Business Statistics A First Course, 5E David M. Levine Timothy C. Krehbiel Test Bank

Type:

Test Bank

Business Statistics A First Resource: Course Edition:

5th Edition

Author(s):

David M. Levine Timothy C. Krehbiel Mark L. Berenson

Table of Contents Preface

vii

Keywords Index

ix

Chapter 1

Introduction and Data Collection

1

Chapter 2

Presenting Data in Tables and Charts

29

Chapter 3

Numerical Descriptive Measures

74

Chapter 4

Basic Probability

101

Chapter 5

Discrete Probability Distributions

136

Chapter 6

The Normal Distribution

162

Chapter 7

Sampling and Sampling Distributions

188

Chapter 8

Confidence Interval Estimation

225

Chapter 9

Fundamentals of Hypothesis Testing: One-Sample Tests

265

Chapter 10

Two-Sample Tests and One-Way ANOVA

300

Chapter 11

Chi-Square Tests

365

Chapter 12

Simple Linear Regression

393

Chapter 13

Multiple Regression

443

Chapter 14

Statistical Applications in Quality and Production Management

500

iii

Preface

The Test Item File contains a variety of multiple-choice, true-false, problem and fill-in questions based on the definitions, concepts, and ideas developed in each chapter. In addition, numerical problems and Microsoft Excel computer output problems are also given with solutions provided in multiple-choice, true-false, problem and fill-in format.

The Test Item File is intended to assist instructors in preparing examinations. The questions included herein highlight the key topics covered throughout each chapter. Keywords are available after each question to help instructors locate questions on a specific topic or concept. Explanation is provided when the rationale of the correct answer to a difficult question is rather obscure. The format for the Test Item File will facilitate grading and should be helpful to instructors who teach very large sections.

The intended difficulty level (easy, moderate, difficult) of each question in the Test Item File is stated in order to facilitate test item selection by instructors wishing to create specific types of exams. However, some words of caution must be given. The classification of question difficulty level is very subjective and each question should be evaluated based on the emphasis the particular topic was given in class and how much emphasis is to be given to numerical results obtained by calculator rather than computerized results obtained from Microsoft Excel. As an operational definition that is used here, items are classified as easy if they pertain directly to definitions and fundamental concepts. Test items are classified as moderate if they require some numerical calculations with more than a minimal number of steps or if they require a broader understanding of the topic. Test items that are classified as difficult are done so because of the level of rigor of the subject, the length of the narrative, the amount of effort required for solution, or for responses that require more thought and analysis.

Instructors are also advised that all answers in the Test Item File are computed using Microsoft Excel or PHStat with no rounding involved in the intermediate steps. If students use rounding with formulae and a calculator, their answers might be different from those provided in the answer keys. Likewise, if students use the statistical tables at the end of the book instead of Microsoft Excel or PHStat, their answers might also differ from those provided in the answer keys due to rounding. Whenever possible, we provide answers obtained using both Microsoft Excel/PHStat and the statistical tables if they are different.

This Test Item File and others that are similar suffer from one major weakness. They do not permit an evaluation of the students’ written communication skill. The authors highly recommend that, if possible, instructors who use this Test Item File supplement it with at least one short essay type question so that an assessment can be made of the students’ understanding of concepts as well as how they can make connections across various topics.

The following tabular display is a breakdown of the number of questions in each chapter by type.

iv

Chapter

Multiple Choice

True/False

Fill in

Problem

1

48

42

40

0

130

2

51

32

83

18

184

3

27

33

31

38

129

4

60

19

38

37

154

5

21

13

47

41

122

6

16

14

43

48

121

7

44

65

39

31

179

8

29

105

27

21

182

9

70

57

16

7

150

10

109

51

57

53

270

11

45

31

31

6

113

12

78

33

59

28

198

13

102

51

42

18

213

14

44

24

35

10

113

Total

744

582

588

368

2282

v

Total

Keywords Index A a priori classical probability A2 factor addition rule adjusted coefficient of determination adjusted r-square assumption autocorrelation

control limit convenience sample counting rule coverage error critical value cumulative frequency distribution cumulative percentage distribution cumulative percentage polygon (ogive) cumulative relative frequency

B

D

bar chart Bayes' theorem beta-risk binomial distribution box-and-whisker plot

d2 factor D3 factor D4 factor data decision degrees of freedom Deming's 14 points descriptive statistics deviance statistic difference between two means difference between two proportions difference between two variances discrete random variable dummy variable Durbin-Watson statistic

C categorical random variable center line central limit theorem Chebyshev rule chi-square test Chi-square test for difference in proportions Chi-square test of independence chunk sample class boundaries class interval class midpoint cluster sample coefficient of correlation coefficient of determination coefficient of multiple determination coefficient of variation collective exhaustive column percentages combination common causes of variation complement completely randomized design conclusion conditional probability confidence coefficient confidence interval contingency table continuity adjustment continuous random variable control chart

E empirical classical probability empirical rule estimation estimation of mean values ethical issues

F F distribution F test F test for factor F test on slope F test on the entire regression fitted value five-number summary form of hypothesis frame frequency distribution

vi

H

one-sided one-tailed test one-way analysis of variance ordinal scale outcomes

histogram homoscedasticity

I P inferential statistics interaction intercept interpretation interquartile range interval scale

p chart Pareto diagram parameter percentage distribution percentage polygon permutation pie chart point estimate Poisson distribution polygon pooled variance population power prediction interval prediction of individual values primary data sources probability probability distribution probability sample properties proportion p-value

J joint probability judgment sample

L law of large number learning statistical programs least squares level of significance Levene's test

M marginal probability mean (expected value) mean difference mean of the sum mean squares measure of variation measurement error measure of central tendency median mode multiplication rule mutually exclusive

Q quadratic regression quartile quota sample

R R chart random number range ratio scale reasons for learning statistics reasons for sampling red bead experiment rejection region relative frequency distribution residual residual plot resistant to outliers risk

N nonprobability sample nonresponse error normal distribution normal probability plot number of classes

O vii

robust test row percentages

two-tailed test type I error type II error types of data

S

U

sample sample size sample size determination sample space sampling sampling distribution sampling error sampling method sampling with replacement sampling without replacement scatter plot secondary data sources selection bias shape Shewhart Deming cycle simple random sample six sigma management slope sources of data special causes of variation standard deviation standard error standard error of estimate standard normal quantile standardized normal distribution statistic statistical control statistical independence statistical package statistics stem-and-leaf display stratified sample subjective probability sum of squares survey worthiness systematic sample

unbiased

V value variance variance of sum variation

W width

X XBar chart

Z Z scores Z test

T t distribution t test t test for correlation coefficient t test on slope test statistic testing total percentages Tukey-Kramer procedure viii

Introduction and Data Collection

1

CHAPTER 1: INTRODUCTION AND DATA COLLECTION 1. The process of using sample statistics to draw conclusions about true population parameters is called a) statistical inference. b) the scientific method. c) sampling. d) descriptive statistics.

ANSWER: a TYPE: MC DIFFICULTY: Easy KEYWORDS: inferential statistics 2. Those methods involving the collection, presentation, and characterization of a set of data in order to properly describe the various features of that set of data are called a) statistical inference. b) the scientific method. c) sampling. d) descriptive statistics.

ANSWER: d TYPE: MC DIFFICULTY: Easy KEYWORDS: descriptive statistics 3. The collection and summarization of the socioeconomic and physical characteristics of the employees of a particular firm is an example of a) inferential statistics. b) descriptive statistics. c) a parameter. d) a statistic.

ANSWER: b TYPE: MC DIFFICULTY: Easy KEYWORDS: descriptive statistics 4. The estimation of the population average family expenditure on food based on the sample average expenditure of 1,000 families is an example of a) inferential statistics. b) descriptive statistics. c) a parameter. d) a statistic.

ANSWER: a TYPE: MC DIFFICULTY: Easy KEYWORDS: inferential statistics

.

2

Introduction and Data Collection

5. The universe or "totality of items or things" under consideration is called a) a sample. b) a population. c) a parameter. d) a statistic.

ANSWER: b TYPE: MC DIFFICULTY: Easy KEYWORDS: population 6.

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

ANSWER: a TYPE: MC DIFFICULTY: Easy KEYWORDS: sample 7. A summary measure that is computed to describe a characteristic from only a sample of the population is called a) a parameter. b) a census. c) a statistic. d) the scientific method.

ANSWER: c TYPE: MC DIFFICULTY: Easy KEYWORDS: statistic 8. A summary measure that is computed to describe a characteristic of an entire population is called a) a parameter. b) a census. c) a statistic. d) the scientific method.

ANSWER: a TYPE: MC DIFFICULTY: Easy KEYWORDS: parameter

.

Introduction and Data Collection

9. Which of the following is most likely a population as opposed to a sample? a) respondents to a newspaper survey. b) the first 5 students completing an assignment. c) every third person to arrive at the bank. d) registered voters in a county.

ANSWER: d TYPE: MC DIFFICULTY: Moderate KEYWORDS: population, sample 10. Which of the following is most likely a parameter as opposed to a statistic? a) The average score of the first five students completing an assignment. b) The proportion of females registered to vote in a county. c) The average height of people randomly selected from a database. d) The proportion of trucks stopped yesterday that were cited for bad brakes.

ANSWER: b TYPE: MC DIFFICULTY: Moderate KEYWORDS: parameter, statistic 11. Which of the following is not an element of descriptive statistical problems? a) An inference made about the population based on the sample. b) The population or sample of interest. c) Tables, graphs, or numerical summary tools. d) Identification of patterns in the data.

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

ANSWER: b TYPE: MC DIFFICULTY: Moderate KEYWORDS: population, sample

.

3

4

Introduction and Data Collection

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

ANSWER: b TYPE: MC DIFFICULTY: Easy KEYWORDS: data, sampling 14. A study is under way in Yosemite National Forest to determine the adult height of American pine trees. Specifically, the study is attempting to determine what factors aid a tree in reaching heights greater than 60 feet tall. It is estimated that the forest contains 25,000 adult American pines. The study involves collecting heights from 250 randomly selected adult American pine trees and analyzing the results. Identify the sample in the study. a) The 250 randomly selected adult American pine trees. b) The 25,000 adult American pine trees in the forest. c) All the adult American pine trees taller than 60 feet. d) All American pine trees, of any age, in the forest.

ANSWER: a TYPE: MC DIFFICULTY: Easy KEYWORDS: population, sample 15. Most analysts focus on the cost of tuition as the way to measure the cost of a college education.

But incidentals, such as textbook costs, are rarely considered. A researcher at Drummand University wishes to estimate the textbook costs of first-year students at Drummand. To do so, she monitored the textbook cost of 250 first-year students and found that their average textbook cost was $300 per semester. Identify the population of interest to the researcher. a) All Drummand University students. b) All college students. c) All first-year Drummand University students. d) The 250 students that were monitored.

ANSWER: c TYPE: MC DIFFICULTY: Easy KEYWORDS: population, sample

.

Introduction and Data Collection

16. Most analysts focus on the cost of tuition as the way to measure the cost of a college education.

But incidentals, such as textbook costs, are rarely considered. A researcher at Drummand University wishes to estimate the textbook costs of first-year students at Drummand. To do so, she monitored the textbook cost of 250 first-year students and found that their average textbook cost was $300 per semester. Identify the variable of interest to the researcher. a) The textbook cost of first-year Drummand University students. b) The year in school of Drummand University students. c) The age of Drummand University students. d) The cost of incidental expenses of Drummand University students.

ANSWER: a TYPE: MC DIFFICULTY: Easy KEYWORDS: data, sampling 17. Most analysts focus on the cost of tuition as the way to measure the cost of a college education.

But incidentals, such as textbook costs, are rarely considered. A researcher at Drummand University wishes to estimate the textbook costs of first-year students at Drummand. To do so, she monitored the textbook cost of 250 first-year students and found that their average textbook cost was $300 per semester. Identify the sample in the study. a) All Drummand University students. b) All college students. c) All first-year Drummand University students. d) The 250 students that were monitored.

ANSWER: d TYPE: MC DIFFICULTY: Easy KEYWORDS: population, sample 18. Researchers suspect that the average number of units earned per semester by college students is rising. A researcher at Calendula College wishes to estimate the number of units earned by students during the spring semester at Calendula. To do so, he randomly selects 100 student transcripts and records the number of units each student earned in the spring term. He found that the average number of semester units completed was 12.96 units per student. Identify the population of interest to the researcher. a) All Calendula College students. b) All college students. c) All Calendula College students enrolled in the spring. d) All college students enrolled in the spring.

ANSWER: c TYPE: MC DIFFICULTY: Moderate KEYWORDS: population, sample

.

5

6

Introduction and Data Collection

19. The average number of units earned per semester by college students is suspected to be rising. A researcher at Calendula College wishes to estimate the number of units earned by students during the spring semester at Calendula. To do so, he randomly selects 100 student transcripts and records the number of units each student earned in the spring term. Identify the variable of interest to the researcher. a) The number of students enrolled at Calendula College during the spring term. b) The average indebtedness of Calendula College students enrolled in the spring. c) The age of Calendula College students enrolled in the spring. d) The number of units earned by Calendula College students during the spring term.

ANSWER: d TYPE: MC DIFFICULTY: Easy KEYWORDS: data, sampling 20. Jared was working on a project to look at global warming and accessed an Internet site where he captured average global surface temperatures from 1866. Which of the four methods of data collection was he using? a) Published sources b) Experimentation c) Surveying d) Observation ANSWER: a TYPE: MC DIFFICULTY: Easy KEYWORDS: sources of data 21. The British Airways Internet site provides a questionnaire instrument that can be answered electronically. Which of the 4 methods of data collection is involved when people complete the questionnaire? a) Published sources b) Experimentation c) Surveying d) Observation ANSWER: c TYPE: MC DIFFICULTY: Easy KEYWORDS: sources of data

.

Introduction and Data Collection

7

22. A marketing research firm, in conducting a comparative taste test, provided three types of peanut butter to a sample of households randomly selected within the state. Which of the 4 methods of data collection is involved when people are asked to compare the three types of peanut butter? a) Published sources b) Experimentation c) Surveying d) Observation ANSWER: b TYPE: MC DIFFICULTY: Easy KEYWORDS: sources of data 23. Tim was planning for a meeting with his boss to discuss a raise in his annual salary. In preparation, he wanted to use the Consumer Price Index to determine the percentage increase in his real (inflation-adjusted) salary over the last three years. Which of the 4 methods of data collection was involved when he used the Consumer Price Index? a) Published sources b) Experimentation c) Surveying d) Observation ANSWER: a TYPE: MC DIFFICULTY: Easy KEYWORDS: sources of data 24. Which of the 4 methods of data collection is involved when a person counts the number of cars passing designated locations on the Los Angeles freeway system? a) Published sources b) Experimentation c) Surveying d) Observation ANSWER: d TYPE: MC DIFFICULTY: Moderate KEYWORDS: sources of data 25. A statistics student found a reference in the campus library that contained the median family incomes for all 50 states. She would report her data as being collected using a) a designed experiment. b) observational data. c) a random sample. d) a published source.

ANSWER: d TYPE: MC DIFFICULTY: Easy KEYWORDS: sources of data

.

8

Introduction and Data Collection 26. The personnel director at a large company studied the eating habits of the company’s employees.

The director noted whether employees brought their own lunches to work, ate at the company cafeteria, or went out to lunch. The goal of the study was to improve the food service at the company cafeteria. This type of data collection would best be considered as a) an observational study. b) a designed experiment. c) a random sample. d) a quota sample.

ANSWER: a TYPE: MC DIFFICULTY: Easy KEYWORDS: sources of data 27. A study attempted to estimate the proportion of Florida residents who were willing to spend more tax dollars on protecting the beaches from environmental disasters. Twenty-five hundred Florida residents were surveyed. What type of data collection procedure was most likely used to collect the data for this study? a) A designed experiment b) A published source c) A random sample d) Observational data ANSWER: c TYPE: MC DIFFICULTY: Easy KEYWORDS: sources of data 28. Which of the following is a discrete quantitative variable? a) The Dow Jones Industrial average b) The volume of water released from a dam c) The distance you drove yesterday. d) The number of employees of an insurance company ANSWER: d TYPE: MC DIFFICULTY: Easy KEYWORDS: discrete random variable, types of data 29. Which of the following is a continuous quantitative variable? a) The color of a student’s eyes b) The number of employees of an insurance company c) The amount of milk produced by a cow in one 24-hour period d) The number of gallons of milk sold at the local grocery store yesterday ANSWER: c TYPE: MC DIFFICULTY: Easy KEYWORDS: continuous random variable, types of data

.

Introduction and Data Collection

9

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

ANSWER: b TYPE: MC DIFFICULTY: Moderate KEYWORDS: discrete random variable, types of data 31. Researchers are concerned that the weight of the average American school child is increasing implying, among other things, that children’s clothing should be manufactured and marketed in larger sizes. If X is the weight of school children sampled in a nationwide study, then X is an example of a) a categorical random variable. b) a discrete random variable. c) a continuous random variable. d) a parameter.

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

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

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

.

10

Introduction and Data Collection

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

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

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

.

Introduction and Data Collection

11

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

ANSWER: c TYPE: MC DIFFICULTY: Difficult KEYWORDS: continuous random variable, types of data

TABLE 1-1 The manager of the customer service division of a major consumer electronics company is interested in determining whether the customers who have purchased a videocassette recorder made by the company over the past 12 months are satisfied with their products.

38. Referring to Table 1-1, the population of interest is a) all the customers who have bought a videocassette recorder made by the company over the past 12 months. b) all the customers who have bought a videocassette recorder made by the company and brought it in for repair over the past 12 months. c) all the customers who have used a videocassette recorder over the past 12 months. d) all the customers who have ever bought a videocassette recorder made by the company.

ANSWER: a TYPE: MC DIFFICULTY: Difficult KEYWORDS: population 39. Referring to Table 1-1, which of the following will be a good frame for drawing a sample? a) Telephone directory. b) Voting registry. c) The list of customers who returned the registration card. d) A list of potential customers purchased from a database marketing company.

ANSWER: c TYPE: MC DIFFICULTY: Easy KEYWORDS: frame

.

12

Introduction and Data Collection

40. Referring to Table 1-1, the possible responses to the question "How many videocassette recorders made by other manufacturers have you used?" are values from a a) discrete random variable. b) continuous random variable. c) categorical random variable. d) parameter.

ANSWER: a TYPE: MC DIFFICULTY: Easy KEYWORDS: discrete random variable, types of data 41. Referring to Table 1-1, the possible responses to the question "Are you happy, indifferent, or unhappy with the performance per dollar spent on the videocassette recorder?" are values from a a) discrete numerical random variable. b) continuous numerical random variable. c) categorical random variable. d) parameter.

ANSWER: c TYPE: MC DIFFICULTY: Easy KEYWORDS: categorical random variable, types of data 42. Referring to Table 1-1, the possible responses to the question "What is your annual income rounded to the nearest thousands?" are values from a a) discrete numerical random variable. b) continuous numerical random variable. c) categorical random variable. d) parameter.

ANSWER: a TYPE: MC DIFFICULTY: Easy KEYWORDS: discrete random variable, types of data 43. Referring to Table 1-1, the possible responses to the question "How much time do you use the videocassette recorder every week on the average?" are values from a a) discrete numerical random variable. b) continuous numerical random variable. c) categorical random variable. d) parameter.

ANSWER: b TYPE: MC DIFFICULTY: Easy KEYWORDS: continuous random variable, types of data

.

Introduction and Data Collection

13

44. Referring to Table 1-1, the possible responses to the question "How many people are there in your household?" are values from a a) discrete numerical random variable. b) continuous numerical random variable. c) categorical random variable. d) parameter.

ANSWER: a TYPE: MC DIFFICULTY: Easy KEYWORDS: discrete random variable, types of data 45. Referring to Table 1-1, the possible responses to the question "How would you rate the quality of your purchase experience with 1 = excellent, 2 = good, 3 = decent, 4 = poor, 5 = terrible?" are values from a a) discrete numerical random variable. b) continuous numerical random variable. c) categorical random variable. d) parameter.

ANSWER: c TYPE: MC DIFFICULTY: Easy KEYWORDS: categorical random variable, types of data 46. Referring to Table 1-1, the possible responses to the question "What brand of videocassette recorder did you purchase?" are values from a a) discrete numerical random variable. b) continuous numerical random variable. c) categorical random variable. d) parameter.

ANSWER: c TYPE: MC DIFFICULTY: Easy KEYWORDS: categorical random variable, types of data 47. Referring to Table 1-1, the possible responses to the question "Out of a 100 point score with 100 being the highest and 0 being the lowest, what is your satisfaction level on the videocassette recorder that you purchased?" are values from a a) discrete numerical random variable. b) continuous numerical random variable. c) categorical random variable. d) parameter.

ANSWER: a TYPE: MC DIFFICULTY: Moderate KEYWORDS: discrete random variable, types of data

.

14

Introduction and Data Collection

48. Referring to Table 1-1, the possible responses to the question "In which year were you born?" are values from a a) discrete numerical random variable. b) continuous numerical random variable. c) categorical random variable. d) parameter.

ANSWER: a TYPE: MC DIFFICULTY: Moderate KEYWORDS: discrete random variable, types of data 49. True or False: A population is the totality of items or things under consideration.

ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: population 50. True or False: A sample is the portion of the universe that is selected for analysis.

ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: sample 51. True or False: Problems may arise when statistically unsophisticated users who do not understand the assumptions behind the statistical procedures or their limitations are misled by results obtained from computer software.

ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: statistical package 52. True or False: Managers need an understanding of statistics to be able to present and describe information accurately, draw conclusions about large populations based on small samples, improve processes, and make reliable forecasts.

ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: reasons for learning statistics 53. True or False: The possible responses to the question “How long have you been living at your current residence?” are values from a continuous variable.

ANSWER: True TYPE: TF DIFFICULTY: Moderate KEYWORDS: continuous random variable, types of data

.

Introduction and Data Collection

15

54. True or False: The possible responses to the question “How many times in the past three months have you visited a city park?” are values from a discrete variable.

ANSWER: True TYPE: TF DIFFICULTY: Moderate KEYWORDS: discrete random variable, types of data 55. True or False: A continuous variable may take on any value within its relevant range even though the measurement device may not be precise enough to record it.

ANSWER: True TYPE: TF DIFFICULTY: Moderate KEYWORDS: continuous random variable, types of data 56. True or False: Faculty rank (professor to lecturer) is an example of discrete numerical data.

ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: categorical random variable, types of data 57. True or False: Student grades (A to F) are an example of continuous numerical data.

ANSWER: False TYPE: TF DIFFICULTY: Moderate KEYWORDS: categorical random variables, types of data 58. True or False: The amount of coffee consumed by an individual in a day is an example of a discrete numerical variable.

ANSWER: False TYPE: TF DIFFICULTY: Moderate KEYWORDS: continuous random variables, types of data 59. True or False: A statistic is usually used to provide an estimate for a usually unobserved parameter.

ANSWER: True TYPE: TF DIFFICULTY: Moderate KEYWORDS: statistic, parameter, inferential statistics 60. True or False: A statistic is usually unobservable while a parameter is usually observable.

ANSWER: False TYPE: TF DIFFICULTY: Moderate KEYWORDS: statistic, parameter, inferential statistic .

16

Introduction and Data Collection 61. True or False: The answer to the question “What is your favorite color?” is an example of an ordinal scaled variable.

ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: nominal scale 62. True or False: The answer to the question “How do you rate the quality of your business statistics course” is an example of an ordinal scaled variable.

ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: ordinal scale 63. True or False: A professor computed the sample average exam score of 20 students and used it to estimate the average exam score of the 1,500 students taking the exam was an example of inferential statistics.

ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: descriptive statistics, inferential statistics 64. True or False: Using the number of registered voters who turned out to vote for the primary in Iowa to predict the number of registered voters who will turn out to vote in Vermont’s primary is an example of descriptive statistics.

ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: descriptive statistics, inferential statistics 65. True or False: Compiling the number of registered voters who turned out to vote for the primary in Iowa is an example of descriptive statistics.

ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: descriptive statistics, inferential statistics 66. The Human Resources Director of a large corporation wishes to develop an employee benefits package and decides to select 500 employees from a list of all (N = 40,000) workers in order to study their preferences for the various components of a potential package. All the employees in the corporation constitute the _______.

ANSWER: population TYPE: FI DIFFICULTY: Easy KEYWORDS: population

.

Introduction and Data Collection

17

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

ANSWER: sample TYPE: FI DIFFICULTY: Easy KEYWORDS: sample 68. The Human Resources Director of a large corporation wishes to develop an employee benefits package and decides to select 500 employees from a list of all (N = 40,000) workers in order to study their preferences for the various components of a potential package. The Director will use the data from the sample to compute _______.

ANSWER: statistics TYPE: FI DIFFICULTY: Easy KEYWORDS: statistic 69. The Human Resources Director of a large corporation wishes to develop an employee benefits package and decides to select 500 employees from a list of all (N = 40,000) workers in order to study their preferences for the various components of a potential package. Information obtained from the sample will be used to draw conclusions about the true population _______.

ANSWER: parameters TYPE: FI DIFFICULTY: Easy KEYWORDS: parameter 70. The Human Resources Director of a large corporation wishes to develop an employee benefits package and decides to select 500 employees from a list of all (N = 40,000) workers in order to study their preferences for the various components of a potential package. In this study, methods involving the collection, presentation, and characterization of the data are called _______.

ANSWER: descriptive statistics/methods TYPE: FI DIFFICULTY: Easy KEYWORDS: descriptive statistics 71. The Human Resources Director of a large corporation wishes to develop an employee benefits package and decides to select 500 employees from a list of all (N = 40,000) workers in order to study their preferences for the various components of a potential package. In this study, methods that result in decisions concerning population characteristics based only on the sample results are called _______.

ANSWER: inferential statistics/methods TYPE: FI DIFFICULTY: Easy KEYWORDS: inferential statistics

.

18

Introduction and Data Collection

72. Mediterranean fruit flies were discovered in California a few years ago and badly damaged the oranges grown in that state. Suppose the manager of a large farm wanted to study the impact of the fruit flies on the orange crops on a daily basis over a 6-week period. On each day a random sample of orange trees were selected from within a random sample of acres. The daily average number of damaged oranges per tree and the proportion of trees having damaged oranges were calculated. The two main measures calculated each day (i.e., average number of damaged oranges per tree and proportion of trees having damaged oranges) are called _______.

ANSWER: statistics TYPE: FI DIFFICULTY: Moderate KEYWORDS: statistic 73. Mediterranean fruit flies were discovered in California a few years ago and badly damaged the oranges grown in that state. Suppose the manager of a large farm wanted to study the impact of the fruit flies on the orange crops on a daily basis over a 6-week period. On each day a random sample of orange trees were selected from within a random sample of acres. The daily average number of damaged oranges per tree and the proportion of trees having damaged oranges were calculated. The two main measures calculated each day (i.e., average number of damaged oranges per tree and proportion of trees having damaged oranges) may be used on a daily basis to estimate the respective true population _______.

ANSWER: parameters TYPE: FI DIFFICULTY: Easy KEYWORDS: parameters 74. Mediterranean fruit flies were discovered in California a few years ago and badly damaged the oranges grown in that state. Suppose the manager of a large farm wanted to study the impact of the fruit flies on the orange crops on a daily basis over a 6-week period. On each day a random sample of orange trees were selected from within a random sample of acres. The daily average number of damaged oranges per tree and the proportion of trees having damaged oranges were calculated. In this study, drawing conclusions on any one day about the true population characteristics based on information obtained from the sample is called _______.

ANSWER: inferential statistics/methods TYPE: FI DIFFICULTY: Moderate KEYWORDS: inferential statistics

.

Introduction and Data Collection

19

75. Mediterranean fruit flies were discovered in California a few years ago and badly damaged the oranges grown in that state. Suppose the manager of a large farm wanted to study the impact of the fruit flies on the orange crops on a daily basis over a 6-week period. On each day a random sample of orange trees were selected from within a random sample of acres. The daily average number of damaged oranges per tree and the proportion of trees having damaged oranges were calculated. In this study, the presentation and characterization of the two main measures calculated each day (i.e., average number of damaged oranges per tree and proportion of trees having damaged oranges) is called _______ .

ANSWER: descriptive statistics/methods TYPE: FI DIFFICULTY: Moderate KEYWORDS: descriptive statistics 76. The Quality Assurance Department of a large urban hospital is attempting to monitor and evaluate patient satisfaction with hospital services. Prior to discharge, a random sample of patients is asked to fill out a questionnaire to rate such services as medical care, nursing, therapy, laboratory, food, and cleaning. The Quality Assurance Department prepares weekly reports that are presented at the Board of Directors meetings and extraordinary/atypical ratings are easy to flag. Values computed from the sample results each week are called _______.

ANSWER: statistics TYPE: FI DIFFICULTY: Easy KEYWORDS: statistic 77. The Quality Assurance Department of a large urban hospital is attempting to monitor and evaluate patient satisfaction with hospital services. Prior to discharge, a random sample of patients is asked to fill out a questionnaire to rate such services as medical care, nursing, therapy, laboratory, food, and cleaning. The Quality Assurance Department prepares weekly reports that are presented at the Board of Directors meetings and extraordinary/atypical ratings are easy to flag. True population characteristics estimated from the sample results each week are called _______.

ANSWER: parameters TYPE: FI DIFFICULTY: Easy KEYWORDS: parameter 78. The Commissioner of Health in New York State wanted to study malpractice litigation in New York. A sample of 31 thousand medical records was drawn from a population of 2.7 million patients who were discharged during the year 1997. The proportion of malpractice claims filed from the sample of 31 thousand patients is a _______.

ANSWER: statistic TYPE: FI DIFFICULTY: Moderate KEYWORDS: statistic

.

20

Introduction and Data Collection

79. The Commissioner of Health in New York State wanted to study malpractice litigation in New York. A sample of 31 thousand medical records was drawn from a population of 2.7 million patients who were discharged during the year 1997. The true proportion of malpractice claims filed from the population of 2.7 million patients is a _______.

ANSWER: parameter TYPE: FI DIFFICULTY: Easy KEYWORDS: parameter 80. The Commissioner of Health in New York State wanted to study malpractice litigation in New York. A sample of 31 thousand medical records was drawn from a population of 2.7 million patients who were discharged during the year 1997. Using the information obtained from the sample to predict population characteristics with respect to malpractice litigation is an example of _______.

ANSWER: inferential statistics TYPE: FI DIFFICULTY: Moderate KEYWORDS: inferential statistics 81. The Commissioner of Health in New York State wanted to study malpractice litigation in New York. A sample of 31 thousand medical records was drawn from a population of 2.7 million patients who were discharged during the year 1997. The collection, presentation, and characterization of the data from patient medical records are examples of _______.

ANSWER: descriptive statistics/methods TYPE: FI DIFFICULTY: Easy KEYWORDS: descriptive statistics 82. An insurance company evaluates many numerical variables about a person before deciding on an appropriate rate for automobile insurance. The number of claims a person has made in the last 3 years is an example of a _______ numerical variable.

ANSWER: discrete TYPE: FI DIFFICULTY: Easy KEYWORDS: discrete random variable, types of data 83. An insurance company evaluates many numerical variables about a person before deciding on an appropriate rate for automobile insurance. The distance a person drives in a year is an example of a _______ variable.

ANSWER: continuous TYPE: FI DIFFICULTY: Easy KEYWORDS: continuous random variable, types of data

.

Introduction and Data Collection

21

84. An insurance company evaluates many numerical variables about a person before deciding on an appropriate rate for automobile insurance. A person's age is an example of a _______ numerical variable.

ANSWER: continuous TYPE: FI DIFFICULTY: Easy KEYWORDS: continuous random variable, types of data 85. An insurance company evaluates many numerical variables about a person before deciding on an appropriate rate for automobile insurance. How long a person has been a licensed driver is an example of a _______ numerical variable.

ANSWER: continuous TYPE: FI DIFFICULTY: Moderate KEYWORDS: continuous random variable, types of data 86. An insurance company evaluates many numerical variables about a person before deciding on an appropriate rate for automobile insurance. The number of tickets a person has received in the last 3 years is an example of a _______ numerical variable.

ANSWER: discrete TYPE: FI DIFFICULTY: Easy KEYWORDS: discrete random variable, types of data 87. In purchasing an automobile, there are a number of variables to consider. The body style of the car (sedan, coupe, wagon, etc.) is an example of a _______ variable.

ANSWER: categorical TYPE: FI DIFFICULTY: Easy KEYWORDS: categorical random variable, types of data 88. In purchasing an automobile, there are a number of variables to consider. The classification of the car as a subcompact, compact, standard, or luxury size is an example of a _______ variable.

ANSWER: categorical TYPE: FI DIFFICULTY: Easy KEYWORDS: categorical random variable, types of data 89. In purchasing an automobile, there are a number of variables to consider. The color of the car is an example of a _______ variable.

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

.

22

Introduction and Data Collection

90. Most colleges admit students based on their achievements in a number of different areas. Whether a student has taken any advanced placement courses is an example of a _______ variable.

ANSWER: categorical TYPE: FI DIFFICULTY: Easy KEYWORDS: categorical random variable, types of data 91. Most colleges admit students based on their achievements in a number of different areas. The grade obtained in senior level English. (A, B, C, D, or F) is an example of a _______ variable.

ANSWER: categorical TYPE: FI DIFFICULTY: Moderate KEYWORDS: categorical random variable, types of data 92. Most colleges admit students based on their achievements in a number of different areas. The total SAT score achieved by a student is an example of a _______ numerical variable.

ANSWER: discrete TYPE: FI DIFFICULTY: Moderate KEYWORDS: discrete random variable, types of data 93. The Dean of Students conducted a survey on campus. The gender of the student is an example of a _______ variable.

ANSWER: categorical TYPE: FI DIFFICULTY: Easy KEYWORDS: categorical random variable, types of data 94. The Dean of Students conducted a survey on campus. Class designation (Freshman, Sophomore, Junior, Senior) is an example of a _______ variable.

ANSWER: categorical TYPE: FI DIFFICULTY: Easy KEYWORDS: categorical random variable, types of data 95. The Dean of Students conducted a survey on campus. Major area of study is an example of a _______ variable.

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

.

Introduction and Data Collection

23

96. The Dean of Students conducted a survey on campus. Average SAT score in mathematics is an example of a _______ numerical variable.

ANSWER: continuous TYPE: FI DIFFICULTY: Easy KEYWORDS: continuous random variable, types of data 97. The Dean of Students conducted a survey on campus. Grade point average (GPA) is an example of a _______ numerical variable.

ANSWER: continuous TYPE: FI DIFFICULTY: Easy KEYWORDS: continuous random variable, types of data 98. The Dean of Students conducted a survey on campus. Number of credits currently enrolled for is an example of a _______ numerical variable.

ANSWER: discrete TYPE: FI DIFFICULTY: Easy KEYWORDS: discrete random variable, types of data 99. The Dean of Students conducted a survey on campus. Number of clubs, groups, teams, and organizations affiliated with on campus is an example of a _______ numerical variable.

ANSWER: discrete TYPE: FI DIFFICULTY: Easy KEYWORDS: discrete random variable, types of data 100. A personal computer user survey was conducted. Computer brand primarily used is an example of a _______ variable.

ANSWER: categorical TYPE: FI DIFFICULTY: Easy KEYWORDS: categorical random variable, types of data 101. A personal computer user survey was conducted. Number of personal computers owned is an example of a _______ numerical variable.

ANSWER: discrete TYPE: FI DIFFICULTY: Easy KEYWORDS: discrete random variable, types of data

.

24

Introduction and Data Collection

102. A personal computer user survey was conducted. The number of years using a personal computer is an example of a _______ numerical variable.

ANSWER: continuous TYPE: FI DIFFICULTY: Moderate KEYWORDS: continuous random variable, types of data 103. A personal computer user survey was conducted. Hours of personal computer use per week is an example of a _______ numerical variable ANSWER: continuous TYPE: FI DIFFICULTY: Moderate KEYWORDS: continuous random variable, types of data 104. A personal computer user survey was conducted. Primary word processing package used is an example of a _______ variable ANSWER: categorical TYPE: FI DIFFICULTY: Easy KEYWORDS: categorical random variable, types of data 105. A personal computer user survey was conducted. The number of computer magazine subscriptions is an example of a _______ numerical variable.

ANSWER: discrete TYPE: FI DIFFICULTY: Moderate KEYWORDS: discrete random variable, types of data 106.

The type of TV one owns is an example of an ordinal scaled variable.

ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: nominal scale, ordinal scale 107.

The type of TV one owns is an example of a numerical variable.

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

.

Introduction and Data Collection

108.

25

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

ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: nominal scale 109. Whether the university is private or public is an example of a categorical variable.

ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: categorical random variable 110. Marital status is an example of an ordinal scaled variable.

ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: nominal scale, ordinal scale 111. Marital status is an example of a numerical variable.

ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: categorical random variable 112. The grade level (K-12) of a student is an example of a nominal scaled variable.

ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: nominal scale, ordinal scale 113. The grade level (K-12) of a student is an example of a numerical variable.

ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: categorical random variable 114. The level of satisfaction (“Very unsatisfied”, “Fairly unsatisfied”, “Fairly satisfied”, and “Very satisfied”) in a class is an example of an ordinal scaled variable.

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

.

26

Introduction and Data Collection 115. The level of satisfaction (“Very unsatisfied”, “Fairly unsatisfied”, “Fairly satisfied”, and “Very satisfied”) in a class is an example of a categorical variable.

ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: categorical random variable 116. The quality (“terrible”, “poor”, “fair”, “acceptable”, “very good” and “excellent”) of a day care center is an example of a nominal scaled variable.

ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: nominal scale, ordinal scale 117. The quality (“terrible”, “poor”, “fair”, “acceptable”, “very good” and “excellent”) of a day care center is an example of a numerical variable.

ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: categorical random variable 118. The amount of alcohol consumed by a person per week is an example of a continuous variable.

ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: continuous random variable 119. The number of defective apples in a single box is an example of a continuous variable.

ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: discrete random variable, continuous random variable 120. The amount of calories contained in a pack of 12-ounce cheese is an example of a discrete variable.

ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: discrete random variable, continuous random variable

.

Introduction and Data Collection

27

121. The amount of time a student spent studying for an exam is an example of a continuous variable.

ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: continuous random variable 122. Data sources are considered primary sources when they are primarily used by the people who analyze them.

ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: primary source 123. Data sources are considered primary sources when they are analyzed by the people who also collect them.

ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: primary source 124. Data sources are considered secondary sources when they are not primarily used by the people who analyze them.

ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: secondary source 125. Data sources are considered secondary sources when they are analyzed by the people who do not collect them.

ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: secondary source 126. To learn to use statistical programs, you need to understand the underlying statistical concepts.

ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: learning statistical programs

.

28

Introduction and Data Collection

127. To learn to use statistical programs, you need to know how to review the results for potential errors.

ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: learning statistical programs 128. To learn to use statistical programs, you need to know how to organize and present information.

ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: learning statistical programs 129. To learn to use statistical programs, you need to know how to write computer program.

ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: learning statistical programs 130. To learn to use statistical programs, you need to be a computer scientist.

ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: learning statistical programs

.

Presenting Data in Tables and Charts 29

CHAPTER 2: PRESENTING DATA IN TABLES AND CHARTS TABLE 2-1 An insurance company evaluates many numerical variables about a person before deciding on an appropriate rate for automobile insurance. A representative from a local insurance agency selected a random sample of insured drivers and recorded, X, the number of claims each made in the last 3 years, with the following results.

X f 1 14 2 18 3 12 4 5 5 1 1. Referring to Table 2-1, how many drivers are represented in the sample? a) 5 b) 15 c) 18 d) 50 ANSWER: d TYPE: MC DIFFICULTY: Easy KEYWORDS: frequency distribution 2. Referring to Table 2-1, how many total claims are represented in the sample? a) 15 b) 50 c) 111 d) 250 ANSWER: c TYPE: MC DIFFICULTY: Moderate KEYWORDS: interpretation, frequency distribution

3. A type of vertical bar chart in which the categories are plotted in the descending rank order of the magnitude of their frequencies is called a .

30

Presenting Data in Tables and Charts

a) b) c) d)

contingency table.

Pareto diagram. dot plot. pie chart.

ANSWER: b TYPE: MC DIFFICULTY: Easy KEYWORDS: Pareto diagram

TABLE 2-2 At a meeting of information systems officers for regional offices of a national company, a survey was taken to determine the number of employees the officers supervise in the operation of their departments, where X is the number of employees overseen by each information systems officer.

X f_ 1 7 2 5 3 11 4 8 5 9 4. Referring to Table 2-2, how many regional offices are represented in the survey results? a) 5 b) 11 c) 15 d) 40 ANSWER: d TYPE: MC DIFFICULTY: Easy KEYWORDS: interpretation, frequency distribution 5. Referring to Table 2-2, across all of the regional offices, how many total employees were supervised by those surveyed? a) 15 b) 40 c) 127 d) 200 ANSWER: c TYPE: MC DIFFICULTY: Moderate KEYWORDS: interpretation, frequency distribution

6. The width of each bar in a histogram corresponds to the a) differences between the boundaries of the class. .

Presenting Data in Tables and Charts 31

b) number of observations in each class. c) midpoint of each class. d) percentage of observations in each class.

ANSWER: a TYPE: MC DIFFICULTY: Easy KEYWORDS: frequency distribution

TABLE 2-3 Every spring semester, the School of Business coordinates with local business leaders a luncheon for graduating seniors, their families, and friends. Corporate sponsorship pays for the lunches of each of the seniors, but students have to purchase tickets to cover the cost of lunches served to guests they bring with them. The following histogram represents the attendance at the senior luncheon, where X is the number of guests each graduating senior invited to the luncheon and f is the number of graduating seniors in each category.

160

152

140 120 100 Frequency

85

80 60 40 20

18

17

3

0

4

5

0 0

1

2 Guests per Student

3

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

KEYWORDS: interpretation, histogram

8. Referring to the histogram from Table 2-3, if all the tickets purchased were used, how many guests attended the luncheon? .

32

Presenting Data in Tables and Charts

a) b) c) d)

4 152 275 388

ANSWER: d TYPE: MC DIFFICULTY: Difficult EXPLANATION: The total number of guests is

 Xf 6

i =1

i i

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

Seniors 14%

Freshmen 10%

Juniors 30% Sophomores 46%

a) b) c) d)

14% 44% 54% 86%

ANSWER: d TYPE: MC DIFFICULTY: Easy KEYWORDS: interpretation, pie chart 10. When polygons or histograms are constructed, which axis must show the true zero or "origin"? a) The horizontal axis. b) The vertical axis. c) Both the horizontal and vertical axes. d) Neither the horizontal nor the vertical axis.

ANSWER: b TYPE: MC DIFFICULTY: Easy KEYWORDS: polygon, histogram 11. When constructing charts, the following is plotted at the class midpoints: a) frequency histograms. .

Presenting Data in Tables and Charts 33

b) percentage polygons. c) cumulative relative frequency ogives. d) All of the above.

ANSWER: b TYPE: MC DIFFICULTY: Easy KEYWORDS: percentage polygon

TABLE 2-4 A survey was conducted to determine how people rated the quality of programming available on television. Respondents were asked to rate the overall quality from 0 (no quality at all) to 100 (extremely good quality). The stem-and-leaf display of the data is shown below.

Stem Leaves 3 24 4 03478999 5 0112345 6 12566 7 01 8 9 2 12. Referring to Table 2-4, what percentage of the respondents rated overall television quality with a rating of 80 or above? a) 0 b) 4 c) 96 d) 100 ANSWER: b TYPE: MC DIFFICULTY: Easy KEYWORDS: stem-and-leaf display, interpretation 13. Referring to Table 2-4, what percentage of the respondents rated overall television quality with a rating of 50 or below? a) 11 b) 40 c) 44 d) 56 ANSWER: c TYPE: MC DIFFICULTY: Moderate KEYWORDS: stem-and-leaf display, interpretation

14. Referring to Table 2-4, what percentage of the respondents rated overall television quality with a rating between 50 and 75? a) 11 .

34

Presenting Data in Tables and Charts

b) 40 c) 44 d) 56 ANSWER: d TYPE: MC DIFFICULTY: Moderate KEYWORDS: stem-and-leaf display, interpretation

TABLE 2-5 The following are the durations in minutes of a sample of long-distance phone calls made within the continental United States reported by one long-distance carrier.

Time (in Minutes) 0 but less than 5 5 but less than 10 10 but less than 15 15 but less than 20 20 but less than 25 25 but less than 30 30 or more

Relative Frequency 0.37 0.22 0.15 0.10 0.07 0.07 0.02

15. Referring to Table 2-5, what is the width of each class? a) 1 minute b) 5 minutes c) 2% d) 100% ANSWER: b TYPE: MC DIFFICULTY: Easy KEYWORDS: class interval, relative frequency distribution 16. Referring to Table 2-5, if 1,000 calls were randomly sampled, how many calls lasted under 10 minutes? a) 220 b) 370 c) 410 d) 590 ANSWER: d TYPE: MC DIFFICULTY: Moderate KEYWORDS: relative frequency distribution, interpretation

17. Referring to Table 2-5, if 100 calls were randomly sampled, how many calls lasted 15 minutes or longer? a) 10 b) 14 .

Presenting Data in Tables and Charts 35

c) 26 d) 74 ANSWER: c TYPE: MC DIFFICULTY: Moderate KEYWORDS: relative frequency distribution, interpretation 18. Referring to Table 2-5, if 10 calls lasted 30 minutes or more, how many calls lasted less than 5 minutes? a) 10 b) 185 c) 295 d) 500 ANSWER: b TYPE: MC DIFFICULTY: Moderate KEYWORDS: relative frequency distribution, interpretation 19. Referring to Table 2-5, what is the cumulative relative frequency for the percentage of calls that lasted under 20 minutes? a) 0.10 b) 0.59 c) 0.76 d) 0.84 ANSWER: d TYPE: MC DIFFICULTY: Easy KEYWORDS: cumulative relative frequency 20. Referring to Table 2-5, what is the cumulative relative frequency for the percentage of calls that lasted 10 minutes or more? a) 0.16 b) 0.24 c) 0.41 d) 0.90 ANSWER: c TYPE: MC DIFFICULTY: Moderate KEYWORDS: cumulative relative frequency

21. Referring to Table 2-5, if 100 calls were randomly sampled, _______ of them would have lasted at least 15 minutes but less than 20 minutes a) 0.10 b) 0.16 c) 10 .

36

Presenting Data in Tables and Charts

d) 16 ANSWER: c TYPE: MC DIFFICULTY: Easy KEYWORDS: relative frequency distribution, interpretation 22. Referring to Table 2-5, if 100 calls were sampled, _______ of them would have lasted less than 15 minutes. a) 26 b) 74 c) 10 d) None of the above.

ANSWER: b TYPE: MC DIFFICULTY: Moderate KEYWORDS: relative frequency distribution, interpretation 23. Referring to Table 2-5, if 100 calls were sampled, _______of them would have lasted 20 minutes or more. a) 26 b) 16 c) 74 d) None of the above.

ANSWER: b TYPE: MC DIFFICULTY: Moderate KEYWORDS: relative frequency distribution, interpretation 24. Referring to Table 2-5, if 100 calls were sampled, _______ of them would have lasted less than 5 minutes or at least 30 minutes or more. a) 35 b) 37 c) 39 d) None of the above.

ANSWER: c TYPE: MC DIFFICULTY: Difficult KEYWORDS: relative frequency distribution, interpretation

25. When studying the simultaneous responses to two categorical questions, we should set up a a) contingency table. b) frequency distribution table. c) cumulative percentage distribution table. d) histogram. .

Presenting Data in Tables and Charts 37

ANSWER: a TYPE: MC DIFFICULTY: Easy KEYWORDS: contingency table 26. Data on 1,500 students’ height were collected at a larger university in the East Coast. Which of the following is the best chart for presenting the information? a) A pie chart. b) A Pareto diagram. c) A scatter plot. d) A histogram.

ANSWER: d TYPE: MC DIFFICULTY: Easy KEYWORDS: choice of chart, histogram 27. Data on the number of part-time hours students at a public university worked in a week were collected. Which of the following is the best chart for presenting the information? a) A pie chart. b) A Pareto diagram. c) A percentage table. d) A percentage polygon.

ANSWER: d TYPE: MC DIFFICULTY: Easy KEYWORDS: choice of chart, percentage polygon 28. Data on the number of credit hours of 20,000 students at a public university enrolled in a Spring semester were collected. Which of the following is the best for presenting the information? a) A pie chart. b) A Pareto diagram. c) A stem-and-leaf display. d) A contingency table.

ANSWER: c TYPE: MC DIFFICULTY: Easy KEYWORDS: choice of chart, stem-and-leaf

29. A survey of 150 executives were asked what they think is the most common mistake candidates make during job interviews. Six different mistakes were given. Which of the following is the best for presenting the information? a) A bar chart. b) A histogram c) A stem-and-leaf display. d) A contingency table. .

38

Presenting Data in Tables and Charts

ANSWER: a TYPE: MC DIFFICULTY: Easy KEYWORDS: choice of chart, bar chart 30. You have collected information on the market share of 5 different search engines used by U.S. Internet users in May 2007. Which of the following is the best for presenting the information? a) A pie chart. b) A histogram c) A stem-and-leaf display. d) A contingency table.

ANSWER: a TYPE: MC DIFFICULTY: Easy KEYWORDS: choice of chart, pie chart 31. You have collected information on the market share of 5 different search engines used by U.S. Internet users in May 2007. Which of the following is the best for presenting the information? a) A pie chart. b) A histogram c) A stem-and-leaf display. d) A contingency table.

ANSWER: a TYPE: MC DIFFICULTY: Easy KEYWORDS: choice of chart, pie chart 32. You have collected information on the consumption by the 15 largest coffee-consuming nations.

Which of the following is the best for presenting the share of the consumption? a) A pie chart. b) A Pareto diagram c) A scatter plot. d) A contingency table.

ANSWER: b TYPE: MC DIFFICULTY: Moderate KEYWORDS: choice of chart, Pareto diagram NOTE: Even though a pie chart can also be used, the Pareto diagram is preferable for separating the “vital few” from the “trivial many”.

33. You have collected data on the approximate retail price (in $) and the energy cost per year (in $) of 15 refrigerators. Which of the following is the best for presenting the data? a) A pie chart. b) A scatter diagram. c) A Pareto diagram. d) A contingency table.

ANSWER: .

Presenting Data in Tables and Charts 39

b TYPE: MC DIFFICULTY: Easy KEYWORDS: choice of chart, scatter diagram 34. You have collected data on the number of U.S. households actively using online banking and/or online bill payment from 1995 to 2007. Which of the following is the best for presenting the data? a) A pie chart. b) A stem-and-leaf display c) A Pareto diagram. d) A time-series plot.

ANSWER: d TYPE: MC DIFFICULTY: Easy KEYWORDS: choice of chart, time-series plot 35. You have collected data on the monthly seasonally adjusted civilian unemployment rate for the United States from 1998 to 2007. Which of the following is the best for presenting the data? a) A contingency table. b) A stem-and-leaf display c) A time-series plot. d) A Pareto diagram.

ANSWER: c TYPE: MC DIFFICULTY: Easy KEYWORDS: choice of chart, time-series plot 36. You have collected data on the responses to two questions asked in a survey of 40 college students majoring in business—What is your gender (Male = M; Female = F) and What is your major (Accountancy = A; Computer Information Systems = C; Marketing = M). Which of the following is the best for presenting the data? a) A contingency table. b) A stem-and-leaf display c) A time-series plot. d) A Pareto diagram.

ANSWER: a TYPE: MC DIFFICULTY: Moderate KEYWORDS: choice of chart, contingency table

TABLE 2-6 A sample of 200 students at a Big-Ten university was taken after the midterm to ask them whether they went bar hopping the weekend before the midterm or spent the weekend studying, and whether they did well or poorly on the midterm. The following table contains the result.

Studying for Exam Went Bar Hopping .

Did Well in Midterm 80 30

Did Poorly in Midterm 20 70

40

Presenting Data in Tables and Charts

37. Referring to Table 2-6, of those who went bar hopping the weekend before the midterm in the sample, _______ percent of them did well on the midterm. a) 15 b) 27.27 c) 30 d) 55 ANSWER: c TYPE: MC DIFFICULTY: Easy KEYWORDS: contingency table, interpretation 38. Referring to Table 2-6, of those who did well on the midterm in the sample, _______ percent of them went bar hopping the weekend before the midterm. a) 15 b) 27.27 c) 30 d) 50 ANSWER: b TYPE: MC DIFFICULTY: Easy KEYWORDS: contingency table, interpretation 39. Referring to Table 2-6, _______ percent of the students in the sample went bar hopping the weekend before the midterm and did well on the midterm. a) 15 b) 27.27 c) 30 d) 50 ANSWER: a TYPE: MC DIFFICULTY: Easy KEYWORDS: contingency table, interpretation

40. Referring to Table 2-6, _______ percent of the students in the sample spent the weekend studying and did well on the midterm. a) 40 b) 50 c) 72.72 d) 80 ANSWER: a TYPE: MC DIFFICULTY: Easy .

Presenting Data in Tables and Charts 41

KEYWORDS: contingency table, interpretation 41. Referring to Table 2-6, if the sample is a good representation of the population, we can expect _______ percent of the students in the population to spend the weekend studying and do poorly on the midterm. a) 10 b) 20 c) 45 d) 50 ANSWER: a TYPE: MC DIFFICULTY: Easy KEYWORDS: contingency table, interpretation 42. Referring to Table 2-6, if the sample is a good representation of the population, we can expect _______ percent of those who spent the weekend studying to do poorly on the midterm. a) 10 b) 20 c) 45 d) 50 ANSWER: b TYPE: MC DIFFICULTY: Moderate KEYWORDS: contingency table, interpretation 43. Referring to Table 2-6, if the sample is a good representation of the population, we can expect _______ percent of those who did poorly on the midterm to have spent the weekend studying. a) 10 b) 22.22 c) 45 d) 50 ANSWER: b TYPE: MC DIFFICULTY: Moderate KEYWORDS: contingency table, interpretation

44. In a contingency table, the number of rows and columns a) must always be the same. b) must always be 2. c) must add to 100%. d) None of the above.

ANSWER: d TYPE: MC DIFFICULTY: Moderate KEYWORDS: contingency table .

42

Presenting Data in Tables and Charts

45. Retailers are always interested in determining why a customer selected their store to make a purchase. A sporting goods retailer conducted a customer survey to determine why its customers shopped at the store. The results are shown in the bar chart below. What proportion of the customers responded that they shopped at the store because of the merchandise or the convenience?

Prices

20%

Merchandise

Convenience

15%

Other

15%

0%

a) b) c) d)

50%

10%

20%

30% Responses

40%

50%

60%

35% 50% 65% 85%

ANSWER: c TYPE: MC DIFFICULTY: Easy KEYWORDS: bar chart, interpretation

TABLE 2-7 The Stem-and-Leaf display below contains data on the number of months between the date a civil suit is filed and when the case is actually adjudicated for 50 cases heard in superior court.

Stem Leaves 1L23444 1H 7899 2L222234 2H 55678889 3L001113 .

Presenting Data in Tables and Charts 43

3H 5778 4L0234 4H 5579 5L1124 5H 66 6L15 6H 8 Note: 1L means the “low teens” – 10, 11, 12, 13, or 14; 1H means the “high teens” – 15, 16, 17, 18, or 19; 2L means the “low twenties” – 20, 21, 22, 23, or 24, etc.

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

ANSWER: 12 TYPE: FI DIFFICULTY: 1 Easy KEYWORDS: stem-and-leaf display, interpretation 47. Referring to Table 2-7, the civil suit with the longest wait between when the suit was filed and when it was adjudicated had a wait of ________ months.

ANSWER: 68 TYPE: FI DIFFICULTY: Easy KEYWORDS: stem-and-leaf display, interpretation 48. Referring to Table 2-7, the civil suit with the fourth shortest waiting time between when the suit was filed and when it was adjudicated had a wait of ________ months.

ANSWER: 14 TYPE: FI DIFFICULTY: Moderate KEYWORDS: stem-and-leaf display, interpretation 49. Referring to Table 2-7, ________ percent of the cases were adjudicated within the first 2 years.

ANSWER: 30 TYPE: FI DIFFICULTY: Moderate KEYWORDS: stem-and-leaf display, interpretation 50. Referring to Table 2-7, ________ percent of the cases were not adjudicated within the first 4 years.

ANSWER: 20 TYPE: FI DIFFICULTY: Moderate KEYWORDS: stem-and-leaf display, interpretation 51. Referring to Table 2-7, if a frequency distribution with equal sized classes was made from this data, and the first class was "10 but less than 20," the frequency of that class would be ________. .

44

Presenting Data in Tables and Charts

ANSWER: 9 TYPE: FI DIFFICULTY: Easy KEYWORDS: stem-and-leaf display, interpretation 52. Referring to Table 2-7, if a frequency distribution with equal sized classes was made from this data, and the first class was "10 but less than 20," the relative frequency of the third class would be ________.

ANSWER: 0.20 or 20% or 10/50 TYPE: FI DIFFICULTY: Moderate KEYWORDS: stem-and-leaf display, relative frequency distribution 53. Referring to Table 2-7, if a frequency distribution with equal sized classes was made from this data, and the first class was "10 but less than 20," the cumulative percentage of the second class would be ________.

ANSWER: 46% or 0.46 or 23/50 TYPE: FI DIFFICULTY: Moderate KEYWORDS: stem-and-leaf display, cumulative percentage distribution

TABLE 2-8 The Stem-and-Leaf display represents the number of times in a year that a random sample of 100 "lifetime" members of a health club actually visited the facility.

Stem Leaves 0 012222233333344566666667789999 1 1111222234444455669999 2 00011223455556889 3 0000446799 4 011345567 .

Presenting Data in Tables and Charts 45

5 6 7 8 9

0077 8 67 3 0247

54. Referring to Table 2-8, the person who has the largest leaf associated with the smallest stem visited the facility ________ times.

ANSWER: 9 TYPE: FI DIFFICULTY: Moderate KEYWORDS: stem-and-leaf display, interpretation 55. Referring to Table 2-8, the person who visited the health club less than anyone else in the sample visited the facility ________ times.

ANSWER: 0 or no TYPE: FI DIFFICULTY: Easy KEYWORDS: stem-and-leaf display, interpretation 56. Referring to Table 2-8, the person who visited the health club more than anyone else in the sample visited the facility ________ times.

ANSWER: 97 TYPE: FI DIFFICULTY: Easy KEYWORDS: stem-and-leaf display, interpretation 57. Referring to Table 2-8, ________ of the 100 members visited the health club at least 52 times in a year.

ANSWER: 10 TYPE: FI DIFFICULTY: Moderate KEYWORDS: stem-and-leaf display, interpretation

58. Referring to Table 2-8, ________ of the 100 members visited the health club no more than 12 times in a year.

ANSWER: 38 TYPE: FI DIFFICULTY: Moderate KEYWORDS: stem-and-leaf display, interpretation 59. Referring to Table 2-8, if a frequency distribution with equal sized classes was made from this data, and the first class was "0 but less than 10," the frequency of the fifth class would be ________. .

46

Presenting Data in Tables and Charts

ANSWER: 9 TYPE: FI DIFFICULTY: Moderate KEYWORDS: stem-and-leaf display, frequency distribution

60. Referring to Table 2-8, if a frequency distribution with equal sized classes was made from this data, and the first class was "0 but less than 10," the relative frequency of the last class would be ________.

ANSWER: 4% or 0.04 or 4/100 TYPE: FI DIFFICULTY: Moderate KEYWORDS: stem-and-leaf display, relative frequency distribution 61. Referring to Table 2-8, if a frequency distribution with equal sized classes was made from this data, and the first class was "0 but less than 10," the cumulative percentage of the next-to-last class would be ________.

ANSWER: 96% or 0.96 or 96/100 TYPE: FI DIFFICULTY: Moderate KEYWORDS: stem-and-leaf display, cumulative percentage distribution 62. Referring to Table 2-8, if a frequency distribution with equal sized classes was made from this data, and the first class was "0 but less than 10," the class midpoint of the third class would be ________.

ANSWER: 25 or (20+30)/2 TYPE: FI DIFFICULTY: Moderate KEYWORDS: stem-and-leaf display, class midpoint

TABLE 2-9 The frequency distribution below represents the rents of 250 randomly selected federally subsidized apartments in Minneapolis.

Rent in $ Frequency 300 but less than 400 113 400 but less than 500 85 500 but less than 600 32 600 but less than 700 16 700 but less than 800 4 .

Presenting Data in Tables and Charts 47

63. Referring to Table 2-9, ________ apartments rented for at least $400 but less than $600.

ANSWER: 117 TYPE: FI DIFFICULTY: Easy KEYWORDS: frequency distribution 64. Referring to Table 2-9, ________ percent of the apartments rented for no less than $600.

ANSWER: 8% or 20/250 TYPE: FI DIFFICULTY: Easy KEYWORDS: frequency distribution, cumulative percentage distribution 65. Referring to Table 2-9, ________ percent of the apartments rented for at least $500.

ANSWER: 20.8% or 52/250 TYPE: FI DIFFICULTY: Moderate KEYWORDS: frequency distribution, cumulative percentage distribution 66. Referring to Table 2-9, the class midpoint of the second class is ________.

ANSWER: 450 TYPE: FI DIFFICULTY: Easy KEYWORDS: frequency distribution, class midpoint 67. Referring to Table 2-9, the relative frequency of the second class is ________.

ANSWER: 85/250 or 17/50 or 34% or 0.34 TYPE: FI DIFFICULTY: Easy KEYWORDS: frequency distribution, relative frequency distribution

68. Referring to Table 2-9, the percentage of apartments renting for less than $600 is ________.

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

.

48

Presenting Data in Tables and Charts

0.30 Rel.Freq.

0.20

0.20

0.20

0.20

0.10

0.10

0.10

0.10

0.10

0.00 0

10

20

30

40

50

60

70

69. Referring to the histogram from Table 2-10, ________ percent of the job applicants scored between 10 and 20.

ANSWER: 20% TYPE: FI DIFFICULTY: Easy KEYWORDS: histogram, percentage distribution 70. Referring to the histogram from Table 2-10, ________ percent of the job applicants scored below 50.

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

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

ANSWER: 80 TYPE: FI DIFFICULTY: Moderate KEYWORDS: histogram 72. Referring to the histogram from Table 2-10, the number of job applicants who scored 50 or above is _______.

ANSWER: 40 .

Presenting Data in Tables and Charts 49

TYPE: FI DIFFICULTY: Moderate KEYWORDS: histogram 73. Referring to the histogram from Table 2-10, 90% of the job applicants scored above or equal to ________.

ANSWER: 10 TYPE: FI DIFFICULTY: Moderate KEYWORDS: histogram, cumulative percentage distribution 74. Referring to the histogram from Table 2-10, half of the job applicants scored below ________.

ANSWER: 30 TYPE: FI DIFFICULTY: Moderate KEYWORDS: histogram, cumulative percentage distribution

75. Referring to the histogram from Table 2-10, _______ percent of the applicants scored below 20 or at least 50.

ANSWER: 50% TYPE: FI DIFFICULTY: Moderate KEYWORDS: histogram, cumulative percentage distribution 76. Referring to the histogram from Table 2-10, _______ percent of the applicants scored between 20 and below 50.

ANSWER: 50% TYPE: FI DIFFICULTY: Moderate KEYWORDS: histogram, cumulative percentage distribution

TABLE 2-11 The ordered array below resulted from taking a sample of 25 batches of 500 computer chips and determining how many in each batch were defective.

Defects 1 2 4 17 20 21

4 23

5 23

5 25

6 26

7 27

9 27

9 28

12 29

12 29

15

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

.

50

Presenting Data in Tables and Charts

ANSWER: 4 TYPE: FI DIFFICULTY: Easy KEYWORDS: frequency distribution 78. Referring to Table 2-11, if a frequency distribution for the defects data is constructed, using "0 but less than 5" as the first class, the relative frequency of the “15 but less than 20” class would be ________.

ANSWER: 0.08 or 8% or 2/25 TYPE: FI DIFFICULTY: Moderate KEYWORDS: relative frequency distribution 79. Referring to Table 2-11, construct a frequency distribution for the defects data, using "0 but less than 5" as the first class.

ANSWER: Defects Frequency 0 but less than 5 4 5 but less than 10 6 10 but less than 15 2 15 but less than 20 2 20 but less than 25 4 25 but less than 30 7 TYPE: PR DIFFICULTY: Easy KEYWORDS: frequency distribution

80. Referring to Table 2-11, construct a relative frequency or percentage distribution for the defects data, using "0 but less than 5" as the first class.

ANSWER: Defects Percentage 0 but less than 5 16 5 but less than 10 24 10 but less than 15 8 15 but less than 20 8 20 but less than 25 16 25 but less than 30 28 TYPE: PR DIFFICULTY: Moderate KEYWORDS: relative frequency distribution, percentage distribution

.

Presenting Data in Tables and Charts 51

81. Referring to Table 2-11, construct a cumulative percentage distribution for the defects data if the corresponding frequency distribution uses "0 but less than 5" as the first class.

