLE 16-5 A contractor developed a multiplicative time-series model to forecast the number of contracts in future quarters, using quarterly data on number of contracts during the 3-year period from 1996 to 1998. The following is the resulting regression equation: ln Yˆ = 3.37 + 0.117 X – 0.083 Q1 + 1.28 Q2 + 0.617 Q3 where Yˆis the estimated number of contracts in a quarter X is the coded quarterly value with X = 0 in the first quarter of 1996. Q1 is a dummy variable equal to 1 in the first quarter of a year and 0 otherwise. Q2 is a dummy variable equal to 1 in the second quarter of a year and 0 otherwise. Q3 is a dummy variable equal to 1 in the third quarter of a year and 0 otherwise. 40. Referring to Table 16-5 , the best interpretation of the constant 3.37 in the regression equation is: a) the fitted value for the first quarter of 1996, prior to seasonal adjustment, is log10 3.37. b) the fitted value for the first quarter of 1996, after to seasonal adjustment, is log10 3.37. c) the fitted value for the first quarter of 1996, prior to seasonal adjustment, is 103.37. d) the fitted value for the first quarter of 1996, after to seasonal adjustment, is 103.37. ANSWER: c TYPE: MC DIFFICULTY: Moderate KEYWORDS: exponential model, intercept, interpretation

Time-Series Analysis and Index Numbers

41. Referring to Table 16-5, the best interpretation of the coefficient of X (0.117) in the regression equation is: a) the quarterly compound growth rate in contracts is around 30.92%. b) the annually compound growth rate in contracts is around 30.92%. c) the quarterly compound growth rate in contracts is around 11.7%. d) the annually compound growth rate in contracts is around 11.7%. ANSWER: a TYPE: MC DIFFICULTY: Moderate KEYWORDS: exponential model, slope, interpretation

42. Referring to Table 16-5, the best interpretation of the coefficient of Q3 (0.617) in the regression equation is: a) the number of contracts in the third quarter of a year is approximately 62% higher than the average over all 4 quarters. b) the number of contracts in the third quarter of a year is approximately 62% higher than it would be during the fourth quarter. c) the number of contracts in the third quarter of a year is approximately 314% higher than the average over all 4 quarters. d) the number of contracts in the third quarter of a year is approximately 314% higher than it would be during the fourth quarter. ANSWER: d TYPE: MC DIFFICULTY: Moderate KEYWORDS: exponential model, slope, interpretation 43. Referring to Table 16-5, to obtain a forecast for the first quarter of 1999 using the model, which of the following sets of values should be used in the regression equation? a) X = 12, Q1 = 0, Q2 = 0, Q3 = 0 b) X = 12, Q1 = 1, Q2 = 0, Q3 = 0 c) X = 13, Q1 = 0, Q2 = 0, Q3 = 0 d) X = 13, Q1 = 1, Q2 = 0, Q3 = 0 ANSWER: b TYPE: MC DIFFICULTY: Moderate KEYWORDS: exponential model, forecasting 44. Referring to Table 16-5, to obtain a forecast for the fourth quarter of 1999 using the model, which of the following sets of values should be used in the regression equation? a) X = 15, Q1 = 0, Q2 = 0, Q3 = 0 b) X = 15, Q1 = 1, Q2 = 0, Q3 = 0 c) X = 16, Q1 = 0, Q2 = 0, Q3 = 0 d) X = 16, Q1 = 1, Q2 = 0, Q3 = 0 ANSWER: a TYPE: MC DIFFICULTY: Moderate KEYWORDS: exponential model, forecasting

60 Time-Series Analysis and Index Numbers

45. Referring to Table 16-5, using the regression equation, which of the following values is the best forecast for the number of contracts in the third quarter of 1999? a) 49091 b) 133352 c) 421697 d) 1482518 ANSWER: c TYPE: MC DIFFICULTY: Moderate KEYWORDS: exponential model, forecasting 46. Referring to Table 16-5, using the regression equation, which of the following values is the best forecast for the number of contracts in the second quarter of 2000? a) 144212 b) 391742 c) 1238797 d) 4355119 ANSWER: d TYPE: MC DIFFICULTY: Moderate KEYWORDS: exponential model, forecasting

47. Referring to Table 16-5, in testing the coefficient of X in the regression equation (0.117) the results were a t-statistic of 9.08 and an associated p-value of 0.0000. Which of the following is the best interpretation of this result? a) The quarterly growth rate in the number of contracts is significantly different than 0% ( = 0.05). b) The quarterly growth rate in the number of contracts is not significantly different than 0% ( = 0.05). c) The quarterly growth rate in the number of contracts is significantly different than 100% ( = 0.05). d) The quarterly growth rate in the number of contracts is not significantly different than 100% ( = 0.05). ANSWER: a TYPE: MC DIFFICULTY: Moderate KEYWORDS: exponential model, t test on slope, decision, conclusion, interpretation

48. Referring to Table 16-5, in testing the coefficient for Q1 in the regression equation (– 0.083),

the results were a t-statistic of – 0.66 and an associated p-value of 0.530. Which of the following is the best interpretation of this result? a) The number of contracts in the first quarter of the year is significantly different than the number of contracts in an average quarter ( = 0.05). b) The number of contracts in the first quarter of the year is not significantly different than the number of contracts in an average quarter ( = 0.05). c) The number of contracts in the first quarter of the year is significantly different than the number of contracts in the fourth quarter for a given coded quarterly value of X (  = 0.05).

Time-Series Analysis and Index Numbers

d) The number of contracts in the first quarter of the year is not significantly different than the number of contracts in the fourth quarter for a given coded quarterly value of X ( = 0.05). ANSWER: d TYPE: MC DIFFICULTY: Moderate KEYWORDS: exponential model, t test on slope, decision, conclusion, interpretation 49. True or False: A trend is a persistent pattern in annual time-series data that has to be followed for several years. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: component factor 50. True or False: Given a data set with 15 yearly observations, a 3-year moving average will have fewer observations than a 5-year moving average. ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: moving averages, properties 51. True or False: Given a data set with 15 yearly observations, there are only thirteen 3-year moving averages. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: moving averages, properties 52. True or False: Given a data set with 15 yearly observations, there are only seven 9-year moving averages. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: moving averages, properties

53. True or False: MAD is the summation of the residuals divided by the sample size. ANSWER: False TYPE: TF DIFFICULTY: Moderate KEYWORDS: model selection 54. True or False: A least squares linear trend line is just a simple regression line with the years recoded. ANSWER: True

62 Time-Series Analysis and Index Numbers TYPE: TF DIFFICULTY: Easy KEYWORDS: least-squares trend fitting 55. True or False: If a time series does not exhibit a long-term trend, the method of exponential smoothing may be used to obtain short-term predictions about the future. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: exponential smoothing 56. True or False: The method of least squares may be used to estimate both linear and curvilinear trends. ANSWER: True TYPE: TF DIFFICULTY: Moderate KEYWORDS: least-squares trend fitting 57. True or False: The principle of parsimony indicates that the simplest model that gets the job done adequately should be used. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: model selection 58. True or False: In selecting a forecasting model, we should perform a residual analysis. ANSWER: True TYPE: TF DIFFICULTY: Moderate KEYWORDS: model selection 59. True or False: Each forecast using the method of exponential smoothing depends on all the previous observations in the time series. ANSWER: True TYPE: TF DIFFICULTY: Moderate KEYWORDS: exponential smoothing, properties

60. True or False: The MAD is a measure of the average of the absolute discrepancies between the actual and the fitted values in a given time series. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: model selection TABLE 16-6 The number of cases of merlot wine sold by a Paso Robles winery in an 8-year period follows. Year Cases of Wine

Time-Series Analysis and Index Numbers

1991 1992 1993 1994 1995 1996 1997 1998

270 356 398 456 358 500 410 376

61. Referring to Table 16-6, a centered 3-year moving average is to be constructed for the wine sales. The result of this process will lead to a total of moving averages. ANSWER: 6 TYPE: FI DIFFICULTY: Easy KEYWORDS: moving averages, properties 62. Referring to Table 16-6, a centered 3-year moving average is to be constructed for the wine sales. The moving average for 1992 is . ANSWER: 341.33 TYPE: FI DIFFICULTY: Easy KEYWORDS: moving averages, properties 63. Referring to Table 16-6, a centered 3-year moving average is to be constructed for the wine sales. The moving average for 1995 is . ANSWER: 438 TYPE: FI DIFFICULTY: Easy KEYWORDS: moving averages, properties 64. Referring to Table 16-6, construct a centered 3-year moving average for the wine sales. ANSWER: Period Cases MA 1 270 * 2 356 341.333 3 398 403.333 4 456 404.000 5 358 438.000 6 500 422.667 7 410 428.667 8 376 * TYPE: PR DIFFICULTY: Moderate KEYWORDS: moving averages 65. Referring to Table 16-6, a centered 5-year moving average is to be constructed for the wine sales. The number of moving averages that will be calculated is . ANSWER: 4 TYPE: FI DIFFICULTY: Easy

64 Time-Series Analysis and Index Numbers KEYWORDS: moving averages 66. Referring to Table 16-6, a centered 5-year moving average is to be constructed for the wine sales. The moving average for 1993 is . ANSWER: 367.6 TYPE: FI DIFFICULTY: Moderate KEYWORDS: moving averages 67. Referring to Table 16-6, a centered 5-year moving average is to be constructed for the wine sales. The moving average for 1996 is . ANSWER: 420.0 TYPE: FI DIFFICULTY: Moderate KEYWORDS: moving averages 68. Referring to Table 16-6, construct a centered 5-year moving average for the wine sales. ANSWER: Period Cases MA 1 270 * 2 356 * 3 398 367.6 4 456 413.6 5 358 424.4 6 500 420.0 7 410 * 8 376 * TYPE: PR DIFFICULTY: Moderate KEYWORDS: moving averages

69. Referring to Table 16-6, exponential smoothing with a weight or smoothing constant of 0.2 will be used to smooth the wine sales. The value of E2, the smoothed value for 1992 is . ANSWER: 287.2 TYPE: FI DIFFICULTY: Moderate KEYWORDS: exponential smoothing

70. Referring to Table 16-6, exponential smoothing with a weight or smoothing constant of 0.2 will be used to smooth the wine sales. The value of E4, the smoothed value for 1994 is . ANSWER: 338.7 TYPE: FI DIFFICULTY: Moderate KEYWORDS: exponential smoothing 71. Referring to Table 16-6, exponential smoothing with a weight or smoothing constant of 0.2 will be used to forecast wine sales. The forecast for 1999 is .

Time-Series Analysis and Index Numbers

ANSWER: 380.2 TYPE: FI DIFFICULTY: Moderate KEYWORDS: exponential smoothing, forecasting 72. Referring to Table 16-6, exponentially smooth the wine sales with a weight or smoothing constant of 0.2. ANSWER: Time CaseWine Smooth 1 270 270.000 2 356 287.200 3 398 309.360 4 456 338.688 5 358 342.550 6 500 374.040 7 410 381.232 8 376 380.186 TYPE: PR DIFFICULTY: Difficult KEYWORDS: exponential smoothing

73. Referring to Table 16-6, exponential smoothing with a weight or smoothing constant of 0.4 will be used to smooth the wine sales. The value of E2, the smoothed value for 1992 is . ANSWER: 304.4 TYPE: FI DIFFICULTY: Moderate KEYWORDS: exponential smoothing

74. Referring to Table 16-6, exponential smoothing with a weight or smoothing constant of 0.4 will be used to smooth the wine sales. The value of E5, the smoothed value for 1995 is . ANSWER: 375.7 TYPE: FI DIFFICULTY: Moderate KEYWORDS: exponential smoothing 75. Referring to Table 16-6, exponential smoothing with a weight or smoothing constant of 0.4 will be used to forecast wine sales. The forecast for 1999 is . ANSWER: 401.95 TYPE: FI DIFFICULTY: Moderate KEYWORDS: exponential smoothing, forecasting 76. Referring to Table 16-6, exponentially smooth the wine sales with a weight or smoothing constant of 0.4. ANSWER: Time 1

CaseWine Smooth 270 270.000

66 Time-Series Analysis and Index Numbers 2 356 304.400 3 398 341.840 4 456 387.504 5 358 375.702 6 500 425.421 7 410 419.253 8 376 401.952 TYPE: PR DIFFICULTY: Difficult KEYWORDS: exponential smoothing TABLE 16-7 The number of passengers arriving at San Francisco on the Amtrak cross-country express on 6 successive Mondays were: 60, 72, 96, 84, 36, and 48. 77. Referring to Table 16-7, the number of arrivals will be smoothed with a 3-term moving average. There will be a total of smoothed values. ANSWER: 4 TYPE: FI DIFFICULTY: Easy KEYWORDS: moving averages, properties 78. Referring to Table 16-7, the number of arrivals will be smoothed with a 3-term moving average. The first smoothed value will be . ANSWER: 76 TYPE: FI DIFFICULTY: Easy KEYWORDS: moving averages 79. Referring to Table 16-7, the number of arrivals will be smoothed with a 3-term moving average. The last smoothed value will be . ANSWER: 56 TYPE: FI DIFFICULTY: Easy KEYWORDS: moving averages 80. Referring to Table 16-7, the number of arrivals will be smoothed with a 5-term moving average. The first smoothed value will be . ANSWER: 69.6 TYPE: FI DIFFICULTY: Moderate KEYWORDS: moving averages 81. Referring to Table 16-7, the number of arrivals will be smoothed with a 5-term moving average. The last smoothed value will be . ANSWER: 67.2 TYPE: FI DIFFICULTY: Moderate KEYWORDS: moving averages

Time-Series Analysis and Index Numbers

82. Referring to Table 16-7, the number of arrivals will be exponentially smoothed with a smoothing constant of 0.1. The smoothed value for the second Monday will be

