TEST BANK for Applied Statistics in Business and Economics 6e (Indian Edition) David Doane, Lori Sew by StudyGuide

Applied Statistics in Business and Economics, 6e (Doane) Chapter 1 Overview of Statistics 1) Statistics is the science of collecting, organizing, analyzing, interpreting, and presenting data. Answer: TRUE Explanation: This is one of many good definitions of statistics. Difficulty: 1 Easy Topic: 01.01 What Is Statistics? Learning Objective: 01-01 Define statistics and explain some of its uses. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 2) Inferential statistics refers to generalizing from a sample to a population, estimating unknown parameters, drawing conclusions, and making decisions. Answer: TRUE Explanation: We can use statistics either to describe data or to infer something about a population. Difficulty: 1 Easy Topic: 01.03 Statistics in Business Learning Objective: 01-01 Define statistics and explain some of its uses. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 3) Descriptive statistics refers to summarizing data rather than generalizing about the population. Answer: TRUE Explanation: When we do not infer, we are only describing the available sample data. Difficulty: 1 Easy Topic: 01.03 Statistics in Business Learning Objective: 01-01 Define statistics and explain some of its uses. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation

4) Estimating parameters and testing hypotheses are important aspects of descriptive statistics. Answer: FALSE Explanation: When we generalize to a population, we are using inferential statistics. Difficulty: 2 Medium Topic: 01.03 Statistics in Business Learning Objective: 01-01 Define statistics and explain some of its uses. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 5) Inconsistent treatment of data by a researcher is a symptom of poor survey or research design. Answer: FALSE Explanation: Good survey data can still be misused or misinterpreted. Difficulty: 2 Medium Topic: 01.05 Critical Thinking Learning Objective: 01-04 State the common challenges facing business professionals using statistics. Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation 6) Empirical data are collected through observations and/or experiments. Answer: TRUE Explanation: Empirical data are contrasted with a priori estimates (e.g., expecting 10 heads in 20 coin flips). Difficulty: 2 Medium Topic: 01.05 Critical Thinking Learning Objective: 01-04 State the common challenges facing business professionals using statistics. Bloom's: Remember AACSB: Reflective Thinking Accessibility: Keyboard Navigation 7) Business intelligence refers to collecting, storing, accessing, and analyzing data on the company's operations in order to make better business decisions. Answer: TRUE Explanation: See Wikipedia for similar definitions of business intelligence. Difficulty: 1 Easy Topic: 01.03 Statistics in Business Learning Objective: 01-01 Define statistics and explain some of its uses. Bloom's: Remember AACSB: Reflective Thinking Accessibility: Keyboard Navigation 2 Copyright 2019 © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.

8) When a statistician omits data contrary to her findings in a study, she is justified as long as the sample supports her objective. Answer: FALSE Explanation: We do not omit data unless it is proven to be an error. Difficulty: 2 Medium Topic: 01.04 Statistical Challenges Learning Objective: 01-04 State the common challenges facing business professionals using statistics. Bloom's: Understand AACSB: Ethics Accessibility: Keyboard Navigation 9) A strong correlation between A and B would imply that B is caused by A. Answer: FALSE Explanation: Temporal sequence does not prove causation. Difficulty: 1 Easy Topic: 01.05 Critical Thinking Learning Objective: 01-05 List and explain common statistical pitfalls. Bloom's: Understand AACSB: Reflective Thinking Accessibility: Keyboard Navigation 10) The post hoc fallacy says that when B follows A then B is caused by A. Answer: TRUE Explanation: Temporal sequence does not prove causation. Difficulty: 1 Easy Topic: 01.05 Critical Thinking Learning Objective: 01-05 List and explain common statistical pitfalls. Bloom's: Remember AACSB: Reflective Thinking Accessibility: Keyboard Navigation 11) A statistical test may be significant yet have no practical importance. Answer: TRUE Explanation: Large samples sometimes reveal tiny effects that may not matter very much. Difficulty: 1 Easy Topic: 01.05 Critical Thinking Learning Objective: 01-05 List and explain common statistical pitfalls. Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation 3 Copyright 2019 © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.

12) Valid statistical inferences cannot be made when sample sizes are small. Answer: FALSE Explanation: Small samples may be all that we have, and statistics does have rules for them. Difficulty: 2 Medium Topic: 01.04 Statistical Challenges Learning Objective: 01-04 State the common challenges facing business professionals using statistics. Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation 13) Statistics is an essential part of critical thinking because it allows us to transform the empirical evidence from a sample so it will agree with our preferred conclusions. Answer: FALSE Explanation: Ethical analysts challenge their beliefs with data rather than forcing data to fit their beliefs. Difficulty: 1 Easy Topic: 01.04 Statistical Challenges Learning Objective: 01-04 State the common challenges facing business professionals using statistics. Bloom's: Understand AACSB: Ethics Accessibility: Keyboard Navigation 14) Statistical challenges include imperfect data, practical constraints, and ethical dilemmas. Answer: TRUE Explanation: The list is longer, but these three are big challenges. Difficulty: 1 Easy Topic: 01.04 Statistical Challenges Learning Objective: 01-05 List and explain common statistical pitfalls. Bloom's: Understand AACSB: Ethics Accessibility: Keyboard Navigation 15) A business data analyst needs a Ph.D. in statistics. Answer: FALSE Explanation: Every business person does some statistics. Difficulty: 1 Easy Topic: 01.02 Why Study Statistics? Learning Objective: 01-02 List reasons for a business student to study statistics. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 4 Copyright 2019 © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.

16) The science of statistics tells us whether the sample evidence is convincing. Answer: TRUE Explanation: There are clear scientific rules for statistical inference. Difficulty: 1 Easy Topic: 01.01 What Is Statistics? Learning Objective: 01-01 Define statistics and explain some of its uses. Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation 17) Pitfalls to consider in a statistical test include nonrandom samples, small sample size, and lack of causal links. Answer: TRUE Explanation: These are among many other pitfalls. Difficulty: 1 Easy Topic: 01.05 Critical Thinking Learning Objective: 01-05 List and explain common statistical pitfalls. Bloom's: Remember AACSB: Reflective Thinking Accessibility: Keyboard Navigation 18) In business communication, a table of numbers is preferred to a graph because it is more able to convey meaning. Answer: FALSE Explanation: Although tables can show exact numbers, a good graph may be more helpful. Difficulty: 1 Easy Topic: 01.04 Statistical Challenges Learning Objective: 01-03 Explain the uses of statistics in business. Bloom's: Understand AACSB: Communication Accessibility: Keyboard Navigation 19) Statistical data analysis can often distinguish between real versus perceived ethical issues. Answer: TRUE Explanation: Proper framing of a question may reveal that there is no real ethical issue. Difficulty: 1 Easy Topic: 01.04 Statistical Challenges Learning Objective: 01-04 State the common challenges facing business professionals using statistics. Bloom's: Understand AACSB: Ethics Accessibility: Keyboard Navigation 5 Copyright 2019 © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.

20) Excel has limited use in business because advanced statistical software is widely available. Answer: FALSE Explanation: Small businesses may lack advanced software (and the training to use it). Difficulty: 1 Easy Topic: 01.01 What Is Statistics? Learning Objective: 01-01 Define statistics and explain some of its uses. Bloom's: Understand AACSB: Technology Accessibility: Keyboard Navigation 21) Statistics helps surmount language barriers to solve problems in multinational businesses. Answer: TRUE Explanation: Statistics is part of the international language of science. Difficulty: 1 Easy Topic: 01.02 Why Study Statistics? Learning Objective: 01-02 List reasons for a business student to study statistics. Bloom's: Remember AACSB: Diversity Accessibility: Keyboard Navigation 22) Statistics can help you handle either too little or too much information. Answer: TRUE Explanation: Statistical tasks include sampling to obtain more information or finding meaning in large piles of data. Difficulty: 1 Easy Topic: 01.02 Why Study Statistics? Learning Objective: 01-02 List reasons for a business student to study statistics. Bloom's: Remember AACSB: Technology Accessibility: Keyboard Navigation 23) Predicting a presidential candidate's percentage of the statewide vote from a sample of 800 voters would be an example of inferential statistics. Answer: TRUE Explanation: Generalizing from a sample is an inference. Difficulty: 2 Medium Topic: 01.03 Statistics in Business Learning Objective: 01-01 Define statistics and explain some of its uses. Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation 6 Copyright 2019 © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.

24) Surveying electric vehicle owners would provide a representative random sample of Americans' views on global warming policies. Answer: FALSE Explanation: They are not a random sample of all Americans. Difficulty: 1 Easy Topic: 01.05 Critical Thinking Learning Objective: 01-04 State the common challenges facing business professionals using statistics. Bloom's: Apply AACSB: Reflective Thinking Accessibility: Keyboard Navigation 25) An example of descriptive statistics would be reporting the percentage of students in your accounting class that attended the review session for the last exam. Answer: TRUE Explanation: As long as you do not generalize, it is a descriptive statistic. Difficulty: 2 Medium Topic: 01.03 Statistics in Business Learning Objective: 01-01 Define statistics and explain some of its uses. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 26) "Bob must be rich. He's a lawyer, and lawyers make lots of money." This statement best illustrates which fallacy? A) Using poor survey methods B) Confusing significance with importance C) Unconscious bias D) Generalizing from an average to an individual Answer: D Explanation: Many lawyers do not work for big firms. (Remember My Cousin Vinnie?) Difficulty: 2 Medium Topic: 01.05 Critical Thinking Learning Objective: 01-05 List and explain common statistical pitfalls. Bloom's: Apply AACSB: Reflective Thinking Accessibility: Keyboard Navigation

27) Which is not an ethical obligation of a statistician? A) To know and follow accepted procedures B) To ensure data integrity and accurate calculations C) To support client wishes in drawing conclusions from the data D) To acknowledge sources of financial support Answer: C Explanation: The analyst must sometimes present findings that the client does not like. Difficulty: 1 Easy Topic: 01.04 Statistical Challenges Learning Objective: 01-04 State the common challenges facing business professionals using statistics. Bloom's: Understand AACSB: Ethics Accessibility: Keyboard Navigation 28) Which of the following statements is correct? A) A parameter is a measure that is calculated from a sample. B) Statistics is the science of collecting, organizing, analyzing, interpreting, and presenting data. C) For day-to-day business data analysis, most firms rely on a large staff of expert statisticians. D) A statistical test result that is significant also has practical importance. Answer: B Explanation: A parameter is a population characteristic. Firms often lack professional statisticians on staff, so all business graduates need some degree of statistical training to handle day-to-day problems. Sometimes an effect, while not due to chance, is too small to matter. Difficulty: 2 Medium Topic: 01.01 What Is Statistics? Learning Objective: 01-01 Define statistics and explain some of its uses. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 29) Which is least likely to be an application where statistics will be useful? A) Predicting whether an airfare is likely to rise or fall B) Designing the most desirable features for a ski pass C) Deciding whether offering Rice Krispies improves restaurant sales D) Choosing the wording of a corporate policy prohibiting smoking Answer: D Explanation: Policy wording is probably left up to writers, not statisticians. Difficulty: 2 Medium Topic: 01.03 Statistics in Business Learning Objective: 01-02 List reasons for a business student to study statistics. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 8 Copyright 2019 © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.

30) Because 25 percent of the students in my morning statistics class watch eight or more hours of television a week, I conclude that 25 percent of all students at the university watch eight or more hours of television a week. The most important logical weakness of this conclusion would be A) relying on a sample instead of surveying every student. B) using a sample that may not be representative of all students. C) failing to correct for unconscious interviewer bias. D) assuming cause and effect where none exists. Answer: B Explanation: Generalizing from a nonrandom sample is risky. The morning class may not be representative of all students. Difficulty: 3 Hard Topic: 01.05 Critical Thinking Learning Objective: 01-04 State the common challenges facing business professionals using statistics. Bloom's: Apply AACSB: Reflective Thinking Accessibility: Keyboard Navigation 31) Which of the following is not a characteristic of an ideal statistician? A) Technically current (e.g., software) B) Communicates well (both written and oral) C) Advocates client's objectives D) Can deal with imperfect information Answer: C Explanation: There is an unattractive name for a consultant who always agrees with the client. Difficulty: 1 Easy Topic: 01.04 Statistical Challenges Learning Objective: 01-03 Explain the uses of statistics in business. Bloom's: Remember AACSB: Ethics Accessibility: Keyboard Navigation

32) Which of the following statements is not true? A) Statistics helps refine theories through ongoing hypothesis testing. B) Statistics is the science of collecting, organizing, analyzing, interpreting, and presenting data. C) Estimating parameters is an important aspect of descriptive statistics. D) Statistical challenges include imperfect data and practical constraints. Answer: C Explanation: Estimating a population parameter is an inference rather than a description. Difficulty: 2 Medium Topic: 01.03 Statistics in Business Learning Objective: 01-01 Define statistics and explain some of its uses. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 33) Which is not a practical constraint facing the business researcher or data analyst? A) Time and money are always limited. B) The world is no laboratory, so some experiments are impractical. C) Research on human subjects is fraught with danger and ethical issues. D) Survey respondents usually will tell the truth if well compensated. Answer: D Explanation: Paid respondents may try to tell you what you want to hear. Difficulty: 2 Medium Topic: 01.04 Statistical Challenges Learning Objective: 01-04 State the common challenges facing business professionals using statistics. Bloom's: Understand AACSB: Ethics Accessibility: Keyboard Navigation 34) Which is not an essential characteristic of a good business data analyst? A) Effective writer B) Stays current on techniques C) Has a Ph.D. or master's degree in statistics D) Can deal with imperfect information Answer: C Explanation: No advanced degree is required for ordinary data analysis, which is why all business students study it. Difficulty: 1 Easy Topic: 01.04 Statistical Challenges Learning Objective: 01-03 Explain the uses of statistics in business. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 10 Copyright 2019 © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.

35) An ethical statistical consultant would not always A) follow accepted statistical procedures. B) support management's desired conclusions. C) acknowledge sources of financial support. D) report limitations of the data. Answer: B Explanation: There is a nasty name for a consultant who always agrees with management. Difficulty: 1 Easy Topic: 01.04 Statistical Challenges Learning Objective: 01-04 State the common challenges facing business professionals using statistics. Bloom's: Understand AACSB: Ethics Accessibility: Keyboard Navigation 36) GM's experience with ignition switches suggests that A) statistics is not applicable to automotive manufacturing. B) limited data may still contain important clues. C) good engineers can eliminate all risks. D) ignition switches are inherently dangerous. Answer: B Explanation: When small samples are all that we have, we must study them carefully, especially when the consequences are extreme (e.g., car crashes). Difficulty: 1 Easy Topic: 01.03 Statistics in Business Learning Objective: 01-01 Define statistics and explain some of its uses. Bloom's: Apply AACSB: Reflective Thinking Accessibility: Keyboard Navigation 37) Which is not a goal of the ethical data analyst? A) To be an honest broker of data. B) To learn to downplay inconvenient data. C) To understand the firm's code of ethics (or help create one). D) To look for hidden agendas in data collection. Answer: B Explanation: We do not ignore data unless it is an actual error. Difficulty: 2 Medium Topic: 01.04 Statistical Challenges Learning Objective: 01-04 State the common challenges facing business professionals using statistics. Bloom's: Remember AACSB: Ethics Accessibility: Keyboard Navigation 11 Copyright 2019 © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.

38) Which of the following statements is not true? A) A statistic is a single measure (usually numerical) that is calculated from a sample. B) Statistics is the science of collecting, organizing, analyzing, interpreting, and presenting data. C) For day-to-day business data analysis, most firms rely on a large staff of expert statisticians. D) A statistical test may be significant yet have no practical importance. Answer: C Explanation: Few firms have staffs of statistics experts, so all of us need to know the basics. Difficulty: 2 Medium Topic: 01.03 Statistics in Business Learning Objective: 01-01 Define statistics and explain some of its uses. Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation 39) "Smoking is not harmful. My Aunt Harriet smoked, but lived to age 90." This best illustrates which fallacy? A) Unconscious bias B) Significance versus practical importance C) Post hoc reasoning D) Small sample generalization Answer: D Explanation: Individual cases sometimes deviate from the average. Difficulty: 2 Medium Topic: 01.05 Critical Thinking Learning Objective: 01-05 List and explain common statistical pitfalls. Bloom's: Apply AACSB: Reflective Thinking Accessibility: Keyboard Navigation

40) Which best illustrates the distinction between statistical significance and practical importance? A) "In 2016, 240 out of 400 statistics students at Oxnard Technical College sold their textbooks at the end of the semester, compared with 220 out of 330 students in 2015, a significant decrease." B) "Our new manufacturing technique has increased the life of the 80 GB USB AsimoDrive external hard disk significantly, from 240,000 hours to 250,000 hours." C) "In 50,000 births, the new vaccine reduced the incidence of infant mortality in Morrovia significantly from 14.2 deaths per 1000 births to 10.3 deaths per 1000 births." D) "The new Sky Penetrator IV business jet's cruising range has increased significantly from 3,975 miles to 4,000 miles." Answer: B Explanation: Consumers would not notice because 240,000 hours is approximately 27 years. Difficulty: 3 Hard Topic: 01.05 Critical Thinking Learning Objective: 01-05 List and explain common statistical pitfalls. Bloom's: Evaluate AACSB: Reflective Thinking Accessibility: Keyboard Navigation 41) "Circulation fell in the month after the new editor took over the newspaper Oxnard News Herald. The new editor should be fired." Which is not a serious fallacy in this conclusion? A) Generalizing from a small sample B) Applying post hoc reasoning C) Failing to identify causes D) Using a biased sample Answer: D Explanation: There is no apparent bias here, just a shaky inference from a small sample with no apparent causal link. Difficulty: 2 Medium Topic: 01.05 Critical Thinking Learning Objective: 01-05 List and explain common statistical pitfalls. Bloom's: Evaluate AACSB: Reflective Thinking Accessibility: Keyboard Navigation

42) An ethical data analyst would be least likely to A) check data for accuracy. B) cite his/her data sources and their limitations. C) acknowledge sources of financial support. D) rely on consultants for all calculations. Answer: D Explanation: When you farm out your calculations, you have lost control of your work. Difficulty: 1 Easy Topic: 01.04 Statistical Challenges Learning Objective: 01-04 State the common challenges facing business professionals using statistics. Bloom's: Understand AACSB: Ethics Accessibility: Keyboard Navigation 43) "Tom's SUV rolled over. SUVs are dangerous." This best illustrates which fallacy? A) Unconscious bias B) Significance versus practical importance C) Post hoc reasoning D) Small sample generalization Answer: D Explanation: One instance proves little. Difficulty: 2 Medium Topic: 01.05 Critical Thinking Learning Objective: 01-05 List and explain common statistical pitfalls. Bloom's: Apply AACSB: Reflective Thinking Accessibility: Keyboard Navigation 44) "Bob didn't wear his lucky T-shirt to class, so he failed his chemistry exam." This best illustrates which fallacy? A) Small sample generalization B) Poor survey methods C) Post hoc reasoning D) More than one of the above Answer: C Explanation: There is no credible causal link between these two events. Difficulty: 2 Medium Topic: 01.05 Critical Thinking Learning Objective: 01-05 List and explain common statistical pitfalls. Bloom's: Apply AACSB: Reflective Thinking Accessibility: Keyboard Navigation 14 Copyright 2019 © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.

45) Which is not a reason for an average student to study statistics? A) Improve technical writing skills B) Gain information management skills C) Enhance technical literacy D) Learn stock market strategies Answer: D Explanation: Statistics helps improve writing and technical literacy, but to learn about the stock market, you should probably study finance. Difficulty: 1 Easy Topic: 01.02 Why Study Statistics? Learning Objective: 01-02 List reasons for a business student to study statistics. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 46) Which is not a likely area of application of statistics in business? A) Auditing supplier invoices for correct payment B) Questioning the executives' strategic decisions C) Looking for patterns in a large marketing database D) Making forecasts of several key product lines Answer: B Explanation: Business strategy may involve statistics, but not as much as the others listed here. Difficulty: 1 Easy Topic: 01.03 Statistics in Business Learning Objective: 01-03 Explain the uses of statistics in business. Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation 47) Which is not a likely task of descriptive statistics? A) Summarizing a sample B) Describing data numerically C) Estimating unknown parameters D) Making visual displays of data Answer: C Explanation: Estimating a population parameter is an inference. Difficulty: 2 Medium Topic: 01.03 Statistics in Business Learning Objective: 01-01 Define statistics and explain some of its uses. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 15 Copyright 2019 © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.

48) We would associate the term inferential statistics with which task? A) Making visual displays of data B) Estimating unknown parameters C) Describing a sample of data D) Tabulating a survey Answer: B Explanation: Estimating a population parameter is an inference. Difficulty: 1 Easy Topic: 01.03 Statistics in Business Learning Objective: 01-01 Define statistics and explain some of its uses. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 49) A good data analyst A) removes data if so instructed by client. B) works alone to avoid team conflicts. C) communicates with numbers rather than with graphs. D) reports findings that may contradict client's ideas. Answer: D Explanation: Analysts study all the data, work on teams, and use charts to clarify all findings. Difficulty: 1 Easy Topic: 01.04 Statistical Challenges Learning Objective: 01-04 State the common challenges facing business professionals using statistics. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 50) Which is not an analytical method commonly used to improve business decisions? A) Descriptive analytics. B) Predictive analytics. C) Prescriptive analytics. D) Reactive analytics. Answer: D Explanation: See Minicase 1.1 Using Analytics to Improve Business. Difficulty: 1 Easy Topic: 01.03 Statistics in Business Learning Objective: 01-03 Explain the uses of statistics in business. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 16 Copyright 2019 © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.

Applied Statistics in Business and Economics, 6e (Doane) Chapter 2 Data Collection 1) Categorical data have values that are described by words rather than numbers. Answer: TRUE Explanation: Categories are nominal data but may sometimes also be ranked (e.g., sophomore, junior, senior). Difficulty: 1 Easy Topic: 02.01 Variables and Data Learning Objective: 02-02 Explain the difference between numerical and categorical data. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 2) Numerical data can be either discrete or continuous. Answer: TRUE Explanation: Numerical data can be counts (e.g., cars owned) or continuous measures (e.g., height). Difficulty: 1 Easy Topic: 02.01 Variables and Data Learning Objective: 02-02 Explain the difference between numerical and categorical data. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 3) Categorical data are also referred to as nominal or qualitative data. Answer: TRUE Explanation: Categories are nominal data (nonnumerical), sometimes called qualitative data. Difficulty: 1 Easy Topic: 02.01 Variables and Data Learning Objective: 02-02 Explain the difference between numerical and categorical data. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation

4) The number of checks processed at a bank in a day is an example of categorical data. Answer: FALSE Explanation: Integers are numerical data. Difficulty: 1 Easy Topic: 02.01 Variables and Data Learning Objective: 02-02 Explain the difference between numerical and categorical data. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 5) The number of planes per day that land at an airport is an example of discrete data. Answer: TRUE Explanation: Integers are discrete numerical data. Difficulty: 1 Easy Topic: 02.01 Variables and Data Learning Objective: 02-02 Explain the difference between numerical and categorical data. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 6) The weight of a bag of dog food is an example of discrete data. Answer: FALSE Explanation: Weight is measured on a continuous scale. Difficulty: 1 Easy Topic: 02.01 Variables and Data Learning Objective: 02-02 Explain the difference between numerical and categorical data. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 7) In last year's annual report, Thompson Distributors indicated that it had 12 regional warehouses. This is an example of ordinal level data. Answer: FALSE Explanation: "Number of" is ratio data because a zero exists. Difficulty: 1 Easy Topic: 02.02 Level of Measurement Learning Objective: 02-04 Recognize levels of measurement in data and ways of coding data. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

8) Nominal data refer to data that can be ordered in a natural way. Answer: FALSE Explanation: Nominal (categorical) data would be called ordinal only if categories can be ranked. Difficulty: 1 Easy Topic: 02.02 Level of Measurement Learning Objective: 02-04 Recognize levels of measurement in data and ways of coding data. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 9) This year, Oxnard University produced two football All-Americans. This is an example of continuous data. Answer: FALSE Explanation: The "number of" anything is discrete. Difficulty: 1 Easy Topic: 02.01 Variables and Data Learning Objective: 02-02 Explain the difference between numerical and categorical data. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 10) The type of statistical test that we can perform is independent of the level of measurement of the variable of interest. Answer: FALSE Explanation: Some statistical operations are restricted unless you have ratio or interval data. Difficulty: 1 Easy Topic: 02.02 Level of Measurement Learning Objective: 02-04 Recognize levels of measurement in data and ways of coding data. Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation 11) Your weight recorded at your annual physical would not be ratio data, because you cannot have zero weight. Answer: FALSE Explanation: Zero is only a reference point, not necessarily an observable data value. Difficulty: 2 Medium Topic: 02.02 Level of Measurement Learning Objective: 02-04 Recognize levels of measurement in data and ways of coding data. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 3 Copyright 2019 © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.

12) The level of measurement for categorical data is nominal. Answer: TRUE Explanation: Categorical and nominal are equivalent terms. Difficulty: 1 Easy Topic: 02.02 Level of Measurement Learning Objective: 02-04 Recognize levels of measurement in data and ways of coding data. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 13) Temperature measured in degrees Fahrenheit is an example of interval data. Answer: TRUE Explanation: For temperature, scale distances are meaningful (20 to 25 is the same as 50 to 55 degrees), and 0 degrees Fahrenheit does not mean the absence of heat, so it is not a ratio measurement. Difficulty: 2 Medium Topic: 02.02 Level of Measurement Learning Objective: 02-04 Recognize levels of measurement in data and ways of coding data. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 14) The closing price of a stock is an example of ratio data. Answer: TRUE Explanation: True zero exists as a reference point, whether or not it is observed. Difficulty: 2 Medium Topic: 02.02 Level of Measurement Learning Objective: 02-04 Recognize levels of measurement in data and ways of coding data. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 15) The Statistical Abstract of the United States is a huge annual compendium of data for the United States, and it is available online free of charge. Answer: FALSE Explanation: In 2012 the U.S. Census Bureau ceased publishing this free compendium of data, but students can buy it for $199 from a private publisher. Difficulty: 1 Easy Topic: 02.05 Data Sources Learning Objective: 02-08 Find everyday print or electronic data sources. Bloom's: Remember AACSB: Technology 4 Copyright 2019 © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.

Accessibility: Keyboard Navigation 16) Ordinal data can be treated as if it were nominal data but not vice versa. Answer: TRUE Explanation: You can always go back to a lower level of measurement (but not vice versa). Difficulty: 1 Easy Topic: 02.02 Level of Measurement Learning Objective: 02-04 Recognize levels of measurement in data and ways of coding data. Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation 17) Responses on a seven-point Likert scale are usually treated as ratio data. Answer: FALSE Explanation: No true zero point exists on a Likert scale. Difficulty: 1 Easy Topic: 02.02 Level of Measurement Learning Objective: 02-04 Recognize levels of measurement in data and ways of coding data. Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation 18) Likert scales are especially important in opinion polls and marketing surveys. Answer: TRUE Explanation: Likert scales are used in all kinds of surveys. Difficulty: 1 Easy Topic: 02.02 Level of Measurement Learning Objective: 02-05 Recognize a Likert scale and know how to use it. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 19) Ordinal data are data that can be ranked based on some natural characteristic of the items. Answer: TRUE Explanation: For example, the eras Jurassic, Paleozoic, and Mesozoic can be ranked in time. Difficulty: 1 Easy Topic: 02.02 Level of Measurement Learning Objective: 02-04 Recognize levels of measurement in data and ways of coding data. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation

20) Ratio data are distinguished from interval data by the presence of a zero reference point. Answer: TRUE Explanation: The true zero is a reference that need not be observable. Difficulty: 2 Medium Topic: 02.02 Level of Measurement Learning Objective: 02-04 Recognize levels of measurement in data and ways of coding data. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 21) It is better to attempt a census of a large population instead of relying on a sample. Answer: FALSE Explanation: A census may flounder on cost and time, while samples can be quick and accurate. Difficulty: 2 Medium Topic: 02.03 Sampling Concepts Learning Objective: 02-06 Use the correct terminology for samples and populations. Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation 22) Judgment sampling and convenience sampling are nonrandom sampling techniques. Answer: TRUE Explanation: To be random, every item must have the same chance of being chosen. Difficulty: 1 Easy Topic: 02.04 Sampling Methods Learning Objective: 02-07 Explain the common sampling methods and how to implement them. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 23) A problem with judgment sampling is that the sample may not reflect the population. Answer: TRUE Explanation: While better than mere convenience, judgment may still have flaws. Difficulty: 1 Easy Topic: 02.04 Sampling Methods Learning Objective: 02-07 Explain the common sampling methods and how to implement them. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation

24) When the population is large, a sample estimate is usually preferable to a census. Answer: TRUE Explanation: A census may flounder on cost and time, while samples can be quick and accurate. Difficulty: 1 Easy Topic: 02.03 Sampling Concepts Learning Objective: 02-06 Use the correct terminology for samples and populations. Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation 25) Sampling error is avoidable by choosing the sample scientifically. Answer: FALSE Explanation: Sampling error is unavoidable, though it can be reduced by careful sampling. Difficulty: 2 Medium Topic: 02.04 Sampling Methods Learning Objective: 02-07 Explain the common sampling methods and how to implement them. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 26) A sampling frame is used to identify the target population in a statistical study. Answer: TRUE Explanation: Only some portion of the population may be targeted (e.g., independent voters). Difficulty: 2 Medium Topic: 02.03 Sampling Concepts Learning Objective: 02-06 Use the correct terminology for samples and populations. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 27) By taking a systematic sample, in which we select every 50th shopper arriving at a specific store, we are approximating a random sample of shoppers. Answer: TRUE Explanation: There is no bias if this method is implemented correctly. Difficulty: 2 Medium Topic: 02.04 Sampling Methods Learning Objective: 02-07 Explain the common sampling methods and how to implement them. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

28) A worker collecting data from every other shopper who leaves a store is taking a simple random sample of customer opinion. Answer: FALSE Explanation: Not unless the target population is customers who shopped today (cf., all customers). Also, this is a systematic (not simple) random sample. Difficulty: 2 Medium Topic: 02.04 Sampling Methods Learning Objective: 02-07 Explain the common sampling methods and how to implement them. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 29) Creating a list of people by taking the third name listed on every 10th page of the phone book is an example of convenience sampling. Answer: FALSE Explanation: This resembles two-stage cluster sampling combined with systematic sampling. Difficulty: 2 Medium Topic: 02.04 Sampling Methods Learning Objective: 02-07 Explain the common sampling methods and how to implement them. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 30) Internet surveys posted on popular websites have no bias since anyone can reply. Answer: FALSE Explanation: Self-selection bias exists (respondents may be atypical). Difficulty: 2 Medium Topic: 02.06 Surveys Learning Objective: 02-09 Describe basic elements of survey types, survey designs, and response scales. Bloom's: Understand AACSB: Technology Accessibility: Keyboard Navigation

31) Analysis of month-by-month changes in stock market prices during the most recent recession would require the use of time series data. Answer: TRUE Explanation: Data collected and recorded over time would be a time series. Difficulty: 2 Medium Topic: 02.01 Variables and Data Learning Objective: 02-03 Explain the difference between time series and cross-sectional data. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 32) A cluster sample is a type of stratified sample that is based on geographical location. Answer: TRUE Explanation: An example would be sampling voters randomly within random zip codes. Difficulty: 1 Easy Topic: 02.04 Sampling Methods Learning Objective: 02-07 Explain the common sampling methods and how to implement them. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 33) An advantage of a systematic sample is that no list of enumerated data items is required. Answer: TRUE Explanation: Systematic sampling works with a list (like random sampling) but also without one. Difficulty: 1 Easy Topic: 02.04 Sampling Methods Learning Objective: 02-07 Explain the common sampling methods and how to implement them. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 34) Telephone surveys often have a low response rate and fail to reach the desired population. Answer: TRUE Explanation: Phone surveys are cheaper, but suffer from these weaknesses. Difficulty: 1 Easy Topic: 02.06 Surveys Learning Objective: 02-09 Describe basic elements of survey types, survey designs, and response scales. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 9 Copyright 2019 © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.

35) Mail surveys are attractive because of their high response rates. Answer: FALSE Explanation: Mail surveys have low response rates and invite self-selection bias. Difficulty: 1 Easy Topic: 02.06 Surveys Learning Objective: 02-09 Describe basic elements of survey types, survey designs, and response scales. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 36) A problem with convenience sampling is that the target population is not well-defined. Answer: TRUE Explanation: Convenience sampling is quick but not random, and the target population is unclear. Difficulty: 2 Medium Topic: 02.04 Sampling Methods Learning Objective: 02-07 Explain the common sampling methods and how to implement them. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 37) If you randomly sample 50 students about their favorite places to eat, the data collected would be referred to as cross-sectional data. Answer: TRUE Explanation: Data for individuals would be a cross section (not a time series). Difficulty: 2 Medium Topic: 02.01 Variables and Data Learning Objective: 02-03 Explain the difference between time series and cross-sectional data. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 38) The number of FedEx shipping centers in each of 50 cities would be ordinal level data. Answer: FALSE Explanation: The "number of" anything is ratio data because a true zero reference point exists. Difficulty: 2 Medium Topic: 02.02 Level of Measurement Learning Objective: 02-04 Recognize levels of measurement in data and ways of coding data. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 11 Copyright 2019 © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.

39) Internet surveys posted on popular websites suffer from nonresponse bias. Answer: TRUE Explanation: Nonresponse or self-selection bias is rampant in such surveys. Difficulty: 2 Medium Topic: 02.06 Surveys Learning Objective: 02-09 Describe basic elements of survey types, survey designs, and response scales. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 40) Different variables are usually shown as columns of a multivariate data set. Answer: TRUE Explanation: It is customary to use a column for each variable, while each row is an observation. Difficulty: 1 Easy Topic: 02.01 Variables and Data Learning Objective: 02-01 Use basic terminology for describing data and samples. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 41) Each row in a multivariate data matrix is an observation (e.g., an individual response). Answer: TRUE Explanation: It is customary to use a column for each variable, while each row is an observation. Difficulty: 1 Easy Topic: 02.01 Variables and Data Learning Objective: 02-01 Use basic terminology for describing data and samples. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 42) A bivariate data set has only two observations on a variable. Answer: FALSE Explanation: Bivariate refers to the number of variables, not the number of observations. Difficulty: 1 Easy Topic: 02.01 Variables and Data Learning Objective: 02-01 Use basic terminology for describing data and samples. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 12 Copyright 2019 © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.

43) Running times for 3,000 runners in a 5k race would be a multivariate data set. Answer: FALSE Explanation: Regardless of the number of observations, we have only one variable (running time). Difficulty: 1 Easy Topic: 02.01 Variables and Data Learning Objective: 02-01 Use basic terminology for describing data and samples. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 44) Running times for 500 runners in a 5k race would be a univariate data set. Answer: TRUE Explanation: Regardless of the number of observations, we have only one variable (running time). Difficulty: 1 Easy Topic: 02.01 Variables and Data Learning Objective: 02-01 Use basic terminology for describing data and samples. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 45) A list of the salaries, ages, and years of experience for 50 CEOs is a multivariate data set. Answer: TRUE Explanation: We would have a data matrix with 50 rows and 3 columns. Difficulty: 1 Easy Topic: 02.01 Variables and Data Learning Objective: 02-01 Use basic terminology for describing data and samples. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 46) The daily closing price of Apple stock over the past month would be a time series. Answer: TRUE Explanation: Data collected over time is a time series. Difficulty: 2 Medium Topic: 02.01 Variables and Data Learning Objective: 02-03 Explain the difference between time series and cross-sectional data. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 13 Copyright 2019 © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.

47) The number of words on 50 randomly chosen textbook pages would be cross-sectional data. Answer: TRUE Explanation: Data were not collected over time, so we have cross-sectional data. Difficulty: 2 Medium Topic: 02.01 Variables and Data Learning Objective: 02-03 Explain the difference between time series and cross-sectional data. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 48) A Likert scale with an even number of scale points between "Strongly Agree" and "Strongly Disagree" is intended to prevent "neutral" choices. Answer: TRUE Explanation: An even number of scale points (e.g., 4) forces the respondent to "lean" toward one end of the scale or the other. Difficulty: 2 Medium Topic: 02.02 Level of Measurement Learning Objective: 02-05 Recognize a Likert scale and know how to use it. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 49) Private statistical databases (e.g., CRSP) are usually free. Answer: FALSE Explanation: Private research databases generally require a subscription (often expensive). Difficulty: 1 Easy Topic: 02.05 Data Sources Learning Objective: 02-08 Find everyday print or electronic data sources. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation

50) An investment firm rates bonds for Aard Co Inc. as "B+," while bonds of Deva Corp. are rated "AA." Which level of measurement would be appropriate for such data? A) Nominal B) Ordinal C) Interval D) Ratio Answer: B Explanation: Ranks are clear, but interval would require assumed equal scale distances (doubtful). Difficulty: 2 Medium Topic: 02.02 Level of Measurement Learning Objective: 02-04 Recognize levels of measurement in data and ways of coding data. Bloom's: Evaluate AACSB: Analytical Thinking Accessibility: Keyboard Navigation 51) Which variable is least likely to be regarded as ratio data? A) Length of time required for a randomly chosen vehicle to cross a toll bridge (minutes) B) Weight of a randomly chosen student (pounds) C) Number of fatalities in a randomly chosen traffic disaster (persons) D) Student's evaluation of a professor's teaching (Likert scale) Answer: D Explanation: A Likert scale has no true zero. The other examples do. Difficulty: 2 Medium Topic: 02.02 Level of Measurement Learning Objective: 02-04 Recognize levels of measurement in data and ways of coding data. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 52) Which of the following is numerical data? A) Your gender B) The brand of cell phone you own C) Whether you have an American Express card D) The fuel economy (MPG) of your car Answer: D Explanation: Fuel economy is numerical. The others are categorical. Difficulty: 1 Easy Topic: 02.01 Variables and Data Learning Objective: 02-02 Explain the difference between numerical and categorical data. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 15 Copyright 2019 © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.

53) Measurements from a sample are called A) statistics. B) inferences. C) parameters. D) variables. Answer: A Explanation: A measurement calculated from a sample is a statistic. Difficulty: 1 Easy Topic: 02.04 Sampling Methods Learning Objective: 02-06 Use the correct terminology for samples and populations. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 54) Quantitative variables use which two levels of measurement? A) Ordinal and ratio B) Interval and ordinal C) Nominal and ordinal D) Interval and ratio Answer: D Explanation: Numerical (quantitative) data can be interval or ratio. Difficulty: 2 Medium Topic: 02.02 Level of Measurement Learning Objective: 02-04 Recognize levels of measurement in data and ways of coding data. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 55) Temperature in degrees Fahrenheit is an example of a(n) ________ variable. A) nominal B) ordinal C) interval D) ratio Answer: C Explanation: No true zero exists in temperature measurements except on the Kelvin scale. Difficulty: 1 Easy Topic: 02.02 Level of Measurement Learning Objective: 02-04 Recognize levels of measurement in data and ways of coding data. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

56) Using a sample to make generalizations about an aspect of a population is called A) data mining. B) descriptive statistics. C) random sampling. D) statistical inference. Answer: D Explanation: Generalizing from a sample to a population is an inference. Difficulty: 1 Easy Topic: 02.03 Sampling Concepts Learning Objective: 02-06 Use the correct terminology for samples and populations. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 57) Your telephone area code is an example of a(n) ________ variable. A) nominal B) ordinal C) interval D) ratio Answer: A Explanation: Area codes are not ranked, so they are merely nominal (i.e., categorical). Difficulty: 2 Medium Topic: 02.02 Level of Measurement Learning Objective: 02-04 Recognize levels of measurement in data and ways of coding data. Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation 58) Which is least likely to be regarded as a ratio variable? A) A critic's rating of a restaurant on a 1 to 4 scale B) Automobile exhaust emission of nitrogen dioxide (milligrams per mile) C) Number of customer complaints per day at a cable TV company office D) Cost of an eBay purchase Answer: A Explanation: Ratings on a Likert scale have no meaningful zero. Difficulty: 2 Medium Topic: 02.02 Level of Measurement Learning Objective: 02-04 Recognize levels of measurement in data and ways of coding data. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

59) Automobile exhaust emission of CO2 (milligrams per mile) is ________ data. A) nominal B) ordinal C) interval D) ratio Answer: D Explanation: Meaningful zero emissions are possible (e.g., electric car) so ratio. Difficulty: 2 Medium Topic: 02.02 Level of Measurement Learning Objective: 02-04 Recognize levels of measurement in data and ways of coding data. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 60) Your rating of the food served at a local restaurant using a three-point scale of 0 = gross, 1 = decent, 2 = yummy is ________ data. A) nominal B) ordinal C) interval D) ratio Answer: B Explanation: Only rankings are implied (not equal scale distances). Difficulty: 2 Medium Topic: 02.02 Level of Measurement Learning Objective: 02-04 Recognize levels of measurement in data and ways of coding data. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 61) The number of passengers "bumped" on a particular airline flight is ________ data. A) nominal B) ordinal C) interval D) ratio Answer: D Explanation: A true zero point exists (no passengers might be bumped). Difficulty: 1 Easy Topic: 02.02 Level of Measurement Learning Objective: 02-04 Recognize levels of measurement in data and ways of coding data. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 18 Copyright 2019 © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.

62) Which should not be regarded as a continuous random variable? A) Tonnage carried by a randomly chosen oil tanker at sea B) Wind velocity at 7 o'clock this morning C) Number of personal fouls by the Miami Heat in a game D) Length of time to play a Wimbledon tennis match Answer: C Explanation: Counting things yields integer (discrete) data. Difficulty: 2 Medium Topic: 02.01 Variables and Data Learning Objective: 02-02 Explain the difference between numerical and categorical data. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 63) Which of the following is not true? A) Categorical data have values that are described by words rather than numbers. B) Categorical data are also referred to as nominal or qualitative data. C) The number of checks processed at a bank in a day is categorical data. D) Numerical data can be either discrete or continuous. Answer: C Explanation: The "number of" anything is a discrete numerical variable. Difficulty: 2 Medium Topic: 02.01 Variables and Data Learning Objective: 02-02 Explain the difference between numerical and categorical data. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 64) Which of the following is true? A) The type of charge card used by a customer (Visa, MasterCard, AmEx) is ordinal data. B) The duration (minutes) of a flight from Boston to Minneapolis is ratio data. C) The number of Nobel Prize–winning faculty at Oxnard University is continuous data. D) The number of regional warehouses owned by Jankord Industries is ordinal data. Answer: B Explanation: A true zero exists as a reference point (even if not observed), so ratios have meaning. Difficulty: 2 Medium Topic: 02.02 Level of Measurement Learning Objective: 02-04 Recognize levels of measurement in data and ways of coding data. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

65) Which statement is correct? A) Judgment sampling is preferred to systematic sampling. B) Sampling without replacement introduces bias in our estimates of parameters. C) Cluster sampling is useful when strata characteristics are unknown. D) Focus groups usually work best without a moderator. Answer: C Explanation: Review the characteristics of each sampling method. Difficulty: 2 Medium Topic: 02.04 Sampling Methods Learning Objective: 02-07 Explain the common sampling methods and how to implement them. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 66) A Likert scale A) yields interval data if scale distances are equal. B) must have an odd number of scale points. C) must have a verbal label on each scale point. D) is rarely used in marketing surveys. Answer: A Explanation: Marketers use Likert scales and try to make scales with meaningful intervals. Difficulty: 2 Medium Topic: 02.02 Level of Measurement Learning Objective: 02-05 Recognize a Likert scale and know how to use it. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 67) Which is most nearly correct regarding sampling error? A) It can be eliminated by increasing the sample size. B) It cannot be eliminated by any statistical sampling method. C) It can be eliminated by using Excel's =RANDBETWEEN() function. D) It can be eliminated by utilizing systematic random sampling. Answer: B Explanation: Sampling involves error, though it can be minimized by proper methodology. Difficulty: 2 Medium Topic: 02.03 Sampling Concepts Learning Objective: 02-06 Use the correct terminology for samples and populations. Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation

68) Which statement is false? A) Random dialing phone surveys have low response and are poorly targeted. B) Selection bias means that many respondents dislike the interviewer. C) Simple random sampling requires a list of the population. D) Web surveys are economical but suffer from nonresponse bias. Answer: B Explanation: Selection bias occurs when respondents are atypical. Difficulty: 2 Medium Topic: 02.06 Surveys Learning Objective: 02-09 Describe basic elements of survey types, survey designs, and response scales. Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation 69) Judgment sampling is sometimes preferred over random sampling, for example, when A) the desired sample size is much larger than the population. B) the sampling budget is large and the population is conveniently located. C) time is short and the sampling budget is limited. D) the population is readily accessible and sampling is nondestructive. Answer: C Explanation: Judgment sampling can save time and may be better than mere convenience. Difficulty: 2 Medium Topic: 02.04 Sampling Methods Learning Objective: 02-07 Explain the common sampling methods and how to implement them. Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation 70) An advantage of convenience samples is that A) the required sample size is easier to calculate. B) sampling error can be reduced. C) computation of statistics is easier. D) they are often quicker and cheaper. Answer: D Explanation: Convenience samples are quick, with a possible trade-off of accuracy. Difficulty: 1 Easy Topic: 02.04 Sampling Methods Learning Objective: 02-07 Explain the common sampling methods and how to implement them. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation

71) Before deciding whether to assess heavy fines against noisy airlines, which sampling method would the Federal Aviation Administration probably use to measure the peak noise from departing jets as measured by a ground-level observer at a point one mile from the end of the departure runway? A) Radio survey of pilots. B) Simple random sample. C) Judgment sample. D) Stratified sample. Answer: D Explanation: From the cockpit, pilots can't assess external noise levels, so a radio survey of pilots is not useful. Measurements must be taken from the ground. No list is available for the unpredictable mix of departing flights, so we cannot use a simple random sample. A judgment sample would not provide an objective basis for assessing fines. A reasonable option would be for ground observers to record the aircraft size, type, and carrier (airline) for each departing flight for a week and use this information to construct a stratified sample. Difficulty: 3 Hard Topic: 02.04 Sampling Methods Learning Objective: 02-07 Explain the common sampling methods and how to implement them. Bloom's: Evaluate AACSB: Analytical Thinking Accessibility: Keyboard Navigation 72) Professor Hardtack chose a sample of 7 students from his statistics class of 35 students by picking every student who was wearing red that day. Which kind of sample is this? A) Simple random sample B) Judgment sample C) Systematic sample D) Convenience sample Answer: D Explanation: It may be quick but no judgment is involved and may not be representative of all students. Difficulty: 2 Medium Topic: 02.04 Sampling Methods Learning Objective: 02-07 Explain the common sampling methods and how to implement them. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

73) Thirty work orders are selected from a filing cabinet containing 500 work order folders by choosing every 15th folder. Which sampling method is this? A) Simple random sample B) Systematic sample C) Stratified sample D) Cluster sample Answer: B Explanation: This is a classic systematic sample from an accessible but unlisted population. Difficulty: 2 Medium Topic: 02.04 Sampling Methods Learning Objective: 02-07 Explain the common sampling methods and how to implement them. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 74) Which of the following is not a likely reason for sampling? A) The destructive nature of certain tests B) The physical impossibility of checking all the items in the population C) Prohibitive cost of studying the entire population D) The expense of obtaining random numbers Answer: D Explanation: Random numbers are cheap (e.g., Excel's =RANDBETWEEN function). Difficulty: 2 Medium Topic: 02.03 Sampling Concepts Learning Objective: 02-06 Use the correct terminology for samples and populations. Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation 75) Comparing a census of a large population to a sample drawn from it, we expect that the A) sample is usually a more practical method of obtaining the desired information. B) accuracy of the observations in the census is surely higher than in the sample. C) sample must be a large fraction of the population to be accurate. Answer: A Explanation: Census is often impractical, while samples can be extremely accurate. Difficulty: 2 Medium Topic: 02.03 Sampling Concepts Learning Objective: 02-06 Use the correct terminology for samples and populations. Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation

76) A stratified sample is sometimes recommended when A) the sample size is very large. B) the population is small compared to the sample. C) distinguishable strata can be identified in the populations. D) the population is spread out geographically. Answer: C Explanation: Identifiable strata such as gender, ethnicity, or region can be used. Difficulty: 2 Medium Topic: 02.04 Sampling Methods Learning Objective: 02-07 Explain the common sampling methods and how to implement them. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 77) A random sample is one in which the A) probability that an item is selected for the sample is the same for all population items. B) population items are selected haphazardly by experienced workers. C) items to be selected from the population are specified based on expert judgment. D) probability of selecting a population item depends on the item's data value. Answer: A Explanation: Each item must have the same chance of being picked if the sample is random. Difficulty: 2 Medium Topic: 02.04 Sampling Methods Learning Objective: 02-07 Explain the common sampling methods and how to implement them. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 78) An advantage of convenience samples over random samples is that A) they are easy to analyze. B) it is easier to determine the sample size needed. C) it is easier to calculate the sampling errors involved. D) data collection cost is reduced. Answer: D Explanation: Convenience samples are often used because they are quick (but maybe not accurate). Difficulty: 1 Easy Topic: 02.04 Sampling Methods Learning Objective: 02-07 Explain the common sampling methods and how to implement them. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation

79) To measure satisfaction with its cell phone service, AT&T takes a stratified sample of its customers by age, gender, and location. Which is an advantage of this type of sampling, as opposed to other sampling methods? A) It is less intrusive on customers' privacy. B) It does not require random numbers. C) It gives faster results. D) It can give more accurate results. Answer: D Explanation: Stratified sampling can yield more complete and accurate information. Difficulty: 2 Medium Topic: 02.04 Sampling Methods Learning Objective: 02-07 Explain the common sampling methods and how to implement them. Bloom's: Analyze AACSB: Diversity Accessibility: Keyboard Navigation 80) An accounting professor wishing to know how many MBA students would take a summer elective in international accounting did a survey of the class she was teaching. Which kind of sample is this? A) Simple random sample B) Cluster sample C) Systematic sample D) Convenience sample Answer: D Explanation: She may bias the estimate because only accounting students were surveyed. Difficulty: 2 Medium Topic: 02.04 Sampling Methods Learning Objective: 02-07 Explain the common sampling methods and how to implement them. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

81) A binary variable (also called a dichotomous variable or dummy variable) has A) only two possible values. B) continuous scale values. C) rounded data values. D) ordinal or interval values. Answer: A Explanation: Binary variables are used in every field of business to code qualitative (nominal) data. Difficulty: 1 Easy Topic: 02.01 Variables and Data Learning Objective: 02-02 Explain the difference between numerical and categorical data. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 82) A population has groups that have a small amount of variation within them, but large variation among or between the groups themselves. The proper sampling technique is A) simple random. B) stratified. C) cluster. D) judgment. Answer: B Explanation: Identifiable strata call for stratified sampling if you can afford the extra time and cost. Difficulty: 3 Hard Topic: 02.04 Sampling Methods Learning Objective: 02-07 Explain the common sampling methods and how to implement them. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

83) A manager chose two people from her team of eight to give an oral presentation because she felt they were representative of the whole team's views. What sampling technique did she use in choosing these two people? A) Convenience B) Simple random C) Judgment D) Cluster Answer: C Explanation: Expert judgment may be better than just pointing a finger (we hope). Difficulty: 2 Medium Topic: 02.04 Sampling Methods Learning Objective: 02-07 Explain the common sampling methods and how to implement them. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 84) Sampling bias can best be reduced by A) using appropriate data coding. B) having a computer tabulate the results. C) utilizing random sampling. D) taking a judgment sample. Answer: C Explanation: Sampling error cannot be eliminated, but sampling bias can be avoided. Difficulty: 2 Medium Topic: 02.04 Sampling Methods Learning Objective: 02-07 Explain the common sampling methods and how to implement them. Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation 85) A sampling technique used when groups are defined by their geographical location is A) cluster sampling. B) convenience sampling. C) judgment sampling. D) random sampling. Answer: A Explanation: Strata based on location can be targeted through cluster sampling. Difficulty: 2 Medium Topic: 02.04 Sampling Methods Learning Objective: 02-07 Explain the common sampling methods and how to implement them. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 27 Copyright 2019 © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.

86) If we choose 500 random numbers using Excel's function =RANDBETWEEN(1,99), we would most likely find that A) numbers near the mean (50) would tend to occur more frequently. B) numbers near 1 and 99 would tend to occur less frequently. C) some numbers would occur more than once. D) the numbers would have a clear pattern. Answer: C Explanation: On average, we would expect each number to occur around five times. Difficulty: 2 Medium Topic: 02.04 Sampling Methods Learning Objective: 02-07 Explain the common sampling methods and how to implement them. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 87) A problem with nonrandom sampling is that A) larger samples need to be taken to reduce the sampling error inherent in this approach. B) not every item in the population has the same chance of being selected, as it should. C) it is usually more expensive than random sampling. D) it generally provides lower response rates than random sampling. Answer: B Explanation: Only random sampling gives every item the same chance to be picked. Difficulty: 2 Medium Topic: 02.04 Sampling Methods Learning Objective: 02-07 Explain the common sampling methods and how to implement them. Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation 88) From its 32 regions, the FAA selects 6 regions, and then randomly audits 25 departing commercial flights in each region for compliance with legal fuel and weight requirements. This is an example of A) simple random sampling. B) stratified random sampling. C) cluster sampling. D) judgment sampling. Answer: C Explanation: Two-stage cluster sampling is being used (a special form of stratification). Difficulty: 2 Medium Topic: 02.04 Sampling Methods Learning Objective: 02-07 Explain the common sampling methods and how to implement them. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 28 Copyright 2019 © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.

89) Which of the following is a correct statement? A) Choosing the third person listed on every fifth page of the phone book is stratified sampling. B) An advantage of a systematic sample is that no list of enumerated data items is required. C) Convenience sampling is used to study shoppers in convenience stores. D) Judgment sampling is an example of true random sampling. Answer: B Explanation: Review the sampling methods and their characteristics. Difficulty: 2 Medium Topic: 02.04 Sampling Methods Learning Objective: 02-07 Explain the common sampling methods and how to implement them. Bloom's: Analyze AACSB: Analytical Thinking Accessibility: Keyboard Navigation 90) Which of the following is false? A) Sampling error is the difference between the true parameter and the sample estimate. B) Sampling error is a result of unavoidable random variation in a sample. C) A sampling frame is chosen from the target population in a statistical study. D) The target population must first be defined by a full list or data file of all individuals. Answer: D Explanation: Review the terminology of sampling. Difficulty: 2 Medium Topic: 02.03 Sampling Concepts Learning Objective: 02-06 Use the correct terminology for samples and populations. Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation 91) When we are choosing a random sample and we do not place chosen units back into the population, we are A) sampling with replacement. B) sampling without replacement. C) using a systematic sample. D) using a voluntary sample. Answer: B Explanation: In sampling without replacement the items chosen are not independent. Statistics formulas usually assume replacement to ensure unbiased estimates. Difficulty: 1 Easy Topic: 02.04 Sampling Methods Learning Objective: 02-07 Explain the common sampling methods and how to implement them. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 29 Copyright 2019 © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.

92) Which method is likely to be used by a journalism student who is casually surveying opinions of students about the university's cafeteria food for an article that she is writing? A) Simple random sample B) Systematic random sample C) Cluster sample D) Convenience sample Answer: D Explanation: Quick and easy may trump true random sampling for a busy journalist. Difficulty: 2 Medium Topic: 02.04 Sampling Methods Learning Objective: 02-07 Explain the common sampling methods and how to implement them. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 93) Which of the following is false? A) Mail surveys are cheap but have low response rates. B) Coverage error is when respondents give untruthful answers. C) Focus groups are nonrandom but can probe issues more deeply. D) Surveys posted on popular websites suffer from selection bias. Answer: B Explanation: Coverage error is when you miss some segment of the target population. Difficulty: 2 Medium Topic: 02.06 Surveys Learning Objective: 02-09 Describe basic elements of survey types, survey designs, and response scales. Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation

94) Which is a time series variable? A) VISA balances of 30 students on December 31 of this year B) Net earnings reported by Xena Corp. for the last 10 quarters C) Dollar exchange rates yesterday against 10 other world currencies D) Titles of the top 10 movies in total revenue last week Answer: B Explanation: If x1, x2, . . ., xn do not refer to n time periods, it isn't a time series. Difficulty: 2 Medium Topic: 02.02 Level of Measurement Learning Objective: 02-03 Explain the difference between time series and cross-sectional data. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 95) An observation in a data set would refer to A) only a variable whose value is recorded by visual inspection. B) a data item whose value is numerical (as opposed to categorical). C) a single row that contains one or more observed variables. D) the values of all the variables in the entire data set. Answer: C Explanation: We usually put observations in rows on a spreadsheet, while each column is a variable. Difficulty: 1 Easy Topic: 02.01 Variables and Data Learning Objective: 02-01 Use basic terminology for describing data and samples. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 96) A multivariate data set contains A) more than two observations. B) more than two categorical variables. C) more than two variables. D) more than two levels of measurement. Answer: C Explanation: When you have more than two variables, it is multivariate data. Difficulty: 1 Easy Topic: 02.01 Variables and Data Learning Objective: 02-01 Use basic terminology for describing data and samples. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 31 Copyright 2019 © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.

97) The Centers for Disease Control and Prevention (CDC) wants to estimate the average extra hospital stay that occurs when heart surgery patients experience postoperative atrial fibrillation. They divide the United States into nine regions. In each region, hospitals are selected at random within each hospital size group (small, medium, large). In each hospital, heart surgery patients are sampled according to known percentages by age group (under 50, 50 to 64, 65 and over) and gender (male, female). This procedure combines which sampling methods? A) Systematic, simple random, and convenience B) Convenience, systematic, and judgment C) Cluster, stratified, and simple random D) Judgment, systematic, and simple random Answer: C Explanation: Identifiable strata were sampled, but also random sampling within strata and regional clusters was used. Difficulty: 3 Hard Topic: 02.04 Sampling Methods Learning Objective: 02-07 Explain the common sampling methods and how to implement them. Bloom's: Evaluate AACSB: Analytical Thinking Accessibility: Keyboard Navigation 98) Which statement is correct? A) Selecting every fifth shopper arriving at a store will approximate a random sample of shoppers. B) Selecting only shoppers who drive SUVs is a stratified sampling method. C) A census is preferable to a sample for most business problems. D) Stratified samples are usually cheaper than other methods. Answer: A Explanation: Done carefully, systematic sampling is close to random when there is no list. Difficulty: 2 Medium Topic: 02.04 Sampling Methods Learning Objective: 02-07 Explain the common sampling methods and how to implement them. Bloom's: Evaluate AACSB: Analytical Thinking Accessibility: Keyboard Navigation

99) Which is a categorical variable? A) The brand of jeans you usually wear B) The price you paid for your last pair of jeans C) The distance to the store where you purchased your last pair of jeans D) The number of pairs of jeans that you own Answer: A Explanation: Categories have only names (e.g., Calvin Klein). Difficulty: 1 Easy Topic: 02.01 Variables and Data Learning Objective: 02-02 Explain the difference between numerical and categorical data. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 100) Which is a discrete variable? A) The time it takes to put on a pair of jeans B) The price you paid for your last pair of jeans C) The distance to the store where you purchased your last pair of jeans D) The number of pairs of jeans that you own Answer: D Explanation: The "number of" anything is discrete numerical data. Difficulty: 1 Easy Topic: 02.01 Variables and Data Learning Objective: 02-02 Explain the difference between numerical and categorical data. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 101) A section of the population we have targeted for analysis is A) a statistic. B) a frame. C) a sample. D) a coven. Answer: B Explanation: We must define the segment we want to look at (e.g., independent voters). Difficulty: 1 Easy Topic: 02.03 Sampling Concepts Learning Objective: 02-06 Use the correct terminology for samples and populations. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation

102) Which is not a time series variable? A) Closing checkbook balances of 30 students on December 31 of this year B) Net earnings reported by Xena Corp. for the last 10 quarters C) Dollar/euro exchange rates at 12 noon GMT for the last 30 days D) Movie attendance at a certain theater for each Saturday last year Answer: A Explanation: If x1, x2, . . ., xn do not refer to n time periods, it is not a time series. Difficulty: 2 Medium Topic: 02.02 Level of Measurement Learning Objective: 02-03 Explain the difference between time series and cross-sectional data. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 103) A good Likert scale may not have A) unequal distances between scale points. B) an odd number of scale points. C) a verbal label on each scale point. D) verbal anchors at its end points. Answer: A Explanation: Surveys try to create scales with meaningful intervals. Difficulty: 2 Medium Topic: 02.02 Level of Measurement Learning Objective: 02-02 Explain the difference between numerical and categorical data. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 104) A Likert scale with an odd number of scale points between "Strongly Agree" and "Strongly Disagree" A) cannot have equal scale distances. B) cannot have a neutral middle point. C) must have a verbal label on each scale point. D) is often used in marketing surveys. Answer: D Explanation: Likert scales should have arguably equal intervals. A middle neutral response is possible with an odd number of scale points (e.g., 5 or 7). Difficulty: 2 Medium Topic: 02.02 Level of Measurement Learning Objective: 02-05 Recognize a Likert scale and know how to use it. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 34 Copyright 2019 © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.

105) A Likert scale with an even number of scale points between "Strongly Agree" and "Strongly Disagree" A) cannot have equal scale distances. B) is intended to prevent "neutral" choices. C) must have a verbal label on each scale point. D) is rarely used in surveys. Answer: B Explanation: Likert scales should have arguably equal intervals. An even number of scale points (e.g., 4) forces the respondent to "lean" toward one end of the scale or the other. Difficulty: 2 Medium Topic: 02.02 Level of Measurement Learning Objective: 02-05 Recognize a Likert scale and know how to use it. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 106) Which statement is correct? A) Analysts rarely consult business periodicals (e.g., Bloomberg Businessweek). B) Web searches (e.g., Google) often yield unverifiable data. C) Government data sources (e.g., www.bls.gov) are often costly. D) Private statistical databases (e.g., CRSP) are usually free. Answer: B Explanation: Periodicals are often up-to-date and readily available data sources. Web data may be unreliable, and searches may be directed toward obtaining payment for data. Private research databases generally require a subscription, while government data sources generally are free. Difficulty: 1 Easy Topic: 02.05 Data Sources Learning Objective: 02-08 Find everyday print or electronic data sources. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation

107) Which statement is correct? A) Analysts avoid business periodicals (e.g., Bloomberg Businessweek). B) Web searches (e.g., Google) yield reliable and easily verified data. C) Government data sources (e.g., www.bls.gov) usually are free. D) Private statistical databases (e.g., CRSP) usually are free. Answer: C Explanation: Periodicals are often up-to-date and readily available data sources. Web data may be unreliable, and searches may be directed toward obtaining payment for data. Private research databases generally require a subscription, while government data sources generally are free. Difficulty: 1 Easy Topic: 02.05 Data Sources Learning Objective: 02-08 Find everyday print or electronic data sources. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 108) A valid survey is one that A) measures what the researcher wants to measure. B) has been approved by top management. C) is administered by a professional statistician. D) has a large number of questions. Answer: A Explanation: There is no need for a large number of questions, although the survey must be tested to be sure it measures what the researcher wants to know. Difficulty: 1 Easy Topic: 02.06 Surveys Learning Objective: 02-09 Describe basic elements of survey types, survey designs, and response scales. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation

109) A reliable survey is one that A) is administered by trusted employees. B) has been approved by quality engineers. C) gives consistent measurements. D) has many easy questions. Answer: A Explanation: Over time and across groups of similar respondents, a reliable survey should give consistent results (within expected ranges of statistical variation). Difficulty: 1 Easy Topic: 02.06 Surveys Learning Objective: 02-09 Describe basic elements of survey types, survey designs, and response scales. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation

Applied Statistics in Business and Economics, 6e (Doane) Chapter 3 Describing Data Visually 1) It is easier to read the data values on a 3D column chart than on a 2D column chart. Answer: FALSE Explanation: Height is harder to judge on a 3D chart. Difficulty: 1 Easy Topic: 03.05 Column and Bar Charts Learning Objective: 03-06 Make an effective column chart or bar chart. Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation 2) The column chart should be avoided if you are plotting time series data. Answer: FALSE Explanation: Line charts are more common, but column charts also work for a time series. Difficulty: 2 Medium Topic: 03.05 Column and Bar Charts Learning Objective: 03-06 Make an effective column chart or bar chart. Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation 3) The line chart is appropriate for categorical (qualitative) data. Answer: FALSE Explanation: Only numerical data can be plotted on a line chart. Difficulty: 1 Easy Topic: 03.04 Line Charts Learning Objective: 03-05 Make an effective line chart. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 4) The Pareto chart is used to display the "vital few" causes of problems. Answer: TRUE Explanation: Causes are displayed in a column or bar chart sorted in order of frequency. Difficulty: 1 Easy Topic: 03.05 Column and Bar Charts Learning Objective: 03-06 Make an effective column chart or bar chart. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 1 Copyright 2019 © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.

5) Excel's pyramid chart is generally preferred to a plain 2D column chart. Answer: FALSE Explanation: Avoid novelty charts when a plain 2D bar chart will do. Difficulty: 1 Easy Topic: 03.09 Deceptive Graphs Learning Objective: 03-06 Make an effective column chart or bar chart. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 6) Excel's pyramid charts make it easier to read the data values. Answer: FALSE Explanation: Avoid novelty charts when a plain 2D column chart will do. Difficulty: 1 Easy Topic: 03.09 Deceptive Graphs Learning Objective: 03-06 Make an effective column chart or bar chart. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 7) Dot plots are similar to histograms with many bins (classes). Answer: TRUE Explanation: Height of the dot stack is analogous to bar height. Difficulty: 2 Medium Topic: 03.01 Stem-and-Leaf Displays and Dot Plots Learning Objective: 03-01 Make a stem-and-leaf or dot plot. Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation 8) Compared to a dot plot, we lose some detail when we present data in a frequency distribution. Answer: TRUE Explanation: Individual data values are easier to see on a dot plot. Difficulty: 2 Medium Topic: 03.02 Frequency Distributions and Histograms Learning Objective: 03-02 Create a frequency distribution for a data set. Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation

9) Stacked dot plots are useful in understanding the association between two paired quantitative variables (X, Y). Answer: FALSE Explanation: You would prefer a scatter plot for X-Y data. Difficulty: 3 Hard Topic: 03.01 Stem-and-Leaf Displays and Dot Plots Learning Objective: 03-01 Make a stem-and-leaf or dot plot. Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation 10) Log scales are common because most people are familiar with them. Answer: FALSE Explanation: Log scales seem to be harder for the average business person to interpret Difficulty: 2 Medium Topic: 03.04 Line Charts Learning Objective: 03-05 Make an effective line chart. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 11) Sturges' Rule should override judgment about the "right" number of histogram bins. Answer: FALSE Explanation: Sturges' Rule is only a starting point or guideline. Difficulty: 2 Medium Topic: 03.02 Frequency Distributions and Histograms Learning Objective: 03-03 Make a histogram with appropriate bins. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 12) Sturges' Rule is merely a suggestion, not an ironclad requirement. Answer: TRUE Explanation: Sturges' Rule is only a starting point or guideline. Difficulty: 1 Easy Topic: 03.02 Frequency Distributions and Histograms Learning Objective: 03-03 Make a histogram with appropriate bins. Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation

13) Excel's 3D pie charts are usually clearer than 2D pie charts. Answer: FALSE Explanation: Many people like 3D better, but it is harder to judge slice size. Difficulty: 1 Easy Topic: 03.06 Pie Charts Learning Objective: 03-07 Make an effective pie chart. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 14) A common error with pie charts is using too few "slices." Answer: FALSE Explanation: The opposite is true (too many slices are often seen). Difficulty: 2 Medium Topic: 03.06 Pie Charts Learning Objective: 03-07 Make an effective pie chart. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 15) A pie chart can generally be used instead of a bar chart. Answer: FALSE Explanation: No, a pie chart is specialized (only for parts of a whole). Difficulty: 2 Medium Topic: 03.06 Pie Charts Learning Objective: 03-07 Make an effective pie chart. Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation 16) A column chart can sometimes be used instead of a line chart for time series data. Answer: TRUE Explanation: Line charts and column charts may be used to display time series data. Difficulty: 2 Medium Topic: 03.05 Column and Bar Charts Learning Objective: 03-06 Make an effective column chart or bar chart. Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation

17) Pie charts are attractive to statisticians, but are rarely used in business or general media. Answer: FALSE Explanation: Pie charts make it hard to judge data values precisely, but often are colorful. Difficulty: 1 Easy Topic: 03.06 Pie Charts Learning Objective: 03-07 Make an effective pie chart. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 18) Pie charts are useful in displaying frequencies that sum to a total. Answer: TRUE Explanation: That is exactly what pie charts are for (e.g., industry market shares). Difficulty: 1 Easy Topic: 03.06 Pie Charts Learning Objective: 03-07 Make an effective pie chart. Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation 19) Dot plots may not reveal the shape of a distribution when the sample is small. Answer: TRUE Explanation: You need a fairly large sample size to assess shape on a dot plot. Difficulty: 1 Easy Topic: 03.01 Stem-and-Leaf Displays and Dot Plots Learning Objective: 03-01 Make a stem-and-leaf or dot plot. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 20) Scatter plots are used to visualize association in samples of paired data (X, Y). Answer: TRUE Explanation: That is exactly what a scatter plot is for. Difficulty: 1 Easy Topic: 03.07 Scatter Plots Learning Objective: 03-08 Make and interpret a scatter plot. Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation

21) The zero origin rule may be waived for column or line charts if the objective is merely to visualize relative change over time. Answer: FALSE Explanation: A nonzero origin is always potentially deceptive if we want to know relative size. Difficulty: 2 Medium Topic: 03.09 Deceptive Graphs Learning Objective: 03-10 Recognize deceptive graphing techniques. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 22) In a bimodal histogram, the two highest bars will have the same height. Answer: FALSE Explanation: Not necessarily, though they might be the same. Difficulty: 2 Medium Topic: 03.02 Frequency Distributions and Histograms Learning Objective: 03-04 Identify skewness, modal classes, and outliers in a histogram. Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation 23) A frequency distribution is a tabulation of n data values into classes called bins. Answer: TRUE Explanation: This is the definition of a frequency distribution. Difficulty: 2 Medium Topic: 03.02 Frequency Distributions and Histograms Learning Objective: 03-02 Create a frequency distribution for a data set. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 24) A dot plot would be useful in visualizing scores on an exam in a class of 30 students. Answer: TRUE Explanation: Because the sample is small and data values are discrete, a dot plot would be good. Difficulty: 2 Medium Topic: 03.01 Stem-and-Leaf Displays and Dot Plots Learning Objective: 03-01 Make a stem-and-leaf or dot plot. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

25) A frequency distribution usually has equal bin widths. Answer: TRUE Explanation: Unequal bins are possible, but rare (software default is equal bins). Difficulty: 1 Easy Topic: 03.02 Frequency Distributions and Histograms Learning Objective: 03-02 Create a frequency distribution for a data set. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 26) Line charts are not used for cross-sectional data. Answer: TRUE Explanation: Line charts are for time series data. Difficulty: 2 Medium Topic: 03.04 Line Charts Learning Objective: 03-05 Make an effective line chart. Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation 27) A scatter plot is useful in visualizing trends over time. Answer: FALSE Explanation: Use a line chart to visualize trends. Difficulty: 1 Easy Topic: 03.07 Scatter Plots Learning Objective: 03-08 Make and interpret a scatter plot. Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation 28) A scatter plot requires two quantitative variables (i.e., not categorical data). Answer: TRUE Explanation: An X-Y plot makes no sense for nonnumerical data. Difficulty: 2 Medium Topic: 03.07 Scatter Plots Learning Objective: 03-08 Make and interpret a scatter plot. Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation

29) The number of bins in this histogram (caffeine content in mg/oz for 65 soft drinks) is consistent with Sturges' Rule.

Answer: FALSE Explanation: Sturges' Rule suggests k = 1 + 3.3 log (65) = 6.98, or about 7 bins. Difficulty: 2 Medium Topic: 03.02 Frequency Distributions and Histograms Learning Objective: 03-03 Make a histogram with appropriate bins. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

30) Because most data values are on the left, we would say that this dot plot (burglary rates per 100,000 persons in 350 U.S. cities) shows a distribution that is skewed to the left (negatively skewed).

Answer: FALSE Explanation: The long right tail suggests positive skewness. Difficulty: 2 Medium Topic: 03.01 Stem-and-Leaf Displays and Dot Plots Learning Objective: 03-04 Identify skewness, modal classes, and outliers in a histogram. Bloom's: Evaluate AACSB: Analytical Thinking Accessibility: Keyboard Navigation 31) It is possible to construct a histogram or frequency polygon with open-ended classes. Answer: FALSE Explanation: Without limits, we cannot mark bin limits on a graph. Difficulty: 2 Medium Topic: 03.02 Frequency Distributions and Histograms Learning Objective: 03-02 Create a frequency distribution for a data set. Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation 32) Except for the Y-axis scaling, a histogram will look the same if we use relative frequencies instead of raw frequencies (with the same bin limits). Answer: TRUE Explanation: Relative frequencies are just raw frequencies divided by the sample size. Difficulty: 2 Medium Topic: 03.02 Frequency Distributions and Histograms Learning Objective: 03-02 Create a frequency distribution for a data set. Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation 9 Copyright 2019 © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.

33) The ________ can be used to differentiate the "vital few" causes of quality problems from the "trivial many" causes of quality problems. A) histogram B) scatter plot C) pareto chart D) box plot Answer: C Explanation: A Pareto chart shows the causes in descending order of frequency. Difficulty: 1 Easy Topic: 03.05 Column and Bar Charts Learning Objective: 03-06 Make an effective column chart or bar chart. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 34) Which is not a characteristic of a dot plot? A) Simplicity B) Legibility C) Wide bins D) Dot stacking Answer: C Explanation: In a simple dot plot, "bins" are really individual data values (not a range). If two dots appear at the same axis location, they must be stacked. Difficulty: 1 Easy Topic: 03.01 Stem-and-Leaf Displays and Dot Plots Learning Objective: 03-01 Make a stem-and-leaf or dot plot. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 35) Which display is most likely to reveal association between X and Y? A) Dot plot B) Scatter plot C) Histogram D) Pareto chart Answer: B Explanation: Scatter plots reveal the degree of covariance between X and Y. Difficulty: 1 Easy Topic: 03.07 Scatter Plots Learning Objective: 03-08 Make and interpret a scatter plot. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 10 Copyright 2019 © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.

36) Which criterion is least likely to be used in choosing bins (classes) in a frequency distribution? A) Following Sturges' Rule B) Selecting "nice" class (bin) limits C) Using aesthetic judgment D) Always starting at zero Answer: D Explanation: Bins can start at any value. Judgment is often used to select "nice" bins. Difficulty: 2 Medium Topic: 03.02 Frequency Distributions and Histograms Learning Objective: 03-02 Create a frequency distribution for a data set. Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation 37) Which of the following is true? A) Line charts are not used for cross-sectional data. B) Line charts are useful for visualizing categorical data. C) Pyramid charts are generally preferred instead of column charts. D) Pie charts can generally be used instead of bar charts. Answer: A Explanation: Line charts require numerical data over time (not categorical data). Difficulty: 2 Medium Topic: 03.04 Line Charts Learning Objective: 03-05 Make an effective line chart. Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation 38) Histograms generally do not reveal the A) exact data range. B) modal classes (bins). C) degree of skewness. D) relative frequencies. Answer: A Explanation: "Nice" bin limits may not end exactly at xmin and xmax. Difficulty: 2 Medium Topic: 03.02 Frequency Distributions and Histograms Learning Objective: 03-04 Identify skewness, modal classes, and outliers in a histogram. Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation 11 Copyright 2019 © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.

39) A column chart would be least suitable to display which data? A) Annual compensation of 500 company CEOs B) U.S. exports to its six largest trading partners C) Exxon-Mobil's quarterly sales for the last four years D) One-year CD interest rates paid by the eight largest U.S. banks Answer: A Explanation: With 500 data values, a column chart would reveal little. Make a histogram instead. Difficulty: 2 Medium Topic: 03.05 Column and Bar Charts Learning Objective: 03-06 Make an effective column chart or bar chart. Bloom's: Evaluate AACSB: Analytical Thinking Accessibility: Keyboard Navigation 40) A line chart would not be suitable to display which data? A) U.S. oil imports from OPEC nations for the last 20 years B) Annual compensation of the top 50 CEOs C) Exxon-Mobil's quarterly sales data for the last five years D) Daily stock market closing prices of Microsoft for the past month Answer: B Explanation: Line charts are for time series data (not cross-sectional data). Difficulty: 2 Medium Topic: 03.04 Line Charts Learning Objective: 03-05 Make an effective line chart. Bloom's: Evaluate AACSB: Analytical Thinking Accessibility: Keyboard Navigation 41) Which is not a tip for effective column charts? A) Time usually goes on the horizontal axis. B) Column height should be proportional to the quantity displayed. C) Label data values at the top of each column unless graphing lots of data. D) The zero origin rule may be waived for financial reports. Answer: D Explanation: Especially in financial reports, the zero origin rule is vital. Difficulty: 2 Medium Topic: 03.05 Column and Bar Charts Learning Objective: 03-06 Make an effective column chart or bar chart. Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation

42) Which is not a tip for effective line charts? A) Line charts are better than bar charts to display cross-sectional data. B) Numerical labels are omitted on a line chart if there are many data values. C) Omit data markers (e.g., squares, triangles) when there are many data values. D) Thick lines make it harder to see exact data values. Answer: A Explanation: Line charts are for numerical time series data. Difficulty: 2 Medium Topic: 03.04 Line Charts Learning Objective: 03-05 Make an effective line chart. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 43) Which is a reason for using a log scale for time series data? A) It helps compare growth in time series of dissimilar magnitude. B) General business audiences find it easier to interpret a log scale. C) On a log scale, equal distances represent equal dollar amounts. D) The axis labels are usually easier to read in log units. Answer: A Explanation: While less familiar to some people, changing magnitude may necessitate a log scale. Difficulty: 2 Medium Topic: 03.03 Effective Excel Charts Learning Objective: 03-05 Make an effective line chart. Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation 44) Which is a not a characteristic of pie charts? A) Pie charts can only convey a general idea of the data values. B) Pie charts are ineffective when they have too many slices. C) Exploded and 3D pie charts will allow more "slices." D) Pie chart data always represent parts of a whole (e.g., market share). Answer: C Explanation: Pie charts with too many slices are hard to read, regardless whether they are 2D or 3D. Difficulty: 1 Easy Topic: 03.06 Pie Charts Learning Objective: 03-07 Make an effective pie chart. Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation 14 Copyright 2019 © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.

45) Excel's pyramid charts A) are generally preferred to pie charts. B) should be avoided despite their visual appeal. C) are generally preferred to line charts. D) show trends more clearly than column charts. Answer: B Explanation: Avoid novelty charts in business presentations. They are fun but unclear. Difficulty: 2 Medium Topic: 03.09 Deceptive Graphs Learning Objective: 03-10 Recognize deceptive graphing techniques. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 46) Which is not a reason why pie charts are popular in business? A) They can convey a general idea of the data to a nontechnical audience. B) They can display major changes in parts of a whole (e.g., market share). C) They are more precise than line charts, despite their low visual impact. D) They can be labeled with data values to facilitate interpretation. Answer: C Explanation: Pie charts are attractive to display parts of a whole, but can be hard to read precisely despite their strong visual impact. Difficulty: 2 Medium Topic: 03.06 Pie Charts Learning Objective: 03-07 Make an effective pie chart. Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation 47) Which data would be suitable for a pie chart? A) Whirlpool Corporation's sales revenue for the last five years B) Oxnard University student category (undergraduate, masters, doctoral) C) Average SAT scores for entering freshmen at 10 major U.S. universities D) U.S. toy imports from China over the past decade Answer: B Explanation: Pie charts are only used to display parts of a whole (not time series data). Difficulty: 2 Medium Topic: 03.06 Pie Charts Learning Objective: 03-07 Make an effective pie chart. Bloom's: Analyze AACSB: Analytical Thinking Accessibility: Keyboard Navigation

48) Which data would be suitable for a pie chart? A) Percent vote in the last election by party (Democrat, Republican, Other) B) Retail prices of six major brands of color laser printers C) Labor cost per vehicle for 10 major world automakers D) Prices paid by 10 students for their accounting textbooks Answer: A Explanation: Pie charts are only used to display parts of a whole. Difficulty: 2 Medium Topic: 03.06 Pie Charts Learning Objective: 03-07 Make an effective pie chart. Bloom's: Analyze AACSB: Analytical Thinking Accessibility: Keyboard Navigation 49) Which data would be suitable for a pie chart? A) Average starting salary of MBA graduates from six ivy-league universities B) APR interest rates charged by the top five U.S. credit cards C) Last semester's average GPA for students in seven majors in a business school D) The number of U.S. primary care clinics by type (urban, suburban, rural) Answer: D Explanation: Pie charts are only used to display parts of a whole. Difficulty: 2 Medium Topic: 03.06 Pie Charts Learning Objective: 03-07 Make an effective pie chart. Bloom's: Analyze AACSB: Analytical Thinking Accessibility: Keyboard Navigation 50) Scatter plots are A) useful in visualizing trends over time. B) useful in identifying causal relationships. C) often fitted with a linear equation in Excel. D) applicable mainly to discrete data. Answer: C Explanation: Scatter plots reveal association (not cause-and-effect). Excel makes it easy to fit a linear equation to a time series on a scatter plot. Difficulty: 2 Medium Topic: 03.07 Scatter Plots Learning Objective: 03-08 Make and interpret a scatter plot. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation

51) Which is not a characteristic of an effective summary table? A) The main point should be clear within 10 seconds. B) Data to be compared should be displayed in rows, not columns. C) Data should be rounded to three or four significant digits. D) Use space instead of lines to separate columns. Answer: B Explanation: Research suggests that side-by-side comparisons are easier. Rounding also helps, as well as not trying to show too much in one table. Difficulty: 2 Medium Topic: 03.08 Tables Learning Objective: 03-08 Make and interpret a scatter plot. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 52) Effective summary tables generally A) have very detailed column headings and footnotes. B) round their data to three or four significant digits. C) use plenty of heavy vertical lines to separate the columns. D) have the raw data listed in a footnote for clarity. Answer: B Explanation: We round because too much accuracy may make it harder to assess magnitudes (e.g., 5.01873 mm and 5.02016 mm both round to 5.02 mm). Modern table style usually omits vertical lines. Difficulty: 2 Medium Topic: 03.08 Tables Learning Objective: 03-09 Make simple tables and pivot tables. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation

53) Pivot tables A) are similar in purpose to simple 2D pie charts. B) show how the data "pivot" around a point. C) show cross-tabulations of data. D) are not really tables at all. Answer: C Explanation: A pivot table shows frequency counts (or sums or averages) in a row-column format. Difficulty: 2 Medium Topic: 03.08 Tables Learning Objective: 03-09 Make simple tables and pivot tables. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 54) Which of the following is least useful in visualizing categorical data? A) Bar chart B) Pie chart C) Line chart D) Pareto chart Answer: C Explanation: Line charts are used to display numerical data over time. Difficulty: 1 Easy Topic: 03.04 Line Charts Learning Objective: 03-05 Make an effective line chart. Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation 55) Which is considered a novelty chart in Excel? A) Pie chart B) Column chart C) Pyramid chart D) Line chart Answer: C Explanation: Pyramid charts utilize the area trick and are hard to read. Difficulty: 1 Easy Topic: 03.09 Deceptive Graphs Learning Objective: 03-10 Recognize deceptive graphing techniques. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation

56) We would use a pivot table to A) cross-tabulate frequencies of occurrence of two variables. B) rotate the data around a central point. C) establish cause-and-effect between X and Y. D) display the data in a 3D scatter plot. Answer: A Explanation: A pivot table shows frequency counts (or sums or averages) in a row-column format. Difficulty: 1 Easy Topic: 03.08 Tables Learning Objective: 03-09 Make simple tables and pivot tables. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 57) Which is not considered a deceptive graphical technique? A) Nonzero origin B) Elastic graph proportions C) Dramatic title D) Axis demarcations Answer: D Explanation: Axis demarcations are helpful on graphs with numerical scales. We try to avoid nonzero origin and leading titles. Difficulty: 1 Easy Topic: 03.09 Deceptive Graphs Learning Objective: 03-10 Recognize deceptive graphing techniques. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 58) Which is not considered a deceptive graphical technique? A) Undefined units B) 2D graphs C) Authority figures D) Distracting pictures Answer: B Explanation: 2D graphs are generally acceptable. We avoid distracting pictures and unclear units of measurement. Difficulty: 1 Easy Topic: 03.09 Deceptive Graphs Learning Objective: 03-10 Recognize deceptive graphing techniques. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 19 Copyright 2019 © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.

59) Which is the most serious deceptive graphical technique? A) Vague source B) Using bold colors C) Nonzero origin D) Unlabeled data points Answer: C Explanation: A nonzero origin can distort relative size. Difficulty: 2 Medium Topic: 03.09 Deceptive Graphs Learning Objective: 03-10 Recognize deceptive graphing techniques. Bloom's: Evaluate AACSB: Analytical Thinking Accessibility: Keyboard Navigation 60) Which is not a poor graphing technique? A) Gratuitous pictures B) Labeled axis scales C) 3D bar charts D) Rotated axis Answer: B Explanation: Labeled axis scales are desirable for numerical data. Rotated graphs are harder to read, and we avoid unnecessary pictures. Difficulty: 1 Easy Topic: 03.09 Deceptive Graphs Learning Objective: 03-10 Recognize deceptive graphing techniques. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 61) Which of these deficiencies would be considered a major graphical deception? A) Vague or unclear source B) Using more than one color or font C) Bar widths proportional to bar height D) Using a dramatic graph title Answer: C Explanation: The area trick occurs when bar width increases along with bar height. Difficulty: 2 Medium Topic: 03.09 Deceptive Graphs Learning Objective: 03-10 Recognize deceptive graphing techniques. Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation 20 Copyright 2019 © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.

62) Which is not a characteristic of a log scale for time series data? A) Log scales are useful when data change by an order of magnitude. B) The distance from 5 to 50 is the same as the distance from 50 to 500. C) On a log scale, equal distances represent equal ratios. D) Log scales are generally familiar to the average reader. Answer: D Explanation: Some readers lack training to understand a simple log scale, such as the fact that equal distances represent equal ratios. Difficulty: 2 Medium Topic: 03.04 Line Charts Learning Objective: 03-05 Make an effective line chart. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 63) Which is not a characteristic of using a log scale to display time series data? A) A log scale helps if we are comparing changes in two time series of dissimilar magnitude. B) General business audiences find it easier to interpret a log scale. C) If you display data on a log scale, equal distances represent equal ratios. Answer: B Explanation: Some business audiences lack training to understand a simple log scale, such as the fact that equal distances represent equal ratios. Difficulty: 2 Medium Topic: 03.04 Line Charts Learning Objective: 03-05 Make an effective line chart. Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation

64) This histogram shows Chris's golf scores in his last 77 rounds at Devil's Ridge. Which is not a correct statement?

A) The number of bins is consistent with Sturges' Rule. B) The histogram has a noticeable bimodal shape. C) The modal class is 78 < 80. D) About 15 percent of his scores are in the interval 74 < 76. Answer: A Explanation: Sturges' Rule suggests k = 1 + 3.3 log (77) = 7.22, or about 7 bins. Difficulty: 2 Medium Topic: 03.02 Frequency Distributions and Histograms Learning Objective: 03-03 Make a histogram with appropriate bins. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 65) Which is not revealed on a scatter plot? A) Pairs of observed (xi, yi) data values B) Nonlinear relationships between X and Y C) Missing data values due to nonresponses D) Unusual data values (outliers) Answer: C Explanation: Excel simply omits missing data for a scatter plot. Difficulty: 2 Medium Topic: 03.07 Scatter Plots Learning Objective: 03-07 Make an effective pie chart. Bloom's: Analyze AACSB: Analytical Thinking Accessibility: Keyboard Navigation 22 Copyright 2019 © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.

66) The distribution pictured below is

A) bimodal and skewed right. B) bimodal and skewed left. C) skewed right. D) skewed left. Answer: A Explanation: You can see two modes and a longer right tail. Difficulty: 2 Medium Topic: 03.02 Frequency Distributions and Histograms Learning Objective: 03-04 Identify skewness, modal classes, and outliers in a histogram. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 67) The distribution pictured below is

A) bimodal and skewed right. B) bimodal and skewed left. C) skewed right. D) skewed left. Answer: D Explanation: You can see one mode and a long left tail. Difficulty: 2 Medium Topic: 03.02 Frequency Distributions and Histograms Learning Objective: 03-04 Identify skewness, modal classes, and outliers in a histogram. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 23 Copyright 2019 © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.

68) The graph below illustrates which deceptive technique?

A) Poor Y axis scale B) Area trick C) Unclear grid lines D) Dramatic title Answer: B Explanation: Area trick, because area increases along with height. Difficulty: 2 Medium Topic: 03.09 Deceptive Graphs Learning Objective: 03-10 Recognize deceptive graphing techniques. Bloom's: Analyze AACSB: Analytical Thinking Accessibility: Keyboard Navigation 69) Which is a characteristic of a histogram's bars? A) The bar widths reveal the cumulative frequencies of data values. B) The bar widths indicate class intervals, and their areas indicate frequencies. C) The bar widths show class intervals, and their heights indicate frequencies. D) The bar widths are an exact multiple of the sample size. Answer: C Explanation: Histogram bar height shows frequencies within each interval. Difficulty: 2 Medium Topic: 03.02 Frequency Distributions and Histograms Learning Objective: 03-03 Make a histogram with appropriate bins. Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation

70) Below is a frequency distribution of earnings of 50 contractors in a country. Earnings (thousands) 1–10 11–20 20–30 31–40 41–50 50–60

Number of Contractors 2 7 12 15 8 6

Regarding this distribution, which of the following is correct? A) The frequency distribution contains too many class intervals. B) The class interval limits are ambiguous. C) Too few classes were chosen. D) The class intervals are too wide. Answer: B Explanation: There are overlapping bin limits. Difficulty: 2 Medium Topic: 03.02 Frequency Distributions and Histograms Learning Objective: 03-02 Create a frequency distribution for a data set. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

71) Bob found an error in the following frequency distribution. What is it? Class 1–10 11–20 25–30 31–40 44–50

Frequency 2 6 8 12 6

A) The class limits are overlapping too much. B) The classes are not collectively exhaustive. C) There are too many classes by Sturges' Rule. D) The first class must start at 0. Answer: B Explanation: Bin limits overlap. Where would you put a data value of 23? Difficulty: 2 Medium Topic: 03.02 Frequency Distributions and Histograms Learning Objective: 03-02 Create a frequency distribution for a data set. Bloom's: Analyze AACSB: Analytical Thinking Accessibility: Keyboard Navigation 72) The point halfway between the bin limits in a frequency distribution is known as the A) bin midpoint. B) bin limit. C) bin frequency. D) bin width. Answer: A Explanation: Bin midpoint is the average of the bin limits. Difficulty: 1 Easy Topic: 03.02 Frequency Distributions and Histograms Learning Objective: 03-02 Create a frequency distribution for a data set. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation

73) When using a dot plot with a small sample, which is least apparent? A) Dispersion of data values within the array B) The overall shape of the distribution C) Central tendency of data in the data set D) Location of data values within the array Answer: B Explanation: Dot plots reveal center and variation, but shape cannot be judged very well in small samples. Difficulty: 2 Medium Topic: 03.01 Stem-and-Leaf Displays and Dot Plots Learning Objective: 03-01 Make a stem-and-leaf or dot plot. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 74) If you have 256 data points, how many classes (bins) would Sturges' Rule suggest? A) 6 B) 7 C) 8 D) 9 Answer: D Explanation: Sturges' Rule suggests k = 1 + 3.3 log (256) = 9 bins. Difficulty: 2 Medium Topic: 03.02 Frequency Distributions and Histograms Learning Objective: 03-02 Create a frequency distribution for a data set. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 75) If you have 32 data points, how many classes (bins) would Sturges' Rule suggest? A) 5 B) 6 C) 7 D) 8 Answer: B Explanation: Sturges' Rule suggests k = 1 + 3.3 log (32) = 6 bins. Difficulty: 2 Medium Topic: 03.02 Frequency Distributions and Histograms Learning Objective: 03-02 Create a frequency distribution for a data set. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 27 Copyright 2019 © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.

76) Which statement is not true concerning Sturges' Rule? A) It proposes adding one class (bin) to the histogram for each extra observation. B) If you double the sample size, you should add one class. C) Its purpose is to tell how many classes (bins) to use in a frequency distribution. D) It is only a guideline and may be overruled by other considerations. Answer: A Explanation: Sturges' Rule suggests about k = 1 + 3.3 log (n) bins. Difficulty: 3 Hard Topic: 03.02 Frequency Distributions and Histograms Learning Objective: 03-02 Create a frequency distribution for a data set. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 77) To classify prices from 62 recent home sales, Sturges' Rule would recommend A) 7 classes. B) 8 classes. C) 9 classes. D) 10 classes. Answer: A Explanation: Sturges' Rule suggests k = 1 + 3.3 log (62) = 7 bins. Difficulty: 2 Medium Topic: 03.02 Frequency Distributions and Histograms Learning Objective: 03-02 Create a frequency distribution for a data set. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 78) A histogram can be defined as A) a chart whose bar widths show the cumulative frequencies of data values. B) a chart whose bar widths indicate class intervals and whose areas indicate frequencies. C) a chart whose bar widths show class intervals and whose heights indicate frequencies. D) a chart whose bar heights represent the value of each data point. Answer: C Explanation: Check the textbook definition of histogram. Difficulty: 1 Easy Topic: 03.02 Frequency Distributions and Histograms Learning Objective: 03-03 Make a histogram with appropriate bins. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation

79) An open-ended bin (e.g., "50 and over") might be seen in a frequency distribution when A) some data values are not integers. B) data values are nonnumerical. C) extremely large data values exist. D) some data are missing. Answer: C Explanation: For example, General Electric's CEO earned $15.2 million in 2010, which would not fit the bins of a histogram of incomes for ordinary taxpayers. Difficulty: 1 Easy Topic: 03.02 Frequency Distributions and Histograms Learning Objective: 03-03 Make a histogram with appropriate bins. Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation 80) The width of a class in a frequency distribution is known as the A) midpoint. B) class limit. C) bin frequency. D) class interval. Answer: D Explanation: Bin width and class interval have the same meaning. Difficulty: 1 Easy Topic: 03.02 Frequency Distributions and Histograms Learning Objective: 03-03 Make a histogram with appropriate bins. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 81) A population is of size 5,500 observations. When the data are represented in a relative frequency distribution, the relative frequency of a given interval is 0.15. The frequency in this interval is equal to A) 4,675 B) 800 C) 675 D) 825 Answer: D Explanation: 15 percent of 5,500 is 825. Difficulty: 1 Easy Topic: 03.02 Frequency Distributions and Histograms Learning Objective: 03-02 Create a frequency distribution for a data set. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 29 Copyright 2019 © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.

82) A population has 75 observations. One class interval has a frequency of 15 observations. The relative frequency in this category is A) 0.20 B) 0.10 C) 0.15 D) 0.75 Answer: A Explanation: 15/75 = 0.20, or 20 percent. Difficulty: 1 Easy Topic: 03.02 Frequency Distributions and Histograms Learning Objective: 03-02 Create a frequency distribution for a data set. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 83) Below is a sorted stem-and-leaf diagram for the measured speeds (miles per hour) of 49 randomly chosen vehicles on highway I-80 in Nebraska. How many vehicles were traveling exactly the speed limit (70 mph)? Stem unit Leaf unit

= =

10 1

Frequency Stem 1 4 1 5 17 6 19 7 7 8 4 9 49

Leaf 9 5 01222455556688999 0111233345666778999 0133557 0122

A) 0 B) 1 C) 19 D) Impossible to tell Answer: B Explanation: The fourth stem has only one value of 70. Difficulty: 1 Easy Topic: 03.01 Stem-and-Leaf Displays and Dot Plots Learning Objective: 03-01 Make a stem-and-leaf or dot plot. Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation 30 Copyright 2019 © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.

84) Below is a sorted stem-and-leaf diagram for the measured speeds (miles per hour) of 49 randomly chosen vehicles on highway I-80 in Nebraska. What is the highest observed speed? Stem unit Leaf unit

= =

10 1

Frequency Stem 1 4 1 5 17 6 19 7 7 8 4 9 49

Leaf 9 5 01222455556688999 0111233345666778999 0133557 0122

A) 92 B) 90 C) 87 D) Impossible to tell Answer: A Explanation: The maximum data value in the top stem is 92. Difficulty: 1 Easy Topic: 03.01 Stem-and-Leaf Displays and Dot Plots Learning Objective: 03-01 Make a stem-and-leaf or dot plot. Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation

85) Below is a sorted stem-and-leaf diagram for the measured speeds (miles per hour) of 49 randomly chosen vehicles on highway I-80 in Nebraska. What is the mode? Stem unit Leaf unit

= =

10 1

Frequency Stem 1 4 1 5 17 6 19 7 7 8 4 9 49

Leaf 9 5 01222455556688999 0111233345666778999 0133557 0122

A) 62 B) 79 C) 65 D) Impossible to tell Answer: C Explanation: The value 65 occurs 4 times. Some other data values occur 3 times. Difficulty: 2 Medium Topic: 03.01 Stem-and-Leaf Displays and Dot Plots Learning Objective: 03-01 Make a stem-and-leaf or dot plot. Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation

86) Below is a sorted stem-and-leaf diagram for the measured speeds (miles per hour) of 49 randomly chosen vehicles on highway I-80 in Nebraska. What is the fourth slowest speed in the sorted data array? Stem unit Leaf unit

= =

10 1

Frequency Stem 1 4 1 5 17 6 19 7 7 8 4 9 49

Leaf 9 5 01222455556688999 0111233345666778999 0133557 0122

A) 61 B) 60 C) 55 D) Impossible to tell Answer: A Explanation: The second data value in the third stem is 61 (this is the fourth item in the sorted data array). Difficulty: 2 Medium Topic: 03.01 Stem-and-Leaf Displays and Dot Plots Learning Objective: 03-01 Make a stem-and-leaf or dot plot. Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation

87) Below is a sorted stem-and-leaf diagram for the measured speeds (miles per hour) of 49 randomly chosen vehicles on highway I-80 in Nebraska. Which is the modal class? Stem unit Leaf unit

= =

10 1

Frequency Stem 1 4 1 5 17 6 19 7 7 8 4 9 49

Leaf 9 5 01222455556688999 0111233345666778999 0133557 0122

A) 60 but less than 70 B) 70 but less than 80 C) 80 but less than 90 D) impossible to determine Answer: B Explanation: All the raw data are available, so we can calculate any statistic. Difficulty: 1 Easy Topic: 03.01 Stem-and-Leaf Displays and Dot Plots Learning Objective: 03-01 Make a stem-and-leaf or dot plot. Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation 88) A statistician prepared a bar chart showing, in descending order, the frequency of six underlying causes of general aviation accidents (pilot error, mechanical problems, disorientation, miscommunication, controller error, other). What would we call this type of chart? A) Pivot table B) Pareto chart C) Log scale chart D) Frequency polygon Answer: B Explanation: A Pareto chart displays the causes in descending order of frequency. Difficulty: 2 Medium Topic: 03.05 Column and Bar Charts Learning Objective: 03-06 Make an effective column chart or bar chart. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 34 Copyright 2019 © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.

Applied Statistics in Business and Economics, 6e (Doane) Chapter 4 Descriptive Statistics 1) A data set with two values that are tied for the highest number of occurrences is called bimodal. Answer: TRUE Explanation: "Bimodal" means two modes. Difficulty: 1 Easy Topic: 04.02 Measures of Center Learning Objective: 04-01 Explain the concepts of center, variability, and shape. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 2) The midrange is not greatly affected by outliers. Answer: FALSE Explanation: Extremes distort the midrange (average of highest and lowest data values). Difficulty: 1 Easy Topic: 04.02 Measures of Center Learning Objective: 04-02 Calculate and interpret common measures of center. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 3) The second quartile is the same as the median. Answer: TRUE Explanation: The second quartile, the median, and the 50th percentile are the same thing. Difficulty: 1 Easy Topic: 04.05 Percentiles, Quartiles, and Box Plots Learning Objective: 04-07 Calculate quartiles and other percentiles. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 4) A trimmed mean may be preferable to a mean when a data set has extreme values. Answer: TRUE Explanation: Trimming diminishes the effect of outliers. Difficulty: 1 Easy Topic: 04.02 Measures of Center Learning Objective: 04-02 Calculate and interpret common measures of center. Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation 1 Copyright 2019 © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.

5) One benefit of the box plot is that it clearly displays the standard deviation. Answer: FALSE Explanation: A box plot shows quartiles. Difficulty: 2 Medium Topic: 04.05 Percentiles, Quartiles, and Box Plots Learning Objective: 04-08 Make and interpret box plots. Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation 6) It is inappropriate to apply the Empirical Rule to a population that is right-skewed. Answer: TRUE Explanation: The Empirical Rule applies to normal populations. Difficulty: 2 Medium Topic: 04.04 Standardized Data Learning Objective: 04-05 Apply the Empirical Rule and recognize outliers. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 7) Given the data set 10, 5, 2, 6, 3, 4, 20, the median value is 5. Answer: TRUE Explanation: Sort and find the middle value. Difficulty: 2 Medium Topic: 04.02 Measures of Center Learning Objective: 04-02 Calculate and interpret common measures of center. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 8) Given the data set 2, 5, 10, 6, 3, the median value is 3. Answer: FALSE Explanation: Sort and find the middle value: 2 3 5 6 10. Difficulty: 2 Medium Topic: 04.02 Measures of Center Learning Objective: 04-02 Calculate and interpret common measures of center. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

9) When data are right-skewed, we expect the median to be greater than the mean. Answer: FALSE Explanation: It is the other way around, as the mean will be pulled up by extremes. Difficulty: 1 Easy Topic: 04.02 Measures of Center Learning Objective: 04-01 Explain the concepts of center, variability, and shape. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 10) The sum of the deviations around the mean is always zero. Answer: TRUE Explanation: The mean is the fulcrum (balancing point), so deviations must sum to zero. Difficulty: 2 Medium Topic: 04.02 Measures of Center Learning Objective: 04-02 Calculate and interpret common measures of center. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 11) The midhinge is a robust measure of center when there are outliers. Answer: TRUE Explanation: Outliers have little effect on the midhinge (average of the 25th and 75th percentiles). Difficulty: 2 Medium Topic: 04.05 Percentiles, Quartiles, and Box Plots Learning Objective: 04-07 Calculate quartiles and other percentiles. Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation 12) Chebyshev's Theorem says that at most 50 percent of the data lie within 2 standard deviations of the mean. Answer: FALSE Explanation: At least 75 percent by Chebyshev. Difficulty: 2 Medium Topic: 04.04 Standardized Data Learning Objective: 04-04 Apply Chebyshev's theorem. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

13) Chebyshev's Theorem says that at least 95 percent of the data lie within 2 standard deviations of the mean. Answer: FALSE Explanation: At least 75 percent by Chebyshev. Difficulty: 2 Medium Topic: 04.04 Standardized Data Learning Objective: 04-04 Apply Chebyshev's theorem. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 14) If there are 19 data values, the median will have 10 values above it and 9 below it since n is odd. Answer: FALSE Explanation: When n is odd, the median is the middle member of the sorted data set. In this case, the median is x10 and there will be 9 below x10 (x1, . . . , x9) and 9 above x10 (x11, . . . , x19). Difficulty: 2 Medium Topic: 04.02 Measures of Center Learning Objective: 04-02 Calculate and interpret common measures of center. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 15) If there are 20 data values, the median will be halfway between two data values. Answer: TRUE Explanation: Median is between two data values when n is even. Difficulty: 2 Medium Topic: 04.02 Measures of Center Learning Objective: 04-02 Calculate and interpret common measures of center. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 16) In a left-skewed distribution, we expect that the median will be greater than the mean. Answer: TRUE Explanation: The mean is likely to be pulled down by low extremes. Difficulty: 1 Easy Topic: 04.02 Measures of Center Learning Objective: 04-01 Explain the concepts of center, variability, and shape. Bloom's: Understand AACSB: Analytical Thinking 4 Copyright 2019 © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.

Accessibility: Keyboard Navigation 17) If the standard deviations of two samples are the same, so are their coefficients of variation. Answer: FALSE Explanation: The means may differ, which affects their coefficients of variation. Difficulty: 2 Medium Topic: 04.03 Measures of Variability Learning Objective: 04-03 Calculate and interpret common measures of variability. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 18) A certain health maintenance organization (HMO) examined the number of office visits by its members in the last year. This data set would probably be skewed to the left due to low outliers. Answer: FALSE Explanation: The lower bound is zero, but high extremes are likely for sicker individuals. Difficulty: 3 Hard Topic: 04.02 Measures of Center Learning Objective: 04-01 Explain the concepts of center, variability, and shape. Bloom's: Evaluate AACSB: Analytical Thinking Accessibility: Keyboard Navigation 19) A certain health maintenance organization examined the number of office visits by its members in the last year. For this data set, the mean is probably not a very good measure of a "typical" person's office visits. Answer: TRUE Explanation: The lower bound is zero, but high extremes are likely for sicker individuals. Difficulty: 3 Hard Topic: 04.02 Measures of Center Learning Objective: 04-02 Calculate and interpret common measures of center. Bloom's: Evaluate AACSB: Analytical Thinking Accessibility: Keyboard Navigation

20) Referring to this box plot of ice cream fat content, the median seems more "typical" of fat content than the midrange as a measure of center.

Answer: TRUE Explanation: The midrange (average of low and high) will be pulled down by the left-tail minimum in this left-skewed distribution. Difficulty: 2 Medium Topic: 04.05 Percentiles, Quartiles, and Box Plots Learning Objective: 04-08 Make and interpret box plots. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 21) Referring to this box plot of ice cream fat content, the mean would exceed the median.

Answer: FALSE Explanation: The data are skewed left. Difficulty: 2 Medium Topic: 04.05 Percentiles, Quartiles, and Box Plots Learning Objective: 04-08 Make and interpret box plots. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

22) Referring to this box plot of ice cream fat content, the skewness would be negative.

Answer: TRUE Explanation: The data are skewed left (negative skewness) as indicated by the long left tail. Difficulty: 2 Medium Topic: 04.05 Percentiles, Quartiles, and Box Plots Learning Objective: 04-08 Make and interpret box plots. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 23) Referring to this graph of ice cream fat content, the second quartile is between 60 and 61.

Answer: TRUE Explanation: Drop a line to lower axis to hit between 60 and 61. Difficulty: 2 Medium Topic: 04.05 Percentiles, Quartiles, and Box Plots Learning Objective: 04-08 Make and interpret box plots. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 24) The range as a measure of variability is very sensitive to extreme data values. Answer: TRUE Explanation: The range depends only on highest and lowest data values, so it is easily distorted. Difficulty: 1 Easy Topic: 04.03 Measures of Variability Learning Objective: 04-03 Calculate and interpret common measures of variability. Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation

25) In calculating the sample variance, the sum of the squared deviations around the mean is divided by n − 1 to avoid underestimating the unknown population variance. Answer: TRUE Explanation: Check the definition. You lose one piece of information because the mean is estimated. Difficulty: 2 Medium Topic: 04.03 Measures of Variability Learning Objective: 04-03 Calculate and interpret common measures of variability. Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation 26) Outliers are data values that fall beyond ±2 standard deviations from the mean. Answer: FALSE Explanation: Outliers are 3 standard deviations from the mean Difficulty: 1 Easy Topic: 04.04 Standardized Data Learning Objective: 04-05 Apply the Empirical Rule and recognize outliers. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 27) The Empirical Rule assumes that the distribution of data follows a normal curve. Answer: TRUE Explanation: Unlike Chebyshev, the Empirical Rule assumes a normal population. Difficulty: 1 Easy Topic: 04.04 Standardized Data Learning Objective: 04-05 Apply the Empirical Rule and recognize outliers. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 28) The Empirical Rule can be applied to any distribution, unlike Chebyshev's theorem. Answer: FALSE Explanation: The Empirical Rule assumes a normal population, while Chebyshev applies to any population. Difficulty: 2 Medium Topic: 04.04 Standardized Data Learning Objective: 04-05 Apply the Empirical Rule and recognize outliers. Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation 8 Copyright 2019 © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.

29) When applying the Empirical Rule to a distribution of grades, if a student scored one standard deviation below the mean, then she would be at the 25th percentile of the distribution. Answer: FALSE Explanation: About 15.87 percent (not 25 percent) are less than one standard deviation below the mean (in a normal distribution). Difficulty: 2 Medium Topic: 04.04 Standardized Data Learning Objective: 04-05 Apply the Empirical Rule and recognize outliers. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 30) Kurtosis cannot be judged accurately by looking at a histogram. Answer: TRUE Explanation: Histograms are affected by scaling, so peakedness is hard to judge. Difficulty: 2 Medium Topic: 04.08 Skewness and Kurtosis Learning Objective: 04-11 Assess skewness and kurtosis in a sample. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 31) A platykurtic distribution is more sharply peaked (i.e., thinner tails) than a normal distribution. Answer: FALSE Explanation: A platykurtic distribution is flatter than a normal distribution (thicker tails). Difficulty: 2 Medium Topic: 04.08 Skewness and Kurtosis Learning Objective: 04-11 Assess skewness and kurtosis in a sample. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 32) A leptokurtic distribution is more sharply peaked (i.e., thinner tails) than a normal distribution. Answer: TRUE Explanation: A leptokurtic distribution is more sharply peaked and has thinner tails. Difficulty: 2 Medium Topic: 04.08 Skewness and Kurtosis Learning Objective: 04-11 Assess skewness and kurtosis in a sample. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 9 Copyright 2019 © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.

33) A positive kurtosis coefficient in Excel indicates a leptokurtic condition in a distribution. Answer: TRUE Explanation: The sign of Excel's kurtosis coefficient indicates the kurtosis direction relative to a normal distribution. Difficulty: 2 Medium Topic: 04.08 Skewness and Kurtosis Learning Objective: 04-11 Assess skewness and kurtosis in a sample. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 34) A sample consists of the following data: 7, 11, 12, 18, 20, 22, 43. Using the "three standard deviation" criterion, the last observation (X = 43) would be considered an outlier. Answer: FALSE Explanation: The observation 43 is not more than three standard deviations above the mean for this data set. The sample mean is 19.00 and the sample standard deviation is 11.86. Difficulty: 3 Hard Topic: 04.04 Standardized Data Learning Objective: 04-06 Transform a data set into standardized values. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 35) The coefficient of variation is A) measured on a scale from 0 to 100. B) a unit-free statistic. C) helpful when the sample means are zero. D) a measure of correlation for two variables. Answer: B Explanation: The coefficient of variation is unit free. It is the standard deviation as a percentage of the mean. But it cannot be used when the mean is zero because the mean is in the denominator. Difficulty: 1 Easy Topic: 04.03 Measures of Variability Learning Objective: 04-03 Calculate and interpret common measures of variability. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation

36) Which is not an advantage of the method of medians to find Q1 and Q3? A) Ease of interpolating quartile positions B) Ease of application in small data sets C) Intuitive definitions without complex formulas D) Same method as Excel's =QUARTILE.EXC function. Answer: D Explanation: When the quartiles lie between two data values, the method of medians goes halfway between the values (very simple), while Excel interpolates between them in a more complex way. Difficulty: 2 Medium Topic: 04.05 Percentiles, Quartiles, and Box Plots Learning Objective: 04-07 Calculate quartiles and other percentiles. Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation 37) Which is a characteristic of the mean as a measure of center? A) Deviations do not sum to zero when there are extreme values. B) It is less reliable than the mode when the data are continuous. C) It utilizes all the information in a sample. D) It is usually equal to the median in business data. Answer: C Explanation: The mean utilizes all n data values. Deviations always sum to zero around the mean. The mean works for continuous data (unlike the mode). The mean often differs from the median in business data. Difficulty: 2 Medium Topic: 04.02 Measures of Center Learning Objective: 04-02 Calculate and interpret common measures of center. Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation

38) The position of the median is A) n/2 in any sample. B) n/2 if n is even. C) n/2 if n is odd. D) (n+1)/2 in any sample. Answer: D Explanation: This formula always works for the median position. For example, if n = 10 (even) the median is at position (10+1)/2 = 5.5, or halfway between x5 and x6. But if n = 11 (odd) the median is at position (11+1)/2 = 6, which is observation x6. Difficulty: 3 Hard Topic: 04.02 Measures of Center Learning Objective: 04-02 Calculate and interpret common measures of center. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 39) Which is a characteristic of the trimmed mean as a measure of center? A) It is similar to the mean if there are offsetting high and low extremes. B) It is especially helpful in a small sample. C) It does not require sorting the sample. D) It is basically the same as the midrange. Answer: A Explanation: After sorting, we can trim unusual values to stabilize the mean. The trimmed mean may be similar to the mean if the extremes on either end offset each other. Unlike the trimmed mean, the midrange is affected by outliers. Difficulty: 2 Medium Topic: 04.02 Measures of Center Learning Objective: 04-02 Calculate and interpret common measures of center. Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation

40) Which is not a characteristic of the geometric mean as a measure of center? A) It is similar to the mean if the data are skewed right. B) It mitigates the effects of large data values. C) It is useful in business data to calculate average growth rates. D) It cannot be calculated when the data contain negative or zero values. Answer: A Explanation: Although both the mean and the geometric mean are affected by high extremes in skewed data, the geometric mean tends to reduce their influence. The geometric mean cannot be used when any data values are zero or negative. Difficulty: 2 Medium Topic: 04.02 Measures of Center Learning Objective: 04-02 Calculate and interpret common measures of center. Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation 41) Which is not a characteristic of the standard deviation? A) It is always the square root of the variance. B) It is not applicable when data are continuous. C) It can be calculated when the data contain negative or zero values. D) Its physical interpretation is not as easy as the MAD. Answer: B Explanation: The standard deviation applies to any data measured on a ratio or interval scale. Because it is a square root, its visual interpretation may be less clear than the MAD. Difficulty: 2 Medium Topic: 04.03 Measures of Variability Learning Objective: 04-03 Calculate and interpret common measures of variability. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation

42) Chebyshev's Theorem A) applies to all samples. B) applies only to samples from a normal population. C) gives a narrower range of predictions than the Empirical Rule. D) is based on Sturges' Rule for data classification. Answer: A Explanation: The strength of Chebyshev's Theorem is that it makes no assumption about normality, while the Empirical Rule only works for normal populations. Difficulty: 2 Medium Topic: 04.04 Standardized Data Learning Objective: 04-04 Apply Chebyshev's theorem. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 43) Which of the following is not a valid description of an outlier? A) A data value beyond the outer fences B) A data value that is very unusual C) A data value that lies below Q1 or above Q3 D) A data value several standard deviations from the mean. Answer: C Explanation: Data values outside the quartiles (top or bottom 25 percent) are not very unusual. Difficulty: 2 Medium Topic: 04.04 Standardized Data Learning Objective: 04-05 Apply the Empirical Rule and recognize outliers. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 44) If samples are from a normal distribution with μ = 100 and σ = 10, we expect A) about 68 percent of the data within 90 to 110. B) almost all the data within 80 to 120. C) about 95 percent of the data within 70 to 130. D) about half the data to exceed 75. Answer: A Explanation: Review the Empirical Rule. For example, the interval 90 to 110 is the μ ± 1σ range. Difficulty: 2 Medium Topic: 04.04 Standardized Data Learning Objective: 04-05 Apply the Empirical Rule and recognize outliers. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 15 Copyright 2019 © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.