ANSWER: Defects CumPct 0 0 5 16 10 40 15 48 20 56 25 72 30 100 TYPE: PR DIFFICULTY: Moderate KEYWORDS: cumulative percentage distribution 82. Referring to Table 2-11, construct a histogram for the defects data, using "0 but less than 5" as the first class.

ANSWER: 7

7

Frequency 6

6 5 4

4

4 3 2

2

2 1 0

0

5

10

15 20 Number of Defects

25

30

TYPE: PR DIFFICULTY: Easy KEYWORDS: histogram, frequency distribution 83. Referring to Table 2-11, construct a cumulative percentage polygon for the defects data if the corresponding frequency distribution uses "0 but less than 5" as the first class.

ANSWER:

.

52

Presenting Data in Tables and Charts

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

5

10

15

20

25

30

Number of Defects

TYPE: PR DIFFICULTY: Moderate KEYWORDS: cumulative percentage polygon 84. The point halfway between the boundaries of each class interval in a grouped frequency distribution is called the _______.

ANSWER: class midpoint TYPE: FI DIFFICULTY: Easy KEYWORDS: cumulative percentage polygon, frequency distribution 85. A _______ is a vertical bar chart in which the rectangular bars are constructed at the boundaries of each class interval.

ANSWER: histogram TYPE: FI DIFFICULTY: Easy KEYWORDS: histogram 86. It is essential that each class grouping or interval in a frequency distribution be ________.

ANSWER: non-overlapping and of equal width TYPE: FI DIFFICULTY: Moderate KEYWORDS: frequency distribution, class interval 87. In order to compare one large batch of numerical data to another, a ________ distribution must be developed from the frequency distribution.

ANSWER: relative frequency or percentage TYPE: FI DIFFICULTY: Easy KEYWORDS: relative frequency distribution, percentage distribution .

Presenting Data in Tables and Charts 53

88. When comparing two or more large batches of numerical data, the distributions being developed should use the same ________.

ANSWER: class boundaries.

TYPE: FI DIFFICULTY: Easy KEYWORDS: class boundaries 89. The width of each class grouping or interval in a frequency distribution should be ________.

ANSWER: the same or equal TYPE: FI DIFFICULTY: Easy KEYWORDS: class interval, frequency distribution 90. In constructing a polygon, each class grouping is represented by its _______ and then these are consecutively connected to one another.

ANSWER: midpoint TYPE: FI DIFFICULTY: Easy KEYWORDS: polygon, class interval, midpoint 91. A _______ is a summary table in which numerical data are tallied into class intervals or categories.

ANSWER: frequency distribution TYPE: FI DIFFICULTY: Easy KEYWORDS: frequency distribution, class interval 92. True or False: In general, grouped frequency distributions should have between 5 and 15 class intervals.

ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: frequency distribution, number of classes

93. True or False: The sum of relative frequencies in a distribution always equals 1.

ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: relative frequency

.

54

Presenting Data in Tables and Charts

94. True or False: The sum of cumulative frequencies in a distribution always equals 1.

ANSWER: False TYPE: TF DIFFICULTY: Moderate KEYWORDS: cumulative frequency distribution 95. True or False: When constructing a frequency distribution, classes should be selected in such a way that they are of equal width.

ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: frequency distribution 96. True or False: A research analyst was directed to arrange raw data collected on the yield of wheat, ranging from 40 to 93 bushels per acre, in a frequency distribution. He should choose 30 as the class interval width.

ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: frequency distribution, class interval 97. True or False: If the values of the seventh and eighth class in a cumulative frequency distribution are the same, we know that there are no observations in the eighth class.

ANSWER: True TYPE: TF DIFFICULTY: Moderate KEYWORDS: cumulative frequency distribution 98. True or False: Research on Human perception concludes that the bar chart is preferred to the pie chart, because the human eye can more accurately judge length comparisons against a fixed scale (as in a bar chart) than angular measures (as in a pie chart).

ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: bar chart, pie chart

99. True or False: One of the advantages of a pie chart is that it clearly shows that the total of all the categories of the pie adds to 100%.

ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: pie chart

.

Presenting Data in Tables and Charts 55

100. True or False: The larger the number of observations in a numerical data set, the larger the number of class intervals needed for a grouped frequency distribution.

ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: class interval, frequency distribution 101. True or False: Determining the class boundaries of a frequency distribution is highly subjective.

ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: class boundaries, frequency distribution

102. True or False: The original data values cannot be assessed once they are grouped into a frequency distribution table.

ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: frequency distribution 103. True or False: The percentage distribution cannot be constructed from the frequency distribution directly.

ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: percentage distribution, frequency distribution 104. True or False: The stem-and-leaf display is often superior to the frequency distribution in that it maintains the original values for further analysis.

ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: stem-and-leaf display, frequency distribution

105. True or False: The relative frequency is the frequency in each class divided by the total number of observations.

ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: relative frequency distribution 106. True or False: Ogives are plotted at the midpoints of the class groupings. .

56

Presenting Data in Tables and Charts

ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: ogives, midpoint 107. True or False: Percentage polygons are plotted at the boundaries of the class groupings.

ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: percentage polygons 108. True or False: The main principle behind the Pareto diagram is the ability to track the "vital few" from the "trivial many."

ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: Pareto diagram 109. True or False: A histogram can have gaps between the bars, whereas bar charts cannot have gaps.

ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: histogram, bar chart 110. True or False: Histograms are used for numerical data while bar charts are suitable for categorical data.

ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: histogram, bar chart

111. True or False: A Wal-Mart store in a small town monitors customer complaints and organizes these complaints into six distinct categories. Over the past year, the company has received 534 complaints. One possible graphical method for representing these data would be a Pareto chart.

ANSWER: True TYPE: TF DIFFICULTY: Moderate KEYWORDS: Pareto diagram

.

Presenting Data in Tables and Charts 57

112. True or False: Apple Computer, Inc. collected information on the age of their customers. The youngest customer was 12 and the oldest was 72. To study the distribution of the age among its customers, it can use a Pareto diagram.

ANSWER: False TYPE: TF DIFFICULTY: Moderate KEYWORDS: Pareto diagram 113. True or False: Apple Computer, Inc. collected information on the age of their customers. The youngest customer was 12 and the oldest was 72. To study the distribution of the age among its customers, it is best to use a pie chart.

ANSWER: False TYPE: TF DIFFICULTY: Moderate KEYWORDS: pie chart 114. True or False: Apple Computer, Inc. collected information on the age of their customers. The youngest customer was 12 and the oldest was 72. To study the distribution of the age among its customers, it can use a percentage polygon.

ANSWER: True TYPE: TF DIFFICULTY: Moderate KEYWORDS: percentage polygon 115. True or False: Apple Computer, Inc. collected information on the age of their customers. The youngest customer was 12 and the oldest was 72. To study the percentage of their customers who are below a certain age, it can use an ogive.

ANSWER: True TYPE: TF DIFFICULTY: Moderate KEYWORDS: ogive 116. True or False: If you wish to construct a graph of a relative frequency distribution, you would most likely construct an ogive first.

ANSWER: False TYPE: TF DIFFICULTY: Moderate KEYWORDS: Ogive 117. True or False: An ogive is a cumulative percentage polygon.

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

.

58

Presenting Data in Tables and Charts

118. True or False: A good choice for the number of class groups to use in constructing frequency distribution is to have at least 5 but no more than 15 class groups.

ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: number of classes 119. True or False: In general, a frequency distribution should have at least 8 class groups but no more than 20.

ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: number of classes 120. True of False: To determine the width of class interval, divide the number of class groups by the range of the data.

ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: class interval 121. True or False: The percentage polygon is formed by having the lower boundary of each class represent the data in that class and then connecting the sequence of lower boundaries at their respective class percentages.

ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: percentage polygon 122. True or False: A polygon can be constructed from a bar chart.

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

123. To evaluate two categorical variables at the same time, a _______ could be developed.

ANSWER: contingency or cross-classification table TYPE: FI DIFFICULTY: Easy KEYWORDS: contingency table, cross-classification table

.

Presenting Data in Tables and Charts 59

124. Relationships in a contingency table can be examined more fully if the frequencies are converted into _______ .

ANSWER: percentages or proportions TYPE: FI DIFFICULTY: Easy KEYWORDS: contingency table

TABLE 2-12 The table below contains the opinions of a sample of 200 people broken down by gender about the latest congressional plan to eliminate anti-trust exemptions for professional baseball.

For 38 12 50

Female Male Totals

Neutral 54 36 90

Against Totals 12 104 48 96 60 200

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

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

Totals 100.00 100.00 100.00

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

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

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

ANSWER: For Neutral Against Totals Female 19.00 27.00 6.00 52.00 Male 6.00 18.00 24.00 48.00 Totals 25.00 45.00 30.00 100.00 .

60

Presenting Data in Tables and Charts

TYPE: PR DIFFICULTY: Easy KEYWORDS: total percentages 128. Referring to Table 2-12, of those for the plan in the sample, ________ percent were females.

ANSWER: 76% TYPE: FI DIFFICULTY: Moderate KEYWORDS: contingency table, column percentages 129. Referring to Table 2-12, of those neutral in the sample, ________ percent were males.

ANSWER: 40% TYPE: FI DIFFICULTY: Moderate KEYWORDS: contingency table, column percentages 130. Referring to Table 2-12, of the males in the sample, ________ percent were for the plan.

ANSWER: 12.50% TYPE: FI DIFFICULTY: Moderate KEYWORDS: contingency table 131. Referring to Table 2-12, of the females in the sample, ________ percent were against the plan.

ANSWER: 11.54% TYPE: FI DIFFICULTY: Moderate KEYWORDS: contingency table 132. Referring to Table 2-12, of the females in the sample, ________ percent were either neutral or against the plan.

ANSWER: 63.46% or (51.92+11.54)% TYPE: FI DIFFICULTY: Moderate KEYWORDS: contingency table 133. Referring to Table 2-12, ________ percent of the 200 were females who were against the plan.

ANSWER: 6% TYPE: FI DIFFICULTY: Moderate KEYWORDS: contingency table 134. Referring to Table 2-12, ________ percent of the 200 were males who were neutral.

ANSWER: 18% TYPE: FI DIFFICULTY: Moderate KEYWORDS: contingency table

.

Presenting Data in Tables and Charts 61

135. Referring to Table 2-12, ________ percent of the 200 were females who were either neutral or against the plan.

ANSWER: 33% TYPE: FI DIFFICULTY: Difficult KEYWORDS: contingency table 136. Referring to Table 2-12, _______ percent of the 200 were males who were not against the plan.

ANSWER: 24% TYPE: FI DIFFICULTY: Difficult KEYWORDS: contingency table 137. Referring to Table 2-12, _______ percent of the 200 were not neutral.

ANSWER: 55% TYPE: FI DIFFICULTY: Difficult KEYWORDS: contingency table, row percentages 138. Referring to Table 2-12, _______ percent of the 200 were against the plan.

ANSWER: 30% TYPE: FI DIFFICULTY: Moderate KEYWORDS: contingency table, row percentages 139. Referring to Table 2-12, ________ percent of the 200 were males.

ANSWER: 48% TYPE: FI DIFFICULTY: Easy KEYWORDS: contingency table, column percentages 140. Referring to Table 2-12, if the sample is a good representation of the population, we can expect _______ percent of the population will be for the plant.

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

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

ANSWER: 48% TYPE: FI DIFFICULTY: Moderate KEYWORDS: contingency table, column percentages

.

62

Presenting Data in Tables and Charts

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

ANSWER: 24% TYPE: FI DIFFICULTY: Moderate KEYWORDS: contingency table 143. Referring to Table 2-12, if the sample is a good representation of the population, we can expect _______ percent of the males in the population will be against the plan.

ANSWER: 50% TYPE: FI DIFFICULTY: Moderate KEYWORDS: contingency table 144. Referring to Table 2-12, if the sample is a good representation of the population, we can expect _______ percent of the females in the population will not be against the plan.

ANSWER: 88.46% or (36.54+51.92) TYPE: FI DIFFICULTY: Moderate KEYWORDS: contingency table

TABLE 2-13 Given below is the stem-and-leaf display representing the amount of detergent used in gallons (with leaves in 10ths of gallons) in a month by 25 drive-through car wash operations in Phoenix.

9 | 147 10 | 02238 11 | 135566777 12 | 223489 13 | 02 145. Referring to Table 2-13, if a frequency distribution for the amount of detergent used is constructed, using "9.0 but less than 10.0 gallons" as the first class, the frequency of the “11.0 but less than 12.0 gallons” class would be ________.

ANSWER: 9 TYPE: FI DIFFICULTY: Easy KEYWORDS: frequency distribution 146. Referring to Table 2-13, if a percentage histogram for the detergent data is constructed, using "9.0 but less than 10.0 gallons" as the first class, the percentage of drive-through car wash operations that use “12.0 but less than 13.0 gallons” of detergent would be ________.

ANSWER: 24% TYPE: FI DIFFICULTY: Moderate KEYWORDS: relative frequency distribution, percentage distribution .

Presenting Data in Tables and Charts 63

147. Referring to Table 2-13, if a percentage histogram for the detergent data is constructed, using "9.0 but less than 10.0 gallons" as the first class, what percentage of drive-through car wash operations use less than 12 gallons of detergent in a month?

ANSWER: 68% TYPE: FI DIFFICULTY: Easy KEYWORDS: percentage distribution, cumulative relative frequency 148. Referring to Table 2-13, if a relative frequency or percentage distribution for the detergent data is constructed, using "9.0 but less than 10.0 gallons" as the first class, what percentage of drivethrough car wash operations use at least 10 gallons of detergent in a month?

ANSWER: 88% TYPE: FI DIFFICULTY: Easy KEYWORDS: relative frequency distribution, percentage distribution 149. Referring to Table 2-13, if a relative frequency or percentage distribution for the detergent data is constructed, using "9.0 but less than 10.0 gallons" as the first class, what percentage of drivethrough car wash operations use at least 10 gallons but less than 13 gallons of detergent in a month?

ANSWER: 80% TYPE: FI DIFFICULTY: Easy KEYWORDS: relative frequency distribution, percentage distribution 150. Referring to Table 2-13, construct a frequency distribution for the detergent data, using "9.0 but less than 10.0 gallons" as the first class.

ANSWER: Purchases (gals) Frequency 9.0 but less than 10.0 3 10.0 but less than 11.0 5 11.0 but less than 12.0 9 12.0 but less than 13.0 6 13.0 but less than 14.0 2 TYPE: PR DIFFICULTY: Moderate KEYWORDS: frequency distribution

151. Referring to Table 2-13, construct a relative frequency or percentage distribution for the detergent data, using "9.0 but less than 10.0" as the first class.

ANSWER: Gasoline Purchases (gals) 9.0 but less than 10.0 10.0 but less than 11.0

Percentage 12% 20 .

Presenting Data in Tables and Charts 11.0 but less than 12.0 36 12.0 but less than 13.0 24 13.0 but less than 14.0 8 TYPE: PR DIFFICULTY: Moderate KEYWORDS: relative frequency distribution, percentage distribution 152. Referring to Table 2-13, construct a cumulative percentage distribution for the detergent data if the corresponding frequency distribution uses "9.0 but less than 10.0" as the first class.

ANSWER: Gasoline Frequency Percentage Purchases (gals) Less Than Less Than 9.0 but less than 10.0 3 12 10.0 but less than 11.0 8 32 11.0 but less than 12.0 17 68 12.0 but less than 13.0 23 92 13.0 but less than 14.0 25 100 TYPE: PR DIFFICULTY: Moderate KEYWORDS: cumulative percentage distribution 153. Referring to Table 2-13, construct a percentage histogram for the detergent data, using "9.0 but less than 10.0" as the first class.

ANSWER: % Histogram 40 35 30 25 %

64

20 15 10 5 0 9.0 -- 9.9

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

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

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

ANSWER:

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Presenting Data in Tables and Charts 65

Ogive 100 80 %

60 40 20 0 9

10

11

12

13

14

Purchase

TYPE: PR DIFFICULTY: Moderate KEYWORDS: cumulative percentage polygon 155. Referring to Table 2-13, construct a percentage polygon for the detergent data if the corresponding frequency distribution uses "9.0 but less than 10.0" as the first class.

ANSWER:

%

% Polygon 40 35 30 25 20 15 10 5 0

%

8.5

9.5

10.5

11.5

12.5

13.5

14.5

Purchase

TYPE: PR DIFFICULTY: Moderate KEYWORDS: percentage distribution

TABLE 2-14 The table below contains the number of people who own a portable DVD player in a sample of 600 broken down by gender.

Own a Portable .

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Presenting Data in Tables and Charts

DVD Player Yes No

Male 96 224

Female 40 240

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

ANSWER: Own Yes No Total

Male 70.59% 48.28% 53.33%

Female 29.41% 51.72% 46.67%

Total 100.00% 100.00% 100.00%

TYPE: PR DIFFICULTY: Easy KEYWORDS: row percentages 157. Referring to Table 2-14, construct a table of column percentages.

ANSWER: Own Yes No Total

Male 30.00% 70.00% 100.00%

Female 14.29% 85.71% 100.00%

Total 22.67% 77.33% 100.00%

TYPE: PR DIFFICULTY: Easy KEYWORDS: column percentages 158. Referring to Table 2-12, construct a table of total percentages.

ANSWER: Own Yes No Total

Male 16.00% 37.33% 53.33%

Female 6.67% 40.00% 46.67%

Total 22.67% 77.33% 100.00%

TYPE: PR DIFFICULTY: Easy KEYWORDS: total percentages 159. Referring to Table 2-14, of those who owned a portable DVD in the sample, ________ percent were females.

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

160. Referring to Table 2-14, of those who did not own a portable DVD in the sample, ________ percent were males.

ANSWER: 48.28% TYPE: FI DIFFICULTY: Moderate KEYWORDS: contingency table, row percentages .

Presenting Data in Tables and Charts 67

161. Referring to Table 2-14, of the males in the sample, ________ percent owned a portable DVD. ANSWER: 30% TYPE: FI DIFFICULTY: Moderate KEYWORDS: contingency table, column percentages 162. Referring to Table 2-14, of the females in the sample, ________ percent did not own a portable DVD. ANSWER: 85.71% TYPE: FI DIFFICULTY: Moderate KEYWORDS: contingency table, column percentages 163. Referring to Table 2-14, of the females in the sample, ________ percent owned a portable DVD. ANSWER: 14.29% TYPE: FI DIFFICULTY: Moderate KEYWORDS: contingency table, column percentages 164. Referring to Table 2-14, ________ percent of the 600 were females who owned a portable DVD. ANSWER: 6.67% TYPE: FI DIFFICULTY: Moderate KEYWORDS: contingency table, total percentage 165. Referring to Table 2-14, ________ percent of the 600 were males who owned a portable DVD. ANSWER: 16% TYPE: FI DIFFICULTY: Moderate KEYWORDS: contingency table, total percentage

166. Referring to Table 2-14, ________ percent of the 600 were females who either owned or did not own a portable DVD. ANSWER: 46.67% TYPE: FI DIFFICULTY: Moderate KEYWORDS: contingency table, total percentage

.

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Presenting Data in Tables and Charts

167. Referring to Table 2-14, _______ percent of the 600 were males who did not owned a portable DVD. ANSWER: 37.33% TYPE: FI DIFFICULTY: Moderate KEYWORDS: contingency table, total percentage 168. Referring to Table 2-14, _______ percent of the 600 owned a portable DVD. ANSWER: 22.67% TYPE: FI DIFFICULTY: Moderate KEYWORDS: contingency table, column percentages 169. Referring to Table 2-14, _______ percent of the 600 did not owned a portable DVD. ANSWER: 77.33% TYPE: FI DIFFICULTY: Moderate KEYWORDS: contingency table, column percentages 170. Referring to Table 2-14, ________ percent of the 600 were females.

ANSWER: 46.67% TYPE: FI DIFFICULTY: Easy KEYWORDS: contingency table, row percentages 171. Referring to Table 2-14, if the sample is a good representation of the population, we can expect _______ percent of the population will own a portable DVD. ANSWER: 22.67% TYPE: FI DIFFICULTY: Moderate KEYWORDS: contingency table, column percentages

172. Referring to Table 2-14, if the sample is a good representation of the population, we can expect _______ percent of the population will be males.

ANSWER: 53.33% TYPE: FI DIFFICULTY: Moderate KEYWORDS: contingency table, column percentages

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Presenting Data in Tables and Charts 69

173. Referring to Table 2-14, if the sample is a good representation of the population, we can expect _______ percent of those who own a portable DVD in the population will be males.

ANSWER: 70.59% TYPE: FI DIFFICULTY: Moderate KEYWORDS: contingency table, row percentages 174. Referring to Table 2-14, if the sample is a good representation of the population, we can expect _______ percent of the males in the population will own a portable DVD. ANSWER: 30% TYPE: FI DIFFICULTY: Moderate KEYWORDS: contingency table, column percentages 175. Referring to Table 2-14, if the sample is a good representation of the population, we can expect _______ percent of the females in the population will not own a portable DVD. ANSWER: 85.71% TYPE: FI DIFFICULTY: Moderate KEYWORDS: contingency table, column percentages

TABLE 2-15 The figure below is the ogive for the amount of fat (in grams) for a sample of 36 pizzas products where the upper boundaries of the intervals are: 5, 10, 15, 20, 25, and 30.

.

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Presenting Data in Tables and Charts

Cumulative Percentage Polygon for Fat 120% 100% 80% 60% 40% 20% 0% -0.01

4.99

9.99 14.99 19.99 24.99 29.99 34.99

Fat (grams)

176. Referring to Table 2-14, roughly what percentage of pizza products contains less than 10 grams of fat? a) 3% b) 14% c) 50% d) 75% ANSWER: b TYPE: MC DIFFICULTY: Easy KEYWORDS: cumulative percentage polygon, ogive, interpretation 177. Referring to Table 2-14, what percentage of pizza products contains at least 20 grams of fat? a) 5% b) 25% c) 75% d) 96% ANSWER: b TYPE: MC DIFFICULTY: Easy KEYWORDS: cumulative percentage polygon, ogive, interpretation

178. Referring to Table 2-14, what percentage of pizza products contains between 10 and 25 grams of fat? a) 14% b) 44% c) 62% d) 81% .

Presenting Data in Tables and Charts 71

ANSWER: d TYPE: MC DIFFICULTY: Easy KEYWORDS: cumulative percentage polygon, ogive, interpretation TABLE 2-16 The figure below is the percentage polygon for the amount of calories for a sample of 36 pizzas products where the upper limits of the intervals are: 310, 340, 370, 400 and 430.

Percentage Polygon for Calories 30% 25% 20% 15% 10% 5% 0% 264.99 294.99 324.99 354.99 384.99 414.99 444.99

Calories

179. Referring to Table 2-15, roughly what percentage of pizza products contains between 400 and 430 calories? a) 0% b) 11% c) 89% d) 100% ANSWER: b TYPE: MC DIFFICULTY: Easy KEYWORDS: percentage polygon, interpretation 180. Referring to Table 2-15, roughly what percentage of pizza products contains between 340 and 400 calories? a) 22% b) 25% c) 28% d) 50% ANSWER: d TYPE: MC DIFFICULTY: Moderate KEYWORDS: percentage polygon, interpretation 181. Referring to Table 2-15, roughly what percentage of pizza products contains at least 340 calories? a) 25% b) 28% c) 39% d) 61% .

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ANSWER: d TYPE: MC DIFFICULTY: Moderate KEYWORDS: percentage polygon, interpretation

TABLE 2-17 The following table presents total retail sales in millions of dollars for the leading apparel companies during April 2001 and April 2002.

APPAREL COMPANY Gap TJX Limited Kohl’s Nordstrom Talbots AnnTaylor

April 01 April 02 1,159.00 962 781.7 899 596.5 620.4 544.9 678.9 402.6 418.3 139.9 130.1 114.2 124.8

182. Referring to Table 2-17, construct a table of column percentages.

ANSWER: Apparel Company April 2001 Gap 31.00% TJX 20.91% Limited 15.95% Kohl's 14.57% Nordstrom 10.77% Talbots 3.74% AnnTaylor 3.05% Total 100.00% TYPE: PR DIFFICULTY: Moderate KEYWORDS: column percentages

April 2002 25.09% 23.45% 16.18% 17.71% 10.91% 3.39% 3.26% 100.00%

183. True or False: Referring to Table 2-17, in general, retail sales for the apparel industry have seen a modest growth between April 2001 and April 2002.

ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: column percentages, interpretation 184. Referring to Table 2-17, among the 8 stores, _______ saw a sales decline.

ANSWER: Gap and Talbots TYPE: FI DIFFICULTY: Easy KEYWORDS: column percentages, interpretation .

Presenting Data in Tables and Charts 73

.

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CHAPTER 3: NUMERICAL DESCRIPTIVE MEASURES 1. Which of the following statistics is not a measure of central tendency? a) Arithmetic mean. b) Median. c) Mode. d) Q3.

ANSWER: d TYPE: MC DIFFICULTY: Easy KEYWORDS: measure of central tendency, arithmetic mean, median, mode, quartiles 2. Which measure of central tendency can be used for both numerical and categorical variables? a) Arithmetic mean. b) Median. c) Mode. d) Standard deviation.

ANSWER: c TYPE: MC DIFFICULTY: Moderate KEYWORDS: measure of central tendency, mode, arithmetic mean, median 3. Which of the arithmetic mean, median, and mode are resistant measures of central tendency? a) The arithmetic mean and median only. b) The median and mode only. c) The mode and arithmetic mean only. d) All the three are resistant measures.

ANSWER: b TYPE: MC DIFFICULTY: Moderate KEYWORDS: measure of central tendency, resistant to outliers, mean, median, mode 4. In a right-skewed distribution a) the median equals the arithmetic mean. b) the median is less than the arithmetic mean. c) the median is larger than the arithmetic mean. d) none of the above.

ANSWER: b TYPE: MC DIFFICULTY: Easy KEYWORDS: shape

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Numerical Descriptive Measures

5. Which of the following statements about the median is not true? a) It is more affected by extreme values than the arithmetic mean. b) It is a measure of central tendency. c) It is equal to Q2. d) It is equal to the mode in bell-shaped "normal" distributions.

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

ANSWER: d TYPE: MC DIFFICULTY: Easy KEYWORDS: shape, normal distribution 7. In a perfectly symmetrical distribution a) the range equals the interquartile range. b) the interquartile range equals the arithmetic mean. c) the median equals the arithmetic mean. d) the variance equals the standard deviation.

ANSWER: c TYPE: MC DIFFICULTY: Easy KEYWORDS: shape 8. When extreme values are present in a set of data, which of the following descriptive summary measures are most appropriate: a) CV and range. b) arithmetic mean and standard deviation. c) interquartile range and median. d) variance and interquartile range.

ANSWER: c TYPE: MC DIFFICULTY: Easy KEYWORDS: resistant to outliers, coefficient of variation, range, arithmetic mean, standard deviation, interquartile range, median, variance

.

75

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Numerical Descriptive Measures

9. In general, which of the following descriptive summary measures cannot be easily approximated from a boxplot? a) The variance. b) The range. c) The interquartile range. d) The median.

ANSWER: a TYPE: MC DIFFICULTY: Easy KEYWORDS: boxplot, variance, range, interquartile range, median 10. The smaller the spread of scores around the arithmetic mean, a) the smaller the interquartile range. b) the smaller the standard deviation. c) the smaller the coefficient of variation. d) All the above.

ANSWER: d TYPE: MC DIFFICULTY: Moderate KEYWORDS: variation, arithmetic mean, interquartile range, standard deviation, coefficient of variation 11. Which descriptive summary measures are considered to be resistant statistics? a) The arithmetic mean and standard deviation. b) The interquartile range and range. c) The mode and variance. d) The median and interquartile range.

ANSWER: d TYPE: MC DIFFICULTY: Moderate KEYWORDS: resistant to outliers, arithmetic mean, standard deviation, interquartile range, range, mode, variance, median 12. In right-skewed distributions, which of the following is the correct statement? a) The distance from Q1 to Q2 is larger than the distance from Q2 to Q3. b) The distance from Q1 to Q2 is smaller than the distance from Q2 to Q3. c) The arithmetic mean is smaller than the median. d) The mode is larger than the arithmetic mean.

ANSWER: b TYPE: MC DIFFICULTY: Easy KEYWORDS: shape, quartiles, arithmetic mean, mode

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Numerical Descriptive Measures

77

13. In perfectly symmetrical distributions, which of the following is NOT a correct statement? a) The distance from Q1 to Q2 equals to the distance from Q2 to Q3. b) The distance from the smallest observation to Q1 is the same as the distance from Q3 to the largest observation. c) The distance from the smallest observation to Q2 is the same as the distance from Q2 to the largest observation. d) The distance from Q1 to Q3 is half of the distance from the smallest to the largest observation.

ANSWER: d TYPE: MC DIFFICULTY: Easy KEYWORDS: shape, quartiles, five-number summary, quartiles 14. In left-skewed distributions, which of the following is the correct statement? a) The distance from Q1 to Q2 is smaller than the distance from Q2 to Q3. b) The distance from the smallest observation to Q1 is larger than the distance from Q3 to the largest observation. c) The distance from the smallest observation to Q2 is smaller than the distance from Q2 to the largest observation. d) The distance from Q1 to Q3 is twice the distance from the Q1 to Q2.

ANSWER: b TYPE: MC DIFFICULTY: Easy KEYWORDS: shape, quartiles, five-number summary, quartiles 15. According to the empirical rule, if the data form a "bell-shaped" normal distribution, _______ percent of the observations will be contained within 2 standard deviations around the arithmetic mean. a) 68.26 b) 88.89 c) 93.75 d) 95.44 ANSWER: d TYPE: MC DIFFICULTY: Moderate KEYWORDS: empirical rule, normal distribution 16. According to the empirical rule, if the data form a "bell-shaped" normal distribution, _______ percent of the observations will be contained within 1 standard deviation around the arithmetic mean. a) 68.26 b) 75.00 c) 88.89 d) 93.75 ANSWER: a TYPE: MC DIFFICULTY: Moderate KEYWORDS: empirical rule, normal distribution .

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Numerical Descriptive Measures

17. According to the empirical rule, if the data form a "bell-shaped" normal distribution, _______ percent of the observations will be contained within 3 standard deviations around the arithmetic mean. a) 68.26 b) 75.00 c) 95.0 d) 99.7 ANSWER: d TYPE: MC DIFFICULTY: Moderate KEYWORDS: empirical rule, normal distribution 18. Which of the following is NOT a measure of central tendency? a) The arithmetic mean. b) The median. c) The mode. d) The interquartile range.

ANSWER: d TYPE: MC DIFFICULTY: Easy KEYWORDS: measure of central tendency, arithmetic mean, median, mode, interquartile range 19. Which of the following is NOT sensitive to extreme values? a) The range. b) The standard deviation. c) The interquartile range. d) The coefficient of variation.

ANSWER: c TYPE: MC DIFFICULTY: Moderate KEYWORDS: resistant to outliers, range, standard deviation, interquartile range, coefficient of variation 20. Which of the following is sensitive to extreme values? a) The median. b) The interquartile range. c) The arithmetic mean. d) The 1st quartile.

ANSWER: c TYPE: MC DIFFICULTY: Moderate KEYWORDS: resistant to outliers, median, interquartile range, arithmetic mean, quartiles

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Numerical Descriptive Measures

79

21. Which of the following is the easiest to compute? a) The arithmetic mean. b) The median. c) The mode. d) The standard deviation.

ANSWER: c TYPE: MC DIFFICULTY: Easy KEYWORDS: mode, arithmetic mean, median 22. According to the Chebyshev rule, at least 75% of all observations in any data set are contained within a distance of how many standard deviations around the mean? a) 1 b) 2 c) 3 d) 4 ANSWER: b TYPE: MC DIFFICULTY: Easy KEYWORDS: Chebyshev rule 23. According to the Chebyshev rule, at least 93.75% of all observations in any data set are contained within a distance of how many standard deviations around the mean? a) 1 b) 2 c) 3 d) 4 ANSWER: d TYPE: MC DIFFICULTY: Moderate KEYWORDS: Chebyshev rule 24. According to the Chebyshev rule, at least what percentage of the observations in any data set are contained within a distance of 3 standard deviations around the mean? a) 67% b) 75% c) 88.89% d) 99.7% ANSWER: c TYPE: MC DIFFICULTY: Easy KEYWORDS: Chebyshev rule

.

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Numerical Descriptive Measures

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

ANSWER: 74.04 years TYPE: PR DIFFICULTY: Easy KEYWORDS: arithmetic mean 27. Referring to Table 3-1, identify the median age of the uninsured senior citizens.

ANSWER: 73 years TYPE: PR DIFFICULTY: Easy KEYWORDS: median 28. Referring to Table 3-1, identify the first quartile of the ages of the uninsured senior citizens.

ANSWER: 65.5 years TYPE: PR DIFFICULTY: Moderate KEYWORDS: quartiles 29. Referring to Table 3-1, identify the third quartile of the ages of the uninsured senior citizens.

ANSWER: 81.5 years TYPE: PR DIFFICULTY: Moderate KEYWORDS: quartiles

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Numerical Descriptive Measures

81

30. Referring to Table 3-1, identify the interquartile range of the ages of the uninsured senior citizens.

ANSWER: 16 years TYPE: PR DIFFICULTY: Moderate KEYWORDS: interquartile range 31. Referring to Table 3-1, identify which of the following is the correct statement. a) One fourth of the senior citizens sampled are below 65.5 years of age. b) The middle 50% of the senior citizens sampled are between 65.5 and 73.0 years of age. c) The average age of senior citizens sampled is 73.5 years of age. d) All of the above are correct.

ANSWER: a TYPE: MC DIFFICULTY: Moderate KEYWORDS: quartiles, arithmetic mean 32. Referring to Table 3-1, identify which of the following is the correct statement. a) One fourth of the senior citizens sampled are below 64 years of age. b) The middle 50% of the senior citizens sampled are between 65.5 and 73.0 years of age. c) 25% of the senior citizens sampled are older than 81.5 years of age. d) All of the above are correct.

ANSWER: c TYPE: MC DIFFICULTY: Moderate KEYWORDS: quartiles 33. Referring to Table 3-1, what type of shape does the distribution of the sample appear to have?

ANSWER: Slightly positive or right-skewed.

TYPE: PR DIFFICULTY: Moderate KEYWORDS: shape 34. Referring to Table 3-1, calculate the variance of the ages of the uninsured senior citizens correct to the nearest hundredth of a year squared.

ANSWER: 94.96 years2 TYPE: PR DIFFICULTY: Moderate KEYWORDS: variance 35. Referring to Table 3-1, calculate the standard deviation of the ages of the uninsured senior citizens correct to the nearest hundredth of a year.

ANSWER: 9.74 years TYPE: PR DIFFICULTY: Moderate KEYWORDS: standard deviation

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Numerical Descriptive Measures

36. Referring to Table 3-1, calculate the coefficient of variation of the ages of the uninsured senior citizens.

ANSWER: 13.16% TYPE: PR DIFFICULTY: Moderate KEYWORDS: coefficient of variation 37. True or False: The median of the values 3.4, 4.7, 1.9, 7.6, and 6.5 is 1.9.

ANSWER: False TYPE: TF DIFFICULTY: Moderate KEYWORDS: median 38. True or False: The median of the values 3.4, 4.7, 1.9, 7.6, and 6.5 is 4.05.

ANSWER: False TYPE: TF DIFFICULTY: Moderate KEYWORDS: median 39. True or False: In a set of numerical data, the value for Q3 can never be smaller than the value for Q1.

ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: quartiles 40. True or False: In a set of numerical data, the value for Q2 is always halfway between Q1 and Q3.

ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: quartiles 41. True or False: If the distribution of a data set were perfectly symmetrical, the distance from Q1 to the median would always equal the distance from Q3 to the median in a boxplot.

ANSWER: True TYPE: TF DIFFICULTY: Moderate KEYWORDS: boxplot, quartiles, shape 42. True or False: In right-skewed distributions, the distance from Q3 to the largest observation exceeds the distance from the smallest observation to Q1.

ANSWER: True TYPE: TF DIFFICULTY: Moderate KEYWORDS: quartiles, shape .

Numerical Descriptive Measures

83

43. True or False: In left-skewed distributions, the distance from the smallest observation to Q1 exceeds the distance from Q3 to the largest observation.

ANSWER: True TYPE: TF DIFFICULTY: Moderate KEYWORDS: quartiles, shape 44. True or False: A boxplot is a graphical representation of a 5-number summary.

ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: boxplot, five-number summary 45. True or False: The 5-number summary consists of the smallest observation, the first quartile, the median, the third quartile, and the largest observation.

ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: five-number summary 46. True or False: In a boxplot, the box portion represents the data between the first and third quartile values.

ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: boxplot 47. True or False: The line drawn within the box of the boxplot always represents the arithmetic mean.

ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: boxplot, arithmetic mean

48. True or False: The line drawn within the box of the boxplot always represents the median.

ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: boxplot, median

49. True or False: In a sample of size 40, the sample mean is 15. In this case, the sum of all observations in the sample is  X i = 600. .

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Numerical Descriptive Measures

ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: arithmetic mean 50. True or False: A population with 200 elements has an arithmetic mean of 10. From this information, it can be shown that the population standard deviation is 15.

ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: arithmetic mean, standard deviation 51. True or False: In exploratory data analysis, a boxplot can be used to illustrate the median, quartiles, and extreme values.

ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: boxplot, five-number summary 52. True or False: An economics professor bases his final grade on homework, two midterm examinations, and a final examination. The homework counts 10% toward the final grade, while each midterm examination counts 25%. The remaining portion consists of the final examination.

If a student scored 95% in homework, 70% on the first midterm examination, 96% on the second midterm examination, and 72% on the final, his final average is 79.8%.

ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: arithmetic mean 53. True or False: The median of a data set with 20 items would be the average of the 10th and the 11th items in the ordered array.

ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: median, arithmetic mean 54. True or False: The coefficient of variation measures variability in a data set relative to the size of the arithmetic mean.

ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: coefficient of variation 55. True or False: The coefficient of variation is expressed as a percentage.

ANSWER: .

Numerical Descriptive Measures

85

True TYPE: TF DIFFICULTY: Easy KEYWORDS: coefficient of variation 56. True or False: The coefficient of variation is a measure of central tendency in the data.

ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: coefficient of variation, measure of variation 57. True or False: The interquartile range is a measure of variation or dispersion in a set of data.

ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: interquartile range, measure of variation 58. True or False: The interquartile range is a measure of central tendency in a set of data.

ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: interquartile range, measure of variation 59. True or False: If a set of data is perfectly symmetrical, the arithmetic mean must be identical to the median.

ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: shape, arithmetic mean, median 60. True or False: The coefficient of variation is a measure of relative variation.

ANSWER: True TYPE: TF DIFFICULTY: Moderate KEYWORDS: coefficient of variation 61. True or False: If the data set is approximately bell-shaped, the empirical rule will more accurately reflect the greater concentration of data close to the mean as compared to the Chebyshev rule.

ANSWER: True TYPE: TF DIFFICULTY: Moderate KEYWORDS: empirical rule, Chebyshev rule, normal distribution

62. If the arithmetic mean of a numerical data set exceeds the median, the data are considered to be _______ skewed.

ANSWER: .

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Numerical Descriptive Measures

positive or right TYPE: FI DIFFICULTY: Easy KEYWORDS: shape, arithmetic mean, median TABLE 3-2 The data below represent the amount of grams of carbohydrates in a serving of breakfast cereal in a sample of 11 different servings.

11 15 23 29 19 22 21 20 15 25 17 63. Referring to Table 3-2, the arithmetic mean carbohydrates in this sample is ________ grams.

ANSWER: 217/11 = 19.73 TYPE: FI DIFFICULTY: Moderate KEYWORDS: arithmetic mean 64. Referring to Table 3-2, the median carbohydrate amount in the cereal is ________ grams.

ANSWER: 20 TYPE: FI DIFFICULTY: Moderate KEYWORDS: median 65. Referring to Table 3-2, the first quartile of the carbohydrate amounts is ________ grams.

ANSWER: 15 TYPE: FI DIFFICULTY: Moderate KEYWORDS: quartiles 66. Referring to Table 3-2, the third quartile of the carbohydrate amounts is ________ grams.

ANSWER: 23 TYPE: FI DIFFICULTY: Moderate KEYWORDS: quartiles 67. Referring to Table 3-2, the range in the carbohydrate amounts is ________ grams.

ANSWER: 18 TYPE: FI DIFFICULTY: Easy KEYWORDS: range

68. Referring to Table 3-2, the interquartile range in the carbohydrate amounts is ________ grams.

ANSWER: 8 .

Numerical Descriptive Measures

TYPE: FI DIFFICULTY: Moderate KEYWORDS: interquartile range 69. Referring to Table 3-2, the variance of the carbohydrate amounts is ________ (grams squared).

ANSWER: 26.02 TYPE: FI DIFFICULTY: Moderate KEYWORDS: variance 70. Referring to Table 3-2, the standard deviation of the carbohydrate amounts is ________ grams.

ANSWER: 5.10 TYPE: FI DIFFICULTY: Moderate KEYWORDS: standard deviation 71. Referring to Table 3-2, the coefficient of variation of the carbohydrate amounts is ________ percent.

ANSWER: 25.86% TYPE: FI DIFFICULTY: Moderate KEYWORDS: coefficient of variation 72. Referring to Table 3-2, the five-number summary of the carbohydrate amounts consists of ________, ________, ________, ________, ________.

ANSWER: 11, 15, 20, 23, 29 TYPE: FI DIFFICULTY: Moderate KEYWORDS: five-number summary

73. Referring to Table 3-2, construct a boxplot for the carbohydrate amounts.

.

87

88

Numerical Descriptive Measures

ANSWER: Box-and-whisker Plot

10

15

20

25

30

TYPE: PR DIFFICULTY: Moderate KEYWORDS: boxplot 74. Referring to Table 3-2, what type of shape does the distribution of the sample appear to have?

ANSWER: Slightly positive or right-skewed.

TYPE: PR DIFFICULTY: Moderate EXPLANATION: The distribution is considered as slightly right-skewed because the length of the right whisker (6) is slighter longer than the length of the left whisker (4). The distribution could also be considered as slight left-skewed since the mean (19.73) is slightly lower than the median (20).

KEYWORDS: shape

TABLE 3-3

.

Numerical Descriptive Measures

89

The stem-and-leaf display below represents the number of vitamin supplements sold by a health food store in a sample of 16 days.

Stem Leaves 1H 99 2L 0023 2H 567 3L 034 3H 568 4L 1 Note (1): 1H means the “high teens” – 15, 16, 17, 18, or 19; 2L means the “low twenties” – 20, 21, 22, 23, or 24; 2H means the “high twenties” – 25, 26, 27, 28, or 29, etc.

Note (2): For this sample, the sum of the observations is 448, the sum of the squares of the observations is 13,356, and the sum of the squared differences between each observation and the mean is 812.

75. Referring to Table 3-3, the arithmetic mean of the number of vitamin supplements sold in this sample is ________.

ANSWER: 28 TYPE: FI DIFFICULTY: Easy KEYWORDS: arithmetic mean 76. Referring to Table 3-3, the first quartile of the number of vitamin supplements sold in this sample is ________.

ANSWER: 20 TYPE: FI DIFFICULTY: Moderate KEYWORDS: quartiles 77. Referring to Table 3-3, the third quartile of the number of vitamin supplements sold in this sample is ________.

ANSWER: 35 TYPE: FI DIFFICULTY: Moderate KEYWORDS: quartiles 78. Referring to Table 3-3, the median number of vitamin supplements sold in this sample is ________.

ANSWER: 26.5 TYPE: FI DIFFICULTY: Moderate KEYWORDS: median

79. Referring to Table 3-3, the range of the number of vitamin supplements sold in this sample is ________.

.

90

Numerical Descriptive Measures

ANSWER: 22 TYPE: FI DIFFICULTY: Easy KEYWORDS: range 80. Referring to Table 3-3, the interquartile range of the number of vitamin supplements sold in this sample is ________.

ANSWER: 15 TYPE: FI DIFFICULTY: Moderate KEYWORDS: interquartile range 81. Referring to Table 3-3, the variance of the number of vitamin supplements sold in this sample is ________.

ANSWER: 54.1 TYPE: FI DIFFICULTY: Easy KEYWORDS: variance 82. Referring to Table 3-3, the standard deviation of the number of vitamin supplements sold in this sample is ________.

ANSWER: 7.4 TYPE: FI DIFFICULTY: Easy KEYWORDS: standard deviation 83. Referring to Table 3-3, the coefficient of variation of the number of vitamin supplements sold in this sample is ________ percent.

ANSWER: 26.3% TYPE: FI DIFFICULTY: Moderate KEYWORDS: coefficient of variation 84. Referring to Table 3-3, the five-number summary of the data in this sample consists of ________, ________, ________, ________, ________.

ANSWER: 19, 20, 26.5, 35, 41 TYPE: FI DIFFICULTY: Moderate KEYWORDS: five-number summary

85. Referring to Table 3-3, construct a boxplot for the data in this sample.

.

Numerical Descriptive Measures

91

ANSWER: Box-and-whisker Plot

10

20

30

40

TYPE: PR DIFFICULTY: Moderate KEYWORDS: boxplot 86. Referring to Table 3-3, what type of shape does the distribution of the sample appear to have?

ANSWER: Slightly positive or right-skewed.

TYPE: PR DIFFICULTY: Moderate KEYWORDS: shape

TABLE 3-4 The stem-and-leaf display below represents the number of cargo manifests approved by customs inspectors of the Port of New York in a sample of 35 days.

STEM LEAVES 1H 67889 2L 0011122223333444 2H 5566678899 3L 1122 Note (1): 1H means the “high teens” – 15, 16, 17, 18, or 19; 2L means the “low twenties” – 20, 21, 22, 23, or 24; 2H means the “high twenties” – 25, 26, 27, 28, or 29, etc.

Note (2): For this sample, the sum of the observations is 838, the sum of the squares of the observations is 20,684, and the sum of the squared differences between each observation and the mean is 619.89.

87. Referring to Table 3-4, the arithmetic mean of the customs data is ________.

ANSWER: 23.9 TYPE: FI DIFFICULTY: Easy KEYWORDS: arithmetic mean

88. Referring to Table 3-4, the median of the customs data is ________.

.

92

Numerical Descriptive Measures

ANSWER: 23 TYPE: FI DIFFICULTY: Moderate KEYWORDS: median 89. Referring to Table 3-4, the first quartile of the customs data is ________.

ANSWER: 21 TYPE: FI DIFFICULTY: Moderate KEYWORDS: quartiles 90. Referring to Table 3-4, the third quartile of the customs data is ________.

ANSWER: 27 TYPE: FI DIFFICULTY: Moderate KEYWORDS: quartiles 91. Referring to Table 3-4, the range of the customs data is ________.

ANSWER: 16 TYPE: FI DIFFICULTY: Easy KEYWORDS: range 92. Referring to Table 3-4, the interquartile range of the customs data is ________.

ANSWER: 6 TYPE: FI DIFFICULTY: Moderate KEYWORDS: interquartile range 93. Referring to Table 3-4, the variance of the customs data is ________.

ANSWER: 18.2 TYPE: FI DIFFICULTY: Easy KEYWORDS: variance 94. Referring to Table 3-4, the standard deviation of the customs data is ________.

ANSWER: 4.3 TYPE: FI DIFFICULTY: Easy KEYWORDS: standard deviation

95. Referring to Table 3-4, the coefficient of variation of the customs data is ________ percent.

.

Numerical Descriptive Measures

93

ANSWER: 17.8% or 18% TYPE: FI DIFFICULTY: Moderate KEYWORDS: coefficient of variation 96. Referring to Table 3-4, the five-number summary for the data in the customs sample consists of ________, ________, ________, ________, ________.

ANSWER: 16, 21, 23, 27, 32 TYPE: FI DIFFICULTY: Moderate KEYWORDS: five-number summary 97. Referring to Table 3-4, construct a boxplot of this sample.

ANSWER: Box-and-whisker Plot

10

15

20

25

30

35

TYPE: PR DIFFICULTY: Moderate KEYWORDS: boxplot

TABLE 3-5 The rate of return of a Fortune 500 company over the past 15 years are: 3.17%, 4.43%, 5.93%, 5.43%, 7.29%, 8.21%, 6.23%, 5.23%, 4.34%, 6.68%, 7.14%, -5.56%, -5.23%, -5.73%, -10.34% 98. Referring to Table 3-5, compute the arithmetic mean rate of return.

ANWER: 2.48% TYPE: PR DIFFICULTY: Easy KEYWORDS: arithmetic mean

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

.

94

Numerical Descriptive Measures

ANSWER: 18.55% TYPE: PR DIFFICULTY: Easy KEYWORDS: range 100. Referring to Table 3-5, construct a boxplot for the rate of return.

ANSWER: Box-and-whisker Plot

-0.15

-0.1

-0.05

0

0.05

0.1

TYPE: PR DIFFICULTY: Easy KEYWORDS: boxplot 101. Referring to Table 3-5, what is the shape of the distribution for the rate of return?

ANSWER: Left-skewed TYPE: PR DIFFICULTY: Easy KEYWORDS: shape

TABLE 3-6 The rate of return of an Internet Service Provider over a 10 year period are: 10.25%, 12.64%, 8.37%, 9.29%, 6.23%, 42.53%, 29.23%, 15.25%, 21.52%, -2.35%.

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

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

103. Referring to Table 3-6, construct a boxplot for the rate of return

.

Numerical Descriptive Measures

95

ANWER: Box-and-whisker Plot

-0.1

0

0.1

0.2

0.3

0.4

0.5

TYPE: PR DIFFICULTY: Moderate KEYWORDS: boxplot 104. Referring to Table 3-6, what is the shape of the distribution for the rate of return?

ANSWER: Right-skewed TYPE: PR DIFFICULTY: Moderate KEYWORDS: shape 105. The Z scores can be used to identify outliers.

ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: Z scores 106. The larger the Z score, the farther is the distance from the observation to the median.

ANSWER: False TYPE: TF DIFFICULTY: Moderate KEYWORDS: Z scores 107. As a general rule, an observation is considered an extreme value if its Z score is greater than −3.

ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: Z score 108. As a general rule, an observation is considered an extreme value if its Z score is greater than 3.

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

109. As a general rule, an observation is considered an extreme value if its Z score is less than 3.

ANSWER: .

96

Numerical Descriptive Measures

False TYPE: TF DIFFICULTY: Easy KEYWORDS: Z score 110. As a general rule, an observation is considered an extreme value if its Z score is less than −3.

ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: Z score 111. The Z score of an observation can never be negative.

ANSWER: False TYPE: TF DIFFICULTY: Moderate KEYWORDS: Z score 112. The Z score of an observation measures how many standard deviations is the value from the mean.

ANSWER: True TYPE: TF DIFFICULTY: Moderate KEYWORDS: Z score

TABLE 3-7 Given below is the frequency distribution of the 12-month rate of return achieved by 30 stocks.

12-month Rate of Return −0.10 but less than 0.0 0.0 but less than 0.1 0.1 but less than 0.2 0.2 but less than 0.3 0.3 but less than 0.4

Number of Funds 3 5 6 10 6

113. Referring to Table 3-7, what is the approximate mean rate of return?

ANSWER: 0.1867 TYPE: PR DIFFICULTY: Moderate KEYWORDS: approximate mean

114. Referring to Table 3-7, what is the approximate variance?

ANSWER: 0.0162 .

Numerical Descriptive Measures

97

TYPE: PR DIFFICULTY: Moderate KEYWORDS: approximate variance 115. Referring to Table 3-7, what is the approximate standard deviation?

ANSWER: 0.1273 TYPE: PR DIFFICULTY: Moderate KEYWORDS: approximate standard deviation

TABLE 3-8 Given below is the frequency distribution of the number of defective tomatoes in 50 different shipments.

Defective Items 0 but less than 5 5 but less than 10 10 but less than 15 15 but less than 20 20 but less than 25 25 but less than 30

Frequency 35 7 5 2 1 0

116. Referring to Table 3-8, what is the approximate mean number of defective tomatoes?

ANSWER: 5.2 TYPE: PR DIFFICULTY: Moderate KEYWORDS: approximate mean 117. Referring to Table 3-8, what is the approximate variance?

ANSWER: 23.6837 TYPE: PR DIFFICULTY: Moderate KEYWORDS: approximate variance 118. Referring to Table 3-8, what is the approximate standard deviation?

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

TABLE 3-9 In the 2002–2003 academic year, many public universities in the United States raised tuition and fees due to a decrease in state subsidies. The change in the cost of tuition, a shared dormitory room, and .

98

Numerical Descriptive Measures

the most popular meal plan from the 2001–2002 academic year for a sample of 10 public universities were as follows: $1589, $593, $1223, $869, $423, $1720, $708, $1425, $922 and $308.

119. Referring to Table 3-9, what is the mean and median change in the cost?

ANSWER: $978, $895.5 TYPE: PR DIFFICULTY: Moderate KEYWORDS: mean 120. Referring to Table 3-9, what are the five-number summary of the change in the cost?

ANSWER: $308, $593, $895.5, $1425, and $1720 TYPE: PR DIFFICULTY: Moderate KEYWORDS: five-number summary 121. Referring to Table 3-9, what is the standard deviation of the change in the cost?

ANSWER: $491.8 TYPE: PR DIFFICULTY: Moderate KEYWORDS: standard deviation 122. Referring to Table 3-9, the middle 50% of the change in the cost is spread over what value?

ANSWER: $832 TYPE: PR DIFFICULTY: Moderate KEYWORDS: interquartile range 123. Referring to Table 3-9, what is the coefficient of variation of the change in cost?

ANSWER: 50.29% TYPE: PR DIFFICULTY: Moderate KEYWORDS: coefficient of variation 124. Referring to Table 3-9, what are the (absolute values of) the Z scores of the change in cost?

ANSWER: 1.24, 0.78, 0.50, 0.22, 1.13, 1.51, 0.55, 0.91, 0.11, 1.36 TYPE: PR DIFFICULTY: Moderate KEYWORDS: Z scores

125. Referring to Table 3-9, are the data skewed? If so, how?

ANSWER: skewed to the right .

Numerical Descriptive Measures

TYPE: PR DIFFICULTY: Moderate KEYWORDS: shape

TABLE 3-10 The following table represents the assets in billions of dollars of the five largest bond funds.

Bond Fund Vanguard GNMA Vanguard Total Bond Mkt. Index Bond Fund of America A Franklin Calif. Tax-Free Inc. A Vanguard Short-Term Corp.

Assets (Billions $) 19.5 16.8 13.7 12.8 10.9

126. Referring to Table 3-10, what is the mean for this population of the five largest bond funds?

ANSWER: $14.74 billions TYPE: PR DIFFICULTY: Easy KEYWORDS: population mean 127. Referring to Table 3-10, what are the variance and standard deviation for this population?

ANSWER: 9.2984 squared billion dollars, $3.0493 billions, respectively TYPE: PR DIFFICULTY: Moderate KEYWORDS: population variance, population standard deviation

TABLE 3-11 The population of eight analysts at a software firm were asked to estimate the reuse rate when developing a new software system. The following data are given as a percentage of the total code written for a software system that is part of the reuse database.

50, 62.5, 37.5, 75.0, 45.0, 47.5, 15.0, 25.0 128. Referring to Table 3-11, what is the mean percentage of the total code that is part of the reuse database?

ANSWER: 44.69 TYPE: PR DIFFICULTY: Easy KEYWORDS: population mean 129. Referring to Table 3-11, what are the variance and standard deviation of the total code that is part of the reuse database?

ANSWER: 324.12 and 18.00, respectively TYPE: PR DIFFICULTY: Moderate .

99

100

Numerical Descriptive Measures

KEYWORDS: population variance, population standard deviation

.

Basic Probability

101

CHAPTER 4: BASIC PROBABILITY 1. If two events are collectively exhaustive, what is the probability that one or the other occurs? a) 0. b) 0.50. c) 1.00. d) Cannot be determined from the information given.

ANSWER: c TYPE: MC DIFFICULTY: Easy KEYWORDS: collectively exhaustive 2. If two events are collectively exhaustive, what is the probability that both occur at the same time? a) 0. b) 0.50. c) 1.00. d) Cannot be determined from the information given.

ANSWER: d TYPE: MC DIFFICULTY: Moderate KEYWORDS: collectively exhaustive, mutually exclusive EXPLANATION: We do not know if they are mutually exclusive.

3. If two events are mutually exclusive, what is the probability that one or the other occurs? a) 0. b) 0.50. c) 1.00. d) Cannot be determined from the information given.

ANSWER: d TYPE: MC DIFFICULTY: moderate KEYWORDS: collectively exhaustive, mutually exclusive EXPLANATION: We do not know if they are collectively exhaustive.

4. If two events are mutually exclusive, what is the probability that both occur at the same time? a) 0. b) 0.50. c) 1.00. d) Cannot be determined from the information given.

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

.

102

Basic Probability

5. If two events are mutually exclusive and collectively exhaustive, what is the probability that both occur? a) 0. b) 0.50. c) 1.00. d) Cannot be determined from the information given.

ANSWER: a TYPE: MC DIFFICULTY: Easy KEYWORDS: collective exhaustive, mutually exclusive 6. If two events are mutually exclusive and collectively exhaustive, what is the probability that one or the other occurs? a) 0. b) 0.50. c) 1.00. d) Cannot be determined from the information given.

ANSWER: c TYPE: MC DIFFICULTY: Easy KEYWORDS: collectively exhaustive, mutually exclusive 7. If events A and B are mutually exclusive and collectively exhaustive, what is the probability that event A occurs? a) 0. b) 0.50. c) 1.00. d) Cannot be determined from the information given.

ANSWER: d TYPE: MC DIFFICULTY: moderate KEYWORDS: collectively exhaustive, mutually exclusive EXPLANATION: We do not know if they are equally likely events.

8. If two equally likely events A and B are mutually exclusive and collectively exhaustive, what is the probability that event A occurs? a) 0. b) 0.50. c) 1.00. d) Cannot be determined from the information given.

ANSWER: b TYPE: MC DIFFICULTY: easy KEYWORDS: collectively exhaustive, mutually exclusive

.

Basic Probability

103

9. If two equally likely events A and B are mutually exclusive, what is the probability that event A occurs? a) 0. b) 0.50. c) 1.00. d) Cannot be determined from the information given.

ANSWER: d TYPE: MC DIFFICULTY: moderate KEYWORDS: collectively exhaustive, mutually exclusive EXPLANATION: We do not know if they are collectively exhaustive.

10. If two equally likely events A and B are collectively exhaustive, what is the probability that event A occurs? a) 0. b) 0.50. c) 1.00. d) Cannot be determined from the information given.

ANSWER: d TYPE: MC DIFFICULTY: moderate KEYWORDS: collectively exhaustive, mutually exclusive EXPLANATION: We do not know if they are mutually exclusive.

11. Selection of raffle tickets from a large bowl is an example of a) sampling with replacement. b) sampling without replacement. c) subjective probability. d) None of the above.

ANSWER: b TYPE: MC DIFFICULTY: Easy KEYWORDS: sampling with replacement, sampling without replacement 12. If two events are independent, what is the probability that they both occur? a) 0. b) 0.50. c) 1.00. d) Cannot be determined from the information given.

ANSWER: d TYPE: MC DIFFICULTY: Moderate KEYWORDS: statistical independence

.

104

Basic Probability

13. If the outcome of event A is not affected by event B, then events A and B are said to be a) mutually exclusive. b) statistically independent. c) collectively exhaustive. d) None of the above.

ANSWER: b TYPE: MC DIFFICULTY: Easy KEYWORDS: statistical independence 14. If event A and event B cannot occur at the same time, then events A and B are said to be a) mutually exclusive. b) statistically independent. c) collectively exhaustive. d) None of the above.

ANSWER: a TYPE: MC DIFFICULTY: Easy KEYWORDS: mutually exclusive 15. If either event A or event B must occur, then events A and B are said to be a) mutually exclusive. b) statistically independent. c) collectively exhaustive. d) None of the above.

ANSWER: c TYPE: MC DIFFICULTY: Moderate KEYWORDS: collectively exhaustive 16. The collection of all possible events is called a) a simple probability. b) a sample space. c) a joint probability. d) the null set.

ANSWER: b TYPE: MC DIFFICULTY: Moderate KEYWORDS: sample space

.

Basic Probability

105

17. All the events in the sample space that are not part of the specified event are called a) simple events. b) joint events. c) the sample space. d) the complement of the event.

ANSWER: d TYPE: MC DIFFICULTY: Moderate KEYWORDS: sample space, complement 18. Simple probability is also called a) marginal probability. b) joint probability. c) conditional probability. d) Bayes' theorem.

ANSWER: a TYPE: MC DIFFICULTY: Easy KEYWORDS: marginal probability 19. When using the general multiplication rule, P(A and B) is equal to a) P(A|B)P(B). b) P(A)P(B). c) P(B)/P(A). d) P(A)/P(B).

ANSWER: a TYPE: MC DIFFICULTY: Moderate KEYWORDS: multiplication rule 20. A business venture can result in the following outcomes (with their corresponding chance of occurring in parentheses): Highly Successful (10%), Successful (25%), Break Even (25%), Disappointing (20%), and Highly Disappointing (?). If these are the only outcomes possible for the business venture, what is the chance that the business venture will be considered Highly Disappointing? a) 10% b) 15% c) 20% d) 25% ANSWER: c TYPE: MC DIFFICULTY: Easy KEYWORDS: marginal probability

.

106

Basic Probability

21. A recent survey of banks revealed the following distribution for the interest rate being charged on a home loan (based on a 30-year mortgage with a 10% down payment).

Interest Rate Probability

7.0% 0.12

7.5% 0.23

8.0% 0.24

8.5% 0.35

> 8.5% 0.06

If a bank is selected at random from this distribution, what is the chance that the interest rate charged on a home loan will exceed 8.0%? a) 0.06 b) 0.41 c) 0.59 d) 1.00 ANSWER: b TYPE: MC DIFFICULTY: Easy KEYWORDS: marginal probability, addition rule 22. The employees of a company were surveyed on questions regarding their educational background and marital status. Of the 600 employees, 400 had college degrees, 100 were single, and 60 were single college graduates. The probability that an employee of the company is single or has a college degree is: a) 0.10 b) 0.25 c) 0.667 d) 0.733 ANSWER: d TYPE: MC DIFFICULTY: Moderate KEYWORDS: addition rule 23. The employees of a company were surveyed on questions regarding their educational background and marital status. Of the 600 employees, 400 had college degrees, 100 were single, and 60 were single college graduates. The probability that an employee of the company is married and has a college degree is: a) 40/600 b) 340/600 c) 400/600 d) 500/600 ANSWER: b TYPE: MC DIFFICULTY: Moderate KEYWORDS: joint probability

.

Basic Probability

107

24. The employees of a company were surveyed on questions regarding their educational background and marital status. Of the 600 employees, 400 had college degrees, 100 were single, and 60 were single college graduates. The probability that an employee of the company does not have a college degree is: a) 0.10 b) 0.33 c) 0.67 d) 0.75 ANSWER: b TYPE: MC DIFFICULTY: Easy KEYWORDS: complement 25. The probability that house sales will increase in the next 6 months is estimated to be 0.25. The probability that the interest rates on housing loans will go up in the same period is estimated to be 0.74. The probability that house sales or interest rates will go up during the next 6 months is estimated to be 0.89. The probability that both house sales and interest rates will increase during the next 6 months is: a) 0.10 b) 0.185 c) 0.705 d) 0.90 ANSWER: a TYPE: MC DIFFICULTY: Moderate KEYWORDS: joint probability 26. The probability that house sales will increase in the next 6 months is estimated to be 0.25. The probability that the interest rates on housing loans will go up in the same period is estimated to be 0.74. The probability that house sales or interest rates will go up during the next 6 months is estimated to be 0.89. The probability that neither house sales nor interest rates will increase during the next 6 months is: a) 0.11 b) 0.195 c) 0.89 d) 0.90 ANSWER: a TYPE: MC DIFFICULTY: Moderate KEYWORDS: joint probability, complement

.

108

Basic Probability

27. The probability that house sales will increase in the next 6 months is estimated to be 0.25. The probability that the interest rates on housing loans will go up in the same period is estimated to be 0.74. The probability that house sales or interest rates will go up during the next 6 months is estimated to be 0.89. The probability that house sales will increase but interest rates will not during the next 6 months is: a) 0.065 b) 0.15 c) 0.51 d) 0.89 ANSWER: b TYPE: MC DIFFICULTY: Moderate KEYWORDS: joint probability, complement 28. The probability that house sales will increase in the next 6 months is estimated to be 0.25. The probability that the interest rates on housing loans will go up in the same period is estimated to be 0.74. The probability that house sales or interest rates will go up during the next 6 months is estimated to be 0.89. The events of increase in house sales and increase in interest rates in the next 6 months are a) statistically independent. b) mutually exclusive. c) collectively exhaustive. d) None of the above.

ANSWER: d TYPE: MC DIFFICULTY: Moderate KEYWORDS: joint probability, statistical independence EXPLANATION: They are not statistically independent.