ANSWER: 61.2 TYPE: FI DIFFICULTY: Moderate KEYWORDS: exponential smoothing 83. Referring to Table 16-7, the number of arrivals will be exponentially smoothed with a smoothing constant of 0.1. The smoothed value for the sixth Monday will be

ANSWER: 62.0 TYPE: FI DIFFICULTY: Moderate KEYWORDS: exponential smoothing 84. Referring to Table 16-7, the number of arrivals will be exponentially smoothed with a smoothing constant of 0.1. Then the forecast for the seventh Monday will be

ANSWER: 62.0 TYPE: FI DIFFICULTY: Moderate KEYWORDS: exponential smoothing, forecasting

85. Referring to Table 16.7, exponentially smooth the number of arrivals using a smoothing constant of 0.1. ANSWER: Time Arrivals Smooth 1 60 60.0000 2 72 61.2000 3 96 64.6800 4 84 66.6120 5 36 63.5508 6 48 61.9957 TYPE: PR DIFFICULTY: Difficult KEYWORDS: exponential smoothing 86. Referring to Table 16-7, the number of arrivals will be exponentially smoothed with a smoothing constant of 0.25. The smoothed value for the second Monday will be . ANSWER: 63.0 TYPE: FI DIFFICULTY: Moderate KEYWORDS: exponential smoothing 87. Referring to Table 16-7, the number of arrivals will be exponentially smoothed with a smoothing constant of 0.25. The smoothed value for the third Monday will be ANSWER: 71.25 TYPE: FI DIFFICULTY: Moderate

68 Time-Series Analysis and Index Numbers KEYWORDS: exponential smoothing 88. Referring to Table 16-7, the number of arrivals will be exponentially smoothed with a smoothing constant of 0.25. The forecast of the number of arrivals on the seventh Monday will be . ANSWER: 60.6 TYPE: FI DIFFICULTY: Moderate KEYWORDS: exponential smoothing, forecasting 89. Referring to Table 16-7, exponentially smooth the number of arrivals using a smoothing constant of 0.25. ANSWER: Time Arrivals Smooth 1 60 60.0000 2 72 63.0000 3 96 71.2500 4 84 74.4375 5 36 64.8281 6 48 60.6211 TYPE: PR DIFFICULTY: Difficult KEYWORDS: exponential smoothing TABLE 16-8 The president of a chain of department stores believes that her stores' total sales have been showing a linear trend since 1980. She uses Microsoft Excel to obtain the partial output below. The dependent variable is sales (in millions of dollars), while the independent variable is coded years, where 1980 is coded as 0, 1981 is coded as 1, etc. SUMMARY OUTPUT Regression Statistics Multiple R R Square Adjusted R Square Standard Error Observations Intercept Coded Year

0.604 0.365 0.316 4.800 17

Coefficients 31.2 0.78

90. Referring to Table 16-8, the fitted trend value (in millions of dollars) for 1980 is . ANSWER: 31.2 TYPE: FI DIFFICULTY: Moderate KEYWORDS: least-squares trend fitting, fitted value

Time-Series Analysis and Index Numbers

91. Referring to Table 16-8, the fitted trend value (in millions of dollars) for 1985 is . ANSWER: 35.1 TYPE: FI DIFFICULTY: Moderate KEYWORDS: least-squares trend fitting, fitted value 92. Referring to Table 16-8, the estimate of the amount by which sales (in millions of dollars) is increasing each year is . ANSWER: 0.78 TYPE: FI DIFFICULTY: Moderate KEYWORDS: least-squares trend fitting, slope, interpretation 93. Referring to Table 16-8, the forecast for sales (in millions of dollars) in 2000 is

ANSWER: 46.8 TYPE: FI DIFFICULTY: Moderate KEYWORDS: least-squares trend fitting, forecasting 94. Referring to Table 16-8, the forecast for sales (in millions of dollars) in 2005 is

ANSWER: 50.7 TYPE: FI DIFFICULTY: Moderate KEYWORDS: least-squares trend fitting, forecasting TABLE 16-9 The executive vice-president of a drug manufacturing firm believes that the demand for the firm's most popular drug has been evidencing an exponential trend since 1985. She uses Microsoft Excel to obtain the partial output below. The dependent variable is the log base 10 of the demand for the drug, while the independent variable is years, where 1985 is coded as 0, 1986 is coded as 1, etc. SUMMARY OUTPUT Regression Statistics Multiple R R Square Adjusted R Square Standard Error Observations

Intercept Coded Year

0.996 0.992 0.991 0.02831 12

Coefficients 1.44 0.068

95. Referring to Table 16-9, the fitted trend value for 1985 is

70 Time-Series Analysis and Index Numbers ANSWER: 27.54 TYPE: FI DIFFICULTY: Moderate KEYWORDS: least-squares trend fitting, fitted value 96. Referring to Table 16-9, the fitted trend value for 1990 is

ANSWER: 60.26 TYPE: FI DIFFICULTY: Moderate KEYWORDS: least-squares trend fitting, fitted value

97. Referring to Table 16-9, the fitted exponential trend equation to predict Y is

ANSWER: 27.54(1.17)X TYPE: FI DIFFICULTY: Difficult KEYWORDS: least-squares trend fitting, fitted value 98. Referring to Table 16-9, the forecast for the demand in 1999 is

ANSWER: 246.6 TYPE: FI DIFFICULTY: Moderate KEYWORDS: least-squares trend fitting, forecasting 99. Referring to Table 16-9, the forecast for the demand in 2002 is

ANSWER: 394.46 TYPE: FI DIFFICULTY: Moderate KEYWORDS: least-squares trend fitting, forecasting TABLE 16-10 The manager of a marketing consulting firm has been examining his company's yearly profits. He believes that these profits have been showing a quadratic trend since 1980. He uses Microsoft Excel to obtain the partial output below. The dependent variable is profit (in thousands of dollars), while the independent variables are coded years and squared of coded years, where 1980 is coded as 0, 1981 is coded as 1, etc. SUMMARY OUTPUT Regression Statistics Multiple R R Square Adjusted R Square Standard Error Observations

Intercept Coded Year

0.998 0.996 0.996 4.996 17

Coefficients 35.5 0.45

Time-Series Analysis and Index Numbers

Year Squared 100.

1.00

Referring to Table 16-10, the fitted value for 1980 is

ANSWER: 35.5 TYPE: FI DIFFICULTY: Moderate KEYWORDS: least-squares trend fitting, fitted value 101.

Referring to Table 16-10, the fitted value for 1985 is

ANSWER: 62.75 TYPE: FI DIFFICULTY: Moderate KEYWORDS: least-squares trend fitting, fitted value 102.

Referring to Table 16-10, the forecast for profits in 2000 is

ANSWER: 444.5 TYPE: FI DIFFICULTY: Moderate KEYWORDS: least-squares trend fitting, forecasting 103.

Referring to Table 16-10, the forecast for profits in 2005 is

ANSWER: 671.75 TYPE: FI DIFFICULTY: Moderate KEYWORDS: least-squares trend fitting, forecasting 104. Microsoft Excel was used to obtain the following quadratic trend equation: Sales = 100 – 10X + 15X2. The data used was from 1989 through 1998, coded 0 to 9. The forecast for 1999 is . ANSWER: 1,500 TYPE: FI DIFFICULTY: Moderate KEYWORDS: least-squares trend fitting, forecasting 105. The manager of a company believed that her company's profits were following an exponential trend. She used Microsoft Excel to obtain a prediction equation for the logarithm (base 10) of profits: log10(Profits) = 2 + 0.3X The data she used were from 1993 through 1998, coded 0 to 5. The forecast for 1999 profits is . ANSWER: 6,309.57 TYPE: FI DIFFICULTY: Moderate KEYWORDS: least-squares trend fitting, forecasting 106.

A first-order autoregressive model for stock sales is: Salesi = 800 + 1.2(Sales)i-1.

72 Time-Series Analysis and Index Numbers If sales in 1998 is 6000, the forecast of sales for 1999 is

ANSWER: 8,000 TYPE: FI DIFFICULTY: Moderate KEYWORDS: autoregressive model, forecasting 107.

A second-order autoregressive model for average mortgage rate is: Ratei = – 2.0 + 1.8(Rate)i-1 – 0.5 (Rate)i-2. If the average mortgage rate in 1998 was 7.0, and in 1997 was 6.4, the forecast for 1999 is .

ANSWER: 7.4 TYPE: FI DIFFICULTY: Moderate KEYWORDS: autoregressive model, forecasting 108.

ANSWER: 7.82 TYPE: FI DIFFICULTY: Difficult KEYWORDS: autoregressive model, forecasting TABLE 16-11 Business closures in Laramie, Wyoming from 1989 to 1994 were: 1993 10 1994 11 1995 13 1996 19 1997 24 1998 35 Microsoft Excel was used to fit both first-order and second-order autoregressive models, resulting in the following partial outputs: SUMMARY OUTPUT – 2nd Order Model Intercept X Variable 1 X Variable 2

Coefficients -5.77 0.80 1.14

SUMMARY OUTPUT – 1st Order Model Intercept X Variable 1

Coefficients -4.16 1.59

Time-Series Analysis and Index Numbers

109.

Referring to Table 16-11, the fitted values for the first-order autoregressive model are , , , , and .

ANSWER: 11.74, 13.33, 16.51, 26.05, and 34.00 TYPE: FI DIFFICULTY: Moderate KEYWORDS: autoregressive model, fitted value 110.

Referring to Table 16-11, the residuals for the first-order autoregressive model are , , , , and .

ANSWER: -.74, -.33, 2.49, -2.05, and 1.00 TYPE: FI DIFFICULTY: Moderate KEYWORDS: autoregressive model, residual

111. Referring to Table 16-11, the value of the MAD for the first-order autoregressive model is . ANSWER: 1.322 TYPE: FI DIFFICULTY: Moderate KEYWORDS: autoregressive model, model selection 112.

Referring to Table 16-11, the fitted values for the second-order autoregressive model are , , , and .

ANSWER: 14.43, 17.17, 24.25, and 35.09 TYPE: FI DIFFICULTY: Moderate KEYWORDS: autoregressive model, fitted value 113.

Referring to Table 16-11, the residuals for the second-order autoregressive model are , , , and .

ANSWER: – 1.43, 1.83, – 0.25, and – 0.09 TYPE: FI DIFFICULTY: Moderate KEYWORDS: autoregressive model, residual

114. Referring to Table 16-11, the value of the MAD for the second-order autoregressive model is . ANSWER: 0.90 TYPE: FI DIFFICULTY: Moderate KEYWORDS: autoregressive model, model selection

115. True or False: Referring to Table 16-11, the values of the MAD for the two models indicate that the first-order model should be used for forecasting. ANSWER: False

74 Time-Series Analysis and Index Numbers TYPE: TF DIFFICULTY: Moderate KEYWORDS: autoregressive model, model selection TABLE 16-12 The manager of a health club has recorded average attendance in newly introduced step classes over the last 15 months: 32.1, 39.5, 40.3, 46.0, 65.2, 73.1, 83.7, 106.8, 118.0, 133.1, 163.3, 182.8, 205.6, 249.1, and 263.5. She then used Microsoft Excel to obtain the following partial output for both a first- and second-order autoregressive model. SUMMARY OUTPUT – 2nd Order Model Regression Statistics Multiple R R Square Adjusted R Square Standard Error Observations

Intercept X Variable 1 X Variable 2

0.993 0.987 0.985 9.276 15

Coefficients 5.86 0.37 0.85

SUMMARY OUTPUT – 1st Order Model Regression Statistics Multiple R R Square Adjusted R Square Standard Error Observations

Intercept X Variable 1

0.993 0.987 0.985 9.150 15

Coefficients 5.66 1.10

116. Referring to Table 16-12, using the first-order model, the forecast of average attendance for month 16 is . ANSWER: 295.51 TYPE: FI DIFFICULTY: Moderate KEYWORDS: autoregressive model, forecasting 117. Referring to Table 16-12, using the first-order model, the forecast of average attendance for month 17 is . ANSWER: 330.72 TYPE: FI DIFFICULTY: Moderate Y

Time-Series Analysis and Index Numbers

KEYWORDS: autoregressive model, forecasting 118. Referring to Table 16-12, using the second-order model, the forecast of average attendance for month 16 is . ANSWER: 315.09 TYPE: FI DIFFICULTY: Moderate KEYWORDS: autoregressive model, forecasting 119. Referring to Table 16-12, using the second-order model, the forecast of average attendance for month 17 is . ANSWER: 346.42 TYPE: FI DIFFICULTY: Moderate KEYWORDS: autoregressive model, forecasting 120. True or False: Referring to Table 16-12, based on the parsimony principle, the secondorder model is the better model for making forecasts. ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: autoregressive model, model selection TABLE 16-13 A local store developed a multiplicative time-series model to forecast its revenues in future quarters, using quarterly data on its revenues during the 4-year period from 1998 to 2002. The following is the resulting regression equation: log10 Yµ = 6.102 + 0.012 X – 0.129 Q1 – 0.054 Q2 + 0.098 Q3 where Yµ is the estimated number of contracts in a quarter X is the coded quarterly value with X = 0 in the first quarter of 1998. Q1 is a dummy variable equal to 1 in the first quarter of a year and 0 otherwise. Q2 is a dummy variable equal to 1 in the second quarter of a year and 0 otherwise. Q3 is a dummy variable equal to 1 in the third quarter of a year and 0 otherwise. 121. Referring to Table 16-13, the best interpretation of the constant 6.102 in the regression equation is: a) the fitted value for the first quarter of 1998, prior to seasonal adjustment, is log10(6.102). b) the fitted value for the first quarter of 1998, after to seasonal adjustment, is log10(6.102). c) the fitted value for the first quarter of 1998, prior to seasonal adjustment, is 106.102. d) the fitted value for the first quarter of 1998, after to seasonal adjustment, is 106.102. ANSWER: c TYPE: MC DIFFICULTY: Moderate KEYWORDS: exponential model, intercept, interpretation