45) In a sample of 10,000 observations from a normal population, how many would you expect to lie beyond three standard deviations of the mean? A) None of them B) About 27 C) About 100 D) About 127 Answer: B Explanation: Within μ ± 3σ we would expect 99.73 percent of 10,000, or 9,973 data values. Difficulty: 2 Medium Topic: 04.04 Standardized Data Learning Objective: 04-05 Apply the Empirical Rule and recognize outliers. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 46) Which is the Excel formula for the standard deviation of a sample array named Data? A) =STDEV.S(Data) B) =STANDEV(Data) C) =STDEV.P(Data) D) =SUM(Data)/(COUNT(Data)-1) Answer: A Explanation: STDEV.S(Data) denotes a sample standard deviation. Difficulty: 2 Medium Topic: 04.03 Measures of Variability Learning Objective: 04-03 Calculate and interpret common measures of variability. Bloom's: Remember AACSB: Technology Accessibility: Keyboard Navigation 47) Which is not true of an outlier? A) It is likely to be from a different population. B) It suggests an error in recording the data. C) It is best discarded to get a better mean. D) It is an anomaly that may tell the researcher something. Answer: C Explanation: We are reluctant to delete outliers, as they may tell us something important. Difficulty: 1 Easy Topic: 04.04 Standardized Data Learning Objective: 04-05 Apply the Empirical Rule and recognize outliers. Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation 16 Copyright 2019 © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.

48) Estimating the mean from grouped data will tend to be most accurate when A) observations are distributed uniformly within classes. B) there are few classes with wide class limits. C) the sample is not very large and bins are wide. D) the standard deviation is large relative to the mean. Answer: A Explanation: Many bins and uniform data distribution within bins would give a result closest to the ungrouped mean μ. Difficulty: 1 Easy Topic: 04.07 Grouped Data Learning Objective: 04-10 Calculate the mean and standard deviation from grouped data. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 49) Which is true of the kurtosis of a distribution? A) A distribution that is flatter than a normal distribution (i.e., thicker tails) is mesokurtic. B) A distribution that is more peaked than a normal distribution (i.e., thinner tails) is platykurtic. C) It is risky to assess kurtosis if the sample size is less than 50. D) The expected range of the kurtosis coefficient increases as n increases. Answer: C Explanation: Shape is hard to judge in small samples. Excel computes kurtosis for samples of any size, but tables of critical values may not go down below n = 50. Difficulty: 3 Hard Topic: 04.08 Skewness and Kurtosis Learning Objective: 04-11 Assess skewness and kurtosis in a sample. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 50) Which is true of skewness? A) In business data, positive skewness is unusual. B) In a negatively skewed distribution, the mean is likely to exceed the median. C) Skewness often is evidenced by one or more outliers. D) The expected range of Excel's skewness coefficient increases as n increases. Answer: C Explanation: Skewness due to extreme data values is common in business data. Right skewness is common, which increases the mean relative to the median. Difficulty: 3 Hard Topic: 04.08 Skewness and Kurtosis Learning Objective: 04-11 Assess skewness and kurtosis in a sample. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 17 Copyright 2019 © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.

51) Which is not true of the Empirical Rule? A) It applies to any distribution. B) It can be applied to fewer distributions than Chebyshev's Theorem. C) It assumes that the distribution of data follows a bell-shaped, normal curve. D) It predicts more observations within μ ± kσ than Chebyshev's Theorem. Answer: A Explanation: The Empirical Rule applies only to normal populations, while Chebyshev's Theorem is general. Difficulty: 2 Medium Topic: 04.04 Standardized Data Learning Objective: 04-05 Apply the Empirical Rule and recognize outliers. Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation 52) Which is a correct statement concerning the median? A) In a left-skewed distribution, we expect that the median will exceed the mean. B) The sum of the deviations around the median is zero. C) The median is an observed data value in any data set. D) The median is halfway between Q1 and Q3 on a box plot. Answer: A Explanation: The mean is pulled down in left-skewed data, but deviations around it sum to zero in any data set. The median may be between two data values and may not be in the middle of the box plot. Difficulty: 2 Medium Topic: 04.02 Measures of Center Learning Objective: 04-02 Calculate and interpret common measures of center. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

53) Which statement is true? A) With nominal data, we can find the mode. B) Outliers distort the mean, but not the standard deviation. C) Business and economic data are rarely skewed to the right. D) If we sample a normal population, the sample skewness coefficient is exactly 0. Answer: A Explanation: The mode (most frequent data value) works for nominal data. Outliers affect both the mean and the standard deviation. Skewness will be near zero in samples from a normal population, but not exactly due to sample variation. Difficulty: 2 Medium Topic: 04.02 Measures of Center Learning Objective: 04-02 Calculate and interpret common measures of center. Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation 54) Exam scores in a small class were 10, 10, 20, 20, 40, 60, 80, 80, 90, 100, 100. For this data set, which statement is incorrect concerning measures of center? A) The median is 60.00. B) The mode is not helpful. C) The 5 percent trimmed mean would be awkward. D) The geometric mean is 35.05. Answer: D Explanation: To find the geometric mean, multiply the data values and take the 11th root to get G = 41.02. Outliers affect both the mean and the standard deviation. There are multiple modes in this example. Difficulty: 3 Hard Topic: 04.02 Measures of Center Learning Objective: 04-02 Calculate and interpret common measures of center. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

55) Exam scores in a small class were 0, 50, 50, 70, 70, 80, 90, 90, 100, 100. For this data set, which statement is incorrect concerning measures of center? A) The median is 70. B) The mode is not helpful. C) The geometric mean is useless. D) The mean is 70. Answer: A Explanation: The median is 75 (halfway between x5 = 70 and x6 = 80 in the sorted array). The zeros render the geometric mean useless. The modes in this case are not unique. Difficulty: 3 Hard Topic: 04.02 Measures of Center Learning Objective: 04-02 Calculate and interpret common measures of center. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 56) Exam scores in a random sample of students were 0, 50, 50, 70, 70, 80, 90, 90, 90, 100. Which statement is incorrect? A) The standard deviation is 29.61. B) The data are slightly left-skewed. C) The midrange and mean are almost the same. D) The third quartile is 90. Answer: C Explanation: The midrange is (0 + 100)/2 = 50, while the mean is 69. Q3 falls between 90 and 90. Difficulty: 2 Medium Topic: 04.02 Measures of Center Learning Objective: 04-02 Calculate and interpret common measures of center. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

57) For U.S. adult males, the mean height is 178 cm with a standard deviation of 8 cm and the mean weight is 84 kg with a standard deviation of 8 kg. Elmer is 170 cm tall and weighs 70 kg. It is most nearly correct to say that A) Elmer's weight is more unusual than his height. B) Elmer is heavier than he is tall. C) Height and weight have the same degree of variation. D) Height has more variation than weight. Answer: A Explanation: Convert Elmer's height and weight to z-scores. For Elmer's weight, z = (x − μ)/σ = (70 − 84)/8 = −1.75, while for Elmer's height, z = (x − μ)/σ = (170 − 178)/8 = −1.00. Therefore, Elmer is farther from the mean weight than from the mean height. Difficulty: 3 Hard Topic: 04.04 Standardized Data Learning Objective: 04-06 Transform a data set into standardized values. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 58) John scored 85 on Prof. Hardtack's exam (Q1 = 40 and Q3 = 60). Based on the fences, which is correct? A) John is an extreme outlier. B) John is an outlier. C) John is not an outlier. D) John is in the 85th percentile. Answer: C Explanation: The upper inner fence is 60 + 1.5(60 − 40) = 90 so John's score does not quite qualify as an outlier. He is in the top quartile, but without more information we do not know his percentile. Difficulty: 2 Medium Topic: 04.05 Percentiles, Quartiles, and Box Plots Learning Objective: 04-07 Calculate quartiles and other percentiles. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

59) John scored 35 on Prof. Johnson's exam (Q1 = 70 and Q3 = 80). Based on the fences, which is correct? A) John is unusual but not an outlier. B) John is an outlier. C) John is neither unusual nor an outlier. D) John is in the 30th percentile. Answer: B Explanation: The lower inner fence is 70 − 1.5(80 − 70) = 55 so John is an outlier. Actually, John is an extreme outlier because the lower outer fence is 70 − 3.0(80 − 70) = 40. Difficulty: 2 Medium Topic: 04.05 Percentiles, Quartiles, and Box Plots Learning Objective: 04-07 Calculate quartiles and other percentiles. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 60) A population consists of the following data: 7, 11, 12, 18, 20, 22, 25. The population variance is A) 6.07 B) 36.82 C) 5.16 D) 22.86 Answer: B Explanation: Use the population formula or Excel's =VAR.P(Data). Difficulty: 2 Medium Topic: 04.03 Measures of Variability Learning Objective: 04-03 Calculate and interpret common measures of variability. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

61) Consider the following data: 6, 7, 17, 51, 3, 17, 23, and 69. The range and the median are A) 69 and 17.5. B) 66 and 17.5. C) 66 and 17. D) 69 and 17. Answer: C Explanation: The range = xmax − xmin = 69 − 3 = 66. The median is halfway between 17 and 17. Difficulty: 2 Medium Topic: 04.02 Measures of Center Learning Objective: 04-02 Calculate and interpret common measures of center. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 62) When a sample has an odd number of observations, the median is the A) observation in the center of the data array. B) average of the two observations in the center of the data array. C) value of the most frequent observation. D) average of Q1 and Q3. Answer: A Explanation: Median position is always (n+1)/2. It need not be halfway between the quartiles. Difficulty: 1 Easy Topic: 04.02 Measures of Center Learning Objective: 04-02 Calculate and interpret common measures of center. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 63) As a measure of variability, compared to the range, an advantage of the standard deviation is that it A) is calculated easily through the use of a formula. B) considers only the data values in the middle of the data array. C) describes the distance between the highest and lowest values. D) considers all data values. Answer: D Explanation: The range is easy to calculate but utilizes only two data values, which may be unusual. Difficulty: 2 Medium Topic: 04.03 Measures of Variability Learning Objective: 04-03 Calculate and interpret common measures of variability. Bloom's: Remember AACSB: Analytical Thinking 23 Copyright 2019 © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.

Accessibility: Keyboard Navigation 64) Which two statistics offer robust measures of center when outliers are present? A) Mean and mode. B) Median and trimmed mean. C) Midrange and geometric mean. D) Variance and standard deviation. Answer: B Explanation: Extremes are excluded from the trimmed mean and do not affect the median. Difficulty: 1 Easy Topic: 04.02 Measures of Center Learning Objective: 04-02 Calculate and interpret common measures of center. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 65) Which Excel function is designed to calculate z = (x − μ)/σ for a column of data? A) =STANDARDIZE B) =NORM.DIST C) =STDEV.P D) =AVEDEV Answer: A Explanation: You need the sample mean and sample standard deviation to find the z-score. Difficulty: 1 Easy Topic: 04.04 Standardized Data Learning Objective: 04-06 Transform a data set into standardized values. Bloom's: Remember AACSB: Technology Accessibility: Keyboard Navigation 66) Which Excel function would be least useful to calculate the quartiles for a column of data? A) =STANDARDIZE B) =PERCENTILE.EXC C) =QUARTILE.EXC D) =RANK Answer: A Explanation: Check Appendix J for function definitions. Difficulty: 2 Medium Topic: 04.05 Percentiles, Quartiles, and Box Plots Learning Objective: 04-07 Calculate quartiles and other percentiles. Bloom's: Remember AACSB: Technology Accessibility: Keyboard Navigation 24 Copyright 2019 © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.

67) Shown the spending by a sample of 50 breakfast customers of McDonald's. Which statement is least likely to be correct?

A) The median is very close to the midhinge. B) The median purchase is slightly less than $5. C) About 75 percent of the customers spend less than $7. D) The mean is a reasonable measure of center. Answer: D Explanation: Outliers and right-skewness would affect the mean. Difficulty: 3 Hard Topic: 04.05 Percentiles, Quartiles, and Box Plots Learning Objective: 04-08 Make and interpret box plots. Bloom's: Evaluate AACSB: Analytical Thinking Accessibility: Keyboard Navigation

68) VenalCo Market Research surveyed 50 individuals who recently purchased a certain CD, revealing the age distribution shown below. Which statement is least defensible?

A) The mean age probably exceeds the median age. B) The mode would be a reasonable measure of center. C) The data are somewhat skewed to the left. D) The CD is unlikely to appeal to retirees. Answer: A Explanation: The mean would be lower than the median due to left-skewness. Difficulty: 3 Hard Topic: 04.02 Measures of Center Learning Objective: 04-02 Calculate and interpret common measures of center. Bloom's: Evaluate AACSB: Analytical Thinking Accessibility: Keyboard Navigation 69) Given a sample of three items (X = 4, 6, 5), which statement is incorrect? A) The geometric mean is 5.2. B) The standard deviation is 1. C) The coefficient of variation is 20 percent. D) The quartiles are useless. Answer: A Explanation: Multiply and take the 3rd root to get the geometric mean of 4.932. With only three data values, the quartiles cannot be calculated (we cannot divide three items into four groups). Difficulty: 1 Easy Topic: 04.02 Measures of Center Learning Objective: 04-02 Calculate and interpret common measures of center. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 27 Copyright 2019 © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.

70) A sample of customers from Barnsboro National Bank shows an average account balance of $315 with a standard deviation of $87. A sample of customers from Wellington Savings and Loan shows an average account balance of $8350 with a standard deviation of $1800. Which statement about account balances is correct? A) Barnsboro Bank has more variation. B) Wellington S&L has more variation. C) Both have the same variation. Answer: A Explanation: Calculate the coefficient of variation for each bank. For Barnsboro, CV = 100 × s/ = 100 × 87/315 = 27.62, while for Wellington CV = 100 × s/ = 100 × 1800/8350 = 21.56. Difficulty: 1 Easy Topic: 04.03 Measures of Variability Learning Objective: 04-03 Calculate and interpret common measures of variability. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 71) Histograms are best used to A) provide a visual estimate of the standard deviation. B) show the quartiles of the data set. C) assess the shape of the distribution. D) reveal the interquartile range of the data set. Answer: C Explanation: A histogram does not show quartiles or standard deviation. Difficulty: 1 Easy Topic: 04.02 Measures of Center Learning Objective: 04-01 Explain the concepts of center, variability, and shape. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation

72) The ________ shows the relationship between two variables. A) box plot B) bar chart C) histogram D) scatter plot Answer: D Explanation: The X-Y scatter plot measures association in a visual way. Difficulty: 1 Easy Topic: 04.06 Covariance and Correlation Learning Objective: 04-09 Calculate and interpret a correlation coefficient and covariance. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 73) If the mean and median of a population are the same, then its distribution is A) normal. B) skewed. C) symmetric. D) uniform. Answer: C Explanation: A symmetric population need not be normal or uniform. Difficulty: 2 Medium Topic: 04.02 Measures of Center Learning Objective: 04-02 Calculate and interpret common measures of center. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 74) In the following data set {7, 5, 0, 2, 7, 15, 5, 2, 7, 18, 7, 3, 0}, the value 7 is A) the mean. B) the mode. C) both the mode and median. D) both the mean and mode. Answer: B Explanation: The value 7 occurs four times. The median is 5. The mean is 6. Difficulty: 1 Easy Topic: 04.02 Measures of Center Learning Objective: 04-02 Calculate and interpret common measures of center. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

75) The median of 600, 800, 1000, 1200 is A) 800 B) 1000 C) 900 D) 950 Answer: C Explanation: Median is halfway between 800 and 1000. Difficulty: 1 Easy Topic: 04.02 Measures of Center Learning Objective: 04-02 Calculate and interpret common measures of center. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 76) The 25th percentile for waiting time in a doctor's office is 19 minutes. The 75th percentile is 31 minutes. The interquartile range (IQR) is A) 12 minutes. B) 16 minutes. C) 22 minutes. D) impossible to determine without knowing n. Answer: A Explanation: The IQR is 31 − 19 = 12. Difficulty: 2 Medium Topic: 04.05 Percentiles, Quartiles, and Box Plots Learning Objective: 04-07 Calculate quartiles and other percentiles. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 77) The 25th percentile for waiting time in a doctor's office is 19 minutes. The 75th percentile is 31 minutes. Which is incorrect regarding the fences? A) The upper inner fence is 49 minutes. B) The upper outer fence is 67 minutes. C) A waiting time of 45 minutes exceeds the upper inner fence. D) A waiting time of 70 minutes would be an outlier. Answer: C Explanation: Apply definitions of fences. For example, the upper inner fence is 31 + 1.5(31 − 19) = 49. Difficulty: 3 Hard Topic: 04.05 Percentiles, Quartiles, and Box Plots Learning Objective: 04-07 Calculate quartiles and other percentiles. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 30 Copyright 2019 © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.

78) When using Chebyshev's Theorem, the minimum percentage of sample observations that will fall within two standard deviations of the mean will be ________ the percentage within two standard deviations if a normal distribution is assumed (Empirical Rule). A) smaller than B) greater than C) the same as Answer: A Explanation: Chebyshev guarantees fewer observations within two standard deviations than the Empirical Rule. Difficulty: 2 Medium Topic: 04.04 Standardized Data Learning Objective: 04-04 Apply Chebyshev's theorem. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 79) Which distribution is least likely to be skewed to the right by high values? A) Annual incomes of n passengers on a flight from New York to London B) Weekend gambling losses of n customers at a major casino C) Accident damage losses by n renters of an auto rental company D) Cost of a plain McDonald's hamburger in n U.S. cities Answer: D Explanation: A few high values would skew the data badly in all but the hamburger example, because a McDonald's hamburger is a standard menu item. Difficulty: 3 Hard Topic: 04.02 Measures of Center Learning Objective: 04-02 Calculate and interpret common measures of center. Bloom's: Analyze AACSB: Analytical Thinking Accessibility: Keyboard Navigation

80) Based on daily measurements, Bob's weight has a mean of 200 pounds with a standard deviation of 16 pounds, while Mary's weight has a mean of 125 pounds with a standard deviation of 15 pounds. Who has the smaller relative variation? A) Bob B) Mary C) They are the same. Answer: A Explanation: Calculate the coefficients of variation for Bob and Mary. Bob's CV = 100 × s/ = 100 × 16/200 = 8.00, while Mary's CV = 100 × s/ = 100 × 15/125 = 12.00. Therefore, Bob's weight varies less than Mary's weight in relative terms. Difficulty: 2 Medium Topic: 04.03 Measures of Variability Learning Objective: 04-03 Calculate and interpret common measures of variability. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 81) Frieda is 67 inches tall and weighs 135 pounds. Women her age have a mean height of 65 inches with a standard deviation of 2.5 inches and a mean weight of 125 pounds with a standard deviation of 10 pounds. In relative terms, it is correct to say that A) Frieda is taller and thinner than women in her age group. B) for this group of women, weight has greater variation than height. C) Frieda's height is more unusual than her weight. D) the variation coefficient exceeds 10 percent for both height and weight. Answer: C Explanation: Calculate the z-scores for Frieda's weight and Frieda's height. For Frieda's height, z = (x − μ)/σ = (67 − 65)/(2.5) = 0.80, while for Frieda's weight, z = (x − μ)/σ = (135 − 125)/10 = 1.00. Therefore, Frieda's weight is farther from the mean than her height. For heights, the CV = 100 × σ/μ = 100 × (2.5)/(65) = 3.8%, while for weights, CV = 100 × σ/μ = 100 × 10/125 = 8.0% (both CVs are below 10%). Difficulty: 3 Hard Topic: 04.03 Measures of Variability Learning Objective: 04-03 Calculate and interpret common measures of variability. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

82) Which statement is false? A) The coefficient of variation cannot be used when the mean is zero. B) The standard deviation is in the same units as the mean (e.g., kilograms). C) The mean from a frequency tabulation may differ from the mean from raw data. D) The skewness coefficient is zero in a sample from any normal distribution. Answer: D Explanation: We cannot calculate a coefficient of variation when the mean is zero. The mean and standard deviation have the same units (e.g., pounds). Information is lost when data are tabulated so the group mean may differ from the raw mean. Normal populations are symmetric, but a sample may differ from the population. Difficulty: 2 Medium Topic: 04.08 Skewness and Kurtosis Learning Objective: 04-11 Assess skewness and kurtosis in a sample. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 83) The values of xmin and xmax can be inferred accurately except in a A) box plot. B) dot plot. C) histogram. D) scatter plot. Answer: C Explanation: The bin limits in a histogram may be rounded, so the values of xmin and xmax may be unclear. Difficulty: 2 Medium Topic: 04.02 Measures of Center Learning Objective: 04-01 Explain the concepts of center, variability, and shape. Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation

84) Which of the following statements is likely to be true? A) The median personal income of California taxpayers would probably be near the mean. B) The interquartile range offers a measure of income inequality among California residents. C) For income, the sum of squared deviations about the mean is negative about half the time. D) For personal incomes in California, outliers in either tail would be equally likely. Answer: B Explanation: Incomes are likely to be skewed due to high extremes, while income is bounded on the low end by zero. A wider interquartile range (IQR) would suggest greater inequality of incomes. Difficulty: 3 Hard Topic: 04.05 Percentiles, Quartiles, and Box Plots Learning Objective: 04-07 Calculate quartiles and other percentiles. Bloom's: Analyze AACSB: Analytical Thinking Accessibility: Keyboard Navigation 85) Which statistics offer robust (resistant to outliers) measures of center? A) Mean, midrange, mode. B) Median, midhinge, trimmed mean. C) Trimmed mean, midrange, midhinge. D) Mean, mode, quartiles. Answer: B Explanation: Any measure of center using the mean is subject to the influence of outliers. Difficulty: 2 Medium Topic: 04.02 Measures of Center Learning Objective: 04-02 Calculate and interpret common measures of center. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 86) The Empirical Rule says that A) most business data sets are normally distributed. B) outliers are within three standard deviations of the mean. C) in most business data we expect the mean and median to be similar. D) about 32 percent of the data are beyond one standard deviation from the mean. Answer: D Explanation: The Empirical Rule says that about 68 percent of the observations are within one standard deviation of the mean. Business data often are skewed. Difficulty: 2 Medium Topic: 04.04 Standardized Data Learning Objective: 04-05 Apply the Empirical Rule and recognize outliers. Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation 34 Copyright 2019 © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.

87) Three randomly chosen Seattle students were asked how many round trips they made to Canada last year. Their replies were 3, 4, 5. The geometric mean is A) 3.877 B) 4.000 C) 3.915 D) 4.422 Answer: C Explanation: Multiply the three numbers and take the 3rd root of 60 to get 3.915. Difficulty: 2 Medium Topic: 04.02 Measures of Center Learning Objective: 04-02 Calculate and interpret common measures of center. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 88) Three randomly chosen California students were asked how many times they drove to Mexico last year. Their replies were 4, 5, 6. The geometric mean is A) 3.87 B) 5.00 C) 5.42 D) 4.93 Answer: D Explanation: Multiply the three numbers and take the 3rd root of 120 to get 4.932. Difficulty: 2 Medium Topic: 04.02 Measures of Center Learning Objective: 04-02 Calculate and interpret common measures of center. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

89) Three randomly chosen Colorado students were asked how many times they went rock climbing last month. Their replies were 5, 6, 7. The standard deviation is A) 1.212 B) 0.816 C) 1.000 D) 1.056 Answer: C Explanation: Use the sample formula or Excel's =STDEV.S(Data). Difficulty: 2 Medium Topic: 04.03 Measures of Variability Learning Objective: 04-03 Calculate and interpret common measures of variability. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 90) Patient survival times after a certain type of surgery have a very right-skewed distribution due to a few high outliers. Consequently, which statement is most likely to be correct? A) Median > Midrange B) Mean < Median C) Mean > Midrange D) Mean > Trimmed Mean Answer: D Explanation: Mean is pulled up by high outliers. Difficulty: 2 Medium Topic: 04.02 Measures of Center Learning Objective: 04-02 Calculate and interpret common measures of center. Bloom's: Evaluate AACSB: Analytical Thinking Accessibility: Keyboard Navigation

91) So far this year, stock A has had a mean price of $6.58 per share with a standard deviation of $1.88, while stock B has had a mean price of $10.57 per share with a standard deviation of $3.02. Which stock is more volatile? A) Stock A B) Stock B C) They are the same. Answer: C Explanation: They are the same: CVA = 100(1.88/6.58) = 28.57%. CVB = 100(3.02/10.57) = 28.57%. Difficulty: 2 Medium Topic: 04.03 Measures of Variability Learning Objective: 04-03 Calculate and interpret common measures of variability. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 92) Outliers are indicated using fences on a A) box plot. B) dot plot. C) histogram. D) Pareto chart. Answer: A Explanation: On a boxplot, outliers are identified by their distance from the median. Data values outside the inner fences (median ± 1.5 IQR) are outliers. Data values beyond the outer fences (median ± 3.0 IQR) are extreme outliers. This definition of "outlier" is not the same as the Empirical Rule, which is based on the distance from the mean. Difficulty: 1 Easy Topic: 04.05 Percentiles, Quartiles, and Box Plots Learning Objective: 04-08 Make and interpret box plots. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation

93) Which is not a measure of variability? A) Mean absolute deviation (MAD) B) Range C) Coefficient of variation D) Trimmed mean Answer: D Explanation: The trimmed mean is a measure of center. Difficulty: 1 Easy Topic: 04.03 Measures of Variability Learning Objective: 04-03 Calculate and interpret common measures of variability. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 94) Twelve randomly chosen students were asked how many times they had missed class during a certain semester, with this result: 3, 2, 1, 2, 1, 5, 9, 1, 2, 3, 3, 10. The geometric mean is A) B) 2.604 C) 1.517 D) Answer: B Explanation: Take the 12th root of the product (97,200) to get 2.604. Difficulty: 3 Hard Topic: 04.02 Measures of Center Learning Objective: 04-02 Calculate and interpret common measures of center. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

95) Twelve randomly chosen students were asked how many times they had missed class during a certain semester, with this result: 3, 2, 1, 2, 1, 5, 9, 1, 2, 3, 3, 10. The median is A) 7.0 B) 3.0 C) 3.5 D) 2.5 Answer: D Explanation: Sort the data and go halfway between the two middle data values. Difficulty: 2 Medium Topic: 04.02 Measures of Center Learning Objective: 04-02 Calculate and interpret common measures of center. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 96) One disadvantage of the range is that A) only extreme values are used in its calculation. B) it is expressed in different units than the mean. C) it does not exist for some data sets. D) it is undefined if any X values are 0 or negative. Answer: A Explanation: Range ignores all the sample except the extremes xmax and xmin. Difficulty: 2 Medium Topic: 04.03 Measures of Variability Learning Objective: 04-03 Calculate and interpret common measures of variability. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation

97) Which is a characteristic of the standard deviation? A) It is not greatly affected by outliers. B) It is measured in the same units as the mean. C) It measures dispersion around the median. D) It has a natural, concrete meaning. Answer: B Explanation: Although we square the deviations around the mean, we take the square root of the sum to get back to the original units of X. However, the standard deviation is affected by outliers and its interpretation may be nonintuitive. Difficulty: 2 Medium Topic: 04.03 Measures of Variability Learning Objective: 04-03 Calculate and interpret common measures of variability. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 98) Twelve randomly chosen students were asked how many times they had missed class during a certain semester, with this result: 2, 1, 5, 1, 1, 3, 4, 3, 1, 1, 5, 18. For this sample, the geometric mean is A) 2.158 B) 1.545 C) 2.376 D) 3.017 Answer: C Explanation: Take the 12th root of the product (32,400) to get 2.376. Difficulty: 3 Hard Topic: 04.02 Measures of Center Learning Objective: 04-02 Calculate and interpret common measures of center. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

99) Twelve randomly chosen students were asked how many times they had missed class during a certain semester, with this result: 2, 1, 5, 1, 1, 3, 4, 3, 1, 1, 5, 18. For this sample, the median is A) 2 B) 3 C) 3.5 D) 2.5 Answer: D Explanation: Sort the data and look halfway between the two middle data values. Difficulty: 2 Medium Topic: 04.02 Measures of Center Learning Objective: 04-02 Calculate and interpret common measures of center. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 100) Twelve randomly chosen students were asked how many times they had missed class during a certain semester, with this result: 2, 1, 5, 1, 1, 3, 4, 3, 1, 1, 5, 18. For this sample, which measure of center is least representative of the "typical" student? A) Mean B) Median C) Mode D) Midrange Answer: D Explanation: The unusual data value pulls up the mean (3.75) but affects the midrange (1 + 18)/2 = 9.5 even more noticeably. Difficulty: 2 Medium Topic: 04.02 Measures of Center Learning Objective: 04-02 Calculate and interpret common measures of center. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

101) Here are statistics on order sizes of Megalith Construction Supply's shipments of two kinds of construction materials last year. Mean Std. Dev.

Girders 160 48

Rivets 2800 702

Which order sizes have greater variability? A) Girders B) Rivets C) They are the same. D) Cannot be determined without knowing n Answer: A Explanation: Calculate the coefficient of variation. For girders, the CV = 100 × s/ = 100 × (48)/(160) = 30.00%, while for rivets, CV = 100 × s/ = 100 × 702/2800 = 25.07%. Difficulty: 2 Medium Topic: 04.03 Measures of Variability Learning Objective: 04-03 Calculate and interpret common measures of variability. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 102) The quartiles of a distribution are most clearly revealed in which display? A) Box plot B) Scatter plot C) Histogram D) Dot plot Answer: A Explanation: The histogram, scatter plot, or dot plot will not directly show quartiles. Difficulty: 1 Easy Topic: 04.05 Percentiles, Quartiles, and Box Plots Learning Objective: 04-08 Make and interpret box plots. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation

103) The sum of the deviations around the mean is A) greater than zero if data are right-skewed. B) smaller when the units are smaller (e.g., milligrams versus kilograms). C) always zero. D) dependent on the sample size. Answer: C Explanation: The sum must be zero unless you calculated the mean incorrectly. Difficulty: 2 Medium Topic: 04.03 Measures of Variability Learning Objective: 04-03 Calculate and interpret common measures of variability. Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation 104) What does the graph below (profit/sales ratios for 25 Fortune 500 companies) reveal?

A) That the median exceeds the mean. B) That the data are slightly left-skewed. C) That the interquartile range is about 8. D) That the distribution is bell-shaped. Answer: C Explanation: The box is skewed right, so mean probably exceeds the median. The interquartile range (IQR) is about 12 − 4 = 8. Difficulty: 2 Medium Topic: 04.05 Percentiles, Quartiles, and Box Plots Learning Objective: 04-08 Make and interpret box plots. Bloom's: Analyze AACSB: Analytical Thinking Accessibility: Keyboard Navigation

105) Find the sample correlation coefficient for the following data. X 3 7 5 9 11 13 19 21

Y 8 12 13 10 17 23 39 38

A) .8911 B) .9132 C) .9822 D) .9556 Answer: D Explanation: You could use Excel's =CORREL(XData,YData). Difficulty: 2 Medium Topic: 04.06 Covariance and Correlation Learning Objective: 04-09 Calculate and interpret a correlation coefficient and covariance. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 106) The heights of male students in a certain statistics class range from Xmin = 61 to Xmax = 79. Applying the Empirical Rule, a reasonable estimate of σ would be A) 2.75 B) 3.00 C) 3.25 D) 3.50 Answer: B Explanation: One-sixth of the presumed 6σ range would be (79 − 61)/6 = 3. Difficulty: 2 Medium Topic: 04.04 Standardized Data Learning Objective: 04-05 Apply the Empirical Rule and recognize outliers. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

107) A reporter for the campus paper asked five randomly chosen students how many occupants, including the driver, ride to school in their cars. The responses were 1, 1, 1, 1, 6. The coefficient of variation is A) 25 percent. B) 250 percent. C) 112 percent. D) 100 percent. Answer: C Explanation: You first have to calculate the mean and standard deviation. Difficulty: 2 Medium Topic: 04.03 Measures of Variability Learning Objective: 04-03 Calculate and interpret common measures of variability. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 108) A smooth distribution with one mode is negatively skewed (skewed to the left). The median of the distribution is $65. Which of the following is a reasonable value for the distribution mean? A) $76 B) $54 C) $81 D) $65 Answer: B Explanation: The mean is expected to be less than the median due to left-skewness. Difficulty: 1 Easy Topic: 04.02 Measures of Center Learning Objective: 04-02 Calculate and interpret common measures of center. Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation

109) In a positively skewed distribution, the percentage of observations that fall below the median is: A) about 50 percent. B) less than 50 percent. C) more than 50 percent. D) cannot tell without knowing n. Answer: A Explanation: The median always puts about half above and half below. Difficulty: 2 Medium Topic: 04.02 Measures of Center Learning Objective: 04-02 Calculate and interpret common measures of center. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 110) Which is a weakness of the mode? A) It does not apply to qualitative data. B) It is inappropriate for continuous data. C) It is hard to calculate when n is small. D) It is usually about the same as the median. Answer: B Explanation: The mode is helpful for categorical data and is easy to calculate in small samples, but requires sorting the sample. Continuous (decimal) data generally have no mode, or, if a mode exists, it is often not near the center. Difficulty: 2 Medium Topic: 04.02 Measures of Center Learning Objective: 04-02 Calculate and interpret common measures of center. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 111) The mode is least appropriate for A) continuous data. B) categorical data. C) discrete data. D) Likert scale data. Answer: A Explanation: The mode is good for discrete or categorical data but fails for continuous data. Difficulty: 2 Medium Topic: 04.02 Measures of Center Learning Objective: 04-02 Calculate and interpret common measures of center. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 46 Copyright 2019 © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.

112) Craig operates a part-time snow-plowing business using a 2002 GMC 2500 HD extended cab short box truck. This box plot of Craig's MPG on 195 tanks of gas does not support which statement?

A) There are several outliers. B) This is a very right-skewed distribution. C) Most MPG values are concentrated in a narrow range. D) The interquartile range is less than 2 MPG. Answer: B Explanation: It is a narrow box. With outliers in both tails, it is unclear which way skewness would be. Difficulty: 2 Medium Topic: 04.05 Percentiles, Quartiles, and Box Plots Learning Objective: 04-08 Make and interpret box plots. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

113) Estimate the mean exam score for the 50 students in Prof. Axolotl's class. Score 40 but less than 50 50 but less than 60 60 but less than 70 70 but less than 80 80 but less than 90 Total

f 6 18 14 9 3 50

A) 59.2 B) 62.0 C) 63.5 D) 64.1 Answer: B Explanation: Apply the formulas for weighted average using the interval midpoint multiplied by the interval frequency. Difficulty: 2 Medium Topic: 04.07 Grouped Data Learning Objective: 04-10 Calculate the mean and standard deviation from grouped data. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

114) A survey of salary increases received during a recent year by 44 working MBA students is shown. Find the approximate mean percent raise. Percent Increase 0 but less than 2 2 but less than 4 4 but less than 6 6 but less than 10 Total

Number 3 4 10 27 44

A) 6.56 B) 6.74 C) 5.90 D) 6.39 Answer: D Explanation: Apply the formulas for weighted average using the interval midpoint multiplied by the interval frequency. Difficulty: 2 Medium Topic: 04.07 Grouped Data Learning Objective: 04-10 Calculate the mean and standard deviation from grouped data. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

115) The following frequency distribution shows the amount earned yesterday by employees of a large Las Vegas casino. Estimate the mean daily earnings. Earnings (dollars) 50 < 75 75 < 100 100 < 125 125 < 150 150 < 175

Frequency 10 15 60 40 10

A) $112.50 B) $125.01 C) $105.47 D) $117.13 Answer: D Explanation: Apply the formula for the weighted average using each interval midpoint multiplied by its relative frequency. Difficulty: 2 Medium Topic: 04.07 Grouped Data Learning Objective: 04-10 Calculate the mean and standard deviation from grouped data. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

116) The following table is the frequency distribution of parking fees for a day: Fee (dollars) 6.00 < 6.50 6.50 < 7.00 7.00 < 7.50 7.50 < 8.00

Number of Garages 5 8 10 7

What is the mean parking fee? A) $7.07 B) $6.95 C) $7.00 D) $7.25 Answer: A Explanation: Apply the formulas for weighted average using each interval midpoint multiplied by its relative frequency. Difficulty: 2 Medium Topic: 04.07 Grouped Data Learning Objective: 04-10 Calculate the mean and standard deviation from grouped data. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 117) Find the standard deviation of this sample: 4, 7, 9, 12, 15. A) 4.550 B) 3.798 C) 4.278 D) 2.997 Answer: C Explanation: Use the sample formula or Excel's =STDEV.S(Data). Difficulty: 2 Medium Topic: 04.03 Measures of Variability Learning Objective: 04-03 Calculate and interpret common measures of variability. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

118) The 25th percentile for waiting time in a doctor's office is 10 minutes. The 75th percentile is 30 minutes. Which is incorrect regarding the fences? A) The upper inner fence is 60 minutes. B) The upper outer fence is 90 minutes. C) A waiting time of 45 minutes would be an outlier. D) The lower fences are not relevant in this problem. Answer: C Explanation: Add 1.5 times the interquartile range to the third quartile to get the upper inner fence. Add 3.0 times the interquartile range to the third quartile to get the upper outer fence. An outlier is beyond the inner upper fence. Difficulty: 3 Hard Topic: 04.05 Percentiles, Quartiles, and Box Plots Learning Objective: 04-07 Calculate quartiles and other percentiles. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 119) Five homes were recently sold in Oxnard Acres. Four of the homes sold for $400,000, while the fifth home sold for $2.5 million. Which measure of central tendency best represents a typical home price in Oxnard Acres? A) The mean or median. B) The median or mode. C) The mean or mode. D) The midrange or mean. Answer: B Explanation: Outliers will affect the mean or midrange. Difficulty: 2 Medium Topic: 04.02 Measures of Center Learning Objective: 04-02 Calculate and interpret common measures of center. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

120) In Tokyo, construction workers earn an average of ¥420,000 (yen) per month with a standard deviation of ¥20,000, while in Hamburg, Germany, construction workers earn an average of €3,200 (euros) per month with a standard deviation of €57. Who is earning relatively more, a worker making ¥460,000 per month in Tokyo or one earning €3,300 per month in Hamburg? A) The workers are the same in relative terms. B) The Tokyo worker is relatively better off. C) The Hamburg worker is relatively better off. Answer: B Explanation: Calculate and compare the z-score for each nation's workers. For Tokyo, z = (x − μ)/σ = (460000 − 420000)/(20000) = 2.00, while for Hamburg, z = (x − μ)/σ = (3300 − 3200)/57 = 1.75. Therefore, wages for this worker are relatively higher in Tokyo. Difficulty: 2 Medium Topic: 04.04 Standardized Data Learning Objective: 04-06 Transform a data set into standardized values. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 121) Which statement is false? Explain. A) If μ = 52 and σ = 15, then X = 81 would be an outlier. B) If the data are from a normal population, about 68 percent of the values will be within μ ± σ. C) If μ = 640 and σ = 128, then the coefficient of variation is 20 percent. Answer: A Explanation: Calculate the z-score to detect outliers: z = (x − μ)/σ = (81 − 52)/(15) = 1.93, which is not an outlier, while the CV is 100 × σ/μ = 100 × 128/640 = 20%. Difficulty: 2 Medium Topic: 04.04 Standardized Data Learning Objective: 04-06 Transform a data set into standardized values. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

122) Which is not a measure of variability? A) Mean absolute deviation (MAD) B) Standard deviation C) Midhinge D) Interquartile range Answer: C Explanation: The midhinge measures center, not variability. Difficulty: 2 Medium Topic: 04.03 Measures of Variability Learning Objective: 04-03 Calculate and interpret common measures of variability. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 123) If Q1 = 150 and Q3 = 250, the upper fences (inner and outer) are A) 450 and 600. B) 350 and 450. C) 400 and 550. D) impossible to determine without more information. Answer: C Explanation: Add 1.5 times the interquartile range to the third quartile to get the upper inner fence. Add 3.0 times the interquartile range to the third quartile to get the upper outer fence. Difficulty: 2 Medium Topic: 04.05 Percentiles, Quartiles, and Box Plots Learning Objective: 04-07 Calculate quartiles and other percentiles. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

124) Variables X and Y have the strongest correlation in which scatter plot?

A) Figure A B) Figure B C) They are about the same. Answer: C Explanation: Except for sign, the correlations appear roughly the same. Difficulty: 1 Easy Topic: 04.06 Covariance and Correlation Learning Objective: 04-09 Calculate and interpret a correlation coefficient and covariance. Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation 125) Which of the following statements is likely to apply to the incomes of 50 randomly chosen taxpayers in California? A) The median income would probably be near the mean. B) The midhinge would be a robust measure of center. C) The sum of squared deviations about the mean would be negative. D) Outliers in either tail would be equally likely. Answer: B Explanation: The data will be extremely skewed by a few very rich taxpayers. Difficulty: 3 Hard Topic: 04.02 Measures of Center Learning Objective: 04-02 Calculate and interpret common measures of center. Bloom's: Evaluate AACSB: Analytical Thinking Accessibility: Keyboard Navigation

126) A certain health maintenance organization (HMO) examined the number of office visits by each of its members in the last year. For this data set, we would anticipate that the geometric mean would be A) a reasonable measure of center. B) zero because some HMO members would not have an office visit. C) too high because the distribution is likely to be skewed to the left. D) negative because some data values would be below the mean. Answer: B Explanation: Zeros would exist for those who had no office visits, so the geometric mean would be zero. Difficulty: 3 Hard Topic: 04.02 Measures of Center Learning Objective: 04-02 Calculate and interpret common measures of center. Bloom's: Evaluate AACSB: Analytical Thinking Accessibility: Keyboard Navigation 127) Three randomly chosen Colorado students were asked how many times they went rock climbing last month. Their replies were 5, 6, 7. The coefficient of variation is A) 16.7 percent. B) 13.6 percent. C) 20.0 percent. D) 35.7 percent. Answer: A Explanation: Calculate the mean and standard deviation first. The mean is 6 and the sample standard deviation is 1 so CV = 100 × 1/6 = 16.7%. Difficulty: 2 Medium Topic: 04.03 Measures of Variability Learning Objective: 04-03 Calculate and interpret common measures of variability. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

128) The mean of a population is 50 and the median is 40. Which histogram is most likely for samples from this population?

A) Sample A. B) Sample B. C) Sample C. Answer: A Explanation: The mean exceeds the median, so the data are skewed right. Difficulty: 1 Easy Topic: 04.02 Measures of Center Learning Objective: 04-02 Calculate and interpret common measures of center. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 129) If Excel's sample skewness coefficient is positive, we conclude that A) the population is skewed to the right, if the sample size is small. B) the population is symmetric, as long as the sample size is very large. C) the coefficient is within the range of chance for a symmetric population. D) we should consult a table of percentiles that takes sample size into consideration. Answer: D Explanation: We have tables that show the expected range of expected variation for a sample skewness coefficient for various sample sizes from a symmetric, normal population. Difficulty: 2 Medium Topic: 04.08 Skewness and Kurtosis Learning Objective: 04-11 Assess skewness and kurtosis in a sample. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

130) If Excel's sample kurtosis coefficient is negative, we conclude that A) the population is platykurtic, as long as the sample size is small. B) the population is leptokurtic, as long as the sample size is large. C) the coefficient is within the range of chance for a symmetric population. D) we should consult a table of percentiles that takes sample size into consideration. Answer: D Explanation: We have tables that show the expected range of expected variation for a sample kurtosis coefficient for various sample sizes from a normal population. Difficulty: 2 Medium Topic: 04.08 Skewness and Kurtosis Learning Objective: 04-11 Assess skewness and kurtosis in a sample. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 131) The midhinge lies halfway between A) xmin and xmax. B) Q1 and Q3. C) the mean and the median. D) the inner fences. Answer: B Explanation: The midhinge is the average of Q1 and Q3. Difficulty: 2 Medium Topic: 04.05 Percentiles, Quartiles, and Box Plots Learning Objective: 04-07 Calculate quartiles and other percentiles. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation

132) Which is not a characteristic of the midhinge? A) It lies halfway between Q1 and Q3. B) It is used to detect asymmetry. C) It is equal to the median in a symmetric data set. D) It is strongly affected by outliers. Answer: D Explanation: The midhinge is the average of Q1 and Q3 and is a robust measure of center when outliers exist. Difficulty: 2 Medium Topic: 04.05 Percentiles, Quartiles, and Box Plots Learning Objective: 04-07 Calculate quartiles and other percentiles. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 133) Which measure of association is unit-free? A) =CORREL(Xdata, YData) B) =COVAR.P(XData,YData) C) =COVAR.S(Xdata,YData) Answer: A Explanation: Covariance depends on the units of measurement of X and Y (e.g., dollars) while the correlation coefficient is always on a scale from −1 to 1. Difficulty: 1 Easy Topic: 04.06 Covariance and Correlation Learning Objective: 04-09 Calculate and interpret a correlation coefficient and covariance. Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation 134) To calculate the Pearson 2 coefficient of skewness Sk2 for a sample, we need the A) mean, median, and mode. B) mean, median, and standard deviation. C) mean, mode, and standard deviation. D) mode and quartiles. Answer: D Explanation: The Pearson 2 coefficient Sk2 = 3(mean − median)/s is zero if the mean and median are equal, its sign indicates the direction of skewness. Its appeal is its simplicity. Difficulty: 2 Medium Topic: 04.03 Measures of Variability Learning Objective: 04-11 Assess skewness and kurtosis in a sample. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 59 Copyright 2019 © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.

Applied Statistics in Business and Economics, 6e (Doane) Chapter 5 Probability 1) A sample space is the set of all possible outcomes in an experiment. Answer: TRUE Explanation: Review the definition of sample space. Difficulty: 1 Easy Topic: 05.01 Random Experiments Learning Objective: 05-01 Describe the sample space of a random experiment. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 2) The sum of all the probabilities of simple events in a sample space equals one. Answer: TRUE Explanation: Simple events are non-overlapping, so they sum to unity. Difficulty: 1 Easy Topic: 05.03 Rules of Probability Learning Objective: 05-03 Apply the definitions and rules of probability. Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation 3) The sum of the probabilities of all compound events in a sample space equals one. Answer: FALSE Explanation: Compound events may overlap, so you cannot simply add their probabilities. Difficulty: 1 Easy Topic: 05.02 Probability Learning Objective: 05-03 Apply the definitions and rules of probability. Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation 4) Probability is the measure of the relative likelihood that an event will occur. Answer: TRUE Explanation: This is one of three ways to view probability. Difficulty: 1 Easy Topic: 05.02 Probability Learning Objective: 05-03 Apply the definitions and rules of probability. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 1 Copyright 2019 © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.

5) The probability of the union of two events P(A or B) can exceed one. Answer: FALSE Explanation: Review the General Rule of Addition. No event probability can exceed one. Difficulty: 2 Medium Topic: 05.03 Rules of Probability Learning Objective: 05-03 Apply the definitions and rules of probability. Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation 6) The empirical view of probability is based on relative frequencies. Answer: TRUE Explanation: This is one view of probability, calculated from experience. Difficulty: 1 Easy Topic: 05.02 Probability Learning Objective: 05-02 Distinguish among the three views of probability. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 7) Grandma's predicting rain based on how much her arthritis is acting up is an example of the classical view of probability. Answer: FALSE Explanation: This would be a subjective probability, where no data are available. Difficulty: 2 Medium Topic: 05.02 Probability Learning Objective: 05-02 Distinguish among the three views of probability. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 8) The odds of an event can be calculated by dividing the event's probability by the probability of its complement. Answer: TRUE Explanation: Review the definition of odds. Difficulty: 2 Medium Topic: 05.02 Probability Learning Objective: 05-04 Calculate odds from given probabilities. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 2 Copyright 2019 © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.

9) The union of two events A and B is the event consisting of all outcomes in the sample space that are contained in both event A and event B. Answer: FALSE Explanation: The union is all outcomes in either A or B. Difficulty: 2 Medium Topic: 05.03 Rules of Probability Learning Objective: 05-03 Apply the definitions and rules of probability. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 10) The general law of addition for probabilities says P(A or B) = P(A) + P(B) − P(A ∩ B). Answer: TRUE Explanation: We must subtract the probability of the event intersection from the sum. Difficulty: 1 Easy Topic: 05.03 Rules of Probability Learning Objective: 05-03 Apply the definitions and rules of probability. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 11) Events A and B are mutually exclusive if P(A ∩ B) = 0. Answer: TRUE Explanation: Nonoverlapping events cannot both occur (intersection is the empty set). Difficulty: 2 Medium Topic: 05.03 Rules of Probability Learning Objective: 05-03 Apply the definitions and rules of probability. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 12) Independent events are mutually exclusive. Answer: FALSE Explanation: Only if P(A ∩ B) = 0 would they be mutually exclusive. Difficulty: 2 Medium Topic: 05.04 Independent Events Learning Objective: 05-05 Determine when events are independent. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation

13) If events A and B are mutually exclusive, then P(A) + P(B) = 0. Answer: FALSE Explanation: Their intersection would have zero probability, but not their sum. Difficulty: 2 Medium Topic: 05.03 Rules of Probability Learning Objective: 05-03 Apply the definitions and rules of probability. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 14) P(A | B) is the joint probability of events A and B divided by the probability of A. Answer: FALSE Explanation: You would divide the joint probability by P(B) rather than by P(A). Difficulty: 2 Medium Topic: 05.03 Rules of Probability Learning Objective: 05-03 Apply the definitions and rules of probability. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 15) Two events A and B are independent if P(A | B) is the same as P(A). Answer: TRUE Explanation: This is the definition of independence. Event B does not affect event A. Difficulty: 1 Easy Topic: 05.04 Independent Events Learning Objective: 05-05 Determine when events are independent. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 16) If events A and B are dependent, it can be concluded that one event causes the other. Answer: FALSE Explanation: Maybe there is causation, but both events could be affected by something else. Difficulty: 2 Medium Topic: 05.04 Independent Events Learning Objective: 05-05 Determine when events are independent. Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation

17) For any event A, the probability of A is always 0 ≤ P(A) ≤ 1. Answer: TRUE Explanation: Any probability must be between 0 and 1. Difficulty: 1 Easy Topic: 05.02 Probability Learning Objective: 05-03 Apply the definitions and rules of probability. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 18) If events A and B are mutually exclusive, the joint probability of the events is zero. Answer: TRUE Explanation: The intersection of nonoverlapping events is the empty set. Difficulty: 2 Medium Topic: 05.03 Rules of Probability Learning Objective: 05-03 Apply the definitions and rules of probability. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 19) When the outcome of a random experiment is a continuous measurement, the sample space is described by a rule instead of listing the possible simple events. Answer: TRUE Explanation: For example, the speeds of vehicles on a freeway are not listable (nonintegers). Difficulty: 1 Easy Topic: 05.01 Random Experiments Learning Objective: 05-01 Describe the sample space of a random experiment. Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation 20) If A and B are independent events, then P(A or B) = P(A) P(B). Answer: FALSE Explanation: Review the definition of independence. The proffered statement would have been correct if "and" were substituted for "or." The correct statement would be P(A or B) = P(A) + P(B) − P(A ∩ B). Difficulty: 2 Medium Topic: 05.04 Independent Events Learning Objective: 05-05 Determine when events are independent. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 5 Copyright 2019 © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.

21) If A and B are mutually exclusive events, then P(A ∩ B) = P(A) + P(B). Answer: FALSE Explanation: Their intersection is empty so it has zero probability. Difficulty: 2 Medium Topic: 05.03 Rules of Probability Learning Objective: 05-03 Apply the definitions and rules of probability. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 22) The probability of A and its complement (A´) will always sum to one. Answer: TRUE Explanation: An event and its complement comprise the sample space. Difficulty: 2 Medium Topic: 05.03 Rules of Probability Learning Objective: 05-03 Apply the definitions and rules of probability. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 23) If event A occurs, then its complement (A´) will also occur. Answer: FALSE Explanation: An event and its complement are nonoverlapping. Difficulty: 1 Easy Topic: 05.03 Rules of Probability Learning Objective: 05-03 Apply the definitions and rules of probability. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 24) The sum of the probabilities of two mutually exclusive events is one. Answer: FALSE Explanation: Review definitions for event probabilities. Their probabilities would sum to one only if these two events comprise the entire sample space. Difficulty: 2 Medium Topic: 05.03 Rules of Probability Learning Objective: 05-03 Apply the definitions and rules of probability. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation

25) P(A ∩ B) = .50 is an example of a joint probability. Answer: TRUE Explanation: Intersection of two events yields a joint probability. Difficulty: 1 Easy Topic: 05.03 Rules of Probability Learning Objective: 05-03 Apply the definitions and rules of probability. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 26) The general law of addition for probabilities says P(A or B) = P(A) P(B). Answer: FALSE Explanation: Review the General Law of Addition: P(A or B) = P(A) + P(B) − P(A ∩ B). Difficulty: 1 Easy Topic: 05.03 Rules of Probability Learning Objective: 05-03 Apply the definitions and rules of probability. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 27) If P(A) = .50, P(B) = .30, and P(A ∩ B) = .15, then A and B are independent events. Answer: TRUE Explanation: This is true because for independent events: P(A ∩ B) = P(A) P(B) and in this example P(A) P(B) = (.50)(.30) = .15 = P(A ∩ B). Difficulty: 2 Medium Topic: 05.04 Independent Events Learning Objective: 05-05 Determine when events are independent. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 28) Insurance company life tables are an example of the classical (a priori) approach to probability. Answer: FALSE Explanation: Actuarial data are based on relative frequencies (empirical probability). Difficulty: 2 Medium Topic: 05.02 Probability Learning Objective: 05-02 Distinguish among the three views of probability. Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation

29) When two events cannot occur at the same time, they are said to be mutually exclusive. Answer: TRUE Explanation: This is the definition of mutually exclusive events. Difficulty: 2 Medium Topic: 05.03 Rules of Probability Learning Objective: 05-03 Apply the definitions and rules of probability. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 30) The probability of events A or B occurring can be found by summing their probabilities. Answer: FALSE Explanation: Event probabilities cannot be summed unless their intersection is the empty set. Difficulty: 2 Medium Topic: 05.03 Rules of Probability Learning Objective: 05-03 Apply the definitions and rules of probability. Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation 31) When two events A and B are independent, the probability of their intersection can be found by multiplying their probabilities. Answer: TRUE Explanation: This is true by the definition of independent events. Difficulty: 2 Medium Topic: 05.04 Independent Events Learning Objective: 05-05 Determine when events are independent. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 32) Two events are mutually exclusive when they contain no outcomes in common. Answer: TRUE Explanation: This is the definition of mutually exclusive events. Difficulty: 1 Easy Topic: 05.03 Rules of Probability Learning Objective: 05-03 Apply the definitions and rules of probability. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation

33) In a contingency table, the probability of the union of two events is found by taking the frequency of the intersection of the two events and dividing by the total. Answer: FALSE Explanation: Review the general law of addition P(A or B) = P(A) + P(B) − P(A ∩ B). Difficulty: 2 Medium Topic: 05.05 Contingency Tables Learning Objective: 05-06 Apply the concepts of probability to contingency tables. Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation 34) Bayes' Theorem shows how to revise a prior probability to obtain a conditional or posterior probability when another event's occurrence is known. Answer: TRUE Explanation: This is a rather general statement about Bayes' contribution to statistics. Difficulty: 2 Medium Topic: 05.07 Bayes' Theorem Learning Objective: 05-08 Use Bayes' Theorem to calculate revised probabilities. Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation 35) The union of two events is all outcomes in either or both, while the intersection is only those events in both. Answer: TRUE Explanation: Review the definitions of union and intersection. Difficulty: 2 Medium Topic: 05.03 Rules of Probability Learning Objective: 05-03 Apply the definitions and rules of probability. Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation 36) A contingency table is a cross-tabulation of frequencies for two categorical variables. Answer: TRUE Explanation: The categories define the rows and columns. Difficulty: 1 Easy Topic: 05.05 Contingency Tables Learning Objective: 05-06 Apply the concepts of probability to contingency tables. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 9 Copyright 2019 © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.

37) The number of arrangements of sampled items drawn from a population is found with the formula for permutations (if order is important) or combinations (if order does not matter). Answer: TRUE Explanation: Review the definitions of permutations and combinations. Difficulty: 2 Medium Topic: 05.08 Counting Rules Learning Objective: 05-09 Apply counting rules to calculate possible event arrangements. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 38) If P(A) = .20 then the odds against event A's occurrence are 4 to 1. Answer: TRUE Explanation: Review the definition of odds. Difficulty: 2 Medium Topic: 05.02 Probability Learning Objective: 05-04 Calculate odds from given probabilities. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 39) The value of 7! is 5040. Answer: TRUE Explanation: Factorial n (written as n!) is 7! = 7 × 6 × 5 × 4 × 3 × 2 × 1 = 5040. Difficulty: 1 Easy Topic: 05.08 Counting Rules Learning Objective: 05-09 Apply counting rules to calculate possible event arrangements. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

40) Events A and B are mutually exclusive when A) their joint probability is zero. B) they are independent events. C) P(A)P(B) = 0 D) P(A)P(B) = P(A | B) Answer: A Explanation: By the definition of mutually exclusive P(A and B) = 0. That is, both events cannot happen. Difficulty: 2 Medium Topic: 05.03 Rules of Probability Learning Objective: 05-03 Apply the definitions and rules of probability. Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation 41) If two events are complementary, then we know that A) the sum of their probabilities is one. B) the joint probability of the two events is one. C) their intersection has a nonzero probability. D) they are independent events. Answer: A Explanation: Together, complementary events comprise the sample space. Difficulty: 2 Medium Topic: 05.03 Rules of Probability Learning Objective: 05-03 Apply the definitions and rules of probability. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation

42) Regarding probability, which of the following is correct? A) When events A and B are mutually exclusive, then P(A ∩ B) = P(A) + P(B). B) The union of events A and B consists of all outcomes in the sample space that are contained in both event A and event B. C) When two events A and B are independent, the joint probability of the events can be found by multiplying the probabilities of the individual events. D) The probability of the union of two events can exceed one. Answer: C Explanation: Review the rules of probability. When events are independent, their joint probability is their product. However, this is not true in general. Difficulty: 2 Medium Topic: 05.04 Independent Events Learning Objective: 05-05 Determine when events are independent. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 43) Independent events A and B would be consistent with which of the following statements? A) P(A) = .3, P(B) = .5, P(A ∩ B) = .4. B) P(A) = .4, P(B) = .5, P(A ∩ B) = .2. C) P(A) = .5, P(B) = .4, P(A ∩ B) = .3. D) P(A) = .4, P(B) = .3, P(A ∩ B) = .5. Answer: B Explanation: For independence, the product P(A)P(B) must equal P(A ∩ B). Difficulty: 1 Easy Topic: 05.04 Independent Events Learning Objective: 05-05 Determine when events are independent. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

44) Find the probability that either event A or B occurs if the chance of A occurring is .5, the chance of B occurring is .3, and events A and B are independent. A) .80 B) .15 C) .65 D) .85 Answer: C Explanation: Given that the events are independent, the product P(A)P(B) must equal P(A ∩ B). Thus, P(A or B) = P(A) + P(B) − P(A ∩ B) = .50 + .30 − (.50)(.30) = .80 − .15 = .65 using the General Law of Addition. Difficulty: 2 Medium Topic: 05.04 Independent Events Learning Objective: 05-05 Determine when events are independent. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 45) Regarding the rules of probability, which of the following statements is correct? A) If A and B are independent events, then P(B) = P(A)P(B). B) The sum of two mutually exclusive events is one. C) The probability of A and its complement will sum to one. D) If event A occurs, then its complement will also occur. Answer: C Explanation: By definition P(A and A') = 1. Difficulty: 2 Medium Topic: 05.03 Rules of Probability Learning Objective: 05-03 Apply the definitions and rules of probability. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation

46) Within a given population, 22 percent of the people are smokers, 57 percent of the people are males, and 12 percent are males who smoke. If a person is chosen at random from the population, what is the probability that the selected person is either a male or a smoker? A) .67 B) .79 C) .22 D) .43 Answer: A Explanation: Use the General Law of Addition P(A or B) = P(A) + P(B) − P(A ∩ B). Difficulty: 2 Medium Topic: 05.03 Rules of Probability Learning Objective: 05-03 Apply the definitions and rules of probability. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 47) Information was collected on those who attended the opening of a new movie. The analysis found that 56 percent of the moviegoers were female, 26 percent were under age 25, and 17 percent were females under the age of 25. Find the probability that a moviegoer is either female or under age 25. A) .79 B) .82 C) .65 D) .50 Answer: C Explanation: Use the General Law of Addition P(A or B) = P(A) + P(B) − P(A ∩ B). Difficulty: 3 Hard Topic: 05.03 Rules of Probability Learning Objective: 05-03 Apply the definitions and rules of probability. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

48) Given the contingency table shown here, find P(V). County Macomb (M) Oakland (O) Wayne (W) Col Total

Cell Phone Service Provider Sprint (S) AT&T (A) Verizon (V) 17 25 8 19 38 13 24 37 19 60 100 40

Row Total 50 70 80 200

A) .20 B) .40 C) .50 D) .80 Answer: A Explanation: This is a marginal probability P(V) = 40/200 = .20. Difficulty: 2 Medium Topic: 05.05 Contingency Tables Learning Objective: 05-06 Apply the concepts of probability to contingency tables. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 49) Given the contingency table shown here, find P(V | W). County Macomb (M) Oakland (O) Wayne (W) Col Total

Cell Phone Service Provider Sprint (S) AT&T (A) Verizon (V) 17 25 8 19 38 13 24 37 19 60 100 40

Row Total 50 70 80 200

A) .4000 B) .0950 C) .2375 D) .5875 Answer: C Explanation: This is a conditional probability P(V | W) = 19/80. Difficulty: 2 Medium Topic: 05.05 Contingency Tables Learning Objective: 05-06 Apply the concepts of probability to contingency tables. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 15 Copyright 2019 © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.

50) Given the contingency table shown here, find the probability P(V'), that is, the probability of the complement of V. County Macomb (M) Oakland (O) Wayne (W) Col Total

Cell Phone Service Provider Sprint (S) AT&T (A) Verizon (V) 17 25 8 19 38 13 24 37 19 60 100 40

Row Total 50 70 80 200

A) .30 B) .50 C) .80 D) .15 Answer: C Explanation: Calculate the probability of the complement of V by subtracting from its marginal probability P(V ') = 40/200 to get P(V ') = 1 − P(V) = 1 − 40/200. Difficulty: 3 Hard Topic: 05.05 Contingency Tables Learning Objective: 05-06 Apply the concepts of probability to contingency tables. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

51) Given the contingency table shown here, find P(W ∩ S). County Macomb (M) Oakland (O) Wayne (W) Col Total

Cell Phone Service Provider Sprint (S) AT&T (A) Verizon (V) 17 25 8 19 38 13 24 37 19 60 100 40

Row Total 50 70 80 200

A) .12 B) .30 C) .40 D) .58 Answer: A Explanation: This is a joint probability P(W and S) = 24/200. Difficulty: 2 Medium Topic: 05.05 Contingency Tables Learning Objective: 05-06 Apply the concepts of probability to contingency tables. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 52) Given the contingency table shown here, find P(A or M). County Macomb (M) Oakland (O) Wayne (W) Col Total

Cell Phone Service Provider Sprint (S) AT&T (A) Verizon (V) 17 25 8 19 38 13 24 37 19 60 100 40

Row Total 50 70 80 200

A) .2500 B) .7500 C) .6250 D) .1250 Answer: C Explanation: Use the General Law of Addition P(A or M) = 100/200 + 50/200 − 25/200. Difficulty: 3 Hard Topic: 05.05 Contingency Tables Learning Objective: 05-06 Apply the concepts of probability to contingency tables. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 17 Copyright 2019 © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.

53) Given the contingency table shown here, find P(A2).

B1 B2 B3 Col Total

A1 12 14 18 44

26 28 32 86

42 44 47 133

68 64 72 204

Row Total 148 150 169 467

A) .1842 B) .1766 C) .8163 D) .0578 Answer: A Explanation: This is a marginal probability: P(A2) = 86/467. Difficulty: 2 Medium Topic: 05.05 Contingency Tables Learning Objective: 05-06 Apply the concepts of probability to contingency tables. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 54) Given the contingency table shown here, find P(A3 ∩ B2).

B1 B2 B3 Col Total

A1 12 14 18 44

26 28 32 86

42 44 47 133

68 64 72 204

Row Total 148 150 169 467

A) .3212 B) .2933 C) .0942 D) .1006 Answer: C Explanation: This is a joint probability: P(A3 and B2) = 44/467. Difficulty: 2 Medium Topic: 05.05 Contingency Tables Learning Objective: 05-06 Apply the concepts of probability to contingency tables. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 18 Copyright 2019 © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.

55) Given the contingency table shown here, find P(A2 | B3).

B1 B2 B3 Col Total

A1 12 14 18 44

26 28 32 86

42 44 47 133

68 64 72 204

Row Total 148 150 169 467

A) .0685 B) .1893 C) .3721 D) .1842 Answer: B Explanation: This is a conditional probability: P(A2 | B3) = 32/169. Difficulty: 2 Medium Topic: 05.05 Contingency Tables Learning Objective: 05-06 Apply the concepts of probability to contingency tables. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

56) Given the contingency table shown here, find P(A1 or B2).

B1 B2 B3 Col Total

A1 12 14 18 44

26 28 32 86

42 44 47 133

68 64 72 204

Row Total 148 150 169 467

A) .0933 B) .3182 C) .0300 D) .3854 Answer: D Explanation: Apply the General Law of Addition: P(A1 or B2) = 44/467 + 150/467 − 14/467. Difficulty: 3 Hard Topic: 05.05 Contingency Tables Learning Objective: 05-06 Apply the concepts of probability to contingency tables. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 57) Given the contingency table shown here, find P(A1 ∩ A2).

B1 B2 B3 Col Total

A1 12 14 18 44

26 28 32 86

42 44 47 133

68 64 72 204

Row Total 148 150 169 467

A) .00 B) .09 C) .28 D) .38 Answer: A Explanation: This is a joint probability. The important thing here is that events A1 and A2 are mutually exclusive and so both events cannot occur. Difficulty: 2 Medium Topic: 05.05 Contingency Tables Learning Objective: 05-06 Apply the concepts of probability to contingency tables. Bloom's: Apply AACSB: Analytical Thinking 20 Copyright 2019 © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.

Accessibility: Keyboard Navigation 58) Given the contingency table shown here, find the probability that either event A2 or event B2 will occur.

B1 B2 B3 Col Total

A1 12 14 18 44

26 28 32 86

42 44 47 133

68 64 72 204

Row Total 148 150 169 467

A) .4454 B) .5054 C) .0600 Answer: A Explanation: Use the General Law of Addition: P(A2 or B2) =86/467 + 150/467 − 28/467. Difficulty: 3 Hard Topic: 05.05 Contingency Tables Learning Objective: 05-06 Apply the concepts of probability to contingency tables. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 59) Given the contingency table shown here, find P(B). Age Absences Under 2 days (B) 2 or more days (B') Column Total

Under 25 (A) 50 30 80

25 or More (A') 40 80 120

Row Total 90 110 200

A) .85 B) .25 C) .45 D) .22 Answer: C Explanation: This is a marginal probability: P(B) = 90/200. Difficulty: 1 Easy Topic: 05.05 Contingency Tables Learning Objective: 05-06 Apply the concepts of probability to contingency tables. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 21 Copyright 2019 © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.

60) Given the contingency table shown here, find P(A or B). Age Absences Under 2 days (B) 2 or more days (B') Column Total

Under 25 (A) 50 30 80

25 or More (A') 40 80 120

Row Total 90 110 200

A) .25 B) .85 C) .60 D) .42 Answer: C Explanation: Use the General Law of Addition: P(A or B) = 80/200 + 90/200 − 50/200. Difficulty: 2 Medium Topic: 05.05 Contingency Tables Learning Objective: 05-06 Apply the concepts of probability to contingency tables. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 61) Given the contingency table shown here, find P(B | A). Age Absences Under 2 days (B) 2 or more days (B') Column Total

Under 25 (A) 50 30 80

25 or More (A') 40 80 120

Row Total 90 110 200

A) .250 B) .555 C) .855 D) .625 Answer: D Explanation: This is a conditional probability: P(B | A) = 50/80. Difficulty: 2 Medium Topic: 05.05 Contingency Tables Learning Objective: 05-06 Apply the concepts of probability to contingency tables. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

62) Given the contingency table shown here, what is the probability that a randomly chosen employee who is under age 25 would be absent 2 or more days? Age Absences Under 2 days (B) 2 or more days (B') Column Total

Under 25 (A) 50 30 80

25 or More (A') 40 80 120

Row Total 90 110 200

A) .625 B) .375 C) .150 D) .273 Answer: B Explanation: This is a conditional probability: P(B ' | A) = 30/80. Difficulty: 2 Medium Topic: 05.05 Contingency Tables Learning Objective: 05-06 Apply the concepts of probability to contingency tables. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 63) Oxnard Casualty wants to ensure that their e-mail server has 99.98 percent reliability. They will use several independent servers in parallel, each of which is 95 percent reliable. What is the smallest number of independent file servers that will accomplish the goal? A) 1 B) 2 C) 3 D) 4 Answer: C Explanation: 1 − P(F1 ∩ F2 ∩ F3) = 1 − (.05)(.05)(.05) = 1 − .000125 = .999875, so 3 servers will do. Difficulty: 3 Hard Topic: 05.04 Independent Events Learning Objective: 05-05 Determine when events are independent. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

64) Given the contingency table shown here, does the decision to retire appear independent of the employee type? Survey question: Do you plan on retiring or keep working when you turn 65? Employee Management (M) Line worker (L) Total

Retire (R) 13 39 52

Work (W) 18 54 72

Total 31 93 124

A) Yes B) No Answer: A Explanation: Does the product of the marginal probabilities equal their joint probability? This can be checked by asking whether P(M and R) = P(M) P(R). In this example, because (31/124) (52/124) = 13/124, we can see that M and R are independent events. Difficulty: 2 Medium Topic: 05.05 Contingency Tables Learning Objective: 05-06 Apply the concepts of probability to contingency tables. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

65) Given the contingency table shown here, find the probability that a randomly chosen employee is a line worker who plans to retire at age 65. Survey question: Do you plan on retiring or keep working when you turn 65? Employee Management (M) Line worker (L) Total

Retire (R) 13 39 52

Work (W) 18 54 72

Total 31 93 124

A) .227 B) .419 C) .750 D) .315 Answer: D Explanation: This is a joint probability: P(L and R) = 39/124. Difficulty: 2 Medium Topic: 05.05 Contingency Tables Learning Objective: 05-06 Apply the concepts of probability to contingency tables. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 66) Given the contingency table shown here, find P(R ∩ L). Survey question: Do you plan on retiring or keep working when you turn 65? Employee Management (M) Line worker (L) Total

Retire (R) 13 39 52

Work (W) 18 54 72

Total 31 93 124

A) .250 B) .315 C) .425 D) .850 Answer: B Explanation: This is a joint probability: P(R ∩ L) = 39/124. Difficulty: 2 Medium Topic: 05.05 Contingency Tables Learning Objective: 05-06 Apply the concepts of probability to contingency tables. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 25 Copyright 2019 © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.

67) Given the contingency table shown here, find P(W | M). Survey question: Do you plan on retiring or keep working when you turn 65? Employee Management (M) Line worker (L) Total

Retire (R) 13 39 52

Work (W) 18 54 72

Total 31 93 124

A) .145 B) .250 C) .581 D) .687 Answer: C Explanation: This is a conditional probability: P(W | M) = 18/31. Difficulty: 2 Medium Topic: 05.05 Contingency Tables Learning Objective: 05-06 Apply the concepts of probability to contingency tables. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 68) Given the contingency table shown here, find P(L or W). Survey question: Do you plan on retiring or keep working when you turn 65? Employee Management (M) Line worker (L) Total

Retire (R) 13 39 52

Work (W) 18 54 72

Total 31 93 124

A) .750 B) .588 C) .435 D) .895 Answer: D Explanation: Use the General Law of Addition: P(L or W) = 93/124 + 72/124 − 54/124. Difficulty: 2 Medium Topic: 05.05 Contingency Tables Learning Objective: 05-06 Apply the concepts of probability to contingency tables. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 26 Copyright 2019 © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.

69) Ramjac Company wants to set up k independent file servers, each capable of running the company's intranet. Each server has average "uptime" of 98 percent. What must k be to achieve 99.999 percent probability that the intranet will be "up"? A) 1 B) 2 C) 3 D) 4 Answer: C Explanation: 1 − P(F1∩F2∩F3) = 1 − (.02)(.02)(.02) = 1 − .000008 = .999992, so 3 servers will do. Difficulty: 3 Hard Topic: 05.04 Independent Events Learning Objective: 05-05 Determine when events are independent. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 70) Given the contingency table shown here, what is the probability that a mother in the study smoked during pregnancy? Mother's Education Below High School High School Some College College Degree Col Total

Smoked during Pregnancy 393 560 121 48 1,122

Didn't Smoke during Pregnancy 640 1,370 635 550 3,195

Row Total 1,033 1,930 756 598 4,317

A) .2599 B) .3174 C) .5000 D) .7401 Answer: A Explanation: This is a marginal probability: P(smoked) = 1122/4317. Difficulty: 1 Easy Topic: 05.05 Contingency Tables Learning Objective: 05-06 Apply the concepts of probability to contingency tables. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

71) Given the contingency table shown here, what is the probability that a mother smoked during pregnancy if her education level was below high school? Mother's Education Below High School High School Some College College Degree Col Total

Smoked during Pregnancy 393 560 121 48 1,122

Didn't Smoke during Pregnancy 640 1,370 635 550 3,195

Row Total 1,033 1,930 756 598 4,317

A) .2385 B) .0907 C) .3503 D) .3804 Answer: D Explanation: This is a conditional probability: P(smoked | below high school) = 393/1033. Difficulty: 2 Medium Topic: 05.05 Contingency Tables Learning Objective: 05-06 Apply the concepts of probability to contingency tables. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

72) Given the contingency table shown here, what is the probability that a mother smoked during pregnancy and had a college degree? Mother's Education Below High School High School Some College College Degree Col Total

Smoked during Pregnancy 393 560 121 48 1,122

Didn't Smoke during Pregnancy 640 1,370 635 550 3,195

Row Total 1,033 1,930 756 598 4,317

A) .0111 B) .0428 C) .0803 D) .2385 Answer: A Explanation: This is a joint probability: P(smoked and college) = 48/4317. Difficulty: 2 Medium Topic: 05.05 Contingency Tables Learning Objective: 05-06 Apply the concepts of probability to contingency tables. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

73) Given the contingency table shown here, what is the probability that a mother smoked during pregnancy or that she graduated from college? Mother's Education Below High School High School Some College College Degree Col Total

Smoked during Pregnancy 393 560 121 48 1,122

Didn't Smoke during Pregnancy 640 1,370 635 550 3,195

Row Total 1,033 1,930 756 598 4,317

A) .0111 B) .2591 C) .3873 D) .7850 Answer: C Explanation: Use the General Law of Addition: 1122/4317 + 598/4317 − 48/4317. Difficulty: 3 Hard Topic: 05.05 Contingency Tables Learning Objective: 05-06 Apply the concepts of probability to contingency tables. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

74) Given the contingency table shown here, if a mother attended some college but did not have a degree, what is the probability that she did not smoke during her pregnancy? Mother's Education Below High School High School Some College College Degree Col Total

Smoked during Pregnancy 393 560 121 48 1,122

Didn't Smoke during Pregnancy 640 1,370 635 550 3,195

Row Total 1,033 1,930 756 598 4,317

A) .2736 B) .8399 C) .8752 D) .9197 Answer: B Explanation: This is a conditional probability: 635/756. Difficulty: 3 Hard Topic: 05.05 Contingency Tables Learning Objective: 05-06 Apply the concepts of probability to contingency tables. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

75) Given the contingency table shown here, find the probability that a mother with some college smoked during pregnancy. Mother's Education Below High School High School Some College College Degree Col Total

Smoked during Pregnancy 393 560 121 48 1,122

Didn't Smoke during Pregnancy 640 1,370 635 550 3,195

Row Total 1,033 1,930 756 598 4,317

A) .1078 B) .1746 C) .1601 D) .1117 Answer: C Explanation: This is a conditional probability: 121/756. Difficulty: 2 Medium Topic: 05.05 Contingency Tables Learning Objective: 05-06 Apply the concepts of probability to contingency tables. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

76) Given the contingency table shown here, if a survey participant is selected at random, what is the probability he/she is an undergrad who favors the change to a quarter system? Opinion: Oppose Change (N) Favor Change (S) Col Total

Group Surveyed Undergrads (U) Graduates (G) 73 27 27 23 100 50

Faculty (F) 20 30 50

Row Total 120 80 200

A) .270 B) .135 C) .338 D) .756 Answer: B Explanation: This is a joint probability: P(U and S) =27/200. Difficulty: 2 Medium Topic: 05.05 Contingency Tables Learning Objective: 05-06 Apply the concepts of probability to contingency tables. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 77) Given the contingency table shown here, if a faculty member is chosen at random, what is the probability he/she opposes the change to a quarter system? Opinion: Oppose Change (N) Favor Change (S) Col Total

Group Surveyed Undergrads (U) Graduates (G) 73 27 27 23 100 50

Faculty (F) 20 30 50

Row Total 120 80 200

A) .10 B) .25 C) .40 D) .60 Answer: C Explanation: This is a marginal probability: P(N | F) = 20/50 = .40. Difficulty: 2 Medium Topic: 05.05 Contingency Tables Learning Objective: 05-06 Apply the concepts of probability to contingency tables. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 33 Copyright 2019 © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.

78) Given the contingency table shown here, what is the probability that a participant selected at random is a graduate student who opposes the change to a quarter system? Opinion: Oppose Change (N) Favor Change (S) Col Total

Group Surveyed Undergrads (U) Graduates (G) 73 27 27 23 100 50

Faculty (F) 20 30 50

Row Total 120 80 200

A) .135 B) .250 C) .375 D) .540 Answer: A Explanation: This is a joint probability: P(G and N) = 27/200. Difficulty: 2 Medium Topic: 05.05 Contingency Tables Learning Objective: 05-06 Apply the concepts of probability to contingency tables. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

79) Given the contingency table shown here, what is the probability that a student attends a public school in a rural area? What type of school do you attend? Location Inner City (I) Suburban (S) Rural (R) Col Total

Public (P) 35 45 25 105

Religious (R) 15 10 5 30

Other Private (O) 20 25 5 50

Row Total 70 80 35 185

A) .238 B) .714 C) .135 D) .567 Answer: C Explanation: This is a joint probability: P(P and R) = 25/185 = .135. Difficulty: 1 Easy Topic: 05.05 Contingency Tables Learning Objective: 05-06 Apply the concepts of probability to contingency tables. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

80) Given the contingency table shown here, if a randomly chosen student attends a religious school, what is the probability the location is rural? What type of school do you attend? Location Inner City (I) Suburban (S) Rural (R) Col Total

Public (P) 35 45 25 105

Religious (R) 15 10 5 30

Other Private (O) 20 25 5 50

Row Total 70 80 35 185

A) .142 B) .162 C) .167 D) .333 Answer: C Explanation: This is a conditional probability: P(R | L) = 5/30 = .167. Difficulty: 2 Medium Topic: 05.05 Contingency Tables Learning Objective: 05-06 Apply the concepts of probability to contingency tables. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

81) Given the contingency table shown here, if a randomly chosen student attends school in an inner-city location, what is the probability that it is a public school? What type of school do you attend? Location Inner City (I) Suburban (S) Rural (R) Col Total

Public (P) 35 45 25 105

Religious (R) 15 10 5 30

Other Private (O) 20 25 5 50

Row Total 70 80 35 185

A) .189 B) .333 C) .500 D) .567 Answer: C Explanation: This is a conditional probability: P(P | I) = 35/70 = .500. Difficulty: 2 Medium Topic: 05.05 Contingency Tables Learning Objective: 05-06 Apply the concepts of probability to contingency tables. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 82) Given the contingency table shown here, find P(E). Gender Male (M) Female (F) Col Total

Accounting (A) 210 150 360

Major Gen. Mgmt. (G) 180 160 340

Economics (E) 140 160 300

Row Total 530 470 1000

A) .180 B) .300 C) .529 D) .641 Answer: B Explanation: This is a marginal probability: P(E) = 300/1000 = .300. Difficulty: 1 Easy Topic: 05.05 Contingency Tables Learning Objective: 05-06 Apply the concepts of probability to contingency tables. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 37 Copyright 2019 © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.

83) Given the contingency table shown here, find P(E | F). Gender Male (M) Female (F) Col Total

Accounting (A) 210 150 360

Major Gen. Mgmt. (G) 180 160 340

Economics (E) 140 160 300

Row Total 530 470 1000

A) .160 B) .300 C) .340 D) .533 Answer: C Explanation: This is a conditional probability: P(E | F) = 160/470 = .340. Difficulty: 2 Medium Topic: 05.05 Contingency Tables Learning Objective: 05-06 Apply the concepts of probability to contingency tables. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 84) Given the contingency table shown here, find P(A ∩ M). Gender Male (M) Female (F) Col Total

Accounting (A) 210 150 360

Major Gen. Mgmt. (G) 180 160 340

Economics (E) 140 160 300

Row Total 530 470 1000

A) .210 B) .360 C) .396 D) .583 Answer: A Explanation: This is a joint probability: P(A ∩ M) = 210/1000 = .210. Difficulty: 1 Easy Topic: 05.05 Contingency Tables Learning Objective: 05-06 Apply the concepts of probability to contingency tables. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

85) Given the contingency table shown here, find P(F or G). Gender Male (M) Female (F) Col Total

Accounting (A) 210 150 360

Major Gen. Mgmt. (G) 180 160 340

Economics (E) 140 160 300

Row Total 530 470 1000

A) .160 B) .470 C) .650 D) .810 Answer: C Explanation: Use the General Law of Addition: P(F or G) = 470/1000 + 340/1000 − 160/1000. Difficulty: 3 Hard Topic: 05.05 Contingency Tables Learning Objective: 05-06 Apply the concepts of probability to contingency tables. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 86) Given the contingency table shown here, find the probability that a randomly chosen individual is a female economics major. Gender Male (M) Female (F) Col Total

Accounting (A) 210 150 360

Major Gen. Mgmt. (G) 180 160 340

Economics (E) 140 160 300

Row Total 530 470 1000

A) .3404 B) .4700 C) .1600 D) .5333 Answer: C Explanation: This is a joint probability: P(F and E) = 160/1000 = .16. Difficulty: 2 Medium Topic: 05.05 Contingency Tables Learning Objective: 05-06 Apply the concepts of probability to contingency tables. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

87) Debbie has two stocks, X and Y. Consider the following events: X = the event that the price of stock X has increased Y = the event that the price of stock Y has increased The event "the price of stock X has increased and the price of stock Y has not increased" may be written as A) X ' ∩ Y B) X or Y ′ C) X ∩ Y ′ D) X or Y Answer: C Explanation: This is a joint probability that also entails the notation for an event's complement. Difficulty: 1 Easy Topic: 05.03 Rules of Probability Learning Objective: 05-03 Apply the definitions and rules of probability. Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation 88) If P(A | B) = .40 and P(B) = .30, find P(A ∩ B). A) .171 B) .525 C) .571 D) .120 Answer: D Explanation: Use the definition for conditional probability. Difficulty: 3 Hard Topic: 05.03 Rules of Probability Learning Objective: 05-03 Apply the definitions and rules of probability. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

89) A company is producing two types of ski goggles. Thirty percent of the production is of type A, and the rest is of type B. Five percent of all type A goggles are returned within 10 days after the sale, whereas only two percent of type B are returned. If a pair of goggles is returned within the first 10 days after the sale, the probability that the goggles returned are of type B is A) .014 B) .140 C) .070 D) .483 Answer: D Explanation: Review Bayes' Theorem, and perhaps make a table or tree. Difficulty: 3 Hard Topic: 05.07 Bayes' Theorem Learning Objective: 05-08 Use Bayes' Theorem to calculate revised probabilities. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 90) Given the contingency table shown here, find the joint probability that a call sampled at random out of this population is local and 2−5 minutes long. Type of Phone Call Local Long Distance Total

0−1 150 170 320

Call Length (minutes) 2− 5 6+ 250 100 120 10 370 110

Total 500 300 800

A) .5000 B) .3125 C) .4000 D) .4625 Answer: B Explanation: This is a joint probability: 250/800 = .3125 Difficulty: 2 Medium Topic: 05.05 Contingency Tables Learning Objective: 05-06 Apply the concepts of probability to contingency tables. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

91) Given the contingency table shown here, if a call is sampled at random, find the marginal probability that the call is long distance. Type of Phone Call Local Long Distance Total

0−1 150 170 320

Call Length (minutes) 2− 5 6+ 250 100 120 10 370 110

Total 500 300 800

A) .3750 B) .6250 C) .4000 D) 300/500 Answer: A Explanation: 500/800 = .6250. Difficulty: 2 Medium Topic: 05.05 Contingency Tables Learning Objective: 05-06 Apply the concepts of probability to contingency tables. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 92) If a call is sampled at random, the conditional probability that the call is not "6+" minutes long given that it is a long distance call is Type of Phone Call Local Long Distance Total

0−1 150 170 320

Call Length (minutes) 2− 5 6+ 250 100 120 10 370 110

Total 500 300 800

A) 120/300 B) 10/300 C) .9667 D) .6667 Answer: C Explanation: Calculate the conditional probability 1 − 10/300 = .9667. Difficulty: 3 Hard Topic: 05.05 Contingency Tables Learning Objective: 05-06 Apply the concepts of probability to contingency tables. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 42 Copyright 2019 © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.

93) The following table gives a classification of the 10,000 shareholders of Oxnard Xylophone Distributors, Inc. A few numbers are missing from the table. Given that a shareholder holding 500−999 shares is picked, there is a .625 probability that the shareholder will be a woman. Consequently, what is the number of men holding 1000 or more shares? Shareholders Men Women Joint Accounts Total

0−499 3,000 2,000 0 5,000

Number of Shares Held 500−999 1,000+ 1,000 4,000

200 1,000

Total 4,000 4,800 1,200 10,000

A) 1,000 B) 250 C) 7,500 D) 500 Answer: D Explanation: Multiply by the column total and subtract to fill in the remaining frequencies. Difficulty: 3 Hard Topic: 05.05 Contingency Tables Learning Objective: 05-06 Apply the concepts of probability to contingency tables. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 94) In any sample space P(A | B) and P(B | A) A) are always equal to one another. B) are never equal to one another. C) are reciprocals of one another. D) are equal only if P(A) = P(B). Answer: D Explanation: Use the definition of conditional probability. Difficulty: 3 Hard Topic: 05.03 Rules of Probability Learning Objective: 05-03 Apply the definitions and rules of probability. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation

95) If P(A ∩ B) = .50, can P(A) = .20? A) Only if P(A | B) = .10 B) Not unless P(B) = .30 C) Only if P(B ∩ A) = .60 D) If P(A) = .20, then P(A ∩ B) cannot equal .50. Answer: D Explanation: The given information contains a contradiction, because P(A ∩ B) cannot exceed P(A). Difficulty: 2 Medium Topic: 05.03 Rules of Probability Learning Objective: 05-03 Apply the definitions and rules of probability. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 96) The following relationship always holds true for events A and B in a sample space. A) P(A | B) = P(B | A) B) P(A ∩ B) = P(A | B) P(B) C) P(A | B) = P(B | A) P(A) Answer: B Explanation: Use the definition of conditional probability: P(A | B) = P(A ∩ B) / P(B). Difficulty: 3 Hard Topic: 05.03 Rules of Probability Learning Objective: 05-03 Apply the definitions and rules of probability. Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation 97) The following probabilities are given about events A and B in a sample space: P(A) = .30, P(B) = .40, P(A or B) = .60. We can say that A) P(A ∩ B) = .70. B) P(A) = P(A ∩ B). C) P(A ∩ B) = .10. D) A and B are independent events. Answer: C Explanation: Apply the General Rule of Addition: P(A or B) = P(A) + P(B) − P(A ∩ B). Difficulty: 3 Hard Topic: 05.03 Rules of Probability Learning Objective: 05-03 Apply the definitions and rules of probability. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 44 Copyright 2019 © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.

98) If P(A) = .35, P(B) = .60, and P(A or B) = .70, then A) A and B are mutually exclusive. B) P(A ∩ B) = .15. C) P(A ∩ B) = .25. D) P(A ∩ B) = .35. Answer: C Explanation: Apply the General Rule of Addition: P(A or B) = P(A) + P(B) − P(A ∩ B). Difficulty: 3 Hard Topic: 05.03 Rules of Probability Learning Objective: 05-03 Apply the definitions and rules of probability. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 99) The following table shows the survival experience of 1,000 males who retire at age 65: Age 65 70 75 80 85

Number of Males Surviving 1,000 907 775 596 383

Based on these data, the probability that a 75-year-old male will survive to age 80 is A) .596 B) 1 − .596 = .404 C) 1 − .775 = .225 D) .769 Answer: D Explanation: Given that 775 have survived to 75, the probability is 596 divided by 775. Difficulty: 3 Hard Topic: 05.03 Rules of Probability Learning Objective: 05-03 Apply the definitions and rules of probability. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

100) Given the contingency table shown here, find P(G | M). Vehicle Type Car (C) Minivan (M) Full-Size Van (F) SUV (V) Truck (T) Col Total

Somerset (S) 44 21 2 19 14 100

Oakland (O) 49 15 3 27 6 100

Great Lakes (G) 36 18 3 26 17 100

Row Total 129 54 8 72 37 300

A) .1800 B) .0450 C) .3333 D) .1350 Answer: C Explanation: This is a conditional probability: P(G | M) = 18/54. Difficulty: 2 Medium Topic: 05.05 Contingency Tables Learning Objective: 05-06 Apply the concepts of probability to contingency tables. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

101) Given the contingency table shown here, find P(V or S). Vehicle Type Car (C) Minivan (M) Full-Size Van (F) SUV (V) Truck (T) Col Total

Somerset (S) 44 21 2 19 14 100

Oakland (O) 49 15 3 27 6 100

Great Lakes (G) 36 18 3 26 17 100

Row Total 129 54 8 72 37 300

A) .5100 B) .4300 C) .0475 D) .4775 Answer: A Explanation: Use the General Rule of Addition: P(V or S) = 72/300 + 100/300 − 19/300. Difficulty: 3 Hard Topic: 05.05 Contingency Tables Learning Objective: 05-06 Apply the concepts of probability to contingency tables. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

102) Given the contingency table shown here, find P(V | S). Vehicle Type Car (C) Minivan (M) Full-Size Van (F) SUV (V) Truck (T) Col Total

Somerset (S) 44 21 2 19 14 100

Oakland (O) 49 15 3 27 6 100

Great Lakes (G) 36 18 3 26 17 100

Row Total 129 54 8 72 37 300

A) .2639 B) .1900 C) .0475 D) .4144 Answer: B Explanation: This is a conditional probability: P(V | S) = 19/100. Difficulty: 2 Medium Topic: 05.05 Contingency Tables Learning Objective: 05-06 Apply the concepts of probability to contingency tables. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 103) The manager of Ardmore Pharmacy knows that 25 percent of the customers entering the store buy prescription drugs, 65 percent buy over-the-counter drugs, and 18 percent buy both types of drugs. What is the probability that a randomly selected customer will buy at least one of these two types of drugs? A) .90 B) .85 C) .72 D) .65 Answer: C Explanation: Use the General Rule of Addition: P(A or B) = P(A) + P(B) − P(A ∩ B) = .25 + . 65 − .18. Difficulty: 2 Medium Topic: 05.03 Rules of Probability Learning Objective: 05-03 Apply the definitions and rules of probability. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

104) Two events are complementary (i.e., they are complements) if A) the sum of their probabilities equals one. B) they are disjoint and their probabilities sum to one. C) the joint probability of the two events equals one. D) they are independent events with equal probabilities. Answer: B Explanation: Review the rules of probability. Difficulty: 2 Medium Topic: 05.03 Rules of Probability Learning Objective: 05-03 Apply the definitions and rules of probability. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 105) Which statement is false? A) If P(A) = .05, then the odds against event A's occurrence are 19 to 1. B) If A and B are mutually exclusive events, then P(A or B) = 0. C) The number of permutations of five things taken two at a time is 20. Answer: B Explanation: Review rules of probability and counting rules. Difficulty: 3 Hard Topic: 05.03 Rules of Probability Learning Objective: 05-03 Apply the definitions and rules of probability. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 106) The number of unique orders in which five items (A, B, C, D, E) can be arranged is A) 5 B) 840 C) 120 D) 24 Answer: C Explanation: Apply rules of counting: 5 × 4 × 3 × 2 × 1 = 120. Difficulty: 3 Hard Topic: 05.08 Counting Rules Learning Objective: 05-09 Apply counting rules to calculate possible event arrangements. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

107) If four items are chosen at random without replacement from seven items, in how many ways can the four items be arranged, treating each arrangement as a different event (i.e., if order is important)? A) 35 B) 840 C) 5040 D) 24 Answer: B Explanation: This is 7P4. Difficulty: 3 Hard Topic: 05.08 Counting Rules Learning Objective: 05-09 Apply counting rules to calculate possible event arrangements. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 108) How many ways can we choose three items at random without replacement from five items (A, B, C, D, E) if the order of the selected items is not important? A) 60 B) 120 C) 10 D) 24 Answer: C Explanation: This is 5C3. Difficulty: 3 Hard Topic: 05.08 Counting Rules Learning Objective: 05-09 Apply counting rules to calculate possible event arrangements. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

109) The value of 6C2 is A) 15 B) 30 C) 720 D) 12 Answer: A Explanation: Apply the formula for combinations. Difficulty: 2 Medium Topic: 05.08 Counting Rules Learning Objective: 05-09 Apply counting rules to calculate possible event arrangements. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 110) The value of 4P2 is A) 8 B) 6 C) 24 D) 12 Answer: D Explanation: Apply the formula for permutations. Difficulty: 2 Medium Topic: 05.08 Counting Rules Learning Objective: 05-09 Apply counting rules to calculate possible event arrangements. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 111) The probability that event A occurs, given that event B has occurred, is an example of: A) a marginal probability. B) a conditional probability. C) a joint probability. D) more than one of the above. Answer: B Explanation: Review the definition of conditional probability P(A | B). Difficulty: 1 Easy Topic: 05.03 Rules of Probability Learning Objective: 05-03 Apply the definitions and rules of probability. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation

112) If each of two independent file servers has a reliability of 93 percent and either alone can run the website, then the overall website availability is A) .9951 B) .8649 C) .9300 D) .9522 Answer: A Explanation: Failure probability for either server is 1 − .93 = .07. We are told that failures are independent events, so we can multiply their probabilities. The probability that both will fail is P(F1 and F2) = P(F1) P(F2) = (.07)(.07) = .0049, so the probability that one or the other will operate is 1 − .0049 = .9951 Difficulty: 2 Medium Topic: 05.04 Independent Events Learning Objective: 05-05 Determine when events are independent. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 113) In a certain city, 5 percent of all drivers have expired licenses, 10 percent have an unpaid parking ticket, and 1 percent have both an expired license and an unpaid parking ticket. Are these events independent? A) No B) Yes C) Can't tell from given information Answer: A Explanation: For independence we would require P(A)P(B) = P(A ∩ B). Difficulty: 2 Medium Topic: 05.04 Independent Events Learning Objective: 05-05 Determine when events are independent. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

114) In a certain city, 5 percent of all drivers have expired licenses and 10 percent have an unpaid parking ticket. If these events are independent, what is the probability that a driver has both an expired license and an unpaid parking ticket? A) .010 B) .005 C) .001 D) Cannot be determined Answer: B Explanation: Because they are independent events then P(A ∩ B) = P(A)P(B) = (.05)(.10). Difficulty: 2 Medium Topic: 05.04 Independent Events Learning Objective: 05-05 Determine when events are independent. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 115) If two events are collectively exhaustive, what is the probability that one or the other will occur? A) 1.00 B) .00 C) .50 D) Cannot tell from given information Answer: A Explanation: Review definition of probabilities (collectively exhaustive covers all the possibilities). Difficulty: 2 Medium Topic: 05.03 Rules of Probability Learning Objective: 05-03 Apply the definitions and rules of probability. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

116) Which best exemplifies a subjective probability? A) The probability that a female age 30 will have an accident in a week's car rental at Hertz B) The probability that a pair of dice will come up 7 in a given throw C) The probability that the summer Olympic games will be held in Chicago in 2020 D) The probability that a checked bag on Flight 1872 will weigh more than 40 pounds Answer: C Explanation: Subjective probabilities are not based on empirical frequencies. Difficulty: 2 Medium Topic: 05.02 Probability Learning Objective: 05-02 Distinguish among the three views of probability. Bloom's: Analyze AACSB: Analytical Thinking Accessibility: Keyboard Navigation 117) Which best exemplifies the classical definition of probability? A) The probability that a male age 50 will have an accident in a week's car rental at Alamo B) The probability that a pair of dice will come up 7 when they are rolled C) The probability that the winter Olympic games will be held in Europe in 2022 D) The probability that a checked bag on Flight 1872 will weigh more than 30 pounds Answer: B Explanation: Classical probability is determined a priori by the nature of the experiment. Difficulty: 2 Medium Topic: 05.02 Probability Learning Objective: 05-02 Distinguish among the three views of probability. Bloom's: Analyze AACSB: Analytical Thinking Accessibility: Keyboard Navigation 118) Which best exemplifies the empirical definition of probability? A) The probability that a Chinese athlete will win the diving competition in the next Olympics B) The probability that a fair coin will come up heads when it is flipped C) The probability that your own bank will become insolvent within 12 months D) The probability that a checked bag on Flight 1872 will weigh less than 30 pounds Answer: D Explanation: Empirical probabilities are based on observed frequencies. Difficulty: 2 Medium Topic: 05.02 Probability Learning Objective: 05-02 Distinguish among the three views of probability. Bloom's: Analyze AACSB: Analytical Thinking Accessibility: Keyboard Navigation

119) From the following tree, find the probability that a randomly chosen person will get the flu vaccine and will also get the flu.

A) .10 B) .07 C) .19 D) .70 Answer: B Explanation: Multiply down the branch: .70 × .10 = .07. Difficulty: 2 Medium Topic: 05.06 Tree Diagrams Learning Objective: 05-07 Interpret a tree diagram. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

120) From the following tree, find the probability that a randomly chosen person will not get a vaccination and will not get the flu.

A) .18 B) .60 C) .19 D) .70 Answer: A Explanation: Multiply down the branch: .30 × .60 = .18. Difficulty: 2 Medium Topic: 05.06 Tree Diagrams Learning Objective: 05-07 Interpret a tree diagram. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

121) From the following tree, find the probability that a randomly chosen person will get the flu.

A) .19 B) .07 C) .81 D) .70 Answer: A Explanation: Multiply down two branches and add .07 to .12. That is (.70)(.10) + (.30)(.40). Difficulty: 2 Medium Topic: 05.06 Tree Diagrams Learning Objective: 05-07 Interpret a tree diagram. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 122) At Joe's Restaurant, 80 percent of the diners are new customers (N), while 20 percent are returning customers (R). Fifty percent of the new customers pay by credit card, compared with 70 percent of the regular customers. If a customer pays by credit card, what is the probability that the customer is a new customer? A) .7407 B) .8000 C) .5400 D) .5000 Answer: A Explanation: Review Bayes' Theorem, and perhaps make a table or tree. Difficulty: 3 Hard Topic: 05.07 Bayes' Theorem Learning Objective: 05-08 Use Bayes' Theorem to calculate revised probabilities. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 57 Copyright 2019 © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.

123) At Dolon General Hospital, 30 percent of the patients have Medicare insurance (M) while 70 percent do not have Medicare insurance (M´). Twenty percent of the Medicare patients arrive by ambulance, compared with 10 percent of the non-Medicare patients. If a patient arrives by ambulance, what is the probability that the patient has Medicare insurance? A) .7000 B) .5000 C) .4615 D) .1300 Answer: C Explanation: Review Bayes' Theorem, and perhaps make a table or tree. Difficulty: 3 Hard Topic: 05.07 Bayes' Theorem Learning Objective: 05-08 Use Bayes' Theorem to calculate revised probabilities. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation

Applied Statistics in Business and Economics, 6e (Doane) Chapter 6 Discrete Probability Distributions 1) A random variable is a function or rule that assigns a numerical value to each outcome in the sample space of a stochastic (chance) experiment. Answer: TRUE Explanation: Review the definition of discrete random variable. Difficulty: 1 Easy Topic: 06.01 Discrete Probability Distributions Learning Objective: 06-01 Define a discrete random variable and its probability distribution. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 2) A discrete random variable has a countable number of distinct values. Answer: TRUE Explanation: Review the definition of discrete random variable. But "countable" does not necessarily imply that we know the upper limit (e.g., number of computer virus attacks per year). Difficulty: 1 Easy Topic: 06.01 Discrete Probability Distributions Learning Objective: 06-01 Define a discrete random variable and its probability distribution. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 3) The expected value of a discrete random variable E(X) is the sum of all X values weighted by their respective probabilities. Answer: TRUE Explanation: Review the definition of expected value. The mean is a weighted average. Difficulty: 1 Easy Topic: 06.02 Expected Value and Variance Learning Objective: 06-02 Solve problems using expected value and variance. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation

4) A discrete distribution can be described by its probability density function (PDF) or by its cumulative distribution function (CDF). Answer: TRUE Explanation: Review the definition of PDF (point probability) and CDF (cumulative sum of probabilities). Difficulty: 1 Easy Topic: 06.01 Discrete Probability Distributions Learning Objective: 06-01 Define a discrete random variable and its probability distribution. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 5) A random variable may be discrete or continuous, but not both. Answer: TRUE Explanation: Review the definition of discrete and continuous. Discrete implies enumerable. Difficulty: 2 Medium Topic: 06.01 Discrete Probability Distributions Learning Objective: 06-01 Define a discrete random variable and its probability distribution. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 6) To describe the number of blemishes per sheet of white bond paper, we would use a discrete uniform distribution. Answer: FALSE Explanation: Not all X values would be equally likely and we have no upper limit (Poisson distribution would be better). Difficulty: 2 Medium Topic: 06.05 Poisson Distribution Learning Objective: 06-05 Find Poisson probabilities using tables, formulas, or Excel. Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation 7) The outcomes for the sum of two dice can be described as a discrete uniform distribution. Answer: FALSE Explanation: The sum of two dice follows a triangular distribution, as was shown in Chapter 5. Difficulty: 2 Medium Topic: 06.01 Discrete Probability Distributions Learning Objective: 06-01 Define a discrete random variable and its probability distribution. Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation 2 Copyright 2019 © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.

8) A discrete binomial distribution is skewed right when π > .50. Answer: FALSE Explanation: Most outcomes would be on the right, so a longer left tail exists. Difficulty: 2 Medium Topic: 06.04 Binomial Distribution Learning Objective: 06-04 Find binomial probabilities using tables, formulas, or Excel. Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation 9) When π = .70 the discrete binomial distribution is negatively skewed. Answer: TRUE Explanation: Most outcomes would be on the right, so a longer left tail exists. Difficulty: 2 Medium Topic: 06.04 Binomial Distribution Learning Objective: 06-04 Find binomial probabilities using tables, formulas, or Excel. Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation 10) The Poisson distribution describes the number of occurrences within a randomly chosen unit of time or space. Answer: TRUE Explanation: The Poisson distribution describes events per unit of time. Difficulty: 1 Easy Topic: 06.05 Poisson Distribution Learning Objective: 06-05 Find Poisson probabilities using tables, formulas, or Excel. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 11) The Poisson distribution can be skewed either left or right, depending on λ. Answer: FALSE Explanation: The Poisson distribution is always right-skewed. Difficulty: 2 Medium Topic: 06.05 Poisson Distribution Learning Objective: 06-05 Find Poisson probabilities using tables, formulas, or Excel. Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation

12) Although the shape of the Poisson distribution is positively skewed, it becomes more nearly symmetric as its mean becomes larger. Answer: TRUE Explanation: Although always right-skewed, the Poisson approaches a normal as the mean increases. Difficulty: 2 Medium Topic: 06.05 Poisson Distribution Learning Objective: 06-05 Find Poisson probabilities using tables, formulas, or Excel. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 13) As a rule of thumb, the Poisson distribution can be used to approximate a binomial distribution when n ≥ 20 and π ≤ .05. Answer: TRUE Explanation: The Poisson is a better approximation to a binomial when n is large and π is small. Difficulty: 2 Medium Topic: 06.05 Poisson Distribution Learning Objective: 06-06 Use the Poisson approximation to the binomial (optional). Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 14) The hypergeometric distribution is skewed right. Answer: FALSE Explanation: The hypergeometric is skewed right if s/N < .50 (and conversely). Difficulty: 2 Medium Topic: 06.06 Hypergeometric Distribution Learning Objective: 06-07 Find hypergeometric probabilities using Excel. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation

15) The hypergeometric distribution assumes that the probability of a success remains the same from one trial to the next. Answer: FALSE Explanation: In the hypergeometric, π is not constant because we are sampling without replacement. Difficulty: 1 Easy Topic: 06.06 Hypergeometric Distribution Learning Objective: 06-07 Find hypergeometric probabilities using Excel. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 16) The hypergeometric distribution is not applicable if sampling is done with replacement. Answer: TRUE Explanation: The hypergeometric is used when there is no replacement in sampling from a finite population. Difficulty: 1 Easy Topic: 06.06 Hypergeometric Distribution Learning Objective: 06-07 Find hypergeometric probabilities using Excel. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 17) As a rule of thumb, the binomial distribution can be used to approximate the hypergeometric distribution whenever the population is at least 20 times as large as the sample. Answer: TRUE Explanation: It is safe to use the binomial-hypergeometric approximation if n/N < .05. Difficulty: 1 Easy Topic: 06.06 Hypergeometric Distribution Learning Objective: 06-08 Use the binomial approximation to the hypergeometric (optional). Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation

18) An example of a geometric random variable is the number of pine trees with pine beetle infestation in a random sample of 15 pine trees in Colorado. Answer: FALSE Explanation: This is a binomial experiment, assuming π is constant. Difficulty: 2 Medium Topic: 06.07 Geometric Distribution (Optional) Learning Objective: 06-09 Calculate geometric probabilities (optional). Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation 19) Calculating the probability of getting three aces in a hand of five cards dealt from a deck of 52 cards would require the use of a hypergeometric distribution. Answer: TRUE Explanation: This is a hypergeometric experiment (sampling without replacement). Difficulty: 2 Medium Topic: 06.06 Hypergeometric Distribution Learning Objective: 06-07 Find hypergeometric probabilities using Excel. Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation 20) The Poisson distribution is appropriate to describe the number of babies born in a small hospital on a given day. Answer: TRUE Explanation: Events per unit of time with no clear upper limit suggests a Poisson event. Difficulty: 2 Medium Topic: 06.05 Poisson Distribution Learning Objective: 06-05 Find Poisson probabilities using tables, formulas, or Excel. Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation 21) The gender (M, F) of a randomly chosen unborn child is a Bernoulli event. Answer: TRUE Explanation: Bernoulli events have two outcomes (0 or 1). Difficulty: 2 Medium Topic: 06.04 Binomial Distribution Learning Objective: 06-04 Find binomial probabilities using tables, formulas, or Excel. Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation 6 Copyright 2019 © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.

22) The Poisson distribution has only one parameter. Answer: TRUE Explanation: The one Poisson parameter is its mean. Difficulty: 1 Easy Topic: 06.05 Poisson Distribution Learning Objective: 06-05 Find Poisson probabilities using tables, formulas, or Excel. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 23) The standard deviation of a Poisson random variable is the square root of its mean. Answer: TRUE Explanation: Yes, because the mean and variance of a Poisson are the same. Difficulty: 1 Easy Topic: 06.05 Poisson Distribution Learning Objective: 06-05 Find Poisson probabilities using tables, formulas, or Excel. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 24) Customer arrivals per unit of time would tend to follow a binomial distribution. Answer: FALSE Explanation: This would be a Poisson event (arrivals per unit of time). Difficulty: 1 Easy Topic: 06.05 Poisson Distribution Learning Objective: 06-05 Find Poisson probabilities using tables, formulas, or Excel. Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation 25) The two outcomes (success, failure) in the Bernoulli model are equally likely. Answer: FALSE Explanation: No, the probability of success need not be .50. Difficulty: 2 Medium Topic: 06.04 Binomial Distribution Learning Objective: 06-04 Find binomial probabilities using tables, formulas, or Excel. Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation

26) The expected value of a random variable is its mean. Answer: TRUE Explanation: The mean is another name for expected value. Difficulty: 2 Medium Topic: 06.02 Expected Value and Variance Learning Objective: 06-02 Solve problems using expected value and variance. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 27) A discrete probability distribution A) is a listing of all possible values of the random variable. B) assigns a probability to each possible value of the random variable. C) can assume values between −1 and +1. D) is independent of the parameters of the distribution. Answer: B Explanation: A discrete PDF assigns a probability to each X value and they must sum to 1. The PDF depends on its parameter(s). Difficulty: 2 Medium Topic: 06.01 Discrete Probability Distributions Learning Objective: 06-01 Define a discrete random variable and its probability distribution. Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation 28) The number of male babies in a sample of 10 randomly chosen babies is a A) continuous random variable. B) Poisson random variable. C) binary random variable. D) binomial random variable. Answer: D Explanation: Here, we are counting "successes" with a constant probability of success in n Bernoulli trials. Difficulty: 1 Easy Topic: 06.04 Binomial Distribution Learning Objective: 06-04 Find binomial probabilities using tables, formulas, or Excel. Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation

29) A discrete random variable A) can be treated as continuous when it has a large range of values. B) cannot be treated as continuous. C) is best avoided if at all possible. D) is usually uniformly distributed. Answer: A Explanation: For example, the Sunday vehicle count on a freeway is a discrete (but large) number. Difficulty: 2 Medium Topic: 06.01 Discrete Probability Distributions Learning Objective: 06-01 Define a discrete random variable and its probability distribution. Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation 30) Which is not a discrete random variable? A) The number of defects in a 4 × 8 sheet of plywood B) The number of female passengers who board a plane C) The time until failure of a vehicle headlamp D) The number of correct answers on a statistics exam Answer: C Explanation: Time is continuous. Difficulty: 1 Easy Topic: 06.01 Discrete Probability Distributions Learning Objective: 06-01 Define a discrete random variable and its probability distribution. Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation 31) Which is a not a discrete random variable? A) The number of births in a hospital on a given day B) The number of fives obtained in four rolls of a die C) The hourly earnings of a call center employee in Boston D) The number of applicants applying for a civil service job Answer: C Explanation: Someone's earnings would be more like a continuous measurement (not an integer). Difficulty: 1 Easy Topic: 06.01 Discrete Probability Distributions Learning Objective: 06-01 Define a discrete random variable and its probability distribution. Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation 9 Copyright 2019 © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.