29. The probability that house sales will increase in the next 6 months is estimated to be 0.25. The probability that the interest rates on housing loans will go up in the same period is estimated to be 0.74. The probability that house sales or interest rates will go up during the next 6 months is estimated to be 0.89. The events of increase in house sales and no increase in house sales in the next 6 months are a) statistically independent. b) mutually exclusive. c) collectively exhaustive. d) (b) and (c) ANSWER: d TYPE: MC DIFFICULTY: Moderate KEYWORDS: mutually exclusive, collectively exhaustive, complement

.

Basic Probability

109

30. The probability that a new advertising campaign will increase sales is assessed as being 0.80. The probability that the cost of developing the new ad campaign can be kept within the original budget allocation is 0.40. Assuming that the two events are independent, the probability that the cost is kept within budget and the campaign will increase sales is: a) 0.20 b) 0.32 c) 0.40 d) 0.88 ANSWER: b TYPE: MC DIFFICULTY: Easy KEYWORDS: statistical independence, joint probability, multiplication rule 31. The probability that a new advertising campaign will increase sales is assessed as being 0.80. The probability that the cost of developing the new ad campaign can be kept within the original budget allocation is 0.40. Assuming that the two events are independent, the probability that the cost is kept within budget or the campaign will increase sales is: a) 0.20 b) 0.32 c) 0.68 d) 0.88 ANSWER: d TYPE: MC DIFFICULTY: Easy KEYWORDS: statistical independence, multiplication rule, addition rule 32. The probability that a new advertising campaign will increase sales is assessed as being 0.80. The probability that the cost of developing the new ad campaign can be kept within the original budget allocation is 0.40. Assuming that the two events are independent, the probability that the cost is not kept within budget or the campaign will not increase sales is: a) 0.12 b) 0.32 c) 0.68 d) 0.88 ANSWER: c TYPE: MC DIFFICULTY: Easy KEYWORDS: statistical independence, multiplication rule, addition rule, complement

.

110

Basic Probability

33. The probability that a new advertising campaign will increase sales is assessed as being 0.80. The probability that the cost of developing the new ad campaign can be kept within the original budget allocation is 0.40. Assuming that the two events are independent, the probability that neither the cost is kept within budget nor the campaign will increase sales is: a) 0.12 b) 0.32 c) 0.68 d) 0.88 ANSWER: a TYPE: MC DIFFICULTY: Easy KEYWORDS: statistical independence, multiplication rule, joint probability, complement 34. According to a survey of American households, the probability that the residents own 2 cars if annual household income is over $25,000 is 80%. Of the households surveyed, 60% had incomes over $25,000 and 70% had 2 cars. The probability that the residents of a household own 2 cars and have an income over $25,000 a year is: a) 0.12 b) 0.18 c) 0.22 d) 0.48 ANSWER: d TYPE: MC DIFFICULTY: Moderate KEYWORDS: joint probability, conditional probability 35. According to a survey of American households, the probability that the residents own 2 cars if annual household income is over $25,000 is 80%. Of the households surveyed, 60% had incomes over $25,000 and 70% had 2 cars. The probability that the residents of a household do not own 2 cars and have an income over $25,000 a year is: a) 0.12 b) 0.18 c) 0.22 d) 0.48 ANSWER: a TYPE: MC DIFFICULTY: Difficult KEYWORDS: joint probability, complement, multiplication rule, conditional probability

.

Basic Probability

111

36. According to a survey of American households, the probability that the residents own 2 cars if annual household income is over $25,000 is 80%. Of the households surveyed, 60% had incomes over $25,000 and 70% had 2 cars. The probability that the residents of a household own 2 cars and have an income less than or equal to $25,000 a year is: a) 0.12 b) 0.18 c) 0.22 d) 0.48 ANSWER: c TYPE: MC DIFFICULTY: Difficult KEYWORDS: joint probability, complement, multiplication rule, conditional probability 37. According to a survey of American households, the probability that the residents own 2 cars if annual household income is over $25,000 is 80%. Of the households surveyed, 60% had incomes over $25,000 and 70% had 2 cars. The probability that annual household income is over $25,000 if the residents of a household own 2 cars is: a) 0.42 b) 0.48 c) 0.50 d) 0.69 ANSWER: d TYPE: MC DIFFICULTY: Difficult KEYWORDS: conditional probability, Bayes’ theorem 38. According to a survey of American households, the probability that the residents own 2 cars if annual household income is over $25,000 is 80%. Of the households surveyed, 60% had incomes over $25,000 and 70% had 2 cars. The probability that annual household income is over $25,000 if the residents of a household do not own 2 cars is: a) 0.12 b) 0.18 c) 0.40 d) 0.70 ANSWER: c TYPE: MC DIFFICULTY: Difficult KEYWORDS: conditional probability, complement, Bayes’ theorem

.

112

Basic Probability

39. According to a survey of American households, the probability that the residents own 2 cars if annual household income is over $25,000 is 80%. Of the households surveyed, 60% had incomes over $25,000 and 70% had 2 cars. The probability that the residents do not own 2 cars if annual household income is not over $25,000 is: a) 0.12 b) 0.18 c) 0.45 d) 0.70 ANSWER: c TYPE: MC DIFFICULTY: Difficult KEYWORDS: conditional probability, complement 40. A company has 2 machines that produce widgets. An older machine produces 23% defective widgets, while the new machine produces only 8% defective widgets. In addition, the new machine produces 3 times as many widgets as the older machine does. Given that a widget was produced by the new machine, what is the probability it is not defective? a) 0.06 b) 0.50 c) 0.92 d) 0.94 ANSWER: c TYPE: MC DIFFICULTY: Easy KEYWORDS: conditional probability, complement 41. A company has 2 machines that produce widgets. An older machine produces 23% defective widgets, while the new machine produces only 8% defective widgets. In addition, the new machine produces 3 times as many widgets as the older machine does. What is the probability that a randomly chosen widget produced by the company is defective? a) 0.078 b) 0.1175 c) 0.156 d) 0.310 ANSWER: b TYPE: MC DIFFICULTY: Moderate KEYWORDS: marginal probability

.

Basic Probability

113

42. A company has 2 machines that produce widgets. An older machine produces 23% defective widgets, while the new machine produces only 8% defective widgets. In addition, the new machine produces 3 times as many widgets as the older machine does. Given a randomly chosen widget was tested and found to be defective, what is the probability it was produced by the new machine? a) 0.08 b) 0.15 c) 0.489 d) 0.511 ANSWER: d TYPE: MC DIFFICULTY: Difficult KEYWORDS: conditional probability, Bayes’ theorem

TABLE 4-1 Mothers Against Drunk Driving is a very visible group whose main focus is to educate the public about the harm caused by drunk drivers. A study was recently done that emphasized the problem we all face with drinking and driving. Four hundred accidents that occurred on a Saturday night were analyzed. Two items noted were the number of vehicles involved and whether alcohol played a role in the accident. The numbers are shown below:

Did alcohol play a role?

Yes No Totals

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

Totals 170 230 400

43. Referring to Table 4-1, what proportion of accidents involved more than one vehicle? a) 50/400 or 12.5% b) 75/400 or 18.75% c) 275/400 or 68.75% d) 325/400 or 81.25% ANSWER: d TYPE: MC DIFFICULTY: Easy KEYWORDS: contingency table, empirical classical probability, addition rule

44. Referring to Table 4-1, what proportion of accidents involved alcohol and a single vehicle? .

114

Basic Probability

a) b) c) d)

25/400 or 6.25% 50/400 or 12.5% 195/400 or 48.75% 245/400 or 61.25%

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

48. Referring to Table 4-1, given that 3 vehicles were involved, what proportion of accidents involved alcohol? .

Basic Probability

a) b) c) d)

115

20/30 or 66.67% 20/50 or 40% 20/170 or 11.77% 20/400 or 5%

ANSWER: b TYPE: MC DIFFICULTY: Easy KEYWORDS: contingency table, empirical classical probability, conditional probability 49. Referring to Table 4-1, given that alcohol was not involved, what proportion of the accidents were single vehicle? a) 50/75 or 66.67% b) 25/230 or 10.87% c) 50/170 or 29.41% d) 25/75 or 33.33% ANSWER: b TYPE: MC DIFFICULTY: Easy KEYWORDS: contingency table, empirical classical probability, conditional probability, complement 50. Referring to Table 4-1, given that alcohol was not involved, what proportion of the accidents were multiple vehicle? a) 50/170 or 29.41% b) 120/170 or 70.59% c) 205/230 or 89.13% d) 25/230 or 10.87% ANSWER: c TYPE: MC DIFFICULTY: Easy KEYWORDS: contingency table, empirical classical probability, conditional probability, complement

TABLE 4-2

.

116

Basic Probability An alcohol awareness task force at a Big-Ten university sampled 200 students after the midterm to ask them whether they went bar hopping the weekend before the midterm or spent the weekend studying, and whether they did well or poorly on the midterm. The following result was obtained.

Studying for Exam Went Bar Hopping

Did Well on Midterm 80 30

Did Poorly on Midterm 20 70

51. Referring to Table 4-2, what is the probability that a randomly selected student who went bar hopping will do well on the midterm? a. 30/100 or 30% b. 30/110 or 27.27% c. 30/200 or 15% d. (100/200)*(110/200) or 27.50% ANSWER: a TYPE: MC DIFFICULTY: Easy KEYWORDS: contingency table, empirical classical probability, conditional probability 52. Referring to Table 4-2, what is the probability that a randomly selected student did well on the midterm or went bar hopping the weekend before the midterm? a) 30/200 or 15% b) (80+30)/200 or 55% c) (30+70)/200 or 50% d) (80+30+70)/200 or 90% ANSWER: d TYPE: MC DIFFICULTY: Easy KEYWORDS: contingency table, empirical classical probability, addition rule 53. Referring to Table 4-2, what is the probability that a randomly selected student did well on the midterm and also went bar hopping the weekend before the midterm? a) 30/200 or 15% b) (80+30)/200 or 55% c) (30+70)/200 or 50% d) (80+30+70)/200 or 90% ANSWER: a TYPE: MC DIFFICULTY: Easy KEYWORDS: contingency table, empirical classical probability, joint probability

54. Referring to Table 4-2, the events "Did Well on Midterm" and "Studying for Exam" are a) statistically dependent. .

Basic Probability

117

b) mutually exclusive. c) collective exhaustive. d) None of the above.

ANSWER: a TYPE: MC DIFFICULTY: Easy KEYWORDS: contingency table, empirical classical probability, statistical independence, joint probability 55. Referring to Table 4-2, the events "Did Well on Midterm" and "Studying for Exam" are a) not statistically dependent. b) not mutually exclusive. c) collective exhaustive. d) None of the above.

ANSWER: b TYPE: MC DIFFICULTY: Easy KEYWORDS: contingency table, empirical classical probability, mutually exclusive, joint probability 56. Referring to Table 4-2, the events "Did Well on Midterm" and "Did Poorly on Midterm" are a) statistically dependent. b) mutually exclusive. c) collective exhaustive. d) All of the above.

ANSWER: d TYPE: MC DIFFICULTY: Easy KEYWORDS: contingency table, empirical classical probability, statistical independence, mutually exclusive, collective exhaustive, joint probability 57. True or False: When A and B are mutually exclusive, P(A or B) can be found by adding P(A) and P(B).

ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: mutually exclusive, addition rule 58. True or False: The collection of all the possible events is called a sample space.

ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: sample space

59. True or False: If A and B cannot occur at the same time they are called mutually exclusive.

.

118

Basic Probability

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

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

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

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

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

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

ANSWER: True TYPE: TF DIFFICULTY: Moderate KEYWORDS: collective exhaustive

66. True or False: If P(A and B) = 0, then A and B must be mutually exclusive.

.

Basic Probability

119

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

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

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

ANSWER: False TYPE: TF DIFFICULTY: Difficult KEYWORDS: mutually exclusive, collective exhaustive 70. Suppose A and B are independent events where P(A) = 0.4 and P(B) = 0.5. Then P(A and B) = __________.

ANSWER: 0.2 TYPE: FI DIFFICULTY: Easy KEYWORDS: statistical independence, multiplication rule 71. Suppose A and B are mutually exclusive events where P(A) = 0.4 and P(B) = 0.5. Then P(A and B) = __________.

ANSWER: 0 TYPE: FI DIFFICULTY: Easy KEYWORDS: mutually exclusive, joint probability, multiplication rule 72. Suppose A and B are mutually exclusive events where P(A) = 0.4 and P(B) = 0.5. Then P(A or B) = __________.

ANSWER: 0.9 TYPE: FI DIFFICULTY: Easy KEYWORDS: mutually exclusive, addition rule 73. Suppose A and B are independent events where P(A) = 0.4 and P(B) = 0.5. Then P(A or B) = __________. .

120

Basic Probability

ANSWER: 0.7 TYPE: FI DIFFICULTY: Easy KEYWORDS: statistical independence, addition rule 74. Suppose A and B are events where P(A) = 0.4, P(B) = 0.5, and P(A and B) = 0.1. Then P(A or B) = __________.

ANSWER: 0.8 TYPE: FI DIFFICULTY: Easy KEYWORDS: addition rule 75. Suppose A and B are events where P(A) = 0.4, P(B) = 0.5, and P(A and B) = 0.1. Then P(A|B) = __________.

ANSWER: 0.2 TYPE: FI DIFFICULTY: Moderate KEYWORDS: conditional probability 76. Suppose A and B are events where P(A) = 0.4, P(B) = 0.5, and P(A and B) = 0.1. Then P(B|A) = __________.

ANSWER: 0.25 TYPE: FI DIFFICULTY: Moderate KEYWORDS: conditional probability

TABLE 4-3 A survey is taken among customers of a fast-food restaurant to determine preference for hamburger or chicken. Of 200 respondents selected, 75 were children and 125 were adults. 120 preferred hamburger and 80 preferred chicken. 55 of the children preferred hamburger.

77. Referring to Table 4-3, the probability that a randomly selected individual is an adult is __________.

ANSWER: 125/200 or 62.5% TYPE: FI DIFFICULTY: Moderate KEYWORDS: empirical classical probability, conditional probability, marginal probability

78. Referring to Table 4-3, the probability that a randomly selected individual is an adult or a child is __________.

.

Basic Probability

121

ANSWER: 200/200 or 100% TYPE: FI DIFFICULTY: Moderate KEYWORDS: empirical classical probability, addition rule 79. Referring to Table 4-3, the probability that a randomly selected individual is a child and prefers chicken is __________.

ANSWER: 20/200 or 10% TYPE: FI DIFFICULTY: Moderate KEYWORDS: empirical classical probability, joint probability, multiplication rule 80. Referring to Table 4-3, the probability that a randomly selected individual is an adult and prefers chicken is __________.

ANSWER: 60/200 or 30% TYPE: FI DIFFICULTY: Moderate KEYWORDS: empirical classical probability, joint probability, multiplication rule 81. Referring to Table 4-3, the probability that a randomly selected individual is a child or prefers hamburger is __________.

ANSWER: 140/200 or 70% TYPE: FI DIFFICULTY: Moderate KEYWORDS: empirical classical probability, addition rule 82. Referring to Table 4-3, assume we know the person is a child. The probability that this individual prefers hamburger is __________.

ANSWER: 55/75 or 73.33% TYPE: FI DIFFICULTY: Moderate KEYWORDS: empirical classical probability, conditional probability 83. Referring to Table 4-3, assume we know that a person has ordered chicken. The probability that this individual is an adult is __________.

ANSWER: 60/80 or 75% TYPE: FI DIFFICULTY: Moderate KEYWORDS: empirical classical probability, conditional probability, Bayes’ theorem

84. Referring to Table 4-3, assume we know that a person has ordered hamburger. The probability that this individual is a child is __________.

ANSWER: .

122

Basic Probability

55/120 or 45.83% TYPE: FI DIFFICULTY: Moderate KEYWORDS: empirical classical probability, conditional probability, Bayes’ theorem

TABLE 4-4 Suppose that patrons of a restaurant were asked whether they preferred beer or whether they preferred wine. 70% said that they preferred beer. 60% of the patrons were male. 80% of the males preferred beer.

85. Referring to Table 4-4, the probability a randomly selected patron prefers wine is __________.

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

ANSWER: 0.40 TYPE: FI DIFFICULTY: Moderate KEYWORDS: addition rule, multiplication rule, marginal probability 87. Referring to Table 4-4, the probability a randomly selected patron is a female who prefers wine is __________.

ANSWER: 0.18 TYPE: FI DIFFICULTY: Difficult KEYWORDS: addition rule, multiplication rule, joint probability 88. Referring to Table 4-4, the probability a randomly selected patron is a female who prefers beer is __________.

ANSWER: 0.22 TYPE: FI DIFFICULTY: Difficult KEYWORDS: addition rule, multiplication rule, joint probability 89. Referring to Table 4-4, suppose a randomly selected patron prefers wine. Then the probability the patron is a male is __________.

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

ANSWER: .

Basic Probability

123

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

ANSWER: 0.55 TYPE: FI DIFFICULTY: Difficult KEYWORDS: addition rule, multiplication rule, conditional probability 92. True or False: Referring to Table 4-4, the two events "preferring" beer and "preferring" wine are statistically independent.

ANSWER: False TYPE: TF DIFFICULTY: Moderate KEYWORDS: addition rule, multiplication rule, statistical independence 93. True or False: Referring to Table 4-4, the two events "preferring" beer and being a male are statistically independent.

ANSWER: False TYPE: TF DIFFICULTY: Moderate KEYWORDS: addition rule, multiplication rule, statistical independence

TABLE 4-5 In a meat packaging plant Machine A accounts for 60% of the plant's output, while Machine B accounts for 40% of the plant's output. In total, 4% of the packages are improperly sealed. Also, 3% of the packages are from Machine A and are improperly sealed.

94. Referring to Table 4-5, if a package is selected at random, the probability that it will be properly sealed is ________.

ANSWER: 0.96 TYPE: FI DIFFICULTY: Moderate KEYWORDS: addition rule, marginal probability

95. Referring to Table 4-5, if a package selected at random is improperly sealed, the probability that it came from machine A is ________.

ANSWER: 0.75 .

124

Basic Probability

TYPE: FI DIFFICULTY: Difficult KEYWORDS: conditional probability 96. Referring to Table 4-5, if a package selected at random came from Machine A, the probability that it is improperly sealed is ________.

ANSWER: 0.05 TYPE: FI DIFFICULTY: Difficult KEYWORDS: conditional probability 97. Referring to Table 4-5, if a package selected at random came from Machine B, the probability that it is properly sealed is ________.

ANSWER: 0.975 TYPE: FI DIFFICULTY: Difficult KEYWORDS: conditional probability, complement, addition rule 98. Referring to Table 4-5, if a package selected at random came from Machine B, the probability that it is improperly sealed is ________.

ANSWER: 0.025 TYPE: FI DIFFICULTY: Difficult KEYWORDS: conditional probability, complement, addition rule

TABLE 4-6 At a Texas college, 60% of the students are from the southern part of the state, 30% are from the northern part of the state, and the remaining 10% are from out-of-state. All students must take and pass an Entry Level Math (ELM) test. 60% of the southerners have passed the ELM, 70% of the northerners have passed the ELM, and 90% of the out-of-staters have passed the ELM. 99. Referring to Table 4-6, the probability that a randomly selected student is someone from northern Texas who has not passed the ELM is ________.

ANSWER: 0.09 TYPE: FI DIFFICULTY: Difficult KEYWORDS: joint probability, multiplication rule

100. Referring to Table 4-6, the probability that a randomly selected student has passed the ELM is ________.

ANSWER: 0.66 TYPE: FI DIFFICULTY: Difficult .

Basic Probability

125

KEYWORDS: marginal probability, multiplication rule, addition rule 101. Referring to Table 4-6, if a randomly selected student has passed the ELM, the probability the student is from out-of-state is ________.

ANSWER: 0.136 TYPE: FI DIFFICULTY: Difficult KEYWORDS: conditional probability, Bayes’ theorem 102. Referring to Table 4-6, if a randomly selected student has not passed the ELM, the probability the student is from southern Texas is ________.

ANSWER: 0.706 TYPE: FI DIFFICULTY: Difficult KEYWORDS: conditional probability, complement, Bayes’ theorem 103. Referring to Table 4-6, the probability that a randomly selected student is not from southern Texas and has not passed the ELM is ________.

ANSWER: 0.10 TYPE: FI DIFFICULTY: Difficult KEYWORDS: joint probability, complement, multiplication rule, addition 104. Referring to Table 4-6, if a randomly selected student has not passed the ELM, the probability the student is not from northern Texas is ________.

ANSWER: 0.735 TYPE: FI DIFFICULTY: Difficult KEYWORDS: conditional probability, complement, addition rule, multiplication rule, Bayes’ theorem 105. Referring to Table 4-6, if a randomly selected student is not from southern Texas, the probability the student has not passed the ELM is ________.

ANSWER: 0.25 TYPE: FI DIFFICULTY: Difficult KEYWORDS: conditional probability, complement, addition rule, multiplication rule, Bayes’ theorem

106. Referring to Table 4-6, if a randomly selected student is not from out-of-state, the probability the student has passed the ELM is ________.

ANSWER: 0.633 TYPE: FI DIFFICULTY: Difficult

.

126

Basic Probability KEYWORDS: conditional probability, complement, addition rule, multiplication rule, Bayes’ theorem

TABLE 4-7 The next state lottery will have the following payoffs possible with their associated probabilities.

Payoff Probability $2.00 0.0500 $25.00 0.0100 $100.00 0.0050 $500.00 0.0010 $5,000.00 0.0005 $10,000.00 0.0001 You buy a single ticket.

107. Referring to Table 4-7, the probability that you win any money is ________.

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

ANSWER: 0.0066 TYPE: FI DIFFICULTY: Easy KEYWORDS: addition rule 109. Referring to Table 4-7, if you have a winning ticket, the probability that you win at least $100.00 is ________.

ANSWER: 0.10 TYPE: FI DIFFICULTY: Moderate KEYWORDS: conditional probability, addition rule

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

Basic Probability

127

d) Conditional probability ANSWER: a TYPE: MC DIFFICULTY: Easy KEYWORDS: a priori classical probability, empirical classical probability, subjective probability 111. The closing price of a company’s stock tomorrow can be lower, higher or the same as today’s closing price. Based on the closing price of the stock collected over the last month, 25% of the days the closing price was higher than previous day’s closing price, 45% was lower than previous day’s and 30% was the same as previous day’s. Based on this information, the probability that tomorrow’s closing price will be higher than today’s is 25%. This is an example of using which of the following probability approach? a) A priori classical probability b) Empirical classical probability c) Subjective probability d) Conditional probability ANSWER: b TYPE: MC DIFFICULTY: Easy KEYWORDS: a priori classical probability, empirical classical probability, subjective probability 112. The closing price of a company’s stock tomorrow can be lower, higher or the same as today’s closing price. After evaluating all the information available on the company’s fundamentals and the economic environment, an analyst has determined that the probability that tomorrow’s closing price will be higher than today’s is determined to be 25%. This is an example of using which of the following probability approach? a) A priori classical probability b) Empirical classical probability c) Subjective probability d) Conditional probability ANSWER: c TYPE: MC DIFFICULTY: Easy KEYWORDS: a priori classical probability, empirical classical probability, subjective probability

TABLE 4-8 According to the record of the registrar’s office at a state university, 35% of the students are freshman, 25% are sophomore, 16% are junior and the rest are senior. Among the freshmen, sophomores, juniors and seniors, the portion of students who live in the dormitory are, respectively, 80%, 60%, 30% and 20%.

.

128

Basic Probability 113. Referring to Table 4-8, what is the probability that a randomly selected student is a freshman who lives in a dormitory?

ANSWER: 0.28 TYPE: PR DIFFICULTY: Easy KEYWORDS: joint probability, multiplication rule 114. Referring to Table 4-8, what is the probability that a randomly selected student is a sophomore who does not live in a dormitory?

ANSWER: 0.1 TYPE: PR DIFFICULTY: Easy KEYWORDS: joint probability, multiplication rule, complement 115. Referring to Table 4-8, what is the probability that a randomly selected student is a junior who does not live in a dormitory?

ANSWER: 0.112 TYPE: PR DIFFICULTY: Easy KEYWORDS: joint probability, multiplication rule, complement 116. Referring to Table 4-8, what is the probability that a randomly selected student is a junior or senior who lives in a dormitory?

ANSWER: 0.096 TYPE: PR DIFFICULTY: Moderate KEYWORDS: joint probability, multiplication rule, addition rule 117. Referring to Table 4-8, what percentage of the students live in a dormitory?

ANSWER: 0.526 TYPE: PR DIFFICULTY: Easy KEYWORDS: marginal probability, multiplication rule, addition rule 118. Referring to Table 4-8, what percentage of the students do not live in a dormitory?

ANSWER: 0.474 TYPE: PR DIFFICULTY: Easy KEYWORDS: marginal probability, multiplication rule, addition rule 119. Referring to Table 4-8, if a randomly selected student lives in the dormitory, what is the probability that the student is a freshman?

ANSWER: 0.532 TYPE: PR DIFFICULTY: Moderate KEYWORDS: conditional probability, Bayes’ theorem .

Basic Probability

129

120. Referring to Table 4-8, if a randomly selected student lives in the dormitory, what is the probability that the student is not a freshman?

ANSWER: 0.468 TYPE: PR DIFFICULTY: Moderate KEYWORDS: conditional probability, Bayes’ theorem, complement, addition rule 121. Referring to Table 4-8, if a randomly selected student does not live in the dormitory, what is the probability that the student is a junior or a senior?

ANSWER: 0.641 TYPE: PR DIFFICULTY: Moderate KEYWORDS: conditional probability, Bayes’ theorem, complement, addition rule 122. Referring to Table 4-8, determine whether the class status of a student and whether the student lives in a dormitory are statistically independent.

ANSWER: Not statistical independent TYPE: PR DIFFICULTY: Easy KEYWORDS: statistical independence

TABLE 4-9 A survey conducted by the Segal Company of New York found that in a sample of 189 large companies, 40 offered stock options to their board members as part of their non-cash compensation packages. For small- to mid-sized companies, 43 of the 180 surveyed indicated that they offer stock options as part of their noncash compensation packages to their board members.

123. Referring to Table 4-9, set up a contingency table.

ANSWER: Large Stock Options

Yes No

Company Size Small-to-midsized 40 43 149 137 189 180

83 286 369

TYPE: PR DIFFICULTY: Moderate KEYWORDS: contingency table 124. Referring to Table 4-9, if a company is selected at random, what is the probability that the company offered stock options to their board members?

ANSWER: 0.2249 or 22.49% TYPE: PR DIFFICULTY: Easy KEYWORDS: marginal probability

.

130

Basic Probability 125. Referring to Table 4-9, if a company is selected at random, what is the probability that the company is small to mid-sized and did not offer stock options to their board members?

ANSWER: 0.3713 or 37.13% TYPE: PR DIFFICULTY: Easy KEYWORDS: joint probability 126. Referring to Table 4-9, if a company is selected at random, what is the probability that the company is small to mid-sized or offered stock options to their board members?

ANSWER: 0.5962 or 59.62% TYPE: PR DIFFICULTY: Easy KEYWORDS: addition rule 127. Referring to Table 4-9, if a randomly selected company is a large company, what is the probability that it offered stock options to their board members?

ANSWER: 0.2116 or 21.16% TYPE: PR DIFFICULTY: Easy KEYWORDS: conditional probability 128. Referring to Table 4-9, if a randomly selected company offered stock options to their board member, what is the probability that it is a large company?

ANSWER: 0.4819 or 48.19% TYPE: PR DIFFICULTY: Easy KEYWORDS: conditional probability 129. Referring to Table 4-9, is the size of the company statistically independent of whether stock options are offered to their board members and why?

ANSWER: No, because P(stock option) is not equal to P(stock option | large) TYPE: PR DIFFICULTY: Moderate KEYWORDS: conditional probability, statistical independence

TABLE 4-10 Are whites more likely to claim bias? It was found that 60% of the workers were white, 30% were black and 10% are other races. Given that a worker was white, the probability that the worker had claimed bias was 30%. Given that a worker was black, the probability that the worker had claimed bias was 40%. Given that a worker was other race, the probability that the worker had claimed bias was 0%. .

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130. Referring to Table 4-10, what is the probability that a randomly selected worker had not claimed bias?

ANSWER: 0.7 or 70% TYPE: PR DIFFICULTY: Moderate KEYWORDS: complement, marginal probability, multiplication rule 131. Referring to Table 4-10, if a randomly selected worker had claimed bias, what is the probability that the worker is white?

ANSWER: 0.6 or 60% TYPE: PR DIFFICULTY: Moderate KEYWORDS: conditional probability, Bayes’ theorem 132. Referring to Table 4-10, if a randomly selected worker had not claimed bias, what is the probability that the worker is white?

ANSWER: 0.6 or 60% TYPE: PR DIFFICULTY: Moderate KEYWORDS: conditional probability, Bayes’ theorem 133. Referring to Table 4-10, what is the probability that a randomly selected worker is white and had claimed bias?

ANSWER: 0.18 or 18% TYPE: PR DIFFICULTY: Moderate KEYWORDS: joint probability, multiplication rule 134. Referring to Table 4-10, what is the probability that a randomly selected worker is black and had not claimed bias?

ANSWER: 0.18 or 18% TYPE: PR DIFFICULTY: Moderate KEYWORDS: joint probability, multiplication rule

135. Referring to Table 4-10, what is the probability that a randomly selected worker is black and had not claimed bias or is white and has claimed bias?

ANSWER: 0.36 or 36% TYPE: PR DIFFICULTY: Moderate KEYWORDS: joint probability, multiplication rule, addition rule

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Basic Probability 136. Referring to Table 4-10, what is the probability that a randomly selected worker is not black and had not claimed bias?

ANSWER: 0.52 or 52% TYPE: PR DIFFICULTY: Moderate KEYWORDS: joint probability, multiplication rule, addition rule 137. Referring to Table 4-10, when a randomly selected worker was not white, what is the probability that the worker had not claimed bias?

ANSWER: 0.7 or 70% TYPE: PR DIFFICULTY: Moderate KEYWORDS: conditional probability, multiplication rule, addition rule 138. You know that the probability of a randomly selected student will cheat on an exam is

1%. You also know that the probability of a randomly selected student will cheat on an exam knowing that his/her fellow classmate is cheating on the exam is also 1%. Which of the following is true about the event of "the randomly selected student cheating on an exam" and "his/her classmate is cheating on the exam"? a) b) c) d)

They are mutually exclusive.

They are collectively exhaustive.

They are statistically independent.

None of the above.

ANSWER: c TYPE: MC DIFFICULTY: Easy KEYWORDS: statistical independence 139. It is considered an unethical practice when advertisements try to mislead consumers by providing misinterpretations on claims using probability statements.

ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: ethical practice

140. It is considered an ethical practice when advertisements explain the basis on which their probability claims rest.

ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: ethical practice 141. It is considered an ethical practice when using statements that involve probabilities to provide a correct interpretation of the statements. .

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ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: ethical practice 142. It is not an unethical practice when using statements that involve probabilities in advertisement without providing a correct interpretation of the statements because most people are comfortable with the concept of probability.

ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: ethical practice TABLE 4-11 A survey of 500 men and 500 women, designed to study financial tensions between couples, asked how likely each was to hide purchases, cash, or investments from his or her partner. The results were as follows: Likely to Hide Type Men Women Auto-related 66 36 Cash 126 133 Clothes 62 116 Electronics 79 56 Entertainment 96 76 Food 74 94 Investments 76 52 Travel 53 39 143. Referring to Table 4-11, if a respondent is chosen at random, what is the probability that he or she is likely to hide a clothing purchase?

ANSWER: 0.178 TYPE: PR DIFFICULTY: Easy KEYWORDS: empirical classical probability, marginal probability 144. Referring to Table 4-11, if a respondent is chosen at random, what is the probability that he or she is likely to hide investments?

ANSWER: 0.128 TYPE: PR DIFFICULTY: Easy KEYWORDS: empirical classical probability, marginal probability 145. Referring to Table 4-11, if a respondent is chosen at random, what is the probability that the person is a male and is likely to hide auto-related purchases?

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Basic Probability ANSWER: 0.066 TYPE: PR DIFFICULTY: Easy KEYWORDS: empirical classical probability, joint probability 146. Referring to Table 4-11, if a respondent is chosen at random, what is the probability that the person is a female and is likely to hide clothing purchases?

ANSWER: 0.116 TYPE: PR DIFFICULTY: Easy KEYWORDS: empirical classical probability, joint probability 147. Referring to Table 4-11, given that the person hides investments from his or her partner, what is the probability that the person is a male?

ANSWER: 0.5938 TYPE: PR DIFFICULTY: Moderate KEYWORDS: empirical classical probability, conditional probability 148. Referring to Table 4-11, given that the person hides cash from his or her partner, what is the probability that the person is a male?

ANSWER: 0.4865 TYPE: PR DIFFICULTY: Moderate KEYWORDS: empirical classical probability, conditional probability 149. Referring to Table 4-11, given that the person hides auto-related purchases from his or her partner, what is the probability that the person is a female?

ANSWER: 0.3529 TYPE: PR DIFFICULTY: Moderate KEYWORDS: empirical classical probability, conditional probability

150. Referring to Table 4-11, given that the person is a male, what is the probability that he will hide auto-related purchases?

ANSWER: 0.132 TYPE: PR DIFFICULTY: Moderate KEYWORDS: empirical classical probability, conditional probability 151. Referring to Table 4-11, given that the person is a female, what is the probability that he will hide clothing purchases?

ANSWER: .

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0.232 TYPE: PR DIFFICULTY: Moderate KEYWORDS: empirical classical probability, conditional probability 152. Referring to Table 4-11, is hiding clothing purchases statistically independent of being a female?

ANSWER: No TYPE: PR DIFFICULTY: Moderate KEYWORDS: empirical classical probability, conditional probability, statistical independence 153. Referring to Table 4-11, is hiding an auto-related purchase statistically independent of being a male?

ANSWER: No TYPE: PR DIFFICULTY: Moderate KEYWORDS: empirical classical probability, conditional probability, statistical independence 154. Referring to Table 4-11, is hiding a travel purchase statistically independent of being a male?

ANSWER: No TYPE: PR DIFFICULTY: Moderate KEYWORDS: empirical classical probability, conditional probability, statistical independence

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

ANSWER: a TYPE: MC DIFFICULTY: Moderate KEYWORDS: probability distribution

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Discrete Probability Distributions 4. The connotation "expected value" or "expected gain" from playing roulette at a casino means a) the amount you expect to "gain" on a single play. b) the amount you expect to "gain" in the long run over many plays. c) the amount you need to "break even" over many plays. d) the amount you should expect to gain if you are lucky.

ANSWER: b TYPE: MC DIFFICULTY: Easy KEYWORDS: expected value 5. Which of the following about the binomial distribution is not a true statement? a) The probability of event of interest must be constant from trial to trial. b) Each outcome is independent of the other. c) Each outcome may be classified as either "event of interest" or "not event of interest." d) The random variable of interest is continuous.

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

ANSWER: b TYPE: MC DIFFICULTY: Easy KEYWORDS: binomial distribution, properties 7.

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

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

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Discrete Probability Distributions 8. Whenever p = 0.1 and n is small, the binomial distribution will be a) symmetric. b) right-skewed. c) left-skewed. d) None of the above.

ANSWER: b TYPE: MC DIFFICULTY: Moderate KEYWORDS: binomial distribution, properties 9. If n = 10 and p = 0.70, then the mean of the binomial distribution is a) 0.07 b) 1.45. c) 7.00 d) 14.29 ANSWER: c TYPE: MC DIFFICULTY: Easy KEYWORDS: binomial distribution, mean 10. If n = 10 and p = 0.70, then the standard deviation of the binomial distribution is a) 0.07 b) 1.45 c) 7.00 d) 14.29 ANSWER: b TYPE: MC DIFFICULTY: Easy KEYWORDS: binomial distribution, standard deviation 11. If the outcomes of a random variable follow a Poisson distribution, then their a) mean equals the standard deviation. b) median equals the standard deviation. c) mean equals the variance. d) median equals the variance.

ANSWER: c TYPE: MC DIFFICULTY: Easy KEYWORDS: Poisson distribution, mean, standard deviation, properties

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12. What type of probability distribution will the consulting firm most likely employ to analyze the insurance claims in the following problem?

An insurance company has called a consulting firm to determine if the company has an unusually high number of false insurance claims. It is known that the industry proportion for false claims is 3%. The consulting firm has decided to randomly and independently sample 100 of the company’s insurance claims. They believe the number of these 100 that are false will yield the information the company desires. a) binomial distribution. b) Poisson distribution. c) normal distribution. d) none of the above.

ANSWER: a TYPE: MC DIFFICULTY: Difficult KEYWORDS: binomial distribution, properties 13. What type of probability distribution will most likely be used to analyze warranty repair needs on new cars in the following problem?

The service manager for a new automobile dealership reviewed dealership records of the past 20 sales of new cars to determine the number of warranty repairs he will be called on to perform in the next 90 days. Corporate reports indicate that the probability any one of their new cars needs a warranty repair in the first 90 days is 0.05. The manager assumes that calls for warranty repair are independent of one another and is interested in predicting the number of warranty repairs he will be called on to perform in the next 90 days for this batch of 20 new cars sold. a) binomial distribution. b) Poisson distribution. c) normal distribution. d) none of the above.

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

14. What type of probability distribution will most likely be used to analyze the number of chocolate chip parts per cookie in the following problem? .

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Discrete Probability Distributions The quality control manager of Marilyn’s Cookies is inspecting a batch of chocolate chip cookies. When the production process is in control, the average number of chocolate chip parts per cookie is 6.0. The manager is interested in analyzing the probability that any particular cookie being inspected has fewer than 5.0 chip parts. a) binomial distribution. b) Poisson distribution. c) normal distribution. d) none of the above.

ANSWER: b TYPE: MC DIFFICULTY: Moderate KEYWORDS: Poisson distribution, properties 15. A professor receives, on average, 24.7 e-mails from students the day before the midterm exam.

To compute the probability of receiving at least 10 e-mails on such a day, he will use what type of probability distribution? a) binomial distribution. b) Poisson distribution. c) normal distribution. d) none of the above.

ANSWER: b TYPE: MC DIFFICULTY: Moderate KEYWORDS: Poisson distribution, properties 16. A company has 125 personal computers. The probability that any one of them will require repair on a given day is 0.025. To find the probability that exactly 20 of the computers will require repair on a given day, one will use what type of probability distribution? a) binomial distribution. b) Poisson distribution. c) normal distribution. d) none of the above.

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

17. On the average, 1.8 customers per minute arrive at any one of the checkout counters of a grocery store. What type of probability distribution can be used to find out the probability that there will be no customer arriving at a checkout counter? a) binomial distribution. b) Poisson distribution. .

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c) normal distribution. d) none of the above.

ANSWER: b TYPE: MC DIFFICULTY: Moderate KEYWORDS: Poisson distribution, properties 18. A multiple-choice test has 30 questions. There are 4 choices for each question. A student who has not studied for the test decides to answer all questions randomly. What type of probability distribution can be used to figure out his chance of getting at least 20 questions right? a) binomial distribution. b) Poisson distribution. c) normal distribution. d) none of the above.

ANSWER: a TYPE: MC DIFFICULTY: Moderate KEYWORDS: binomial distribution, properties 19. A lab orders 100 rats a week for each of the 52 weeks in the year for experiments that the lab conducts. Suppose the mean cost of rats used in lab experiments turned out to be $13.00 per week. Interpret this value. a) Most of the weeks resulted in rat costs of $13.00. b) The median cost for the distribution of rat costs is $13.00. c) The expected or average cost for all weekly rat purchases is $13.00. d) The rat cost that occurs more often than any other is $13.00.

ANSWER: c TYPE: MC DIFFICULTY: Easy KEYWORDS: mean

20. A lab orders 100 rats a week for each of the 52 weeks in the year for experiments that the lab conducts. Prices for 100 rats follow the following distribution: Price: $10.00 $12.50 $15.00 Probability: 0.35 0.40 0.25 How much should the lab budget for next year’s rat orders be, assuming this distribution does not change? a) $520 .

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Discrete Probability Distributions b) $637 c) $650 d) $780 ANSWER: b TYPE: MC DIFFICULTY: Moderate KEYWORDS: mean, probability distribution 21. The local police department must write, on average, 5 tickets a day to keep department revenues at budgeted levels. Suppose the number of tickets written per day follows a Poisson distribution with a mean of 6.5 tickets per day. Interpret the value of the mean. a) The number of tickets that is written most often is 6.5 tickets per day. b) Half of the days have less than 6.5 tickets written and half of the days have more than 6.5 tickets written. c) If we sampled all days, the arithmetic average or expected number of tickets written would be 6.5 tickets per day. d) The mean has no interpretation since 0.5 ticket can never be written.

ANSWER: c TYPE: MC DIFFICULTY: Easy KEYWORDS: mean, Poisson distribution 22. True or False: The Poisson distribution can be used to model a continuous random variable.

ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: Poisson distribution, properties 23. True or False: Another name for the mean of a probability distribution is its expected value.

ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: mean

24. True or False: The number of customers arriving at a department store in a 5-minute period has a binomial distribution.

ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: Poisson distribution, properties

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25. True or False: The number of customers arriving at a department store in a 5-minute period has a Poisson distribution.

ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: Poisson distribution, properties 26. True or False: The number of males selected in a sample of 5 students taken without replacement from a class of 9 females and 18 males has a binomial distribution.

ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: binomial distribution, properties 27. True or False: The diameters of 10 randomly selected bolts have a binomial distribution.

ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: binomial distribution, properties 28. True or False: The largest value that a Poisson random variable X can have is n.

ANSWER: False TYPE: TF DIFFICULTY: Moderate KEYWORDS: Poisson distribution, properties 29. True or False: In a Poisson distribution, the mean and standard deviation are equal.

ANSWER: False TYPE: TF DIFFICULTY: Moderate KEYWORDS: Poisson distribution, properties

30. True or False: In a Poisson distribution, the mean and variance are equal.

ANSWER: True TYPE: TF DIFFICULTY: Moderate KEYWORDS: Poisson distribution, properties 31. True or False: If p remains constant in a binomial distribution, an increase in n will increase the variance. .

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ANSWER: True TYPE: TF DIFFICULTY: Moderate KEYWORDS: binomial distribution, properties 32. True or False: If p remains constant in a binomial distribution, an increase in n will not change the mean.

ANSWER: False TYPE: TF DIFFICULTY: Moderate KEYWORDS: binomial distribution, properties 33. True or False: Suppose that a judge’s decisions follow a binomial distribution and that his verdict is correct 90% of the time. In his next 10 decisions, the probability that he makes fewer than 2 incorrect verdicts is 0.736.

ANSWER: True TYPE: TF DIFFICULTY: Moderate KEYWORDS: binomial distribution, probability 34. True or False: Suppose that the number of airplanes arriving at an airport per minute is a Poisson process. The average number of airplanes arriving per minute is 3. The probability that exactly 6 planes arrive in the next minute is 0.0504.

ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: Poisson distribution, probability

TABLE 5-1 The probability that a particular type of smoke alarm will function properly and sound an alarm in the presence of smoke is 0.8. You have 2 such alarms in your home and they operate independently.

35. Referring to Table 5-1, the probability that both sound an alarm in the presence of smoke is ________.

ANSWER: 0.64 .

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TYPE: FI DIFFICULTY: Moderate KEYWORDS: binomial distribution 36. Referring to Table 5-1, the probability that neither sound an alarm in the presence of smoke is ________.

ANSWER: 0.04 TYPE: FI DIFFICULTY: Moderate KEYWORDS: binomial distribution 37. Referring to Table 5-1, the probability that at least one sounds an alarm in the presence of smoke is ________.

ANSWER: 0.96 TYPE: FI DIFFICULTY: Difficult KEYWORDS: binomial distribution TABLE 5-2 A certain type of new business succeeds 60% of the time. Suppose that 3 such businesses open (where they do not compete with each other, so it is reasonable to believe that their relative successes would be independent).

38. Referring to Table 5-2, the probability that all 3 businesses succeed is ________.

ANSWER: 0.216 TYPE: FI DIFFICULTY: Moderate KEYWORDS: binomial distribution 39. Referring to Table 5-2, the probability that all 3 businesses fail is ________.

ANSWER: 0.064 TYPE: FI DIFFICULTY: Moderate KEYWORDS: binomial distribution

40. Referring to Table 5-2, the probability that at least 1 business succeeds is ________.

ANSWER: 0.936 TYPE: FI DIFFICULTY: Moderate KEYWORDS: binomial distribution 41. Referring to Table 5-2, the probability that exactly 1 business succeeds is ________.

ANSWER: 0.288 .

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TYPE: FI DIFFICULTY: Moderate KEYWORDS: binomial distribution 42. If X has a binomial distribution with n = 4 and p = 0.3, then P(X = 1) = ________ .

ANSWER: 0.4116 TYPE: FI DIFFICULTY: Easy KEYWORDS: binomial distribution 43. If X has a binomial distribution with n = 4 and p = 0.3, then P(X > 1) = ________ .

ANSWER: 0.3483 TYPE: FI DIFFICULTY: Moderate KEYWORDS: binomial distribution 44. If X has a binomial distribution with n = 5 and p = 0.1, then P(X = 2) = ________ .

ANSWER: 0.0729 TYPE: FI DIFFICULTY: Easy KEYWORDS: binomial distribution 45. Suppose that past history shows that 60% of college students prefer Brand C cola. A sample of 5 students is to be selected. The probability that exactly 1 prefers brand C is ________.

ANSWER: 0.0768 TYPE: FI DIFFICULTY: Easy KEYWORDS: binomial distribution 46. Suppose that past history shows that 60% of college students prefer Brand C cola. A sample of 5 students is to be selected. The probability that at least 1 prefers brand C is ________.

ANSWER: 0.9898 TYPE: FI DIFFICULTY: Moderate KEYWORDS: binomial distribution 47. Suppose that past history shows that 60% of college students prefer Brand C cola. A sample of 5 students is to be selected. The probability that exactly 3 prefer brand C is ________.

ANSWER: 0.3456 TYPE: FI DIFFICULTY: Easy KEYWORDS: binomial distribution 48. Suppose that past history shows that 60% of college students prefer Brand C cola. A sample of 5 students is to be selected. The probability that exactly 4 prefer brand C is ________. .

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ANSWER: 0.2592 TYPE: FI DIFFICULTY: Easy KEYWORDS: binomial distribution 49. Suppose that past history shows that 60% of college students prefer Brand C cola. A sample of 5 students is to be selected. The probability that at most 2 prefer brand C is ________.

ANSWER: 0.3174 TYPE: FI DIFFICULTY: Moderate KEYWORDS: binomial distribution 50. Suppose that past history shows that 60% of college students prefer Brand C cola. A sample of 5 students is to be selected. The probability that more than 3 prefer brand C is ________.

ANSWER: 0.3370 TYPE: FI DIFFICULTY: Moderate KEYWORDS: binomial distribution 51. Suppose that past history shows that 60% of college students prefer Brand C cola. A sample of 5 students is to be selected. The probability that less than 2 prefer brand C is ________.

ANSWER: 0.0870 TYPE: FI DIFFICULTY: Moderate KEYWORDS: binomial distribution 52. Suppose that past history shows that 60% of college students prefer Brand C cola. A sample of 5 students is to be selected. The average number that you would expect to prefer brand C is ________.

ANSWER: 3 TYPE: FI DIFFICULTY: Easy KEYWORDS: binomial distribution, mean 53. Suppose that past history shows that 60% of college students prefer Brand C cola. A sample of 5 students is to be selected. The variance of the number that prefer brand C is ________.

ANSWER: 1.2 TYPE: FI DIFFICULTY: Easy KEYWORDS: binomial distribution, variance TABLE 5-3 The following table contains the probability distribution for X = the number of retransmissions necessary to successfully transmit a 1024K data package through a double satellite media. .

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Discrete Probability Distributions X P(X)

0 0.35

1 0.35

2 0.25

3 0.05

54. Referring to Table 5-3, the probability of no retransmissions is ________.

ANSWER: 0.35 TYPE: FI DIFFICULTY: Easy KEYWORDS: probability distribution 55. Referring to Table 5-3, the probability of at least one retransmission is ________.

ANSWER: 0.65 TYPE: FI DIFFICULTY: Easy KEYWORDS: probability distribution 56. Referring to Table 5-3, the mean or expected value for the number of retransmissions is ________.

ANSWER: 1.0 TYPE: FI DIFFICULTY: Easy KEYWORDS: probability distribution, mean 57. Referring to Table 5-3, the variance for the number of retransmissions is ________.

ANSWER: 0.80 TYPE: FI DIFFICULTY: Easy KEYWORDS: probability distribution, variance

58. Referring to Table 5-3, the standard deviation of the number of retransmissions is ________.

ANSWER: 0.894 TYPE: FI DIFFICULTY: Easy KEYWORDS: probability distribution, standard deviation 59. In a game called Taxation and Evasion, a player rolls a pair of dice. If on any turn the sum is 7, 11, or 12, the player gets audited. Otherwise, she avoids taxes. Suppose a player takes 5 turns at rolling the dice. The probability that she does not get audited is ________.

ANSWER: 0.2373 .

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TYPE: FI DIFFICULTY: Moderate KEYWORDS: binomial distribution 60. In a game called Taxation and Evasion, a player rolls a pair of dice. If on any turn the sum is 7, 11, or 12, the player gets audited. Otherwise, she avoids taxes. Suppose a player takes 5 turns at rolling the dice. The probability that she gets audited once is ________.

ANSWER: 0.3955 TYPE: FI DIFFICULTY: Moderate KEYWORDS: binomial distribution 61. In a game called Taxation and Evasion, a player rolls a pair of dice. If on any turn the sum is 7, 11, or 12, the player gets audited. Otherwise, she avoids taxes. Suppose a player takes 5 turns at rolling the dice. The probability that she gets audited at least once is ________.

ANSWER: 0.7627 TYPE: FI DIFFICULTY: Moderate KEYWORDS: binomial distribution 62. In a game called Taxation and Evasion, a player rolls a pair of dice. If on any turn the sum is 7, 11, or 12, the player gets audited. Otherwise, she avoids taxes. Suppose a player takes 5 turns at rolling the dice. The probability that she gets audited no more than 2 times is ________.

ANSWER: 0.8965 TYPE: FI DIFFICULTY: Moderate KEYWORDS: binomial distribution

63. In a game called Taxation and Evasion, a player rolls a pair of dice. If on any turn the sum is 7, 11, or 12, the player gets audited. Otherwise, she avoids taxes. Suppose a player takes 5 turns at rolling the dice. The expected number of times she will be audited is ________.

ANSWER: 1.25 TYPE: FI DIFFICULTY: Easy KEYWORDS: binomial distribution, mean 64. In a game called Taxation and Evasion, a player rolls a pair of dice. If on any turn the sum is 7, 11, or 12, the player gets audited. Otherwise, she avoids taxes. Suppose a player takes 5 turns at rolling the dice. The variance of the number of times she will be audited is ________.

ANSWER: 0.9375 TYPE: FI DIFFICULTY: Moderate .

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KEYWORDS: binomial distribution, variance 65. In a game called Taxation and Evasion, a player rolls a pair of dice. If on any turn the sum is 7, 11, or 12, the player gets audited. Otherwise, she avoids taxes. Suppose a player takes 5 turns at rolling the dice. The standard deviation of the number of times she will be audited is ________.

ANSWER: 0.968 TYPE: FI DIFFICULTY: Moderate KEYWORDS: binomial distribution, standard deviation TABLE 5-4 The following table contains the probability distribution for X = the number of traffic accidents reported in a day in Corvallis, Oregon.

X 0 1 2 3 4 5 P(X) 0.10 0.20 0.45 0.15 0.05 0.05 66. Referring to Table 5-4, the probability of 3 accidents is ________.

ANSWER: 0.15 TYPE: FI DIFFICULTY: Easy KEYWORDS: probability distribution 67. Referring to Table 5-4, the probability of at least 1 accident is ________.

ANSWER: 0.90 TYPE: FI DIFFICULTY: Easy KEYWORDS: probability distribution 68. Referring to Table 5-4, the mean or expected value of the number of accidents is ________.

ANSWER: 2.0 TYPE: FI DIFFICULTY: Easy KEYWORDS: probability distribution, mean, 69. Referring to Table 5-4, the variance of the number of accidents is ________.

ANSWER: 1.4 TYPE: FI DIFFICULTY: Easy KEYWORDS: probability distribution, variance 70. Referring to Table 5-4, the standard deviation of the number of accidents is ________.

.

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ANSWER: 1.18 TYPE: FI DIFFICULTY: Easy KEYWORDS: probability distribution, standard deviation 71. The number of power outages at a nuclear power plant has a Poisson distribution with a mean of 6 outages per year. The probability that there will be exactly 3 power outages in a year is ____________.

ANSWER: 0.0892 TYPE: FI DIFFICULTY: Easy KEYWORDS: Poisson distribution 72. The number of power outages at a nuclear power plant has a Poisson distribution with a mean of 6 outages per year. The probability that there will be at least 3 power outages in a year is ____________.

ANSWER: 0.9380 TYPE: FI DIFFICULTY: Difficult KEYWORDS: Poisson distribution 73. The number of power outages at a nuclear power plant has a Poisson distribution with a mean of 6 outages per year. The probability that there will be at least 1 power outage in a year is ____________.

ANSWER: 0.9975 TYPE: FI DIFFICULTY: Moderate KEYWORDS: Poisson distribution 74. The number of power outages at a nuclear power plant has a Poisson distribution with a mean of 6 outages per year. The probability that there will be no more than 1 power outage in a year is ____________.

ANSWER: 0.0174 TYPE: FI DIFFICULTY: Moderate KEYWORDS: Poisson distribution 75. The number of power outages at a nuclear power plant has a Poisson distribution with a mean of 6 outages per year. The probability that there will be between 1 and 3 inclusive power outages in a year is ____________.

ANSWER: 0.1487 TYPE: FI DIFFICULTY: Difficult KEYWORDS: Poisson distribution .

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Discrete Probability Distributions 76. The number of power outages at a nuclear power plant has a Poisson distribution with a mean of 6 outages per year. The variance of the number of power outages is ____________.

ANSWER: 6 TYPE: FI DIFFICULTY: Easy KEYWORDS: Poisson distribution 77. The number of 911 calls in Butte, Montana, has a Poisson distribution with a mean of 10 calls a day. The probability of seven 911 calls in a day is ____________.

ANSWER: 0.0901 TYPE: FI DIFFICULTY: Easy KEYWORDS: Poisson distribution 78. The number of 911 calls in Butte, Montana, has a Poisson distribution with a mean of 10 calls a day. The probability of seven or eight 911 calls in a day is ____________.

ANSWER: 0.2027 TYPE: FI DIFFICULTY: Moderate KEYWORDS: Poisson distribution 79. The number of 911 calls in Butte, Montana, has a Poisson distribution with a mean of 10 calls a day. The probability of 2 or more 911 calls in a day is ____________.

ANSWER: 0.9995 TYPE: FI DIFFICULTY: Moderate KEYWORDS: Poisson distribution 80. The number of 911 calls in Butte, Montana, has a Poisson distribution with a mean of 10.0 calls a day. The standard deviation of the number of 911 calls in a day is ____________ .

ANSWER: 3.16 TYPE: FI DIFFICULTY: Easy KEYWORDS: Poisson distribution 81. The Department of Commerce in a particular state has determined that the number of small businesses that declare bankruptcy per month is approximately a Poisson distribution with a mean of 6.4. Find the probability that more than 3 bankruptcies occur next month.

ANSWER: 0.881 TYPE: PR DIFFICULTY: Moderate KEYWORDS: Poisson distribution

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82. The Department of Commerce in a particular state has determined that the number of small businesses that declare bankruptcy per month is approximately a Poisson distribution with a mean of 6.4. Find the probability that exactly 5 bankruptcies occur next month.

ANSWER: 0.149 TYPE: PR DIFFICULTY: Easy KEYWORDS: Poisson distribution 83. The on-line access computer service industry is growing at an extraordinary rate. Current estimates suggest that only 20% of the home-based computers have access to on-line services.

This number is expected to grow quickly over the next 5 years. Suppose 20 people with homebased computers were randomly and independently sampled. Find the probability that fewer than 10 of those sampled currently have access to on-line services.

ANSWER: 0.9974 TYPE: PR DIFFICULTY: Moderate KEYWORDS: binomial distribution 84. The on-line access computer service industry is growing at an extraordinary rate. Current estimates suggest that only 20% of the home-based computers have access to on-line services.

This number is expected to grow quickly over the next 5 years. Suppose 20 people with homebased computers were randomly and independently sampled. Find the probability that more than 15 of those sampled currently do not have access to on-line services.

ANSWER: 0.6296 TYPE: PR DIFFICULTY: Difficult KEYWORDS: binomial distribution 85. A national trend predicts that women will account for half of all business travelers in the next 3 years. To attract these women business travelers, hotels are providing more amenities that women particularly like. A recent survey of American hotels found that 70% offer hairdryers in the bathrooms. Consider a random and independent sample of 20 hotels. Find the probability all of the hotels in the sample offered hairdryers in the bathrooms.

ANSWER: 0.0008 TYPE: PR DIFFICULTY: Moderate KEYWORDS: binomial distribution 86. A national trend predicts that women will account for half of all business travelers in the next 3 years. To attract these women business travelers, hotels are providing more amenities that women particularly like. A recent survey of American hotels found that 70% offer hairdryers in the bathrooms. Consider a random and independent sample of 20 hotels. Find the probability that more than 7 but less than 13 of the hotels in the sample offered hairdryers in the bathrooms.

ANSWER: 0.2264 (using PHStat) or 0.2265 (using Table E.60 .

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Discrete Probability Distributions TYPE: PR DIFFICULTY: Moderate KEYWORDS: binomial distribution 87. A national trend predicts that women will account for half of all business travelers in the next 3 years. To attract these women business travelers, hotels are providing more amenities that women particularly like. A recent survey of American hotels found that 70% offer hairdryers in the bathrooms. Consider a random and independent sample of 20 hotels. Find the probability that at least 9 of the hotels in the sample do not offer hairdryers in the bathrooms.

ANSWER: 0.1133 TYPE: PR DIFFICULTY: Difficult KEYWORDS: binomial distribution 88. The local police department must write, on average, 5 tickets a day to keep department revenues at budgeted levels. Suppose the number of tickets written per day follows a Poisson distribution with a mean of 6.4 tickets per day. Find the probability that less than 6 tickets are written on a randomly selected day from this population.

ANSWER: 0.384 TYPE: PR DIFFICULTY: Easy KEYWORDS: Poisson distribution

89. The local police department must write, on average, 5 tickets a day to keep department revenues at budgeted levels. Suppose the number of tickets written per day follows a Poisson distribution with a mean of 6.4 tickets per day. Find the probability that exactly 6 tickets are written on a randomly selected day from this population.

ANSWER: 0.159 TYPE: PR DIFFICULTY: Easy KEYWORDS: Poisson distribution TABLE 5-5 The U.S. Department of Transportation maintains statistics for mishandled bags per 1,000 airline passengers. In 2005, Jet Blue had 4.06 mishandled bags per 1,000 passengers. Assume that the number of mishandled bags has a Poisson distribution.

90. Referring to Table 5-5, what is the probability that in the next 1,000 passengers, Jet Blue will have no mishandled bags?

ANSWER: 0.0172 .

Discrete Probability Distributions TYPE: PR DIFFICULTY: Easy KEYWORDS: Poisson distribution 91. Referring to Table 5-5, what is the probability that in the next 1,000 passengers, Jet Blue will have at least one mishandled bags?

ANSWER: 0.9828 TYPE: PR DIFFICULTY: Easy KEYWORDS: Poisson distribution 92. Referring to Table 5-5, what is the probability that in the next 1,000 passengers, Jet Blue will have at least two mishandled bags?

ANSWER: 0.9127 TYPE: PR DIFFICULTY: Easy KEYWORDS: Poisson distribution 93. Referring to Table 5-5, what is the probability that in the next 1,000 passengers, Jet Blue will have no more than three mishandled bags?

ANSWER: 0.4218 TYPE: PR DIFFICULTY: Easy KEYWORDS: Poisson distribution

94. Referring to Table 5-5, what is the probability that in the next 1,000 passengers, Jet Blue will have no more than four mishandled bags?

ANSWER: 0.6171 TYPE: PR DIFFICULTY: Easy KEYWORDS: Poisson distribution 95. Referring to Table 5-5, what is the probability that in the next 1,000 passengers, Jet Blue will have fewer than six mishandled bags?

ANSWER: 0.7757 TYPE: PR DIFFICULTY: Easy KEYWORDS: Poisson distribution 96. Referring to Table 5-5, what is the probability that in the next 1,000 passengers, Jet Blue will have fewer than eight mishandled bags?

ANSWER: 0.9452 TYPE: PR DIFFICULTY: Easy .

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Discrete Probability Distributions KEYWORDS: Poisson distribution 97. Referring to Table 5-5, what is the probability that in the next 1,000 passengers, Jet Blue will have more than eight mishandled bags?

ANSWER: 0.02320 TYPE: PR DIFFICULTY: Easy KEYWORDS: Poisson distribution 98. Referring to Table 5-5, what is the probability that in the next 1,000 passengers, Jet Blue will have more than ten mishandled bags?

ANSWER: 0.003172 TYPE: PR DIFFICULTY: Easy KEYWORDS: Poisson distribution 99. Referring to Table 5-5, what is the probability that in the next 1,000 passengers, Jet Blue will have between two and four mishandled bags?

ANSWER: 0.5298 TYPE: PR DIFFICULTY: Moderate KEYWORDS: Poisson distribution 100. Referring to Table 5-5, what is the probability that in the next 1,000 passengers, Jet Blue will have more than five but less than eight mishandled bags?

ANSWER: 0.1695 TYPE: PR DIFFICULTY: Moderate KEYWORDS: Poisson distribution 101. Referring to Table 5-5, what is the probability that in the next 1,000 passengers, Jet Blue will have less than two or more than eight mishandled bags?

ANSWER: 0.1105 TYPE: PR DIFFICULTY: Moderate KEYWORDS: Poisson distribution 102. Referring to Table 5-5, what is the probability that in the next 1,000 passengers, Jet Blue will have no more than two or at least eight mishandled bags?

ANSWER: 0.2842 TYPE: PR DIFFICULTY: Moderate KEYWORDS: Poisson distribution .

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103. Referring to Table 5-5, what is the probability that in the next 1,000 passengers, Jet Blue will have less than two and more than eight mishandled bags?

ANSWER: 0 TYPE: PR DIFFICULTY: Difficult KEYWORDS: Poisson distribution 104. Referring to Table 5-5, what is the probability that in the next 1,000 passengers, Jet Blue will have no more than two and at least eight mishandled bags?

ANSWER: 0 TYPE: PR DIFFICULTY: Difficult KEYWORDS: Poisson distribution

TABLE 5-6 The quality control manager of Marilyn’s Cookies is inspecting a batch of chocolate chip cookies.

When the production process is in control, the average number of chocolate chip parts per cookie is 6.0.

105. Referring to Table 5-6, what is the probability that any particular cookie being inspected has 4.0 chip parts.

ANSWER: 0.1339 TYPE: PR DIFFICULTY: Easy KEYWORDS: Poisson distribution 106. Referring to Table 5-6, what is the probability that any particular cookie being inspected has fewer than 5.0 chip parts.

ANSWER: 0.2851 TYPE: PR DIFFICULTY: Moderate KEYWORDS: Poisson distribution 107. Referring to Table 5-6, what is the probability that any particular cookie being inspected has at least 6.0 chip parts. .

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Discrete Probability Distributions ANSWER: 0.5543 TYPE: PR DIFFICULTY: Moderate KEYWORDS: Poisson distribution 108. Referring to Table 5-6, what is the probability that any particular cookie being inspected has between 5.0 and 8.0 inclusive chip parts.

ANSWER: 0.5622 TYPE: PR DIFFICULTY: Moderate KEYWORDS: Poisson distribution 109. Referring to Table 5-6, what is the probability that any particular cookie being inspected has less than 5.0 or more than 8.0 chip parts.

ANSWER: 0.4378 TYPE: PR DIFFICULTY: Difficult KEYWORDS: Poisson distribution

TABLE 5-7 There are two houses with almost identical characteristics available for investment in two different neighborhoods with drastically different demographic composition. The anticipated gain in value when the houses are sold in 10 years has the following probability distribution:

Probability .25 .40 .35

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

110. Referring to Table 5-7, what is the expected value gain for the house in neighborhood A? ANSWER: $ 12,550 TYPE: PR DIFFICULTY: Easy KEYWORDS: mean 111. Referring to Table 5-7, what is the expected value gain for the house in neighborhood B? ANSWER: $ 21,300 TYPE: PR DIFFICULTY: Easy KEYWORDS: mean .