76 Time-Series Analysis and Index Numbers

122. Referring to Table 16-13, the best interpretation of the coefficient of X (0.012) in the regression equation is: a) the quarterly compound growth rate in revenues is around 2.8%. b) the annual growth rate in revenues is around 2.8%. c) the quarterly growth rate in revenues is around 1.2%. d) the annual growth rate in revenues is around 1.2%. ANSWER: a TYPE: MC DIFFICULTY: Moderate KEYWORDS: exponential model, slope, interpretation

123. Referring to Table 16-13, the estimated quarterly compound growth rate in revenues is around: a) 1.2%. b) 2.8%. c) 12%. d) 28%. ANSWER: b TYPE: MC DIFFICULTY: Moderate KEYWORDS: exponential model, slope, interpretation

124. Referring to Table 16-13, the best interpretation of the coefficient of Q2 (–0.054) in the regression equation is: a) the revenues in the second quarter of a year is approximately 5.4% lower than the average over all 4 quarters. b) the revenues in the second quarter of a year is approximately 5.4% lower than it would be during the fourth quarter. c) the revenues in the second quarter of a year is approximately 11.69% lower than the average over all 4 quarters. d) the revenues in the second quarter of a year is approximately 11.69% lower than it would be during the fourth quarter. ANSWER: d TYPE: MC DIFFICULTY: Moderate KEYWORDS: exponential model, slope, interpretation

125. Referring to Table 16-13, the best interpretation of the coefficient of Q3 (0.098) in the regression equation is: a) the revenues in the third quarter of a year is approximately 9.8% higher than the average over all 4 quarters. b) the revenues in the third quarter of a year is approximately 9.8% higher than it would be during the fourth quarter. c) the revenues in the third quarter of a year is approximately 25.31% higher than the average over all 4 quarters. d) the revenues in the third quarter of a year is approximately 25.31% higher than it would be during the fourth quarter. ANSWER: d

Time-Series Analysis and Index Numbers

TYPE: MC DIFFICULTY: Moderate KEYWORDS: exponential model, slope, interpretation 126. Referring to Table 16-13, to obtain a forecast for the first quarter of 2002 using the model, which of the following sets of values should be used in the regression equation? a) X = 16, Q1 = 1, Q2 = 0, Q3 = 0 b) X = 16, Q1 = 0, Q2 = 1, Q3 = 0 c) X = 17, Q1 = 1, Q2 = 0, Q3 = 0 d) X = 17, Q1 = 0, Q2 = 1, Q3 = 0 ANSWER: a TYPE: MC DIFFICULTY: Moderate KEYWORDS: exponential model, forecasting 127. Referring to Table 16-13, to obtain a forecast for the fourth quarter of 1999 using the model, which of the following sets of values should be used in the regression equation? a) X = 7, Q1 = 0, Q2 = 0, Q3 = 0 b) X = 7, Q1 = 1, Q2 = 0, Q3 = 0 c) X = 8, Q1 = 0, Q2 = 0, Q3 = 0 d) X = 8, Q1 = 1, Q2 = 0, Q3 = 0 ANSWER: a TYPE: MC DIFFICULTY: Moderate KEYWORDS: exponential model, forecasting 128. Referring to Table 16-13, to obtain a forecast for the third quarter of 2003 using the model, which of the following sets of values should be used in the regression equation? a) X = 22, Q1 = 0, Q2 = 0, Q3 = 0 b) X = 22, Q1 = 0, Q2 = 0, Q3 = 1 c) X = 23, Q1 = 0, Q2 = 0, Q3 = 0 d) X = 23, Q1 = 0, Q2 = 0, Q3 = 1 ANSWER: b TYPE: MC DIFFICULTY: Moderate KEYWORDS: exponential model, forecasting 129. Referring to Table 16-13, using the regression equation, what is the forecast for the revenues in the third quarter of 2003? ANSWER: 2,910,717.12 TYPE: PR DIFFICULTY: Moderate KEYWORDS: exponential model, forecasting 130. Referring to Table 16-13, using the regression equation, what is the forecast for the revenues in the first quarter of 2005? ANSWER: 2,037,042.08

78 Time-Series Analysis and Index Numbers TYPE: PR DIFFICULTY: Moderate KEYWORDS: exponential model, forecasting 131. Referring to Table 16-13, using the regression equation, what is the forecast for the revenues in the fourth quarter of 2004? ANSWER: 2,666,858.67 TYPE: PR DIFFICULTY: Moderate KEYWORDS: exponential model, forecasting

132. Referring to Table 16-13, in testing the significance of the coefficient of X in the regression equation (0.012) which has a p-value of 0.0000. Which of the following is the best interpretation of this result? a) The quarterly growth rate in revenues is significantly different from 0% ( = 0.05). b) The quarterly growth rate in revenues is not significantly different from 0% (  = 0.05). c) The quarterly growth rate in revenues is significantly different from 1.2% ( = 0.05). d) The quarterly growth rate in revenues is not significantly different from 1.2% ( = 0.05). ANSWER: a TYPE: MC DIFFICULTY: Moderate KEYWORDS: exponential model, t test on slope, decision, conclusion, interpretation

133. Referring to Table 16-13, in testing the significance of the coefficient for Q1 in the regression equation (– 0.129) which has a p-value of 0.492. Which of the following is the best interpretation of this result? a) The revenues in the first quarter of the year are significantly different than the revenues in an average quarter ( = 0.05). b) The revenues in the first quarter of the year are not significantly different than the revenues in an average quarter ( = 0.05). c) The revenues in the first quarter of the year are significantly different than the revenues in the fourth quarter ( = 0.05). d) The revenues in the first quarter of the year are not significantly different than the revenues in the fourth quarter ( = 0.05). ANSWER: d TYPE: MC DIFFICULTY: Moderate KEYWORDS: exponential model, t test on slope, decision, conclusion, interpretation 134. True or False: A simple price index tracks the price of a group of commodities at a given period of time to the price paid for that group of commodities at a particular point of time in the past. ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: simple price index, aggregate price index

Time-Series Analysis and Index Numbers

135. True or False: For a price index, it is preferable to select the base period in a period of economic stability. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: base period

136. True or False: The base period should be recent so that a price index is not severely affected by change in technology, and consumer attitudes and habits. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: base period 137. True or False: The more expensive commodities are overly influential in an unweighted aggregate price index. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: unweighted aggregate price index, properties 138. True or False: The changes in the price of the least consumed commodities are overly influential in a weighted aggregate price index. ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: unweighted aggregate price index, properties

139. True or False: Unweighted aggregate price indexes account for differences in the magnitude of prices per unit and differences in the consumption levels of the items in the market basket. ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: unweighted aggregate price index, weighted aggregate price index, properties

140. True or False: The Laspeyres price index is a form of weighted aggregate price index. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: weighted aggregate price index, Laspeyres price index

141. True or False: The Paasche price index is a form of unweighted aggregate price index. ANSWER: False TYPE: TF DIFFICULTY: Easy

80 Time-Series Analysis and Index Numbers KEYWORDS: weighted aggregate price index, Paasche price index 142. True or False: The Laspeyres price index uses the initial consumption quantities as the weights. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: Laspeyres price index, properties 143. True or False: The Paasche price index uses the consumption quantities in the year of interest as the weights. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: Paasche price index, properties 144. True or False: The Paasche price index reflects more accurately the consumption cost at a point in time because it uses the consumption quantities in the initial year as the base. ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: Paasche price index, properties 145. True or False: The Paasche price index has the disadvantage that current consumption quantities are usually hard to obtain. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: Paasche price index, properties

146. True or False: The Laspeyres price index has the disadvantage that the consumption pattern in the initial period might be quite different from that in the current period and, hence, does not reflect accurately the current consumption cost. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: Laspeyres price index, properties 147.

True of False: The consumer price index is a Paasche price index.

ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: Laspeyres price index, Paasche price index TABLE 16-14 Given below are the average prices for three types of energy products in the United States from 1992 to 1995.

Time-Series Analysis and Index Numbers

Year Electricity 1992 43.205 1993 16.959 1994 47.202 1995 48.874 1996

Natural Gas

48.693

25.893 28.749 28.933 29.872 28.384

Fuel Oil 0.892 0.969 1.034 0.913 0.983

148. Referring to Table 16-14, what are the simple price indexes for electricity, natural gas and fuel oil, respectively, in 1994 using 1992 as the base year? ANSWER: 109.25, 111.74 and 115.92 TYPE: PR DIFFICULTY: Easy KEYWORDS: simple price index 149. Referring to Table 16-14, what are the simple price indexes for electricity, natural gas and fuel oil, respectively, in 1995 using 1992 as the base year? ANSWER: 113.12, 115.37 and 102.35 TYPE: PR DIFFICULTY: Easy KEYWORDS: simple price index 150. Referring to Table 16-14, what are the simple price indexes for electricity, natural gas and fuel oil, respectively, in 1996 using 1992 as the base year? ANSWER: 112.70, 109.62 and 110.20 TYPE: PR DIFFICULTY: Easy KEYWORDS: simple price index

151. Referring to Table 16-14, , what are the simple price indexes for electricity, natural gas and fuel oil, respectively, in 1992 using 1996 as the base year? ANSWER: 88.73, 91.22 and 90.74 TYPE: PR DIFFICULTY: Easy KEYWORDS: simple price index 152. Referring to Table 16-14, , what are the simple price indexes for electricity, natural gas and fuel oil, respectively, in 1993 using 1996 as the base year? ANSWER: 34.83, 101.29 and 98.58 TYPE: PR DIFFICULTY: Easy KEYWORDS: simple price index 153. Referring to Table 16-14, , what are the simple price indexes for electricity, natural gas and fuel oil, respectively, in 1994 using 1996 as the base year? ANSWER: 96.94, 101.93 and 105.19 TYPE: PR DIFFICULTY: Easy

82 Time-Series Analysis and Index Numbers KEYWORDS: simple price index

154. Referring to Table 16-14, , what is the unweighted price index for the group of three energy items in 1994 using 1992 as the base year? ANSWER: 110.26 TYPE: PR DIFFICULTY: Moderate KEYWORDS: unweighted price index 155. Referring to Table 16-14, , what is the unweighted price index for the group of three energy items in 1995 using 1992 as the base year? ANSWER: 113.81 TYPE: PR DIFFICULTY: Moderate KEYWORDS: unweighted price index 156. Referring to Table 16-14, , what is the unweighted price index for the group of three energy items in 1996 using 1992 as the base year? ANSWER: 111.53 TYPE: PR DIFFICULTY: Moderate KEYWORDS: unweighted price index 157. Referring to Table 16-14, what is the Paasche price index for the group of three energy items in 1994 for a family that consumed 13 units of electricity, 26 unites of natural gas and 235 units of fuel oil in 1994 using 1992 as the base year? ANSWER: 111.38 TYPE: PR DIFFICULTY: Difficult KEYWORDS: Paasche price index

158. Referring to Table 16-14, what is the Paasche price index for the group of three energy items in 1995 for a family that consumed 13 units of electricity, 26 unites of natural gas and 235 units of fuel oil in 1995 using 1992 as the base year? ANSWER: 112.61 TYPE: PR DIFFICULTY: Difficult KEYWORDS: Paasche price index 159. Referring to Table 16-14, what is the Paasche price index for the group of three energy items in 1996 for a family that consumed 13 units of electricity, 26 unites of natural gas and 235 units of fuel oil in 1996 using 1992 as the base year? ANSWER: 110.90 TYPE: PR DIFFICULTY: Difficult KEYWORDS: Paasche price index

Time-Series Analysis and Index Numbers

160. Referring to Table 16-14, what is the Laspeyres price index for the group of three energy items in 1994 for a family that consumed 15 units of electricity, 24 unites of natural gas and 200 units of fuel oil in 1992 using 1992 as the base year? ANSWER: 111.14 TYPE: PR DIFFICULTY: Difficult KEYWORDS: Laspeyres price index 161. Referring to Table 16-14, what is the Laspeyres price index for the group of three energy items in 1995 for a family that consumed 15 units of electricity, 24 unites of natural gas and 200 units of fuel oil in 1992 using 1992 as the base year? ANSWER: 112.76 TYPE: PR DIFFICULTY: Difficult KEYWORDS: Laspeyres price index 162. Referring to Table 16-14, what is the Laspeyres price index for the group of three energy items in 1996 for a family that consumed 15 units of electricity, 24 unites of natural gas and 200 units of fuel oil in 1992 using 1992 as the base year? ANSWER: 111.07 TYPE: PR DIFFICULTY: Difficult KEYWORDS: Laspeyres price index

Decision Making

CHAPTER 17: DECISION MAKING

1. A tabular presentation that shows the outcome for each decision alternative under the various states of nature is called: a) a payback period matrix. b) a decision matrix. c) a decision tree. d) a payoff table. ANSWER: d TYPE: MC DIFFICULTY: Easy KEYWORDS: payoff 2. The difference between expected payoff under certainty and expected value of the best act without certainty is the: a) expected monetary value. b) expected net present value. c) expected value of perfect information. d) expected rate of return. ANSWER: c TYPE: MC DIFFICULTY: Easy KEYWORDS: expected value of perfect information 3. A medical doctor is involved in a $1 million malpractice suit. He can either settle out of court for $250,000 or go to court. If he goes to court and loses, he must pay $825,000 plus $175,000 in court costs. If he wins in court the plaintiffs pay the court costs. Identify the actions of this decision-making problem. a) Two choices: (1) go to court and (2) settle out of court. b) Two possibilities: (1) win the case in court and (2) lose the case in court. c) Four consequences resulting from Go/Settle and Win/Lose combinations. d) The amount of money paid by the doctor. ANSWER: a TYPE: MC DIFFICULTY: Easy KEYWORDS: courses of action 4. A medical doctor is involved in a $1 million malpractice suit. He can either settle out of court for $250,000 or go to court. If he goes to court and loses, he must pay $825,000 plus $175,000 in court costs. If he wins in court the plaintiffs pay the court costs. Identify the states of nature of this decision-making problem. a) Two choices: (1) go to court and (2) settle out of court. b) Two possibilities: (1) win the case in court and (2) lose the case in court. c) Four consequences resulting from Go/Settle and Win/Lose combinations. d) The amount of money paid by the doctor. ANSWER:

Decision Making b TYPE: MC DIFFICULTY: Easy KEYWORDS: states of the world 5. A medical doctor is involved in a $1 million malpractice suit. He can either settle out of court for $250,000 or go to court. If he goes to court and loses, he must pay $825,000 plus $175,000 in court costs. If he wins in court the plaintiffs pay the court costs. Identify the outcomes of this decision-making problem. a) Two choices: (1) go to court and (2) settle out of court. b) Two possibilities: (1) win the case in court and (2) lose the case in court. c) Four consequences resulting from Go/Settle and Win/Lose combinations. d) The amount of money paid by the doctor. ANSWER: c TYPE: MC DIFFICULTY: Easy KEYWORDS: outcomes 6. A company that manufactures designer jeans is contemplating whether to increase its advertising budget by $1 million for next year. If the expanded advertising campaign is successful, the company expects sales to increase by $1.6 million next year. If the advertising campaign fails, the company expects sales to increase by only $400,000 next year. If the advertising budget is not increased, the company expects sales to increase by $200,000. Identify the states of nature in this decision-making problem. a) Two choices: (1) increase the budget and (2) do not increase the budget. b) Two possibilities: (1) campaign is successful and (2) campaign is not successful. c) Four consequences resulting from the Increase/Do Not Increase and Successful/Not Successful combinations. d) The increase in sales dollars next year. ANSWER: b TYPE: MC DIFFICULTY: Easy KEYWORDS: states of the world 7. A company that manufactures designer jeans is contemplating whether to increase its advertising budget by $1 million for next year. If the expanded advertising campaign is successful, the company expects sales to increase by $1.6 million next year. If the advertising campaign fails, the company expects sales to increase by only $400,000 next year. If the advertising budget is not increased, the company expects sales to increase by $200,000. Identify the actions in this decision-making problem. a) Two choices: (1) increase the budget and (2) do not increase the budget. b) Two possibilities: (1) campaign is successful and (2) campaign is not successful. c) Four consequences resulting from the Increase/Do Not Increase and Successful/Not Successful combinations. d) The increase in sales dollars next year. ANSWER: a TYPE: MC DIFFICULTY: Easy KEYWORDS: courses of action 8. A company that manufactures designer jeans is contemplating whether to increase its advertising budget by $1 million for next year. If the expanded advertising campaign is successful, the

Decision Making

company expects sales to increase by $1.6 million next year. If the advertising campaign fails, the company expects sales to increase by only $400,000 next year. If the advertising budget is not increased, the company expects sales to increase by $200,000. Identify the outcomes in this decision-making problem. a) Two choices: (1) increase the budget and (2) do not increase the budget. b) Two possibilities: (1) campaign is successful and (2) campaign is not successful. c) Four consequences resulting from the Increase/Do Not Increase and Successful/Not Successful combinations. d) The increase in sales dollars next year. ANSWER: c TYPE: MC DIFFICULTY: Easy KEYWORDS: outcomes 9. A company that manufactures designer jeans is contemplating whether to increase its advertising budget by $1 million for next year. If the expanded advertising campaign is successful, the company expects sales to increase by $1.6 million next year. If the advertising campaign fails, the company expects sales to increase by only $400,000 next year. If the advertising budget is not increased, the company expects sales to increase by $200,000. Identify the objective variable in this decision-making problem. a) Two choices: (1) increase the budget and (2) do not increase the budget. b) Two possibilities: (1) campaign is successful and (2) campaign is not successful. c) Four consequences resulting from the Increase/Do Not Increase and Successful/Not Successful combinations. d) The increase in sales dollars next year. ANSWER: d TYPE: MC DIFFICULTY: Easy KEYWORDS: objective TABLE 17-1 The following payoff table shows profits associated with a set of 3 alternatives under 2 possible states of nature. States A1 A2 A3 1 12 –2 8 2 4 10 5 where: S1 is state of nature 1 A1 is action alternative 1 S2 is state of nature 2 A2 is action alternative 2 A3 is action alternative 3

10. Referring to Table 17-1, the opportunity loss for A3 when S2 occurs is a) 0 b) 4 c) 5 d) 6

ANSWER: c TYPE: MC DIFFICULTY: Moderate KEYWORDS: opportunity loss

Decision Making

11. Referring to Table 17-1, the opportunity loss for A2 when S1 occurs is a) b) c) d)

–2 0 5 14

ANSWER: d TYPE: MC DIFFICULTY: Moderate KEYWORDS: opportunity loss

12. Referring to Table 17-1, if the probability of S1 is 0.4, then the probability of S2 is a) 0.4 b) 0.5 c) 0.6 d) 1.0 ANSWER: c TYPE: MC DIFFICULTY: Moderate KEYWORDS: probability

13. Referring to Table 17-1, if the probability of S1 is 0.2 and S2 is 0.8, then the expected monetary value of A1 is a) 2.4 b) 5.6 c) 8 d) 16 ANSWER: b TYPE: MC DIFFICULTY: Moderate KEYWORDS: expected monetary value

14. Referring to Table 17-1, if the probability of S1 is 0.2 and S2 is 0.8, then the expected opportunity loss (EOL) for A1 is a) 0 b) 1.2 c) 4.8 d) 5.6 ANSWER: c TYPE: MC DIFFICULTY: Moderate KEYWORDS: expected opportunity loss

15. Referring to Table 17-1, if the probability of S1 is 0.2, what is the optimal alternative using EOL?

a) A1. b) A2. c) A3. d) It cannot be determined.

Decision Making

ANSWER: b TYPE: MC DIFFICULTY: Moderate KEYWORDS: decision making, expected opportunity loss

16. Referring to Table 17-1, if the probability of S1 is 0.5, then the expected monetary value (EMV ) for A1 is a) 3 b) 4 c) 6.5 d) 8 ANSWER: d TYPE: MC DIFFICULTY: Moderate KEYWORDS: expected monetary value

17. Referring to Table 17-1, if the probability of S1 is 0.5, then the expected monetary value (EMV ) for A2 is a) 3 b) 4 c) 6.5 d) 8 ANSWER: b TYPE: MC DIFFICULTY: Moderate KEYWORDS: expected monetary value

18. Referring to Table 17-1, if the probability of S1 is 0.5, then the expected opportunity loss (EOL) for A1 is a) 3 b) 4.5 c) 7 d) 8 ANSWER: a TYPE: MC DIFFICULTY: Moderate KEYWORDS: expected opportunity loss

19. Referring to Table 17-1, if the probability of S1 is 0.5, then the expected opportunity loss (EOL) for A3 is a) 3 b) 4.5 c) 7 d) 8 ANSWER: b TYPE: MC DIFFICULTY: Moderate KEYWORDS: expected opportunity loss

100 Decision Making

20. Referring to Table 17-1, if the probability of S1 is 0.5, what is the optimal alternative using EMV?

a) A1 b) A2 c) A3 d) It cannot be determined. ANSWER: a TYPE: MC DIFFICULTY: Moderate KEYWORDS: decision making, expected monetary value

21. Referring to Table 17-1, if the probability of S1 is 0.5, then the EVPI for the payoff table is a) b) c) d)

–3 3 8 11

ANSWER: b TYPE: MC DIFFICULTY: Moderate KEYWORDS: expected value of perfect information

22. Referring to Table 17-1, if the probability of S1 is 0.5, then the coefficient of variation for A1 is a) 0.231 b) 0.5 c) 1.5 d) 2 ANSWER: b TYPE: MC DIFFICULTY: Moderate KEYWORDS: coefficient of variation

23. Referring to Table 17-1, if the probability of S1 is 0.5, then the coefficient of variation for A2 is a) 0.231 b) 0.5 c) 1.5 d) 2 ANSWER: c TYPE: MC DIFFICULTY: Moderate KEYWORDS: coefficient of variation

24. Referring to Table 17-1, if the probability of S1 is 0.5, then the return to risk ratio for A1 is a) 0.667 b) 1.5 c) 2 d) 4.333 ANSWER:

Decision Making

101

c TYPE: MC DIFFICULTY: Moderate KEYWORDS: return to risk ratio

25. Referring to Table 17-1, if the probability of S1 is 0.5, then the return to risk ratio for A3 is a) 0.667 b) 1.5 c) 2 d) 4.333 ANSWER: d TYPE: MC DIFFICULTY: Moderate KEYWORDS: return to risk ratio

26. Referring to Table 17-1, if the probability of S1 is 0.5, then the expected profit under certainty (EPUC ) is a) 3 b) 5 c) 8 d) 11 ANSWER: d TYPE: MC DIFFICULTY: Moderate KEYWORDS: expected profit under certainty 27. Blossom’s Flowers purchases roses for sale for Valentine’s Day. The roses are purchased for $10 a dozen and are sold for $20 a dozen. Any roses not sold on Valentine’s Day can be sold for $5 per dozen. The owner will purchase 1 of 3 amounts of roses for Valentine’s Day: 100, 200, or 400 dozen roses. The number of alternatives for the payoff table is a) 2 b) 3 c) 4 d) It cannot be determined. ANSWER: b TYPE: MC DIFFICULTY: Moderate KEYWORDS: courses of action 28. Blossom’s Flowers purchases roses for sale for Valentine’s Day. The roses are purchased for $10 a dozen and are sold for $20 a dozen. Any roses not sold on Valentine’s Day can be sold for $5 per dozen. The owner will purchase 1 of 3 amounts of roses for Valentine’s Day: 100, 200, or 400 dozen roses. The number of states of nature for the payoff table is a) 2 b) 3 c) 4 d) It cannot be determined. ANSWER: b TYPE: MC DIFFICULTY: Moderate

102 Decision Making KEYWORDS: states of the world 29. Blossom’s Flowers purchases roses for sale for Valentine’s Day. The roses are purchased for $10 a dozen and are sold for $20 a dozen. Any roses not sold on Valentine’s Day can be sold for $5 per dozen. The owner will purchase 1 of 3 amounts of roses for Valentine’s Day: 100, 200, or 400 dozen roses. The payoff for buying 200 dozen roses and selling 100 dozen roses at the full price is a) $2,000 b) $1,000 c) $500 d) – $500 ANSWER: c TYPE: MC DIFFICULTY: Moderate KEYWORDS: payoff 30. Blossom’s Flowers purchases roses for sale for Valentine’s Day. The roses are purchased for $10 a dozen and are sold for $20 a dozen. Any roses not sold on Valentine’s Day can be sold for $5 per dozen. The owner will purchase 1 of 3 amounts of roses for Valentine’s Day: 100, 200, or 400 dozen roses. The payoff for buying and selling 400 dozen roses at the full price is a) $12,000 b) $6,000 c) $4,000 d) It cannot be determined. ANSWER: c TYPE: MC DIFFICULTY: Moderate KEYWORDS: payoff 31. Blossom’s Flowers purchases roses for sale for Valentine’s Day. The roses are purchased for $10 a dozen and are sold for $20 a dozen. Any roses not sold on Valentine’s Day can be sold for $5 per dozen. The owner will purchase 1 of 3 amounts of roses for Valentine’s Day: 100, 200, or 400 dozen roses. The opportunity loss for buying 200 dozen roses and selling 100 dozen roses at the full price is a) $1,000 b) $500 c) – $500 d) – $2,000 ANSWER: b TYPE: MC DIFFICULTY: Moderate KEYWORDS: opportunity loss 32. Blossom’s Flowers purchases roses for sale for Valentine’s Day. The roses are purchased for $10 a dozen and are sold for $20 a dozen. Any roses not sold on Valentine’s Day can be sold for $5 per dozen. The owner will purchase 1 of 3 amounts of roses for Valentine’s Day: 100, 200, or 400 dozen roses. The opportunity loss for buying 400 dozen roses and selling 200 dozen roses at the full price is a) – $2,000 b) $1,000 c) $500

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d) $0 ANSWER: b TYPE: MC DIFFICULTY: Moderate KEYWORDS: opportunity loss 33. Blossom’s Flowers purchases roses for sale for Valentine’s Day. The roses are purchased for $10 a dozen and are sold for $20 a dozen. Any roses not sold on Valentine’s Day can be sold for $5 per dozen. The owner will purchase 1 of 3 amounts of roses for Valentine’s Day: 100, 200, or 400 dozen roses. If the probability of selling 100 dozen roses is 0.2 and 200 dozen roses is 0.5, then the probability of selling 400 dozen roses is a) 0.7 b) 0.5 c) 0.3 d) 0.2 ANSWER: c TYPE: MC DIFFICULTY: Moderate KEYWORDS: probability

34. Blossom’s Flowers purchases roses for sale for Valentine’s Day. The roses are purchased for $10 a dozen and are sold for $20 a dozen. Any roses not sold on Valentine’s Day can be sold for $5 per dozen. The owner will purchase 1 of 3 amounts of roses for Valentine’s Day: 100, 200, or 400 dozen roses. Given 0.2, 0.4, and 0.4 are the probabilities for the sale of 100, 200, or 400 dozen roses, respectively, then the EMV for buying 200 dozen roses is a) $4,500 b) $2,500 c) $1,700 d) $1,000