32) Which statement is incorrect? A) The Poisson distribution is always skewed right. B) The binomial distribution may be skewed left or right. C) The discrete uniform distribution is always symmetric. D) The hypergeometric distribution is symmetric. Answer: D Explanation: A hypergeometric distribution is symmetric only if s/N = .50. Difficulty: 2 Medium Topic: 06.06 Hypergeometric Distribution Learning Objective: 06-07 Find hypergeometric probabilities using Excel. Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation 33) The random variable X is the number of shots it takes before you make the first free throw in basketball. Assuming the probability of success (making a free throw) is constant from trial to trial, what type of distribution does X follow? A) Binomial B) Poisson C) Hypergeometric D) Geometric Answer: D Explanation: The geometric model describes the number of trials until the first success. Difficulty: 2 Medium Topic: 06.07 Geometric Distribution (Optional) Learning Objective: 06-09 Calculate geometric probabilities (optional). Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

34) Which probability model is most nearly appropriate to describe the number of burned-out fluorescent tubes in a classroom with 12 fluorescent tubes, assuming a constant probability of a burned-out tube? A) Binomial B) Poisson C) Hypergeometric D) Geometric Answer: A Explanation: n = 12 Bernoulli trials with fixed probability of success would be a binomial model. Difficulty: 2 Medium Topic: 06.04 Binomial Distribution Learning Objective: 06-04 Find binomial probabilities using tables, formulas, or Excel. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 35) Which distribution is most nearly appropriate to describe the number of fatalities in Texas in a given year due to poisonous snakebites? A) Binomial B) Poisson C) Hypergeometric D) Geometric Answer: B Explanation: Events per unit of time with no clear upper limit would resemble a Poisson distribution. Difficulty: 2 Medium Topic: 06.05 Poisson Distribution Learning Objective: 06-05 Find Poisson probabilities using tables, formulas, or Excel. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

36) Which model would you use to describe the probability that a call-center operator will make the first sale on the third call, assuming a constant probability of making a sale? A) Binomial B) Poisson C) Hypergeometric D) Geometric Answer: D Explanation: Geometric describes the number of trials to first success. Difficulty: 2 Medium Topic: 06.07 Geometric Distribution (Optional) Learning Objective: 06-09 Calculate geometric probabilities (optional). Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 37) In a randomly chosen week, which probability model would you use to describe the number of accidents at the intersection of two streets? A) Binomial B) Poisson C) Hypergeometric D) Geometric Answer: B Explanation: Events per unit of time with no clear upper limit would resemble a Poisson distribution. Difficulty: 2 Medium Topic: 06.05 Poisson Distribution Learning Objective: 06-05 Find Poisson probabilities using tables, formulas, or Excel. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

38) Which model best describes the number of nonworking web URLs ("This page cannot be displayed") you encounter in a randomly chosen minute while surfing websites for Florida vacation rental condos? A) Binomial B) Poisson C) Hypergeometric D) Geometric Answer: B Explanation: Events per unit of time with no clear upper limit would resemble a Poisson distribution. Difficulty: 2 Medium Topic: 06.05 Poisson Distribution Learning Objective: 06-05 Find Poisson probabilities using tables, formulas, or Excel. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 39) Which probability model would you use to describe the number of damaged printers in a random sample of 4 printers taken from a shipment of 28 printers that contains 3 damaged printers? A) Poisson B) Hypergeometric C) Binomial D) Uniform Answer: B Explanation: Sampling (n = 4 printers) without replacement with known number of "successes" (s = 3 damaged printers) in the population (N = 28 printers). Difficulty: 2 Medium Topic: 06.06 Hypergeometric Distribution Learning Objective: 06-07 Find hypergeometric probabilities using Excel. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

40) Which model best describes the number of incorrect fare quotations by a well-trained airline ticket agent between 2 p.m. and 3 p.m. on a particular Thursday. A) Binomial B) Poisson C) Hypergeometric D) Geometric Answer: B Explanation: Events per unit of time with no clear upper limit would resemble a Poisson distribution. Difficulty: 2 Medium Topic: 06.05 Poisson Distribution Learning Objective: 06-05 Find Poisson probabilities using tables, formulas, or Excel. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 41) Which model best describes the number of blemishes per sheet of white bond paper? A) Binomial B) Poisson C) Hypergeometric D) Geometric Answer: B Explanation: Events per unit of area with no clear upper limit would resemble a Poisson distribution. Difficulty: 2 Medium Topic: 06.05 Poisson Distribution Learning Objective: 06-05 Find Poisson probabilities using tables, formulas, or Excel. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

42) To ensure quality, customer calls for airline fare quotations are monitored at random. On a particular Thursday afternoon, ticket agent Bob gives 40 fare quotations, of which 4 are incorrect. In a random sample of 8 of these customer calls, which model best describes the number of incorrect quotations Bob will make? A) Binomial B) Poisson C) Hypergeometric D) Geometric Answer: C Explanation: Sampling (n = 8 calls selected) without replacement with known number of "successes" (s = 4 incorrect quotes) in the population (N = 40 quotes). Difficulty: 2 Medium Topic: 06.06 Hypergeometric Distribution Learning Objective: 06-07 Find hypergeometric probabilities using Excel. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 43) The number of people injured in rafting expeditions on the Colorado River on a randomly chosen Thursday in August is best described by which model? A) Binomial B) Poisson C) Hypergeometric D) Geometric Answer: B Explanation: Independent events per unit of time with no clear upper limit would be Poisson. Difficulty: 1 Easy Topic: 06.05 Poisson Distribution Learning Objective: 06-05 Find Poisson probabilities using tables, formulas, or Excel. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

44) On a particular Thursday in August, 40 Grand Canyon tourists enter a drawing for a free mule ride. Ten of the entrants are European tourists. Five entrants are selected at random to get the free mule ride. Which model best describes the number of European tourists in the random sample? A) Binomial B) Poisson C) Hypergeometric D) Geometric Answer: C Explanation: Sampling (n = 5 tourists selected) without replacement with known number of "successes" (s = 10 Europeans) in the population (N = 40). Difficulty: 2 Medium Topic: 06.06 Hypergeometric Distribution Learning Objective: 06-07 Find hypergeometric probabilities using Excel. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 45) Which model best describes the number of births in a hospital until the first twins are delivered? A) Binomial B) Poisson C) Hypergeometric D) Geometric Answer: D Explanation: The geometric distribution describes the number of trials until the first success. Difficulty: 2 Medium Topic: 06.07 Geometric Distribution (Optional) Learning Objective: 06-09 Calculate geometric probabilities (optional). Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

46) On a randomly chosen Wednesday, which probability model would you use to describe the number of convenience store robberies in Los Angeles? A) Binomial B) Poisson C) Hypergeometric D) Geometric Answer: B Explanation: Events per unit of time with no clear upper limit would be Poisson. Difficulty: 2 Medium Topic: 06.05 Poisson Distribution Learning Objective: 06-05 Find Poisson probabilities using tables, formulas, or Excel. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 47) Which probability model would you use to describe the number of customers served at a certain California Pizza Kitchen until the first customer orders split pea soup? A) Binomial B) Geometric C) Uniform D) Poisson Answer: B Explanation: The geometric distribution describes the number of trials until the first success. Difficulty: 2 Medium Topic: 06.07 Geometric Distribution (Optional) Learning Objective: 06-09 Calculate geometric probabilities (optional). Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 48) Which distribution has a mean of 5? A) Poisson with λ = 25. B) Binomial with n = 200, π = .05 C) Hypergeometric with N = 100, n = 10, s = 50 Answer: C Explanation: Review model parameters. The hypergeometric mean is ns/N = (10)(50)/100 = 5. Difficulty: 2 Medium Topic: 06.06 Hypergeometric Distribution Learning Objective: 06-07 Find hypergeometric probabilities using Excel. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 18 Copyright 2019 © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.

49) Of the following, the one that most resembles a Poisson random variable is the number of A) heads in 200 flips of a fair coin. B) annual power failures at your residence. C) face cards in a bridge hand of 13 cards. D) defective CDs in a spool containing 15 CDs. Answer: B Explanation: Independent arrivals per unit of time with no clear upper limit would be Poisson. Difficulty: 1 Easy Topic: 06.05 Poisson Distribution Learning Objective: 06-05 Find Poisson probabilities using tables, formulas, or Excel. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 50) A charity raffle prize is $1,000. The charity sells 4,000 raffle tickets. One winner will be selected at random. At what ticket price would a ticket buyer expect to break even? A) $0.50 B) $0.25 C) $0.75 D) $1.00 Answer: B Explanation: Expected winning is (1/4000) × $1000 = $0.25. Difficulty: 3 Hard Topic: 06.02 Expected Value and Variance Learning Objective: 06-02 Solve problems using expected value and variance. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 51) A die is rolled. If it rolls a 1, 2, or 3, you win $2. If it rolls to a 4, 5, or 6, you lose $1. Calculate the expected winnings. A) $0.50 B) $3.00 C) $1.50 D) $1.00 Answer: A Explanation: E(X) = (3/6) × $2 + (3/6) × (−$1) = $0.50. Difficulty: 2 Medium Topic: 06.02 Expected Value and Variance Learning Objective: 06-02 Solve problems using expected value and variance. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 19 Copyright 2019 © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.

52) A fair die is rolled. If it comes up 1 or 2 you win $2. If it comes up 3, 4, 5, or 6, you lose $1. Calculate the expected winnings. A) $0.00 B) $1.00 C) $0.50 D) $0.25 Answer: A Explanation: E(X) = (2/6) × $2 + (4/6) × (−$1) = $0.6667 − $0.6667 = 0. Difficulty: 2 Medium Topic: 06.02 Expected Value and Variance Learning Objective: 06-02 Solve problems using expected value and variance. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 53) A carnival has a game of chance: a fair coin is tossed. If it lands heads you win $1.00, and if it lands tails you lose $0.50. How much should a ticket to play this game cost if the carnival wants to break even? A) $0.25 B) $0.50 C) $0.75 D) $1.00 Answer: A Explanation: E(X) = (.5) × $1 + (.5) × (−$0.50) = $0.50 − $0.25 = $0.25. Difficulty: 3 Hard Topic: 06.02 Expected Value and Variance Learning Objective: 06-02 Solve problems using expected value and variance. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

54) Ephemeral Services Corporation (ESCO) knows that nine other companies besides ESCO are bidding for a $900,000 government contract. Each company has an equal chance of being awarded the contract. If ESCO has already spent $100,000 in developing its bidding proposal, what is its expected net profit? A) $100,000 B) $90,000 C) −$10,000 D) $0 Answer: D Explanation: E(X) = (1/9) × $900,000 = $100,000. ESCO only can expect to cover its sunk cost (no profit). Difficulty: 3 Hard Topic: 06.02 Expected Value and Variance Learning Objective: 06-02 Solve problems using expected value and variance. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 55) The discrete random variable X is the number of students that show up for Professor Smith's office hours on Monday afternoons. The table below shows the probability distribution for X. What is the expected value E(X) for this distribution? X P(X)

0 .40

1 .30

2 .20

3 .10

Total 1.00

A) 1.2 B) 1.0 C) 1.5 D) 2.0 Answer: B Explanation: For each X, multiply X by P(X) and sum the values. Difficulty: 2 Medium Topic: 06.02 Expected Value and Variance Learning Objective: 06-02 Solve problems using expected value and variance. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

56) The discrete random variable X is the number of students that show up for Professor Smith's office hours on Monday afternoons. The table below shows the probability distribution for X. What is the probability that at least 1 student comes to office hours on any given Monday? X P(X)

0 .40

1 .30

2 .20

3 .10

Total 1.00

A) .30 B) .40 C) .50 D) .60 Answer: D Explanation: P(X ≥ 1) = 1 − P(X = 0) = 1 − .40 = .60. Difficulty: 1 Easy Topic: 06.01 Discrete Probability Distributions Learning Objective: 06-01 Define a discrete random variable and its probability distribution. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 57) The discrete random variable X is the number of students that show up for Professor Smith's office hours on Monday afternoons. The table below shows the probability distribution for X. What is the probability that fewer than 2 students come to office hours on any given Monday? X P(X)

0 .40

1 .30

2 .20

3 .10

Total 1.00

A) .10 B) .40 C) .70 D) .90 Answer: C Explanation: P(X < 2) = P(X = 0) + P(X = 1) = .40 + .30 = .70. Difficulty: 1 Easy Topic: 06.01 Discrete Probability Distributions Learning Objective: 06-01 Define a discrete random variable and its probability distribution. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

58) The discrete random variable X is the number of passengers waiting at a bus stop. The table below shows the probability distribution for X. What is the expected value E(X) for this distribution? X P(X)

0 .40

1 .30

2 .20

3 .10

Total 1.00

A) 1.1 B) 1.3 C) 1.7 D) 1.9 Answer: B Explanation: For each X, multiply X times P(X) and sum the values. Difficulty: 2 Medium Topic: 06.02 Expected Value and Variance Learning Objective: 06-02 Solve problems using expected value and variance. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 59) Given the following probability distribution, what is the expected value of the random variable X? X 100 150 200 250 300 Sum

P(X) .10 .20 .30 .30 .10 1.00

A) 175 B) 150 C) 200 D) 205 Answer: D Explanation: For each X, multiply X by P(X) and sum the values. Difficulty: 2 Medium Topic: 06.02 Expected Value and Variance Learning Objective: 06-02 Solve problems using expected value and variance. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 23 Copyright 2019 © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.

60) Which of the following characterizes a Bernoulli process? A) A random experiment that has only two outcomes. B) The probability of "success" varies with each trial. C) Either outcome has the same chance of occurrence. D) The "success" must be a desirable outcome. Answer: A Explanation: Review the characteristics of the Bernoulli (binary) process. The two outcomes need not be equiprobable (e.g., "bumped" on your flight or not) nor must the event of interest be desirable ("bumped" is not desirable). Difficulty: 1 Easy Topic: 06.04 Binomial Distribution Learning Objective: 06-04 Find binomial probabilities using tables, formulas, or Excel. Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation 61) The binomial distribution describes the number of A) trials to obtain the first "success" in a Bernoulli process. B) trials to obtain n "successes" in a Bernoulli process. C) "successes" or "failures" in a Bernoulli process. D) "successes" in n Bernoulli trials. Answer: D Explanation: In a binomial distribution, we count the number of "successes" in n independent binary trials. Difficulty: 1 Easy Topic: 06.04 Binomial Distribution Learning Objective: 06-04 Find binomial probabilities using tables, formulas, or Excel. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation

62) Which of the following is not a requirement of a binomial distribution? A) Constant probability of success B) Only two possible Bernoulli outcomes C) Fixed number of trials D) Equally likely outcomes Answer: D Explanation: In a binomial distribution, we count the number of "successes" in n independent binary trials. Difficulty: 2 Medium Topic: 06.04 Binomial Distribution Learning Objective: 06-04 Find binomial probabilities using tables, formulas, or Excel. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 63) The binomial distribution is symmetrical when A) π = 1 and 1 − π = 0. B) π = ¼ and 1 − π = ¾. C) π = ½ and 1 − π = ½. D) π = 0 and 1 − π = 1. Answer: C Explanation: The binomial distribution is skewed unless π = .50. Difficulty: 2 Medium Topic: 06.04 Binomial Distribution Learning Objective: 06-04 Find binomial probabilities using tables, formulas, or Excel. Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation 64) The variance will reach a maximum in a binomial distribution when A) π = 1 and 1 − π = 0. B) π = ¼ and 1 − π = ¾. C) π = ½ and 1 − π = ½. D) π = 0 and 1 − π = 1. Answer: C Explanation: Review the formula for the binomial distribution standard deviation. Difficulty: 3 Hard Topic: 06.04 Binomial Distribution Learning Objective: 06-04 Find binomial probabilities using tables, formulas, or Excel. Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation

65) Which distribution is most strongly right-skewed? A) Binomial with n = 50, π = .70 B) Binomial with n = 50, π = .90 C) Binomial with n = 50, π = .40 D) Binomial with n = 50, π = .10 Answer: D Explanation: The binomial is right-skewed when π < .50. Difficulty: 2 Medium Topic: 06.04 Binomial Distribution Learning Objective: 06-04 Find binomial probabilities using tables, formulas, or Excel. Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation 66) A random variable is binomially distributed with n = 16 and π = .40. The expected value and standard deviation of the variables are A) 2.00 and 1.24 B) 4.80 and 4.00 C) 6.40 and 1.96 D) 2.00 and 1.20 Answer: C Explanation: Review formulas for the binomial distribution mean and standard deviation. Difficulty: 2 Medium Topic: 06.04 Binomial Distribution Learning Objective: 06-04 Find binomial probabilities using tables, formulas, or Excel. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 67) The expected value (mean) of a binomial variable is 15. The number of trials is 20. The probability of "success" is A) .25 B) .50 C) .75 D) .30 Answer: C Explanation: Set E(X) = nπ = (20)π = 15 and solve for π. Difficulty: 2 Medium Topic: 06.04 Binomial Distribution Learning Objective: 06-04 Find binomial probabilities using tables, formulas, or Excel. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 26 Copyright 2019 © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.

68) If 90 percent of automobiles in Orange County have both headlights working, what is the probability that in a sample of eight automobiles, at least seven will have both headlights working? A) .6174 B) .3826 C) .8131 D) .1869 Answer: C Explanation: Use Appendix A with n = 8 and π = .90 to find P(X ≥ 7) or else use the Excel function =1-BINOM.DIST(6,8,.90,1) = .8131. Difficulty: 2 Medium Topic: 06.04 Binomial Distribution Learning Objective: 06-04 Find binomial probabilities using tables, formulas, or Excel. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 69) In Quebec, 90 percent of the population subscribes to the Roman Catholic religion. In a random sample of eight Quebecois, find the probability that the sample contains at least five Roman Catholics. A) .0050 B) .0331 C) .9950 D) .9619 Answer: C Explanation: Use Appendix A with n = 8 and π = .90 to find P(X ≥ 5) or else use the Excel function =1-BINOM.DIST(4,8,.90,1) = .99498. Difficulty: 2 Medium Topic: 06.04 Binomial Distribution Learning Objective: 06-04 Find binomial probabilities using tables, formulas, or Excel. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

70) Hardluck Harry has a batting average of .200 (i.e., a 20 percent chance of a hit each time he's at bat). Scouts for a rival baseball club secretly observe Harry's performance in 12 random times at bat. What is the probability that Harry will get more than 2 hits? A) .2055 B) .2362 C) .7946 D) .4417 Answer: D Explanation: Use Appendix A with n = 12 and π = .20 to find P(X ≥ 3) or else use the Excel function =1-BINOM.DIST(2,12,.20,1) = .44165. Difficulty: 2 Medium Topic: 06.04 Binomial Distribution Learning Objective: 06-04 Find binomial probabilities using tables, formulas, or Excel. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 71) The probability that a visitor to an animal shelter will adopt a dog is .20. Out of nine visits, what is the probability that at least one dog will be adopted? A) .8658 B) .3020 C) .5639 D) .1342 Answer: A Explanation: Use Appendix A with n = 9 and π = .20 to find P(X ≥ 1) or else use the Excel function =1-BINOM.DIST(0,9,.20,1) = .865778. Difficulty: 2 Medium Topic: 06.04 Binomial Distribution Learning Objective: 06-04 Find binomial probabilities using tables, formulas, or Excel. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

72) Based on experience, 60 percent of the women who request a pregnancy test at a certain clinic are actually pregnant. In a random sample of 12 women, what is the probability that at least 10 are pregnant? A) .0639 B) .1424 C) .0196 D) .0835 Answer: D Explanation: Use Appendix A with n = 12 and π = .60 to find P(X ≥ 10) or else use the Excel function =1-BINOM.DIST(9,12,.60,1) = .08344. Difficulty: 2 Medium Topic: 06.04 Binomial Distribution Learning Objective: 06-04 Find binomial probabilities using tables, formulas, or Excel. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 73) If 5 percent of automobiles in Oakland County have one burned-out headlight, what is the probability that, in a sample of 10 automobiles, none will have a burned-out headlight? A) .5987 B) .3151 C) .0116 D) .1872 Answer: A Explanation: Use Appendix A with n = 10 and π = .05 find P(X = 0) or else use the Excel function =BINOM.DIST(0,10,.05,0) = .59874. Difficulty: 1 Easy Topic: 06.04 Binomial Distribution Learning Objective: 06-04 Find binomial probabilities using tables, formulas, or Excel. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

74) Jankord Jewelers permits the return of their diamond wedding rings, provided the return occurs within two weeks. Typically, 10 percent are returned. If eight rings are sold today, what is the probability that fewer than three will be returned? A) .9950 B) .9619 C) .0331 D) .1488 Answer: B Explanation: Use Appendix A with n = 8 and π = .10 to find P(X < 3) or else use the Excel function =BINOM.DIST(2,8,.1,1) = .96191. Difficulty: 2 Medium Topic: 06.04 Binomial Distribution Learning Objective: 06-04 Find binomial probabilities using tables, formulas, or Excel. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 75) The probability that an Oxnard University student is carrying a backpack is .70. If 10 students are observed at random, what is the probability that fewer than 7 will be carrying backpacks? A) .3504 B) .2001 C) .6177 D) .2668 Answer: A Explanation: Use Appendix A with n = 10 and π = .70 to find P(X < 7) or else use the Excel function =BINOM.DIST(6,10,.7,1) = .35039. Difficulty: 2 Medium Topic: 06.04 Binomial Distribution Learning Objective: 06-04 Find binomial probabilities using tables, formulas, or Excel. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

76) An insurance company is issuing 16 car insurance policies. Suppose the probability for a claim during a year is 15 percent. If the binomial probability distribution is applicable, then the probability that there will be at least two claims during the year is equal to A) .5615 B) .2775 C) .7161 D) .0388 Answer: C Explanation: Use Appendix A with n = 16 and π = .15 to find P(X ≥ 2) or else use the Excel function =1-BINOM.DIST(1,16,.15,1) = .7161. Difficulty: 2 Medium Topic: 06.04 Binomial Distribution Learning Objective: 06-04 Find binomial probabilities using tables, formulas, or Excel. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 77) A random variable X is distributed binomially with n = 8 and π = .70. The standard deviation of the variable X is approximately A) 0.458 B) 2.828 C) 1.680 D) 1.296 Answer: D Explanation: Use the formula for the binomial standard deviation. Difficulty: 1 Easy Topic: 06.04 Binomial Distribution Learning Objective: 06-04 Find binomial probabilities using tables, formulas, or Excel. Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation

78) Suppose X is binomially distributed with n = 12 and π = .20. The probability that X will be less than or equal to 3 is A) .5584 B) .7946 C) .2362 D) .7638 Answer: B Explanation: Use Appendix A with n = 12 and π = .20 to find P(X ≤ 3) or else use the Excel function =BINOM.DIST(3,12,.2,1) = .79457. Difficulty: 2 Medium Topic: 06.04 Binomial Distribution Learning Objective: 06-04 Find binomial probabilities using tables, formulas, or Excel. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 79) Which Excel function would generate a single random X value for a binomial random variable with parameters n = 16 and π = .25? A) =BINOM.DIST(RAND(),16,.25,0) B) =BINOM.DIST(0,16,.25,RAND()) C) =BINOM.INV(16,.25,RAND()) D) =BINOM.INV(0,16,.25,RAND()) Answer: C Explanation: This is the Excel function for the inverse of a binomial. Difficulty: 3 Hard Topic: 06.04 Binomial Distribution Learning Objective: 06-04 Find binomial probabilities using tables, formulas, or Excel. Bloom's: Remember AACSB: Technology Accessibility: Keyboard Navigation

80) A network has three independent file servers, each with 90 percent reliability. The probability that the network will be functioning correctly (at least one server is working) at a given time is A) 99.9 percent. B) 97.2 percent. C) 95.9 percent. D) 72.9 percent. Answer: A Explanation: Use Appendix A with n = 3 and π = .90. Difficulty: 3 Hard Topic: 06.04 Binomial Distribution Learning Objective: 06-04 Find binomial probabilities using tables, formulas, or Excel. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 81) Which statement concerning the binomial distribution is correct? A) Its PDF covers all integer values of X from 0 to n. B) Its PDF is the same as its CDF when π = .50. C) Its CDF shows the probability of each value of X. D) Its CDF is skewed right when π < .50. Answer: A Explanation: Review the definitions of the binomial distribution. The binomial domain is X = 0, 1, . . . , n. Difficulty: 2 Medium Topic: 06.04 Binomial Distribution Learning Objective: 06-04 Find binomial probabilities using tables, formulas, or Excel. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation

82) Historically, 2 percent of the stray dogs in Southfield are unlicensed. On a randomly chosen day, the Southfield city animal control officer picks up seven stray dogs. What is the probability that fewer than two will be unlicensed? A) .8681 B) .9921 C) .3670 D) .0076 Answer: B Explanation: Use Appendix A with n = 7 and π = .02. Difficulty: 2 Medium Topic: 06.04 Binomial Distribution Learning Objective: 06-04 Find binomial probabilities using tables, formulas, or Excel. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 83) The domain of X in a Poisson probability distribution is discrete and can include: A) any real X value. B) any integer X value. C) any nonnegative integer X value. D) any X value except zero. Answer: C Explanation: For a Poisson random variable, X = 0, 1, 2, . . . (no upper limit). Difficulty: 1 Easy Topic: 06.05 Poisson Distribution Learning Objective: 06-05 Find Poisson probabilities using tables, formulas, or Excel. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 84) On Saturday morning, calls arrive at TicketMaster at a rate of 108 calls per hour. What is the probability of fewer than three calls in a randomly chosen minute? A) .1607 B) .8913 C) .2678 D) .7306 Answer: D Explanation: Use Appendix B with λ = 108/60 = 1.8. Difficulty: 3 Hard Topic: 06.05 Poisson Distribution Learning Objective: 06-05 Find Poisson probabilities using tables, formulas, or Excel. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 34 Copyright 2019 © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.

85) On average, a major earthquake (Richter scale 6.0 or above) occurs three times a decade in a certain California county. Find the probability that at least one major earthquake will occur within the next decade. A) .7408 B) .1992 C) .1494 D) .9502 Answer: D Explanation: Use Appendix B with λ = 3.0. Difficulty: 2 Medium Topic: 06.05 Poisson Distribution Learning Objective: 06-05 Find Poisson probabilities using tables, formulas, or Excel. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 86) On average, an IRS auditor discovers 4.7 fraudulent income tax returns per day. On a randomly chosen day, what is the probability that she discovers fewer than two? A) .0518 B) .0427 C) .1005 D) .1523 Answer: A Explanation: Use Appendix B with λ = 4.7. Difficulty: 2 Medium Topic: 06.05 Poisson Distribution Learning Objective: 06-05 Find Poisson probabilities using tables, formulas, or Excel. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

87) On a Sunday in April, dog bite victims arrive at Carver Memorial Hospital at a historical rate of 0.6 victim per day. On a given Sunday in April, what is the probability that exactly two dog bite victims will arrive? A) .0875 B) .0902 C) .0988 D) .0919 Answer: C Explanation: Use Appendix B with λ = 0.6. Difficulty: 2 Medium Topic: 06.05 Poisson Distribution Learning Objective: 06-05 Find Poisson probabilities using tables, formulas, or Excel. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 88) If tubing averages 16 defects per 100 meters, what is the probability of finding exactly 2 defects in a randomly chosen 10-meter piece of tubing? A) .8795 B) .2674 C) .3422 D) .2584 Answer: D Explanation: Use Appendix B with λ = 16/10 = 1.6. Difficulty: 3 Hard Topic: 06.05 Poisson Distribution Learning Objective: 06-05 Find Poisson probabilities using tables, formulas, or Excel. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

89) Cars are arriving at a toll booth at a rate of four per minute. What is the probability that exactly eight cars will arrive in the next two minutes? A) .0349 B) .1396 C) .9666 D) .0005 Answer: B Explanation: Use Appendix B with λ = 4.0. Difficulty: 3 Hard Topic: 06.05 Poisson Distribution Learning Objective: 06-05 Find Poisson probabilities using tables, formulas, or Excel. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 90) Arrival of cars per minute at a toll booth may be characterized by the Poisson distribution if A) the arrivals are independent. B) no more than one arrival can occur in a minute. C) there is only one lane leading to the booth. D) the mean arrival rate is at least 30. Answer: A Explanation: Events per unit of time with no clear upper limit. Difficulty: 1 Easy Topic: 06.05 Poisson Distribution Learning Objective: 06-05 Find Poisson probabilities using tables, formulas, or Excel. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 91) The coefficient of variation for a Poisson distribution with λ = 5 is A) 35.2 percent. B) 58.9 percent. C) 44.7 percent. D) 31.1 percent. Answer: C Explanation: Use the coefficient of variation with standard deviation equal to the square root of the mean. Difficulty: 2 Medium Topic: 06.05 Poisson Distribution Learning Objective: 06-05 Find Poisson probabilities using tables, formulas, or Excel. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 37 Copyright 2019 © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.

92) The coefficient of variation for a Poisson distribution with λ = 4 is A) 35.2 percent. B) 58.9 percent. C) 50.0 percent. D) 26.4 percent. Answer: C Explanation: The Poisson standard deviation is the square root of the mean. Difficulty: 2 Medium Topic: 06.05 Poisson Distribution Learning Objective: 06-05 Find Poisson probabilities using tables, formulas, or Excel. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 93) For which binomial distribution would a Poisson approximation be unacceptable? A) n = 30, π = .02 B) n = 50, π = .03 C) n = 200, π = .10 D) n = 500, π = .01 Answer: C Explanation: We want n ≥ 20 and π ≤ .05. Difficulty: 2 Medium Topic: 06.05 Poisson Distribution Learning Objective: 06-06 Use the Poisson approximation to the binomial (optional). Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 94) For which binomial distribution would a Poisson approximation be acceptable? A) n = 60, π = .08 B) n = 100, π = .15 C) n = 40, π = .03 D) n = 20, π = .20 Answer: C Explanation: We want n ≥ 20 and π ≤ .05 for an acceptable Poisson approximation. Difficulty: 2 Medium Topic: 06.05 Poisson Distribution Learning Objective: 06-06 Use the Poisson approximation to the binomial (optional). Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

95) For which binomial distribution would a Poisson approximation not be acceptable? A) n = 35, π = .07 B) n = 95, π = .01 C) n = 80, π = .02 D) n = 50, π = .03 Answer: A Explanation: We want n ≥ 20 and π ≤ .05 for an acceptable Poisson approximation. Difficulty: 2 Medium Topic: 06.05 Poisson Distribution Learning Objective: 06-06 Use the Poisson approximation to the binomial (optional). Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 96) The true proportion of accounts receivable with some kind of error is .02 for Venal Enterprises. If an auditor randomly samples 200 accounts receivable, what is the approximate Poisson probability that fewer than two will contain errors? A) .1038 B) .0916 C) .1465 D) .0015 Answer: B Explanation: Since n ≥ 20 and π ≤ .05 we can set λ = nπ = (200)(.02) = 4.0 and use Appendix B to find P(X ≤ 1), or else use the Excel cumulative distribution function =POISSON.DIST(1,4.0,1) = .09158. Difficulty: 3 Hard Topic: 06.05 Poisson Distribution Learning Objective: 06-06 Use the Poisson approximation to the binomial (optional). Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

97) The probability that a rental car will be stolen is .0004. If 3500 cars are rented, what is the approximate Poisson probability that 2 or fewer will be stolen? A) .3452 B) .2417 C) .5918 D) .8335 Answer: D Explanation: Since n ≥ 20 and π ≤ .05 we can set λ = nπ = (3500)(.0004) = 1.4 and use Appendix B to find P(X ≤ 2), or else use the Excel cumulative distribution function =POISSON.DIST(2,1.4,1) = .8335. Difficulty: 3 Hard Topic: 06.05 Poisson Distribution Learning Objective: 06-06 Use the Poisson approximation to the binomial (optional). Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 98) The probability that a customer will use a stolen credit card to make a purchase at a certain Target store is .003. If 400 purchases are made in a given day, what is the approximate Poisson probability that 4 or fewer will be with stolen cards? A) .0053 B) .0076 C) .9923 D) .0555 Answer: C Explanation: Since n ≥ 20 and π ≤ .05 we can set λ = nπ = (400)(.003) = 1.2 and use Appendix B, or else use the Excel cumulative distribution function =POISSON.DIST(4,.003*400,1) = . 9923. Difficulty: 3 Hard Topic: 06.05 Poisson Distribution Learning Objective: 06-06 Use the Poisson approximation to the binomial (optional). Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

99) The probability that a ticket holder will miss a flight is .005. If 180 passengers take the flight, what is the approximate Poisson probability that at least 2 will miss the flight? A) .9372 B) .0628 C) .1647 D) .2275 Answer: D Explanation: Since n ≥ 20 and π ≤ .05 we can set λ = nπ = (.005)(180) = 0.9 and use Appendix B to find P(X ≥ 2), or else use the Excel cumulative distribution function =1POISSON.DIST(1,0.9,1) = .2275. Difficulty: 3 Hard Topic: 06.05 Poisson Distribution Learning Objective: 06-06 Use the Poisson approximation to the binomial (optional). Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 100) The probability that a certain daily flight's departure from ORD to LAX is delayed is .02. Over six months, this flight departs 180 times. What is the approximate Poisson probability that it will be delayed fewer than 2 times? A) .4471 B) .3028 C) .1257 D) .1771 Answer: C Explanation: Since n ≥ 20 and π ≤ .05 we can set λ = nπ = (180)(.02) = 3.6 and use Appendix B to find P(X ≤ 1) or else use the Excel cumulative distribution function =POISSON.DIST(1,3.6,1) = .12569. Difficulty: 3 Hard Topic: 06.05 Poisson Distribution Learning Objective: 06-06 Use the Poisson approximation to the binomial (optional). Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

101) If X is a discrete uniform random variable ranging from 0 to 12, find P(X ≥ 10). A) .1126 B) .1666 C) .2308 D) .2500 Answer: C Explanation: 3 out of 13 outcomes (don't forget to count 0 as an outcome). Difficulty: 2 Medium Topic: 06.03 Uniform Distribution Learning Objective: 06-03 Define and apply the uniform discrete model. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 102) If X is a discrete uniform random variable ranging from one to eight, find P(X < 6). A) .6250 B) .5000 C) .7500 D) .3750 Answer: A Explanation: We count five out of eight outcomes that meet this requirement. Difficulty: 2 Medium Topic: 06.03 Uniform Distribution Learning Objective: 06-03 Define and apply the uniform discrete model. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 103) If X is a discrete uniform random variable ranging from one to eight, what is its mean? A) 4.0 B) 4.5 C) 5.0 D) 5.5 Answer: B Explanation: The mean is halfway between the lower and upper limits 1 and 8. Difficulty: 1 Easy Topic: 06.03 Uniform Distribution Learning Objective: 06-03 Define and apply the uniform discrete model. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

104) If X is a discrete uniform random variable ranging from 12 to 24, what is its mean? A) 18.5 B) 16.0 C) 18.0 D) 19.5 Answer: C Explanation: The mean is halfway between the lower and upper limits 12 and 24. Difficulty: 1 Easy Topic: 06.03 Uniform Distribution Learning Objective: 06-03 Define and apply the uniform discrete model. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 105) At Ersatz University, the graduating class of 480 includes 96 guest students from Latvia. A sample of 10 students is selected at random to attend a dinner with the Board of Governors. Use the binomial model to obtain the approximate hypergeometric probability that the sample contains at least three Latvian students. A) .3222 B) .1209 C) .8791 D) .6778 Answer: A Explanation: Since n/N < .05 we can use Appendix A with n = 10 and π = 96/480 = .20 to find P(X ≥ 3). Difficulty: 2 Medium Topic: 06.06 Hypergeometric Distribution Learning Objective: 06-08 Use the binomial approximation to the hypergeometric (optional). Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

106) There are 90 passengers on a commuter flight from SFO to LAX, of whom 27 are traveling on business. In a random sample of five passengers, use the binomial model to find the approximate hypergeometric probability that there is at least one business passenger. A) .3087 B) .1681 C) .3602 D) .8319 Answer: D Explanation: Since n/N < .05 we can use Appendix A with n = 5 and π = 27/90 = .30 to find P(X ≥ 1). Difficulty: 3 Hard Topic: 06.06 Hypergeometric Distribution Learning Objective: 06-07 Find hypergeometric probabilities using Excel. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 107) Use the binomial model to find the approximate hypergeometric probability of at least two damaged flash drives in a sample of five taken from a shipment of 150 that contains 30 damaged flash drives. A) .9421 B) .0579 C) .7373 D) .2627 Answer: D Explanation: Since n/N < .05 we can use Appendix A with n = 5 and π = 30/150 = .20 to find P(X ≥ 2). Difficulty: 2 Medium Topic: 06.06 Hypergeometric Distribution Learning Objective: 06-08 Use the binomial approximation to the hypergeometric (optional). Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

108) On a particular day, 112 of 280 passengers on a particular DTW-LAX flight used the eticket check-in kiosk to obtain boarding passes. In a random sample of eight passengers, use the binomial model to find the approximate hypergeometric probability that four will have used the e-ticket check-in kiosk to obtain boarding passes. A) .2322 B) .8263 C) .2926 D) .5613 Answer: A Explanation: Since n/N < .05 we can use Appendix A with n = 8 and π = 112/280 = .40 to find P(X = 4). Difficulty: 3 Hard Topic: 06.06 Hypergeometric Distribution Learning Objective: 06-08 Use the binomial approximation to the hypergeometric (optional). Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 109) A clinic employs nine physicians. Five of the physicians are female. Four patients arrive at once. Assuming the doctors are assigned randomly to patients, what is the probability that all of the assigned physicians are female? A) .0397 B) .0295 C) .0808 D) .0533 Answer: A Explanation: You cannot use the binomial approximation because we have sampled more than 5% of the population (n/N = 4/9 = .444) so we use the hypergeometric formula with x = 4, n = 4, s = 5, N = 9, or else use the Excel function =HYPGEOM.DIST(4,4,5,9,0) = .03938. Difficulty: 3 Hard Topic: 06.06 Hypergeometric Distribution Learning Objective: 06-08 Use the binomial approximation to the hypergeometric (optional). Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

110) There is a .02 probability that a customer's Visa charge will be rejected at a certain Target store because the transaction exceeds the customer's credit limit. What is the probability that the first such rejection occurs on the third Visa transaction? A) .0192 B) .0025 C) .0247 D) .0200 Answer: A Explanation: Use the formulas for the geometric PDF (not the CDF) with π = .02 to find P(X = 3) = .02(1 − .02)3−1 = .02(.98)2 = .02(.9604) = .019208. Difficulty: 3 Hard Topic: 06.07 Geometric Distribution (Optional) Learning Objective: 06-09 Calculate geometric probabilities (optional). Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 111) Ten percent of the corporate managers at Axolotl Industries majored in humanities. If you start interviewing Axolotl managers, what is the probability that the first humanities major is the fifth manager that you interview? A) .0656 B) .8561 C) .5904 D) .4095 Answer: A Explanation: Use the formulas for the geometric PDF (not the CDF) with π = .10 to find P(X = 5) = .10(1 − .10)5−1 = .10(.90)4 = .10(.6561) = .06561. Difficulty: 3 Hard Topic: 06.07 Geometric Distribution (Optional) Learning Objective: 06-09 Calculate geometric probabilities (optional). Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

112) Ten percent of the corporate managers at Axolotl Industries majored in humanities. What is the expected number of managers to be interviewed before finding the first one with a humanities major? A) 15 B) 20 C) 10 D) 17 Answer: C Explanation: The geometric mean is 1/π = 1/(.10) = 10. Difficulty: 2 Medium Topic: 06.07 Geometric Distribution (Optional) Learning Objective: 06-09 Calculate geometric probabilities (optional). Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 113) When you send out a resume, the probability of being called for an interview is .20. What is the probability that the first interview occurs on the fourth resume that you send out? A) .4096 B) .1024 C) .2410 D) .0016 Answer: B Explanation: Use the formulas for the geometric PDF (not the CDF) with π = .20 to find P(X = 4) = .20(1 − .20)4−1 = .20(.80)3 = .20(.512) = .1024. Difficulty: 3 Hard Topic: 06.07 Geometric Distribution (Optional) Learning Objective: 06-09 Calculate geometric probabilities (optional). Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

114) When you send out a resume, the probability of being called for an interview is .20. What is the expected number of resumes you send out before you get the first interview? A) 5 B) 7 C) 10 D) 12 Answer: A Explanation: The geometric mean is 1/π = 1/(.20) = 5. Difficulty: 2 Medium Topic: 06.07 Geometric Distribution (Optional) Learning Objective: 06-09 Calculate geometric probabilities (optional). Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 115) When you send out a resume, the probability of being called for an interview is .20. What is the probability that you get your first interview within the first five resumes that you send out? A) .6723 B) .1024 C) .2410 D) .0016 Answer: A Explanation: Use the formulas for the geometric CDF (not the PDF) with π = .20 to find P(X ≤ 5) = 1 − (1 − .20)5 = 1 − (.80)5 = 1 − .32678 = .67232. Difficulty: 3 Hard Topic: 06.07 Geometric Distribution (Optional) Learning Objective: 06-09 Calculate geometric probabilities (optional). Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

116) There is a .02 probability that a customer's Visa charge will be rejected at a certain Target store because the transaction exceeds the customer's credit limit. What is the probability that the first such rejection occurs within the first 20 Visa transactions? A) .1362 B) .4000 C) .3324 D) .4538 Answer: C Explanation: Use the formulas for the geometric CDF (not the PDF) with π = .02 to find P(X ≤ 20) = 1 − (1 − .02)20 = 1 − (.98)20 = 1 − .6676 = .3324. Difficulty: 2 Medium Topic: 06.07 Geometric Distribution (Optional) Learning Objective: 06-09 Calculate geometric probabilities (optional). Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 117) There is a .02 probability that a customer's Visa charge will be rejected at a certain Target store because the transaction exceeds the customer's credit limit. What is the expected number of Visa transactions until the first one is rejected? A) 10 B) 20 C) 50 D) 98 Answer: C Explanation: The geometric mean is 1/π = 1/(.02) = 50. Difficulty: 2 Medium Topic: 06.07 Geometric Distribution (Optional) Learning Objective: 06-09 Calculate geometric probabilities (optional). Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

118) The geometric distribution best describes A) the number of successes in a sample of n trials. B) the number of trials until the first success. C) the number of events in a given unit of time. D) the process of sampling without replacement. Answer: B Explanation: Review the definition of geometric distribution. Difficulty: 2 Medium Topic: 06.07 Geometric Distribution (Optional) Learning Objective: 06-09 Calculate geometric probabilities (optional). Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 119) The CDF for the geometric distribution shows A) the probability of success in a random experiment consisting of n independent trials. B) the probability that the first success will occur within a given number of trials. C) the probability that no success will be obtained in a given Bernoulli trial. D) the probability of more than one success in the first n trials. Answer: B Explanation: Review the definition of geometric distribution. Difficulty: 2 Medium Topic: 06.07 Geometric Distribution (Optional) Learning Objective: 06-09 Calculate geometric probabilities (optional). Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation 120) If the probability of success is .25, what is the probability of obtaining the first success within the first three trials? A) .4218 B) .5781 C) .1406 D) .2228 Answer: B Explanation: Use the formulas for the geometric CDF (not the PDF) with π = .25 to find P(X ≤ 3) = 1 − (1 − .25)3 = 1 − (.75)3 = 1 − .421875 = .578125. Difficulty: 3 Hard Topic: 06.07 Geometric Distribution (Optional) Learning Objective: 06-09 Calculate geometric probabilities (optional). Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 50 Copyright 2019 © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.

121) If the probability of success is .30, what is the probability of obtaining the first success within the first five trials? A) .0024 B) .8319 C) .1681 D) .9976 Answer: B Explanation: Use the formulas for the geometric CDF (not the PDF) with π = .30 to find P(X ≤ 5) = 1 − (1 − .30)5 = 1 − (.70)5 = 1 − .16807 = .83193. Difficulty: 3 Hard Topic: 06.07 Geometric Distribution (Optional) Learning Objective: 06-09 Calculate geometric probabilities (optional). Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 122) A project has three independent stages that must be completed in sequence. The time to complete each stage is a random variable. The expected times to complete the stages are μ1 = 23, μ2 = 11, μ3 = 17. The expected project completion time is A) 51 B) 23 C) 40 D) 32 Answer: A Explanation: The means can be summed because the stages are independent. Difficulty: 1 Easy Topic: 06.08 Transformations of Random Variables (Optional) Learning Objective: 06-10 Apply rules for transformations of random variables (optional). Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

123) A project has 3 independent stages that must be completed in sequence. The time to complete each stage is a random variable. The standard deviations of the completion times for the stages are σ1 = 5, σ2 = 4, σ3 = 6. The standard deviation of the overall project completion time is A) 8.77. B) 15.0. C) 14.2. D) 9.24. Answer: A Explanation: The variances can be summed because the stages are independent (Rule 4). You have to square the standard deviations to get the variances σ12 = 25, σ22 = 16, σ32 = 36; then add them and take the square root of the sum. Be careful—the standard deviations cannot be summed. Difficulty: 3 Hard Topic: 06.08 Transformations of Random Variables (Optional) Learning Objective: 06-10 Apply rules for transformations of random variables (optional). Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 124) A stock portfolio consists of two stocks X and Y. Their daily closing prices are independent random variables with standard deviations σX = 2.51 and σY = 5.22. What is the standard deviation of the sum of the closing prices of these two stocks? A) 33.55 B) 6.48 C) 7.73 D) 5.79 Answer: D Explanation: The variances can be summed because the stages are independent (Rule 4). You have to square the standard deviations to get the variances σX2 = 6.3001 and σY2 = 27.2484; then add them and take the square root of the sum. Be careful—the standard deviations cannot be summed. Difficulty: 2 Medium Topic: 06.08 Transformations of Random Variables (Optional) Learning Objective: 06-10 Apply rules for transformations of random variables (optional). Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

125) A stock portfolio consists of two stocks X and Y. Their daily closing prices are correlated random variables with variances σX2 = 3.51 and σY2= 5.22, and covariance σXY = −1.55. What is the standard deviation of the sum of the closing prices of these two stocks? A) 5.63 B) 7.18 C) 8.73 D) 2.68 Answer: D Explanation: Use the formula for the variance of correlated (nonindependent) events. We sum the variances and covariance, and then take the square root: σX+Y = [σX2 + σY2 + σXY]1/2 = [3.51 + 5.22 − 1.55]1/2 = [7.18]1/2 = 2.67955. Difficulty: 3 Hard Topic: 06.08 Transformations of Random Variables (Optional) Learning Objective: 06-10 Apply rules for transformations of random variables (optional). Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 126) The expected value of a random variable X is 140 and the standard deviation is 14. The standard deviation of the random variable Y = 3X − 10 is A) 42 B) 6.48 C) 14 D) 32 Answer: A Explanation: Use the rule for functions of a random variable (Rule 2) to get σY = 3σX = (3)(14) = 42. The constant −10 merely shifts the distribution and has no effect on the standard deviation. The mean of Y is not requested. Difficulty: 3 Hard Topic: 06.08 Transformations of Random Variables (Optional) Learning Objective: 06-10 Apply rules for transformations of random variables (optional). Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

127) The expected value of a random variable X is 10 and the standard deviation is 2. The standard deviation of the random variable Y = 2X − 10 is A) 2 B) 4 C) −10 D) −6 Answer: B Explanation: Use the rule for functions of a random variable (Rule 2) to get σY = 2σX = (2)(2) = 4. The constant −10 merely shifts the distribution and has no effect on the standard deviation. The mean of Y is not requested. Difficulty: 3 Hard Topic: 06.08 Transformations of Random Variables (Optional) Learning Objective: 06-10 Apply rules for transformations of random variables (optional). Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

Applied Statistics in Business and Economics, 6e (Doane) Chapter 7 Continuous Probability Distributions 1) A continuous uniform distribution is always symmetric. Answer: TRUE Explanation: The PDF is the same height for all X values. Difficulty: 1 Easy Topic: 07.02 Uniform Continuous Distribution Learning Objective: 07-02 Calculate uniform probabilities. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 2) The height and width of a continuous uniform distribution's PDF are the same. Answer: FALSE Explanation: The PDF height must be 1/(b − a) so that the total area is unity. Difficulty: 1 Easy Topic: 07.02 Uniform Continuous Distribution Learning Objective: 07-02 Calculate uniform probabilities. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 3) A continuous uniform distribution U(0,800) will have μ = 400 and σ = 230.94. Answer: TRUE Explanation: Apply the formulas for the uniform distribution mean and standard deviation. The standard deviation is [(800 − 0)2/12]1/2 = 230.94. Difficulty: 2 Medium Topic: 07.02 Uniform Continuous Distribution Learning Objective: 07-02 Calculate uniform probabilities. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

4) A continuous uniform distribution U(100,200) will have the same standard deviation as a continuous uniform distribution U(200,300). Answer: TRUE Explanation: In the standard deviation formula, (b − a)2 is the same for both these examples. Difficulty: 2 Medium Topic: 07.02 Uniform Continuous Distribution Learning Objective: 07-02 Calculate uniform probabilities. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 5) For a continuous uniform distribution U(200,400), the parameters are μ = 300 and σ = 100. Answer: FALSE Explanation: The standard deviation is [(400 − 200)2/12]1/2 = 57.7. Difficulty: 2 Medium Topic: 07.02 Uniform Continuous Distribution Learning Objective: 07-02 Calculate uniform probabilities. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 6) The exponential distribution describes the number of arrivals per unit of time. Answer: FALSE Explanation: Arrivals per unit of time would be Poisson (but waiting time is exponential). Difficulty: 1 Easy Topic: 07.06 Exponential Distribution Learning Objective: 07-07 Find the exponential probability for a given x. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 7) The exponential distribution is always skewed right. Answer: TRUE Explanation: The PDF clearly shows extreme right-skewness. Difficulty: 1 Easy Topic: 07.06 Exponential Distribution Learning Objective: 07-07 Find the exponential probability for a given x. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation

8) If arrivals follow a Poisson distribution, waiting times follow the exponential distribution. Answer: TRUE Explanation: Poisson arrivals (discrete) imply exponential waiting times (continuous). Difficulty: 1 Easy Topic: 07.06 Exponential Distribution Learning Objective: 07-07 Find the exponential probability for a given x. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 9) The triangular distribution is used in "what-if" analysis for business planning. Answer: TRUE Explanation: Simplicity in visualizing planning scenarios is an attraction of the triangular distribution. We only need to specify the low, high, and mode. Difficulty: 1 Easy Topic: 07.07 Triangular Distribution (Optional) Learning Objective: 07-09 Use the triangular distribution for "what-if" analysis (optional). Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 10) The triangular distribution is symmetric. Answer: FALSE Explanation: The triangular distribution is symmetric only if the mode is at the axis midpoint. Difficulty: 1 Easy Topic: 07.07 Triangular Distribution (Optional) Learning Objective: 07-09 Use the triangular distribution for "what-if" analysis (optional). Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 11) The triangular distribution T(0,10,20) is skewed left. Answer: FALSE Explanation: It is left-skewed only if the mode is right of the axis midpoint. Difficulty: 2 Medium Topic: 07.07 Triangular Distribution (Optional) Learning Objective: 07-09 Use the triangular distribution for "what-if" analysis (optional). Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

12) A triangular distribution can be skewed either left or right. Answer: TRUE Explanation: It is left-skewed only the mode is right of the axis midpoint, and vice versa. Difficulty: 1 Easy Topic: 07.07 Triangular Distribution (Optional) Learning Objective: 07-09 Use the triangular distribution for "what-if" analysis (optional). Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 13) For a continuous random variable, the total area beneath the PDF will be greater than zero but less than one. Answer: FALSE Explanation: If it is a PDF, the total area must be 1. Difficulty: 1 Easy Topic: 07.01 Continuous Probability Distributions Learning Objective: 07-01 Define a continuous random variable. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 14) The exponential distribution is continuous and the Poisson distribution is discrete, yet the two distributions are closely related. Answer: TRUE Explanation: Poisson arrivals (discrete) imply exponential waiting times (continuous). Difficulty: 2 Medium Topic: 07.06 Exponential Distribution Learning Objective: 07-07 Find the exponential probability for a given x. Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation 15) The mean, median, and mode of a normal distribution will always be the same. Answer: TRUE Explanation: A normal distribution is perfectly symmetric. Difficulty: 2 Medium Topic: 07.03 Normal Distribution Learning Objective: 07-03 Know the form and parameters of the normal distribution. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation

16) There is a simple formula for normal areas, but we prefer a table for greater accuracy. Answer: FALSE Explanation: We have a formula for the PDF, but there is no exact formula for areas under the curve. Difficulty: 2 Medium Topic: 07.03 Normal Distribution Learning Objective: 07-03 Know the form and parameters of the normal distribution. Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation 17) Normal distributions differ only in their means and variances. Answer: TRUE Explanation: All normal distributions look the same except for scaling. Difficulty: 1 Easy Topic: 07.03 Normal Distribution Learning Objective: 07-03 Know the form and parameters of the normal distribution. Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation 18) Any normal distribution has a mean of 0 and a standard deviation of 1. Answer: FALSE Explanation: Only the standardized normal is N(0,1). Difficulty: 1 Easy Topic: 07.03 Normal Distribution Learning Objective: 07-03 Know the form and parameters of the normal distribution. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 19) We would use a normal distribution to model the waiting time until the next Florida hurricane strike. Answer: FALSE Explanation: Hurricane arrivals might be regarded as Poisson events, so waiting times are exponential. Difficulty: 2 Medium Topic: 07.06 Exponential Distribution Learning Objective: 07-07 Find the exponential probability for a given x. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 5 Copyright 2019 © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.

20) Experience suggests that 4 percent of all college students have had a tonsillectomy. In a sample of 300 college students, we need to find the probability that at least 10 have had a tonsillectomy. It is acceptable to use the normal distribution to estimate this probability. Answer: TRUE Explanation: The quick rule is nπ ≥ 10 and n(1 − π) ≥ 10, which is the case in this example. Difficulty: 2 Medium Topic: 07.05 Normal Approximations Learning Objective: 07-06 Use the normal approximation to a binomial or a Poisson. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 21) The normal is a good approximation to the binomial when n is greater than or equal to 10. Answer: FALSE Explanation: Without knowing π we cannot be sure. The quick rule is nπ ≥ 10 and n(1 − π) ≥ 10. Difficulty: 1 Easy Topic: 07.05 Normal Approximations Learning Objective: 07-06 Use the normal approximation to a binomial or a Poisson. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 22) The true proportion of accounts receivable with some kind of error is 4 percent for Venal Enterprises. If an auditor randomly samples 50 accounts receivable, it is acceptable to use the normal approximation to estimate the probability that fewer than two will contain errors. Answer: FALSE Explanation: The quick rule is nπ ≥ 10 and n(1 − π) ≥ 10, which is not fulfilled in this case. Difficulty: 2 Medium Topic: 07.05 Normal Approximations Learning Objective: 07-06 Use the normal approximation to a binomial or a Poisson. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 23) The normal distribution is a good approximation to the binomial if n ≥ 30. Answer: FALSE Explanation: Without knowing π we cannot be sure. The quick rule is nπ ≥ 10 and n(1 − π) ≥ 10. Difficulty: 1 Easy Topic: 07.05 Normal Approximations Learning Objective: 07-06 Use the normal approximation to a binomial or a Poisson. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 6 Copyright 2019 © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.

24) The normal distribution is a good approximation to the binomial if n = 200 and π = .03. Answer: FALSE Explanation: The quick rule is nπ ≥ 10 and n(1 − π) ≥ 10, which is not fulfilled in this case. Difficulty: 1 Easy Topic: 07.05 Normal Approximations Learning Objective: 07-06 Use the normal approximation to a binomial or a Poisson. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 25) The normal distribution is a good approximation to the binomial if n = 25 and π = .50. Answer: TRUE Explanation: The quick rule is nπ ≥ 10 and n(1 − π) ≥ 10, which is fulfilled in this case. Difficulty: 1 Easy Topic: 07.05 Normal Approximations Learning Objective: 07-06 Use the normal approximation to a binomial or a Poisson. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 26) The exponential distribution can be either right-skewed or left-skewed, depending on λ. Answer: FALSE Explanation: The PDF of the exponential shows that it is always right-skewed. Difficulty: 1 Easy Topic: 07.06 Exponential Distribution Learning Objective: 07-07 Find the exponential probability for a given x. Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation 27) The number of lightning strikes in a day in Miami is a continuous random variable. Answer: FALSE Explanation: The "number of" anything is discrete. Difficulty: 1 Easy Topic: 07.01 Continuous Probability Distributions Learning Objective: 07-01 Define a continuous random variable. Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation

28) The area under a normal curve is 1 only if the distribution is standardized N(0,1). Answer: FALSE Explanation: Any normal distribution has a total area of one under the PDF. Difficulty: 1 Easy Topic: 07.04 Standard Normal Distribution Learning Objective: 07-03 Know the form and parameters of the normal distribution. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 29) The area under an exponential curve can exceed 1 because the distribution is right-skewed. Answer: FALSE Explanation: If it is a PDF, the total area under the PDF is one. Difficulty: 2 Medium Topic: 07.06 Exponential Distribution Learning Objective: 07-07 Find the exponential probability for a given x. Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation 30) A machine dispenses water into a glass. Assuming that the amount of water dispensed follows a continuous uniform distribution from 10 ounces to 16 ounces, the average amount of water dispensed by the machine is A) 12 ounces. B) 13 ounces. C) 14 ounces. D) 16 ounces. Answer: B Explanation: The mean is halfway between the end points of the distribution. Difficulty: 1 Easy Topic: 07.02 Uniform Continuous Distribution Learning Objective: 07-02 Calculate uniform probabilities. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

31) A machine dispenses water into a glass. Assuming that the amount of water dispensed follows a continuous uniform distribution from 10 ounces to 16 ounces, the standard deviation of the amount of water dispensed is about A) 1.73 ounces. B) 3.00 ounces. C) 0.57 ounce. D) 3.51 ounces. Answer: A Explanation: The standard deviation is [(16 − 10)2/12]1/2 = 1.73. Difficulty: 2 Medium Topic: 07.02 Uniform Continuous Distribution Learning Objective: 07-02 Calculate uniform probabilities. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 32) A machine dispenses water into a glass. Assuming that the amount of water dispensed follows a continuous uniform distribution from 10 ounces to 16 ounces, what is the probability that 13 or more ounces will be dispensed in a given glass? A) .1666 B) .3333 C) .5000 D) .6666 Answer: C Explanation: Half the area is above 13. Difficulty: 2 Medium Topic: 07.02 Uniform Continuous Distribution Learning Objective: 07-02 Calculate uniform probabilities. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

33) A random variable X is best described by a continuous uniform distribution from 20 to 45 inclusive. The mean of this distribution is A) 30.5 B) 31.5 C) 32.5 D) 33.5 Answer: C Explanation: The mean is halfway between the end points of the distribution. Difficulty: 1 Easy Topic: 07.02 Uniform Continuous Distribution Learning Objective: 07-02 Calculate uniform probabilities. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 34) A random variable X is best described by a continuous uniform distribution from 20 to 45 inclusive. The standard deviation of this distribution is approximately A) 52.1 B) 32.5 C) 6.85 D) 7.22 Answer: D Explanation: The standard deviation is [(45 − 20)2/12]1/2 = 7.22. Difficulty: 2 Medium Topic: 07.02 Uniform Continuous Distribution Learning Objective: 07-02 Calculate uniform probabilities. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

35) A random variable X is best described by a continuous uniform distribution from 20 to 45 inclusive. What is P(30 ≤ X ≤ 40)? A) .20 B) .40 C) .60 D) .80 Answer: B Explanation: The desired area is 10/25 = .40. Difficulty: 2 Medium Topic: 07.02 Uniform Continuous Distribution Learning Objective: 07-02 Calculate uniform probabilities. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 36) The Excel function =800*RAND() would generate random numbers with standard deviation approximately equal to A) 200 B) 188 C) 231 D) 400 Answer: C Explanation: The standard deviation is [(800 − 0)2/12]1/2 = 230.94. Difficulty: 2 Medium Topic: 07.02 Uniform Continuous Distribution Learning Objective: 07-02 Calculate uniform probabilities. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

37) The Excel function =40*RAND() would generate random numbers with standard deviation approximately equal to A) 13.33 B) 20.00 C) 11.55 D) 19.27 Answer: C Explanation: The standard deviation is [(40 − 0)2/12]1/2 = 11.55. Difficulty: 2 Medium Topic: 07.02 Uniform Continuous Distribution Learning Objective: 07-02 Calculate uniform probabilities. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 38) If arrivals occur at a mean rate of 3.6 events per hour, the exponential probability of waiting more than 0.5 hour for the next arrival is A) .2407 B) .1653 C) .1222 D) .5000 Answer: B Explanation: P(X > .50) = exp(−3.6 × 0.50) = .1653. Difficulty: 2 Medium Topic: 07.06 Exponential Distribution Learning Objective: 07-07 Find the exponential probability for a given x. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

39) If arrivals occur at a mean rate of 3.6 events per hour, the exponential probability of waiting less than 0.5 hour for the next arrival is A) .7122 B) .8105 C) .8347 D) .7809 Answer: C Explanation: P(X < .50) = 1 − exp(−3.6 × 0.50) = 1 − 0.1653 = .8347. Difficulty: 2 Medium Topic: 07.06 Exponential Distribution Learning Objective: 07-07 Find the exponential probability for a given x. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 40) If arrivals occur at a mean rate of 2.6 events per minute, the exponential probability of waiting more than 1.5 minutes for the next arrival is A) .0202 B) .0122 C) .0535 D) .2564 Answer: A Explanation: P(X > 1.5) = exp(−2.6 × 1.50) = .0202. Difficulty: 2 Medium Topic: 07.06 Exponential Distribution Learning Objective: 07-07 Find the exponential probability for a given x. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 41) If arrivals occur at a mean rate of 1.6 events per minute, the exponential probability of waiting less than 1 minute for the next arrival is A) .2019 B) .7104 C) .8812 D) .7981 Answer: D Explanation: (X < 1) = 1 − exp(−1.6 × 1) = 1 − .2019 = .7981. Difficulty: 2 Medium Topic: 07.06 Exponential Distribution Learning Objective: 07-07 Find the exponential probability for a given x. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 13 Copyright 2019 © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.

42) Bob's z-score for the last exam was 1.52 in Prof. Axolotl's class BIO 417, "Life Cycle of the Ornithorhynchus." Bob said, "Oh, good, my score is in the top 10 percent." Assuming a normal distribution of scores, is Bob right? A) Yes B) No C) Must have n to answer. Answer: A Explanation: P(Z < 1.52) = .9357. Difficulty: 2 Medium Topic: 07.04 Standard Normal Distribution Learning Objective: 07-04 Find the normal probability for a given z or x using tables or Excel. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 43) The lengths of brook trout caught in a certain Colorado stream are normally distributed with a mean of 14 inches and a standard deviation of 3 inches. What proportion of brook trout caught will be between 12 and 18 inches in length? A) .6563 B) .6826 C) .2486 D) .4082 Answer: A Explanation: P(12 < X < 18) = P(−.67 < Z < 1.33) = .6568 (from Appendix C) or .6563 using Excel. Difficulty: 2 Medium Topic: 07.04 Standard Normal Distribution Learning Objective: 07-04 Find the normal probability for a given z or x using tables or Excel. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

44) The lengths of brook trout caught in a certain Colorado stream are normally distributed with a mean of 14 inches and a standard deviation of 3 inches. The first quartile for the lengths of brook trout would be A) 16.01 inches. B) 11.00 inches. C) 11.98 inches. D) 10.65 inches. Answer: C Explanation: Using Excel =NORM.INV(.25,14,3) = 11.98, or Q1 = 14 − 0.675(3) = 11.975 using Appendix C. Difficulty: 3 Hard Topic: 07.04 Standard Normal Distribution Learning Objective: 07-05 Solve for z or x for a given normal probability using tables or Excel. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 45) The lengths of brook trout caught in a certain Colorado stream are normally distributed with a mean of 14 inches and a standard deviation of 3 inches. What lower limit should the State Game Commission set on length if it is desired that 80 percent of the catch may be kept by fishers? A) 12.80 inches B) 11.48 inches C) 12.00 inches D) 9.22 inches Answer: B Explanation: Using Excel =NORM.INV(.20,14,3) = 11.475, or X = 14 − 0.84(3) = 11.48 using Appendix C. Difficulty: 3 Hard Topic: 07.04 Standard Normal Distribution Learning Objective: 07-05 Solve for z or x for a given normal probability using tables or Excel. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

46) In Melanie's Styling Salon, the time to complete a simple haircut is normally distributed with a mean of 25 minutes and a standard deviation of 4 minutes. What percentage of customers require less than 32 minutes for a simple haircut? A) 95.99 percent B) 99.45 percent C) 97.72 percent D) 45.99 percent Answer: A Explanation: Using Excel =NORMDIST(32,25,4,1) = 0.9599, or use z = (32 − 25)/4 = 1.75 with Appendix C. Difficulty: 2 Medium Topic: 07.04 Standard Normal Distribution Learning Objective: 07-04 Find the normal probability for a given z or x using tables or Excel. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 47) In Melanie's Styling Salon, the time to complete a simple haircut is normally distributed with a mean of 25 minutes and a standard deviation of 4 minutes. The slowest quartile of customers will require more than how many minutes for a simple haircut? A) 3(n + 1)/4 minutes B) 26 minutes C) 25.7 minutes D) 27.7 minutes Answer: D Explanation: Using Excel =NORM.INV(.75,25,4) = 27.698, or Q3 = 25 + 0.675(4) = 27.7 using Appendix C. Difficulty: 2 Medium Topic: 07.04 Standard Normal Distribution Learning Objective: 07-05 Solve for z or x for a given normal probability using tables or Excel. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

48) In Melanie's Styling Salon, the time to complete a simple haircut is normally distributed with a mean of 25 minutes and a standard deviation of 4 minutes. For a simple haircut, the middle 90 percent of the customers will require A) between 18.4 and 31.6 minutes. B) between 19.9 and 30.1 minutes. C) between 20.0 and 30.0 minutes. D) between 17.2 and 32.8 minutes. Answer: A Explanation: The 90 percent range is μ ± 1.645σ. Difficulty: 2 Medium Topic: 07.04 Standard Normal Distribution Learning Objective: 07-05 Solve for z or x for a given normal probability using tables or Excel. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 49) The area under the normal curve between z = 0 and z = 1 is ________ the area under the normal curve between z = 1 and z = 2. A) less than B) greater than C) equal to Answer: B Explanation: The standard normal PDF grows closer to the axis as z increases to the right of zero. Difficulty: 1 Easy Topic: 07.04 Standard Normal Distribution Learning Objective: 07-04 Find the normal probability for a given z or x using tables or Excel. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

50) The price-earnings ratio for firms in a given industry follows the normal distribution. In this industry, a firm whose price-earnings ratio has a standardized value of z = 1.00 is approximately in the highest ________ percent of firms in the industry. A) 16 percent B) 34 percent C) 68 percent D) 75 percent Answer: A Explanation: About 15.86 percent of the area is above one standard deviation. Difficulty: 2 Medium Topic: 07.04 Standard Normal Distribution Learning Objective: 07-04 Find the normal probability for a given z or x using tables or Excel. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 51) A student's grade on an examination was transformed to a z value of 0.67. Assuming a normal distribution, we know that she scored approximately in the top A) 15 percent. B) 50 percent. C) 40 percent. D) 25 percent. Answer: D Explanation: P(Z > 0.67) = 1 − P(Z < 0.67) = 1 − .2514 = .7486. Difficulty: 2 Medium Topic: 07.04 Standard Normal Distribution Learning Objective: 07-04 Find the normal probability for a given z or x using tables or Excel. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

52) The MPG (miles per gallon) for a certain compact car is normally distributed with a mean of 31 and a standard deviation of 0.8. What is the probability that the MPG for a randomly selected compact car would be less than 32? A) .3944 B) .8944 C) .1056 D) .5596 Answer: B Explanation: P(X < 32) = P(Z < 1.25) = .8944. Difficulty: 2 Medium Topic: 07.04 Standard Normal Distribution Learning Objective: 07-04 Find the normal probability for a given z or x using tables or Excel. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 53) The probability is .80 that a standard normal random variable is between −z and +z. The value of z is approximately A) 1.28 B) 1.35 C) 1.96 D) 1.45 Answer: A Explanation: For tail areas of .1000 we would use z = 1.282. Difficulty: 2 Medium Topic: 07.04 Standard Normal Distribution Learning Objective: 07-05 Solve for z or x for a given normal probability using tables or Excel. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

54) The time required for a citizen to complete the 2010 U.S. Census "long" form is normally distributed with a mean of 40 minutes and a standard deviation of 10 minutes. What proportion of the citizens will require less than one hour? A) .4772 B) .9772 C) .9974 D) .9997 Answer: B Explanation: P(X < 60) = P(Z < 2.00) = .9772. Difficulty: 2 Medium Topic: 07.04 Standard Normal Distribution Learning Objective: 07-04 Find the normal probability for a given z or x using tables or Excel. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 55) The time required for a citizen to complete the 2010 U.S. Census "long" form is normally distributed with a mean of 40 minutes and a standard deviation of 10 minutes. The slowest 10 percent of the citizens would need at least how many minutes to complete the form? A) 27.2 B) 35.8 C) 52.8 D) 59.6 Answer: C Explanation: Using Excel =NORM.INV(.90,40,10) = 52.82, or 40 + 1.282(10) = 52.82 using Appendix C. Difficulty: 3 Hard Topic: 07.04 Standard Normal Distribution Learning Objective: 07-05 Solve for z or x for a given normal probability using tables or Excel. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

56) The time required for a citizen to complete the 2010 U.S. Census "long" form is normally distributed with a mean of 40 minutes and a standard deviation of 10 minutes. What is the third quartile (in minutes) for the time required to complete the form? A) 44.75 B) 46.75 C) 47.50 D) 52.50 Answer: B Explanation: Using Excel =NORM.INV(.75,40,10) = 46.75, or Q3 = 40 + 0.675(10) = 46.75 using Appendix C. Difficulty: 3 Hard Topic: 07.04 Standard Normal Distribution Learning Objective: 07-05 Solve for z or x for a given normal probability using tables or Excel. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 57) Exam scores were normal in BIO 200. Jason's exam score was one standard deviation above the mean. What percentile is he in? A) 68th B) 75th C) 78th D) 84th Answer: D Explanation: About 15.87 percent of the area lies above one standard deviation. Difficulty: 1 Easy Topic: 07.04 Standard Normal Distribution Learning Objective: 07-04 Find the normal probability for a given z or x using tables or Excel. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

58) Compared to the area between z = 1.00 and z = 1.25, the area between z = 2.00 and z = 2.25 in the standard normal distribution will be A) smaller. B) larger. C) the same. D) impossible to compare without knowing μ and σ. Answer: A Explanation: The normal PDF approaches the axis as z increases beyond zero. Difficulty: 2 Medium Topic: 07.04 Standard Normal Distribution Learning Objective: 07-04 Find the normal probability for a given z or x using tables or Excel. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 59) A large number of applicants for admission to graduate study in business are given an aptitude test. Scores are normally distributed with a mean of 460 and standard deviation of 80. What fraction of applicants would you expect to have scores of 600 or above? A) .0401 B) .4599 C) .5401 D) .0852 Answer: A Explanation: P(X > 600) = P(Z > 1.75) = .0401. Difficulty: 2 Medium Topic: 07.04 Standard Normal Distribution Learning Objective: 07-04 Find the normal probability for a given z or x using tables or Excel. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

60) A large number of applicants for admission to graduate study in business are given an aptitude test. Scores are normally distributed with a mean of 460 and standard deviation of 80. What fraction of the applicants would you expect to have a score of 400 or above? A) .2734 B) .7734 C) .7266 D) .7500 Answer: B Explanation: P(X > 400) = P(Z > −0.75) = .7734. Difficulty: 2 Medium Topic: 07.04 Standard Normal Distribution Learning Objective: 07-04 Find the normal probability for a given z or x using tables or Excel. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 61) A large number of applicants for admission to graduate study in business are given an aptitude test. Scores are normally distributed with a mean of 460 and standard deviation of 80. The top 2.5 percent of the applicants would have a score of at least (choose the nearest integer) A) 606 B) 617 C) 600 D) 646 Answer: B Explanation: Using z = 1.96, we get X = 460 + 1.96 × 80 = 616.8. Difficulty: 2 Medium Topic: 07.04 Standard Normal Distribution Learning Objective: 07-05 Solve for z or x for a given normal probability using tables or Excel. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

62) If the random variable Z has a standard normal distribution, then P(1.25 ≤ Z ≤ 2.17) is A) .0906 B) .9200 C) .4700 D) .3944 Answer: A Explanation: P(Z ≤ 2.17) − P(Z ≤ 1.25) = .0906. Difficulty: 2 Medium Topic: 07.04 Standard Normal Distribution Learning Objective: 07-04 Find the normal probability for a given z or x using tables or Excel. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 63) If the random variable Z has a standard normal distribution, then P(Z ≤ −1.37) is A) .9147 B) .4147 C) .5016 D) .0853 Answer: D Explanation: From Appendix C, we get the left-tail area of .0853. Difficulty: 1 Easy Topic: 07.04 Standard Normal Distribution Learning Objective: 07-04 Find the normal probability for a given z or x using tables or Excel. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 64) Assume that X is normally distributed with a mean μ = $64. Given that P(X ≥ $75) = .2981, we can calculate that the standard deviation of X is approximately A) $20.76 B) $13.17 C) $5.83 D) $7.05 Answer: A Explanation: For a right-tail area of .2981, we need z = −.53. So with x = 75, we set z = (x − μ)/σ and solve for σ. Difficulty: 3 Hard Topic: 07.04 Standard Normal Distribution Learning Objective: 07-05 Solve for z or x for a given normal probability using tables or Excel. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 24 Copyright 2019 © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.

65) The standard deviation of a normal random variable X is $20. Given that P(X ≤ $10) = .1841 we can determine that the mean of the distribution is equal to A) $13 B) $26 C) $20 D) $28 Answer: D Explanation: For a left-tail area of .1841, we need z = −.90. With x = 10, we set z = (x − μ)/σ and solve for μ. Difficulty: 3 Hard Topic: 07.04 Standard Normal Distribution Learning Objective: 07-05 Solve for z or x for a given normal probability using tables or Excel. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 66) The random variable X is normally distributed with mean of 80 and variance of 36. The 67th percentile of the distribution is A) 72.00 B) 95.84 C) 90.00 D) 82.64 Answer: D Explanation: Since P(Z < 0.44) = .6700 (from Appendix C), we get 80 + 0.44(6) = 82.64. Difficulty: 2 Medium Topic: 07.04 Standard Normal Distribution Learning Objective: 07-05 Solve for z or x for a given normal probability using tables or Excel. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 67) The area under the normal curve between the 20th and 70th percentiles is equal to A) .7000 B) .5000 C) .9193 Answer: B Explanation: Logically, this must be .70 − .20 = .50, as you can verify from Appendix C. Difficulty: 3 Hard Topic: 07.04 Standard Normal Distribution Learning Objective: 07-04 Find the normal probability for a given z or x using tables or Excel. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 25 Copyright 2019 © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.

68) The variable in a normal distribution can assume any value between A) −3 and +3 B) −4 and +4 C) −1 and +1 D) −∞ and +∞ Answer: D Explanation: Almost all the area is within −3 and +3, but the curve never quite touches the zaxis. Difficulty: 1 Easy Topic: 07.03 Normal Distribution Learning Objective: 07-03 Know the form and parameters of the normal distribution. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 69) What are the mean and standard deviation for the standard normal distribution? A) μ = 0, σ = 0 B) μ = 1, σ = 1 C) μ = 1, σ = 0 D) μ = 0, σ = 1 Answer: D Explanation: It's a standardized normal distribution, so the mean must be zero. Difficulty: 1 Easy Topic: 07.04 Standard Normal Distribution Learning Objective: 07-03 Know the form and parameters of the normal distribution. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 70) Any two normal curves are the same except for their A) standard deviations. B) means. C) standard deviations and means. D) standard deviations, means, skewness, and kurtosis. Answer: C Explanation: We write N(μ,σ) to show the similar form of all normals. Difficulty: 1 Easy Topic: 07.03 Normal Distribution Learning Objective: 07-03 Know the form and parameters of the normal distribution. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation

71) Light bulbs are normally distributed with an average lifetime of 1000 hours and a standard deviation of 250 hours. The probability that a light bulb picked at random will last less than 1500 hours is about A) 97.72 percent. B) 95.44 percent. C) 75.00 percent. D) 68.00 percent. Answer: A Explanation: P(Z < 1500) = P(Z < 2.00) = .9772 from Appendix C (or from Excel). Difficulty: 2 Medium Topic: 07.04 Standard Normal Distribution Learning Objective: 07-04 Find the normal probability for a given z or x using tables or Excel. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 72) To convert a normally distributed variable X into a standard Z score we would A) subtract the mean from the original observation and divide the result by the variance. B) subtract the mean from the original observation and divide the result by the standard deviation. C) add the mean and the original observation, then divide by the variance. D) subtract the mean from the standard deviation and divide by the variance. Answer: B Explanation: Review the z-score transformation. Difficulty: 1 Easy Topic: 07.04 Standard Normal Distribution Learning Objective: 07-03 Know the form and parameters of the normal distribution. Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation 73) Regarding continuous probability distributions, which statement is incorrect? A) The triangular distribution may be skewed left or right. B) The uniform distribution is never skewed. C) The normal distribution is sometimes skewed. D) The exponential distribution is always skewed right. Answer: C Explanation: Review the characteristics of these four distributions. Difficulty: 2 Medium Topic: 07.03 Normal Distribution Learning Objective: 07-03 Know the form and parameters of the normal distribution. Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation 27 Copyright 2019 © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.

74) Which model best describes your waiting time until you get the next nonworking web URL ("This page cannot be displayed") as you click on various websites for Florida condo rentals? A) Triangular B) Uniform C) Normal D) Exponential Answer: D Explanation: The waiting time until the next event resembles an exponential distribution. Difficulty: 1 Easy Topic: 07.06 Exponential Distribution Learning Objective: 07-07 Find the exponential probability for a given x. Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation 75) On average, a major earthquake (Richter scale 6.0 or above) occurs 3 times a decade in a certain California county. What is the probability that less than six months will pass before the next earthquake? A) .1393 B) .8607 C) .0952 D) .9048 Answer: A Explanation: Set λ = 3/120 = 0.025 earthquake/month so P(X < 6) = 1 − exp(−0.025 × 6) = 1 − . 8607 = .1393. Difficulty: 3 Hard Topic: 07.06 Exponential Distribution Learning Objective: 07-07 Find the exponential probability for a given x. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

76) If the mean time between in-flight aircraft engine shutdowns is 12,500 operating hours, the 90th percentile of waiting times to the next shutdown will be A) 20,180 hours. B) 28,782 hours. C) 23,733 hours. D) 18,724 hours. Answer: B Explanation: Set λ = 1/12500. To solve for x, set the left-tail area of 1 − exp(−λx) equal to .90. Then solve for x by taking logs of both sides. Difficulty: 3 Hard Topic: 07.06 Exponential Distribution Learning Objective: 07-08 Solve for x for a given exponential probability. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 77) On average, 15 minutes elapse between discoveries of fraudulent corporate tax returns in a certain IRS office. What is the probability that less than 30 minutes will elapse before the next fraudulent corporate tax return is discovered? A) .1353 B) .6044 C) .7389 D) .8647 Answer: D Explanation: P(X < 30) = 1 − exp(−λx) = 1 − exp(−(1/15) × 30) = 1 − .1353 = .8647. Difficulty: 3 Hard Topic: 07.06 Exponential Distribution Learning Objective: 07-07 Find the exponential probability for a given x. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

78) If the mean time between unscheduled maintenance of LCD displays in a hospital's CT scan facility is 4,000 operating hours, what is the probability of unscheduled maintenance in the next 5,000 hours? A) .8000 B) .7135 C) .2865 D) .5000 Answer: B Explanation: P(X < 5000) = 1 − exp(−λx) = 1 − exp(−(1/4000) × 5000) = 1 − .2865 = .7135. Difficulty: 2 Medium Topic: 07.06 Exponential Distribution Learning Objective: 07-07 Find the exponential probability for a given x. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 79) A certain assembly line at Vexing Manufacturing Company averages 30 minutes between breakdowns. What is the probability that less than 6 minutes will elapse before the next breakdown? A) .8187 B) .0488 C) .1813 D) .2224 Answer: C Explanation: (X < 6) = 1 − exp(−λx) = 1 − exp(−(1/30) × 6) = 1 − .8187 = .1813. Difficulty: 3 Hard Topic: 07.06 Exponential Distribution Learning Objective: 07-07 Find the exponential probability for a given x. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

80) A certain assembly line at Vexing Manufacturing Company averages 30 minutes between breakdowns. The median time between breakdowns is A) 30.0 minutes. B) 35.7 minutes. C) 25.4 minutes. D) 20.8 minutes. Answer: D Explanation: Set λ = 1/30. To solve for x, set the right-tail area exp(−λx) = .50. Then solve for x by taking logs of both sides. Difficulty: 3 Hard Topic: 07.06 Exponential Distribution Learning Objective: 07-08 Solve for x for a given exponential probability. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 81) Which probability model is most appropriate to describe the waiting time (working days) until an office photocopier breaks down (i.e., requires unscheduled maintenance)? A) Normal B) Uniform C) Exponential D) Poisson Answer: C Explanation: Poisson breakdowns suggest exponential waiting time. Difficulty: 2 Medium Topic: 07.06 Exponential Distribution Learning Objective: 07-07 Find the exponential probability for a given x. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 82) On the last exam in FIN 417, "Capital Budgeting Strategies" Bob's z-score was −1.15. Bob said, "Yipe! My score is within the bottom quartile." Assuming a normal distribution, is Bob right? A) Yes B) No C) Must know the class size to answer Answer: A Explanation: The bottom quartile would be below z = −.675 so Bob is indeed below that point. Difficulty: 2 Medium Topic: 07.04 Standard Normal Distribution Learning Objective: 07-04 Find the normal probability for a given z or x using tables or Excel. Bloom's: Apply AACSB: Analytical Thinking 31 Copyright 2019 © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.

Accessibility: Keyboard Navigation 83) Exam scores were normal in MIS 200. Jason's exam score was 1.41 standard deviations above the mean. What percentile is he in? A) 68th B) 75th C) 84th D) 92nd Answer: D Explanation: P(Z < 1.41) = .9207. Difficulty: 1 Easy Topic: 07.04 Standard Normal Distribution Learning Objective: 07-04 Find the normal probability for a given z or x using tables or Excel. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 84) Compared to the area between z = 0.50 and z = 0.75, the area between z = 1.50 and z = 1.75 in the standard normal distribution will be A) smaller B) larger C) the same Answer: A Explanation: The normal PDF approaches the axis as z increases beyond zero, so areas get smaller. Difficulty: 2 Medium Topic: 07.04 Standard Normal Distribution Learning Objective: 07-04 Find the normal probability for a given z or x using tables or Excel. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

85) If GMAT scores for applicants at Oxnard Graduate School of Business are N(500, 50), then the top 5 percent of the applicants would have a score of at least (choose the nearest integer) A) 575 B) 582 C) 601 D) 608 Answer: B Explanation: The top 5 percent would require z = 1.645, so x = 500 + 1.645(50) = 582.25. Difficulty: 2 Medium Topic: 07.04 Standard Normal Distribution Learning Objective: 07-05 Solve for z or x for a given normal probability using tables or Excel. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 86) If the random variable Z has a standard normal distribution, then P(1.17 ≤ Z ≤ 2.26) is A) .1091 B) .1203 C) .2118 D) .3944 Answer: A Explanation: Subtract P(Z ≤ 2.26) − P(Z ≤ 1.17) = .9881 − .8790 = .1091. Difficulty: 2 Medium Topic: 07.04 Standard Normal Distribution Learning Objective: 07-04 Find the normal probability for a given z or x using tables or Excel. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 87) If the random variable Z has a standard normal distribution, then P(Z ≤ − 1.72) is A) .9573 B) .0446 C) .5016 D) .0427 Answer: D Explanation: Use Appendix C or Excel =NORM.S.DIST(−1.72,1). Difficulty: 1 Easy Topic: 07.04 Standard Normal Distribution Learning Objective: 07-04 Find the normal probability for a given z or x using tables or Excel. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

88) Excel's =100*RAND() function produces continuous random numbers that are uniformly distributed between 0 and 100. The standard deviation of this distribution is approximately A) 50.00 B) 28.87 C) 33.33 D) 25.00 Answer: B Explanation: The standard deviation of a uniform distribution is [(b - a)2/12]1/2 = [(100 − 0)2/12]1/2. Difficulty: 2 Medium Topic: 07.02 Uniform Continuous Distribution Learning Objective: 07-02 Calculate uniform probabilities. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 89) Excel's =RAND() function produces random numbers that are uniformly distributed between 0 and 1. The mean of this distribution is approximately A) .5000 B) .2500 C) .3333 D) .2887 Answer: A Explanation: The mean is halfway between the end points. Difficulty: 1 Easy Topic: 07.02 Uniform Continuous Distribution Learning Objective: 07-02 Calculate uniform probabilities. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

90) Excel's =RAND() function produces random numbers that are uniformly distributed from 0 to 1. What is the probability that the random number exceeds .75? A) 75 percent B) 50 percent C) 25 percent Answer: C Explanation: This is the upper 25 percent. Difficulty: 1 Easy Topic: 07.02 Uniform Continuous Distribution Learning Objective: 07-02 Calculate uniform probabilities. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 91) Which is the correct Excel formula for the 80th percentile of a distribution that is N(475, 33)? A) =NORM.DIST(80, 475, 33,1) B) =NORM.INV(0.80, 475, 33) C) =NORM.S.INV((80 − 475)/33) Answer: B Explanation: Review the Excel functions in Appendix J. Difficulty: 2 Medium Topic: 07.04 Standard Normal Distribution Learning Objective: 07-05 Solve for z or x for a given normal probability using tables or Excel. Bloom's: Remember AACSB: Technology Accessibility: Keyboard Navigation 92) If arrivals follow a Poisson distribution with mean 1.2 arrivals per minute, find the 75th percentile of waiting times until the next arrival (i.e., 75 percent below). A) 1.155 minutes (69.3 seconds) B) 0.240 minute (14.4 seconds) C) 1.919 minutes (115.1 seconds) Answer: A Explanation: Set λ = 1.2. To solve for x, set the right-tail area exp(−λx) equal to .25. Then solve for x by taking logs of both sides. Difficulty: 3 Hard Topic: 07.06 Exponential Distribution Learning Objective: 07-08 Solve for x for a given exponential probability. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

93) A software developer makes 175 phone calls to its current customers. There is an 8 percent chance of reaching a given customer (instead of a busy signal, no answer, or answering machine). The normal approximation of the probability of reaching at least 20 customers is A) .022 B) .007 C) .063 D) .937 Answer: C Explanation: Set n = 175 and π = .08. Calculate μ = nπ = (175)(.08) = 14 and σ = [nπ(1 − π)]1/2 = [175(.08)(1 − .08)]1/2 = 3.588872. Use x = 19.5 (with the continuity correction) and calculate the binomial P(X ≥ 20) ≈ P(z ≥ 1.532515) using z = (x − μ)/σ = 1.532515. Difficulty: 3 Hard Topic: 07.05 Normal Approximations Learning Objective: 07-06 Use the normal approximation to a binomial or a Poisson. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 94) For Gardyloo Manufacturing, the true proportion of accounts receivable with some kind of error is .20. If an auditor randomly samples 225 accounts receivable, what is the approximate normal probability that 39 or fewer will contain errors? A) .1797 B) .2097 C) .1587 D) .0544 Answer: A Explanation: Set μ = nπ and σ = [nπ(1 − π)]1/2 and convert x = 39.5 (using the continuity correction) to a z score with z = (x − μ)/σ. Set n = 225 and π = .20. Calculate μ = nπ = (225)(.20) = 45 and σ = [nπ(1 − π)]1/2 = [225(.20)(1 − .20)]1/2 = 6.000. Use x = 39.5 (with the continuity correction) and calculate the binomial P(X ≤ 39) ≈ P(z ≤ −.916667) using z = (x − μ)/σ = −.916667. Difficulty: 3 Hard Topic: 07.05 Normal Approximations Learning Objective: 07-06 Use the normal approximation to a binomial or a Poisson. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

95) A letter is mailed to a sample of 500 homeowners. Based on past experience, the probability of an undeliverable letter is 0.06. The normal approximation to the binomial probability of 40 or more undeliverable letters is A) .9632 B) .0368 C) .2305 D) .7695 Answer: B Explanation: Set n = 500 and π = .06. Calculate μ = nπ = (500)(.06) = 30 and σ = [nπ(1 − π)]1/2 = [500(.06)(1 − .06)]1/2 = 5.31037. Use x = 39.5 (with the continuity correction) and calculate the binomial P(X ≥ 40) ≈ P(z ≥ 1.78895) using z = (x − μ)/σ = 1.78895. Difficulty: 3 Hard Topic: 07.05 Normal Approximations Learning Objective: 07-06 Use the normal approximation to a binomial or a Poisson. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 96) In a True-False exam with 100 questions, passing requires a score of at least 60. What is the approximate normal probability that a "guesser" will score at least 60 points? A) .0287 B) .4713 C) .0251 D) .0377 Answer: A Explanation: A guesser would have a 50 percent chance of a correct answer, so we set π = .50. There are n = 100 questions, so we calculate μ = nπ = (100)(.50) = 50 and σ = [nπ(1 − π)]1/2 = [100(.50)(1 − .50)]1/2 = 5. Use x = 59.5 (with the continuity correction) and calculate the binomial P(X ≥ 60) ≈ P(z ≥ 1.90) using z = (x − μ)/σ = 1.90. Difficulty: 3 Hard Topic: 07.05 Normal Approximations Learning Objective: 07-06 Use the normal approximation to a binomial or a Poisson. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

97) A multiple choice exam has 100 questions. Each question has five choices. What would be the approximate probability that a "guesser" could achieve a score of 30 or more? A) .0088 B) .0062 C) .0015 D) .4913 Answer: A Explanation: A guesser would have a 20 percent chance of a correct answer (1 out of 5), so we set π = .20. There are n = 100 questions, so we calculate μ = nπ = (100)(.20) = 20 and σ = [nπ(1 − π)]1/2 = [100(.20)(1 − .20)]1/2 = 4. Use x = 29.5 (with the continuity correction) and calculate the binomial P(X ≥ 30) ≈ P(z ≥ 2.375) using z = (x − μ)/σ = 2.375. Difficulty: 3 Hard Topic: 07.05 Normal Approximations Learning Objective: 07-06 Use the normal approximation to a binomial or a Poisson. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 98) For which binomial distribution would a normal approximation be most acceptable? A) n = 50, π = .05 B) n = 100, π = .04 C) n = 40, π = .25 D) n = 400, π = .02 Answer: C Explanation: We want nπ ≥ 10 and n(1 − π) ≥ 10. Difficulty: 2 Medium Topic: 07.05 Normal Approximations Learning Objective: 07-06 Use the normal approximation to a binomial or a Poisson. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

99) Historically, the default rate on a certain type of commercial loan is 20 percent. If a bank makes 100 of these loans, what is the approximate probability that at least 26 will result in default? A) .2000 B) .0668 C) .0846 D) .0336 Answer: C Explanation: Set n = 100 and π = .20. Calculate μ = nπ = (100)(.20) = 20 and σ = [nπ(1 − π)]1/2 = [100(.20)(1 − .20)]1/2 = 4. Use x = 25.5 (with the continuity correction) and calculate the binomial P(X ≥ 26) ≈ P(z ≥ 1.375) using z = (x − μ)/σ = 1.375. Difficulty: 3 Hard Topic: 07.05 Normal Approximations Learning Objective: 07-06 Use the normal approximation to a binomial or a Poisson. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 100) A company employs 300 employees. Each year, there is a 30 percent turnover rate for employees. We want to do a normal approximation to the binomial distribution of the number of employees who leave each year. For this normal approximation, the mean is ________ and the standard deviation is ________. A) 90, 63 B) 90, 7.937 C) 90, 30 D) 90, 15 Answer: B Explanation: Use n = 300 and π = .30, and then calculate μ = nπ and σ = [nπ(1 − π)]1/2. Difficulty: 2 Medium Topic: 07.05 Normal Approximations Learning Objective: 07-06 Use the normal approximation to a binomial or a Poisson. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

101) The probability that a rental car will be stolen is .001. If 25,000 cars are rented from Hertz, what is the normal approximation to the probability that fewer than 20 will be stolen? A) .2577 B) .1335 C) .1128 D) .8335 Answer: B Explanation: Set n = 25,000 and π = .001. Calculate μ = nπ = (25000)(.001) = 25 and σ = [nπ(1 − π)]1/2 = [25000(.001)(1 − .001)]1/2 = 4.9975. Use x = 19.5 (with the continuity correction) and calculate the binomial P(X < 20) ≈ P(z < − 1.10055) using z = (x − μ)/σ = −1.10055. Difficulty: 3 Hard Topic: 07.05 Normal Approximations Learning Objective: 07-06 Use the normal approximation to a binomial or a Poisson. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 102) If adult male heights are normally distributed with a mean of 180 cm and a standard deviation of 7 cm, how high should an aircraft lavatory door be to ensure that 99.9 percent of adult males will not have to stoop as they enter? A) 195.7 cm B) 201.6 cm C) 207.3 cm D) 201.4 cm Answer: B Explanation: With Excel, we get =NORM.INV(.999,180,7) = 201.63, or Appendix C with z = 3.09. Difficulty: 2 Medium Topic: 07.04 Standard Normal Distribution Learning Objective: 07-05 Solve for z or x for a given normal probability using tables or Excel. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

103) TotCo is developing a new deluxe baby bassinet. If the length of a newborn baby is normally distributed with a mean of 50 cm and a standard deviation of 5 cm, what should be the interior length of the bassinet to ensure that 99 percent of newborn babies will fit, with a safety margin of 15 cm on each end of the bassinet? A) 95.45 cm B) 85.22 cm C) 91.63 cm D) 98.92 cm Answer: C Explanation: With Excel, we get =NORM.INV(.99,50,5)+30 = 91.63, or Appendix C with z = 2.33. Difficulty: 2 Medium Topic: 07.04 Standard Normal Distribution Learning Objective: 07-05 Solve for z or x for a given normal probability using tables or Excel. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 104) The triangular distribution T(4, 12, 26) has a mean of A) 14 B) 18 C) 12 D) 13 Answer: A Explanation: Mean is (4 + 12 + 26)/3. Difficulty: 1 Easy Topic: 07.07 Triangular Distribution (Optional) Learning Objective: 07-09 Use the triangular distribution for "what-if" analysis (optional). Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

105) The triangular distribution T(0, 10, 20) has a standard deviation of A) 4.082 B) 3.775 C) 3.024 D) 2.994 Answer: A Explanation: Set a = 0, b = 10, c = 20, and use the triangular standard deviation formula σ = [(a2 + b2 + c2 − ab − ac − bc)/18]1/2 = 4.082. Difficulty: 2 Medium Topic: 07.07 Triangular Distribution (Optional) Learning Objective: 07-09 Use the triangular distribution for "what-if" analysis (optional). Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 106) The triangular distribution T(5, 23, 62) has a mean of A) 23 B) 30 C) 33 D) 35 Answer: B Explanation: Mean is (5 + 23 + 62)/3. Difficulty: 1 Easy Topic: 07.07 Triangular Distribution (Optional) Learning Objective: 07-09 Use the triangular distribution for "what-if" analysis (optional). Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 107) The triangular distribution T(10, 20, 50) has a standard deviation of A) 9.498 B) 9.225 C) 8.498 D) 7.710 Answer: C Explanation: Set a = 10, b = 20, c = 50, and use the triangular standard deviation formula σ = [(a2 + b2 + c2 − ab − ac − bc)/18]1/2 = 8.498. Difficulty: 2 Medium Topic: 07.07 Triangular Distribution (Optional) Learning Objective: 07-09 Use the triangular distribution for "what-if" analysis (optional). Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 42 Copyright 2019 © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.

108) Which statement is incorrect? A) The triangular distribution always has a single mode. B) The mean of the triangular distribution is (a + b + c)/3. C) The triangular distribution is right-skewed. Answer: C Explanation: The triangular distribution may be skewed right or left, and is symmetric only if the mode is halfway between a and c. Review the properties of the triangular distribution. Difficulty: 2 Medium Topic: 07.07 Triangular Distribution (Optional) Learning Objective: 07-09 Use the triangular distribution for "what-if" analysis (optional). Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 109) Bob used a triangular distribution of T(20, 30, 61) to represent his daily commute time (minutes). Which statement is incorrect? A) The distribution is right-skewed. B) The mode of the distribution exceeds the mean. C) The mean of the distribution is 37. D) The midrange of the distribution is 40.5. Answer: B Explanation: This triangular distribution is right-skewed. The mean is (20 + 30 + 61)/3 = 37, which exceeds the mode b = 30. It would be helpful to sketch a graph of the PDF. Difficulty: 3 Hard Topic: 07.07 Triangular Distribution (Optional) Learning Objective: 07-09 Use the triangular distribution for "what-if" analysis (optional). Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

110) Phyllis used a triangular distribution of T(10, 15, 20) to represent her daily commute time (minutes). Which statement is incorrect? A) The distribution is right-skewed. B) The mode of the distribution is at the mean. C) The mean of the distribution is 15. D) The midrange of the distribution is 15. Answer: A Explanation: The distribution is symmetric if the mode b lies halfway between the end points. In this example, (a + c)/2 = (10 + 20)/2 = 15, so the mode b = 15 is halfway between the minimum and maximum. Difficulty: 2 Medium Topic: 07.07 Triangular Distribution (Optional) Learning Objective: 07-09 Use the triangular distribution for "what-if" analysis (optional). Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 111) In a continuous distribution, A) P(X < 5) is the same as P(X ≤ 5). B) P(X < 5) is less than P(X ≤ 5). C) P(X < 5) is more than P(X ≤ 5). Answer: A Explanation: A point has no area in a continuous CDF. Difficulty: 2 Medium Topic: 07.01 Continuous Probability Distributions Learning Objective: 07-01 Define a continuous random variable. Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation 112) In a continuous distribution the A) PDF is usually higher than the CDF. B) CDF is used to find left-tail probabilities. C) PDF shows the area under the curve. D) CDF is usually the same as the PDF. Answer: B Explanation: The CDF shows P(X ≤ x). Review the definitions of PDF and CDF. Difficulty: 2 Medium Topic: 07.01 Continuous Probability Distributions Learning Objective: 07-01 Define a continuous random variable. Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation 44 Copyright 2019 © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.

113) If the mean waiting time for the next arrival is 12 minutes, what is the median waiting time? A) 7.2 minutes B) 8.3 minutes C) 9.1 minutes D) 12 minutes Answer: B Explanation: Set λ = 1/12 minute per arrival and take logs of both sides of exp(−λx) = .50 to solve for x. Difficulty: 3 Hard Topic: 07.06 Exponential Distribution Learning Objective: 07-08 Solve for x for a given exponential probability. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 114) If the mean waiting time for the next arrival is 18 minutes, what is the first quartile (25th percentile) for waiting times? A) 13 minutes B) 7.9 minutes C) 5.2 minutes D) 3.1 minutes Answer: C Explanation: Set λ = 1/18 minute per arrival and take logs of both sides of exp(−λx) = .75 to solve for x. Difficulty: 3 Hard Topic: 07.06 Exponential Distribution Learning Objective: 07-08 Solve for x for a given exponential probability. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

115) Could this function be a PDF?

A) Yes B) No C) It depends on x. Answer: B Explanation: Area =1/2 × base × height = .500, which is not 1, so it cannot be a PDF. Difficulty: 2 Medium Topic: 07.01 Continuous Probability Distributions Learning Objective: 07-01 Define a continuous random variable. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

116) Could this function be a PDF?

A) Yes B) No C) It depends on x. Answer: B Explanation: Area = base × height = 2, which is not 1, so it cannot be a PDF. Difficulty: 2 Medium Topic: 07.01 Continuous Probability Distributions Learning Objective: 07-01 Define a continuous random variable. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 117) The ages of job applicants for a security guard position are uniformly distributed between 25 and 65. Could a 25-year-old job applicant be two standard deviations below the mean (or more than two standard deviations)? A) Yes B) No C) Impossible to determine from given information Answer: B Explanation: Since σ = [(65 − 25)2/12]1/2 = 11.54, we can see that 25 is not 2σ below the mean of 45. Difficulty: 3 Hard Topic: 07.02 Uniform Continuous Distribution Learning Objective: 07-02 Calculate uniform probabilities. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

118) The figure shows a standard normal N(0,1) distribution. Find the shaded area.

A) .6444 B) .7514 C) .9245 D) .9850 Answer: D Explanation: Appendix C.2 gives 1 − P(z < − 2.17) = 1 − .0150 = 0.9750 = 1NORM.S.DIST(−2.17,1). Difficulty: 2 Medium Topic: 07.04 Standard Normal Distribution Learning Objective: 07-04 Find the normal probability for a given z or x using tables or Excel. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

119) The figure shows a standard normal N(0,1) distribution. Find the shaded area.

A) .4400 B) .3300 C) .2998 D) .2502 Answer: B Explanation: Appendix C.2. gives 1 − P(z < .44) = 1 − .6700 = .3300 = 1NORM.S.DIST(0.44,1). Difficulty: 2 Medium Topic: 07.04 Standard Normal Distribution Learning Objective: 07-04 Find the normal probability for a given z or x using tables or Excel. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

120) The figure shows a standard normal N(0,1) distribution. Find the z value for the shaded area.

A) −1.98 B) −1.87 C) −1.75 D) −1.62 Answer: C Explanation: Appendix C.2 gives P(z < −1.75) = .0401, or use Excel = NORM.S.INV(0.04) = −1.75. Difficulty: 2 Medium Topic: 07.04 Standard Normal Distribution Learning Objective: 07-05 Solve for z or x for a given normal probability using tables or Excel. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

121) The figure shows a standard normal N(0,1) distribution. Find the z value for the shaded area.

A) −2.17 B) −2.09 C) −1.99 D) −1.94 Answer: A Explanation: Appendix C.2 gives P(z < −2.17) = .0150, or use Excel =NORM.S.INV(0.015) = −2.17. Difficulty: 2 Medium Topic: 07.04 Standard Normal Distribution Learning Objective: 07-05 Solve for z or x for a given normal probability using tables or Excel. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

122) The figure shows a normal N(400,23) distribution. Find the approximate shaded area.

A) .0410 B) .0501 C) .0724 D) .0838 Answer: A Explanation: z = (440 − 400)/23 = 1.739, so Appendix C.2 gives 1 − P(z < 1.74) = 1 − .9591 = . 0409, or from Excel =1-NORM.DIST(440,400,23,1) = .0410. Difficulty: 3 Hard Topic: 07.04 Standard Normal Distribution Learning Objective: 07-04 Find the normal probability for a given z or x using tables or Excel. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

123) The figure shows a normal N(400,23) distribution. Find the approximate shaded area.

A) .3811 B) .3527 C) .2299 D) .1940 Answer: C Explanation: z = (417 − 400)/23 = 0.7391, so Appendix C.2 gives 1 − P(z < 0.74) = 1 − .7704 = .2296, or from Excel =1-NORM.DIST(417,400,23,1) = .2299. Difficulty: 3 Hard Topic: 07.04 Standard Normal Distribution Learning Objective: 07-04 Find the normal probability for a given z or x using tables or Excel. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

124) The figure shows a normal N(400,23) distribution. Find the x value for the shaded area.

A) 379.1 B) 362.2 C) 355.7 D) 347.6 Answer: B Explanation: From Appendix C.2, we get P(z < −1.645) = .05, so x = µ + zσ = 400 − 1.645(23) = 362.2, or from Excel =NORM.INV(0.05,400,23) = 362.2. Difficulty: 3 Hard Topic: 07.03 Normal Distribution Learning Objective: 07-05 Solve for z or x for a given normal probability using tables or Excel. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

125) The figure shows a normal N(400,23) distribution. Find the x value for the shaded area.

A) 412.9 B) 426.7 C) 436.2 D) 440.3 Answer: D Explanation: From Appendix C.2, we get P(z > 1.75) = .0401, so x = µ + zσ = 400 + 1.75(23) = 440.3, or from Excel =NORM.INV(0.96,400,23) = 440.3. Difficulty: 3 Hard Topic: 07.04 Standard Normal Distribution Learning Objective: 07-05 Solve for z or x for a given normal probability using tables or Excel. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

Applied Statistics in Business and Economics, 6e (Doane) Chapter 8 Sampling Distributions and Estimation 1) The expected value of an unbiased estimator is equal to the parameter whose value is being estimated. Answer: TRUE Explanation: An unbiased estimator's expected value is the true parameter value. Difficulty: 1 Easy Topic: 08.01 Sampling and Estimation Learning Objective: 08-02 Explain the desirable properties of estimators. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 2) All estimators are biased since sampling error always exists to some extent. Answer: FALSE Explanation: An estimator is unbiased if its expected value is the true parameter, even though there may be sampling error. Difficulty: 1 Easy Topic: 08.01 Sampling and Estimation Learning Objective: 08-02 Explain the desirable properties of estimators. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 3) An estimator must be unbiased if you are to use it for statistical analysis. Answer: FALSE Explanation: An estimator can be useful as long as its bias is known. Difficulty: 2 Medium Topic: 08.01 Sampling and Estimation Learning Objective: 08-02 Explain the desirable properties of estimators. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation

4) The efficiency of an estimator depends on the variance of the estimator's sampling distribution. Answer: TRUE Explanation: Efficiency is measured by the variance of the estimator's sampling distribution. Difficulty: 1 Easy Topic: 08.01 Sampling and Estimation Learning Objective: 08-02 Explain the desirable properties of estimators. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 5) In comparing estimators, the more efficient estimator will have a smaller standard error. Answer: TRUE Explanation: Efficiency is measured by the variance of the estimator's sampling distribution. Difficulty: 1 Easy Topic: 08.01 Sampling and Estimation Learning Objective: 08-02 Explain the desirable properties of estimators. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 6) A 90 percent confidence interval will be wider than a 95 percent confidence interval, ceteris paribus. Answer: FALSE Explanation: For example, z.025 = 1.960 (for 95 percent confidence) gives a wider interval than z.05 = 1.645 (for 90 percent confidence). The proffered statement would also hold true for the Student's t distribution. Difficulty: 1 Easy Topic: 08.04 Confidence Interval for a Mean (mew) with Known sigma Learning Objective: 08-05 Construct a confidence interval for a population mean using z. Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation

7) In constructing a confidence interval for the mean, the z distribution provides a result nearly identical to the t distribution when n is large. Answer: TRUE Explanation: Student's t approaches z as the sample size increases. Difficulty: 1 Easy Topic: 08.05 Confidence Interval for a Mean (mew) with Unknown sigma Learning Objective: 08-06 Know when and how to use Student's t instead of z to estimate a mean. Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation 8) The Central Limit Theorem (CLT) says that, if n exceeds 30, the population will be normal. Answer: FALSE Explanation: The CLT refers to asymptotic properties of the mean. The population cannot be changed. Difficulty: 2 Medium Topic: 08.02 Central Limit Theorem Learning Objective: 08-03 State and apply the Central Limit Theorem for a mean. Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation 9) The Central Limit Theorem says that a histogram of the sample means will have a bell shape, even if the population is skewed and the sample is small. Answer: FALSE Explanation: The CLT says that the histogram of sample means will have a smaller range than the population. However, a large sample size may be required to achieve bell shape if the population is skewed. Difficulty: 2 Medium Topic: 08.02 Central Limit Theorem Learning Objective: 08-03 State and apply the Central Limit Theorem for a mean. Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation

10) The confidence level refers to the procedure used to construct the confidence interval, rather than to the particular confidence interval we have constructed. Answer: TRUE Explanation: A particular interval either does or does not contain the true parameter. Difficulty: 2 Medium Topic: 08.04 Confidence Interval for a Mean (mew) with Known sigma Learning Objective: 08-05 Construct a confidence interval for a population mean using z. Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation 11) The Central Limit Theorem guarantees an approximately normal sampling distribution when n is sufficiently large. Answer: TRUE Explanation: Yes, although a large sample size may be required if the population is skewed. Difficulty: 2 Medium Topic: 08.02 Central Limit Theorem Learning Objective: 08-03 State and apply the Central Limit Theorem for a mean. Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation 12) A sample of size 5 shows a mean of 45.2 and a sample standard deviation of 6.4. The standard error of the sample mean is approximately 2.86. Answer: TRUE Explanation: The standard error is the standard deviation divided by the square root of the sample size. We would use Student's t instead of z to construct a confidence interval for the population mean, but this problem did not ask for a confidence interval. Difficulty: 1 Easy Topic: 08.05 Confidence Interval for a Mean (mew) with Unknown sigma Learning Objective: 08-06 Know when and how to use Student's t instead of z to estimate a mean. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

13) As n increases, the width of the confidence interval will decrease, ceteris paribus. Answer: TRUE Explanation: The standard error is the standard deviation divided by the square root of the sample size. Difficulty: 1 Easy Topic: 08.03 Sample Size and Standard Error Learning Objective: 08-04 Explain how sample size affects the standard error. Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation 14) As n increases, the standard error decreases. Answer: TRUE Explanation: The standard error is the standard deviation divided by the square root of the sample size. Difficulty: 2 Medium Topic: 08.03 Sample Size and Standard Error Learning Objective: 08-04 Explain how sample size affects the standard error. Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation 15) A higher confidence level leads to a narrower confidence interval, ceteris paribus. Answer: FALSE Explanation: Higher confidence requires more uncertainty (a wider interval). For example, z.025 = 1.960 (for 95 percent confidence) gives a wider interval than z.05 = 1.645 (for 90 percent confidence). The proffered statement would also hold true for the Student's t distribution. Difficulty: 2 Medium Topic: 08.04 Confidence Interval for a Mean (mew) with Known sigma Learning Objective: 08-05 Construct a confidence interval for a population mean using z. Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation

16) When the sample standard deviation is used to construct a confidence interval for the mean, we would use the Student's t distribution instead of the normal distribution. Answer: TRUE Explanation: We should use the t distribution when the population variance is unknown. Difficulty: 1 Easy Topic: 08.05 Confidence Interval for a Mean (mew) with Unknown sigma Learning Objective: 08-06 Know when and how to use Student's t instead of z to estimate a mean. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 17) As long as the sample is more than one item, the standard error of the sample mean will be smaller than the standard deviation of the population. Answer: TRUE Explanation: The standard error is the standard deviation divided by the square root of the sample size. Difficulty: 1 Easy Topic: 08.03 Sample Size and Standard Error Learning Objective: 08-04 Explain how sample size affects the standard error. Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation 18) For a sample size of 20, a 95 percent confidence interval using the t distribution would be wider than one constructed using the z distribution. Answer: TRUE Explanation: Student's t is always larger than z for the same level of confidence. Difficulty: 2 Medium Topic: 08.05 Confidence Interval for a Mean (mew) with Unknown sigma Learning Objective: 08-06 Know when and how to use Student's t instead of z to estimate a mean. Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation

19) In constructing a confidence interval for a mean, the width of the interval is dependent on the sample size, the confidence level, and the population standard deviation. Answer: TRUE Explanation: The confidence interval depends on all of these. Difficulty: 2 Medium Topic: 08.04 Confidence Interval for a Mean (mew) with Known sigma Learning Objective: 08-05 Construct a confidence interval for a population mean using z. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 20) In constructing confidence intervals, it is conservative to use the z distribution when n ≥ 30. Answer: FALSE Explanation: While t and z values may be similar for large samples, it is more conservative to use t values. Using z would tend to make the interval narrower than is justified. Difficulty: 1 Easy Topic: 08.05 Confidence Interval for a Mean (mew) with Unknown sigma Learning Objective: 08-06 Know when and how to use Student's t instead of z to estimate a mean. Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation 21) The Central Limit Theorem (CLT) can be applied to the sample proportion. Answer: TRUE Explanation: We are sampling a Bernoulli population (1 = success, 0 = failure), but the CLT still applies. Difficulty: 1 Easy Topic: 08.06 Confidence Interval for a Proportion (pi) Learning Objective: 08-07 Construct a confidence interval for a population proportion. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 22) The distribution of the sample proportion p = x/n is normal when n ≥ 30. Answer: FALSE Explanation: We want at least 10 successes and 10 failures to assume that p is normally distributed. Difficulty: 1 Easy Topic: 08.06 Confidence Interval for a Proportion (pi) Learning Objective: 08-07 Construct a confidence interval for a population proportion. Bloom's: Understand AACSB: Analytical Thinking 7 Copyright 2019 © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.

Accessibility: Keyboard Navigation 23) The standard deviation of the sample proportion p = x/n increases as n increases. Answer: FALSE Explanation: The proffered statement is backward because n is in the denominator of [p(1 − p)/n]1/2. Difficulty: 1 Easy Topic: 08.06 Confidence Interval for a Proportion (pi) Learning Objective: 08-07 Construct a confidence interval for a population proportion. Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation 24) A 95 percent confidence interval constructed around p will be wider than a 90 percent confidence interval. Answer: TRUE Explanation: The interval for 95 percent confidence will be wider (z = 1.96) than for 90 percent confidence (z = 1.645). Difficulty: 1 Easy Topic: 08.06 Confidence Interval for a Proportion (pi) Learning Objective: 08-07 Construct a confidence interval for a population proportion. Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation 25) The sample proportion is always the midpoint of a confidence interval for the population proportion. Answer: TRUE Explanation: The interval is p ± z[p(1 − p)/n]1/2. Difficulty: 2 Medium Topic: 08.06 Confidence Interval for a Proportion (pi) Learning Objective: 08-07 Construct a confidence interval for a population proportion. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation

26) The standard error of the sample proportion is largest when π = .50. Answer: TRUE Explanation: The value of [π(1 − π)/n]1/2 is smaller for any value other than π = .50. Difficulty: 2 Medium Topic: 08.06 Confidence Interval for a Proportion (pi) Learning Objective: 08-07 Construct a confidence interval for a population proportion. Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation 27) The standard error of the sample proportion does not depend on the confidence level. Answer: TRUE Explanation: The standard error of p is [π(1 − π)/n]1/2. Difficulty: 2 Medium Topic: 08.06 Confidence Interval for a Proportion (pi) Learning Objective: 08-07 Construct a confidence interval for a population proportion. Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation 28) To narrow the confidence interval for π, we can either increase n or decrease the level of confidence. Answer: TRUE Explanation: The interval is p ± z[p(1 − p)/n]1/2. For example, z.025 = 1.96 for 95 percent confidence versus z.05 = 1.645 for 90 percent confidence. Difficulty: 2 Medium Topic: 08.06 Confidence Interval for a Proportion (pi) Learning Objective: 08-07 Construct a confidence interval for a population proportion. Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation 29) Ceteris paribus, the narrowest confidence interval for π is achieved when p = .50. Answer: FALSE Explanation: The value of [p(1 − p)/n]1/2 is smaller for any value other than π = .50. Difficulty: 2 Medium Topic: 08.06 Confidence Interval for a Proportion (pi) Learning Objective: 08-07 Construct a confidence interval for a population proportion. Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation 9 Copyright 2019 © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.