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112. Referring to Table 5-7, what is the variance of the gain in value for the house in neighborhood A? ANSWER: 583,147,500 TYPE: PR DIFFICULTY: Easy KEYWORDS: variance 113. Referring to Table 5-7, what is the variance of the gain in value for the house in neighborhood B? ANSWER: 67,460,000 TYPE: PR DIFFICULTY: Easy KEYWORDS: variance 114. Referring to Table 5-7, what is the standard deviation of the value gain for the house in neighborhood A? ANSWER: $24,148.45 TYPE: PR DIFFICULTY: Easy KEYWORDS: standard deviation 115. Referring to Table 5-7, what is the standard deviation of the value gain for the house in neighborhood B? ANSWER: $8,213.40 TYPE: PR DIFFICULTY: Easy KEYWORDS: standard deviation TABLE 5-8 Two different designs on a new line of winter jackets for the coming winter are available for your manufacturing plants. Your profit (in thousands of dollars) will depend on the taste of the consumers when winter arrives. The probability of the three possible different tastes of the consumers and the corresponding profits are presented in the following table.

Probability 0.2 0.5 0.3

Taste more conservative no change more liberal

Design A 180 230 350

Design B 520 310 270

116. Referring to Table 5-8, the table above is called the ______________ for the two designs.

ANSWER: probability distribution .

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Discrete Probability Distributions TYPE: FI DIFFICULTY: Easy KEYWORDS: probability distribution 117. Referring to Table 5-8, what is your expected profit when Design A is chosen?

ANSWER: $256 thousands or $256,000 TYPE: PR DIFFICULTY: Easy KEYWORDS: mean 118. Referring to Table 5-8, what is your expected profit when Design B is chosen?

ANSWER: $340 thousands or $340,000 TYPE: PR DIFFICULTY: Easy KEYWORDS: mean 119. Referring to Table 5-8, what is the variance of your profit when Design A is chosen?

ANSWER: 4,144 millions or 4,144,000,000 TYPE: PR DIFFICULTY: Moderate KEYWORDS: variance 120. Referring to Table 5-8, what is the variance of your profit when Design B is chosen?

ANSWER: 8,400 millions or 8,400,000,000 TYPE: PR DIFFICULTY: Moderate KEYWORDS: variance 121. Referring to Table 5-8, what is the standard deviation of your profit when Design A is chosen?

ANSWER: $64.37391 thousands or $64,373.91 TYPE: PR DIFFICULTY: Easy KEYWORDS: standard deviation 122. Referring to Table 5-8, what is the standard deviation of your profit when Design B is chosen?

ANSWER: $91.65151 thousands or $91,651.51 TYPE: PR DIFFICULTY: Easy KEYWORDS: standard deviation

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The Normal Distribution

CHAPTER 6: THE NORMAL DISTRIBUTION 1. In its standardized form, the normal distribution a) has a mean of 0 and a standard deviation of 1. b) has a mean of 1 and a variance of 0. c) has an area equal to 0.5. d) cannot be used to approximate discrete probability distributions.

ANSWER: a TYPE: MC DIFFICULTY: Easy KEYWORDS: standardized normal distribution, properties 2. Which of the following about the normal distribution is not true? a) Theoretically, the mean, median, and mode are the same. b) About 2/3 of the observations fall within  1 standard deviation from the mean. c) It is a discrete probability distribution. d) Its parameters are the mean,  , and standard deviation,  .

ANSWER: c TYPE: MC DIFFICULTY: Easy KEYWORDS: normal distribution, properties 3. If a particular batch of data is approximately normally distributed, we would find that approximately a) 2 of every 3 observations would fall between  1 standard deviation around the mean. b) 4 of every 5 observations would fall between  1.28 standard deviations around the mean. c) 19 of every 20 observations would fall between  2 standard deviations around the mean. d) All the above.

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

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The Normal Distribution 163 5. For some value of Z, the probability that a standard normal variable is below Z is 0.2090. The value of Z is a) – 0.81 b) – 0.31 c) 0.31 d) 1.96 ANSWER: a TYPE: MC DIFFICULTY: Moderate KEYWORDS: standardized normal distribution, value 6. For some positive value of Z, the probability that a standard normal variable is between 0 and Z is 0.3340. The value of Z is a) 0.07 b) 0.37 c) 0.97 d) 1.06 ANSWER: c TYPE: MC DIFFICULTY: Moderate KEYWORDS: standardized normal distribution, value 7. For some positive value of X, the probability that a standard normal variable is between 0 and +2X is 0.1255. The value of X is a) 0.99 b) 0.40 c) 0.32 d) 0.16 ANSWER: d TYPE: MC DIFFICULTY: Difficult KEYWORDS: normal distribution, value 8. For some positive value of X, the probability that a standard normal variable is between 0 and +1.5X is 0.4332. The value of X is a) 0.10 b) 0.50 c) 1.00 d) 1.50 ANSWER: c TYPE: MC DIFFICULTY: Difficult KEYWORDS: normal distribution, value

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The Normal Distribution 9. Given that X is a normally distributed random variable with a mean of 50 and a standard deviation of 2, find the probability that X is between 47 and 54.

ANSWER: 0.9104 TYPE: PR DIFFICULTY: Easy KEYWORDS: normal distribution, probability 10. A company that sells annuities must base the annual payout on the probability distribution of the length of life of the participants in the plan. Suppose the probability distribution of the lifetimes of the participants is approximately a normal distribution with a mean of 68 years and a standard deviation of 3.5 years. What proportion of the plan recipients would receive payments beyond age 75?

ANSWER: 0.0228 TYPE: PR DIFFICULTY: Easy KEYWORDS: normal distribution, probability 11. A company that sells annuities must base the annual payout on the probability distribution of the length of life of the participants in the plan. Suppose the probability distribution of the lifetimes of the participants is approximately a normal distribution with a mean of 68 years and a standard deviation of 3.5 years. What proportion of the plan recipients die before they reach the standard retirement age of 65?

ANSWER: 0.1957 using Excel or 0.1949 using Table E.2 TYPE: PR DIFFICULTY: Moderate KEYWORDS: normal distribution, probability 12. A company that sells annuities must base the annual payout on the probability distribution of the length of life of the participants in the plan. Suppose the probability distribution of the lifetimes of the participants is approximately a normal distribution with a mean of 68 years and a standard deviation of 3.5 years. Find the age at which payments have ceased for approximately 86% of the plan participants.

ANSWER: 71.78 years old TYPE: PR DIFFICULTY: Difficult KEYWORDS: normal distribution, value

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13. If we know that the length of time it takes a college student to find a parking spot in the library parking lot follows a normal distribution with a mean of 3.5 minutes and a standard deviation of 1 minute, find the probability that a randomly selected college student will find a parking spot in the library parking lot in less than 3 minutes. a) 0.3551 b) 0.3085 c) 0.2674 d) 0.1915 ANSWER: b TYPE: MC DIFFICULTY: Easy KEYWORDS: normal distribution, probability 14. If we know that the length of time it takes a college student to find a parking spot in the library parking lot follows a normal distribution with a mean of 3.5 minutes and a standard deviation of 1 minute, find the probability that a randomly selected college student will take between 2 and 4.5 minutes to find a parking spot in the library parking lot. a) 0.0919 b) 0.2255 c) 0.4938 d) 0.7745 ANSWER: d TYPE: MC DIFFICULTY: Easy KEYWORDS: normal distribution, probability 15. If we know that the length of time it takes a college student to find a parking spot in the library parking lot follows a normal distribution with a mean of 3.5 minutes and a standard deviation of 1 minute, find the point in the distribution in which 75.8% of the college students exceed when trying to find a parking spot in the library parking lot. a) 2.8 minutes b) 3.2 minutes c) 3.4 minutes d) 4.2 minutes ANSWER: a TYPE: MC DIFFICULTY: Moderate KEYWORDS: normal distribution, value 16. The owner of a fish market determined that the average weight for a catfish is 3.2 pounds with a standard deviation of 0.8 pound. Assuming the weights of catfish are normally distributed, the probability that a randomly selected catfish will weigh more than 4.4 pounds is _______?

ANSWER: 0.0668 .

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The Normal Distribution TYPE: FI DIFFICULTY: Easy KEYWORDS: normal distribution, probability 17. The owner of a fish market determined that the average weight for a catfish is 3.2 pounds with a standard deviation of 0.8 pound. Assuming the weights of catfish are normally distributed, the probability that a randomly selected catfish will weigh between 3 and 5 pounds is _______?

ANSWER: 0.5865 TYPE: FI DIFFICULTY: Easy KEYWORDS: normal distribution, probability 18. The owner of a fish market determined that the average weight for a catfish is 3.2 pounds with a standard deviation of 0.8 pound. A citation catfish should be one of the top 2% in weight.

Assuming the weights of catfish are normally distributed, at what weight (in pounds) should the citation designation be established? a) 1.56 pounds b) 4.84 pounds c) 5.20 pounds d) 7.36 pounds ANSWER: b TYPE: MC DIFFICULTY: Moderate KEYWORDS: normal distribution, value 19. The owner of a fish market determined that the average weight for a catfish is 3.2 pounds with a standard deviation of 0.8 pound. Assuming the weights of catfish are normally distributed, above what weight (in pounds) do 89.80% of the weights occur?

ANSWER: 2.184 pounds TYPE: PR DIFFICULTY: Moderate KEYWORDS: normal distribution, value 20. The owner of a fish market determined that the average weight for a catfish is 3.2 pounds with a standard deviation of 0.8 pound. Assuming the weights of catfish are normally distributed, the probability that a randomly selected catfish will weigh less than 2.2 pounds is _______?

ANSWER: 0.1056 TYPE: FI DIFFICULTY: Easy KEYWORDS: normal distribution, probability 21. A food processor packages orange juice in small jars. The weights of the filled jars are approximately normally distributed with a mean of 10.5 ounces and a standard deviation of 0.3 ounce. Find the proportion of all jars packaged by this process that have weights that fall below 10.875 ounces.

ANSWER: 0.8944 .

The Normal Distribution 167 TYPE: PR DIFFICULTY: Easy KEYWORDS: normal distribution, probability 22. A food processor packages orange juice in small jars. The weights of the filled jars are approximately normally distributed with a mean of 10.5 ounces and a standard deviation of 0.3 ounce. Find the proportion of all jars packaged by this process that have weights that fall above 10.95 ounces.

ANSWER: 0.0668 TYPE: PR DIFFICULTY: Easy KEYWORDS: normal distribution, probability 23. True or False: The probability that a standard normal random variable, Z, falls between – 1.50 and 0.81 is 0.7242.

ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: standardized normal distribution, probability 24. True or False: The probability that a standard normal random variable, Z, is between 1.50 and 2.10 is the same as the probability Z is between – 2.10 and – 1.50.

ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: standardized normal distribution, probability 25. True or False: The probability that a standard normal random variable, Z, is below 1.96 is 0.4750.

ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: standardized normal distribution, probability 26. True or False: The probability that a standard normal random variable, Z, is between 1.00 and 3.00 is 0.1574.

ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: standardized normal distribution, probability 27. True or False: The probability that a standard normal random variable, Z, falls between –2.00 and –0.44 is 0.6472.

ANSWER: False TYPE: TF DIFFICULTY: Easy .

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The Normal Distribution KEYWORDS: standardized normal distribution, probability 28. True or False: The probability that a standard normal random variable, Z, is less than 5.0 is approximately 0.

ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: standardized normal distribution, probability 29. True or False: A worker earns $15 per hour at a plant and is told that only 2.5% of all workers make a higher wage. If the wage is assumed to be normally distributed and the standard deviation of wage rates is $5 per hour, the average wage for the plant is $7.50 per hour.

ANSWER: False TYPE: TF DIFFICULTY: Moderate KEYWORDS: normal distribution, mean 30. True or False: Theoretically, the mean, median, and the mode are all equal for a normal distribution.

ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: normal distribution, properties 31. True or False: Any set of normally distributed data can be transformed to its standardized form.

ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: normal distribution, properties 32. True or False: The "middle spread," that is the middle 50% of the normal distribution, is equal to one standard deviation.

ANSWER: False TYPE: TF DIFFICULTY: Moderate KEYWORDS: normal distribution, probability, value 33. True or False: A normal probability plot may be used to assess the assumption of normality for a particular batch of data.

ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: normal probability plot

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The Normal Distribution 169

34. True or False: If a data batch is approximately normally distributed, its normal probability plot would be S-shaped.

ANSWER: False TYPE: TF DIFFICULTY: Moderate KEYWORDS: normal probability plot 35. The probability that a standard normal variable Z is positive is ________.

ANSWER: 0.50 TYPE: FI DIFFICULTY: Easy KEYWORDS: standardized normal distribution 36. The amount of pyridoxine (in grams) in a multiple vitamin is normally distributed with  = 110 grams and  = 25 grams. What is the probability that a randomly selected vitamin will contain between 100 and 110 grams of pyridoxine?

ANSWER: 0.1554 TYPE: PR DIFFICULTY: Easy KEYWORDS: normal distribution, probability 37. The amount of pyridoxine (in grams) in a multiple vitamin is normally distributed with  = 110 grams and  = 25 grams. What is the probability that a randomly selected vitamin will contain between 82 and 100 grams of pyridoxine?

ANSWER: 0.2132 TYPE: PR DIFFICULTY: Easy KEYWORDS: normal distribution, probability 38. The amount of pyridoxine (in grams) in a multiple vitamin is normally distributed with  = 110 grams and  = 25 grams. What is the probability that a randomly selected vitamin will contain at least 100 grams of pyridoxine?

ANSWER: 0.6554 TYPE: PR DIFFICULTY: Easy KEYWORDS: normal distribution, probability 39. The amount of pyridoxine (in grams) in a multiple vitamin is normally distributed with  = 110 grams and  = 25 grams. What is the probability that a randomly selected vitamin will contain between 100 and 120 grams of pyridoxine?

ANSWER: 0.3108 TYPE: PR DIFFICULTY: Moderate .

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The Normal Distribution KEYWORDS: normal distribution, probability 40. The amount of pyridoxine (in grams) in a multiple vitamin is normally distributed with  = 110 grams and  = 25 grams. What is the probability that a randomly selected vitamin will contain less than 100 grams of pyridoxine?

ANSWER: 0.3446 TYPE: PR DIFFICULTY: Easy KEYWORDS: normal distribution, probability 41. The amount of pyridoxine (in grams) in a multiple vitamin is normally distributed with  = 110 grams and  = 25 grams. What is the probability that a randomly selected vitamin will contain less than 100 grams or more than 120 grams of pyridoxine?

ANSWER: 0.6892 TYPE: PR DIFFICULTY: Easy KEYWORDS: normal distribution, probability 42. The amount of pyridoxine (in grams) in a multiple vitamin is normally distributed with  = 110 grams and  = 25 grams. Approximately 83% of the vitamins will have at least how many grams of pyridoxine?

ANSWER: 86.15 using Excel or 86.25 using Table E.2 TYPE: PR DIFFICULTY: Moderate KEYWORDS: normal distribution, value 43. The true length of boards cut at a mill with a listed length of 10 feet is normally distributed with a mean of 123 inches and a standard deviation of 1 inch. What proportion of the boards will be between 121 and 124 inches?

ANSWER: 0.8186 using Excel or 0.8185 using Table E.2 TYPE: PR DIFFICULTY: Easy KEYWORDS: normal distribution, probability 44. The true length of boards cut at a mill with a listed length of 10 feet is normally distributed with a mean of 123 inches and a standard deviation of 1 inch. What proportion of the boards will be over 125 inches in length?

ANSWER: 0.0228 TYPE: PR DIFFICULTY: Easy KEYWORDS: normal distribution, probability

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The Normal Distribution 171 45. The true length of boards cut at a mill with a listed length of 10 feet is normally distributed with a mean of 123 inches and a standard deviation of 1 inch. What proportion of the boards will be less than 124 inches?

ANSWER: 0.8413 TYPE: PR DIFFICULTY: Easy KEYWORDS: normal distribution, probability 46. You were told that the amount of time lapsed between consecutive trades on the New York Stock Exchange followed a normal distribution with a mean of 15 seconds. You were also told that the probability that the time lapsed between two consecutive trades to fall between 16 to 17 seconds was 13%. The probability that the time lapsed between two consecutive trades would fall below 13 seconds was 7%. What is the probability that the time lapsed between two consecutive trades will be longer than 17 seconds?

ANSWER: 7% or 0.07 TYPE: PR DIFFICULTY: Easy KEYWORDS: normal distribution, probability 47. You were told that the amount of time lapsed between consecutive trades on the New York Stock Exchange followed a normal distribution with a mean of 15 seconds. You were also told that the probability that the time lapsed between two consecutive trades to fall between 16 to 17 seconds was 13%. The probability that the time lapsed between two consecutive trades would fall below 13 seconds was 7%. What is the probability that the time lapsed between two consecutive trades will be between 13 and 14 seconds?

ANSWER: 13% or 0.13 TYPE: PR DIFFICULTY: Easy KEYWORDS: normal distribution, probability 48. You were told that the amount of time lapsed between consecutive trades on the New York Stock Exchange followed a normal distribution with a mean of 15 seconds. You were also told that the probability that the time lapsed between two consecutive trades to fall between 16 to 17 seconds was 13%. The probability that the time lapsed between two consecutive trades would fall below 13 seconds was 7%. What is the probability that the time lapsed between two consecutive trades will be between 15 and 16 seconds?

ANSWER: 30% or 0.30 TYPE: PR DIFFICULTY: Easy KEYWORDS: normal distribution, probability

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The Normal Distribution 49. You were told that the amount of time lapsed between consecutive trades on the New York Stock Exchange followed a normal distribution with a mean of 15 seconds. You were also told that the probability that the time lapsed between two consecutive trades to fall between 16 to 17 seconds was 13%. The probability that the time lapsed between two consecutive trades would fall below 13 seconds was 7%. What is the probability that the time lapsed between two consecutive trades will be between 14 and 15 seconds?

ANSWER: 30% or 0.30 TYPE: PR DIFFICULTY: Easy KEYWORDS: normal distribution, probability 50. You were told that the amount of time lapsed between consecutive trades on the New York Stock Exchange followed a normal distribution with a mean of 15 seconds. You were also told that the probability that the time lapsed between two consecutive trades to fall between 16 to 17 seconds was 13%. The probability that the time lapsed between two consecutive trades would fall below 13 seconds was 7%. What is the probability that the time lapsed between two consecutive trades will be between 13 and 16 seconds?

ANSWER: 73% or 0.73 TYPE: PR DIFFICULTY: Moderate KEYWORDS: normal distribution, probability 51. You were told that the amount of time lapsed between consecutive trades on the New York Stock Exchange followed a normal distribution with a mean of 15 seconds. You were also told that the probability that the time lapsed between two consecutive trades to fall between 16 to 17 seconds was 13%. The probability that the time lapsed between two consecutive trades would fall below 13 seconds was 7%. What is the probability that the time lapsed between two consecutive trades will be between 14 and 17 seconds?

ANSWER: 73% or 0.73 TYPE: PR DIFFICULTY: Moderate KEYWORDS: normal distribution, probability 52. You were told that the amount of time lapsed between consecutive trades on the New York Stock Exchange followed a normal distribution with a mean of 15 seconds. You were also told that the probability that the time lapsed between two consecutive trades to fall between 16 to 17 seconds was 13%. The probability that the time lapsed between two consecutive trades would fall below 13 seconds was 7%. The probability is 20% that the time lapsed will be shorter how many seconds?

ANSWER: 14 seconds TYPE: PR DIFFICULTY: Easy KEYWORDS: normal distribution, probability

.

The Normal Distribution 173 53. You were told that the amount of time lapsed between consecutive trades on the New York Stock Exchange followed a normal distribution with a mean of 15 seconds. You were also told that the probability that the time lapsed between two consecutive trades to fall between 16 to 17 seconds was 13%. The probability that the time lapsed between two consecutive trades would fall below 13 seconds was 7%. The probability is 80% that the time lapsed will be longer than how many seconds?

ANSWER: 14 seconds TYPE: PR DIFFICULTY: Easy KEYWORDS: normal distribution, probability 54. You were told that the amount of time lapsed between consecutive trades on the New York Stock Exchange followed a normal distribution with a mean of 15 seconds. You were also told that the probability that the time lapsed between two consecutive trades to fall between 16 to 17 seconds was 13%. The probability that the time lapsed between two consecutive trades would fall below 13 seconds was 7%. The middle 60% of the time lapsed will fall between which two numbers?

ANSWER: 14 seconds and 16 seconds TYPE: PR DIFFICULTY: Moderate KEYWORDS: normal distribution, probability 55. You were told that the amount of time lapsed between consecutive trades on the New York Stock Exchange followed a normal distribution with a mean of 15 seconds. You were also told that the probability that the time lapsed between two consecutive trades to fall between 16 to 17 seconds was 13%. The probability that the time lapsed between two consecutive trades would fall below 13 seconds was 7%. The middle 86% of the time lapsed will fall between which two numbers?

ANSWER: 13 seconds and 17 seconds TYPE: PR DIFFICULTY: Moderate KEYWORDS: normal distribution, probability 56. You were told that the mean score on a statistics exam is 75 with the scores normally distributed.

In addition, you know the probability of a score between 55 and 60 is 4.41% and that the probability of a score greater than 90 is 6.68%. What is the probability of a score between 90 and 95?

ANSWER: 4.41% or 0.0441 TYPE: PR DIFFICULTY: Easy KEYWORDS: normal distribution, probability

.

174

The Normal Distribution 57. You were told that the mean score on a statistics exam is 75 with the scores normally distributed.

In addition, you know the probability of a score between 55 and 60 is 4.41% and that the probability of a score greater than 90 is 6.68%. What is the probability of a score greater than 95?

ANSWER: 2.27% or 0.0227 TYPE: PR DIFFICULTY: Moderate KEYWORDS: normal distribution, probability 58. You were told that the mean score on a statistics exam is 75 with the scores normally distributed.

In addition, you know the probability of a score between 55 and 60 is 4.41% and that the probability of a score greater than 90 is 6.68%. What is the probability of a score lower than 55?

ANSWER: 2.27% or 0.0227 TYPE: PR DIFFICULTY: Easy KEYWORDS: normal distribution, probability 59. You were told that the mean score on a statistics exam is 75 with the scores normally distributed.

In addition, you know the probability of a score between 55 and 60 is 4.41% and that the probability of a score greater than 90 is 6.68%. What is the probability of a score between 75 and 90?

ANSWER: 43.32% or 0.4332 TYPE: PR DIFFICULTY: Easy KEYWORDS: normal distribution, probability 60. You were told that the mean score on a statistics exam is 75 with the scores normally distributed.

In addition, you know the probability of a score between 55 and 60 is 4.41% and that the probability of a score greater than 90 is 6.68%. What is the probability of a score between 60 and 75?

ANSWER: 43.32% or 0.4332 TYPE: PR DIFFICULTY: Easy KEYWORDS: normal distribution, probability 61. You were told that the mean score on a statistics exam is 75 with the scores normally distributed.

In addition, you know the probability of a score between 55 and 60 is 4.41% and that the probability of a score greater than 90 is 6.68%. What is the probability of a score between 60 and 95?

ANSWER: 91.05% or 0.9105 TYPE: PR DIFFICULTY: Moderate KEYWORDS: normal distribution, probability

.

The Normal Distribution 175 62. You were told that the mean score on a statistics exam is 75 with the scores normally distributed.

In addition, you know the probability of a score between 55 and 60 is 4.41% and that the probability of a score greater than 90 is 6.68%. What is the probability of a score between 55 and 90?

ANSWER: 91.05% or 0.9105 TYPE: PR DIFFICULTY: Moderate KEYWORDS: normal distribution, probability 63. You were told that the mean score on a statistics exam is 75 with the scores normally distributed.

In addition, you know the probability of a score between 55 and 60 is 4.41% and that the probability of a score greater than 90 is 6.68%. What is the probability of a score between 55 and 95?

ANSWER: 95.46% or 0.9546 TYPE: PR DIFFICULTY: Moderate KEYWORDS: normal distribution, probability 64. You were told that the mean score on a statistics exam is 75 with the scores normally distributed.

In addition, you know the probability of a score between 55 and 60 is 4.41% and that the probability of a score greater than 90 is 6.68%. The middle 86.64% of the students will score between which two scores?

ANSWER: 60 and 90 TYPE: PR DIFFICULTY: Moderate KEYWORDS: normal distribution, probability 65. You were told that the mean score on a statistics exam is 75 with the scores normally distributed.

In addition, you know the probability of a score between 55 and 60 is 4.41% and that the probability of a score greater than 90 is 6.68%. The middle 95.46% of the students will score between which two scores?

ANSWER: 55 and 95 TYPE: PR DIFFICULTY: Moderate KEYWORDS: normal distribution, probability 66. Suppose Z has a standard normal distribution with a mean of 0 and standard deviation of 1. The probability that Z is less than 1.15 is __________.

ANSWER: 0.8749 TYPE: FI DIFFICULTY: Easy KEYWORDS: standardized normal distribution, probability

.

176

The Normal Distribution 67. Suppose Z has a standard normal distribution with a mean of 0 and standard deviation of 1. The probability that Z is more than 0.77 is __________.

ANSWER: 0.2206 TYPE: FI DIFFICULTY: Easy KEYWORDS: standardized normal distribution, probability 68. Suppose Z has a standard normal distribution with a mean of 0 and standard deviation of 1. The probability that Z is less than -2.20 is __________.

ANSWER: 0.0139 TYPE: FI DIFFICULTY: Easy KEYWORDS: standardized normal distribution, probability 69. Suppose Z has a standard normal distribution with a mean of 0 and standard deviation of 1. The probability that Z is more than -0.98 is __________.

ANSWER: 0.8365 TYPE: FI DIFFICULTY: Easy KEYWORDS: standardized normal distribution, probability 70. Suppose Z has a standard normal distribution with a mean of 0 and standard deviation of 1. The probability that Z is between -2.33 and 2.33 is __________.

ANSWER: 0.9802 TYPE: FI DIFFICULTY: Easy KEYWORDS: standardized normal distribution, probability 71. Suppose Z has a standard normal distribution with a mean of 0 and standard deviation of 1. The probability that Z is between -2.89 and -1.03 is __________.

ANSWER: 0.1496 TYPE: FI DIFFICULTY: Easy KEYWORDS: standardized normal distribution, probability 72. Suppose Z has a standard normal distribution with a mean of 0 and standard deviation of 1. The probability that Z is between -0.88 and 2.29 is __________.

ANSWER: 0.7996 TYPE: FI DIFFICULTY: Easy KEYWORDS: standardized normal distribution, probability

.

The Normal Distribution 177 73. Suppose Z has a standard normal distribution with a mean of 0 and standard deviation of 1. The probability that Z values are larger than __________ is 0.3483.

ANSWER: 0.39 TYPE: FI DIFFICULTY: Moderate KEYWORDS: standardized normal distribution, probability 74. Suppose Z has a standard normal distribution with a mean of 0 and standard deviation of 1. The probability that Z values are larger than __________ is 0.6985.

ANSWER: -0.52 TYPE: FI DIFFICULTY: Moderate KEYWORDS: standardized normal distribution, probability 75. Suppose Z has a standard normal distribution with a mean of 0 and standard deviation of 1. So 27% of the possible Z values are smaller than __________.

ANSWER: -0.61 TYPE: FI DIFFICULTY: Moderate KEYWORDS: standardized normal distribution, value 76. Suppose Z has a standard normal distribution with a mean of 0 and standard deviation of 1. So 85% of the possible Z values are smaller than __________.

ANSWER: 1.04 TYPE: FI DIFFICULTY: Moderate KEYWORDS: standardized normal distribution, value 77. Suppose Z has a standard normal distribution with a mean of 0 and standard deviation of 1. So 96% of the possible Z values are between __________ and __________ (symmetrically distributed about the mean).

ANSWER: -2.05 and 2.05 or -2.06 and 2.06 TYPE: FI DIFFICULTY: Moderate KEYWORDS: standardized normal distribution, value 78. Suppose Z has a standard normal distribution with a mean of 0 and standard deviation of 1. So 50% of the possible Z values are between __________ and __________ (symmetrically distributed about the mean).

ANSWER: -0.67 and 0.67 or -0.68 and 0.68 TYPE: FI DIFFICULTY: Moderate KEYWORDS: standardized normal distribution, value

.

178

The Normal Distribution 79. The amount of time necessary for assembly line workers to complete a product is a normal random variable with a mean of 15 minutes and a standard deviation of 2 minutes. The probability is __________ that a product is assembled in less than 12 minutes.

ANSWER: 0.0668 TYPE: FI DIFFICULTY: Easy KEYWORDS: normal distribution, probability 80. The amount of time necessary for assembly line workers to complete a product is a normal random variable with a mean of 15 minutes and a standard deviation of 2 minutes. The probability is __________ that a product is assembled in between 14 and 16 minutes.

ANSWER: 0.3829 using Excel or 0.3830 using Table E.2 TYPE: FI DIFFICULTY: Easy KEYWORDS: normal distribution, probability 81. The amount of time necessary for assembly line workers to complete a product is a normal random variable with a mean of 15 minutes and a standard deviation of 2 minutes. The probability is __________ that a product is assembled in between 10 and 12 minutes.

ANSWER: 0.0606 TYPE: FI DIFFICULTY: Easy KEYWORDS: normal distribution, probability 82. The amount of time necessary for assembly line workers to complete a product is a normal random variable with a mean of 15 minutes and a standard deviation of 2 minutes. The probability is __________ that a product is assembled in between 15 and 21 minutes.

ANSWER: 0.49865 TYPE: FI DIFFICULTY: Easy KEYWORDS: normal distribution, probability 83. The amount of time necessary for assembly line workers to complete a product is a normal random variable with a mean of 15 minutes and a standard deviation of 2 minutes. The probability is __________ that a product is assembled in between 16 and 21 minutes.

ANSWER: 0.30719 TYPE: FI DIFFICULTY: Easy KEYWORDS: normal distribution, probability

.

The Normal Distribution 179 84. The amount of time necessary for assembly line workers to complete a product is a normal random variable with a mean of 15 minutes and a standard deviation of 2 minutes. The probability is __________ that a product is assembled in more than 11 minutes.

ANSWER: 0.9772 TYPE: FI DIFFICULTY: Easy KEYWORDS: normal distribution, probability 85. The amount of time necessary for assembly line workers to complete a product is a normal random variable with a mean of 15 minutes and a standard deviation of 2 minutes. The probability is __________ that a product is assembled in more than 19 minutes.

ANSWER: 0.0228 TYPE: FI DIFFICULTY: Easy KEYWORDS: normal distribution, probability 86. The amount of time necessary for assembly line workers to complete a product is a normal random variable with a mean of 15 minutes and a standard deviation of 2 minutes. The probability is __________ that a product is assembled in less than 20 minutes.

ANSWER: 0.9938 TYPE: FI DIFFICULTY: Easy KEYWORDS: normal distribution, probability 87. The amount of time necessary for assembly line workers to complete a product is a normal random variable with a mean of 15 minutes and a standard deviation of 2 minutes. So, 15% of the products require more than __________ minutes for assembly.

ANSWER: 17.0729 using Excel or 17.08 using Table E.2 TYPE: FI DIFFICULTY: Moderate KEYWORDS: normal distribution, value 88. The amount of time necessary for assembly line workers to complete a product is a normal random variable with a mean of 15 minutes and a standard deviation of 2 minutes. So, 90% of the products require more than __________ minutes for assembly.

ANSWER: 12.44 TYPE: FI DIFFICULTY: Moderate KEYWORDS: normal distribution, value

.

180

The Normal Distribution 89. The amount of time necessary for assembly line workers to complete a product is a normal random variable with a mean of 15 minutes and a standard deviation of 2 minutes. So, 60% of the products would be assembled within __________ and __________ minutes (symmetrically distributed about the mean).

ANSWER: 13.32 and 16.68 or 13.31 and 16.69 TYPE: FI DIFFICULTY: Difficult KEYWORDS: normal distribution, value 90. The amount of time necessary for assembly line workers to complete a product is a normal random variable with a mean of 15 minutes and a standard deviation of 2 minutes. So, 17% of the products would be assembled within __________ minutes.

ANSWER: 13.1 TYPE: FI DIFFICULTY: Moderate KEYWORDS: normal distribution, value 91. The amount of time necessary for assembly line workers to complete a product is a normal random variable with a mean of 15 minutes and a standard deviation of 2 minutes. So, 70% of the products would be assembled within __________ minutes.

ANSWER: 16.0488 using Excel or 16.04 using Table E.2 TYPE: FI DIFFICULTY: Moderate KEYWORDS: normal distribution, value

TABLE 6-1 The manager of a surveying company believes that the average number of phone surveys completed per hour by her employees has a normal distribution. She takes a sample of 15 days output from her employees and determines the average number of surveys per hour on these days. The ordered array for this data is: 10.0, 10.1, 10.3, 10.5, 10.7, 11.2, 11.4, 11.5, 11.7, 11.8, 11.8, 12.0, 12.2, 12.2, 12.5.

92. Referring to Table 6-1, the first standard normal quantile is ________.

ANSWER: -1.5341 TYPE: FI DIFFICULTY: Moderate KEYWORDS: standard normal quantile, normal probability plot 93. Referring to Table 6-1, the fourth standard normal quantile is ________.

ANSWER: -0.6745 TYPE: FI DIFFICULTY: Moderate KEYWORDS: standard normal quantile, normal probability plot

.

The Normal Distribution 181 94. Referring to Table 6-1, the ninth standard normal quantile is ________.

ANSWER: +0.1573 TYPE: FI DIFFICULTY: Moderate KEYWORDS: standard normal quantile, normal probability plot 95. Referring to Table 6-1, the fourteenth standard normal quantile is ________.

ANSWER: +1.1503 TYPE: FI DIFFICULTY: Moderate KEYWORDS: standard normal quantile, normal probability plot 96. Referring to Table 6-1, the last standard normal quantile is ________.

ANSWER: +1.5341 TYPE: FI DIFFICULTY: Moderate KEYWORDS: standard normal quantile, normal probability plot 97. Referring to Table 6-1, construct a normal probability plot for the data.

ANSWER: 13

Surveys

12

11

10 -2

-1.5

-1

-0.5

0

0.5

1

1.5

2

Z -Score

TYPE: PR DIFFICULTY: Moderate KEYWORDS: normal probability plot 98. True or False: Referring to Table 6-1, the data appear reasonably normal but not perfectly normal.

ANSWER: True TYPE: TF DIFFICULTY: Moderate KEYWORDS: normal probability plot .

182

The Normal Distribution TABLE 6-2 The city manager of a large city believes that the number of reported accidents on any weekend has a normal distribution. She takes a sample of nine weekends and determines the number of reported accidents during each. The ordered array for this data is: 15, 46, 53, 54, 55, 76, 82, 256, 407.

99. Referring to Table 6-2, the first standard normal quantile is ________.

ANSWER: -1.28 TYPE: FI DIFFICULTY: Moderate KEYWORDS: standard normal quantile, normal probability plot 100. Referring to Table 6-2, the fifth standard normal quantile is ________.

ANSWER: 0 TYPE: FI DIFFICULTY: Moderate KEYWORDS: standard normal quantile, normal probability plot 101. Referring to Table 6-2, the sixth standard normal quantile is ________.

ANSWER: +0.25 TYPE: FI DIFFICULTY: Moderate KEYWORDS: standard normal quantile, normal probability plot 102. Referring to Table 6-2, the second standard normal quantile is ________.

ANSWER: -0.84 TYPE: FI DIFFICULTY: Moderate KEYWORDS: standard normal quantile, normal probability plot 103. Referring to Table 6-2, the seventh standard normal quantile is ________.

ANSWER: +0.52 TYPE: FI DIFFICULTY: Moderate KEYWORDS: standard normal quantile, normal probability plot

.

The Normal Distribution 183 104. Referring to Table 6-2, construct a normal probability plot.

ANSWER: 450

Accidents

400 350 300 250 200 150 100 50 0 -1.5

-1

-0.5

0

0.5

1

1.5

Z -Scores

TYPE: PR DIFFICULTY: Moderate KEYWORDS: normal probability plot 105. True or False: Referring to Table 6-2, the data appear normal.

ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: normal probability plot

TABLE 6-3 The number of column inches of classified advertisements appearing on Mondays in a certain daily newspaper is normally distributed with population mean 320 and population standard deviation 20 inches.

106. Referring to Table 6-3, for a randomly chosen Monday, what is the probability there will be less than 340 column inches of classified advertisement?

ANSWER: 0.8413 TYPE: PR DIFFICULTY: Easy KEYWORDS: normal distribution, probability

.

184

The Normal Distribution 107. Referring to Table 6-3, for a randomly chosen Monday, what is the probability there will be between 280 and 360 column inches of classified advertisement?

ANSWER: 0.9545 using Excel or 0.9544 using Table E.2 TYPE: PR DIFFICULTY: Moderate KEYWORDS: normal distribution, probability 108. Referring to Table 6-3, for a randomly chosen Monday the probability is 0.1 that there will be less than how many column inches of classified advertisements?

ANSWER: 294.4 TYPE: PR DIFFICULTY: Moderate KEYWORDS: normal distribution, value 109. Referring to Table 6-3, a single Monday is chosen at random. State in which of the following ranges the number of column inches of classified advertisement is most likely to be: a) 300 --320 b) 310 --330 c) 320 -- 340 d) 330 -- 350 ANSWER: b TYPE: MC Difficulty: Moderate KEYWORDS: normal distribution, probability

TABLE 6-4 John has two jobs. For daytime work at a jewelry store he is paid $200 per month, plus a commission. His monthly commission is normally distributed with mean $600 and standard deviation $40. At night he works as a waiter, for which his monthly income is normally distributed with mean $100 and standard deviation $30. John's income levels from these two sources are independent of each other.

110. Referring to Table 6-4, for a given month, what is the probability that John's commission from the jewelry store is less than $640?

ANSWER: 0.8413 TYPE: PR DIFFICULTY: Easy KEYWORDS: normal distribution, probability

.

The Normal Distribution 185 111. Referring to Table 6-4, for a given month, what is the probability that John's income as a waiter is between $70 and $160?

ANSWER: 0.8186 using Excel or 0.8185 using Table E.2 TYPE: PR DIFFICULTY: Easy KEYWORDS: normal distribution, probability 112. Referring to Table 6-4, the probability is 0.9 that John's income as a waiter is less than how much in a given month?

ANSWER: $138.45 using Excel or $138.40 using Table E.2 TYPE: PR DIFFICULTY: Moderate KEYWORDS: normal distribution, value 113. Referring to Table 6-4, find the mean and standard deviation of John's total income from these two jobs for a given month.

ANSWER: $900; $50 TYPE: PR DIFFICULTY: Difficult EXPLANATION: Total mean = $200 + $600 + $100 = $900; Total variance = 402 + 302 = 2,500 KEYWORDS: normal distribution, mean, standard deviation 114. Referring to Table 6-4, for a given month, what is the probability that John's total income from these two jobs is less than $825?

ANSWER: 0.0668 TYPE: PR DIFFICULTY: Difficult EXPLANATION: Total mean = $900, Total standard deviation = $50 KEYWORDS: normal distribution, probability 115. Referring to Table 6-4, the probability is 0.2 that John's total income from these two jobs in a given month is less than how much?

ANSWER: $857.92 using Excel or $858 using Table E.2 TYPE: PR DIFFICULTY: Difficult KEYWORDS: normal distribution, value

.

186

The Normal Distribution TABLE 6-5 A company has 125 personal computers. The probability that any one of them will require repair on a given day is 0.15.

116. Referring to Table 6-5, which of the following is one of the properties required so that the binomial distribution can be used to compute the probability that no more than 2 computers will require repair on a given day? a) The number of personal computers the company owns on a given day is fixed. b) The probability that a computer that will require repair in the morning is the same as that in the afternoon. c) The number of computers that will require repair in the morning is independent of the number of computers that will require repair in the afternoon. d) The probability that two or more computers that will require repair in a given day approaches zero.

ANSWER: a TYPE: MC DIFFICULTY: Easy KEYWORDS: binomial distribution, properties 117. Referring to Table 6-5, which of the following is one of the properties required so that the binomial distribution can be used to compute the probability that no more than 2 computers will require repair on a given day? a) The probability that a computer that will require repair in the morning is the same as that in the afternoon. b) A randomly selected computer on a given day will either require a repair or will not. c) The number of computers that will require repair in the morning is independent of the number of computers that will require repair in the afternoon. d) The probability that two or more computers that will require repair in a given day approaches zero.

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

.

The Normal Distribution 187 118. Referring to Table 6-5, which of the following is one of the properties required so that the binomial distribution can be used to compute the probability that no more than 2 computers will require repair on a given day? a) The probability that a computer that will require repair in the morning is the same as that in the afternoon. b) The number of computers that will require repair in the morning is independent of the number of computers that will require repair in the afternoon. c) The probability that any one of the computers that will require repair on a given day will not affect or change the probability that any other computers that will require repair on the same day. d) The probability that two or more computers that will require repair in a given day approaches zero.

ANSWER: c TYPE: MC DIFFICULTY: Easy KEYWORDS: binomial distribution, properties 119. The owner of a fish market determined that the average weight for a catfish is 3.2

pounds. He also knew that the probability of a randomly selected catfish that would weigh more than 3.8 pounds is 20% and the probability that a randomly selected catfish that would weigh less than 2.8 pounds is 30%. The probability that a randomly selected catfish will weigh less than 3.6 pounds is ______.

ANSWER: 70% or 0.7 TYPE: FI DIFFICULTY: Easy KEYWORDS: normal distribution, probability, properties 120. The owner of a fish market determined that the average weight for a catfish is 3.2

pounds. He also knew that the probability of a randomly selected catfish that would weigh more than 3.8 pounds is 20% and the probability that a randomly selected catfish that would weigh less than 2.8 pounds is 30%. The probability that a randomly selected catfish will weigh between 2.6 and 3.6 pounds is ______.

ANSWER: 50% or 0.5 TYPE: FI DIFFICULTY: Easy KEYWORDS: normal distribution, probability, properties 121. The owner of a fish market determined that the average weight for a catfish is 3.2

pounds. He also knew that the probability of a randomly selected catfish that would weigh more than 3.8 pounds is 20% and the probability that a randomly selected catfish that would weigh less than 2.8 pounds is 30%. The middle 40% of the catfish will weigh between ______ pounds and ______ pounds.

ANSWER: 2.8 and 3.6 TYPE: FI DIFFICULTY: Easy KEYWORDS: normal distribution, probability, properties .

188

Sampling and Sampling Distributions

CHAPTER 7: SAMPLING AND SAMPLING DISTRIBUTIONS 1. Sampling distributions describe the distribution of a) parameters. b) statistics. c) both parameters and statistics. d) neither parameters nor statistics.

ANSWER: b TYPE: MC DIFFICULTY: Easy KEYWORDS: statistics, sampling distribution 2. The standard error of the mean a) is never larger than the standard deviation of the population. b) decreases as the sample size increases. c) measures the variability of the mean from sample to sample. d) All of the above.

ANSWER: d TYPE: MC DIFFICULTY: Easy KEYWORDS: standard error, mean 3. The Central Limit Theorem is important in statistics because a) for a large n, it says the population is approximately normal. b) for any population, it says the sampling distribution of the sample mean is approximately normal, regardless of the sample size. c) for a large n, it says the sampling distribution of the sample mean is approximately normal, regardless of the shape of the population. d) for any sized sample, it says the sampling distribution of the sample mean is approximately normal.

ANSWER: c TYPE: MC DIFFICULTY: Difficult KEYWORDS: central limit theorem

.

Sampling and Sampling Distributions 189

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

ANSWER: a TYPE: MC DIFFICULTY: Moderate KEYWORDS: unbiased 5. For air travelers, one of the biggest complaints is of the waiting time between when the airplane taxis away from the terminal until the flight takes off. This waiting time is known to have a skewed-right distribution with a mean of 10 minutes and a standard deviation of 8 minutes.

Suppose 100 flights have been randomly sampled. Describe the sampling distribution of the mean waiting time between when the airplane taxis away from the terminal until the flight takes off for these 100 flights. a) Distribution is skewed-right with mean = 10 minutes and standard error = 0.8 minutes. b) Distribution is skewed-right with mean = 10 minutes and standard error = 8 minutes. c) Distribution is approximately normal with mean = 10 minutes and standard error = 0.8 minutes. d) Distribution is approximately normal with mean = 10 minutes and standard error = 8 minutes.

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

ANSWER: d TYPE: MC DIFFICULTY: Easy KEYWORDS: sampling distribution, properties

.

190

Sampling and Sampling Distributions

7. Which of the following is true about the sampling distribution of the sample mean? a) The mean of the sampling distribution is always  . b) The standard deviation of the sampling distribution is always  . c) The shape of the sampling distribution is always approximately normal. d) All of the above are true.

ANSWER: a TYPE: MC DIFFICULTY: Moderate KEYWORDS: sampling distribution, properties 8. True or False: The amount of time it takes to complete an examination has a skewed-left distribution with a mean of 65 minutes and a standard deviation of 8 minutes. If 64 students were randomly sampled, the probability that the sample mean of the sampled students exceeds 71 minutes is approximately 0.

ANSWER: True TYPE: TF DIFFICULTY: Moderate KEYWORDS: sampling distribution, central limit theorem 9. Suppose the ages of students in Statistics 101 follow a skewed-right distribution with a mean of 23 years and a standard deviation of 3 years. If we randomly sampled 100 students, which of the following statements about the sampling distribution of the sample mean age is incorrect? a) The mean of the sampling distribution is equal to 23 years. b) The standard deviation of the sampling distribution is equal to 3 years. c) The shape of the sampling distribution is approximately normal. d) The standard error of the sampling distribution is equal to 0.3 years.

ANSWER: b TYPE: MC DIFFICULTY: Easy KEYWORDS: sampling distribution, central limit theorem 10. Why is the Central Limit Theorem so important to the study of sampling distributions? a) It allows us to disregard the size of the sample selected when the population is not normal. b) It allows us to disregard the shape of the sampling distribution when the size of the population is large. c) It allows us to disregard the size of the population we are sampling from. d) It allows us to disregard the shape of the population when n is large.

ANSWER: d TYPE: MC DIFFICULTY: Moderate KEYWORDS: central limit theorem

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

Sampling and Sampling Distributions 191

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

ANSWER: True TYPE: TF DIFFICULTY: Moderate KEYWORDS: central limit theorem 13. Suppose a sample of n = 50 items is drawn from a population of manufactured products and the weight, X, of each item is recorded. Prior experience has shown that the weight has a probability distribution with  = 6 ounces and  = 2.5 ounces. Which of the following is true about the sampling distribution of the sample mean if a sample of size 15 is selected? a) The mean of the sampling distribution is 6 ounces. b) The standard deviation of the sampling distribution is 2.5 ounces. c) The shape of the sample distribution is approximately normal. d) All of the above are correct.

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

ANSWER: 0.0228 TYPE: PR DIFFICULTY: Moderate KEYWORDS: sampling distribution, mean, probability, central limit theorem 15. The distribution of the number of loaves of bread sold per day by a large bakery over the past 5 years has a mean of 7,750 and a standard deviation of 145 loaves. Suppose a random sample of n = 40 days has been selected. What is the approximate probability that the average number of loaves sold in the sampled days exceeds 7,895 loaves?

ANSWER: Approximately 0 TYPE: PR DIFFICULTY: Moderate KEYWORDS: sampling distribution, mean, probability, central limit theorem 16. Sales prices of baseball cards from the 1960s are known to possess a skewed-right distribution with a mean sale price of $5.25 and a standard deviation of $2.80. Suppose a random sample of .

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100 cards from the 1960s is selected. Describe the sampling distribution for the sample mean sale price of the selected cards. a) Skewed-right with a mean of $5.25 and a standard error of $2.80 b) Normal with a mean of $5.25 and a standard error of $0.28 c) Skewed-right with a mean of $5.25 and a standard error of $0.28 d) Normal with a mean of $5.25 and a standard error of $2.80 ANSWER: b TYPE: MC DIFFICULTY: Easy KEYWORDS: sampling distribution, central limit theorem 17. Major league baseball salaries averaged $1.5 million with a standard deviation of $0.8 million in 1994. Suppose a sample of 100 major league players was taken. Find the approximate probability that the average salary of the 100 players exceeded $1 million. a) Approximately 0 b) 0.2357 c) 0.7357 d) Approximately 1 ANSWER: d TYPE: MC DIFFICULTY: Moderate KEYWORDS: sampling distribution, mean, probability, central limit theorem 18. At a computer manufacturing company, the actual size of computer chips is normally distributed with a mean of 1 centimeter and a standard deviation of 0.1 centimeter. A random sample of 12 computer chips is taken. What is the standard error for the sample mean? a) 0.029 b) 0.050 c) 0.091 d) 0.120 ANSWER: a TYPE: MC DIFFICULTY: Easy KEYWORDS: standard error, mean 19. At a computer manufacturing company, the actual size of computer chips is normally distributed with a mean of 1 centimeter and a standard deviation of 0.1 centimeter. A random sample of 12 computer chips is taken. What is the probability that the sample mean will be between 0.99 and 1.01 centimeters?

ANSWER: 0.2710 using Excel or 0.2736 using Table E.2 TYPE: PR DIFFICULTY: Moderate KEYWORDS: sampling distribution, mean, probability 20. At a computer manufacturing company, the actual size of computer chips is normally distributed with a mean of 1 centimeter and a standard deviation of 0.1 centimeter. A random sample of 12 .

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computer chips is taken. What is the probability that the sample mean will be below 0.95 centimeters?

ANSWER: 0.0416 using Excel or 0.0418 using Table E.2 TYPE: PR DIFFICULTY: Moderate KEYWORDS: sampling distribution, mean, probability 21. At a computer manufacturing company, the actual size of computer chips is normally distributed with a mean of 1 centimeter and a standard deviation of 0.1 centimeter. A random sample of 12 computer chips is taken. Above what value do 2.5% of the sample means fall?

ANSWER: 1.057 TYPE: PR DIFFICULTY: Difficult KEYWORDS: sampling distribution, mean, value 22. The owner of a fish market has an assistant who has determined that the weights of catfish are normally distributed, with mean of 3.2 pounds and standard deviation of 0.8 pound. If a sample of 16 fish is taken, what would the standard error of the mean weight equal? a) 0.003 b) 0.050 c) 0.200 d) 0.800 ANSWER: c TYPE: MC DIFFICULTY: Easy KEYWORDS: standard error, mean 23. The owner of a fish market has an assistant who has determined that the weights of catfish are normally distributed, with mean of 3.2 pounds and standard deviation of 0.8 pound. If a sample of 25 fish yields a mean of 3.6 pounds, what is the Z-score for this observation? a) 18.750 b) 2.500 c) 1.875 d) 0.750 ANSWER: b TYPE: MC DIFFICULTY: Easy KEYWORDS: sampling distribution, mean

24. The owner of a fish market has an assistant who has determined that the weights of catfish are normally distributed, with mean of 3.2 pounds and standard deviation of 0.8 pound. If a sample of 64 fish yields a mean of 3.4 pounds, what is probability of obtaining a sample mean this large or larger? .

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a) b) c) d)

0.0001 0.0013 0.0228 0.4987

ANSWER: c TYPE: MC DIFFICULTY: Moderate KEYWORDS: sampling distribution, mean, probability 25. The owner of a fish market has an assistant who has determined that the weights of catfish are normally distributed, with mean of 3.2 pounds and standard deviation of 0.8 pound. What percentage of samples of 4 fish will have sample means between 3.0 and 4.0 pounds? a) 84% b) 67% c) 29% d) 16% ANSWER: b TYPE: MC DIFFICULTY: Moderate KEYWORDS: sampling distribution, mean, probability 26. For sample size 16, the sampling distribution of the mean will be approximately normally distributed a) regardless of the shape of the population. b) if the shape of the population is symmetrical. c) if the sample standard deviation is known. d) if the sample is normally distributed.

ANSWER: b TYPE: MC DIFFICULTY: Moderate KEYWORDS: sampling distribution, mean, central limit theorem 27. The standard error of the mean for a sample of 100 is 30. In order to cut the standard error of the mean to 15, we would a) increase the sample size to 200. b) increase the sample size to 400. c) decrease the sample size to 50. d) decrease the sample to 25.

ANSWER: b TYPE: MC DIFFICULTY: Moderate KEYWORDS: standard error, mean 28. Which of the following is true regarding the sampling distribution of the mean for a large sample size? a) It has the same shape, mean, and standard deviation as the population. b) It has a normal distribution with the same mean and standard deviation as the population. c) It has the same shape and mean as the population, but has a smaller standard deviation. .

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d) It has a normal distribution with the same mean as the population but with a smaller standard deviation.

ANSWER: d TYPE: MC DIFFICULTY: Moderate KEYWORDS: sampling distribution, mean, central limit theorem 29. For sample sizes greater than 30, the sampling distribution of the mean will be approximately normally distributed a) regardless of the shape of the population. b) only if the shape of the population is symmetrical. c) only if the standard deviation of the samples are known. d) only if the population is normally distributed.

ANSWER: a TYPE: MC DIFFICULTY: Easy KEYWORDS: sampling distribution, mean, central limit theorem 30. For sample size 1, the sampling distribution of the mean will be normally distributed a) regardless of the shape of the population. b) only if the shape of the population is symmetrical. c) only if the population values are positive. d) only if the population is normally distributed.

ANSWER: d TYPE: MC DIFFICULTY: Moderate KEYWORDS: sampling distribution, mean, central limit theorem 31. The standard error of the population proportion will become larger a) as population proportion approaches 0. b) as population proportion approaches 0.50. c) as population proportion approaches 1.00. d) as the sample size increases.

ANSWER: b TYPE: MC DIFFICULTY: Moderate KEYWORDS: standard error, proportion

32. True or False: As the sample size increases, the standard error of the mean increases.

ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: standard error, mean .

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33. True or False: If the population distribution is symmetric, the sampling distribution of the mean can be approximated by the normal distribution if the samples contain 15 observations.

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

ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: sampling distribution, mean, central limit theorem 35. True or False: If the amount of gasoline purchased per car at a large service station has a population mean of $15 and a population standard deviation of $4, then 99.73% of all cars will purchase between $3 and $27.

ANSWER: False TYPE: TF DIFFICULTY: Moderate KEYWORDS: sampling distribution, mean, probability 36. True or False: If the amount of gasoline purchased per car at a large service station has a population mean of $15 and a population standard deviation of $4 and a random sample of 4 cars is selected, there is approximately a 68.26% chance that the sample mean will be between $13 and $17.

ANSWER: False TYPE: TF DIFFICULTY: Moderate EXPLANATION: The sample is too small for the normal approximation.

KEYWORDS: sampling distribution, mean, probability

37. True or False: If the amount of gasoline purchased per car at a large service station has a population mean of $15 and a population standard deviation of $4 and it is assumed that the amount of gasoline purchased per car is symmetric, there is approximately a 68.26% chance that a random sample of 16 cars will have a sample mean between $14 and $16.

ANSWER: True .

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TYPE: TF DIFFICULTY: Moderate KEYWORDS: sampling distribution, mean, probability 38. True or False: If the amount of gasoline purchased per car at a large service station has a population mean of $15 and a population standard deviation of $4 and a random sample of 64 cars is selected, there is approximately a 95.44% chance that the sample mean will be between $14 and $16.

ANSWER: True TYPE: TF DIFFICULTY: Difficult KEYWORDS: sampling distribution, mean, probability 39. True or False: As the sample size increases, the effect of an extreme value on the sample mean becomes smaller.

ANSWER: True TYPE: TF DIFFICULTY: Moderate KEYWORDS: sampling distribution, law of large numbers 40. True or False: If the population distribution is skewed, in most cases the sampling distribution of the mean can be approximated by the normal distribution if the samples contain at least 30 observations.

ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: sampling distribution, mean, central limit theorem 41. True or False: A sampling distribution is a distribution for a statistic.

ANSWER: True TYPE: TF DIFFICULTY: Moderate KEYWORDS: sampling distribution

42. True or False: Suppose  = 50 and  = 100 for a population. In a sample where n = 100 is randomly taken, 95% of all possible sample means will fall between 48.04 and 51.96.

2

ANSWER: True TYPE: TF DIFFICULTY: Moderate KEYWORDS: sampling distribution, mean, probability

.

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Sampling and Sampling Distributions 43. True or False: Suppose  = 80 and  = 400 for a population. In a sample where n = 100 is randomly taken, 95% of all possible sample means will fall above 76.71.

2

ANSWER: True TYPE: TF DIFFICULTY: Moderate KEYWORDS: sampling distribution, mean, probability 44. True or False: Suppose  = 50 and  = 100 for a population. In a sample where n = 100 is randomly taken, 90% of all possible sample means will fall between 49 and 51.

2

ANSWER: False TYPE: TF DIFFICULTY: Moderate KEYWORDS: sampling distribution, mean, probability 45. True or False: The Central Limit Theorem ensures that the sampling distribution of the sample mean approaches normal as the sample size increases.

ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: central limit theorem 46. True or False: The standard error of the mean is also known as the standard deviation of the sampling distribution of the sample mean.

ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: standard error, mean 47. True or False: A sampling distribution is defined as the probability distribution of possible sample sizes that can be observed from a given population.

ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: sampling distribution

48. True or False: As the size of the sample is increased, the standard deviation of the sampling distribution of the sample mean for a normally distributed population will stay the same.

ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: standard error, properties 49. True or False: For distributions such as the normal distribution, the arithmetic mean is considered more stable from sample to sample than other measures of central tendency. .

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ANSWER: True TYPE: TF DIFFICULTY: Moderate KEYWORDS: sampling distribution, mean 50. True or False: The fact that the sample means are less variable than the population data can be observed from the standard error of the mean.

ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: sampling distribution, mean, standard error 51. The amount of pyridoxine (in grams) per multiple vitamin is normally distributed with  = 110 grams and  = 25 grams. A sample of 25 vitamins is to be selected. What is the probability that the sample mean will be between 100 and 120 grams?

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

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

ANSWER: 0.9772 TYPE: PR DIFFICULTY: Moderate KEYWORDS: sampling distribution, mean, probability 54. The amount of pyridoxine (in grams) per multiple vitamin is normally distributed with  = 110 grams and  = 25 grams. A sample of 25 vitamins is to be selected. So, 95% of all sample means will be greater than how many grams?

ANSWER: 101.7757 TYPE: PR DIFFICULTY: Difficult KEYWORDS: sampling distribution, mean, value

.

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Sampling and Sampling Distributions 55. The amount of pyridoxine (in grams) per multiple vitamin is normally distributed with  = 110 grams and  = 25 grams. A sample of 25 vitamins is to be selected. So, the middle 70% of all sample means will fall between what two values?

ANSWER: 104.8 and 115.2 TYPE: PR DIFFICULTY: Difficult KEYWORDS: sampling distribution, mean, value 56. The amount of time required for an oil and filter change on an automobile is normally distributed with a mean of 45 minutes and a standard deviation of 10 minutes. A random sample of 16 cars is selected. What would you expect the standard error of the mean to be?

ANSWER: 2.5 minutes TYPE: PR DIFFICULTY: Easy KEYWORDS: standard error, mean 57. The amount of time required for an oil and filter change on an automobile is normally distributed with a mean of 45 minutes and a standard deviation of 10 minutes. A random sample of 16 cars is selected. What is the probability that the sample mean is between 45 and 52 minutes?

ANSWER: 0.4974 TYPE: PR DIFFICULTY: Moderate KEYWORDS: sampling distribution, mean, probability 58. The amount of time required for an oil and filter change on an automobile is normally distributed with a mean of 45 minutes and a standard deviation of 10 minutes. A random sample of 16 cars is selected. What is the probability that the sample mean will be between 39 and 48 minutes?

ANSWER: 0.8767 TYPE: PR DIFFICULTY: Moderate KEYWORDS: sampling distribution, mean, probability

59. The amount of time required for an oil and filter change on an automobile is normally distributed with a mean of 45 minutes and a standard deviation of 10 minutes. A random sample of 16 cars is selected. So, 95% of all sample means will fall between what two values?

ANSWER: 40.1 and 49.9 minutes TYPE: PR DIFFICULTY: Difficult KEYWORDS: sampling distribution, mean, probability

.

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60. The amount of time required for an oil and filter change on an automobile is normally distributed with a mean of 45 minutes and a standard deviation of 10 minutes. A random sample of 16 cars is selected. So, 90% of the sample means will be greater than what value?

ANSWER: 41.8 minutes TYPE: PR DIFFICULTY: Difficult KEYWORDS: sampling distribution, mean, probability 61. True or False: The amount of bleach a machine pours into bottles has a mean of 36 oz. with a standard deviation of 0.15 oz. Suppose we take a random sample of 36 bottles filled by this machine. The sampling distribution of the sample mean has a mean of 36 oz.

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

ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: sampling distribution, mean, standard error 63. True or False: The amount of bleach a machine pours into bottles has a mean of 36 oz. with a standard deviation of 0.15 oz. Suppose we take a random sample of 36 bottles filled by this machine. The sampling distribution of the sample mean will be approximately normal only if the population sampled is normal.

ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: sampling distribution, mean, central limit theorem

64. The amount of bleach a machine pours into bottles has a mean of 36 oz. with a standard deviation of 0.15 oz. Suppose we take a random sample of 36 bottles filled by this machine. The probability that the mean of the sample exceeds 36.01 oz. is __________.

ANSWER: 0.3446 TYPE: FI DIFFICULTY: Moderate KEYWORDS: sampling distribution, mean, probability, central limit theorem 65. The amount of bleach a machine pours into bottles has a mean of 36 oz. with a standard deviation of 0.15 oz. Suppose we take a random sample of 36 bottles filled by this machine. The probability that the mean of the sample is less than 36.03 is __________. .

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ANSWER: 0.8849 TYPE: FI DIFFICULTY: Moderate KEYWORDS: sampling distribution, mean, probability, central limit theorem 66. The amount of bleach a machine pours into bottles has a mean of 36 oz. with a standard deviation of 0.15 oz. Suppose we take a random sample of 36 bottles filled by this machine. The probability that the mean of the sample is between 35.94 and 36.06 oz. is __________.

ANSWER: 0.9836 TYPE: FI DIFFICULTY: Moderate KEYWORDS: sampling distribution, mean, probability, central limit theorem 67. The amount of bleach a machine pours into bottles has a mean of 36 oz. with a standard deviation of 0.15 oz. Suppose we take a random sample of 36 bottles filled by this machine. The probability that the mean of the sample is between 35.95 and 35.98 oz. is __________.

ANSWER: 0.1891 TYPE: FI DIFFICULTY: Moderate KEYWORDS: sampling distribution, mean, probability, central limit theorem 68. The amount of bleach a machine pours into bottles has a mean of 36 oz. with a standard deviation of 0.15 oz. Suppose we take a random sample of 36 bottles filled by this machine. So, 95% of the sample means based on samples of size 36 will be between __________ and __________.

ANSWER: 35.951 and 36.049 ounces TYPE: FI DIFFICULTY: Difficult KEYWORDS: sampling distribution, mean, value, central limit theorem

69. A manufacturer of power tools claims that the average amount of time required to assemble their top-of-the-line table saw is 80 minutes with a standard deviation of 40 minutes. Suppose a random sample of 64 purchasers of this table saw is taken. The mean of the sampling distribution of the sample mean is __________ minutes.

ANSWER: 80 TYPE: FI DIFFICULTY: Easy KEYWORDS: sampling distribution, mean, unbiased 70. A manufacturer of power tools claims that the average amount of time required to assemble their top-of-the-line table saw is 80 minutes with a standard deviation of 40 minutes. Suppose a random sample of 64 purchasers of this table saw is taken. The standard deviation of the sampling distribution of the sample mean is __________ minutes. .

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ANSWER: 5 TYPE: FI DIFFICULTY: Easy KEYWORDS: sampling distribution, mean, standard error 71. A manufacturer of power tools claims that the average amount of time required to assemble their top-of-the-line table saw is 80 minutes with a standard deviation of 40 minutes. Suppose a random sample of 64 purchasers of this table saw is taken. The probability that the sample mean will be less than 82 minutes is __________.

ANSWER: 0.6554 TYPE: FI DIFFICULTY: Moderate KEYWORDS: sampling distribution, mean, probability, central limit theorem 72. A manufacturer of power tools claims that the average amount of time required to assemble their top-of-the-line table saw is 80 minutes with a standard deviation of 40 minutes. Suppose a random sample of 64 purchasers of this table saw is taken. The probability that the sample mean will be between 77 and 89 minutes is __________.

ANSWER: 0.6898 TYPE: FI DIFFICULTY: Moderate KEYWORDS: sampling distribution, mean, probability, central limit theorem 73. A manufacturer of power tools claims that the average amount of time required to assemble their top-of-the-line table saw is 80 minutes with a standard deviation of 40 minutes. Suppose a random sample of 64 purchasers of this table saw is taken. The probability that the sample mean will be greater than 88 minutes is __________.

ANSWER: 0.0548 TYPE: FI DIFFICULTY: Moderate KEYWORDS: sampling distribution, mean, probability, central limit theorem 74. A manufacturer of power tools claims that the average amount of time required to assemble their top-of-the-line table saw is 80 minutes with a standard deviation of 40 minutes. Suppose a random sample of 64 purchasers of this table saw is taken. So, 95% of the sample means based on samples of size 64 will be between __________ and __________.

ANSWER: 70.2 and 89.8 minutes TYPE: FI DIFFICULTY: Difficult KEYWORDS: sampling distribution, mean, value, central limit theorem 75. To use the normal distribution to approximate the binomial distribution, we need ______ and ______ to be at least 5.