ANSWER: c TYPE: MC DIFFICULTY: Moderate KEYWORDS: expected monetary value

35. Blossom’s Flowers purchases roses for sale for Valentine’s Day. The roses are purchased for $10 a dozen and are sold for $20 a dozen. Any roses not sold on Valentine’s Day can be sold for $5 per dozen. The owner will purchase 1 of 3 amounts of roses for Valentine’s Day: 100, 200, or 400 dozen roses. Given 0.2, 0.4, and 0.4 are the probabilities for the sale of 100, 200, or 400 dozen roses, respectively, then the EOL for buying 200 dozen roses is a) $700 b) $900 c) $1,500 d) $1,600

ANSWER: b TYPE: MC DIFFICULTY: Moderate KEYWORDS: expected opportunity loss

104 Decision Making

36. Blossom’s Flowers purchases roses for sale for Valentine’s Day. The roses are purchased for $10 a dozen and are sold for $20 a dozen. Any roses not sold on Valentine’s Day can be sold for $5 per dozen. The owner will purchase 1 of 3 amounts of roses for Valentine’s Day: 100, 200, or 400 dozen roses. Given 0.2, 0.4, and 0.4 are the probabilities for the sale of 100, 200, or 400 dozen roses, respectively, then the optimal EOL for buying roses is a) $700 b) $900 c) $1,500 d) $1,600

ANSWER: a TYPE: MC DIFFICULTY: Moderate KEYWORDS: decision making, expected opportunity loss

37. Blossom’s Flowers purchases roses for sale for Valentine’s Day. The roses are purchased for $10 a dozen and are sold for $20 a dozen. Any roses not sold on Valentine’s Day can be sold for $5 per dozen. The owner will purchase 1 of 3 amounts of roses for Valentine’s Day: 100, 200, or 400 dozen roses. Given 0.2, 0.4, and 0.4 are the probabilities for the sale of 100, 200, or 400 dozen roses, respectively, then the optimal alternative using EMV for selling roses is to buy dozen roses. a) 100 b) 200 c) 400 d) 600

ANSWER: c TYPE: MC DIFFICULTY: Moderate KEYWORDS: decision making, expected monetary value

38. Blossom’s Flowers purchases roses for sale for Valentine’s Day. The roses are purchased for $10 a dozen and are sold for $20 a dozen. Any roses not sold on Valentine’s Day can be sold for $5 per dozen. The owner will purchase 1 of 3 amounts of roses for Valentine’s Day: 100, 200, or 400 dozen roses. Given 0.2, 0.4, and 0.4 are the probabilities for the sale of 100, 200, or 400 dozen roses, respectively, then the optimal EMV for buying roses is a) $700 b) $900 c) $1,700 d) $1,900

ANSWER: d TYPE: MC DIFFICULTY: Moderate KEYWORDS: decision making, expected monetary value

39. Blossom’s Flowers purchases roses for sale for Valentine’s Day. The roses are purchased for $10 a dozen and are sold for $20 a dozen. Any roses not sold on Valentine’s Day can be sold for $5 per dozen. The owner will purchase 1 of 3 amounts of roses for Valentine’s Day: 100, 200, or 400 dozen roses. Given 0.2, 0.4, and 0.4 are the probabilities for the sale of 100, 200, or 400 dozen roses, respectively, then the EVPI for buying roses is a) $700 b) $1,500

Decision Making

c) $1,900 d) $2,600 ANSWER: a TYPE: MC DIFFICULTY: Moderate KEYWORDS: expected value of perfect information TABLE 17-2 The following payoff matrix is given in dollars. Action Event A B 1 400 700 2 200 500 Suppose the probability of Event 1 is 0.5 and Event 2 is 0.5.

40. Referring to Table 17-2, the EMV for Action A is a) $300 b) $550 c) $600 d) $700 ANSWER: a TYPE: MC DIFFICULTY: Moderate KEYWORDS: expected monetary value

41. Referring to Table 17-2, the EOL for Action A is a) 0 b) 100 c) 200 d) 300 ANSWER: d TYPE: MC DIFFICULTY: Moderate KEYWORDS: expected opportunity loss

42. Referring to Table 17-2, the coefficient of variation for Action A is a) 12.8% b) 33.3% c) 133.33% d) 333.3% ANSWER: b TYPE: MC DIFFICULTY: Moderate KEYWORDS: coefficient of variation

43. Referring to Table 17-2, the return to risk ratio for Action B is a) 0.167 b) 3.0

105

106 Decision Making c) 6.0 d) 9.0 ANSWER: c TYPE: MC DIFFICULTY: Moderate KEYWORDS: return to risk ratio 44. Referring to Table 17-2, what is the action with the preferable return to risk ratio? a) Action A b) Action B c) Either Action A or Action B d) It cannot be determined. ANSWER: b TYPE: MC DIFFICULTY: Moderate KEYWORDS: decision making, return to risk action 45. Referring to Table 17-2, what is the action with the preferable coefficient of variation? a) Action A b) Action B c) Either Action A or Action B d) It cannot be determined. ANSWER: b TYPE: MC DIFFICULTY: Moderate KEYWORDS: decision making, coefficient of variation

46. Referring to Table 17-2, what is the optimal action using the EMV criterion? a) Action A b) Action B c) Either Action A or Action B d) It cannot be determined. ANSWER: b TYPE: MC DIFFICULTY: Moderate KEYWORDS: decision making, expected monetary value

47. Referring to Table 17-2, what is the optimal action using the EOL criterion? a) Action A b) Action B c) Either Action A or Action B d) It cannot be determined. ANSWER: b TYPE: MC DIFFICULTY: Moderate KEYWORDS: decision making, expected opportunity loss

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48. Referring to Table 17-2, the EVPI is a) 0 b) 300 c) 400 d) 600 ANSWER: a TYPE: MC DIFFICULTY: Moderate KEYWORDS: expected value of perfect information

49. Referring to Table 17-2, the expected profit under certainty (EPUC ) is a) 0 b) 300 c) 500 d) 600 ANSWER: d TYPE: MC DIFFICULTY: Moderate KEYWORDS: expected profit under certainty

50. For a potential investment of $5,000, a portfolio has an EMV of $1,000 and a standard deviation of $100. What is the rate of return? a) 5% b) 10% c) 20% d) 50% ANSWER: c TYPE: MC DIFFICULTY: Moderate KEYWORDS: rate of return

51. For a potential investment of $5,000, a portfolio has an EMV of $1,000 and a standard deviation of $100. What is the coefficient of variation? a) 10% b) 20% c) 50% d) 100% ANSWER: a TYPE: MC DIFFICULTY: Moderate KEYWORDS: coefficient of variation

52. For a potential investment of $5,000, a portfolio has an EMV of $1,000 and a standard deviation of $100. The return to risk ratio is a) 50 b) 20 c) 10 d) 5

108 Decision Making ANSWER: c TYPE: MC DIFFICULTY: Moderate KEYWORDS: return to risk ratio 53. The minimum expected opportunity loss is also equal to a. expected profit under certainty. b. expected value of perfect information. c. coefficient of variation. d. expected value under certainty minus the expected monetary value of the worst alternative. ANSWER: b TYPE: MC DIFFICULTY: Easy KEYWORDS: expected opportunity loss, expected value of perfect information TABLE 17-3 The following information is from 2 investment opportunities. A B Expected monetary value $900 $600 Standard deviation 100 50

54. Referring to Table 17-3, what is the coefficient of variation for investment A? a) 90.0% b) 11.1% c) 8.3% d) 5.0% ANSWER: b TYPE: MC DIFFICULTY: Moderate KEYWORDS: coefficient of variation 55. Referring to Table 17-3, which investment has the optimal coefficient of variation? a) Investment A b) Investment B c) The investments are equal. d) It cannot be determined. ANSWER: b TYPE: MC DIFFICULTY: Moderate KEYWORDS: decision making, coefficient of variation 56. Referring to Table 17-3, which investment has the optimal return to risk ratio? a) Investment A b) Investment B c) The investments are equal. d) It cannot be determined. ANSWER:

Decision Making

b TYPE: MC DIFFICULTY: Moderate KEYWORDS: decision making, return to risk ratio 57. Referring to Table 17-3, what is the return to risk ratio for Investment B? a) 8 b) 10 c) 12 d) 24 ANSWER: c TYPE: MC DIFFICULTY: Moderate KEYWORDS: return to risk ratio TABLE 17-4 A stock portfolio has the following returns under the market conditions listed below. Market Condition Probability Return Bull 0.4 $200 Stable 0.3 $100 Bear 0.3 – $100

58. Referring to Table 17-4, what is the EMV? a) $180 b) $130 c) $90 d) $80 ANSWER: d TYPE: MC DIFFICULTY: Moderate KEYWORDS: expected monetary value 59. Referring to Table 17-4, what is the standard deviation? a) 4,890 b) 4,840 c) 124.9 d) 69.6 ANSWER: c TYPE: MC DIFFICULTY: Moderate KEYWORDS: standard deviation 60. Referring to Table 17-4, what is the coefficient of variation? a) 88.8% b) 90.3% c) 100% d) 156.1% ANSWER: d TYPE: MC DIFFICULTY: Moderate

109

110 Decision Making KEYWORDS: coefficient of variation 61. Referring to Table 17-4, what is the return to risk ratio? a) 0.64 b) 1.08 c) 1.18 d) 2.00 ANSWER: a TYPE: MC DIFFICULTY: Moderate KEYWORDS: return to risk ratio

62. The curve for the

will show a rapid increase in utility for initial amounts of money followed by a gradual leveling off for increasing dollar amounts. a) risk taker b) risk averter c) risk neutral d) profit seeker

ANSWER: b TYPE: MC DIFFICULTY: Moderate KEYWORDS: utility 63. Look at the utility function graphed below and select the type of decision maker that corresponds to the graph. a) b) c) d)

Risk averter Risk neutral Risk taker Risk player Utility

Monetary Outcome

ANSWER: c TYPE: MC DIFFICULTY: Moderate KEYWORDS: utility

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64. The risk seeker’s curve represents the utility of one who enjoys taking risks. Therefore, the slope of the utility curve becomes a) smaller b) stable c) larger d) uncertain

for large dollar amounts.

ANSWER: c TYPE: MC DIFFICULTY: Moderate KEYWORDS: utility 65. a) b) c) d)

is a procedure for revising probabilities based upon additional information. Utility theory Bernoulli’s theorem Beckman’s theorem Bayes’ theorem

ANSWER: d TYPE: MC DIFFICULTY: Moderate KEYWORDS: Bayes’ theorem 66. Look at the utility function graphed below and select the type of decision maker that corresponds to the graph. a) b) c) d)

Risk averter Risk neutral Risk taker Risk player Utility

Monetary Outcome

ANSWER: b TYPE: MC DIFFICULTY: Moderate KEYWORDS: utility

67. The

curve represents the expected monetary value approach. a) risk averter’s b) risk taker’s

112 Decision Making c) risk neutral d) Bernoulli ANSWER: c TYPE: MC DIFFICULTY: Moderate KEYWORDS: utility 68. Look at the utility function graphed below and select the type of decision-maker that corresponds to the graph. a) Risk averter b) Risk neutral c) Risk taker d) Risk player Utility

Monetary Outcome

ANSWER: a TYPE: MC DIFFICULTY: Moderate KEYWORDS: utility 69. In a local cellular phone area, company A accounts for 60% of the cellular phone market, while company B accounts for the remaining 40% of the market. Of the cellular calls made with company A, 1% of the calls will have some sort of interference, while 2% of the cellular calls with company B will have interference. If a cellular call is selected at random, the probability that it will have interference is a) 0.014 b) 0.028 c) 0.14 d) 0.986 ANSWER: a TYPE: MC DIFFICULTY: Moderate KEYWORDS: probability 70. In a local cellular phone area, company A accounts for 60% of the cellular phone market, while company B accounts for the remaining 40% of the market. Of the cellular calls made with company A, 1% of the calls will have some sort of interference, while 2% of the cellular calls with company B will have interference. If a cellular call is selected at random, the probability that it will not have interference is

Decision Making

a) b) c) d)

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0.014 0.028 0.14 0.986

ANSWER: d TYPE: MC DIFFICULTY: Difficult KEYWORDS: probability 71. In a local cellular phone area, company A accounts for 60% of the cellular phone market, while company B accounts for the remaining 40% of the market. Of the cellular calls made with company A, 1% of the calls will have some sort of interference, while 2% of the cellular calls with company B will have interference. If a cellular call is selected at random and has interference, what is the probability that it was with company A? a) 0.071 b) 0.429 c) 0.571 d) It cannot be determined. ANSWER: b TYPE: MC DIFFICULTY: Difficult KEYWORDS: probability 72. At Eastern University, 60% of the students are from suburban areas, 30% are from rural areas, and 10% are from urban areas. Of the students from the suburban areas, 60% are nonbusiness majors. Of the students from the rural areas, 70% are nonbusiness majors. Of the students from the urban areas, 90% are nonbusiness majors. The probability that a randomly selected student is a business major is a) 0.66 b) 0.54 c) 0.44 d) 0.34 ANSWER: d TYPE: MC DIFFICULTY: Moderate KEYWORDS: probability 73. At Eastern University, 60% of the students are from suburban areas, 30% are from rural areas, and 10% are from urban areas. Of the students from the suburban areas, 60% are nonbusiness majors. Of the students from the rural areas, 70% are nonbusiness majors. Of the students from the urban areas, 90% are nonbusiness majors. If a randomly selected student is not a business major, the probability that the student is from the urban area is a) 0.136 b) 0.214 c) 0.666 d) 0.706 ANSWER: a TYPE: MC DIFFICULTY: Moderate KEYWORDS: probability

114 Decision Making

74. True or False: Removal of uncertainty from a decision-making problem leads to a case referred to as perfect information. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: expected value of perfect information, expected profit under certainty 75. True or False: A firm wishing to maximize profits will increase production as long as expected marginal profit exceeds expected marginal loss. ANSWER: True TYPE: TF DIFFICULTY: Moderate KEYWORDS: payoff 76. True or False: Opportunity loss is the difference between the lowest profit for an event and the actual profit obtained for an action taken. ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: opportunity loss 77. True or False: To calculate expected profit under certainty, we need to have perfect information about which event will occur. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: expected profit under certainty 78. In portfolio analysis, the

is the reciprocal of the return to risk ratio.