30) The statistic p = x/n may be assumed normally distributed when np ≥ 10 and n(1 − p) ≥ 10. Answer: TRUE Explanation: We want at least 10 successes and 10 failures in the sample to assume normality of p. Difficulty: 1 Easy Topic: 08.06 Confidence Interval for a Proportion (pi) Learning Objective: 08-07 Construct a confidence interval for a population proportion. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 31) The Student's t distribution is always symmetric and bell-shaped, but its tails lie above the normal. Answer: TRUE Explanation: Student's t resembles a normal, but its PDF is above the normal PDF in the tails. Difficulty: 2 Medium Topic: 08.05 Confidence Interval for a Mean (mew) with Unknown sigma Learning Objective: 08-06 Know when and how to use Student's t instead of z to estimate a mean. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 32) The confidence interval half-width when π = .50 is called the margin of error. Answer: TRUE Explanation: Pollsters use this conservative definition, which reflects both the sample size and the confidence level. Difficulty: 1 Easy Topic: 08.06 Confidence Interval for a Proportion (pi) Learning Objective: 08-07 Construct a confidence interval for a population proportion. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation

33) Based on the Rule of Three, if no events occur in n independent trials, we can set the upper 95 percent confidence bound at 3/n. Answer: TRUE Explanation: We need a special rule because when p = 0, we cannot apply the usual formula p ± z[p(1 − p)/n]1/2. Difficulty: 2 Medium Topic: 08.06 Confidence Interval for a Proportion (pi) Learning Objective: 08-07 Construct a confidence interval for a population proportion. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 34) The sample standard deviation s is halfway between the lower and upper confidence limits for the population σ (i.e., the confidence interval is symmetric around s). Answer: FALSE Explanation: The chi-square distribution is not symmetric, so neither is the confidence interval for σ. Difficulty: 3 Hard Topic: 08.10 Confidence Interval for a Population Variance, sigma2 (Optional) Learning Objective: 08-11 Construct a confidence interval for a variance (optional). Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation 35) In a sample size calculation, if the confidence level decreases, the size of the sample needed will increase. Answer: FALSE Explanation: Reduced confidence yields a smaller required sample because z will be smaller. For example, z.025 = 1.96 for 95 percent confidence versus z.05 = 1.645 for 90 percent confidence. Difficulty: 2 Medium Topic: 08.08 Sample Size Determination for a Mean Learning Objective: 08-09 Calculate sample size to estimate a mean. Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation

36) To calculate the sample size needed for a survey to estimate a proportion, the population standard deviation σ must be known. Answer: FALSE Explanation: For a proportion, the sample size formula requires π not σ. Difficulty: 2 Medium Topic: 08.09 Sample Size Determination for a Proportion Learning Objective: 08-10 Calculate sample size to estimate a proportion. Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation 37) Assuming that π = .50 is a quick and conservative approach to use in a sample size calculation for a proportion. Answer: TRUE Explanation: Assuming that π = .50 is quick and safe (but may give a larger sample than is needed). Difficulty: 2 Medium Topic: 08.09 Sample Size Determination for a Proportion Learning Objective: 08-10 Calculate sample size to estimate a proportion. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 38) To estimate the required sample size for a proportion, one method is to take a small pilot sample to estimate π and then apply the sample size formula. Answer: TRUE Explanation: Taking a pilot sample is a common method, but assuming that π = .50 is quick and safe (maybe too conservative). Difficulty: 2 Medium Topic: 08.09 Sample Size Determination for a Proportion Learning Objective: 08-10 Calculate sample size to estimate a proportion. Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation

39) To estimate π, you typically need a sample size equal to at least 5 percent of your population. Answer: FALSE Explanation: The required sample size n bears no necessary proportionality to N. Difficulty: 2 Medium Topic: 08.09 Sample Size Determination for a Proportion Learning Objective: 08-10 Calculate sample size to estimate a proportion. Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation 40) To estimate a proportion with a 4 percent margin of error and a 95 percent confidence level, the required sample size is over 800. Answer: FALSE Explanation: n = (z/E)2(π)(1 − π) = (1.96/.04)2(.50)(1 − .50) = 600.25. Difficulty: 2 Medium Topic: 08.09 Sample Size Determination for a Proportion Learning Objective: 08-10 Calculate sample size to estimate a proportion. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 41) Approximately 95 percent of the population X values will lie within the 95 percent confidence interval for the mean. Answer: FALSE Explanation: The confidence interval is for the population mean, not for individual X values. Difficulty: 2 Medium Topic: 08.04 Confidence Interval for a Mean (mew) with Known sigma Learning Objective: 08-05 Construct a confidence interval for a population mean using z. Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation 42) A 99 percent confidence interval has higher confidence but less precision than a 95 percent confidence interval. Answer: TRUE Explanation: The higher confidence level widens the interval, so it is less precise. Difficulty: 2 Medium Topic: 08.04 Confidence Interval for a Mean (mew) with Known sigma Learning Objective: 08-05 Construct a confidence interval for a population mean using z. Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation 13 Copyright 2019 © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.

43) Sampling variation is not controllable by the statistician. Answer: TRUE Explanation: Sampling variation is inevitable. All we can do is use best practices in estimation. Difficulty: 1 Easy Topic: 08.01 Sampling and Estimation Learning Objective: 08-01 Define sampling error, parameter, and estimator. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 44) The sample mean is not a random variable when the population parameters are known. Answer: FALSE Explanation: The sample mean is a random variable regardless of what we know about the population. Difficulty: 1 Easy Topic: 08.01 Sampling and Estimation Learning Objective: 08-01 Define sampling error, parameter, and estimator. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 45) The finite population correction factor (FPCF) can be ignored if n = 7 and N = 700. Answer: TRUE Explanation: The FPCF has a negligible effect when the sample is less than 5 percent of the population. Difficulty: 2 Medium Topic: 08.07 Estimating from Finite Populations Learning Objective: 08-08 Know how to modify confidence intervals when the population is finite. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

46) In constructing a confidence interval, the finite population correction factor (FPCF) can be ignored if samples of 12 items are drawn from a population of 300 items. Answer: TRUE Explanation: The FPCF has a negligible effect when the sample is less than 5 percent of the population. Difficulty: 2 Medium Topic: 08.07 Estimating from Finite Populations Learning Objective: 08-08 Know how to modify confidence intervals when the population is finite. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 47) The finite population correction factor (FPCF) can be ignored when the sample size is large relative to the population size. Answer: FALSE Explanation: The FPCF has a negligible effect when n is small relative to N. Difficulty: 2 Medium Topic: 08.07 Estimating from Finite Populations Learning Objective: 08-08 Know how to modify confidence intervals when the population is finite. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 48) A sampling distribution describes the distribution of A) a parameter. B) a statistic. C) either a parameter or a statistic. D) neither a parameter nor a statistic. Answer: B Explanation: A statistic is a random variable. Its sampling distribution describes its behavior. Difficulty: 1 Easy Topic: 08.01 Sampling and Estimation Learning Objective: 08-01 Define sampling error, parameter, and estimator. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation

49) As the sample size increases, the standard error of the mean A) increases. B) decreases. C) may increase or decrease. Answer: B Explanation: The standard error of the mean is σ/(n1/2). Difficulty: 1 Easy Topic: 08.03 Sample Size and Standard Error Learning Objective: 08-04 Explain how sample size affects the standard error. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 50) Which statement is most nearly correct, other things being equal? A) Doubling the sample size will cut the standard error of the mean in half. B) The standard error of the mean depends on the population size. C) Quadrupling the sample size roughly halves the standard error of the mean. D) The standard error of the mean depends on the confidence level. Answer: C Explanation: The standard error of the mean is σ/(n1/2) so replacing n by 4n would cut the standard error of the mean in half. Difficulty: 2 Medium Topic: 08.03 Sample Size and Standard Error Learning Objective: 08-04 Explain how sample size affects the standard error. Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation 51) The width of a confidence interval for μ is not affected by A) the sample size. B) the confidence level. C) the standard deviation. D) the sample mean. Answer: D Explanation: The mean is not used in calculating the width of the confidence interval zσ/(n1/2). Difficulty: 2 Medium Topic: 08.04 Confidence Interval for a Mean (mew) with Known sigma Learning Objective: 08-05 Construct a confidence interval for a population mean using z. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation

52) The Central Limit Theorem implies that A) the population will be approximately normal if n ≥ 30. B) repeated samples must be taken to obtain normality. C) the distribution of the mean is approximately normal for large n. D) the mean follows the same distribution as the population. Answer: C Explanation: The sampling distribution of the mean is asymptotically normal for any population. Difficulty: 2 Medium Topic: 08.02 Central Limit Theorem Learning Objective: 08-03 State and apply the Central Limit Theorem for a mean. Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation 53) The owner of Limp Pines Resort wanted to know the average age of its clients. A random sample of 25 tourists is taken. It shows a mean age of 46 years with a standard deviation of 5 years. The width of a 98 percent confidence interval for the true mean client age is approximately A) ±1.711 years. B) ±2.326 years. C) ±2.492 years. D) ±2.797 years. Answer: C Explanation: The width is ts/(n1/2) = (2.492)(5)/(25)1/2 = 2.492. Difficulty: 2 Medium Topic: 08.05 Confidence Interval for a Mean (mew) with Unknown sigma Learning Objective: 08-06 Know when and how to use Student's t instead of z to estimate a mean. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

54) In constructing a confidence interval for a mean with unknown variance with a sample of 25 items, Beth used z instead of t. "Well, at least my interval will be wider than necessary, so it was a conservative error," said she. Is Beth's statement correct? A) Yes. B) No. C) It depends on μ. Answer: B Explanation: z is always smaller than t (ceteris paribus), so the interval would be narrower than is justified. Difficulty: 2 Medium Topic: 08.05 Confidence Interval for a Mean (mew) with Unknown sigma Learning Objective: 08-06 Know when and how to use Student's t instead of z to estimate a mean. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 55) A random sample of 16 ATM transactions at the Last National Bank of Flat Rock revealed a mean transaction time of 2.8 minutes with a standard deviation of 1.2 minutes. The width (in minutes) of the 95 percent confidence interval for the true mean transaction time is A) ±0.639 B) ±0.588 C) ±0.300 D) ±2.131 Answer: A Explanation: The width is ts/(n1/2) = (2.131)(1.2)/(16)1/2 = 0.639. Difficulty: 2 Medium Topic: 08.05 Confidence Interval for a Mean (mew) with Unknown sigma Learning Objective: 08-06 Know when and how to use Student's t instead of z to estimate a mean. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

56) We could narrow a 95 percent confidence interval by A) using 99 percent confidence. B) using a larger sample. C) raising the standard error. Answer: B Explanation: A larger sample would narrow the interval width zs/(n1/2). Difficulty: 1 Easy Topic: 08.04 Confidence Interval for a Mean (mew) with Known sigma Learning Objective: 08-05 Construct a confidence interval for a population mean using z. Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation 57) The owner of Torpid Oaks B&B wanted to know the average distance its guests had traveled. A random sample of 16 guests showed a mean distance of 85 miles with a standard deviation of 32 miles. The 90 percent confidence interval (in miles) for the mean is approximately A) (71.0, 99.0) B) (71.8, 98.2) C) (74.3, 95.7) D) (68.7, 103.2) Answer: A Explanation: The interval is 85 ± ts/(n1/2) or 85 ± (1.753)(32)/(16)1/2 with d.f. = 15 (do not use z). Difficulty: 2 Medium Topic: 08.05 Confidence Interval for a Mean (mew) with Unknown sigma Learning Objective: 08-06 Know when and how to use Student's t instead of z to estimate a mean. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

58) A highway inspector needs an estimate of the mean weight of trucks crossing a bridge on the interstate highway system. She selects a random sample of 49 trucks and finds a mean of 15.8 tons with a sample standard deviation of 3.85 tons. The 90 percent confidence interval for the population mean is A) 14.72 to 16.88 tons. B) 14.90 to 16.70 tons. C) 14.69 to 16.91 tons. D) 14.88 to 16.72 tons. Answer: D Explanation: The interval is 15.8 ± ts/(n1/2) or 15.8 ± (1.677)(3.85)/(49)1/2 using d.f. = 48 (do not use z). Difficulty: 2 Medium Topic: 08.05 Confidence Interval for a Mean (mew) with Unknown sigma Learning Objective: 08-06 Know when and how to use Student's t instead of z to estimate a mean. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 59) To determine a 72 percent level of confidence for a proportion, the value of z is approximately A) ±1.65 B) ±0.77 C) ±1.08 D) ±1.55 Answer: C Explanation: Look up the z value that puts 14 percent in each tail. Difficulty: 2 Medium Topic: 08.06 Confidence Interval for a Proportion (pi) Learning Objective: 08-07 Construct a confidence interval for a population proportion. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

60) To estimate the average annual expenses of students on books and class materials, a sample of size 36 is taken. The sample mean is $850 and the sample standard deviation is $54. A 99 percent confidence interval for the population mean is A) $823.72 to $876.28 B) $832.36 to $867.64 C) $826.82 to $873.18 D) $825.48 to $874.52 Answer: D Explanation: The interval is 850 ± ts/(n1/2), or 850 ± (2.724)(54)/(36)1/2 with d.f. = 35 (do not use z). Difficulty: 2 Medium Topic: 08.05 Confidence Interval for a Mean (mew) with Unknown sigma Learning Objective: 08-06 Know when and how to use Student's t instead of z to estimate a mean. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 61) In constructing a 95 percent confidence interval, if you increase n to 4n, the width of your confidence interval will be (assuming other things remain the same) A) about 25 percent of its former width. B) about two times wider. C) about 50 percent of its former width. D) about four times wider. Answer: C Explanation: The standard error of the mean is σ/(n1/2) so replacing n by 4n would cut the standard error of the mean in half. Difficulty: 2 Medium Topic: 08.04 Confidence Interval for a Mean (mew) with Known sigma Learning Objective: 08-05 Construct a confidence interval for a population mean using z. Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation

62) Which of the following is not a characteristic of the t distribution? A) It is a continuous distribution. B) It has a mean of 0. C) It is a symmetric distribution. D) It approaches z as degrees of freedom decrease. Answer: D Explanation: Student's t approaches z as the degrees of freedom increase. Difficulty: 1 Easy Topic: 08.05 Confidence Interval for a Mean (mew) with Unknown sigma Learning Objective: 08-06 Know when and how to use Student's t instead of z to estimate a mean. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 63) Which statement is incorrect? Explain. A) If p = .50 and n = 100, the standard error of the sample proportion is .05. B) In a sample size calculation for estimating π, it is conservative to assume π = .50. C) If n = 250 and p = .06, we cannot assume normality in a confidence interval for π. Answer: C Explanation: Normality of p may be assumed because np = 15 and n(1 − p) = 235 (at least 10 "successes" and 10 "failures"). Difficulty: 2 Medium Topic: 08.06 Confidence Interval for a Proportion (pi) Learning Objective: 08-07 Construct a confidence interval for a population proportion. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 64) What is the approximate width of a 90 percent confidence interval for the true population proportion if there are 12 successes in a sample of 25? A) ±.196 B) ±.164 C) ±.480 D) ±.206 Answer: B Explanation: The interval width is ±z[p(1 − p)/n]1/2 = ±(1.645)[(.48)(.52)/25]1/2. Difficulty: 2 Medium Topic: 08.06 Confidence Interval for a Proportion (pi) Learning Objective: 08-07 Construct a confidence interval for a population proportion. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 22 Copyright 2019 © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.

65) A poll showed that 48 out of 120 randomly chosen graduates of California medical schools last year intended to specialize in family practice. What is the width of a 90 percent confidence interval for the proportion that plan to specialize in family practice? A) ±.0447 B) ±.0736 C) ±.0876 D) ±.0894 Answer: B Explanation: The interval width is ±z[p(1 − p)/n]1/2 = ±(1.645)[(.40)(.60)/120]1/2. Difficulty: 2 Medium Topic: 08.06 Confidence Interval for a Proportion (pi) Learning Objective: 08-07 Construct a confidence interval for a population proportion. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 66) What is the approximate width of an 80 percent confidence interval for the true population proportion if there are 12 successes in a sample of 80? A) ±.078 B) ±.066 C) ±.051 D) ±.094 Answer: C Explanation: The interval width is ±z[p(1 − p)/n]1/2 = ±(1.282)[(.15)(.85)/80]1/2. Difficulty: 2 Medium Topic: 08.06 Confidence Interval for a Proportion (pi) Learning Objective: 08-07 Construct a confidence interval for a population proportion. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

67) A random sample of 160 commercial customers of PayMor Lumber revealed that 32 had paid their accounts within a month of billing. The 95 percent confidence interval for the true proportion of customers who pay within a month would be A) 0.148 to 0.252 B) 0.138 to 0.262 C) 0.144 to 0.256 D) 0.153 to 0.247 Answer: B Explanation: The interval is p ± z[p(1 − p)/n]1/2 = .20 ± (1.960)[(.20)(.80)/160]1/2. Difficulty: 2 Medium Topic: 08.06 Confidence Interval for a Proportion (pi) Learning Objective: 08-07 Construct a confidence interval for a population proportion. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 68) A random sample of 160 commercial customers of PayMor Lumber revealed that 32 had paid their accounts within a month of billing. Can normality be assumed for the sample proportion? A) Yes B) No C) Need more information to say Answer: A Explanation: Yes, because there were at least 10 "successes" and at least 10 "failures" in the sample. This is a conservative rule of thumb. Difficulty: 1 Easy Topic: 08.06 Confidence Interval for a Proportion (pi) Learning Objective: 08-07 Construct a confidence interval for a population proportion. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

69) The conservative sample size required for a 95 percent confidence interval for π with an error of ±0.04 is A) 271 B) 423 C) 385 D) 601 Answer: D Explanation: n = (z/E)2(π)(1 − π) = (1.96/.04)2(.50)(1 − .50) = 600.25 (round up). Difficulty: 2 Medium Topic: 08.09 Sample Size Determination for a Proportion Learning Objective: 08-10 Calculate sample size to estimate a proportion. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 70) Last week, 108 cars received parking violations in the main university parking lot. Of these, 27 had unpaid parking tickets from a previous violation. Assuming that last week was a random sample of all parking violators, find the 95 percent confidence interval for the percentage of parking violators that have prior unpaid parking tickets. A) 18.1 to 31.9 percent. B) 16.8 to 33.2 percent. C) 15.3 to 34.7 percent. D) 19.5 to 30.5 percent. Answer: B Explanation: The interval is p ± z[p(1 − p)/n]1/2 = .25 ± (1.960)[(.25)(.75)/108]1/2. Difficulty: 2 Medium Topic: 08.06 Confidence Interval for a Proportion (pi) Learning Objective: 08-07 Construct a confidence interval for a population proportion. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

71) In a random sample of 810 women employees, it is found that 81 would prefer working for a female boss. The width of the 95 percent confidence interval for the proportion of women who prefer a female boss is A) ±.0288 B) ±.0105 C) ±.0207 D) ±.0196 Answer: C Explanation: The width is ±z[p(1 − p)/n]1/2 or ±(1.960)[(.10)(.90)/810]1/2 or ±.0207. Difficulty: 2 Medium Topic: 08.06 Confidence Interval for a Proportion (pi) Learning Objective: 08-07 Construct a confidence interval for a population proportion. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 72) Jolly Blue Giant Health Insurance (JBGHI) is concerned about rising lab test costs and would like to know what proportion of the positive lab tests for prostate cancer are actually proven correct through subsequent biopsy. JBGHI demands a sample large enough to ensure an error of ±2 percent with 90 percent confidence. What is the necessary sample size? A) 4,148 B) 2,401 C) 1,692 D) 1,604 Answer: C Explanation: n = (z/E)2(π)(1 − π) = (1.645/.02)2(.50)(1 − .50) = 1691.3 (round up). Difficulty: 2 Medium Topic: 08.09 Sample Size Determination for a Proportion Learning Objective: 08-10 Calculate sample size to estimate a proportion. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

73) A university wants to estimate the average distance that commuter students travel to get to class with an error of ±3 miles and 90 percent confidence. What sample size would be needed, assuming that travel distances are normally distributed with a range of X = 0 to X = 50 miles, using the Empirical Rule μ ± 3σ to estimate σ. A) About 28 students B) About 47 students C) About 30 students D) About 21 students Answer: D Explanation: Using σ = (50 − 0)/6 = 8.333, we get n = [zσ/E]2 = [(1.645)(8.333)/3]2 = 20.9 (round up). Difficulty: 3 Hard Topic: 08.08 Sample Size Determination for a Mean Learning Objective: 08-09 Calculate sample size to estimate a mean. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 74) A financial institution wishes to estimate the mean balances owed by its credit card customers. The population standard deviation is $300. If a 99 percent confidence interval is used and an interval of ±$75 is desired, how many cardholders should be sampled? A) 3382 B) 629 C) 87 D) 107 Answer: D Explanation: n = [zσ/E]2 = [(2.576)(300)/75]2 = 106.2 (round up). Difficulty: 2 Medium Topic: 08.08 Sample Size Determination for a Mean Learning Objective: 08-09 Calculate sample size to estimate a mean. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

75) A company wants to estimate the time its trucks take to drive from city A to city B. The standard deviation is known to be 12 minutes. What sample size is required so that the error does not exceed ±2 minutes, with 95 percent confidence? A) 12 observations B) 139 observations C) 36 observations D) 129 observations Answer: B Explanation: n = [zσ/E]2 = [(1.960)(12)/2]2 = 138.3 (round up). Difficulty: 2 Medium Topic: 08.08 Sample Size Determination for a Mean Learning Objective: 08-09 Calculate sample size to estimate a mean. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 76) In a large lecture class, the professor announced that the scores on a recent exam were normally distributed with a range from 51 to 87. Using the Empirical Rule μ ± 3σ to estimate σ, how many students would you need to sample to estimate the true mean score for the class with 90 percent confidence and an error of ±2? A) About 17 students B) About 35 students C) About 188 students D) About 25 students Answer: D Explanation: Using σ = (87 − 51)/6 = 6, we get n = [zσ/E]2 = [(1.645)(6)/2]2 = 24.35 (round up). Difficulty: 2 Medium Topic: 08.08 Sample Size Determination for a Mean Learning Objective: 08-09 Calculate sample size to estimate a mean. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

77) Using the conventional polling definition, find the margin of error for a customer satisfaction survey of 225 customers who have recently dined at Applebee's. A) ±5.0 percent B) ±4.2 percent C) ±7.1 percent D) ±6.5 percent Answer: D Explanation: The margin of error is ±z[π(1 − π)/n]1/2 or ±(1.960)[(.50)(.50)/225]1/2 or ±.065. Difficulty: 2 Medium Topic: 08.09 Sample Size Determination for a Proportion Learning Objective: 08-10 Calculate sample size to estimate a proportion. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 78) A marketing firm is asked to estimate the percentage of existing customers who would purchase a "digital upgrade" to their basic cable TV service. The firm wants 99 percent confidence and an error of ±5 percent. What is the required sample size (to the next higher integer)? A) 664 B) 625 C) 801 D) 957 Answer: A Explanation: n = (z/E)2(π)(1 − π) = (2.576/.05)2(.50)(1 − .50) = 663.6 (round up). Difficulty: 2 Medium Topic: 08.09 Sample Size Determination for a Proportion Learning Objective: 08-10 Calculate sample size to estimate a proportion. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

79) An airport traffic analyst wants to estimate the proportion of daily takeoffs by small business jets (as opposed to commercial passenger jets or other aircraft) with an error of ±4 percent with 90 percent confidence. What sample size should the analyst use? A) 385 B) 601 C) 410 D) 423 Answer: D Explanation: n = (z/E)2(π)(1 − π) = (1.645/.04)2(.50)(1 − .50) = 422.8 (round up). Difficulty: 2 Medium Topic: 08.09 Sample Size Determination for a Proportion Learning Objective: 08-10 Calculate sample size to estimate a proportion. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 80) Ersatz Beneficial Insurance wants to estimate the cost of damage to cars due to accidents. The standard deviation of the cost is known to be $200. They want to estimate the mean cost using a 95 percent confidence interval within ±$10. What is the minimum sample size n? A) 1083 B) 4002 C) 1537 D) 2301 Answer: C Explanation: n = [zσ/E]2 = [(1.960)(200)/10]2 = 1536.6 (round up). Difficulty: 2 Medium Topic: 08.08 Sample Size Determination for a Mean Learning Objective: 08-09 Calculate sample size to estimate a mean. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

81) Professor York randomly surveyed 240 students at Oxnard University and found that 150 of the students surveyed watch more than 10 hours of television weekly. Develop a 95 percent confidence interval to estimate the true proportion of students who watch more than 10 hours of television each week. The confidence interval is A) .533 to .717 B) .564 to .686 C) .552 to .698 D) .551 to .739 Answer: B Explanation: The interval is p ± z[p(1 − p)/n]1/2 = .625 ± (1.960)[(.625)(.375)/240]1/2. Difficulty: 2 Medium Topic: 08.06 Confidence Interval for a Proportion (pi) Learning Objective: 08-07 Construct a confidence interval for a population proportion. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 82) Professor York randomly surveyed 240 students at Oxnard University and found that 150 of the students surveyed watch more than 10 hours of television weekly. How many additional students would Professor York have to sample to estimate the proportion of all Oxnard University students who watch more than 10 hours of television each week within ±3 percent with 99 percent confidence? A) 761 B) 1001 C) 1489 D) 1728 Answer: C Explanation: Using p = .625, we get n = (z/E)2(π)(1 − π) = (2.576/.03)2(.625)(.375) = 1728.06 (round up). Difficulty: 3 Hard Topic: 08.09 Sample Size Determination for a Proportion Learning Objective: 08-10 Calculate sample size to estimate a proportion. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

83) The sample proportion is in the middle of the confidence interval for the population proportion A) in any sample. B) only if the samples are large. C) only if π is not too far from .50. Answer: A Explanation: The interval is p ± z[p(1 − p)/n]1/2. Difficulty: 2 Medium Topic: 08.06 Confidence Interval for a Proportion (pi) Learning Objective: 08-07 Construct a confidence interval for a population proportion. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 84) For a sample of size 16, the critical values of chi-square for a 95 percent confidence interval for the population variance are A) 6.262, 27.49 B) 6.908, 28.85 C) 5.629, 26.12 D) 7.261, 25.00 Answer: A Explanation: Using d.f. = n − 1 = 15, we get χ2L = 6.262 and χ2U = 27.49 from Appendix E. Difficulty: 2 Medium Topic: 08.10 Confidence Interval for a Population Variance, sigma2 (Optional) Learning Objective: 08-11 Construct a confidence interval for a variance (optional). Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 85) For a sample of size 11, the critical values of chi-square for a 90 percent confidence interval for the population variance are A) 6.262, 27.49 B) 6.908, 28.85 C) 3.940, 18.31 D) 3.247, 20.48 Answer: C Explanation: d.f. = n − 1 = 10, we get χ2L = 3.940 and χ2U = 18.31 from Appendix E. Difficulty: 2 Medium Topic: 08.10 Confidence Interval for a Population Variance, sigma2 (Optional) Learning Objective: 08-11 Construct a confidence interval for a variance (optional). Bloom's: Apply AACSB: Analytical Thinking 32 Copyright 2019 © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.

Accessibility: Keyboard Navigation 86) For a sample of size 18, the critical values of chi-square for a 99 percent confidence interval for the population variance are A) 6.262, 27.49 B) 5.697, 35.72 C) 5.629, 26.12 D) 7.261, 25.00 Answer: B Explanation: d.f. = n − 1 = 17, we get χ2L = 5.697 and χ2U = 35.72 from Appendix E. Difficulty: 2 Medium Topic: 08.10 Confidence Interval for a Population Variance, sigma2 (Optional) Learning Objective: 08-11 Construct a confidence interval for a variance (optional). Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 87) Which of the following statements is most nearly correct, other things being equal? A) Using Student's t instead of z makes a confidence interval narrower. B) The table values of z and t are about the same when the mean is large. C) For a given confidence level, the z value is always smaller than the t value. D) Student's t is rarely used because it is more conservative to use z. Answer: C Explanation: As n increases, t approaches z, but t is always larger than z. It is conservative to use Student's t if the population variance is unknown. Difficulty: 2 Medium Topic: 08.05 Confidence Interval for a Mean (mew) with Unknown sigma Learning Objective: 08-06 Know when and how to use Student's t instead of z to estimate a mean. Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation 88) The Central Limit Theorem (CLT) A) applies only to samples from normal populations. B) applies to any population. C) applies best to populations that are skewed. D) applies only when μ and σ are known. Answer: B Explanation: The appeal of the CLT is that it applies to populations of any shape. Difficulty: 2 Medium Topic: 08.02 Central Limit Theorem Learning Objective: 08-03 State and apply the Central Limit Theorem for a mean. 33 Copyright 2019 © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.

Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 89) In which situation may the sample proportion safely be assumed to follow a normal distribution? A) 12 successes in a sample of 72 items B) 8 successes in a sample of 40 items C) 6 successes in a sample of 200 items D) 4 successes in a sample of 500 items Answer: A Explanation: We would like at least 10 "successes" and at least 10 "failures" to assume that p is normal. (This is a conservative rule.) Difficulty: 2 Medium Topic: 08.06 Confidence Interval for a Proportion (pi) Learning Objective: 08-07 Construct a confidence interval for a population proportion. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 90) In which situation may the sample proportion safely be assumed to follow a normal distribution? A) n = 100, π = .06 B) n = 250, π = .02 C) n = 30, π = .50 D) n = 500, π = .01 Answer: C Explanation: We want nπ ≥ 10 and n(1 − π) ≥ 10 to assume that p is normal. Difficulty: 2 Medium Topic: 08.06 Confidence Interval for a Proportion (pi) Learning Objective: 08-07 Construct a confidence interval for a population proportion. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

91) If σ = 12, find the sample size to estimate the mean with an error of ±4 and 95 percent confidence (rounded to the next higher integer). A) 75 B) 35 C) 58 D) 113 Answer: B Explanation: n = [zσ/E]2 = [(1.960)(12)/4]2 = 34.6 (round up). Difficulty: 2 Medium Topic: 08.08 Sample Size Determination for a Mean Learning Objective: 08-09 Calculate sample size to estimate a mean. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 92) If σ = 25, find the sample size to estimate the mean with an error of ±3 and 90 percent confidence (rounded to the next higher integer). A) 426 B) 512 C) 267 D) 188 Answer: D Explanation: n = [zσ/E]2 = [(1.645)(25)/3]2 = 187.9 (round up). Difficulty: 2 Medium Topic: 08.08 Sample Size Determination for a Mean Learning Objective: 08-09 Calculate sample size to estimate a mean. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 93) Sampling error can be avoided A) by using an unbiased estimator. B) by eliminating nonresponses (e.g., older people). C) by no method under the statistician's control. D) either by using an unbiased estimator or by eliminating nonresponse. Answer: C Explanation: Sampling error occurs in any random sample. Difficulty: 1 Easy Topic: 08.01 Sampling and Estimation Learning Objective: 08-01 Define sampling error, parameter, and estimator. Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation 35 Copyright 2019 © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.

94) A consistent estimator for the mean A) converges on the true parameter μ as the variance increases. B) converges on the true parameter μ as the sample size increases. C) consistently follows a normal distribution. D) is impossible to obtain using real sample data. Answer: B Explanation: A consistent estimator's variance becomes smaller and the estimator approaches the true parameter as n increases. Difficulty: 2 Medium Topic: 08.01 Sampling and Estimation Learning Objective: 08-02 Explain the desirable properties of estimators. Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation 95) Concerning confidence intervals, which statement is most nearly correct? A) We should use z instead of t when n is large. B) We use the Student's t distribution when σ is unknown. C) We use the Student's t distribution to narrow the confidence interval. Answer: B Explanation: Student's t distribution widens the confidence interval (compare with z) and should be used whenever σ is unknown. Difficulty: 2 Medium Topic: 08.05 Confidence Interval for a Mean (mew) with Unknown sigma Learning Objective: 08-06 Know when and how to use Student's t instead of z to estimate a mean. Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation 96) The standard error of the mean decreases when A) the sample size decreases. B) the standard deviation increases. C) the standard deviation decreases or n increases. D) the population size decreases. Answer: C Explanation: The standard error of the mean σ/(n1/2) depends on n and σ. Difficulty: 2 Medium Topic: 08.03 Sample Size and Standard Error Learning Objective: 08-04 Explain how sample size affects the standard error. Bloom's: Understand AACSB: Analytical Thinking 36 Copyright 2019 © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.

Accessibility: Keyboard Navigation 97) For a given sample size, the higher the confidence level, the A) more accurate the point estimate. B) smaller the standard error. C) smaller the interval width. D) greater the interval width. Answer: D Explanation: To have more confidence, we must widen the interval. For example, z.025 = 1.960 (for 95 percent confidence) gives a wider interval than z.05 = 1.645 (for 90 percent confidence). The proffered statement would also be true for the Student's t distribution. Difficulty: 2 Medium Topic: 08.04 Confidence Interval for a Mean (mew) with Known sigma Learning Objective: 08-05 Construct a confidence interval for a population mean using z. Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation 98) A sample is taken and a confidence interval is constructed for the mean of the distribution. Which value is always found at the center of the interval? A) The sample mean B) The population mean μ C) Neither nor μ since with a sample anything can happen D) Both and μ as long as there are not too many outliers Answer: A Explanation: Any confidence interval for a mean is symmetric around the sample mean. Difficulty: 2 Medium Topic: 08.04 Confidence Interval for a Mean (mew) with Known sigma Learning Objective: 08-05 Construct a confidence interval for a population mean using z. Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation

99) If a normal population has parameters μ = 40 and σ = 8, then for a sample size n = 4 A) the standard error of the sample mean is approximately 2. B) the standard error of the sample mean is approximately 4. C) the standard error of the sample mean is approximately 8. D) the standard error of the sample mean is approximately 10. Answer: B Explanation: The standard error is σ/(n1/2) = (8)/(41/2) = 4. Difficulty: 1 Easy Topic: 08.03 Sample Size and Standard Error Learning Objective: 08-04 Explain how sample size affects the standard error. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 100) In a confidence interval, the finite population correction factor (FPCF) can be ignored when A) n = 100 and N = 700. B) n = 50 and N = 200. C) n = 6 and N = 500. D) n = 8 and N = 80. Answer: C Explanation: The FPCF has a negligible effect when the sample is less than 5 percent of the population. Difficulty: 2 Medium Topic: 08.07 Estimating from Finite Populations Learning Objective: 08-08 Know how to modify confidence intervals when the population is finite. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

101) In a confidence interval, the finite population correction factor (FPCF) can be ignored when A) n = 12 and N = 96. B) n = 80 and N = 400. C) n = 6 and N = 60. D) n = 8 and N = 800. Answer: D Explanation: The FPCF has a negligible effect when the sample is less than 5 percent of the population. Difficulty: 2 Medium Topic: 08.07 Estimating from Finite Populations Learning Objective: 08-08 Know how to modify confidence intervals when the population is finite. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

Applied Statistics in Business and Economics, 6e (Doane) Chapter 9 One-Sample Hypothesis Tests 1) The level of significance refers to the probability of making a Type II error. Answer: FALSE Explanation: The level of significance is the desired probability of Type I error. Difficulty: 1 Easy Topic: 09.02 Type I and Type II Error Learning Objective: 09-02 Define Type I error, Type II error, and power. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 2) The level of significance refers to the probability of making a Type I error. Answer: TRUE Explanation: The level of significance is the desired probability of Type I error. Difficulty: 1 Easy Topic: 09.02 Type I and Type II Error Learning Objective: 09-02 Define Type I error, Type II error, and power. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 3) A simultaneous reduction in both α and β will require a larger sample size. Answer: TRUE Explanation: In general, there is a trade-off between α and β, but with a larger n we can reduce both Type I and Type II error. Difficulty: 2 Medium Topic: 09.02 Type I and Type II Error Learning Objective: 09-02 Define Type I error, Type II error, and power. Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation

4) The probability of rejecting a false null hypothesis increases as the sample size increases, other things being equal. Answer: TRUE Explanation: Larger samples cut the chance of Type II error (β) and increase power (1 − β). Difficulty: 2 Medium Topic: 09.02 Type I and Type II Error Learning Objective: 09-02 Define Type I error, Type II error, and power. Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation 5) When the probability of a Type I error increases, the probability of a Type II error must decrease, ceteris paribus. Answer: TRUE Explanation: For a given sample size, there is a trade-off between α and β. Difficulty: 2 Medium Topic: 09.02 Type I and Type II Error Learning Objective: 09-02 Define Type I error, Type II error, and power. Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation 6) A false positive in a drug test for steroids is a Type II error. Answer: FALSE Explanation: A false positive is a Type I error. Difficulty: 2 Medium Topic: 09.02 Type I and Type II Error Learning Objective: 09-02 Define Type I error, Type II error, and power. Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation 7) If a judge acquits every defendant, the judge will never commit a Type I error. (H0 is the hypothesis of innocence.) Answer: TRUE Explanation: If no one is convicted, there is no Type I error (but there can be Type II error). Difficulty: 2 Medium Topic: 09.02 Type I and Type II Error Learning Objective: 09-02 Define Type I error, Type II error, and power. Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation 2 Copyright 2019 © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.

8) When your sample size increases, the chance of both Type I and Type II error will increase. Answer: FALSE Explanation: There is a trade-off between α and β unless we can increase n. Difficulty: 2 Medium Topic: 09.02 Type I and Type II Error Learning Objective: 09-02 Define Type I error, Type II error, and power. Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation 9) A Type II error can only occur when you fail to reject H0. Answer: TRUE Explanation: If you do not reject H0, you may commit a Type II error. Difficulty: 1 Easy Topic: 09.02 Type I and Type II Error Learning Objective: 09-02 Define Type I error, Type II error, and power. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 10) A Type I error can only occur if you reject H0. Answer: TRUE Explanation: If you reject H0, a false positive can occur. Difficulty: 1 Easy Topic: 09.02 Type I and Type II Error Learning Objective: 09-02 Define Type I error, Type II error, and power. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 11) John rejected H0, so we know definitely that he did not commit a Type II error. Answer: TRUE Explanation: If you reject H0, you may commit a Type I error but not a Type II error. Difficulty: 2 Medium Topic: 09.02 Type I and Type II Error Learning Objective: 09-02 Define Type I error, Type II error, and power. Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation 3 Copyright 2019 © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.

12) In hypothesis testing, we cannot prove a null hypothesis is true. Answer: TRUE Explanation: The null hypothesis could be falsified by a different sample. Difficulty: 1 Easy Topic: 09.01 Logic of Hypothesis Testing Learning Objective: 09-01 Know the steps in testing hypotheses and define H0 and H1. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 13) For a given level of significance (α), increasing the sample size will increase the probability of Type II error because there are more ways to make an incorrect decision. Answer: FALSE Explanation: A large sample size is beneficial in reducing error of either type. Difficulty: 2 Medium Topic: 09.02 Type I and Type II Error Learning Objective: 09-02 Define Type I error, Type II error, and power. Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation 14) For a given sample size, reducing the level of significance will decrease the probability of making a Type II error. Answer: FALSE Explanation: For fixed n, reducing α would tend to increase β. Difficulty: 2 Medium Topic: 09.02 Type I and Type II Error Learning Objective: 09-02 Define Type I error, Type II error, and power. Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation 15) The probability of a false positive is decreased if we reduce α. Answer: TRUE Explanation: By definition, α is the chance of a false positive. However, reducing α would tend to increase β. Difficulty: 2 Medium Topic: 09.02 Type I and Type II Error Learning Objective: 09-02 Define Type I error, Type II error, and power. Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation 4 Copyright 2019 © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.

16) A hypothesis test may be statistically significant, yet have little practical importance. Answer: TRUE Explanation: Small effects may be unimportant in some applications. Difficulty: 1 Easy Topic: 09.04 Testing a Mean: Known Population Variance Learning Objective: 09-05 Perform a hypothesis test for a mean with known σ using z. Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation 17) Compared to using α = .01, choosing α = .001 will make it less likely that a true null hypothesis will be rejected. Answer: TRUE Explanation: Smaller α makes it harder to reject the null hypothesis (but may increase β). Difficulty: 1 Easy Topic: 09.03 Decision Rules and Critical Values Learning Objective: 09-04 Explain decision rules, critical values, and rejection regions. Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation 18) A two-tailed hypothesis test for H0: μ = 15 at α = .10 is analogous to asking if a 90 percent confidence interval for μ contains 15. Answer: TRUE Explanation: Only in a two-tailed hypothesis test is this statement true. Difficulty: 2 Medium Topic: 09.04 Testing a Mean: Known Population Variance Learning Objective: 09-05 Perform a hypothesis test for a mean with known σ using z. Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation 19) For a given sample size and α level, the Student's t value always exceeds the z value. Answer: TRUE Explanation: As n increases, t approaches z, but t is always larger. Difficulty: 2 Medium Topic: 09.05 Testing a Mean: Unknown Population Variance Learning Objective: 09-07 Perform a hypothesis test for a mean with unknown σ using t. Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation 5 Copyright 2019 © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.

20) For a given level of significance, the critical value of Student's t increases as n increases. Answer: FALSE Explanation: As n increases, t declines and approaches the corresponding z. Difficulty: 2 Medium Topic: 09.05 Testing a Mean: Unknown Population Variance Learning Objective: 09-07 Perform a hypothesis test for a mean with unknown σ using t. Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation 21) For a sample of nine items, the critical value of Student's t for a left-tailed test of a mean at α = .05 is −1.860. Answer: TRUE Explanation: Use Appendix D or Excel's function =T.INV(.05,8). Difficulty: 2 Medium Topic: 09.05 Testing a Mean: Unknown Population Variance Learning Objective: 09-07 Perform a hypothesis test for a mean with unknown σ using t. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 22) Holding other factors constant, it is harder to reject the null hypothesis for a mean when conducting a two-tailed test rather than a one-tailed test. Answer: TRUE Explanation: For a two-tailed test, the critical value is farther out in the tail. Difficulty: 2 Medium Topic: 09.03 Decision Rules and Critical Values Learning Objective: 09-04 Explain decision rules, critical values, and rejection regions. Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation 23) If we desire α = .10, then a p-value of .13 would lead us to reject the null hypothesis. Answer: FALSE Explanation: Reject the null if the p-value is less than α. Difficulty: 1 Easy Topic: 09.04 Testing a Mean: Known Population Variance Learning Objective: 09-05 Perform a hypothesis test for a mean with known σ using z. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

24) The p-value is the probability of the sample result (or one more extreme), assuming H0 is true. Answer: TRUE Explanation: This is the definition of a p-value. Difficulty: 1 Easy Topic: 09.04 Testing a Mean: Known Population Variance Learning Objective: 09-05 Perform a hypothesis test for a mean with known σ using z. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 25) The probability of rejecting a true null hypothesis is the significance level of the test. Answer: TRUE Explanation: This is the definition of α. Difficulty: 2 Medium Topic: 09.02 Type I and Type II Error Learning Objective: 09-02 Define Type I error, Type II error, and power. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 26) A null hypothesis is rejected when the calculated p-value is less than the critical value of the test statistic. Answer: FALSE Explanation: No, the p-value is compared with α (not with the critical value from a table). Difficulty: 2 Medium Topic: 09.04 Testing a Mean: Known Population Variance Learning Objective: 09-05 Perform a hypothesis test for a mean with known σ using z. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 27) In a right-tailed test, the null hypothesis is rejected when the value of the test statistic exceeds the critical value. Answer: TRUE Explanation: For example, we would reject H0 if zcalc > 1.645 at α = .05. Difficulty: 1 Easy Topic: 09.03 Decision Rules and Critical Values Learning Objective: 09-04 Explain decision rules, critical values, and rejection regions. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 7 Copyright 2019 © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.

28) The critical value of a hypothesis test is based on the researcher's selected level of significance. Answer: TRUE Explanation: The level of significance is the desired tail area, which dictates the critical value. Difficulty: 1 Easy Topic: 09.03 Decision Rules and Critical Values Learning Objective: 09-03 Formulate a null and alternative hypothesis for µ or π. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 29) If the null and alternative hypotheses are H0: μ ≤ 100 and H1: μ > 100, the test is right-tailed. Answer: TRUE Explanation: The direction of the test is always revealed by the direction of the inequality in H1. Difficulty: 1 Easy Topic: 09.03 Decision Rules and Critical Values Learning Objective: 09-03 Formulate a null and alternative hypothesis for µ or π. Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation 30) The null hypothesis is rejected when the p-value exceeds the level of significance. Answer: FALSE Explanation: Reject the null if the p-value is less than α. Difficulty: 1 Easy Topic: 09.04 Testing a Mean: Known Population Variance Learning Objective: 09-05 Perform a hypothesis test for a mean with known σ using z. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 31) For a given null hypothesis and level of significance, the critical value for a two-tailed test is greater than the critical value for a one-tailed test. Answer: TRUE Explanation: For a two-tailed test, we have to go farther into the tails to put α/2 in the tail. Difficulty: 2 Medium Topic: 09.03 Decision Rules and Critical Values Learning Objective: 09-04 Explain decision rules, critical values, and rejection regions. Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation 8 Copyright 2019 © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.

32) For a given H0 and level of significance, if you reject the H0 for a one-tailed test, you would also reject H0 for a two-tailed test. Answer: FALSE Explanation: The opposite is true because the two-tailed critical value is bigger. Difficulty: 2 Medium Topic: 09.03 Decision Rules and Critical Values Learning Objective: 09-04 Explain decision rules, critical values, and rejection regions. Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation 33) If the hypothesized proportion is π0 = .025 in a sample of size 120, it is safe to assume normality of the sample proportion p. Answer: FALSE Explanation: We can assume normality of p if nπ0 ≥ 10 and n(1 − π0) ≥ 10, which is not true here. Difficulty: 1 Easy Topic: 09.06 Testing a Proportion Learning Objective: 09-09 Check whether normality may be assumed in testing a proportion. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 34) For a mean, we would expect the test statistic to be near zero if the null hypothesis is true. Answer: TRUE Explanation: The difference between the sample mean and the hypothesized mean would be small. Difficulty: 1 Easy Topic: 09.03 Decision Rules and Critical Values Learning Objective: 09-04 Explain decision rules, critical values, and rejection regions. Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation 35) In the hypothesis H0: π = π0, the value of π0 is derived from the sample. Answer: FALSE Explanation: The hypothesized proportion is a target or historical benchmark. Difficulty: 2 Medium Topic: 09.06 Testing a Proportion Learning Objective: 09-08 Perform a hypothesis test for a proportion and find the p-value. Bloom's: Understand 10 Copyright 2019 © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.

AACSB: Analytical Thinking Accessibility: Keyboard Navigation 36) In testing the hypotheses H0: π ≤ π0, H1: π > π0, we would use a right-tailed test. Answer: TRUE Explanation: The direction of the test is always revealed by the direction of the inequality in H1. Difficulty: 1 Easy Topic: 09.03 Decision Rules and Critical Values Learning Objective: 09-03 Formulate a null and alternative hypothesis for µ or π. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 37) To test the hypothesis H0: π = .0125 using n = 160, it is safe to assume normality of p. Answer: FALSE Explanation: We can assume normality of p if nπ0 ≥ 10 and n(1 − π0) ≥ 10, which is not true here. Difficulty: 1 Easy Topic: 09.06 Testing a Proportion Learning Objective: 09-09 Check whether normality may be assumed in testing a proportion. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 38) In testing a proportion, normality of p can be assumed if nπ0 ≥ 10 and n(1 − π0) ≥ 10. Answer: TRUE Explanation: This is a conservative rule of thumb. Difficulty: 2 Medium Topic: 09.06 Testing a Proportion Learning Objective: 09-09 Check whether normality may be assumed in testing a proportion. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 39) Power is the probability of rejecting the null hypothesis when it is false and is equal to 1 − β. Answer: TRUE Explanation: High power (small chance of Type II error) is desirable. Difficulty: 2 Medium Topic: 09.02 Type I and Type II Error Learning Objective: 09-02 Define Type I error, Type II error, and power. Bloom's: Remember 11 Copyright 2019 © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.

AACSB: Analytical Thinking Accessibility: Keyboard Navigation

40) Other things being equal, a smaller standard deviation implies higher power. Answer: TRUE Explanation: Higher variance makes it harder to detect a departure from H0. Difficulty: 3 Hard Topic: 09.07 Power Curves and OC Curves (Optional) Learning Objective: 09-10 Interpret a power curve or OC curve (optional). Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation 41) The power of a test is the probability that the test will reject a false null hypothesis. Answer: TRUE Explanation: High power (small chance of Type II error) is desirable. Difficulty: 2 Medium Topic: 09.02 Type I and Type II Error Learning Objective: 09-02 Define Type I error, Type II error, and power. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 42) The height of the power curve shows the probability of accepting a true null hypothesis. Answer: FALSE Explanation: Power is the chance of correctly rejecting a false null hypothesis. Difficulty: 3 Hard Topic: 09.07 Power Curves and OC Curves (Optional) Learning Objective: 09-10 Interpret a power curve or OC curve (optional). Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation 43) The power curve plots β on the Y-axis and the test statistic on the X-axis. Answer: FALSE Explanation: A power curve plots the true parameter value on the X-axis and 1 − β on the Y-axis. Difficulty: 2 Medium Topic: 09.07 Power Curves and OC Curves (Optional) Learning Objective: 09-10 Interpret a power curve or OC curve (optional). Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation

44) A smaller probability of Type II error implies higher power of the test. Answer: TRUE Explanation: By definition, power is 1 − β. Difficulty: 2 Medium Topic: 09.02 Type I and Type II Error Learning Objective: 09-02 Define Type I error, Type II error, and power. Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation 45) Varying the true mean is a movement along the power curve, not a shift in the curve. Answer: TRUE Explanation: The power curve shows how power varies with the true mean. Difficulty: 2 Medium Topic: 09.07 Power Curves and OC Curves (Optional) Learning Objective: 09-10 Interpret a power curve or OC curve (optional). Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation 46) Increasing the sample size shifts the power curve upward, ceteris paribus. Answer: TRUE Explanation: A larger n would raise the power curve at all points along the X-axis. Difficulty: 3 Hard Topic: 09.07 Power Curves and OC Curves (Optional) Learning Objective: 09-10 Interpret a power curve or OC curve (optional). Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation 47) Increasing the level of significance shifts the power curve upward, ceteris paribus. Answer: TRUE Explanation: For a given n, increasing α would decrease β and hence raise power (1 − β). Difficulty: 3 Hard Topic: 09.07 Power Curves and OC Curves (Optional) Learning Objective: 09-10 Interpret a power curve or OC curve (optional). Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation

48) A power curve for a mean is at its lowest point when the true μ is very near μ0. Answer: TRUE Explanation: This is why it is hard to detect small departures from H0. Difficulty: 2 Medium Topic: 09.07 Power Curves and OC Curves (Optional) Learning Objective: 09-10 Interpret a power curve or OC curve (optional). Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation 49) Larger samples lead to increased power, ceteris paribus. Answer: TRUE Explanation: A larger n would raise the power curve at all points along the X-axis. Difficulty: 2 Medium Topic: 09.07 Power Curves and OC Curves (Optional) Learning Objective: 09-10 Interpret a power curve or OC curve (optional). Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation 50) In graphing power curves, there is a different power curve for each sample size n. Answer: TRUE Explanation: A larger n would raise the power curve at all points along the X-axis. Difficulty: 2 Medium Topic: 09.07 Power Curves and OC Curves (Optional) Learning Objective: 09-10 Interpret a power curve or OC curve (optional). Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation 51) In hypothesis testing, we are trying to reject the alternative hypothesis. Answer: FALSE Explanation: We are trying to reject the null hypothesis H0. Difficulty: 1 Easy Topic: 09.01 Logic of Hypothesis Testing Learning Objective: 09-01 Know the steps in testing hypotheses and define H0 and H1. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation

52) In hypothesis testing, we are trying to prove the null hypothesis. Answer: FALSE Explanation: We cannot prove the null hypothesis, for H0 could be falsified by a different sample. Difficulty: 1 Easy Topic: 09.01 Logic of Hypothesis Testing Learning Objective: 09-01 Know the steps in testing hypotheses and define H0 and H1. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 53) When σ is unknown, it is more conservative to use z instead of t for the critical value. Answer: FALSE Explanation: Because z is smaller than t, we would reject too often if we use z (not conservative). Difficulty: 1 Easy Topic: 09.05 Testing a Mean: Unknown Population Variance Learning Objective: 09-07 Perform a hypothesis test for a mean with unknown σ using t. Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation 54) For a given sample size, when we increase the probability of a Type I error, the probability of a Type II error A) remains unchanged. B) increases. C) decreases. D) is impossible to determine without more information. Answer: C Explanation: For a given sample size, there is a trade-off between α and β. Difficulty: 2 Medium Topic: 09.02 Type I and Type II Error Learning Objective: 09-02 Define Type I error, Type II error, and power. Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation

55) After testing a hypothesis regarding the mean, we decided not to reject H0. Thus, we are exposed to A) Type I error. B) Type II error. C) either Type I or Type II error. D) neither Type I nor Type II error. Answer: B Explanation: Failure to reject H0 could lead to Type II error (but not Type I error). Difficulty: 2 Medium Topic: 09.02 Type I and Type II Error Learning Objective: 09-02 Define Type I error, Type II error, and power. Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation 56) After testing a hypothesis, we decided to reject the null hypothesis. Thus, we are exposed to A) Type I error. B) Type II error. C) either Type I or Type II error. D) neither Type I nor Type II error. Answer: A Explanation: Rejecting H0 could lead to Type I error (but not Type II error). Difficulty: 2 Medium Topic: 09.02 Type I and Type II Error Learning Objective: 09-02 Define Type I error, Type II error, and power. Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation 57) Which statement about α is not correct? A) It is the probability of committing a Type I error. B) It is the test's significance level. C) It is the probability of rejecting a true H0. D) It is equal to 1 − β. Answer: D Explanation: There is an inverse relationship between α and β, but it is not a simple equation. Difficulty: 2 Medium Topic: 09.02 Type I and Type II Error Learning Objective: 09-02 Define Type I error, Type II error, and power. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 17 Copyright 2019 © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.

58) Which of the following is correct? A) When the sample size increases, both α and β may decrease. B) Type II error can only occur when you reject H0. C) Type I error can only occur if you fail to reject H0. D) The level of significance is the probability of Type II error. Answer: A Explanation: Only a larger sample size can allow a reduction in both α and β (ceteris paribus). Difficulty: 2 Medium Topic: 09.02 Type I and Type II Error Learning Objective: 09-02 Define Type I error, Type II error, and power. Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation 59) Which of the following is incorrect? A) The level of significance is the probability of making a Type I error. B) Lowering both α and β at once will require a larger sample size. C) The probability of rejecting a true null hypothesis increases as n increases. D) When Type I error increases, Type II error must decrease, ceteris paribus. Answer: C Explanation: The critical value for the desired α takes the sample size into consideration. Difficulty: 2 Medium Topic: 09.02 Type I and Type II Error Learning Objective: 09-02 Define Type I error, Type II error, and power. Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation 60) John rejected his null hypothesis in a right-tailed test for a mean at α = .025 because his critical t value was 2.000 and his calculated t value was 2.345. We can be sure that A) John did not commit a Type I error. B) John did not commit a Type II error. C) John committed neither a Type I nor Type II error. D) John committed both a Type I and a Type II error. Answer: B Explanation: John could have committed Type II error only if he failed to reject H0. Difficulty: 3 Hard Topic: 09.05 Testing a Mean: Unknown Population Variance Learning Objective: 09-07 Perform a hypothesis test for a mean with unknown σ using t. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 18 Copyright 2019 © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.

61) "My careful physical examination shows no evidence of any serious problem," said Doctor Morpheus. "However, a very costly lab test can be performed to check for the rare condition known as estomalgia fatalis. The test is almost invariably negative for persons with your age and symptoms. My personal hypothesis is that the occasional stomach pain you reported is due to indigestion caused by eating tacos with too much hot sauce. But you must decide for yourself." As you consider your doctor's hypothesis, what would be the consequence of Type I error on your part? A) It cannot be determined without knowing the type of test. B) Your estomalgia fatalis will go undetected. C) You will waste money on an unnecessary lab test. D) Your survivors will enjoy a sizeable malpractice award. Answer: C Explanation: Type I error is rejecting the doctor's advice when it was correct. Difficulty: 3 Hard Topic: 09.02 Type I and Type II Error Learning Objective: 09-02 Define Type I error, Type II error, and power. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 62) Which of the following statements is correct? A) Increasing α will make it more likely that we will reject H0, ceteris paribus. B) Doubling the sample size roughly doubles the test statistic, ceteris paribus. C) A higher standard deviation would increase the power of a test for a mean. D) The p-value shows the probability that the null hypothesis is false. Answer: A Explanation: A larger α will make it easier to reject H0 (e.g., z.05 = 1.645 versus z.01 = 2.326). Difficulty: 3 Hard Topic: 09.04 Testing a Mean: Known Population Variance Learning Objective: 09-05 Perform a hypothesis test for a mean with known σ using z. Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation

63) "I believe your airplane's engine is sound," states the mechanic. "I've been over it carefully, and can't see anything wrong. I'd be happy to tear the engine down completely for an internal inspection at a cost of $1,500. But I believe that roughness you heard in the engine on your last flight was probably just a bit of water in the fuel, which passed harmlessly through the engine and is now gone." As the pilot considers the mechanic's hypothesis, the cost of Type I error is A) the pilot will experience the thrill of no-engine flight. B) the pilot will be out $1,500 unnecessarily. C) the mechanic will lose a good customer. D) impossible to determine without knowing α. Answer: B Explanation: Type I error is rejecting the mechanic's advice when it was correct. Difficulty: 3 Hard Topic: 09.02 Type I and Type II Error Learning Objective: 09-02 Define Type I error, Type II error, and power. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 64) A study over a 10-year period showed that a certain mammogram test had a 50 percent rate of false positives. This indicates that A) about half the tests indicated cancer. B) about half the tests missed a cancer that exists. C) about half the tests showed a cancer that did not exist. D) about half the women tested actually had no cancer. Answer: C Explanation: This is a 50 percent chance of Type I error. Difficulty: 2 Medium Topic: 09.02 Type I and Type II Error Learning Objective: 09-02 Define Type I error, Type II error, and power. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

65) You are driving a van packed with camping gear (total weight 3,500 pounds including yourself and family) into a northern wilderness area. You take a "short cut" that turns into a onelane road, with no room to turn around. After 11 miles you come to a narrow bridge with a faded sign saying, "Safe Up to 2 Tons." About a half-mile ahead, you can see that your road rejoins the main highway. You consider the sign's hypothesis carefully before making a decision. The cost of Type I error is A) that you pass safely over the bridge and everyone's happy. B) about $23,900, not including medical bills. C) that you will find out just how cold that river actually is. D) that your kids will think you are a chicken. Answer: D Explanation: Type I error is rejecting the sign's message when it was correct. Difficulty: 3 Hard Topic: 09.02 Type I and Type II Error Learning Objective: 09-02 Define Type I error, Type II error, and power. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 66) After lowering the landing gear, the pilot notices that the "gear down and locked" light is not illuminated. "It's probably just a burned out light bulb," she says, as she proceeds on final approach for landing. Considering the pilot's hypothesis, which is the result of Type I error? A) The sound of metal scraping on concrete will be heard upon landing. B) The landing is delayed unnecessarily while the bulb and gear are checked. C) We cannot be sure without knowing whether or not the bulb is actually faulty. Answer: B Explanation: Type I error is concluding there is a problem when there was not. Difficulty: 2 Medium Topic: 09.02 Type I and Type II Error Learning Objective: 09-02 Define Type I error, Type II error, and power. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

67) As you are crossing a field at the farm, your country cousin Jake assures you, "Don't worry about that old bull coming toward us. He's harmless." As you consider Jake's hypothesis, what would be Type I error on your part? A) You will soon feel the bull's horns. B) You will run away for no good reason. C) Jake will not have any more visits from you. Answer: B Explanation: Type I error is rejecting Jake's advice when he was right. Difficulty: 2 Medium Topic: 09.02 Type I and Type II Error Learning Objective: 09-02 Define Type I error, Type II error, and power. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 68) Which is not true of p-values? A) When they are small, we want to reject H0. B) They measure the probability of an incorrect decision. C) They show the chance of Type I error if we reject H0. D) They do not require α to be specified a priori. Answer: B Explanation: The p-value tells the likelihood of the observed sample result (or one that is more extreme) assuming that H0 is true. Difficulty: 2 Medium Topic: 09.04 Testing a Mean: Known Population Variance Learning Objective: 09-05 Perform a hypothesis test for a mean with known σ using z. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation

69) For a test of a mean, which of the following is incorrect? A) H0 is rejected when the calculated p-value is less than the critical value of the test statistic. B) In a right-tailed test, we reject H0 when the test statistic exceeds the critical value. C) The critical value is based on the researcher's chosen level of significance. D) If H0: μ ≤ 100 and H1: μ > 100, then the test is right-tailed. Answer: A Explanation: We compare the p-value with α (not with the critical value). Difficulty: 2 Medium Topic: 09.04 Testing a Mean: Known Population Variance Learning Objective: 09-05 Perform a hypothesis test for a mean with known σ using z. Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation 70) Guidelines for the Jolly Blue Giant Health Insurance Company say that the average hospitalization for a triple hernia operation should not exceed 30 hours. A diligent auditor studied records of 16 randomly chosen triple hernia operations at Hackmore Hospital and found a mean hospital stay of 40 hours with a standard deviation of 20 hours. "Aha!" she cried, "the average stay exceeds the guideline." At α = .025, the critical value for a right-tailed test of her hypothesis is A) 1.753 B) 2.131 C) 1.645 D) 1.960 Answer: B Explanation: Using Appendix D with d.f. = 16 − 1 = 15, we get t.025 = 2.131. Alternatively, use the Excel function =T.INV(0.975,15) = 2.1345. Difficulty: 2 Medium Topic: 09.05 Testing a Mean: Unknown Population Variance Learning Objective: 09-07 Perform a hypothesis test for a mean with unknown σ using t. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

71) Guidelines for the Jolly Blue Giant Health Insurance Company say that the average hospitalization for a triple hernia operation should not exceed 30 hours. A diligent auditor studied records of 16 randomly chosen triple hernia operations at Hackmore Hospital and found a mean hospital stay of 40 hours with a standard deviation of 20 hours. "Aha!" she cried, "the average stay exceeds the guideline." The value of the test statistic for her hypothesis is A) 2.080 B) 0.481 C) 1.866 D) 2.000 Answer: D Explanation: tcalc = (40 − 30)/[(20)/161/2] = 2.000. Difficulty: 2 Medium Topic: 09.05 Testing a Mean: Unknown Population Variance Learning Objective: 09-07 Perform a hypothesis test for a mean with unknown σ using t. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 72) Guidelines for the Jolly Blue Giant Health Insurance Company say that the average hospitalization for a triple hernia operation should not exceed 30 hours. A diligent auditor studied records of 16 randomly chosen triple hernia operations at Hackmore Hospital and found a mean hospital stay of 40 hours with a standard deviation of 20 hours. "Aha!" she cried, "the average stay exceeds the guideline." The p-value for a right-tailed test of her hypothesis is A) between .05 and .10. B) between .025 and .05. C) between .01 and .025. D) less than .01. Answer: B Explanation: Use Appendix D with tcalc = 2.000 or Excel =T.DIST.RT(2.000,15) = .0320. Difficulty: 3 Hard Topic: 09.05 Testing a Mean: Unknown Population Variance Learning Objective: 09-06 Use tables or Excel to find the p-value in tests of µ. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

73) For a right-tailed test of a hypothesis for a population mean with n = 14, the value of the test statistic was t = 1.863. The p-value is A) between .05 and .025. B) between .10 and .05. C) greater than .10. D) less than .01. Answer: A Explanation: For d.f. = 13, t.025 = 2.160 and t.05 = 1.771 or Excel =T.DIST.RT(1.863,13) = . 0426. Difficulty: 2 Medium Topic: 09.05 Testing a Mean: Unknown Population Variance Learning Objective: 09-06 Use tables or Excel to find the p-value in tests of µ. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 74) Hypothesis tests for a mean using the critical value method require A) using a two-tailed test. B) sampling a normal population. C) knowing the true population mean. D) specifying α in advance. Answer: D Explanation: You cannot find the critical value without specifying α. Difficulty: 2 Medium Topic: 09.03 Decision Rules and Critical Values Learning Objective: 09-04 Explain decision rules, critical values, and rejection regions. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 75) The level of significance is not A) the probability of a "false rejection." B) a value between 0 and 1. C) the likelihood of rejecting the null hypothesis when it is true. D) the chance of accepting a true null hypothesis. Answer: D Explanation: The level of significance is the risk of rejecting a true null hypothesis. Difficulty: 2 Medium Topic: 09.03 Decision Rules and Critical Values Learning Objective: 09-04 Explain decision rules, critical values, and rejection regions. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 25 Copyright 2019 © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.

76) The critical value in a hypothesis test A) is calculated from the sample data. B) usually is .05 or .01 in most statistical tests. C) separates the acceptance and rejection regions. D) depends on the value of the test statistic. Answer: C Explanation: We can specify whatever α we wish to set the desired tail area(s). Difficulty: 2 Medium Topic: 09.03 Decision Rules and Critical Values Learning Objective: 09-04 Explain decision rules, critical values, and rejection regions. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 77) Which is not a likely reason to choose the z distribution for a hypothesis test of a mean? A) The value of σ is known. B) The sample size n is very large. C) The population is normal. D) The value of σ is very large. Answer: D Explanation: We use z any time σ is known. Difficulty: 2 Medium Topic: 09.04 Testing a Mean: Known Population Variance Learning Objective: 09-05 Perform a hypothesis test for a mean with known σ using z. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation

78) Dullco Manufacturing claims that its alkaline batteries last at least 40 hours on average in a certain type of portable CD player. But tests on a random sample of 18 batteries from a day's large production run showed a mean battery life of 37.8 hours with a standard deviation of 5.4 hours. To test DullCo's hypothesis, the test statistic is A) −1.980 B) −1.728 C) −2.101 D) −1.960 Answer: B Explanation: tcalc = (37.8 − 40)/[(5.4)/181/2] = −1.72848. Difficulty: 2 Medium Topic: 09.05 Testing a Mean: Unknown Population Variance Learning Objective: 09-07 Perform a hypothesis test for a mean with unknown σ using t. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 79) Dullco Manufacturing claims that its alkaline batteries last at least 40 hours on average in a certain type of portable CD player. But tests on a random sample of 18 batteries from a day's large production run showed a mean battery life of 37.8 hours with a standard deviation of 5.4 hours. In a left-tailed test at α = .05, which is the most accurate statement? A) We would strongly reject the claim. B) We would clearly fail to reject the claim. C) We would face a rather close decision. D) We would switch to α = .01 for a more powerful test. Answer: C Explanation: tcalc = (37.8 − 40)/[(5.4)/181/2] = −1.728, while for d.f. = 18 − 1 = 17 we get t.05 = −1.740, so it is a close decision. Difficulty: 2 Medium Topic: 09.05 Testing a Mean: Unknown Population Variance Learning Objective: 09-07 Perform a hypothesis test for a mean with unknown σ using t. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

80) Dullco Manufacturing claims that its alkaline batteries last at least 40 hours on average in a certain type of portable CD player. But tests on a random sample of 18 batteries from a day's large production run showed a mean battery life of 37.8 hours with a standard deviation of 5.4 hours. To test DullCo's hypothesis, the p-value is A) slightly less than .05. B) exactly equal to .05. C) slightly greater than .05. D) uncertain without knowing α. Answer: C Explanation: tcalc = −1.728, t.05 = −1.740 or Excel =T.DIST(-1.72848,17,1) = .0511. Difficulty: 2 Medium Topic: 09.05 Testing a Mean: Unknown Population Variance Learning Objective: 09-07 Perform a hypothesis test for a mean with unknown σ using t. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 81) For tests of a mean, if other factors are held constant, which statement is correct? A) The critical value of Student's t increases as n increases. B) A test statistic tcalc = 1.853 with n = 16 leads to rejection at α = .05 in a one-tailed test. C) It is harder to reject the null hypothesis in a two-tailed test rather than a one-tailed test. D) If we desire α = .10, then a p-value of .13 would lead us to reject the null hypothesis. Answer: C Explanation: Rejection in a two-tailed test implies rejection in a one-tailed test, but not vice versa. Difficulty: 3 Hard Topic: 09.05 Testing a Mean: Unknown Population Variance Learning Objective: 09-07 Perform a hypothesis test for a mean with unknown σ using t. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

82) For a sample size of n = 100, and σ = 10, we want to test the hypothesis H0: μ = 100. The sample mean is 103. The test statistic is A) 1.645 B) 1.960 C) 3.000 D) 0.300 Answer: C Explanation: zcalc = (103 − 100)/[(10)/1001/2] = 3.000. Difficulty: 2 Medium Topic: 09.04 Testing a Mean: Known Population Variance Learning Objective: 09-05 Perform a hypothesis test for a mean with known σ using z. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 83) When testing the hypothesis H0: μ = 100 with n = 100 and σ2 = 100, we find that the sample mean is 97. The test statistic is A) −3.000 B) −10.00 C) −0.300 D) −0.030 Answer: A Explanation: zcalc = (97 − 100)/[(10)/1001/2] = −3.000. Difficulty: 2 Medium Topic: 09.04 Testing a Mean: Known Population Variance Learning Objective: 09-05 Perform a hypothesis test for a mean with known σ using z. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

84) Given a normal distribution with σ = 3, we want to test the hypothesis H0: μ = 20. We find that the sample mean is 21. The test statistic is A) 1.000 B) 1.645 C) 1.960 D) impossible to find without more information. Answer: D Explanation: The sample size is needed to calculate the z test statistic. Difficulty: 2 Medium Topic: 09.04 Testing a Mean: Known Population Variance Learning Objective: 09-05 Perform a hypothesis test for a mean with known σ using z. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 85) In testing a proportion, which of the following statements is incorrect? A) Using α = .05 rather than α = .01 would make it more likely that H0 will be rejected. B) When the sample proportion is p = .02 and n = 150, it is safe to assume normality. C) An 80 percent confidence interval is narrower than the 90 percent confidence interval, ceteris paribus. D) The sample proportion may be assumed approximately normal if the sample is large enough. Answer: B Explanation: We want at least 10 "successes," but np = 3 in this example. Difficulty: 2 Medium Topic: 09.06 Testing a Proportion Learning Objective: 09-09 Check whether normality may be assumed in testing a proportion. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 86) Which of the following is not a characteristic of the t distribution? A) It is a continuous distribution. B) It has a mean of zero. C) It a symmetric distribution. D) It is similar to the z distribution when n is small. Answer: D Explanation: Student's t resembles z most closely for a large sample size. Difficulty: 1 Easy Topic: 09.05 Testing a Mean: Unknown Population Variance Learning Objective: 09-07 Perform a hypothesis test for a mean with unknown σ using t. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 30 Copyright 2019 © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.

87) Which of the following is not a valid null hypothesis? A) H0: μ ≥ 0 B) H0: μ ≤ 0 C) H0: μ ≠ 0 D) H0: μ = 0 Answer: C Explanation: The null hypothesis cannot contain a two-tailed inequality. Difficulty: 2 Medium Topic: 09.03 Decision Rules and Critical Values Learning Objective: 09-03 Formulate a null and alternative hypothesis for µ or π. Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation 88) Given that in a one-tailed test you cannot reject H0, can you reject H0 in a two-tailed test at the same α? A) Yes. B) No. C) Maybe. Answer: B Explanation: Rejection in a two-tailed test implies rejection in a one-tailed test, but not vice versa. Difficulty: 2 Medium Topic: 09.03 Decision Rules and Critical Values Learning Objective: 09-04 Explain decision rules, critical values, and rejection regions. Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation

89) The process that produces Sonora Bars (a type of candy) is intended to produce bars with a mean weight of 56 gm. The process standard deviation is known to be 0.77 gm. A random sample of 49 candy bars yields a mean weight of 55.82 gm. Which are the hypotheses to test whether the mean is smaller than it is supposed to be? A) H0: μ ≤ 56, H1: μ > 56 B) H0: μ ≥ 56, H1: μ < 56 C) H0: μ = 56, H1: μ ≠ 56 D) H0: μ < 56, H1: μ ≥ 56 Answer: B Explanation: We want a left-tailed alternative hypothesis. Difficulty: 1 Easy Topic: 09.04 Testing a Mean: Known Population Variance Learning Objective: 09-03 Formulate a null and alternative hypothesis for µ or π. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 90) The process that produces Sonora Bars (a type of candy) is intended to produce bars with a mean weight of 56 gm. The process standard deviation is known to be 0.77 gm. A random sample of 49 candy bars yields a mean weight of 55.82 gm. Find the test statistic to see whether the candy bars are smaller than they are supposed to be. A) −1.636 B) −1.645 C) −1.677 Answer: A Explanation: zcalc = (55.82 − 56)/[(0.77)/491/2] = −1.63636. Difficulty: 2 Medium Topic: 09.04 Testing a Mean: Known Population Variance Learning Objective: 09-05 Perform a hypothesis test for a mean with known σ using z. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

91) The process that produces Sonora Bars (a type of candy) is intended to produce bars with a mean weight of 56 gm. The process standard deviation is known to be 0.77 gm. A random sample of 49 candy bars yields a mean weight of 55.82 gm. Find the p-value for a test to see whether the candy bars are smaller than they are supposed to be. A) Between .05 and .10 B) Between .025 and .05 C) Between .01 and .025 D) Less than .01 Answer: A Explanation: zcalc = (55.82 − 56)/[(0.77)/491/2] = −1.63636 and z.05 = −1.645, or find the exact p-value as =NORM.S.DIST(-1.63636,1) = .0509. Difficulty: 3 Hard Topic: 09.04 Testing a Mean: Known Population Variance Learning Objective: 09-06 Use tables or Excel to find the p-value in tests of µ. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 92) A sample of 16 ATM transactions shows a mean transaction time of 67 seconds with a standard deviation of 12 seconds. Find the test statistic to decide whether the mean transaction time exceeds 60 seconds. A) 1.457 B) 2.037 C) 2.333 D) 1.848 Answer: C Explanation: tcalc = (67 − 60)/[(12)/161/2] = 2.333. Difficulty: 2 Medium Topic: 09.05 Testing a Mean: Unknown Population Variance Learning Objective: 09-07 Perform a hypothesis test for a mean with unknown σ using t. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

93) A sample of 16 ATM transactions shows a mean transaction time of 67 seconds with a standard deviation of 12 seconds. State the hypotheses to test whether the mean transaction time exceeds 60 seconds. A) H0: μ ≤ 60, H1: μ > 60 B) H0: μ ≥ 60, H1: μ < 60 C) H0: μ = 60, H1: μ ≠ 60 D) H0: μ < 60, H1: μ ≥ 60 Answer: A Explanation: We want a right-tailed test in this case. Difficulty: 1 Easy Topic: 09.05 Testing a Mean: Unknown Population Variance Learning Objective: 09-03 Formulate a null and alternative hypothesis for µ or π. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 94) A sample of 16 ATM transactions shows a mean transaction time of 67 seconds with a standard deviation of 12 seconds. Find the critical value to test whether the mean transaction time exceeds 60 seconds at α = .01. A) 2.947 B) 2.602 C) 2.583 D) 2.333 Answer: B Explanation: For d.f. = 15, use Appendix D to find t.01 = 2.602. Difficulty: 1 Easy Topic: 09.05 Testing a Mean: Unknown Population Variance Learning Objective: 09-07 Perform a hypothesis test for a mean with unknown σ using t. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

95) Given H0: μ ≥ 18 and H1: μ < 18, we would commit a Type I error if we A) conclude that μ ≥ 18 when the truth is that μ < 18. B) conclude that μ < 18 when the truth is that μ ≥ 18. C) fail to reject μ ≥ 18 when the truth is that μ < 18. Answer: B Explanation: Rejecting a true null hypothesis is a Type I error. Difficulty: 2 Medium Topic: 09.02 Type I and Type II Error Learning Objective: 09-02 Define Type I error, Type II error, and power. Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation 96) For a right-tailed test of a hypothesis for a single population mean with n = 10, the value of the test statistic was t = 1.411. The p-value is A) between .05 and .025. B) between .10 and .05. C) greater than .10. D) less than .001. Answer: B Explanation: From Appendix D with d.f. = 9, we get t.05 = 1.833 and t.10 = 1.383. Difficulty: 2 Medium Topic: 09.05 Testing a Mean: Unknown Population Variance Learning Objective: 09-06 Use tables or Excel to find the p-value in tests of µ. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 97) Last year, 10 percent of all teenagers purchased a new iPhone. This year, a sample of 260 randomly chosen teenagers showed that 39 had purchased a new iPhone. The test statistic to find out whether the percentage has risen would be A) 2.687 B) 2.758 C) .0256 D) 2.258 Answer: A Explanation: p = 39/260 = .15, π0 = .10, zcalc = (.15 − .10)/[(.10)(1 − .10)/260]1/2 = 2.68742. Difficulty: 2 Medium Topic: 09.06 Testing a Proportion Learning Objective: 09-08 Perform a hypothesis test for a proportion and find the p-value. Bloom's: Apply AACSB: Analytical Thinking 35 Copyright 2019 © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.

Accessibility: Keyboard Navigation 98) Last year, 10 percent of all teenagers purchased a new iPhone. This year, a sample of 260 randomly chosen teenagers showed that 39 had purchased a new iPhone. To test whether the percentage has risen, the critical value at α = .05 is A) 1.645 B) 1.658 C) 1.697 D) 1.960 Answer: A Explanation: z.05 = 1.645. Difficulty: 1 Easy Topic: 09.06 Testing a Proportion Learning Objective: 09-08 Perform a hypothesis test for a proportion and find the p-value. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 99) Last year, 10 percent of all teenagers purchased a new iPhone. This year, a sample of 260 randomly chosen teenagers showed that 39 had purchased a new iPhone. To test whether the percentage has risen, the p-value is approximately A) .0501 B) .0314 C) .0492 D) .0036 Answer: D Explanation: p = 39/260 = .15, π0 = .10, zcalc = (.15 − .10)/[(.10)(1 − .10)/260]1/2 = 2.68742. From Appendix C we get P(Z > 2.69) = .0036, or from Excel =1-NORM.S.DIST(2.68742,1) = . 0036. Difficulty: 3 Hard Topic: 09.06 Testing a Proportion Learning Objective: 09-08 Perform a hypothesis test for a proportion and find the p-value. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

100) Ajax Peanut Butter's quality control allows 2 percent of the jars to exceed the quality standard for insect fragments. A sample of 150 jars from the current day's production reveals that 30 exceed the quality standard for insect fragments. Which is incorrect? A) Normality of p may safely be assumed in the hypothesis test. B) A right-tailed test would be appropriate. C) We strongly suspect that quality control standards are not met. D) Type II error is more of a concern in this case than Type I error. Answer: A Explanation: nπ0 = (150)(.02) = 3, so normality of p is doubtful. Difficulty: 2 Medium Topic: 09.06 Testing a Proportion Learning Objective: 09-09 Check whether normality may be assumed in testing a proportion. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 101) In the nation of Gondor, the EPA requires that half the new cars sold will meet a certain particulate emission standard a year later. A sample of 64 one-year-old cars revealed that only 24 met the particulate emission standard. The test statistic to see whether the proportion is below the requirement is A) −1.645 B) −2.066 C) −2.000 D) −1.960 Answer: C Explanation: p = 24/64 = .375, π0 = .50, zcalc = (.375 − .50)/[(.50)(1 − .50)/64]1/2 = −2.000. Difficulty: 2 Medium Topic: 09.06 Testing a Proportion Learning Objective: 09-08 Perform a hypothesis test for a proportion and find the p-value. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

102) The hypotheses H0: π ≥ .40, H1: π < .40 would require A) a left-tailed test. B) a right-tailed test. C) a two-tailed test. Answer: A Explanation: The inequality in the alternative hypothesis points to the direction of the test. Difficulty: 1 Easy Topic: 09.06 Testing a Proportion Learning Objective: 09-03 Formulate a null and alternative hypothesis for µ or π. Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation 103) At α = .05, the critical value to test the hypotheses H0: π ≥ .40, H1: π < .40 would be A) −1.645 B) −1.960 C) −2.326 D) impossible to determine without more information. Answer: A Explanation: z.05 = −1.645. Difficulty: 1 Easy Topic: 09.06 Testing a Proportion Learning Objective: 09-08 Perform a hypothesis test for a proportion and find the p-value. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 104) In a test of a mean, the reported p-value is .025. Using α =.05 the conclusion would be to A) accept the null hypothesis. B) reject the null hypothesis. C) fail to reject the null hypothesis. D) gather more evidence due to inconclusive results. Answer: B Explanation: Reject the null hypothesis if the p-value is smaller than α. Difficulty: 1 Easy Topic: 09.04 Testing a Mean: Known Population Variance Learning Objective: 09-05 Perform a hypothesis test for a mean with known σ using z. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

105) Which of the following decisions could result in a Type II error for a test? A) Reject the alternative hypothesis. B) Reject the null hypothesis. C) Fail to reject the null hypothesis. D) Make no decision. Answer: C Explanation: Failing to reject H0 could lead to a Type II error (but not a Type I error). Difficulty: 1 Easy Topic: 09.02 Type I and Type II Error Learning Objective: 09-02 Define Type I error, Type II error, and power. Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation 106) The Melodic Kortholt Company will change its current health plan if at least half the employees are dissatisfied with it. A trial sample of 25 employees shows that 16 are dissatisfied. In this problem A) normality of the sample proportion should not be assumed. B) normality of the sample proportion can be assumed. C) normality of the sample proportion cannot be judged without knowing π. Answer: B Explanation: nπ0 = (25)(.50) = 12.5, so we expect at least 10 "successes" and 10 "failures" (be careful to use π0 instead of p to check for normality). Difficulty: 2 Medium Topic: 09.06 Testing a Proportion Learning Objective: 09-09 Check whether normality may be assumed in testing a proportion. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

107) The Melodic Kortholt Company will change its current health plan if at least half the employees are dissatisfied with it. A trial sample of 25 employees shows that 16 are dissatisfied. The p-value for a right-tailed test is A) .1337 B) .4192 C) .0901 D) .0808 Answer: D Explanation: p = 16/25 = .64, π0 = .50, zcalc = (.64 − .50)/[(.50)(1 − .50)/25]1/2 = 1.400. From Appendix C we get P(Z > 1.40) = .0808, or from Excel =1-NORM.S.DIST(1.400,1) = .08076. Difficulty: 3 Hard Topic: 09.06 Testing a Proportion Learning Objective: 09-08 Perform a hypothesis test for a proportion and find the p-value. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 108) The Melodic Kortholt Company will change its current health plan if at least half the employees are dissatisfied with it. A trial sample of 25 employees shows that 16 are dissatisfied. For a right-tailed test, the test statistic would be A) 1.227 B) 1.375 C) 1.400 D) 1.115 Answer: C Explanation: p = 16/25 = .64, π0 = .50, zcalc = (.64 − .50)/[(.50)(1 − .50)/25]1/2 = 1.400. Difficulty: 2 Medium Topic: 09.06 Testing a Proportion Learning Objective: 09-08 Perform a hypothesis test for a proportion and find the p-value. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

109) If the sample size increases from 25 to 100 and the level of significance stays the same, then A) the risk of a Type II error would decrease. B) the risk of a Type I error would decrease. C) the risk of both Type I and Type II errors would decrease. D) the risk of neither Type I nor Type II error would decrease. Answer: A Explanation: We are holding α constant, so the larger sample size will reduce β. Difficulty: 2 Medium Topic: 09.02 Type I and Type II Error Learning Objective: 09-02 Define Type I error, Type II error, and power. Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation 110) "Currently, only 20 percent of arrested drug pushers are convicted," cried candidate Courageous Calvin in a campaign speech. "Elect me and you'll see a big increase in convictions." A year after his election, a random sample of 144 case files of arrested drug pushers showed 36 convictions. For a right-tailed test, the p-value is approximately A) .9332 B) .0668 C) .0435 D) .0250 Answer: B Explanation: p = 36/144 = .25, π0 = .20, zcalc = (.25 − .20)/[(.20)(1 − .20)/144]1/2 = 1.500. From Appendix C, we get P(Z > 1.50) = .0668, or from Excel =1-NORM.S.DIST(1.500,1) = . 0668. Difficulty: 3 Hard Topic: 09.06 Testing a Proportion Learning Objective: 09-08 Perform a hypothesis test for a proportion and find the p-value. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

111) In a right-tailed test, a statistician got a z test statistic of 1.47. What is the p-value? A) .4292 B) .0708 C) .0874 D) .9292 Answer: B Explanation: From Appendix C we get P(Z > 1.47) = .0708 or from Excel =1NORM.S.DIST(1.47,1) = .0708. Difficulty: 2 Medium Topic: 09.04 Testing a Mean: Known Population Variance Learning Objective: 09-06 Use tables or Excel to find the p-value in tests of µ. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 112) In a left-tailed test, a statistician got a z test statistic of −1.720. What is the p-value? A) .4292 B) .0709 C) .0427 D) .0301 Answer: C Explanation: From Appendix C, we get P(Z < −1.72) = .0427 or from the Excel function =NORM.S.DIST(-1.720,1) = .0427. Difficulty: 2 Medium Topic: 09.04 Testing a Mean: Known Population Variance Learning Objective: 09-06 Use tables or Excel to find the p-value in tests of µ. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 113) In a two-tailed test, a statistician got a z test statistic of 1.47. What is the p-value? A) .0708 B) .1416 C) .0874 D) .0301 Answer: B Explanation: From Appendix C, we get 2 × P(Z > 1.47) = 2 × .0708 = .1416. The Excel version of this calculation is =2*(1-NORM.S.DIST(1.47,1)) = .14156. Difficulty: 2 Medium Topic: 09.04 Testing a Mean: Known Population Variance Learning Objective: 09-06 Use tables or Excel to find the p-value in tests of µ. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 42 Copyright 2019 © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.

114) Which of the following statements is true? A) Decreasing α will increase the power of the test. B) Doubling the sample size will double the power of the test. C) A higher standard deviation would increase the power if we are testing a mean. D) Power of the test rises if the true mean is farther from the hypothesized mean. Answer: D Explanation: A test becomes more sensitive (greater power) when the truth differs greatly from H0. Difficulty: 3 Hard Topic: 09.07 Power Curves and OC Curves (Optional) Learning Objective: 09-10 Interpret a power curve or OC curve (optional). Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation 115) High power in a hypothesis test about one sample mean is likely to be associated with A) small sample size. B) low α. C) large β. D) small σ. Answer: D Explanation: Less variation in the population makes the test more sensitive (greater power). Difficulty: 2 Medium Topic: 09.07 Power Curves and OC Curves (Optional) Learning Objective: 09-10 Interpret a power curve or OC curve (optional). Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation 116) The power of a test is the probability of A) concluding H1 when H1 is true. B) concluding H1 when H0 is true. C) concluding H0 when H0 is true. D) concluding H0 when H1 is true. Answer: A Explanation: Review the definition of power. Difficulty: 2 Medium Topic: 09.02 Type I and Type II Error Learning Objective: 09-02 Define Type I error, Type II error, and power. Bloom's: Remember AACSB: Analytical Thinking 43 Copyright 2019 © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.

Accessibility: Keyboard Navigation 117) Which is not a step in hypothesis testing? A) Formulate the hypotheses. B) Specify the desired Type I error. C) Find the test statistic from a table. D) Formulate a decision rule. Answer: C Explanation: Review the steps in hypothesis testing. Difficulty: 2 Medium Topic: 09.01 Logic of Hypothesis Testing Learning Objective: 09-01 Know the steps in testing hypotheses and define H0 and H1. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 118) Which is an invalid alternative hypothesis? A) H1: μ ≠ 18 B) H1: μ = 18 C) H1: μ > 18 D) H1: μ < 18 Answer: B Explanation: You cannot have an equality in the alternative hypothesis. Difficulty: 2 Medium Topic: 09.03 Decision Rules and Critical Values Learning Objective: 09-03 Formulate a null and alternative hypothesis for µ or π. Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation 119) Which is a valid null hypothesis? A) H0: μ ≠ 18 B) H0: μ = 18 C) H0: μ > 18 D) H0: μ < 18 Answer: B Explanation: The null hypothesis cannot have < or > or ≠. Difficulty: 2 Medium Topic: 09.03 Decision Rules and Critical Values Learning Objective: 09-04 Explain decision rules, critical values, and rejection regions. Bloom's: Understand AACSB: Analytical Thinking 44 Copyright 2019 © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.

Accessibility: Keyboard Navigation 120) A two-tailed hypothesis test for H0: π = .30 at α = .05 is analogous to A) asking if the 90 percent confidence interval for π contains .30. B) asking if the 95 percent confidence interval for π contains .30. C) asking if the p-value (area in both tails combined) is less than .025. D) asking if the p-value (area in both tails combined) is less than .10. Answer: B Explanation: This statement is true for a two-tailed test only. Difficulty: 2 Medium Topic: 09.06 Testing a Proportion Learning Objective: 09-08 Perform a hypothesis test for a proportion and find the p-value. Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation 121) For a right-tailed test of hypothesis for a population mean with known σ, the test statistic was z = 1.79. The p-value is A) .0367 B) .9633 C) .1186 D) .0179 Answer: A Explanation: From Appendix C, we get P(Z > 1.79) = .0367 or from Excel =1NORM.S.DIST(1.79,1) = .0367. Difficulty: 1 Easy Topic: 09.04 Testing a Mean: Known Population Variance Learning Objective: 09-06 Use tables or Excel to find the p-value in tests of µ. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

122) If n = 25 and α = .05 in a right-tailed test of a mean with unknown σ, the critical value is A) 1.960 B) 1.645 C) 1.711 D) .0179 Answer: C Explanation: Using d.f. = 24, t.05 = 1.711 from Appendix D. Alternatively, use the Excel function =T.INV(0.95,24) = 1.71088. Difficulty: 1 Easy Topic: 09.05 Testing a Mean: Unknown Population Variance Learning Objective: 09-06 Use tables or Excel to find the p-value in tests of µ. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 123) The researcher's null hypothesis is H0: σ2 ≤ 22. A sample of n = 25 items yields a sample variance of s2 = 28.5. The critical value of chi-square for a right-tailed test at α = .05 is A) 1.960 B) 1.645 C) 13.85 D) 36.42 Answer: D Explanation: From Appendix E with d.f. = 24 we get χ2.05 = 36.42. Alternatively, use the Excel function =CHISQ.INV.RT(0.05,24) = 36.4150. Difficulty: 2 Medium Topic: 09.08 Tests for One Variance (Optional) Learning Objective: 09-11 Perform a hypothesis test for a variance (optional). Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

124) The researcher's null hypotheses is H0: σ2 ≤ 22. A sample of n = 25 items yields a sample variance of s2 = 28.5. The test statistic is A) 31.09 B) 26.42 C) impossible to determine unless we know whether it is a one-tailed test. D) impossible to determine without knowing α. Answer: A Explanation: χ2calc = (n − 1)s2/σ2 = (25 − 1)(28.5)/(22) = 31.09. Difficulty: 3 Hard Topic: 09.08 Tests for One Variance (Optional) Learning Objective: 09-11 Perform a hypothesis test for a variance (optional). Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 125) The researcher's null hypothesis is H0: σ2 = 420. A sample of n = 18 items yields a sample variance of s2 = 512. The critical values of chi-square for a two-tailed test at α = .05 are: A) 8.672 and 27.59 B) 7.564 and 30.19 C) −1.960 and +1.960 D) 9.390 and 28.87 Answer: B Explanation: From Appendix E with d.f. = 17 we get χ2.025 = 7.564 (left tail) and 30.19 (right tail). Difficulty: 2 Medium Topic: 09.08 Tests for One Variance (Optional) Learning Objective: 09-11 Perform a hypothesis test for a variance (optional). Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

126) The researcher's null hypotheses is H0: σ2 = 420. A sample of n = 18 items yields a sample variance of s2 = 512. The test statistic is A) 34.09 B) 20.72 C) 14.77 D) must know α to answer. Answer: B Explanation: χ2calc = (n − 1)s2/σ2 = (18 − 1)(512)/(420) = 20.72. Difficulty: 3 Hard Topic: 09.08 Tests for One Variance (Optional) Learning Objective: 09-11 Perform a hypothesis test for a variance (optional). Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 127) In hypothesis testing, Type I error is A) always set at 5 percent. B) smaller than or equal to 5 percent. C) the probability of rejecting H0 when H0 is true. D) the probability of rejecting H0 when H1 is true. Answer: C Explanation: Rejecting a true null hypothesis is a Type I error. Difficulty: 2 Medium Topic: 09.02 Type I and Type II Error Learning Objective: 09-02 Define Type I error, Type II error, and power. Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation 128) In hypothesis testing, the value of β is A) equal to 1 minus the probability of committing a Type I error. B) the probability of concluding H0 when H0 is true. C) the probability of concluding H0 when H1 is true. Answer: C Explanation: Failing to reject a false null hypothesis is a Type II error. Difficulty: 2 Medium Topic: 09.02 Type I and Type II Error Learning Objective: 09-02 Define Type I error, Type II error, and power. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 48 Copyright 2019 © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.

129) Regarding the probability of Type I error (α) and Type II error (β), which statement is true? A) β > α B) β < α C) α + β = 1 D) Power = 1 − β. Answer: D Explanation: Although α and β are related, there is no simple equation relating α and β. Difficulty: 3 Hard Topic: 09.02 Type I and Type II Error Learning Objective: 09-02 Define Type I error, Type II error, and power. Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation 130) In the hypothesis H0: μ = μ0, the value of μ0 is not derived from A) the sample. B) past experience. C) a target or benchmark. D) a scientific theory. Answer: A Explanation: The hypothesized value of the mean is a target or is based on past experience. Difficulty: 2 Medium Topic: 09.03 Decision Rules and Critical Values Learning Objective: 09-04 Explain decision rules, critical values, and rejection regions. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 131) In testing the hypotheses H0: π ≤ π0, H1: π > π0, we would use a A) two-tailed test. B) left-tailed test. C) right-tailed test. D) breathalyzer test. Answer: C Explanation: The inequality in H1 always points to the direction of the test. Difficulty: 1 Easy Topic: 09.03 Decision Rules and Critical Values Learning Objective: 09-03 Formulate a null and alternative hypothesis for µ or π. Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation 49 Copyright 2019 © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.

132) We can assume that the sample proportion is normally distributed if A) we have 10 successes in the sample. B) we have 10 failures in the sample. C) we have both 10 successes and 10 failures in the sample. D) the population is known. Answer: C Explanation: As a guideline, we want at least 10 successes and 10 failures to assume a normal p. Difficulty: 1 Easy Topic: 09.06 Testing a Proportion Learning Objective: 09-09 Check whether normality may be assumed in testing a proportion. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 133) The critical value in a hypothesis test A) is derived from the sample. B) is determined by α and the type of test. C) is another name for the null hypothesis. D) is a dissenting view by scientific critics. Answer: B Explanation: The critical value is a cutoff between the rejection and nonrejection regions. Difficulty: 2 Medium Topic: 09.03 Decision Rules and Critical Values Learning Objective: 09-04 Explain decision rules, critical values, and rejection regions. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 134) The critical value in a hypothesis test A) is a test statistic calculated from the sample. B) is a contradictory result from a different test. C) is a cutoff between the rejection and nonrejection regions. D) is another name for level of significance. Answer: C Explanation: The critical value is a cutoff between the rejection and nonrejection regions. Difficulty: 2 Medium Topic: 09.03 Decision Rules and Critical Values Learning Objective: 09-04 Explain decision rules, critical values, and rejection regions. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation

135) The rejection region in a hypothesis test A) is an area in the tail(s) of a sampling distribution. B) is a consensus formed by trained statisticians. C) is always set to correspond to α = .05 in the test. D) is always equal to the p-value in the test. Answer: A Explanation: The rejection region is depicted visually as a tail area. Difficulty: 1 Easy Topic: 09.03 Decision Rules and Critical Values Learning Objective: 09-04 Explain decision rules, critical values, and rejection regions. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 136) A two-tailed hypothesis test A) would not be used if the test statistic could be negative. B) will double the probability of rejecting the null hypothesis. C) is used when the direction of the test is of no research interest. D) would lead to rejection more often than a one-tailed test at the same α. Answer: C Explanation: The rejection region is depicted visually as a tail area. Difficulty: 2 Medium Topic: 09.03 Decision Rules and Critical Values Learning Objective: 09-04 Explain decision rules, critical values, and rejection regions. Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation 137) The rejection region in a hypothesis test A) is an area in the tail(s) of a sampling distribution. B) is a consensus formed by trained statisticians. C) is always set to correspond to α = .05 in the test. D) is always equal to the p-value in the test. Answer: A Explanation: The rejection region is depicted visually as a tail area. Difficulty: 1 Easy Topic: 09.03 Decision Rules and Critical Values Learning Objective: 09-04 Explain decision rules, critical values, and rejection regions. Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation

138) In a statistical test, we A) try to reject the null hypothesis. B) try to prove the alternative hypothesis. C) choose a null hypothesis that is easy to reject. D) choose a large level of significance if possible. Answer: A Explanation: We assume that H0 is true and see if the sample contradicts that assumption. Difficulty: 1 Easy Topic: 09.01 Logic of Hypothesis Testing Learning Objective: 09-01 Know the steps in testing hypotheses and define H0 and H1. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 139) The null hypothesis is A) often from a benchmark or historical value. B) chosen after looking at the sample data. C) what the researcher is trying to prove. D) dependent on the level of significance. Answer: A Explanation: H0 is often a prevailing view that we are testing against sample evidence. Difficulty: 1 Easy Topic: 09.03 Decision Rules and Critical Values Learning Objective: 09-04 Explain decision rules, critical values, and rejection regions. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 140) The decision rule is A) based on the sampling distribution and chosen level of significance. B) specified so as to support the researcher's alternative hypothesis. C) chosen after selecting the sample and calculating the test statistic. D) specified by a panel of expert statisticians elected annually. Answer: A Explanation: The decision rule depends on the type of test and the level of significance. Difficulty: 1 Easy Topic: 09.03 Decision Rules and Critical Values Learning Objective: 09-04 Explain decision rules, critical values, and rejection regions. Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation

141) In a left-tailed test, a statistician got a z test statistic of −1.61. What is the p-value? A) .4292 B) .0709 C) .0537 D) .0301 Answer: C Explanation: From Appendix C, we get P(Z < −1.61) = .0537 or from the Excel function =NORM.S.DIST(-1.61,1) = .0537. Difficulty: 2 Medium Topic: 09.04 Testing a Mean: Known Population Variance Learning Objective: 09-06 Use tables or Excel to find the p-value in tests of µ. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 142) In a two-tailed test, a statistician got a z test statistic of 1.82. What is the p-value? A) .0708 B) .0688 C) .0874 D) .0301 Answer: B Explanation: From Appendix C we get 2 × P(Z > 1.82) = 2 × .0344 = .0688. The Excel version of this calculation is =2*(1-NORM.S.DIST(1.82,1)) = 0.0688. Difficulty: 2 Medium Topic: 09.04 Testing a Mean: Known Population Variance Learning Objective: 09-06 Use tables or Excel to find the p-value in tests of µ. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

Applied Statistics in Business and Economics, 6e (Doane) Chapter 10 Two-Sample Hypothesis Tests 1) A pooled proportion is calculated by giving each sample proportion an equal weight. Answer: FALSE Explanation: In pc = (x1 + x2)/(n1 + n2) the sample successes are combined, as are the sample sizes. Difficulty: 2 Medium Topic: 10.05 Comparing Two Proportions Learning Objective: 10-05 Perform a test to compare two proportions using z. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 2) The difference between two sample proportions p1 − p2 may be assumed normally distributed if each sample has at least 10 "successes" and 10 "failures." Answer: TRUE Explanation: This rule of thumb assures near-normality of the difference p1 − p2. Difficulty: 1 Easy Topic: 10.05 Comparing Two Proportions Learning Objective: 10-06 Check whether normality may be assumed for two proportions. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 3) When testing the difference between two population proportions, it is necessary to use the same size sample from each population. Answer: FALSE Explanation: The formula for the test statistic does not require equal sample sizes. Difficulty: 1 Easy Topic: 10.05 Comparing Two Proportions Learning Objective: 10-05 Perform a test to compare two proportions using z. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation

4) When using independent samples to test the difference between two population means, a pooled variance is used if the population variances are unknown and assumed equal. Answer: TRUE Explanation: Sample variances are combined using weights equal to their respective degrees of freedom. Difficulty: 1 Easy Topic: 10.02 Comparing Two Means: Independent Samples Learning Objective: 10-02 Explain the assumptions underlying the two-sample test of means. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 5) In comparing the means of two independent samples, if the test statistic indicates a significant difference at α = .05, it will also be significant at α = .10. Answer: TRUE Explanation: It is easier to reject at α = .10 because the critical value for α = .10 is smaller than for α = .05. Difficulty: 1 Easy Topic: 10.02 Comparing Two Means: Independent Samples Learning Objective: 10-01 Recognize and perform a test for two means. Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation 6) The degrees of freedom for the t-test used to compare two population means (independent samples) with unknown variances (assumed equal) will be n1 + n2 − 2. Answer: TRUE Explanation: For the pooled variance case, we can add the degrees of freedom n1 − 1 and n2 − 1. Difficulty: 2 Medium Topic: 10.02 Comparing Two Means: Independent Samples Learning Objective: 10-02 Explain the assumptions underlying the two-sample test of means. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation

7) When using independent samples to test the difference between two population means, it is desirable but not necessary for the sample sizes to be the same. Answer: TRUE Explanation: Equal sample sizes may improve power, but are not required. Difficulty: 2 Medium Topic: 10.02 Comparing Two Means: Independent Samples Learning Objective: 10-02 Explain the assumptions underlying the two-sample test of means. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 8) The Welch-Satterthwaite test is more conservative than the pooled variance test to compare two population means with unknown variances in independent samples. Answer: TRUE Explanation: When in doubt, assume unequal variances (but power may be lost if variances were actually equal). Difficulty: 2 Medium Topic: 10.02 Comparing Two Means: Independent Samples Learning Objective: 10-02 Explain the assumptions underlying the two-sample test of means. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 9) When sample data occur in pairs, an advantage of choosing a paired t-test is that it tends to increase the power of a test, as compared to treating each sample independently. Answer: TRUE Explanation: Assuming independent samples loses information, so power may be lost. Difficulty: 2 Medium Topic: 10.04 Comparing Two Means: Paired Samples Learning Objective: 10-04 Recognize paired data and be able to perform a paired t test. Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation 10) A paired t-test with two columns of 10 observations in each column would use d.f. = 18. Answer: FALSE Explanation: We have 10 differences, so we would use d.f. = n − 1 = 10 − 1 = 9. Difficulty: 1 Easy Topic: 10.04 Comparing Two Means: Paired Samples Learning Objective: 10-04 Recognize paired data and be able to perform a paired t test. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 3 Copyright 2019 © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.

11) In conducting a paired t-test of the difference between two population means, the usual null hypothesis is that the mean of the population of paired differences is zero. Answer: TRUE Explanation: Testing for a zero difference is common (to ask whether a difference exists or does not exist). Difficulty: 1 Easy Topic: 10.04 Comparing Two Means: Paired Samples Learning Objective: 10-04 Recognize paired data and be able to perform a paired t test. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 12) The t-test for two samples of paired data will use n differences, making it a one-sample t-test. Answer: TRUE Explanation: There are n data pairs, so it is a one-sample test. Difficulty: 1 Easy Topic: 10.04 Comparing Two Means: Paired Samples Learning Objective: 10-04 Recognize paired data and be able to perform a paired t test. Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation 13) The F test is used to test for the equality of two population variances. Answer: TRUE Explanation: The ratio of sample variances should be near 1 if the population variances are the same. Difficulty: 1 Easy Topic: 10.07 Comparing Two Variances Learning Objective: 10-08 Carry out a test of two variances using the F distribution. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 14) The F distribution is never negative and is always skewed right. Answer: TRUE Explanation: The ratio of variances has no upper limit but is bounded at zero on the low end. Difficulty: 2 Medium Topic: 10.07 Comparing Two Variances Learning Objective: 10-08 Carry out a test of two variances using the F distribution. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 4 Copyright 2019 © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.