ANSWER: n and n( − 1) .

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TYPE: FI DIFFICULTY: Easy KEYWORDS: sampling distribution, proportion, central limit theorem 76. True or False: The sample mean is an unbiased estimate of the population mean.

ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: mean, unbiased 77. True or False: The sample proportion is an unbiased estimate of the population proportion.

ANSWER: True TYPE: TF DIFFICULTY: Difficult KEYWORDS: proportion, unbiased 78. True or False: The mean of the sampling distribution of a sample proportion is the population proportion,  .

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

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

 (1 −  ) n

where  is the population proportion.

ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: sampling distribution, proportion 81. True or False: A sample of size 25 provides a sample variance of 400. The standard error, in this case equal to 4, is best described as the estimate of the standard deviation of means calculated from samples of size 25.

ANSWER: True .

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TYPE: TF DIFFICULTY: Moderate KEYWORDS: standard error 82. True or False: An unbiased estimator will have a value, on average across samples, equal to the population parameter value.

ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: unbiased 83. True or False: In inferential statistics, the standard error of the sample mean assesses the uncertainty or error of estimation.

ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: mean, standard error 84. Assume that house prices in a neighborhood are normally distributed with standard deviation $20,000. A random sample of 16 observations is taken. What is the probability that the sample mean differs from the population mean by more than $5,000?

ANSWER: 0.3173 using Excel or 0.3174 using Table E.2 TYPE: PR DIFFICULTY: Moderate KEYWORDS: sampling distribution, mean, probability

TABLE 7-1 Times spent studying by students in the week before final exams follow a normal distribution with standard deviation 8 hours. A random sample of 4 students was taken in order to estimate the mean study time for the population of all students.

85. Referring to Table 7-1, what is the probability that the sample mean exceeds the population mean by more than 2 hours?

ANSWER: 0.3085 TYPE: PR DIFFICULTY: Moderate KEYWORDS: sampling distribution, mean, probability 86. Referring to Table 7-1, what is the probability that the sample mean is more than 3 hours below the population mean? .

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ANSWER: 0.2266 TYPE: PR DIFFICULTY: Moderate KEYWORDS: sampling distribution, mean, probability 87. Referring to Table 7-1, what is the probability that the sample mean differs from the population mean by less than 2 hours?

ANSWER: 0.3829 using Excel or 0.3830 using Table E.2 TYPE: PR DIFFICULTY: Moderate KEYWORDS: sampling distribution, mean, probability 88. Referring to Table 7-1, what is the probability that the sample mean differs from the population mean by more than 3 hours?

ANSWER: 0.4533 using Excel or 0.4532 using Table E.2 TYPE: PR DIFFICULTY: Moderate KEYWORDS: sampling distribution, mean, probability TABLE 7-2 The mean selling price of new homes in a city over a year was $115,000. The population standard deviation was $25,000. A random sample of 100 new home sales from this city was taken.

89. Referring to Table 7-2, what is the probability that the sample mean selling price was more than $110,000?

ANSWER: 0.9772 TYPE: PR DIFFICULTY: Easy KEYWORDS: sampling distribution, mean, probability, central limit theorem 90. Referring to Table 7-2, what is the probability that the sample mean selling price was between $113,000 and $117,000?

ANSWER: 0.5763 using Excel or 0.5762 using Table E.2 TYPE: PR DIFFICULTY: Easy KEYWORDS: sampling distribution, mean, probability, central limit theorem 91. Referring to Table 7-2, what is the probability that the sample mean selling price was between $114,000 and $116,000?

ANSWER: 0.3108 TYPE: PR DIFFICULTY: Easy KEYWORDS: sampling distribution, mean, probability, central limit theorem

.

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92. Referring to Table 7-2, without doing the calculations, state in which of the following ranges the sample mean selling price is most likely to lie? a) $113,000 -- $115,000 b) $114,000 -- $116,000 c) $115,000 -- $117,000 d) $116,000 -- $118,000 ANSWER: b TYPE: MC DIFFICULTY: Easy KEYWORDS: sampling distribution, mean, probability, central limit theorem

TABLE 7-3 The lifetimes of a certain brand of light bulbs are known to be normally distributed with a mean of 1,600 hours and a standard deviation of 400 hours. A random sample of 64 of these light bulbs is taken.

93. Referring to Table 7-3, what is the probability that the sample mean lifetime is more than 1,550 hours?

ANSWER: 0.8413 TYPE: PR DIFFICULTY: Easy KEYWORDS: sampling distribution, mean, probability 94. Referring to Table 7-3, the probability is 0.15 that the sample mean lifetime is more than how many hours?

ANSWER: 1,651.82 hours using Excel or 1,652 hours using Table E.2 TYPE: PR DIFFICULTY: Moderate KEYWORDS: sampling distribution, mean, probability 95. Referring to Table 7-3, the probability is 0.20 that the sample mean lifetime differs from the population mean lifetime by at least how many hours?

ANSWER: 64.08 hours using Excel or 64 hours using Table E.2 TYPE: PR DIFFICULTY: Difficult KEYWORDS: sampling distribution, mean, value TABLE 7-4 According to a survey, only 15% of customers who visited the web site of a major retail store made a purchase. Random samples of size 50 are selected.

96. Referring to Table 7-4, the average of all the sample proportions of customers who will make a purchase after visiting the web site is _______.

ANSWER: .

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0.15 or 15% TYPE: FI DIFFICULTY: Moderate KEYWORDS: sampling distribution, proportion, mean 97. Referring to Table 7-4, the standard deviation of all the sample proportions of customers who will make a purchase after visiting the web site is ________.

ANSWER: 0.05050 TYPE: FI DIFFICULTY: Moderate KEYWORDS: sampling distribution, proportion, standard error 98. True of False: Referring to Table 7-4, the requirements for using a normal distribution to approximate a binomial distribution is fulfilled.

ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: sampling distribution, proportion, central limit theorem 99. Referring to Table 7-4, what proportion of the samples will have between 20% and 30% of customers who will make a purchase after visiting the web site?

ANSWER: 0.1596 TYPE: PR DIFFICULTY: Easy KEYWORDS: sampling distribution, proportion, probability

100. Referring to Table 7-4, what proportion of the samples will have less than 15% of customers who will make a purchase after visiting the web site?

ANSWER: 0.5 TYPE: PR DIFFICULTY: Easy KEYWORDS: sampling distribution, proportion, probability 101. Referring to Table 7-4, what is the probability that a random sample of 50 will have at least 30% of customers who will make a purchase after visiting the web site?

ANSWER: 0.0015 TYPE: PR DIFFICULTY: Easy KEYWORDS: sampling distribution, proportion, probability 102. Referring to Table 7-4, 90% of the samples will have less than what percentage of customers who will make a purchase after visiting the web site? .

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ANSWER: 21.47% TYPE: PR DIFFICULTY: Difficult KEYWORDS: sampling distribution, proportion, value 103. Referring to Table 7-4, 90% of the samples will have more than what percentage of customers who will make a purchase after visiting the web site?

ANSWER: 8.528% using Excel or 8.536 using Table E.2 TYPE: PR DIFFICULTY: Difficult KEYWORDS: sampling distribution, proportion, value 104. A study at a college in the west coast reveals that, historically, 45% of their students are minority students. The expected percentage of minority students in their next batch of freshmen is _______.

ANSWER: 45% TYPE: FI DIFFCULTY: Moderate KEYWORDS: sampling distribution, proportion, mean 105. A study at a college in the west coast reveals that, historically, 45% of their students are minority students. If random samples of size 75 are selected, the standard error of the proportions of students in the samples who are minority students is _________.

ANSWER: 0.05745 TYPE: FI DIFFCULTY: Moderate KEYWORDS: sampling distribution, proportion, standard error 106. A study at a college in the west coast reveals that, historically, 45% of their students are minority students. If a random sample of size 75 is selected, the probability is _______ that between 30% and 50% of the students in the sample will be minority students.

ANSWER: 0.8034 using Excel or 0.8033 using Table E.2 TYPE: FI DIFFCULTY: Easy KEYWORDS: sampling distribution, proportion, probability 107. A study at a college in the west coast reveals that, historically, 45% of their students are minority students. If a random sample of size 75 is selected, the probability is _______ that more than half of the students in the sample will be minority students.

ANSWER: 0.1920 using Excel or 0.1922 using Table E.2 TYPE: FI DIFFCULTY: Easy KEYWORDS: sampling distribution, proportion, probability

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108. A study at a college in the west coast reveals that, historically, 45% of their students are minority students. If random samples of size 75 are selected, 80% of the samples will have less than ______% of minority students.

ANSWER: 49.83 TYPE: FI DIFFCULTY: Difficult KEYWORDS: sampling distribution, proportion, value 109. A study at a college in the west coast reveals that, historically, 45% of their students are minority students. If random samples of size 75 are selected, 95% of the samples will have more than ______% of minority students.

ANSWER: 35.55 TYPE: FI DIFFCULTY: Difficult KEYWORDS: sampling distribution, proportion, value 110. Which of the following is NOT a reason for the need for sampling? a) It is usually too costly to study the whole population. b) It is usually too time consuming to look at the whole population. c) It is sometimes destructive to observe the entire population. d) It is always more informative by investigating a sample than the entire population.

ANSWER: d TYPE: MC DIFFICULTY: Moderate KEYWORDS: reasons for sampling

111. Which of the following is NOT a reason for drawing a sample? a) A sample is less time consuming than a census. b) A sample is less costly to administer than a census. c) A sample is usually not a good representation of the target population. d) A sample is less cumbersome and more practical to administer.

ANSWER: c TYPE: MC DIFFICULTY: Easy KEYWORDS: reasons for sampling 112. Which of the following sampling methods is a probability sample? a) Chunk b) Quota sample c) Stratified sample d) Judgment sample ANSWER: .

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c TYPE: MC DIFFICULTY: Easy KEYWORDS: probability sample 113. A sample of 300 subscribers to a particular magazine is selected from a population frame of 9,000 subscribers. If, upon examining the data, it is determined that no subscriber had been selected in the sample more than once, a) the sample could not have been random. b) the sample may have been selected without replacement or with replacement. c) the sample had to have been selected with replacement. d) the sample had to have been selected without replacement.

ANSWER: b TYPE: MC DIFFICULTY: Moderate KEYWORDS: sampling method 114. For a population frame containing N = 1,007 individuals, what code number should you assign to the first person on the list in order to use a table of random numbers? a) 0 b) 1 c) 01 d) 0001 ANSWER: d TYPE: MC DIFFICULTY: Easy KEYWORDS: random number

115. Which of the following types of samples can you use if you want to make valid statistical inferences from a sample to a population? a) A judgment sample b) A quota sample c) A chunk d) A probability sample ANSWER: d TYPE: MC DIFFICULTY: Easy KEYWORDS: probability sample, sampling method 116. The evening host of a dinner dance reached into a bowl, mixed all the tickets around, and selected the ticket to award the grand door prize. What sampling method was used? a) Simple random sample b) Systematic sample c) Stratified sample d) Cluster sample

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ANSWER: a TYPE: MC DIFFICULTY: Easy KEYWORDS: simple random sample, probability sample, sampling method 117. The Dean of Students mailed a survey to a total of 400 students. The sample included 100 students randomly selected from each of the freshman, sophomore, junior, and senior classes on campus last term. What sampling method was used? a) Simple random sample b) Systematic sample c) Stratified sample d) Cluster sample ANSWER: c TYPE: MC DIFFICULTY: Easy KEYWORDS: stratified sample, probability sample, sampling method 118. A telemarketer set the company’s computerized dialing system to contact every 25th person listed in the local telephone directory. What sampling method was used? a) Simple random sample b) Systematic sample c) Stratified sample d) Cluster sample ANSWER: b TYPE: MC DIFFICULTY: Easy KEYWORDS: systematic sample, probability sample, sampling method

119. Since a _______ is not a randomly selected probability sample, there is no way to know how well it represents the overall population. a) Simple random sample b) Quota sample c) Stratified sample d) Cluster sample ANSWER: b TYPE: MC DIFFICULTY: Easy KEYWORDS: quota sample, nonprobability sample, sampling method 120. A population frame for a survey contains a listing of 72,345 names. Using a table of random numbers, how many digits will the code numbers for each member of your population contain? a) 3 b) 4 c) 5 d) 6 ANSWER: .

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c TYPE: MC DIFFICULTY: Easy KEYWORDS: random number 121. A population frame for a survey contains a listing of 6,179 names. Using a table of random numbers, which of the following code numbers will appear on your list? a) 06 b) 0694 c) 6946 d) 61790 ANSWER: b TYPE: MC DIFFICULTY: Easy KEYWORDS: random number 122. Which of the following can be reduced by proper interviewer training? a) Sampling error b) Measurement error c) Both of the above d) None of the above ANSWER: b TYPE: MC DIFFICULTY: Difficult KEYWORDS: measurement error, survey worthiness

123. Which of the following sampling methods will more likely be susceptible to ethical violation? a) Simple random sample b) Cluster sample c) Convenience sample d) Stratified sample ANSWER: c TYPE: MC DIFFICULTY: Easy KEYWORDS: ethical issues, sampling method 124. Which of the following sampling methods will more likely be susceptible to ethical violation when used to form conclusions about the entire population? a) Simple random sample b) Cluster sample c) Judgement sample d) Stratified sample ANSWER: c .

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TYPE: MC DIFFICULTY: Easy KEYWORDS: judgment sample, ethical issues, sampling method 125. Which of the following sampling methods will more likely be susceptible to ethical violation when used to form conclusions about the entire population? a) Simple random sample b) Cluster sample c) Systematic sample d) Convenient sample ANSWER: d TYPE: MC DIFFICULTY: Easy KEYWORDS: convenience sample, ethical issues, sampling method

TABLE 7-5 The manager of the customer service division of a major consumer electronics company is interested in determining whether the customers who have purchased a videocassette recorder made by the company over the past 12 months are satisfied with their products.

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

ANSWER: c TYPE: MC DIFFICULTY: Difficult KEYWORDS: coverage error, survey worthiness

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127. Referring to Table 7-5, if there are 4 different brands of videocassette recorders made by the company, the best sampling strategy would be to use a a) a simple random sample. b) a stratified sample. c) a cluster sample. d) a systematic sample.

ANSWER: b TYPE: MC DIFFICULTY: Difficult KEYWORDS: stratified sample, probability sample, sampling method 128. Referring to Table 7-5, which of the following questions in the survey will NOT likely induce a measurement error? a) How many times have you illegally copied copyrighted sporting events? b) What is your exact annual income? c) How many times have you brought the videocassette recorder back for service? d) How many times have you failed to set the time on the videocassette recorder?

ANSWER: c TYPE: MC DIFFICULTY: Difficult KEYWORDS: measurement error, survey worthiness

129. Referring to Table 7-5, if a customer survey questionnaire is included in all the videocassette recorders made and sold by the company over the past 12 months, this method of collecting data will most like suffer from a) nonresponse error. b) measurement error. c) coverage error. d) nonprobability sampling.

ANSWER: a TYPE: MC DIFFICULTY: Difficult KEYWORDS: nonresponse error, survey worthiness 130. True or False: As a population becomes large, it is usually better to obtain statistical information from the entire population.

ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: population, sample, reasons for samplings

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131. True or False: If a simple random sample is chosen with replacement, each individual has the same chance of selection on every draw.

ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: simple random sample, probability sample, sampling method 132. True or False: When dealing with human surveys, we are usually interested in sampling with replacement.

ANSWER: False TYPE: TF DIFFICULTY: Moderate KEYWORDS: sampling with replacement, sampling method, survey worthiness 133. True or False: The only reliable way a researcher can make statistical inferences from a sample to a population is to use nonprobability sampling methods.

ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: nonprobability, probability sample, sampling method 134. True or False: A sample is always a good representation of the target population.

ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: sample, population, sampling method 135. True or False: There can be only one sample drawn from a population.

ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: sample, sampling method 136. True or False: Using different frames to generate data can lead to totally different conclusions.

ANSWER: True TYPE:TF DIFFICULTY: Easy KEYWORDS: frame, sampling method 137. True or False: Sampling error can be completely eliminated by taking larger sample sizes.

ANSWER: False TYPE: TF DIFFICULTY: Difficult KEYWORDS: sampling error

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138. True or False: Sampling error can be reduced by taking larger sample sizes.

ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: sampling error 139. True or False: Chunk sample is a type of probability sample.

ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: chunk sample 140. True or False: Items or individuals in a judgment sample are chosen with regard to their probability of occurrence.

ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: judgment sample, nonprobability sample 141. True or False: When participants are allowed to self-select into the sample, you have a nonprobability sample.

ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: nonprobability sample

142. True or False: Systematic samples are less efficient than stratified sample.

ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: systematic sample, stratified sample 143. True or False: The professor of a business statistics class wanted to find out the average amount of time per week her students spent studying for the class. Among the 50 students in her class, 20% were freshmen, 50% were sophomores and 30% were juniors. She decided to draw 2 students randomly from the freshmen, 5 randomly from the sophomores and 3 randomly from the juniors. This is an example of a systematic sample.

ANSWER: False TYPE: TF DIFFICULTY: Moderate KEYWORDS: stratified sample 144. ________ results from the exclusion of certain groups of subjects from a population frame.

ANSWER: .

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Coverage error TYPE: FI DIFFICULTY: Difficult KEYWORDS: coverage error, survey worthiness, frame 145. Coverage error results in a ________.

ANSWER: selection bias TYPE: FI DIFFICULTY: Difficult KEYWORDS: selection bias, survey worthiness 146. ________ results from the failure to collect data on all subjects in the sample.

ANSWER: Nonresponse error or bias TYPE: FI DIFFICULTY: Moderate KEYWORDS: nonresponse error, survey worthiness 147. The sampling process begins by locating appropriate data sources called ___________.

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

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

ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: chunk sample 149. True or False: You stand at the main entrance to a departmental store and pick the first 20 customers that enter the store after it has opened its door for business on a single day. This is an example of a systematic sample.

ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: systematic sample 150. True or False: An electronic appliance chain gathered customer opinions on their services using the customer feedback forms that are attached to the product registration forms. This is an example of a convenience sample.

ANSWER: .

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True TYPE: TF DIFFICULTY: Easy KEYWORDS: convenience sample 151. True or False: To gather opinions on the efficacy of U.S. foreign policies, a sample of 50 faculty members is selected from the pool of university professors who have taught political science at the graduate level. This is an example of a judgment sample.

ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: judgment sample

152. True or False: A sample is selected by including everybody who sits in the first row of a business statistics class. This is an example of a cluster sample.

ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: cluster sample 153. True or False: The question “How many times have you abused your spouse in the last 6 months?” will most likely result in nonresponse error.

ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: nonresponse error, survey worthiness 154. True or False: The question “Is your household income last year somewhere in between $25,000 and $35,000?” will most likely result in coverage error.

ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: coverage error, survey worthiness 155. True or False: The only way one can eliminate sampling error is to take the whole population as the sample.

ANSWER: True TYPE: TF DIFFICULTY: Moderte KEYWORDS: sampling error, survey worthiness 156. True or False: Coverage error can become an ethical issue if a particular group is intentionally excluded from the frame.

ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: ethical issue, coverage error, survey worthiness, frame .

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157. True or False: Measurement error will become an ethical issue when the findings are presented without reference to sample size and margin of error.

ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: ethical issue, measurement error, sampling error, survey worthiness 158. True or False: Measurement error can become an ethical issue when a survey sponsor chooses leading questions that guide the responses in a particular direction.

ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: ethical issue, measurement error, survey worthiness 159. True or False: Measurement error can become an ethical issue when an interviewer purposely creates a Hawthorne effect or otherwise guides the responses in a particular direction.

ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: ethical issue, measurement error, survey worthiness

160. True or False: Sampling error becomes an ethical issue if the findings are purposely presented without reference to sample size and margin of error so that the sponsor can promote a viewpoint that might otherwise be truly insignificant.

ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: ethical issue, sampling error, survey worthiness 161. True or False: The professor of a business statistics class wanted to find out the average amount of time per week her students spent studying for the class. She divided the students into the left, right and center groups according to the location they sat in the class that day. One of these 3 groups was randomly selected and everyone in the group was asked the average amount of time per week he/she spent studying for the class. This is an example of a cluster sample.

ANSWER: True TYPE: TF DIFFICULTY: Moderate KEYWORDS: cluster sample 162. True or False: The professor of a business statistics class wanted to find out the average amount of time per week her students spent studying for the class. She divided the fifty students on her roster into ten groups starting from the first student on the roster. The first student was randomly selected from the first group. Then every tenth student was selected from the remaining students.

This is an example of a cluster sample. .

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ANSWER: False TYPE: TF DIFFICULTY: Moderate KEYWORDS: systematic sample 163. True or False: Selection bias occurs more frequently in systematic samples than in simple random samples.

ANSWER: True TYPE: TF DIFFICULTY: easy KEYWORDS: simple random sample, systematic sample 164. True or False: The question: “Have you used any form of illicit drugs over the past 2 months?” will most likely result in measurement error if the question is answered.

ANSWER: True TYPE: TF DIFFICULTY: Moderate KEYWORDS: measurement error, survey worthiness

165. True or False: The question: “How much did you make last year rounded to the nearest hundreds of dollars?” will most likely result in measurement error.

ANSWER: True TYPE: TF DIFFICULTY: Moderate KEYWORDS: measurement error, survey worthiness

TABLE 7-6 According to an article, 19% of the entire U.S. population have high-speed access to the Internet.

Random samples of size 200 are selected from the U.S. population.

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

ANSWER: 19% or 0.19 TYPE: FI DIFFICULTY: Easy KEYWORDS: mean, sampling distribution, proportion, central limit theorem 167. Referring to Table 7-6, the standard error of all the sample proportions is ______.

ANSWER: 0.0277 .

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TYPE: FI DIFFICULTY: Easy KEYWORDS: standard error, sampling distribution, proportion, central limit theorem 168. Referring to Table 7-6, among all the random samples of size 200, ______ % will have between 14% and 24% who have high-speed access to the Internet.

ANSWER: 92.85 using Excel or 92.82 using Table E.2 TYPE: FI DIFFICULTY: Easy KEYWORDS: sampling distribution, proportion, central limit theorem 169. Referring to Table 7-6, among all the random samples of size 200, ______ % will have between 9% and 29% who have high-speed access to the Internet.

ANSWER: 99.97 TYPE: FI DIFFICULTY: Easy KEYWORDS: sampling distribution, proportion, central limit theorem 170. Referring to Table 7-6, among all the random samples of size 200, ______ % will have more than 30% who have high-speed access to the Internet.

ANSWER: 0.0000 or virtually zero TYPE: FI DIFFICULTY: Easy KEYWORDS: sampling distribution, proportion, central limit theorem 171. Referring to Table 7-6, among all the random samples of size 200, ______ % will have less than 20% who have high-speed access to the Internet.

ANSWER: 64.08 using Excel or 64.06 using Table E.2 TYPE: FI DIFFICULTY: Easy KEYWORDS: sampling distribution, proportion, central limit theorem 172. Referring to Table 7-6, among all the random samples of size 200, 90 % will have less than _____% who have high-speed access to the Internet.

ANSWER: 22.56 using Excel or 22.55 using Table E.2 TYPE: FI DIFFICULTY: Moderate KEYWORDS: sampling distribution, proportion, central limit theorem 173. Referring to Table 7-6, among all the random samples of size 200, 90 % will have more than _____% who have high-speed access to the Internet.

ANSWER: 15.45 TYPE: FI DIFFICULTY: Moderate KEYWORDS: sampling distribution, proportion, central limit theorem TABLE 7-7 .

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Online customer service is a key element to successful online retailing. According to a marketing survey, 37.5% of online customers take advantage of the online customer service. Random samples of 200 customers are selected.

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

ANSWER: 0.375 or 37.5% TYPE: FI DIFFICULTY: Easy KEYWORDS: mean, sampling distribution, proportion, central limit theorem 175. Referring to Table 7-7, the standard error of all possible sample proportions is ______.

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

176. Referring to Table 7-7, ____ % of the samples are likely to have between 35% and 40% who take advantage of online customer service.

ANSWER: 53.48 using Excel or 53.46 using Table E.2 TYPE: FI DIFFICULTY: Easy KEYWORDS: sampling distribution, proportion, central limit theorem 177. Referring to Table 7-7, ____ % of the samples are likely to have less than 37.5% who take advantage of online customer service.

ANSWER: 50 TYPE: FI DIFFICULTY: Easy KEYWORDS: sampling distribution, proportion, central limit theorem 178. Referring to Table 7-7, 90% of the samples proportions symmetrically around the population proportion will have between _____% and _____% of the customers who take advantage of online customer service.

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

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179. Referring to Table 7-7, 95% of the samples proportions symmetrically around the population proportion will have between _____% and _____% of the customers who take advantage of online customer service.

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

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

ANSWER: c TYPE: MC DIFFICULTY: Moderate KEYWORDS: confidence interval, proportion, properties, width 2. When determining the sample size for a proportion for a given level of confidence and sampling error, the closer to 0.50 that  is estimated to be, the sample size required __________. a) is smaller b) is larger c) is not affected d) can be smaller, larger or unaffected ANSWER: b TYPE: MC DIFFICULTY: Moderate KEYWORDS: confidence interval, proportion, properties 3. A 99% confidence interval estimate can be interpreted to mean that a) if all possible samples are taken and confidence interval estimates are developed, 99% of them would include the true population mean somewhere within their interval. b) we have 99% confidence that we have selected a sample whose interval does include the population mean. c) Both of the above. d) None of the above.

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

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Confidence Interval Estimation 5. Which of the following is not true about the Student’s t distribution? a) It has more area in the tails and less in the center than does the normal distribution. b) It is used to construct confidence intervals for the population mean when the population standard deviation is known. c) It is bell shaped and symmetrical. d) As the number of degrees of freedom increases, the t distribution approaches the normal distribution.

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

ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: confidence interval, mean, standard deviation unknown 7. The t distribution a) assumes the population is normally distributed. b) approaches the normal distribution as the sample size increases. c) has more area in the tails than does the normal distribution. d) All of the above.

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

ANSWER: d TYPE: MC DIFFICULTY: Difficult KEYWORDS: confidence interval, interpretation

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9. It is desired to estimate the average total compensation of CEOs in the Service industry. Data were randomly collected from 18 CEOs and the 97% confidence interval was calculated to be ($2,181,260, $5,836,180). Based on the interval above, do you believe the average total compensation of CEOs in the Service industry is more than $3,000,000? a) Yes, and I am 97% confident of it. b) Yes, and I am 78% confident of it. c) I am 97% confident that the average compensation is $3,000,000. d) I cannot conclude that the average exceeds $3,000,000 at the 97% confidence level.

ANSWER: d TYPE: MC DIFFICULTY: Difficult KEYWORDS: confidence interval, interpretation 10. A confidence interval was used to estimate the proportion of statistics students that are females. A random sample of 72 statistics students generated the following 90% confidence interval: (0.438, 0.642). Based on the interval above, is the population proportion of females equal to 0.60? a) No, and we are 90% sure of it. b) No. The proportion is 54.17%. c) Maybe. 0.60 is a believable value of the population proportion based on the information above. d) Yes, and we are 90% sure of it.

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

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

ANSWER: a TYPE: MC DIFFICULTY: Easy KEYWORDS: mean, sample size determination 13. Suppose a 95% confidence interval for  turns out to be (1,000, 2,100). Give a definition of what it means to be “95% confident” as an inference. a) In repeated sampling, the population parameter would fall in the given interval 95% of the time. b) In repeated sampling, 95% of the intervals constructed would contain the population mean. c) 95% of the observations in the entire population fall in the given interval. d) 95% of the observations in the sample fall in the given interval.

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

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

15. Suppose a 95% confidence interval for  has been constructed. If it is decided to take a larger sample and to decrease the confidence level of the interval, then the resulting interval width .

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would . (Assume that the sample statistics gathered would not change very much for the new sample.) a) be larger than the current interval width b) be narrower than the current interval width c) be the same as the current interval width d) be unknown until actual sample sizes and reliability levels were determined ANSWER: b TYPE: MC DIFFICULTY: Moderate KEYWORDS: confidence interval, properties, width 16. In the construction of confidence intervals, if all other quantities are unchanged, an increase in the sample size will lead to a interval. a) narrower b) wider c) less significant d) biased ANSWER: a TYPE: MC DIFFICULTY: Easy KEYWORDS: confidence interval, properties, width 17. A major department store chain is interested in estimating the average amount its credit card customers spent on their first visit to the chain’s new store in the mall. Fifteen credit card accounts were randomly sampled and analyzed with the following results: X = $50.50 and s 2 = 400 . Assuming the distribution of the amount spent on their first visit is approximately normal, what is the shape of the sampling distribution of the sample mean that will be used to create the desired confidence interval for  ? a) Approximately normal with a mean of $50.50 b) A standard normal distribution c) A t distribution with 15 degrees of freedom d) A t distribution with 14 degrees of freedom ANSWER: d TYPE: MC DIFFICULTY: Easy KEYWORDS: confidence interval, mean, t distribution

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

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

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

19. Private colleges and universities rely on money contributed by individuals and corporations for their operating expenses. Much of this money is put into a fund called an endowment, and the college spends only the interest earned by the fund. A recent survey of 8 private colleges in the United States revealed the following endowments (in millions of dollars): 60.2, 47.0, 235.1, 490.0, 122.6, 177.5, 95.4, and 220.0. What value will be used as the point estimate for the mean endowment of all private colleges in the United States? a) $1,447.8 b) $180.975 c) $143.042 d) $8 ANSWER: b TYPE: MC DIFFICULTY: Easy KEYWORDS: point estimate, mean 20. Private colleges and universities rely on money contributed by individuals and corporations for their operating expenses. Much of this money is put into a fund called an endowment, and the college spends only the interest earned by the fund. A recent survey of 8 private colleges in the United States revealed the following endowments (in millions of dollars): 60.2, 47.0, 235.1, 490.0, 122.6, 177.5, 95.4, and 220.0. Summary statistics yield X = 180.975 and s = 143.042 .

Calculate a 95% confidence interval for the mean endowment of all the private colleges in the United States assuming a normal distribution for the endowments. a) $180.975  $94.066 b) $180.975  $99.123 c) $180.975  $116.621 d) $180.975  $119.586 ANSWER: d TYPE: MC DIFFICULTY: Moderate KEYWORDS: confidence interval, mean, t distribution

21. A university system enrolling hundreds of thousands of students is considering a change in the way students pay for their education. Currently, the students pay $55 per credit hour. The university system administrators are contemplating charging each student a set fee of $750 per quarter, regardless of how many credit hours each takes. To see if this proposal would be economically feasible, the administrators would like to know how many credit hours, on the .

Confidence Interval Estimation 231

average, each student takes per quarter. A random sample of 250 students yields a mean of 14.1 credit hours per quarter and a standard deviation of 2.3 credit hours per quarter. Suppose the administration wanted to estimate the mean to within 0.1 hours at 95% reliability and assumed that the sample standard deviation provided a good estimate for the population standard deviation.

How large a total sample would they need to take?

ANSWER: n = 2033 TYPE: PR DIFFICULTY: Easy KEYWORDS: mean, sample size determination 22. As an aid to the establishment of personnel requirements, the director of a hospital wishes to estimate the mean number of people who are admitted to the emergency room during a 24-hour period. The director randomly selects 64 different 24-hour periods and determines the number of admissions for each. For this sample, X = 19.8 and s2 = 25. Which of the following assumptions is necessary in order for a confidence interval to be valid? a) The population sampled from has an approximate normal distribution. b) The population sampled from has an approximate t distribution. c) The mean of the sample equals the mean of the population. d) None of these assumptions are necessary.

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

ANSWER: 19.8  1.249 TYPE: PR DIFFICULTY: Moderate KEYWORDS: confidence interval, mean, t distribution

24. As an aid to the establishment of personnel requirements, the director of a hospital wishes to estimate the mean number of people who are admitted to the emergency room during a 24-hour period. The director randomly selects 64 different 24-hour periods and determines the number of admissions for each. For this sample, X = 19.8 and s2 = 25. Using the sample standard deviation as an estimate for the population standard deviation, what size sample should the director choose if she wishes to estimate the mean number of admissions per 24-hour period to within 1 admission with 99% reliability,? .

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ANSWER: n = 166 TYPE: PR DIFFICULTY: Moderate KEYWORDS: mean, sample size determination 25. A university dean is interested in determining the proportion of students who receive some sort of financial aid. Rather than examine the records for all students, the dean randomly selects 200 students and finds that 118 of them are receiving financial aid. Use a 90% confidence interval to estimate the true proportion of students who receive financial aid.

ANSWER: 0.59  0.057 or 0.533    0.647 TYPE: PR DIFFICULTY: Moderate KEYWORDS: confidence interval, proportion 26. A university dean is interested in determining the proportion of students who receive some sort of financial aid. Rather than examine the records for all students, the dean randomly selects 200 students and finds that 118 of them are receiving financial aid. The 95% confidence interval for p is 0.59  0.07. Interpret this interval. a) We are 95% confident that the true proportion of all students receiving financial aid is between 0.52 and 0.66. b) 95% of the students get between 52% and 66% of their tuition paid for by financial aid. c) We are 95% confident that between 52% and 66% of the sampled students receive some sort of financial aid. d) We are 95% confident that 59% of the students are on some sort of financial aid.

ANSWER: a TYPE: MC DIFFICULTY: Moderate KEYWORDS: confidence interval, proportion, interpretation

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

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30. An economist is interested in studying the incomes of consumers in a particular region. The population standard deviation is known to be $1,000. A random sample of 50 individuals resulted in an average income of $15,000. What total sample size would the economist need to use for a 95% confidence interval if the width of the interval should not be more than $100? a) n = 1537 b) n = 385 c) n = 40 d) n = 20 ANSWER: a TYPE: MC DIFFICULTY: Easy KEYWORDS: mean, sample size determination 31. The head librarian at the Library of Congress has asked her assistant for an interval estimate of the mean number of books checked out each day. The assistant provides the following interval estimate: from 740 to 920 books per day. What is an efficient, unbiased point estimate of the number of books checked out each day at the Library of Congress? a) 740 b) 830 c) 920 d) 1,660 ANSWER: b TYPE: MC DIFFICULTY: Easy KEYWORDS: point estimate, mean 32. The head librarian at the Library of Congress has asked her assistant for an interval estimate of the mean number of books checked out each day. The assistant provides the following interval estimate: from 740 to 920 books per day. If the head librarian knows that the population standard deviation is 150 books checked out per day, approximately how large a sample did her assistant use to determine the interval estimate? a) 2 b) 3 c) 12 d) It cannot be determined from the information given.

ANSWER: d TYPE: MC DIFFICULTY: Difficult KEYWORDS: mean, sample size determination

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Confidence Interval Estimation 235

33. The head librarian at the Library of Congress has asked her assistant for an interval estimate of the mean number of books checked out each day. The assistant provides the following interval estimate: from 740 to 920 books per day. If the head librarian knows that the population standard deviation is 150 books checked out per day, and she asked her assistant for a 95% confidence interval, approximately how large a sample did her assistant use to determine the interval estimate? a) 125 b) 13 c) 11 d) 4 ANSWER: c TYPE: MC DIFFICULTY: Difficult KEYWORDS: mean, sample size determination 34. The head librarian at the Library of Congress has asked her assistant for an interval estimate of the mean number of books checked out each day. The assistant provides the following interval estimate: from 740 to 920 books per day. If the head librarian knows that the population standard deviation is 150 books checked out per day, and she asked her assistant to use 25 days of data to construct the interval estimate, what confidence level can she attach to the interval estimate? a) 99.7% b) 99.0% c) 98.0% d) 95.4% ANSWER: a TYPE: MC DIFFICULTY: Difficult KEYWORDS: mean, sample size determination 35. True or False: A race car driver tested his car for time from 0 to 60 mph, and in 20 tests obtained an average of 4.85 seconds with a standard deviation of 1.47 seconds. A 95% confidence interval for the 0 to 60 time is 4.52 seconds to 5.18 seconds.

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

ANSWER: False TYPE: TF DIFFICULTY: Moderate KEYWORDS: confidence interval, mean, t distribution

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37. True or False: Given a sample mean of 2.1 and a population standard deviation of 0.7 from a sample of 10 data points, a 90% confidence interval will have a width of 2.36.

ANSWER: False TYPE: TF DIFFICULTY: Moderate KEYWORDS: confidence interval, mean, standardized normal distribution 38. True or False: A sample size of 5 provides a sample mean of 9.6. If the population variance is known to be 5 and the population distribution is assumed to be normal, the lower limit for a 92% confidence interval is 7.85.

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

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

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

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

ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: confidence interval, mean, t distribution

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43. True or False: The t distribution allows the calculation of confidence intervals for means for small samples when the population variance is not known, regardless of the shape of the distribution in the population.

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

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

ANSWER: False TYPE: TF DIFFICULTY: Moderate KEYWORDS: confidence interval, proportion 46. True or False: The sample mean is a point estimate of the population mean.

ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: point estimate, mean 47. True or False: The confidence interval estimate of the population mean is constructed around the sample mean.

ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: confidence interval, mean 48. True or False: The confidence interval estimate of the population proportion is constructed around the sample proportion.

ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: confidence interval, proportion

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49. True or False: A point estimate consists of a single sample statistic that is used to estimate the true population parameter.

ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: point estimate 50. True or False: The confidence interval obtained will always correctly estimate the population parameter.

ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: confidence interval, interpretation 51. True or False: Other things being equal, the confidence interval for the mean will be wider for 95% confidence than for 90% confidence.

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

ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: confidence interval, mean, t distribution 53. True or False: The t distribution is used to develop a confidence interval estimate of the population proportion when the population standard deviation is unknown.

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

ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: confidence interval, proportion, standardized normal distribution

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Confidence Interval Estimation 239

55. True or False: The standardized normal distribution is used to develop a confidence interval estimate of the population proportion when the sample size is sufficiently large.

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

ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: t distribution, standardized normal distribution 57. True or False: The t distribution is used to develop a confidence interval estimate of the population mean when the population standard deviation is unknown.

ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: t distribution, standardized normal distribution 58. True or False: In estimating the population mean with the population standard deviation unknown, if the sample size is 12, there will be 6 degrees of freedom.

ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: confidence interval, mean, t distribution 59. True or False: The difference between the sample mean and the population mean is called the sampling error.

ANSWER: False TYPE: TF DIFFICULTY: Difficult KEYWORDS: sampling error 60. True or False: The difference between the sample proportion and the population proportion is called the sampling error.

ANSWER: False TYPE: TF DIFFICULTY: Moderate KEYWORDS: sampling error

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61. True or False: The difference between the sample size and the population size is called the sampling error.

ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: sampling error 62. True or False: The difference between the upper limit of a confidence interval and the point estimate used in constructing the confidence interval is called the sampling error.

ANSWER: True TYPE: TF DIFFICULTY: Moderate KEYWORDS: sampling error 63. True or False: The difference between the lower limit of a confidence interval and the point estimate used in constructing the confidence interval is called the sampling error.

ANSWER: True TYPE: TF DIFFICULTY: Moderate KEYWORDS: sampling error 64. True or False: Sampling error equals half the width of a confidence interval.

ANSWER: True TYPE: TF DIFFICULTY: Moderate KEYWORDS: sampling error 65. True or False: The width of a confidence interval equals twice the sampling error.

ANSWER: True TYPE: TF DIFFICULTY: Moderate KEYWORDS: sampling error 66. True or False: The sampling error can either be positive or negative.

ANSWER: True TYPE: TF DIFFICULTY: Difficult KEYWORDS: sampling error 67. True or False: A population parameter is used to estimate a confidence interval.

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

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Confidence Interval Estimation 241

68. True or False: For a given data set, the confidence interval will be wider for 95% confidence than for 90% confidence.

ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: confidence interval, properties, width 69. True or False: Holding the sample size fixed, increasing the level of confidence in a confidence interval will necessarily lead to a wider confidence interval.

ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: confidence interval, properties, width, trade-off 70. True or False: Holding the level of confidence fixed, increasing the sample size will lead to a wider confidence interval.

ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: confidence interval, properties, width, trade-off 71. True or False: Holding the width of a confidence interval fixed, increasing the level of confidence can be achieved with a lower sample size.

ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: confidence interval, properties, width, trade-off TABLE 8-1 A random sample of 100 stores from a large chain of 1,000 garden supply stores was selected to determine the average number of lawnmowers sold at an end-of-season clearance sale. The sample results indicated an average of 6 and a standard deviation of 2 lawnmowers sold. A 95% confidence interval (5.623 to 6.377) was established based on these results.

72. True or False: Referring to Table 8-1, if the population had consisted of 10,000 stores, the confidence interval estimate of the mean would have been wider in range.

ANSWER: False TYPE: TF DIFFICULTY: Moderate KEYWORDS: confidence interval, mean, properties, width

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73. True or False: Referring to Table 8-1, of all possible samples of 100 stores drawn from the population of 1,000 stores, 95% of the sample means will fall between 5.623 and 6.377 lawnmowers.

ANSWER: False TYPE: TF DIFFICULTY: Moderate KEYWORDS: confidence interval, mean, interpretation 74. True or False: Referring to Table 8-1, of all possible samples of 100 stores taken from the population of 1,000 stores, 95% of the confidence intervals developed will contain the true population mean within the interval.

ANSWER: True TYPE: TF DIFFICULTY: Moderate KEYWORDS: confidence interval, mean, interpretation 75. True or False: Referring to Table 8-1, there are 10 possible samples of 100 stores that can be selected out of the population of 1,000 stores.

ANSWER: False TYPE: TF DIFFICULTY: Moderate KEYWORDS: confidence interval, mean, interpretation 76. True or False: Referring to Table 8-1, 95% of the stores have sold between 5.623 and 6.377 lawnmowers.

ANSWER: False TYPE: TF DIFFICULTY: Moderate KEYWORDS: confidence interval, mean, interpretation 77. True or False: Referring to Table 8-1, we do not know for sure whether the true population mean is between 5.623 and 6.377 lawnmowers sold.

ANSWER: True TYPE: TF DIFFICULTY: Moderate KEYWORDS: confidence interval, mean, interpretation

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Confidence Interval Estimation 243

TABLE 8-2 The managers of a company are worried about the morale of their employees. In order to determine if a problem in this area exists, they decide to evaluate the attitudes of their employees with a standardized test. They select the Fortunato test of job satisfaction, which has a known standard deviation of 24 points.

78. Referring to Table 8-2, they should sample ________ employees if they want to estimate the mean score of the employees within 5 points with 90% confidence.

ANSWER: 63 TYPE: FI DIFFICULTY: Easy KEYWORDS: mean, sample size determination 79. Referring to Table 8-2, due to financial limitations, the managers decide to take a sample of 45 employees. This yields a mean score of 88.0 points. A 90% confidence interval would go from ________ to ________.

ANSWER: 82.12 to 93.88 TYPE: FI DIFFICULTY: Moderate KEYWORDS: confidence interval, mean, standardized normal distribution 80. True or False: Referring to Table 8-2, this confidence interval is only valid if the scores on the Fortunato test are normally distributed.

ANSWER: False TYPE: TF DIFFICULTY: Difficult EXPLANATION: With a sample size of 45, this confidence interval will still be valid if the scores are not normally distributed due to the central limit theorem.

KEYWORDS: confidence interval, mean, standardized normal distribution, central limit theorem TABLE 8-3 A quality control engineer is interested in the mean length of sheet insulation being cut automatically by machine. The desired length of the insulation is 12 feet. It is known that the standard deviation in the cutting length is 0.15 feet. A sample of 70 cut sheets yields a mean length of 12.14 feet. This sample will be used to obtain a 99% confidence interval for the mean length cut by machine.

81. Referring to Table 8-3, the critical value to use in obtaining the confidence interval is ________.

ANSWER: 2.58 TYPE: FI DIFFICULTY: Easy KEYWORDS: confidence interval, mean, standardized normal distribution

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82. Referring to Table 8-3, the confidence interval goes from ________ to ________.

ANSWER: 12.09 to 12.19 TYPE: FI DIFFICULTY: Moderate KEYWORDS: confidence interval, mean, standardized normal distribution 83. True or False: Referring to Table 8-3, the confidence interval indicates that the machine is not working properly.

ANSWER: True TYPE: TF DIFFICULTY: Moderate KEYWORDS: confidence interval, mean, standardized normal distribution, interpretation 84. True or False: Referring to Table 8-3, the confidence interval is valid only if the lengths cut are normally distributed.

ANSWER: False TYPE: TF DIFFICULTY: Moderate EXPLANATION: With a sample size of 70, this confidence interval will still be valid if the lengths cut are not normally distributed due to the central limit theorem.

KEYWORDS: confidence interval, mean, standardized normal distribution, central limit theorem 85. Referring to Table 8-3, suppose the engineer had decided to estimate the mean length to within 0.03 with 99% confidence. Then the sample size would be ________.

ANSWER: 165.8724 rounds up to 166 TYPE: FI DIFFICULTY: Moderate KEYWORDS: mean, sample size determination TABLE 8-4 To become an actuary, it is necessary to pass a series of 10 exams, including the most important one, an exam in probability and statistics. An insurance company wants to estimate the mean score on this exam for actuarial students who have enrolled in a special study program. They take a sample of 8 actuarial students in this program and determine that their scores are: 2, 5, 8, 8, 7, 6, 5, and 7. This sample will be used to calculate a 90% confidence interval for the mean score for actuarial students in the special study program.

86. Referring to Table 8-4, the mean of the sample is __________, while the standard deviation of the sample is __________.

ANSWER: 6.0; 2.0 TYPE: FI DIFFICULTY: Moderate KEYWORDS: confidence interval

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Confidence Interval Estimation 245

87. Referring to Table 8-4, the confidence interval will be based on __________ degrees of freedom.

ANSWER: 7 TYPE: FI DIFFICULTY: Easy KEYWORDS: confidence interval, mean, t distribution 88. Referring to Table 8-4, the critical value used in constructing a 90% confidence interval is __________.

ANSWER: 1.8946 TYPE: FI DIFFICULTY: Easy KEYWORDS: confidence interval, mean, t distribution 89. Referring to Table 8-4, a 90% confidence interval for the mean score of actuarial students in the special program is from __________ to __________.

ANSWER: 4.66 to 7.34 TYPE: FI DIFFICULTY: Moderate KEYWORDS: confidence interval, mean, t distribution 90. True or False: Referring to Table 8-4, for the confidence interval to be valid, it is necessary that test scores of students in the special study program on the actuarial exam be normally distributed.

ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: confidence interval, mean, t distribution 91. True or False: Referring to Table 8-4, it is possible that the confidence interval obtained will not contain the mean score for all actuarial students in the special class.

ANSWER: True TYPE: TF DIFFICULTY: Moderate KEYWORDS: confidence interval, mean, interpretation 92. True or False: Referring to Table 8-4, if we use the same sample information to obtain a 95% confidence interval, the resulting interval would be narrower than the one obtained here with 90% confidence.

ANSWER: False TYPE: TF DIFFICULTY: Moderate KEYWORDS: confidence interval, mean, properties, width

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93. Suppose a department store wants to estimate the average age of the customers of its contemporary apparel department, correct to within 2 years, with level of confidence equal to 95%. Management believes that the standard deviation is 8 years. The sample size they should take is ________.

ANSWER: 62 TYPE: FI DIFFICULTY: Easy KEYWORDS: mean, sample size determination TABLE 8-5 The actual voltages of power packs labeled as 12 volts are as follows: 11.77, 11.90, 11.64, 11.84, 12.13, 11.99, and 11.77.

94. Referring to Table 8-5, a confidence interval for this sample would be based on the t distribution with __________ degrees of freedom.

ANSWER: 6 TYPE: FI DIFFICULTY: Easy KEYWORDS: confidence interval, mean, t distribution 95. Referring to Table 8-5, the critical value for a 99% confidence interval for this sample is __________.

ANSWER: 3.7074 TYPE: FI DIFFICULTY: Easy KEYWORDS: confidence interval, mean, t distribution 96. Referring to Table 8-5, a 99% confidence interval for the mean voltage of the power packs is from __________ to __________.

ANSWER: 11.6367 to 12.0891 TYPE: FI DIFFICULTY: Moderate KEYWORDS: confidence interval, mean, t distribution 97. True or False: Referring to Table 8-5, a 95% confidence interval for the mean voltage of the power pack is wider than a 99% confidence interval.

ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: confidence interval, mean, properties, width

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98. True or False: Referring to Table 8-5, a 99% confidence interval will contain 99% of the voltages for all such power packs.

ANSWER: False TYPE: TF DIFFICULTY: Moderate KEYWORDS: confidence interval, mean, interpretation 99. True or False: Referring to Table 8-5, a confidence interval estimate of the population mean would only be valid if the distribution of voltages is normal.

ANSWER: True TYPE: TF DIFFICULTY: Moderate KEYWORDS: confidence interval, mean, t distribution 100.

True or False: Referring to Table 8-5, a 90% confidence interval calculated from the same data would be narrower than a 99% confidence interval.

ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: confidence interval, mean, properties, width 101.

True or False: Referring to Table 8-5, it is possible that the 99% confidence interval calculated from the data will not contain the mean voltage for the sample.

ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: confidence interval, mean, interpretation 102.

True or False: Referring to Table 8-5, it is possible that the 99% confidence interval calculated from the data will not contain the mean voltage for the entire population.

ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: confidence interval, mean, interpretation

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TABLE 8-6 A sample of salary offers (in thousands of dollars) given to management majors is: 28, 31, 26, 32, 27, 28, 27, 30, 31, and 29. Using this data to obtain a 95% confidence interval resulted in an interval from 27.5 to 30.3.

103. True or False: Referring to Table 8-6, 95% of the time, the sample mean salary offer to management majors will fall between 27.5 and 30.3.

ANSWER: False TYPE: TF DIFFICULTY: Moderate KEYWORDS: confidence interval, mean, interpretation 104. True or False: Referring to Table 8-6, 95% of the salary offers are between 27.5 and 30.3.

ANSWER: False TYPE: TF DIFFICULTY: Moderate KEYWORDS: confidence interval, mean, interpretation 105. True or False: Referring to Table 8-6, it is possible that the mean of the population is between 27.5 and 30.3.

ANSWER: True TYPE: TF DIFFICULTY: Moderate KEYWORDS: confidence interval, mean, interpretation 106. True or False: Referring to Table 8-6, it is possible that the mean of the population is not between 27.5 and 30.3.

ANSWER: True TYPE: TF DIFFICULTY: Moderate KEYWORDS: confidence interval, mean, interpretation 107. True or False: Referring to Table 8-6, 95% of the sample means will fall between 27.5 and 30.3.

ANSWER: False TYPE: TF DIFFICULTY: Moderate KEYWORDS: confidence interval, mean, interpretation 108. True or False: Referring to Table 8-6, 95% of all confidence intervals constructed similarly to this one with a sample size of 10 will contain the mean of the population.

ANSWER: True TYPE: TF DIFFICULTY: Moderate KEYWORDS: confidence interval, mean, interpretation .

Confidence Interval Estimation 249

109. True or False: Referring to Table 8-6, a 99% confidence interval for the mean of the population from the same sample would be wider than 27.5 to 30.3.

ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: confidence interval, mean, properties, width 110. True or False: Referring to Table 8-6, the confidence interval obtained is valid only if the distribution of the population of salary offers is normal.

ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: confidence interval, mean, t distribution 111. A prison official wants to estimate the proportion of cases of recidivism. Examining the records of 250 convicts, the official determines that there are 65 cases of recidivism. A confidence interval will be obtained for the proportion of cases of recidivism. Part of this calculation includes the estimated standard error of the sample proportion. For this sample, the estimated standard error is __________.

ANSWER: 0.028 TYPE: FI DIFFICULTY: Easy KEYWORDS: confidence interval, proportion 112. A prison official wants to estimate the proportion of cases of recidivism. Examining the records of 250 convicts, the official determines that there are 65 cases of recidivism. A 99% confidence interval for the proportion of cases of recidivism would go from __________ to __________.

ANSWER: 0.189 to 0.331 TYPE: FI DIFFICULTY: Moderate KEYWORDS: confidence interval, proportion TABLE 8-7 After an extensive advertising campaign, the manager of a company wants to estimate the proportion of potential customers that recognize a new product. She samples 120 potential consumers and finds that 54 recognize this product. She uses this sample information to obtain a 95% confidence interval that goes from 0.36 to 0.54.

113. True or False: Referring to Table 8-7, the parameter of interest to the manager is the proportion of potential customers in this sample that recognize the new product.

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

.

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Confidence Interval Estimation

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

ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: confidence interval, proportion, parameter 115. True or False: Referring to Table 8-7, this interval requires the use of the t distribution to obtain the confidence coefficient.

ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: confidence interval, proportion, t distribution 116. True or False: Referring to Table 8-7, this interval requires the assumption that the distribution of the number of people recognizing the product has a normal distribution.

ANSWER: False TYPE: TF DIFFICULTY: Moderate KEYWORDS: confidence interval, proportion 117. True or False: Referring to Table 8-7, 95% of the time, the proportion of people that recognize the product will fall between 0.36 and 0.54.

ANSWER: False TYPE: TF DIFFICULTY: Moderate KEYWORDS: confidence interval, proportion, interpretation 118. True or False: Referring to Table 8-7, 95% of the time, the sample proportion of people that recognize the product will fall between 0.36 and 0.54.

ANSWER: False TYPE: TF DIFFICULTY: Moderate KEYWORDS: confidence interval, proportion, interpretation 119. True or False: Referring to Table 8-7, 95% of the people will recognize the product between 36% and 54% of the time.

ANSWER: False TYPE: TF DIFFICULTY: Difficult KEYWORDS: confidence interval, proportion, interpretation

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Confidence Interval Estimation 251

120. True or False: Referring to Table 8-7, it is possible that the true proportion of people that recognize the product is between 0.36 and 0.54.

ANSWER: True TYPE: TF DIFFICULTY: Moderate KEYWORDS: confidence interval, proportion, interpretation 121. True or False: Referring to Table 8-7, it is possible that the true proportion of people that recognize the product is not between 0.36 and 0.54.

ANSWER: True TYPE: TF DIFFICULTY: Moderate KEYWORDS: confidence interval, proportion, interpretation 122. The head of a computer science department is interested in estimating the proportion of students entering the department who will choose the new computer engineering option. Suppose there is no information about the proportion of students who might choose the option. What size sample should the department head take if she wants to be 95% confident that the estimate is within 0.10 of the true proportion?

ANSWER: 97 TYPE: PR DIFFICULTY: Moderate KEYWORDS: proportion, sample size determination 123. The head of a computer science department is interested in estimating the proportion of students entering the department who will choose the new computer engineering option. A preliminary sample indicates that the proportion will be around 0.25. Therefore, what size sample should the department head take if she wants to be 95% confident that the estimate is within 0.10 of the true proportion?

ANSWER: 73 TYPE: PR DIFFICULTY: Easy KEYWORDS: proportion, sample size determination

.

252

Confidence Interval Estimation

TABLE 8-8 A hotel chain wants to estimate the average number of rooms rented daily in each month. The population of rooms rented daily is assumed to be normally distributed for each month with a standard deviation of 24 rooms.

124. Referring to Table 8-8, during January, a sample of 16 days has a sample mean of 48 rooms.

This information is used to calculate an interval estimate for the population mean to be from 40 to 56 rooms. What is the level of confidence of this interval?

ANSWER: 81.76% TYPE: PR DIFFICULTY: Difficult KEYWORDS: confidence interval, mean 125. Referring to Table 8-8, during February, a sample of 25 days has a sample mean of 37 rooms.

Use this information to calculate a 92% confidence interval for the population mean.

ANSWER: 28.60 to 45.40 TYPE: PR DIFFICULTY: Easy KEYWORDS: confidence interval, mean 126. The county clerk wants to estimate the proportion of retired voters who will need special election facilities. The clerk wants to find a 95% confidence interval for the population proportion which extends at most 0.07 to either side of the sample proportion. How large a sample must be taken to assure these conditions are met?

ANSWER: 196 TYPE: PR DIFFICULTY: Easy KEYWORDS: proportion, sample size determination 127. The county clerk wants to estimate the proportion of retired voters who will need special election facilities. Suppose a sample of 400 retired voters was taken. If 150 need special election facilities, calculate an 80% confidence interval for the population proportion.

ANSWER: 0.344 to 0.406 TYPE: PR DIFFICULTY: Easy KEYWORDS: confidence interval, proportion

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Confidence Interval Estimation 253

TABLE 8-9 A wealthy real estate investor wants to decide whether it is a good investment to build a high-end shopping complex in a suburban county in Chicago. His main concern is the total market value of the 3,605 houses in the suburban county. He commissioned a statistical consulting group to take a sample of 200 houses and obtained a sample average market price of $225,000 and a sample standard deviation of $38,700. The consulting group also found out that the average differences between market prices and appraised prices was $125,000 with a standard deviation of $3,400. Also the proportion of houses in the sample that are appraised for higher than the market prices is 0.24.

128. A quality control engineer is interested in estimating the proportion of defective items coming off a production line. In a sample of 300 items, 27 are defective. A 90% confidence interval for the proportion of defectives from this production line would go from __________ to __________.

ANSWER: 0.063 to 0.117 TYPE: FI DIFFICULTY: Moderate KEYWORDS: confidence interval, proportion

TABLE 8-10 The president of a university would like to estimate the proportion of the student population that owns a personal computer. In a sample of 500 students, 417 own a personal computer.

129. True or False: Referring to Table 8-10, the parameter of interest is the average number of students in the population who own a personal computer.

ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: confidence interval, proportion, parameter 130. True or False: Referring to Table 8-10, the parameter of interest is the proportion of student population who own a personal computer.

ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: confidence interval, proportion, parameter 131. Referring to Table 8-10, the critical value for a 99% confidence interval for this sample is __________.

ANSWER: 2.5758 TYPE: FI DIFFICULTY: Easy KEYWORDS: confidence interval, proportion

132. Referring to Table 8-10, a 99% confidence interval for the proportion of student population who own a personal computer is from __________ to __________. .

254

Confidence Interval Estimation

ANSWER: 0.7911 to 0.8769 TYPE: FI DIFFICULTY: Moderate KEYWORDS: confidence interval, proportion 133. Referring to Table 8-10, the sampling error of a 99% confidence interval for the proportion of student population who own a personal computer is __________.

ANSWER: 0.04286 TYPE: FI DIFFICULTY: Moderate KEYWORDS: confidence interval, proportion, sampling error 134. True or False: Referring to Table 8-10, a 95% confidence interval for the proportion of student population who own a personal computer is narrower than a 99% confidence interval.

ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: confidence interval, proportion, properties, width 135. True or False: Referring to Table 8-10, a 99% confidence interval will contain 99% of the student population who own a personal computer.

ANSWER: False TYPE: TF DIFFICULTY: Moderate KEYWORDS: confidence interval, proportion, interpretation 136. True or False: Referring to Table 8-10, a confidence interval estimate of the population proportion would only be valid if the distribution of the number of students who own a personal computer is normal.

ANSWER: False TYPE: TF DIFFICULTY: Moderate KEYWORDS: confidence interval, proportion 137. True or False: Referring to Table 8-10, a 90% confidence interval calculated from the same data would be narrower than a 99% confidence interval.

ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: confidence interval, proportion, properties, width

138. True or False: Referring to Table 8-10, it is possible that the 99% confidence interval calculated from the data will not contain the sample proportion of students who own a personal computer.

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Confidence Interval Estimation 255

ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: confidence interval, proportion, interpretation 139. True or False: Referring to Table 8-10, it is possible that the 99% confidence interval calculated from the data will not contain the proportion of student population who own a personal computer.

ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: confidence interval, proportion, interpretation 140. True or False: Referring to Table 8-10, we are 99% confident that the average numbers of student population who own a personal computer is between 0.7911 and 0.8769.

ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: confidence interval, proportion, interpretation 141. True or False: Referring to Table 8-10, we are 99% confident that between 79.11% and 87.69% of the student population own a personal computer.

ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: confidence interval, proportion, interpretation

TABLE 8-11 The superintendent of a unified school district of a small town wants to make sure that no more than 5% of the students skip more than 10 days of school in a year. A random sample of 145 students from a population of 800 showed that 12 students skipped more than 10 days of school last year.

142. Referring to Table 8-11, what is the critical value for the 95% one-sided confidence interval for the proportion of students who skipped more than 10 days of school last year?

ANSWER: 1.6449 (using PHStat) or 1.645 (using Table E.2) TYPE: PR DIFFICULTY: Easy KEYWORDS: confidence interval, proportion, one-sided, critical value

143. Referring to Table 8-11, what is the upper bound of the 95% one-sided confidence interval for the proportion of students who skipped more than 10 days of school last year?

ANSWER: .

256

Confidence Interval Estimation

0.1168 TYPE: PR DIFFICULTY: Easy KEYWORDS: confidence interval, proportion, one-sided 144. True or False: Referring to Table 8-11, the superintendent can conclude with 95% level of confidence that no more than 5% of the students in the unified school district skipped more than 10 days of school last year.

ANSWER: False TYPE: TF DIFFICULTY: Moderate KEYWORDS: confidence interval, proportion, one-sided, interpretation

TABLE 8-12 The president of a university is concerned that illicit drug use on campus is higher than the 5% acceptable level. A random sample of 250 students from a population of 2000 revealed that 7 of them had used illicit drug during the last 12 months.

145. Referring to Table 8-12, what is the critical value for the 90% one-sided confidence interval for the proportion of students who had used illicit drug during the last 12 months?

ANSWER: 1.2816 (using PHStat) or 1.28 (using Table E.2) TYPE: PR DIFFICULTY: Easy KEYWORDS: confidence interval, proportion, one-sided, critical value 146. Referring to Table 8-12, what is the upper bound of the 90% one-sided confidence interval for the proportion of students who had used illicit drug during the last 12 months?

ANSWER: 0.0405 TYPE: PR DIFFICULTY: Easy KEYWORDS: confidence interval, proportion, one-sided 147. True or False: Referring to Table 8-12, the president can be 90% confident that no more than 5% of the students at the university had used illicit drug during the last 12 months.

ANSWER: True TYPE: TF DIFFICULTY: Moderate KEYWORDS: confidence interval, proportion, one-sided, interpretation

148. True or False: Referring to Table 8-12, using the 90% one-sided confidence interval, the president can be 95% confident that no more than 5% of the students at the university had used illicit drug during the last 12 months.

ANSWER: False .

Confidence Interval Estimation 257

TYPE: TF DIFFICULTY: Difficult EXPLANATION: Since the upper limit of the 95% one-sided confidence interval will be larger than 0.0405, the president cannot be 95% confident that no more than 5% of the students at the university had used illicit drug during the last 12 months.

KEYWORDS: confidence interval, proportion, one-sided, interpretation 149. True or False: Referring to Table 8-12, using the 90% one-sided confidence interval, the president can be 85% confident that no more than 5% of the students at the university had used illicit drug during the last 12 months.

ANSWER: True TYPE: TF DIFFICULTY: Difficult EXPLANATION: Since the upper limit of the 85% one-sided confidence interval will be smaller than 0.0405, the president can be 85% confident that no more than 5% of the students at the university had used illicit drug during the last 12 months.

KEYWORDS: confidence interval, proportion, one-sided, interpretation

TABLE 8-13 A university wanted to find out the percentage of students who felt comfortable reporting cheating by their fellow students. A surveyed of 2,800 students was conducted and the students were asked if they felt comfortable reporting cheating by their fellow students. The results were 1,344 answered “Yes” and 1,456 answered “no”.

150. True or False: Referring to Table 8-13, the parameter of interest is the total number of students in the population who feel comfortable reporting cheating by their fellow students.

ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: confidence interval, proportion, parameter 151. True or False: Referring to Table 8-13, the parameter of interest is the proportion of student population who feel comfortable reporting cheating by their fellow students.

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

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

ANSWER: 2.5758 TYPE: FI DIFFICULTY: Easy KEYWORDS: confidence interval, proportion .

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Confidence Interval Estimation

153. Referring to Table 8-13, a 99% confidence interval for the proportion of student population who feel comfortable reporting cheating by their fellow students is from __________ to __________.

ANSWER: 0.4557 to 0.5043 TYPE: FI DIFFICULTY: Moderate KEYWORDS: confidence interval, proportion 154. Referring to Table 8-13, the sampling error of a 99% confidence interval for the proportion of student population who feel comfortable reporting cheating by their fellow students is __________.

ANSWER: 0.0243 TYPE: FI DIFFICULTY: Moderate KEYWORDS: confidence interval, proportion, sampling error

155. True or False: Referring to Table 8-13, a 95% confidence interval for the proportion of student population who feel comfortable reporting cheating by their fellow students is narrower than a 99% confidence interval.

ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: confidence interval, proportion, properties, width 156. True or False: Referring to Table 8-13, a 99% confidence interval will contain 99% of the student population who feel comfortable reporting cheating by their fellow students.

ANSWER: False TYPE: TF DIFFICULTY: Moderate KEYWORDS: confidence interval, proportion, interpretation 157. True or False: Referring to Table 8-13, a confidence interval estimate of the population proportion would only be valid if the distribution of the number of students who feel comfortable reporting cheating by their fellow students is normal.

ANSWER: False TYPE: TF DIFFICULTY: Moderate KEYWORDS: confidence interval, proportion 158. True or False: Referring to Table 8-13, a 90% confidence interval calculated from the same data would be narrower than a 99% confidence interval.

ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: confidence interval, proportion, properties, width .

Confidence Interval Estimation 259

159. True or False: Referring to Table 8-13, it is possible that the 99% confidence interval calculated from the data will not contain the sample proportion of students who feel comfortable reporting cheating by their fellow students.

ANSWER: False TYPE: TF DIFFICULTY: Moderate EXPLANATION: The confidence interval always contains the sample proportion.

KEYWORDS: confidence interval, proportion, interpretation 160. True or False: Referring to Table 8-13, it is possible that the 99% confidence interval calculated from the data will not contain the proportion of student population who feel comfortable reporting cheating by their fellow students.

ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: confidence interval, proportion, interpretation 161. True or False: Referring to Table 8-13, we are 99% confident that the total numbers of student population who feel comfortable reporting cheating by their fellow students is between 0.4557 and 0.5043.

ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: confidence interval, proportion, interpretation 162. True or False: Referring to Table 8-13, we are 99% confident that between 45.57% and 50.43% of the student population feel comfortable reporting cheating by their fellow students.

ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: confidence interval, proportion, interpretation

TABLE 8-14 The president of a university is concerned that the percentage of students who have cheated on an exam is higher than the 1% acceptable level. A confidential random sample of 1000 students from a population of 7000 revealed that 6 of them said that they had cheated on an exam during the last semester.

.

260

Confidence Interval Estimation

163. Referring to Table 8-14, what is the critical value for the 90% one-sided confidence interval for the proportion of students who had cheated on an exam during the last 12 months?

ANSWER: 1.2816 TYPE: PR DIFFICULTY: Easy KEYWORDS: confidence interval, proportion, one-sided, critical value 164. Referring to Table 8-14, what is the upper bound of the 90% one-sided confidence interval for the proportion of students who had cheated on an exam during the last 12 months?

ANSWER: 0.0089 TYPE: PR DIFFICULTY: Easy KEYWORDS: confidence interval, proportion, one-sided 165. True or False: Referring to Table 8-14, the president can be 90% confident that no more than 1% of the students at the university had cheated on an exam during the last semester.

ANSWER: True TYPE: TF DIFFICULTY: Moderate KEYWORDS: confidence interval, proportion, one-sided, interpretation 166. True or False: Referring to Table 8-14, using on the 90% one-sided confidence interval, the president can be 95% confident that no more than 1% of the students at the university had cheated on an exam during the last semester.

ANSWER: False TYPE: TF DIFFICULTY: Difficult EXPLANATION: Since the upper limit of the 95% one-sided confidence interval will be larger than 0.0089, the president cannot be 95% confident that no more than 1% of the students at the university had cheated on an exam during the last semester.

KEYWORDS: confidence interval, proportion, one-sided, interpretation

167. True or False: Referring to Table 8-14, using on the 90% one-sided confidence interval, the superintendent can be 85% confident that no more than 1% of the students at the university had cheated on an exam during the last semester.

ANSWER: True TYPE: TF DIFFICULTY: Difficult

.

Confidence Interval Estimation 261

EXPLANATION: Since the upper limit of the 85% one-sided confidence interval will be smaller than 0.0089, the president can be 85% confident that no more than 1% of the students at the university had cheated on an exam during the last semester.

KEYWORDS: confidence interval, proportion, one-sided, interpretation

TABLE 8-15 A sales and marketing management magazine conducted a survey on salespeople cheating on their expense reports and other unethical conduct. In the survey on 200 managers, managers have caught salespeople cheating on an expense report 58% of the time, working a second job on company time 50% of the time, listing a “strip bar” as a restaurant on an expense report 22% of the time, and giving a kickback to a customer 19% of the time.

168. Referring to Table 8-15, construct a 95% confidence interval estimate of the population proportion of managers who have caught salespeople cheating on an expense report.

ANSWER: 0.5116 to 0.6484 TYPE: PR DIFFICULTY: Moderate KEYWORDS: confidence interval, proportion 169. Referring to Table 8-15, the critical value for a 95% confidence interval estimate of the population proportion of managers who have caught salespeople cheating on an expense report is __________.

ANSWER: 1.96 TYPE: FI DIFFICULTY: Easy KEYWORDS: confidence interval, proportion

170. Referring to Table 8-15, the sampling error of a 95% confidence interval estimate of the population proportion of managers who have caught salespeople cheating on an expense report is __________.

ANSWER: 0.0684 TYPE: FI DIFFICULTY: Moderate KEYWORDS: confidence interval, proportion, sampling error

171. True or False: Referring to Table 8-15, a 95% confidence interval will contain 95% of the population proportion of managers who have caught salespeople cheating on an expense report.

ANSWER: False TYPE: TF DIFFICULTY: Moderate KEYWORDS: confidence interval, proportion, interpretation

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262

Confidence Interval Estimation

172. True or False: Referring to Table 8-15, it is possible that the 95% confidence interval calculated from the data will not contain the sample proportion of managers who have caught salespeople cheating on an expense report.

ANSWER: False TYPE: TF DIFFICULTY: Moderate EXPLANATION: The confidence interval always contains the sample proportion.

KEYWORDS: confidence interval, proportion, interpretation 173. True or False: Referring to Table 8-15, it is possible that the 95% confidence interval calculated from the data will not contain the population proportion of managers who have caught salespeople cheating on an expense report.

ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: confidence interval, proportion, interpretation 174. True or False: Referring to Table 8-15, we are 95% confident that the population average number of managers who have caught salespeople cheating on an expense report is between 0.5116 to 0.6484.

ANSWER: False TYPE: TF DIFFICULTY: Difficult KEYWORDS: confidence interval, proportion, interpretation 175. True or False: Referring to Table 8-15, we are 95% confident that between 51.16% and 64.84% of managers in the population have caught salespeople cheating on an expense report.

ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: confidence interval, proportion, interpretation 176. Referring to Table 8-15, construct a 95% confidence interval estimate of the population proportion of managers who have caught salespeople working a second job on company time.

ANSWER: 0.4307 to 0.5693 TYPE: PR DIFFICULTY: Moderate KEYWORDS: confidence interval, proportion 177. Referring to Table 8-15, the sampling error of a 95% confidence interval estimate of the population proportion of managers who have caught salespeople working a second job on company time is __________.

ANSWER: 0.0693 TYPE: FI DIFFICULTY: Moderate KEYWORDS: confidence interval, proportion, sampling error

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Confidence Interval Estimation 263

178. Referring to Table 8-15, construct a 95% confidence interval estimate of the population proportion of managers who have caught salespeople listing a “strip bar” as a restaurant on an expense report.

ANSWER: 0.1626 to 0.2774 TYPE: PR DIFFICULTY: Moderate KEYWORDS: confidence interval, proportion 179. Referring to Table 8-15, the sampling error of a 95% confidence interval estimate of the population proportion of managers who have caught salespeople listing a “strip bar” as a restaurant on an expense report is __________.

ANSWER: 0.05741 TYPE: FI DIFFICULTY: Moderate KEYWORDS: confidence interval, proportion, sampling error 180. Referring to Table 8-15, construct a 95% confidence interval estimate of the population proportion of managers who have caught salespeople giving a kickback to a customer.

ANSWER: 0.1356 to 0.2444 TYPE: PR DIFFICULTY: Moderate KEYWORDS: confidence interval, proportion 181. Referring to Table 8-15, the sampling error of a 95% confidence interval estimate of the population proportion of managers who have caught salespeople giving a kickback to a customer is __________.

ANSWER: 0.05437 TYPE: FI DIFFICULTY: Moderate KEYWORDS: confidence interval, proportion, sampling error

182. Referring to Table 8-15, determine the sample size needed to estimate the proportion of managers who have caught salespeople working a second job on company time to within ±0.02 with 95% confidence.

ANSWER: 2401 TYPE: PR DIFFICULTY: Moderate KEYWORDS: confidence interval, proportion, sample size determination

.

Fundamentals of Hypothesis Testing: One-Sample Tests

CHAPTER 9: FUNDAMENTALS OF HYPOTHESIS TESTING: ONE-SAMPLE TESTS 1. Which of the following would be an appropriate null hypothesis? a) The mean of a population is equal to 55. b) The mean of a sample is equal to 55. c) The mean of a population is greater than 55. d) Only (a) and (c) are true.

ANSWER: a TYPE: MC DIFFICULTY: Easy KEYWORDS: form of hypothesis 2. Which of the following would be an appropriate null hypothesis? a) The population proportion is less than 0.65. b) The sample proportion is less than 0.65. c) The population proportion is no less than 0.65. d) The sample proportion is no less than 0.65.

ANSWER: c TYPE: MC DIFFICULTY: Easy KEYWORDS: form of hypothesis 3. Which of the following would be an appropriate alternative hypothesis? a) The mean of a population is equal to 55. b) The mean of a sample is equal to 55. c) The mean of a population is greater than 55. d) The mean of a sample is greater than 55.

ANSWER: c TYPE: MC DIFFICULTY: Easy KEYWORDS: form of hypothesis 4. Which of the following would be an appropriate alternative hypothesis? a) The population proportion is less than 0.65. b) The sample proportion is less than 0.65. c) The population proportion is no less than 0.65. d) The sample proportion is no less than 0.65.

ANSWER: a TYPE: MC DIFFICULTY: Easy KEYWORDS: form of hypothesis

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265

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Fundamentals of Hypothesis Testing: One-Sample Tests

5. A Type II error is committed when a) we reject a null hypothesis that is true. b) we don't reject a null hypothesis that is true. c) we reject a null hypothesis that is false. d) we don't reject a null hypothesis that is false.

ANSWER: d TYPE: MC DIFFICULTY: Easy KEYWORDS: type II error 6. A Type I error is committed when a) we reject a null hypothesis that is true. b) we don't reject a null hypothesis that is true. c) we reject a null hypothesis that is false. d) we don't reject a null hypothesis that is false.

ANSWER: a TYPE: MC DIFFICULTY: Easy KEYWORDS: type I error 7. The power of a test is measured by its capability of a) rejecting a null hypothesis that is true. b) not rejecting a null hypothesis that is true. c) rejecting a null hypothesis that is false. d) not rejecting a null hypothesis that is false.

ANSWER: c TYPE: MC DIFFICULTY: Easy KEYWORDS: power 8. True or False: For a given level of significance, if the sample size is increased, the power of the test will increase.

ANSWER: True TYPE: TF DIFFICULTY: Moderate KEYWORDS: power, level of significance, sample size 9. True or False: For a given level of significance, if the sample size is increased, the probability of committing a Type I error will increase.

ANSWER: False TYPE: TF DIFFICULTY: Moderate KEYWORDS: level of significance, sample size, type I error

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Fundamentals of Hypothesis Testing: One-Sample Tests

267

10. True or False: For a given level of significance, if the sample size is increased, the probability of committing a Type II error will increase.

ANSWER: False TYPE: TF DIFFICULTY: Moderate KEYWORDS: level of significance, sample size, type II error 11. True or False: For a given sample size, the probability of committing a Type II error will increase when the probability of committing a Type I error is reduced.

ANSWER: True TYPE: TF DIFFICULTY: Moderate KEYWORDS: sample size, type I error, type II error 12. If an economist wishes to determine whether there is evidence that average family income in a community exceeds $25,000 a) either a one-tailed or two-tailed test could be used with equivalent results. b) a one-tailed test should be utilized. c) a two-tailed test should be utilized. d) None of the above.

ANSWER: b TYPE: MC DIFFICULTY: Easy KEYWORDS: one-tailed test 13. If an economist wishes to determine whether there is evidence that average family income in a community equals $25,000 a) either a one-tailed or two-tailed test could be used with equivalent results. b) a one-tailed test should be utilized. c) a two-tailed test should be utilized. d) None of the above.

ANSWER: c TYPE: MC DIFFICULTY: Easy KEYWORDS: two-tailed test 14. If the p-value is less than  in a two-tailed test, a) the null hypothesis should not be rejected. b) the null hypothesis should be rejected. c) a one-tailed test should be used. d) no conclusion should be reached.

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

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268

Fundamentals of Hypothesis Testing: One-Sample Tests 15. If a test of hypothesis has a Type I error probability (  ) of 0.01, we mean a) if the null hypothesis is true, we don't reject it 1% of the time. b) if the null hypothesis is true, we reject it 1% of the time. c) if the null hypothesis is false, we don't reject it 1% of the time. d) if the null hypothesis is false, we reject it 1% of the time.

ANSWER: b TYPE: MC DIFFICULTY: Moderate KEYWORDS: type I error, level of significance 16. If the Type I error (  ) for a given test is to be decreased, then for a fixed sample size n a) the Type II error (  ) will also decrease. b) the Type II error (  ) will increase. c) the power of the test will increase. d) a one-tailed test must be utilized.

ANSWER: b TYPE: MC DIFFICULTY: Moderate KEYWORDS: type I error, type II error, sample size 17. For a given sample size n, if the level of significance (  ) is decreased, the power of the test a) will increase. b) will decrease. c) will remain the same. d) cannot be determined.

ANSWER: b TYPE: MC DIFFICULTY: Moderate KEYWORDS: level of significance, power, sample size 18. For a given level of significance (  ), if the sample size n is increased, the probability of a Type II error (  ) a) will decrease. b) will increase. c) will remain the same. d) cannot be determined.

ANSWER: a TYPE: MC DIFFICULTY: Moderate KEYWORDS: level of significance, beta-risk, sample size

19. You know that the level of significance (  ) of a test is 5%, you can tell that the probability of committing a Type II error (  ) is .

Fundamentals of Hypothesis Testing: One-Sample Tests

a) b) c) d)

2.5% 95%.

97.5% unknown

ANSWER: d TYPE: MC DIFFICULTY: Moderate KEYWORDS: level of significance, beta-risk 20. You know that the probability of committing a Type II error (  ) is 5%, you can tell that the power of the test is a) 2.5% b) 95%. c) 97.5% d) unknown ANSWER: b TYPE: MC DIFFICULTY: Easy KEYWORDS: level of significance, beta-risk, power 21. If a researcher rejects a true null hypothesis, she has made a _______error.

ANSWER: Type I TYPE: FI DIFFICULTY: Easy KEYWORDS: type I error 22. If a researcher accepts a true null hypothesis, she has made a _______decision.

ANSWER: correct TYPE: FI DIFFICULTY: Easy KEYWORDS: decision 23. If a researcher rejects a false null hypothesis, she has made a _______decision.

ANSWER: correct TYPE: FI DIFFICULTY: Easy KEYWORDS: decision 24. If a researcher accepts a false null hypothesis, she has made a _______error.

ANSWER: Type II TYPE: FI DIFFICULTY: Easy KEYWORDS: type II error 25. It is possible to directly compare the results of a confidence interval estimate to the results obtained by testing a null hypothesis if a) a two-tailed test for  is used. .

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Fundamentals of Hypothesis Testing: One-Sample Tests b) a one-tailed test for  is used. c) Both of the previous statements are true. d) None of the previous statements is true.

ANSWER: a TYPE: MC DIFFICULTY: Moderate KEYWORDS: confidence interval, two-tailed test 26. The power of a statistical test is a) the probability of not rejecting H0 when it is false. b) the probability of rejecting H0 when it is true. c) the probability of not rejecting H0 when it is true. d) the probability of rejecting H0 when it is false.

ANSWER: d TYPE: MC DIFFICULTY: Moderate KEYWORDS: power 27. The symbol for the power of a statistical test is a)  . b) 1 –  . c)  . d) 1 –  .

ANSWER: d TYPE: MC DIFFICULTY: Easy KEYWORDS: power 28. Suppose we wish to test H0:   47 versus H1:  > 47. What will result if we conclude that the mean is greater than 47 when its true value is really 52? a) We have made a Type I error. b) We have made a Type II error. c) We have made a correct decision d) None of the above are correct.

ANSWER: c TYPE: MC DIFFICULTY: Moderate KEYWORDS: one-tailed test, conclusion

29. How many Kleenex should the Kimberly Clark Corporation package of tissues contain?

Researchers determined that 60 tissues is the average number of tissues used during a cold.

Suppose a random sample of 100 Kleenex users yielded the following data on the number of

.

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tissues used during a cold: X = 52, s = 22. Give the null and alternative hypotheses to determine if the number of tissues used during a cold is less than 60. a) H0 :   60 and H1 :   60. b) H0 :   60 and H1 :   60.

H0 : X  60 and H1 : X  60. d) H0 : X = 52 and H1 : X  52. c)

ANSWER: b TYPE: MC DIFFICULTY: Moderate KEYWORDS: one-tailed test, form of hypothesis, form of hypothesis, mean, t test 30. How many Kleenex should the Kimberly Clark Corporation package of tissues contain?

Researchers determined that 60 tissues is the average number of tissues used during a cold.

Suppose a random sample of 100 Kleenex users yielded the following data on the number of tissues used during a cold: X = 52, s = 22. Using the sample information provided, calculate the value of the test statistic. a) t = ( 52 − 60 ) / 22 b) t = ( 52 − 60 ) / ( 22 /100 ) c) t = ( 52 − 60 ) / ( 22 /1002 ) d) t = ( 52 − 60 ) / ( 22 /10 ) ANSWER: d TYPE: MC DIFFICULTY: Easy KEYWORDS: one-tailed test, mean, t test 31. How many Kleenex should the Kimberly Clark Corporation package of tissues contain?

Researchers determined that 60 tissues is the average number of tissues used during a cold.

Suppose a random sample of 100 Kleenex users yielded the following data on the number of tissues used during a cold: X = 52, s = 22. Suppose the alternative we wanted to test was H1 :   60. State the correct rejection region for  = 0.05. a) Reject H0 if t > 1.6604. b) Reject H0 if t < – 1.6604. c) Reject H0 if t > 1.9842 or Z < – 1.9842. d) Reject H0 if t < – 1.9842.

ANSWER: b TYPE: MC DIFFICULTY: Moderate KEYWORDS: one-tailed test, mean, t test, rejection region

32. How many Kleenex should the Kimberly Clark Corporation package of tissues contain?

Researchers determined that 60 tissues is the average number of tissues used during a cold.

Suppose a random sample of 100 Kleenex users yielded the following data on the number of .

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tissues used during a cold: X = 52, s = 22. Suppose the test statistic does fall in the rejection region at  = 0.05. Which of the following decision is correct? a) At  = 0.05, we do not reject H0. b) At  = 0.05, we reject H0. c) At  = 0.05, we accept H0. d) At  = 0.10, we do not reject H0.

ANSWER: b TYPE: MC DIFFICULTY: Easy KEYWORDS: one-tailed test, mean, t test, decision 33. How many Kleenex should the Kimberly Clark Corporation package of tissues contain?

Researchers determined that 60 tissues is the average number of tissues used during a cold.

Suppose a random sample of 100 Kleenex users yielded the following data on the number of tissues used during a cold: X = 52, s = 22. Suppose the test statistic does fall in the rejection region at  = 0.05. Which of the following conclusion is correct? a) At  = 0.05, there is not sufficient evidence to conclude that the average number of tissues used during a cold is 60 tissues. b) At  = 0.05, there is sufficient evidence to conclude that the average number of tissues used during a cold is 60 tissues. c) At  = 0.05, there is not sufficient evidence to conclude that the average number of tissues used during a cold is not 60 tissues. d) At  = 0.10, there is sufficient evidence to conclude that the average number of tissues used during a cold is not 60 tissues.

ANSWER: d TYPE: MC DIFFICULTY: Moderate KEYWORDS: one-tailed test, mean, t test, conclusion 34. We have created a 95% confidence interval for  with the result (10, 15). What decision will we make if we test H0 :  = 16 versus H1 :   16 at  = 0.05? a) Reject H0 in favor of H1. b) Accept H0 in favor of H1. c) Fail to reject H0 in favor of H1. d) We cannot tell what our decision will be from the information given.

ANSWER: a TYPE: MC DIFFICULTY: Easy KEYWORDS: two-tailed test, confidence interval, mean, t test, decision

35. We have created a 95% confidence interval for  with the result (10, 15). What decision will we make if we test H0 :  = 16 versus H1 :   16 at  = 0.10? a) Reject H0 in favor of H1. b) Accept H0 in favor of H1. .

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c) Fail to reject H0 in favor of H1. d) We cannot tell what our decision will be from the information given.

ANSWER: a TYPE: MC DIFFICULTY: Moderate EXPLANATION: The 90% confidence interval is narrower than (10, 15), which still does not contain 16.

KEYWORDS: two-tailed test, confidence interval, mean, t test, decision 36. We have created a 95% confidence interval for  with the result (10, 15). What decision will we make if we test H0 :  = 16 versus H1 :   16 at  = 0.025? a) Reject H0 in favor of H1. b) Accept H0 in favor of H1. c) Fail to reject H0 in favor of H1. d) We cannot tell what our decision will be from the information given.

ANSWER: d TYPE: MC DIFFICULTY: Moderate EXPLANATION: The 97.5% confidence interval is wider than (10, 15), which could have contained 16 or not have contained 16.

KEYWORDS: two-tailed test, confidence interval, mean, t test, decision 37. Suppose we want to test H0 :   30 versus H1 :   30 . Which of the following possible sample results based on a sample of size 36 gives the strongest evidence to reject H0 in favor of H1? a) X = 28, s = 6 b) X = 27, s = 4 c) X = 32, s = 2 d) X = 26, s = 9 ANSWER: b TYPE: MC DIFFICULTY: Moderate KEYWORDS: one-tailed test, rejection region

38. Which of the following statements is not true about the level of significance in a hypothesis test? a) The larger the level of significance, the more likely you are to reject the null hypothesis. b) The level of significance is the maximum risk we are willing to accept in making a Type I error. c) The significance level is also called the  level. .

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d) The significance level is another name for Type II error.

ANSWER: d TYPE: MC DIFFICULTY: Moderate KEYWORDS: level of significance 39. If, as a result of a hypothesis test, we reject the null hypothesis when it is false, then we have committed a) a Type II error. b) a Type I error. c) no error. d) an acceptance error.

ANSWER: c TYPE: MC DIFFICULTY: Moderate KEYWORDS: decision, type I error, type II error 40. The value that separates a rejection region from a non-rejection region is called the _______.

ANSWER: critical value TYPE: FI DIFFICULTY: Easy KEYWORDS: critical value, rejection region 41. A is a numerical quantity computed from the data of a sample and is used in reaching a decision on whether or not to reject the null hypothesis. a) significance level b) critical value c) test statistic d) parameter ANSWER: c TYPE: MC DIFFICULTY: Easy KEYWORDS: test statistic

42. The owner of a local nightclub has recently surveyed a random sample of n = 250 customers of the club. She would now like to determine whether or not the mean age of her customers is over 30. If so, she plans to alter the entertainment to appeal to an older crowd. If not, no entertainment changes will be made. The appropriate hypotheses to test are: a) H0 :   30 versus H1 :   30 . b) H0 :   30 versus H1 :   30 . .

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H0 : X  30 versus H1 : X  30 . d) H0 : X  30 versus H1 : X  30 . c)

ANSWER: b TYPE: MC DIFFICULTY: Easy KEYWORDS: one-tailed test, form of hypothesis, form of hypothesis, mean 43. The owner of a local nightclub has recently surveyed a random sample of n = 250 customers of the club. She would now like to determine whether or not the mean age of her customers is over 30. If so, she plans to alter the entertainment to appeal to an older crowd. If not, no entertainment changes will be made. If she wants to be 99% confident in her decision, what rejection region should she use? a) Reject H0 if t < – 2.34. b) Reject H0 if t < – 2.55. c) Reject H0 if t > 2.34. d) Reject H0 if t > 2.58.

ANSWER: c TYPE: MC DIFFICULTY: Moderate KEYWORDS: one-tailed test, mean, Z test, t test, rejection region 44. The owner of a local nightclub has recently surveyed a random sample of n = 250 customers of the club. She would now like to determine whether or not the mean age of her customers is over 30. If so, she plans to alter the entertainment to appeal to an older crowd. If not, no entertainment changes will be made. Suppose she found that the sample mean was 30.45 years and the sample standard deviation was 5 years. If she wants to be 99% confident in her decision, what decision should she make? a) Reject H0. b) Accept H0. c) Fail to reject H0. d) We cannot tell what her decision should be from the information given.

ANSWER: c TYPE: MC DIFFICULTY: Moderate KEYWORDS: one-tailed test, mean, t test, decision

45. The owner of a local nightclub has recently surveyed a random sample of n = 250 customers of the club. She would now like to determine whether or not the mean age of her customers is over 30. If so, she plans to alter the entertainment to appeal to an older crowd. If not, no entertainment changes will be made. Suppose she found that the sample mean was 30.45 years and the sample standard deviation was 5 years. If she wants to be 99% confident in her decision, what conclusion can she make? a) There is not sufficient evidence that the mean age of her customers is over 30. .

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b) There is sufficient evidence that the mean age of her customers is over 30. c) There is not sufficient evidence that the mean age of her customers is not over 30. d) There is sufficient evidence that the mean age of her customers is not over 30.

ANSWER: a TYPE: MC DIFFICULTY: Moderate KEYWORDS: one-tailed test, mean, t test, conclusion 46. The owner of a local nightclub has recently surveyed a random sample of n = 250 customers of the club. She would now like to determine whether or not the mean age of her customers is over 30. If so, she plans to alter the entertainment to appeal to an older crowd. If not, no entertainment changes will be made. Suppose she found that the sample mean was 30.45 years and the sample standard deviation was 5 years. What is the p-value associated with the test statistic? a) 0.3577 b) 0.1423 c) 0.0780 d) 0.02 ANSWER: c TYPE: MC DIFFICULTY: Moderate KEYWORDS: one-tailed test, mean, t test, p-value 47. A survey claims that 9 out of 10 doctors recommend aspirin for their patients with headaches. To test this claim against the alternative that the actual proportion of doctors who recommend aspirin is less than 0.90, a random sample of 100 doctors results in 83 who indicate that they recommend aspirin. The value of the test statistic in this problem is approximately equal to: a) – 4.12 b) – 2.33 c) – 1.86 d) – 0.07 ANSWER: b TYPE: MC DIFFICULTY: Easy KEYWORDS: one-tailed test, proportion, test statistic

48. A survey claims that 9 out of 10 doctors recommend aspirin for their patients with headaches. To test this claim against the alternative that the actual proportion of doctors who recommend aspirin is less than 0.90, a random sample of 100 doctors was selected. Suppose that the test statistic is – 2.20. Can we conclude that H0 should be rejected at the (a)  = 0.10, (b)  = 0.05, and (c)  = 0.01 level of Type I error? a) (a) yes; (b) yes; (c) yes b) (a) no; (b) no; (c) no c) (a) no; (b) no; (c) yes .

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d) (a) yes; (b) yes; (c) no ANSWER: d TYPE: MC DIFFICULTY: Moderate KEYWORDS: one-tailed test, proportion, decision 49. A survey claims that 9 out of 10 doctors recommend aspirin for their patients with headaches. To test this claim against the alternative that the actual proportion of doctors who recommend aspirin is less than 0.90, a random sample of 100 doctors was selected. Suppose you reject the null hypothesis. What conclusion can you draw? a) There is not sufficient evidence that the proportion of doctors who recommend aspirin is not less than 0.90. b) There is sufficient evidence that the proportion of doctors who recommend aspirin is not less than 0.90. c) There is not sufficient evidence that the proportion of doctors who recommend aspirin is less than 0.90. d) There is sufficient evidence that the proportion of doctors who recommend aspirin is less than 0.90.

ANSWER: d TYPE: MC DIFFICULTY: Moderate KEYWORDS: one-tailed test, proportion, conclusion 50. A major videocassette rental chain is considering opening a new store in an area that currently does not have any such stores. The chain will open if there is evidence that more than 5,000 of the 20,000 households in the area are equipped with videocassette recorders (VCRs). It conducts a telephone poll of 300 randomly selected households in the area and finds that 96 have VCRs. State the test of interest to the rental chain. a) H 0 :   0.32 versus H1 :   0.32 b) H 0 :   0.25 versus H1 :   0.25 c) H 0 :   5,000 versus H1 :   5,000 d) H0 :   5,000 versus H1 :   5,000 ANSWER: b TYPE: MC DIFFICULTY: Easy KEYWORDS: one-tailed test, proportion, form of hypothesis, form of hypothesis

51. A major videocassette rental chain is considering opening a new store in an area that currently does not have any such stores. The chain will open if there is evidence that more than 5,000 of the 20,000 households in the area are equipped with videocassette recorders (VCRs). It conducts a telephone poll of 300 randomly selected households in the area and finds that 96 have VCRs. The value of the test statistic in this problem is approximately equal to: a) 2.80 b) 2.60 c) 1.94 d) 1.30 .

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ANSWER: a TYPE: MC DIFFICULTY: Easy KEYWORDS: one-tailed test, proportion, Z test, test statistic 52. A major videocassette rental chain is considering opening a new store in an area that currently does not have any such stores. The chain will open if there is evidence that more than 5,000 of the 20,000 households in the area are equipped with videocassette recorders (VCRs). It conducts a telephone poll of 300 randomly selected households in the area and finds that 96 have VCRs. The p-value associated with the test statistic in this problem is approximately equal to: a) 0.0100 b) 0.0051 c) 0.0026 d) 0.0013 ANSWER: c TYPE: MC DIFFICULTY: Easy KEYWORDS: one-tailed test, proportion, Z test, p-value 53. A major videocassette rental chain is considering opening a new store in an area that currently does not have any such stores. The chain will open if there is evidence that more than 5,000 of the 20,000 households in the area are equipped with videocassette recorders (VCRs). It conducts a telephone poll of 300 randomly selected households in the area and finds that 96 have VCRs. The decision on the hypothesis test using a 3% level of significance is: a) to reject H0 in favor of H1. b) to accept H0 in favor of H1. c) to fail to reject H0 in favor of H1. d) we cannot tell what the decision should be from the information given.

ANSWER: a TYPE: MC DIFFICULTY: Easy KEYWORDS: one-tailed test, proportion, Z test, decision

54. A major videocassette rental chain is considering opening a new store in an area that currently does not have any such stores. The chain will open if there is evidence that more than 5,000 of the 20,000 households in the area are equipped with videocassette recorders (VCRs). It conducts a telephone poll of 300 randomly selected households in the area and finds that 96 have VCRs. The rental chain's conclusion from the hypothesis test using a 3% level of significance is: a) to open a new store. b) not to open a new store. c) to delay opening a new store until additional evidence is collected. d) we cannot tell what the decision should be from the information given.

ANSWER: .

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a TYPE: MC DIFFICULTY: Easy KEYWORDS: one-tailed test, proportion, Z test, conclusion 55. An entrepreneur is considering the purchase of a coin-operated laundry. The current owner claims that over the past 5 years, the average daily revenue was $675 with a standard deviation of $75. A sample of 30 days reveals a daily average revenue of $625. If you were to test the null hypothesis that the daily average revenue was $675, which test would you use? a) Z-test of a population mean b) Z-test of a population proportion c) t-test of population mean d) t-test of a population proportion ANSWER: a TYPE: MC DIFFICULTY: Moderate KEYWORDS: two-tailed test, mean, Z test 56. An entrepreneur is considering the purchase of a coin-operated laundry. The current owner claims that over the past 5 years, the average daily revenue was $675 with a standard deviation of $75. A sample of 30 days reveals a daily average revenue of $625. If you were to test the null hypothesis that the daily average revenue was $675 and decide not to reject the null hypothesis, what can you conclude? a) There is not enough evidence to conclude that the daily average revenue was $675. b) There is not enough evidence to conclude that the daily average revenue was not $675. c) There is enough evidence to conclude that the daily average revenue was $675. d) There is enough evidence to conclude that the daily average revenue was not $675.

ANSWER: b TYPE: MC DIFFICULTY: Moderate KEYWORDS: two-tailed test, mean, Z test, conclusion

57. A manager of the credit department for an oil company would like to determine whether the average monthly balance of credit card holders is equal to $75. An auditor selects a random sample of 100 accounts and finds that the average owed is $83.40 with a sample standard deviation of $23.65. If you were to conduct a test to determine whether the auditor should conclude that there is evidence that the average balance is different from $75, which test would you use? a) Z-test of a population mean b) Z-test of a population proportion c) t-test of population mean d) t-test of a population proportion ANSWER: .

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c TYPE: MC DIFFICULTY: Easy KEYWORDS: two-tailed test, mean, t test 58. A manager of the credit department for an oil company would like to determine whether the average monthly balance of credit card holders is equal to $75. An auditor selects a random sample of 100 accounts and finds that the average owed is $83.40 with a sample standard deviation of $23.65. If you wanted to test whether the average balance is different from $75 and decided to reject the null hypothesis, what conclusion could you draw? a) There is not evidence that the average balance is $75. b) There is not evidence that the average balance is not $75. c) There is evidence that the average balance is $75. d) There is evidence that the average balance is not $75.

ANSWER: d TYPE: MC DIFFICULTY: moderate KEYWORDS: two-tailed test, mean, t test, conclusion 59. The marketing manager for an automobile manufacturer is interested in determining the proportion of new compact-car owners who would have purchased a passenger-side inflatable air bag if it had been available for an additional cost of $300. The manager believes from previous information that the proportion is 0.30. Suppose that a survey of 200 new compact-car owners is selected and 79 indicate that they would have purchased the inflatable air bags. If you were to conduct a test to determine whether there is evidence that the proportion is different from 0.30, which test would you use? a) Z-test of a population mean b) Z-test of a population proportion c) t-test of population mean d) t-test of a population proportion ANSWER: b TYPE: MC DIFFICULTY: Easy KEYWORDS: two-tailed test, proportion

60. The marketing manager for an automobile manufacturer is interested in determining the proportion of new compact-car owners who would have purchased a passenger-side inflatable air bag if it had been available for an additional cost of $300. The manager believes from previous information that the proportion is 0.30. Suppose that a survey of 200 new compact-car owners is selected and 79 indicate that they would have purchased the inflatable air bags. If you were to conduct a test to determine whether there is evidence that the proportion is different from 0.30 and decided not to reject the null hypothesis, what conclusion could you draw? a) There is sufficient evidence that the proportion is 0.30. b) There is not sufficient evidence that the proportion is 0.30. c) There is sufficient evidence that the proportion is 0.30. d) There is not sufficient evidence that the proportion is not 0.30.

ANSWER: .

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d TYPE: MC DIFFICULTY: Easy KEYWORDS: two-tailed test, proportion, conclusion

TABLE 9-1 Microsoft Excel was used on a set of data involving the number of parasites found on 46 Monarch butterflies captured in Pismo Beach State Park. A biologist wants to know if the mean number of parasites per butterfly is over 20. She will make her decision using a test with a level of significance of 0.10. The following information was extracted from the Microsoft Excel output for the sample of 46 Monarch butterflies: n = 46; Arithmetic Mean = 28.00; Standard Deviation = 25.92; Standard Error = 3.82; Null Hypothesis: H0 :   20.000 ;  = 0.10; df = 45; T Test Statistic = 2.09; One-Tailed Test Upper Critical Value = 1.3006; p-value = 0.021; Decision = Reject.

61. Referring to Table 9-1, the parameter the biologist is interested in is: a) the mean number of butterflies in Pismo Beach State Park. b) the mean number of parasites on these 46 butterflies. c) the mean number of parasites on Monarch butterflies in Pismo Beach State Park. d) the proportion of butterflies with parasites.

ANSWER: c TYPE: MC DIFFICULTY: Easy KEYWORDS: mean, t test, parameter 62. Referring to Table 9-1, state the alternative hypothesis for this study.

ANSWER:

H 1 :   20.000

TYPE: PR DIFFICULTY: Easy KEYWORDS: one-tailed test, mean, t test, form of hypothesis

63. Referring to Table 9-1, what critical value should the biologist use to determine the rejection region? a) 1.6794 b) 1.3011 c) 1.3006 d) 0.6800 ANSWER: c TYPE: MC DIFFICULTY: Easy KEYWORDS: one-tailed test, mean, t test, critical value 64. True or False: Referring to Table 9-1, the null hypothesis would be rejected. .

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ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: one-tailed test, mean, t test, decision 65. True or False: Referring to Table 9-1, the null hypothesis would be rejected if a 4% probability of committing a Type I error is allowed.

ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: one-tailed test, mean, t test, decision 66. True or False: Referring to Table 9-1, the null hypothesis would be rejected if a 1% probability of committing a Type I error is allowed.

ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: one-tailed test, mean, t test, decision 67. Referring to Table 9-1, the lowest level of significance at which the null hypothesis can be rejected is ______.

ANSWER: 0.021 TYPE: FI DIFFICULTY: Moderate KEYWORDS: one-tailed test, mean, t test, p-value 68. True or False: Referring to Table 9-1, the evidence proves beyond a doubt that the mean number of parasites on butterflies in Pismo Beach State Park is over 20.

ANSWER: False TYPE: TF DIFFICULTY: Moderate KEYWORDS: one-tailed test, mean, t test, conclusion 69. True of False: Referring to Table 9-1, the biologist can conclude that there is sufficient evidence to show that the average number of parasites per Monarch butterfly in Pismo Beach State Park is over 20 using a level of significance of 0.10.

ANSWER: True TYPE: TF DIFFICULTY: Moderate KEYWORDS: one-tailed test, mean, t test, conclusion 70. True or False: Referring to Table 9-1, the biologist can conclude that there is sufficient evidence to show that the average number of parasites per Monarch butterfly in Pismo Beach State Park is over 20 with no more than a 5% probability of incorrectly rejecting the true null hypothesis.

ANSWER: .

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True TYPE: TF DIFFICULTY: Moderate KEYWORDS: one-tailed test, mean, t test, decision 71. True or False: Referring to Table 9-1, the biologist can conclude that there is sufficient evidence to show that the average number of parasites per Monarch butterfly in Pismo Beach State Park is over 20 with no more than a 1% probability of incorrectly rejecting the true null hypothesis.

ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: one-tailed test, mean, t test, decision 72. True or False: Referring to Table 9-1, the value of  is 0.90.

ANSWER: False TYPE: TF DIFFICULTY: Moderate KEYWORDS: one-tailed test, mean, t test, beta-risk 73. True or False: Referring to Table 9-1, if these data were used to perform a two-tailed test, the pvalue would be 0.042.

ANSWER: True TYPE: TF DIFFICULTY: Moderate KEYWORDS: one-tailed test, mean, t test, p-value 74. True or False: Suppose, in testing a hypothesis about a proportion, the p-value is computed to be 0.043. The null hypothesis should be rejected if the chosen level of significance is 0.05.

ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: mean, t test, p-value, level of significance, decision

75. True or False: Suppose, in testing a hypothesis about a proportion, the p-value is computed to be 0.034. The null hypothesis should be rejected if the chosen level of significance is 0.01.

ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: p-value, level of significance, decision 76. True or False: Suppose, in testing a hypothesis about a proportion, the Z test statistic is computed to be 2.04. The null hypothesis should be rejected if the chosen level of significance is 0.01 and a two-tailed test is used.

ANSWER: False TYPE: TF DIFFICULTY: Easy .

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KEYWORDS: proportion, Z test, test statistic, critical value, decision 77. True or False: In testing a hypothesis, statements for the null and alternative hypotheses as well as the selection of the level of significance should precede the collection and examination of the data.

ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: ethical issues 78. True or False: It is not considered an unethical practice to change the direction of a test after the data are collected.

ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: ethical issues 79. True or False: It is not considered an unethical practice to not allow potential respondents be permitted to self-select for a study or be purposely selected.

ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: ethical issues 80. True or False: It is not considered an unethical practice to arbitrarily change or discard extreme or unusual values to alter the results of the hypothesis tests.

ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: ethical issues 81. True or False: It is not considered an unethical practice to discard extreme or unusual values if you can determine that a measurement is incomplete or grossly in error because of some equipment problem or unusual behavioral occurrence unrelated to the study.

ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: ethical issues 82. True or False: In instances in which there is insufficient evidence to reject the null hypothesis, making it clear that this does not prove that the null hypothesis is true is considered an ethical practice.

ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: ethical issues .

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83. True or False: Just reporting the results of hypothesis tests that show statistical significance but omitting those for which there is insufficient evidence in the findings is considered an ethical practice.

ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: ethical issues 84. True or False: The test statistic measures how close the computed sample statistic has come to the hypothesized population parameter.

ANSWER: True TYPE: TF DIFFICULTY: Moderate KEYWORDS: test statistic 85. True or False: The statement of the null hypothesis always contains an equality.

ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: form of null hypoehsis 86. True or False: The larger is the p-value, the more likely one is to reject the null hypothesis.

ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: p-value

87. True or False: The smaller is the p-value, the stronger is the evidence against the null hypothesis.

ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: p-value 88. True or False: A sample is used to obtain a 95% confidence interval for the mean of a population.

The confidence interval goes from 15 to 19. If the same sample had been used to test the null hypothesis that the mean of the population is equal to 20 versus the alternative hypothesis that the mean of the population differs from 20, the null hypothesis could be rejected at a level of significance of 0.05.

ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: confidence interval, two-tailed test, decision .

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89. True or False: A sample is used to obtain a 95% confidence interval for the mean of a population.

The confidence interval goes from 15 to 19. If the same sample had been used to test the null hypothesis that the mean of the population is equal to 18 versus the alternative hypothesis that the mean of the population differs from 18, the null hypothesis could be rejected at a level of significance of 0.05.

ANSWER: False TYPE: TF DIFFICULTY: Moderate KEYWORDS: confidence interval, two-tailed test, decision 90. True or False: A sample is used to obtain a 95% confidence interval for the mean of a population.

The confidence interval goes from 15 to 19. If the same sample had been used to test the null hypothesis that the mean of the population is equal to 20 versus the alternative hypothesis that the mean of the population differs from 20, the null hypothesis could be rejected at a level of significance of 0.10.

ANSWER: True TYPE: TF DIFFICULTY: Moderate KEYWORDS: confidence interval, two-tailed test, decision 91. True or False: A sample is used to obtain a 95% confidence interval for the mean of a population.

The confidence interval goes from 15 to 19. If the same sample had been used to test the null hypothesis that the mean of the population is equal to 20 versus the alternative hypothesis that the mean of the population differs from 20, the null hypothesis could be rejected at a level of significance of 0.02.

ANSWER: False TYPE: TF DIFFICULTY: Moderate EXPLANATION: We are not sure if 20 will be in the wider confidence interval.

KEYWORDS: confidence interval, two-tailed test, decision 92. True or False: A sample is used to obtain a 95% confidence interval for the mean of a population.

The confidence interval goes from 15 to 19. If the same sample had been used to test the null hypothesis that the mean of the population is equal to 20 versus the alternative hypothesis that the mean of the population differs from 20, the null hypothesis could be accepted at a level of significance of 0.02.

ANSWER: False TYPE: TF DIFFICULTY: Moderate EXPLANATION: We are not sure if 20 will be in the wider confidence interval.

KEYWORDS: confidence interval, two-tailed test, decision

TABLE 9-2 A student claims that he can correctly identify whether a person is a business major or an agriculture major by the way the person dresses. Suppose in actuality that if someone is a business major, he can .

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correctly identify that person as a business major 87% of the time. When a person is an agriculture major, the student will incorrectly identify that person as a business major 16% of the time. Presented with one person and asked to identify the major of this person (who is either a business or agriculture major), he considers this to be a hypothesis test with the null hypothesis being that the person is a business major and the alternative that the person is an agriculture major.

93. Referring to Table 9-2, what would be a Type I error? a) Saying that the person is a business major when in fact the person is a business major. b) Saying that the person is a business major when in fact the person is an agriculture major. c) Saying that the person is an agriculture major when in fact the person is a business major. d) Saying that the person is an agriculture major when in fact the person is an agriculture major.

ANSWER: c TYPE: MC DIFFICULTY: Difficult KEYWORDS: form of hypothesis, form of hypothesis 94. Referring to Table 9-2, what would be a Type II error? a) Saying that the person is a business major when in fact the person is a business major. b) Saying that the person is a business major when in fact the person is an agriculture major. c) Saying that the person is an agriculture major when in fact the person is a business major. d) Saying that the person is an agriculture major when in fact the person is an agriculture major.

ANSWER: b TYPE: MC DIFFICULTY: Difficult KEYWORDS: type II error

95. Referring to Table 9-2, what is the “actual level of significance” of the test? a) 0.13 b) 0.16 c) 0.84 d) 0.87 ANSWER: a TYPE: MC DIFFICULTY: Difficult KEYWORDS: level of significance 96. Referring to Table 9-2, what is the “actual confidence coefficient”? a) 0.13 b) 0.16 c) 0.84 d) 0.87

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ANSWER: d TYPE: MC DIFFICULTY: Difficult KEYWORDS: confidence coefficient 97. Referring to Table 9-2, what is the value of  ? a) 0.13 b) 0.16 c) 0.84 d) 0.87 ANSWER: a TYPE: MC DIFFICULTY: Difficult KEYWORDS: level of significance 98. Referring to Table 9-2, what is the value of  ? a) 0.13 b) 0.16 c) 0.84 d) 0.87 ANSWER: b TYPE: MC DIFFICULTY: Difficult KEYWORDS: beta-risk

TABLE 9-3 An appliance manufacturer claims to have developed a compact microwave oven that consumes an average of no more than 250 W. From previous studies, it is believed that power consumption for microwave ovens is normally distributed with a standard deviation of 15 W. A consumer group has decided to try to discover if the claim appears true. They take a sample of 20 microwave ovens and find that they consume an average of 257.3 W. 99. Referring to Table 9-3, the population of interest is a) the power consumption in the 20 microwave ovens. b) the power consumption in all such microwave ovens. c) the mean power consumption in the 20 microwave ovens. d) the mean power consumption in all such microwave ovens.

ANSWER: b TYPE: MC DIFFICULTY: Moderate .

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KEYWORDS: one-tailed test, mean, Z test 100. Referring to Table 9-3, the parameter of interest is a) the mean power consumption of the 20 microwave ovens. b) the mean power consumption of all such microwave ovens. c) 250 d) 257.3 ANSWER: b TYPE: MC DIFFICULTY: Easy KEYWORDS: one-tailed test, mean, Z test, parameter 101. Referring to Table 9-3, the appropriate hypotheses to determine if the manufacturer's claim appears reasonable are: a) H 0 :  = 250 versus H1 :   250 b) H 0 :   250 versus H1 :   250

H 0 :   250 versus H1 :   250 d) H 0 :   257.3 versus H1 :   257.3 c)

ANSWER: c TYPE: MC DIFFICULTY: Moderate KEYWORDS: one-tailed test, mean, Z test, form of hypothesis, form of hypothesis 102. Referring to Table 9-3, for a test with a level of significance of 0.05, the critical value would be ________.

ANSWER: Z = 1.645 TYPE: FI DIFFICULTY: Moderate KEYWORDS: one-tailed test, mean, Z test, critical value 103. Referring to Table 9.3, the value of the test statistic is ________.

ANSWER: 2.18 TYPE: FI DIFFICULTY: Easy KEYWORDS: one-tailed test, mean, Z test, test statistic 104. Referring to Table 9-3, the p-value of the test is ________.

ANSWER: 0.0148 using Excel or 0.0146 using Table E.2 TYPE: FI DIFFICULTY: Moderate KEYWORDS: one-tailed test, mean, Z test, p-value 105. True or False: Referring to Table 9-3, for this test to be valid, it is necessary that the power consumption for microwave ovens has a normal distribution.

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ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: one-tailed test, mean, Z test, assumptions 106. True or False: Referring to Table 9-3, the null hypothesis will be rejected at 5% level of significance.

ANSWER: True TYPE: TF DIFFICULTY: Moderate KEYWORDS: one-tailed test, mean, Z test, decision 107. True or False: Referring to Table 9-3, the null hypothesis will be rejected at 1% level of significance.

ANSWER: False TYPE: TF DIFFICULTY: Moderate KEYWORDS: one-tailed test, mean, Z test, decision 108. True or False: Referring to Table 9-3, the consumer group can conclude that there is enough evidence to prove that the manufacturer’s claim is not true when allowing for a 5% probability of committing a type I error.

ANSWER: True TYPE: TF DIFFICULTY: Difficult KEYWORDS: one-tailed test, mean, Z test, conclusion

TABLE 9-4 A drug company is considering marketing a new local anesthetic. The effective time of the anesthetic the drug company is currently producing has a normal distribution with an average of 7.4 minutes with a standard deviation of 1.2 minutes. The chemistry of the new anesthetic is such that the effective time should be normal with the same standard deviation, but the mean effective time may be lower. If it is lower, the drug company will market the new anesthetic; otherwise, they will continue to produce the older one. A sample of size 36 results in a sample mean of 7.1. A hypothesis test will be done to help make the decision.

109. Referring to Table 9-4, the appropriate hypotheses are: a) H 0 :  = 7.4 versus H1 :   7.4 b) H 0 :   7.4 versus H1 :   7.4

H 0 :   7.4 versus H1 :   7.4 d) H 0 :   7.4 versus H1 :   7.4 c)

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ANSWER: c TYPE: MC DIFFICULTY: Easy KEYWORDS: one-tailed test, mean, Z test, form of hypothesis, form of hypothesis 110. Referring to Table 9-4, for a test with a level of significance of 0.10, the critical value would be ________.

ANSWER: -1.28 TYPE: FI DIFFICULTY: Easy KEYWORDS: one-tailed test, mean, Z test, critical value 111. Referring to Table 9-4, the value of the test statistic is ________.

ANSWER: -1.50 TYPE: FI DIFFICULTY: Easy KEYWORDS: one-tailed test, mean, Z test, test statistic 112. Referring to Table 9-4, the p-value of the test is ________.

ANSWER: 0.0668 TYPE: FI DIFFICULTY: Moderate KEYWORDS: one-tailed test, mean, Z test, p-value 113. True or False: Referring to Table 9-4, the null hypothesis will be rejected with a level of significance of 0.10.

ANSWER: True TYPE: TF DIFFICULTY: Moderate KEYWORDS: one-tailed test, mean, Z test, decision 114. True or False: Referring to Table 9-4, if the level of significance had been chosen as 0.05, the null hypothesis would be rejected.

ANSWER: False TYPE: TF DIFFICULTY: Moderate KEYWORDS: one-tailed test, mean, Z test, decision 115. True or False: Referring to Table 9-4, if the level of significance had been chosen as 0.05, the company would market the new anesthetic.

ANSWER: False TYPE: TF DIFFICULTY: Moderate KEYWORDS: one-tailed test, mean, Z test, conclusion

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TABLE 9-5 A bank tests the null hypothesis that the mean age of the bank's mortgage holders is less than or equal to 45, versus an alternative that the mean age is greater than 45. They take a sample and calculate a pvalue of 0.0202.

116. True or False: Referring to Table 9-5, the null hypothesis would be rejected at a significance level of  = 0.05.

ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: one-tailed test, mean, decision 117. True or False: Referring to Table 9-5, the null hypothesis would be rejected at a significance level of  = 0.01.

ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: one-tailed test, mean, decision 118. True or False: Referring to Table 9-5, the bank can conclude that the average age is greater than 45 at a significance level of  = 0.01.

ANSWER: False TYPE: TF DIFFICULTY: Moderate KEYWORDS: one-tailed test, mean, conclude

119. Referring to Table 9-5, if the same sample was used to test the opposite one-tailed test, what would be that test's p-value? a) 0.0202 b) 0.0404 c) 0.9596 d) 0.9798 ANSWER: d TYPE: MC DIFFICULTY: Moderate KEYWORDS: one-tailed test, mean, p-value

TABLE 9-6 The quality control engineer for a furniture manufacturer is interested in the mean amount of force necessary to produce cracks in stressed oak furniture. She performs a two-tailed test of the null

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hypothesis that the mean for the stressed oak furniture is 650. The calculated value of the Z test statistic is a positive number that leads to a p-value of 0.080 for the test.

120. True or False: Referring to Table 9-6, if the test is performed with a level of significance of 0.10, the null hypothesis would be rejected.

ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: one-tailed test, mean, Z test, decision 121. True or False: Referring to Table 9-6, if the test is performed with a level of significance of 0.10, the engineer can conclude that the mean amount of force necessary to produce cracks in stressed oak furniture is 650.

ANSWER: False TYPE: TF DIFFICULTY: Moderate KEYWORDS: one-tailed test, mean, Z test, conclusion 122. True or False: Referring to Table 9-6, if the test is performed with a level of significance of 0.05, the null hypothesis would be rejected.

ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: one-tailed test, mean, Z test, decision

123. True or False: Referring to Table 9-6, if the test is performed with a level of significance of 0.05, the engineer can conclude that the mean amount of force necessary to produce cracks in stressed oak furniture is 650.

ANSWER: False TYPE: TF DIFFICULTY: Moderate EXPLANATION: the engineer can conclude that there is insufficient evidence to show that the mean amount of force needed is not 650 but cannot conclude that there is evidence to show that the force needed is 650.

KEYWORDS: one-tailed test, mean, Z test, conclusion 124. True or False: Referring to Table 9-6, suppose the engineer had decided that the alternative hypothesis to test was that the mean was greater than 650. Then if the test is performed with a level of significance of 0.10, the null hypothesis would be rejected.

ANSWER: True TYPE: TF DIFFICULTY: Easy .

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KEYWORDS: one-tailed test, mean, Z test, decision 125. Referring to Table 9-6, suppose the engineer had decided that the alternative hypothesis to test was that the mean was greater than 650. What would be the p-value of this one-tailed test? a) 0.040 b) 0.160 c) 0.840 d) 0.960 ANSWER: a TYPE: MC DIFFICULTY: Moderate KEYWORDS: one-tailed test, mean, Z test, p-value 126. Referring to Table 9-6, suppose the engineer had decided that the alternative hypothesis to test was that the mean was less than 650. What would be the p-value of this one-tailed test? a) 0.040 b) 0.160 c) 0.840 d) 0.960 ANSWER: d TYPE: MC DIFFICULTY: Moderate KEYWORDS: one-tailed test, mean, Z test, p-value

127. True or False: Referring to Table 9-6, suppose the engineer had decided that the alternative hypothesis to test was that the mean was less than 650. Then if the test is performed with a level of significance of 0.10, the null hypothesis would be rejected.

ANSWER: False TYPE: TF DIFFICULTY: Moderate KEYWORDS: one-tailed test, mean, Z test, decision

TABLE 9-7 A major home improvement store conducted its biggest brand recognition campaign in the company’s history. A series of new television advertisements featuring well-known entertainers and sports figures were launched. A key metric for the success of television advertisements is the proportion of viewers who “like the ads a lot”. A study of 1,189 adults who viewed the ads reported that 230 indicated that they “like the ads a lot.” The percentage of a typical television advertisement receiving the “like the ads a lot” score is believed to be 22%. Company officials wanted to know if there is evidence that the series of television advertisements are less successful than the typical ad (i.e. if there is evidence that the population proportion of “like the ads a lot” for the company’s ads is less than 0.22) at a 0.01 level of significance.

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128. Referring to Table 9-7, the parameter the company officials is interested in is: a) the mean number of viewers who “like the ads a lot”. b) the total number of viewers who “like the ads a lot”. c) the mean number of company officials who “like the ads a lot”. d) the proportion of viewers who “like the ads a lot”.

ANSWER: d TYPE: MC DIFFICULTY: Easy KEYWORDS: proportion, Z test, parameter 129. Referring to Table 9-7, state the null hypothesis for this study.

ANSWER:

H 0 :   0.22

TYPE: PR DIFFICULTY: Easy KEYWORDS: one-tailed test, proportion, Z test, form of hypothesis 130. Referring to Table 9-7, state the alternative hypothesis for this study.

ANSWER:

H 1 :   0.22

TYPE: PR DIFFICULTY: Easy KEYWORDS: one-tailed test, proportion, Z test, form of hypothesis

131. Referring to Table 9-7, what critical value should the company officials use to determine the rejection region?

ANSWER: −2.3263 TYPE: PR DIFFICULTY: Easy KEYWORDS: one-tailed test, proportion, Z test, critical value 132. Referring to Table 9-7, the null hypothesis will be rejected if the test statistics is a) greater than 2.3263 b) less than 2.3263 c) greater than −2.3263 d) less than −2.3263 ANSWER: d TYPE: MC DIFFICULTY: Easy KEYWORDS: one-tailed test, proportion, Z test, rejection region 133. True or False: Referring to Table 9-7, the null hypothesis would be rejected. .

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ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: one-tailed test, proportion, Z test, decision 134. Referring to Table 9-7, the lowest level of significance at which the null hypothesis can be rejected is ______.

ANSWER: 0.0135 TYPE: FI DIFFICULTY: Moderate KEYWORDS: one-tailed test, proportion, Z test, p-value 135. Referring to Table 9-7, the largest level of significance at which the null hypothesis will not be rejected is ______.

ANSWER: 0.0135 TYPE: FI DIFFICULTY: Moderate KEYWORDS: one-tailed test, proportion, Z test, p-value 136. True of False: Referring to Table 9-7, the company officials can conclude that there is sufficient evidence to show that the series of television advertisements are less successful than the typical ad using a level of significance of 0.01.

ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: one-tailed test, proportion, Z test, conclusion 137. True or False: Referring to Table 9-7, the company officials can conclude that there is sufficient evidence to show that the series of television advertisements are less successful than the typical ad using a level of significance of 0.05.

ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: one-tailed test, proportion, Z test, decision 138. True or False: Referring to Table 9-7, the value of  is 0.90.

ANSWER: False TYPE: TF DIFFICULTY: Moderate KEYWORDS: one-tailed test, mean, Z test, beta-risk 139. Referring to Table 9-7, what will be the p-value if these data were used to perform a two-tailed test?

ANSWER: 0.027 TYPE: TF DIFFICULTY: Moderate .

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KEYWORDS: one-tailed test, proportion, Z test, p-value TABLE 9-8 One of the biggest issues facing e-retailers is the ability to turn browsers into buyers. This is measured by the conversion rate, the percentage of browsers who buy something in their visit to a site. The conversion rate for a company’s website was 10.1%. The website at the company was redesigned in an attempt to increase its conversion rates. Samples of 200 browsers at the redesigned site were selected. Suppose that 24 browsers made a purchase. The company officials would like to know if there is evidence of an increase in conversion rate at the 5% level of significance.

140. Referring to Table 9-8, the parameter the company officials is interested in is: a) the mean number of browsers who buy something in their visit to the company’s website. b) the total number of browsers who buy something in their visit to the company’s website. c) the mean number of company officials who buy something in their visit to the company’s website. d) the proportion of browsers who buy something in their visit to the company’s website.

ANSWER: d TYPE: MC DIFFICULTY: Easy KEYWORDS: proportion, Z test, parameter 141. Referring to Table 9-8, state the null hypothesis for this study.

ANSWER: H0:   0.101 TYPE: PR DIFFICULTY: Easy KEYWORDS: one-tailed test, proportion, Z test, form of hypothesis 142. Referring to Table 9-8, state the alternative hypothesis for this study.

ANSWER: H1:  > 0.101 TYPE: PR DIFFICULTY: Easy KEYWORDS: one-tailed test, proportion, Z test, form of hypothesis 143. Referring to Table 9-8, what critical value should the company officials use to determine the rejection region?

ANSWER:  TYPE: PR DIFFICULTY: Easy KEYWORDS: one-tailed test, proportion, Z test, critical value 144. Referring to Table 9-8, the null hypothesis will be rejected if the test statistics is a) greater than 1.645 b) less than 1.645 c) greater than −1.645 d) less than −1.645 ANSWER: .

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a TYPE: MC DIFFICULTY: Easy KEYWORDS: one-tailed test, proportion, Z test, rejection region 145. True or False: Referring to Table 9-8, the null hypothesis would be rejected.

ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: one-tailed test, proportion, Z test, decision 146. Referring to Table 9-8, the lowest level of significance at which the null hypothesis can be rejected is ______.

ANSWER: 0.1863 TYPE: FI DIFFICULTY: Moderate KEYWORDS: one-tailed test, proportion, Z test, p-value 147. Referring to Table 9-8, the largest level of significance at which the null hypothesis will not be rejected is ______.

ANSWER: 0.1863 TYPE: FI DIFFICULTY: Moderate KEYWORDS: one-tailed test, proportion, Z test, p-value 148. True of False: Referring to Table 9-8, the company officials can conclude that there is sufficient evidence that the conversion rate at the company’s website has increased using a level of significance of 0.05.

ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: one-tailed test, proportion, Z test, conclusion 149. True or False: Referring to Table 9-8, the value of the probability of committing a type II error  is 0.95.

ANSWER: False TYPE: TF DIFFICULTY: Moderate KEYWORDS: one-tailed test, mean, Z test, beta-risk 150. Referring to Table 9-8, what will be the p-value if these data were used to perform a two-tailed test?

ANSWER: 0.3725 TYPE: TF DIFFICULTY: Moderate KEYWORDS: one-tailed test, proportion, Z test, p-value

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CHAPTER 10: TWO-SAMPLE TESTS AND ONE-WAY ANOAVA 1. The t test for the difference between the means of 2 independent populations assumes that the respective a) sample sizes are equal. b) sample variances are equal. c) populations are approximately normal. d) All of the above.

ANSWER: c TYPE: MC DIFFICULTY: Moderate KEYWORDS: pooled variance, t test, assumption 2. The t test for the mean difference between 2 related populations assumes that the a) population sizes are equal. b) sample variances are equal. c) population of differences is approximately normal or sample sizes are large enough. d) All of the above.

ANSWER: c TYPE: MC DIFFICULTY: Moderate KEYWORDS: t test, mean difference, assumption 3. If we are testing for the difference between the means of 2 related populations with samples of n1 = 20 and n2 = 20, the number of degrees of freedom is equal to a) 39. b) 38. c) 19. d) 18.

ANSWER: c TYPE: MC DIFFICULTY: Easy KEYWORDS: t test, mean difference, degrees of freedom 4. If we are testing for the difference between the means of 2 independent populations presuming equal variances with samples of n1 = 20 and n2 = 20, the number of degrees of freedom is equal to a) 39. b) 38. c) 19. d) 18.

ANSWER: b .

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TYPE: MC DIFFICULTY: Easy KEYWORDS: pooled variance, t test, degrees of freedom 5. In what type of test is the variable of interest the difference between the values of the observations rather than the observations themselves? a) A test for the equality of variances from 2 independent populations. b) A test for the difference between the means of 2 related populations. c) A test for the difference between the means of 2 independent populations. d) All of the above.

ANSWER: b TYPE: MC DIFFICULTY: Moderate KEYWORDS: t test, mean difference 6. In testing for the differences between the means of 2 independent populations where the variances in each population are unknown but assumed equal, the degrees of freedom are a) n – 1. b) n1 + n2 – 1. c) n1 + n2 – 2. d) n – 2.

ANSWER: c TYPE: MC DIFFICULTY: Easy KEYWORDS: pooled variance, t test, degrees of freedom 7. In testing for differences between the means of 2 related populations where the variance of the differences is unknown, the degrees of freedom are a) n – 1. b) n1 + n2 – 1. c) n1 + n2 – 2. d) n – 2.

ANSWER: a TYPE: MC DIFFICULTY: Easy KEYWORDS: t test, mean difference, degrees of freedom 8. In testing for differences between the means of two related populations, the null hypothesis is a) H 0 :  D = 2 . b) H 0 :  D = 0 . c)

H0 : D  0 .

d) H 0 :  D  0 .

ANSWER: b TYPE: MC DIFFICULTY: Easy KEYWORDS: t test, mean difference, form of hypothesis

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9. In testing for differences between the means of two independent populations, the null hypothesis is: a) H 0 : 1 − 2 = 2. b) H0 : 1 – 2 = 0. c) H0 : 1 – 2 > 0. d) H0 : 1 – 2 < 2.

ANSWER: b TYPE: MC DIFFICULTY: Easy KEYWORDS: pooled variance, t test, form of hypothesis 10. When testing for the difference between 2 population variances with sample sizes of n1 = 8 and n2 = 10, the number of degrees of freedom are a) 8 and 10. b) 7 and 9. c) 18. d) 16.

ANSWER: b TYPE: MC DIFFICULTY: Easy KEYWORDS: F test, difference between two variances, degrees of freedom 11. The statistical distribution used for testing the difference between two population variances is the ___ distribution. a) t b) standardized normal c) binomial d) F ANSWER: d TYPE: MC DIFFICULTY: Easy KEYWORDS: F test, difference between two variances, degrees of freedom 12. The test for the equality of two population variances is based on a) the difference between the 2 sample variances. b) the ratio of the 2 sample variances. c) the difference between the 2 population variances. d) the difference between the sample variances divided by the difference between the sample means.

ANSWER: b TYPE: MC DIFFICULTY: Easy KEYWORDS: F test, difference between two variances

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13. True or False: The F test used for testing the difference in two population variances is always a one-tailed test.

ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: F test, difference between two variances, rejection region 14. In testing for the differences between the means of two related populations, the _______ hypothesis is the hypothesis of "no differences."

ANSWER: null TYPE: FI DIFFICULTY: Easy KEYWORDS: t test, mean difference, form of hypothesis 15. In testing for the differences between the means of two related populations, we assume that the differences follow a _______ distribution.

ANSWER: normal TYPE: FI DIFFICULTY: Easy KEYWORDS: t test, mean difference, assumption 16. In testing for the differences between the means of two independent populations, we assume that the 2 populations each follow a _______ distribution.