ANSWER: coefficient of variation TYPE: FI DIFFICULTY: Easy KEYWORDS: coefficient of variation 79. The riskcurve shows a rapid increase in utility for initial amounts of money followed by a gradual leveling off for increasing dollar amounts. ANSWER: averter’s TYPE: FI DIFFICULTY: Easy KEYWORDS: utility 80. The risk-

curve represents the expected monetary value approach.

ANSWER: neutral’s TYPE: FI DIFFICULTY: Easy KEYWORDS: utility

Decision Making

TABLE 17-5 The following payoff table shows profits associated with a set of 2 alternatives under 3 possible events. Action Event A B 1 1000 1200 2 500 700 3 300 – 200 Suppose that the probability of Event 1 is 0.2, Event 2 is 0.5, and Event 3 is 0.3.

81. Referring to Table 17-5, what is the EMV for Action A? ANSWER: $540 TYPE: PR DIFFICULTY: Moderate KEYWORDS: expected monetary value

82. Referring to Table 17-5, what is the EMV for Action B? ANSWER: $530 TYPE: PR DIFFICULTY: Moderate KEYWORDS: expected monetary value

83. Referring to Table 17-5, what is the opportunity loss for Action B with Event 3? ANSWER: $500 TYPE: PR DIFFICULTY: Moderate KEYWORDS: opportunity loss

84. Referring to Table 17-5, what is the opportunity loss for Action B with Event 1? ANSWER: $0 TYPE: PR DIFFICULTY: Moderate KEYWORDS: opportunity loss

85. Referring to Table 17-5, what is the opportunity loss for Action A with Event 1? ANSWER: $200 TYPE: PR DIFFICULTY: Moderate KEYWORDS: opportunity loss

86. Referring to Table 17-5, what is the opportunity loss for Action A with Event 2? ANSWER: $200 TYPE: PR DIFFICULTY: Moderate KEYWORDS: opportunity loss

115

116 Decision Making

87. Referring to Table 17-5, what is the EOL for Action A? ANSWER: $140 TYPE: PR DIFFICULTY: Moderate KEYWORDS: expected opportunity loss

88. Referring to Table 17-5, what is the EOL for Action B? ANSWER: $150 TYPE: PR DIFFICULTY: Moderate KEYWORDS: expected opportunity loss

89. Referring to Table 17-5, what is the optimal action using EMV? ANSWER: Action A TYPE: PR DIFFICULTY: Moderate KEYWORDS: decision making, expected monetary value

90. Referring to Table 17-5, what is the optimal action using EOL? ANSWER: Action A TYPE: PR DIFFICULTY: Moderate KEYWORDS: decision making, expected opportunity loss

91. Referring to Table 17-5, what is the EVPI for this problem? ANSWER: $140 TYPE: PR DIFFICULTY: Moderate KEYWORDS: expected value of perfect information

92. Referring to Table 17-5, what is the expected profit under certainty (EPUC ) for this problem? ANSWER: $680 TYPE: PR DIFFICULTY: Moderate KEYWORDS: expected profit under certainty

93. Referring to Table 17-5, what is the standard deviation for Action A? ANSWER: 245.8 TYPE: PR DIFFICULTY: Moderate KEYWORDS: standard deviation

94. Referring to Table 17-5, what is the coefficient of variation for Action A? ANSWER:

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117

0.4551 TYPE: PR DIFFICULTY: Moderate KEYWORDS: coefficient of variation

95. Referring to Table 17-5, what is the return to risk ratio for Action B? ANSWER: 1.0313 TYPE: PR DIFFICULTY: Moderate KEYWORDS: return to risk ratio 96. Referring to Table 17-5, what is the optimal action using the coefficient of variation? ANSWER: Action A TYPE: PR DIFFICULTY: Moderate KEYWORDS: decision making, coefficient of variation 97. Referring to Table 17-5, what is the optimal action using the return to risk ratio? ANSWER: Action A TYPE: PR DIFFICULTY: Moderate KEYWORDS: decision making, return to risk ratio TABLE 17-6 A student wanted to find out the optimal strategy to study for a Business Statistics exam. He constructed the following payoff table based on the average amount of time he needed to put in every week studying for the course and the degree of difficulty of the exam. From the information that he gathered from students who had taken the course, he concluded that there was a 40% probability that the exam would be easy. Easy Exam Difficult Exam

16 hours 40 100

8 hours 60 50

4 hours 80 0

98. Referring to Table 17-6, how many possible courses of action are there? ANSWER: 3 TYPE: PR DIFFICULTY: Easy KEYWORDS: courses of action 99. Referring to Table 17-6, how many possible states of the world are there? ANSWER: 2 TYPE: PR DIFFICULTY: Easy KEYWORDS: states of the world 100. Referring to Table 17-6, what is the opportunity loss of spending 4 hours per week on average studying for the exam when the exam turns out to be easy?

118 Decision Making

ANSWER: 0 TYPE: PR DIFFICULTY: Moderate KEYWORDS: opportunity loss 101. Referring to Table 17-6, what is the opportunity loss of spending 16 hours per week on average studying for the exam when the exam turns out to be easy? ANSWER: 40 TYPE: PR DIFFICULTY: Moderate KEYWORDS: opportunity loss 102. Referring to Table 17-6, what is the opportunity loss of spending 8 hours per week on average studying for the exam when the exam turns out to be difficult? ANSWER: 50 TYPE: PR DIFFICULTY: Moderate KEYWORDS: opportunity loss 103. Referring to Table 17-6, what is the expected monetary value of spending 8 hours per week on average studying for the exam? ANSWER: 54 TYPE: PR DIFFICULTY: Moderate KEYWORDS: expected monetary value 104. Referring to Table 17-6, what is the expected opportunity loss of spending 8 hours per week on average studying for the exam? ANSWER: 38 TYPE: PR DIFFICULTY: Moderate KEYWORDS: expected opportunity loss 105. Referring to Table 17-6, what is the coefficient of variation of spending 8 hours per week on average studying for the exam? ANSWER: 0.0907 TYPE: PR DIFFICULTY: Moderate KEYWORDS: coefficient of variation 106. Referring to Table 17-6, what is the return-to-risk ratio of spending 8 hours per week on average studying for the exam? ANSWER: 11.0227 TYPE: PR DIFFICULTY: Moderate KEYWORDS: return to risk ratio

Decision Making

107.

119

Referring to Table 17-6, what is the expected value of perfect information?

ANSWER: 16 TYPE: PR DIFFICULTY: Moderate KEYWORDS: expected value of perfect information 108. Referring to Table 17-6, what is the maximum amount that the student is willing to pay to obtain perfect information? ANSWER: 16 TYPE: PR DIFFICULTY: Difficult KEYWORDS: expected value of perfect information 109.

Referring to Table 17-6, what is the expected profit under certainty?

ANSWER: 92 TYPE: PR DIFFICULTY: Moderate KEYWORDS: expected profit under certainty 110. Referring to Table 17-6, what would be the expected profit if the student had perfect information on whether the exam will be easy or difficult? ANSWER: 92 TYPE: PR DIFFICULTY: Difficult KEYWORDS: expected profit under certainty 111. True or False: Referring to Table 17-6, the optimal strategy using the expected monetary value criterion is to study 8 hours per week on average for the exam. ANSWER: False TYPE: TF DIFFICULTY: Moderate KEYWORDS: decision making, expected monetary value 112. True or False: Referring to Table 17-6, the optimal strategy using the expected monetary value criterion is to study 16 hours per week on average for the exam. ANSWER: True TYPE: TF DIFFICULTY: Moderate KEYWORDS: decision making, expected monetary value 113. True or False: Referring to Table 17-6, the optimal strategy using the return-to-risk ratio criterion is to study 8 hours per week on average for the exam. ANSWER: True TYPE: TF DIFFICULTY: Moderate KEYWORDS: decision making, return to risk ratio

120 Decision Making 114. True or False: Referring to Table 17-6, the optimal strategy using the coefficient of variation criterion is to study 8 hours per week on average for the exam. ANSWER: True TYPE: TF DIFFICULTY: Moderate KEYWORDS: decision making, coefficient of variation 115. True or False: Referring to Table 17-6, the optimal strategy using the expected opportunity loss criterion is to study 8 hours per week on average for the exam. ANSWER: False TYPE: TF DIFFICULTY: Moderate KEYWORDS: decision making, expected opportunity loss 116. True or False: Referring to Table 17-6, the optimal strategy using the expected opportunity loss criterion is to study 16 hours per week on average for the exam. ANSWER: True TYPE: TF DIFFICULTY: Moderate KEYWORDS: decision making, expected opportunity loss

117. Which of the following is NOT a decision making criterion? a. b. c. d.

Maximizing the expected monetary value of an action. Minimizing the expected opportunity loss of an action. Minimizing expected profit under certainty. Maximizing the return-to-risk ratio.

ANSWER: c TYPE: MC DIFFICULTY: Easy KEYWORDS: decision making

124 Statistical Applications in Quality and Production Management

CHAPTER 18: STATISTICAL APPLICATIONS IN QUALITY MANAGEMENT 1. The control chart a) focuses on the time dimension of a system. b) captures the natural variability in the system. c) can be used for categorical, discrete, or continuous variables. d) All of the above. ANSWER: d TYPE: MC DIFFICULTY: Moderate KEYWORDS: control chart 2. Variation signaled by individual fluctuations or patterns in the data is called a) special or assignable causes. b) common or chance causes. c) explained variation. d) the standard deviation. ANSWER: a TYPE: MC DIFFICULTY: Easy KEYWORDS: special causes of variation 3. Variation due to the inherent variability in a system of operation is called a) special or assignable causes. b) common or chance causes. c) explained variation. d) the standard deviation. ANSWER: b TYPE: MC DIFFICULTY: Easy KEYWORDS: common causes of variation

4. Which of the following is not one of Deming's 14 points? a) b) c) d)

Belief in mass inspection. Create constancy of purpose for improvement of product or service. Adopt and institute leadership. Drive out fear.

ANSWER: a TYPE: MC DIFFICULTY: Moderate KEYWORDS: Deming’s 14 points

125 Statistical Applications in Quality and Production Management

5. Which of the following is not one of Deming's 14 points? a) b) c) d)

Create constancy of purpose for improvement of product or service. Award business on the basis of price tag alone. Break down barriers between staff areas. Drive out fear.

ANSWER: b TYPE: MC DIFFICULTY: Moderate KEYWORDS: Deming’s 14 points 6. The principal focus of the control chart is the attempt to separate special or assignable causes of variation from common causes of variation. What cause of variation can be reduced only by changing the system? a) Special or assignable causes b) Common causes c) Total causes d) None of the above ANSWER: b TYPE: MC DIFFICULTY: Easy KEYWORDS: common causes of variation 7. Once the control limits are set for a control chart, one attempts to a) discern patterns that might exist in values over time. b) determine whether any points fall outside the control limits. c) Both of the above. d) None of the above. ANSWER: c TYPE: MC DIFFICULTY: Easy KEYWORDS: control chart 8. Which of the following is not part of the Shewhart-Deming cycle? a) Plan b) Do c) React d) Act ANSWER: c TYPE: MC DIFFICULTY: Moderate KEYWORDS: Shewhart-Deming cycle 9. The Shewhart-Deming cycle plays an important role in which of the following Deming’s 14 points for management? a) Adopt the new philosophy. b) Break down barriers between staff areas. c) Create constancy of purpose for improvement of product and services. d) Eliminate slogans, exhortation, and targets for the workforce. ANSWER:

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c TYPE: MC DIFFICULTY: Moderate KEYWORDS: Shewhart-Deming cycle 10. Which of the following situations suggests a process that appears to be operating in a state of statistical control? a) A control chart with a series of consecutive points that are above the center line and a series of consecutive points that are below the center line. b) A control chart in which no points fall outside either the upper control limit or the lower control limit and no patterns are present. c) A control chart in which several points fall outside the upper control limit. d) All of the above. ANSWER: b TYPE: MC DIFFICULTY: Easy KEYWORDS: control chart, statistical control 11. Which of the following situations suggests a process that appears to be operating out of statistical control? a) A control chart with a series of consecutive points that are above the center line and a series of consecutive points that are below the center line. b) A control chart in which points fall outside the lower control limit. c) A control chart in which several points fall outside the upper control limit. d) All of the above. ANSWER: d TYPE: MC DIFFICULTY: Moderate KEYWORDS: control chart, statistical control 12. A process is said to be out of control if a) a point falls above the upper or below the lower control lines. b) a run of 8 of more points is observed. c) Either of the above. d) None of the above. ANSWER: c TYPE: MC DIFFICULTY: Moderate KEYWORDS: statistical control 13. One of the morals of the red bead experiment is: a) variation is part of the process. b) only management can change the system. c) it is the system that primarily determines performance. d) All of the above. ANSWER: d TYPE: MC DIFFICULTY: Moderate KEYWORDS: red bead experiment 14. Which famous statistician developed the 14 Points of Quality?