15) In an F test for the ratio of two population variances, the degrees of freedom in both the numerator and the denominator must be equal. Answer: FALSE Explanation: Degrees of freedom are equal only if the sample sizes are the same. Difficulty: 2 Medium Topic: 10.07 Comparing Two Variances Learning Objective: 10-08 Carry out a test of two variances using the F distribution. Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation 16) The critical value in an F test for equal variances is the ratio of the sample variances. Answer: FALSE Explanation: The test statistic is the ratio of variances, but the critical value is from a table (or Excel). Difficulty: 2 Medium Topic: 10.07 Comparing Two Variances Learning Objective: 10-08 Carry out a test of two variances using the F distribution. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 17) The test statistic in an F test for equal variances is the ratio of the sample variances. Answer: TRUE Explanation: The ratio of sample variances should be near 1 if the population variances are the same. Difficulty: 1 Easy Topic: 10.07 Comparing Two Variances Learning Objective: 10-08 Carry out a test of two variances using the F distribution. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation

18) We could use the same data set for two independent samples (i.e., two columns of data) either to compare the means (t-test) or to compare the variances (F test). Answer: TRUE Explanation: It depends on whether we are interested in means or variances, but both tests use the same data. Difficulty: 1 Easy Topic: 10.07 Comparing Two Variances Learning Objective: 10-08 Carry out a test of two variances using the F distribution. Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation 19) In general, the Welch-Satterthwaite t-test for two means has the same degrees of freedom as the pooled samples t-test for two means. Answer: FALSE Explanation: The statement is true if the sample variances are identical, but it is not true in general. Difficulty: 2 Medium Topic: 10.02 Comparing Two Means: Independent Samples Learning Objective: 10-02 Explain the assumptions underlying the two-sample test of means. Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation 20) When the variances are known, a test comparing two independent sample means would use the normal distribution. Answer: TRUE Explanation: In this (rare) case, we would utilize the normal distribution. Difficulty: 1 Easy Topic: 10.02 Comparing Two Means: Independent Samples Learning Objective: 10-02 Explain the assumptions underlying the two-sample test of means. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation

21) When the variances are unknown, a test comparing two independent sample means would use the Student's t distribution. Answer: TRUE Explanation: Student's t is needed when we do not know the population variances. Difficulty: 1 Easy Topic: 10.02 Comparing Two Means: Independent Samples Learning Objective: 10-02 Explain the assumptions underlying the two-sample test of means. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 22) If the sample proportions are p1 = 15/60 and p2 = 20/90, normality may be assumed in a test comparing the two population proportions. Answer: TRUE Explanation: In each sample, we have at least 10 successes (x1 = 15, x2 = 20) and 10 failures (n1 − x1 = 45, n2 − x2 = 70). Difficulty: 2 Medium Topic: 10.05 Comparing Two Proportions Learning Objective: 10-06 Check whether normality may be assumed for two proportions. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 23) If the sample proportions are p1 = 6/90 and p2 =4/100, normality may be assumed in a test comparing the two population proportions. Answer: FALSE Explanation: We do not have at least 10 successes (x1 = 6, x2 = 4) in each sample. Difficulty: 2 Medium Topic: 10.05 Comparing Two Proportions Learning Objective: 10-06 Check whether normality may be assumed for two proportions. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

24) A paired t-test with two columns of 8 observations in each column would use d.f. = 7. Answer: TRUE Explanation: There are 8 differences so d.f. = n − 1 = 7. Difficulty: 1 Easy Topic: 10.04 Comparing Two Means: Paired Samples Learning Objective: 10-04 Recognize paired data and be able to perform a paired t test. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 25) If the population variances are exactly equal, the sample F test statistic will be zero. Answer: FALSE Explanation: The ratio of variances should be near 1 (not 0) if the population variances are the same. Difficulty: 1 Easy Topic: 10.07 Comparing Two Variances Learning Objective: 10-08 Carry out a test of two variances using the F distribution. Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation 26) In a right-tailed test comparing two proportions, the test statistic was zcalc = +1.81. The pvalue is: A) .9649 B) .0351 C) .4649 D) Must know n to answer. Answer: B Explanation: Appendix C gives P(Z ≥ 1.81) = .0351 or Excel =1-NORM.S.DIST(1.81,1). Difficulty: 2 Medium Topic: 10.05 Comparing Two Proportions Learning Objective: 10-05 Perform a test to compare two proportions using z. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

27) In a left-tailed test comparing two means with unknown variances assumed to be equal, the test statistic was t = −1.81 with sample sizes of n1 = 8 and n2 = 12. The p-value would be A) between .025 and .05. B) between .01 and .025. C) between .05 and .10. D) Must know α to answer. Answer: A Explanation: For d.f. = 18, Appendix D gives t.05 = 1.734 and t.025 = 2.101. For an exact answer, you can use the Excel function =T.DIST(-1.81,8+12-2,1) = .04351. Difficulty: 2 Medium Topic: 10.02 Comparing Two Means: Independent Samples Learning Objective: 10-01 Recognize and perform a test for two means. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 28) In a left-tailed test comparing two means with variances unknown but assumed to be equal, the sample sizes were n1 = 8 and n2 = 12. At α = .05, the critical value would be A) −1.960 B) −2.101 C) −1.734 D) −1.645 Answer: C Explanation: For d.f. = 18, Appendix D gives t.05 = −1.734. Difficulty: 2 Medium Topic: 10.02 Comparing Two Means: Independent Samples Learning Objective: 10-02 Explain the assumptions underlying the two-sample test of means. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

29) In a right-tailed test comparing two means with known variances, the sample sizes were n1 = 8 and n2 = 12. At α = .05, the critical value would be A) 1.960 B) 1.645 C) 1.734 D) 1.282 Answer: B Explanation: For a right-tailed test with known variances, we would use z.05 = 1.645. Difficulty: 1 Easy Topic: 10.02 Comparing Two Means: Independent Samples Learning Objective: 10-01 Recognize and perform a test for two means. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 30) In a test for equality of two proportions, the sample proportions were p1 = 12/50 and p2 = 18/50. The test statistic is A) −1.44 B) −1.31 C) −1.67 D) Impossible to determine without knowing α. Answer: B Explanation: Use combined proportion pc = (x1 + x2)/(n1 + n2) = (12 + 18)/(50 + 50) = .30 in zcalc. Difficulty: 3 Hard Topic: 10.05 Comparing Two Proportions Learning Objective: 10-05 Perform a test to compare two proportions using z. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

31) In a test for equality of two proportions, the sample proportions were p1 = 12/50 and p2 = 18/50. The pooled proportion is A) .20 B) .24 C) .36 D) .30 Answer: D Explanation: Use combined proportion pc = (x1 + x2)/(n1 + n2) = (12 + 18)/(50 + 50) = .30 in the calculation. Difficulty: 1 Easy Topic: 10.05 Comparing Two Proportions Learning Objective: 10-05 Perform a test to compare two proportions using z. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 32) If the sample proportions were p1 = 12/50 and p2 = 18/50, what is the approximate 95 percent confidence interval for the difference of the population proportions? A) [−.144, +.244] B) [−.120, +.120] C) [−.298, +.058] D) [−.011, .214] Answer: C Explanation: Enter the proportions separately in the formula. (We are not combining the samples.) Difficulty: 3 Hard Topic: 10.06 Confidence Interval for the Difference of Two Proportions, pi1 - pi2 Learning Objective: 10-07 Construct a confidence interval for pi1 - pi2. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

33) John wants to compare two means. His sample statistics were 1 = 22.7, s12 = 5.4 n1 = 9 and 2 2 = 20.5, s2 = 3.6, n2 = 9. Assuming equal variances, which is the approximate 95 percent confidence interval for the difference of the population means? A) [2.44, 6.19] B) [1.17, 5.08] C) [0.08, 4.32] D) [−0.09, 3.19] Answer: C Explanation: Pool the variances and add the degrees of freedom, because equal variances are assumed. With d.f. = (9 − 1) + (9 − 1) = 16 we find t.025 = 2.120 from Appendix D or from the Excel function =T.INV.2T(0.05,16) = 2.119905. Apply formula 10.5 to find the confidence interval. Difficulty: 3 Hard Topic: 10.03 Confidence Interval for the Difference of Two Means, mu1 - mu2 Learning Objective: 10-03 Construct a confidence interval for mu1 - mu2. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 34) Kate wants to compare two means. Her sample statistics were 1 = 22.7, s12 = 5.4, n1 = 9 and 2 2 = 20.5, s2 = 3.6, n2 = 9. Assuming equal variances, the pooled variance is A) 4.5 B) 4.9 C) 5.1 D) 3.8 Answer: A Explanation: The pooled variance is [(n1 − 1)s12 + (n2 − 1)s22]/[(n1 − 1) + (n2 − 1)] = 4.5. Difficulty: 2 Medium Topic: 10.02 Comparing Two Means: Independent Samples Learning Objective: 10-02 Explain the assumptions underlying the two-sample test of means. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

35) John wants to compare two means. His sample statistics were 1 = 22.7, s12 = 5.4, n1 = 9 and 2 2 = 20.5, s2 = 3.6, n2 = 9. Assuming equal variances, the test statistic is A) 2.37 B) 2.20 C) 1.96 D) Must know α to answer. Answer: B Explanation: The pooled variance is (n1 − 1)s12 + (n2 − 1)s22 = (9 − 1)(5.4) + (9 − 1)(3.6) = 4.5. Therefore, the test statistic Difficulty: 2 Medium Topic: 10.02 Comparing Two Means: Independent Samples Learning Objective: 10-02 Explain the assumptions underlying the two-sample test of means. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 36) Kate wants to compare two means. Her sample statistics were 1 = 22.7, s12 = 5.4, n1 = 9 and 2 2 = 20.5, s2 = 3.6, n2 = 9. Assuming equal variances, the degrees of freedom for his test will be A) 16 B) 18 C) 9 D) 8 Answer: A Explanation: With pooled samples, we add the d.f. = (n1 − 1) + (n2 − 1) = (9 − 1) + (9 − 1) = 16. Difficulty: 1 Easy Topic: 10.02 Comparing Two Means: Independent Samples Learning Objective: 10-02 Explain the assumptions underlying the two-sample test of means. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

37) In a random sample of patient records in Cutter Memorial Hospital, six-month postoperative exams were given in 90 out of 200 prostatectomy patients, while in Paymor Hospital such exams were given in 110 out of 200 cases. In comparing these two proportions, normality of the difference may be assumed because A) the populations are large enough to be assumed normal. B) the probability of success can reasonably be assumed constant. C) the samples are random, so the proportions are unbiased estimates. D) nπ ≥ 10 and n(1 − π) ≥ 10 for each sample taken separately. Answer: D Explanation: We have at least 10 successes (x1 = 90, x2 = 110) and 10 failures (n1 − x1 = 110, n2 − x2 = 90). Difficulty: 2 Medium Topic: 10.05 Comparing Two Proportions Learning Objective: 10-06 Check whether normality may be assumed for two proportions. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 38) In a random sample of patient records in Cutter Memorial Hospital, six-month postoperative exams were given in 90 out of 200 prostatectomy patients, while in Paymor Hospital such exams were given in 110 out of 200 cases. In a left-tailed test for equality of proportions, the test statistic is A) −1.96 B) −2.00 C) −4.00 D) −3.48 Answer: B Explanation: Combined pc = (90 + 110)/(200 + 200) = .50, so zcalc = (p1 − p2)/[pc(1 − pc)/n1 + pc(1 − pc)/n2]1/2 = −2.000. Difficulty: 3 Hard Topic: 10.05 Comparing Two Proportions Learning Objective: 10-05 Perform a test to compare two proportions using z. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

39) In a random sample of patient records in Cutter Memorial Hospital, six-month postoperative exams were given in 90 out of 200 prostatectomy patients, while in Paymor Hospital such exams were given in 110 out of 200 cases. In a left-tailed test for equality of proportions, the p-value is A) .9772 B) .0228 C) .4772 D) .0517 Answer: B Explanation: Combined pc = (90 + 110)/(200 + 200) = .50, so zcalc = (p1 − p2)/[pc(1 − pc)/n1 + pc(1 − pc)/n2]1/2 = −2.000 and using Appendix C we get P(Z ≤ −2.000) = .0228. Alternatively, use the Excel function =NORM.S.DIST(−2.000,1) = 0.0227501. Difficulty: 3 Hard Topic: 10.05 Comparing Two Proportions Learning Objective: 10-05 Perform a test to compare two proportions using z. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 40) In a random sample of patient records in Cutter Memorial Hospital, six-month postoperative exams were given in 90 out of 200 cases, while in Paymor Hospital such exams were given in 110 out of 200 cases. The pooled proportion is A) .50 B) .40 C) .30 D) .20 Answer: A Explanation: Use combined proportion pc = (x1 + x2)/(n1 + n2) = (90 + 110)/(200 + 200) = .50 in the calculation. Difficulty: 1 Easy Topic: 10.05 Comparing Two Proportions Learning Objective: 10-05 Perform a test to compare two proportions using z. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

41) Management of Melodic Kortholt Company compared absenteeism rates in two plants on the third Monday in November. Of Plant A's 800 employees, 120 were absent. Of Plant B's 1200 employees, 144 were absent. To compare the two proportions, the pooled proportion is A) .130 B) .140 C) .132 D) .135 Answer: C Explanation: Combined proportion is pc = (x1 + x2)/(n1 + n2) = (120 + 144)/(800 + 1200) = . 132. Difficulty: 1 Easy Topic: 10.05 Comparing Two Proportions Learning Objective: 10-05 Perform a test to compare two proportions using z. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

42) Management of Melodic Kortholt Company compared absenteeism rates in two plants on the third Monday in November. Of Plant A's 800 employees, 120 were absent. Of Plant B's 1200 employees, 144 were absent. MegaStat's results for a two-tailed test are shown below. p1 0.15 120/800 120. 800 0.03 0.00 0.01545 x.xx .0522

p2 0.12 144/1200 144. 1200

p (as decimal) p (as fraction) X n

sample difference hypothesized difference std. error z p-value (two-tailed)

The test statistic (shown as z = x.xx) is approximately A) 2.022 B) 1.960 C) 1.942 D) 1.645 Answer: C Explanation: Combined proportion is pc = (120 + 144)/(800 + 1200) = .132, so zcalc = (.15 − . 12)/[.132(1 − .132)/800 + .132(1 − .132)/1200]1/2 = 1.942. Difficulty: 2 Medium Topic: 10.05 Comparing Two Proportions Learning Objective: 10-05 Perform a test to compare two proportions using z. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

43) Management of Melodic Kortholt Company compared absenteeism rates in two plants on the third Monday in November. Of Plant A's 800 employees, 120 were absent. Of Plant B's 1200 employees, 144 were absent. MegaStat's results for a two-tailed test are shown below. p1 0.15 120/800 120. 800 0.03 0.00 0.01545 x.xx .0522

p2 0.12 144/1200 144. 1200

p (as decimal) p (as fraction) X n

sample difference hypothesized difference std. error z p-value (two-tailed)

At α = .05, the two-tailed test for a difference in proportions is A) just barely significant. B) not quite significant. C) not feasible due to nonnormality. Answer: B Explanation: Because the p-value is slightly greater than .05, we cannot reject H0. Difficulty: 1 Easy Topic: 10.05 Comparing Two Proportions Learning Objective: 10-05 Perform a test to compare two proportions using z. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

44) A new policy of "flex hours" is proposed. Random sampling showed that 28 of 50 female workers favored the change, while 22 of 50 male workers favored the change. Management wonders if there is a difference between the two groups. For a test comparing the two proportions, the assumption of normality for the difference of proportions is A) clearly justified. B) clearly unjustified. C) a borderline call. Answer: A Explanation: In each sample, we have at least 10 successes (x1 = 28, x2 = 22) and 10 failures (n1 − x1 = 22, n2 − x2 = 28). Difficulty: 2 Medium Topic: 10.05 Comparing Two Proportions Learning Objective: 10-06 Check whether normality may be assumed for two proportions. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 45) A new policy of "flex hours" is proposed. Random sampling showed that 28 of 50 female workers favored the change, while 22 of 50 male workers favored the change. Management wonders if there is a difference between the two groups. What is the test statistic to test for a zero difference in the population proportions? A) 1.321 B) 1.287 C) 1.200 D) −1.255 Answer: C Explanation: Combined proportion is pc = (28 + 22)/(50 + 50) = .50, so zcalc = (.56 − .44)/ [.50(1 − .50)/50 + .50(1 − .50)/50]1/2 = −1.255. Difficulty: 3 Hard Topic: 10.05 Comparing Two Proportions Learning Objective: 10-05 Perform a test to compare two proportions using z. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

46) A new policy of "flex hours" is proposed. Random sampling showed that 28 of 50 female workers favored the change, while 22 of 50 male workers favored the change. Management wonders if there is a difference between the two groups. What is the p-value for a two-tailed test? A) .3849 B) .1151 C) .2301 D) .3453 Answer: C Explanation: Combined proportion is pc = (28 + 22)/(50 + 50) = .50, so zcalc = (.56 − .44)/ [.50(1 − .50)/50 + 50(1 − .50)/50]1/2 = −1.20 and 2 × P(Z < −1.20) = 2 × .1151 = .2302 (or .2301 using Excel). Difficulty: 3 Hard Topic: 10.05 Comparing Two Proportions Learning Objective: 10-05 Perform a test to compare two proportions using z. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 47) At Huge University, a sample of 200 business school seniors showed that 26 planned to pursue an MBA degree, compared with 120 of 800 arts and sciences seniors. We want to know if the proportion is higher in the arts and sciences group. The pooled proportion for this test is A) .130 B) .140 C) .145 D) .146 Answer: D Explanation: Combined proportion is pc = (26 + 120)/(200 + 800) = .146. Difficulty: 1 Easy Topic: 10.05 Comparing Two Proportions Learning Objective: 10-05 Perform a test to compare two proportions using z. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

48) At Huge University, a sample of 200 business school seniors showed that 26 planned to pursue an MBA degree, compared with 120 of 800 arts and sciences seniors. We want to know if the proportion is higher in the arts and sciences group. For this test, the assumption of normality for the difference of proportions is A) clearly unjustified. B) clearly justified. C) a borderline call. Answer: B Explanation: In each sample, we have at least 10 successes (x1 = 26, x2 = 120) and 10 failures (n1 − x1 = 174, n2 − x2 = 680). Difficulty: 1 Easy Topic: 10.05 Comparing Two Proportions Learning Objective: 10-06 Check whether normality may be assumed for two proportions. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 49) At Huge University, a sample of 200 business school seniors showed that 26 planned to pursue an MBA degree, compared with 120 of 800 arts and sciences seniors. We want to know if the proportion is higher in the arts and sciences group. What is the z test statistic? A) −1.322. B) −1.122. C) −0.716. D) We must first know α. Answer: C Explanation: Combined proportion is pc = (26 + 120)/(200 + 800) = .146, so zcalc = (.13 − .15)/ [.146(1 − .146)/200 + .146(1 − .146)/800]1/2 = −0.716. Difficulty: 3 Hard Topic: 10.05 Comparing Two Proportions Learning Objective: 10-05 Perform a test to compare two proportions using z. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

50) At Huge University, a sample of 200 business school seniors showed that 26 planned to pursue an MBA degree, compared with 120 of 800 arts and sciences seniors. We want to know if the proportion is higher in the arts and sciences group. The p-value for a left-tailed test is approximately A) .38 B) .48 C) .24 D) .51 Answer: C Explanation: Combined proportion is pc = (26 + 120)/(200 + 800) = .146, so zcalc = (.13 − .15)/ [.146(1 − .146)/200 + .146(1 − .146)/800]1/2 = −0.716. From Appendix C, we get P(Z ≤ −0.72) = .2358. Alternatively, use the Excel function =NORM.S.DIST(−0.72,1) = 0.235762. Difficulty: 3 Hard Topic: 10.05 Comparing Two Proportions Learning Objective: 10-05 Perform a test to compare two proportions using z. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 51) Two well-known aviation training schools are being compared using random samples of their graduates. It is found that 70 of 140 graduates of Fly-More Academy passed their FAA exams on the first try, compared with 104 of 260 graduates of Blue Yonder Institute. To compare the pass rates, the pooled proportion would be A) .500 B) .435 C) .400 D) .345 Answer: B Explanation: Combined proportion is pc = (70 + 104)/(140 + 260) = .435. Difficulty: 1 Easy Topic: 10.05 Comparing Two Proportions Learning Objective: 10-05 Perform a test to compare two proportions using z. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

52) Two well-known aviation training schools are being compared using random samples of their graduates. It is found that 70 of 140 graduates of Fly-More Academy passed their FAA exams on the first try, compared with 104 of 260 graduates of Blue Yonder Institute. To compare the two proportions, the assumption of normality of the test statistic is A) justified, but it is a borderline case. B) clearly justified. C) clearly not justified. Answer: B Explanation: We have at least 10 successes (x1 = 70, x2 = 104) and 10 failures (n1 − x1 = 70, n2 − x2 = 156). Difficulty: 1 Easy Topic: 10.05 Comparing Two Proportions Learning Objective: 10-06 Check whether normality may be assumed for two proportions. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 53) Two well-known aviation training schools are being compared using random samples of their graduates. It is found that 70 of 140 graduates of Fly-More Academy passed their FAA exams on the first try, compared with 104 of 260 graduates of Blue Yonder Institute. The test statistic to test the pass rates for equality is A) 2.141 B) 1.298 C) 1.227 D) 1.924 Answer: D Explanation: Combined proportion is pc = (70 + 104)/(140 + 260) = .435, so zcalc = (.50 − .40)/ [.435(1 − .435)/140 +.435(1 − .435)/260]1/2 = 1.924. Difficulty: 3 Hard Topic: 10.05 Comparing Two Proportions Learning Objective: 10-05 Perform a test to compare two proportions using z. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

54) Two well-known aviation training schools are being compared using random samples of their graduates. It is found that 70 of 140 graduates of Fly-More Academy passed their FAA exams on the first try, compared with 104 of 260 graduates of Blue Yonder Institute. To compare the pass rates, find the critical value for a right-tailed test at α = .05. A) 1.960 B) 1.645 C) 2.326 Answer: B Explanation: z.05 = 1.645. Difficulty: 1 Easy Topic: 10.05 Comparing Two Proportions Learning Objective: 10-05 Perform a test to compare two proportions using z. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 55) Two well-known aviation training schools are being compared using random samples of their graduates. It is found that 70 of 140 graduates of Fly-More Academy passed their FAA exams on the first try, compared with 104 of 260 graduates of Blue Yonder Institute. To compare the pass rates, the p-value for a right-tailed test is approximately A) .054 B) .027 C) .155 D) .013 Answer: B Explanation: Combined proportion is pc = (70 + 104)/(140 + 260) = .435, so zcalc = (.50 − .40)/ [.435(1 − .435)/140 + .435(1 − .435)/260]1/2 = 1.924. From Appendix C we get P(Z ≥ 1.92) = . 0274. Alternatively, use the Excel function =1–NORM.S.DIST(1.92,1) = 0.027429. Difficulty: 3 Hard Topic: 10.05 Comparing Two Proportions Learning Objective: 10-05 Perform a test to compare two proportions using z. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

56) Two well-known aviation training schools are being compared using random samples of their graduates. It is found that 70 of 140 graduates of Fly-More Academy passed their FAA exams on the first try, compared with 104 of 260 graduates of Blue Yonder Institute. In a right-tailed test, the p-value is .0275, so at α = .025 we should A) reject the hypothesis of equal proportions. B) not reject the hypothesis of equal proportions. C) change the α to .05 to get a rejection. Answer: A Explanation: The p-value is less than the chosen level of significance. Difficulty: 1 Easy Topic: 10.05 Comparing Two Proportions Learning Objective: 10-05 Perform a test to compare two proportions using z. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 57) Of 200 youthful gamers (under 18) who tried the new Z-Box-Plus game, 160 rated it "excellent," compared with only 144 of 200 adult gamers (18 or over). The pooled proportion for a test to compare the two proportions would be A) .76 B) .72 C) .77 D) Must know α to answer. Answer: A Explanation: Combined proportion is pc = (160 + 144)/(200 + 200) = .76. Difficulty: 1 Easy Topic: 10.05 Comparing Two Proportions Learning Objective: 10-05 Perform a test to compare two proportions using z. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

58) Of 200 youthful gamers (under 18) who tried the new Z-Box-Plus game, 160 rated it "excellent," compared with only 144 of 200 adult gamers (18 or over). The test statistic to test the two proportions for equality would be A) 1.645 B) 1.960 C) 1.873 D) 1.448 Answer: C Explanation: Combined proportion is pc = (160 + 144)/(200 + 200) = .76, so zcalc = (.80 − .72)/ [.76(1 − .76)/200 + .76(1 − .76)/200]1/2 = 1.873. Difficulty: 3 Hard Topic: 10.05 Comparing Two Proportions Learning Objective: 10-05 Perform a test to compare two proportions using z. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 59) Of 200 youthful gamers (under 18) who tried the new Z-Box-Plus game, 160 rated it "excellent," compared with only 144 of 200 adult gamers (18 or over). The p-value for a righttailed test to compare the two proportions would be approximately A) .042 B) .031 C) .054 D) .095 Answer: B Explanation: Combined proportion is pc = (160 + 144)/(200 + 200) = .76, so zcalc = (.80 − .72)/ [.76(1 − .76)/200 + .76(1 − .76)/200]1/2 = 1.873. From Appendix C we get P(Z ≥ 1.87) = .0307. Difficulty: 3 Hard Topic: 10.05 Comparing Two Proportions Learning Objective: 10-05 Perform a test to compare two proportions using z. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

60) Of 200 youthful gamers (under 18) who tried the new Z-Box-Plus game, 160 rated it "excellent," compared with only 144 of 200 adult gamers (18 or over). Calculate the 95 percent confidence interval for the difference of proportions. A) [+.013, +.263] B) [−.014, +.188] C) [−.003, +.163] D) [+.057, +.261] Answer: C Explanation: Do not pool the proportions when you calculate the standard error of p1 − p2. Difficulty: 3 Hard Topic: 10.06 Confidence Interval for the Difference of Two Proportions, pi1 - pi2 Learning Objective: 10-07 Construct a confidence interval for pi1 - pi2. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 61) Carver Memorial Hospital's surgeons have a new procedure that they think will decrease the time to perform an appendectomy. A sample of 8 appendectomies using the old method had a mean of 38 minutes with a variance of 36 minutes, while a sample of 10 appendectomies using the experimental method had a mean of 29 minutes with a variance of 16 minutes. For a righttailed test for equal means (assume equal variances), the critical value at α = .10 is A) 1.746 B) 1.337 C) 2.120 D) 2.754 Answer: B Explanation: For d.f. = (n1 − 1) + (n2 − 1) = 7 + 9 = 16, we get t.10 = 1.337. Difficulty: 2 Medium Topic: 10.02 Comparing Two Means: Independent Samples Learning Objective: 10-02 Explain the assumptions underlying the two-sample test of means. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

62) Carver Memorial Hospital's surgeons have a new procedure that they think will decrease the time to perform an appendectomy. A sample of 8 appendectomies using the old method had a mean of 38 minutes with a variance of 36 minutes, while a sample of 10 appendectomies using the experimental method had a mean of 29 minutes with a variance of 16 minutes. For a righttailed test of equal means (assume equal variances), the pooled variance is A) 14.76 B) 26.00 C) 24.75 D) 27.54 Answer: C Explanation: The pooled variance is [(8 − 1)36 + (10 − 1)16]/[(8 − 1) + (10 − 1)] = 24.75. Difficulty: 2 Medium Topic: 10.02 Comparing Two Means: Independent Samples Learning Objective: 10-02 Explain the assumptions underlying the two-sample test of means. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 63) Carver Memorial Hospital's surgeons have a new procedure that they think will decrease the time to perform an appendectomy. A sample of 8 appendectomies using the old method had a mean of 38 minutes with a variance of 36 minutes, while a sample of 10 appendectomies using the experimental method had a mean of 29 minutes with a variance of 16 minutes. For a righttailed test of means (assume equal variances), the test statistic is A) 3.814 B) 2.365 C) 3.000 D) 1.895 Answer: A Explanation: The pooled variance is [(n1 − 1)s12 + (n2 − 1)s22]/[(n1 − 1) + (n2 − 1)] = [(8 − 1)36 + (10 − 1)16]/[(8 − 1) + (10 − 1)] = 24.75, so tcalc = (38 − 29)/(24.75/8 + 24.75/10)1/2 = 3.814. Difficulty: 3 Hard Topic: 10.02 Comparing Two Means: Independent Samples Learning Objective: 10-02 Explain the assumptions underlying the two-sample test of means. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

64) Based on a random sample of 13 tire changes, the mean time to change a tire on a Boeing 777 has a mean of 59.5 minutes with a standard deviation of 8.4 minutes. For 10 tire changes on a Boeing 787, the mean time was 64.3 minutes with a standard deviation of 12.4 minutes. To test for equal variances in a two-tailed test at α = .10, the critical values are A) 3.73 and 0.228 B) 2.51 and 3.67 C) 3.07 and 0.398 D) 3.07 and 0.357 Answer: D Explanation: Using α/2 in Appendix F, we get F12,9 = 3.07 and 1/F9,12 = 1/(2.80) = 0.357 (or else use Excel's functions =F.INV.RT(.05,12,9) and =F.INV(.05,12,9). Difficulty: 3 Hard Topic: 10.07 Comparing Two Variances Learning Objective: 10-08 Carry out a test of two variances using the F distribution. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

65) A certain psychological theory predicts that men want bigger families than women. Kate asked each student in her psychology class how many children he or she considered ideal for a married couple and obtained the Excel results shown below at α = .05. Desired Number of Children t-Test: Two-sample Assuming Equal t-Test: Two-sample Assuming Equal Variances Variances Men Women Men Women 2.133 Mean 2.4571 2.1333 Mean 2.4571 3 0.266 Variance 0.7261 0.2667 Variance 0.7261 7 Observations 35 15 Observations 35 15 Pooled 0.59206 Variance Hypothesized Hypothesized 0 0 Diff Diff df 48 df 42 t Stat 1.364 t Stat 1.650 P(T <= t) oneP(T <= t) one0.090 0.053 tail tail t Critical onet Critical one1.677 1.682 tail tail P(T <= t) twoP(T <= t) two0.179 0.106 tail tail t Critical twot Critical two2.011 2.018 tail tail What conclusion can you draw in a two-tailed test at α = .05? A) Men want larger families on average than women. B) Women want larger families on average than men. C) We cannot reject the hypothesis of equal population means. D) The decision depends on whether or not the variances are equal. Answer: C Explanation: Neither assumption about variances leads to a p-value smaller than .05. Difficulty: 2 Medium Topic: 10.02 Comparing Two Means: Independent Samples Learning Objective: 10-02 Explain the assumptions underlying the two-sample test of means. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

66) Nacirema Airlines is buying a fleet of new fuel-efficient planes. The HogJet and the LitheJet both meet their price and performance needs, and both planes meet EPA noise guidelines. However, the quieter plane is preferred. Each plane is flown through a typical takeoff and landing sequence 10 times, while remote sensors at ground level record the noise levels (in decibels). The table below summarizes the sound level tests using Excel's default level of significance (α = .05). t-Test Assuming Equal Variances LitheJet HogJet Mean

80.3368

82.4669

Variance Observations Pooled Variance Hypothesized Diff df t Stat P(T <= t) onetail t Critical onetail P(T <= t) twotail t Critical twotail

0.7178 10

4.7385 10

2.7282

t-Test Assuming Equal Variances LitheJet HogJet 82.466 Mean 80.3368 9 Variance 0.7178 4.7385 Observations 10 10 Hypothesized 0 Diff

0.0000 18 −2.8837 0.0049 1.7341 0.0099 2.1009

df 12 t Stat −2.8837 P(T <= t) one0.0069 tail t Critical one1.7823 tail P(T <= t) two0.0137 tail t Critical two2.1788 tail

In a left-tailed test comparing the means at α = .05, we would A) not reject H0. B) reject H0. C) have insufficient information to make a decision. Answer: B Explanation: Either assumption about variances leads to a p-value smaller than .05. Difficulty: 2 Medium Topic: 10.02 Comparing Two Means: Independent Samples Learning Objective: 10-02 Explain the assumptions underlying the two-sample test of means. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

67) Nacirema Airlines is buying a fleet of new fuel-efficient planes. The HogJet and the LitheJet both meet their price and performance needs, and both planes meet EPA noise guidelines. However, the quieter plane is preferred. Each plane is flown through a typical takeoff and landing sequence 10 times, while remote sensors at ground level record the noise levels (in decibels). The table below summarizes the sound level tests using Excel's default level of significance (α = .05). t-Test Assuming Equal Variances LitheJet HogJet Mean

80.3368

82.4669

Variance Observations Pooled Variance Hypothesized Diff df t Stat P(T <= t) onetail t Critical onetail P(T <= t) twotail t Critical twotail

0.7178 10

4.7385 10

2.7282

t-Test Assuming Equal Variances LitheJet HogJet 82.466 Mean 80.3368 9 Variance 0.7178 4.7385 Observations 10 10 Hypothesized 0 Diff

0.0000 18 −2.8837 0.0049 1.7341 0.0099 2.1009

df 12 t Stat −2.8837 P(T <= t) one0.0069 tail t Critical one1.7823 tail P(T <= t) two0.0137 tail t Critical two2.1788 tail

After inspecting this table, we would most likely A) use the test assuming unequal variances. B) use the test for equal variances. C) perform another test to determine if the variances are equal before proceeding. D) realize that the decision is not affected by our assumptions concerning the variance. Answer: D Explanation: Fcalc = 4.7385/0.7178 = 6.60 > F.05 = 5.35 for d.f. = (9, 9), so we would want to assume unequal variances. However, we notice that our decision is the same under either assumption about variances (one-tailed p-values are less than .05 in either test). Difficulty: 2 Medium Topic: 10.02 Comparing Two Means: Independent Samples Learning Objective: 10-02 Explain the assumptions underlying the two-sample test of means. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

68) Nacirema Airlines is buying a fleet of new fuel-efficient planes. The HogJet and the LitheJet both meet their price and performance needs, and both planes meet EPA noise guidelines. However, the quieter plane is preferred. Each plane is flown through a typical takeoff and landing sequence 10 times, while remote sensors at ground level record the noise levels (in decibels). The table below summarizes the sound level tests using Excel's default level of significance (α = .05). t-Test Assuming Equal Variances LitheJet HogJet Mean

80.3368

82.4669

Variance Observations Pooled Variance Hypothesized Diff df t Stat P(T <= t) onetail t Critical onetail P(T <= t) twotail t Critical twotail

0.7178 10

4.7385 10

2.7282

t-Test Assuming Equal Variances LitheJet HogJet 82.466 Mean 80.3368 9 Variance 0.7178 4.7385 Observations 10 10 Hypothesized 0 Diff

0.0000 18 −2.8837 0.0049 1.7341 0.0099 2.1009

df 12 t Stat −2.8837 P(T <= t) one0.0069 tail t Critical one1.7823 tail P(T <= t) two0.0137 tail t Critical two2.1788 tail

If we switched from α = .05 to α = .005 in a two-tailed test of means, our assumption about variances (assumed equal or assumed unequal) would A) affect the decision. B) not affect the decision. C) require a new analysis. Answer: A Explanation: For equal variances the p-value is .0049 (reject equal means), while for unequal variances the p-value = .0069 (do not reject equal means), so the choice of tests does matter. Difficulty: 2 Medium Topic: 10.02 Comparing Two Means: Independent Samples Learning Objective: 10-02 Explain the assumptions underlying the two-sample test of means. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

69) A psychology researcher has a theory that predicts women will tend to carry more cash than men. A random sample of Ersatz University students revealed that 16 females had a mean of $22.30 in their wallets with a standard deviation of $3.20, while 16 males had a mean of $17.30 with a standard deviation of $9.60.The researcher's hypothesis would lead us to perform a A) right-tailed test. B) left-tailed test. C) two-tailed test. Answer: A Explanation: The a priori theory requires a right-tailed test. Difficulty: 1 Easy Topic: 10.02 Comparing Two Means: Independent Samples Learning Objective: 10-02 Explain the assumptions underlying the two-sample test of means. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 70) A psychology researcher has a theory that predicts women will tend to carry more cash than men. A random sample of Ersatz University students revealed that 16 females had a mean of $22.30 in their wallets with a standard deviation of $3.20, while 16 males had a mean of $17.30 with a standard deviation of $9.60. The test statistic for the researcher's hypothesis is A) impossible to determine without knowing α. B) 1.250 C) 1.504 D) 1.976 Answer: D Explanation: Using a folded F test at α = .05, we get Fcalc = 9.602/3.202 = 9.00 > F.025 = 2.86 for d.f. = (15, 15), so we would want to assume unequal variances. For unequal variances tcalc = (22.30 − 17.30)/[(3.2)2/16 + (9.6)2/16]1/2 = (5)/(6.4)1/2 = 1.976. Difficulty: 3 Hard Topic: 10.02 Comparing Two Means: Independent Samples Learning Objective: 10-02 Explain the assumptions underlying the two-sample test of means. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

71) A random sample of Ersatz University students revealed that 16 females had a mean of $22.30 in their wallets with a standard deviation of $3.20, while 16 males had a mean of $17.30 with a standard deviation of $9.60. In comparing the population variances at α = .10 in a twotailed test, we conclude that A) the variances are equal. B) the variances are unequal. C) the variances are incomparable (different sample sizes). Answer: B Explanation: Folded Fcalc = (9.60)2/(3.20)2 = 9.00 > F.01 = 3.52 for d.f. = (15, 15), so we conclude that the variances are unequal. Difficulty: 3 Hard Topic: 10.07 Comparing Two Variances Learning Objective: 10-08 Carry out a test of two variances using the F distribution. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

72) Randomly chosen MBA students were asked their opinions about the ideal number of children for a married couple. The sample data were entered into MegaStat, and the following results were produced.

Men 2.812 1.2505 11 0.6582 0.40722 0 1.62 0.065

Hypothesis Test: independent Groups (t-test, unequal variance) Women 2.1538 0.5547 13 13 difference (Men - Women) standard error of difference hypothesized difference t p-value (one-tailed, upper)

F-test for equality of variance 1.56375 0.307692 5.08 0.01

mean std. dev. sample size pooled df

variance: Men variance: Women F p-value

To compare the means, would it be appropriate to use a test that assumes equal variances? A) Yes, because by pooling our variances, we can simplify our calculations. B) Yes, we should assume equal variances when comparing independent samples. C) No, because the sample statistics show that the variances are unequal at α = .05. D) No, because the variances will differ because the means differ significantly. Answer: D Explanation: Folded Fcalc = (1.2505)2/(0.5547)2 = 5.08 > F.025 = 3.37 for d.f. = (10, 12), so we conclude that the variances are unequal. Difficulty: 2 Medium Topic: 10.07 Comparing Two Variances Learning Objective: 10-08 Carry out a test of two variances using the F distribution. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

73) Litter sizes (number of pups) for randomly chosen dogs from two breeds were compared. The sample data were entered into Excel, and the following results were produced. t-Test: Two-Sample Assuming Unequal Variances Dalmatian Mean 4.8125 Variance 3.4958333 Observations 16 Hypothesized Mean Diff 0 df 21 t Stat −1.261183

Labrador 5.461538 0.602564 13

What is the critical value for a left-tailed test comparing the means at α = .05? A) −1.645 B) −1.721 C) −1.703 D) −1.699 Answer: B Explanation: t.05 = −1.721 for Welch's test with d.f. = 21. Difficulty: 1 Easy Topic: 10.02 Comparing Two Means: Independent Samples Learning Objective: 10-02 Explain the assumptions underlying the two-sample test of means. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

74) Randomly chosen MBA students were asked their opinions about the ideal number of children for a married couple. The sample data were entered into MegaStat, and the following results were produced.

Men 2.812 1.2505 11 0.6582 0.40722 0 1.62 0.065

Hypothesis Test: independent Groups (t-test, unequal variance) Women 2.1538 0.5547 13 13 difference (Men - Women) standard error of difference hypothesized difference t p-value (one-tailed, upper)

F-test for equality of variance 1.56375 0.307692 5.08 0.01

mean std. dev. sample size pooled df

variance: Men variance: Women F p-value

What conclusion can you draw from this analysis at α = .05? A) Men want larger families on average than women. B) Women want larger families on average than men. C) This is insufficient evidence to suggest a difference in means. D) We could conclude that men want larger families if we used a two-tailed test. Answer: C Explanation: Cannot reject equal means because the p-value (.065) is not less than α (.05). Difficulty: 2 Medium Topic: 10.02 Comparing Two Means: Independent Samples Learning Objective: 10-02 Explain the assumptions underlying the two-sample test of means. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

75) Litter sizes (number of pups) for randomly chosen dogs from two breeds were compared. The sample data were entered into Excel, and the following results were produced. t-Test: Two-Sample Assuming Unequal Variances Dalmatian Mean 4.8125 Variance 3.4958333 Observations 16 Hypothesized Mean Diff 0 df 21 t Stat −1.261183

Labrador 5.461538 0.602564 13

What is the p-value for a left-tailed test comparing the means at α = .05? A) Less than .10 B) More than .10 C) Between .10 and .05 D) Between .05 and .01 Answer: B Explanation: The test statistic (−1.261) is not less than t.10 (−1.372) for d.f. = 21. For a more precise answer, we can use the Excel function =T.DIST(-1.261183,21,1) = .11054. Difficulty: 2 Medium Topic: 10.02 Comparing Two Means: Independent Samples Learning Objective: 10-01 Recognize and perform a test for two means. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 76) During a test period, an experimental group of 10 vehicles using an 85 percent ethanolgasoline mixture showed mean CO2 emissions of 667 pounds per 1000 miles, with a standard deviation of 20 pounds. A control group of 14 vehicles using regular gasoline showed mean CO2 emissions of 679 pounds per 1000 miles with a standard deviation of 15 pounds. At α = .05, in a left-tailed test, the critical value to compare the means (assuming equal variances) is A) −2.508 B) −2.074 C) −1.321 D) −1.717 Answer: D Explanation: Assuming equal variances, d.f. = (10 − 1) + (14 − 1) = 22, so t.05 = −1.717. Difficulty: 2 Medium Topic: 10.02 Comparing Two Means: Independent Samples Learning Objective: 10-02 Explain the assumptions underlying the two-sample test of means. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 39 Copyright 2019 © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.

77) During a test period, an experimental group of 10 vehicles using an 85 percent ethanolgasoline mixture showed mean CO2 emissions of 667 pounds per 1000 miles, with a standard deviation of 20 pounds. A control group of 14 vehicles using regular gasoline showed mean CO2 emissions of 679 pounds per 1000 miles with a standard deviation of 15 pounds. Assuming equal variances, the pooled variance is A) 296.59 B) 225.00 C) 400.00 D) 522.16 Answer: A Explanation: The pooled variance is [(10 − 1)202 + (14 − 1)152]/[(10 − 1) + (14 − 1)] = 296.59. Difficulty: 2 Medium Topic: 10.02 Comparing Two Means: Independent Samples Learning Objective: 10-02 Explain the assumptions underlying the two-sample test of means. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 78) During a test period, an experimental group of 10 vehicles using an 85 percent ethanolgasoline mixture showed mean CO2 emissions of 667 pounds per 1000 miles, with a standard deviation of 20 pounds. A control group of 14 vehicles using regular gasoline showed mean CO2 emissions of 679 pounds per 1000 miles with a standard deviation of 15 pounds. At α = .05, in a left-tailed test (assuming equal variances) the test statistic is A) −1.310 B) −2.042 C) −1.645 D) −1.683 Answer: D Explanation: The pooled variance is [(10 − 1)202 + (14 − 1)152]/[(10 − 1) + (14 − 1)] = 296.591, so tcalc = (667 − 679)/(296.591/10 + 296.591/14)1/2 = −1.683. Difficulty: 3 Hard Topic: 10.02 Comparing Two Means: Independent Samples Learning Objective: 10-02 Explain the assumptions underlying the two-sample test of means. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

79) During a test period, an experimental group of 10 vehicles using an 85 percent ethanolgasoline mixture showed mean CO2 emissions of 240 pounds per 100 miles, with a standard deviation of 20 pounds. A control group of 14 vehicles using regular gasoline showed mean CO2 emissions of 252 pounds per 100 miles with a standard deviation of 15 pounds. A quick comparison of the sample variances suggests that the population variances are A) probably equal. B) probably unequal. C) incomparable. Answer: A Explanation: Folded Fcalc = (20)2/(15)2 = 1.78 < F.025 = 3.31 for d.f. = (9, 13), so we conclude that the variances are equal. Difficulty: 2 Medium Topic: 10.07 Comparing Two Variances Learning Objective: 10-08 Carry out a test of two variances using the F distribution. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 80) Mary did an analysis of acute care variances in samples of occupancy rates at two community hospitals and obtained the following results: F-test for equality of variance 108.243 variance: group 1 98.371 variance: group 1 1.100 F 0.905 p-value Can Mary conclude that the variances are unequal at α = .05? A) No, there is not enough evidence to believe the variances are unequal. B) Yes, the analysis shows that the variances are unequal. C) You cannot tell without knowing the sample sizes. Answer: A Explanation: The p-value (.905) is far greater than .05, so we cannot reject equal variances. Difficulty: 2 Medium Topic: 10.07 Comparing Two Variances Learning Objective: 10-08 Carry out a test of two variances using the F distribution. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

81) Mary analyzed occupancy rates at two community hospitals and obtained the following Excel results.

Mean Variance Observations Hypothesized Diff df t Stat P(T<=t) one-tail t Critical one-tail P(T<=t) two-tail t Critical two-tail

t-Test for Acue Care Occupancy Rates MaxHealth 62.5462 108.2377 13 0 20 −1.3923 0.0896 1.7247 0.1791 2.0860

HealthPro 68.4800 98.3707 10

Which conclusion is correct in a two-tailed test at α = .05? A) There appears to be no difference in the mean occupancy rates. B) HealthPro has a significantly higher mean occupancy rate. C) There is a significant difference in the mean occupancy rates. Answer: A Explanation: The p-value (.1791) is greater than .05, so we cannot reject equal means. Difficulty: 2 Medium Topic: 10.02 Comparing Two Means: Independent Samples Learning Objective: 10-02 Explain the assumptions underlying the two-sample test of means. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

82) A medical researcher wondered if there is a significant difference between the mean birth weight of boy and girl babies. Random samples of 5 babies' weights (pounds) for each gender showed the following: Boys Girls

8.0 5.3

4.7 2.8

7.3 6.4

6.2 6.8

3.4 7.4

To test the researcher's hypothesis, we should use the A) paired (dependent) samples t-test. B) independent samples t-test. C) large-sample z-test. D) t-test for correlation. Answer: B Explanation: Although arranged side by side, these are unrelated data (independent samples). Difficulty: 1 Easy Topic: 10.02 Comparing Two Means: Independent Samples Learning Objective: 10-02 Explain the assumptions underlying the two-sample test of means. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 83) A medical researcher wondered if there is a significant difference between the mean birth weight of boy and girl babies. A random sample of babies' weights (pounds) showed the following: Boys Girls

8.0 5.3

4.7 2.8

7.3 6.4

6.2 6.8

3.4 7.4

How many degrees of freedom would be used to test for a zero difference in means? A) 4 B) 8 C) 10 D) Must know α to say. Answer: B Explanation: Although arranged side by side, these are unrelated data so it would be inappropriate to treat them as paired data. Using an independent samples t-test with n1 = 5 and n2 = 5 we have df = n1 + n2 − 2 = 5 + 5 − 2 = 8. Difficulty: 2 Medium Topic: 10.02 Comparing Two Means: Independent Samples Learning Objective: 10-02 Explain the assumptions underlying the two-sample test of means. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 43 Copyright 2019 © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.

84) In a test of a new surgical procedure, the five most respected surgeons in FlatBroke Township were invited to Carver Hospital. Each surgeon was assigned two patients of the same age, gender, and overall health. One patient was operated upon in the old way, and the other in the new way. Both procedures are considered equally safe. The surgery times are shown below:

Old way New way

Allen 36 31

Surgeon Bob 55 45

Chloe 28 28

Daphne 40 35

Edgar 62 57

Which test should we use to test for zero difference in mean times? A) Use the paired t-test. B) Use the independent samples t-test. C) Use the independent samples z test. D) Cannot be sure which test to use without knowing α. Answer: A Explanation: There are five pairs of data, so we would calculate the differences. Difficulty: 1 Easy Topic: 10.04 Comparing Two Means: Paired Samples Learning Objective: 10-04 Recognize paired data and be able to perform a paired t test. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

85) In a test of a new surgical procedure, the five most respected surgeons in FlatBroke Township were invited to Carver Hospital. Each surgeon was assigned two patients of the same age, gender, and overall health. One patient was operated upon in the old way, and the other in the new way. Both procedures are considered equally safe. The surgery times are shown below:

Old way New way

Allen 36 31

Surgeon Bob 55 45

Chloe 28 28

Daphne 40 35

Edgar 62 57

The time (in minutes) to complete each procedure was carefully recorded. In a right-tailed test for a difference of means, the test statistic is A) 3.162 B) 1.645 C) 1.860 D) 2.132 Answer: A Explanation: The test statistic is tcalc = (5 − 0)/[(3.5355)/51/2] = 3.162. Difficulty: 3 Hard Topic: 10.04 Comparing Two Means: Paired Samples Learning Objective: 10-04 Recognize paired data and be able to perform a paired t test. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

86) A corporate analyst is testing whether mean inventory turnover has increased. Inventory turnover in six randomly chosen product distribution centers (PDCs) is shown. PDC 1 PDC 2 PDC 3 PDC 4 PDC 5 PDC 6

This Year 5.1 3.9 4.8 3.4 4.6 7.7

Last Year 4.1 2.9 2.8 3.4 2.6 4.7

The degrees of freedom for the appropriate test would be A) 6 B) 5 C) 4 D) 12 Answer: B Explanation: These are paired samples, so d.f. = 6 − 1 = 5. Difficulty: 1 Easy Topic: 10.04 Comparing Two Means: Paired Samples Learning Objective: 10-04 Recognize paired data and be able to perform a paired t test. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

87) A corporate analyst is testing whether mean inventory turnover has increased. Inventory turnover in six randomly chosen product distribution centers (PDCs) is shown. PDC 1 PDC 2 PDC 3 PDC 4 PDC 5 PDC 6

This Year 5.1 3.9 4.8 3.4 4.6 7.7

Last Year 4.1 2.9 2.8 3.4 2.6 4.7

The right-tailed critical value at α = .005 is A) 1.645 B) 1.479 C) 4.032 D) 2.015 Answer: C Explanation: Paired data with d.f. = 6 − 1 = 5 gives us t.005 = 4.032. Difficulty: 1 Easy Topic: 10.04 Comparing Two Means: Paired Samples Learning Objective: 10-04 Recognize paired data and be able to perform a paired t test. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

88) A corporate analyst is testing whether mean inventory turnover has increased. Inventory turnover in six randomly chosen product distribution centers (PDCs) is shown. PDC 1 PDC 2 PDC 3 PDC 4 PDC 5 PDC 6

This Year 5.1 3.9 4.8 3.4 4.6 7.7

Last Year 4.1 2.9 2.8 3.4 2.6 4.7

The value of the test statistic is A) 3.798 B) 2.449 C) 1.225 D) 3.503 Answer: D Explanation: Paired data test statistic is tcalc = (1.5 − 0)/[(1.0488)/61/2] = 3.503. Difficulty: 3 Hard Topic: 10.04 Comparing Two Means: Paired Samples Learning Objective: 10-04 Recognize paired data and be able to perform a paired t test. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

89) The table below shows the mean number of daily errors by seven air traffic controller trainees during the first two weeks on the job. We want to perform a paired t-test at α = .05 to see if the mean daily errors have decreased from Week 1 to Week 2.

Week 1 Week 2

T1 5.1 3.2

T2 3.0 2.2

Trainee T3 12.1 8.7

T4 6.2 7.7

T5 11.5 9.4

T6 7.8 7.8

T7 2.2 3.1

The right-tailed critical value at α = .05 is A) 1.895 B) 1.943 C) 2.447 D) 2.365 Answer: B Explanation: Paired data with d.f. = 7 − 1 = 6 gives us t.05 = 1.943. Difficulty: 1 Easy Topic: 10.04 Comparing Two Means: Paired Samples Learning Objective: 10-04 Recognize paired data and be able to perform a paired t test. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

90) The table below shows the mean number of daily errors by air traffic controller trainees during the first two weeks on the job. We want to perform a paired t-test at α = .05 to see if the mean daily errors decreased significantly.

Week 1 Week 2

T1 5.1 3.2

T2 3.0 2.2

Trainee T3 12.1 8.7

T4 6.2 7.7

T5 11.5 9.4

T6 7.8 7.8

T7 2.2 3.1

The test statistic is A) 1.25 B) 1.75 C) 0.87 D) 0.79 Answer: A Explanation: Paired data test statistic is tcalc = (0.8286 − 0)/[(1.7547)/71/2] = 1.249. Difficulty: 3 Hard Topic: 10.04 Comparing Two Means: Paired Samples Learning Objective: 10-04 Recognize paired data and be able to perform a paired t test. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

91) The table below shows the mean number of daily errors by air traffic controller trainees during the first two weeks on the job. We want to perform a paired t-test at α = .05 to see if the mean daily errors decreased significantly.

Week 1 Week 2

T1 5.1 3.2

T2 3.0 2.2

Trainee T3 12.1 8.7

T4 6.2 7.7

T5 11.5 9.4

T6 7.8 7.8

T7 2.2 3.1

What would be the degrees of freedom for the appropriate test? A) 14 B) 12 C) 7 D) 6 Answer: D Explanation: Paired data with d.f. = 7 − 1 = 6. Difficulty: 1 Easy Topic: 10.04 Comparing Two Means: Paired Samples Learning Objective: 10-04 Recognize paired data and be able to perform a paired t test. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 92) The F-test for equality of variances assumes A) normal populations. B) equal means. C) equal sample sizes. D) equal means and sample sizes. Answer: A Explanation: Severely skewed populations could pose a problem for the F-test. Difficulty: 1 Easy Topic: 10.07 Comparing Two Variances Learning Objective: 10-08 Carry out a test of two variances using the F distribution. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation

93) Which is not true of the two-tailed F-test for equality of variances? A) It requires reversing the numerator and denominator d.f. to obtain the left-tail critical value. B) It can be avoided by "folding" the larger variance into the numerator and adjusting α. C) It is fairly robust to the presence of nonnormality in the populations being sampled. Answer: C Explanation: Simulation studies suggest that the F-test is robust to modest departures from normality. Difficulty: 2 Medium Topic: 10.07 Comparing Two Variances Learning Objective: 10-08 Carry out a test of two variances using the F distribution. Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation 94) Which of the following is not a characteristic of the F distribution? A) It is a continuous distribution. B) It is always a positive number. C) It is a family based on two sets of degrees of freedom. D) It describes the ratio of two sample means. Answer: D Explanation: The F-test is a ratio of two variances. Difficulty: 1 Easy Topic: 10.07 Comparing Two Variances Learning Objective: 10-08 Carry out a test of two variances using the F distribution. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation

95) Carver Memorial Hospital's surgeons have a new procedure that they think will decrease the variance in the time it takes to perform an appendectomy. A sample of 8 appendectomies using the old method had a variance of 36 minutes, while a sample of 10 appendectomies using the experimental method had a variance of 16 minutes. At α = .10 in a two-tailed test for equal variances, the critical values are A) 0.272 and 3.29 B) 0.299 and 3.07 C) 0.368 and 2.51 D) −1.645 and +1.645 Answer: A Explanation: Using α/2 = .05 in each tail, from Appendix F, F7,9 = 3.29 and 1/F9,7 = 1/(3.68) = 0.272, or else use Excel's functions =F.INV.RT(.05,7,9) and =F.INV(.05,7,9). Difficulty: 3 Hard Topic: 10.07 Comparing Two Variances Learning Objective: 10-08 Carry out a test of two variances using the F distribution. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 96) The folded F-test for equality of variances A) is rarely used because of its complexity. B) requires looking up the critical value for α/2. C) puts the smaller variance in the numerator. D) requires looking up two critical values of F instead of one. Answer: B Explanation: Putting the larger variance in the numerator makes it into a right-tailed test using α/2. Difficulty: 2 Medium Topic: 10.07 Comparing Two Variances Learning Objective: 10-08 Carry out a test of two variances using the F distribution. Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation

97) An F-test for equality of variances gives a p-value of .003. At α = .05, what conclusion can be made about the preferred test to compare the means for the same sample? A) We would prefer a pooled variance t-test for equality of means. B) We would not wish to pool the variances in a t-test for equality of means. C) We would prefer a paired t-test for equality of means. D) The variances have nothing to do with the t-test for equality of means. Answer: B Explanation: The results of the F-test suggest that we should assume unequal variances in the ttest. Difficulty: 2 Medium Topic: 10.07 Comparing Two Variances Learning Objective: 10-08 Carry out a test of two variances using the F distribution. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 98) A random sample of Ersatz University students revealed that 16 females had a mean of $22.30 in their wallets with a standard deviation of $3.20, while 6 males had a mean of $17.30 with a standard deviation of $9.60. At α = .10, to test for equal variances in a two-tailed test, the critical values are A) 0.441 and 3.24 B) 0.556 and 2.27 C) 0.345 and 4.62 D) 0.387 and 2.90 Answer: C Explanation: Using α/2 = .05 in each tail, from Appendix F, F15,5 = 4.62 and 1/F5,15 = 1/(2.90) = 0.345, or else use Excel's functions =F.INV.RT(.05,15,5) and =F.INV(.05,15,5). Difficulty: 3 Hard Topic: 10.07 Comparing Two Variances Learning Objective: 10-08 Carry out a test of two variances using the F distribution. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

99) A random sample of Ersatz University students revealed that 16 females had a mean of $22.30 in their wallets with a standard deviation of $3.20, while 6 males had a mean of $17.30 with a standard deviation of $9.60. The value of the test statistic for a folded F-test for equal variances is A) 0.333 B) 0.111 C) 9.00 D) 3.00 Answer: C Explanation: Fcalc = (9.60)2/(3.20)2 = 9.00 because a folded F-test puts the larger variance in the numerator. Difficulty: 2 Medium Topic: 10.07 Comparing Two Variances Learning Objective: 10-08 Carry out a test of two variances using the F distribution. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 100) Assuming unequal variances in a t-test for a zero difference of two means, we would A) sum the degrees of freedom for each sample. B) use the larger degrees of freedom for simplicity. C) use a complicated formula for the degrees of freedom. D) use a z-test to be conservative in the calculation. Answer: C Explanation: The formula for Welch's adjusted degrees of freedom is not easy without a computer. Difficulty: 2 Medium Topic: 10.02 Comparing Two Means: Independent Samples Learning Objective: 10-02 Explain the assumptions underlying the two-sample test of means. Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation

101) The z-test for zero difference in two means A) is generally the preferred test for means. B) is rarely suitable for business data. C) is the most powerful test for means. D) is not available in Excel's Data Analysis. Answer: B Explanation: We rarely know the population variances, so we would generally use a t-test. Difficulty: 2 Medium Topic: 10.02 Comparing Two Means: Independent Samples Learning Objective: 10-01 Recognize and perform a test for two means. Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation 102) A confidence interval for the difference of two population means A) must pool the sample variances. B) may or may not pool the sample variances. C) cannot be used to test for equal population means. Answer: B Explanation: Whether to pool or not is a choice the researcher can make. Difficulty: 2 Medium Topic: 10.03 Confidence Interval for the Difference of Two Means, mu1 - mu2 Learning Objective: 10-03 Construct a confidence interval for mu1 - mu2. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

103) A medical researcher compared the variances in birth weights for five randomly chosen babies of each gender, with the MegaStat results shown below. F-test for equality of variance 3.537 variance: Boys 3.288 variance: Girls 1.08 F .9453 p-value The population variances A) may be assumed equal at any customary α. B) should be assumed unequal at any customary α. C) are not relevant to this paired t-test. Answer: A Explanation: The test statistic Fcalc = (3.537)2/(3.288)2 does not differ significantly from 1 at any typical α. Difficulty: 2 Medium Topic: 10.07 Comparing Two Variances Learning Objective: 10-08 Carry out a test of two variances using the F distribution. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

104) A medical researcher wondered if there is a significant difference between the mean birth weight of boy and girl babies. She weighed a random sample of five babies of each gender. Their weights (pounds) are shown below, along with some MegaStat results. Boys 5.920 1.881 5 8 0.1800 3.4125 1.8473 1.1683 0 0.15 .4407

Girls 5.740 mean 1.813 std. dev. 5 n df difference (Boys - Girls) pooled variance pooled std. dev. standard error of difference hypothesized of difference t p-value (one-tailed, upper)

The population means A) may be assumed equal at any customary α. B) should be assumed unequal at any customary α. C) are not relevant to this paired t-test. Answer: A Explanation: The p-value (.4407) does not lead to rejection of equal means at any typical α. Difficulty: 2 Medium Topic: 10.02 Comparing Two Means: Independent Samples Learning Objective: 10-02 Explain the assumptions underlying the two-sample test of means. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

105) The coach of an adult Master's Swim class selected eight swimmers within each of the two age groups shown below. A 50-yard freestyle time is recorded for each swimmer. The resulting times (seconds) are shown below. Which statistical test would you choose to compare the two groups? Obs 1 2 3 4 5 6 7 8 Mean St Dev

Age Group 25-34 Yrs 35-44 Yrs 26.26 31.08 28.44 28.63 26.61 25.65 29.74 31.81 26.94 27.22 26.31 30.56 34.68 33.38 22.55 30.82 27.69 3.50

29.89 2.54

A) t-test for independent samples with known variances B) t-test for independent samples with unknown variances C) t-test for paired samples D) z-test for two independent proportions Answer: B Explanation: Despite being arranged side-by-side, there is no link between the columns. The similar standard deviations suggest that it would be reasonable to "pool" the variances (pun intended) although this question was not posed. Difficulty: 1 Easy Topic: 10.02 Comparing Two Means: Independent Samples Learning Objective: 10-02 Explain the assumptions underlying the two-sample test of means. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

106) Jason wants to perform a two-tailed test for equality between two independent sample proportions. Each sample has at least 10 "successes" and 10 "failures." Jason's test statistic is −1.44. What is his p-value? A) .1498 B) .0749 C) .9251 D) Between .01 and .05 Answer: A Explanation: From Appendix C, we find P(Z ≤ −1.44) = .0749, which we double for a two-tailed test. Difficulty: 2 Medium Topic: 10.05 Comparing Two Proportions Learning Objective: 10-05 Perform a test to compare two proportions using z. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 107) Does the Speedo Fastskin II Male Hi-Neck Bodyskin competition racing swimsuit improve a swimmer's 200-yard individual medley performance times? A test of 100 randomly chosen male varsity swimmers at several different universities showed that 66 enjoyed improved times, compared with only 54 of 100 female varsity swimmers. To test for equality in the proportions of men versus women who experienced improvement, the test statistic is approximately A) 1.73 B) 1.47 C) 2.31 D) We cannot tell without knowing the tail of the test. Answer: A Explanation: The combined proportion is pc = (66 + 54)/(100 + 100) = .60, so zcalc = (.66 − . 54)/[.60(1 − .60)/100 + .60(1 − .60)/100]1/2 = 1.73. Difficulty: 3 Hard Topic: 10.05 Comparing Two Proportions Learning Objective: 10-05 Perform a test to compare two proportions using z. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

108) Does the Speedo Fastskin II Male Hi-Neck Bodyskin competition racing swimsuit improve a swimmer's 200-yard individual medley performance times? A test of 100 randomly chosen male varsity swimmers at several different universities showed that 66 enjoyed improved times, compared with only 54 of 100 female varsity swimmers. In comparing the proportions of males versus females, is it safe to assume normality? A) Yes, clearly B) Yes, but just barely C) No. D) We cannot tell without knowing α. Answer: A Explanation: We have at least 10 successes (x1 = 66, x2 = 54) and 10 failures (n1 − x1 = 34, n2 − x2 = 46). Difficulty: 1 Easy Topic: 10.05 Comparing Two Proportions Learning Objective: 10-06 Check whether normality may be assumed for two proportions. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 109) The table below shows two samples taken to compare the mean age of individuals who purchased the iPhone 3G at two AT&T store locations. Statistic Mean St Dev Sample size

Ann Arbor 25.817 3.389 7

Livonia 31.248 1.874 10

What are the critical values for a two-tailed test for equal variances at α = .05? A) 0.275, 3.14 B) 0.244, 3.37 C) 0.210, 3.95 D) 0.181, 4.32 Answer: D Explanation: Using α/2 = .025 in each tail, from Appendix F, F6,9 = 4.32 and 1/F9,6 = 1/(5.52) = 0.181, or else use Excel's functions =F.INV.RT(.025,6,9) and =F.INV(.025,6,9). Difficulty: 3 Hard Topic: 10.07 Comparing Two Variances Learning Objective: 10-08 Carry out a test of two variances using the F distribution. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

110) The table below shows two samples taken to compare the mean age of individuals who purchased the iPhone 3G at two AT&T store locations. Statistic Mean St Dev Sample size

Ann Arbor 25.817 4.389 7

Livonia 31.248 1.874 10

At α = .05, can you conclude that the first sample has a larger variance than the second sample? A) Yes, clearly. B) Yes, but just barely. C) No. Answer: A Explanation: The test statistic Fcalc = (4.389)2/(1.874)2 = 5.59 exceeds F.05 = 3.37 with d.f. = (6, 9). Difficulty: 1 Easy Topic: 10.07 Comparing Two Variances Learning Objective: 10-08 Carry out a test of two variances using the F distribution. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 111) Group 1 has a mean of 13.4 and group 2 has a mean of 15.2. Both populations are known to have a variance of 9.0 and each sample consists of 18 items. What is the test statistic to test for equality of population means? A) −1.755 B) −1.643 C) −1.800 D) −1.285 Answer: C Explanation: With known variances, zcalc = (13.4 − 15.2)/[9.0/18 + 9.0/18]1/2 = −1.800. Difficulty: 2 Medium Topic: 10.02 Comparing Two Means: Independent Samples Learning Objective: 10-01 Recognize and perform a test for two means. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

112) Which is not a type of comparison for which you would anticipate a two-sample test? A) Before Treatment versus After Treatment B) Old Method versus New Method C) Sample Mean versus Desired Mean D) Experimental Group versus Control Group Answer: C Explanation: The point of comparison is between two samples, not comparison with a benchmark or target. Difficulty: 2 Medium Topic: 10.01 Two-Sample Tests Learning Objective: 10-01 Recognize and perform a test for two means. Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation 113) Which Excel function would give the critical value for a left-tailed F test to compare two sample variances? A) =F.INV(α, df1,df2) B) =F.DIST(s12/s22, df1,df2) C) =F.INV(α/2, df1,df2) D) =1−F.DIST(s12/s22, df1,df2) Answer: A Explanation: The critical value does not depend on the sample variances. The inverse Excel function provides critical values. Difficulty: 2 Medium Topic: 10.07 Comparing Two Variances Learning Objective: 10-08 Carry out a test of two variances using the F distribution. Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation

114) Which Excel function would give a p-value for a left-tailed F test to compare two sample variances? A) =1−F.DIST(s12/s22, df1,df2) B) =F.INV(α, df1,df2) C) =F.DIST(s12/s22, df1,df2) D) =F.INV(α/2, df1,df2) Answer: C Explanation: The =F.DIST function gives a left-tail area. The inverse Excel functions provide critical values. Difficulty: 2 Medium Topic: 10.07 Comparing Two Variances Learning Objective: 10-08 Carry out a test of two variances using the F distribution. Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation

Applied Statistics in Business and Economics, 6e (Doane) Chapter 11 Analysis of Variance 1) ANOVA is a procedure intended to compare the variances of several groups (treatments). Answer: FALSE Explanation: ANOVA compares several means (although its test statistic is based on variances). Difficulty: 1 Easy Topic: 11.01 Overview of ANOVA Learning Objective: 11-01 Use basic ANOVA terminology correctly. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 2) ANOVA is a procedure intended to compare the means of several groups (treatments). Answer: TRUE Explanation: Although its test statistic is based on variances, ANOVA compares several means. Difficulty: 1 Easy Topic: 11.01 Overview of ANOVA Learning Objective: 11-01 Use basic ANOVA terminology correctly. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 3) If you have four factors (call them A, B, C, and D) in an ANOVA experiment with replication, you could have a maximum of four different two-factor interactions. Answer: FALSE Explanation: There could be six two-way interactions: AB, AC, AD, BC, BD, CD. Difficulty: 3 Hard Topic: 11.07 Higher-Order ANOVA Models (Optional) Learning Objective: 11-12 Recognize the need for experimental design and GLM (optional). Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation 4) Hartley's test measures the equality of the means for several groups. Answer: FALSE Explanation: Hartley's test is designed to detect unequal population variances. Difficulty: 2 Medium Topic: 11.04 Tests for Homogeneity of Variances Learning Objective: 11-08 Use Hartley's test for equal variances in c treatment groups. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 1 Copyright 2019 © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.

5) Hartley's test is to check for unequal variances for c groups. Answer: TRUE Explanation: Unequal population variances would violate an ANOVA assumption. Difficulty: 2 Medium Topic: 11.04 Tests for Homogeneity of Variances Learning Objective: 11-08 Use Hartley's test for equal variances in c treatment groups. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 6) Comparison of c means in one-factor ANOVA can equivalently be done by using c individual t-tests on c pairs of means at the same α. Answer: FALSE Explanation: Multiple two-sample t-tests from the same data set would inflate the overall α. Difficulty: 2 Medium Topic: 11.03 Multiple Comparisons Learning Objective: 11-07 Understand and perform Tukey's test for paired means. Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation 7) ANOVA assumes equal variances within each treatment group. Answer: TRUE Explanation: ANOVA checks for unequal means, while assuming homogeneous variances. Difficulty: 1 Easy Topic: 11.01 Overview of ANOVA Learning Objective: 11-02 Explain the assumptions of ANOVA and why they are important. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 8) Three-factor ANOVA is required if we have three treatment groups (i.e., three data columns). Answer: FALSE Explanation: If there are only three columns of data, we only have one factor (with three treatments). The hypothesis is whether the three treatment group means are the same. Difficulty: 2 Medium Topic: 11.07 Higher-Order ANOVA Models (Optional) Learning Objective: 11-12 Recognize the need for experimental design and GLM (optional). Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation 2 Copyright 2019 © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.

9) ANOVA assumes normal populations. Answer: TRUE Explanation: Populations are assumed to be normally distributed and to have equal variances. Difficulty: 1 Easy Topic: 11.01 Overview of ANOVA Learning Objective: 11-02 Explain the assumptions of ANOVA and why they are important. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 10) Tukey's test compares pairs of treatment means in an ANOVA. Answer: TRUE Explanation: Tukey's test is a follow-up to ANOVA to detect which pairs of means differ (if any). Difficulty: 2 Medium Topic: 11.03 Multiple Comparisons Learning Objective: 11-07 Understand and perform Tukey's test for paired means. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 11) Tukey's test is similar to a two-sample t-test except that it pools the variances for all c samples. Answer: TRUE Explanation: There is a strong analogy with the two-sample t-test except that we pool all the variances. Difficulty: 2 Medium Topic: 11.03 Multiple Comparisons Learning Objective: 11-07 Understand and perform Tukey's test for paired means. Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation 12) Tukey's test is not needed if we have the overall F statistic for the ANOVA. Answer: FALSE Explanation: Tukey's test is a follow-up to ANOVA to detect which pairs of means differ (if any). Difficulty: 2 Medium Topic: 11.03 Multiple Comparisons Learning Objective: 11-07 Understand and perform Tukey's test for paired means. Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation 3 Copyright 2019 © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.

13) Interaction plots that show crossing lines indicate likely interactions. Answer: TRUE Explanation: Interaction plots provide an intuitive visual way of seeing possible interactions. Difficulty: 2 Medium Topic: 11.06 Two-Factor ANOVA with Replication (Full Factorial Model) Learning Objective: 11-11 Interpret main effects and interaction effects in two-factor ANOVA. Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation 14) Interaction plots that show parallel lines would suggest interaction effects. Answer: FALSE Explanation: Interaction plots that show crossing lines indicate likely interactions. Difficulty: 2 Medium Topic: 11.06 Two-Factor ANOVA with Replication (Full Factorial Model) Learning Objective: 11-11 Interpret main effects and interaction effects in two-factor ANOVA. Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation 15) In a two-factor ANOVA with three columns and four rows, there can be more than two interaction effects. Answer: FALSE Explanation: There can only be one interaction (row × column). Difficulty: 2 Medium Topic: 11.06 Two-Factor ANOVA with Replication (Full Factorial Model) Learning Objective: 11-11 Interpret main effects and interaction effects in two-factor ANOVA. Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation 16) Sample sizes must be equal in one-factor ANOVA. Answer: FALSE Explanation: Sample sizes often are equal by design, but it is not necessary. Difficulty: 1 Easy Topic: 11.02 One-Factor ANOVA (Completely Randomized Model) Learning Objective: 11-03 Recognize from data format when one-factor ANOVA is appropriate. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation

17) In a 3×4 randomized block (two-factor unreplicated) ANOVA, we have 12 treatment groups. Answer: TRUE Explanation: Each row/column combination is a treatment group. Difficulty: 2 Medium Topic: 11.05 Two-Factor ANOVA without Replication (Randomized Block Model) Learning Objective: 11-10 Interpret results in a two-factor ANOVA without replication. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 18) One-factor ANOVA with two groups is equivalent to a two-tailed t-test. Answer: TRUE Explanation: The p-values will be the same in either test as long as the t-test is two-tailed. Difficulty: 2 Medium Topic: 11.02 One-Factor ANOVA (Completely Randomized Model) Learning Objective: 11-04 Interpret sums of squares and calculations in an ANOVA table. Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation 19) One-factor ANOVA stacked data for five groups will be arranged in five separate columns. Answer: FALSE Explanation: One column will contain the data, while a second column names the group. Difficulty: 2 Medium Topic: 11.02 One-Factor ANOVA (Completely Randomized Model) Learning Objective: 11-03 Recognize from data format when one-factor ANOVA is appropriate. Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation 20) Hartley's test is the largest sample mean divided by the smallest sample mean. Answer: FALSE Explanation: Hartley's test statistic is the ratio of s2max to s2min (ratio of variances not means). Difficulty: 2 Medium Topic: 11.04 Tests for Homogeneity of Variances Learning Objective: 11-08 Use Hartley's test for equal variances in c treatment groups. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation

21) Tukey's test for five groups would require 10 comparisons of means. Answer: TRUE Explanation: The number of possible comparisons is c(c − 1)/2 = 5(4)/2 = 10. Difficulty: 2 Medium Topic: 11.03 Multiple Comparisons Learning Objective: 11-07 Understand and perform Tukey's test for paired means. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 22) ANOVA is robust to violations of the equal-variance assumption as long as group sizes are equal. Answer: TRUE Explanation: Studies suggest that equal group sizes strengthen the ANOVA test. Difficulty: 2 Medium Topic: 11.02 One-Factor ANOVA (Completely Randomized Model) Learning Objective: 11-02 Explain the assumptions of ANOVA and why they are important. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 23) Levene's test for homogeneity of variance is attractive because it does not depend on the assumption of normality. Answer: TRUE Explanation: While Hartley's test is sensitive to nonnormality, Levene's test statistic is not. However, Levene's test is not provided by Excel so you need another computer package. Difficulty: 2 Medium Topic: 11.04 Tests for Homogeneity of Variances Learning Objective: 11-08 Use Hartley's test for equal variances in c treatment groups. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 24) Tukey's test with seven groups would entail 21 comparisons of means. Answer: TRUE Explanation: The number of possible comparisons is c(c − 1)/2 = 7(6)/2 = 21. Difficulty: 2 Medium Topic: 11.03 Multiple Comparisons Learning Objective: 11-07 Understand and perform Tukey's test for paired means. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 7 Copyright 2019 © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.

25) Tukey's test pools all the sample variances. Answer: TRUE Explanation: In a Tukey test, all c sample variances are combined (weighted by their degrees of freedom). Difficulty: 1 Easy Topic: 11.03 Multiple Comparisons Learning Objective: 11-07 Understand and perform Tukey's test for paired means. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 26) It is desirable, but not necessary, that sample sizes be equal in a one-factor ANOVA. Answer: TRUE Explanation: Studies suggest that equal group sizes strengthen the ANOVA test. Difficulty: 1 Easy Topic: 11.02 One-Factor ANOVA (Completely Randomized Model) Learning Objective: 11-02 Explain the assumptions of ANOVA and why they are important. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 27) Which is the Excel function to find the critical value of F for α = .05, df1 = 3, df2 = 25? A) =F.DIST(.05, 2, 24) B) =F.INV.RT(.05, 3, 25) C) =F.DIST(.05, 3, 25) D) =F.INV(.05, 2, 24) Answer: B Explanation: The equivalent Excel 2007 function would be =FINV(.05, 3, 25). Difficulty: 2 Medium Topic: 11.02 One-Factor ANOVA (Completely Randomized Model) Learning Objective: 11-06 Use a table or Excel to find critical values for the F distribution. Bloom's: Apply AACSB: Technology Accessibility: Keyboard Navigation

28) Which Excel function gives the right-tail p-value for an ANOVA test with a test statistic Fcalc = 4.52, n = 29 observations, and c = 4 groups? A) =F.DIST.RT(4.52, 3, 25) B) =F.INV(4.52, 4, 28) C) =F.DIST(4.52, 4, 28) D) =F.INV(4.52, 3, 25) Answer: A Explanation: The equivalent Excel 2007 function would be =FDIST(.05, 3, 25). Difficulty: 3 Hard Topic: 11.02 One-Factor ANOVA (Completely Randomized Model) Learning Objective: 11-06 Use a table or Excel to find critical values for the F distribution. Bloom's: Apply AACSB: Technology Accessibility: Keyboard Navigation 29) Variation "within" the ANOVA treatments represents A) random variation. B) differences between group means. C) differences between group variances. D) the effect of sample size. Answer: A Explanation: Variation within groups is also called error variance or unexplained variance. Difficulty: 1 Easy Topic: 11.01 Overview of ANOVA Learning Objective: 11-01 Use basic ANOVA terminology correctly. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 30) Which is not an assumption of ANOVA? A) Normality of the treatment populations B) Homogeneous treatment variances C) Independent sample observations D) Equal population sizes for groups Answer: D Explanation: It is desirable, but not necessary, that sample sizes be equal in a one-factor ANOVA. Difficulty: 1 Easy Topic: 11.01 Overview of ANOVA Learning Objective: 11-02 Explain the assumptions of ANOVA and why they are important. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 9 Copyright 2019 © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.

31) In an ANOVA, when would the F-test statistic be zero? A) When there is no difference in the variances B) When the treatment means are the same C) When the observations are normally distributed D) The F-test statistic cannot ever be zero. Answer: B Explanation: If each group mean equals the overall mean, then Fcalc could be zero (an unusual situation). Difficulty: 2 Medium Topic: 11.02 One-Factor ANOVA (Completely Randomized Model) Learning Objective: 11-04 Interpret sums of squares and calculations in an ANOVA table. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 32) ANOVA is used to compare A) proportions of several groups. B) variances of several groups. C) means of several groups. D) both means and variances. Answer: C Explanation: Although its test statistic is based on variances, ANOVA compares several means. Difficulty: 1 Easy Topic: 11.01 Overview of ANOVA Learning Objective: 11-01 Use basic ANOVA terminology correctly. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 33) Analysis of variance is a technique used to test for A) equality of two or more variances. B) equality of two or more means. C) equality of a population mean and a given value. D) equality of more than two variances. Answer: B Explanation: Although its test statistic is based on variances, ANOVA compares several means. Difficulty: 1 Easy Topic: 11.01 Overview of ANOVA Learning Objective: 11-01 Use basic ANOVA terminology correctly. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 10 Copyright 2019 © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.

34) Which of the following is not a characteristic of the F distribution? A) It is always right-skewed. B) It describes the ratio of two variances. C) It is a family based on two sets of degrees of freedom. D) It is negative when s12 is smaller than s22. Answer: D Explanation: The F distribution is the ratio of two mean squares, so it cannot be negative. Difficulty: 2 Medium Topic: 11.02 One-Factor ANOVA (Completely Randomized Model) Learning Objective: 11-04 Interpret sums of squares and calculations in an ANOVA table. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 35) In an ANOVA, the SSE (error) sum of squares reflects A) the effect of the combined factor(s). B) the overall variation in Y that is to be explained. C) the variation that is not explained by the factors. D) the combined effect of treatments and sample size. Answer: C Explanation: The error variance or unexplained variance is variation within groups. Difficulty: 1 Easy Topic: 11.02 One-Factor ANOVA (Completely Randomized Model) Learning Objective: 11-04 Interpret sums of squares and calculations in an ANOVA table. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 36) To test the null hypothesis H0: μ1 = μ2 = μ3 using samples from normal populations with unknown but equal variances, we A) cannot safely use ANOVA. B) can safely employ ANOVA. C) would prefer three separate t-tests. D) would need three-factor ANOVA. Answer: B Explanation: As long as the variances are equal, we can safely use ANOVA. Difficulty: 1 Easy Topic: 11.02 One-Factor ANOVA (Completely Randomized Model) Learning Objective: 11-03 Recognize from data format when one-factor ANOVA is appropriate. Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation 11 Copyright 2019 © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.

37) Which is not assumed in ANOVA? A) Observations are independent. B) Populations are normally distributed. C) Variances of all treatment groups are the same. D) Population variances are known. Answer: D Explanation: Population variances are almost never known. Difficulty: 1 Easy Topic: 11.01 Overview of ANOVA Learning Objective: 11-02 Explain the assumptions of ANOVA and why they are important. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 38) In a one-factor ANOVA, the computed value of F will be negative A) when there is no difference in the treatment means. B) when there is no difference within the treatments. C) when the SST (total) is larger than SSE (error). D) under no circumstances. Answer: D Explanation: The F distribution is the ratio of two mean squares, so it cannot be negative. Difficulty: 1 Easy Topic: 11.02 One-Factor ANOVA (Completely Randomized Model) Learning Objective: 11-04 Interpret sums of squares and calculations in an ANOVA table. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 39) Degrees of freedom for the between-group variation in a one-factor ANOVA with n1 = 5, n2 = 6, n3 = 7 would be A) 18 B) 17 C) 6 D) 2 Answer: D Explanation: For between-group degrees of freedom we have df = c − 1 = 3 − 1 = 2. Difficulty: 2 Medium Topic: 11.02 One-Factor ANOVA (Completely Randomized Model) Learning Objective: 11-04 Interpret sums of squares and calculations in an ANOVA table. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 12 Copyright 2019 © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.

40) Degrees of freedom for the between-group variation in a one-factor ANOVA with n1 = 8, n2 = 5, n3 = 7, n4 = 9 would be A) 28 B) 3 C) 29 D) 4 Answer: B Explanation: For between-group degrees of freedom we have df = c − 1 = 4 − 1 = 3. Difficulty: 2 Medium Topic: 11.02 One-Factor ANOVA (Completely Randomized Model) Learning Objective: 11-04 Interpret sums of squares and calculations in an ANOVA table. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 41) Using one-factor ANOVA with 30 observations, we find at α = .05 that we cannot reject the null hypothesis of equal means. We increase the sample size from 30 observations to 60 observations and obtain the same value for the sample F-test statistic. Which is correct? A) We might now be able to reject the null hypothesis. B) We surely must reject H0 for 60 observations C) We cannot reject H0 since we obtained the same F-value. D) It is impossible to get the same F-value for n = 60 as for n = 30. Answer: A Explanation: With more degrees of freedom, the critical value F.05 will be smaller, so we might reject. Difficulty: 2 Medium Topic: 11.02 One-Factor ANOVA (Completely Randomized Model) Learning Objective: 11-04 Interpret sums of squares and calculations in an ANOVA table. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

42) One-factor analysis of variance A) requires that the number of observations in each group be identical. B) has less power when the number of observations per group is not identical. C) is extremely sensitive to slight departures from normality. D) is a generalization of the t-test for paired observations. Answer: B Explanation: Studies suggest that equal group sizes strengthen the power of the ANOVA test. Difficulty: 2 Medium Topic: 11.02 One-Factor ANOVA (Completely Randomized Model) Learning Objective: 11-02 Explain the assumptions of ANOVA and why they are important. Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation 43) In a one-factor ANOVA, the total sum of squares is equal to A) the sum of squares within groups plus the sum of squares between groups. B) the sum of squares within groups times the sum of squares between groups. C) the sum of squares within groups divided by the sum of squares between groups. D) the means of all the groups squared. Answer: A Explanation: The basic identify is SSbetween + SSwithin = SStotal. Difficulty: 1 Easy Topic: 11.02 One-Factor ANOVA (Completely Randomized Model) Learning Objective: 11-04 Interpret sums of squares and calculations in an ANOVA table. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 44) The within-treatment variation reflects A) variation among individuals of the same group. B) variation between individuals in different groups. C) variation explained by factors included in the ANOVA model. D) variation that is not part of the ANOVA model. Answer: A Explanation: Variation within groups is also called error variance or unexplained variance. Difficulty: 2 Medium Topic: 11.02 One-Factor ANOVA (Completely Randomized Model) Learning Objective: 11-04 Interpret sums of squares and calculations in an ANOVA table. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation

45) Given the following ANOVA table (some information is missing), find the F statistic. Source Treatment Error Total

Sum of Squares 744.00 751.50 1,495.50

df 4 15 19

Mean Square

A) 3.71 B) 0.99 C) 0.497 D) 4.02 Answer: A Explanation: MStreatment = 744/4 = 186, MSerror = (751.5)/15 = 50.1, so F = 186/50.1 = 3.71. Difficulty: 2 Medium Topic: 11.02 One-Factor ANOVA (Completely Randomized Model) Learning Objective: 11-04 Interpret sums of squares and calculations in an ANOVA table. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 46) Given the following ANOVA table (some information is missing), find the critical value of F.05. Source Treatment Error Total

Sum of Squares 744.00 751.50 1,495.50

Mean Square

F.05

4 15 19

A) 3.06 B) 2.90 C) 2.36 D) 3.41 Answer: A Explanation: For df = (4, 15) we use Appendix F to get F.05 = 3.06. Difficulty: 2 Medium Topic: 11.02 One-Factor ANOVA (Completely Randomized Model) Learning Objective: 11-06 Use a table or Excel to find critical values for the F distribution. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

47) Identify the degrees of freedom for the treatment and error in this one-factor ANOVA (blanks indicate missing information). Source Treatment Error Total

Sum of Squares 993 1,002 1,995

Mean Square 331.0 50.1

A) 4, 24 B) 3, 20 C) 5, 23 Answer: B Explanation: Since SS/df = MS, we know that df = SS/MS. Hence, 993/331 = 3 and 1002/50.1 = 20. Difficulty: 2 Medium Topic: 11.02 One-Factor ANOVA (Completely Randomized Model) Learning Objective: 11-04 Interpret sums of squares and calculations in an ANOVA table. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 48) For this one-factor ANOVA (some information is missing), how many treatment groups were there? Source Treatment Error Total

Sum of Squares 654 3,456 4,110

Mean Square 218 128

A) Cannot be determined B) 3 C) 4 D) 2 Answer: C Explanation: Since SS/df = MS, we know that df = SS/MS and, hence, 654/218 = 3 = c − 1. Difficulty: 2 Medium Topic: 11.02 One-Factor ANOVA (Completely Randomized Model) Learning Objective: 11-04 Interpret sums of squares and calculations in an ANOVA table. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

49) For this one-factor ANOVA (some information is missing), what is the F-test statistic? Source Treatment Error Total

Sum of Squares 654 3,456 4,110

Mean Square 218 128

A) 0.159 B) 2.833 C) 1.703 D) Cannot be determined Answer: C Explanation: Fcalc = (MStreatment)/(MSerror) = 218/128 = 1.703. Difficulty: 2 Medium Topic: 11.02 One-Factor ANOVA (Completely Randomized Model) Learning Objective: 11-04 Interpret sums of squares and calculations in an ANOVA table. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 50) Refer to the following partial ANOVA results from Excel (some information is missing). Source of Variation Between groups Within groups Total

SS 1483 2113.833

MS 210.2788 74.15

The F-test statistic is A) 2.84 B) 3.56 C) 2.80 D) 2.79 Answer: A Explanation: Fcalc = (MSbetween)/(MSwithin) = (210.2788)/(74.15) = 2.836. Difficulty: 2 Medium Topic: 11.02 One-Factor ANOVA (Completely Randomized Model) Learning Objective: 11-04 Interpret sums of squares and calculations in an ANOVA table. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

51) Refer to the following partial ANOVA results from Excel (some information is missing). Source of Variation Between groups Within groups Total

1483 2113.833

MS 210.2778 74.15

Degrees of freedom for between-groups variation are A) 3 B) 4 C) 5 D) We cannot tell from given information. Answer: A Explanation: SSbetween = 2113.833 − 1483 = 630.833, so df = (630.833)/(210.2778) = 3. Difficulty: 3 Hard Topic: 11.02 One-Factor ANOVA (Completely Randomized Model) Learning Objective: 11-04 Interpret sums of squares and calculations in an ANOVA table. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 52) Refer to the following partial ANOVA results from Excel (some information is missing). Source of Variation Between groups Within groups Total

SS 1483 2113.833

MS 210.2778 74.15

SS for between-groups variation will be A) 129.99 B) 630.83 C) 1233.4 D) We cannot tell from given information. Answer: B Explanation: SSbetween = 2113.833 − 1483 = 630.833. Difficulty: 1 Easy Topic: 11.02 One-Factor ANOVA (Completely Randomized Model) Learning Objective: 11-04 Interpret sums of squares and calculations in an ANOVA table. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

53) Refer to the following partial ANOVA results from Excel (some information is missing). Source of Variation Between groups Within groups Total

1483 2113.833

MS 210.2778 74.15

The number of treatment groups is A) 4 B) 3 C) 2 D) 1 Answer: A Explanation: SSbetween = 2113.833 − 1483 = 630.833, so df = (630.833)/(210.2778) = 3 = c − 1. Difficulty: 3 Hard Topic: 11.02 One-Factor ANOVA (Completely Randomized Model) Learning Objective: 11-04 Interpret sums of squares and calculations in an ANOVA table. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 54) Refer to the following partial ANOVA results from Excel (some information is missing). Source of Variation Between groups Within groups Total

SS 1483 2113.833

MS 210.2778 74.15

The sample size is A) 20 B) 23 C) 24 D) 21 Answer: C Explanation: (630.833)/(210.2778) = 3 and (1483)/(74.15) = 20, so 3 + 20 = 23 = n − 1. Difficulty: 3 Hard Topic: 11.02 One-Factor ANOVA (Completely Randomized Model) Learning Objective: 11-04 Interpret sums of squares and calculations in an ANOVA table. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

55) Refer to the following partial ANOVA results from Excel (some information is missing). Source of Variation Between groups Within groups Total

1483 2113.833

MS 210.2778 74.15

Assuming equal group sizes, the number of observations in each group is A) 2 B) 3 C) 4 D) 6 Answer: D Explanation: (630.833)/(210.2778) = 3 and (1483)/(74.15) = 20, so 3 + 20 = 23 = n − 1 and n/c = 24/4 = 6. Difficulty: 3 Hard Topic: 11.02 One-Factor ANOVA (Completely Randomized Model) Learning Objective: 11-04 Interpret sums of squares and calculations in an ANOVA table. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 56) Refer to the following partial ANOVA results from Excel (some information is missing). Source of Variation Between groups Within groups Total

SS 1483 2113.833

MS 210.2778 74.15

Degrees of freedom for the F-test are A) 5, 22 B) 4, 21 C) 3, 20 D) impossible to determine. Answer: C Explanation: (630.833)/(210.2778) = 3 and (1483)/(74.15) = 20. Difficulty: 3 Hard Topic: 11.02 One-Factor ANOVA (Completely Randomized Model) Learning Objective: 11-04 Interpret sums of squares and calculations in an ANOVA table. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

57) Refer to the following partial ANOVA results from Excel (some information is missing). Source of Variation Between groups Within groups Total

1483 2113.833

MS 210.2778 74.15

P-value 0.064139

F crit

The critical value of F at α = .05 is A) 1.645 B) 2.84 C) 3.10 D) 4.28 Answer: C Explanation: (630.833)/(210.2778) = 3 and (1483)/(74.15) = 20, so F.05 = 3.10 for df = (3, 20). Difficulty: 3 Hard Topic: 11.02 One-Factor ANOVA (Completely Randomized Model) Learning Objective: 11-06 Use a table or Excel to find critical values for the F distribution. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 58) Refer to the following partial ANOVA results from Excel (some information is missing). Source of Variation Between groups Within groups Total

1483 2113.833

MS 210.2778 74.15

P-value 0.064139

At α = .05, the difference between group means is A) highly significant. B) barely significant. C) not quite significant. D) clearly insignificant. Answer: C Explanation: The p-value is not less than .05, so we cannot reject the hypothesis of equal means. Difficulty: 2 Medium Topic: 11.02 One-Factor ANOVA (Completely Randomized Model) Learning Objective: 11-04 Interpret sums of squares and calculations in an ANOVA table. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

59) The Internal Revenue Service wishes to study the time required to process tax returns in three regional centers. A random sample of three tax returns is chosen from each of three centers. The time (in days) required to process each return is recorded as shown below. East 49 39 45

West 47 52 51

Midwest 54 49 56

The test to use to compare the means for all three groups would require: A) three-factor ANOVA. B) one-factor ANOVA. C) repeated two-sample test of means. D) two-factor ANOVA with replication. Answer: B Explanation: One factor ANOVA is required (three group means are to be compared). Difficulty: 1 Easy Topic: 11.02 One-Factor ANOVA (Completely Randomized Model) Learning Objective: 11-03 Recognize from data format when one-factor ANOVA is appropriate. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

60) The Internal Revenue Service wishes to study the time required to process tax returns in three regional centers. A random sample of three tax returns is chosen from each of three centers. The time (in days) required to process each return is recorded as shown below. Subsequently, an ANOVA test was performed. East 49 39 45

West 47 52 51

Midwest 54 49 56

Degrees of freedom for the error sum of squares in the ANOVA would be A) 11 B) 2 C) 4 D) 6 Answer: D Explanation: Error df = n − c = 9 − 3 = 6. Difficulty: 2 Medium Topic: 11.02 One-Factor ANOVA (Completely Randomized Model) Learning Objective: 11-04 Interpret sums of squares and calculations in an ANOVA table. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

61) The Internal Revenue Service wishes to study the time required to process tax returns in three regional centers. A random sample of three tax returns is chosen from each of three centers. The time (in days) required to process each return is recorded as shown below. East 49 39 45

West 47 52 51

Midwest 54 49 56

Degrees of freedom for the between-groups sum of squares in the ANOVA would be A) 11 B) 2 C) 4 D) 6 Answer: B Explanation: Between groups df = c − 1 = 3 − 1 = 2. Difficulty: 2 Medium Topic: 11.02 One-Factor ANOVA (Completely Randomized Model) Learning Objective: 11-04 Interpret sums of squares and calculations in an ANOVA table. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 62) Prof. Gristmill sampled exam scores for five randomly chosen students from each of his two sections of ACC 200. His sample results are shown. Day Class Night Class

93 91

58 81

74 85

85 60

82 73

He could test the population means for equality using A) a t-test for two means from independent samples. B) a t-test for two means from paired (related) samples. C) a one-factor ANOVA. D) either a one-factor ANOVA or a two-tailed t-test. Answer: D Explanation: As there are only two groups, either ANOVA or a two-tailed t-test will give the same p-value. Difficulty: 2 Medium Topic: 11.02 One-Factor ANOVA (Completely Randomized Model) Learning Objective: 11-03 Recognize from data format when one-factor ANOVA is appropriate. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 24 Copyright 2019 © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.

63) Systolic blood pressure of randomly selected HMO patients was recorded on a particular Wednesday, with the results shown here: Under 20 105 113 108 114 123

Patient Age Group 20 to 29 30 to 49 110 122 101 114 112 128 127 124 123 125

50 and Over 139 115 136 124 123

The appropriate hypothesis test is A) one-factor ANOVA. B) two-factor ANOVA. C) three-factor ANOVA. D) four-factor ANOVA. Answer: A Explanation: One factor ANOVA is required (four group means are to be compared). Difficulty: 1 Easy Topic: 11.02 One-Factor ANOVA (Completely Randomized Model) Learning Objective: 11-03 Recognize from data format when one-factor ANOVA is appropriate. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

64) Systolic blood pressure of randomly selected HMO patients was recorded on a particular Wednesday, with the results shown here. An ANOVA test was performed using these data. Under 20 105 113 108 114 123

Patient Age Group 20 to 29 30 to 49 110 122 101 114 112 128 127 124 123 125

50 and Over 139 115 136 124 123

Degrees of freedom for the between-treatments sum of squares would be A) 3 B) 19 C) 17 D) It depends on α. Answer: A Explanation: Between-treatments df = c − 1 = 4 − 1 = 3. Difficulty: 1 Easy Topic: 11.02 One-Factor ANOVA (Completely Randomized Model) Learning Objective: 11-04 Interpret sums of squares and calculations in an ANOVA table. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

65) Systolic blood pressure of randomly selected HMO patients was recorded on a particular Wednesday, with the results shown here. An ANOVA test was performed using these data. Under 20 105 113 108 114 123

Patient Age Group 20 to 29 30 to 49 110 122 101 114 112 128 127 124 123 125

50 and Over 139 115 136 124 123

What are the degrees of freedom for the error sum of squares? A) 3 B) 19 C) 16 D) It depends on α. Answer: C Explanation: Error df = n − c = 20 − 4 = 16. Difficulty: 2 Medium Topic: 11.02 One-Factor ANOVA (Completely Randomized Model) Learning Objective: 11-04 Interpret sums of squares and calculations in an ANOVA table. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

66) Sound levels are measured at random moments under typical driving conditions for various full-size truck models. The Excel ANOVA results are shown below. Big Bruin Gran Conto MaxRanger Oso Grande Overall Source Between Within Total

Mean 57.44 56.5 64.51 54.15 57.73 SS 462.4 934.70 1,397.10

Std Dev 8.944 6.361 3.983 4.166 6.824 d.f 3 27 30

n 6 8 7 10 31 MS 154.1 34.6

The test statistic to compare the five means simultaneously is A) 2.96 B) 15.8 C) 5.56 D) 4.45 Answer: D Explanation: Fcalc = (154.1)/(34.6) = 4.45. Difficulty: 2 Medium Topic: 11.02 One-Factor ANOVA (Completely Randomized Model) Learning Objective: 11-04 Interpret sums of squares and calculations in an ANOVA table. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

67) Sound levels are measured at random moments under typical driving conditions for various full-size truck models. The ANOVA results are shown below. Big Bruin Gran Conto MaxRanger Oso Grande Overall Source Between Within Total

Mean 57.44 56.5 64.51 54.15 57.73 SS 462.4 934.70 1,397.10

Std Dev 8.944 6.361 3.983 4.166 6.824 d.f 3 27 30

n 6 8 7 10 31 MS 154.1 34.6

The test statistic for Hartley's test for homogeneity of variance is A) 2.25 B) 5.04 C) 4.61 D) 4.45 Answer: B Explanation: Hartley's H = s2max/s2min = (8.944)2/(3.983)2 = 5.04. Difficulty: 2 Medium Topic: 11.04 Tests for Homogeneity of Variances Learning Objective: 11-08 Use Hartley's test for equal variances in c treatment groups. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

68) Refer to the following partial ANOVA results from Excel (some information is missing). ANOVA Table Source Treatment Error Total

SS 44,757 89,025 133,782

df 55 59

MS 11,189 1,619

The number of treatment groups is A) 5 B) 4 C) 3 D) impossible to ascertain from given information. Answer: A Explanation: 59 − 55 = 4 = c − 1, so c = 5 Difficulty: 2 Medium Topic: 11.02 One-Factor ANOVA (Completely Randomized Model) Learning Objective: 11-04 Interpret sums of squares and calculations in an ANOVA table. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 69) Refer to the following partial ANOVA results from Excel (some information is missing). ANOVA Table Source Treatment Error Total

SS 44,757 89,025 133,782

df 55 59

MS 11,189 1,619

The F statistic is A) 2.88 B) 4.87 C) 5.93 D) 6.91 Answer: D Explanation: Fcalc = 11,189/1619 = 6.91. Difficulty: 2 Medium Topic: 11.02 One-Factor ANOVA (Completely Randomized Model) Learning Objective: 11-04 Interpret sums of squares and calculations in an ANOVA table. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 30 Copyright 2019 © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.

70) Refer to the following partial ANOVA results from Excel (some information is missing). ANOVA Table Source Treatment Error Total

SS 44,757 89,025 133,782

df 55 59

MS 11,189 1,619

The number of observations in the original sample was A) 59 B) 60 C) 58 D) 54 Answer: B Explanation: n − 1 = 59, so n = 60. Difficulty: 1 Easy Topic: 11.02 One-Factor ANOVA (Completely Randomized Model) Learning Objective: 11-04 Interpret sums of squares and calculations in an ANOVA table. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 71) Refer to the following partial ANOVA results from Excel (some information is missing). ANOVA Table Source Treatment Error Total

SS 44,757 89,025 133,782

df 55 59

MS 11,189 1,619

Using Appendix F, the 5 percent critical value for the F-test is approximately A) 3.24 B) 6.91 C) 2.56 D) 2.06 Answer: C Explanation: Treatment df = 59 − 55 = 4, so F.05 = 2.56 using df = (4, 50) in Appendix F. Difficulty: 3 Hard Topic: 11.02 One-Factor ANOVA (Completely Randomized Model) Learning Objective: 11-06 Use a table or Excel to find critical values for the F distribution. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 31 Copyright 2019 © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.

72) Refer to the following partial ANOVA results from Excel (some information is missing). ANOVA Table Source Treatment Error Total

SS 44,757 89,025 133,782

MS 11,189 1,619

55 59

The p-value for the F-test would be A) much less than .05. B) slightly less than .05. C) slightly greater than .05. D) much greater than .05. Answer: A Explanation: Fcalc = 11,189/1619 = 6.91, while F.05 = 2.56 using df = (4, 50) in Appendix F. Difficulty: 3 Hard Topic: 11.02 One-Factor ANOVA (Completely Randomized Model) Learning Objective: 11-06 Use a table or Excel to find critical values for the F distribution. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 73) Refer to the following partial ANOVA results from Excel (some information is missing). ANOVA table Source Treatment Error Total

SS 717.4

1,848.2

P-value .0442

70.675

The MS (mean square) for the treatments is A) 239.13 B) 106.88 C) 1,130.8 D) impossible to ascertain from the information given. Answer: A Explanation: (717.4)/3 = 239.133. Difficulty: 1 Easy Topic: 11.02 One-Factor ANOVA (Completely Randomized Model) Learning Objective: 11-04 Interpret sums of squares and calculations in an ANOVA table. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 32 Copyright 2019 © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.

74) Refer to the following partial ANOVA results from Excel (some information is missing). ANOVA table Source Treatment Error Total

SS 717.4

1,848.2

P-value .0442

70.675

The F statistic is A) 4.87 B) 3.38 C) 5.93 D) 6.91 Answer: B Explanation: Between-groups MS = (717.4)/3 = 239.133, so Fcalc = (239.133)/(70.675) = 3.383. Difficulty: 3 Hard Topic: 11.02 One-Factor ANOVA (Completely Randomized Model) Learning Objective: 11-04 Interpret sums of squares and calculations in an ANOVA table. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 75) Refer to the following partial ANOVA results from Excel (some information is missing). ANOVA table Source Treatment Error Total

SS 717.4

1,848.2

P-value .0442

70.675

The number of observations in the entire sample is A) 20 B) 19 C) 22 Answer: A Explanation: n − 1 = 19, so n = 20. Difficulty: 1 Easy Topic: 11.02 One-Factor ANOVA (Completely Randomized Model) Learning Objective: 11-04 Interpret sums of squares and calculations in an ANOVA table. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

76) Refer to the following partial ANOVA results from Excel (some information is missing). ANOVA table Source Treatment Error Total

SS 717.4

1,848.2

P-value .0442

70.675

The 5 percent critical value for the F test is A) 2.46 B) 3.24 C) 3.38 D) impossible to ascertain from the given information. Answer: B Explanation: Error df = 19 − 3 = 16, so F.05 = 3.24 using df = (3, 16) in Appendix F. Difficulty: 2 Medium Topic: 11.02 One-Factor ANOVA (Completely Randomized Model) Learning Objective: 11-06 Use a table or Excel to find critical values for the F distribution. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 77) Refer to the following partial ANOVA results from Excel (some information is missing). ANOVA table Source Treatment Error Total

SS 717.40

df 3

1,848.20

P-value .0442

70.675

Our decision about the hypothesis of equal treatment means is that the null hypothesis A) cannot be rejected at α = .05. B) can be rejected at α = .05. C) can be rejected for any typical value of α. D) cannot be assessed from the given information. Answer: B Explanation: The p-value is less than .05, so we conclude unequal population means. Difficulty: 1 Easy Topic: 11.02 One-Factor ANOVA (Completely Randomized Model) Learning Objective: 11-04 Interpret sums of squares and calculations in an ANOVA table. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 34 Copyright 2019 © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.

78) To compare the cost of three shipping methods, a random sample of four shipments is taken for each of three firms. The cost per shipment is shown below. SpeedyShip GetItThere WeRTops

355 342 361

435 441 430

422 402 435

518 488 528

In a one-factor ANOVA, the degrees of freedom for the between-groups sum of squares will be A) 11 B) 3 C) 2 D) 9 Answer: C Explanation: Between-groups df = c − 1 = 3 − 1 = 2. Difficulty: 2 Medium Topic: 11.02 One-Factor ANOVA (Completely Randomized Model) Learning Objective: 11-04 Interpret sums of squares and calculations in an ANOVA table. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 79) To compare the cost of three shipping methods, a random sample of four shipments is taken for each of three firms. The cost per shipment is shown below. SpeedyShip GetItThere WeRTops

355 342 361

435 441 430

422 402 435

518 488 528

In a one-factor ANOVA, the degrees of freedom for the within-groups sum of squares will be A) 11 B) 3 C) 9 D) 2 Answer: C Explanation: Within-groups df = n − c = 12 − 3 = 9. Difficulty: 2 Medium Topic: 11.02 One-Factor ANOVA (Completely Randomized Model) Learning Objective: 11-04 Interpret sums of squares and calculations in an ANOVA table. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

80) To compare the cost of three shipping methods, a random sample of four shipments is taken for each of three firms. The cost per shipment is shown below. SpeedyShip GetItThere WeRTops

355 342 361

435 441 430

422 402 435

518 488 528

The degrees of freedom for the total sum of squares in a one-factor ANOVA would be A) 11 B) 8 C) 2 D) 9 Answer: A Explanation: Total df = n − 1 = 12 − 1 = 11. Difficulty: 1 Easy Topic: 11.02 One-Factor ANOVA (Completely Randomized Model) Learning Objective: 11-04 Interpret sums of squares and calculations in an ANOVA table. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

81) Refer to the following MegaStat output (some information is missing). The sample size was n = 65 in a one-factor ANOVA. Tukey simultaneous comparison t-values Mon Fri Tue Thu Mon Fri Tue Thu Wed

3.02 3.08 4.46 4.64

0.07 1.45 1.62

1.38 1.55

Wed

0.17

At α = .05, which is the critical value of the test statistic for a two-tailed test for a significant difference in means that are to be compared simultaneously? Note: This question requires a Tukey table. A) 2.81 B) 2.54 C) 2.33 D) 1.96 Answer: A Explanation: T.05 = 2.81 for df = (c, n − c) with c = 5 and n = 65. Difficulty: 3 Hard Topic: 11.03 Multiple Comparisons Learning Objective: 11-07 Understand and perform Tukey's test for paired means. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

82) Refer to the following MegaStat output (some information is missing). The sample size was n = 65 in a one-factor ANOVA. Tukey simultaneous comparison t-values Mon Fri Tue Thu Mon Fri Tue Thu Wed

3.02 3.08 4.46 4.64

0.07 1.45 1.62

1.38 1.55

Wed

0.17

Which pairs of days differ significantly? Note: This question requires access to a Tukey table. A) (Mon, Thu) and (Mon, Wed) only. B) (Mon, Wed) only. C) (Mon, Thu) only. D) (Mon, Thu) and (Mon, Wed) and (Mon, Fri) and (Mon, Tue). Answer: D Explanation: Use T.05 = 2.81 for df = (c, n − c) with c = 5 and n = 65. Difficulty: 3 Hard Topic: 11.03 Multiple Comparisons Learning Objective: 11-07 Understand and perform Tukey's test for paired means. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

83) Refer to the following MegaStat output (some information is missing). The sample size was n = 24 in a one-factor ANOVA. Tukey simultaneous comparison t-values Med 4 Med 1 Med 3 Med 4 Med 1 Med 3 Med 2

1.78 2.01 2.84

0.24 1.07

Med 2

0.83

At α = .05, what is the critical value of the Tukey test statistic for a two-tailed test for a significant difference in means that are to be compared simultaneously? Note: This question requires access to a Tukey table. A) 2.07 B) 2.80 C) 2.76 D) 1.96 Answer: B Explanation: T.05 = 2.80 for df = (c, n − c) with c = 4 and n = 24. Difficulty: 2 Medium Topic: 11.03 Multiple Comparisons Learning Objective: 11-07 Understand and perform Tukey's test for paired means. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

84) Refer to the following MegaStat output (some information is missing). The sample size was n = 24 in a one-factor ANOVA. Tukey simultaneous comparison t-values Med 4 Med 1 Med 3 Med 4 Med 1 Med 3 Med 2

1.78 2.01 2.84

0.24 1.07

Med 2

0.83

Which pairs of meds differ at α = .05? Note: This question requires access to a Tukey table. A) Med 1, Med 2 B) Med 2, Med 4 C) Med 3, Med 4 D) None of them. Answer: B Explanation: T.05 = 2.80 for df = (c, n − c) with c = 4 and n = 24. Difficulty: 2 Medium Topic: 11.03 Multiple Comparisons Learning Objective: 11-07 Understand and perform Tukey's test for paired means. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 85) What is the .05 critical value of Hartley's test statistic for a one-factor ANOVA with n1 = 5, n2 = 8, n3 = 7, n4 = 8, n5 = 6, n6 = 8? Note: This question requires access to a Hartley table. A) 10.8 B) 11.8 C) 13.7 D) 15.0 Answer: C Explanation: H.05 = 13.7 for df = (c, (n/c) − 1) where c = 6 and n = 42, so we use df = (6, 6). Difficulty: 2 Medium Topic: 11.04 Tests for Homogeneity of Variances Learning Objective: 11-08 Use Hartley's test for equal variances in c treatment groups. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

86) What is the .05 critical value of Tukey's test statistic for a one-factor ANOVA with n1 = 6, n2 = 6, n3 = 6? Note: This question requires access to a Tukey table. A) 3.67 B) 2.60 C) 3.58 D) 2.75 Answer: B Explanation: T.05 = 2.60 for df = (c, n − c) with c = 3 and n = 18. Difficulty: 2 Medium Topic: 11.03 Multiple Comparisons Learning Objective: 11-07 Understand and perform Tukey's test for paired means. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 87) What are the degrees of freedom for Hartley's test statistic for a one-factor ANOVA with n1 = 5, n2 = 8, n3 = 7, n4 = 8, n5 = 6, n6 = 8? A) 7, 6 B) 6, 6 C) 6, 41 Answer: B Explanation: Use df = (c, (n/c) − 1) where c = 6 and n = 42, or df = (6, 6). Difficulty: 2 Medium Topic: 11.04 Tests for Homogeneity of Variances Learning Objective: 11-08 Use Hartley's test for equal variances in c treatment groups. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

88) What are the degrees of freedom for Tukey's test statistic for a one-factor ANOVA with n1 = 6, n2 = 6, n3 = 6? A) 3, 6 B) 6, 3 C) 6, 15 D) 3, 15 Answer: D Explanation: Use df = (c, n − c) with c = 3 and n = 18, or df = (3, 15). Difficulty: 2 Medium Topic: 11.03 Multiple Comparisons Learning Objective: 11-07 Understand and perform Tukey's test for paired means. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 89) After performing a one-factor ANOVA test, John noticed that the sample standard deviations for his four groups were, respectively, 33, 24, 79, and 35. John should A) feel confident in his ANOVA test. B) use Hartley's test to check his assumptions. C) use an independent samples t-test instead of ANOVA. D) use a paired t-test instead of ANOVA. Answer: B Explanation: The unusually large standard deviation for group 3 suggests unequal variances. Difficulty: 2 Medium Topic: 11.04 Tests for Homogeneity of Variances Learning Objective: 11-08 Use Hartley's test for equal variances in c treatment groups. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 90) Which statement is incorrect? A) We need a Tukey test because ANOVA does not tell which pairs of means differ. B) Hartley's test is needed to determine whether the means of the groups differ. C) ANOVA assumes equal variances in the c groups being compared. Answer: B Explanation: Hartley's test compares variances (not means). Difficulty: 2 Medium Topic: 11.04 Tests for Homogeneity of Variances Learning Objective: 11-08 Use Hartley's test for equal variances in c treatment groups. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 42 Copyright 2019 © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.

91) Which is not an assumption of unreplicated two-factor ANOVA (randomized block)? A) Normality of the population B) Homogeneous variances C) Additive treatment effects D) There is factor interaction. Answer: D Explanation: The usual assumptions apply to a two-factor ANOVA (but no interaction estimate is possible without replication). Difficulty: 2 Medium Topic: 11.05 Two-Factor ANOVA without Replication (Randomized Block Model) Learning Objective: 11-02 Explain the assumptions of ANOVA and why they are important. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 92) Which is correct concerning a two-factor unreplicated (randomized block) ANOVA? A) No interaction effect is estimated. B) The interaction effect would have its own F statistic. C) The interaction would be insignificant unless the main effects were significant. Answer: A Explanation: We cannot estimate the interaction effect without replication in a two-factor ANOVA. Difficulty: 2 Medium Topic: 11.05 Two-Factor ANOVA without Replication (Randomized Block Model) Learning Objective: 11-10 Interpret results in a two-factor ANOVA without replication. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 93) In a two-factor unreplicated (randomized block) ANOVA, what is the F statistic for the treatment effect given that SSA (treatments) = 216, SSB (block) = 126, SSE (error) = 18? A) 12 B) 1.71 C) 7 D) We cannot tell without more information. Answer: D Explanation: We cannot calculate the mean squares without knowing r, c, and n, so no F statistics can be calculated. Difficulty: 2 Medium Topic: 11.05 Two-Factor ANOVA without Replication (Randomized Block Model) Learning Objective: 11-10 Interpret results in a two-factor ANOVA without replication. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 43 Copyright 2019 © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.

94) Three bottles of wine are tasted by three experts. Each rater assigns a rating (scale is from 1=terrible to 10=superb). Which test would you use for the most obvious hypothesis? Bob Fritz Sasha

Wine 1 9 9 8

Wine 2 8 7 8

Wine 3 7 8 6

A) t-test for independent means B) One-factor ANOVA C) Two-factor ANOVA without replication D) Two-factor ANOVA with replication Answer: C Explanation: There are two factors, but there is only one observation per row/column cell (i.e., no replication exists). Difficulty: 1 Easy Topic: 11.05 Two-Factor ANOVA without Replication (Randomized Block Model) Learning Objective: 11-09 Recognize from data format when two-factor ANOVA is needed. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

95) To compare the cost of three shipping methods, a firm ships material to each of four different destinations over a six-month period. The average cost per shipment is shown below. Shipper SpeedyShip GetItThere WeRTops

Toledo 355 342 361

Destination Oshawa 435 441 430

Janesville 422 402 435

Dallas 518 488 528

Which test would be appropriate? A) Independent samples t-test B) Two-factor ANOVA with replication C) Dependent (paired-samples) t-test D) Two-factor ANOVA without replication Answer: D Explanation: There are two factors, but there is only one observation per row/column cell (i.e., no replication exists). Difficulty: 1 Easy Topic: 11.05 Two-Factor ANOVA without Replication (Randomized Block Model) Learning Objective: 11-09 Recognize from data format when two-factor ANOVA is needed. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

96) To compare the cost of three shipping methods, a firm ships material to each of four different destinations over a six-month period. The average cost per shipment is shown below. Shipper SpeedyShip GetItThere WeRTops

Destination Oshawa 435 441 430

Toledo 355 342 361

Janesville 422 402 435

Dallas 518 488 528

For the appropriate type of ANOVA, total degrees of freedom would be A) 11 B) 3 C) 4 D) 12 Answer: A Explanation: df = n − 1 = 12 − 1 = 11. Difficulty: 1 Easy Topic: 11.05 Two-Factor ANOVA without Replication (Randomized Block Model) Learning Objective: 11-10 Interpret results in a two-factor ANOVA without replication. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 97) Here is an Excel ANOVA table that summarizes the results of an experiment to assess the effects of ambient noise level and plant location on worker productivity. The test used α = .05. Source of Variation Plant location Noise level Error Total

3.0075 8.4075 3.5225 14.9375

3 3 9

1.0025 2.8025 0.3914

2.561 7.160

P-value

F crit

0.1199 0.0093

3.862 3.863

Is the effect of plant location significant at α = .05? A) Yes B) No C) Need more information to say Answer: B Explanation: The p-value is not less than .05, so plant location has no significant effect. Difficulty: 1 Easy Topic: 11.05 Two-Factor ANOVA without Replication (Randomized Block Model) Learning Objective: 11-10 Interpret results in a two-factor ANOVA without replication. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 46 Copyright 2019 © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.