ANSWER: normal TYPE: FI DIFFICULTY: Easy KEYWORDS: pooled variance, t test, assumption 17. Given the following information, calculate the degrees of freedom that should be used in the pooled-variance t test. s 12 = 4 s 22 = 6 n1 = 16 n2 = 25 a) df = 41 b) df = 39 c) df = 16 d) df = 25 ANSWER: b TYPE: MC DIFFICULTY: Easy KEYWORDS: pooled variance, t test, degrees of freedom

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Two-Sample Tests and One-Way ANOVA 18. Given the following information, calculate sp2, the pooled sample variance that should be used in the pooled-variance t test. s 12 = 4 s 22 = 6 n1 = 16 n2 = 25 a) sp2 = 6.00 b) sp2 = 5.00 c) sp2 = 5.23 d) sp2 = 4.00 ANSWER: c TYPE: MC DIFFICULTY: Easy KEYWORDS: pooled variance, t test

TABLE 10-1 Are Japanese managers more motivated than American managers? A randomly selected group of each were administered the Sarnoff Survey of Attitudes Toward Life (SSATL), which measures motivation for upward mobility. The SSATL scores are summarized below.

Sample Size Mean SSATL Score Population Std. Dev.

American 211 65.75 11.07

Japanese 100 79.83 6.41

19. Referring to Table 10-1, judging from the way the data were collected, which test would likely be most appropriate to employ? a) Paired t test b) Pooled-variance t test for the difference between two means c) Independent samples Z test for the difference between two means d) Related samples Z test for the mean difference ANSWER: c TYPE: MC DIFFICULTY: Easy KEYWORDS: Z test, difference between two means 20. Referring to Table 10-1, give the null and alternative hypotheses to determine if the average SSATL score of Japanese managers differs from the average SSATL score of American managers. a) H0 :  A –  J  0 versus H1 :  A –  J  0 b) H0 :  A –  J  0 versus H1 :  A –  J  0 c) H0 :  A –  J = 0 versus H1 :  A –  J  0 d) H0 : XA – XJ = 0 versus H1 : X A – XJ  0 ANSWER: c TYPE: MC DIFFICULTY: Easy KEYWORDS: Z test, mean differences, form of hypothesis

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Two-Sample Tests and One-Way ANOVA 305

21. Referring to Table 10-1, assuming the independent samples procedure was used, calculate the value of the test statistic.

65.75 – 79.83 9.82 9.82 + 211 100 65.75 – 79.83 b) Z = 11.07 6.41 + 211 100 65.75 – 79.83 c) Z = 9.82 2 9.82 2 + 211 100 65.75 – 79.83 d) Z = 11.07 2 6.412 + 211 100 a)

Z=

ANSWER: d TYPE: MC DIFFICULTY: Easy KEYWORDS: Z test, mean differences, test statistic 22. Referring to Table 10-1, suppose that the test statistic is Z = 2.45. Find the p-value if we assume that the alternative hypothesis was a two-tailed test ( H1 :  A –  J  0 ). a) 0.0071 b) 0.0142 c) 0.4929 d) 0.9858 ANSWER: b TYPE: MC DIFFICULTY: Moderate KEYWORDS: Z test, mean differences, p-value

TABLE 10-2 .

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Two-Sample Tests and One-Way ANOVA

A researcher randomly sampled 30 graduates of an MBA program and recorded data concerning their starting salaries. Of primary interest to the researcher was the effect of gender on starting salaries.

Analysis of the mean salaries of the females and males in the sample is given below.

Hypothesized Difference Level of Significance Population 1 Sample Sample Size Sample Mean Sample Standard Deviation Population 2 Sample Sample Size Sample Mean Sample Standard Deviation Difference in Sample Means t-Test Statistic Lower-Tail Test Lower Critical Value p-Value

0 0.05 18 48266.7 13577.63 12 55000 11741.29 -6733.3 -1.40193 -1.70113 0.085962

23. Referring to Table 10-2, the researcher was attempting to show statistically that the female MBA graduates have a significantly lower mean starting salary than the male MBA graduates.

According to the test run, which of the following is an appropriate alternative hypothesis? a) H1 : females  males b) H1 : females  males

H1 : females  males d) H1 : females = males c)

ANSWER: b TYPE: MC DIFFICULTY: Moderate KEYWORDS: pooled variance, t test, form of hypothesis 24. Referring to Table 10-2, the researcher was attempting to show statistically that the female MBA graduates have a significantly lower mean starting salary than the male MBA graduates. From the analysis in Table 10-2, the correct test statistic is: a) 0.0860 b) – 1.4019 c) – 1.7011 d) – 6,733.33 ANSWER: b TYPE: MC DIFFICULTY: Moderate KEYWORDS: pooled variance, t test, test statistic

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Two-Sample Tests and One-Way ANOVA 307

25. Referring to Table 10-2, the researcher was attempting to show statistically that the female MBA graduates have a significantly lower mean starting salary than the male MBA graduates. The proper conclusion for this test is: a) At the  = 0.10 level, there is sufficient evidence to indicate a difference in the mean starting salaries of male and female MBA graduates. b) At the  = 0.10 level, there is sufficient evidence to indicate that females have a lower mean starting salary than male MBA graduates. c) At the  = 0.10 level, there is sufficient evidence to indicate that females have a higher mean starting salary than male MBA graduates. d) At the  = 0.10 level, there is insufficient evidence to indicate any difference in the mean starting salaries of male and female MBA graduates.

ANSWER: b TYPE: MC DIFFICULTY: Moderate KEYWORDS: pooled variance, t test, conclusion 26. Referring to Table 10-2, the researcher was attempting to show statistically that the female MBA graduates have a significantly lower mean starting salary than the male MBA graduates. What assumptions were necessary to conduct this hypothesis test? a) Both populations of salaries (male and female) must have approximate normal distributions. b) The population variances are approximately equal. c) The samples were randomly and independently selected. d) All of the above assumptions were necessary.

ANSWER: d TYPE: MC DIFFICULTY: Moderate KEYWORDS: pooled variance, t test, assumption 27. Referring to Table 10-2, what is the 99% confidence interval estimate for the difference between two means?

ANSWER: −$20,004.90 to $6,538.30 TYPE: PR DIFFICULTY: Moderate KEYWORDS: confidence interval, difference between two means 28. Referring to Table 10-2, what is the 95% confidence interval estimate for the difference between two means?

ANSWER: −$16571.55 to $3,104.95 TYPE: PR DIFFICULTY: Moderate KEYWORDS: confidence interval, difference between two means

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29. Referring to Table 10-2, what is the 90% confidence interval estimate for the difference between two means?

ANSWER: −$14,903.61 to $1,437.01 TYPE: PR DIFFICULTY: Moderate KEYWORDS: confidence interval, difference between two means

TABLE 10-3 The use of preservatives by food processors has become a controversial issue. Suppose 2 preservatives are extensively tested and determined safe for use in meats. A processor wants to compare the preservatives for their effects on retarding spoilage. Suppose 15 cuts of fresh meat are treated with preservative I and 15 are treated with preservative II, and the number of hours until spoilage begins is recorded for each of the 30 cuts of meat. The results are summarized in the table below.

Preservative I Preservative II X I = 106.4 hours X II = 96.54 hours S I = 10.3 hours S II = 13.4 hours 30. Referring to Table 10-3, state the null and alternative hypotheses for testing if the population variances differ for preservatives I and II. a) H 0 :  I2 –  II2  0 versus H 1 :  I2 –  II2  0 b)

H 0 :  I2 –  II2  0 versus H 1 :  I2 –  II2  0

c)

H 0 :  I2 –  II2 = 0 versus H1 :  I2 –  II2  0

d) H 0 :  I2 –  II2  0 versus H 1 :  I2 –  II2 = 0 ANSWER: c TYPE: MC DIFFICULTY: Moderate KEYWORDS: F test, difference between two variances, form of hypothesis 31. Referring to Table 10-3, what is the value of the test statistic for testing if the population variances differ for preservatives I and II? ANSWER: 1.6925 TYPE: PR DIFFICULTY: Easy KEYWORDS: F test, difference between two variances, test statistic

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Two-Sample Tests and One-Way ANOVA 309

32. Referring to Table 10-3, what assumptions are necessary for testing if the population variances differ for preservatives I and II? a) Both sampled populations are normally distributed. b) Both samples are random and independent. c) Neither (a) nor (b) is necessary. d) Both (a) and (b) are necessary.

ANSWER: d TYPE: MC DIFFICULTY: Easy KEYWORDS: F test, difference between two variances, assumption 33. Referring to Table 10-3, what is the critical value for testing if the population variances differ for preservatives I and II at the 5% level of significance?

ANSWER: 2.9786 (using Excel) or 2.95 (using Table E.5 with 15 and 14 degrees of freedom) TYPE: PR DIFFICULTY: Easy KEYWORDS: F test, difference between two variances, critical value 34. Referring to Table 10-3, what is the largest level of significance at which a test if the population variances differ for preservatives I and II will not be rejected?

ANSWER: 0.3362 or ‘greater than 0.05’ (using Table E.5 with 15 and 14 degrees of freedom) TYPE: PR DIFFICULTY: Moderate KEYWORDS: F test, difference between two variances, p-value 35. Referring to Table 10-3, suppose  = 0.05. Which of the following represents the result of the relevant hypothesis test? a) The alternative hypothesis is rejected. b) The null hypothesis is rejected. c) The null hypothesis is not rejected. d) Insufficient information exists on which to make a decision.

ANSWER: c TYPE: MC DIFFICULTY: Easy KEYWORDS: F test, difference between two variances, decision

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310

Two-Sample Tests and One-Way ANOVA 36. Referring to Table 10-3, suppose  = 0.05. Which of the following represents the correct conclusion? a) There is no evidence of a difference in the population variances betweeen preservatives I and II. b) There is evidence of a difference in the population variances betweeen preservatives I and II. c) There is no evidence that the population variances betweeen preservatives I and II are the same. d) There is evidence that the population variances betweeen preservatives I and II are the same.

ANSWER: a TYPE: MC DIFFICULTY: Easy KEYWORDS: F test, difference between two variances, conclusion

TABLE 10-4 A real estate company is interested in testing whether, on average, families in Gotham have been living in their current homes for less time than families in Metropolis have. Assume that the two population variances are equal. A random sample of 100 families from Gotham and a random sample of 150 families in Metropolis yield the following data on length of residence in current homes.

Gotham: X G = 35 months, sG2 = 900 Metropolis: X M = 50 months, sM2 = 1050 37. Referring to Table 10-4, which of the following represents the relevant hypotheses tested by the real estate company? a) H0 :  G – M  0 versus H1 :  G –  M  0 b) H0 : G –  M  0 versus H1 :  G –  M  0 c) H0 :  G – M = 0 versus H1 :  G –  M  0 d) H0 : XG – XM  0 versus H1 : XG – XM  0 ANSWER: a TYPE: MC DIFFICULTY: Easy KEYWORDS: pooled variance, t test, form of hypothesis 38. Referring to Table 10-4, what is the estimated standard error of the difference between the 2 sample means? a) 4.06 b) 5.61 c) 8.01 d) 16.00 ANSWER: a TYPE: MC DIFFICULTY: Moderate KEYWORDS: pooled variance, t test, standard error

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Two-Sample Tests and One-Way ANOVA 311

39. Referring to Table 10-4, what is an unbiased point estimate for the mean of the sampling distribution of the difference between the 2 sample means? a) – 22 b) – 10 c) – 15 d) 0 ANSWER: c TYPE: MC DIFFICULTY: Easy KEYWORDS: pooled variance, t test, unbiased, point estimate 40. Referring to Table 10-4, what is(are) the critical value(s) of the relevant hypothesis test if the level of significance is 0.05? a) t  Z = – 1.645 b) t  Z =  1.96 c) t  Z = – 1.96 d) t  Z = – 2.080 ANSWER: a TYPE: MC DIFFICULTY: Easy KEYWORDS: pooled variance, t test, critical value 41. Referring to Table 10-4, what is(are) the critical value(s) of the relevant hypothesis test if the level of significance is 0.01? a) t  Z = – 1.96 b) t  Z =  1.96 c) t  Z = – 2.080 d) t  Z = – 2.33 ANSWER: d TYPE: MC DIFFICULTY: Easy KEYWORDS: pooled variance, t test, critical value 42. Referring to Table 10-4, what is the standardized value of the estimate of the mean of the sampling distribution of the difference between sample means? a) – 8.75 b) – 3.69 c) – 2.33 d) – 1.96 ANSWER: b TYPE: MC DIFFICULTY: Moderate KEYWORDS: pooled variance, t test, test statistic

43. Referring to Table 10-4, suppose  = 0.10. Which of the following represents the result of the relevant hypothesis test? .

312

Two-Sample Tests and One-Way ANOVA

a) b) c) d)

The alternative hypothesis is rejected.

The null hypothesis is rejected.

The null hypothesis is not rejected.

Insufficient information exists on which to make a decision.

ANSWER: b TYPE: MC DIFFICULTY: Easy KEYWORDS: pooled variance, t test, decision 44. Referring to Table 10-4, suppose  = 0.05. Which of the following represents the result of the relevant hypothesis test? a) The alternative hypothesis is rejected. b) The null hypothesis is rejected. c) The null hypothesis is not rejected. d) Insufficient information exists on which to make a decision.

ANSWER: b TYPE: MC DIFFICULTY: Easy KEYWORDS: pooled variance, t test, decision 45. Referring to Table 10-4, suppose  = 0.01. Which of the following represents the result of the relevant hypothesis test? a) The alternative hypothesis is rejected. b) The null hypothesis is rejected. c) The null hypothesis is not rejected. d) Insufficient information exists on which to make a decision.

ANSWER: b TYPE: MC DIFFICULTY: Easy KEYWORDS: pooled variance, t test, decision 46. Referring to Table 10-4, suppose  = 0.1. Which of the following represents the correct conclusion? a) There is not enough evidence that, on average, families in Gotham have been living in their current homes for less time than families in Metropolis have. b) There is enough evidence that, on average, families in Gotham have been living in their current homes for less time than families in Metropolis have. c) There is not enough evidence that, on average, families in Gotham have been living in their current homes for no less time than families in Metropolis have. d) There is enough evidence that, on average, families in Gotham have been living in their current homes for no less time than families in Metropolis have.

ANSWER: b TYPE: MC DIFFICULTY: Easy KEYWORDS: pooled variance, t test, conclusion 47. Referring to Table 10-4, suppose  = 0.05. Which of the following represents the correct conclusion? .

Two-Sample Tests and One-Way ANOVA 313

a) There is not enough evidence that, on average, families in Gotham have been living in their current homes for less time than families in Metropolis have. b) There is enough evidence that, on average, families in Gotham have been living in their current homes for less time than families in Metropolis have. c) There is not enough evidence that, on average, families in Gotham have been living in their current homes for no less time than families in Metropolis have. d) There is enough evidence that, on average, families in Gotham have been living in their current homes for no less time than families in Metropolis have.

ANSWER: b TYPE: MC DIFFICULTY: Easy KEYWORDS: pooled variance, t test, conclusion 48. Referring to Table 10-4, suppose  = 0.01. Which of the following represents the correct conclusion? a) There is not enough evidence that, on average, families in Gotham have been living in their current homes for less time than families in Metropolis have. b) There is enough evidence that, on average, families in Gotham have been living in their current homes for less time than families in Metropolis have. c) There is not enough evidence that, on average, families in Gotham have been living in their current homes for no less time than families in Metropolis have. d) There is enough evidence that, on average, families in Gotham have been living in their current homes for no less time than families in Metropolis have.

ANSWER: b TYPE: MC DIFFICULTY: Easy KEYWORDS: pooled variance, t test, conclusion 49. Referring to Table 10-4, what is the 99% confidence interval estimate for the difference in the two means?

ANSWER: −25.54 to −4.46 months TYPE: PR DIFFICULTY: Moderate KEYWORDS: confidence interval, difference between two means 50. Referring to Table 10-4, what is the 95% confidence interval estimate for the difference in the two means?

ANSWER: −23.00 to −7.00 months TYPE: PR DIFFICULTY: Moderate KEYWORDS: confidence interval, difference between two means

TABLE 10-5

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Two-Sample Tests and One-Way ANOVA

To test the effectiveness of a business school preparation course, 8 students took a general business test before and after the course. The results are given below.

Exam Score Exam Score Student Before Course (1) After Course (2) 1 530 670 2 690 770 3 910 1,000 4 700 710 5 450 550 6 820 870 7 820 770 8 630 610 51. Referring to Table 10-5, the number of degrees of freedom is a) 14. b) 13. c) 8. d) 7.

ANSWER: d TYPE: MC DIFFICULTY: Easy KEYWORDS: t test, mean difference, degrees of freedom 52. Referring to Table 10-5, the value of the sample mean difference is _______ if the difference scores reflect the results of the exam after the course minus the results of the exam before the course. a) 0 b) 50 c) 68 d) 400 ANSWER: b TYPE: MC DIFFICULTY: Easy KEYWORDS: t test, mean difference, test statistic 53. Referring to Table 10-5, the value of the standard error of the difference scores is a) 65.027 b) 60.828 c) 22.991 d) 14.696 ANSWER: c TYPE: MC DIFFICULTY: Easy KEYWORDS: t test, mean difference, standard error

54. Referring to Table 10-5, what is the critical value for testing at the 5% level of significance whether the business school preparation course is effective in improving exam scores? a) 2.365 .

Two-Sample Tests and One-Way ANOVA 315

b) 2.145 c) 1.761 d) 1.895 ANSWER: d TYPE: MC DIFFICULTY: Easy KEYWORDS: t test, mean difference, critical value 55. Referring to Table 10-5, at the 0.05 level of significance, the decision for this hypothesis test would be: a) reject the null hypothesis. b) do not reject the null hypothesis. c) reject the alternative hypothesis. d) It cannot be determined from the information given.

ANSWER: a TYPE: MC DIFFICULTY: Easy KEYWORDS: t test, mean difference, decision 56. Referring to Table 10-5, at the 0.05 level of significance, the conclusion for this hypothesis test would be: a) the business school preparation course does improve exam score. b) the business school preparation course does not improve exam score. c) the business school preparation course has no impact on exam score. d) It cannot be drawn from the information given.

ANSWER: a TYPE: MC DIFFICULTY: Easy KEYWORDS: t test, mean difference, conclusion 57. True or False: Referring to Table 10-5, one must assume that the population of difference scores is normally distributed.

ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: t test, mean difference, assumption 58. Referring to Table 10-5, the calculated value of the test statistic is ________.

ANSWER: 2.175 TYPE: FI DIFFICULTY: Easy KEYWORDS: t test, mean difference, test statistic

59. Referring to Table 10-5, the p-value of the test statistic is ________.

ANSWER: .

316

Two-Sample Tests and One-Way ANOVA 0.0331 (using Excel) or ‘between 0.025 and 0.05’ (using Table E.3 with 7 degrees of freedom) TYPE: FI DIFFICULTY: Moderate EXPLANATION: p-value obtained from Excel KEYWORDS: t test, mean difference, p-value 60. True or False: Referring to Table 10-5, in examining the differences between related samples we are essentially sampling from an underlying population of difference "scores."

ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: t test, mean difference, sampling distribution 61. True or False: The sample size in each independent sample must be the same if we are to test for differences between the means of 2 independent populations.

ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: pooled variance, t test, sample size 62. True or False: When we test for differences between the means of 2 independent populations, we can only use a two-tailed test.

ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: pooled variance, t test, rejection region 63. True or False: When testing for differences between the means of 2 related populations, we can use either a one-tailed or two-tailed test.

ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: t test, mean difference, rejection region 64. True or False: Repeated measurements from the same individuals is an example of data collected from 2 related populations.

ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: t test, mean difference

65. True or False: The test for the equality of 2 population variances assumes that each of the 2 populations is normally distributed.

ANSWER: .

Two-Sample Tests and One-Way ANOVA 317

True TYPE: TF DIFFICULTY: Easy KEYWORDS: F test, difference between two variances, assumption 66. True or False: For all two-sample tests, the sample sizes must be equal in the 2 groups.

ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: sample size 67. True or False: When the sample sizes are equal, the pooled variance of the 2 groups is the average of the 2 sample variances.

ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: pooled variance, t test, sample size 68. True or False: The F distribution is symmetric.

ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: F distribution, properties 69. True or False: The F distribution can only have positive values.

ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: F distribution, properties 70. True or False: All F tests are one-tailed tests.

ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: F distribution, properties

71. True or False: A researcher is curious about the effect of sleep on students’ test performances. He chooses 60 students and gives each 2 tests: one given after 2 hours’ sleep and one after 8 hours’ sleep. The test the researcher should use would be a related samples test.

ANSWER: True .

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TYPE: TF DIFFICULTY: Easy KEYWORDS: t test, mean difference 72. When testing H0 : 1 – 2 = 0 versus H1 : 1 –  2  0 , the observed value of the Z-score was found to be – 2.13. The p-value for this test would be a) 0.0166. b) 0.0332. c) 0.9668. d) 0.9834.

ANSWER: b TYPE: MC DIFFICULTY: Easy KEYWORDS: Z test, difference between two means, p-value 73. When testing H 0 : 1 − 2  0 versus H1 : 1 − 2  0 , the observed value of the Z-score was found to be – 2.13. The p-value for this test would be a) 0.0166. b) 0.0332. c) 0.9668. d) 0.9834.

ANSWER: d TYPE: MC DIFFICULTY: Moderate KEYWORDS: Z test, difference between two means, p-value 74. When testing H 0 : 1 − 2  0 versus H1 : 1 − 2  0 , the observed value of the Z-score was found to be – 2.13. The p-value for this test would be a) 0.0166. b) 0.0332. c) 0.9668. d) 0.9834.

ANSWER: a TYPE: MC DIFFICULTY: Easy KEYWORDS: Z test, difference between two means, p-value

75. True or False: A statistics professor wanted to test whether the grades on a statistics test were the same for upper and lower classmen. The professor took a random sample of size 10 from each, conducted a test and found out that the variances were equal. For this situation, the professor should use a t test with related samples.

ANSWER: False .

Two-Sample Tests and One-Way ANOVA 319

TYPE: TF DIFFICULTY: Easy KEYWORDS: pooled variance, t test 76. True or False: A statistics professor wanted to test whether the grades on a statistics test were the same for upper and lower classmen. The professor took a random sample of size 10 from each, conducted a test and found out that the variances were equal. For this situation, the professor should use a t test with independent samples.

ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: pooled variance, t test 77. True or False: A Marine drill instructor recorded the time in which each of 11 recruits completed an obstacle course both before and after basic training. To test whether any improvement occurred, the instructor would use a t-distribution with 11 degrees of freedom.

ANSWER: False TYPE: TF DIFFICULTY: Moderate KEYWORDS: t test, mean difference, degrees of freedom 78. True or False: A Marine drill instructor recorded the time in which each of 11 recruits completed an obstacle course both before and after basic training. To test whether any improvement occurred, the instructor would use a t-distribution with 10 degrees of freedom.

ANSWER: True TYPE: TF DIFFICULTY: Moderate KEYWORDS: t test, mean difference, degrees of freedom TABLE 10-6 Two samples each of size 25 are taken from independent populations assumed to be normally distributed with equal variances. The first sample has a mean of 35.5 and standard deviation of 3.0 while the second sample has a mean of 33.0 and standard deviation of 4.0.

79. Referring to Table 10-6, the pooled (i.e., combined) variance is _______.

ANSWER: 12.5 TYPE: FI DIFFICULTY: Easy KEYWORDS: pooled variance, t test

80. Referring to Table 10-6, the computed t statistic is _______.

ANSWER: 2.50 TYPE: FI DIFFICULTY: Easy KEYWORDS: pooled variance, t test, test statistic

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320

Two-Sample Tests and One-Way ANOVA

81. Referring to Table 10-6, there are _______ degrees of freedom for this test.

ANSWER: 48 TYPE: FI DIFFICULTY: Easy KEYWORDS: pooled variance, t test, degrees of freedom 82. Referring to Table 10-6, the critical values for a two-tailed test of the null hypothesis of no difference in the population means at the  = 0.05 level of significance are _______.

ANSWER:  2.0106 TYPE: FI DIFFICULTY: Easy KEYWORDS: pooled variance, t test, critical value 83. Referring to Table 10-6, a two-tailed test of the null hypothesis of no difference would _______ (be rejected/not be rejected) at the  = 0.05 level of significance.

ANSWER: be rejected TYPE: FI DIFFICULTY: Easy KEYWORDS: pooled variance, t test, decision 84. Referring to Table 10-6, the p-value for a two-tailed test whose computed t statistic is 2.50 is between _____ and _______ .

ANSWER: 0.01 and 0.02 TYPE: FI DIFFICULTY: Moderate KEYWORDS: pooled variance, t test, p-value 85. Referring to Table 10-6, if we were interested in testing against the one-tailed alternative that 1   2 at the  = 0.01 level of significance, the null hypothesis would _______ .

ANSWER: be rejected TYPE: FI DIFFICULTY: Easy KEYWORDS: pooled variance, t test, decision

86. Referring to Table 10-6, the p-value for a one-tailed test whose computed statistic is 2.50 (in the hypothesized direction) is between _______ .

ANSWER: 0.005 and 0.01 TYPE: FI DIFFICULTY: Moderate KEYWORDS: pooled variance, t test, p-value

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Two-Sample Tests and One-Way ANOVA 321

87. Referring to Table 10-6, what is the 99% confidence interval estimate for the difference in the two means?

ANSWER: −0.1822 to 5.1822 TYPE: PR DIFFICULTY: Moderate KEYWORDS: confidence interval, difference between two means 88. Referring to Table 10-6, what is the 95% confidence interval estimate for the difference in the two means?

ANSWER: 0.4894 to 4.5106 TYPE: PR DIFFICULTY: Moderate KEYWORDS: confidence interval, difference between two means 89. Referring to Table 10-6, what is the 90% confidence interval estimate for the difference in the two means?

ANSWER: 0.8228 to 4.1772 TYPE: PR DIFFICULTY: Moderate KEYWORDS: confidence interval, difference between two means

TABLE 10-7 To investigate the efficacy of a diet, a random sample of 16 male patients is drawn from a population of adult males using the diet. The weight of each individual in the sample is taken at the start of the diet and at a medical follow-up 4 weeks later. Assuming that the population of differences in weight before versus after the diet follow a normal distribution, the t-test for related samples can be used to determine if there was a significant decrease in the mean weight during this period. Suppose the mean decrease in weights over all 16 subjects in the study is 3.0 pounds with the standard deviation of differences computed as 6.0 pounds.

90. Referring to Table 10-7, the t test should be _______-tailed.

ANSWER: one TYPE: FI DIFFICULTY: Easy KEYWORDS: t test, mean difference, rejection region

91. Referring to Table 10-7, the computed t statistic is _______.

ANSWER: 2.00 TYPE: FI DIFFICULTY: Easy KEYWORDS: t test, mean difference, test statistic

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92. Referring to Table 10-7, there are _______ degrees of freedom for this test.

ANSWER: 15 TYPE: FI DIFFICULTY: Easy KEYWORDS: t test, mean difference, degrees of freedom 93. Referring to Table 10-7, the critical value for a one-tailed test of the null hypothesis of no difference at the  = 0.05 level of significance is _______.

ANSWER: 1.7531 TYPE: FI DIFFICULTY: Easy KEYWORDS: t test, mean difference, critical value 94. Referring to Table 10-7, a one-tailed test of the null hypothesis of no difference would _______ (be rejected/not be rejected) at the  = 0.05 level of significance.

ANSWER: be rejected TYPE: FI DIFFICULTY: Easy KEYWORDS: t test, mean difference, decision 95. Referring to Table 10-7, the p-value for a one-tailed test whose computed t statistic is 2.00 is between _______.

ANSWER: 0.025 and 0.05 TYPE: FI DIFFICULTY: Moderate KEYWORDS: t test, mean difference, p-value 96. Referring to Table 10-7, if we were interested in testing against the two-tailed alternative that  D is not equal to zero at the  = 0.05 level of significance, the null hypothesis would _______ (be rejected/not be rejected).

ANSWER: not be rejected TYPE: FI DIFFICULTY: Easy KEYWORDS: t test, mean difference, decision

97. Referring to Table 10-7, the p-value for a two-tailed test whose computed statistic is 2.00 is between ________ .

ANSWER: 0.05 and 0.10 TYPE: FI DIFFICULTY: Moderate KEYWORDS: t test, mean difference, p-value .

Two-Sample Tests and One-Way ANOVA 323

98. Referring to Table 10-7, what is the 95% confidence interval estimate for the mean difference in weight before and after the diet?

ANSWER: −0.20 to 6.20 TYPE: PR DIFFICULTY: Moderate KEYWORDS: confidence interval, mean difference 99. Referring to Table 10-7, what is the 99% confidence interval estimate for the mean difference in weight before and after the diet?

ANSWER: − to 7.42 TYPE: PR DIFFICULTY: Moderate KEYWORDS: confidence interval, mean difference 100.

Referring to Table 10-7, what is the 90% confidence interval estimate for the mean difference in weight before and after the diet?

ANSWER: 0 to 5.63 TYPE: PR DIFFICULTY: Moderate KEYWORDS: confidence interval, mean difference

TABLE 10-8 A buyer for a manufacturing plant suspects that his primary supplier of raw materials is overcharging.

In order to determine if his suspicion is correct, he contacts a second supplier and asks for the prices on various identical materials. He wants to compare these prices with those of his primary supplier.

The data collected is presented in the table below, with some summary statistics presented (all of these might not be necessary to answer the questions which follow). The buyer believes that the differences are normally distributed and will use this sample to perform an appropriate test at a level of significance of 0.01. .

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Primary Material Supplier 1 $55 2 $48 3 $31 4 $83 5 $37 6 $55 Sum: $309 Sum of Squares: $17,573

Secondary Supplier $45 $47 $32 $77 $37 $54 $292 $15,472

Difference $10 $1 – $1 $6 $0 $1 $17 $139

101.

Referring to Table 10-8, the hypotheses that the buyer should test are a null hypothesis that ________ versus an alternative hypothesis that ________.

ANSWER:

H 0 :  D  0, H1 :  D  0

TYPE: FI DIFFICULTY: Moderate KEYWORDS: t test, mean difference, form of hypothesis 102. a) b) c) d)

Referring to Table 10-8, the test to perform is a pooled-variance t test for differences between two means.

F test for the difference among more than two means.

Z test for the difference between two means. paired t-test for the mean difference.

ANSWER: d TYPE: MC DIFFICULTY: Easy KEYWORDS: t test, mean difference 103. Referring to Table 10-8, the decision rule is to reject the null hypothesis if ________.

ANSWER: t > 3.3649 TYPE: FI DIFFICULTY: Easy KEYWORDS: t test, mean difference, decision

104. Referring to Table 10-8, the calculated value of the test statistic is ________.

ANSWER: 1.628 TYPE: FI DIFFICULTY: Easy KEYWORDS: t test, mean difference, test statistic 105. Referring to Table 10-8, the p-value of the test is between ________ and ________.

ANSWER: .

Two-Sample Tests and One-Way ANOVA 325

0.05 and 0.1 TYPE: FI DIFFICULTY: Moderate KEYWORDS: t test, mean difference, p-value 106. True or False: Referring to Table 10-8, the null hypothesis should be rejected.

ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: t test, mean difference, decision 107. Referring to Table 10-8, the buyer should decide that the primary supplier is a) overcharging because there is strong evidence that this is the case. b) overcharging because there is insufficient evidence to prove otherwise. c) not overcharging because there is insufficient evidence to prove otherwise. d) not overcharging because there is strong evidence to prove otherwise.

ANSWER: c TYPE: MC DIFFICULTY: Moderate KEYWORDS: t test, mean difference, conclusion 108. Referring to Table 10-8, if the buyer had decided to perform a two-tailed test, the p-value would have been between ________ and ________.

ANSWER: 0.01 and 0.2 TYPE: FI DIFFICULTY: Moderate KEYWORDS: t test, mean difference, p-value 109. Referring to Table 10-8, what is the 99% confidence interval estimate for the mean difference in prices?

ANSWER: −$4.18 to $9.85 TYPE: PR DIFFICULTY: Moderate KEYWORDS: confidence interval, mean difference

110. Referring to Table 10-8, what is the 95% confidence interval estimate for the mean difference in prices?

ANSWER: −$1.64 to $7.31 TYPE: PR DIFFICULTY: Moderate KEYWORDS: confidence interval, mean difference 111. Referring to Table 10-8, what is the 90% confidence interval estimate for the mean difference in prices?

.

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ANSWER: −$0.67 to $6.34 TYPE: PR DIFFICULTY: Moderate KEYWORDS: confidence interval, mean difference 112. If we wish to determine whether there is evidence that the proportion of items of interest is higher in group 1 than in group 2, the appropriate test to use is a) the Z test for the difference between two proportions. b) the F test for the difference between two variances. c) the pooled-variance t test for the difference between two proportions. d) the F test for the difference between two proportions.

ANSWER: a TYPE: MC DIFFICULTY: Easy KEYWORDS: Z test for the difference between two proportions 113. Moving companies are required by the government to publish a Carrier Performance Report each year. One of the descriptive statistics they must include is the annual percentage of shipments on which a $50 or greater claim for loss or damage was filed. Suppose two companies, Econo-Move and On-the-Move, each decide to estimate this figure by sampling their records, and they report the data shown in the following table.

Econo-Move

On-the-Move

Total shipments sampled

900

750

Number of shipments with a claim  $50

162

60

The owner of On-the-Move is hoping to use these data to show that the company is superior to Econo-Move with regard to the percentage of claims filed. Which test would be used to properly analyze the data in this experiment? a) Z test for the difference between two means b) F test for the difference between two variances c) F test for the difference among more than two means d) Z test for the difference between two proportions ANSWER: d TYPE: MC DIFFICULTY: Moderate KEYWORDS: Z test, difference between two proportions 114. The Wall Street Journal recently ran an article indicating differences in perception of sexual harassment on the job between men and women. The article claimed that women perceived the problem to be much more prevalent than did men. One question asked of both men and women was: “Do you think sexual harassment is a major problem in the American workplace?” Some 24% of the men compared to 62% of the women responded “Yes.” Assuming W designates women’s responses and M designates men’s, what hypothesis should The Wall Street Journal test in order to show that its claim is true? a) H0:  W −  M  0 versus H1:  W −  M < 0 b) H0:  W −  M  0 versus H1:  W −  M > 0 c) H0:  W −  M = 0 versus H1:  W −  M  0 .

Two-Sample Tests and One-Way ANOVA 327 d) H0:  W −  M  0 versus H1:  W −  M = 0 ANSWER: b TYPE: MC DIFFICULTY: Easy KEYWORDS: Z test, difference between two proportions, form of hypothesis 115. The Wall Street Journal recently ran an article indicating differences in perception of sexual harassment on the job between men and women. The article claimed that women perceived the problem to be much more prevalent than did men. One question asked to both men and women was: “Do you think sexual harassment is a major problem in the American workplace?” Some 24% of the men compared to 62% of the women responded “Yes.” Suppose that 150 women and 200 men were interviewed. For a 0.01 level of significance, what is the critical value for the rejection region? a) 7.173 b) 7.106 c) 6.635 d) 2.33 ANSWER: d TYPE: MC DIFFICULTY: Moderate KEYWORDS: Z test, difference between two proportions, critical value 116. The Wall Street Journal recently ran an article indicating differences in perception of sexual harassment on the job between men and women. The article claimed that women perceived the problem to be much more prevalent than did men. One question asked to both men and women was: “Do you think sexual harassment is a major problem in the American workplace?” Some 24% of the men compared to 62% of the women responded “Yes.” Suppose that 150 women and 200 men were interviewed. What is the value of the test statistic? a) 7.173 b) 7.106 c) 6.635 d) 2.33 ANSWER: a TYPE: MC DIFFICULTY: Moderate KEYWORDS: Z test, difference between two proportions, test statistic 117. The Wall Street Journal recently ran an article indicating differences in perception of sexual harassment on the job between men and women. The article claimed that women perceived the problem to be much more prevalent than did men. One question asked to both men and women was: “Do you think sexual harassment is a major problem in the American workplace?” Some 24% of the men compared to 62% of the women responded “Yes.” Suppose that 150 women and 200 men were interviewed. Construct a 99% confidence interval estimate of the difference between the proportion of women and men who think sexual harassment is a major problem in the American workplace.

ANSWER: 0.25 to 0.51 TYPE: PR DIFFICULTY: Moderate .

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KEYWORDS: confidence interval, difference between two proportions 118. The Wall Street Journal recently ran an article indicating differences in perception of sexual harassment on the job between men and women. The article claimed that women perceived the problem to be much more prevalent than did men. One question asked to both men and women was: “Do you think sexual harassment is a major problem in the American workplace?” Some 24% of the men compared to 62% of the women responded “Yes.” Suppose that 150 women and 200 men were interviewed. Construct a 95% confidence interval estimate of the difference between the proportion of women and men who think sexual harassment is a major problem in the American workplace.

ANSWER: 0.28 to 0.48 TYPE: PR DIFFICULTY: Moderate KEYWORDS: confidence interval, difference between two proportions 119. The Wall Street Journal recently ran an article indicating differences in perception of sexual harassment on the job between men and women. The article claimed that women perceived the problem to be much more prevalent than did men. One question asked to both men and women was: “Do you think sexual harassment is a major problem in the American workplace?” Some 24% of the men compared to 62% of the women responded “Yes.” Suppose that 150 women and 200 men were interviewed. Construct a 90% confidence interval estimate of the difference between the proportion of women and men who think sexual harassment is a major problem in the American workplace.

ANSWER: 0.30 to 0.46 TYPE: PR DIFFICULTY: Moderate KEYWORDS: confidence interval, difference between two proportions

120. The Wall Street Journal recently ran an article indicating differences in perception of sexual harassment on the job between men and women. The article claimed that women perceived the problem to be much more prevalent than did men. One question asked to both men and women was: “Do you think sexual harassment is a major problem in the American workplace?” Some 24% of the men compared to 62% of the women responded “Yes.” Suppose that 150 women and 200 men were interviewed. What conclusion should be reached? a) Using a 0.01 level of significance, there is sufficient evidence to conclude that women perceive the problem of sexual harassment on the job as much more prevalent than do men. b) There is insufficient evidence to conclude with at least 99% confidence that women perceive the problem of sexual harassment on the job as much more prevalent than do men.

.

Two-Sample Tests and One-Way ANOVA 329

c) There is no evidence of a significant difference between the men and women in their perception. d) More information is needed to draw any conclusions from the data set.

ANSWER: a TYPE: MC DIFFICULTY: Moderate KEYWORDS: Z test, difference between two proportions, decision, conclusion 121. A powerful women’s group has claimed that men and women differ in attitudes about sexual discrimination. A group of 50 men (group 1) and 40 women (group 2) were asked if they thought sexual discrimination is a problem in the United States. Of those sampled, 11 of the men and 19 of the women did believe that sexual discrimination is a problem. Assuming W designates women’s responses and M designates men’s, which of the following are the appropriate null and alternative hypotheses to test the group’s claim? a) H0:  M −  W  0 versus H1:  M −  W  0 b) H0:  M −  W  0 versus H1:  M −  W  0 c) H0:  M −  W = 0 versus H1:  M −  W  0 d) H0:  M −  W  0  0 versus H1:  M −  W = 0 ANSWER: c TYPE: MC DIFFICULTY: Moderate KEYWORDS: Z test, difference between two proportions, form of hypothesis

122. A powerful women’s group has claimed that men and women differ in attitudes about sexual discrimination. A group of 50 men (group 1) and 40 women (group 2) were asked if they thought sexual discrimination is a problem in the United States. Of those sampled, 11 of the men and 19 of the women did believe that sexual discrimination is a problem. Find the value of the test statistic. a) Z = – 2.55 b) Z = – 0.85 c) Z = – 1.05 d) Z = – 1.20 ANSWER: a TYPE: MC DIFFICULTY: Moderate .

330

Two-Sample Tests and One-Way ANOVA

KEYWORDS: Z test, difference between two proportions, test statistic 123. A powerful women’s group has claimed that men and women differ in attitudes about sexual discrimination. A group of 50 men (group 1) and 40 women (group 2) were asked if they thought sexual discrimination is a problem in the United States. Of those sampled, 11 of the men and 19 of the women did believe that sexual discrimination is a problem. If the p-value turns out to be 0.035 (which is not the real value in this data set), then a) at  = 0.05, we should fail to reject H0 b) at  = 0.04, we should reject H0 c) at  = 0.03, we should reject H0 d) None of the above would be correct statements.

ANSWER: b TYPE: MC DIFFICULTY: Moderate KEYWORDS: Z test, difference between two proportions, decision, conclusion 124. A powerful women’s group has claimed that men and women differ in attitudes about sexual discrimination. A group of 50 men (group 1) and 40 women (group 2) were asked if they thought sexual discrimination is a problem in the United States. Of those sampled, 11 of the men and 19 of the women did believe that sexual discrimination is a problem. Construct a 99% confidence interval estimate of the difference between the proportion of men and women who believe that sexual discrimination is a problem.

ANSWER: −0.508 to −0.002 or 0.002 to 0.508 TYPE: PR DIFFICULTY: Moderate KEYWORDS: confidence interval, difference between two proportions

125. A powerful women’s group has claimed that men and women differ in attitudes about sexual discrimination. A group of 50 men (group 1) and 40 women (group 2) were asked if they thought sexual discrimination is a problem in the United States. Of those sampled, 11 of the men and 19 of the women did believe that sexual discrimination is a problem. Construct a 95% confidence interval estimate of the difference between the proportion of men and women who believe that sexual discrimination is a problem.

ANSWER: −0.448 to −0.062 or 0.062 to 0.448 TYPE: PR DIFFICULTY: Moderate KEYWORDS: confidence interval, difference between two proportions 126. A powerful women’s group has claimed that men and women differ in attitudes about sexual discrimination. A group of 50 men (group 1) and 40 women (group 2) were asked if they thought .

Two-Sample Tests and One-Way ANOVA 331

sexual discrimination is a problem in the United States. Of those sampled, 11 of the men and 19 of the women did believe that sexual discrimination is a problem. Construct a 90% confidence interval estimate of the difference between the proportion of men and women who believe that sexual discrimination is a problem.

ANSWER: −0.417 to −0.093 or 0.093 to 0.417 TYPE: PR DIFFICULTY: Moderate KEYWORDS: confidence interval, difference between two proportions

TABLE 10-9 A few years ago, Pepsi invited consumers to take the “Pepsi Challenge.” Consumers were asked to decide which of two sodas, Coke or Pepsi, they preferred in a blind taste test. Pepsi was interested in determining what factors played a role in people’s taste preferences. One of the factors studied was the gender of the consumer. Below are the results of analyses comparing the taste preferences of men and women with the proportions depicting preference for Pepsi.

Males: n = 109, pM = 0.422018 Females: n = 52, pF = 0.25 pM – pF = 0.172018 Z = 2.11825 127. Referring to Table 10-9, to determine if a difference exists in the taste preferences of men and women, give the correct alternative hypothesis that Pepsi would test. a) H1: M –  F  0 b) H1: M –  F  0 c) H1:  M −  F  0 d) H1:  M −  F = 0

ANSWER: c TYPE: MC DIFFICULTY: Easy KEYWORDS: Z test, difference between two proportions, form of hypothesis

128. Referring to Table 10-9, suppose Pepsi wanted to test to determine if the males preferred Pepsi more than the females. Using the test statistic given, compute the appropriate p-value for the test. a) 0.0171 b) 0.0340 c) 0.2119 d) 0.4681 ANSWER: a TYPE: MC DIFFICULTY: Moderate KEYWORDS: Z test, difference between two proportions, p-value 129. Referring to Table 10-9, suppose Pepsi wanted to test to determine if the males preferred Pepsi less than the females. Using the test statistic given, compute the appropriate p-value for the test. .

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Two-Sample Tests and One-Way ANOVA

a) b) c) d)

0.0170 0.0340 0.9660 0.9830

ANSWER: d TYPE: MC DIFFICULTY: Moderate KEYWORDS: Z test, difference between two proportions, p-value 130. Referring to Table 10-9, suppose that the two-tailed p-value was really 0.0734. State the proper conclusion. a) At  = 0.05, there is sufficient evidence to indicate the proportion of males preferring Pepsi differs from the proportion of females preferring Pepsi. b) At  = 0.10, there is sufficient evidence to indicate the proportion of males preferring Pepsi differs from the proportion of females preferring Pepsi. c) At  = 0.05, there is sufficient evidence to indicate the proportion of males preferring Pepsi equals the proportion of females preferring Pepsi. d) At  = 0.08, there is insufficient evidence to indicate the proportion of males preferring Pepsi differs from the proportion of females preferring Pepsi.

ANSWER: b TYPE: MC DIFFICULTY: Moderate KEYWORDS: Z test, difference between two proportions, decision, conclusion 131. Referring to Table 10-9, construct a 90% confidence interval estimate of the difference between the proportion of males and females who prefer Pepsi.

ANSWER: 0.05 to 0.30 TYPE: PR DIFFICULTY: Moderate KEYWORDS: confidence interval, difference between two proportions

132. Referring to Table 10-9, construct a 95% confidence interval estimate of the difference between the proportion of males and females who prefer Pepsi.

ANSWER: 0.02 to 0.32 TYPE: PR DIFFICULTY: Moderate KEYWORDS: confidence interval, difference between two proportions 133. Referring to Table 10-9, construct a 99% confidence interval estimate of the difference between the proportion of males and females who prefer Pepsi.

ANSWER: −0.03 to 0.37 TYPE: PR DIFFICULTY: Moderate KEYWORDS: confidence interval, difference between two proportions .

Two-Sample Tests and One-Way ANOVA 333

TABLE 10-10 The following EXCEL output contains the results of a test to determine if the proportions of satisfied guests at two resorts are the same or different.

Hypothesized Difference Level of Significance

0 0.05 Group 1

Number of Successes Sample Size Group 2 Number of Successes Sample Size Group 1 Proportion Group 2 Proportion Difference in Two Proportions Average Proportion Test Statistic Two-Tailed Test Lower Critical Value Upper Critical Value p-Value

163 227 154 262 0.718061674 0.58778626 0.130275414 0.648261759 3.00875353 −1.959961082 1.959961082 0.002623357

134. Referring to Table 10-10, allowing for 0.75% probability of committing a Type I error, what are the decision and conclusion on testing whether there is any difference in the proportions of satisfied guests in the two resorts? a) Do not reject the null hypothesis; there is enough evidence to conclude that there is significant difference in the proportions of satisfied guests at the two resorts. b) Do not reject the null hypothesis; there is not enough evidence to conclude that there is significant difference in the proportions of satisfied guests at the two resorts. c) Reject the null hypothesis; there is enough evidence to conclude that there is significant difference in the proportions of satisfied guests at the two resorts. d) Reject the null hypothesis; there is not enough evidence to conclude that there is significant difference in the proportions of satisfied guests at the two resorts.

ANSWER: c TYPE: MC DIFFICULTY: Easy KEYWORDS: Z test, difference between two proportions, decision, conclusion .

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Two-Sample Tests and One-Way ANOVA

135. Referring to Table 10-10, if you want to test the claim that "Resort 1 (Group 1) has a higher proportion of satisfied guests compared to Resort 2 (Group 2)", the p-value of the test will be a) 0.00262 b) 0.00262/2 c) 2*(0.00262) d) 1-(0.00262/2) ANSWER: b TYPE: MC DIFFICULTY: Moderate KEYWORDS: Z test, difference between two proportions, p-value 136. Referring to Table 10-10, if you want to test the claim that "Resort 1 (Group 1) has a lower proportion of satisfied guests compared to Resort 2 (Group 2)", you will use a) a t-test for the difference between two proportions. b) a Z-test for the difference between two proportions. c) an F test for the difference between two proportions. d) a  2 test for the difference between two proportions.

ANSWER: b TYPE: MC DIFFICULTY: Moderate KEYWORDS: Z test, difference between two proportions 137. Referring to Table 10-10, construct a 99% confidence interval estimate of the difference in the population proportion of satisfied guests between the two resorts.

ANSWER: 0.02 to 0.24 TYPE: PR DIFFICULTY: Moderate KEYWORDS: confidence interval, difference between two proportions

138. Referring to Table 10-10, construct a 95% confidence interval estimate of the difference in the population proportion of satisfied guests between the two resorts.

ANSWER: 0.05 to 0.21 TYPE: PR DIFFICULTY: Moderate KEYWORDS: confidence interval, difference between two proportions 139. Referring to Table 10-10, construct a 90% confidence interval estimate of the difference in the population proportion of satisfied guests between the two resorts.

ANSWER: 0.06 to 0.20 TYPE: PR DIFFICULTY: Moderate KEYWORDS: confidence interval, difference between two proportions

TABLE 10-11 .

Two-Sample Tests and One-Way ANOVA 335

A corporation randomly selects 150 salespeople and finds that 66% who have never taken a selfimprovement course would like such a course. The firm did a similar study 10 years ago in which 60% of a random sample of 160 salespeople wanted a self-improvement course. The groups are assumed to be independent random samples. Let  1 and  2 represent the true proportion of workers who would like to attend a self-improvement course in the recent study and the past study, respectively.

140. Referring to Table 10-11, if the firm wanted to test whether this proportion has changed from the previous study, which represents the relevant hypotheses? a) H0:  1 −  2 = 0 versus H1:  1 −  2  0 b) H0:  1 −  2  0 versus H1:  1 −  2 = 0 c) H0:  1 −  2  0 versus H1:  1 −  2 > 0 d) H0:  1 −  2  0 versus H1:  1 −  2 < 0 ANSWER: a TYPE: MC DIFFICULTY: Easy KEYWORDS: Z test, difference between two proportions, form of hypothesis

141. Referring to Table 10-11, if the firm wanted to test whether a greater proportion of workers would currently like to attend a self-improvement course than in the past, which represents the relevant hypotheses? a) H0:  1 −  2 = 0 versus H1:  1 −  2  0 b) H0:  1 −  2  0 versus H1:  1 −  2 = 0 c) H0:  1 −  2  0 versus H1:  1 −  2 > 0 d) H0:  1 −  2  0 versus H1:  1 −  2 < 0 ANSWER: c TYPE: MC DIFFICULTY: Easy KEYWORDS: Z test, difference between two proportions, form of hypothesis 142. Referring to Table 10-11, what is the unbiased point estimate for the difference between the two population proportions? a) 0.06 b) 0.10 c) 0.15 .

336

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d) 0.22 ANSWER: a TYPE: MC DIFFICULTY: Easy KEYWORDS: point estimate 143. Referring to Table 10-11, what is/are the critical value(s) when performing a Z test on whether population proportions are different if  = 0.05? a)  1.645 b)  1.96 c) −1.96 d)  2.08 ANSWER: b TYPE: MC DIFFICULTY: Easy KEYWORDS: Z test, difference between two proportions, critical value 144. Referring to Table 10-11, what is/are the critical value(s) when testing whether population proportions are different if  = 0.10? a)  1.645 b)  1.96 c) -1.96 d)  2.08 ANSWER: a TYPE: MC DIFFICULTY: Easy KEYWORDS: Z test, difference between two proportions, critical value

145. Referring to Table 10-11, what is/are the critical value(s) when testing whether the current population proportion is higher than before if  = 0.05? a)  1.645 b) + 1.645 c)  1.96 d) + 1.96 ANSWER: b TYPE: MC DIFFICULTY: Easy KEYWORDS: Z test for the difference between two proportions, critical value 146. Referring to Table 10-11, what is the estimated standard error of the difference between the two sample proportions? a) 0.629 b) 0.500 c) 0.055 d) 0 ANSWER: .

Two-Sample Tests and One-Way ANOVA 337

c TYPE: MC DIFFICULTY: Moderate KEYWORDS: Z test, difference between two proportions, standard error 147. Referring to Table 10-11, what is the value of the test statistic to use in evaluating the alternative hypothesis that there is a difference in the two population proportions? a) 4.335 b) 1.96 c) 1.093 d) 0 ANSWER: c TYPE: MC DIFFICULTY: Moderate KEYWORDS: Z test, difference between two proportions, test statistic 148. Referring to Table 10-11, the company tests to determine at the 0.05 level whether the population proportion has changed from the previous study. Which of the following is most correct? a) Reject the null hypothesis and conclude that the proportion of employees who are interested in a self-improvement course has changed over the intervening 10 years. b) Do not reject the null hypothesis and conclude that the proportion of employees who are interested in a self-improvement course has not changed over the intervening 10 years. c) Reject the null hypothesis and conclude that the proportion of employees who are interested in a self-improvement course has increased over the intervening 10 years. d) Do not reject the null hypothesis and conclude that the proportion of employees who are interested in a self-improvement course has increased over the intervening 10 years.

ANSWER: b TYPE: MC DIFFICULTY: Moderate KEYWORDS: Z test, difference between two proportions, decision, conclusion 149. Referring to Table 10-11, construct a 99% confidence interval estimate of the difference in proportion of workers who would like to attend a self-improvement course in the recent study and the past study.

ANSWER: −0.08 to 0.20 TYPE: PR DIFFICULTY: Moderate KEYWORDS: confidence interval, difference between two proportions 150. Referring to Table 10-11, construct a 95% confidence interval estimate of the difference in proportion of workers who would like to attend a self-improvement course in the recent study and the past study.

ANSWER: −0.05 to 0.17 TYPE: PR DIFFICULTY: Moderate KEYWORDS: confidence interval, difference between two proportions

.

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151. Referring to Table 10-11, construct a 90% confidence interval estimate of the difference in proportion of workers who would like to attend a self-improvement course in the recent study and the past study.

ANSWER: −0.03 to 0.15 TYPE: PR DIFFICULTY: Moderate KEYWORDS: confidence interval, difference between two proportions 152. True or False: In testing the difference between two proportions using the normal distribution, we may use a two-tailed Z test.

ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: Z test, difference between two proportions 153. If we wish to determine whether there is evidence that the proportion of items of interest is higher in Group 1 than in Group 2, and the test statistic for Z = +2.07 where the difference is defined as Group 1’s proportion minus Group 2’s proportion, the p-value is equal to ______.

ANSWER: 0.0192 TYPE: FI DIFFICULTY: Moderate KEYWORDS: Z test, difference between two proportions, p-value

154. If we wish to determine whether there is evidence that the proportion of items of interest is higher in Group 1 than in Group 2, and the test statistic for Z = −2.07 where the difference is defined as Group 1’s proportion minus Group 2’s proportion, the p-value is equal to ______.

ANSWER: 0.9808 TYPE: FI DIFFICULTY: Moderate KEYWORDS: Z test, difference between two proportions, p-value

TABLE 10-12 The dean of a college is interested in the proportion of graduates from his college who have a job offer on graduation day. He is particularly interested in seeing if there is a difference in this proportion for accounting and economics majors. In a random sample of 100 of each type of major at graduation, he found that 65 accounting majors and 52 economics majors had job offers. If the accounting majors are designated as “Group 1” and the economics majors are designated as “Group 2,” perform the appropriate hypothesis test using a level of significance of 0.05.

155. Referring to Table 10-12, the hypotheses the dean should use are: a) H0:  1 −  2 = 0 versus H1:  1 −  2  0 .

Two-Sample Tests and One-Way ANOVA 339 b) H0:  1 −  2  0 versus H1:  1 −  2 = 0 c) H0:  1 −  2  0 versus H1:  1 −  2 > 0 d) H0:  1 −  2  0 versus H1:  1 −  2 < 0 ANSWER: a TYPE: MC DIFFICULTY: Easy KEYWORDS: Z test, difference between two proportions, form of hypothesis 156. Referring to Table 10-12, the null hypothesis will be rejected if the test statistic is ________.

ANSWER: Z > 1.96 or < -1.96 TYPE: FI DIFFICULTY: Easy KEYWORDS: Z test, difference between two proportions, critical value 157. Referring to Table 10-12, the value of the test statistic is ________.

ANSWER: Z = 1.866 TYPE: FI DIFFICULTY: Moderate KEYWORDS: Z test, difference between two proportions, test statistic 158. Referring to Table 10-12, the p-value of the test is ________.

ANSWER: 0.0621 TYPE: FI DIFFICULTY: Moderate KEYWORDS: Z test, difference between two proportions, p-value 159. True or False: Referring to Table 10-12, the null hypothesis should be rejected.

ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: Z test, difference between two proportions, decision 160. True or False: Referring to Table 10-12, the same decision would be made with this test if the level of significance had been 0.01 rather than 0.05.

ANSWER: True TYPE: TF DIFFICULTY: Moderate KEYWORDS: Z test, difference between two proportions, decision 161. True or False: Referring to Table 10-12, the same decision would be made with this test if the level of significance had been 0.10 rather than 0.05.

ANSWER: False TYPE: TF DIFFICULTY: Moderate KEYWORDS: Z test, difference between two proportions, decision .

340

Two-Sample Tests and One-Way ANOVA

162. Referring to Table 10-12, construct a 99% confidence interval estimate of the difference in proportion between accounting majors and economic majors who have a job offer on graduation day.

ANSWER: −0.05 to 0.31 TYPE: PR DIFFICULTY: Moderate KEYWORDS: confidence interval, difference between two proportions 163. Referring to Table 10-12, construct a 95% confidence interval estimate of the difference in proportion between accounting majors and economic majors who have a job offer on graduation day.

ANSWER: −0.01 to 0.27 TYPE: PR DIFFICULTY: Moderate KEYWORDS: confidence interval, difference between two proportions 164. Referring to Table 10-12, construct a 90% confidence interval estimate of the difference in proportion between accounting majors and economic majors who have a job offer on graduation day.

ANSWER: 0.02 to 0.24 TYPE: PR DIFFICULTY: Moderate KEYWORDS: confidence interval, difference between two proportions

TABLE 10-13 A quality control engineer is in charge of the manufacture of computer disks. Two different processes can be used to manufacture the disks. He suspects that the Kohler method produces a greater proportion of defects than the Russell method. He samples 150 of the Kohler and 200 of the Russell disks and finds that 27 and 18 of them, respectively, are defective. If Kohler is designated as “Group 1” and Russell is designated as “Group 2,” perform the appropriate test at a level of significance of 0.01.

165. Referring to Table 10-13, the hypotheses that should be tested are: a) H0:  1 −  2 = 0 versus H1:  1 −  2  0 b) H0:  1 −  2  0 versus H1:  1 −  2 = 0 c) H0:  1 −  2  0 versus H1:  1 −  2 > 0 d) H0:  1 −  2  0 versus H1:  1 −  2 < 0 ANSWER: c TYPE: MC DIFFICULTY: Easy KEYWORDS: Z test, difference between two proportions, form of hypothesis 166. Referring to Table 10-13, the null hypothesis will be rejected if the test statistic is ________. .

Two-Sample Tests and One-Way ANOVA 341

ANSWER: Z > 2.33 TYPE: FI DIFFICULTY: Easy KEYWORDS: Z test, difference between two proportions, critical value 167. Referring to Table 10-13, the value of the test statistic is ________.

ANSWER: 2.49 TYPE: FI DIFFICULTY: Moderate KEYWORDS: Z test, difference between two proportions, test statistic 168. Referring to Table 10-13, the p-value of the test is ________.

ANSWER: 0.0064 TYPE: FI DIFFICULTY: Moderate KEYWORDS: Z test, difference between two proportions, p-value 169. True or False: Referring to Table 10-13, the null hypothesis should be rejected.

ANSWER: True TYPE: TF DIFFICULTY: Moderate KEYWORDS: Z test, difference between two proportions, decision

170. True or False: Referring to Table 10-13, the same decision would be made with this test if the level of significance had been 0.05 rather than 0.01.

ANSWER: True TYPE: TF DIFFICULTY: Moderate KEYWORDS: Z test, difference between two proportions, decision 171. True or False: Referring to Table 10-13, the same decision would be made if this had been a two-tailed test at a level of significance of 0.01.

ANSWER: False TYPE: TF DIFFICULTY: Moderate KEYWORDS: Z test, difference between two proportions, decision 172. Referring to Table 10-13, construct a 90% confidence interval estimate of the difference in proportion between the Kohler and Russell disks that are defective.

ANSWER: 0.03 to 0.15 TYPE: PR DIFFICULTY: Moderate KEYWORDS: confidence interval, difference between two proportions

.

342

Two-Sample Tests and One-Way ANOVA

173. Referring to Table 10-13, construct a 95% confidence interval estimate of the difference in proportion between the Kohler and Russell disks that are defective.

ANSWER: 0.02 to 0.16 TYPE: PR DIFFICULTY: Moderate KEYWORDS: confidence interval, difference between two proportions 174. Referring to Table 10-13, construct a 99% confidence interval estimate of the difference in proportion between the Kohler and Russell disks that are defective.

ANSWER: -0.01 to 0.19 TYPE: PR DIFFICULTY: Moderate KEYWORDS: confidence interval, difference between two proportions

TABLE 10-14 A problem with a telephone line that prevents a customer from receiving or making calls is disconcerting to both the customer and the telephone company. The data on samples of 20 problems reported to two different offices of a telephone company and the time to clear these problems (in minutes) from the customers’ lines are collected. Below is the Excel output to see whether there is evidence of a difference in the mean waiting time between the two offices assuming that the population variances in the two offices are not equal. t-Test: Two-Sample Assuming Unequal Variances Office 1 Mean 2.214 Variance 2.951657 Observations 20 Hypothesized Mean Difference 0 Df 38 t Stat 0.354386 P(T<=t) one-tail 0.362504 t Critical one-tail 1.685953 P(T<=t) two-tail 0.725009 t Critical two-tail 2.024394

Office 2 2.0115 3.57855 20

175. Referring to Table 10-14, what is the 95% confidence interval estimate for the difference in the two means if the two population variances are assumed to be the same?

.

Two-Sample Tests and One-Way ANOVA 343

ANSWER:

−0.9543  1 − 2  1.3593

TYPE: PR DIFFICULTY: Moderate KEYWORDS: confidence interval, difference between two means 176. Referring to Table 10-14, state the null and alternative hypotheses for testing if there is evidence of a difference in the variability of the waiting time between the two offices. a) H 0 :  I2 –  II2  0 versus H 1 :  I2 –  II2  0 b)

H 0 :  I2 –  II2  0 versus H 1 :  I2 –  II2  0

c)

H 0 :  I2 –  II2 = 0 versus H1 :  I2 –  II2  0

d) H 0 :  I2 –  II2  0 versus H 1 :  I2 –  II2 = 0 ANSWER: c TYPE: MC DIFFICULTY: Moderate KEYWORDS: F test, difference between two variances, form of hypothesis 177. Referring to Table 10-14, what is the value of the test statistic for testing if there is evidence of a difference in the variability of the waiting time between the two offices?

ANSWER: 1.2124 TYPE: PR DIFFICULTY: Easy KEYWORDS: F test, difference between two variances, test statistic 178. Referring to Table 10-14, what assumptions are necessary for testing if there is evidence of a difference in the variability of the waiting time between the two offices to be valid? a) Both sampled populations are normally distributed. b) Both samples are random and independent. c) Neither (a) nor (b) is necessary. d) Both (a) and (b) are necessary.

ANSWER: d TYPE: MC DIFFICULTY: Easy KEYWORDS: F test, difference between two variances, assumption 179. Referring to Table 10-14, what is the upper critical value for testing if there is evidence of a difference in the variability of the waiting time between the two offices at the 5% level of significance?

ANSWER: 2.5264 (using Excel) or 2.51 (using Table E.5 with 20 and 19 degrees of freedom) TYPE: PR DIFFICULTY: Easy KEYWORDS: F test, difference between two variances, critical value 180. Referring to Table 10-14, what is the largest level of significance at which a test on a difference in the variability of the waiting time between the two offices will not be rejected?

ANSWER: 0.6789 (using Excel) or ‘greater than 0.10’ (using Table E.5 with 20 and 19 degrees of freedom) .

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Two-Sample Tests and One-Way ANOVA

TYPE: PR DIFFICULTY: Difficult KEYWORDS: F test, difference between two variances, p-value 181. Referring to Table 10-14, suppose  = 0.05. Which of the following represents the result of the relevant hypothesis test? a) The alternative hypothesis is rejected. b) The null hypothesis is rejected. c) The null hypothesis is not rejected. d) Insufficient information exists on which to make a decision.

ANSWER: c TYPE: MC DIFFICULTY: Easy KEYWORDS: F test, difference between two variances, decision

182. Referring to Table 10-14, suppose  = 0.05. Which of the following represents the correct conclusion? a) There is no evidence of a difference in the variability of the waiting time between the two offices. b) There is evidence of a difference in the variability of the waiting time between the two offices. c) There is no evidence that the variability of the waiting time between the two offices are the same. d) There is evidence that the variability of the waiting time between the two offices are the same.

ANSWER: a TYPE: MC DIFFICULTY: Easy KEYWORDS: F test, difference between two variances, conclusion 183. In a one-way ANOVA, if the computed F statistic exceeds the critical F value we may a) reject H0 since there is evidence all the means differ. b) reject H0 since there is evidence of a treatment effect. c) not reject H0 since there is no evidence of a difference. d) not reject H0 because a mistake has been made.

ANSWER: b TYPE: MC DIFFICULTY: Easy KEYWORDS: one-way analysis of variance, decision

.