127 Statistical Applications in Quality and Production Management a) b) c) d)

Shewhart Deming Chebyshev Taguchi

ANSWER: b TYPE: MC DIFFICULTY: Easy KEYWORDS: Deming’s 14 points TABLE 18-1 A local newspaper has 10 delivery boys who each deliver the morning paper to 50 customers every day. The owner decides to record the number of papers delivered on time for a 10-day period and construct a p chart to see whether the percentage is too erratic. Number of Papers Day Delivered on Time 1 46 2 45 3 46 4 45 5 43 6 48 7 46 8 49 9 48 10 47

15. Referring to Table 18-1, what is the numerical value of the center line for the p chart? a) 0.926 b) 0.911 c) 0.885 d) 0.500 ANSWER: a TYPE: MC DIFFICULTY: Moderate KEYWORDS: p chart, center line

16. Referring to Table 18-1, what is the numerical value of the lower control limit for the p chart? a) 0.920 b) 0.911 c) 0.815 d) 0.798 ANSWER: c TYPE: MC DIFFICULTY: Moderate KEYWORDS: p chart, control limit

17. Referring to Table 18-1, what is the numerical value of the upper control limit for the p chart? a) 0.926 b) 0.961 c) 0.979

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d) 1.000 ANSWER: d TYPE: MC DIFFICULTY: Moderate KEYWORDS: p chart, control limit

18. Referring to Table 18-1, which expression best characterizes the p chart? a) b) c) d)

Cycles Increasing trend In-control Individual outliers

ANSWER: c TYPE: MC DIFFICULTY: Moderate KEYWORDS: p chart TABLE 18-2 A political pollster randomly selects a sample of 100 voters each day for 8 successive days and asks how many will vote for the incumbent. The pollster wishes to construct a p chart to see if the percentage favoring the incumbent candidate is too erratic. Sample Number Favoring (Day) Incumbent Candidate 1 57 2 57 3 53 4 51 5 55 6 60 7 56 8 59

19. Referring to Table 18-2, what is the numerical value of the center line for the p chart? a) 0.53 b) 0.56 c) 0.63 d) 0.66 ANSWER: b TYPE: MC DIFFICULTY: Moderate KEYWORDS: p chart, center line

20. Referring to Table 18-2, what is the numerical value of the lower control limit for the p chart? a) 0.37 b) 0.41 c) 0.50 d) 0.71 ANSWER: b

129 Statistical Applications in Quality and Production Management TYPE: MC DIFFICULTY: Moderate KEYWORDS: p chart, control limit

21. Referring to Table 18-2, what is the numerical value of the upper control limit for the p chart? a) 0.92 b) 0.89 c) 0.71 d) 0.62 ANSWER: c TYPE: MC DIFFICULTY: Moderate KEYWORDS: p chart, control limit

22. Referring to Table 18-2, which expression best characterizes the p chart? a) b) c) d)

Hugging the center line Cycles Hugging the control line Individual outliers

ANSWER: a TYPE: MC DIFFICULTY: Moderate KEYWORDS: p chart TABLE 18-3 A quality control analyst for a light bulb manufacturer is concerned that the time it takes to produce a batch of light bulbs is too erratic. Accordingly, the analyst randomly surveys 10 production periods each day for 14 days and records the sample mean and range for each day. Day R X (in minutes) 1 58.5 5.1 2 47.6 7.8 3 64.3 6.1 4 60.6 5.7 5 63.7 6.2 6 57.5 6.0 7 55.0 5.4 8 54.9 6.1 9 55.0 5.9 10 62.7 5.0 11 61.9 7.1 12 60.0 6.5 13 58.3 5.9 14 52.0 5.2 23. Referring to Table 18-3, suppose the analyst constructs an R chart to see if the variability in production times is in-control. What is the center line of this R chart? a) 4.8 b) 6.0 c) 6.9 d) 7.1 ANSWER:

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b TYPE: MC DIFFICULTY: Moderate KEYWORDS: R chart, center line 24. Referring to Table 18-3, suppose the analyst constructs an R chart to see if the variability in production times is in-control. What is the lower control limit for this R chart? a) 4.84 b) 3.37 c) 2.98 d) 1.34 ANSWER: d TYPE: MC DIFFICULTY: Moderate KEYWORDS: R chart, control limit 25. Referring to Table 18-3, suppose the analyst constructs an R chart to see if the variability in production times is in-control. What is the upper control limit for this R chart? a) 10.66 b) 9.37 c) 7.98 d) 6.34 ANSWER: a TYPE: MC DIFFICULTY: Moderate KEYWORDS: R chart, control limit 26. Referring to Table 18-3, suppose the analyst constructs an R chart to see if the variability in production times is in-control. The R chart is characterized by which of the following? a) Increasing trend b) Jump in the level around which the observations vary c) Hugging the center line d) Hugging the control limits ANSWER: c TYPE: MC DIFFICULTY: Moderate KEYWORDS: R chart 27. Referring to Table 18-3, suppose the sample mean and range data were based on 11 observations per day instead of 10. How would this change affect the lower and upper control limits of the R chart? a) LCL would increase; UCL would decrease. b) LCL would remain the same; UCL would decrease. c) Both LCL and UCL would remain the same. d) LCL would decrease; UCL would increase. ANSWER: a TYPE: MC DIFFICULTY: Moderate KEYWORDS: control limit

131 Statistical Applications in Quality and Production Management

28. Referring to Table 18-3, suppose the analyst constructs an X chart to see if the production process is in-control. What is the center line for this chart? a) 64.3 b) 59.5 c) 58.0 d) 57.1 ANSWER: c TYPE: MC DIFFICULTY: Moderate KEYWORDS: XBar chart, center line

29. Referring to Table 18-3, suppose the analyst constructs an X chart to see if the production process is in-control. What is the lower control limit (LCL) for this chart? a) 47.60 b) 56.15 c) 57.15 d) 58.05 ANSWER: b TYPE: MC DIFFICULTY: Moderate KEYWORDS: XBar chart, control limit

30. Referring to Table 18-3, suppose the analyst constructs an X chart to see if the production process is in-control. What is the upper control limit (UCL) for this chart? a) 62.15 b) 60.95 c) 59.85 d) 58.75 ANSWER: c TYPE: MC DIFFICULTY: Moderate KEYWORDS: XBar chart, control limit

31. Referring to Table 18-3, suppose the analyst constructs an X chart to see if the production process is in-control. Which expression best describes this chart? a) In-control b) Increasing trend c) Hugging the center line d) Individual outliers ANSWER: d TYPE: MC DIFFICULTY: Moderate KEYWORDS: XBar chart TABLE 18-4 A factory supervisor is concerned that the time it takes workers to complete an important production task (measured in seconds) is too erratic and adversely affects expected profits. The supervisor

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proceeds by randomly sampling 5 individuals per hour for a period of 10 hours. The sample mean and range for each hour are listed below. Hour R X 1 18.4 25 2 16.9 27 3 23.0 30 4 21.2 23 5 21.0 24 6 24.0 25 7 19.3 12 8 15.8 14 9 20.0 13 10 23.0 11 She also decides that lower and upper specification limit for the critical-to-quality variable should be 10 and 30 seconds, respectively. 32. Referring to Table 18-4, suppose the supervisor constructs an R chart to see if the variability in collection times is in-control. What is the center line of this R chart? a) 20.00 b) 20.56 c) 20.40 d) 24.00 ANSWER: c TYPE: MC DIFFICULTY: Moderate KEYWORDS: R chart, center line 33. Referring to Table 18-4, suppose the supervisor constructs an R chart to see if the variability in collection times is in-control. What are the lower and upper control limits for this R chart? a) – 2.33, 43.13 b) – 2.28, 42.28 c) 0, 42.28 d) 0, 43.13 ANSWER: d TYPE: MC DIFFICULTY: Moderate KEYWORDS: R chart, control limit 34. Referring to Table 18-4, suppose the supervisor constructs an R chart to see if the variability in collection times is in-control. This R chart is characterized by which of the following? a) Jump in the level around which the observations vary b) Increasing trend c) Cycles d) Hugging the control limits ANSWER: a TYPE: MC DIFFICULTY: Moderate KEYWORDS: R chart

133 Statistical Applications in Quality and Production Management 35. Referring to Table 18-4, suppose the sample mean and range data were based on 6 observations per hour instead of 5. How would this change affect the lower and upper control limits of an R chart? a) LCL would increase; UCL would decrease. b) LCL would remain the same; UCL would decrease. c) Both LCL and UCL would remain the same. d) LCL would decrease; UCL would increase. ANSWER: b TYPE: MC DIFFICULTY: Moderate KEYWORDS: R chart, control limit

36. Referring to Table 18-4, suppose the supervisor constructs an X chart to see if the process is incontrol. What is the center line of the chart? a) 20.00 b) 20.26 c) 21.00 d) 24.26 ANSWER: b TYPE: MC DIFFICULTY: Moderate KEYWORDS: XBar chart, center line

37. Referring to Table 18-4, suppose the supervisor constructs an X chart to see if the process is incontrol. What are the lower and upper control limits of this chart? a) 10.00, 30.00 b) 8.49, 32.03 c) 5.39, 35.13 d) 4.96, 35.56 ANSWER: b TYPE: MC DIFFICULTY: Moderate KEYWORDS: XBar chart, control limit

38. Referring to Table 18-4, suppose the supervisor constructs an X chart to see if the process is incontrol. Which expression best describes this chart? a) Decreasing trend b) Hugging the center line c) Increasing trend d) Cycles ANSWER: b TYPE: MC DIFFICULTY: Moderate KEYWORDS: XBar chart

39. Referring to Table 18-4, what is the value of d2 factor? ANSWER: 2.325

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TYPE: PR DIFFICULTY: Easy KEYWORDS: d2 factor 40. Referring to Table 18-4, what percentage of the time it takes workers to complete an important production task will fall inside the specification limits? ANSWER: 0.75 TYPE: PR DIFFICULTY: Moderate KEYWORDS: process capacity

41. Referring to Table 18-4, what is the value of the C p index? ANSWER: 0.38 TYPE: PR DIFFICULTY: Easy KEYWORDS: Cp index

42. Referring to Table 18-4, what is the value of the CPL index? ANSWER: 0.39 TYPE: PR DIFFICULTY: Easy KEYWORDS: CPL index

43. Referring to Table 18-4, what is the value of the CPU index? ANSWER: 0.37 TYPE: PR DIFFICULTY: Easy KEYWORDS: CPU index

44. Referring to Table 18-4, what is the value of the Cpk index? ANSWER: 0.37 TYPE: PR DIFFICULTY: Easy KEYWORDS: Cpk index 45. True or False: The control limits are based on the standard deviation of the process. ANSWER: True TYPE: TF DIFFICULTY: Moderate KEYWORDS: control limit 46. True or False: The purpose of a control chart is to eliminate common cause variation. ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: control chart, common causes of variation

135 Statistical Applications in Quality and Production Management

47. True or False: Special or assignable causes of variation are signaled by individual fluctuations or patterns in the data. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: special causes of variation 48. True or False: One of the focuses of management by process is continuous improvement of a product or service. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: Deming’s 14 points 49. True or False: Common causes of variation represent variation due to the inherent variability in the system. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: common causes of variation 50. True or False: Common causes of variation are correctable without modifying the system. ANSWER: False TYPE: TF DIFFICULTY: Moderate KEYWORDS: common causes of variation 51. True or False: Changes in the system to reduce common cause variation are the responsibility of management. ANSWER: True TYPE: TF DIFFICULTY: Moderate KEYWORDS: common causes of variation 52. True or False: One of the morals of the red bead experiment is that variation is part of any process. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: red bead experiment 53. True or False: In the United States, the control limits on a control chart are placed so that they are 3 standard deviations above and below a central line. ANSWER: True TYPE: TF DIFFICULTY: Easy

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KEYWORDS: control chart, control limit

54. True or False: The p chart is a control chart used for monitoring the proportion of items in a batch that meet given specifications. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: p chart 55. True or False: The R chart is a control chart used to monitor a process mean. ANSWER: False TYPE: TF DIFFICULTY: Moderate KEYWORDS: R chart

56. True or False: It is not possible for the X chart to be out-of-control when the R chart is in control. ANSWER: False TYPE: TF DIFFICULTY: Moderate KEYWORDS: XBar chart, R chart, statistical control 57. The cause of variation that can be reduced only by changing the system is variation.

cause

ANSWER: common TYPE: FI DIFFICULTY: Easy KEYWORDS: common causes of variation 58.

causes of variation are correctable without modifying the system.