98) Here is an Excel ANOVA table that summarizes the results of an experiment to assess the effects of ambient noise level and plant location on worker productivity. The test used α = .05. Source of Variation Plant location Noise level Error Total

3.0075 8.4075 3.5225 14.9375

3 3 9

1.0025 2.8025 0.3914

2.561 7.160

P-value

F crit

0.1199 0.0093

3.862 3.863

Is the effect of noise level significant at α = .05? A) Yes B) No C) We need more information to say. Answer: A Explanation: The p-value is much less than .05, so noise level has a significant effect. Difficulty: 1 Easy Topic: 11.05 Two-Factor ANOVA without Replication (Randomized Block Model) Learning Objective: 11-10 Interpret results in a two-factor ANOVA without replication. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

99) Here is an Excel ANOVA table that summarizes the results of an experiment to assess the effects of ambient noise level and plant location on worker productivity. The test used α = .05. Source of Variation Plant location Noise level Error Total

3.0075 8.4075 3.5225 14.9375

3 3 9

1.0025 2.8025 0.3914

2.561 7.160

P-value

F crit

0.1199 0.0093

3.862 3.863

The experimental design and ANOVA appear to be A) replicated two factor. B) unreplicated two-factor. C) impossible to determine. Answer: B Explanation: The absence of an interaction suggests an unreplicated two-factor model. Difficulty: 1 Easy Topic: 11.05 Two-Factor ANOVA without Replication (Randomized Block Model) Learning Objective: 11-10 Interpret results in a two-factor ANOVA without replication. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

100) Here is an Excel ANOVA table that summarizes the results of an experiment to assess the effects of ambient noise level and plant location on worker productivity. The test used α = .05. Source of Variation Plant location Noise level Error Total

3.0075 8.4075 3.5225 14.9375

3 3 9

1.0025 2.8025 0.3914

2.561 7.160

P-value

F crit

0.1199 0.0093

3.862 3.863

The sample size is A) 15 B) 10 C) 16 D) impossible to determine. Answer: C Explanation: For unreplicated two-factor ANOVA, total df = 3 + 3 + 9 = 15 = n − 1, so n = 16. Difficulty: 1 Easy Topic: 11.05 Two-Factor ANOVA without Replication (Randomized Block Model) Learning Objective: 11-10 Interpret results in a two-factor ANOVA without replication. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

101) At the Seymour Clinic, the number of patients seen by three doctors over five days is as follows: Day Mon Tue Wed Thu Fri

Dr. Able 19 20 22 19 20

Physician Br. Baker 25 22 24 20 34

Dr. Chow 27 27 32 22 27

This data set would call for A) two-factor ANOVA without replication. B) two-factor ANOVA with replication. C) three-factor ANOVA. D) five-factor ANOVA. Answer: A Explanation: There are two factors, but there is only one observation per row/column cell (i.e., no replication exists). Difficulty: 1 Easy Topic: 11.05 Two-Factor ANOVA without Replication (Randomized Block Model) Learning Objective: 11-09 Recognize from data format when two-factor ANOVA is needed. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

102) At the Seymour Clinic, the number of patients seen by three doctors over five days is as follows: Day Mon Tue Wed Thu Fri

Dr. Able 19 20 22 19 20

Physician Br. Baker 25 22 24 20 34

Dr. Chow 27 27 32 22 27

Degrees of freedom for the error sum of squares would be A) 6 B) 14 C) 8 D) 15 Answer: C Explanation: For unreplicated two-factor ANOVA, the error df = (r − 1)(c − 1) = (5 − 1)(3 − 1) = 8. Difficulty: 3 Hard Topic: 11.05 Two-Factor ANOVA without Replication (Randomized Block Model) Learning Objective: 11-10 Interpret results in a two-factor ANOVA without replication. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

103) Here is an Excel ANOVA table for an experiment that analyzed factors that may affect patients' blood pressure (some information is missing). Source of Variation Medication type Patient age group Interaction Error Total

SS 16.5313 25.0938 1.8438 43.2504 86.7192

df 1 3 24

MS 16.5313 8.3646 0.6146 1.8021

F 9.173 4.642 0.341

P-value 0.006 0.011 0.796

The number of medication types is A) 1 B) 2 C) 3 D) 4 Answer: B Explanation: df = 1 = (number of medications − 1), so there were 2 medications. Difficulty: 2 Medium Topic: 11.06 Two-Factor ANOVA with Replication (Full Factorial Model) Learning Objective: 11-11 Interpret main effects and interaction effects in two-factor ANOVA. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

104) Here is an Excel ANOVA table for an experiment that analyzed two factors that may affect patients' blood pressure (some information is missing). Source of Variation Medication type Patient age group Interaction Error Total

SS 16.5313 25.0938 1.8438 43.2504 86.7192

df 1 3 24

MS 16.5313 8.3646 0.6146 1.8021

F 9.173 4.642 0.341

P-value 0.006 0.011 0.796

The number of patient age groups is A) 1 B) 2 C) 3 D) 4 Answer: D Explanation: For patient age group, df = (25.0938)/(8.3646) = 3 = c − 1 (there are 4 age groups). Difficulty: 3 Hard Topic: 11.06 Two-Factor ANOVA with Replication (Full Factorial Model) Learning Objective: 11-11 Interpret main effects and interaction effects in two-factor ANOVA. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

105) Here is an Excel ANOVA table for an experiment that analyzed two factors that may affect patients' blood pressure (some information is missing). Source of Variation Medication type Patient age group Interaction Error Total

SS 16.5313 25.0938 1.8438 43.2504 86.7192

df 1 3 24

MS 16.5313 8.3646 0.6146 1.8021

F 9.173 4.642 0.341

P-value 0.006 0.011 0.796

The number of patients per replication is A) 1 B) 2 C) 3 D) 4 Answer: D Explanation: c − 1 = (25.0938)/(8.3646) = 3 (so 4 age groups), r − 1 = 1 (so 2 meds), total df = 1 + 3 + 3 + 24 = 31 = n − 1 (so n = 32), 8 treatments (3 × 4) and thus 32/8 = 4 replications per treatment. Difficulty: 3 Hard Topic: 11.06 Two-Factor ANOVA with Replication (Full Factorial Model) Learning Objective: 11-11 Interpret main effects and interaction effects in two-factor ANOVA. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

106) Here is an Excel ANOVA table for an experiment that analyzed two factors that may affect patients' blood pressure (some information is missing). Source of Variation Medication type Patient age group Interaction Error Total

SS 16.5313 25.0938 1.8438 43.2504 86.7192

df 1 3 24

MS 16.5313 8.3646 0.6146 1.8021

F 9.173 4.642 0.341

P-value 0.006 0.011 0.796

The overall sample size is A) 7 B) 25 C) 32 D) impossible to determine as given. Answer: C Explanation: c − 1 = (25.0938)/(8.3646) = 3 (so 4 age groups), r − 1 = 1 (so 2 meds), total df = 1 + 3 + 3 + 24 = 31 = n − 1 (so n = 32). Difficulty: 2 Medium Topic: 11.06 Two-Factor ANOVA with Replication (Full Factorial Model) Learning Objective: 11-11 Interpret main effects and interaction effects in two-factor ANOVA. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

107) Here is an Excel ANOVA table for an experiment that analyzed two factors that may affect patients' blood pressure (some information is missing). Source of Variation Medication type Patient age group Interaction Error Total

SS 16.5313 25.0938 1.8438 43.2504 86.7192

df 1 3 24

MS 16.5313 8.3646 0.6146 1.8021

F 9.173 4.642 0.341

P-value 0.006 0.011 0.796

At α = .05 the effect of medication type is A) significant. B) insignificant. C) borderline. Answer: A Explanation: The p-value is much less than .05, so medication type has a highly significant effect. Difficulty: 1 Easy Topic: 11.06 Two-Factor ANOVA with Replication (Full Factorial Model) Learning Objective: 11-11 Interpret main effects and interaction effects in two-factor ANOVA. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

108) Here is an Excel ANOVA table for an experiment that analyzed two factors that may affect patients' blood pressure (some information is missing). Source of Variation Medication type Patient age group Interaction Error Total

SS 16.5313 25.0938 1.8438 43.2504 86.7192

df 1 3 24

MS 16.5313 8.3646 0.6146 1.8021

F 9.173 4.642 0.341

P-value 0.006 0.011 0.796

At α = .01 the effect of patient age is A) very clearly significant. B) just barely significant. C) not quite significant. Answer: C Explanation: The p-value of .011 is greater than .01, so age group does not have a significant effect at α = .01. However, it is a very close decision. Difficulty: 1 Easy Topic: 11.06 Two-Factor ANOVA with Replication (Full Factorial Model) Learning Objective: 11-11 Interpret main effects and interaction effects in two-factor ANOVA. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

109) Here is an Excel ANOVA table for an experiment that analyzed two factors that may affect patients' blood pressure (some information is missing). Source of Variation Medication type Patient age group Interaction Error Total

SS 16.5313 25.0938 1.8438 43.2504 86.7192

df 1 3 24

MS 16.5313 8.3646 0.6146 1.8021

F 9.173 4.642 0.341

P-value 0.006 0.011 0.796

At α = .05 the interaction is A) significant. B) insignificant. C) borderline. Answer: B Explanation: The p-value is much greater than .05 so there is no significant interaction. Difficulty: 1 Easy Topic: 11.06 Two-Factor ANOVA with Replication (Full Factorial Model) Learning Objective: 11-11 Interpret main effects and interaction effects in two-factor ANOVA. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

110) Three randomly chosen pieces of four types of polyvinyl chloride (PVC) pipe of equal wall thickness are tested to determine the burst strength (in pounds per square inch) under three temperature conditions, yielding the results shown below. Temperature Hot (70° C) Warm (40° C) Cold (10° C)

PVC1 250 273 281 321 322 299 358 363 341

PVC2 301 285 275 342 322 339 375 355 354

PVC3 235 260 279 302 315 301 328 336 342

PVC4 217 255 241 240 260 278 301 333 302

Which test would be appropriate? A) One-factor ANOVA B) Two-factor ANOVA with replication C) Dependent (paired-samples) t-test D) Two-factor ANOVA with no replication Answer: B Explanation: Within each treatment combination we have three replications. Difficulty: 2 Medium Topic: 11.06 Two-Factor ANOVA with Replication (Full Factorial Model) Learning Objective: 11-09 Recognize from data format when two-factor ANOVA is needed. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

111) Three randomly chosen pieces of four types of PVC pipe of equal wall thickness are tested to determine the burst strength (in pounds per square inch) under three temperature conditions, yielding the results shown below. Temperature Hot (70° C) Warm (40° C) Cold (10° C)

PVC1 250 273 281 321 322 299 358 363 341

PVC2 301 285 275 342 322 339 375 355 354

PVC3 235 260 279 302 315 301 328 336 342

PVC4 217 255 241 240 260 278 301 333 302

Total degrees of freedom for the ANOVA would be" A) 19 B) 12 C) 35 D) 59 Answer: C Explanation: Total df = n − 1 = 36 − 1 = 35. Difficulty: 2 Medium Topic: 11.06 Two-Factor ANOVA with Replication (Full Factorial Model) Learning Objective: 11-11 Interpret main effects and interaction effects in two-factor ANOVA. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

112) A firm is studying the effect of work shift and parts supplier on its defect rate (dependent variable is defects per 1000). The resulting ANOVA results are shown below (some information is missing). Source Shift Supplier Shift*Supplier Error Total

Sum of Squares 704.07 19.6 430.75 1062.05 2216.47

df 2 4 36 44

Mean Square 352.04 9.8 107.69 29.5

F 11.93 0.33 3.65

P-Value 0.001 0.72 0.014

How many suppliers were there? A) 1 B) 2 C) 3 D) 4 Answer: C Explanation: 44 − 36 − 4 − 2 = 2 = c − 1, so there were 3 suppliers. Difficulty: 2 Medium Topic: 11.06 Two-Factor ANOVA with Replication (Full Factorial Model) Learning Objective: 11-11 Interpret main effects and interaction effects in two-factor ANOVA. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

113) A firm is studying the effect of work shift and parts supplier on its defect rate (dependent variable is defects per 1000). The resulting ANOVA results are shown below (some information is missing). Source Shift Supplier Shift*Supplier Error Total

Sum of Squares 704.07 19.6 430.75 1062.05 2216.47

df 2 4 36 44

Mean Square 352.04 9.8 107.69 29.5

F 11.93 0.33 3.65

P-Value 0.001 0.72 0.014

How many replications per cell were there? A) 2 B) 3 C) 4 D) 5 Answer: D Explanation: n − 1 = 44 (n = 45), 44 − 36 − 4 − 2 = 2 = c − 1 (3 suppliers), r − 1 = 2 (3 shifts), so 3 × 3 = 9 row/column cells. Hence, there are 45/9 = 5 replications per treatment combination. Difficulty: 2 Medium Topic: 11.06 Two-Factor ANOVA with Replication (Full Factorial Model) Learning Objective: 11-11 Interpret main effects and interaction effects in two-factor ANOVA. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

114) A firm is studying the effect of work shift and parts supplier on its defect rate (dependent variable is defects per 1000). The resulting ANOVA results are shown below (some information is missing). Source Shift Supplier Shift*Supplier Error Total

Sum of Squares 704.07 19.6 430.75 1062.05 2216.47

df 2 4 36 44

Mean Square 352.04 9.8 107.69 29.5

F 11.93 0.33 3.65

P-Value 0.001 0.72 0.014

At α = .05, the effect of supplier is A) clearly significant. B) just barely significant. C) almost but not quite significant. D) clearly insignificant. Answer: D Explanation: The p-value is much greater than .05, so supplier has no significant effect. Difficulty: 1 Easy Topic: 11.06 Two-Factor ANOVA with Replication (Full Factorial Model) Learning Objective: 11-11 Interpret main effects and interaction effects in two-factor ANOVA. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

115) A firm is studying the effect of work shift and parts supplier on its defect rate (dependent variable is defects per 1000). The resulting ANOVA results are shown below (some information is missing). Source Shift Supplier Shift*Supplier Error Total

Sum of Squares 704.07 19.6 430.75 1062.05 2216.47

df 2 4 36 44

Mean Square 352.04 9.8 107.69 29.5

F 11.93 0.33 3.65

P-Value 0.001 0.72 0.014

The number of observations was A) 37 B) 45 C) 44 D) 40 Answer: B Explanation: n − 1 = 44 (n = 45). Difficulty: 1 Easy Topic: 11.06 Two-Factor ANOVA with Replication (Full Factorial Model) Learning Objective: 11-11 Interpret main effects and interaction effects in two-factor ANOVA. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

116) A firm is studying the effect of work shift and parts supplier on its defect rate (dependent variable is defects per 1000). The resulting ANOVA results are shown below (some information is missing). Source Shift Supplier Shift*Supplier Error Total

Sum of Squares 704.07 19.6 430.75 1062.05 2216.47

df 2 4 36 44

Mean Square 352.04 9.8 107.69 29.5

F 11.93 0.33 3.65

P-Value 0.001 0.72 0.014

At α = .01, the interaction effect is A) strongly significant. B) just barely significant. C) not quite significant. Answer: C Explanation: The p-value is slightly greater than .01, so the interaction effect is not quite significant at α = .01. Difficulty: 2 Medium Topic: 11.06 Two-Factor ANOVA with Replication (Full Factorial Model) Learning Objective: 11-11 Interpret main effects and interaction effects in two-factor ANOVA. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

117) A firm is concerned with variability in hourly output at several factories and shifts. Here are the results of an ANOVA using output per hour as the dependent variable (some information is missing). Source Factory Supplier Factory*Shift Error Total

Sum of Squares 19012.5 258.333 80908.333 8633.333 108812.5

df 1 2 2 12 17

Mean Square 19012.5 129.167 40454.167 719.444 6400.735

F Ratio 26.427 0.180 56.230

The original data matrix has how many treatments (rows × columns)? A) 4 B) 6 C) 3 D) 8 Answer: B Explanation: r − 1 = 1 (2 factories), c − 1 = 2 (3 shifts), so 2 × 3 = 6 row/column cells. Difficulty: 3 Hard Topic: 11.06 Two-Factor ANOVA with Replication (Full Factorial Model) Learning Objective: 11-11 Interpret main effects and interaction effects in two-factor ANOVA. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

118) A firm is concerned with variability in hourly output at several factories and shifts. Here are the results of an ANOVA using output per hour as the dependent variable (some information is missing). Source Factory Supplier Factory*Shift Error Total

Sum of Squares 19012.5 258.333 80908.333 8633.333 108812.5

df 1 2 2 12 17

Mean Square 19012.5 129.167 40454.167 719.444 6400.735

F Ratio 26.427 0.180 56.230

The number of observations in each treatment cell (row-column intersection) is A) 1 B) 2 C) 3 D) impossible to determine. Answer: C Explanation: n − 1 = 17 (n = 18), r − 1 = 1 (2 factories), c − 1 = 2 (3 shifts), so 2 × 3 = 6 row/column cells. Hence, 18/6 = 3 replications per treatment combination. Difficulty: 3 Hard Topic: 11.06 Two-Factor ANOVA with Replication (Full Factorial Model) Learning Objective: 11-11 Interpret main effects and interaction effects in two-factor ANOVA. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

119) A firm is concerned with variability in hourly output at several factories and shifts. Here are the results of an ANOVA using output per hour as the dependent variable (some information is missing). Source Factory Supplier Factory*Shift Error Total

Sum of Squares 19012.5 258.333 80908.333 8633.333 108812.5

df 1 2 2 12 17

Mean Square 19012.5 129.167 40454.167 719.444 6400.735

F Ratio 26.427 0.180 56.230

P-value 0.000 0.838

At α = .01 the effect of factory is A) clearly significant. B) clearly insignificant. C) of borderline significance. Answer: A Explanation: The p-value is much less than .01, so factory has a significant effect. Difficulty: 1 Easy Topic: 11.06 Two-Factor ANOVA with Replication (Full Factorial Model) Learning Objective: 11-11 Interpret main effects and interaction effects in two-factor ANOVA. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

120) A firm is concerned with variability in hourly output at several factories and shifts. Here are the results of an ANOVA using output per hour as the dependent variable (some information is missing). Source Factory Supplier Factory*Shift Error Total

Sum of Squares 19012.5 258.333 80908.333 8633.333 108812.5

df 1 2 2 12 17

Mean Square 19012.5 129.167 40454.167 719.444 6400.735

F Ratio 26.427 0.180 56.230

P-value 0.000 0.838

The p-value for the interaction effect is going to be A) very small (near 0). B) very large (near 1). C) impossible to know—could be either large or small. Answer: A Explanation: For interaction, Fcalc = (40454.167)/(719.444) = 56.23, indicating a very small pvalue. Difficulty: 2 Medium Topic: 11.06 Two-Factor ANOVA with Replication (Full Factorial Model) Learning Objective: 11-11 Interpret main effects and interaction effects in two-factor ANOVA. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

121) Sound engineers studied factors that might affect the output (in decibels) of a rock concert speaker system. The results of their ANOVA tests are shown (some information is missing). Source of Variation Amplifier Position Interaction Error Total

SS 99.02344 93.98698 10.15365 155.875 359.0391

3 16 23

P-value

99.02344 31.32899 3.384549 9.742188

3.215807 0.347412

0.005718 0.051003 0.791505

Which is the number of amplifiers and positions tested? A) 1, 3 B) 2, 4 C) 3, 5 D) 4, 1 Answer: B Explanation: r − 1 = 1 (2 amplifiers), c − 1 = 3 (4 positions). Difficulty: 3 Hard Topic: 11.06 Two-Factor ANOVA with Replication (Full Factorial Model) Learning Objective: 11-11 Interpret main effects and interaction effects in two-factor ANOVA. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

122) Sound engineers studied factors that might affect the output (in decibels) of a rock concert speaker system. The results of their ANOVA tests are shown (some information is missing). Source of Variation Amplifier Position Interaction Error Total

SS 99.02344 93.98698 10.15365 155.875 359.0391

df 3 16 23

MS 99.02344 31.32899 3.384549 9.742188

F 3.215807 0.347412

The number of observations per cell was A) 1 B) 2 C) 3 D) 4 Answer: C Explanation: n − 1 = 23 (n = 24), r − 1 = 1 (2 amplifiers), c − 1 = 3 (4 positions), so 2 × 4 = 8 row/column cells and hence 24/8 = 3 replications per treatment combination. Difficulty: 3 Hard Topic: 11.06 Two-Factor ANOVA with Replication (Full Factorial Model) Learning Objective: 11-11 Interpret main effects and interaction effects in two-factor ANOVA. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

123) Sound engineers studied factors that might affect the output (in decibels) of a rock concert speaker system. The desired level of significance was α = .05. The results of their ANOVA tests are shown (some information is missing). Source of Variation Amplifier Position Interaction Error Total

SS 99.02344 93.98698 10.15365 155.875 359.0391

3 16 23

P-value

99.02344 31.32899 3.384549 9.742188

3.215807 0.347412

0.005718 0.051003 0.791505

The most reasonable conclusion at α = .05 about the three sources of variation (amplifier, position, and interaction) would be that their effects are A) significant, significant, insignificant. B) insignificant, significant, significant. C) very significant, almost significant, insignificant. Answer: C Explanation: The p-value is smaller than .05 for amplifier, but not quite for position and definitely not for the interaction term. Difficulty: 1 Easy Topic: 11.06 Two-Factor ANOVA with Replication (Full Factorial Model) Learning Objective: 11-11 Interpret main effects and interaction effects in two-factor ANOVA. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

124) Sound engineers studied factors that might affect the output, in decibels, of a rock concert speaker system. The results of their ANOVA tests are shown (some information is missing). Source of Variation Amplifier Position Interaction Error Total

SS 99.02344 93.98698 10.15365 155.875 359.0391

df 3 16 23

MS 99.02344 31.32899 3.384549 9.742188

F 3.215807 0.347412

The F statistic for amplifier was A) 9.90 B) 10.16 C) 5.72 D) 4.27 Answer: B Explanation: Fcalc = (99.02344)/(9.742188) = 10.16. Difficulty: 2 Medium Topic: 11.06 Two-Factor ANOVA with Replication (Full Factorial Model) Learning Objective: 11-11 Interpret main effects and interaction effects in two-factor ANOVA. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

125) A multinational firm manufactures several types of 1280 × 1024 LCD displays in several locations. They designed a sampling experiment to analyze the number of pixels per screen that have significant color degradation after 52,560 hours (six years of continuous use) using accelerated life testing. The Excel ANOVA table for their experiment is shown below. Some table entries have been obscured. The response variable (Y) is the number of degraded pixels in a given display. Source of Variation Country of origin Display type Interaction Error Total

SS 202.9 233.2333 147.7667 1096.5 1680.4

45 59

MS 101.45 58.30833 18.47084 24.36667

F 4.163475

Degrees of freedom for display type will be A) 1 B) 4 C) 3 D) 5 Answer: B Explanation: For display type, df = (233.2333)/(58.30833) = 4. Difficulty: 2 Medium Topic: 11.06 Two-Factor ANOVA with Replication (Full Factorial Model) Learning Objective: 11-11 Interpret main effects and interaction effects in two-factor ANOVA. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

126) A multinational firm manufactures several types of 1280 × 1024 LCD displays in several locations. They designed a sampling experiment to analyze the number of pixels per screen that have significant color degradation after 52,560 hours (six years of continuous use) using accelerated life testing. The Excel ANOVA table for their experiment is shown below. Some table entries have been obscured. The response variable (Y) is the number of degraded pixels in a given display. Source of Variation Country of origin Display type Interaction Error Total

SS 202.9 233.2333 147.7667 1096.5 1680.4

45 59

MS 101.45 58.30833 18.47084 24.36667

F 4.163475

How many display types were there? A) 1 B) 2 C) 3 D) 5 Answer: D Explanation: For display type, df = (233.2333)/(58.30833) = 4 = c − 1 (so 5 display types). Difficulty: 2 Medium Topic: 11.06 Two-Factor ANOVA with Replication (Full Factorial Model) Learning Objective: 11-11 Interpret main effects and interaction effects in two-factor ANOVA. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

127) A multinational firm manufactures several types of 1280 × 1024 LCD displays in several locations. They designed a sampling experiment to analyze the number of pixels per screen that have significant color degradation after 52,560 hours (six years of continuous use) using accelerated life testing. The Excel ANOVA table for their experiment is shown below. Some table entries have been obscured. The response variable (Y) is the number of degraded pixels in a given display. Source of Variation Country of origin Display type Interaction Error Total

SS 202.9 233.2333 147.7667 1096.5 1680.4

45 59

MS 101.45 58.30833 18.47084 24.36667

F 4.163475

How many countries were studied? A) 1 B) 2 C) 3 D) 4 Answer: C Explanation: For country, df = (202.9)/(101.45) = 2 = r − 1 (so 3 countries). Difficulty: 2 Medium Topic: 11.06 Two-Factor ANOVA with Replication (Full Factorial Model) Learning Objective: 11-11 Interpret main effects and interaction effects in two-factor ANOVA. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

128) A multinational firm manufactures several types of 1280 × 1024 LCD displays in several locations. They designed a sampling experiment to analyze the number of pixels per screen that have significant color degradation after 52,560 hours (six years of continuous use) using accelerated life testing. The Excel ANOVA table for their experiment is shown below. Some table entries have been obscured. The response variable (Y) is the number of degraded pixels in a given display. Source of Variation Country of origin Display type Interaction Error Total

202.9

101.45

233.2333

58.30833

147.7667 1096.5 1680.4

18.47084 24.36667

45 59

P-value

4.163475

0.021927

F crit

The F statistic for display effect is A) 1.78 B) 3.16 C) 2.39 D) 2.94 Answer: C Explanation: Fcalc = (58.30833)/(24.36667) = 2.393. Difficulty: 2 Medium Topic: 11.06 Two-Factor ANOVA with Replication (Full Factorial Model) Learning Objective: 11-11 Interpret main effects and interaction effects in two-factor ANOVA. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

129) A multinational firm manufactures several types of 1280 × 1024 LCD displays in several locations. They designed a sampling experiment to analyze the number of pixels per screen that have significant color degradation after 52,560 hours (six years of continuous use) using accelerated life testing. The Excel ANOVA table for their experiment is shown below. Some table entries have been obscured. The response variable (Y) is the number of degraded pixels in a given display. Source of Variation Country of origin Display type Interaction Error Total

202.9

101.45

233.2333

58.30833

147.7667 1096.5 1680.4

18.47084 24.36667

45 59

P-value

4.163475

0.021927

F crit

At α = .05, the interaction effect is A) clearly significant. B) just barely significant. C) not quite significant. D) clearly insignificant. Answer: D Explanation: Fcalc = (18.47084)/(24.36667) = 0.76, which is far less than F.05 for df = (8, 45). Difficulty: 3 Hard Topic: 11.06 Two-Factor ANOVA with Replication (Full Factorial Model) Learning Objective: 11-11 Interpret main effects and interaction effects in two-factor ANOVA. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

130) A multinational firm manufactures several types of 1280 × 1024 LCD displays in several locations. They designed a sampling experiment to analyze the number of pixels per screen that have significant color degradation after 52,560 hours (six years of continuous use) using accelerated life testing. The Excel ANOVA table for their experiment is shown below. Some table entries have been obscured. The response variable (Y) is the number of degraded pixels in a given display. Source of Variation Country of origin Display type Interaction Error Total

SS 202.9 233.2333 147.7667 1096.5 1680.4

45 59

MS 101.45 58.30833 18.47084 24.36667

F 4.163475

The numerator degrees of freedom for the interaction test would be A) 2 B) 4 C) 8 D) 16 Answer: C Explanation: For interaction, df = (147.7667)/(18.47084) = 8. Difficulty: 3 Hard Topic: 11.06 Two-Factor ANOVA with Replication (Full Factorial Model) Learning Objective: 11-11 Interpret main effects and interaction effects in two-factor ANOVA. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

131) A veterinarian notes the age (months) at which dogs are brought to the clinic to be neutered. Male Female

Collies 10 9 12 15 7 8

Terries 14 10 8 17 11 6

Chows 8 18 11 9 15 8

What kind of test would be used? A) One-factor ANOVA B) Two-factor ANOVA with replication C) Two-factor ANOVA without replication D) Three-factor ANOVA with replication. Answer: B Explanation: There are three replications and two factors. Difficulty: 1 Easy Topic: 11.06 Two-Factor ANOVA with Replication (Full Factorial Model) Learning Objective: 11-09 Recognize from data format when two-factor ANOVA is needed. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

132) A veterinarian notes the age (months) at which dogs are brought in to the clinic to be neutered. Male Female

Collies 10 9 12 15 7 8

Terries 14 10 8 17 11 6

Chows 8 18 11 9 15 8

Numerator degrees of freedom for the ANOVA interaction test would be A) 2 B) 3 C) 6 D) impossible to determine. Answer: A Explanation: Two factor ANOVA with replication, interaction df = (r − 1)(c − 1) = (2 − 1)(3 − 1) = 2. Difficulty: 2 Medium Topic: 11.06 Two-Factor ANOVA with Replication (Full Factorial Model) Learning Objective: 11-11 Interpret main effects and interaction effects in two-factor ANOVA. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

133) A veterinarian notes the age (months) at which dogs are brought in to the clinic to be neutered. Male Female

Collies 10 9 12 15 7 8

Terries 14 10 8 17 11 6

Chows 8 18 11 9 15 8

Total degrees of freedom for a two-factor replicated ANOVA would be A) 6 B) 14 C) 17 D) 11 Answer: C Explanation: n − 1 = 18 − 1 = 17. Difficulty: 2 Medium Topic: 11.06 Two-Factor ANOVA with Replication (Full Factorial Model) Learning Objective: 11-11 Interpret main effects and interaction effects in two-factor ANOVA. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

134) Refer to the following partial ANOVA results from Excel (some information is missing). Source of Variation Nozzle setting Pressure level Interaction Error Total

SS 3.46722 8.07444 2.80779 8.54000 22.8894

df 1 2

MS 3.46722 4.03722 1.40389 0.711667

F 4.87198 5.67291 1.97268

How many nozzle settings were observed? A) 3 B) 2 C) 1 D) Cannot tell Answer: B Explanation: For nozzle, df = 1 = r − 1 (so 2 nozzles). Difficulty: 2 Medium Topic: 11.06 Two-Factor ANOVA with Replication (Full Factorial Model) Learning Objective: 11-11 Interpret main effects and interaction effects in two-factor ANOVA. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 135) Refer to the following partial ANOVA results from Excel (some information is missing). Source of Variation Nozzle setting Pressure level Interaction Error Total

SS 3.46722 8.07444 2.80779 8.54000 22.8894

df 1 2

MS 3.46722 4.03722 1.40389 0.711667

F 4.87198 5.67291 1.97268

Degrees of freedom for pressure level would be A) 2 B) 3 C) 4 D) 6 Answer: A Explanation: For pressure, df = (8.07444)/(4.03722) = 2 = c − 1 (so 3 pressures). Difficulty: 2 Medium Topic: 11.06 Two-Factor ANOVA with Replication (Full Factorial Model) Learning Objective: 11-11 Interpret main effects and interaction effects in two-factor ANOVA. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 83 Copyright 2019 © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.

136) Refer to the following partial ANOVA results from Excel (some information is missing). Source of Variation Nozzle setting Pressure level Interaction Error Total

SS 3.46722 8.07444 2.80779 8.54000 22.8894

df 1 2

MS 3.46722 4.03722 1.40389 0.711667

F 4.87198 5.67291 1.97268

Error degrees of freedom would be A) 24 B) 15 C) 12 D) 13 Answer: C Explanation: For error, df = (8.5400)/(0.711667) =12. Difficulty: 2 Medium Topic: 11.06 Two-Factor ANOVA with Replication (Full Factorial Model) Learning Objective: 11-11 Interpret main effects and interaction effects in two-factor ANOVA. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

137) Refer to the following partial ANOVA results from Excel (some information is missing). Source of Variation Nozzle setting Pressure level Interaction Error Total

SS 3.46722 8.07444 2.80779 8.54000 22.8894

df 1 2

MS 3.46722 4.03722 1.40389 0.711667

F 4.87198 5.67291 1.97268

The overall sample size was A) 24 B) 23 C) 22 D) 18 Answer: D Explanation: Divide each SS by its MS to get 1 + 2 + 2 + 12 = 17 = n − 1 (so n = 18). Difficulty: 3 Hard Topic: 11.06 Two-Factor ANOVA with Replication (Full Factorial Model) Learning Objective: 11-11 Interpret main effects and interaction effects in two-factor ANOVA. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 138) Refer to the following partial ANOVA results from Excel (some information is missing). Source of Variation Nozzle setting Pressure level Interaction Error Total

SS 3.46722 8.07444 2.80779 8.54000 22.8894

df 1 2

MS 3.46722 4.03722 1.40389 0.711667

F 4.87198 5.67291 1.97268

How many pressure levels were observed? A) 4 B) 3 C) 2 D) 1 Answer: B Explanation: For pressure, df = (8.07444)/(4.03722) = 2 = c − 1 (so 3 pressures). Difficulty: 2 Medium Topic: 11.06 Two-Factor ANOVA with Replication (Full Factorial Model) Learning Objective: 11-11 Interpret main effects and interaction effects in two-factor ANOVA. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 85 Copyright 2019 © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.

139) Refer to the following partial ANOVA results from Excel (some information is missing). Source of Variation Nozzle setting Pressure level Interaction Error Total

P-value

3.46722 8.07444 2.80779 8.54000 22.8894

3.46722 4.03722 1.40389 0.711667

4.87198 5.67291 1.97268

0.04751 0.01844 0.18177

F crit

At α = .05, the critical F value for nozzle setting is A) 4.71 B) 4.75 C) 3.68 D) 3.02 Answer: B Explanation: Using Appendix F with df = (1, 12), we get F.05 = 4.75. Difficulty: 3 Hard Topic: 11.06 Two-Factor ANOVA with Replication (Full Factorial Model) Learning Objective: 11-11 Interpret main effects and interaction effects in two-factor ANOVA. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

140) Refer to the following partial ANOVA results from Excel (some information is missing). Source of Variation Nozzle setting Pressure level Interaction Error Total

SS 3.46722 8.07444 2.80779 8.54000 22.8894

df 1 2

MS 3.46722 4.03722 1.40389 0.711667

F 4.87198 5.67291 1.97268

The form of the original data matrix is A) 3 × 1 table. B) 1 × 2 table. C) 4 × 3 table. D) 2 × 3 table. Answer: D Explanation: Divide each SS by its MS to get (r − 1) = 1, (c − 1) = 2, so r × c = 2 × 3 = 6 treatments. Difficulty: 2 Medium Topic: 11.06 Two-Factor ANOVA with Replication (Full Factorial Model) Learning Objective: 11-11 Interpret main effects and interaction effects in two-factor ANOVA. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

141) Refer to the following partial ANOVA results from Excel (some information is missing). Source of Variation Nozzle setting Pressure level Interaction Error Total

SS 3.46722 8.07444 2.80779 8.54000 22.8894

df 1 2

MS 3.46722 4.03722 1.40389 0.711667

F 4.87198 5.67291 1.97268

The number of replications per treatment was A) 4 B) 3 C) 2 D) 1 Answer: B Explanation: Divide each SS by its MS to get total df = 1 + 2 + 2 + 12 = 17 = n − 1, so n = 18 and r × c = 2 × 3 = 6 treatments, giving three replications per treatment. Difficulty: 3 Hard Topic: 11.06 Two-Factor ANOVA with Replication (Full Factorial Model) Learning Objective: 11-11 Interpret main effects and interaction effects in two-factor ANOVA. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

142) Refer to the following partial ANOVA results from Excel (some information is missing). Source of Variation Nozzle setting Pressure level Interaction Error Total

SS 3.46722 8.07444 2.80779 8.54000 22.8894

df 1 2

MS 3.46722 4.03722 1.40389 0.711667

F 4.87198 5.67291 1.97268

P-value 0.04751 0.01844 0.18177

At α = .05, the effect of nozzle setting is A) highly significant. B) just barely significant. C) not quite significant. D) clearly insignificant. Answer: B Explanation: Its p-value is slightly less than .05, so the nozzle effect is barely significant. Difficulty: 2 Medium Topic: 11.06 Two-Factor ANOVA with Replication (Full Factorial Model) Learning Objective: 11-11 Interpret main effects and interaction effects in two-factor ANOVA. Bloom's: Evaluate AACSB: Analytical Thinking Accessibility: Keyboard Navigation

143) As shown below, a hospital recorded the number of minutes spent in post-op recovery by three randomly chosen knee-surgery patients in each category, based on age and type of anesthetic. Which is the most appropriate test? Anesthetic Type Local Spinal

Patient Age Group Under 30 30 to 49 45 60 60 45 30 75 75 75 60 120 120 90

50 and Over 45 75 90 105 120 90

A) One-factor ANOVA B) Two-factor ANOVA without replication C) Two-factor ANOVA with replication D) Rimsky-Korsakov test Answer: C Explanation: There are three replications per cell with two factors. Difficulty: 2 Medium Topic: 11.06 Two-Factor ANOVA with Replication (Full Factorial Model) Learning Objective: 11-09 Recognize from data format when two-factor ANOVA is needed. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

144) Refer to the following partial ANOVA results from Excel (some information is missing). The response variable was Y = maximum amount of water pumped from wells (gallons per minute). Source of Variation Depth of well Age of well Interaction Error Total

SS 2450 364.667 32.667 855.333 3702.667

df 2 4 18 26

MS 1225

F 3.8371 0.1719

47.519

The degrees of freedom for age of well is A) 2 B) 3 C) 4 D) 5 Answer: A Explanation: For age of well, df = 26 − 18 − 4 − 2 = 2 (so 3 ages). Difficulty: 3 Hard Topic: 11.06 Two-Factor ANOVA with Replication (Full Factorial Model) Learning Objective: 11-11 Interpret main effects and interaction effects in two-factor ANOVA. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

145) Refer to the following partial ANOVA results from Excel (some information is missing). The response variable was Y = maximum amount of water pumped from wells (gallons per minute). Source of Variation Depth of well Age of well Interaction Error Total

SS 2450 364.667 32.667 855.333 3702.667

df 2 4 18 26

MS 1225

F 3.8371 0.1719

47.519

The F statistic for depth of well is A) 25.23 B) 25.78 C) 25.31 D) 25.06 Answer: B Explanation: Fcalc = (1225)/(47.519) = 25.779. Difficulty: 3 Hard Topic: 11.06 Two-Factor ANOVA with Replication (Full Factorial Model) Learning Objective: 11-11 Interpret main effects and interaction effects in two-factor ANOVA. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

146) Refer to the following partial ANOVA results from Excel (some information is missing). The response variable was Y = maximum amount of water pumped from wells (gallons per minute). Source of Variation Depth of well Age of well Interaction Error Total

SS 2450 364.667 32.667 855.333 3702.667

df 2 4 18 26

MS 1225

F 3.8371 0.1719

47.519

The MS for interaction is A) 7.25 B) 8.17 C) 8.37 D) 9.28 Answer: B Explanation: For interaction, we have MS = (32.667)/4 = 8.167. Difficulty: 2 Medium Topic: 11.06 Two-Factor ANOVA with Replication (Full Factorial Model) Learning Objective: 11-11 Interpret main effects and interaction effects in two-factor ANOVA. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

147) Refer to the following partial ANOVA results from Excel (some information is missing). The response variable was Y = maximum amount of water pumped from wells (gallons per minute). Source of Variation Depth of well Age of well Interaction Error Total

SS 2450 364.667 32.667 855.333 3702.667

df 2 4 18 26

MS 1225

F 3.8371 0.1719

47.519

The MS for age of well is A) 185.23 B) 179.26 C) 180.25 D) 182.33 Answer: D Explanation: By subtraction, for age of well df = 26 − 18 − 4 − 2 = 2, so MS = (364.667) / (2) = 182.334. Difficulty: 3 Hard Topic: 11.06 Two-Factor ANOVA with Replication (Full Factorial Model) Learning Objective: 11-11 Interpret main effects and interaction effects in two-factor ANOVA. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

Applied Statistics in Business and Economics, 6e (Doane) Chapter 12 Simple Regression 1) A scatter plot is used to visualize the association (or lack of association) between two quantitative variables. Answer: TRUE Explanation: The scatter plot shows association between two quantitative variables. Difficulty: 1 Easy Topic: 12.01 Visual Displays and Correlation Analysis Learning Objective: 12-01 Calculate and test a correlation coefficient for significance. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 2) The correlation coefficient r measures the strength of the linear relationship between two variables. Answer: TRUE Explanation: A correlation coefficient measures linearity between two variables. Difficulty: 1 Easy Topic: 12.01 Visual Displays and Correlation Analysis Learning Objective: 12-01 Calculate and test a correlation coefficient for significance. Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation 3) Pearson's correlation coefficient (r) requires that both variables be interval or ratio data. Answer: TRUE Explanation: Correlation assumes quantitative data with at least interval measurements. Difficulty: 1 Easy Topic: 12.01 Visual Displays and Correlation Analysis Learning Objective: 12-01 Calculate and test a correlation coefficient for significance. Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation

4) If r = .55 and n = 16, then the correlation is significant at α = .05 in a two-tailed test. Answer: TRUE Explanation: tcalc = r[(n − 2)/(1 − r2)]1/2 = (.55)[(16 − 2)/(1 − .552)]1/2 = 2.464 > t.025 = 2.145 for d.f. = 16 − 2 = 14. Difficulty: 2 Medium Topic: 12.01 Visual Displays and Correlation Analysis Learning Objective: 12-01 Calculate and test a correlation coefficient for significance. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 5) A sample correlation r = .40 indicates a stronger linear relationship than r = −.60. Answer: FALSE Explanation: The sign only indicates the direction, not the strength, of the linear relationship. Difficulty: 2 Medium Topic: 12.01 Visual Displays and Correlation Analysis Learning Objective: 12-01 Calculate and test a correlation coefficient for significance. Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation 6) A common source of spurious correlation between X and Y is when a third unspecified variable Z affects both X and Y. Answer: TRUE Explanation: Both X and Y could be influenced by a "lurking" variable Z. Difficulty: 2 Medium Topic: 12.01 Visual Displays and Correlation Analysis Learning Objective: 12-01 Calculate and test a correlation coefficient for significance. Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation 7) The correlation coefficient r always has the same sign as b1 in Y = b0 + b1X. Answer: TRUE Explanation: The t-test for the slope in simple regression gives the same result as the t-test for r. Difficulty: 1 Easy Topic: 12.03 Regression Models Learning Objective: 12-02 Interpret a regression equation and use it to make predictions. Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation 2 Copyright 2019 © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.

8) The fitted intercept in a regression has little meaning if no data values near X = 0 have been observed. Answer: TRUE Explanation: Predicting Y for X = 0 makes little sense if the observed data have no values near X = 0. Difficulty: 1 Easy Topic: 12.02 Simple Regression Learning Objective: 12-02 Interpret a regression equation and use it to make predictions. Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation 9) The least squares regression line is obtained when the sum of the squared residuals is minimized. Answer: TRUE Explanation: The Ordinary Least Squares (OLS) method minimizes the sum of squared residuals. Difficulty: 1 Easy Topic: 12.04 Ordinary Least Squares Formulas Learning Objective: 12-04 Explain the least squares method, apply formulas for coefficients, and interpret R2. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 10) In a simple regression, if the coefficient for X is positive and significantly different from zero, then an increase in X is associated with an increase in the mean (i.e., the expected value) of Y. Answer: TRUE Explanation: The conditional mean of Y depends on X (unless the slope is effectively zero). Difficulty: 2 Medium Topic: 12.02 Simple Regression Learning Objective: 12-02 Interpret a regression equation and use it to make predictions. Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation

11) In least squares regression, the residuals e1, e2, . . . , en will always have a zero mean. Answer: TRUE Explanation: The residuals must sum to zero if the OLS method is used, so their mean is zero. Difficulty: 2 Medium Topic: 12.04 Ordinary Least Squares Formulas Learning Objective: 12-02 Interpret a regression equation and use it to make predictions. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 12) When using the least squares method, the column of residuals always sums to zero. Answer: TRUE Explanation: The residuals must sum to zero if the OLS method is used. Difficulty: 2 Medium Topic: 12.04 Ordinary Least Squares Formulas Learning Objective: 12-02 Interpret a regression equation and use it to make predictions. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 13) In the model Sales = 268 + 7.37 Ads (both variables in dollars) an additional $1 spent on ads will increase sales by 7.37 percent. Answer: FALSE Explanation: The slope coefficient is in the same units as Y (dollars, not percent, in this case). Difficulty: 2 Medium Topic: 12.02 Simple Regression Learning Objective: 12-02 Interpret a regression equation and use it to make predictions. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 14) If R2 = .36 in the model Sales = 268 + 7.37 Ads with n = 50, the two-tailed test for correlation at α = .05 would say that there is a significant correlation between Sales and Ads. Answer: TRUE Explanation: tcalc = r[(n − 2)/(1 − r2)]1/2 = (.60)[(50 − 2)/(1 − .36)]1/2 = 5.196 > t.025 = 2.011 for d.f. = 50 − 2 = 48. Difficulty: 3 Hard Topic: 12.01 Visual Displays and Correlation Analysis Learning Objective: 12-01 Calculate and test a correlation coefficient for significance. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 4 Copyright 2019 © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.

15) If R2 = .36 in the model Sales = 268 + 7.37 Ads, then Ads explains 36 percent of the variation in Sales. Answer: TRUE Explanation: We interpret R2 as the fraction of variation in Y explained by X (expressed as a percentage). Difficulty: 2 Medium Topic: 12.04 Ordinary Least Squares Formulas Learning Objective: 12-04 Explain the least squares method, apply formulas for coefficients, and interpret R2. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 16) The ordinary least squares regression line always passes through the point ( , ). Answer: TRUE Explanation: The OLS formulas require the fitted line to pass through this point. Difficulty: 1 Easy Topic: 12.03 Regression Models Learning Objective: 12-02 Interpret a regression equation and use it to make predictions. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 17) The least squares regression line gives unbiased estimates of β0 and β1. Answer: TRUE Explanation: The expected values of the OLS estimators b0 and b1 are the true parameters β0 and β1. Difficulty: 2 Medium Topic: 12.04 Ordinary Least Squares Formulas Learning Objective: 12-04 Explain the least squares method, apply formulas for coefficients, and interpret R2. Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation

18) In a simple regression, the correlation coefficient r is the square root of R2. Answer: TRUE Explanation: In fact, we could use the notation r2 instead of R2 when talking about simple regression. Difficulty: 2 Medium Topic: 12.04 Ordinary Least Squares Formulas Learning Objective: 12-04 Explain the least squares method, apply formulas for coefficients, and interpret R2. Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation 19) If SSR is 1800 and SSE is 200, then R2 is .90. Answer: TRUE Explanation: R2 = SSR/SST = SSR/(SSR + SSE) = 1800/(1800 + 200) = .90. Difficulty: 2 Medium Topic: 12.04 Ordinary Least Squares Formulas Learning Objective: 12-04 Explain the least squares method, apply formulas for coefficients, and interpret R2. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 20) The width of a prediction interval for an individual value of Y is less than the standard error se. Answer: FALSE Explanation: The formula for the interval width multiplies the standard error by an expression that is greater than one. Difficulty: 2 Medium Topic: 12.07 Confidence and Prediction Intervals for Y Learning Objective: 12-07 Distinguish between confidence and prediction intervals for Y. Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation

21) If SSE is near zero in a regression, the statistician will conclude that the proposed model probably has too poor a fit to be useful. Answer: FALSE Explanation: SSE is the sum of the square residuals, which would be smaller if the fit is good. Difficulty: 2 Medium Topic: 12.04 Ordinary Least Squares Formulas Learning Objective: 12-04 Explain the least squares method, apply formulas for coefficients, and interpret R2. Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation 22) For a regression with 200 observations, we expect that about 10 residuals will exceed two standard errors. Answer: TRUE Explanation: If the residuals are normal, 95.44 percent (190 of 200) will lie within ±2se so we would expect about ten residuals to exceed two standard errors. Difficulty: 2 Medium Topic: 12.09 Unusual Observations Learning Objective: 12-09 Identify unusual residuals and tell when they are outliers. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 23) Confidence intervals for predicted Y are less precise when the residuals are very small. Answer: FALSE Explanation: Small residuals imply a small standard error and thus a narrower prediction interval. Difficulty: 2 Medium Topic: 12.07 Confidence and Prediction Intervals for Y Learning Objective: 12-07 Distinguish between confidence and prediction intervals for Y. Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation

24) Cause-and-effect direction between X and Y may be determined by running the regression twice and seeing whether Y = β0 + β1X or X = β1 + β0Y has the larger R2. Answer: FALSE Explanation: Cause and effect cannot be determined in the context of simple regression models. Difficulty: 3 Hard Topic: 12.02 Simple Regression Learning Objective: 12-02 Interpret a regression equation and use it to make predictions. Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation 25) The ordinary least squares method of estimation minimizes the estimated slope and intercept. Answer: FALSE Explanation: OLS minimizes the sum of squared residuals. Difficulty: 2 Medium Topic: 12.04 Ordinary Least Squares Formulas Learning Objective: 12-04 Explain the least squares method, apply formulas for coefficients, and interpret R2. Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation 26) The ordinary least squares method ensures that the residuals will be normally distributed. Answer: FALSE Explanation: OLS produces unbiased estimates, but cannot ensure normality of the residuals. Difficulty: 2 Medium Topic: 12.08 Residual Tests Learning Objective: 12-08 Calculate residuals and perform tests of regression assumptions. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 27) If you have a strong outlier in the residuals, it may represent a different causal system. Answer: TRUE Explanation: Outliers might come from a different population or causal system. Difficulty: 2 Medium Topic: 12.10 Other Regression Problems (Optional) Learning Objective: 12-11 Improve data conditioning and use transformations if needed (optional). Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation 8 Copyright 2019 © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.

28) A negative correlation between two variables X and Y usually yields a negative p-value for r. Answer: FALSE Explanation: The p-value cannot be negative. Difficulty: 2 Medium Topic: 12.01 Visual Displays and Correlation Analysis Learning Objective: 12-01 Calculate and test a correlation coefficient for significance. Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation 29) In linear regression between two variables, a significant relationship exists when the p-value of the t test statistic for the slope is greater than α. Answer: FALSE Explanation: Reject β1 = 0 if the p-value is less than α. Difficulty: 2 Medium Topic: 12.05 Tests for Significance Learning Objective: 12-05 Construct confidence intervals and test hypotheses for the slope and intercept. Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation 30) The larger the absolute value of the t statistic of the slope in a simple linear regression, the stronger the linear relationship that exists between X and Y. Answer: TRUE Explanation: The correlation coefficient measures linearity regardless of its sign ( + or −). Difficulty: 2 Medium Topic: 12.05 Tests for Significance Learning Objective: 12-05 Construct confidence intervals and test hypotheses for the slope and intercept. Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation

31) In simple linear regression, the coefficient of determination (R2) is estimated from sums of squares in the ANOVA table. Answer: TRUE Explanation: R2 = SSR/SST or R2 = 1 − SSE/SST. Difficulty: 1 Easy Topic: 12.04 Ordinary Least Squares Formulas Learning Objective: 12-04 Explain the least squares method, apply formulas for coefficients, and interpret R2. Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation 32) In simple linear regression, the p-value of the slope will always equal the p-value of the F statistic. Answer: TRUE Explanation: This is true only if there is one predictor (but is no longer true in multiple regression). Difficulty: 2 Medium Topic: 12.06 Analysis of Variance: Overall Fit Learning Objective: 12-06 Interpret the ANOVA table and use it to calculate F, R2, and standard error. Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation 33) An observation with high leverage will have a large residual (usually an outlier). Answer: FALSE Explanation: The concepts are distinct (a high-leverage point could have a good fit). Difficulty: 2 Medium Topic: 12.09 Unusual Observations Learning Objective: 12-10 Define leverage and identify high-leverage observations. Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation

34) A prediction interval for Y is narrower than the corresponding confidence interval for the mean of Y. Answer: FALSE Explanation: Predicting an individual case requires a wider confidence interval than predicting the mean, Difficulty: 2 Medium Topic: 12.07 Confidence and Prediction Intervals for Y Learning Objective: 12-07 Distinguish between confidence and prediction intervals for Y. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 35) When X is farther from its mean, the prediction interval and confidence interval for Y become wider. Answer: TRUE Explanation: The width increases when X differs from its mean. (Review the formula.) Difficulty: 2 Medium Topic: 12.07 Confidence and Prediction Intervals for Y Learning Objective: 12-07 Distinguish between confidence and prediction intervals for Y. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 36) The total sum of squares (SST) will never exceed the regression sum of squares (SSR). Answer: FALSE Explanation: The identity is SSR + SSE = SST. Difficulty: 1 Easy Topic: 12.06 Analysis of Variance: Overall Fit Learning Objective: 12-06 Interpret the ANOVA table and use it to calculate F, R2, and standard error. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation

37) "High leverage" would refer to a data point that is poorly predicted by the model (large residual). Answer: FALSE Explanation: A high-leverage observation may have a good fit. (Only its X value determines its leverage.) Difficulty: 2 Medium Topic: 12.09 Unusual Observations Learning Objective: 12-10 Define leverage and identify high-leverage observations. Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation 38) Studentized (or standardized) residuals permit us to detect cases where the regression predicts poorly. Answer: TRUE Explanation: Studentized residuals resemble a t-distribution. A large studentized t-value (e.g., t < −2.00 or t > +2.00) implies a poor fit. Difficulty: 2 Medium Topic: 12.09 Unusual Observations Learning Objective: 12-09 Identify unusual residuals and tell when they are outliers. Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation 39) A poor prediction (large residual) indicates an observation with high leverage. Answer: FALSE Explanation: High leverage indicates an unusually large or small X value (not a poor prediction). A high-leverage observation may have a good fit or a poor fit. Only its X value determines its leverage. Difficulty: 2 Medium Topic: 12.09 Unusual Observations Learning Objective: 12-10 Define leverage and identify high-leverage observations. Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation

40) Ill-conditioned refers to a variable whose units are too large or too small (e.g., $2,434,567). Answer: TRUE Explanation: In Excel, a symptom of poor data conditioning is exponential notation (e.g., 4.3E+06). Difficulty: 2 Medium Topic: 12.10 Other Regression Problems (Optional) Learning Objective: 12-11 Improve data conditioning and use transformations if needed (optional). Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 41) A simple decimal transformation (e.g., from 18,291 to 18.291) often improves data conditioning. Answer: TRUE Explanation: Keeping data magnitudes similar helps avoid exponential notation (e.g., 4.3E+06). Difficulty: 2 Medium Topic: 12.10 Other Regression Problems (Optional) Learning Objective: 12-11 Improve data conditioning and use transformations if needed (optional). Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation 42) Two-tailed t-tests are often used because any predictor that differs significantly from zero in a two-tailed test will also be significantly greater than zero or less than zero in a one-tailed test at the same α. Answer: TRUE Explanation: This is true because the critical t is larger in the two-tailed test (the default in most software). Difficulty: 2 Medium Topic: 12.05 Tests for Significance Learning Objective: 12-05 Construct confidence intervals and test hypotheses for the slope and intercept. Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation

43) A predictor that is significant in a one-tailed t-test will also be significant in a two-tailed test at the same level of significance α. Answer: FALSE Explanation: This is false because the critical t would be larger in a two-tailed test. Difficulty: 2 Medium Topic: 12.05 Tests for Significance Learning Objective: 12-05 Construct confidence intervals and test hypotheses for the slope and intercept. Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation 44) Omission of a relevant predictor is a common source of model misspecification. Answer: TRUE Explanation: In a multivariate world, simple regression may be inadequate. Difficulty: 2 Medium Topic: 12.10 Other Regression Problems (Optional) Learning Objective: 12-11 Improve data conditioning and use transformations if needed (optional). Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 45) Using the least squares formulas, the regression line must pass through the origin. Answer: FALSE Explanation: The OLS intercept estimate does not, in general, equal zero. We might be unable to reject a zero intercept with a t-test, but the fitted intercept is rarely exactly zero. Difficulty: 2 Medium Topic: 12.04 Ordinary Least Squares Formulas Learning Objective: 12-04 Explain the least squares method, apply formulas for coefficients, and interpret R2. Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation

46) Outliers can be detected by examining the standardized residuals. Answer: TRUE Explanation: A poor fit implies a large t-value (e.g., larger than ±3 would be an outlier). Difficulty: 1 Easy Topic: 12.09 Unusual Observations Learning Objective: 12-09 Identify unusual residuals and tell when they are outliers. Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation 47) In a simple regression, there are n − 2 degrees of freedom associated with the error sum of squares (SSE). Answer: TRUE Explanation: This is true in simple regression because we estimate two parameters (β0 and β1). Difficulty: 1 Easy Topic: 12.06 Analysis of Variance: Overall Fit Learning Objective: 12-06 Interpret the ANOVA table and use it to calculate F, R2, and standard error. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 48) In a simple regression, the F statistic is calculated by taking the ratio of MSR to the MSE. Answer: TRUE Explanation: By definition, Fcalc = MSR/MSE (obtained from the ANOVA table). Difficulty: 2 Medium Topic: 12.06 Analysis of Variance: Overall Fit Learning Objective: 12-06 Interpret the ANOVA table and use it to calculate F, R2, and standard error. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation

49) The coefficient of determination is the percentage of the total variation in the response variable Y that is explained by the predictor X. Answer: TRUE Explanation: R2 = SSR/SST or R2 = 1 − SSE/SST lies between 0 and 1 and often is expressed as a percentage. Difficulty: 1 Easy Topic: 12.04 Ordinary Least Squares Formulas Learning Objective: 12-04 Explain the least squares method, apply formulas for coefficients, and interpret R2. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 50) A different confidence interval exists for the mean value of Y for each different value of X. Answer: TRUE Explanation: Both the interval width and also E(Y|X) = β0 + β1X depend on the value of X. Difficulty: 2 Medium Topic: 12.07 Confidence and Prediction Intervals for Y Learning Objective: 12-07 Distinguish between confidence and prediction intervals for Y. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 51) A prediction interval for Y is widest when X is near its mean. Answer: FALSE Explanation: The prediction interval is narrowest when X is near its mean. Review the formula, which has a term (xi - )2 in the numerator. The minimum would be when xi = . Difficulty: 1 Easy Topic: 12.07 Confidence and Prediction Intervals for Y Learning Objective: 12-07 Distinguish between confidence and prediction intervals for Y. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation

52) In a two-tailed test for correlation at α = .05, a sample correlation coefficient r = .42 with n = 25 is significantly different than zero. Answer: TRUE Explanation: tcalc = r[(n − 2)/(1 − r2)]1/2 = (.42)[(25 − 2)/(1 − .422)]1/2 = 2.219 > t.025 = 2.069 for d.f. = 25 − 2 = 23. Difficulty: 2 Medium Topic: 12.01 Visual Displays and Correlation Analysis Learning Objective: 12-01 Calculate and test a correlation coefficient for significance. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 53) In correlation analysis, neither X nor Y is designated as the independent variable. Answer: TRUE Explanation: In correlation analysis, X and Y covary without designating either as "independent." Difficulty: 1 Easy Topic: 12.01 Visual Displays and Correlation Analysis Learning Objective: 12-01 Calculate and test a correlation coefficient for significance. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 54) A negative value for the correlation coefficient (r) implies a negative value for the slope (b1). Answer: TRUE Explanation: The sign of r must be the same as the sign of the slope estimate b1. Difficulty: 1 Easy Topic: 12.04 Ordinary Least Squares Formulas Learning Objective: 12-04 Explain the least squares method, apply formulas for coefficients, and interpret R2. Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation 55) High leverage for an observation indicates that X is far from its mean. Answer: TRUE Explanation: By definition, observations have higher leverage when X is far from its mean. Difficulty: 2 Medium Topic: 12.09 Unusual Observations Learning Objective: 12-10 Define leverage and identify high-leverage observations. Bloom's: Remember AACSB: Analytical Thinking 17 Copyright 2019 © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.

Accessibility: Keyboard Navigation 56) Autocorrelated errors are not usually a concern for regression models using cross-sectional data. Answer: TRUE Explanation: We more often expect autocorrelated residuals in time-series data. Difficulty: 1 Easy Topic: 12.08 Residual Tests Learning Objective: 12-08 Calculate residuals and perform tests of regression assumptions. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 57) There are usually several possible regression lines that will minimize the sum of squared errors. Answer: FALSE Explanation: The OLS solution for the estimators b0 and b1 is unique. Difficulty: 2 Medium Topic: 12.04 Ordinary Least Squares Formulas Learning Objective: 12-04 Explain the least squares method, apply formulas for coefficients, and interpret R2. Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation 58) When the errors in a regression model are not independent, the regression model is said to have autocorrelation. Answer: TRUE Explanation: For example, in first-order autocorrelation each εt depends on εt−1. Difficulty: 1 Easy Topic: 12.08 Residual Tests Learning Objective: 12-08 Calculate residuals and perform tests of regression assumptions. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation

59) In a simple bivariate regression, Fcalc = tcalc2. Answer: TRUE Explanation: This statement is true only in a simple regression (one predictor). Difficulty: 2 Medium Topic: 12.06 Analysis of Variance: Overall Fit Learning Objective: 12-06 Interpret the ANOVA table and use it to calculate F, R2, and standard error. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 60) Correlation analysis primarily measures the degree of the linear relationship between X and Y. Answer: TRUE Explanation: The sign of r indicates the direction, and its magnitude indicates the degree of linearity. Difficulty: 2 Medium Topic: 12.01 Visual Displays and Correlation Analysis Learning Objective: 12-01 Calculate and test a correlation coefficient for significance. Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation 61) The variable used to predict another variable is called the A) response variable. B) regression variable. C) independent variable. D) dependent variable. Answer: C Explanation: We might also call the independent variable a predictor of Y. Difficulty: 1 Easy Topic: 12.02 Simple Regression Learning Objective: 12-02 Interpret a regression equation and use it to make predictions. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation

62) The standard error of the regression A) is based on squared deviations from the regression line. B) may assume negative values if b1 < 0. C) is in squared units of the dependent variable. D) may be cut in half to get an approximate 95 percent prediction interval. Answer: A Explanation: In a simple regression, the standard error is the square root of the sum of the squared residuals divided by (n − 2). Difficulty: 2 Medium Topic: 12.05 Tests for Significance Learning Objective: 12-05 Construct confidence intervals and test hypotheses for the slope and intercept. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 63) A local trucking company fitted a regression to relate the travel time (days) of its shipments as a function of the distance traveled (miles). The fitted regression is Time = −7.126 + 0.0214 Distance, based on a sample of 20 shipments. The estimated standard error of the slope is 0.0053. Find the value of tcalc to test for zero slope. A) 2.46 B) 5.02 C) 4.04 D) 3.15 Answer: C Explanation: tcalc = b1/sb1 = (0.0214)/(0.0053) = 4.038. Difficulty: 2 Medium Topic: 12.05 Tests for Significance Learning Objective: 12-05 Construct confidence intervals and test hypotheses for the slope and intercept. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

64) A local trucking company fitted a regression to relate the travel time (days) of its shipments as a function of the distance traveled (miles). The fitted regression is Time = −7.126 + .0214 Distance, based on a sample of 20 shipments. The estimated standard error of the slope is 0.0053. Find the critical value for a right-tailed test to see if the slope is positive, using α = .05. A) 2.101 B) 2.552 C) 1.960 D) 1.734 Answer: D Explanation: For d.f. = n − 2 = 20 − 2 = 18, Appendix D gives t.05 = 1.734. Difficulty: 2 Medium Topic: 12.05 Tests for Significance Learning Objective: 12-05 Construct confidence intervals and test hypotheses for the slope and intercept. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 65) If the attendance at a baseball game is to be predicted by the equation Attendance = 16,500 − 75 Temperature, what would be the predicted attendance if Temperature is 90 degrees? A) 6,750 B) 9,750 C) 12,250 D) 10,020 Answer: B Explanation: The predicted Attendance is 16,500 − 75(90) = 9,750. Difficulty: 2 Medium Topic: 12.02 Simple Regression Learning Objective: 12-02 Interpret a regression equation and use it to make predictions. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

66) A hypothesis test is conducted at the 5 percent level of significance to test whether the population correlation is zero. If the sample consists of 25 observations and the correlation coefficient is .60, what is the computed test statistic? A) 2.071 B) 1.960 C) 3.597 D) 1.645 Answer: C Explanation: tcalc = r[(n − 2)/(1 − r2)]1/2 = (.60)[(25 − 2)/(1 − .602)]1/2 = 3.597. Difficulty: 2 Medium Topic: 12.01 Visual Displays and Correlation Analysis Learning Objective: 12-01 Calculate and test a correlation coefficient for significance. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 67) Which of the following is not a characteristic of the F test in a simple regression? A) It is a test for overall fit of the model. B) The test statistic can never be negative. C) It requires a table with numerator and denominator degrees of freedom. D) The F test gives a different p-value than the t-test. Answer: D Explanation: Fcalc is the ratio of two variances (mean squares) that measures overall fit. The test statistic cannot be negative because the variances are nonnegative. In a simple regression, the F test always agrees with the t-test. Difficulty: 2 Medium Topic: 12.06 Analysis of Variance: Overall Fit Learning Objective: 12-06 Interpret the ANOVA table and use it to calculate F, R2, and standard error. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation

68) A researcher's Excel results are shown below using Femlab (labor force participation rate among females) to try to predict Cancer (death rate per 100,000 population due to cancer) in the 50 U.S. states. Regression Statistics Multiple R R Square Adjusted R Square Standard Error Observations Variable Intercept Femlab

0.313422848 0.098233882 0.079447088 32.07003698 50 Coefficients 343.619889 –2.2833659

Standard Error 61.0823514 0.99855319

t Stat 5.62552 –2.28667

Which of the following statements is not true? A) The standard error is too high for this model to be of any predictive use. B) The 95 percent confidence interval for the coefficient of Femlab is −4.29 to −0.28. C) Significant correlation exists between Femlab and Cancer at α = .05. D) The two-tailed p-value for Femlab will be less than .05. Answer: A Explanation: The magnitude of se depends on Y (and, in this case, the tcalc indicates significance). Difficulty: 2 Medium Topic: 12.05 Tests for Significance Learning Objective: 12-05 Construct confidence intervals and test hypotheses for the slope and intercept. Bloom's: Analyze AACSB: Analytical Thinking Accessibility: Keyboard Navigation

69) A researcher's results are shown below using Femlab (labor force participation rate among females) to try to predict Cancer (death rate per 100,000 population due to cancer) in the 50 U.S. states. Regression Statistics Multiple R R Square Adjusted R Square Standard Error Observations Variable Intercept Femlab

0.313422848 0.098233882 0.079447088 32.07003698 50 Coefficients 343.619889 –2.2833659

Standard Error 61.0823514 0.99855319

t Stat 5.62552 –2.28667

Which statement is valid regarding the relationship between Femlab and Cancer? A) A rise in female labor participation rate will cause the cancer rate to decrease within a state. B) This model explains about 10 percent of the variation in state cancer rates. C) At the .05 level of significance, there isn't enough evidence to say the two variables are related. D) If your sister starts working, the cancer rate in your state will decline. Answer: B Explanation: It is customary to express the R2 as a percentage. (Here, the tcalc indicates significance.) Difficulty: 2 Medium Topic: 12.04 Ordinary Least Squares Formulas Learning Objective: 12-04 Explain the least squares method, apply formulas for coefficients, and interpret R2. Bloom's: Analyze AACSB: Analytical Thinking Accessibility: Keyboard Navigation

70) A researcher's results are shown below using Femlab (labor force participation rate among females) to try to predict Cancer (death rate per 100,000 population due to cancer) in the 50 U.S. states. Source of variation Regression Residual Total

df 1 48 49

SS 5377.836 49367.389 54745.225

MS 5377.836 1028.487

F 5.228879

What is the R2 for this regression? A) .9018 B) .0982 C) .8395 D) .1605 Answer: B Explanation: R2 = SSR/SST = (5,377.836)/(54,745.225) = .0982. Difficulty: 2 Medium Topic: 12.06 Analysis of Variance: Overall Fit Learning Objective: 12-06 Interpret the ANOVA table and use it to calculate F, R2, and standard error. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 71) A news network stated that a study had found a positive correlation between the number of children a worker has and his or her earnings last year. You may conclude that A) people should have more children so they can get better jobs. B) the data are erroneous because the correlation should be negative. C) causation is in serious doubt. D) statisticians have small families. Answer: C Explanation: There is no a priori basis for expecting causation. Difficulty: 2 Medium Topic: 12.01 Visual Displays and Correlation Analysis Learning Objective: 12-01 Calculate and test a correlation coefficient for significance. Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation

72) William used a sample of 68 large U.S. cities to estimate the relationship between Crime (annual property crimes per 100,000 persons) and Income (median annual income per capita, in dollars). His estimated regression equation was Crime = 428 + 0.050 Income. We can conclude that A) the slope is small so Income has no effect on Crime. B) crime seems to create additional income in a city. C) wealthy individuals tend to commit more crimes, on average. D) the intercept is irrelevant since zero median income is impossible in a large city. Answer: D Explanation: Zero median income makes no sense. (Significance cannot be assessed from the given facts.) Difficulty: 3 Hard Topic: 12.02 Simple Regression Learning Objective: 12-02 Interpret a regression equation and use it to make predictions. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 73) Mary used a sample of 68 large U.S. cities to estimate the relationship between Crime (annual property crimes per 100,000 persons) and Income (median annual income per capita, in dollars). Her estimated regression equation was Crime = 428 + 0.050 Income. If Income decreases by 1000, we would expect that Crime will A) increase by 428. B) decrease by 50. C) increase by 500. D) remain unchanged. Answer: B Explanation: The constant has no effect, so ΔCrime = 0.050 ΔIncome = 0.050(−1000) = −50. Difficulty: 2 Medium Topic: 12.02 Simple Regression Learning Objective: 12-02 Interpret a regression equation and use it to make predictions. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

74) Amelia used a random sample of 100 accounts receivable to estimate the relationship between Days (number of days from billing to receipt of payment) and Size (size of balance due in dollars). Her estimated regression equation was Days = 22 + 0.0047 Size with a correlation coefficient of .300. From this information we can conclude that A) 9 percent of the variation in Days is explained by Size. B) autocorrelation is likely to be a problem. C) the relationship between Days and Size is significant. D) larger accounts usually take less time to pay. Answer: A Explanation: R2 = .302 = .09. These are not time-series data, so there is no reason to expect autocorrelation. We cannot judge significance without more information. Difficulty: 3 Hard Topic: 12.04 Ordinary Least Squares Formulas Learning Objective: 12-04 Explain the least squares method, apply formulas for coefficients, and interpret R2. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 75) Prediction intervals for Y are narrowest when A) the mean of X is near the mean of Y. B) the value of X is near the mean of X. C) the mean of X differs greatly from the mean of Y. D) the mean of X is small. Answer: B Explanation: Review the formula, which has (xi - )2 in the numerator. The minimum would be when xi = . Difficulty: 2 Medium Topic: 12.07 Confidence and Prediction Intervals for Y Learning Objective: 12-07 Distinguish between confidence and prediction intervals for Y. Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation

76) If n = 15 and r = .4296, the corresponding t statistic to test for zero correlation is A) 1.715 B) 7.862 C) 2.048 D) impossible to determine without α. Answer: A Explanation: tcalc = r[(n − 2)/(1 − r2)]1/2 = (.4296)[(15 − 2)/(1 − .42962)]1/2 = 1.715. Difficulty: 2 Medium Topic: 12.01 Visual Displays and Correlation Analysis Learning Objective: 12-01 Calculate and test a correlation coefficient for significance. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 77) Using a two-tailed test at α = .05 for n = 30, we would reject the hypothesis of zero correlation if the absolute value of r exceeds A) .2992. B) .3609. C) .0250. D) .2004. Answer: B Explanation: Use rcrit = t.025/(t.0252 + n − 2)1/2 = (2.048)/(2.0482 + 30 − 2)1/2 = .3609 for d.f. = 30 − 2 = 28. Difficulty: 2 Medium Topic: 12.01 Visual Displays and Correlation Analysis Learning Objective: 12-01 Calculate and test a correlation coefficient for significance. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

78) The ordinary least squares (OLS) method of estimation will minimize A) neither the slope nor the intercept. B) only the slope. C) only the intercept. D) both the slope and intercept. Answer: A Explanation: The OLS method minimizes the sum of squared residuals. Difficulty: 2 Medium Topic: 12.04 Ordinary Least Squares Formulas Learning Objective: 12-04 Explain the least squares method, apply formulas for coefficients, and interpret R2. Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation 79) A standardized residual equal to −2.205 indicates A) a rather poor prediction. B) an extreme outlier in the residuals. C) an observation with high leverage. D) a likely data entry error. Answer: A Explanation: This residual is beyond ±2se but is not an outlier (and without xi we cannot assess leverage). Difficulty: 2 Medium Topic: 12.09 Unusual Observations Learning Objective: 12-09 Identify unusual residuals and tell when they are outliers. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

80) In a simple regression, which would suggest a significant relationship between X and Y? A) Large p-value for the estimated slope B) Large t statistic for the slope C) Large p-value for the F statistic D) Small t statistic for the slope Answer: B Explanation: The larger the tcalc, the more we feel like rejecting H0: β1 = 0. Difficulty: 2 Medium Topic: 12.05 Tests for Significance Learning Objective: 12-05 Construct confidence intervals and test hypotheses for the slope and intercept. Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation 81) Which is indicative of an inverse relationship between X and Y? A) A negative F statistic B) A negative p-value for the correlation coefficient C) A negative correlation coefficient D) Either a negative F statistic or a negative p-value Answer: C Explanation: Fcalc and the p-value cannot be negative. Difficulty: 2 Medium Topic: 12.06 Analysis of Variance: Overall Fit Learning Objective: 12-06 Interpret the ANOVA table and use it to calculate F, R2, and standard error. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation

82) Which is not correct regarding the estimated slope of the OLS regression line? A) It is divided by its standard error to obtain its t statistic. B) It shows the change in Y for a unit change in X. C) It is chosen so as to minimize the sum of squared errors. D) It may be regarded as zero if its p-value is less than α. Answer: D Explanation: We would reject H0: β1 = 0 if its p-value is less than the level of significance. Difficulty: 2 Medium Topic: 12.05 Tests for Significance Learning Objective: 12-05 Construct confidence intervals and test hypotheses for the slope and intercept. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 83) Simple regression analysis means that A) the data are presented in a simple and clear way. B) we have only a few observations. C) there are only two independent variables. D) we have only one explanatory variable. Answer: D Explanation: Multiple regression has more than one independent variable (predictor). Difficulty: 1 Easy Topic: 12.02 Simple Regression Learning Objective: 12-02 Interpret a regression equation and use it to make predictions. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 84) The sample coefficient of correlation does not have which property? A) It can range from −1.00 up to +1.00. B) It is also sometimes called Pearson's r. C) It is tested for significance using a t-test. D) It assumes that Y is the dependent variable. Answer: D Explanation: Correlation analysis makes no assumption of causation or dependence. Difficulty: 1 Easy Topic: 12.01 Visual Displays and Correlation Analysis Learning Objective: 12-01 Calculate and test a correlation coefficient for significance. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 31 Copyright 2019 © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.

85) When comparing the 90 percent prediction and confidence intervals for a given regression analysis A) the prediction interval is narrower than the confidence interval. B) the prediction interval is wider than the confidence interval. C) there is no difference between the size of the prediction and confidence intervals. D) no generalization is possible about their comparative width. Answer: B Explanation: Individual values of Y vary more than the mean of Y. Difficulty: 2 Medium Topic: 12.07 Confidence and Prediction Intervals for Y Learning Objective: 12-07 Distinguish between confidence and prediction intervals for Y. Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation 86) Which is not true of the coefficient of determination? A) It is the square of the coefficient of correlation. B) It is negative when there is an inverse relationship between X and Y. C) It reports the percentage of the variation in Y explained by X. D) It is calculated using sums of squares (e.g., SSR, SSE, SST). Answer: B Explanation: R2 cannot be negative. Difficulty: 2 Medium Topic: 12.04 Ordinary Least Squares Formulas Learning Objective: 12-04 Explain the least squares method, apply formulas for coefficients, and interpret R2. Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation 87) If the fitted regression is Y = 3.5 + 2.1X (R2 = .25, n = 25), it is incorrect to conclude that A) Y increases 2.1 percent for a 1 percent increase in X. B) the estimated regression line crosses the Y axis at 3.5. C) the sample correlation coefficient must be positive. D) the value of the sample correlation coefficient is .50. Answer: A Explanation: The units of measurement are not known, and may not be measured in percent. Difficulty: 2 Medium Topic: 12.02 Simple Regression Learning Objective: 12-02 Interpret a regression equation and use it to make predictions. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 32 Copyright 2019 © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.

88) In a simple regression Y = b0 + b1X where Y = number of robberies in a city (thousands of robberies), X = size of the police force in a city (thousands of police), and n = 45 randomly chosen large U.S. cities in 2008, we would be least likely to see which problem? A) Autocorrelated residuals (because this is time-series data) B) Heteroscedastic residuals (because we are using totals uncorrected for city size) C) Nonnormal residuals (because a few larger cities may skew the residuals) D) High leverage for some observations (because some cities may be huge) Answer: A Explanation: This is not a time series, so autocorrelation would not be expected. However, the "size effect" is likely to produce heteroscedasticity and nonnormality. Further, some data values with high leverage are likely. Difficulty: 3 Hard Topic: 12.08 Residual Tests Learning Objective: 12-08 Calculate residuals and perform tests of regression assumptions. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 89) When homoscedasticity exists, we would expect that a plot of the residuals versus the fitted Y A) will form approximately a straight line. B) crosses the centerline too many times. C) will yield a Durbin-Watson statistic near 2. D) will show no pattern at all. Answer: D Explanation: Homoscedastic residuals exhibit no pattern (equal variance for all Y). Difficulty: 2 Medium Topic: 12.08 Residual Tests Learning Objective: 12-08 Calculate residuals and perform tests of regression assumptions. Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation

90) Which statement is not correct? A) Spurious correlation can often be reduced by expressing X and Y in per capita terms. B) Autocorrelation is mainly a concern when we are using time-series data. C) Heteroscedastic residuals will have roughly the same variance for any value of X. D) Standardized residuals make it easy to identify outliers or instances of poor fit. Answer: C Explanation: Heteroscedastic residuals exhibit different variance for different X or Y values. Difficulty: 2 Medium Topic: 12.08 Residual Tests Learning Objective: 12-08 Calculate residuals and perform tests of regression assumptions. Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation 91) In a simple bivariate regression with 25 observations, which statement is most nearly correct? A) A nonstandardized residual whose value is ei = 4.22 would be considered an outlier. B) A leverage statistic of 0.16 or more would indicate high leverage. C) Standardizing the residuals will eliminate any heteroscedasticity. D) Nonnormal residuals imply biased coefficient estimates, a major problem. Answer: B Explanation: For simple regression, the "high leverage criterion" is hi > 4/n = 4/25 = 0.16. We cannot judge a residual's magnitude without knowing the standard error se. Standardizing is only a scale shift, so it does not reduce heteroscedasticity. Nonnormal errors do not bias the OLS estimates. Difficulty: 3 Hard Topic: 12.09 Unusual Observations Learning Objective: 12-10 Define leverage and identify high-leverage observations. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

92) A regression was estimated using these variables: Y = annual value of reported bank robbery losses in all U.S. banks ($ millions), X = annual value of currency held by all U.S. banks ($ millions), n = 100 years (1912 through 2011). We would not anticipate A) autocorrelated residuals due to time-series data. B) heteroscedastic residuals due to the wide variation in data magnitudes. C) nonnormal residuals due to skewed data as bank size increases over time. D) a negative slope because banks hold less currency when they are robbed. Answer: D Explanation: It is a time series, so autocorrelation would be expected, and the "size effect" is likely to produce heteroscedasticity and nonnormality, but growth in both X and Y would yield a positive slope. Difficulty: 3 Hard Topic: 12.08 Residual Tests Learning Objective: 12-08 Calculate residuals and perform tests of regression assumptions. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 93) A fitted regression for an exam in Prof. Hardtack's class showed Score = 20 + 7 Study, where Score is the student's exam score and Study is the student's study hours. The regression yielded R2 = 0.50 and SE = 8. Bob studied 9 hours. The quick 95 percent prediction interval for Bob's grade is approximately A) 69 to 97 B) 75 to 91 C) 67 to 99 D) 76 to 90 Answer: C Explanation: The quick interval is ypredicted ± 2se = 83 ± (2)(8) = 83 ± 16. Difficulty: 3 Hard Topic: 12.07 Confidence and Prediction Intervals for Y Learning Objective: 12-07 Distinguish between confidence and prediction intervals for Y. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

94) Which is not an assumption of least squares regression? A) Normal X values B) Nonautocorrelated errors C) Homoscedastic errors D) Normal errors Answer: A Explanation: The predictor X is not assumed to be a random variable. Difficulty: 1 Easy Topic: 12.03 Regression Models Learning Objective: 12-03 Explain the form and assumptions of a simple regression model. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 95) In a simple bivariate regression with 60 observations, there will be ________ residuals. A) 60 B) 59 C) 58 D) 57 Answer: A Explanation: There is one residual for every observation. Difficulty: 1 Easy Topic: 12.03 Regression Models Learning Objective: 12-02 Interpret a regression equation and use it to make predictions. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 96) Which is correct to find the value of the coefficient of determination (R2)? A) SSR/SSE B) SSR/SST C) 1 − SST/SSE Answer: B Explanation: We use the ANOVA sums of squares to calculate R2. Difficulty: 2 Medium Topic: 12.04 Ordinary Least Squares Formulas Learning Objective: 12-04 Explain the least squares method, apply formulas for coefficients, and interpret R2. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation

97) The critical value for a two-tailed test of H0: β1 = 0 at α = .05 in a simple regression with 22 observations is A) ±1.725 B) ±2.086 C) ±2.528 D) ±1.960 Answer: B Explanation: From Appendix D, tcrit = ±2.086 for d.f. = n − 2 = 22 − 2 = 20. Difficulty: 2 Medium Topic: 12.05 Tests for Significance Learning Objective: 12-05 Construct confidence intervals and test hypotheses for the slope and intercept. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 98) In a sample of size n = 23, a sample correlation of r = .400 provides sufficient evidence to conclude that the population correlation coefficient exceeds zero in a right-tailed test at A) α = .01 but not α = .05. B) α = .05 but not α = .01. C) both α = .05 and α = .01. D) neither α = .05 nor α = .01. Answer: B Explanation: tcalc = r[(n − 2)/(1 − r2)]1/2 = (.40)[(23 − 2)/(1 − .402)]1/2 = 2.000 > t.05 = 1.721 for d.f. = 23 − 2 = 21. However, the test would not be significant for t.01 = 2.518 Difficulty: 2 Medium Topic: 12.01 Visual Displays and Correlation Analysis Learning Objective: 12-01 Calculate and test a correlation coefficient for significance. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

99) In a sample of n = 23, the Student's t test statistic for a correlation of r = .500 would be A) 2.559 B) 2.819 C) 2.646 D) impossible to calculate without knowing α. Answer: C Explanation: tcalc = r[(n − 2)/(1 − r2)]1/2 = (.50)[(23 − 2)/(1 − .502)]1/2 = 2.646. Difficulty: 2 Medium Topic: 12.01 Visual Displays and Correlation Analysis Learning Objective: 12-01 Calculate and test a correlation coefficient for significance. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 100) In a sample of n = 23, the critical value of the correlation coefficient for a two-tailed test at α = .05 is A) ±.524 B) ±.412 C) ±.500 D) ±.497 Answer: B Explanation: Use rcrit = t.025/(t.0252 + n − 2)1/2 = (2.069)/(2.0692 + 23 − 2)1/2 = .4115 for d.f. = 23 − 2 = 21. Difficulty: 2 Medium Topic: 12.01 Visual Displays and Correlation Analysis Learning Objective: 12-01 Calculate and test a correlation coefficient for significance. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

101) In a sample of n = 23, the critical value of Student's t for a two-tailed test of significance for a simple bivariate regression at α = .05 is A) ±2.229 B) ±2.819 C) ±2.646 D) ±2.080 Answer: D Explanation: From Appendix D, t.025 = ±2.080 for d.f. = n − 2 = 23 − 2 = 21. Difficulty: 2 Medium Topic: 12.05 Tests for Significance Learning Objective: 12-05 Construct confidence intervals and test hypotheses for the slope and intercept. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 102) In a sample of n = 40, a sample correlation of r = .400 provides sufficient evidence to conclude that the population correlation coefficient exceeds zero in a right-tailed test at A) α = .025 but not α = .05. B) α = .05 but not α = .025. C) both α = .025 and α = .05. D) neither α = .025 nor α = .05. Answer: C Explanation: tcalc = r[(n − 2)/(1 − r2)]1/2 = (.40)[(40 − 2)/(1 – .402)]1/2 = 2.690 > t.025 = 2.024 for d.f. = 40 − 2 = 38. The test would also be significant a fortiori if we used t.05 = 1.686. Difficulty: 3 Hard Topic: 12.01 Visual Displays and Correlation Analysis Learning Objective: 12-01 Calculate and test a correlation coefficient for significance. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

103) In a sample of n = 20, the Student's t test statistic for a correlation of r = .400 would be A) 2.110 B) 1.645 C) 1.852 D) can't say without knowing if it's a two-tailed or one-tailed test. Answer: C Explanation: tcalc = r[(n − 2)/(1 − r2)]1/2 = (.40)[(20 − 2)/(1 − .402)]1/2 = 1.852. Difficulty: 2 Medium Topic: 12.01 Visual Displays and Correlation Analysis Learning Objective: 12-01 Calculate and test a correlation coefficient for significance. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 104) In a sample of n = 20, the critical value of the correlation coefficient for a two-tailed test at α = .05 is A) ±.587 B) ±.412 C) ±.444 D) ±.497 Answer: C Explanation: Use rcrit = t.025/(t.0252 + n − 2)1/2 = (2.101)/(2.1012 + 20 − 2)1/2 = .4437 for d.f. = 20 − 2 = 18. Difficulty: 2 Medium Topic: 12.01 Visual Displays and Correlation Analysis Learning Objective: 12-01 Calculate and test a correlation coefficient for significance. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

105) In a sample of n = 27, the critical value of Student's t for a two-tailed test of significance for a simple bivariate regression at α = .05 is A) ±2.060 B) ±2.052 C) ±2.898 D) ±2.074 Answer: A Explanation: From Appendix D, t.025 = ±2.060 for d.f. = n − 2 = 27 − 2 = 25. Difficulty: 2 Medium Topic: 12.05 Tests for Significance Learning Objective: 12-05 Construct confidence intervals and test hypotheses for the slope and intercept. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 106) In a sample of size n = 36, a sample correlation of r = −.450 provides sufficient evidence to conclude that the population correlation coefficient differs significantly from zero in a two-tailed test at A) α = .01 B) α = .05 C) both α = .01 and α = .05. D) neither α = .01 nor α = .05. Answer: C Explanation: tcalc = r[(n − 2)/(1 − r2)]1/2 = (−.45)[(36 − 2)/(1 − (−.40)2)]1/2 = −2.938 < t.005 = −2.728 for d.f. = 34. The test would also be significant a fortiori if we used t.025 = −2.032 Difficulty: 3 Hard Topic: 12.01 Visual Displays and Correlation Analysis Learning Objective: 12-01 Calculate and test a correlation coefficient for significance. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

107) In a sample of n = 36, the Student's t test statistic for a correlation of r = −.450 would be A) −2.110 B) −2.938 C) −2.030 D) impossible to calculate without knowing α. Answer: B Explanation: tcalc = r[(n − 2)/(1 − r2)]1/2 = (−.45)[(36 − 2)/(1 − (−.40)2)]1/2 = −2.938. Difficulty: 2 Medium Topic: 12.01 Visual Displays and Correlation Analysis Learning Objective: 12-01 Calculate and test a correlation coefficient for significance. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 108) In a sample of n = 36, the critical value of the correlation coefficient for a two-tailed test at α = .05 is A) ±.329 B) ±.387 C) ±.423 D) ±.497 Answer: A Explanation: Use rcrit = t.025/(t.0252 + n − 2)1/2 = (2.032)/(2.0322 + 36 − 2)1/2 = .3291 for d.f. = 36 − 2 = 34. Difficulty: 2 Medium Topic: 12.01 Visual Displays and Correlation Analysis Learning Objective: 12-01 Calculate and test a correlation coefficient for significance. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

109) In a sample of n = 36, the critical value of Student's t for a two-tailed test of significance of the slope for a simple regression at α = .05 is A) 2.938 B) 2.724 C) 2.032 D) 2.074 Answer: C Explanation: From Appendix D, t.025 = ±2.032 for d.f. = n − 2 = 36 − 2 = 34. Difficulty: 2 Medium Topic: 12.05 Tests for Significance Learning Objective: 12-05 Construct confidence intervals and test hypotheses for the slope and intercept. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 110) A local trucking company fitted a regression to relate the travel time (days) of its shipments as a function of the distance traveled (miles). The fitted regression is Time = −7.126 + 0.0214 Distance. If Distance increases by 50 miles, the expected Time would increase by A) 1.07 days. B) 7.13 days. C) 2.14 days. D) 1.73 days. Answer: A Explanation: ΔTime = 0.0214 ΔDistance = (0.0214)(50) = 1.07. Difficulty: 2 Medium Topic: 12.02 Simple Regression Learning Objective: 12-02 Interpret a regression equation and use it to make predictions. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

111) A local trucking company fitted a regression to relate the cost of its shipments as a function of the distance traveled. The Excel fitted regression is shown.

Based on this estimated relationship, when distance increases by 50 miles, the expected shipping cost would increase by how much? A) $286 B) $143 C) $104 D) $301 Answer: B Explanation: ΔCost = 2.866 ΔDistance = 2.8666(50) = $143.33. Difficulty: 2 Medium Topic: 12.02 Simple Regression Learning Objective: 12-02 Interpret a regression equation and use it to make predictions. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

112) If SSR is 2592 and SSE is 608, then A) the slope is likely to be insignificant. B) the coefficient of determination is .81. C) the SST would be smaller than SSR. D) the standard error would be large. Answer: B Explanation: R2 = SSR/SST = SSR/(SSR + SSE) = 2592/(2592 + 608) = .81. SST cannot be smaller than SSR because SST = SSR + SSE. The significance and standard error cannot be judged without more information. Difficulty: 2 Medium Topic: 12.04 Ordinary Least Squares Formulas Learning Objective: 12-04 Explain the least squares method, apply formulas for coefficients, and interpret R2. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 113) Find the sample correlation coefficient for the following data. X

3 7 5 9 11 13 15 17

8 12 13 10 17 23 35 34

A) .8911 B) .9124 C) .9822 D) .9556 Answer: B Explanation: Use Excel =CORREL(XData, YData) to verify your calculation using the formula for r. Difficulty: 3 Hard Topic: 12.01 Visual Displays and Correlation Analysis Learning Objective: 12-01 Calculate and test a correlation coefficient for significance. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

114) Find the slope of the simple regression of Y on X. X

3 7 5 9 11 13 19 21

8 12 13 10 17 23 39 38

A) 1.833 B) 3.294 C) 0.762 D) −2.228 Answer: A Explanation: Use Excel to verify your calculations using the formulas for b0 and b1. Difficulty: 2 Medium Topic: 12.04 Ordinary Least Squares Formulas Learning Objective: 12-04 Explain the least squares method, apply formulas for coefficients, and interpret R2. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

115) Find the sample correlation coefficient for the following data. X

3 5 9 13 15

9 13 10 23 35

A) .7291 B) .8736 C) .9118 D) .9563 Answer: B Explanation: Use Excel =CORREL(XData, YData) to verify your calculation using the formula for r. Difficulty: 3 Hard Topic: 12.01 Visual Displays and Correlation Analysis Learning Objective: 12-01 Calculate and test a correlation coefficient for significance. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

116) Find the slope of the simple regression of Y on X. X

3 5 9 13 15

9 13 10 23 35

A) 2.595 B) 1.109 C) −2.221 D) 1.884 Answer: D Explanation: Use Excel to verify your calculations using the formulas for b0 and b1. Difficulty: 2 Medium Topic: 12.04 Ordinary Least Squares Formulas Learning Objective: 12-04 Explain the least squares method, apply formulas for coefficients, and interpret R2. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 117) A researcher's results are shown below using n = 25 observations. Variable Intercept Slope

Coefficients 343.619889 –2.2833659

Standard Error 61.0823514 0.99855319

Which is the 95 percent confidence interval for the slope? A) [−3.282, −1.284] B) [−4.349, −0.217] C) [1.118, 5.026] D) [−0.998, +0.998] Answer: B Explanation: For d.f. = n − 2 = 25 − 2 = 23, t.025 = 2.069, so −2.2834 ± (2.069)(0.99855). Difficulty: 2 Medium Topic: 12.05 Tests for Significance Learning Objective: 12-05 Construct confidence intervals and test hypotheses for the slope and intercept. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 48 Copyright 2019 © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.

118) A researcher's regression results are shown below using n = 8 observations. Variable Intercept Slope

Coefficients –0.1667 1.8333

Standard Error 2.8912 0.2307

Which is the 95 percent confidence interval for the slope? A) [1.333, 2.284] B) [1.602, 2.064] C) [1.268, 2.398] D) [1.118, 2.449] Answer: C Explanation: For d.f. = n − 2 = 8 − 2 = 6, t.025 = 2.447, so 1.8333 ± (2.447)(0.2307). Difficulty: 2 Medium Topic: 12.05 Tests for Significance Learning Objective: 12-05 Construct confidence intervals and test hypotheses for the slope and intercept. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

119) Bob thinks there is something wrong with Excel's fitted regression. What do you say?

A) The estimated equation is obviously incorrect. B) The R2 looks a little high but otherwise it looks OK. C) Bob needs to increase his sample size to decide. D) The relationship is linear, so the equation is credible. Answer: A Explanation: A visual estimate of the slope is Δy/Δx = (625 − 100)/(200 − 0) = 2.625, so the indicated slope of less than 1 must be wrong, plus the visual intercept is 100 (not 154.61) and the fit seems better than R2 = .2284. Difficulty: 3 Hard Topic: 12.02 Simple Regression Learning Objective: 12-02 Interpret a regression equation and use it to make predictions. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 120) The error term εi in the population regression model yi = β0 + β1 xi + εi is assumed to be A) observable. B) autocorrelated. C) heteroscedastic. D) normally distributed. Answer: D Explanation: In the population regression model, errors are not observable and are assumed to be independent and normally distributed with constant variance. However, residuals may not fulfill these assumptions. Difficulty: 2 Medium Topic: 12.03 Regression Models Learning Objective: 12-03 Explain the form and assumptions of a simple regression model. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 50 Copyright 2019 © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.

121) Which is not an assumed property of the errors in a population regression model? A) They are unobservable. B) They are independent. C) They have zero variance. D) They are normally distributed. Answer: C Explanation: In the population regression model, errors are assumed to be unobservable, independent, and normally distributed with constant variance. Difficulty: 1 Easy Topic: 12.03 Regression Models Learning Objective: 12-03 Explain the form and assumptions of a simple regression model. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 122) If the residuals from a fitted regression violate the assumption of homoscedasticity, we know that A) they are normally distributed. B) they are independent of one another. C) their variance is not constant. D) there are extreme outliers. Answer: C Explanation: In the population regression model, errors are assumed to have constant variance. However, residuals may not fulfill these assumptions. Difficulty: 2 Medium Topic: 12.08 Residual Tests Learning Objective: 12-08 Calculate residuals and perform tests of regression assumptions. Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation

123) If the residuals violate the assumption of autocorrelation, we know that they A) are probably outliers. B) are not independent. C) have constant variance. D) are normally distributed. Answer: C Explanation: In the population regression model, errors are assumed to be independent and normally distributed with constant variance. However, residuals may not fulfill these assumptions. Difficulty: 1 Easy Topic: 12.08 Residual Tests Learning Objective: 12-08 Calculate residuals and perform tests of regression assumptions. Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation 124) If the residuals violate the assumption of normality, we expect that A) the sample was not taken randomly. B) the residuals will not sum to zero. C) least squares formulas will fail. D) confidence intervals may be unreliable. Answer: D Explanation: Confidence and prediction intervals are predicated on a normal error distribution, so they may be less reliable when this assumption is violated. Difficulty: 2 Medium Topic: 12.08 Residual Tests Learning Objective: 12-08 Calculate residuals and perform tests of regression assumptions. Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation

125) Which is not an assumed characteristic of εi in the population model yi = β0 + β1 xi + εi? A) It is a random variable. B) It has zero mean. C) It is a statistic. D) It is unobservable. Answer: C Explanation: The true error is a random variable that cannot be observed. Its assumed distribution is N(0,σ2). Difficulty: 1 Easy Topic: 12.03 Regression Models Learning Objective: 12-03 Explain the form and assumptions of a simple regression model. Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation 126) To calculate a residual for the ith observation, we do not need the A) actual value of yi. B) estimated slope. C) standard error. D) estimated intercept. Answer: C Explanation: By definition, the residual is ei = yobserved − yestimated where yestimated = b0 + b1x. Difficulty: 2 Medium Topic: 12.03 Regression Models Learning Objective: 12-03 Explain the form and assumptions of a simple regression model. Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation

127) In the population model yi = β0 + β1 xi + εi the βj are A) parameters. B) statistics. C) estimated. D) observable. Answer: A Explanation: The population parameters (indicated by Greek letters) cannot be observed, but can be estimated from a sample. Difficulty: 1 Easy Topic: 12.03 Regression Models Learning Objective: 12-03 Explain the form and assumptions of a simple regression model. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 128) The slope of a proposed population regression model yi = β0 + β1 xi + εi is assumed to be A) a random variable. B) distributed normally. C) a statistic. D) a parameter. Answer: D Explanation: The parameters β0 and β1 are constants (not random variables) that can be estimated from a sample. Difficulty: 2 Medium Topic: 12.03 Regression Models Learning Objective: 12-03 Explain the form and assumptions of a simple regression model. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation

129) A fitted regression Profit = 262 + 1.51 Sales (all variables in thousands of dollars) was estimated from a random sample of 15 small coffee kiosks. We can say that A) the slope is too small to be significant. B) the intercept does not seem reasonable. C) the R2 will be low due to small sample size. D) predictions are likely to be underestimated. Answer: B Explanation: How can we have profit with no revenue? Difficulty: 2 Medium Topic: 12.02 Simple Regression Learning Objective: 12-02 Interpret a regression equation and use it to make predictions. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 130) A fitted regression Profit = −570 + 30 Sales (all variables in thousands of dollars) was estimated from a random sample of 20 pharmacies. For a pharmacy with Sales = 10, we predict that Profit will be A) 3570 B) 2430 C) −270 D) 870 Answer: C Explanation: Predicted profit is negative: Profit = −570 + 30 Sales = −570 + 30 (10) = −570 + 300 = −270. Difficulty: 1 Easy Topic: 12.02 Simple Regression Learning Objective: 12-02 Interpret a regression equation and use it to make predictions. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

131) A fitted regression Profit = −570 + 30 Sales (all variables in thousands of dollars) was estimated from a random sample of pharmacies. From this regression, in order to break even (Profit ≥ 0), a pharmacy's Sales would have to be at least A) 19 B) 300 C) 56 D) 100 Answer: A Explanation: We set Profit = 0 and solve for Sales: 0 = −570 + 30 Sales giving us Sales = 570/30 = 19. Difficulty: 3 Hard Topic: 12.02 Simple Regression Learning Objective: 12-02 Interpret a regression equation and use it to make predictions. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 132) A financial regression yielded a standard error of 12 dollars, so a residual of 23 dollars would be A) a rather poor prediction. B) an extreme outlier in the residuals. C) an observation with high leverage. D) an outlier, but not extreme. Answer: A Explanation: This residual is within ±2se, so it is not an outlier, though it is not a very good prediction. Without xi we cannot assess leverage. Difficulty: 2 Medium Topic: 12.09 Unusual Observations Learning Objective: 12-09 Identify unusual residuals and tell when they are outliers. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

133) A variable transformation in a regression (e.g., replacing Y with log(Y)) A) may reduce heteroscedasticity. B) changes the model specification. C) makes the model easier to interpret. D) leads to severe autocorrelation. Answer: A Explanation: A log transform changes the model and may make it harder to interpret (without a reverse transformation) but may alleviate heteroscedasticity. Difficulty: 2 Medium Topic: 12.10 Other Regression Problems (Optional) Learning Objective: 12-11 Improve data conditioning and use transformations if needed (optional). Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation 134) The estimated intercept in a regression A) is often of secondary interest to the researcher. B) should be forced to zero in the estimated model. C) can be omitted when making predictions. D) is greater than zero as long as Y is positive. Answer: A Explanation: Researchers are usually more interested in testing the slope to assess the potential effect of X on Y. You can force the intercept to zero in Excel, but that would rarely be desirable. We use the intercept in making predictions, even if it is negative or lacks a practical interpretation. Difficulty: 2 Medium Topic: 12.05 Tests for Significance Learning Objective: 12-05 Construct confidence intervals and test hypotheses for the slope and intercept. Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation

135) We would use a logistic regression model A) to estimate a log-linear trend in time series data. B) when the residuals are not normally distributed. C) to predict an event that occurs or does not occur. D) with Excel's exponential trend regression model. Answer: C Explanation: We use logistic regression when Y is a binary (0, 1) variable. Excel does not offer a logit model. Difficulty: 2 Medium Topic: 12.11 Logistic Regression (Optional) Learning Objective: 12-12 Identify when logistic regression is appropriate and calculate predictions for a binary response variable. Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation 136) Which is not a characteristic of the logistic regression model? A) It postulates an S-shaped relationship between X and Y. B) It can best be fitted using the maximum likelihood method. C) Predictions from the fitted logit model are either 0 or 1. D) The logistic model cannot yield predictions greater than 1. Answer: C Explanation: The logit function is S-shaped. Its predictions are probabilities within the range 0 to 1, using a logit model fitted by the maximum likelihood method. Difficulty: 2 Medium Topic: 12.11 Logistic Regression (Optional) Learning Objective: 12-12 Identify when logistic regression is appropriate and calculate predictions for a binary response variable. Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation

137) If you were to use Excel to estimate Y = β0 + β1 X + ε with binary Y (0 or 1) you would expect A) an error message because Excel does not allow a binary Y. B) predicted probabilities greater than 1 or less than 0. C) predictions that are either 0 or 1. D) residuals that are homoscedastic and normally distributed. Answer: B Explanation: Excel will fit a linear regression with binary Y, but residuals will be heteroscedastic and non-normal, and predictions may be outside the range from 0 to 1. That's why we prefer a logit model estimated with non-Excel software. Difficulty: 2 Medium Topic: 12.11 Logistic Regression (Optional) Learning Objective: 12-12 Identify when logistic regression is appropriate and calculate predictions for a binary response variable. Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation

Applied Statistics in Business and Economics, 6e (Doane) Chapter 13 Multiple Regression 1) In regression, the dependent variable is referred to as the response variable. Answer: TRUE Explanation: Y is also sometimes called the dependent variable. Difficulty: 1 Easy Topic: 13.01 Multiple Regression Learning Objective: 13-01 Use a fitted multiple regression equation to make predictions. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 2) If a regression model's F test statistic is Fcalc = 43.82, we could say that the explained variance is approximately 44 percent. Answer: FALSE Explanation: The R2 statistic (not the F statistic) shows the percentage of explained variation. Difficulty: 1 Easy Topic: 13.02 Assessing Overall Fit Learning Objective: 13-02 Use the ANOVA table to perform an F test for overall significance. Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation 3) In a regression, the model with the best fit is preferred over all other models. Answer: FALSE Explanation: Occam's Razor says that complexity is justified only if it is necessary for a good model. Difficulty: 2 Medium Topic: 13.01 Multiple Regression Learning Objective: 13-01 Use a fitted multiple regression equation to make predictions. Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation

4) A common misinterpretation of the principle of Occam's Razor is that a simple regression model (rather than a multiple regression model) is always best. Answer: TRUE Explanation: Occam's Razor says that complexity is justified if it is necessary for a good model. Difficulty: 2 Medium Topic: 13.01 Multiple Regression Learning Objective: 13-01 Use a fitted multiple regression equation to make predictions. Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation 5) A predictor whose pairwise correlation with Y is near zero can still have a significant t-value in a multiple regression when other predictors are included. Answer: TRUE Explanation: The t statistic for a predictor depends on which other predictors are in the model. Difficulty: 2 Medium Topic: 13.03 Predictor Significance Learning Objective: 13-03 Construct confidence intervals for coefficients and test predictors for significance. Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation 6) The F statistic in a multiple regression is significant if at least one of the predictors has a significant t statistic at a given α. Answer: TRUE Explanation: At least one predictor coefficient will differ from zero at the same α used in the F test. Difficulty: 1 Easy Topic: 13.02 Assessing Overall Fit Learning Objective: 13-02 Use the ANOVA table to perform an F test for overall significance. Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation

7) R2adj can exceed R2 if there are several weak predictors. Answer: FALSE Explanation: R2adj is smaller than R2, and a large difference suggests unnecessary predictors. Difficulty: 2 Medium Topic: 13.02 Assessing Overall Fit Learning Objective: 13-02 Use the ANOVA table to perform an F test for overall significance. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 8) A binary (categorical) predictor should not be used along with nonbinary (numerical) predictors. Answer: FALSE Explanation: Binary predictors behave like any other except they look weird on a scatter plot. Difficulty: 1 Easy Topic: 13.05 Categorical Variables Learning Objective: 13-05 Incorporate categorical variables into a multiple regression model. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 9) In a multiple regression with 3 predictors in a sample of 25 U.S. cities, we would use F3,21 in a test of overall significance. Answer: TRUE Explanation: For the F test we use d.f. = (k, n − k − 1). Difficulty: 2 Medium Topic: 13.02 Assessing Overall Fit Learning Objective: 13-02 Use the ANOVA table to perform an F test for overall significance. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 10) Evans' Rule says that if n = 50 you need at least 5 predictors to have a good model. Answer: FALSE Explanation: On the contrary, Evans' Rule is intended to prevent having too many predictors. Difficulty: 2 Medium Topic: 13.02 Assessing Overall Fit Learning Objective: 13-02 Use the ANOVA table to perform an F test for overall significance. Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation 3 Copyright 2019 © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.

11) The model Y = β0 + β1X + β2X2 cannot be estimated by Excel because of the nonlinear term. Answer: FALSE Explanation: The X2 predictor is just a data column like any other. Difficulty: 2 Medium Topic: 13.06 Tests for Nonlinearity and Interaction Learning Objective: 13-06 Perform basic tests for nonlinearity and interaction. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 12) The random error term in a regression model reflects all factors omitted from the model. Answer: TRUE Explanation: The errors are assumed normally distributed with zero mean and constant variance. Difficulty: 1 Easy Topic: 13.01 Multiple Regression Learning Objective: 13-01 Use a fitted multiple regression equation to make predictions. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 13) If the probability plot of residuals resembles a straight line, the residuals show a fairly good fit to the normal distribution. Answer: TRUE Explanation: The probability plot is easy to interpret in a general way (linearity suggests normality). Difficulty: 2 Medium Topic: 13.08 Regression Diagnostics Learning Objective: 13-08 Analyze residuals to check for violations of residual assumptions. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 14) Confidence intervals for Y may be unreliable when the residuals are not normally distributed. Answer: TRUE Explanation: If serious nonnormality exists and n is small, confidence intervals may be affected. Difficulty: 2 Medium Topic: 13.08 Regression Diagnostics Learning Objective: 13-08 Analyze residuals to check for violations of residual assumptions. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 4 Copyright 2019 © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.

15) A negative estimated coefficient in a regression usually indicates a weak predictor. Answer: FALSE Explanation: It is the t statistic that indicates the strength of a predictor. Difficulty: 2 Medium Topic: 13.03 Predictor Significance Learning Objective: 13-03 Construct confidence intervals for coefficients and test predictors for significance. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 16) For a certain firm, the regression equation Bonus = 2,000 + 257 Experience + 0.046 Salary describes employee bonuses with a standard error of 125. John has 10 years' experience, earns $50,000, and earned a bonus of $7,000. John is an outlier. Answer: FALSE Explanation: John's standardized residual is (yactual − yestimated)/se = (7000 − 6870)/(125) = 1.04, which is not unusual. Difficulty: 3 Hard Topic: 13.08 Regression Diagnostics Learning Objective: 13-09 Identify unusual residuals and tell when they are outliers. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 17) There is one residual for each predictor in the regression model. Answer: FALSE Explanation: There are k predictors, but there are n residuals e1, e2, . . . , en. Difficulty: 1 Easy Topic: 13.01 Multiple Regression Learning Objective: 13-01 Use a fitted multiple regression equation to make predictions. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation

18) If R2 and R2adj differ greatly, we should probably add a few predictors to improve the fit. Answer: FALSE Explanation: Evidence of unnecessary predictors can be seen when R2adj is much smaller than R2. Difficulty: 2 Medium Topic: 13.02 Assessing Overall Fit Learning Objective: 13-02 Use the ANOVA table to perform an F test for overall significance. Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation 19) The effect of a binary predictor is to shift the regression intercept. Answer: TRUE Explanation: The omitted category becomes part of the intercept. Difficulty: 1 Easy Topic: 13.05 Categorical Variables Learning Objective: 13-05 Incorporate categorical variables into a multiple regression model. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 20) A parsimonious model is one with many weak predictors but a few strong ones. Answer: FALSE Explanation: On the contrary, a lean (parsimonious) model has strong predictors and no weak ones. Difficulty: 2 Medium Topic: 13.01 Multiple Regression Learning Objective: 13-01 Use a fitted multiple regression equation to make predictions. Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation 21) The F statistic and its p-value give a global test of significance for a multiple regression. Answer: TRUE Explanation: The F test tells whether or not at least some predictors are significant. Difficulty: 1 Easy Topic: 13.02 Assessing Overall Fit Learning Objective: 13-02 Use the ANOVA table to perform an F test for overall significance. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 6 Copyright 2019 © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.

22) In a regression model of student grades, we would code the nine categories of business courses taken (ACC, FIN, ECN, MGT, MKT, MIS, ORG, POM, QMM) by including nine binary (0 or 1) predictors in the regression. Answer: FALSE Explanation: We can code c categories with c − 1 predictors (i.e., omit one). Difficulty: 2 Medium Topic: 13.05 Categorical Variables Learning Objective: 13-05 Incorporate categorical variables into a multiple regression model. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 23) A disadvantage of Excel's Data Analysis regression tool is that it expects the independent variables to be in a block of contiguous columns, so you must delete a column if you want to eliminate a predictor from the model. Answer: TRUE Explanation: This is why we might want to use Minitab, MegaStat, SPSS, or Systat. Difficulty: 1 Easy Topic: 13.01 Multiple Regression Learning Objective: 13-01 Use a fitted multiple regression equation to make predictions. Bloom's: Understand AACSB: Technology Accessibility: Keyboard Navigation 24) A disadvantage of Excel's regression is that it does not give as much accuracy in the estimated regression coefficients as a package like Minitab. Answer: FALSE Explanation: Excel's accuracy is good for most of the common regression statistics. Difficulty: 1 Easy Topic: 13.01 Multiple Regression Learning Objective: 13-01 Use a fitted multiple regression equation to make predictions. Bloom's: Understand AACSB: Technology Accessibility: Keyboard Navigation

25) Non-normality of the residuals from a regression can best be detected by looking at the residual plots against the fitted Y values. Answer: FALSE Explanation: Use a probability plot to check for nonnormality. A residual plot is a test for heteroscedasticity. Difficulty: 2 Medium Topic: 13.08 Regression Diagnostics Learning Objective: 13-08 Analyze residuals to check for violations of residual assumptions. Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation 26) A high variance inflation factor (VIF) indicates a significant predictor in the regression. Answer: FALSE Explanation: A high VIF indicates that a predictor is related to the other predictors in the model. Difficulty: 2 Medium Topic: 13.07 Multicollinearity Learning Objective: 13-07 Detect multicollinearity and assess its effects. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 27) Autocorrelation may be detected by looking at a plot of the residuals against time. Answer: TRUE Explanation: Too many or too few crossings of the zero axis suggest nonrandomness. Difficulty: 2 Medium Topic: 13.08 Regression Diagnostics Learning Objective: 13-08 Analyze residuals to check for violations of residual assumptions. Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation 28) A widening pattern of residuals as X increases would suggest heteroscedasticity. Answer: TRUE Explanation: The absence of a pattern would be ideal (homoscedastic). Difficulty: 1 Easy Topic: 13.08 Regression Diagnostics Learning Objective: 13-08 Analyze residuals to check for violations of residual assumptions. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 8 Copyright 2019 © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.

29) Plotting the residuals against a binary predictor (X = 0, 1) reveals nothing about heteroscedasticity. Answer: FALSE Explanation: You can still spot wider or narrower spread at the two points X = 0 and X = 1. Difficulty: 3 Hard Topic: 13.08 Regression Diagnostics Learning Objective: 13-08 Analyze residuals to check for violations of residual assumptions. Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation 30) The regression equation Bonus = 2,812 + 27 Experience + 0.046 Salary says that Experience is the most significant predictor of Bonus. Answer: FALSE Explanation: You need a t statistic to assess significance of a predictor. Difficulty: 2 Medium Topic: 13.03 Predictor Significance Learning Objective: 13-03 Construct confidence intervals for coefficients and test predictors for significance. Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation 31) A multiple regression with 60 observations should not have 13 predictors. Answer: TRUE Explanation: Evans' Rule suggests no more than n/10 = 60/10 = 6 predictors. Difficulty: 1 Easy Topic: 13.02 Assessing Overall Fit Learning Objective: 13-02 Use the ANOVA table to perform an F test for overall significance. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

32) A regression of Y using four independent variables X1, X2, X3, X4 could also have up to four nonlinear terms (X2) and six simple interaction terms (XjXk) if you have enough observations to justify them. Answer: TRUE Explanation: We must count all the possible squares and two-way combinations of four predictors. Difficulty: 3 Hard Topic: 13.06 Tests for Nonlinearity and Interaction Learning Objective: 13-06 Perform basic tests for nonlinearity and interaction. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 33) When autocorrelation is present, the estimates of the coefficients will be unbiased. Answer: TRUE Explanation: There is no bias in the Ordinary Least Squares (OLS) estimates, though variances and t tests may be affected. Difficulty: 2 Medium Topic: 13.08 Regression Diagnostics Learning Objective: 13-08 Analyze residuals to check for violations of residual assumptions. Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation 34) If the residuals in your regression are nonnormal, a larger sample size might help improve the reliability of confidence intervals for Y. Answer: TRUE Explanation: Asymptotic normality and consistency of the OLS estimators may help. Difficulty: 2 Medium Topic: 13.08 Regression Diagnostics Learning Objective: 13-08 Analyze residuals to check for violations of residual assumptions. Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation

35) Multicollinearity can be detected from t tests of the predictor variables. Answer: FALSE Explanation: The t tests only indicate significance. (We use VIFs to detect multicollinearity.) Difficulty: 2 Medium Topic: 13.07 Multicollinearity Learning Objective: 13-07 Detect multicollinearity and assess its effects. Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation 36) When multicollinearity is present, the regression model is of no use for making predictions. Answer: FALSE Explanation: Multicollinearity makes it hard to assess each predictor's role, but predictions may be useful. Difficulty: 2 Medium Topic: 13.07 Multicollinearity Learning Objective: 13-07 Detect multicollinearity and assess its effects. Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation 37) Autocorrelation of the residuals may affect the reliability of the t-values for the estimated coefficients of the predictors X1, X2, . . . , Xk. Answer: TRUE Explanation: Autocorrelation can affect the variances of the estimators, hence their t-values. Difficulty: 2 Medium Topic: 13.08 Regression Diagnostics Learning Objective: 13-08 Analyze residuals to check for violations of residual assumptions. Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation 38) The first differences transformation might be tried if autocorrelation is found in a time-series data set. Answer: TRUE Explanation: First differences may help and is an easily understood transformation. Difficulty: 2 Medium Topic: 13.08 Regression Diagnostics Learning Objective: 13-08 Analyze residuals to check for violations of residual assumptions. Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation 11 Copyright 2019 © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.

39) Statisticians who work with cross-sectional data generally do not anticipate autocorrelation. Answer: TRUE Explanation: We are more likely to see autocorrelation in time-series data. Difficulty: 1 Easy Topic: 13.08 Regression Diagnostics Learning Objective: 13-08 Analyze residuals to check for violations of residual assumptions. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 40) The ill effects of heteroscedasticity might be mitigated by redefining totals (e.g., total number of homicides) as relative values (e.g., homicide rate per 100,000 population). Answer: TRUE Explanation: Large magnitude ranges for X's and Y's (the "size" problem) can induce heteroscedasticity. Difficulty: 2 Medium Topic: 13.09 Other Regression Topics Learning Objective: 13-11 Explain the purpose of data conditioning and stepwise regression. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 41) Nonnormal residuals lead to biased estimates of the coefficients in a regression model. Answer: FALSE Explanation: There is no bias in the estimated coefficients, though confidence intervals may be affected. Difficulty: 2 Medium Topic: 13.08 Regression Diagnostics Learning Objective: 13-08 Analyze residuals to check for violations of residual assumptions. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation

42) A large VIF (e.g., 10 or more) would indicate multicollinearity. Answer: TRUE Explanation: Some multicollinearity is inevitable, but very large VIFs suggest competing predictors. Difficulty: 2 Medium Topic: 13.07 Multicollinearity Learning Objective: 13-07 Detect multicollinearity and assess its effects. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 43) Heteroscedasticity exists when all the errors (residuals) have the same variance. Answer: FALSE Explanation: The statement would be true if we change the first word to "homoscedasticity." Difficulty: 1 Easy Topic: 13.08 Regression Diagnostics Learning Objective: 13-08 Analyze residuals to check for violations of residual assumptions. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 44) Multicollinearity refers to relationships among the independent variables. Answer: TRUE Explanation: When one predictor is predicted by the other predictors, we have multicollinearity. Difficulty: 1 Easy Topic: 13.07 Multicollinearity Learning Objective: 13-07 Detect multicollinearity and assess its effects. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 45) A squared predictor is used to test for nonlinearity in the predictor's relationship to Y. Answer: TRUE Explanation: Including a squared predictor is an easy way to make the relationship nonlinear. Difficulty: 2 Medium Topic: 13.06 Tests for Nonlinearity and Interaction Learning Objective: 13-06 Perform basic tests for nonlinearity and interaction. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation

46) Nonnormality of residuals is not usually considered a major problem unless there are outliers. Answer: TRUE Explanation: Serious nonnormality can make the confidence intervals unreliable. Difficulty: 2 Medium Topic: 13.08 Regression Diagnostics Learning Objective: 13-08 Analyze residuals to check for violations of residual assumptions. Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation 47) In the fitted regression Y = 12 + 3X1 − 5X2 + 27X3 + 2X4 the most significant predictor is X3. Answer: FALSE Explanation: We must have the t statistics (not just the coefficients) to assess each predictor's significance. Difficulty: 2 Medium Topic: 13.03 Predictor Significance Learning Objective: 13-03 Construct confidence intervals for coefficients and test predictors for significance. Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation 48) Given that the fitted regression is Y = 76.40 − 6.388X1 + 0.870X2, the standard error of b1 is 1.453, and n = 63, at α = .05, we can conclude that X1 is a significant predictor of Y. Answer: TRUE Explanation: tcalc = (−6.388)/(1.453) = −4.396, which is < t.025 = −2.000 for d.f. = 60 in a twotailed test. Difficulty: 2 Medium Topic: 13.03 Predictor Significance Learning Objective: 13-03 Construct confidence intervals for coefficients and test predictors for significance. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

49) Unlike other predictors, a binary predictor has a t-value that is either 0 or 1. Answer: FALSE Explanation: The t-value for a binary predictor is like any other t-value. Difficulty: 2 Medium Topic: 13.05 Categorical Variables Learning Objective: 13-05 Incorporate categorical variables into a multiple regression model. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 50) The t test shows the ratio of an estimated coefficient to its standard error. Answer: TRUE Explanation: In a test for zero coefficient (and in computer output) tcalc = bj/sbj. Difficulty: 1 Easy Topic: 13.03 Predictor Significance Learning Objective: 13-03 Construct confidence intervals for coefficients and test predictors for significance. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 51) In a multiple regression with five predictors in a sample of 56 U.S. cities, we would use F5,50 in a test of overall significance. Answer: TRUE Explanation: F.05 = 2.40 for d.f. = (k, n − k − 1) = (5, 56 − 5 − 1) = (5, 50). Difficulty: 2 Medium Topic: 13.02 Assessing Overall Fit Learning Objective: 13-02 Use the ANOVA table to perform an F test for overall significance. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

52) In a multiple regression with six predictors in a sample of 67 U.S. cities, what would be the critical value for an F test of overall significance at α = .05? A) 2.29 B) 2.25 C) 2.37 D) 2.18 Answer: B Explanation: Using Appendix F, we get F.05 = 2.25 for d.f. = (k, n − k − 1) = (6, 67 − 6 − 1) = (6, 60). Alternatively, use the Excel function =F.INV.RT(0.05,6,60) = 2.25405. Difficulty: 2 Medium Topic: 13.02 Assessing Overall Fit Learning Objective: 13-02 Use the ANOVA table to perform an F test for overall significance. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 53) In a multiple regression with five predictors in a sample of 56 U.S. cities, what would be the critical value for an F test of overall significance at α = .05? A) 2.45 B) 2.37 C) 2.40 D) 2.56 Answer: C Explanation: Using Appendix F, we get F.05 = 2.40 for d.f. = (k, n − k − 1) = (5, 56 − 5 − 1) = (5, 50). Alternatively, use the Excel function =F.INV.RT(0.05,5,50) = 2.40041. Difficulty: 2 Medium Topic: 13.02 Assessing Overall Fit Learning Objective: 13-02 Use the ANOVA table to perform an F test for overall significance. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

54) When predictor variables are strongly related to each other, the ________ of the regression estimates is questionable. A) logic B) fit C) parsimony D) stability Answer: D Explanation: High interpredictor correlation affects their variances, so coefficients are less certain. Difficulty: 1 Easy Topic: 13.07 Multicollinearity Learning Objective: 13-07 Detect multicollinearity and assess its effects. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 55) A test is conducted in 22 cities to see if giving away free transit system maps will increase the number of bus riders. In a regression analysis, the dependent variable Y is the increase in bus riders (in thousands of persons) from the start of the test until its conclusion. The independent variables are X1 = the number (in thousands) of free maps distributed and a binary variable X2 = 1 if the city has free downtown parking, 0 otherwise. The estimated regression equation is Y = 1.32 + 0.0345X1 − 1.45X2 . In city 3, the observed Y value is 7.3, X1 = 140, and X2 = 0. The residual for city 3 (in thousands) is A) 6.15 B) 1.15 C) 4.83 D) 1.57 Answer: B Explanation: yestimated = 1.32 + 0.0345(140) − 1.45(0) = 6.15, so the residual is (7.3 − 6.15) = 1.15. Difficulty: 2 Medium Topic: 13.02 Assessing Overall Fit Learning Objective: 13-01 Use a fitted multiple regression equation to make predictions. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

56) If X2 is a binary predictor in Y = β0 + β1X1 + β2X2, then which statement is most nearly correct? A) X2 = 1 should represent the most desirable condition. B) X2 would be a significant predictor if β2 = 423.72. C) X2 = 0, X2 = 1, X2 = 2 would be appropriate if three categories exist. D) X2 will shift the estimated equation either by 0 units or by β2 units. Answer: D Explanation: If X2 = 0, then nothing is added to the equation, while if X2 = 1 we add β2 units. Difficulty: 3 Hard Topic: 13.05 Categorical Variables Learning Objective: 13-05 Incorporate categorical variables into a multiple regression model. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 57) The unexplained sum of squares measures variation in the dependent variable Y about the: A) mean of the Y values. B) estimated Y values. C) mean of the X values. D) Y-intercept. Answer: B Explanation: We are trying to explain variation in the response variable around its mean. Difficulty: 2 Medium Topic: 13.02 Assessing Overall Fit Learning Objective: 13-02 Use the ANOVA table to perform an F test for overall significance. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 58) Which of the following is not true of the standard error of the regression? A) It is a measure of the accuracy of the prediction. B) It is based on squared vertical deviations between the actual and predicted values of Y. C) It would be negative when there is an inverse relationship in the model. D) It is used in constructing confidence and prediction intervals for Y. Answer: C Explanation: The standard error is the square root of a sum of squares, so it cannot be negative. Difficulty: 2 Medium Topic: 13.04 Confidence Intervals for Y Learning Objective: 13-04 Calculate the standard error and construct approximate confidence and prediction intervals for Y. Bloom's: Remember AACSB: Analytical Thinking 18 Copyright 2019 © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.