ANSWER: Special TYPE: FI DIFFICULTY: Easy KEYWORDS: special causes of variation TABLE 18-5 A manufacturer of computer disks took samples of 240 disks on 15 consecutive days. The number of disks with bad sectors was determined for each of these samples. The results are in the table that follows. Day 1 2 3 4 5 6

Bad 9 7 4 6 8 3

% Bad 0.037500 0.029167 0.016667 0.025000 0.033333 0.012500

137 Statistical Applications in Quality and Production Management 7 8 9 10 11 12 13 14 15

6 10 16 24 15 9 4 8 6

0.025000 0.041667 0.066667 0.100000 0.062500 0.037500 0.016667 0.033333 0.025000

59. Referring to Table 18-5, the best estimate of the average proportion of disks with bad sectors is . ANSWER: 0.0375 TYPE: FI DIFFICULTY: Moderate KEYWORDS: p chart, center line

60. Referring to Table 18-5, a p control chart is to be made for these data. The center line of the control chart is

ANSWER: 0.0375 TYPE: FI DIFFICULTY: Moderate KEYWORDS: p chart, center line

61. Referring to Table 18-5, a p control chart is to be made for these data. The estimate of the standard error of the proportion of disks with bad sectors is

ANSWER: 0.03679/3 or 0.01226 TYPE: FI DIFFICULTY: Moderate KEYWORDS: p chart, standard error

62. Referring to Table 18-5, a p control chart is to be made for these data. The upper control limit is , and the lower control limit is

ANSWER: 0.0743; 0.00071 TYPE: FI DIFFICULTY: Moderate KEYWORDS: p chart, control limit

63. Referring to Table 18-5, construct a p control chart for these data. ANSWER: Proportion of Bad Disks in Samples of Size 240

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p Chart 0.1 0.09 0.08 0.07 0.06 0.05 0.04 0.03 0.02 0.01 0 0

Sample Number

TYPE: PR DIFFICULTY: Difficult KEYWORDS: p chart 64. True or False: Referring to Table 18-5, the process seems to be in control. ANSWER: False TYPE: TF DIFFICULTY: Moderate KEYWORDS: p chart, statistical control TABLE 18-6 The maker of a packaged candy wants to evaluate the quality of her production process. On each of 16 consecutive days, she samples 600 bags of candy and determines the number in each day's sample that she considers to be of poor quality. The data that she developed follow. Number Proportion Day Poor Poor 1 33 0.0550000 2 29 0.0483333 3 31 0.0516667 4 32 0.0533333 5 43 0.0716667 6 45 0.0750000 7 46 0.0766667 8 48 0.0800000 9 48 0.0800000 10 46 0.0766667 11 28 0.0466667 12 32 0.0533333 13 28 0.0466667 14 32 0.0533333

139 Statistical Applications in Quality and Production Management 15 16

31 24

0.0516667 0.0400000

65. Referring to Table 18-6, the estimate of the proportion of poor quality bags of candy is . ANSWER: 0.06 TYPE: FI DIFFICULTY: Moderate KEYWORDS: p chart, center line

66. Referring to Table 18-6, a p control chart is to be constructed for these data. The center line for the chart should be located at

ANSWER: 0.06 TYPE: FI DIFFICULTY: Moderate KEYWORDS: p chart, center line

67. Referring to Table 18-6, a p control chart is to be constructed for these data. The estimate of the standard error of the sample proportion is

ANSWER: 0.009695 TYPE: FI DIFFICULTY: Moderate KEYWORDS: p chart, standard error

68. Referring to Table 18-6, a p control chart is to be constructed for these data. The lower control limit is

, while the upper control limit is

ANSWER: 0.0309139; 0.0890861 TYPE: FI DIFFICULTY: Moderate KEYWORDS: p chart, control limit

69. Referring to Table 18-6, construct a p control chart for these data. ANSWER: P Chart: Sample Sizes = 600

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p Chart 0.09 0.08 0.07 0.06 0.05 0.04 0.03 0.02 0.01 0 0

Sample Number

TYPE: PR DIFFICULTY: Difficult KEYWORDS: p chart TABLE 18-7 A supplier of silicone sheets for producers of computer chips wants to evaluate her manufacturing process. She takes samples of size 5 from each day's output and counts the number of blemishes on each silicone sheet. The results from 20 days of such evaluations are presented below. Day 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20

1 8 8 10 5 8 9 10 10 6 6 8 6 7 5 7 7 8 11 6 6

Sheet 2 10 13 12 9 3 7 10 9 10 9 5 4 5 8 12 11 4 4 10 12

3 14 6 7 12 8 9 5 10 6 8 6 7 7 8 10 4 5 11 6 12

4 6 6 7 7 9 6 7 6 9 6 10 7 6 7 6 7 4 11 10 6

5 5 10 9 10 10 9 6 5 9 8 10 12 9 6 10 8 7 10 10 8

Mean 8.6 8.6 9.0 8.6 7.6 8.0 7.6 8.0 8.0 7.4 7.8 7.2 6.8 6.8 9.0 7.4 5.6 9.4 8.4 8.8

Range 9 7 5 7 7 3 5 5 4 3 5 8 4 3 6 7 4 7 4 6

141 Statistical Applications in Quality and Production Management

She also decides that the upper specification limit is 10 blemishes. 70. Referring to Table 18-7, an R chart is to be constructed for the number of blemishes. The center line of this R chart is located at . ANSWER: 5.45 TYPE: FI DIFFICULTY: Moderate KEYWORDS: R chart, center line

71. Referring to Table 18-7, an R chart is to be constructed for the number of blemishes. One way to create the lower control limit involves multiplying the average of the sample ranges by D3. For this data set, the value of D3 is . ANSWER: 0 TYPE: FI DIFFICULTY: Easy KEYWORDS: R chart, D3 factor

72. Referring to Table 18-7, an R chart is to be constructed for the number of blemishes. One way to create the upper control limit involves multiplying the average of the sample ranges by D4. For this data set, the value of D4 is . ANSWER: 2.114 TYPE: FI DIFFICULTY: Easy KEYWORDS: R chart, D4 factor 73. Referring to Table 18-7, an R chart is to be constructed for the number of blemishes. The lower control limit for this data set is . ANSWER: 0 or has no limit TYPE: FI DIFFICULTY: Moderate KEYWORDS: R chart, control limit 74. Referring to Table 18-7, an R chart is to be constructed for the number of blemishes. The upper control limit for this data set is . ANSWER: 11.52 TYPE: FI DIFFICULTY: Moderate KEYWORDS: R chart, control limit 75. Referring to Table 18-7, construct an R chart for the number of blemishes. ANSWER:

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R Chart 12 10 8 6 4 2 0 0

Sample Number

TYPE: PR DIFFICULTY: Difficult KEYWORDS: R chart 76. True or False: Referring to Table 18-7, based on the R chart, it appears that the process is out of control. ANSWER: False TYPE: TF DIFFICULTY: Moderate KEYWORDS: R chart, statistical control

77. Referring to Table 18-7, an X chart is to be used for the number of blemishes. The center line of this chart is located at

ANSWER: 7.930 TYPE: FI DIFFICULTY: Moderate KEYWORDS: XBar chart, center line

78. Referring to Table 18-7, an X chart is to be used for the number of blemishes. One way to obtain the control limits is to take the grand mean and add and subtract the product of A2 times the average of the sample ranges. For this data set, the value of A2 is . ANSWER: 0.577 TYPE: FI DIFFICULTY: Easy KEYWORDS: XBar chart, A2 factor

79. Referring to Table 18-7, an X chart is to be used for the number of blemishes. The lower control limit for this data set is

, while the upper control limit is

143 Statistical Applications in Quality and Production Management

ANSWER: 4.785; 11.075 TYPE: FI DIFFICULTY: Moderate KEYWORDS: XBar chart, control limit

80. Referring to Table 18-7, construct an X chart for the number of blemishes. ANSWER:

Mean Chart 12 10 8 6 4 2 0 0

Sample Number

TYPE: PR DIFFICULTY: Difficult KEYWORDS: XBar chart

81. True or False: Referring to Table 18-7, based on the X chart for the number of blemishes, it appears that the process is out of control. ANSWER: False TYPE: TF DIFFICULTY: Moderate KEYWORDS: XBar chart, statistical control

82. Referring to Table 18-7, what is the value of d2 factor? ANSWER: 2.326 TYPE: PR DIFFICULTY: Easy KEYWORDS: d2 factor 83. Referring to Table 18-7, what percentage of the chips will fall below the upper specification limit?

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ANSWER: 0.81 TYPE: PR DIFFICULTY: Moderate KEYWORDS: process capacity

84. Referring to Table 18-7, what is the value of the CPU index? ANSWER: 0.29 TYPE: PR DIFFICULTY: Easy KEYWORDS: CPU index TABLE 18-8 Recently, a university switched to a new type of computer-based registration. The registrar is concerned with the amount of time students are spending on the computer registering under the new system. She decides to randomly select 8 students on each of the 12 days of the registration and determine the time each spends on the computer registering. The range, mean, and standard deviation of the times required to register are in the table that follows. Day Range Mean Std.Dev. 1 10 5.250 3.4949 2 31 15.250 10.3060 3 13 20.375 4.9262 4 21 22.875 8.3911 5 35 8.500 11.3767 6 18 7.875 6.9372 7 25 11.250 8.5815 8 30 7.875 9.5235 9 17 10.250 6.3640 10 22 9.500 7.8740 11 27 7.875 8.7086 12 26 12.875 9.3723 85. Referring to Table 18-8, an R chart is to be constructed for the time required to register. The center line of this R chart is located at . ANSWER: 22.92 TYPE: FI DIFFICULTY: Moderate KEYWORDS: R chart, center line

86. Referring to Table 18-8, an R chart is to be constructed for the time required to register. One way to create the lower control limit involves multiplying the average of the sample ranges by D3. For this data set, the value of D3 is . ANSWER: 0.136 TYPE: FI DIFFICULTY: Easy KEYWORDS: R chart, D3 factor

145 Statistical Applications in Quality and Production Management

87. Referring to Table 18-8, an R chart is to be constructed for the time required to register. One way to create the upper control limit involves multiplying the average of the sample ranges by D4. For this data set, the value of D4 is . ANSWER: 1.864 TYPE: FI DIFFICULTY: Easy KEYWORDS: R chart, D4 factor 88. Referring to Table 18-8, an R chart is to be constructed for the time required to register. The lower control limit for this data set is . ANSWER: 3.117 TYPE: FI DIFFICULTY: Moderate KEYWORDS: R chart, control limit 89. Referring to Table 18-8, an R chart is to be constructed for the time required to register. The upper control limit for this data set is . ANSWER: 42.72 TYPE: FI DIFFICULTY: Moderate KEYWORDS: R chart, control limit 90. Referring to Table 18-8, construct an R chart for the time required to register. ANSWER:

R Chart 45 40 35 30 25 20 15 10 5 0 0

6 Sample Number

TYPE: PR DIFFICULTY: Difficult KEYWORDS: R chart

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91. True or False: Referring to Table 18-8, based on the R chart, it appears that the process is out of control. ANSWER: False TYPE: TF DIFFICULTY: Moderate KEYWORDS: R chart, statistical control

92. Referring to Table 18-8, an X chart is to be used for the time required to register. The center line of this chart is located at

ANSWER: 11.65 TYPE: FI DIFFICULTY: Moderate KEYWORDS: XBar chart, center line

93. Referring to Table 18-8, an X chart is to be used for the time required to register. One way to obtain the control limits is to take the grand mean and add and subtract the product of A2 times the average of the sample ranges. For this data set, the value of A2 is . ANSWER: 0.373 TYPE: FI DIFFICULTY: Easy KEYWORDS: XBar chart, A2 factor

94. Referring to Table 18-8, an X chart is to be used for the time required to register. The lower control limit for this data set is

, while the upper control limit is

ANSWER: 3.098; 20.194 TYPE: FI DIFFICULTY: Moderate KEYWORDS: XBar chart, control limit

95. Referring to Table 18-8, construct an X chart for the time required to register. ANSWER:

147 Statistical Applications in Quality and Production Management

Mean Chart 25

20 15

10 5

0 0

Sample Number

TYPE: PR DIFFICULTY: Difficulty KEYWORDS: XBar char

96. True or False: Referring to Table 18-8, based on the X chart, it appears that the process is in control. ANSWER: False TYPE: TF DIFFICULTY: Moderate KEYWORDS: XBar chart, statistical control

97. Which of the following is a method for breaking process into a series of steps in order to eliminate defects and produce near perfect results? a) Deming’s 14 points management. b) Six Sigma management. c) Shewhart-Deming cycle. d) R chart and X chart. ANSWER: b TYPE: MC DIFFICULTY: Easy KEYWORDS: six sigma management 98. Which of the following is not part of the DMAIC process in Six Sigma management?

a) Define b) Do c) Analyze d) Control

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ANSWER: b TYPE: MC DIFFICULTY: Easy KEYWORDS: six sigma management 99. Developing operational definitions for each critical-to-quality characteristic involves which part of the DMAIC process? a) Define b) Measure c) Analyze d) Improve e) Control ANSWER: b TYPE: MC DIFFICULTY: Easy KEYWORDS: six sigma management 100. Determining the root causes of why defects can occur along with the variables in the process that cause these defects to occur involves which part of the DMAIC process? a) Define b) Measure c) Analyze d) Improve e) Control ANSWER: c TYPE: MC DIFFICULTY: Easy KEYWORDS: six sigma management 101. Maintaining the gains that have been made with a revised process in the long term by avoiding potential problems that can occur when a process is changed involves which part of the DMAIC process? a) Define b) Measure c) Analyze d) Improve e) Control ANSWER: e TYPE: MC DIFFICULTY: Easy KEYWORDS: six sigma management

102. Which of the following best measures the ability of a process to consistently meet specified customer-driven requirement? a) Process capability. b) Specification limits c) Upper control limit d) Lower control limit ANSWER:

149 Statistical Applications in Quality and Production Management a TYPE: MC DIFFICULTY: Easy KEYWORDS: process capability 103. True or False: An in-control process must be achieved before being able to estimate process capability. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: process capability 104. True or False: A process capability is estimated by the percentage of product or service that fall outside the specification limits. ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: process capability

105. True or False: The smaller the capability index, the more capable a process is of meeting customer requirement. ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: capability index

106. True or False: The Cp index measures the potential of a process, not its actual performance. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: capability index

107. True or False: C p >1 indicates that if the process average can be centered, then more than 99.73% of the observations will fall inside the specification limits. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: capability index

108. True or False: The CPL and CPU indexes are used to measure process’ actual performance rather than its potential. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: capability index

109. True or False: The Cpk is a one-sided specification limit. ANSWER:

Statistical Applications in Quality and Production Management

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False TYPE: TF DIFFICULTY: Easy KEYWORDS: capability index

110. True or False: CPL >1 implies that the process mean is more than 3 standard deviation away from the lower specification limit. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: capability index

111. True or False: Cpk > 1 indicates that the process average is exactly 3 standard deviations away from the closest specification limit. ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: capability index

112. True or False: Larger Cpk indicates larger capability of meeting the requirements. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: capability index