Accessibility: Keyboard Navigation 59) A multiple regression analysis with two independent variables yielded the following results in the ANOVA table: SS(Total) = 798, SS(Regression) = 738, SS(Error) = 60. The multiple correlation coefficient is A) .2742 B) .0752 C) .9248 D) .9617 Answer: D Explanation: R2 = SSR/SST = 738/798 = .9248, so r = (R2)1/2 = .9617. Difficulty: 2 Medium Topic: 13.02 Assessing Overall Fit Learning Objective: 13-02 Use the ANOVA table to perform an F test for overall significance. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 60) A fitted multiple regression equation is Y = 12 + 3X1 − 5X2 + 7X3 + 2X4. When X1 increases 2 units and X2 increases 2 units as well, while X3 and X4 remain unchanged, what change would you expect in your estimate of Y? A) Decrease by 2 B) Decrease by 4 C) Increase by 2 D) No change in Y Answer: B Explanation: The net effect is + 3ΔX1 − 5ΔX2 = 3(2) − 5(2) = 6 − 10 = −4. Difficulty: 1 Easy Topic: 13.01 Multiple Regression Learning Objective: 13-01 Use a fitted multiple regression equation to make predictions. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

61) A fitted multiple regression equation is Y = 28 + 5X1 − 4X2 + 7X3 + 2X4. When X1 increases 2 units and X2 increases 2 units as well, while X3 and X4 remain unchanged, what change would you expect in your estimate of Y? A) Increase by 2 B) Decrease by 4 C) Increase by 4 D) No change in Y Answer: A Explanation: The net effect is + 5ΔX1 − 4ΔX2 =5(2) − 4(2) = 10 − 8 = +2. Difficulty: 1 Easy Topic: 13.01 Multiple Regression Learning Objective: 13-01 Use a fitted multiple regression equation to make predictions. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 62) Which is not a name often given to an independent variable that takes on just two values (0 or 1) according to whether or not a given characteristic is absent or present? A) Absent variable B) Binary variable C) Dummy variable Answer: A Explanation: A two-valued predictor is a binary or dummy variable (special cases of categorical predictors). Difficulty: 1 Easy Topic: 13.05 Categorical Variables Learning Objective: 13-05 Incorporate categorical variables into a multiple regression model. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation

63) Using a sample of 63 observations, a dependent variable Y is regressed against two variables X1 and X2 to obtain the fitted regression equation Y = 76.40 − 6.388X1 + 0.870X2. The standard error of b1 is 3.453 and the standard error of b2 is 0.611. At α = .05, we could A) conclude that both coefficients differ significantly from zero. B) reject H0: β1 ≥ 0 and conclude H0: β1 < 0. C) reject H0: β2 ≤ 0 and conclude H0: β1 > 0. D) conclude that Evans' Rule has been violated. Answer: C Explanation: For β1 we have tcalc = (−6.388)/(3.453) = −1.849, which is less than t.05 = −1.671 for d.f. = 60 in a left-tailed test. For β2 we have tcalc = (0.870)/(0.611) = +1.424, which does not exceed t.05 = +1.671 for d.f. = 60 in a right-tailed test. For a two-tailed test, t.025 = ±2.000, so neither coefficient would differ significantly from zero at α = .05. Evans' Rule is not violated, because n/k = 63/3 = 21. Difficulty: 3 Hard Topic: 13.03 Predictor Significance Learning Objective: 13-03 Construct confidence intervals for coefficients and test predictors for significance. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 64) Refer to this ANOVA table from a regression: Source Regression Residual Total

df 4 45 49

SS 1793.2356 2695.0996 4488.3352

MS 448.3089 59.8911

F 7.48540

Which statement is not accurate? A) The F test is significant at α = .05. B) There were 50 observations. C) There were 5 predictors. D) There would be 50 residuals. Answer: C Explanation: d.f. = (k, n − k − 1) = (4, 45), so k = 4 predictors. Difficulty: 1 Easy Topic: 13.02 Assessing Overall Fit Learning Objective: 13-02 Use the ANOVA table to perform an F test for overall significance. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

65) Refer to this ANOVA table from a regression: Source Regression Residual Total

df 4 45 49

SS 1793.2356 2695.0996 4488.3352

MS 448.3089 59.8911

F 7.48540

For this regression, the R2 is A) .3995 B) .6005 C) .6654 D) .8822 Answer: A Explanation: R2 = SSR/SST = (1793.2356)/(4488.3352) = .3995. Difficulty: 1 Easy Topic: 13.02 Assessing Overall Fit Learning Objective: 13-02 Use the ANOVA table to perform an F test for overall significance. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

66) Refer to the following regression results. The dependent variable is Abort (the number of abortions per 1000 women of childbearing age). The regression was estimated using data for the 50 U.S. states with these predictors: EdSpend = public K − 12 school expenditure per capita, Age = median age of population, Unmar = percent of total births by unmarried women, Infmor = infant mortality rate in deaths per 1000 live births. Variable Intercept EdSpend Age Unmar InfMor

Coefficients −19.244 0.0080040 0.76012 0.98629 −3.7848

Standard Error 25.387 0.0054472 0.66914 0.22132 1.0173

t-Stat −0.758 1.469 1.136 4.456 −3.720

Which statement is not supported by a two-tailed test? A) Unmar is a significant predictor at α = .01. B) EdSpend is a significant predictor at α = .20. C) Infmor is not a significant predictor at α = .05. D) Age is not a significant predictor at α = .05. Answer: C Explanation: For Infmor, tcalc = (−3.7848)/(1.0173) = −3.720, which is < t.025 = −2.014 for d.f. = 45. Difficulty: 3 Hard Topic: 13.03 Predictor Significance Learning Objective: 13-03 Construct confidence intervals for coefficients and test predictors for significance. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

67) Refer to the following correlation matrix that was part of a regression analysis. The dependent variable was Abort (the number of abortions per 1000 women of childbearing age). The regression was estimated using data for the 50 U.S. states with these predictors: EdSpend = public K − 12 school expenditure per capita, Age = median age of population, Unmar = percent of total births by unmarried women, Infmor = infant mortality rate in deaths per 1000 live births. Correlation Matrix Abort EdSpend Age Unmar InfMor

Abort 1.0000 0.2626 0.1610 0.3286 − 0.2513

EdSpend

Age

Unmar

InfMor

1.0000 − 0.0420 − 0.0949 − 0.2826

1.0000 0.0937 0.0389

1.0000 0.5239

1.0000

Using a two-tailed correlation test, which statement is not accurate? A) Age and Infmor are not significantly correlated at α = .05. B) Abort and Unmar are significantly correlated at α = .05. C) Unmar and Infmor are significantly correlated at α = .05. D) The first column of the table shows evidence of multicollinearity. Answer: D Explanation: Use rcrit = t.025/(t.0252 + n − 2)1/2 = (2.011)/(2.0112 + 50 − 2)1/2 = .2788 for d.f. = 50 − 2 = 48 for a two-tailed test at α = .05. Using this criterion, we see that two pairs of predictors, (Abort and Unmar) and (Unmar and Infmor), have correlations that differ significantly from zero. Difficulty: 2 Medium Topic: 13.07 Multicollinearity Learning Objective: 13-07 Detect multicollinearity and assess its effects. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

68) Part of a regression output is provided below. Some of the information has been omitted. Source of variation Regression Residual Total

SS 3177.17 3478.36

df 2 17 19

MS 1588.6 17.717

The approximate value of F is A) 1605.7 B) 0.9134 C) 89.66 D) impossible to calculate with the given information. Answer: C Explanation: Fcalc = MSR/MSE = (1588.6)/(17.717) = 89.66. Difficulty: 1 Easy Topic: 13.02 Assessing Overall Fit Learning Objective: 13-02 Use the ANOVA table to perform an F test for overall significance. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 69) Part of a regression output is provided below. Some of the information has been omitted. Source of variation Regression Residual Total

SS 3177.17 3478.36

df 2 17 19

MS 1588.6 17.717

The SS (residual) is A) 3177.17 B) 301.19 C) 17.71 D) impossible to determine. Answer: B Explanation: SSE = SST − SSR = 3478.36 − 3177.17 = 301.19. Difficulty: 1 Easy Topic: 13.02 Assessing Overall Fit Learning Objective: 13-02 Use the ANOVA table to perform an F test for overall significance. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

70) A realtor is trying to predict the selling price of houses in Greenville (in thousands of dollars) as a function of Size (measured in thousands of square feet) and whether or not there is a fireplace (FP is 0 if there is no fireplace, 1 if there is a fireplace). Part of the regression output is provided below, based on a sample of 20 homes. Some of the information has been omitted. Variable Intercept Size FP

Coefficients 128.93746 6.47601954

Standard Error 2.6205302 1.2072436 1.9803612

t-Stat 49.203 11.439 3.27

The estimated coefficient for Size is approximately A) 9.5 B) 13.8 C) 122.5 D) 1442.6 Answer: A Explanation: Coefficient = (t Stat)/(Std Err) = (11.439)/(1.2072436) = 9.475. Difficulty: 2 Medium Topic: 13.03 Predictor Significance Learning Objective: 13-03 Construct confidence intervals for coefficients and test predictors for significance. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

71) A realtor is trying to predict the selling price of houses in Greenville (in thousands of dollars) as a function of Size (measured in thousands of square feet) and whether or not there is a fireplace (FP is 0 if there is no fireplace, 1 if there is a fireplace). The regression output is provided below. Some of the information has been omitted. Source of variation Regression Residual Total

SS 3177.17 3478.36

df 2 17 19

MS 1588.6 17.717

How many predictors (independent variables) were used in the regression? A) 20 B) 18 C) 3 D) 2 Answer: D Explanation: d.f. = (k, n − k − 1) = (2, 17), so k = 2. Difficulty: 1 Easy Topic: 13.02 Assessing Overall Fit Learning Objective: 13-02 Use the ANOVA table to perform an F test for overall significance. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

72) A realtor is trying to predict the selling price of houses in Greenville (in thousands of dollars) as a function of Size (measured in thousands of square feet) and whether or not there is a fireplace (FP is 0 if there is no fireplace, 1 if there is a fireplace). The regression output is provided below. Some of the information has been omitted. Source of variation Regression Residual Total

SS 3177.17 3478.36

df 2 17 19

MS 1588.584 17.71713

Which of the following conclusions can be made based on the F test? A) The p-value on the F test will be very high. B) At least one of the predictors is useful in explaining Y. C) The model is of no use in predicting the selling prices of houses. D) The estimates were based on a sample of 19 houses. Answer: B Explanation: Fcalc = MSR/MSE = (1588.6)/(17.717) = 89.66, which exceeds F.05 = 3.59 for d.f. = (2, 17). Difficulty: 2 Medium Topic: 13.02 Assessing Overall Fit Learning Objective: 13-02 Use the ANOVA table to perform an F test for overall significance. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

73) A realtor is trying to predict the selling price of houses in Greenville (in thousands of dollars) as a function of Size (measured in thousands of square feet) and whether or not there is a fireplace (FP is 0 if there is no fireplace, 1 if there is a fireplace). Part of the regression output is provided below, based on a sample of 20 homes. Some of the information has been omitted. Variable Intercept Size FP

Coefficients 128.93746 6.47601954

Standard Error 2.6205302 1.2072436 1.9803612

t-Stat 49.203 11.439 3.27

P-value 8.93E-20 2.09E-09 0.004512

Which statement is supported by the regression output? A) At α = .05, FP is not a significant predictor in a two-tailed test. B) A fireplace adds around $6,476 to the selling price of the average house. C) A large house with no fireplace will sell for more than a small house with a fireplace. D) FP is a more significant predictor than Size. Answer: B Explanation: The estimated coefficient of FP is 6.476 (our home prices are in thousands). Difficulty: 2 Medium Topic: 13.03 Predictor Significance Learning Objective: 13-03 Construct confidence intervals for coefficients and test predictors for significance. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 74) A log transformation might be appropriate to alleviate which problem(s)? A) Heteroscedastic residuals B) Multicollinearity C) Autocorrelated residuals Answer: A Explanation: By reducing data magnitudes, a log transform may help equalize variances. Difficulty: 2 Medium Topic: 13.08 Regression Diagnostics Learning Objective: 13-08 Analyze residuals to check for violations of residual assumptions. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation

75) A useful guideline in determining the extent of collinearity in a multiple regression model is A) Sturges' Rule. B) Klein's Rule. C) Occam's Rule. D) Pearson's Rule. Answer: B Explanation: Klein's Rule suggests severe collinearity if any pairwise correlation between predictors exceeds the overall multiple correlation coefficient. Difficulty: 2 Medium Topic: 13.07 Multicollinearity Learning Objective: 13-07 Detect multicollinearity and assess its effects. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 76) In a multiple regression, all of the following are true regarding residuals except A) their sum always equals zero. B) they are the differences between observed and predicted values of the response variable. C) they may be used to detect multicollinearity. D) they may be used to detect heteroscedasticity. Answer: C Explanation: Residuals help in all these except to detect multicollinearity; we need VIFs for that task. Difficulty: 2 Medium Topic: 13.08 Regression Diagnostics Learning Objective: 13-08 Analyze residuals to check for violations of residual assumptions. Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation

77) The residual plot below suggests which violation(s) of regression assumptions?

A) Autocorrelation B) Heteroscedasticity C) Nonnormality D) Multicollinearity Answer: B Explanation: There seems to be a "fan-out" pattern (nonconstant residual variance). Difficulty: 2 Medium Topic: 13.08 Regression Diagnostics Learning Objective: 13-08 Analyze residuals to check for violations of residual assumptions. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 78) Which is not a standard criterion for assessing a regression model? A) Logic of causation B) Overall fit C) Degree of collinearity D) Binary predictors Answer: D Explanation: Binary predictors may be a useful part of any regression model. Difficulty: 1 Easy Topic: 13.01 Multiple Regression Learning Objective: 13-01 Use a fitted multiple regression equation to make predictions. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation

79) If the standard error is 12, the width of a quick prediction interval for Y is A) ±15 B) ±24 C) ±19 D) impossible to determine without an F table. Answer: B Explanation: Double the standard error to get an approximate width of a prediction interval for Y. Difficulty: 2 Medium Topic: 13.04 Confidence Intervals for Y Learning Objective: 13-04 Calculate the standard error and construct approximate confidence and prediction intervals for Y. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 80) Which is a characteristic of the variance inflation factor (VIF)? A) It is insignificant unless the corresponding t statistic is significant. B) It reveals collinearity rather than multicollinearity. C) It measures the degree of significance of each predictor. D) It indicates the predictor's degree of multicollinearity. Answer: D Explanation: The larger the VIFs, the more evidence we have that the predictors are multicollinear. Difficulty: 2 Medium Topic: 13.07 Multicollinearity Learning Objective: 13-07 Detect multicollinearity and assess its effects. Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation

81) Which statement best describes this regression where Y = highway miles per gallon (MPG) in 91 cars? R2 Adjusted R2 R Std. Error Source Regression Residual Total

0.499 0.444 0.707 4.019 SS 1,305.7251 1,308.3848 2,614.1099

n91 k9 Dep. Var.HwyMPG df 9 81 90

MS 145.0806 16.1529

F 8.98

p-value .0000

A) Regression is statistically significant but with large error in the MPG predictions. B) Regression is statistically significant and has quite small MPG prediction errors. C) Regression is not quite significant, but predictions should be very good. D) This is not a significant regression at any customary level of α. Answer: A Explanation: The p-value for the F test indicates significance, but the quick prediction interval is Y ± 2(4.019) or Y ± 8 MPG, which would not permit a very precise prediction. Difficulty: 2 Medium Topic: 13.02 Assessing Overall Fit Learning Objective: 13-02 Use the ANOVA table to perform an F test for overall significance. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

82) Based on these regression results, in your judgment which statement is most nearly correct (Y = highway miles per gallon in 91 cars)? R2 Adjusted R2 R Std. Error Source Regression Residual Total

0.499 0.444 0.707 4.019 SS 1,305.7251 1,308.3848 2,614.1099

n91 k9 Dep. VarHwyMPG df 9 81 90

MS 145.0806 16.1529

F 8.98

p-value .0000

A) The number of predictors is rather small. B) Some predictors are not contributing much. C) Prediction intervals would be fairly narrow in terms of MPG. D) The overall model lacks significance and/or predictive power. Answer: B Explanation: There is a gap between R2 and R2adj, which suggests some superfluous predictors were used. Difficulty: 2 Medium Topic: 13.02 Assessing Overall Fit Learning Objective: 13-02 Use the ANOVA table to perform an F test for overall significance. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

83) In the following regression, which are the three best predictors? Variables Intercept NumCyl HPMax ManTran Length Wheelbase Width RearStRm Weight Domestic

Coefficients 9.8080 − 1.6804 − 0.0369 0.2868 0.1109 − 0.0701 0.4079 − 0.0085 − 0.0025 − 1.2291

Std. Error 16.9900 0.5757 0.0140 1.2802 0.0601 0.1714 0.2922 0.2018 0.0020 1.1391

t (df = 81) 0.577 − 2.919 − 2.630 0.224 1.845 − 0.409 1.396 − 0.042 − 1.266 − 1.079

p-value .5654 .0045 .0102 .8233 .0686 .6836 .1665 .9666 .2090 .2838

A) ManTran, Wheelbase, RearStRm B) ManTran, Length, Width C) NumCyl, HPMax, Length D) Cannot be ascertained from the given information Answer: C Explanation: The absolute values of the t statistics indicate a ranking of significance. Difficulty: 2 Medium Topic: 13.03 Predictor Significance Learning Objective: 13-03 Construct confidence intervals for coefficients and test predictors for significance. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

84) In the following regression, which are the two best predictors? Variables Intercept NumCyl HPMax ManTran Length Wheelbase Width RearStRm Weight Domestic

Coefficients 9.8080 − 1.6804 − 0.0369 0.2868 0.1109 − 0.0701 0.4079 − 0.0085 − 0.0025 − 1.2291

Std. Error 16.9900 0.5757 0.0140 1.2802 0.0601 0.1714 0.2922 0.2018 0.0020 1.1391

A) NumCyl, HpMax B) Intercept, NumCyl C) NumCyl, Domestic D) ManTran, Width Answer: A Explanation: Absolute t statistics indicate a ranking, so find tcalc = (Coef)/(Std Err) for each predictor. Difficulty: 2 Medium Topic: 13.03 Predictor Significance Learning Objective: 13-03 Construct confidence intervals for coefficients and test predictors for significance. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

85) In the following regression (n = 91), which coefficients differ from zero in a two-tailed test at α = .05? Variables Intercept NumCyl HPMax ManTran Length Wheelbase Width RearStRm Weight Domestic

Coefficients 9.8080 − 1.6804 − 0.0369 0.2868 0.1109 − 0.0701 0.4079 − 0.0085 − 0.0025 − 1.2291

Confidence Interval 95% lower 95% upper − 23.9968 43.6129 − 2.8260 − 0.5349 − 0.0648 − 0.0090 − 2.2604 2.8341 − 0.0087 0.2305 − 0.4111 0.2709 − 0.1735 0.9893 − 0.4100 0.3931 − 0.0064 0.0014 − 3.4955 1.0374

A) NumCyl, HPMax B) Intercept, ManTran C) Intercept, NumCyl, Domestic D) Intercept, Domestic Answer: A Explanation: If the confidence interval includes zero, the predictor is not significant in a twotailed test. Difficulty: 2 Medium Topic: 13.03 Predictor Significance Learning Objective: 13-03 Construct confidence intervals for coefficients and test predictors for significance. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

86) Based on the following regression ANOVA table, what is the R2? Source Regression Residual Total

df 4 45 49

SS 1793.2356 2695.0996 4488.3352

MS 448.3089 59.8911

F 7.48540

A) .1336 B) .6005 C) .3995 D) Insufficient information to answer Answer: C Explanation: R2 = SSR/SST = (1793.2356)/(4488.3352) = .3995. Difficulty: 2 Medium Topic: 13.02 Assessing Overall Fit Learning Objective: 13-02 Use the ANOVA table to perform an F test for overall significance. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

87) In the following regression, which statement best describes the degree of multicollinearity? Variables Intercept NumCyl HPMax ManTran Length Wheelbase Width RearStRm Weight Domestic

Coefficients 9.8080 − 1.6804 − 0.0369 0.2868 0.1109 − 0.0701 0.4079 − 0.0085 − 0.0025 − 1.2291

Std. Error 16.9900 0.5757 0.0140 1.2802 0.0601 0.1714 0.2922 0.2018 0.0020 1.1391

t (df = 81) 0.577 − 2.919 − 2.630 0.224 1.845 − 0.409 1.396 − 0.042 − 1.266 − 1.079

p-value .5654 .0045 .0102 .8233 .0686 .6836 .1665 .9666 .2090 .2838

VIF 3.159 3.068 2.105 4.339 7.553 6.857 2.015 7.670 1.825

A) There is very little evidence of multicollinearity. B) There is much evidence of multicollinearity. C) Only NumCyl and HPMax are collinear. D) Only ManTran and RearStRm are collinear. Answer: B Explanation: Many of these predictors have large VIFs. Difficulty: 2 Medium Topic: 13.07 Multicollinearity Learning Objective: 13-07 Detect multicollinearity and assess its effects. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 88) The relationship of Y to four other variables was established as Y = 12 + 3X1 − 5X2 + 7X3 + 2X4. When X1 increases 5 units and X2 increases 3 units, while X3 and X4 remain unchanged, what change would you expect in your estimate of Y? A) Decrease by 15 B) Increase by 15 C) No change D) Increase by 5 Answer: C Explanation: The net effect is + 3ΔX1 − 5ΔX2 = 3(5) − 5(3) = 15 − 15 = 0. Difficulty: 2 Medium Topic: 13.01 Multiple Regression Learning Objective: 13-01 Use a fitted multiple regression equation to make predictions. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

89) Does the picture below show strong evidence of heteroscedasticity against the predictor Wheelbase?

A) Yes, there is strong evidence of heteroscedasticity. B) No, the plot appears mostly random against Wheelbase. C) We need a probability plot to answer. D) We need VIF statistics to answer. Answer: B Explanation: In this case, the residual scatter plot is random (no systematic difference in vertical spread). Difficulty: 2 Medium Topic: 13.08 Regression Diagnostics Learning Objective: 13-08 Analyze residuals to check for violations of residual assumptions. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 90) Which is not a correct way to find the coefficient of determination? A) SSR/SSE B) SSR/SST C) 1 − SSE/SST Answer: A Explanation: R2 = SSR/SST or R2 = 1 − SSE/SST. Difficulty: 2 Medium Topic: 13.02 Assessing Overall Fit Learning Objective: 13-02 Use the ANOVA table to perform an F test for overall significance. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 40 Copyright 2019 © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.

91) If SSR = 3600, SSE = 1200, and SST = 4800, then R2 is A) .5000 B) .7500 C) .3333 D) .2500 Answer: B Explanation: R2 = SSR/SST = 3600/4800 = .7500. Difficulty: 2 Medium Topic: 13.02 Assessing Overall Fit Learning Objective: 13-02 Use the ANOVA table to perform an F test for overall significance. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 92) Which statement is incorrect? A) Positive autocorrelation results in too many centerline crossings in the residual plot over time. B) The R2 statistic can only increase (or stay the same) when you add more predictors to a regression. C) If the F-statistic is insignificant, the t statistics for the predictors also are insignificant at the same α. D) A regression with 60 observations and 5 predictors does not violate Evans' Rule. Answer: A Explanation: Positive autocorrelation results in too few crossings of the zero point on the axis (cycles). Difficulty: 2 Medium Topic: 13.08 Regression Diagnostics Learning Objective: 13-08 Analyze residuals to check for violations of residual assumptions. Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation

93) Which statement about leverage is incorrect? A) Leverage refers to an observation's distance from the mean of X. B) If n = 40 and k = 4 predictors, a leverage statistic of .15 would indicate high leverage. C) If n = 180 and k = 3 predictors, a leverage statistic of .08 would indicate high leverage. Answer: B Explanation: 2(k + 1)/n = 2(4 + 1)/40 = .25, so hi = .15 would not indicate high leverage. Difficulty: 3 Hard Topic: 13.08 Regression Diagnostics Learning Objective: 13-10 Identify high leverage observations and their possible causes. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 94) Which statement is incorrect? A) Binary predictors shift the intercept of the fitted regression. B) If a qualitative variable has c categories, we would use only c − 1 binaries as predictors. C) A binary predictor has the same t test as any other predictor. D) If there is a binary predictor (X = 0, 1) in the model, the residuals may not sum to zero. Answer: D Explanation: Residuals always sum to zero using the OLS method. Difficulty: 2 Medium Topic: 13.05 Categorical Variables Learning Objective: 13-05 Incorporate categorical variables into a multiple regression model. Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation 95) Heteroscedasticity of residuals in regression suggests that there is A) nonconstant variation in the errors. B) multicollinearity among the predictors. C) nonnormality in the errors. D) lack of independence in successive errors. Answer: A Explanation: Heteroscedasticity is the technical name for nonconstant residual variance. Difficulty: 1 Easy Topic: 13.08 Regression Diagnostics Learning Objective: 13-08 Analyze residuals to check for violations of residual assumptions. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation

96) If you rerun a regression, omitting a predictor X5, which would be unlikely? A) The new R2 will decline if X5 was a relevant predictor. B) The new standard error will increase if X5 was a relevant predictor. C) The remaining estimated β's will change if X5 was collinear with other predictors. D) The numerator degrees of freedom for the F test will increase. Answer: D Explanation: The numerator df is the number of predictors, so omitting one would have the opposite effect. Difficulty: 3 Hard Topic: 13.02 Assessing Overall Fit Learning Objective: 13-02 Use the ANOVA table to perform an F test for overall significance. Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation 97) In a multiple regression, which is an incorrect statement about the residuals? A) They may be used to test for multicollinearity. B) They are differences between observed and estimated values of Y. C) Their sum will always equal zero. D) They may be used to detect heteroscedasticity. Answer: A Explanation: To check for multicollinearity, we would look at the VIFs or a correlation matrix. Difficulty: 2 Medium Topic: 13.07 Multicollinearity Learning Objective: 13-07 Detect multicollinearity and assess its effects. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 98) Which of the following is not a characteristic of the F distribution? A) It is a continuous distribution. B) It uses a test statistic Fcalc that can never be negative. C) Its degrees of freedom vary, depending on α. D) It is used to test for overall significance in a regression. Answer: C Explanation: In ANOVA, we use d.f. = (k, n − k − 1). The value of α does not affect d.f. Difficulty: 2 Medium Topic: 13.02 Assessing Overall Fit Learning Objective: 13-02 Use the ANOVA table to perform an F test for overall significance. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 43 Copyright 2019 © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.

99) Which of the following would be most useful in checking the normality assumption of the errors in a regression model? A) The t statistics for the coefficients. B) The F-statistic from the ANOVA table. C) The histogram of residuals. D) The VIF statistics for the predictors. Answer: C Explanation: A histogram could reveal skewness, outliers, or nonnormal shape. Difficulty: 2 Medium Topic: 13.08 Regression Diagnostics Learning Objective: 13-08 Analyze residuals to check for violations of residual assumptions. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 100) The regression equation Salary = 25,000 + 3200 YearsExperience + 1400 YearsCollege describes employee salaries at Axolotl Corporation. The standard error is 2600. John has 10 years' experience and 4 years of college. His salary is $66,500. What is John's standardized residual? A) −1.250 B) −0.240 C) +0.870 D) +1.500 Answer: D Explanation: John's predicted salary is 25,000 + 3200(10) + 1400(4) = 62,600, so his standardized residual is (66,500 − 62,600)/(2600) = 1.500 (i.e., he is somewhat overpaid according to the fitted regression). Difficulty: 3 Hard Topic: 13.08 Regression Diagnostics Learning Objective: 13-09 Identify unusual residuals and tell when they are outliers. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

101) The regression equation Salary = 28,000 + 2700 YearsExperience + 1900 YearsCollege describes employee salaries at Ramjac Corporation. The standard error is 2400. Mary has 10 years' experience and 4 years of college. Her salary is $58,350. What is Mary's standardized residual (approximately)? A) −1.150 B) +2.007 C) −1.771 D) +1.400 Answer: C Explanation: Mary's predicted salary is 28,000 + 2700(10) + 1900(4) = 62,600, so her standardized residual is (58,350 − 62,600)/(2400) = −1.771 (i.e., she is somewhat underpaid according to the fitted regression). Difficulty: 3 Hard Topic: 13.08 Regression Diagnostics Learning Objective: 13-09 Identify unusual residuals and tell when they are outliers. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 102) Which Excel function will give the p-value for overall significance if a regression has 75 observations and 5 predictors and gives an F test statistic Fcalc = 3.67? A) =F.INV(.05, 5, 75) B) =F.DIST(3.67, 4, 74) C) =F.DIST.RT(3.67, 5, 69) D) =F.DIST(.05, 4, 70) Answer: C Explanation: In pre-2010 versions of Excel, the function was =FDIST(3.67, 5, 69) for d.f. = (k, n − k − 1). Difficulty: 2 Medium Topic: 13.02 Assessing Overall Fit Learning Objective: 13-02 Use the ANOVA table to perform an F test for overall significance. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

103) The ScamMore Energy Company is attempting to predict natural gas consumption for the month of January. A random sample of 50 homes was used to fit a regression of gas usage (in CCF) using as predictors Temperature = the thermostat setting (degrees Fahrenheit) and Occupants = the number of household occupants. They obtained the following results: Variable Intercept Temperature Occupants

Coefficient 21.684 0.9142 2.244

Std Err 4.122 0.2918 1.315

In testing each coefficient for a significant difference from zero (two-tailed test at α = .10), which is the most reasonable conclusion about the predictors? A) Temperature is highly significant; Occupants is barely significant. B) Temperature is not significant; Occupants is significant. C) Temperature is less significant than Occupants. D) Temperature is significant; Occupants is not significant. Answer: A Explanation: Find the test statistic tcalc = Coef /StdErr for each predictor and compare with t.05 = 1.678 for d.f. = n − k − 1 = 50 − 2 − 1 = 47. Difficulty: 3 Hard Topic: 13.03 Predictor Significance Learning Objective: 13-03 Construct confidence intervals for coefficients and test predictors for significance. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 104) In a regression with 60 observations and 7 predictors, there will be ________ residuals. A) 60 B) 59 C) 52 D) 6 Answer: A Explanation: There are 60 residuals e1, e2, . . . , e60 (one residual for each observation). Difficulty: 1 Easy Topic: 13.02 Assessing Overall Fit Learning Objective: 13-01 Use a fitted multiple regression equation to make predictions. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

105) A regression with 72 observations and 9 predictors violates A) Evans' Rule. B) Klein's Rule. C) Doane's Rule. D) Sturges' Rule. Answer: A Explanation: Evans' Rule suggests n/k ≥ 10, but in this example n/k = 72/9 = 8. Difficulty: 2 Medium Topic: 13.02 Assessing Overall Fit Learning Objective: 13-02 Use the ANOVA table to perform an F test for overall significance. Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation 106) The F test for ANOVA in a regression model with 4 predictors and 47 observations would have how many degrees of freedom? A) (3, 44) B) (4, 46) C) (4, 42) D) (3, 43) Answer: C Explanation: d.f. = (k, n − k − 1) = (4, 47 − 4 − 1) = (4, 42). Difficulty: 2 Medium Topic: 13.02 Assessing Overall Fit Learning Objective: 13-02 Use the ANOVA table to perform an F test for overall significance. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

107) In a regression with 7 predictors and 62 observations, degrees of freedom for a t test for each coefficient would use how many degrees of freedom? A) 61 B) 60 C) 55 D) 54 Answer: D Explanation: d.f. = n − k − 1 = 62 − 7 − 1 = 54. Difficulty: 2 Medium Topic: 13.03 Predictor Significance Learning Objective: 13-03 Construct confidence intervals for coefficients and test predictors for significance. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 108) In a multiple regression with k independent variables, the standard error is A) the sum of the squared residuals. B) the sum of the residuals. C) the sum of the residuals divided by n − k − 1. D) the square root of SSE divided by n − k − 1. Answer: D Explanation: The standard error is the square root of SSE divided by its degrees of freedom. Difficulty: 2 Medium Topic: 13.04 Confidence Intervals for Y Learning Objective: 13-04 Calculate the standard error and construct approximate confidence and prediction intervals for Y. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation

109) If the standard error is 18, an approximate prediction interval width for Y is A) ±36 B) ±24 C) ±18 D) impossible to determine without an F table. Answer: A Explanation: Double the standard error to get an approximate width of a prediction interval for Y. Difficulty: 2 Medium Topic: 13.04 Confidence Intervals for Y Learning Objective: 13-04 Calculate the standard error and construct approximate confidence and prediction intervals for Y. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 110) For a given set of values for x1, x2, . . . , xk the confidence interval for the conditional mean of Y is A) narrower than the prediction interval for the individual Y value. B) wider than the prediction interval for the individual Y value. C) usually about the same as the prediction interval for the individual Y value. Answer: A Explanation: A prediction interval for an individual Y is wider than the confidence interval for the conditional mean of Y. Difficulty: 2 Medium Topic: 13.04 Confidence Intervals for Y Learning Objective: 13-04 Calculate the standard error and construct approximate confidence and prediction intervals for Y. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation

111) Which estimated multiple regression contains an interaction term? A) y = 47 − 12x1 + 8x1x2 − 5x2 B) y = 47 − 12x + 8x 2 − 5x + 25x 2 1 1 2 2 C) y = 47 − 12x 2 + 5x 2 1

Answer: A Explanation: An interaction term is the product of two individual predictors. Difficulty: 3 Hard Topic: 13.06 Tests for Nonlinearity and Interaction Learning Objective: 13-06 Perform basic tests for nonlinearity and interaction. Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation 112) Which estimated multiple regression contains an interaction term? A) y = − 92 + 6x 2 − 12x 2 1 2 B) y = 47 − 18x − 3x 2 + 6x + 36x 2 1

C) y = 88 + 11x1 + 7x1x2 + 5x2 Answer: C Explanation: An interaction term is the product of two individual predictors. Difficulty: 3 Hard Topic: 13.06 Tests for Nonlinearity and Interaction Learning Objective: 13-06 Perform basic tests for nonlinearity and interaction. Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation 113) Which estimated multiple regression allows a test for nonlinearity? A) y = 47 − 12x1 + 8x1x2 − 5x2 B) y = 47 − 12x + 8x 2 − 5x + 25x 2 1 1 2 2 y = 47 − 12x + 5x + 13x 1 2 3 C) Answer: B Explanation: Including a squared predictor allows a test for nonlinearity (quadratic). Difficulty: 3 Hard Topic: 13.06 Tests for Nonlinearity and Interaction Learning Objective: 13-06 Perform basic tests for nonlinearity and interaction. Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation

114) Which estimated multiple regression has nonlinearity tests? A) y = − 92 − 5x + 6x 2 + 18x − 12x 2 1 1 2 2 y = 47 − 18x − 3x + 6x + 36x 1 2 3 4 B) y = 88 + 11x + 7x x + 5x 1 1 2 2 C) Answer: A Explanation: Including a squared predictor allows a test for nonlinearity (quadratic). Difficulty: 3 Hard Topic: 13.06 Tests for Nonlinearity and Interaction Learning Objective: 13-06 Perform basic tests for nonlinearity and interaction. Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation 115) Simple tests for nonlinearity in a regression model can be performed by A) squaring the standard error. B) including squared predictors. C) deleting predictors one at a time. Answer: B Explanation: Including a squared predictor allows a test for (quadratic) nonlinearity. Difficulty: 2 Medium Topic: 13.06 Tests for Nonlinearity and Interaction Learning Objective: 13-06 Perform basic tests for nonlinearity and interaction. Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation

116) The regression equation Salary = 35,000 + 3500 YearsExperience + 1200 YearsCollege describes employee salaries at Streeling Research Labs. The standard error is 2500. Doris has 20 years' experience and 4 years of college. Her salary is $113,000. What is Doris's standardized residual? A) +1.28 B) −0.24 C) −1.28 D) +3.20 Answer: A Explanation: Doris's predicted salary is 35,000 + 3500(20) + 1200(4) = 109,800, so her standardized residual is (113000 − 109800)/(2500) = 1.28. She is somewhat overpaid according to the fitted regression but is not an outlier. Difficulty: 3 Hard Topic: 13.08 Regression Diagnostics Learning Objective: 13-09 Identify unusual residuals and tell when they are outliers. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 117) The regression equation Salary = 45,000 + 1500 YearsExperience + 2800 YearsCollege describes employee salaries at Terminus Fissile Labs. The standard error is 2500. Lars has 15 years' experience and 4 years of college. His salary is $70,500. If this regression is valid, we conclude that A) Lars is underpaid, but is not an outlier. B) Lars is underpaid, and his salary is an outlier. C) Lars is overpaid, but is not an outlier. D) Lars is overpaid, and his salary is an outlier. Answer: B Explanation: Lars's predicted salary is 45,000 + 1500(15) + 2800(4) = 78,700, so his standardized residual is (70,500 − 78,700)/(2500) = −3.28. Lars is underpaid according to the fitted regression, and his salary is an outlier. Difficulty: 3 Hard Topic: 13.08 Regression Diagnostics Learning Objective: 13-09 Identify unusual residuals and tell when they are outliers. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

118) In a regression with n = 50 observations and k = 3 predictors, the criterion for high leverage is A) hi ≥ .33 B) hi ≥ .25 C) hi ≥ .21 D) hi ≥ .16 Answer: D Explanation: 2(k + 1)/n = 2(3 + 1)/50 = .16, so hi = .16 or more would indicate high leverage. Difficulty: 3 Hard Topic: 13.08 Regression Diagnostics Learning Objective: 13-10 Identify high leverage observations and their possible causes. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 119) In a regression with n = 100 observations and k = 5 predictors, the criterion for high leverage is A) hi ≥ .06 B) hi ≥ .12 C) hi ≥ .18 D) hi ≥ .24 Answer: B Explanation: 2(k + 1)/n = 2(5 + 1)/100 = .12, so hi = .12 or more would indicate high leverage. Difficulty: 3 Hard Topic: 13.08 Regression Diagnostics Learning Objective: 13-10 Identify high leverage observations and their possible causes. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

120) A high leverage observation will have A) an unusual value of the observed Y. B) unusual values of one or more X values. C) a large standardized residual. D) high correlations between the X values. Answer: B Explanation: Influential observations are those with unusual values of one or more independent variables. Difficulty: 3 Hard Topic: 13.08 Regression Diagnostics Learning Objective: 13-10 Identify high leverage observations and their possible causes. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 121) An observation with extreme values in one or more independent variables (predictors) A) has reduced influence on the estimated coefficients. B) has increased influence on the estimated coefficients. C) causes an overstated estimate of the standard error. D) leads to high correlations between the X values. Answer: B Explanation: Influential observations are those with unusual values of one or more predictors. Difficulty: 3 Hard Topic: 13.08 Regression Diagnostics Learning Objective: 13-10 Identify high leverage observations and their possible causes. Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation 122) A logistic regression is appropriate when A) the dependent variable is binary (0, 1). B) an independent variable is binary (0, 1). C) the dependent variable is log(Y). D) the independent variables have the form log(X). Answer: A Explanation: A logistic regression is needed when the dependent variable is binary (0, 1). Difficulty: 2 Medium Topic: 13.09 Other Regression Topics Learning Objective: 13-11 Explain the purpose of data conditioning and stepwise regression. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 54 Copyright 2019 © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.

123) When the dependent variable is binary (0 or 1), we need A) stepwise regression. B) data transformation to improve conditioning. C) logistic regression. D) best subsets regression. Answer: C Explanation: A logistic regression is needed when the dependent variable is binary (0, 1). Difficulty: 2 Medium Topic: 13.09 Other Regression Topics Learning Objective: 13-11 Explain the purpose of data conditioning and stepwise regression. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 124) When the predictor units of measurement differ greatly in magnitude, which action might be useful? A) Stepwise regression to detect which predictors are best B) Decimal transformations to improve data conditioning C) Logistic regression to produce probability predictions D) Best subsets regression to find the best model Answer: B Explanation: Simple data decimal transformations can improve accuracy and legibility of output. Difficulty: 2 Medium Topic: 13.09 Other Regression Topics Learning Objective: 13-11 Explain the purpose of data conditioning and stepwise regression. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation

125) When we have no prior guidance on which combination of predictors is best, we might consider A) log transformations. B) decimal transformations. C) logistic regression. D) best subsets regression. Answer: D Explanation: Best subsets regression tries all possible combinations of predictors. Many statisticians are uncomfortable with this "model shopping." Difficulty: 2 Medium Topic: 13.09 Other Regression Topics Learning Objective: 13-11 Explain the purpose of data conditioning and stepwise regression. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 126) To find which predictors are most helpful in increasing R2, we might consider A) Log transformations. B) Stepwise regression. C) Logistic regression. D) Nonlinear regression. Answer: B Explanation: Stepwise regression either adds predictors one at a time in order of R2 improvement or deletes predictors one at a time to minimize R2 loss. Many statisticians are uncomfortable with this "model shopping." Difficulty: 2 Medium Topic: 13.09 Other Regression Topics Learning Objective: 13-11 Explain the purpose of data conditioning and stepwise regression. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation

127) The forward selection method of stepwise regression A) starts with all the predictors in the model and deletes the weaker ones. B) adds predictors one at a time starting with the best single predictor. C) runs all possible models and then chooses the best fitting one. D) requires nonlinear estimation using maximum likelihood. Answer: B Explanation: Stepwise regression adds predictors one at a time in order of R2 improvement. Many statisticians are uncomfortable with this "model shopping." Difficulty: 2 Medium Topic: 13.09 Other Regression Topics Learning Objective: 13-11 Explain the purpose of data conditioning and stepwise regression. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 128) The backward elimination of stepwise regression A) sometimes misses the best model for a given number of predictors. B) adds predictors one at a time starting with the best single predictor. C) runs all possible models and then chooses the best one. D) requires nonlinear estimation using maximum likelihood. Answer: A Explanation: By deleting predictors one at a time, the backward elimination method may not choose the best model for any specified number of predictors. Many statisticians are uncomfortable with this "model shopping." Difficulty: 2 Medium Topic: 13.09 Other Regression Topics Learning Objective: 13-11 Explain the purpose of data conditioning and stepwise regression. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation

129) Which is not true of the logistic regression model? A) It is nonlinear. B) It can best be fitted using the maximum likelihood method. C) Its predictions are either 0 or 1. D) It cannot yield predictions greater than 1. Answer: C Explanation: The logit model's predictions are probabilities that vary within the range 0 to 1. It is a nonlinear model that can be fitted using the maximum likelihood method. Difficulty: 2 Medium Topic: 13.10 Logistic Regression (Optional) Learning Objective: 13-12 Identify when logistic regression is appropriate and calculate predictions for a binary response variable. Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation

Applied Statistics in Business and Economics, 6e (Doane) Chapter 14 Time-Series Analysis 1) A multiplicative time-series model Y = T × C × S × I becomes an additive model if we take the logarithm of each side of the equation. Answer: TRUE Explanation: By the additive property of the log function, log(Y) = log(T) + log(C) + log(S) + log(I). Difficulty: 2 Medium Topic: 14.01 Time-Series Components Learning Objective: 14-01 Define time-series data and its components. Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation 2) A firm's income statement contains data measured over a period of time, as opposed to being measured at a point in time. Answer: TRUE Explanation: Income is a flow over time (it does not exist at a point in time). Difficulty: 1 Easy Topic: 14.01 Time-Series Components Learning Objective: 14-01 Define time-series data and its components. Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation 3) A firm's balance sheet contains data measured over a period of time, as opposed to being measured at a point in time. Answer: FALSE Explanation: We measure assets and liabilities at a given moment in time (not over an interval of time). Difficulty: 1 Easy Topic: 14.01 Time-Series Components Learning Objective: 14-01 Define time-series data and its components. Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation

4) Additive models are most appropriate over longer periods of time or when the data magnitude is growing rapidly over time. Answer: FALSE Explanation: Multiplicative models are used when data magnitudes are changing significantly. Difficulty: 1 Easy Topic: 14.01 Time-Series Components Learning Objective: 14-01 Define time-series data and its components. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 5) Multiplicative models are most appropriate over longer periods of time or when the data magnitude is growing rapidly over time. Answer: TRUE Explanation: The additive model only works for short periods or small changes in data size. Difficulty: 1 Easy Topic: 14.01 Time-Series Components Learning Objective: 14-01 Define time-series data and its components. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 6) Multiplicative models are avoided in business because they are too complicated. Answer: FALSE Explanation: Businesses generally use the multiplicative model (e.g., it is the default in Minitab). Difficulty: 1 Easy Topic: 14.01 Time-Series Components Learning Objective: 14-01 Define time-series data and its components. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 7) Cycles are usually ignored because there is no general theory to describe them. Answer: TRUE Explanation: Standard software packages have no provision to estimate cycles for this reason. Difficulty: 1 Easy Topic: 14.01 Time-Series Components Learning Objective: 14-01 Define time-series data and its components. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 2 Copyright 2019 © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.

8) Seasonality is usually ignored because there is no statistical way to describe it. Answer: FALSE Explanation: Seasonality is well understood, and common software packages can deal with it. Difficulty: 1 Easy Topic: 14.06 Seasonality Learning Objective: 14-07 Interpret seasonal factors and use them to make forecasts. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 9) Moving average models are causal models (as opposed to time-series models). Answer: FALSE Explanation: A moving average model is merely a descriptive tool to smooth data. Difficulty: 1 Easy Topic: 14.04 Moving Averages Learning Objective: 14-05 Interpret a moving average and use Excel to create it. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 10) Exponential smoothing and moving averages are especially useful for data with a clear trend. Answer: FALSE Explanation: We would need to use trend models if there is a trend (moving averages are useful for erratic data). Difficulty: 1 Easy Topic: 14.05 Exponential Smoothing Learning Objective: 14-06 Use exponential smoothing to forecast trendless data. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 11) An exponential trend can have a turning point (peak or trough). Answer: FALSE Explanation: An exponential trend either grows or declines monotonically. Difficulty: 1 Easy Topic: 14.02 Trend Forecasting Learning Objective: 14-02 Interpret a linear, exponential, or quadratic trend model. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 3 Copyright 2019 © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.

12) An exponential trend allows growth but cannot be used if the time series is declining. Answer: FALSE Explanation: The exponential trend can fit a time series that declines at a declining rate. Difficulty: 1 Easy Topic: 14.02 Trend Forecasting Learning Objective: 14-02 Interpret a linear, exponential, or quadratic trend model. Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation 13) Over the short run, exponential and linear trends may look alike. Answer: TRUE Explanation: Nonlinearity may not be apparent over a short time period. Difficulty: 1 Easy Topic: 14.02 Trend Forecasting Learning Objective: 14-02 Interpret a linear, exponential, or quadratic trend model. Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation 14) The quadratic model can never have more than one turning point (peaks or troughs). Answer: TRUE Explanation: A quadratic trend (second-order polynomial) can have only one turning point. Difficulty: 1 Easy Topic: 14.02 Trend Forecasting Learning Objective: 14-02 Interpret a linear, exponential, or quadratic trend model. Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation 15) The quadratic model's R2 is always at least as high as the linear model fitted to the same data. Answer: TRUE Explanation: Because it has an extra squared term, the quadratic R2 is at least as high as the linear R2. Difficulty: 2 Medium Topic: 14.02 Trend Forecasting Learning Objective: 14-02 Interpret a linear, exponential, or quadratic trend model. Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation

16) The linear model's R2 may exceed the quadratic model's R2 fitted to the same data. Answer: FALSE Explanation: The linear R2 cannot be higher because the quadratic model has an extra squared predictor. Difficulty: 2 Medium Topic: 14.02 Trend Forecasting Learning Objective: 14-02 Interpret a linear, exponential, or quadratic trend model. Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation 17) The shape of the fitted quadratic model yt = 544 + 62t − 3.2t2 first is declining at first, then rising. Answer: FALSE Explanation: For yt = a + bt + ct2, if c < 0, the trend must eventually decline. Difficulty: 2 Medium Topic: 14.02 Trend Forecasting Learning Objective: 14-02 Interpret a linear, exponential, or quadratic trend model. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 18) The shape of the fitted quadratic model yt = 324 − 42t − 1.3t2 is declining at first, then rising. Answer: FALSE Explanation: For yt = a + bt + ct2, if c < 0, the trend must eventually decline. In this example, the series will decline steadily because both b < 0 and c < 0. Difficulty: 2 Medium Topic: 14.02 Trend Forecasting Learning Objective: 14-02 Interpret a linear, exponential, or quadratic trend model. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

19) The shape of the fitted quadratic model yt = 823 + 72t − 5.1t2 is rising at first, then declining. Answer: TRUE Explanation: For yt = a + bt + ct2, if c < 0, the trend must eventually decline. In this example, the trend will initially rise because b > 0, but will then decline because c < 0. Difficulty: 2 Medium Topic: 14.02 Trend Forecasting Learning Objective: 14-02 Interpret a linear, exponential, or quadratic trend model. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 20) The shape of the fitted exponential model yt = 256e−.07t is always declining. Answer: TRUE Explanation: For yt = aebt the trend will decline as t increases, starting from y0 = a. In this example, the rate of decline (b = −.07) is −7 percent. Difficulty: 1 Easy Topic: 14.02 Trend Forecasting Learning Objective: 14-02 Interpret a linear, exponential, or quadratic trend model. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 21) The shape of the fitted exponential model yt = 256e−.07t is rising at first, then declining. Answer: FALSE Explanation: Because b < 0 in yt = aebt the value of yt will decline as t increases, starting from y0 = a. Difficulty: 2 Medium Topic: 14.02 Trend Forecasting Learning Objective: 14-02 Interpret a linear, exponential, or quadratic trend model. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

22) Moving averages are most useful for irregular data with no clear trend. Answer: TRUE Explanation: The moving average is most helpful when trend models fail. Difficulty: 2 Medium Topic: 14.04 Moving Averages Learning Objective: 14-05 Interpret a moving average and use Excel to create it. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 23) Moving averages are often used for making long-term forecasts (e.g., five periods ahead). Answer: FALSE Explanation: The moving average is a descriptive tool to smooth the data. (It does not make forecasts.) Difficulty: 1 Easy Topic: 14.04 Moving Averages Learning Objective: 14-05 Interpret a moving average and use Excel to create it. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 24) The smoothing constant α indicates the weight assigned to the most recent forecast. Answer: FALSE Explanation: In Ft+1 = αyt + (1 − α)Ft, the value α is the weight on recent data (not on the forecast). Difficulty: 1 Easy Topic: 14.05 Exponential Smoothing Learning Objective: 14-06 Use exponential smoothing to forecast trendless data. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 25) Increasing the smoothing constant α increases the weight assigned to the most recent observation. Answer: TRUE Explanation: In Ft+1 = αyt + (1 − α)Ft, the value α is the weight on recent data. Difficulty: 1 Easy Topic: 14.05 Exponential Smoothing Learning Objective: 14-06 Use exponential smoothing to forecast trendless data. Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation 8 Copyright 2019 © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.

26) A higher value of the smoothing constant α makes the forecast less responsive to recent data. Answer: FALSE Explanation: In Ft+1 = αyt + (1 − α)Ft, a higher value α gives more weight to recent data. Difficulty: 1 Easy Topic: 14.05 Exponential Smoothing Learning Objective: 14-06 Use exponential smoothing to forecast trendless data. Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation 27) Using the first observed data value is a common way of initializing the forecasts in the exponential smoothing model. Answer: TRUE Explanation: This method is simple and widely used. Difficulty: 1 Easy Topic: 14.05 Exponential Smoothing Learning Objective: 14-06 Use exponential smoothing to forecast trendless data. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 28) In exponential smoothing, using α = .20 will give a smoother series than using α = .05. Answer: FALSE Explanation: In Ft+1 = αyt + (1 − α)Ft, using α = .05 gives .95 weight to prior forecasts (more smoothing). Difficulty: 1 Easy Topic: 14.05 Exponential Smoothing Learning Objective: 14-06 Use exponential smoothing to forecast trendless data. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation

29) In exponential smoothing, using α = .20 will provide less smoothing than using α = .10. Answer: TRUE Explanation: In Ft+1 = αyt + (1 − α)Ft, using α = .20 would give more weight to recent data (less smoothing). Difficulty: 1 Easy Topic: 14.05 Exponential Smoothing Learning Objective: 14-06 Use exponential smoothing to forecast trendless data. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 30) Monthly seasonal factors should be adjusted so they sum to 12. Answer: TRUE Explanation: The initial estimated seasonal factors might need to be tweaked so they sum to 12. Difficulty: 1 Easy Topic: 14.06 Seasonality Learning Objective: 14-07 Interpret seasonal factors and use them to make forecasts. Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation 31) Quarterly seasonal factors will sum to 4. Answer: TRUE Explanation: The four seasonal factors must have an average of 1.00, so they must sum to 4.00. Difficulty: 1 Easy Topic: 14.06 Seasonality Learning Objective: 14-07 Interpret seasonal factors and use them to make forecasts. Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation 32) Occam's Razor says we should always choose the simplest model. Answer: FALSE Explanation: We choose the simpler of two models if the both models are adequate. But the simplest model is not preferred if it is inadequate (a more complex model may be required). The linear model, for example, is simple, but not always appropriate. Difficulty: 2 Medium Topic: 14.03 Assessing Fit Learning Objective: 14-04 Know the definitions of common fit measures. Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation 10 Copyright 2019 © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.

33) An attraction of MAPE as a measure of fit is its simple interpretation. Answer: TRUE Explanation: Mean average percent error (MAPE) is easy to understand. Difficulty: 2 Medium Topic: 14.03 Assessing Fit Learning Objective: 14-04 Know the definitions of common fit measures. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 34) The MAD measures the average absolute size of the forecast error. Answer: TRUE Explanation: Taking the mean absolute deviation (MAD) gives the size of the errors (regardless of their signs). Difficulty: 2 Medium Topic: 14.03 Assessing Fit Learning Objective: 14-04 Know the definitions of common fit measures. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 35) A centered moving average provides good forecasts when there is a strong upward trend in the data. Answer: FALSE Explanation: A trend model would be better because a moving average will lag the data when there is an upward trend. Difficulty: 2 Medium Topic: 14.04 Moving Averages Learning Objective: 14-05 Interpret a moving average and use Excel to create it. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation

36) Over long periods of time, multiplicative time-series models may be favored over additive time-series models, because the data magnitude often changes over time. Answer: TRUE Explanation: The multiplicative model is attractive because it handles changing magnitudes. Difficulty: 2 Medium Topic: 14.01 Time-Series Components Learning Objective: 14-01 Define time-series data and its components. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 37) Regression analysis can be used for forecasting monthly time-series data using a trend variable and 11 binary predictors (one for each month except omitting one month). Answer: TRUE Explanation: Use c − 1 binary predictors for c categories. (See Chapter 13 for categorical predictors.) Difficulty: 1 Easy Topic: 14.06 Seasonality Learning Objective: 14-08 Use regression with seasonal binaries to make forecasts. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 38) The shape of the fitted quadratic model yt = 248 − 84t + 2.4t2 is declining at first, then rising. Answer: TRUE Explanation: For yt = a + bt + ct2, if c > 0 the trend must eventually rise despite an early decline (b < 0). Difficulty: 2 Medium Topic: 14.02 Trend Forecasting Learning Objective: 14-02 Interpret a linear, exponential, or quadratic trend model. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

39) Averaging the first six data values is a way of initializing the forecasts in an exponential smoothing process. Answer: TRUE Explanation: This method is less likely to give a weird start than using only the first data value. Difficulty: 2 Medium Topic: 14.05 Exponential Smoothing Learning Objective: 14-06 Use exponential smoothing to forecast trendless data. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 40) The exponential model would be attractive for analyzing a growing company's revenues over time. Answer: TRUE Explanation: Many business time-series variables can be best modeled using an exponential trend. Difficulty: 2 Medium Topic: 14.02 Trend Forecasting Learning Objective: 14-02 Interpret a linear, exponential, or quadratic trend model. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 41) If Y1 = 216 and Y5 = 332, then the simple index number for period 5 is I5 = 153.7. Answer: TRUE Explanation: I5 = 100 × Y5/Y1 = 100 × 332/216 = 153.7. Difficulty: 1 Easy Topic: 14.07 Index Numbers Learning Objective: 14-09 Interpret index numbers. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

42) The time-series model Y = T × C × S × I A) is an additive model. B) is a multiplicative model. C) is an exponential model. D) is a polynomial model. Answer: B Explanation: The four components are multiplied. Difficulty: 1 Easy Topic: 14.01 Time-Series Components Learning Objective: 14-01 Define time-series data and its components. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 43) If we fit a linear trend to 10 observations on time-series data that are growing exponentially, then it is most likely that A) the fitted trend will be too high at t = 1 and t = 10. B) the fitted trend will be too low in the middle. C) the forecasts (if extrapolated) will be too low. D) the residuals will show a pattern like −−− + + + + −−−. Answer: C Explanation: The linear trend will be too low at the ends and too high in the middle. (Make a sketch.) Difficulty: 2 Medium Topic: 14.02 Trend Forecasting Learning Objective: 14-02 Interpret a linear, exponential, or quadratic trend model. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

44) The implied turning point (peak or trough) of yt = 516 − 42t +3t2 would be in which period? Hint: Use calculus to solve for the value of t that would maximize or minimize yt. A) t = 7 B) t = −7 C) t = 6 D) t = −42 Answer: A Explanation: See the quadratic templates in Figure 14.12. The signs of b and c in yt = a + bt + ct2 indicate whether it is a peak or trough. In this example, b < 0 and c > 0 so the series declines and then rises (a trough exists). To find the value of t that minimizes yt, we can take the first derivative of yt with respect to t, set the derivative equal to zero, and solve for t. This gives dy/dt = −42 + 6t = 0 which implies t = 7 (a trough because the series eventually rises). Difficulty: 3 Hard Topic: 14.02 Trend Forecasting Learning Objective: 14-02 Interpret a linear, exponential, or quadratic trend model. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 45) In the model yt = 516 − 42t + 3t2 the turning point A) is a peak. B) is a trough. C) could be either a peak or a trough. D) is neither a peak nor a trough. Answer: B Explanation: The signs of b and c in yt = a + bt + ct2 indicate whether it is a peak or trough. Here, it is a trough because the series eventually rises (the t2 term has a positive coefficient). See the quadratic templates in Figure 14.12. Difficulty: 2 Medium Topic: 14.02 Trend Forecasting Learning Objective: 14-02 Interpret a linear, exponential, or quadratic trend model. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

46) A computer analysis reveals that the best-fitting trend model is yt = 4.12e0.987t. The trend was fitted using year-end common stock prices for Melodic Kortholt Outlet for the last six years. The R2 is .8571. Which conclusion is not supportable? A) The absolute annual growth (in dollars per share) is increasing. B) Few investments could match the astounding growth rate. C) At the end of year 3, the stock price would be nearly $80. D) The exponential model is inappropriate for financial data. Answer: D Explanation: The growth rate is 98.7 percent per year (increasing at an increasing rate) and y3 = 79.59. Difficulty: 3 Hard Topic: 14.02 Trend Forecasting Learning Objective: 14-02 Interpret a linear, exponential, or quadratic trend model. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 47) Suppose the estimated quadratic model yt = 500 + 20t − t2 is the best-fitting trend of sales of XYZ Inc. using data for the past 20 years (t = 1, 2, . . . , 20). Which statement is incorrect? A) Sales are increasing by about 20 units per year. B) The turning point would be in period 10. C) Latest year sales are no better than in year zero. D) The trend was higher in year 10 than in year 20. Answer: A Explanation: We can just try a few different values of t (such as t = 0 and t = 10 and t = 20) to see what happens to yt. Or we can take the first derivative of yt with respect to t, set the derivative equal to zero, and solve for t to get dy/dt = 20 − 2t = 0 which implies t = 10 (a peak). When t = 10 we get yt = 500 + 20t − t2 = 500 + 20(10) − 102 = 600. Similarly, if we insert t = 0 and t = 20, we see that y0 = y20 = 500 (both y0 and y20 are lower than y10). Difficulty: 3 Hard Topic: 14.02 Trend Forecasting Learning Objective: 14-03 Fit any common trend model and use it to make forecasts. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

48) Which statement is most defensible regarding the time series shown below?

A) There appears to be strong seasonality. B) There appears to be a six-month seasonal cycle. C) The trend appears to be exponential. D) The quadratic trend would be required. Answer: A Explanation: A clear monthly pattern exists over 12 months, and the trend appears to be linear. Difficulty: 2 Medium Topic: 14.06 Seasonality Learning Objective: 14-07 Interpret seasonal factors and use them to make forecasts. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

49) If the trend model yt = a + bt + ct2 is fitted to a time series, we would get A) R2 that could be lower than the linear model. B) R2 that could be either higher or lower than the linear model. C) R2 that is at least as high as the linear model. D) no R2 for this type of model because it is nonlinear. Answer: C Explanation: The quadratic R2 always is at least as high as the linear R2 because of the extra t2 predictor. Difficulty: 2 Medium Topic: 14.02 Trend Forecasting Learning Objective: 14-02 Interpret a linear, exponential, or quadratic trend model. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 50) The fitted annual sales trend is yt = 187e−.047t. The sales forecast for year 5 would be A) 236.9 B) 178.7 C) 175.3 D) 147.8 Answer: D Explanation: y5 = 187e−.047(5) = 187(0.79057) = 147.83 (the growth rate is negative). Difficulty: 2 Medium Topic: 14.02 Trend Forecasting Learning Objective: 14-03 Fit any common trend model and use it to make forecasts. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 51) The fitted annual sales trend is yt = 227e.037t. The values of yt are A) rising by an increasing amount each period. B) rising by a declining amount each period. C) declining by a declining amount each period. D) declining by an increasing amount each period. Answer: A Explanation: The growth rate is positive (3.7 percent) so the series increases at an increasing rate. Difficulty: 2 Medium Topic: 14.02 Trend Forecasting Learning Objective: 14-02 Interpret a linear, exponential, or quadratic trend model. Bloom's: Apply AACSB: Analytical Thinking 18 Copyright 2019 © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.

Accessibility: Keyboard Navigation 52) If a fitted trend equation is yt = 120 − 40t + 2.5t2, then the turning point will be A) a peak in period 40. B) a trough in period 5. C) a peak in period 4. D) a trough in period 8. Answer: D Explanation: See the quadratic templates in Figure 14.12. The signs of b and c in yt = a + bt + ct2 indicate whether it is a peak or trough. In this example, b < 0 and c > 0, so the series first declines and then rises (a trough exists). To find the value of t that minimizes yt, we can take the first derivative of yt with respect to t, set the derivative equal to zero, and solve for t. This gives dy/dt = −40 + 5t = 0, which implies t = 8 (a trough because the coefficient of t2 is positive). Difficulty: 2 Medium Topic: 14.02 Trend Forecasting Learning Objective: 14-02 Interpret a linear, exponential, or quadratic trend model. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 53) If the fitted annual trend for a stock price is yt = 27e0.213t, then A) the slow growth rate is not very attractive. B) the stock price seems to be approaching an asymptote. C) the stock is probably undervalued in today's markets. D) the stock would more than double every four years. Answer: D Explanation: The growth rate is 21.3 percent per year (nice) and y5/y1 = (27e1.065)/(27e0.213) = 2.34. Difficulty: 3 Hard Topic: 14.02 Trend Forecasting Learning Objective: 14-02 Interpret a linear, exponential, or quadratic trend model. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

54) What is the approximate slope of a linear trend for the value of Bob's beer can collection?

A) −25 B) −50 C) −15 D) −10 Answer: D Explanation: Visually the estimated slope is Δy/Δx = (50 − 250)/(2009 − 1989) = −200/20 = −10. Difficulty: 2 Medium Topic: 14.02 Trend Forecasting Learning Objective: 14-03 Fit any common trend model and use it to make forecasts. Bloom's: Analyze AACSB: Analytical Thinking Accessibility: Keyboard Navigation

55) Which trend would you choose to forecast the 2013 value of Bob's beer can collection?

A) Linear model is preferred. B) Exponential model is preferred. C) They are equivalent. Answer: B Explanation: The exponential model gives more credible forecasts, with only slightly lower R2 (see Trend-Fitting Criteria). Difficulty: 3 Hard Topic: 14.02 Trend Forecasting Learning Objective: 14-03 Fit any common trend model and use it to make forecasts. Bloom's: Evaluate AACSB: Reflective Thinking Accessibility: Keyboard Navigation

56) Which trend would you choose to forecast the 2013 value of Bob's beer can collection?

A) Exponential model is preferred. B) Quadratic model is preferred. C) They are equivalent. Answer: A Explanation: Despite its higher R2, the quadratic model gives unbelievable forecasts and is more complex. (Occam's Razor favors the simpler exponential model, which has only two parameters and is easy to interpret.) Difficulty: 3 Hard Topic: 14.02 Trend Forecasting Learning Objective: 14-03 Fit any common trend model and use it to make forecasts. Bloom's: Apply AACSB: Reflective Thinking Accessibility: Keyboard Navigation

57) Might it be acceptable to use exponential smoothing to forecast 2010 for Bob's data?

A) No, under no circumstances. B) Yes, if you use only the most recent 10 years. C) Yes, if you need a long-term forecast (e.g., to 2020). D) No, because the data appear to be erratic. Answer: B Explanation: The early years suggest a downward trend, but smoothing would work well for recent data. Difficulty: 2 Medium Topic: 14.05 Exponential Smoothing Learning Objective: 14-06 Use exponential smoothing to forecast trendless data. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

58) Which trend would you choose to forecast the number of tractors sold in 2010?

A) Linear model is best. B) Polynomial model is best. C) Either gives equivalent forecasts. Answer: A Explanation: Avoid higher-order polynomials. (Extrapolate this one and you would get wild forecasts.) Difficulty: 2 Medium Topic: 14.02 Trend Forecasting Learning Objective: 14-03 Fit any common trend model and use it to make forecasts. Bloom's: Evaluate AACSB: Analytical Thinking Accessibility: Keyboard Navigation

59) Which trend would you choose to forecast the number of water damage claims in 2010?

A) Exponential model is best. B) Polynomial model is best. C) Models give equivalent forecasts. Answer: A Explanation: Avoid higher-order polynomials. (Extrapolate this one and you would get wild forecasts.) Difficulty: 2 Medium Topic: 14.02 Trend Forecasting Learning Objective: 14-03 Fit any common trend model and use it to make forecasts. Bloom's: Evaluate AACSB: Analytical Thinking Accessibility: Keyboard Navigation 60) A quadratic trend equation yt = 900 + 80t − 5t2 was fitted to a company's sales. This result implies that the sales trend A) hit a peak in period 8. B) hit a peak in period 5. C) hit a trough in period 8. D) hit a trough in period 5. Answer: A Explanation: See the quadratic templates in Figure 14.12. The signs of b and c in yt = a + bt + ct2 indicate whether it is a peak or trough. In this example, b > 0 and c < 0, so the series rises and then declines (a peak exists). To find the value of t that maximizes yt, we can take the first derivative of yt with respect to t, set the derivative equal to zero, and solve for t. This gives dy/dt = 80 − 10t = 0 which implies t = 8 (a peak because the coefficient of t2 is negative). Difficulty: 3 Hard Topic: 14.02 Trend Forecasting Learning Objective: 14-02 Interpret a linear, exponential, or quadratic trend model. Bloom's: Analyze AACSB: Analytical Thinking 25 Copyright 2019 © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.

Accessibility: Keyboard Navigation 61) If we fit a linear trend to data that are growing exponentially, which is least likely? A) The fitted trend will be too low at the beginning and end. B) The fitted trend will be too high in the middle. C) The forecasts (if extrapolated) will be too low. D) The fit will be poor to the most recent data. Answer: D Explanation: When data are growing exponentially, a linear trend would give forecasts that eventually are too low (poor fit to recent data). The linear trend would also be too low at the beginning and too high in the middle. (Make a sketch.) Difficulty: 2 Medium Topic: 14.02 Trend Forecasting Learning Objective: 14-02 Interpret a linear, exponential, or quadratic trend model. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 62) For the fitted time-series trend model yt = 9.23e−0.0867t, it is correct to say that A) the series is growing by 9.23 percent. B) the series is growing by 8.67 percent. C) the model probably shows a low R2. D) the time series would be declining. Answer: D Explanation: The negative coefficient in yt = aebt indicates a rate of decline of 8.67 percent (b < 0). Difficulty: 2 Medium Topic: 14.02 Trend Forecasting Learning Objective: 14-02 Interpret a linear, exponential, or quadratic trend model. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

63) If a trend is given by yt = aebt, then A) the fitted trend value in period 0 is b. B) the trend is exponentially increasing. C) the trend is exponentially decreasing. D) the trend is increasing if b > 0 and decreasing if b < 0. Answer: D Explanation: The parameter b is the compound rate of growth (if b > 0) or decline (if b < 0). Difficulty: 2 Medium Topic: 14.02 Trend Forecasting Learning Objective: 14-02 Interpret a linear, exponential, or quadratic trend model. Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation 64) Suppose you fit a (linear or nonlinear) trend regression to a monthly time series and discover that the R2 is only 18 percent. The best conclusion is that A) adding seasonal factors might make things worse. B) fitting a seasonal component could raise the R2. C) there is no seasonality in the time series. D) the overall trend is probably negative. Answer: B Explanation: Seasonal factors might help. (They won't reduce the R2.) Difficulty: 2 Medium Topic: 14.06 Seasonality Learning Objective: 14-07 Interpret seasonal factors and use them to make forecasts. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

65) A trend line has been fitted to a company's annual sales. The trend is given by yt = 50 + 5t, where t is the time index (t = 1, 2, . . . , n) and yt is annual sales (in millions of dollars). The implication of this trend line is that sales are expected to increase A) by $5 million every year. B) by exactly 5 percent every year. C) by an average of 5 percent per year. D) by an average of $5 million per year. Answer: D Explanation: It is a simple linear model with slope dy/dt = 5. Difficulty: 1 Easy Topic: 14.02 Trend Forecasting Learning Objective: 14-02 Interpret a linear, exponential, or quadratic trend model. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 66) Consider the following linear trend equation of an industry's sales: yt = 120 + 12t, where t is measured in years and sales are measured in millions of dollars. Which is the most reasonable conclusion? A) We would forecast that sales will increase $12 million in the next year. B) We would forecast that sales will increase 12 percent in the next year. C) On the average, sales will increase 12/(120 + 12 × 10) = 0.05, or 5 percent next year. D) The year-to-year change will depend on the value of t. Answer: A Explanation: It is a simple linear model with slope dy/dt = 12. Sales are in millions so 12 represents $12 million. Difficulty: 2 Medium Topic: 14.02 Trend Forecasting Learning Objective: 14-03 Fit any common trend model and use it to make forecasts. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

67) In a multiplicative model, if the seasonal factor is 1.15 for a particular season, then we expect that the time series in that season would be (other things being equal) A) 115 percent above the trend. B) 115 percent below the trend. C) 15 percent above the trend. D) 15 percent below the trend. Answer: C Explanation: If S = 1.15 in Y = T × C × S × I, then Y would increase by 15 percent. Difficulty: 2 Medium Topic: 14.06 Seasonality Learning Objective: 14-07 Interpret seasonal factors and use them to make forecasts. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 68) A quadratic trend equation was estimated from monthly sales of trucks in the United States from July 2006 to July 2011. The estimated trend is yt = 106 + 1.03t + 0.048t2, where yt units are in thousands. From this trend, how many trucks would be sold in July 2012? A) About 308,419 B) About 524,889 C) About 436,982 D) About 223,831 Answer: C Explanation: Using t = 73 for July 2012, yt = 106 + 1.03(73) + 0.048(73)2 = 436.982. Difficulty: 3 Hard Topic: 14.02 Trend Forecasting Learning Objective: 14-03 Fit any common trend model and use it to make forecasts. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

69) A quadratic trend equation yt = 900 + 30t − 2.5t2 was fitted to a company's sales. This result implies that the sales trend A) hit a maximum in period t = 30. B) hit a maximum in period t = 6. C) hit a minimum in period t = 5. D) hit a minimum in period t = 6. Answer: B Explanation: See the quadratic templates in Figure 14.12. The signs of b and c in yt = a + bt + ct2 indicate whether it is a peak or trough. In this example, b > 0 and c < 0, so the series rises and then declines (a peak exists). To find the value of t that maximizes yt, we can take the first derivative of yt with respect to t, set the derivative equal to zero, and solve for t. dy/dt = 30 − 5t = 0 implies t = 6 (a peak because the coefficient of t2 is negative). Difficulty: 3 Hard Topic: 14.02 Trend Forecasting Learning Objective: 14-03 Fit any common trend model and use it to make forecasts. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 70) The fitted sales trend over the last 12 years is yt = 14.7e0.063t. We can say that A) sales grew (on average) at 63 percent per annum. B) sales in year 10 would be about 39. C) a continuously compounded model was used. D) sales in year 1 were 14.7. Answer: C Explanation: The compound growth rate is 6.3 percent (b > 0) and y10 = 14.7e0.063(10) = 27.6. Difficulty: 2 Medium Topic: 14.02 Trend Forecasting Learning Objective: 14-02 Interpret a linear, exponential, or quadratic trend model. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

71) Which data would be measured over an interval of time as opposed to at a point in time? A) Toyota's accounts receivable on December 31, 2017. B) Costco's sales for fiscal year 2017. C) The Canadian unemployment rate on December 1, 2017. D) The closing price of Wal-Mart's stock last Friday. Answer: B Explanation: Sales are a flow of dollars over a period of time (income statement). Difficulty: 1 Easy Topic: 14.01 Time-Series Components Learning Objective: 14-01 Define time-series data and its components. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 72) Which data would be measured over an interval of time as opposed to at a point in time? A) Toyota's total revenue for fiscal 2017. B) Toyota's salaried employee head count at the end of June 2017. C) Toyota's short-term indebtedness at the end of fiscal 2017. D) Toyota's inventory of unsold vehicles on December 31, 2017. Answer: A Explanation: Revenue is a flow of dollars over a period of time (income statement). Difficulty: 1 Easy Topic: 14.01 Time-Series Components Learning Objective: 14-01 Define time-series data and its components. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 73) Which is a characteristic of an additive (as opposed to multiplicative) time-series model? A) It is appropriate for financial data over a longer time period. B) It is appropriate for rapidly growing financial data. C) It is appropriate for data that vary by an order of magnitude. D) It is appropriate for short-term data with a steady dollar growth. Answer: D Explanation: A linear model may suffice in the short term if magnitudes do not change sharply. Difficulty: 2 Medium Topic: 14.01 Time-Series Components Learning Objective: 14-01 Define time-series data and its components. Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation

74) If yt = 50e0.07t, which forecast for period 10 is correct? A) 100.7 B) 90.7 C) 80.7 D) 70.7 Answer: A Explanation: y10 = 50e0.07(10) = 100.69. Difficulty: 2 Medium Topic: 14.02 Trend Forecasting Learning Objective: 14-03 Fit any common trend model and use it to make forecasts. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 75) If yt = 544e0.07t, which forecast for period 7 is correct? A) 989 B) 888 C) 678 D) 589 Answer: B Explanation: y7 = 544e0.07(7) = 887.98. Difficulty: 2 Medium Topic: 14.02 Trend Forecasting Learning Objective: 14-03 Fit any common trend model and use it to make forecasts. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 76) If yt = 256e−0.07t, which forecast for period 6 is correct? A) 390 B) 203 C) 179 D) 168 Answer: D Explanation: y6 = 256e−0.07(6) = 168.2. Difficulty: 2 Medium Topic: 14.02 Trend Forecasting Learning Objective: 14-03 Fit any common trend model and use it to make forecasts. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 32 Copyright 2019 © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.

77) MegaStat's seasonal factor of 1.073 applied to a monthly trend forecast of 125.820 would give which seasonally adjusted forecast? A) 135.005 B) 126.893 C) 137.044 D) 124.228 Answer: A Explanation: Trend × Seasonal = 125.820 × 1.073 = 135.005 (7.3 percent above trend). Difficulty: 1 Easy Topic: 14.06 Seasonality Learning Objective: 14-07 Interpret seasonal factors and use them to make forecasts. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 78) Which statement is correct regarding forecasting using an exponential smoothing model? A) A model with low α is more responsive to new data than a model with high α. B) Increasing α will typically increase the forecast accuracy. C) As α increases, more weight is put on recent data. D) When data have an upward trend, the forecasts will have an upward bias. Answer: C Explanation: Larger α gives more weight to yt in Ft+1 = αyt + (1 − α)Ft. Difficulty: 2 Medium Topic: 14.05 Exponential Smoothing Learning Objective: 14-06 Use exponential smoothing to forecast trendless data. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 79) Which of the following best describes the decomposition modeling approach to forecasting? A) The values of certain variables are used to predict the values of others. B) Series are separated into trend, seasonal, irregular, and cyclical components. C) Forecasts are based on an average of recent values. D) Sophisticated models such as regression with binaries are preferred. Answer: B Explanation: Decomposition begins with the view that there are four components (although cycles generally are ignored). Difficulty: 1 Easy Topic: 14.01 Time-Series Components Learning Objective: 14-01 Define time-series data and its components. Bloom's: Understand AACSB: Analytical Thinking 33 Copyright 2019 © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.

Accessibility: Keyboard Navigation 80) Which of the following is the least useful time-series forecasting model when there is a strong upward trend in the data? A) Estimated exponential trend model. B) A five-period centered moving average. C) Exponential smoothing with a high α. D) Regression model with trend and seasonal binaries. Answer: B Explanation: Moving averages will lag when there is a strong upward trend. Difficulty: 2 Medium Topic: 14.04 Moving Averages Learning Objective: 14-05 Interpret a moving average and use Excel to create it. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 81) Which model assumes a constant percentage rate of growth? A) Linear B) Quadratic C) Cubic D) Exponential Answer: D Explanation: The exponent b in yt = aebt is the constant compound growth rate. Difficulty: 1 Easy Topic: 14.02 Trend Forecasting Learning Objective: 14-02 Interpret a linear, exponential, or quadratic trend model. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 82) The compound growth rate in the fitted trend equation yt = 228e−.0982t is A) 98.2 percent. B) −9.82 percent. C) 198.2 percent. D) 228 percent. Answer: B Explanation: The negative coefficient b in yt = aebt indicates a 9.8 percent rate of decline. Difficulty: 2 Medium Topic: 14.02 Trend Forecasting Learning Objective: 14-02 Interpret a linear, exponential, or quadratic trend model. Bloom's: Apply 34 Copyright 2019 © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.

AACSB: Analytical Thinking Accessibility: Keyboard Navigation 83) Which statement is most nearly correct regarding time-series trend models? A) The quadratic model is best for a series that is growing by a constant percentage. B) A linear model may yield a higher R2 than a quadratic model. C) The exponential model is rarely used for financial data. D) The exponential model would be linear if we take the natural log of yt. Answer: D Explanation: Log of a product is the sum of the logs, so ln(aebt) = ln(a) + ln(ebt) = ln(a) + bt = a1 + bt. Difficulty: 1 Easy Topic: 14.02 Trend Forecasting Learning Objective: 14-02 Interpret a linear, exponential, or quadratic trend model. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 84) Which is a time series? A) Last year's GDP in 27 developing countries. B) Year-end unemployment rates in the United States, 2000–2010. C) Passenger miles flown by each of 10 major airlines in 2010. D) The number of "hits" on each of 10 websites yesterday. Answer: B Explanation: A time series is measured over t = 1, 2, . . . , n periods. Difficulty: 1 Easy Topic: 14.01 Time-Series Components Learning Objective: 14-01 Define time-series data and its components. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 85) Which is not a component of a time series? A) Trend B) Seasonal C) Irregular D) Linearity Answer: D Explanation: Linearity is not a component. Difficulty: 1 Easy Topic: 14.01 Time-Series Components Learning Objective: 14-01 Define time-series data and its components. Bloom's: Remember 35 Copyright 2019 © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.

AACSB: Analytical Thinking Accessibility: Keyboard Navigation 86) The "up-and-down" component of a time series that represents periods of prosperity followed by recession over extended periods of time longer than one year is called A) trend variation. B) seasonal variation. C) cyclical variation. D) irregular variation. Answer: C Explanation: Regular variation beyond one year would be a cycle. Difficulty: 1 Easy Topic: 14.01 Time-Series Components Learning Objective: 14-01 Define time-series data and its components. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 87) Fluctuations caused by strikes and floods are A) cyclical fluctuations. B) irregular fluctuations. C) residual fluctuations. D) seasonal fluctuations. Answer: B Explanation: Unpredictable events in Y = T × C × S × I cannot be modeled as T, C, or S, so they must be I. Difficulty: 2 Medium Topic: 14.01 Time-Series Components Learning Objective: 14-01 Define time-series data and its components. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation

88) The four components of a time series are which of the following? A) Cycle, season, month, day. B) Month, cycle, seasonal, irregular. C) Cycle, seasonal, irregular, regular. D) Seasonal, cycle, irregular, trend. Answer: D Explanation: Y = Trend × Cycle × Seasonal × Irregular. Difficulty: 1 Easy Topic: 14.01 Time-Series Components Learning Objective: 14-01 Define time-series data and its components. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 89) Which measure of fit is measured in the same units as Y? A) MAPE B) MAD C) R2 D) MSD Answer: B Explanation: Absolute deviations from the mean are simply averaged, so MAD is in dollars (or whatever). Difficulty: 2 Medium Topic: 14.03 Assessing Fit Learning Objective: 14-04 Know the definitions of common fit measures. Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation 90) The fitted annual sales trend is Yt = 187.3e−.047t. On average, sales are A) rising by an increasing absolute amount each year. B) rising by a declining absolute amount each year. C) falling by a declining absolute amount each year. D) falling by an increasing absolute amount each year. Answer: C Explanation: There is a 4.7 percent annual decline (but less each year in dollars). Difficulty: 3 Hard Topic: 14.02 Trend Forecasting Learning Objective: 14-02 Interpret a linear, exponential, or quadratic trend model. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 37 Copyright 2019 © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.

91) If Y1 = 116 and Y7 = 255, which is the simple index number for period 7 (denoted I7)? A) 215.3 B) 219.8 C) 222.7 D) 242.4 Answer: B Explanation: I7 = 100 × Y7/Y1 = 100 × 255/116 = 219.82. Difficulty: 1 Easy Topic: 14.07 Index Numbers Learning Objective: 14-09 Interpret index numbers. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 92) Concerning a seasonal index for monthly data, which statement is incorrect? A) A multiplicative index value of 1.000 indicates no seasonal deviation from trend. B) Additive indexes are adjusted so they always sum to 12. C) To make forecasts, we multiply the trend by each month's seasonal index. D) Seasonal indexes are obtained by the process called decomposition of a time series. Answer: B Explanation: Additive seasonal indexes are in the same units as yt, so they need not add to anything specific. Difficulty: 3 Hard Topic: 14.06 Seasonality Learning Objective: 14-07 Interpret seasonal factors and use them to make forecasts. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 93) Which statement is correct for a simple index number? A) For the base year, the index is set to 0.000. B) We cannot use index numbers to compare two time series measured in different units. C) The simple relative index for period t = 5 is calculated as Y5/Y1. D) Weighted index numbers have few practical applications because of their complexity. Answer: C Explanation: The base period index would be Y1/Y1 = 1.000. Difficulty: 2 Medium Topic: 14.07 Index Numbers Learning Objective: 14-09 Interpret index numbers. Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation 38 Copyright 2019 © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.

94) Which is a time series? A) The M1 component of the money supply for the United States (n = 20 quarters). B) The unemployment rates for the U.S. states (n = 50 states). C) The gross domestic product for E.U. members (n = 27 nations). D) The inflation rate for housing in U.S. metropolitan areas (n = 46 cities). Answer: A Explanation: A time series is measured over t = 1, 2, . . . , n periods of time (not cities, etc.). Difficulty: 1 Easy Topic: 14.01 Time-Series Components Learning Objective: 14-01 Define time-series data and its components. Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation 95) To initialize the forecasts in an exponential smoothing process, it is acceptable to A) use the average of the first six observed data values. B) use the most recent data value in the observed data. C) apply the smoothing constant α to the mean of the data. D) ask a panel of experts to make a guess at the initial forecast. Answer: A Explanation: To get started, it is common to set F1 equal to the average of periods 1 through 6. Difficulty: 1 Easy Topic: 14.05 Exponential Smoothing Learning Objective: 14-06 Use exponential smoothing to forecast trendless data. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 96) If a fitted trend equation is yt = 220 − 40t + 2.5t2, which is the forecast for period 6? A) 225 B) 70 C) 304 D) 110 Answer: B Explanation: yt = 220 − 40(6) + 2.5(6)2 = 70. Difficulty: 2 Medium Topic: 14.02 Trend Forecasting Learning Objective: 14-03 Fit any common trend model and use it to make forecasts. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 39 Copyright 2019 © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.

97) If a fitted trend equation is yt = 184e−.047t, which is the forecast for period 4? A) 195 B) 222 C) 152 D) 138 Answer: C Explanation: y4 = 184e−.047(4) = 184(.8286) = 152.47 (negative growth rate, so it is a declining series). Difficulty: 2 Medium Topic: 14.02 Trend Forecasting Learning Objective: 14-03 Fit any common trend model and use it to make forecasts. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 98) If a fitted trend equation is yt = 816e0.065t, which is the forecast for period 7? A) 1286 B) 895 C) 944 D) 1018 Answer: A Explanation: y7 = 816e0.065(7) = 816(1.57617) = 1286.16. Difficulty: 2 Medium Topic: 14.02 Trend Forecasting Learning Objective: 14-03 Fit any common trend model and use it to make forecasts. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

99) If a fitted trend equation is yt = 227e−.098t, which is the forecast for period 5? A) 372 B) −87 C) −216 D) 139 Answer: D Explanation: y5 = 227e−.098(5) = 227(.61263) = 139.07 (negative growth rate, so it is a declining series). Difficulty: 2 Medium Topic: 14.02 Trend Forecasting Learning Objective: 14-03 Fit any common trend model and use it to make forecasts. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 100) Using exponential smoothing, if Ft = 33, yt = 41, and α = .20, what is the new forecast Ft+1? A) 34.6 B) 36.2 C) 38.1 D) 40.2 Answer: A Explanation: Ft+1 = αyt + (1 − α)Ft = (.20)(41) + (.80)(33) = 34.6. Difficulty: 2 Medium Topic: 14.05 Exponential Smoothing Learning Objective: 14-06 Use exponential smoothing to forecast trendless data. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 101) Using exponential smoothing, if Ft = 12, yt = 15, and α = .10, what is the new forecast Ft+1? A) 14.1 B) 12.6 C) 12.3 D) 13.7 Answer: C Explanation: Ft+1 = αyt + (1 − α)Ft = (.10)(15) + (.90)(12) = 12.3. Difficulty: 2 Medium Topic: 14.05 Exponential Smoothing Learning Objective: 14-06 Use exponential smoothing to forecast trendless data. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 41 Copyright 2019 © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.

102) Using exponential smoothing, if Ft = 220, yt = 240, and α = .20, what is the new forecast Ft+1? A) 217 B) 224 C) 232 D) 243 Answer: B Explanation: Ft+1 = αyt + (1 − α)Ft = (.20)(240) + (.80)(220) = 224. Difficulty: 2 Medium Topic: 14.05 Exponential Smoothing Learning Objective: 14-06 Use exponential smoothing to forecast trendless data. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 103) Which of the following measures of fit is unit free? A) MAD (mean absolute deviation) B) SE (standard error) C) MSD (mean squared deviation) D) R2 (coefficient of determination) Answer: D Explanation: The R2 statistic has no units of measure such as pounds or dollars. Difficulty: 2 Medium Topic: 14.03 Assessing Fit Learning Objective: 14-04 Know the definitions of common fit measures. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 104) Which of the following measures of fit is expressed in the same units as yt? A) SE (standard error) B) MAPE (mean absolute percent error) C) MSD (mean squared error) D) R2 (coefficient of determination) Answer: A Explanation: The standard error is the square root of a sum of squares, so same units as yt. Difficulty: 2 Medium Topic: 14.03 Assessing Fit Learning Objective: 14-04 Know the definitions of common fit measures. Bloom's: Remember AACSB: Analytical Thinking 42 Copyright 2019 © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.

Accessibility: Keyboard Navigation 105) Which of the following measures of fit is expressed in percent? A) SE (standard error) B) MAPE C) MSD D) MAD Answer: B Explanation: MAPE is a mean percent error. Difficulty: 2 Medium Topic: 14.03 Assessing Fit Learning Objective: 14-04 Know the definitions of common fit measures. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 106) For these data, what is the three-period centered moving average for period 4? t yt

1 22

2 33

3 27

4 34

5 38

6 26

7 28

8 35

9 41

A) 33.33 B) 31.33 C) 33.00 D) 34.00 Answer: C Explanation: (27 + 34 + 38)/3 = 33. Difficulty: 1 Easy Topic: 14.04 Moving Averages Learning Objective: 14-05 Interpret a moving average and use Excel to create it. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

107) For these data, what is the three-period trailing moving average for period 6? t yt

1 22

2 33

3 27

4 34

5 38

6 26

7 28

8 35

9 41

A) 32.67 B) 31.33 C) 33.00 D) 34.67 Answer: A Explanation: (26 + 38 + 34)/3 = 32.67. Difficulty: 2 Medium Topic: 14.04 Moving Averages Learning Objective: 14-05 Interpret a moving average and use Excel to create it. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 108) Use the estimated regression equation yt = 448 + 12t + 18 Qtr1 − 26 Qtr2 + 3 Qtr3 to make a forecast for period 13. The regression model has three quarterly binaries (0 or 1). The model was fitted to 12 periods of quarterly data, starting with the first quarter. A) 588 B) 476 C) 622 D) No forecast is possible without knowing R2. Answer: C Explanation: If Qtr1 = 1, Qtr2 = 0, Qtr3 = 0, then y13 = 448 + 12(13) + 18(1) − 26(0) + 3(0) = 622. Difficulty: 3 Hard Topic: 14.06 Seasonality Learning Objective: 14-08 Use regression with seasonal binaries to make forecasts. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

109) The estimated regression equation is yt = 448 + 12t + 18 Qtr1 − 26 Qtr2 + 3 Qtr3. The regression model has three quarterly binaries. The model was fitted to 12 periods of quarterly data starting with the first quarter. Why is there no fourth quarterly binary for Qtr4? A) Because the researcher made a mistake. (We must include a binary for each quarter.) B) Because it is unnecessary. (Its value is implied by the other three binaries.) C) Because the fourth quarter binary is assumed to be the same as the first quarter. D) Because there is no seasonality in the fourth quarter in most time series. Answer: B Explanation: Since Qtr1 + Qtr2 + Qtr3 + Qtr4 = 1, we know that Qtr4 = 1 − Qtr1 − Qtr2 − Qtr3. Difficulty: 3 Hard Topic: 14.06 Seasonality Learning Objective: 14-08 Use regression with seasonal binaries to make forecasts. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

Applied Statistics in Business and Economics, 6e (Doane) Chapter 15 Chi-Square Tests 1) In a chi-square test of a 5 × 5 contingency table at α = .05, the critical value is 37.65. Answer: FALSE Explanation: χ2.05 = 26.30 for d.f. = (5 − 1)(5 − 1) = 16. The Excel function is =CHISQ.INV.RT(0.05,16) = 26.29623. Difficulty: 1 Easy Topic: 15.01 Chi-Square Test for Independence Learning Objective: 15-02 Find degrees of freedom and use the chi-square table of critical values. Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation 2) If two variables are independent, we would anticipate a chi-square test statistic close to zero. Answer: TRUE Explanation: The difference between the observed and expected frequencies should be near zero. Difficulty: 1 Easy Topic: 15.01 Chi-Square Test for Independence Learning Objective: 15-03 Perform a chi-square test for independence on a contingency table. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 3) The null hypothesis for a chi-square test on a contingency table is that the variables are dependent. Answer: FALSE Explanation: The null hypothesis is independence (not dependence). Difficulty: 2 Medium Topic: 15.01 Chi-Square Test for Independence Learning Objective: 15-03 Perform a chi-square test for independence on a contingency table. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation

4) The shape of the chi-square distribution depends only on its degrees of freedom. Answer: TRUE Explanation: The chi-square distribution has only one parameter (called degrees of freedom). Difficulty: 2 Medium Topic: 15.01 Chi-Square Test for Independence Learning Objective: 15-02 Find degrees of freedom and use the chi-square table of critical values. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 5) In a chi-square test for independence, expected frequencies must be integers (or rounded to the nearest integer). Answer: FALSE Explanation: The expected frequencies are integers only in unusual situations (if the frequencies are "nice"). Difficulty: 2 Medium Topic: 15.01 Chi-Square Test for Independence Learning Objective: 15-03 Perform a chi-square test for independence on a contingency table. Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation 6) In a chi-square test for independence, observed frequencies must be at least 5 in every cell. Answer: FALSE Explanation: Small expected (not observed) frequencies are to be avoided. Difficulty: 2 Medium Topic: 15.01 Chi-Square Test for Independence Learning Objective: 15-03 Perform a chi-square test for independence on a contingency table. Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation

7) In a chi-square test for independence, observed and expected frequencies must sum across to the same row totals and down to the same column totals. Answer: TRUE Explanation: Expected frequencies reallocate the row (or column) total, so they must sum to the total. Difficulty: 1 Easy Topic: 15.01 Chi-Square Test for Independence Learning Objective: 15-03 Perform a chi-square test for independence on a contingency table. Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation 8) In samples drawn from a population in which the row and column categories are independent, the value of the chi-square test statistic will be zero. Answer: FALSE Explanation: Sampling variation exists even if the null hypothesis is true for the population. Difficulty: 2 Medium Topic: 15.01 Chi-Square Test for Independence Learning Objective: 15-03 Perform a chi-square test for independence on a contingency table. Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation 9) In a hypothesis test using chi-square, if the null hypothesis is true, the sample value of the sample chi-square test statistic will be exactly zero. Answer: FALSE Explanation: Sampling variation exists even if the null hypothesis is true for the population. Difficulty: 2 Medium Topic: 15.01 Chi-Square Test for Independence Learning Objective: 15-03 Perform a chi-square test for independence on a contingency table. Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation

10) The chi-square test for independence is a nonparametric test (no parameters are estimated). Answer: TRUE Explanation: The chi-square test does not estimate a parameter. Difficulty: 1 Easy Topic: 15.01 Chi-Square Test for Independence Learning Objective: 15-01 Recognize a contingency table and understand how it is created. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 11) Cochran's Rule requires observed frequencies of 5 or more in each cell of a contingency table. Answer: FALSE Explanation: Small expected (not observed) frequencies are to be avoided. Difficulty: 2 Medium Topic: 15.01 Chi-Square Test for Independence Learning Objective: 15-03 Perform a chi-square test for independence on a contingency table. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 12) A large negative chi-square test statistic would indicate that the null hypothesis should be rejected. Answer: FALSE Explanation: It is a sum of squares divided by a positive expected frequency, so it cannot be negative. Difficulty: 2 Medium Topic: 15.01 Chi-Square Test for Independence Learning Objective: 15-03 Perform a chi-square test for independence on a contingency table. Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation

13) The degrees of freedom in a 3 × 4 chi-square contingency table would equal 11. Answer: FALSE Explanation: d.f. = (3 − 1)(4 − 1) = 6. Difficulty: 1 Easy Topic: 15.01 Chi-Square Test for Independence Learning Objective: 15-02 Find degrees of freedom and use the chi-square table of critical values. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 14) The null hypothesis for a chi-square contingency test of independence for two variables always assumes that the variables are independent. Answer: TRUE Explanation: The null hypothesis must be phrased like this so there is only one way it can be true. Difficulty: 2 Medium Topic: 15.01 Chi-Square Test for Independence Learning Objective: 15-03 Perform a chi-square test for independence on a contingency table. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 15) The chi-square test is unreliable when there are any cells with small observed frequency counts. Answer: FALSE Explanation: Small expected (not observed) frequencies are to be avoided. Difficulty: 2 Medium Topic: 15.01 Chi-Square Test for Independence Learning Objective: 15-03 Perform a chi-square test for independence on a contingency table. Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation

16) The chi-square test can only be used to assess independence between two variables. Answer: FALSE Explanation: Chi-square tests can be used to test for goodness of fit, for example. Difficulty: 1 Easy Topic: 15.01 Chi-Square Test for Independence Learning Objective: 15-01 Recognize a contingency table and understand how it is created. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 17) The chi-square test is based on an analysis of frequencies. Answer: TRUE Explanation: Its attraction is that the test can be performed on categorical data (counts). Difficulty: 1 Easy Topic: 15.01 Chi-Square Test for Independence Learning Objective: 15-03 Perform a chi-square test for independence on a contingency table. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 18) A chi-square distribution is always skewed right. Answer: TRUE Explanation: Especially for small degrees of freedom, the chi-square distribution is rightskewed. Difficulty: 1 Easy Topic: 15.01 Chi-Square Test for Independence Learning Objective: 15-02 Find degrees of freedom and use the chi-square table of critical values. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 19) A chi-square test for independence is called a distribution-free test since the test is based on categorical data rather than on populations that follow any particular distribution. Answer: TRUE Explanation: The lack of assumed population shape is an attraction of this test. Difficulty: 1 Easy Topic: 15.01 Chi-Square Test for Independence Learning Objective: 15-01 Recognize a contingency table and understand how it is created. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 6 Copyright 2019 © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.

20) Observed frequencies in a chi-square goodness-of-fit test for normality may be less than 5, or even 0, in some cells, as long as the expected frequencies are large enough. Answer: TRUE Explanation: Small expected (not observed) frequencies are to be avoided. Difficulty: 2 Medium Topic: 15.05 Normal Chi-Square Goodness-of-Fit Test Learning Objective: 15-07 Explain the chi-square GOF test for normality. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 21) In a chi-square goodness-of-fit test, a small p-value would indicate a good fit to the hypothesized distribution. Answer: FALSE Explanation: When the p-value is small, we are inclined to reject the hypothesized distribution. Difficulty: 2 Medium Topic: 15.02 Chi-Square Tests for Goodness-of-Fit Learning Objective: 15-04 Perform a goodness-of-fit (GOF) test for a multinomial distribution. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 22) For a chi-square goodness-of-fit test for a uniform distribution with 5 categories, we would use the critical value for 4 degrees of freedom. Answer: TRUE Explanation: d.f. = k − 1 − m = 5 − 1 − 0 = 4, where m = 0 parameters are estimated and k = 5 categories. Difficulty: 1 Easy Topic: 15.03 Uniform Goodness-of-Fit Test Learning Objective: 15-05 Perform a GOF test for a uniform distribution. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

23) For a chi-square goodness-of-fit test for a uniform distribution with 7 categories, we would use the critical value for 6 degrees of freedom. Answer: TRUE Explanation: d.f. = k − 1 − m = 7 − 1 − 0 = 6, where m = 0 parameters are estimated and k = 7 categories. Difficulty: 1 Easy Topic: 15.03 Uniform Goodness-of-Fit Test Learning Objective: 15-05 Perform a GOF test for a uniform distribution. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 24) For a chi-square goodness-of-fit test for a normal distribution using 8 categories with estimated mean and standard deviation, we would use the critical value for 7 degrees of freedom. Answer: FALSE Explanation: d.f. = k − 1 − m = 8 − 1 − 2 = 5, where m = 2 parameters are estimated and k = 8 categories. Difficulty: 2 Medium Topic: 15.05 Normal Chi-Square Goodness-of-Fit Test Learning Objective: 15-07 Explain the chi-square GOF test for normality. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 25) For a chi-square goodness-of-fit test for a normal distribution using 7 categories with estimated mean and standard deviation, we would use the critical value for 4 degrees of freedom. Answer: TRUE Explanation: d.f. = k − 1 − m = 7 − 1 − 2 = 4, where m = 2 parameters are estimated and k = 7 categories. Difficulty: 2 Medium Topic: 15.05 Normal Chi-Square Goodness-of-Fit Test Learning Objective: 15-07 Explain the chi-square GOF test for normality. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

26) A probability plot usually allows outliers to be detected. Answer: TRUE Explanation: Outliers will be seen as unusual (tail) points far from the main body of data. Difficulty: 1 Easy Topic: 15.06 ECDF Tests (Optional) Learning Objective: 15-08 Interpret ECDF tests and know their advantages compared to chisquare GOF tests. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 27) In a goodness-of-fit test, a linear probability plot suggests that the null hypothesis should be rejected. Answer: FALSE Explanation: Data that follow a straight line would support the null hypothesis (and conversely). Difficulty: 2 Medium Topic: 15.06 ECDF Tests (Optional) Learning Objective: 15-08 Interpret ECDF tests and know their advantages compared to chisquare GOF tests. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 28) In a chi-square goodness-of-fit test, we lose one degree of freedom for each parameter estimated. Answer: TRUE Explanation: d.f. = k − 1 − m, where m = number of parameters that are estimated and k = number of categories. Difficulty: 1 Easy Topic: 15.02 Chi-Square Tests for Goodness-of-Fit Learning Objective: 15-04 Perform a goodness-of-fit (GOF) test for a multinomial distribution. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation

29) In a chi-square goodness-of-fit test, we gain one degree of freedom if n increases by 1. Answer: FALSE Explanation: d.f. = k − 1 − m, for m parameters and k categories (n does not enter this formula). Difficulty: 2 Medium Topic: 15.03 Uniform Goodness-of-Fit Test Learning Objective: 15-04 Perform a goodness-of-fit (GOF) test for a multinomial distribution. Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation 30) In a chi-square goodness-of-fit test, a sample of n observations has n − 1 degrees of freedom. Answer: FALSE Explanation: d.f. = k − 1 − m for m parameters and k categories (n does not enter this formula). Difficulty: 2 Medium Topic: 15.02 Chi-Square Tests for Goodness-of-Fit Learning Objective: 15-04 Perform a goodness-of-fit (GOF) test for a multinomial distribution. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 31) The Poisson goodness-of-fit test is inappropriate for continuous data. Answer: TRUE Explanation: Poisson data are integers. For example, X = number of arrivals. Difficulty: 2 Medium Topic: 15.04 Poisson Goodness-of-Fit Test Learning Objective: 15-06 Explain the GOF test for a Poisson distribution. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 32) The Kolmogorov-Smirnov and Anderson-Darling tests are based on the ECDF (Empirical Cumulative Distribution Function). Answer: TRUE Explanation: The ECDF provides the basis for several such tests. Difficulty: 2 Medium Topic: 15.06 ECDF Tests (Optional) Learning Objective: 15-08 Interpret ECDF tests and know their advantages compared to chisquare GOF tests. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 10 Copyright 2019 © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.

33) Goodness-of-fit tests using the ECDF (Empirical Cumulative Distribution Function) compare the actual cumulative frequencies with expected cumulative frequencies for each observation under the assumption that the data came from the hypothesized distribution. Answer: TRUE Explanation: Yes, the test considers each data value separately (there is no grouping into categories). Difficulty: 2 Medium Topic: 15.06 ECDF Tests (Optional) Learning Objective: 15-08 Interpret ECDF tests and know their advantages compared to chisquare GOF tests. Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation 34) An attraction of the Kolmogorov-Smirnov test is that it is fairly easy to do without a computer. Answer: FALSE Explanation: The Kolmogorov-Smirnov test is done with a computer because the calculations are difficult. Difficulty: 2 Medium Topic: 15.06 ECDF Tests (Optional) Learning Objective: 15-08 Interpret ECDF tests and know their advantages compared to chisquare GOF tests. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 35) An attraction of the Anderson-Darling test is that it is fairly easy to do without a computer. Answer: FALSE Explanation: The Anderson-Darling test is done with a computer. It requires an inverse distribution function which is done using computer software. Difficulty: 2 Medium Topic: 15.06 ECDF Tests (Optional) Learning Objective: 15-08 Interpret ECDF tests and know their advantages compared to chisquare GOF tests. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation

36) ECDF tests have an advantage over the chi-square goodness-of-fit test on frequencies because an ECDF test treats observations individually. Answer: TRUE Explanation: Each data value is considered separately (no grouping into categories), so more power. Difficulty: 2 Medium Topic: 15.06 ECDF Tests (Optional) Learning Objective: 15-08 Interpret ECDF tests and know their advantages compared to chisquare GOF tests. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 37) In an ECDF test for goodness-of-fit, the n observations are grouped into categories rather than being treated individually. Answer: FALSE Explanation: In ECDF tests, each data value is considered separately (no grouping into categories). Difficulty: 2 Medium Topic: 15.06 ECDF Tests (Optional) Learning Objective: 15-08 Interpret ECDF tests and know their advantages compared to chisquare GOF tests. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 38) When raw data are available, ECDF tests usually surpass the chi-square test in their ability to detect departures from the distribution specified in the null hypothesis. Answer: TRUE Explanation: Each data value is considered separately (no grouping into categories), so more power. Difficulty: 2 Medium Topic: 15.06 ECDF Tests (Optional) Learning Objective: 15-08 Interpret ECDF tests and know their advantages compared to chisquare GOF tests. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation

39) The Anderson-Darling test is used to test the assumption of normality. Answer: TRUE Explanation: Most software packages offer the Anderson-Darling normality test because normality tests are frequently needed. Difficulty: 1 Easy Topic: 15.06 ECDF Tests (Optional) Learning Objective: 15-08 Interpret ECDF tests and know their advantages compared to chisquare GOF tests. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 40) Probability plots are used to test the assumption of normality. Answer: TRUE Explanation: Most software packages have the normal probability plot, because normality tests are frequently needed. Difficulty: 1 Easy Topic: 15.06 ECDF Tests (Optional) Learning Objective: 15-08 Interpret ECDF tests and know their advantages compared to chisquare GOF tests. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 41) In a test for a uniform distribution with k categories, the expected frequency is n/k in each cell. Answer: TRUE Explanation: For uniformity, we expect n/k in each category. Difficulty: 1 Easy Topic: 15.03 Uniform Goodness-of-Fit Test Learning Objective: 15-05 Perform a GOF test for a uniform distribution. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation

42) If samples are drawn from a population that is normal, a goodness-of-fit test for normality could yield A) Type I error but not Type II error. B) Type II error but not Type I error. C) either Type I error or Type II error. D) both Type I and Type II errors. Answer: A Explanation: If the hypothesis (H0: population is normal) is true, we cannot commit Type II error (failing to reject a false hypothesis). But in reality, we would not know that H0 is true. Difficulty: 2 Medium Topic: 15.05 Normal Chi-Square Goodness-of-Fit Test Learning Objective: 15-07 Explain the chi-square GOF test for normality. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 43) The number of cars waiting at a certain residential neighborhood stoplight is observed at 6:00 a.m. on 160 different days. The observed sample frequencies are shown here. Number of cars waiting Number of days

0 60

1 50

2 30

3 20

Under the null hypothesis of a uniform distribution, the expected number of days we would see 0 cars is A) 10 B) 20 C) 30 D) 40 Answer: D Explanation: n/k = 160/4 = 40. Difficulty: 2 Medium Topic: 15.03 Uniform Goodness-of-Fit Test Learning Objective: 15-05 Perform a GOF test for a uniform distribution. Bloom's: Analyze AACSB: Analytical Thinking Accessibility: Keyboard Navigation

44) A chi-square goodness of fit test for a normal distribution used 40 observations, and the mean and standard deviation were estimated from the sample. The test used six categories. We would use how many degrees of freedom in looking up the critical value for the test? A) 39 B) 37 C) 5 D) 3 Answer: D Explanation: d.f. = k − 1 − m = 6 − 1 − 2 = 3, where m = 2 parameters are estimated and k = 6 categories. (n is not in the formula.) Difficulty: 2 Medium Topic: 15.05 Normal Chi-Square Goodness-of-Fit Test Learning Objective: 15-07 Explain the chi-square GOF test for normality. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 45) A chi-square goodness of fit test for a normal distribution used 60 observations, and the mean and standard deviation were estimated from the sample. The test used seven categories. We would use how many degrees of freedom in looking up the critical value for the test? A) 6 B) 4 C) 59 D) 57 Answer: B Explanation: d.f. = k − 1 − m = 7 − 1 − 2 = 4, where m = 2 parameters are estimated and k = 7 categories. (n is not in the formula.) Difficulty: 2 Medium Topic: 15.05 Normal Chi-Square Goodness-of-Fit Test Learning Objective: 15-07 Explain the chi-square GOF test for normality. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

46) Which of these statements concerning a chi-square goodness-of-fit test is correct? A) Data could be ratio or interval measurements. B) Population must be normally distributed. C) All the expected frequencies must be equal. Answer: A Explanation: Distances between data values must be meaningful (ratio or interval). Difficulty: 1 Easy Topic: 15.02 Chi-Square Tests for Goodness-of-Fit Learning Objective: 15-04 Perform a goodness-of-fit (GOF) test for a multinomial distribution. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 47) Which of the following is not a potential solution to the problem that arises when not all expected frequencies are 5 or more in a chi-square test for independence? A) Combine some of the columns. B) Combine some of the rows. C) Increase the sample size. D) Add more rows or columns. Answer: D Explanation: Subdividing rows or columns would make the expected frequencies smaller. Difficulty: 2 Medium Topic: 15.01 Chi-Square Test for Independence Learning Objective: 15-01 Recognize a contingency table and understand how it is created. Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation 48) Which of these statements concerning a chi-square goodness-of-fit test is correct? A) It is inapplicable to test for a normal distribution with open-ended top and bottom classes. B) It is generally a better test than the chi-square test of independence. C) There is no way to get the degrees of freedom since the right tail goes to infinity. D) It can be used to test whether a sample follows a specified distribution. Answer: D Explanation: A goodness-of-fit test asks whether the sample contradicts a proposed population distribution. Difficulty: 2 Medium Topic: 15.02 Chi-Square Tests for Goodness-of-Fit Learning Objective: 15-04 Perform a goodness-of-fit (GOF) test for a multinomial distribution. Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation 16 Copyright 2019 © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.

49) A proofreader checked 160 ads for grammatical errors. The sample frequency distribution is shown below. Number of Errors Observed Frequency

0 10

1 65

2 71

3 14

Under the null hypothesis of a uniform distribution, the expected number of times we would get 0 errors is A) 10 B) 20 C) 30 D) 40 Answer: D Explanation: n/k = 160/4 = 40. Difficulty: 2 Medium Topic: 15.03 Uniform Goodness-of-Fit Test Learning Objective: 15-05 Perform a GOF test for a uniform distribution. Bloom's: Analyze AACSB: Analytical Thinking Accessibility: Keyboard Navigation 50) A proofreader checked 160 ads for grammatical errors. The sample frequency distribution is shown below. Number of Errors Observed Frequency

0 10

1 65

2 71

3 14

A goodness-of-fit test to determine whether this sample came from a uniform distribution would result in a chi-square test statistic of approximately A) 55 B) 79 C) 85 D) 161 Answer: B Explanation: (10 − 40)2/40 + (65 − 40)2/40 + (71 − 40)2/40 + (14 − 40)2/40 = 79.05. Difficulty: 2 Medium Topic: 15.03 Uniform Goodness-of-Fit Test Learning Objective: 15-05 Perform a GOF test for a uniform distribution. Bloom's: Analyze AACSB: Analytical Thinking Accessibility: Keyboard Navigation

51) A proofreader checked 160 ads for grammatical errors. The distribution obtained is shown below. Number of Errors Observed Frequency

0 10

1 65

2 71

3 14

At α = .01, what decision would we reach in a goodness-of-fit test to see whether this sample came from a uniform distribution? A) Reject the null and conclude the distribution is not uniform. B) Conclude that there is insufficient evidence to reject the null. C) No conclusion can be made due to small expected frequencies. D) No conclusion can be made due to inadequate sample size. Answer: A Explanation: (10 − 40)2/40 + (65 − 40)2/40 + (71 − 40)2/40 + (14 − 40)2/40 = 79.05 > χ2.01 = 11.34 for d.f. = 3. Difficulty: 3 Hard Topic: 15.03 Uniform Goodness-of-Fit Test Learning Objective: 15-05 Perform a GOF test for a uniform distribution. Bloom's: Analyze AACSB: Analytical Thinking Accessibility: Keyboard Navigation 52) A chi-square test of independence is a one-tailed test. The reason is that A) we are testing whether the frequencies exceed their expected values. B) we square the deviations, so the test statistic lies at or above zero. C) hypothesis tests are one-tailed tests when dealing with sample data. D) the chi-square distribution is positively skewed. Answer: B Explanation: The chi-square test statistic sums the terms (Obs − Exp)2, so differences in either direction will yield a positive contribution to the sum. Difficulty: 2 Medium Topic: 15.01 Chi-Square Test for Independence Learning Objective: 15-03 Perform a chi-square test for independence on a contingency table. Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation

53) We sometimes combine two row or column categories in a chi-square test when the A) observed frequencies are more than 5. B) observed frequencies are less than 5. C) expected frequencies are more than 5. D) expected frequencies are less than 5. Answer: D Explanation: Consolidating two rows (or columns) would increase expected frequencies (but would decrease d.f.). Difficulty: 2 Medium Topic: 15.01 Chi-Square Test for Independence Learning Objective: 15-03 Perform a chi-square test for independence on a contingency table. Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation 54) To determine how well an observed set of frequencies fits an expected set of frequencies from a Poisson distribution we must estimate A) no parameters. B) one parameter (λ). C) two parameters (μ, σ). D) three parameters (μ, σ, n). Answer: B Explanation: We lose one degree of freedom when we estimate the Poisson mean λ. Difficulty: 1 Easy Topic: 15.04 Poisson Goodness-of-Fit Test Learning Objective: 15-06 Explain the GOF test for a Poisson distribution. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation

55) The critical value in a chi-square test for independence depends on A) the normality of the data. B) the variance of the data. C) the number of categories. D) the expected frequencies. Answer: C Explanation: χ2α depends on d.f. = k − 1 − m for m estimated parameters and k categories. Difficulty: 1 Easy Topic: 15.01 Chi-Square Test for Independence Learning Objective: 15-02 Find degrees of freedom and use the chi-square table of critical values. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 56) In a chi-square test of independence, the number of degrees of freedom equals the A) number of observations minus one. B) number of categories minus one. C) number of rows minus one times the number of columns minus one. D) number of sample observations minus the missing observations. Answer: C Explanation: χ2α depends on d.f. = (r − 1)(c − 1). Difficulty: 1 Easy Topic: 15.01 Chi-Square Test for Independence Learning Objective: 15-02 Find degrees of freedom and use the chi-square table of critical values. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 57) In order to apply the chi-square test of independence, we prefer to have A) at least 5 observed frequencies in each cell. B) at least 5 expected observations in each cell. C) at least 5 percent of the observations in each cell. D) not more than 5 observations in each cell. Answer: B Explanation: Larger expected frequencies are desirable (at least 5 according to Cochran's Rule). Difficulty: 1 Easy Topic: 15.01 Chi-Square Test for Independence Learning Objective: 15-03 Perform a chi-square test for independence on a contingency table. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 20 Copyright 2019 © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.

58) Kortholts that fail to meet certain precise specifications must be reworked on the next day until they are within the desired specifications. A sample of one day's output of kortholts from the Melodic Kortholt Company showed the frequencies shown below. Specification Met Specification Not Met Col Total

Plant A 185 15 200

Plant B 85 15 100

Row Total 270 30 300

Find the chi-square test statistic for a hypothesis of independence. A) 7.22 B) 4.17 C) 5.13 D) 6.08 Answer: B Explanation: The test statistic is χ2calc = Σ(Obs − Exp)2/Exp where Exp = [(row sum) × (col sum)]/n. Difficulty: 2 Medium Topic: 15.01 Chi-Square Test for Independence Learning Objective: 15-03 Perform a chi-square test for independence on a contingency table. Bloom's: Analyze AACSB: Analytical Thinking Accessibility: Keyboard Navigation

59) Kortholts that fail to meet certain precise specifications must be reworked on the next day until they are within the desired specifications. A sample of one day's output of kortholts from the Melodic Kortholt Company showed the frequencies shown below. Specification Met Specification Not Met Col Total

Plant A 185 15 200

Plant B 85 15 100

Row Total 270 30 300

Find the p-value for the chi-square test statistic for a hypothesis of independence. A) Less than .01 B) Between .01 and .025 C) Between .025 and .05 D) Greater than .05 Answer: C Explanation: χ2calc = 4.167 with d.f. = 1 is between χ2.05 = 3.841 and χ2.025 = 5.024. Using Excel, the p-value is =CHISQ.DIST.RT(4.167,1) = .0412. Difficulty: 3 Hard Topic: 15.01 Chi-Square Test for Independence Learning Objective: 15-03 Perform a chi-square test for independence on a contingency table. Bloom's: Analyze AACSB: Analytical Thinking Accessibility: Keyboard Navigation 60) An operations analyst counted the number of arrivals per minute at a bank ATM in each of 30 randomly chosen minutes. The results were: 0, 3, 3, 2, 1, 0, 1, 0, 0, 1, 1, 1, 2, 1, 0, 1, 0, 1, 2, 1, 1, 2, 1, 0, 1, 2, 0, 1, 0, 1. Which goodness-of-fit test would you recommend? A) Uniform. B) Poisson. C) Normal. D) Binomial. Answer: B Explanation: Arrivals per unit of time with a small mean would resemble a Poisson distribution. Difficulty: 2 Medium Topic: 15.04 Poisson Goodness-of-Fit Test Learning Objective: 15-06 Explain the GOF test for a Poisson distribution. Bloom's: Analyze AACSB: Analytical Thinking Accessibility: Keyboard Navigation

61) An operations analyst counted the number of arrivals per minute at an ATM in each of 30 randomly chosen minutes. The results were: 0, 3, 3, 2, 1, 0, 1, 0, 0, 1, 1, 1, 2, 1, 0, 1, 0, 1, 2, 1, 1, 2, 1, 0, 1, 2, 0, 1, 0, 1. For the Poisson goodness-of-fit test, what is the expected frequency of the data value X = 1? A) Impossible to determine B) 11.04 C) 1.00 D) 2.47 Answer: B Explanation: The sample mean is 1.00 so n × P(X = 1 | λ = 1.00) = (30)(.3679) = 11.037. Difficulty: 3 Hard Topic: 15.04 Poisson Goodness-of-Fit Test Learning Objective: 15-06 Explain the GOF test for a Poisson distribution. Bloom's: Analyze AACSB: Analytical Thinking Accessibility: Keyboard Navigation 62) The table below is a tabulation of opinions of employees of Axolotl Corporation, who were sampled at random from pay records and asked to complete an anonymous job satisfaction survey. Pay Type Salaried Hourly Col Total

Degree of Satisfaction Satisfied Neutral Dissatisfied 40 10 10 80 50 50 120 60 60

Row Total 60 180 240

For a chi-square test of independence, the degrees of freedom would be A) 2 B) 3 C) 4 D) 6 Answer: A Explanation: d.f. = (2 − 1)(3 − 1) = 2. Difficulty: 1 Easy Topic: 15.01 Chi-Square Test for Independence Learning Objective: 15-02 Find degrees of freedom and use the chi-square table of critical values. Bloom's: Analyze AACSB: Analytical Thinking Accessibility: Keyboard Navigation

63) The table below is a tabulation of opinions of employees of Axolotl Corporation, who were sampled at random from pay records and asked to complete an anonymous job satisfaction survey. Pay Type Salaried Hourly Col Total

Degree of Satisfaction Satisfied Neutral Dissatisfied 40 10 10 80 50 50 120 60 60

Row Total 60 180 240

For a chi-square test of independence, which is the critical value for α = .01? A) 9.210 B) 4.605 C) 11.34 D) 16.81 Answer: A Explanation: χ2.01 = 9.210 for d.f. = (2 − 1)(3 − 1) = 2. The Excel function is =CHISQ.INV.RT(0.01,2) = 9.21034. Difficulty: 2 Medium Topic: 15.01 Chi-Square Test for Independence Learning Objective: 15-02 Find degrees of freedom and use the chi-square table of critical values. Bloom's: Analyze AACSB: Analytical Thinking Accessibility: Keyboard Navigation

64) The table below is a tabulation of opinions of employees of Axolotl Corporation, who were sampled at random from pay records and asked to complete an anonymous job satisfaction survey, with the results shown below. Pay Type Salaried Hourly Col Total

Degree of Satisfaction Satisfied Neutral Dissatisfied 40 10 10 80 50 50 120 60 60

Row Total 60 180 240

Assuming independence, the expected frequency of satisfied hourly employees is A) 80 B) 90 C) 75 D) 60 Answer: B Explanation: e21 = (R2)(C1)/n = (180)(120)/240 = 90. Difficulty: 2 Medium Topic: 15.01 Chi-Square Test for Independence Learning Objective: 15-03 Perform a chi-square test for independence on a contingency table. Bloom's: Analyze AACSB: Analytical Thinking Accessibility: Keyboard Navigation 65) To carry out a chi-square goodness-of-fit test for normality you need at least A) 5 categories altogether. B) 5 observations in each category. C) 5 expected observations in each category. D) 50 samples or more. Answer: C Explanation: Because d.f. = k − 1 − m = k − 3 since m = 2, we need at least k = 4 groups each with e ≥ 5. Difficulty: 1 Easy Topic: 15.05 Normal Chi-Square Goodness-of-Fit Test Learning Objective: 15-07 Explain the chi-square GOF test for normality. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation

66) Students in an introductory college economics class were asked how many credits they had earned in college and how certain they were about their choice of major. Their replies are summarized below. Credits Earned Under 10 10 through 59 60 or more Col Total

Degree of Certainty Very Certain Somewhat Certain Very Certain 24 16 6 16 8 20 2 14 22 42 38 48

Row Tot 46 44 38 128

Under the assumption of independence, the expected frequency in the upper left cell is A) 15.09 B) 24.00 C) 19.72 D) 20.22 Answer: A Explanation: e11 = (R1)(C1)/n = (46)(42)/128 = 15.09. Difficulty: 2 Medium Topic: 15.01 Chi-Square Test for Independence Learning Objective: 15-03 Perform a chi-square test for independence on a contingency table. Bloom's: Analyze AACSB: Analytical Thinking Accessibility: Keyboard Navigation

67) Students in an introductory college economics class were asked how many credits they had earned in college and how certain they were about their choice of major. Their replies are summarized below. Credits Earned Under 10 10 through 59 60 or more Col Total

Degree of Certainty Very Certain Somewhat Certain Very Certain 24 16 6 16 8 20 2 14 22 42 38 48

Row Tot 46 44 38 128

For a chi-square test of independence, degrees of freedom would be A) 2 B) 9 C) 4 D) 127 Answer: C Explanation: d.f. = (3 − 1)(3 − 1) = 4. Difficulty: 1 Easy Topic: 15.01 Chi-Square Test for Independence Learning Objective: 15-02 Find degrees of freedom and use the chi-square table of critical values. Bloom's: Analyze AACSB: Analytical Thinking Accessibility: Keyboard Navigation

68) Students in an introductory college economics class were asked how many credits they had earned in college and how certain they were about their choice of major. Their replies are summarized below. Credits Earned Under 10 10 through 59 60 or more Col Total

Degree of Certainty Very Certain Somewhat Certain Very Certain 24 16 6 16 8 20 2 14 22 42 38 48

Row Tot 46 44 38 128

For a chi-square test of independence, which is the critical value for α = .05? A) 5.991 B) 7.815 C) 9.488 D) 16.92 Answer: C Explanation: χ2.05 = 9.488 for d.f. = (3 − 1)(3 − 1) = 4. The Excel function is =CHISQ.INV.RT(0.05,4) = 9.48773. Difficulty: 2 Medium Topic: 15.01 Chi-Square Test for Independence Learning Objective: 15-02 Find degrees of freedom and use the chi-square table of critical values. Bloom's: Analyze AACSB: Analytical Thinking Accessibility: Keyboard Navigation

69) Students in an introductory college economics class were asked how many credits they had earned in college, and how certain they were about their choice of major. Their replies are summarized below. Credits Earned Under 10 10 through 59 60 or more Col Total

Degree of Certainty Very Certain Somewhat Certain Very Certain 24 16 6 16 8 20 2 14 22 42 38 48

Row Tot 46 44 38 128

Assuming independence, which is the expected frequency of very uncertain students with 60 credits or more? A) 12.47 B) 2.00 C) 14.56 D) 11.09 Answer: A Explanation: e31 = (R3)(C1)/n = (38)(42)/128 = 12.47. Difficulty: 2 Medium Topic: 15.01 Chi-Square Test for Independence Learning Objective: 15-03 Perform a chi-square test for independence on a contingency table. Bloom's: Analyze AACSB: Analytical Thinking Accessibility: Keyboard Navigation

70) Students in an introductory college economics class were asked how many credits they had earned in college, and how certain they were about their choice of major. Their replies are summarized below. Credits Earned Under 10 10 through 59 60 or more Col Total

Degree of Certainty Very Certain Somewhat Certain Very Certain 24 16 6 16 8 20 2 14 22 42 38 48

Row Tot 46 44 38 128

Which statement is most nearly correct? A) The contingency table violates Cochran's Rule. B) Visual inspection of column frequencies suggests independence. C) At α = .05 we would easily reject the null hypothesis of independence. D) At α = .05 we cannot reject he null hypothesis of independence. Answer: C Explanation: χ2calc = 29.528 > χ2.05 = 9.488 for d.f. = (3 − 1)(3 − 1) = 4. The Excel function is =CHISQ.INV.RT(0.05,4) = 9.48773. Difficulty: 3 Hard Topic: 15.01 Chi-Square Test for Independence Learning Objective: 15-03 Perform a chi-square test for independence on a contingency table. Bloom's: Analyze AACSB: Analytical Thinking Accessibility: Keyboard Navigation

71) As an independent project, a team of statistics students tabulated the types of vehicles that were parked in four different suburban shopping malls. Vehicle Type Car Minivan Full-size Van SUV Truck Col Total

Somerset 44 21 2 19 14 100

Mall Location Oakland Great Lakes 49 36 15 18 3 3 27 26 6 17 100 100

Jamestown Row Total 64 193 13 67 2 10 12 84 9 46 100 400

For a chi-square test of independence, the degrees of freedom would be A) 20 B) 12 C) 399 D) 6 Answer: B Explanation: d.f. = (5 − 1)(4 − 1) = 12. Difficulty: 1 Easy Topic: 15.01 Chi-Square Test for Independence Learning Objective: 15-02 Find degrees of freedom and use the chi-square table of critical values. Bloom's: Analyze AACSB: Analytical Thinking Accessibility: Keyboard Navigation

72) As an independent project, a team of statistics students tabulated the types of vehicles that were parked in four different suburban shopping malls. Vehicle Type Car Minivan Full-size Van SUV Truck Col Total

Somerset 44 21 2 19 14 100

Mall Location Oakland Great Lakes 49 36 15 18 3 3 27 26 6 17 100 100

Jamestown Row Total 64 193 13 67 2 10 12 84 9 46 100 400

For a chi-square test of independence, the critical value for α = .10 is A) 10.64 B) 14.68 C) 28.41 D) 18.55 Answer: D Explanation: χ2.10 = 18.55 for d.f. = (5 − 1)(4 − 1) = 12. The Excel function is =CHISQ.INV.RT(0.10,12) =18.54935. Difficulty: 1 Easy Topic: 15.01 Chi-Square Test for Independence Learning Objective: 15-02 Find degrees of freedom and use the chi-square table of critical values. Bloom's: Analyze AACSB: Analytical Thinking Accessibility: Keyboard Navigation

73) As an independent project, a team of statistics students tabulated the types of vehicles that were parked in four different suburban shopping malls. Vehicle Type Car Minivan Full-size Van SUV Truck Col Total

Somerset 44 21 2 19 14 100

Mall Location Oakland Great Lakes 49 36 15 18 3 3 27 26 6 17 100 100

Jamestown Row Total 64 193 13 67 2 10 12 84 9 46 100 400

Assuming independence, the expected frequency of SUVs in Jamestown is A) 12 B) 21 C) 75 D) 60 Answer: B Explanation: e44 = (R4)(C4)/n = (84)(100)/400 = 21. Difficulty: 2 Medium Topic: 15.01 Chi-Square Test for Independence Learning Objective: 15-03 Perform a chi-square test for independence on a contingency table. Bloom's: Analyze AACSB: Analytical Thinking Accessibility: Keyboard Navigation

74) Employees of OxCo Mfg. were surveyed to evaluate the company's pension plan. The table below displays some of the results of the survey. Years Employed Under 10 10 to 25 25 or more Col Total

Rating of Pension Plan Excellent Good Fair 26 92 92 37 159 133 17 69 55 80 320 280

Poor 30 71 19 120

Row Total 240 400 160 800

6.206chi-square The expected frequency for the shaded cell in row 2 and column 2 in the table would be A) 163 B) 158 C) 165 D) 160 Answer: D Explanation: e22 = (R2)(C2)/n = (400)(320)/800 = 160. Difficulty: 2 Medium Topic: 15.01 Chi-Square Test for Independence Learning Objective: 15-03 Perform a chi-square test for independence on a contingency table. Bloom's: Analyze AACSB: Analytical Thinking Accessibility: Keyboard Navigation

75) Employees of OxCo Mfg. were surveyed to evaluate the company's pension plan. The table below displays some of the results of the survey. Years Employed Under 10 10 to 25 25 or more Col Total

Rating of Pension Plan Excellent Good Fair 26 92 92 37 159 133 17 69 55 80 320 280

Poor 30 71 19 120

Row Total 240 400 160 800

6.206chi-square Degrees of freedom for this test would be A) 6 B) 7 C) 799 D) 12 Answer: A Explanation: d.f. = (3 − 1)(4 − 1) = 6. Difficulty: 1 Easy Topic: 15.01 Chi-Square Test for Independence Learning Objective: 15-02 Find degrees of freedom and use the chi-square table of critical values. Bloom's: Analyze AACSB: Analytical Thinking Accessibility: Keyboard Navigation

76) Employees of OxCo Mfg. were surveyed to evaluate the company's pension plan. The table below displays some of the results of the survey. Years Employed Under 10 10 to 25 25 or more Col Total

Rating of Pension Plan Excellent Good Fair 26 92 92 37 159 133 17 69 55 80 320 280

Poor 30 71 19 120

Row Total 240 400 160 800

6.206chi-square Which is the most appropriate conclusion? A) Do not reject H0. B) Reject H0 at α = .10. C) Reject H0 at α = .05. D) Reject H0 at α = .01. Answer: A Explanation: χ2calc = 6.206 does not even exceed χ2.10 = 10.64 for d.f. = (3 − 1)(4 − 1) = 6. Difficulty: 2 Medium Topic: 15.01 Chi-Square Test for Independence Learning Objective: 15-03 Perform a chi-square test for independence on a contingency table. Bloom's: Analyze AACSB: Analytical Thinking Accessibility: Keyboard Navigation 77) You test a hypothesis of independence of two variables. The number of observations is 500 and you have classified the data into a 4×4 contingency table. The test statistic has ________ degrees of freedom. A) 16 B) 9 C) 499 D) 498 Answer: B Explanation: Sample size does not enter into the calculation: d.f. = (4 − 1)(4 − 1) = 9. Difficulty: 1 Easy Topic: 15.01 Chi-Square Test for Independence Learning Objective: 15-02 Find degrees of freedom and use the chi-square table of critical values. Bloom's: Analyze AACSB: Analytical Thinking Accessibility: Keyboard Navigation 36 Copyright 2019 © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.

78) You want to test the hypothesis that the prime rate and inflation are independent. The following table is prepared for the test on the basis of the results of a random sample, collected in various countries and various time periods shown below. Inflation Rates Under 5% 5% or more Col Tot

6-Feb 40 5 45

Prime Rate 10-Jul 30 30 60

20-Nov 5 40 45

Row Total 75 75 150

Using α = .05, which is the critical value of the test statistic that you would use? A) 3.841 B) 12.59 C) 5.991 D) 7.815 Answer: C Explanation: χ2.05 = 5.991 for d.f. = (2 − 1)(3 − 1) = 2. The Excel function is =CHISQ.INV.RT(0.05,2) = 5.99146. Difficulty: 1 Easy Topic: 15.01 Chi-Square Test for Independence Learning Objective: 15-02 Find degrees of freedom and use the chi-square table of critical values. Bloom's: Analyze AACSB: Analytical Thinking Accessibility: Keyboard Navigation

79) You want to test the hypothesis that the prime rate and inflation are independent. The following table is prepared for the test on the basis of the results of a random sample, collected in various countries and various time periods shown below. Inflation Rates Under 5% 5% or more Col Tot

6-Feb 40 5 45

Prime Rate 10-Jul 30 30 60

20-Nov 5 40 45

Row Total 75 75 150

The expected frequency for the cell in row 2 and column 3 is A) 22.5 B) 30 C) 40 D) 40.5 Answer: A Explanation: e23 = (R2)(C3)/n = (45)(75)/150 = 22.5. Difficulty: 1 Easy Topic: 15.01 Chi-Square Test for Independence Learning Objective: 15-03 Perform a chi-square test for independence on a contingency table. Bloom's: Analyze AACSB: Analytical Thinking Accessibility: Keyboard Navigation

80) You want to test the hypothesis that the prime rate and inflation are independent. The following table is prepared for the test on the basis of the results of a random sample, collected in various countries and various time periods shown below. Inflation Rates Under 5% 5% or more Col Tot

6-Feb 40 5 45

Prime Rate 10-Jul 30 30 60

20-Nov 5 40 45

Row Total 75 75 150

What is the value of the test statistic? A) 306.25 B) 0.00 C) 54.44 D) 13.61 Answer: C Explanation: χ2calc = (40 − 22.5)2/22.5 + (30 − 30)2/30 + (5 − 22.5)2/22.5 + (5 − 22.5)2/22.5 + (30 − 30)2/30 + (40 − 22.5)2/22.5 = 54.444. Difficulty: 3 Hard Topic: 15.01 Chi-Square Test for Independence Learning Objective: 15-03 Perform a chi-square test for independence on a contingency table. Bloom's: Analyze AACSB: Analytical Thinking Accessibility: Keyboard Navigation

81) You want to test the hypothesis that the prime rate and inflation are independent. The following table of frequencies is prepared from a random sample, collected in various countries and various time periods shown below. Inflation Rates Under 5% 5% or more Col Tot

6-Feb 40 5 45

Prime Rate 10-Jul 30 30 60

20-Nov 5 40 45

Row Total 75 75 150

Based on an analysis of the data in this table, which conclusion can be made at α = .01? A) The prime rate and inflation rate are independent. B) The prime rate and inflation rate are not independent. C) Small observed frequencies in some cells suggest that no reliable conclusion can be made. D) Small expected frequencies in some cells suggest that no reliable conclusion can be made. Answer: B Explanation: χ2calc = 54.444 greatly exceeds χ2.01 = 9.210 for d.f. = (2 − 1)(3 − 1) = 2. Difficulty: 3 Hard Topic: 15.01 Chi-Square Test for Independence Learning Objective: 15-03 Perform a chi-square test for independence on a contingency table. Bloom's: Analyze AACSB: Analytical Thinking Accessibility: Keyboard Navigation

82) You want to sell your house, and you decide to obtain an appraisal on it. Looking at past data, you discover that actual prices obtained for houses and the appraisals given for them prior to their sale were as shown below. Appraisal Up to $500,000 $500,001 or more Col Tot

Actual Selling Price Up to $500,000 $500,001 or more 21 9 14 6 35 15

Row Total 30 20 50

Based on these data we can say that A) no conclusion is possible without knowing α. B) appraisal and actual price are not independent at α = .05. C) appraisal and actual price are independent at any α. D) the degrees of freedom are insufficient for a decision. Answer: C Explanation: Column frequencies are all in the same ratio 3:2, so perfect independence exists (e = f). Difficulty: 3 Hard Topic: 15.01 Chi-Square Test for Independence Learning Objective: 15-03 Perform a chi-square test for independence on a contingency table. Bloom's: Analyze AACSB: Analytical Thinking Accessibility: Keyboard Navigation

83) Preferences for the type of diet drink from a random sample of 121 shoppers are in the table below. A researcher is interested in determining if there is a relationship between the type of diet drink preferred and the age of the shoppers. Diet Pepsi

Observed Expected % of chisq

Under 25 10 21.40 28.6%

25 and older 60 48.60 12.6%

Total 70 70.00 41.3%

Diet Coke

Observed Expected % of chisq

18 9.79 32.5%

14 22.21 14.3%

32 32.00 46.8%

Diet Sprite

Observed Expected % of chisq

9 5.81 8.3%

10 13.19 3.6%

19 19.00 11.9%

Total

Observed Expected % of chisq

37 37.00 69.4%

84 84.00 30.6%

121 121.00 100.0%

21.21chi-square In performing a chi-square test of independence on these data, how many degrees of freedom will the test statistic have? A) 1 B) 2 C) 4 D) 6 Answer: B Explanation: d.f. = (3 − 1)(2 − 1) = 2. Difficulty: 1 Easy Topic: 15.01 Chi-Square Test for Independence Learning Objective: 15-02 Find degrees of freedom and use the chi-square table of critical values. Bloom's: Analyze AACSB: Analytical Thinking Accessibility: Keyboard Navigation

84) Preferences for the type of diet drink from a random sample of 121 shoppers are in the table below. A researcher is interested in determining if there is a relationship between the type of diet drink preferred and the age of the shoppers. Diet Pepsi

Observed Expected % of chisq

Under 25 10 21.40 28.6%

25 and older 60 48.60 12.6%

Total 70 70.00 41.3%

Diet Coke

Observed Expected % of chisq

18 9.79 32.5%

14 22.21 14.3%

32 32.00 46.8%

Diet Sprite

Observed Expected % of chisq

9 5.81 8.3%

10 13.19 3.6%

19 19.00 11.9%

Total

Observed Expected % of chisq

37 37.00 69.4%

84 84.00 30.6%

121 121.00 100.0%

21.21chi-square Using α = .025, what is the critical value of the test statistic that you would use in a decision rule to test an appropriate hypothesis? A) 5.02 B) 5.99 C) 7.38 D) 14.45 Answer: C Explanation: χ2.025 = 7.378 for d.f. = (3 − 1)(2 − 1) = 2. The Excel function is =CHISQ.INV.RT(0.025,2) = 7.37776. Difficulty: 2 Medium Topic: 15.01 Chi-Square Test for Independence Learning Objective: 15-02 Find degrees of freedom and use the chi-square table of critical values. Bloom's: Analyze AACSB: Analytical Thinking Accessibility: Keyboard Navigation

85) Preferences for the type of diet drink from a random sample of 121 shoppers are in the table below. A researcher is interested in determining if there is a relationship between the type of diet drink preferred and the age of the shoppers. Diet Pepsi

Observed Expected % of chisq

Under 25 10 21.40 28.6%

25 and older 60 48.60 12.6%

Total 70 70.00 41.3%

Diet Coke

Observed Expected % of chisq

18 9.79 32.5%

14 22.21 14.3%

32 32.00 46.8%

Diet Sprite

Observed Expected % of chisq

9 5.81 8.3%

10 13.19 3.6%

19 19.00 11.9%

Total

Observed Expected % of chisq

37 37.00 69.4%

84 84.00 30.6%

121 121.00 100.0%

21.21chi-square What can you conclude for the data analysis at α = .05? A) The means are equal for all three groups. B) There is insufficient evidence to conclude that the type of drink and age are dependent. C) The type of drink and age are dependent. D) No conclusion is possible without knowing more information. Answer: C Explanation: χ2calc = 21.21 greatly exceeds χ2.05 = 5.991 for d.f. = (3 − 1)(2 − 1) = 2. Difficulty: 3 Hard Topic: 15.01 Chi-Square Test for Independence Learning Objective: 15-03 Perform a chi-square test for independence on a contingency table. Bloom's: Analyze AACSB: Analytical Thinking Accessibility: Keyboard Navigation

86) A taste test of randomly selected students was conducted to see if there was a difference in preferences among four popular drinks. The following table shows the frequency of responses. Beverage Frequency

Coke 51

Pepsi 66

A&W Root Beer 43

Dr Pepper 40

The expected number of students preferring Dr. Pepper is A) 25 B) 40 C) 50 D) 60 Answer: C Explanation: n = 51 + 66 + 43 + 40 = 200 so, assuming a uniform distribution, e = n/k = 200/4 = 50. Difficulty: 2 Medium Topic: 15.03 Uniform Goodness-of-Fit Test Learning Objective: 15-05 Perform a GOF test for a uniform distribution. Bloom's: Analyze AACSB: Analytical Thinking Accessibility: Keyboard Navigation 87) A taste test of randomly selected students was conducted to see if there was a difference in preferences among four popular drinks. The following table shows the frequency of responses. Beverage Frequency

Coke 51

Pepsi 66

A&W Root Beer 43

Dr Pepper 40

Using α = .025, the critical value of the test you would use in determining whether the preferences are the same among the drinks is A) 5.991 B) 7.378 C) 9.348 D) 11.07 Answer: C Explanation: χ2.025 = 9.348 for d.f. = k − 1 = 4 − 1 = 3. The Excel function is =CHISQ.INV.RT(0.025,3) = 9.34840. Difficulty: 1 Easy Topic: 15.03 Uniform Goodness-of-Fit Test Learning Objective: 15-05 Perform a GOF test for a uniform distribution. Bloom's: Analyze AACSB: Analytical Thinking Accessibility: Keyboard Navigation

88) A taste test of randomly selected students was conducted to see if there was a difference in preferences among four popular drinks. The following table shows the frequency of responses. Beverage Frequency

Coke 51

Pepsi 66

A&W Root Beer 43

Dr Pepper 40

The value of the chi-square test statistic you would use in testing whether the preferences are the same among the drinks is A) 7.54 B) 8.12 C) 10.76 D) 12.56 Answer: B Explanation: (51 − 50)2/50 + (66 − 50)2/50 + (43 − 50)2/50 + (40 − 50)2/50 = 8.12. Difficulty: 2 Medium Topic: 15.03 Uniform Goodness-of-Fit Test Learning Objective: 15-05 Perform a GOF test for a uniform distribution. Bloom's: Analyze AACSB: Analytical Thinking Accessibility: Keyboard Navigation 89) A taste test of randomly selected students was conducted to see if there was a difference in preferences among four popular drinks. The following table shows the frequency of responses. Beverage Frequency

Coke 51

Pepsi 66

A&W Root Beer 43

Dr Pepper 40

Using α = .025, what can you conclude from your analysis? A) Reject the null and conclude some drinks are preferred more than others. B) There is not enough evidence to say that a preference exists. C) Pepsi is the preferred drink. D) Form no conclusion because Cochran's Rule is violated. Answer: B Explanation: χ2calc = (51 − 50)2/50 + (66 − 50)2/50 + (43 − 50)2/50 + (40 − 50)2/50 = 8.12 does not exceed χ2.025 = 9.348 for d.f. = k − 1 = 4 − 1 = 3. Difficulty: 3 Hard Topic: 15.03 Uniform Goodness-of-Fit Test Learning Objective: 15-05 Perform a GOF test for a uniform distribution. Bloom's: Analyze AACSB: Analytical Thinking Accessibility: Keyboard Navigation

90) The Oxnard Retailers Anti-Theft Alliance (ORATA) published a study that claimed the causes of disappearance of inventory in retail stores were 30 percent shoplifting, 50 percent employee theft, and 20 percent faulty paperwork. The manager of the Melodic Kortholt Outlet performed an audit of the disappearance of 80 items and found the frequencies shown below. She would like to know if her store's experience follows the same pattern as other retailers. Reason Frequency

Shoplifting 32

Employee Theft 38

Poor Paperwork 10

Under the null hypothesis that her store follows the published pattern, the expected number of items that disappeared due to shoplifting is A) 16 B) 40 C) 24 D) 27 Answer: C Explanation: n = 32 + 38 + 11 = 80, so for shoplifting e = .30 × 80 = 24. Difficulty: 2 Medium Topic: 15.02 Chi-Square Tests for Goodness-of-Fit Learning Objective: 15-04 Perform a goodness-of-fit (GOF) test for a multinomial distribution. Bloom's: Analyze AACSB: Analytical Thinking Accessibility: Keyboard Navigation

91) The Oxnard Retailers Anti-Theft Alliance (ORATA) published a study that claimed the causes of disappearance of inventory in retail stores were 30 percent shoplifting, 50 percent employee theft, and 20 percent faulty paperwork. The manager of the Melodic Kortholt Outlet performed an audit of the disappearance of 80 items and found the frequencies shown below. She would like to know if her store's experience follows the same pattern as other retailers. Reason Frequency

Shoplifting 32

Employee Theft 38

Poor Paperwork 10

Using α = .05, the critical value you would use in determining whether the Melodic Kortholt Outlet's pattern differs from the published study is A) 7.815 B) 5.991 C) 1.960 D) 1.645 Answer: B Explanation: χ2.05 = 5.991 for d.f. = k − 1 = 3 − 1 = 2. The Excel function is =CHISQ.INV.RT(0.05,2) = 5.99146. Difficulty: 2 Medium Topic: 15.02 Chi-Square Tests for Goodness-of-Fit Learning Objective: 15-04 Perform a goodness-of-fit (GOF) test for a multinomial distribution. Bloom's: Analyze AACSB: Analytical Thinking Accessibility: Keyboard Navigation

92) The Oxnard Retailers Anti-Theft Alliance (ORATA) published a study that claimed the causes of disappearance of inventory in retail stores were 30 percent shoplifting, 50 percent employee theft, and 20 percent faulty paperwork. The manager of the Melodic Kortholt Outlet performed an audit of the disappearance of 80 items and found the frequencies shown below. She would like to know if her store's experience follows the same pattern as other retailers. Reason Frequency

Shoplifting 32

Employee Theft 38

Poor Paperwork 10

The value of the chi-square test statistic you would use in testing whether there is a difference from the published pattern is A) 7.54 B) 5.02 C) 9.76 D) 9.22 Answer: B Explanation: (32 − 24)2/24 + (38 − 40)2/40 + (10 − 16)2/16 = 5.0167 with π1 = .30, π2 = .50, π3 = .20 and n = 80. Difficulty: 3 Hard Topic: 15.02 Chi-Square Tests for Goodness-of-Fit Learning Objective: 15-04 Perform a goodness-of-fit (GOF) test for a multinomial distribution. Bloom's: Analyze AACSB: Analytical Thinking Accessibility: Keyboard Navigation

93) The Oxnard Retailers Anti-Theft Alliance (ORATA) published a study that claimed the causes of disappearance of inventory in retail stores were 30 percent shoplifting, 50 percent employee theft, and 20 percent faulty paperwork. The manager of the Melodic Kortholt Outlet performed an audit of the disappearance of 80 items and found the frequencies shown below. She would like to know if her store's experience follows the same pattern as other retailers. Reason Frequency

Shoplifting 32

Employee Theft 38

Poor Paperwork 10

Using α = .05, what can you conclude from your analysis? A) The store's pattern is clearly significantly different from the published data. B) The store's pattern is almost, but not quite, significantly different from the published data. C) The store's pattern is very close to the published data. D) We can form no conclusion because Cochran's Rule is violated. Answer: B Explanation: χ2calc = (32 − 24)2/24 + (38 − 40)2/40 + (10 − 16)2/16 = 5.0167 with π1 = .30, π2 = .50, π3 = .20 and n = 80, and χ2.05 = 5.991 for d.f. = k − 1 = 3 − 1 = 2. Thus, we cannot quite reject H0: π1 = .30, π2 = .50, π3 = .20. Difficulty: 2 Medium Topic: 15.02 Chi-Square Tests for Goodness-of-Fit Learning Objective: 15-04 Perform a goodness-of-fit (GOF) test for a multinomial distribution. Bloom's: Analyze AACSB: Analytical Thinking Accessibility: Keyboard Navigation 94) A contingency table shows A) frequency counts. B) means of the data. C) event probabilities. D) chi-square values. Answer: A Explanation: Contingency tables contain count data. Difficulty: 1 Easy Topic: 15.01 Chi-Square Test for Independence Learning Objective: 15-01 Recognize a contingency table and understand how it is created. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation

95) We would create a contingency table by A) summing the probabilities of two variables. B) cross-tabulating frequencies of two variables. C) applying the chi-square distribution to a sample. D) using Cochran's Rule to estimate frequencies. Answer: B Explanation: A contingency table is a two-way frequency distribution. Difficulty: 1 Easy Topic: 15.01 Chi-Square Test for Independence Learning Objective: 15-01 Recognize a contingency table and understand how it is created. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation

96) Which data set is consistent with the hypothesis of a normal population?

A) Data Set A B) Data Set C C) Neither data set D) Both data sets Answer: A

Explanation: Data Set A has a linear probability plot. Data Set C is nonlinear and has a small pvalue for the Anderson-Darling test (shown as AD in the graphs) which suggests rejection of the hypothesis of normality. Difficulty: 2 Medium Topic: 15.06 ECDF Tests (Optional) Learning Objective: 15-08 Interpret ECDF tests and know their advantages compared to chisquare GOF tests. Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation 97) Which statement is most nearly correct regarding ECDF tests? A) An attraction of the Anderson-Darling test is that it is fairly easy to do without a computer. B) In an ECDF test for goodness-of-fit, the n observations are grouped into categories rather than being treated individually. C) When raw data are available, ECDF tests usually surpass the chi-square test in their ability to detect departures from the distribution specified in the null hypothesis. Answer: C Explanation: ECDF tests (e.g., Anderson-Darling, Kolmogorov-Smirnov, probability plot) generally gain power by considering each data point separately. However, these tests are not easy without a computer. Difficulty: 2 Medium Topic: 15.06 ECDF Tests (Optional) Learning Objective: 15-08 Interpret ECDF tests and know their advantages compared to chisquare GOF tests. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation

98) Sam wants to perform a goodness-of-fit test for a Poisson distribution using the following sample data on the frequency of arrivals per minute. What value should Sam use to estimate the required parameter (λ)? x 0 1 2 3 4 5 6 Total

f 3 9 14 11 7 4 2 50

A) 1.9 B) 2.1 C) 2.6 D) 3.1 Answer: C Explanation: The estimated mean is [3(0) + 9(1) + 14(2) + 11(3) + 7(4) + 4(5) + 2(6)]/50 = 130/50 = 2.60. Difficulty: 2 Medium Topic: 15.04 Poisson Goodness-of-Fit Test Learning Objective: 15-06 Explain the GOF test for a Poisson distribution. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

Applied Statistics in Business and Economics, 6e (Doane) Chapter 16 Nonparametric Tests 1) Nonparametric tests can be used in small samples. Answer: TRUE Explanation: Special tables (or suitable software) are required when sample sizes are small. Difficulty: 1 Easy Topic: 16.01 Why Use Nonparametric Tests? Learning Objective: 16-01 Define nonparametric tests and explain when they may be desirable. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 2) Nonparametric tests may require special tables for small samples. Answer: TRUE Explanation: Special tables (or suitable computer software) will be required for small samples. Difficulty: 1 Easy Topic: 16.01 Why Use Nonparametric Tests? Learning Objective: 16-01 Define nonparametric tests and explain when they may be desirable. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 3) Nonparametric tests generally are more powerful than parametric tests when normality cannot be assumed. Answer: TRUE Explanation: When justified, parametric tests generally have more power (but at the cost of making more restrictive assumptions). Difficulty: 2 Medium Topic: 16.01 Why Use Nonparametric Tests? Learning Objective: 16-01 Define nonparametric tests and explain when they may be desirable. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation

4) Parametric tests generally are more powerful than nonparametric tests when normality can be assumed. Answer: TRUE Explanation: When justified, parametric tests generally have more power. Difficulty: 2 Medium Topic: 16.01 Why Use Nonparametric Tests? Learning Objective: 16-01 Define nonparametric tests and explain when they may be desirable. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 5) Rejection of a hypothesis using a nonparametric test is less convincing than using an equivalent parametric test, since nonparametric tests generally make fewer assumptions. Answer: FALSE Explanation: Rejection in a nonparametric test is persuasive because fewer assumptions are made. Difficulty: 2 Medium Topic: 16.01 Why Use Nonparametric Tests? Learning Objective: 16-01 Define nonparametric tests and explain when they may be desirable. Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation 6) Rejection of a hypothesis using a nonparametric test is more convincing than using an equivalent parametric test when the data are badly skewed. Answer: TRUE Explanation: Rejection in a nonparametric test is persuasive when normality is in doubt. Difficulty: 2 Medium Topic: 16.01 Why Use Nonparametric Tests? Learning Objective: 16-01 Define nonparametric tests and explain when they may be desirable. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation

7) If the population is normal, we would usually prefer a nonparametric test. Answer: FALSE Explanation: When justified, parametric tests generally have more power. Difficulty: 1 Easy Topic: 16.01 Why Use Nonparametric Tests? Learning Objective: 16-01 Define nonparametric tests and explain when they may be desirable. Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation 8) Most nonparametric tests assume ordinal data. Answer: TRUE Explanation: Many nonparametric tests convert ratio or interval data into ranks (ordinal data). Difficulty: 1 Easy Topic: 16.01 Why Use Nonparametric Tests? Learning Objective: 16-01 Define nonparametric tests and explain when they may be desirable. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 9) Most nonparametric tests require data measured on a ratio scale. Answer: FALSE Explanation: Many nonparametric tests convert ratio or interval data into ranks (ordinal data). Difficulty: 1 Easy Topic: 16.01 Why Use Nonparametric Tests? Learning Objective: 16-01 Define nonparametric tests and explain when they may be desirable. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 10) The one-sample runs test compares medians of two or more groups. Answer: FALSE Explanation: The runs test checks for randomness. Difficulty: 1 Easy Topic: 16.02 One-Sample Runs Test Learning Objective: 16-02 Use the one-sample runs test. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation

11) The one-sample runs test uses binary data (only two possible values). Answer: TRUE Explanation: Data may be any sequence of two values (e.g., + or −). Difficulty: 1 Easy Topic: 16.02 One-Sample Runs Test Learning Objective: 16-02 Use the one-sample runs test. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 12) The Spearman rank correlation test compares medians of two paired data sets. Answer: FALSE Explanation: To compare medians, we use a Mann-Whitney test (Wilcoxon rank sum test). Difficulty: 1 Easy Topic: 16.07 Spearman Rank Correlation Test Learning Objective: 16-07 Use the Spearman rank correlation test. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 13) The one-sample runs test uses a test statistic that is normally distributed as long as the number of runs of each type is large enough. Answer: TRUE Explanation: As long as we have enough runs of each type, we can use a normal z test. Difficulty: 1 Easy Topic: 16.02 One-Sample Runs Test Learning Objective: 16-02 Use the one-sample runs test. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 14) The one-sample runs test is useful for detecting nonrandom patterns in time-series data. Answer: TRUE Explanation: Random pattern is essentially the null hypothesis. Difficulty: 1 Easy Topic: 16.02 One-Sample Runs Test Learning Objective: 16-02 Use the one-sample runs test. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation

15) The one-sample runs test is similar to a test for autocorrelation. Answer: TRUE Explanation: We can use the runs test to check for autocorrelation in regression residuals. Difficulty: 1 Easy Topic: 16.02 One-Sample Runs Test Learning Objective: 16-02 Use the one-sample runs test. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 16) The one-sample runs test is also called the Wald-Wolfowitz test after its inventors. Answer: TRUE Explanation: Wald and Wolfowitz first demonstrated the runs test, so it goes by either name. Difficulty: 1 Easy Topic: 16.02 One-Sample Runs Test Learning Objective: 16-02 Use the one-sample runs test. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 17) The Wilcoxon signed-rank test is useful when comparing one sample median with a benchmark. Answer: TRUE Explanation: The null hypothesis is for one population median versus a target or benchmark. Difficulty: 2 Medium Topic: 16.03 Wilcoxon Signed-Rank Test Learning Objective: 16-03 Use the Wilcoxon signed-rank test. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 18) The Wilcoxon signed-rank test is useful when comparing more than two sample medians. Answer: FALSE Explanation: For two medians, we would use the Wilcoxon rank sum test (Mann-Whitney test). Difficulty: 2 Medium Topic: 16.03 Wilcoxon Signed-Rank Test Learning Objective: 16-03 Use the Wilcoxon signed-rank test. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation

19) The Wilcoxon signed-rank test is analogous to a one-sample parametric test of a mean. Answer: TRUE Explanation: The idea is to test a measure of center (in this case, a median) against a benchmark. Difficulty: 2 Medium Topic: 16.03 Wilcoxon Signed-Rank Test Learning Objective: 16-03 Use the Wilcoxon signed-rank test. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 20) The Wilcoxon signed-rank test is analogous to a parametric two-sample t test of means from independent samples. Answer: FALSE Explanation: For two medians, we would use the Wilcoxon rank sum test (Mann-Whitney test). Difficulty: 2 Medium Topic: 16.03 Wilcoxon Signed-Rank Test Learning Objective: 16-03 Use the Wilcoxon signed-rank test. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 21) The Wilcoxon signed-rank test is analogous to a parametric t test comparing three or more medians. Answer: FALSE Explanation: The null hypothesis is for one population median versus a target or benchmark. Difficulty: 2 Medium Topic: 16.03 Wilcoxon Signed-Rank Test Learning Objective: 16-03 Use the Wilcoxon signed-rank test. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 22) The Wilcoxon signed-rank test is robust to outliers in the data if the population is at least somewhat symmetric. Answer: TRUE Explanation: The test does require reasonably symmetric populations. Difficulty: 2 Medium Topic: 16.03 Wilcoxon Signed-Rank Test Learning Objective: 16-03 Use the Wilcoxon signed-rank test. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 6 Copyright 2019 © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.

23) The Wilcoxon signed-rank test does not assume a normal population, but it does require a roughly symmetric population. Answer: TRUE Explanation: The test does require reasonably symmetric populations. Difficulty: 2 Medium Topic: 16.03 Wilcoxon Signed-Rank Test Learning Objective: 16-03 Use the Wilcoxon signed-rank test. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 24) The Wilcoxon signed-rank test is less powerful than a one-sample test of a mean when the population is actually normal. Answer: TRUE Explanation: When justified, a t test of a mean would generally be more powerful. Difficulty: 2 Medium Topic: 16.03 Wilcoxon Signed-Rank Test Learning Objective: 16-03 Use the Wilcoxon signed-rank test. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 25) The Wilcoxon signed-rank test has good power over a range of nonnormal populations. Answer: TRUE Explanation: Studies show good power when populations are not too skewed. Difficulty: 2 Medium Topic: 16.03 Wilcoxon Signed-Rank Test Learning Objective: 16-03 Use the Wilcoxon signed-rank test. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation

26) The Wilcoxon signed-rank test is an alternative to the one-sample t test for paired observations. Answer: TRUE Explanation: One version of the Wilcoxon signed-rank test utilizes paired data. Difficulty: 2 Medium Topic: 16.03 Wilcoxon Signed-Rank Test Learning Objective: 16-03 Use the Wilcoxon signed-rank test. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 27) The Wilcoxon rank sum test (Mann-Whitney test) can be used as a test for equality of two population medians. Answer: TRUE Explanation: This test for identical populations is a test of medians if the variances are equal. Difficulty: 2 Medium Topic: 16.04 Wilcoxon Rank Sum Test Learning Objective: 16-04 Use the Wilcoxon rank sum test for two samples. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 28) The Wilcoxon rank sum test (Mann-Whitney test) requires two independent samples. Answer: TRUE Explanation: The samples must be independent. Difficulty: 2 Medium Topic: 16.04 Wilcoxon Rank Sum Test Learning Objective: 16-04 Use the Wilcoxon rank sum test for two samples. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation

29) The Wilcoxon rank sum test (Mann-Whitney test) can be used even when the population variances are unequal. Answer: FALSE Explanation: Wilcoxon rank sum test (Mann-Whitney test) is a test for identical populations, but it is a test for equal medians if the variances are equal. Difficulty: 2 Medium Topic: 16.04 Wilcoxon Rank Sum Test Learning Objective: 16-04 Use the Wilcoxon rank sum test for two samples. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 30) The Wilcoxon rank sum test (Mann-Whitney test) utilizes the ranks of two independent samples. Answer: TRUE Explanation: Raw data in each sample are converted into ranks. Difficulty: 2 Medium Topic: 16.04 Wilcoxon Rank Sum Test Learning Objective: 16-04 Use the Wilcoxon rank sum test for two samples. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 31) The Mann-Whitney test is sometimes called the "Wilcoxon rank sum test" because it was formulated independently by different statisticians. Answer: TRUE Explanation: This test has two names. Difficulty: 2 Medium Topic: 16.04 Wilcoxon Rank Sum Test Learning Objective: 16-04 Use the Wilcoxon rank sum test for two samples. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation

32) We sometimes call the Wilcoxon rank sum test the Mann/Whitney test, but the two are the same. Answer: TRUE Explanation: This test has two names. Difficulty: 2 Medium Topic: 16.04 Wilcoxon Rank Sum Test Learning Objective: 16-04 Use the Wilcoxon rank sum test for two samples. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 33) The Mann-Whitney test is analogous to a one-sample t test comparing a mean with a benchmark. Answer: FALSE Explanation: The Wilcoxon rank sum test (Mann-Whitney test) compares two populations. Difficulty: 2 Medium Topic: 16.04 Wilcoxon Rank Sum Test Learning Objective: 16-04 Use the Wilcoxon rank sum test for two samples. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 34) The Kruskal-Wallis test requires c independent samples (where usually c > 2). Answer: TRUE Explanation: Kruskal-Wallis compares several populations. Difficulty: 2 Medium Topic: 16.05 Kruskal-Wallis Test for Independent Samples Learning Objective: 16-05 Use the Kruskal-Wallis test for c independent samples. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 35) The Kruskal-Wallis test applies even if the c samples to be compared are not independent. Answer: FALSE Explanation: Samples must be independent in the Kruskal-Wallis test. Difficulty: 2 Medium Topic: 16.05 Kruskal-Wallis Test for Independent Samples Learning Objective: 16-05 Use the Kruskal-Wallis test for c independent samples. Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation 10 Copyright 2019 © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.

36) The Kruskal-Wallis test is analogous to the one-sample parametric t test for a mean. Answer: FALSE Explanation: Kruskal-Wallis compares several populations. Difficulty: 2 Medium Topic: 16.05 Kruskal-Wallis Test for Independent Samples Learning Objective: 16-05 Use the Kruskal-Wallis test for c independent samples. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 37) The Kruskal-Wallis test is analogous to the parametric one-factor ANOVA. Answer: TRUE Explanation: Kruskal-Wallis compares c population medians (similar to comparing c means). Difficulty: 2 Medium Topic: 16.05 Kruskal-Wallis Test for Independent Samples Learning Objective: 16-05 Use the Kruskal-Wallis test for c independent samples. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 38) The Kruskal-Wallis test does not require normal populations, but it does require them to be of similar shape. Answer: TRUE Explanation: Kruskal-Wallis is a test for identical populations but becomes a test for equal medians if population shapes are the same. Difficulty: 2 Medium Topic: 16.05 Kruskal-Wallis Test for Independent Samples Learning Objective: 16-05 Use the Kruskal-Wallis test for c independent samples. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 39) The Kruskal-Wallis test is equivalent to comparing medians in c groups. Answer: TRUE Explanation: Kruskal-Wallis is a test for identical populations, but it becomes a test for equal medians if population shapes are the same. Difficulty: 1 Easy Topic: 16.05 Kruskal-Wallis Test for Independent Samples Learning Objective: 16-05 Use the Kruskal-Wallis test for c independent samples. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 11 Copyright 2019 © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.

40) The Kruskal-Wallis test is a test for randomness in sequential data. Answer: FALSE Explanation: The Kruskal-Wallis compares c population medians. To test for randomness, you would use a runs test rather than a Kruskal-Wallis test. Difficulty: 1 Easy Topic: 16.05 Kruskal-Wallis Test for Independent Samples Learning Objective: 16-05 Use the Kruskal-Wallis test for c independent samples. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 41) The Kruskal-Wallis test is less useful in finance or marketing, because normal populations can usually be assumed for financial or marketing data. Answer: FALSE Explanation: Normal populations are rare in many business applications, so the Kruskal-Wallis test is often useful. Difficulty: 1 Easy Topic: 16.05 Kruskal-Wallis Test for Independent Samples Learning Objective: 16-05 Use the Kruskal-Wallis test for c independent samples. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 42) The Kruskal-Wallis test is almost as powerful as one-factor ANOVA even when normality can be assumed. Answer: TRUE Explanation: Studies suggest that the Kruskal-Wallis test is about as powerful as the F test (ANOVA). Difficulty: 1 Easy Topic: 16.05 Kruskal-Wallis Test for Independent Samples Learning Objective: 16-05 Use the Kruskal-Wallis test for c independent samples. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation

43) The Friedman test is a nonparametric equivalent to two-factor ANOVA without replication. Answer: TRUE Explanation: Unreplicated ANOVA data can be tested using the Friedman test. Difficulty: 2 Medium Topic: 16.06 Friedman Test for Related Samples Learning Objective: 16-06 Use the Friedman test for related samples. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 44) The Friedman test is a nonparametric equivalent to a parametric one-sample t test comparing a mean with a benchmark. Answer: FALSE Explanation: The Friedman test is a nonparametric equivalent to two-factor ANOVA without replication. Difficulty: 2 Medium Topic: 16.06 Friedman Test for Related Samples Learning Objective: 16-06 Use the Friedman test for related samples. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 45) The Friedman test requires groups of equal size. Answer: TRUE Explanation: By definition, there must be one observation per cell. Difficulty: 2 Medium Topic: 16.06 Friedman Test for Related Samples Learning Objective: 16-06 Use the Friedman test for related samples. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 46) The Friedman test compares medians when there are two grouping factors (rows, columns). Answer: TRUE Explanation: We can think of the Friedman test as comparing medians if the population shape is similar. Difficulty: 1 Easy Topic: 16.06 Friedman Test for Related Samples Learning Objective: 16-06 Use the Friedman test for related samples. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 13 Copyright 2019 © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.

47) The Friedman test resembles the Kruskal-Wallis test except that there are two grouping factors (rows, columns) instead of one grouping factor (columns). Answer: TRUE Explanation: The Friedman test is analogous to unreplicated two-factor ANOVA. Difficulty: 2 Medium Topic: 16.06 Friedman Test for Related Samples Learning Objective: 16-06 Use the Friedman test for related samples. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 48) The Friedman test is often almost as powerful as two-factor ANOVA without replication (randomized block design). Answer: TRUE Explanation: The Friedman test has good power versus two-factor unreplicated ANOVA. Difficulty: 2 Medium Topic: 16.06 Friedman Test for Related Samples Learning Objective: 16-06 Use the Friedman test for related samples. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 49) Spearman's rank correlation always lies within the range −1.00 to +1.00. Answer: TRUE Explanation: Spearman's rho lies between −1 and +1. Difficulty: 1 Easy Topic: 16.07 Spearman Rank Correlation Test Learning Objective: 16-07 Use the Spearman rank correlation test. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 50) Spearman's rank correlation coefficient lies within the range 0 to 1.00 inclusive. Answer: FALSE Explanation: Spearman's rho lies between −1 and +1. Difficulty: 1 Easy Topic: 16.07 Spearman Rank Correlation Test Learning Objective: 16-07 Use the Spearman rank correlation test. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 14 Copyright 2019 © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.

51) Spearman's rank correlation is named for a British brewer who was testing beer samples. Answer: FALSE Explanation: Charles Spearman was a behavioral psychologist. Difficulty: 1 Easy Topic: 16.07 Spearman Rank Correlation Test Learning Objective: 16-07 Use the Spearman rank correlation test. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 52) Spearman's rank correlation is named for a British behavioral psychologist who was studying human intelligence. Answer: TRUE Explanation: Charles Spearman was a behavioral psychologist studying human IQs. Difficulty: 1 Easy Topic: 16.07 Spearman Rank Correlation Test Learning Objective: 16-07 Use the Spearman rank correlation test. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 53) Spearman's rank correlation is used to measure agreement in rankings of paired (x, y) data. Answer: TRUE Explanation: Spearman's rho is similar in this regard to the usual (Pearson) correlation. Difficulty: 1 Easy Topic: 16.07 Spearman Rank Correlation Test Learning Objective: 16-07 Use the Spearman rank correlation test. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation

54) All of these are nonparametric tests except A) Spearman rank correlation test. B) Friedman test. C) Student's t test. D) Kruskal-Wallis test. Answer: C Explanation: Student's t test (Chapter 10) is a parametric test. Difficulty: 2 Medium Topic: 16.01 Why Use Nonparametric Tests? Learning Objective: 16-01 Define nonparametric tests and explain when they may be desirable. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 55) Which nonparametric test is analogous to a parametric two-sample t test for means? A) Wald-Wolfowitz one-sample runs test B) Wilcoxon signed-rank test C) Wilcoxon rank sum (Mann-Whitney) test D) Kruskal-Wallis test Answer: C Explanation: The Wilcoxon rank sum (Mann-Whitney) test compares two medians if the populations have similar shape. Difficulty: 2 Medium Topic: 16.04 Wilcoxon Rank Sum Test Learning Objective: 16-04 Use the Wilcoxon rank sum test for two samples. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 56) Which nonparametric test is analogous to a parametric k-sample test for means? A) Kruskal-Wallis test B) Wilcoxon signed-rank test C) Wilcoxon rank sum (Mann-Whitney) test D) Wald-Wolfowitz one-sample runs test Answer: A Explanation: The Kruskal-Wallis test compares k medians if the populations have similar shape. Difficulty: 2 Medium Topic: 16.05 Kruskal-Wallis Test for Independent Samples Learning Objective: 16-05 Use the Kruskal-Wallis test for c independent samples. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 16 Copyright 2019 © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.

57) Which nonparametric test is used to compare a one-sample median with a benchmark? A) Wald-Wolfowitz one-sample runs test B) Wilcoxon signed-rank test C) Wilcoxon rank sum (Mann-Whitney) test D) Kruskal-Wallis test Answer: B Explanation: The Wilcoxon signed-rank test compares a median with a benchmark or target. Difficulty: 2 Medium Topic: 16.03 Wilcoxon Signed-Rank Test Learning Objective: 16-03 Use the Wilcoxon signed-rank test. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 58) Which nonparametric test has no parametric counterpart? A) Wald-Wolfowitz one-sample runs test B) Spearman rank correlation test C) Wilcoxon rank sum (Mann-Whitney) test D) Kruskal-Wallis test Answer: A Explanation: The runs test is not quite like any parametric test. Difficulty: 3 Hard Topic: 16.02 One-Sample Runs Test Learning Objective: 16-02 Use the one-sample runs test. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 59) Which nonparametric test is analogous to a parametric one-sample t test for differences in paired data? A) Wald-Wolfowitz one-sample runs test B) Wilcoxon signed-rank test C) Wilcoxon rank sum (Mann-Whitney) test D) Kruskal-Wallis test Answer: B Explanation: One version of the Wilcoxon signed-rank test is for paired data. Difficulty: 2 Medium Topic: 16.03 Wilcoxon Signed-Rank Test Learning Objective: 16-03 Use the Wilcoxon signed-rank test. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 17 Copyright 2019 © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.

60) Which nonparametric test would we use to compare ratings assigned to n pairs of bonds by two different rating agencies if the bonds are rated on an ordinal scale (Aaa, Aa, A, Baa, Ba B, etc.)? A) Wald-Wolfowitz one-sample runs test B) Wilcoxon signed-rank test C) Wilcoxon rank sum (Mann-Whitney) test D) Kruskal-Wallis test Answer: B Explanation: One version of the Wilcoxon signed-rank test is for paired data. Difficulty: 2 Medium Topic: 16.03 Wilcoxon Signed-Rank Test Learning Objective: 16-03 Use the Wilcoxon signed-rank test. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 61) Which is not true of the one-sample runs test? A) It is also called the Wald-Wolfowitz test after its inventors. B) Its purpose is to detect nonrandomness. C) It cannot be applied to sequential observations. D) It is similar to a test for autocorrelation. Answer: C Explanation: The runs test is for randomness in a sequence of two-valued data (e.g., + and −). Difficulty: 2 Medium Topic: 16.02 One-Sample Runs Test Learning Objective: 16-02 Use the one-sample runs test. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 62) Which nonparametric test is used to detect nonrandomness in sequential observations? A) Wald-Wolfowitz one-sample runs test B) Wilcoxon signed-rank test C) Wilcoxon rank sum (Mann-Whitney) test D) Kruskal-Wallis test Answer: A Explanation: The runs test is for randomness in a sequence of two-valued data (e.g., + and −). Difficulty: 2 Medium Topic: 16.02 One-Sample Runs Test Learning Objective: 16-02 Use the one-sample runs test. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 18 Copyright 2019 © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.

63) Which is a nonparametric test for runs (autocorrelation) in a sequence of binary data? A) Kruskal-Wallis test B) Wilcoxon signed-rank test C) Wilcoxon rank sum (Mann-Whitney) test D) Wald-Wolfowitz one-sample runs test Answer: D Explanation: The runs test is for randomness in a sequence of two-valued data (e.g., + and −). Difficulty: 2 Medium Topic: 16.02 One-Sample Runs Test Learning Objective: 16-02 Use the one-sample runs test. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 64) Which parametric test resembles the nonparametric Spearman's rank test? A) t test of a correlation coefficient B) t test of two sample means C) t test of one sample mean D) F test of variances Answer: A Explanation: The t test of a correlation coefficient can be used to test for zero correlation. Difficulty: 1 Easy Topic: 16.07 Spearman Rank Correlation Test Learning Objective: 16-07 Use the Spearman rank correlation test. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 65) Which nonparametric test is used to test for agreement in ranks of paired data? A) Spearman test B) Friedman test C) Wilcoxon rank sum (Mann-Whitney) test D) Kruskal-Wallis test Answer: A Explanation: To see if the ranks agree, we could use Spearman's rho. Difficulty: 1 Easy Topic: 16.07 Spearman Rank Correlation Test Learning Objective: 16-07 Use the Spearman rank correlation test. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 19 Copyright 2019 © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.

66) Twenty customers are randomly chosen, and each is given a sample of Nut Butter ice cream to taste. The customers rank the taste of the ice cream on a 10-point scale. Each is then given a sample of Chewy Gooey ice cream to taste and rank on a 10-point scale. Based on the 20 customers' ratings of each flavor, the research analyst conducting the survey wishes to determine if one flavor is preferred over the other. Which nonparametric test would you use to determine if there is a difference in the ratings? A) Wald-Wolfowitz one-sample runs test B) Wilcoxon signed-rank test C) Wilcoxon rank sum (Mann-Whitney) test D) Kruskal-Wallis test Answer: C Explanation: For two independent samples, use the Wilcoxon rank sum test (Mann-Whitney). Difficulty: 3 Hard Topic: 16.04 Wilcoxon Rank Sum Test Learning Objective: 16-04 Use the Wilcoxon rank sum test for two samples. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 67) Which is not a characteristic of the Wilcoxon rank sum (Mann-Whitney) test? A) It can be used as a test for equality of two population medians. B) It is analogous to a one-sample t test comparing a mean with a benchmark. C) It requires independent samples from populations with equal variances. D) It is similar to one-factor ANOVA when we have c independent samples. Answer: D Explanation: The Wilcoxon/Mann-Whitney test uses two samples (not c samples). Difficulty: 2 Medium Topic: 16.04 Wilcoxon Rank Sum Test Learning Objective: 16-04 Use the Wilcoxon rank sum test for two samples. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

68) Delta Air Lines wants to determine if the number of no-shows for flights originating from Detroit is higher than from Minneapolis. A sample of 20 flights is taken from each city, and the number of no-shows is determined for each flight. Which nonparametric test would you use to determine whether the number of no-shows is greater in Detroit? A) Wald-Wolfowitz one-sample runs test B) Wilcoxon signed-rank test C) Wilcoxon rank sum (Mann-Whitney) test D) Kruskal-Wallis test Answer: C Explanation: For two independent samples, use the Wilcoxon rank sum test (Mann-Whitney). Difficulty: 3 Hard Topic: 16.04 Wilcoxon Rank Sum Test Learning Objective: 16-04 Use the Wilcoxon rank sum test for two samples. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 69) The dean of Oxnard business school wants to know if there is a difference in computer skills for students in four majors (marketing, finance, operations, accounting). A 20-question skills test is given to 10 randomly chosen students in each major. Which nonparametric test could be used by the dean to see if there is a difference in computer skills of students in the various majors? A) Wald-Wolfowitz test B) Wilcoxon signed-rank test C) Mann-Whitney test D) Kruskal-Wallis test Answer: D Explanation: Because we have four samples, we would choose the Kruskal-Wallis test to compare the medians. Difficulty: 3 Hard Topic: 16.05 Kruskal-Wallis Test for Independent Samples Learning Objective: 16-05 Use the Kruskal-Wallis test for c independent samples. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

70) Which is not a characteristic of the Kruskal-Wallis test? A) It is analogous to the parametric one-factor ANOVA. B) It does not make any assumptions about the populations. C) It is useful in comparing the medians in c groups. D) It does not require normal distributions of populations. Answer: B Explanation: The Kruskal-Wallis test requires populations of similar shape if we want to compare their medians. Difficulty: 2 Medium Topic: 16.05 Kruskal-Wallis Test for Independent Samples Learning Objective: 16-05 Use the Kruskal-Wallis test for c independent samples. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 71) Individuals in four different age groups are asked to rate, on a scale of 1 to 10, three flavors of ice creams. Each group has the same number of individuals. Median ratings by the individuals are grouped by the flavor of ice cream and by age group. Which nonparametric test could be used to see if there is a difference in median ratings among the four age groups and three flavors? A) Wilcoxon signed-rank test B) Friedman test C) Mann-Whitney test D) Kruskal-Wallis test Answer: B Explanation: This is a two-factor comparison using medians, so we would use Friedman's test. Difficulty: 3 Hard Topic: 16.06 Friedman Test for Related Samples Learning Objective: 16-06 Use the Friedman test for related samples. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

72) Which nonparametric test is analogous to a two-factor ANOVA without replication? A) Wilcoxon signed-rank test B) Friedman test C) Mann-Whitney test D) Kruskal-Wallis test Answer: B Explanation: Friedman's test uses a two-factor experiment with one observed median per cell. Difficulty: 1 Easy Topic: 16.06 Friedman Test for Related Samples Learning Objective: 16-06 Use the Friedman test for related samples. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 73) If there are three or more populations with ordinal data, what is the appropriate test to determine whether the distributions are equal? A) Friedman test B) t test C) ANOVA D) Kruskal-Wallis test Answer: D Explanation: Kruskal-Wallis is a test for identical populations (or a test for equal medians, if populations differ only in center). Difficulty: 2 Medium Topic: 16.05 Kruskal-Wallis Test for Independent Samples Learning Objective: 16-05 Use the Kruskal-Wallis test for c independent samples. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

74) Hoping to reduce the waiting time, a doctor's office tried a new way of scheduling its appointments. Populations of waiting times are believed to be similar except in central tendency. However, waiting times may not be normally distributed, so a nonparametric test was chosen to compare the waiting times. The Wilcoxon/Mann-Whitney test results are shown below. n 21 25 46

Mann/Whitney Test sum of ranks 600.5 Old Way 480.5 New Way 1081 Total

493.50 expected value 45.31 standard deviation 2.35 z, corrected for ties .0094 p-value (one-tailed, upper) Which is the best conclusion? A) The medians differ at α = .01. B) The means differ at α = .01. C) The samples differ at α = .01. D) There are no differences at α = .01. Answer: A Explanation: The p-value is less than .01, so we reject equal medians Difficulty: 1 Easy Topic: 16.04 Wilcoxon Rank Sum Test Learning Objective: 16-04 Use the Wilcoxon rank sum test for two samples. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

75) At the Food Barn, children can order from the adult menu (A) or the child's menu (C). Below is the pattern of menu orders for 24 children last Wednesday evening. A, A, A, A, C, A, A, C, C, C, A, A, A, A, A, A, C, C, C, A, C, C, C, C Which test would you choose to see if this pattern is random? A) Kruskal-Wallis test B) Wilcoxon rank sum (Mann-Whitney) test C) Wald-Wolfowitz one-sample runs test D) Wilcoxon test Answer: C Explanation: Use the runs test for sequences that indicate a nonrandom pattern. Difficulty: 2 Medium Topic: 16.02 One-Sample Runs Test Learning Objective: 16-02 Use the one-sample runs test. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 76) At the Food Barn, children can order from the adult menu (A) or the child's menu (C). Below is the pattern of menu orders for 24 children last Wednesday evening. A, A, A, A, C, A, A, C, C, C, A, A, A, A, A, A, C, C, C, A, C, C, C, C Which statement is most accurate? A) Neither sample size (for outcomes A or C) is large enough for a runs test using z. B) Both sample sizes (for outcomes A and C) are large enough for a runs test using z. C) Only one outcome has a large enough sample for a runs test using z. D) This type of data would not be suitable for a runs test using z. Answer: B Explanation: We need 10 outcomes of each type to use a large-sample z test. Difficulty: 2 Medium Topic: 16.02 One-Sample Runs Test Learning Objective: 16-02 Use the one-sample runs test. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

77) A clinic has four doctors. They wish to compare the amount of time that doctors spend with their patients. It is suspected that times may not be normally distributed but are otherwise assumed similar except in center, so a nonparametric test was chosen. The Kruskal-Wallis test results are shown below. Median 27.00 28.50 46.00 31.00 31.00

n 11 14 9 11 45

Kruskal-Wallis Test Avg. Rank 19.18 15.93 36.39 24.86

Doc A Doc B Doc C Doc D Total

14.599 H (corrected for ties) 3 d.f .0022 p-value Which is the best conclusion? A) The doctors have the same median times at α = .01. B) The doctors have different median times at α = .01. C) The sample size is insufficient for this kind of test. D) A one-factor ANOVA would be a better kind of test. Answer: B Explanation: The p-value is less than .01. Difficulty: 2 Medium Topic: 16.05 Kruskal-Wallis Test for Independent Samples Learning Objective: 16-05 Use the Kruskal-Wallis test for c independent samples. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

78) Returns on an investor's stock portfolio (n = 19 stocks) are compared for the same stock in each of two consecutive quarters. Since the returns are not normally distributed (normality test pvalues were .005 and .126 respectively), a nonparametric test was chosen. The test results are shown below. n 19 19 38

Wilcoxon Paired Data Test sum of ranks 247 Qtr 1 494 Qtr 2 741 total 370.50 expected value 34.25 standard deviation −3.59 z .0003 p-value (two-tailed)

Which is the best conclusion? A) The means are the same at α = .01. B) The medians are the same at α = .01. C) The means differ at α = .01. D) The medians differ at α = .01. Answer: D Explanation: The p-value is less than .01. Difficulty: 2 Medium Topic: 16.03 Wilcoxon Signed-Rank Test Learning Objective: 16-03 Use the Wilcoxon signed-rank test. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

79) Returns on an investor's stock portfolio (n = 19 stocks) are compared for the same stock in each of two consecutive quarters. Since the returns are not normally distributed (normality test pvalues were .005 and .126 respectively), a nonparametric test was chosen. The test results are shown below.

Qtr1 Qtr2

Spearman Rank Correlation Qtr1 Qtr2 1.000 .577 1.000 19 sample size ±.456 critical value .05 (two-tail) ±.575 critical value 0.15 (two-tail)

Which is the best conclusion? A) The returns are correlated neither at α = .05 nor at α = .01. B) The returns are correlated at α = .05 but not at α = .01. C) The returns are correlated at α = .01 but not at α = .05. D) The returns are correlated both at α = .05 and at α = .01. Answer: D Explanation: The correlation differs from zero at α =.05 and .01 (close decision). Difficulty: 2 Medium Topic: 16.07 Spearman Rank Correlation Test Learning Objective: 16-07 Use the Spearman rank correlation test. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

80) Attendees at an outdoor concert can buy pavilion tickets (A) or lawn tickets (B). Below is the output for the one-sample runs to test whether there is a pattern in 32 consecutive ticket purchases. n 21 11 32 15.44 2.50 −1.17 .2403

Runs Test runs 6 6 12

A B total

expected value standard deviation z p-value (two-tailed)

The best conclusion at α = .05 would be A) the pattern is random. B) there is a difference in medians. C) the sample sizes are too small. D) the sample sizes must be equal. Answer: A Explanation: The p-value is greater than .05 so we cannot reject randomness. Difficulty: 2 Medium Topic: 16.02 One-Sample Runs Test Learning Objective: 16-02 Use the one-sample runs test. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

81) To compare the cost of three shipping methods, a firm shipped five orders to each of four different destinations over a six-month period. Their total shipping costs are shown below. Shipper Speedy Ship GetItThere WeR Tops

Toledo 355 342 361

Destination Oshawa Janesville 435 422 441 402 430 435

Dallas 518 488 528

Which nonparametric test would you use to compare the median shipping costs among destinations and among shippers? A) Friedman test B) Kruskal-Wallis test C) Wilcoxon rank sum (Mann-Whitney) test D) Lebesgue-Stieltjes test Answer: A Explanation: The Friedman test is like an unreplicated two-factor ANOVA except that data are converted to ranks. Difficulty: 2 Medium Topic: 16.06 Friedman Test for Related Samples Learning Objective: 16-06 Use the Friedman test for related samples. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

82) At the Seymour Clinic, the total number of patients seen by three doctors over three days is shown. Day Mon Tue Wed Thu Fri

Dr. Able 19 20 22 19 20

Physician Dr. Baker 25 22 24 20 34

Dr. Chow 27 27 32 22 27

Which nonparametric test would you use to compare the median number of patients seen by the doctors on each day? A) Wilcoxon test B) Kruskal-Wallis test C) Anderson-Darling test D) Friedman test Answer: D Explanation: The Friedman test is like an unreplicated two-factor ANOVA except that data are converted to ranks. Difficulty: 2 Medium Topic: 16.06 Friedman Test for Related Samples Learning Objective: 16-06 Use the Friedman test for related samples. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

83) Systolic blood pressure of randomly selected HMO patients was recorded on a particular Wednesday, with the results shown below. Under 20 105 113 108 114 123

Patient Age Group 20 to 29 30 to 49 110 122 101 114 112 128 127 124 123 125

50 and Over 139 115 136 124 123

Which is the appropriate hypothesis test to compare the medians? A) Wilcoxon test B) Kruskal-Wallis test C) Levene's test D) Friedman test Answer: B Explanation: The Kruskal-Wallis test is like a one-factor ANOVA except that data are converted to ranks. Difficulty: 1 Easy Topic: 16.05 Kruskal-Wallis Test for Independent Samples Learning Objective: 16-05 Use the Kruskal-Wallis test for c independent samples. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

84) You do not wish to assume normality in the population, yet you wish to compare central tendency in c samples. You decide to utilize MegaStat, whose menu is shown below.

Which menu would you choose? A) Hypothesis Tests B) Nonparametric Tests C) Correlation/Regression D) Analysis of Variance Answer: B Explanation: We would need the Kruskal-Wallis nonparametric test for c medians. Difficulty: 1 Easy Topic: 16.01 Why Use Nonparametric Tests? Learning Objective: 16-01 Define nonparametric tests and explain when they may be desirable. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

85) You do not wish to assume normality in the population, yet you wish to compare central tendency in c samples. You decide to utilize Minitab, whose menu is shown below.

Which menu would you choose? A) ANOVA B) Regression C) Nonparametrics D) EDA Answer: C Explanation: We would need the Kruskal-Wallis nonparametric test for c medians. Difficulty: 1 Easy Topic: 16.01 Why Use Nonparametric Tests? Learning Objective: 16-01 Define nonparametric tests and explain when they may be desirable. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

Applied Statistics in Business and Economics, 6e (Doane) Chapter 17 Quality Management 1) Quality control refers to methods used by organizations to ensure that their products and services meet customer expectations. Answer: TRUE Explanation: Quality control is a broad area encompassing various statistical and behavioral tools. Difficulty: 1 Easy Topic: 17.01 Quality and Variation Learning Objective: 17-01 Define quality and explain how it may be measured. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 2) An important attribute of quality is conformance to specifications. Answer: TRUE Explanation: Variation in products and services should lie within a desired target range. Difficulty: 1 Easy Topic: 17.01 Quality and Variation Learning Objective: 17-01 Define quality and explain how it may be measured. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 3) A major goal in statistical quality control is the reduction of variation in a process. Answer: TRUE Explanation: Services and products with less variation perform more consistently. Difficulty: 1 Easy Topic: 17.01 Quality and Variation Learning Objective: 17-01 Define quality and explain how it may be measured. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation

4) The Japanese invented and implemented quality control techniques prior to World War II. Answer: FALSE Explanation: U.S. experts pioneered quality control, but the Japanese adopted and refined its methods. Difficulty: 1 Easy Topic: 17.02 Pioneers in Quality Management Learning Objective: 17-02 Name key individuals and their contributions to the quality movement. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 5) Process control charts were adopted by the Japanese after World War II. Answer: TRUE Explanation: After WWII the Japanese became experts in quality control. Difficulty: 1 Easy Topic: 17.02 Pioneers in Quality Management Learning Objective: 17-02 Name key individuals and their contributions to the quality movement. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 6) Pareto charts show the frequency of problems that affect a process in descending order. Answer: TRUE Explanation: The Pareto chart is like a histogram showing frequencies of problems. Difficulty: 1 Easy Topic: 17.03 Quality Improvement Learning Objective: 17-03 List steps and common analytical tools for quality improvement. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation

7) Statistical process control charts (SPC charts) are attributed to Shewhart. Answer: TRUE Explanation: Walter Shewhart invented and refined control charts. Difficulty: 1 Easy Topic: 17.02 Pioneers in Quality Management Learning Objective: 17-02 Name key individuals and their contributions to the quality movement. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 8) The development of acceptance sampling is attributed to the works of Taguchi and Ishikawa. Answer: FALSE Explanation: Harold F. Dodge and Harry G. Romig created tables for acceptance sampling. Difficulty: 2 Medium Topic: 17.02 Pioneers in Quality Management Learning Objective: 17-02 Name key individuals and their contributions to the quality movement. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 9) An in-control process will always exhibit some common cause variation. Answer: TRUE Explanation: "In-control" processes have variation that is within predictable normal limits. Difficulty: 1 Easy Topic: 17.04 Control Charts: Overview Learning Objective: 17-04 Define a control chart and the types of variables displayed. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 10) Common cause variation does not indicate an out-of-control process. Answer: TRUE Explanation: "In-control" processes have variation, but stay within established limits. Difficulty: 1 Easy Topic: 17.04 Control Charts: Overview Learning Objective: 17-04 Define a control chart and the types of variables displayed. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 3 Copyright 2019 © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.

11) The presence of common cause variation is an indication that the process is out of control. Answer: FALSE Explanation: "In-control" processes have variation, but stay within established limits. Difficulty: 1 Easy Topic: 17.04 Control Charts: Overview Learning Objective: 17-04 Define a control chart and the types of variables displayed. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 12) Special cause variation exists when the process produces observations that are not from the same population as the majority of the observations. Answer: TRUE Explanation: Contrast this with common cause variation, which is within predictable normal limits. Difficulty: 1 Easy Topic: 17.01 Quality and Variation Learning Objective: 17-01 Define quality and explain how it may be measured. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 13) Control charts are used to monitor the quality of a product before it is produced. Answer: FALSE Explanation: Control charts are based on measurements taken from actual output or services. Difficulty: 1 Easy Topic: 17.04 Control Charts: Overview Learning Objective: 17-04 Define a control chart and the types of variables displayed. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation

14) A process may be in a state of control even if one sample mean is more than two standard deviations above the centerline. Answer: TRUE Explanation: We expect a sample mean beyond two standard errors about 5 percent of the time. One sample beyond two standard deviations is not unusual enough to trigger a warning. Difficulty: 2 Medium Topic: 17.05 Control Charts for a Mean Learning Objective: 17-05 Make and interpret control charts for a mean. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 15) A p-chart is a type of process control chart that can be used for plotting the proportion of nonconforming sampled items. Answer: TRUE Explanation: With a p-chart, we can track the proportion of nonconforming items. Difficulty: 2 Medium Topic: 17.07 Other Control Charts Learning Objective: 17-07 Make and interpret control charts for attribute data. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 16) The Cpk index may indicate a capable process even though the Cp index is unacceptable. Answer: FALSE Explanation: The Cpk index may indicate problems even when the Cp index is okay. Difficulty: 2 Medium Topic: 17.09 Process Capability Learning Objective: 17-09 Assess the capability of a process. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation

17) A moving range (MR) chart is appropriate to monitor variation when every single item is being inspected (n = 1) since the range (R) cannot be calculated. Answer: TRUE Explanation: There is no range or standard deviation when n = 1, there, so we use a moving range. Difficulty: 2 Medium Topic: 17.06 Control Charts for a Range Learning Objective: 17-06 Make and interpret control charts for a range. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 18) If a single sample mean is 2.1 standard deviations above the centerline, the process is not in control. Answer: FALSE Explanation: Sample means only lie within two standard errors about 95 percent of the time. A single point outside 2 sigmas does not indicate a problem. Difficulty: 2 Medium Topic: 17.05 Control Charts for a Mean Learning Objective: 17-05 Make and interpret control charts for a mean. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 19) ISO 9000 standards were first developed in the United States under the leadership of Joseph Juran. Answer: FALSE Explanation: The ISO standards were mostly developed in Europe. Difficulty: 1 Easy Topic: 17.10 Additional Quality Topics (Optional) Learning Objective: 17-10 Identify topics commonly associated with quality management (optional). Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation

20) ISO 9000 specifies quality processes rather than defect rates. Answer: TRUE Explanation: The focus in ISO is on systems that prevent quality problems and defects. Difficulty: 1 Easy Topic: 17.10 Additional Quality Topics (Optional) Learning Objective: 17-10 Identify topics commonly associated with quality management (optional). Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 21) Deming stressed identifying the workers who contributed the most to poor quality. Answer: FALSE Explanation: Deming felt that management must take responsibility for ensuring quality. Difficulty: 1 Easy Topic: 17.02 Pioneers in Quality Management Learning Objective: 17-02 Name key individuals and their contributions to the quality movement. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 22) Deming thought that the majority of quality problems were traceable to faulty equipment. Answer: FALSE Explanation: Deming felt that poor management was mainly to blame for low quality. Difficulty: 1 Easy Topic: 17.02 Pioneers in Quality Management Learning Objective: 17-02 Name key individuals and their contributions to the quality movement. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 23) Fishbone diagrams were developed for the Japanese fishing industry. Answer: FALSE Explanation: Fishbone (cause-and-effect) diagrams display likely sources of quality problems. Difficulty: 1 Easy Topic: 17.03 Quality Improvement Learning Objective: 17-03 List steps and common analytical tools for quality improvement. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 7 Copyright 2019 © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.

24) In statistical process control, the Cpk index measures the separate distances between the centerline μ and the USL and LSL. Answer: TRUE Explanation: The Cpk index remedies a weakness of the Cp index by showing poor centering. Difficulty: 2 Medium Topic: 17.09 Process Capability Learning Objective: 17-09 Assess the capability of a process. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 25) As a rule of thumb, if a process Cpk index is less than 1.00, the level of process capability is usually judged acceptable. Answer: FALSE Explanation: The Cpk index must be greater than 1 for the process to have a safety margin. Difficulty: 1 Easy Topic: 17.09 Process Capability Learning Objective: 17-09 Assess the capability of a process. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 26) The Cp index equals the Cpk index if USL = 550, LSL = 540, μ = 545, and σ = 0.75. Answer: TRUE Explanation: Because μ is halfway between the USL and LSL, we know that Cp = Cpk. Difficulty: 2 Medium Topic: 17.09 Process Capability Learning Objective: 17-09 Assess the capability of a process. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

27) The Cp index equals the Cpk index if USL = 550, LSL = 540, μ = 543, and σ = 0.75. Answer: FALSE Explanation: Because μ is not centered between the USL and LSL, we know that Cp ≠ Cpk. Difficulty: 2 Medium Topic: 17.09 Process Capability Learning Objective: 17-09 Assess the capability of a process. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 28) The Cp index equals the Cpk index if USL = 550, LSL = 540, μ = 546, and σ = 1.25. Answer: FALSE Explanation: Because μ is not centered between the USL and LSL, we know that Cp ≠ Cpk. Difficulty: 2 Medium Topic: 17.09 Process Capability Learning Objective: 17-09 Assess the capability of a process. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 29) If USL = 550, LSL = 540, μ = 545, and σ = 1.00, the Cp index is 1.67. Answer: TRUE Explanation: Cp = (USL − LSL)/(6σ) = (550 − 540)/[(6)(1)] = 1.67. Difficulty: 2 Medium Topic: 17.09 Process Capability Learning Objective: 17-09 Assess the capability of a process. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 30) If USL = 550, LSL = 540, μ = 545, and σ = 0.4, the process is highly capable. Answer: TRUE Explanation: Cp = (USL − LSL)/(6σ) = (550 − 540)/[(6)(0.4)] = 4.17. Difficulty: 2 Medium Topic: 17.09 Process Capability Learning Objective: 17-09 Assess the capability of a process. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

31) If USL = 550, LSL = 540, μ = 545, and σ = 1.75, the process is highly capable. Answer: FALSE Explanation: Cp = (USL − LSL)/(6σ) = (550 − 540)/[(6)(1.75)] = 0.95. This is below the minimum acceptable Cp = 1.33. Difficulty: 2 Medium Topic: 17.09 Process Capability Learning Objective: 17-09 Assess the capability of a process. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 32) In general terms, a capable process is one whose variability (σ) is small in relation to the distance between the centerline μ and the upper and lower specification limits. Answer: TRUE Explanation: A highly capable process has low variation relative to the spec range (safety margin). Difficulty: 1 Easy Topic: 17.09 Process Capability Learning Objective: 17-09 Assess the capability of a process. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 33) Management wants a process to be in control and have a capability index at least equal to 1.33 (and ideally much more than 1.33). Answer: TRUE Explanation: A highly capable process has a safety margin of ±1σ or more. Difficulty: 1 Easy Topic: 17.09 Process Capability Learning Objective: 17-09 Assess the capability of a process. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation

34) If you increase the size of the samples taken when using an chart, it is necessary to recalculate your control limits because the limits will be narrower. Answer: TRUE Explanation: The control limits are ±3σ/n1/2 from the centerline. Difficulty: 2 Medium Topic: 17.05 Control Charts for a Mean Learning Objective: 17-05 Make and interpret control charts for a mean. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 35) If you increase the size of the samples taken when using an chart, it is necessary to recalculate your control limits because the limits will be wider. Answer: FALSE Explanation: Control limits are ±3σ/n1/2 from the centerline, so the limits will be narrower. Difficulty: 2 Medium Topic: 17.05 Control Charts for a Mean Learning Objective: 17-05 Make and interpret control charts for a mean. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 36) The upper and lower control limits of an chart are typically set at plus or minus three standard errors from the centerline. Answer: TRUE Explanation: The control limits are ±3σ/n1/2 from the centerline. Difficulty: 1 Easy Topic: 17.05 Control Charts for a Mean Learning Objective: 17-05 Make and interpret control charts for a mean. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation

37) A p-chart is a type of process control chart that is used for plotting the number of defects per unit produced. Answer: FALSE Explanation: The p-chart plots the proportion of defective (or conforming) products or services. Difficulty: 2 Medium Topic: 17.07 Other Control Charts Learning Objective: 17-07 Make and interpret control charts for attribute data. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 38) A c-chart is based on the Poisson distribution. Answer: TRUE Explanation: The c-chart is used when the number of defects follows a Poisson distribution. Difficulty: 2 Medium Topic: 17.07 Other Control Charts Learning Objective: 17-07 Make and interpret control charts for attribute data. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 39) Quality management is characterized by focus on the customer and continual improvement. Answer: TRUE Explanation: The idea is to improve quality by improving consistency and reducing variation. Difficulty: 1 Easy Topic: 17.03 Quality Improvement Learning Objective: 17-03 List steps and common analytical tools for quality improvement. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 40) A process whose output distribution is stable over time is said to be in statistical control, regardless of whether the desired specifications are being met. Answer: TRUE Explanation: A process can be stable and in control while producing nonconforming results. Difficulty: 2 Medium Topic: 17.09 Process Capability Learning Objective: 17-09 Assess the capability of a process. Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation 12 Copyright 2019 © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.

41) A control chart for the mean tells whether the product conforms to specifications. Answer: FALSE Explanation: A process can be stable and in control while producing nonconforming results. Difficulty: 2 Medium Topic: 17.09 Process Capability Learning Objective: 17-09 Assess the capability of a process. Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation 42) In statistical process control, control charts are used to ensure that a process is stable and in control by detecting special cause variation. Answer: TRUE Explanation: Control limits and pattern detection rules will reveal unusual (special cause) variation. Difficulty: 1 Easy Topic: 17.04 Control Charts: Overview Learning Objective: 17-04 Define a control chart and the types of variables displayed. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 43) Walter Shewhart was an American who studied the control charts that the Japanese had invented after World War II and brought those methods back to the United States during the 1980s. Answer: FALSE Explanation: Shewhart invented the control charts that the Japanese adopted after WWII. Difficulty: 1 Easy Topic: 17.02 Pioneers in Quality Management Learning Objective: 17-02 Name key individuals and their contributions to the quality movement. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation

44) Quality experts would probably not recommend A) reducing variation. B) using control charts. C) placing blame for poor work. D) identifying sources of variation. Answer: C Explanation: Deming especially thought that assigning blame distracts from finding solutions. Difficulty: 1 Easy Topic: 17.03 Quality Improvement Learning Objective: 17-03 List steps and common analytical tools for quality improvement. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 45) Quality is ultimately best assessed by A) trained statisticians. B) quality control inspectors. C) management. D) customers. Answer: D Explanation: Customers are the final judges of quality. Difficulty: 2 Medium Topic: 17.01 Quality and Variation Learning Objective: 17-01 Define quality and explain how it may be measured. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 46) Which is not a true statement about the life and philosophy of W. Edwards Deming? A) He taught quality control techniques to Japanese companies during the 1950s. B) He lived a very long life and was a highly paid consultant past age 80. C) He invented control charts and proposed the ISO 9000 standard. D) He believed that poor quality is not primarily the fault of the workers. Answer: C Explanation: Deming lived long and taught the Japanese, but ISO was a European innovation. Difficulty: 2 Medium Topic: 17.02 Pioneers in Quality Management Learning Objective: 17-02 Name key individuals and their contributions to the quality movement. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 14 Copyright 2019 © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.

47) Control charts (SPC charts) are attributed to A) Deming. B) Shewhart. C) Juran. D) Taguchi. Answer: B Explanation: Shewhart is credited with inventing control charts. Difficulty: 1 Easy Topic: 17.02 Pioneers in Quality Management Learning Objective: 17-02 Name key individuals and their contributions to the quality movement. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 48) The development of acceptance sampling is attributed to the works of A) Deming and Shewhart. B) Potter and Granger. C) Dodge and Romig. D) Taguchi and Ishikawa. Answer: C Explanation: Harold F. Dodge and Harry G. Romig created tables for acceptance sampling. Difficulty: 2 Medium Topic: 17.02 Pioneers in Quality Management Learning Objective: 17-02 Name key individuals and their contributions to the quality movement. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation

49) ________ and ________ are well known for their statistical work related to customer satisfaction and the cost of quality. A) Deming and Shewhart B) Harold and Kumar C) Dodge and Romig D) Taguchi and Ishikawa Answer: D Explanation: Genichi Taguchi and Kaoru Ishikawa developed new approaches that focused on customer satisfaction and costs of quality. Difficulty: 2 Medium Topic: 17.02 Pioneers in Quality Management Learning Objective: 17-02 Name key individuals and their contributions to the quality movement. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 50) Which of the following is not a characteristic of Total Quality Management (TQM)? A) Employee empowerment B) Reduction of waste C) Continuous improvement D) Reducing the Cp index Answer: D Explanation: Total Quality Management (TQM) covers a variety of tools but does not deal with capability. Difficulty: 1 Easy Topic: 17.03 Quality Improvement Learning Objective: 17-03 List steps and common analytical tools for quality improvement. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation

51) Instability is most readily apparent on the A) chart. B) R chart. C) np chart. D) I chart. Answer: B Explanation: The R chart reveals variability. Difficulty: 1 Easy Topic: 17.08 Patterns in Control Charts Learning Objective: 17-08 Recognize abnormal patterns in control charts and their potential causes. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 52) The problem that is probably the hardest to identify from a control chart is A) mixture. B) oscillation. C) cycle. D) trend. Answer: A Explanation: Mixing two processes can create nonrandom patterns, but its symptoms can be subtle. Difficulty: 2 Medium Topic: 17.08 Patterns in Control Charts Learning Objective: 17-08 Recognize abnormal patterns in control charts and their potential causes. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation

53) Which is not a tool of statistical quality control? A) Fishbone diagram B) Pareto chart C) Attribute control chart D) Deming chart Answer: D Explanation: There is no Deming chart. Difficulty: 2 Medium Topic: 17.03 Quality Improvement Learning Objective: 17-03 List steps and common analytical tools for quality improvement. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 54) Control charts were an innovation attributed to A) Deming in the 1950s. B) Shewhart in the 1920s. C) Westinghouse in the 1960s. D) Pacioli in the 1490s. Answer: B Explanation: Walter Shewhart created process control charts almost 100 years ago. Difficulty: 1 Easy Topic: 17.02 Pioneers in Quality Management Learning Objective: 17-02 Name key individuals and their contributions to the quality movement. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 55) Which is an appropriate step in continuous quality improvement? A) Taking measurements on a variable and keeping careful records. B) Posting quality banners or company flags where they are visible to all. C) Castigating the lazy employees for their shoddy workmanship. D) Sending employees to Motivation Camp taught by expensive consultants. Answer: A Explanation: No control chart can be created without a record of accurate measurements. Difficulty: 1 Easy Topic: 17.03 Quality Improvement Learning Objective: 17-03 List steps and common analytical tools for quality improvement. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 18 Copyright 2019 © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.

56) Likely reasons for inaccurate control limits would include which of the following? A) The engineering parameter for variance is unknown. B) The engineers were underpaid for their work. C) There was insufficient preliminary sampling. D) Process variation was not zero, as expected. Answer: C Explanation: A control chart requires a long enough record of accurate measurements. Difficulty: 2 Medium Topic: 17.05 Control Charts for a Mean Learning Objective: 17-05 Make and interpret control charts for a mean. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 57) Attribute control charts would not be used to display the A) proportion of nonconforming parts. B) sample range for a measured variable. C) total number of nonconforming parts. D) average number of nonconforming parts. Answer: B Explanation: Calculating a range R = xmax − xmin requires numerical data. Difficulty: 2 Medium Topic: 17.07 Other Control Charts Learning Objective: 17-07 Make and interpret control charts for attribute data. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 58) The R chart is likely to reveal which problem? A) Instability B) Cycles C) Level shift D) Trend Answer: A Explanation: The range R = xmax − xmin measures variability. Difficulty: 2 Medium Topic: 17.08 Patterns in Control Charts Learning Objective: 17-08 Recognize abnormal patterns in control charts and their potential causes. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 19 Copyright 2019 © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.

59) Which of the following is most likely the cause of a level shift in a SPC chart? A) Tool wear B) A new worker C) Temperature fluctuations D) Alternating samples from two machines Answer: B Explanation: Sudden change in a process could be caused by new or untrained personnel. Difficulty: 2 Medium Topic: 17.08 Patterns in Control Charts Learning Objective: 17-08 Recognize abnormal patterns in control charts and their potential causes. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 60) Instability in a process is indicated when samples A) tend to alternate between high and low values. B) drift slowly either upward or downward. C) vary more than expected. D) shift abruptly either above or below the centerline. Answer: C Explanation: Increased variation in a process (relative to the past) is called instability. Difficulty: 1 Easy Topic: 17.08 Patterns in Control Charts Learning Objective: 17-08 Recognize abnormal patterns in control charts and their potential causes. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation

61) A level shift in a process is indicated when samples A) tend to alternate between high and low values. B) drift slowly either upward or downward. C) vary more than expected. D) shift abruptly to a new mean. Answer: D Explanation: A process may stabilize at a new level (but this is a control chart violation). Difficulty: 1 Easy Topic: 17.08 Patterns in Control Charts Learning Objective: 17-08 Recognize abnormal patterns in control charts and their potential causes. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 62) A slow drift of measurements either up or down from the process centerline suggests a(n) A) mixed process. B) trend. C) instability. D) cycle. Answer: B Explanation: Steady drifting of a mean or proportion is called a trend. Difficulty: 1 Easy Topic: 17.08 Patterns in Control Charts Learning Objective: 17-08 Recognize abnormal patterns in control charts and their potential causes. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation

63) Which is not characteristic of a trend? A) Variance is essentially unchanged from sample to sample. B) Trend is often due to mixing two batches of materials. C) Trend is detectable on a control chart if enough samples are taken. D) Rules of thumb can be established to detect trend. Answer: B Explanation: Mixing two processes might produce excessive variation but not steady change. Difficulty: 2 Medium Topic: 17.08 Patterns in Control Charts Learning Objective: 17-08 Recognize abnormal patterns in control charts and their potential causes. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 64) Which is not a rule of thumb to indicate an out-of-control process on the chart? A) Single point outside three sigma. B) Three of four successive points outside two sigma on the same side of the centerline. C) Four of five successive points outside one sigma on the same side of the centerline. D) Nine successive points on the same side of the centerline. Answer: B Explanation: Learn the textbook rules (Rule 3 is four of five successive points beyond one sigma). Difficulty: 2 Medium Topic: 17.05 Control Charts for a Mean Learning Objective: 17-04 Define a control chart and the types of variables displayed. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation

65) Which is not a characteristic of instability? A) Larger than normal amount of variation B) Higher-than-expected frequencies in tails of the distribution of means C) Often caused by untrained operators D) Specification limits that are too narrow Answer: D Explanation: Specification limits are based on customer requirements (achievable or not). Difficulty: 2 Medium Topic: 17.08 Patterns in Control Charts Learning Objective: 17-08 Recognize abnormal patterns in control charts and their potential causes. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 66) Refer to the diagram below.

For this process, the Cp index would be A) less than one. B) equal to one. C) greater than one. Answer: C Explanation: The process is not centered, but there is a safety margin. In this diagram the safety margin appears to be approximately 1σ on the left and 3σ on the right. Difficulty: 2 Medium Topic: 17.09 Process Capability Learning Objective: 17-09 Assess the capability of a process. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 23 Copyright 2019 © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.

67) Find the Cp index for a process with USL = 550, LSL = 540, μ = 545, and σ = 0.75. A) 1.25 B) 1.33 C) 2.22 D) 1.75 Answer: C Explanation: Cp = (USL − LSL)/(6σ) = (550 − 540)/[(6)(0.75)] = 2.22. Difficulty: 2 Medium Topic: 17.09 Process Capability Learning Objective: 17-09 Assess the capability of a process. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 68) Find the Cpk index for a process with USL = 550, LSL = 540, μ = 545, and σ = 0.75. A) 1.33 B) 2.22 C) 1.25 D) 1.75 Answer: B Explanation: Cpk = min(USL − μ, μ − LSL)/(3σ) = min(550 − 545, 545 − 540)/[(3)(0.75)] = 2.22. Difficulty: 3 Hard Topic: 17.09 Process Capability Learning Objective: 17-09 Assess the capability of a process. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 69) Find the Cp index for a process with USL = 550, LSL = 540, μ = 543, and σ = 0.75. A) 1.25 B) 1.33 C) 2.22 D) 1.75 Answer: C Explanation: Cp = (USL − LSL)/(6σ) = (550 − 540)/[(6)(0.75)] = 2.22. Difficulty: 2 Medium Topic: 17.09 Process Capability Learning Objective: 17-09 Assess the capability of a process. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 24 Copyright 2019 © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.

70) Find the Cpk index for a process with USL = 550, LSL = 540, μ = 543, and σ = 0.75. A) 1.33 B) 2.22 C) 1.25 D) 1.75 Answer: A Explanation: Cpk = min(USL − μ, μ − LSL)/(3σ) = min(550 − 543, 543 − 540)/[(3)(0.75)] = 1.33. Difficulty: 3 Hard Topic: 17.09 Process Capability Learning Objective: 17-09 Assess the capability of a process. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 71) Find the Cpk index for a process with USL = 550, LSL = 540, μ = 544, and σ = 1.25. A) 1.33 B) 2.22 C) 1.07 D) 1.75 Answer: C Explanation: Cpk = min(USL − μ, μ − LSL)/(3σ) = min(550 − 544, 544 − 540)/[(3)(1.25)] = 1.07. Difficulty: 3 Hard Topic: 17.09 Process Capability Learning Objective: 17-09 Assess the capability of a process. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

72) Is a process capable if USL = 550, LSL = 540, μ = 542, and σ = 1.25? A) No, clearly not B) No, but very close C) Yes, just barely D) Yes, highly capable Answer: A Explanation: Cpk = min(USL − μ, μ − LSL)/(3σ) = min(550 − 542, 542 − 540)/[(3)(1.25)] = 0.53. This is below the minimum acceptable Cp = 1.33. Difficulty: 3 Hard Topic: 17.09 Process Capability Learning Objective: 17-09 Assess the capability of a process. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 73) Statistical process control charts can measure A) the stability of the process. B) the capability of a process. C) both stability and capability. Answer: A Explanation: Capability cannot be assessed using a control chart (need the Cp or Cpk index). Difficulty: 2 Medium Topic: 17.09 Process Capability Learning Objective: 17-09 Assess the capability of a process. Bloom's: Understand AACSB: Analytical Thinking Accessibility: Keyboard Navigation 74) In manufacturing, if workers readjust the equipment after each sample, it would typically A) increase variation. B) decrease variation. C) widen the specification limits. D) improve conformance to specifications. Answer: A Explanation: Attempting to adjust the process after each sample increases variation. Difficulty: 2 Medium Topic: 17.08 Patterns in Control Charts Learning Objective: 17-08 Recognize abnormal patterns in control charts and their potential causes. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 26 Copyright 2019 © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.

75) Which statistician developed the 14 Points of Quality? A) Juran B) Deming C) Taguchi D) Ishikawa Answer: B Explanation: Deming's 14 Points are still an influential guide to thinking about quality. Difficulty: 1 Easy Topic: 17.02 Pioneers in Quality Management Learning Objective: 17-02 Name key individuals and their contributions to the quality movement. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 76) If the specification subgroup size is n = 4 and the known process parameters are μ = 2.75 and σ = 0.044, which are the control limits for the A) LCL = 2.684, UCL = 2.816 B) LCL = 2.728, UCL = 2.772 C) LCL = 2.618, UCL = 2.882 D) LCL = 2.518, UCL = 2.998

chart?

Answer: A Explanation: The control limits are μ ± 3σ/n1/2, so 2.75 ± 3(0.044)/41/2. Difficulty: 2 Medium Topic: 17.05 Control Charts for a Mean Learning Objective: 17-05 Make and interpret control charts for a mean. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 77) Which is not a characteristic of a p-chart? A) It shows the number of defects per item being inspected. B) It measures the fraction of nonconforming items in a sample. C) It is based on the binomial distribution (or its normal approximation). D) It will have varying control limits if the sample size is changing. Answer: A Explanation: The p-chart shows the proportion of nonconforming items p = x/n. Difficulty: 2 Medium Topic: 17.07 Other Control Charts Learning Objective: 17-07 Make and interpret control charts for attribute data. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 27 Copyright 2019 © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.

78) Ten samples of n = 5 were collected to construct an each sample are shown in the table below. Sample 1 2 3 4 5

Mean 215 204 160 226 232

Range 15 20 10 19 24

Calculate the empirical centerline for the A) 210.5 B) 206.9 C) 205.3 D) 208.2

Sample 6 7 8 9 10

chart. The sample mean and range for Mean 224 205 184 207 212

Range 21 27 12 16 17

chart.

Answer: B Explanation: The control limits are μ ± 3σ/n1/2, so our estimate of μ is (215 + 204 + · · · + 212)/10. Difficulty: 2 Medium Topic: 17.05 Control Charts for a Mean Learning Objective: 17-05 Make and interpret control charts for a mean. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

79) Ten samples of n = 5 were collected to construct an each sample are shown in the table below. Sample 1 2 3 4 5

Mean 215 204 160 226 232

Range 15 20 10 19 24

chart. The sample mean and range for

Sample 6 7 8 9 10

Calculate the empirical lower and upper control limits for the control chart factors.) Subgroup Size 2 3 4 5 6 7 8 9

d2 1.128 1.693 2.059 2.326 2.534 2.704 2.847 2.970

D3 0 0 0 0 0 0.076 0.136 0.184

Mean 224 205 184 207 212

Range 21 27 12 16 17

chart. (You will need a table of

D4 3.267 2.574 2.282 2.114 2.004 1.924 1.864 1.816

A) LCL = 196.46, UCL = 217.34 B) LCL = 171.81, UCL = 241.39 C) LCL = 188.03, UCL = 225.17 D) LCL = 163.64, UCL = 250.56 Answer: A Explanation: Subgroup size is n = 5, so d2 = 2.326, mean of 10 means is 206.9, mean of 10 ranges is 18.1, and the control limits are 206.9 ± 3(18.1)/[(2.326)(51/2)]. Difficulty: 3 Hard Topic: 17.05 Control Charts for a Mean Learning Objective: 17-05 Make and interpret control charts for a mean. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

80) Ten samples of n = 5 were collected to construct an each sample are shown in the table below. Sample 1 2 3 4 5

Mean 215 204 160 226 232

Range 15 20 10 19 24

Sample 6 7 8 9 10

chart. The sample mean and range for Mean 224 205 184 207 212

Range 21 27 12 16 17

Calculate the empirical centerline for the R chart. A) 20.8 B) 17.2 C) 18.1 D) 19.4 Answer: C Explanation: The average range is (15 + 20 + · · · + 17)/10. Difficulty: 2 Medium Topic: 17.06 Control Charts for a Range Learning Objective: 17-06 Make and interpret control charts for a range. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

81) Ten samples of n = 5 were collected to construct an each sample are shown in the table below. Sample 1 2 3 4 5

Mean 215 204 160 226 232

Range 15 20 10 19 24

chart. The sample mean and range for

Sample 6 7 8 9 10

Mean 224 205 184 207 212

Range 21 27 12 16 17

Calculate empirical lower and upper control limits for the R chart. (You will need a table of control chart factors.) Subgroup Size 2 3 4 5 6 7 8 9

d2 1.128 1.693 2.059 2.326 2.534 2.704 2.847 2.970

D3 0 0 0 0 0 0.076 0.136 0.184

D4 3.267 2.574 2.282 2.114 2.004 1.924 1.864 1.816

A) LCL = 0, UCL = 45.86 B) LCL = 0, UCL = 42.49 C) LCL = 0, UCL = 38.26 D) LCL = 4.48, UCL = 35.58 Answer: C Explanation: For n = 5, D3 = 0 and D4 = 2.114, and the average range is (15 + 20 + · · · + 17)/10 = 18.1, so LCL = D3(R-bar) = 0(18.1) = 0 and UCL = D4(R-bar) = (2.114)(18.1) = 38.26. Difficulty: 3 Hard Topic: 17.06 Control Charts for a Range Learning Objective: 17-06 Make and interpret control charts for a range. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

82) Professor Murphy wants to set up a control chart to monitor the percentage of absenteeism in his introductory statistics course (50 students are registered). Absences per period for the last 15 class sessions are in the table below. Session Absent

1 5

2 0

3 2

4 2

5 3

6 7 8 10

8 3

9 5

10 11 12 13 5 1 2 0

14 15 2 3

Calculate the empirical centerline for a p-chart to track absences. A) 0.068 B) 0.072 C) 0.146 D) 0.202 Answer: A Explanation: The average is 3.4 absences per class, so pavg = (avg absences)/(class size) = (3.4)/50 = .068, or you could average the 15 sample proportions [5/50, 0/50, 2/50, . . . , 3/50]. Difficulty: 2 Medium Topic: 17.07 Other Control Charts Learning Objective: 17-07 Make and interpret control charts for attribute data. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

83) Professor Murphy wants to set up a control chart to monitor the percentage of absenteeism in his introductory statistics course (50 students are registered). Absences per period for the last 15 class sessions are in the table below. Session Absent

1 5

2 0

3 2

4 2

5 3

6 7 8 10

8 3

9 5

10 11 12 13 5 1 2 0

14 15 2 3

Using 3 sigma limits, calculate lower and upper control limits for a p-chart to track absences. A) 0, 0.252 B) 0, 0.175 C) 0, 0.114 D) −0.038, 0.272 Answer: B Explanation: The average is 3.4 absences per class of n = 50 students, so pavg = (avg absences)/ (class size) = (3.4)/50 = 0.068, so the control limits are pavg ± 3[pavg(1 − pavg)/n]1/2. Difficulty: 3 Hard Topic: 17.07 Other Control Charts Learning Objective: 17-07 Make and interpret control charts for attribute data. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

84) Given the following control chart, which problem is most likely?

A) Cycle B) Instability C) Trend D) Level shift E) Cycle Answer: D Explanation: Sudden jump to a new level (not a continuing trend) after sample 15. Difficulty: 2 Medium Topic: 17.08 Patterns in Control Charts Learning Objective: 17-08 Recognize abnormal patterns in control charts and their potential causes. Bloom's: Analyze AACSB: Analytical Thinking Accessibility: Keyboard Navigation

85) Given the following control chart, which problem is most likely?

A) Instability B) Trend C) Level shift D) Cycle Answer: B Explanation: Steady drift downward after sample 20. Difficulty: 1 Easy Topic: 17.08 Patterns in Control Charts Learning Objective: 17-08 Recognize abnormal patterns in control charts and their potential causes. Bloom's: Analyze AACSB: Analytical Thinking Accessibility: Keyboard Navigation

86) Given the following control chart, which problem is most likely?

A) Instability B) Trend C) Level shift D) Cycle Answer: A Explanation: Sudden onset of high variability after sample 20. Difficulty: 1 Easy Topic: 17.08 Patterns in Control Charts Learning Objective: 17-08 Recognize abnormal patterns in control charts and their potential causes. Bloom's: Analyze AACSB: Analytical Thinking Accessibility: Keyboard Navigation

87) Given the following control chart, which problem is most likely?

A) Instability B) Trend C) Level shift D) Cycle Answer: D Explanation: Crosses the centerline less than 50/2 = 25 times. Difficulty: 2 Medium Topic: 17.08 Patterns in Control Charts Learning Objective: 17-08 Recognize abnormal patterns in control charts and their potential causes. Bloom's: Analyze AACSB: Analytical Thinking Accessibility: Keyboard Navigation

88) Given the following control chart, which problem is most likely?

A) Instability B) Oscillation C) Level shift D) Cycle Answer: B Explanation: Crosses the centerline more than 50/2 = 25 times. Difficulty: 2 Medium Topic: 17.08 Patterns in Control Charts Learning Objective: 17-08 Recognize abnormal patterns in control charts and their potential causes. Bloom's: Analyze AACSB: Analytical Thinking Accessibility: Keyboard Navigation

89) What does the first letter mean in the Six-Sigma DMAIC acronym? A) Design B) Distribute C) Describe D) Define Answer: D Explanation: Review the Six Sigma steps. Difficulty: 1 Easy Topic: 17.10 Additional Quality Topics (Optional) Learning Objective: 17-10 Identify topics commonly associated with quality management (optional). Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 90) What does the second letter mean in the Six-Sigma DMAIC acronym? A) Maximize B) Measure C) Mentor D) Mobilize Answer: B Explanation: Review the Six Sigma steps. Difficulty: 1 Easy Topic: 17.10 Additional Quality Topics (Optional) Learning Objective: 17-10 Identify topics commonly associated with quality management (optional). Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation

91) What does the third letter mean in the Six-Sigma DMAIC acronym? A) Analyze B) Action C) Absolve D) Attack Answer: A Explanation: Review the Six Sigma steps. Difficulty: 1 Easy Topic: 17.10 Additional Quality Topics (Optional) Learning Objective: 17-10 Identify topics commonly associated with quality management (optional). Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 92) What does the fourth letter mean in the Six-Sigma DMAIC acronym? A) Integrate B) Investigate C) Improve D) Interact Answer: C Explanation: Review the Six Sigma steps. Difficulty: 1 Easy Topic: 17.10 Additional Quality Topics (Optional) Learning Objective: 17-10 Identify topics commonly associated with quality management (optional). Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation

93) What does the fifth letter mean in the Six-Sigma DMAIC acronym? A) Cooperate B) Correlate C) Coordinate D) Control Answer: D Explanation: Review the Six Sigma steps. Difficulty: 1 Easy Topic: 17.10 Additional Quality Topics (Optional) Learning Objective: 17-10 Identify topics commonly associated with quality management (optional). Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 94) Which is not primarily intended to detect excessive variation in a measurement? A) s-chart B) MR-chart C) R-chart D) p-chart Answer: D Explanation: The p-chart tracks the proportion of nonconforming items. Difficulty: 1 Easy Topic: 17.07 Other Control Charts Learning Objective: 17-07 Make and interpret control charts for attribute data. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 95) Which is most applicable when continuous inspection is used? A) I-chart B) c-chart C) p-chart D)

chart

Answer: A Explanation: If every item is inspected, we can use the I-chart (n = 1). Difficulty: 2 Medium Topic: 17.07 Other Control Charts Learning Objective: 17-07 Make and interpret control charts for attribute data. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 41 Copyright 2019 © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.

96) If LSL = 500, USL = 518, μ = 509, and σ = 3, the "safety margin" to maintain the product specifications would be A) nonexistent (i.e., zero). B) 1σ on each side. C) 2σ on each side. D) 3σ on each side. Answer: A Explanation: (USL − LSL)/(6σ) = (518 − 500)/[6(3)] = 1, so there is no safety margin. Difficulty: 2 Medium Topic: 17.09 Process Capability Learning Objective: 17-09 Assess the capability of a process. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation 97) If LSL = 50.00, USL = 56.00, μ = 53.00, and σ = 0.50, the "safety margin" for product specifications would be A) nonexistent (i.e., zero). B) 1σ on each side. C) 2σ on each side. D) 3σ on each side. Answer: D Explanation: The lower end of the expected range of X would be μ − 3σ = 53 − 3(.50) = 51.5, while the upper end of the expected range of X would be μ + 3σ = 53 + 3(.50) = 54.5. This leaves a safety margin of 1.5 on either end, which is 3σ because 3 × 0.5 = 1.5. Difficulty: 2 Medium Topic: 17.09 Process Capability Learning Objective: 17-09 Assess the capability of a process. Bloom's: Apply AACSB: Analytical Thinking Accessibility: Keyboard Navigation

98) Which is a rule of thumb to indicate an out-of-control process on the chart? A) Single point outside one sigma. B) Two of three successive points outside one sigma on the same side of the centerline. C) Four of five successive points on the same side of the centerline. D) Nine successive points alternating in sign. Answer: A Explanation: Review the rules (choices B,C, D are incorrectly stated). Difficulty: 2 Medium Topic: 17.05 Control Charts for a Mean Learning Objective: 17-05 Make and interpret control charts for a mean. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 99) The Malcolm Baldrige award is awarded annually by the A) American Statistical Association to universities. B) American Society for Quality to individual managers. C) Japanese Society for Quality Control to nonprofit agencies. D) President of the United States to firms. Answer: D Explanation: Review the Malcolm Baldrige Award criteria. Difficulty: 1 Easy Topic: 17.10 Additional Quality Topics (Optional) Learning Objective: 17-10 Identify topics commonly associated with quality management (optional). Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 100) The control chart for a range is intended to measure A) variation. B) capability. C) skewness. D) trends. Answer: A Explanation: A range indicates variability in the metric that is being tracked. Difficulty: 1 Easy Topic: 17.06 Control Charts for a Range Learning Objective: 17-06 Make and interpret control charts for a range. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation 43 Copyright 2019 © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.

101) Control limits for a range chart A) are inappropriate for tracking variation. B) vary cyclically when the mean changes. C) are asymmetric around the centerline. D) show whether the mean has changed. Answer: C Explanation: Control limits for the range chart are based on the chi-square distribution, so they are asymmetric around the centerline. Difficulty: 1 Easy Topic: 17.06 Control Charts for a Range Learning Objective: 17-06 Make and interpret control charts for a range. Bloom's: Remember AACSB: Analytical Thinking Accessibility: Keyboard Navigation