Two-Sample Tests and One-Way ANOVA 345

184. Which of the following components in an ANOVA table are not additive? a) Sum of squares. b) Degrees of freedom. c) Mean squares. d) It is not possible to tell.

ANSWER: c TYPE: MC DIFFICULTY: Easy KEYWORDS: one-way analysis of variance, sum of squares, properties 185. Why would you use the Tukey-Kramer procedure? a) To test for normality. b) To test for homogeneity of variance. c) To test independence of errors. d) To test for differences in pairwise means.

ANSWER: d TYPE: MC DIFFICULTY: Easy KEYWORDS: Tukey-Kramer procedure

186. The F test statistic in a one-way ANOVA is a) MSW/MSA. b) SSW/SSA. c) MSA/MSW. d) SSA/SSW. ANSWER: c TYPE: MC DIFFICULTY: Easy KEYWORDS: one-way analysis of variance, F test for factor 187. The degrees of freedom for the F test in a one-way ANOVA are a) (n – c) and (c – 1). b) (c – 1) and (n – c). c) (c – n) and (n – 1). d) (n – 1) and (c – n).

ANSWER: b TYPE: MC DIFFICULTY: Easy KEYWORDS: one-way analysis of variance, F test for factor, degrees of freedom 188. In a one-way ANOVA, the null hypothesis is always a) there is no treatment effect. b) there is some treatment effect. c) all the population means are different. d) some of the population means are different. .

346

Two-Sample Tests and One-Way ANOVA

ANSWER: a TYPE: MC DIFFICULTY: Easy KEYWORDS: one-way analysis of variance, form of hypothesis 189. In a one-way ANOVA a) an interaction term is present. b) an interaction effect can be tested. c) there is no interaction term. d) the interaction term has (c – 1)(n – 1) degrees of freedom.

ANSWER: c TYPE: MC DIFFICULTY: Moderate KEYWORDS: one-way analysis of variance, properties, interaction

190. An airline wants to select a computer software package for its reservation system. Four software packages (1, 2, 3, and 4) are commercially available. The airline will choose the package that bumps as few passengers, on the average, as possible during a month. An experiment is set up in which each package is used to make reservations for 5 randomly selected weeks. (A total of 20 weeks was included in the experiment.) The number of passengers bumped each week is given below. How should the data be analyzed?

Package 1: 12, 14, 9, 11, 16 Package 2: 2, 4, 7, 3, 1 Package 3: 10, 9, 6, 10, 12 Package 4: 7, 6, 6, 15, 12 a) b) c) d)

F test for differences in variances.

One-way ANOVA F test. t test for the differences in means. t test for the mean difference.

ANSWER: b TYPE: MC DIFFICULTY: Easy KEYWORDS: one-way analysis of variance, F test for factor

TABLE 10-15 An airline wants to select a computer software package for its reservation system. Four software packages (1, 2, 3, and 4) are commercially available. The airline will choose the package that bumps as few passengers, on the average, as possible during a month. An experiment is set up in which each package is used to make reservations for 5 randomly selected weeks. (A total of 20 weeks was .

Two-Sample Tests and One-Way ANOVA 347

included in the experiment.) The number of passengers bumped each week is obtained, which gives rise to the following Excel output: ANOVA Source of Variation Between Groups Within Groups

SS 212.4 136.4

Total

348.8

df

MS 3

F 8.304985

P-value 0.001474

F crit 3.238867

8.525

191. Referring to Table 10-15, the within groups degrees of freedom is a) 3 b) 4 c) 16 d) 19 ANSWER: c TYPE: MC DIFFICULTY: Easy KEYWORDS: one-way analysis of variance, degrees of freedom

192. Referring to Table 10-15, the total degrees of freedom is a) 3 b) 4 c) 16 d) 19 ANSWER: d TYPE: MC DIFFICULTY: Easy KEYWORDS: one-way analysis of variance, degrees of freedom 193. Referring to Table 10-15, the among-group (between-group) mean squares is a) 8.525 b) 70.8 c) 212.4 d) 637.2 ANSWER: b TYPE: MC DIFFICULTY: Easy KEYWORDS: one-way analysis of variance, mean squares 194. Referring to Table 10-15, at a significance level of 1%, a) there is insufficient evidence to conclude that the average numbers of customers bumped by the 4 packages are not all the same. b) there is insufficient evidence to conclude that the average numbers of customers bumped by the 4 packages are all the same.

.

348

Two-Sample Tests and One-Way ANOVA

c) there is sufficient evidence to conclude that the average numbers of customers bumped by the 4 packages are not all the same. d) there is sufficient evidence to conclude that the average numbers of customers bumped by the 4 packages are all the same.

ANSWER: c TYPE: MC DIFFICULTY: Easy KEYWORDS: one-way analysis of variance, F test for factor, decision, conclusion

TABLE 10-16 A realtor wants to compare the average sales-to-appraisal ratios of residential properties sold in four neighborhoods (A, B, C, and D). Four properties are randomly selected from each neighborhood and the ratios recorded for each, as shown below.

A: 1.2, 1.1, 0.9, 0.4 C: 1.0, 1.5, 1.1, 1.3 B: 2.5, 2.1, 1.9, 1.6 D: 0.8, 1.3, 1.1, 0.7 Interpret the results of the analysis summarized in the following table: Source Neighborhoods Error Total

df

SS 3.1819

MS 1.0606

F 10.76

PR > F 0.001

12 4.3644

195. Referring to Table 10-16, the among group degrees of freedom is a) 3 b) 4 c) 12 d) 16 ANSWER: a TYPE: MC DIFFICULTY: Easy KEYWORDS: one-way analysis of variance, degrees of freedom 196. Referring to Table 10-16, the within group sum of squares is .

Two-Sample Tests and One-Way ANOVA 349

a) b) c) d)

1.0606 1.1825 3.1819 4.3644

ANSWER: b TYPE: MC DIFFICULTY: Easy KEYWORDS: one-way analysis of variance, sum of squares 197. Referring to Table 10-16, the within group mean squares is a) 0.10 b) 0.29 c) 1.06 d) 1.18 ANSWER: a TYPE: MC DIFFICULTY: Easy KEYWORDS: one-way analysis of variance, mean squares

198. Referring to Table 10-16, a) at the 0.05 level of significance, the mean ratios for the 4 neighborhoods are not all the same. b) at the 0.01 level of significance, the mean ratios for the 4 neighborhoods are all the same. c) at the 0.10 level of significance, the mean ratios for the 4 neighborhoods are not significantly different. d) at the 0.05 level of significance, the mean ratios for the 4 neighborhoods are not significantly different from 0.

ANSWER: a TYPE: MC DIFFICULTY: Easy KEYWORDS: one-way analysis of variance, F test for factor, decision, conclusion 199. Referring to Table 10-16, the null hypothesis for Levene’s test for homogeneity of variances is a) H 0 :  A = B = C = D b)

H0 : M A = M B = M C = M D

H 0 :  A2 =  B2 =  C2 =  D2 d) H 0 :  A =  B =  C =  D c)

ANSWER: c TYPE: MC DIFFICULTY: Easy KEYWORDS: Levene’s test, form of hypothesis

.

350

Two-Sample Tests and One-Way ANOVA 200. Referring to Table 10-16, the value of the test statistic for Levene’s test for homogeneity of variances is a) 0.25 b) 0.37 c) 4.36 d) 10.76 ANSWER: a TYPE: MC DIFFICULTY: Moderate KEYWORDS: Levene’s test, test statistic 201. Referring to Table 10-16, the numerator and denominator degrees of freedom for Levene’s test for homogeneity of variances at a 5% level of significance are, respectively, a) 3, 12 b) 12, 3 c) 3, 15 d) 15, 3 ANSWER: a TYPE: MC DIFFICULTY: Moderate KEYWORDS: Levene’s test, degrees of freedom 202. Referring to Table 10-16, the critical value of Levene’s test for homogeneity of variances at a 5% level of significance is a) 0.64 b) 2.48 c) 3.29 d) 3.49 ANSWER: d TYPE: MC DIFFICULTY: Moderate KEYWORDS: Levene’s test, critical value 203. Referring to Table 10-16, the p-value of the test statistic for Levene’s test for homogeneity of variances is a) 0.25 b) 0.64 c) 0.86 d) 3.49 ANSWER: c TYPE: MC DIFFICULTY: Moderate KEYWORDS: Levene’s test, p-value

.

Two-Sample Tests and One-Way ANOVA 351 204. Referring to Table 10-16, what should be the decision for the Levene’s test for homogeneity of variances at a 5% level of significance? a) Reject the null hypothesis because the p-value is smaller than the level of significance. b) Reject the null hypothesis because the p-value is larger than the level of significance. c) Do not eject the null hypothesis because the p-value is smaller than the level of significance. d) Do not reject the null hypothesis because the p-value is larger than the level of significance.

ANSWER: d TYPE: MC DIFFICULTY: Easy KEYWORDS: Levene’s test, decision 205. Referring to Table 10-16, what should be the conclusion for the Levene’s test for homogeneity of variances at a 5% level of significance? a) There is not sufficient evidence that the variances are all the same. b) There is sufficient evidence that the variances are all the same. c) There is not sufficient evidence that the variances are not all the same. d) There is sufficient evidence that the variances are not all the same.

ANSWER: c TYPE: MC DIFFICULTY: Easy KEYWORDS: Levene’s test, conclusion 206. True or False: The analysis of variance (ANOVA) tests hypotheses about the population variance.

ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: one-way analysis of variance 207. True or False: When the F test is used for ANOVA, the rejection region is always in the right tail.

ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: F test for factor, rejection region 208. True or False: The MSA must always be positive.

ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: mean squares, properties 209. True or False: In a one-factor ANOVA analysis, the among sum of squares and within sum of squares must add up to the total sum of squares.

.

352

Two-Sample Tests and One-Way ANOVA

ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: one-way analysis of variance, sum of squares, properties TABLE 10-17 As part of an evaluation program, a sporting goods retailer wanted to compare the downhill coasting speeds of 4 brands of bicycles. She took 3 of each brand and determined their maximum downhill speeds. The results are presented in miles per hour in the table below.

Trial Barth Tornado Reiser Shaw 1 43 37 41 43 2 46 38 45 45 3 43 39 42 46 210. Referring to Table 10-17, the sporting goods retailer decided to perform an ANOVA F test.

The amount of total variation or SST is __________.

ANSWER: 102.67 TYPE: FI DIFFICULTY: Moderate KEYWORDS: one-way analysis of variance, sum of squares, interpretation

211. Referring to Table 10-17, the among group variation or SSA is __________.

ANSWER: 81.33 TYPE: FI DIFFICULTY: Moderate KEYWORDS: one-way analysis of variance, sum of squares 212. Referring to Table 10-17, the within group variation or SSW is __________.

ANSWER: 21.33 TYPE: FI DIFFICULTY: Moderate KEYWORDS: one-way analysis of variance, sum of squares 213. Referring to Table 10-17, the value of MSA is __________, while MSW is __________.

ANSWER: 27.11; 2.67 TYPE: FI DIFFICULTY: Moderate KEYWORDS: one-way analysis of variance, mean squares 214. Referring to Table 10-17, the null hypothesis that the average downhill coasting speeds of the 4 brands of bicycles are equal will be rejected at a level of significance of 0.05 if the value of the test statistic is greater than __________.

ANSWER: 4.07 .

Two-Sample Tests and One-Way ANOVA 353

TYPE: FI DIFFICULTY: Easy KEYWORDS: one-way analysis of variance, F test for factor, decision 215. Referring to Table 10-17, in testing the null hypothesis that the average downhill coasting speeds of the 4 brands of bicycles are equal, the value of the test statistic is __________.

ANSWER: 10.17 TYPE: FI DIFFICULTY: Moderate KEYWORDS: one-way analysis of variance, F test for factor, test statistic 216. Referring to Table 10-17, construct the ANOVA table from the sample data.

ANSWER: Analysis of Variance Source df SS MS F Bicycle Brands 3 81.33 27.11 10.17 Error 8 21.33 2.67 Total 11 102.67 * or p < 0.005, tabular value TYPE: PR DIFFICULTY: Moderate KEYWORDS: one-way analysis of variance, properties

p 0.004*

217. True or False: Referring to Table 10-17, the null hypothesis should be rejected at a 5% level of significance.

ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: one-way analysis of variance, decision 218. True or False: Referring to Table 10-17, the decision made implies that all 4 means are significantly different.

ANSWER: False TYPE: TF DIFFICULTY: Moderate KEYWORDS: one-way analysis of variance, conclusion, interpretation 219. True or False: Referring to Table 10-17, the test is valid only if the population of speeds has the same variance for the 4 brands.

ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: one-way analysis of variance, assumption 220. True or False: Referring to Table 10-17, the test is less sensitive to the assumption that the population of speeds has the same variance for the 4 brands if the sample sizes of the 4 brands are equal.

ANSWER: .

354

Two-Sample Tests and One-Way ANOVA

True TYPE: TF DIFFICULTY: Easy KEYWORDS: one-way analysis of variance, assumption 221. True or False: Referring to Table 10-17, the test is valid only if the population of speeds is normally distributed.

ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: one-way analysis of variance, assumption 222. True or False: Referring to Table 10-17, the test is robust to the violation of the assumption that the population of speeds is normally distributed.

ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: one-way analysis of variance, assumption

223. Referring to Table 10-17, the sporting goods retailer decided to compare the 4 treatment means by using the Tukey-Kramer procedure with an overall level of significance of 0.05. There are ________ pairwise comparisons that can be made.

ANSWER: 6 TYPE: FI DIFFICULTY: Moderate KEYWORDS: one-way analysis of variance, Tukey-Kramer procedure

224. Referring to Table 10-17, using an overall level of significance of 0.05, the critical value of the Studentized range Q used in calculating the critical range for the Tukey-Kramer procedure is ________.

ANSWER: 4.53 TYPE: FI DIFFICULTY: Moderate KEYWORDS: one-way analysis of variance, Tukey-Kramer procedure, critical value 225. Referring to Table 10-17, using an overall level of significance of 0.05, the critical range for the Tukey-Kramer procedure is ________.

ANSWER: 4.27 TYPE: FI DIFFICULTY: Easy KEYWORDS: one-way analysis of variance,, Tukey-Kramer procedure, critical value 226. True or False: Referring to Table 10-17, based on the Tukey-Kramer procedure with an overall level of significance of 0.05, the retailer would decide that there is a significant difference between all pairs of mean speeds.

.

Two-Sample Tests and One-Way ANOVA 355

ANSWER: False TYPE: TF DIFFICULTY: Moderate KEYWORDS: one-way analysis of variance, Tukey-Kramer procedure, decision, conclusion 227. True or False: Referring to Table 10-17, based on the Tukey-Kramer procedure with an overall level of significance of 0.05, the retailer would decide that there is no significant difference between any pair of mean speeds.

ANSWER: False TYPE: TF DIFFICULTY: Moderate KEYWORDS: one-way analysis of variance, Tukey-Kramer procedure, decision, conclusion

228. True or False: Referring to Table 10-17, based on the Tukey-Kramer procedure with an overall level of significance of 0.05, the retailer would decide that the mean speed for the Tornado brand is significantly different from each of the mean speeds for other brands.

ANSWER: True TYPE: TF DIFFICULTY: Moderate KEYWORDS: one-way analysis of variance, Tukey-Kramer procedure, decision, conclusion 229. True or False: Referring to Table 10-17, based on the Tukey-Kramer procedure with an overall level of significance of 0.05, the retailer would decide that the 3 means other than the mean for Tornado are not significantly different from each other.

ANSWER: True TYPE: TF DIFFICULTY: Moderate KEYWORDS: one-way analysis of variance, Tukey-Kramer procedure, decision, conclusion 230. Referring to Table 10-17, the null hypothesis for Levene’s test for homogeneity of variances is a) H 0 :  A = B = C = D b)

H0 : M A = M B = M C = M D

H 0 :  A2 =  B2 =  C2 =  D2 d) H 0 :  A =  B =  C =  D c)

ANSWER: c TYPE: MC DIFFICULTY: Easy KEYWORDS: Levene’s test, form of hypothesis

.

356

Two-Sample Tests and One-Way ANOVA 231. Referring to Table 10-17, what is the value of the test statistic for Levene’s test for homogeneity of variances?

ANSWER: 0.1333 TYPE: PR DIFFICULTY: Moderate KEYWORDS: Levene’s test, test statistic 232. Referring to Table 10-17, what are the numerator and denominator degrees of freedom for Levene’s test for homogeneity of variances respectively?

ANSWER: 3, 8 TYPE: PR DIFFICULTY: Moderate KEYWORDS: Levene’s test, degrees of freedom

233. Referring to Table 10-17, what is the critical value of Levene’s test for homogeneity of variances at a 5% level of significance?

ANSWER: 4.07 TYPE: PR DIFFICULTY: Moderate KEYWORDS: Levene’s test, critical value 234. Referring to Table 10-17, what is the p-value of the test statistic for Levene’s test for homogeneity of variances?

ANSWER: 0.94 TYPE: PR DIFFICULTY: Moderate KEYWORDS: Levene’s test, p-value 235. Referring to Table 10-17, what should be the decision for the Levene’s test for homogeneity of variances at a 5% level of significance? a) Reject the null hypothesis because the p-value is smaller than the level of significance. b) Reject the null hypothesis because the p-value is larger than the level of significance. c) Do not reject the null hypothesis because the p-value is smaller than the level of significance. d) Do not reject the null hypothesis because the p-value is larger than the level of significance.

ANSWER: d TYPE: MC DIFFICULTY: Easy KEYWORDS: Levene’s test, decision 236. Referring to Table 10-17, what should be the conclusion for the Levene’s test for homogeneity of variances at a 5% level of significance? .

Two-Sample Tests and One-Way ANOVA 357

a) b) c) d)

There is not sufficient evidence that the variances are all the same.

There is sufficient evidence that the variances are all the same.

There is not sufficient evidence that the variances are not all the same.

There is sufficient evidence that the variances are not all the same.

ANSWER: c TYPE: MC DIFFICULTY: Easy KEYWORDS: Levene’s test, conclusion

TABLE 10-18 An agronomist wants to compare the crop yield of 3 varieties of chickpea seeds. She plants 15 fields, 5 with each variety. She then measures the crop yield in bushels per acre. Treating this as a completely randomized design, the results are presented in the table that follows.

Trial Smith Walsh Trevor 1 11.1 19.0 14.6 2 13.5 18.0 15.7 3 15.3 19.8 16.8 4 14.6 19.6 16.7 5 9.8 16.6 15.2 237. Referring to Table 10-18, the agronomist decided to perform an ANOVA F test. The amount of total variation or SST is __________.

ANSWER: 114.82 TYPE: FI DIFFICULTY: Moderate KEYWORDS: one-way analysis of variance, sum of squares 238. Referring to Table 10-18, the among-group variation or SSA is __________.

ANSWER: 82.39 TYPE: FI DIFFICULTY: Moderate KEYWORDS: one-way analysis of variance, sum of squares 239. Referring to Table 10-18, the within-group variation or SSW is __________.

ANSWER: 32.43 TYPE: FI DIFFICULTY: Moderate .

358

Two-Sample Tests and One-Way ANOVA

KEYWORDS: one-way analysis of variance, sum of squares 240. Referring to Table 10-18, the value of MSA is __________, while MSW is __________.

ANSWER: 41.19; 2.70 TYPE: FI DIFFICULTY: Moderate KEYWORDS: one-way analysis of variance, mean squares 241. Referring to Table 10-18, the null hypothesis will be rejected at a level of significance of 0.01 if the value of the test statistic is greater than __________.

ANSWER: 6.93 TYPE: FI DIFFICULTY: Easy KEYWORDS: one-way analysis of variance, F test for factor, critical value

242.

Referring to Table 10-18, the value of the test statistic is __________.

ANSWER: 15.24 TYPE: FI DIFFICULTY: Moderate KEYWORDS: one-way analysis of variance, F test for factor, test statistic 243. Referring to Table 10-18, construct the ANOVA table from the sample data.

ANSWER: Analysis of Variance Source df SS MS F p Seed Varieties 2 82.39 41.19 15.24 0.000508* Error 12 32.43 2.70 Total 14 114.82 * or p < 0.005, tabular value TYPE: PR DIFFICULTY: Moderate KEYWORDS: one-way analysis of variance, F test for factor, properties 244. Referring to Table 10-18, state the null hypothesis that can be tested.

ANSWER: H0: Smith = Walsh = Trevor TYPE: PR DIFFICULTY: Easy KEYWORDS: one-way analysis of variance, F test for factor, form of hypothesis 245. True or False: Referring to Table 10-18, the null hypothesis should be rejected at 0.005 level of significance.

ANSWER: True .

Two-Sample Tests and One-Way ANOVA 359

TYPE: TF DIFFICULTY: Moderate KEYWORDS: one-way analysis of variance, F test for factor, decision 246. True or False: Referring to Table 10-18, the decision made at 0.005 level of significance implies that all 3 means are significantly different.

ANSWER: False TYPE: TF DIFFICULTY: Moderate KEYWORDS: one-way analysis of variance, F test for factor, conclusion 247. True or False: Referring to Table 10-18, the test is valid only if the population of crop yields has the same variance for the 3 varieties.

ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: one-way analysis of variance, F test for factor, assumption

248. True or False: Referring to Table 10-18, the test is valid only if the population of crop yields is normally distributed for the 3 varieties.

ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: one-way analysis of variance, F test for factor, assumption 249. Referring to Table 10-18, the agronomist decided to compare the 3 treatment means by using the Tukey-Kramer procedure with an overall level of significance of 0.01. There are ________ pairwise comparisons that can be made.

ANSWER: 3 TYPE: FI DIFFICULTY: Moderate KEYWORDS: one-way analysis of variance, Tukey-Kramer procedure, properties 250. Referring to Table 10-18, using an overall level of significance of 0.01, the critical value of the Studentized range Q used in calculating the critical range for the Tukey-Kramer procedure is ________.

ANSWER: 5.04 TYPE: FI DIFFICULTY: Easy KEYWORDS: one-way analysis of variance, Tukey-Kramer procedure, critical value 251. Referring to Table 10-18, using an overall level of significance of 0.01, the critical range for the Tukey-Kramer procedure is ________.

ANSWER: 3.70 TYPE: FI DIFFICULTY: Moderate .

360

Two-Sample Tests and One-Way ANOVA

KEYWORDS: one-way analysis of variance, Tukey-Kramer procedure, critical value 252. True or False: Referring to Table 10-18, based on the Tukey-Kramer procedure with an overall level of significance of 0.01, the agronomist would decide that there is a significant difference between the crop yield of Smith and Walsh seeds.

ANSWER: True TYPE: TF DIFFICULTY: Moderate KEYWORDS: one-way analysis of variance, Tukey-Kramer procedure, decision, conclusion 253. True or False: Referring to Table 10-18, based on the Tukey-Kramer procedure with an overall level of significance of 0.01, the agronomist would decide that there is a significant difference between the crop yield of Smith and Trevor seeds.

ANSWER: False TYPE: TF DIFFICULTY: Moderate KEYWORDS: one-way analysis of variance, Tukey-Kramer procedure, decision, conclusion 254. True or False: Referring to Table 10-18, based on the Tukey-Kramer procedure with an overall level of significance of 0.01, the agronomist would decide that there is a significant difference between the crop yield of Walsh and Trevor seeds.

ANSWER: False TYPE: TF DIFFICULTY: Moderate KEYWORDS: one-way analysis of variance, Tukey-Kramer procedure, decision, conclusion

TABLE 10-19 A hotel chain has identically sized resorts in 5 locations. The data that follow resulted from analyzing the hotel occupancies on randomly selected days in the 5 locations.

ROW Caymen Pennkamp California Mayaguez Maui 1 28 40 21 37 22 2 33 35 21 47 19 3 41 33 27 45 25 Analysis of Variance Source df SS Location 4 963.6 Error 10 210.0 Total

MS

F 11.47

p 0.001

255. True or False: Referring to Table 10-19, if a level of significance of 0.05 is chosen, the null hypothesis should be rejected.

ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: one-way analysis of variance, F test for factor, decision .

Two-Sample Tests and One-Way ANOVA 361

256. True or False: Referring to Table 10-19, if a level of significance of 0.05 is chosen, the decision made indicates that all 5 locations have different mean occupancy rates.

ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: one-way analysis of variance, F test for factor, conclusion 257. True or False: Referring to Table 10-19, if a level of significance of 0.05 is chosen, the decision made indicates that at least 2 of the 5 locations have different mean occupancy rates.

ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: one-way analysis of variance, F test for factor, conclusion

258. Referring to Table 10-19, the among-group variation or SSA is _________.

ANSWER: 963.6 TYPE: FI DIFFICULTY: Easy KEYWORDS: one-way analysis of variance, sum of squares 259. Referring to Table 10-19, the within-group variation or SSW is _________.

ANSWER: 210.0 TYPE: FI DIFFICULTY: Easy KEYWORDS: one-way analysis of variance, sum of squares 260. Referring to Table 10-19, the total variation or SST is ________.

ANSWER: 1,173.6 TYPE: FI DIFFICULTY: Easy KEYWORDS: one-way analysis of variance, sum of squares 261. Referring to Table 10-19, the value of MSA is ______ while MSW is _______.

ANSWER: 240.9; 21.0 TYPE: FI DIFFICULTY: Easy KEYWORDS: one-way analysis of variance, mean squares 262. Referring to Table 10-19, the numerator and denominator degrees of freedom of the test ratio are ________ and ________, respectively.

ANSWER: .

362

Two-Sample Tests and One-Way ANOVA

4 and 10 TYPE: FI DIFFICULTY: Easy KEYWORDS: one-way analysis of variance, degrees of freedom 263. True or False: Referring to Table 10-19, the total mean squares is 261.90.

ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: one-way analysis of variance, sum of squares

264. Referring to Table 10-19, the null hypothesis for Levene’s test for homogeneity of variances is a) H 0 :  A = B = C = D b)

H0 : M A = M B = M C = M D

H 0 :  A2 =  B2 =  C2 =  D2 d) H 0 :  A =  B =  C =  D c)

ANSWER: c TYPE: MC DIFFICULTY: Easy KEYWORDS: Levene’s test, form of hypothesis 265. Referring to Table 10-19, what is the value of the test statistic for Levene’s test for homogeneity of variances?

ANSWER: 0.28 TYPE: PR DIFFICULTY: Moderate KEYWORDS: Levene’s test, test statistic 266. Referring to Table 10-19, what are the numerator and denominator degrees of freedom for Levene’s test for homogeneity of variances respectively?

ANSWER: 4, 10 TYPE: PR DIFFICULTY: Moderate KEYWORDS: Levene’s test, degrees of freedom 267. Referring to Table 10-19, what is the critical value of Levene’s test for homogeneity of variances at a 5% level of significance?

.

Two-Sample Tests and One-Way ANOVA 363

ANSWER: 3.48 TYPE: PR DIFFICULTY: Moderate KEYWORDS: Levene’s test, critical value 268. Referring to Table 10-19, what is the p-value of the test statistic for Levene’s test for homogeneity of variances?

ANSWER: 0.88 TYPE: PR DIFFICULTY: Moderate KEYWORDS: Levene’s test, p-value

269. Referring to Table 10-19, what should be the decision for the Levene’s test for homogeneity of variances at a 5% level of significance? a) Reject the null hypothesis because the p-value is smaller than the level of significance. b) Reject the null hypothesis because the p-value is larger than the level of significance. c) Do not reject the null hypothesis because the p-value is smaller than the level of significance. d) Do not reject the null hypothesis because the p-value is larger than the level of significance.

ANSWER: d TYPE: MC DIFFICULTY: Easy KEYWORDS: Levene’s test, decision 270. Referring to Table 10-19, what should be the conclusion for the Levene’s test for homogeneity of variances at a 5% level of significance? a) There is not sufficient evidence that the variances are all the same. b) There is sufficient evidence that the variances are all the same. c) There is not sufficient evidence that the variances are not all the same. d) There is sufficient evidence that the variances are not all the same.

ANSWER: c TYPE: MC DIFFICULTY: Easy KEYWORDS: Levene’s test, conclusion

.

Chi-Square Tests

365

CHAPTER 11: CHI-SQUARE TESTS 1. When testing for independence in a contingency table with 3 rows and 4 columns, there are ________ degrees of freedom. a) 5 b) 6 c) 7 d) 12 ANSWER: b TYPE: MC DIFFICULTY: Easy KEYWORDS: Chi-square test of independence, degrees of freedom 2. If we use the  method of analysis to test for the differences among 4 proportions, the degrees of freedom are equal to: a) 3 b) 4 c) 5 d) 1 2

ANSWER: a TYPE: MC DIFFICULTY: Easy KEYWORDS: Chi-square test for difference in proportions, degrees of freedom 3. If we wish to determine whether there is evidence that the proportion of items of interest is the same in group 1 as in group 2, the appropriate test to use is a) the Z test.

2 b) the  test. c) Both a) and b). d) Neither of a) nor b).

ANSWER: c TYPE: MC DIFFICULTY: Easy KEYWORDS: Chi-square test for difference in proportions

.

366

Chi-Square Tests 4. In testing a hypothesis using the  test, the theoretical frequencies are based on the a) null hypothesis. b) alternative hypothesis. c) normal distribution. d) None of the above.

2

ANSWER: a TYPE: MC DIFFICULTY: Easy KEYWORDS: Chi-square test for difference in proportions, Chi-square test of independence, properties

TABLE 11-1 A study published in the American Journal of Public Health was conducted to determine whether the use of seat belts in motor vehicles depends on ethnic status in San Diego County. A sample of 792 children treated for injuries sustained from motor vehicle accidents was obtained, and each child was classified according to (1) ethnic status (Hispanic or non-Hispanic) and (2) seat belt usage (worn or not worn) during the accident. The number of children in each category is given in the table below.

Seat belts worn Seat belts not worn

Hispanic 31 283

Non-Hispanic 148 330

5. Referring to Table 11-1, the calculated test statistic is a) -0.9991 b) -0.1368 c) 48.1849 d) 72.8063 ANSWER: c TYPE: MC DIFFICULTY: Easy KEYWORDS: Chi-square test of independence, test statistic 6. Referring to Table 11-1, at 5% level of significance, the critical value of the test statistic is a) 3.8415 b) 5.9914 c) 9.4877 d) 13.2767 ANSWER: a TYPE: MC DIFFICULTY: Easy KEYWORDS: Chi-square test of independence, critical value

.

Chi-Square Tests

367

7. Referring to Table 11-1, at 5% level of significance, there is sufficient evidence to conclude that a) use of seat belts in motor vehicles is related to ethnic status in San Diego County. b) use of seat belts in motor vehicles depends on ethnic status in San Diego County. c) use of seat belts in motor vehicles is associated with ethnic status in San Diego County. d) All of the above.

ANSWER: d TYPE: MC DIFFICULTY: Moderate KEYWORDS: Chi-square test of independence, decision, conclusion

TABLE 11-2 Many companies use well-known celebrities as spokespersons in their TV advertisements. A study was conducted to determine whether brand awareness of female TV viewers and the gender of the spokesperson are independent. Each in a sample of 300 female TV viewers was asked to identify a product advertised by a celebrity spokesperson. The gender of the spokesperson and whether or not the viewer could identify the product was recorded. The numbers in each category are given below.

Identified product Could not identify

Male Celebrity 41 109

Female Celebrity 61 89

8. Referring to Table 11-2, the calculated test statistic is a) -0.1006 b) 0.00 c) 5.9418 d) 6.1194 ANSWER: c TYPE: MC DIFFICULTY: Easy KEYWORDS: Chi-square test of independence, test statistic 9. Referring to Table 11-2, at 5% level of significance, the critical value of the test statistic is a) 3.8415 b) 5.9914 c) 9.4877 d) 13.2767 ANSWER: a TYPE: MC DIFFICULTY: Easy KEYWORDS: Chi-square test of independence, critical value

.

368

Chi-Square Tests

10. Referring to Table 11-2, the degrees of freedom of the test statistic are a) 1 b) 2 c) 4 d) 299 ANSWER: a TYPE: MC DIFFICULTY: Easy KEYWORDS: Chi-square test of independence, degrees of freedom 11. Referring to Table 11-2, at 5% level of significance, the conclusion is that a) brand awareness of female TV viewers and the gender of the spokesperson are independent. b) brand awareness of female TV viewers and the gender of the spokesperson are not independent. c) brand awareness of female TV viewers and the gender of the spokesperson are related. d) Both (b) and (c).

ANSWER: d TYPE: MC DIFFICULTY: Moderate KEYWORDS: Chi-square test of independence, decision, conclusion TABLE 11-3 A computer used by a 24-hour banking service is supposed to randomly assign each transaction to one of 5 memory locations. A check at the end of a day’s transactions gave the counts shown in the table to each of the 5 memory locations, along with the number of reported errors.

Memory Location:

1

2

3

4

5

Number of Transactions:

82

100

74

92

102

Number of Reported Errors

11

12

6

9

10

The bank manager wanted to test whether the proportion of errors in transactions assigned to each of the 5 memory locations differ.

12. Referring to Table 11-3, the degrees of freedom of the test statistic is a) 4 b) 8 c) 10 d) 448 ANSWER: a TYPE: MC DIFFICULTY: Easy KEYWORDS: Chi-square test for difference in proportions, degrees of freedom 13. Referring to Table 11-3, the critical value of the test statistic at 1% level of significance is .

Chi-Square Tests

a) b) c) d)

7.7794 13.2767 20.0902 23.2093

ANSWER: b TYPE: MC DIFFICULTY: Easy KEYWORDS: Chi-square test for difference in proportions, critical value 14. Referring to Table 11-3, the calculated value of the test statistic is a) -0.1777 b) -0.0185 c) 1.4999 d) 1.5190 ANSWER: c TYPE: MC DIFFICULTY: Moderate KEYWORDS: Chi-square test for difference in proportions, test statistic 15. Referring to Table 11-3, at 1% level of significance a) there is sufficient evidence to conclude that the proportions of errors in transactions assigned to each of the 5 memory locations are all different. b) there is insufficient evidence to conclude that the proportions of errors in transactions assigned to each of the 5 memory locations are all different. c) there is sufficient evidence to conclude that the proportion of errors in transactions assigned to each of the 5 memory locations are not all the same. d) there is insufficient evidence to conclude that the proportion of errors in transactions assigned to each of the 5 memory locations are not all the same.

ANSWER: d TYPE: MC DIFFICULTY: Moderate KEYWORDS: Chi-square test for difference in proportions, decision, conclusion

TABLE 11-4 .

369

370

Chi-Square Tests

One criterion used to evaluate employees in the assembly section of a large factory is the number of defective pieces per 1,000 parts produced. The quality control department wants to find out whether there is a relationship between years of experience and defect rate. Since the job is repetitious, after the initial training period any improvement due to a learning effect might be offset by a loss of motivation. A defect rate is calculated for each worker in a yearly evaluation. The results for 100 workers are given in the table below.

Years Since Training Period < 1 Year 1 – 4 Years 5 – 9 Years High 6 9 9 Defect Rate: Average 9 19 23 Low 7 8 10 16. Referring to Table 11-4, find the rejection region necessary for testing at the 0.05 level of significance whether there is a relationship between defect rate and years of experience.

2 a) Reject H0 if  > 16.919 2 b) Reject H0 if  > 15.507 2 c) Reject H0 if  > 11.143 2 d) Reject H0 if  > 9.488 ANSWER: d TYPE: MC DIFFICULTY: Easy KEYWORDS: Chi-square test of independence, critical value 17. Referring to Table 11-4, what is the expected number of employees with less than 1 year of training time and a high defect rate? a) 4.17 b) 4.60 c) 5.28 d) 9.17 ANSWER: c TYPE: MC DIFFICULTY: Moderate KEYWORDS: Chi-square test of independence, contingency table, properties 18. Referring to Table 11-4, what is the expected number of employees with 1 to 4 years of training time and a high defect rate? a) 12.00 b) 8.64 c) 6.67 d) 6.00 ANSWER: b TYPE: MC DIFFICULTY: Moderate KEYWORDS: Chi-square test of independence, contingency table, properties 19. Referring to Table 11-4, of the cell for 1 to 4 years of training time and a high defect rate, what is 2 the contribution to the overall  statistic for the independence test? .

Chi-Square Tests

a) b) c) d)

371

0.36 0.1296 0.015 0.0144

ANSWER: c TYPE: MC DIFFICULTY: Moderate KEYWORDS: Chi-square test of independence, contingency table, properties 20. Referring to Table 11-4, a test was conducted to determine if a relationship exists between defect rate and years of experience. Which of the following p-values would indicate that defect rate and years of experience are dependent? Assume you are testing at  = 0.05. a) 0.045 b) 0.055 c) 0.074 d) 0.080 ANSWER: a TYPE: MC DIFFICULTY: Easy KEYWORDS: Chi-square test of independence, p-value, decision

TABLE 11-5 A corporation randomly selects 150 salespeople and finds that 66% who have never taken a selfimprovement course would like such a course. The firm did a similar study 10 years ago in which 60% of a random sample of 160 salespeople wanted a self-improvement course. The groups are assumed to be independent random samples. Let  1 and  2 represent the true proportion of workers who would like to attend a self-improvement course in the recent study and the past study, respectively.

21. Referring to Table 11-5, if the firm wanted to test whether this proportion has changed from the previous study, which represents the relevant hypotheses? a) H0:  1 −  2 = 0 versus H1:  1 −  2  0 b) H0:  1 −  2  0 versus H1:  1 −  2 = 0 c) H0:  1 −  2  0 versus H1:  1 −  2 > 0 d) H0:  1 −  2  0 versus H1:  1 −  2 < 0 ANSWER: a TYPE: MC DIFFICULTY: Easy KEYWORDS: Chi-square test for difference in proportions

22. Referring to Table 11-5, what is the critical value when performing a chi-square test on whether population proportions are different if  = 0.05? a)  1.645 .

372

Chi-Square Tests b)  1.96 c) 3.841 d) 2.706 ANSWER: c TYPE: MC DIFFICULTY: Easy KEYWORDS: Chi-square test for difference in proportions, critical value 23. Referring to Table 11-5, what is the critical value when testing whether population proportions are different if  = 0.10? a)  1.645 b)  1.96 c) 3.842 d) 2.706 ANSWER: d TYPE: MC DIFFICULTY: Easy KEYWORDS: Chi-square test for difference in proportions, critical value 24. Referring to Table 11-5, what is the value of the test statistic to use in evaluating the alternative hypothesis that there is a difference in the two population proportions using  = 0.10? a) 4.335 b) 2.706 c) 1.194 d) 0.274 ANSWER: c TYPE: MC DIFFICULTY: Moderate KEYWORDS: Chi-square test for difference in proportions, test statistic

25. Referring to Table 11-5, the company tests to determine at the 0.05 level whether the population proportion has changed from the previous study. Which of the following is most correct? a) Reject the null hypothesis and conclude that the proportion of employees who are interested in a self-improvement course has changed over the intervening 10 years. .

Chi-Square Tests

373

b) Do not reject the null hypothesis and conclude that the proportion of employees who are interested in a self-improvement course has not changed over the intervening 10 years. c) Reject the null hypothesis and conclude that the proportion of employees who are interested in a self-improvement course has increased over the intervening 10 years. d) Do not reject the null hypothesis and conclude that the proportion of employees who are interested in a self-improvement course has increased over the intervening 10 years.

ANSWER: b TYPE: MC DIFFICULTY: Moderate KEYWORDS: Chi-square test for difference in proportions, decision, conclusion 26. True or False: In testing the difference between two proportions using the normal distribution, we may use either a one-tailed Chi-square test or two-tailed Z test.

ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: Z test for difference in two proportions, Chi-square test for difference in proportions 27. True or False: The squared difference between the observed and theoretical frequencies should be large if there is no significant difference between the proportions.

ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: Chi-square test for difference in proportions, properties 28. True or False: A test for the difference between two proportions can be performed using the chisquare distribution.

ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: Chi-square test for difference in proportions 29. True or False: A test for whether one proportion is higher than the other can be performed using the chi-square distribution.

ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: Chi-square test for difference in proportions 30. True or False: When using the  tests for independence, one should be aware that expected frequencies that are too small will lead to too big a type I error.

2

ANSWER: True .

374

Chi-Square Tests

TYPE: TF DIFFICULTY: Difficult KEYWORDS: Chi-square test of independence, properties, assumption 31. True or False: If we use the chi-square method of analysis to test for the difference between proportions, we must assume that there are at least 5 observed frequencies in each cell of the contingency table.

ANSWER: False TYPE: TF DIFFICULTY: Moderate KEYWORDS: Chi-square test of independence, properties, assumption

TABLE 11-6 The dean of a college is interested in the proportion of graduates from his college who have a job offer on graduation day. He is particularly interested in seeing if there is a difference in this proportion for accounting and economics majors. In a random sample of 100 of each type of major at graduation, he found that 65 accounting majors and 52 economics majors had job offers. If the accounting majors are designated as “Group 1” and the economics majors are designated as “Group 2,” perform the appropriate hypothesis test using a level of significance of 0.05.

32. Referring to Table 11-6, the hypotheses the dean should use are: a) H0:  1 −  2 = 0 versus H1:  1 −  2  0 b) H0:  1 −  2  0 versus H1:  1 −  2 = 0 c) H0:  1 −  2  0 versus H1:  1 −  2 > 0 d) H0:  1 −  2  0 versus H1:  1 −  2 < 0 ANSWER: a TYPE: MC DIFFICULTY: Easy KEYWORDS: Chi-square test for difference in proportions, form of hypothesis 33. Referring to Table 11-6, the null hypothesis will be rejected if the test statistic is ________.

ANSWER:  2 > 3.841 TYPE: FI DIFFICULTY: Easy KEYWORDS: Chi-square test for difference in proportions, critical value

34. Referring to Table 11-6, the value of the test statistic is ________.

ANSWER:  2 = 3.4806 TYPE: FI DIFFICULTY: Moderate KEYWORDS: Chi-square test for difference in proportions, test statistic .

Chi-Square Tests

375

35. Referring to Table 11-6, the p-value of the test is ________.

ANSWER: 0.0621 TYPE: FI DIFFICULTY: Moderate KEYWORDS: Chi-square test for difference in proportions, p-value 36. True or False: Referring to Table 11-6, the null hypothesis should be rejected.

ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: Chi-square test for difference in proportions, decision 37. True or False: Referring to Table 11-6, the same decision would be made with this test if the level of significance had been 0.01 rather than 0.05.

ANSWER: True TYPE: TF DIFFICULTY: Moderate KEYWORDS: Chi-square test for difference in proportions, decision 38. True or False: Referring to Table 11-6, the same decision would be made with this test if the level of significance had been 0.10 rather than 0.05.

ANSWER: False TYPE: TF DIFFICULTY: Moderate KEYWORDS: Chi-square test for difference in proportions, decision

TABLE 11-7 The director of transportation of a large company is interested in the usage of her van pool. She considers her routes to be divided into local and non-local. She is particularly interested in learning if there is a difference in the proportion of males and females who use the local routes. She takes a sample of a day's riders and finds the following: .

376

Chi-Square Tests

Local Non-Local Total

Male 27 33 60

Female 44 25 69

Total 71 58 129

She will use this information to perform a chi-square hypothesis test using a level of significance of 0.05.

39. Referring to Table 11-7, the test will involve _________ degree(s) of freedom.

ANSWER: 1 TYPE: FI DIFFICULTY: Easy KEYWORDS: Chi-square test for difference in proportions, degrees of freedom 40. Referring to Table 11-7, the overall or average proportion of local riders is __________.

ANSWER: 0.550 TYPE: FI DIFFICULTY: Easy KEYWORDS: Chi-square test for difference in proportions, properties 41. Referring to Table 11-7, the expected cell frequency in the Male/Local cell is __________.

ANSWER: 33.02 TYPE: FI DIFFICULTY: Moderate KEYWORDS: Chi-square test for difference in proportions, contingency table, properties 42. Referring to Table 11-7, the expected cell frequency in the Female/Non-Local cell is __________.

ANSWER: 31.02 TYPE: FI DIFFICULTY: Moderate KEYWORDS: Chi-square test for difference in proportions, contingency table, properties 43. Referring to Table 11-7, the critical value of the test is _________.

ANSWER: 3.841 TYPE: FI DIFFICULTY: Easy KEYWORDS: Chi-square test for difference in proportions, critical value 44. Referring to Table 11-7, the value of the test statistic is _________.

ANSWER: 4.568 TYPE: FI DIFFICULTY: Moderate KEYWORDS: Chi-square test for difference in proportions, test statistic .

Chi-Square Tests

377

45. True or False: Referring to Table 11-7, the null hypothesis will be rejected.

ANSWER: True TYPE: TF DIFFICULTY: Moderate KEYWORDS: Chi-square test for difference in proportions, decision 46. True or False: Referring to Table 11-7, the decision made suggests that there is a difference between the proportion of males and females who ride local versus non-local routes.

ANSWER: True TYPE: TF DIFFICULTY: Moderate KEYWORDS: Chi-square test for difference in proportions, conclusion

TABLE 11-8

A market researcher investigated consumer preferences for Coca-Cola and Pepsi before a taste test and after a taste test. The following table summarizes the results from a sample of 200 consumers: PREFERENCE Before taste test After taste test

SOFT DRINK Coca-Cola Pepsi 110 118

90 82

She wanted to perform a chi-square test to see if there is evidence of a difference in preference for Coca-Cola before and after the taste test by allowing a 10% probability of incorrectly concluding that there is a difference when there is in fact no difference.

47. Referring to Table 11-8, the test will involve _________ degree(s) of freedom.

ANSWER: 1 TYPE: FI DIFFICULTY: Easy KEYWORDS: Chi-square test for difference in proportions, degrees of freedom

48. Referring to Table 11-8, the overall or average proportion of consumers who prefer Coca-Cola is __________.

ANSWER: 0.57 TYPE: FI DIFFICULTY: Easy KEYWORDS: Chi-square test for difference in proportions, properties

.

378

Chi-Square Tests

49. Referring to Table 11-8, the expected cell frequency in the Pepsi/After taste test cell is __________.

ANSWER: 86 TYPE: FI DIFFICULTY: Moderate KEYWORDS: Chi-square test for difference in proportions, contingency table, properties 50. Referring to Table 11-8, the expected cell frequency in the Coca-Cola/Before taste test cell is __________.

ANSWER: 114 TYPE: FI DIFFICULTY: Moderate KEYWORDS: Chi-square test for difference in proportions, contingency table, properties 51. Referring to Table 11-8, the critical value of the test is _________.

ANSWER: 2.7055 TYPE: FI DIFFICULTY: Easy KEYWORDS: Chi-square test for difference in proportions, critical value 52. Referring to Table 11-8, the value of the test statistic is _________.

ANSWER: 0.6528 TYPE: FI DIFFICULTY: Moderate KEYWORDS: Chi-square test for difference in proportions, test statistic 53. Referring to Table 11-8, the p-value of the test is _________.

ANSWER: 0.4191 TYPE: FI DIFFICULTY: Moderate KEYWORDS: Chi-square test for difference in proportions, p-value

54. True or False: Referring to Table 11-8, the null hypothesis will be rejected.

ANSWER: False TYPE: TF DIFFICULTY: Moderate KEYWORDS: Chi-square test for difference in proportions, decision 55. True or False: Referring to Table 11-8, the test concludes that there is a difference in preference for Coca-Cola before and after the taste test. .

Chi-Square Tests

379

ANSWER: False TYPE: TF DIFFICULTY: Moderate KEYWORDS: Chi-square test for difference in proportions, conclusion

TABLE 11-9 Four surgical procedures currently are used to install pacemakers. If the patient does not need to return for follow-up surgery, the operation is called a "clear" operation. A heart center wants to compare the proportion of clear operations for the 4 procedures, and collects the following numbers of patients from their own records:

Clear Return Total

A 27 11 38

B 41 15 56

Procedure C 21 9 30

D 7 11 18

Total 96 46 142

They will use this information to test for a difference among the proportion of clear operations using a chi-square test with a level of significance of 0.05.

56. Referring to Table 11-9, the test will involve _________ degrees of freedom.

ANSWER: 3 TYPE: FI DIFFICULTY: Easy KEYWORDS: Chi-square test for difference in proportions, degrees of freedom 57. Referring to Table 11-9, the overall or average proportion of clear operations is __________.

ANSWER: 0.676 TYPE: FI DIFFICULTY: Easy KEYWORDS: Chi-square test for difference in proportions, properties

58. Referring to Table 11-9, the expected cell frequency for the Procedure A/Clear cell is __________.

ANSWER: 25.69 TYPE: FI DIFFICULTY: Moderate KEYWORDS: Chi-square test for difference in proportions, contingency table, properties 59. Referring to Table 11-9, the expected cell frequency for the Procedure D/Return cell is __________. .

380

Chi-Square Tests

ANSWER: 5.83 TYPE: FI DIFFICULTY: Moderate KEYWORDS: Chi-square test for difference in proportions, contingency table, properties 60. Referring to Table 11-9, the critical value of the test is ________.

ANSWER: 7.815 TYPE: FI DIFFICULTY: Easy KEYWORDS: Chi-square test for difference in proportions, critical value 61. Referring to Table 11-9, the value of the test statistic is _________.

ANSWER: 7.867 TYPE: FI DIFFICULTY: Moderate KEYWORDS: Chi-square test for difference in proportions, test statistic 62. True or False: Referring to Table 11-9, the null hypothesis will be rejected.

ANSWER: True TYPE: TF DIFFICULTY: Moderate KEYWORDS: Chi-square test for difference in proportions, decision 63. True or False: Referring to Table 11-9, the decision made suggests that the 4 procedures all have different proportions of clear operations.

ANSWER: False TYPE: TF DIFFICULTY: Moderate KEYWORDS: Chi-square test for difference in proportions, conclusion

64. True or False: Referring to Table 11-9, the decision made suggests that the 4 procedures do not all have the same proportion of clear operations.

ANSWER: True TYPE: TF DIFFICULTY: Moderate KEYWORDS: Chi-square test for difference in proportions, conclusion

TABLE 11-10

.

Chi-Square Tests

381

The director of admissions at a state college is interested in seeing if admissions status (admitted, waiting list, denied admission) at his college is independent of the type of community in which an applicant resides. He takes a sample of recent admissions decisions and forms the following table:

Urban Rural Suburban Total

Admitted 45 33 34 112

Wait List 21 13 12 46

Denied 17 24 39 80

Total 83 70 85 238

He will use this table to do a chi-square test of independence with a level of significance of 0.01.

65. Referring to Table 11-10, the test will involve _________ degrees of freedom.

ANSWER: 4 TYPE: FI DIFFICULTY: Easy KEYWORDS: Chi-square test of independence, degrees of freedom 66. Referring to Table 11-10, the critical value of the test is _________.

ANSWER: 13.277 TYPE: FI DIFFICULTY: Easy KEYWORDS: Chi-square test of independence, critical value 67. Referring to Table 11-10, the expected cell frequency for the Admitted/Urban cell is _________.

ANSWER: 39.06 TYPE: FI DIFFICULTY: Moderate KEYWORDS: Chi-square test of independence, contingency table, properties 68. Referring to Table 11-10, the value of the test statistic is _________.

ANSWER: 12.624 TYPE: FI DIFFICULTY: Moderate KEYWORDS: Chi-square test of independence, test statistic 69. True or False: Referring to Table 11-10, the null hypothesis will be rejected.

ANSWER: False TYPE: TF DIFFICULTY: Moderate KEYWORDS: Chi-square test of independence, decision 70. True or False: Referring to Table 11-10, the p-value of this test is greater than 0.01.

ANSWER: .

382

Chi-Square Tests

True TYPE: TF DIFFICULTY: Moderate KEYWORDS: Chi-square test of independence, p-value 71. True or False: Referring to Table 11-10, the decision made suggests that admissions status at the college is independent of the type of community in which an applicant resides.

ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: Chi-square test of independence, conclusion 72. True or False: Referring to Table 11-10, the same decision would be made with this test if the level of significance had been 0.005.

ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: Chi-square test of independence, decision 73. True or False: Referring to Table 11-10, the same decision would be made with this test if the level of significance had been 0.05.

ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: Chi-square test of independence, decision 74. True or False: Referring to Table 11-10, the null hypothesis claims that "there is no association between admission status at the college and the type of community in which an applicant resides."

ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: Chi-square test of independence, form of hypothesis, conclusion 75. True or False: Referring to Table 11-10, the alternative hypothesis claims that "there is some connection between admission status at the college and the type of community in which an applicant resides."

ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: Chi-square test of independence, form of hypothesis, conclusion 76. True or False: The chi-square test of independence requires that the number of expected frequency in each cell to be at least 5.

ANSWERS: False .

Chi-Square Tests

383

TYPE: TF DIFFICULTY: Easy KEYWORDS: Chi-square test of independence, assumption 77. True or False: The chi-square test of independence requires that the number of expected frequency in each cell to be at least 1.

ANSWERS: True TYPE: TF DIFFICULTY: Easy KEYWORDS: Chi-square test of independence, assumption

TABLE 11-11 Parents complain that children read too few storybooks and watch too much television nowadays. A survey of 1,000 children reveals the following information on average time spent watching TV and average time spent reading storybooks

Average time spent watching TV Less than 2 hours More than 2 hours

Average time spent reading story books Less than 1 hour Between 1and 2 hours

More than 2 hours

90 655

130 8

85 32

78. Referring to Table 11-11, how many children in the survey spent less than 2 hours watching TV and more than 2 hours reading story books on average? a) 8 b) 130 c) 175 d) 687 ANSWER: b TYPE: MC DIFFICULTY: Easy KEYWORDS: contingency table, properties 79. Referring to Table 11-11, how many children in the survey spent less than 2 hours watching TV and no more than 2 hours reading storybooks on average? a) 8 b) 130 c) 175 d) 687 ANSWER: c TYPE: MC DIFFICULTY: Easy KEYWORDS: contingency table, properties

.

384

Chi-Square Tests

80. Referring to Table 11-11, if the null hypothesis of no connection between time spent watching TV and time spent reading story books is true, how many children watching less than 2 hours of TV and reading no more than 2 hours of story books on average can we expect? a) 35.69 b) 227.23 c) 262.91 d) 969.75 ANSWER: c TYPE: MC DIFFICULTY: Difficult KEYWORDS: Chi-square test of independence, contingency table, properties 81. Referring to Table 11-11, if the null hypothesis of no connection between time spent watching TV and time spent reading story books is true, how many children watching less than 2 hours of TV and reading more than 2 hours of story books on average can we expect? a) 42.09 b) 155.25 c) 262.92 d) 987.75 ANSWER: a TYPE: MC DIFFICULTY: Moderate KEYWORDS: Chi-square test of independence, contingency table, properties 82. Referring to Table 11-11, to test whether there is any relationship between average time spent watching TV and average time spent reading story books, the value of the measured test statistic is a) -12.59 b) 1.61 c) 481.49 d) 1,368.06 ANSWER: c TYPE: MC DIFFICULTY: Moderate KEYWORDS: Chi-square test of independence, test statistic 83. Referring to Table 11-11, suppose we want to constrain the probability of committing a Type I error to 5% when testing whether there is any relationship between average time spent watching TV and average time spent reading story books, the critical value will be a) 5.991 b) 7.378 c) 12.592 d) 14.449 ANSWER: a TYPE: MC DIFFICULTY: Easy KEYWORDS: Chi-square test of independence, critical value .

Chi-Square Tests

385

84. Referring to Table 11-11, we want to test whether there is any relationship between average time spent watching TV and average time spent reading storybooks. Suppose the value of the test statistic was 164 (which is not the correct answer) and the critical value was 19.00 (which is not the correct answer), then we could conclude that a) There is connection between time spent reading storybooks and time spent watching TV. b) There is no connection between time spent reading storybooks and time spent watching TV. c) More time spent reading storybooks leads to less time spent watching TV. d) More time spent watching TV leads to less time spent reading storybooks.

ANSWER: a TYPE: MC DIFFICULTY: Moderate KEYWORDS: Chi-square test of independence, decision, conclusion

TABLE 11-12 Recent studies have found that American children are more obese than in the past. The amount of time children spent watching television has received much of the blame. A survey of 100 ten-yearolds revealed the following with regards to weights and average number of hours a day spent watching television. We are interested in testing whether the average number of hours spent watching TV and weights are independent at 1% level of significance.

Weights More than 10 lbs. overweight Within 10 lbs. of normal weight More than 10 lbs. underweight .

0-3 1 20 10

TV Hours 3-6 9 15 5

6+ 20 15 5

Total 30 50 20

386

Chi-Square Tests Total

31

29

40

100

85. Referring to Table 11-12, if there is no connection between weights and average number of hours spent watching TV, we should expect how many children to be spending 3-6 hours on average watching TV and are more than 10 lbs. underweight? a) 5 b) 5.8 c) 6.2 d) 8 ANSWER: b TYPE: MC DIFFICULTY: Easy KEYWORDS: Chi-square test of independence, contingency table, properties 86. Referring to Table 11-12, if there is no connection between weights and average number of hours spent watching TV, we should expect how many children to be spending no more than 6 hours on average watching TV and are more than 10 lbs. underweight? a) 5.8 b) 6.2 c) 8 d) 12 ANSWER: d TYPE: MC DIFFICULTY: Moderate KEYWORDS: Chi-square test of independence, contingency table, properties

87. Referring to Table 11-12, how many children in the survey spend more than 6 hours watching TV and are more than 10 lbs. overweight? a) 1 b) 9 c) 20 d) 40 ANSWER: c TYPE: MC DIFFICULTY: Easy KEYWORDS: Chi-square test of independence, contingency table, properties 88. Referring to Table 11-12, how many children in the survey spend no more than 6 hours watching TV and are more than 10 lbs. underweight? a) 5 .

Chi-Square Tests

387

b) 10 c) 15 d) 20 ANSWER: c TYPE: MC DIFFICULTY: Easy KEYWORDS: Chi-square test of independence, contingency table, properties 89. Referring to Table 11-12, the value of the test statistic is a) 8.532 b) 15.483 c) 18.889 d) 69.744 ANSWER: c TYPE: MC DIFFICULTY: moderate KEYWORDS: Chi-square test of independence, test statistic 90. Referring to Table 11-12, the critical value of the test will be a) 6.635 b) 13.277 c) 14.860 d) 21.666 ANSWER: b TYPE: MC DIFFICULTY: Easy KEYWORDS: Chi-square test of independence, critical value

91. Referring to Table 11-12, suppose the value of the test statistic was 30.00 (which is not the correct value) and the critical value at 1% level of significance was 10.00 (which is not the correct value), which of the following conclusions would be correct? a) We will accept the null and conclude that average number of hours spent watching TV and weights are independent. b) We will reject the null and conclude that average number of hours spent watching TV and weights are independent. c) We will accept the null and conclude that average number of hours spent watching TV and weights are not independent. d) We will reject the null and conclude that average number of hours spent watching TV and weights are not independent.

ANSWER: d TYPE: MC DIFFICULTY: Moderate KEYWORDS: Chi-square test of independence, decision, conclusion

.

388

Chi-Square Tests 92. Referring to Table 11-12, which of the following statements is correct? a) We can accept the null for any level of significance greater than 0.005 b) We can reject the null for any level of significance greater than 0.005. c) We can accept the null for any level of significance smaller than 0.005 d) We can reject the null for any level of significance smaller than 0.005.

ANSWER: b TYPE: MC DIFFICULTY: Difficult KEYWORDS: Chi-square test of independence, decision 93. Referring to Table 11-12, the degrees of freedom of the test statistic are a) 1 b) 2 c) 4 d) 9 ANSWER: c TYPE: MC DIFFICULTY: Easy KEYWORDS: Chi-square test of independence, degrees of freedom 94. True or False: Referring to Table 11-12, the test is always a one-tailed test.

ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: Chi-square test of independence, properties

Table 11-13 According to an article in Marketing News, fewer checks are being written at grocery store checkout stands than in the past. To determine whether there is a difference in the proportion of shoppers who pay by check among three consecutive years at a 0.05 level of significance, the results of a survey of 500 shoppers in three consecutive years are obtained and presented below.

Check Written Yes No 95.

Year 1 225 275

Year Year 2 175 325

Year 3 125 375

Referring to Table 11-13, what is the expected number of shoppers who pay by check in year 1 if there is no difference in the proportion of shoppers who pay by check among the three years?

ANSER: 175 .

Chi-Square Tests

389

TYPE: PR DIFFICULTY: Easy KEYWORDS: Chi-square test for difference in proportions, properties 96.

Referring to Table 11-13, what is the expected number of shoppers who do not pay by check in year 3 if there is no difference in the proportion of shoppers who pay by check among the three years?

ANSER: 325 TYPE: PR DIFFICULTY: Easy KEYWORDS: Chi-square test for difference in proportions, properties 97.

Referring to Table 11-13, what is the form of the null hypothesis? a) H 0 :  1 =  2 =  3 b) H 0 :  1 =  2   3 c)

H 0 : 1   2 =  3

d) H 0 :  1   2   3 ANSWER: a TYPE: MC DIFFICULTY: Easy KEYWORDS: Chi-square test for difference in proportions, forms of hypothesis

98.

Referring to Table 11-13, what is the form of the alternative hypothesis? a) H1 : 1   2   3 b) H1 : 1   2 =  3

H1 : 1 =  2   3 d) H1 : not all  j are the same c)

ANSWER: d TYPE: MC DIFFICULTY: Easy KEYWORDS: Chi-square test for difference in proportions, forms of hypothesis 99.

True or False: Referring to Table 11-13, the assumptions needed to perform the test are satisfied.

ANSWER: True .

390

Chi-Square Tests TYPE: TF DIFFICULTY: Moderate KEYWORDS: Chi-square test for difference in proportions, assumption 100. Referring to Table 11-13, what are the degrees of freedom of the test statistic?

ANSWER: 2 TYPE: PR DIFFICULTY: Easy KEYWORDS: Chi-square test for difference in proportions, degrees of freedom 101. Referring to Table 11-13, what is the value of the test statistic?

ANSWER: 43.96 TYPE: PR DIFFICULTY: Moderate KEYWORDS: Chi-square test for difference in proportions, test statistic 102. Referring to Table 11-13, what is the critical value?

ANSWER: 5.991 TYPE: PR DIFFICULTY: Easy KEYWORDS: Chi-square test for difference in proportions, critical value 103. Referring to Table 11-13, what is the p-value of the test statistic?

ANSWER: 2.9E-10 or smaller than 0.005 TYPE: PR DIFFICULTY: Moderate KEYWORDS: Chi-square test for difference in proportions, p-value

104. True or False: Referring to Table 11-13, the null hypothesis cannot be rejected.

ANSWER: False TYPE: TF DIFFICULTY: Moderate KEYWORDS: Chi-square test for difference in proportions, decision 105. Referring to Table 11-13, what is the correct conclusion? a) There is enough evidence that the proportions are all different in the 3 years. b) There is not enough evidence that the proportions are all different in the 3 years. c) There is enough evidence that at least two of the proportions are not equal. d) There is not enough evidence that at least two of the proportions are not equal.

ANSWER: c TYPE: MC DIFFICULTY: Moderate KEYWORDS: Chi-square test for difference in proportions, conclusion

.

Chi-Square Tests

TABLE 11-14 Data on the percentage of 200 hotels in each of the three large cities across the world on whether minibar charges are correctly posted at checkout are given below.

Yes No

Hong Kong 86% 14%

New York 76% 24%

Paris 78% 22%

At the 0.05 level of significance, you want to know if there is evidence of a difference in the proportion of hotels that correctly post minibar charges among the three cities.

106. Referring to Table 11-14, the test will involve _________ degrees of freedom.

ANSWER: 2 TYPE: FI DIFFICULTY: Easy KEYWORDS: Chi-square test for difference in proportions, degrees of freedom 107. Referring to Table 11-14, the expected cell frequency for the Hong Kong/Yes cell is __________.

ANSWER: 160 TYPE: FI DIFFICULTY: Moderate KEYWORDS: Chi-square test for difference in proportions, contingency table, properties

108. Referring to Table 11-14, the expected cell frequency for the Paris/No cell is __________.

ANSWER: 40 TYPE: FI DIFFICULTY: Moderate KEYWORDS: Chi-square test for difference in proportions, contingency table, properties 109. Referring to Table 11-14, the critical value of the test is ________.

ANSWER: 5.9915 TYPE: FI DIFFICULTY: Easy KEYWORDS: Chi-square test for difference in proportions, critical value 110. Referring to Table 11-14, the value of the test statistic is _________.

ANSWER: .

391

392

Chi-Square Tests

7.00 TYPE: FI DIFFICULTY: Moderate KEYWORDS: Chi-square test for difference in proportions, test statistic 111. True or False: Referring to Table 11-14, the null hypothesis will be rejected.

ANSWER: True TYPE: TF DIFFICULTY: Moderate KEYWORDS: Chi-square test for difference in proportions, decision 112. True or False: Referring to Table 11-14, the decision made suggests that the 3 cities all have different proportions of hotels that correctly post minibar charges.

ANSWER: False TYPE: TF DIFFICULTY: Moderate KEYWORDS: Chi-square test for difference in proportions, conclusion 113. True or False: Referring to Table 11-14, the decision made suggests that the 3 cities do not all have the same proportion of hotels that correctly post minibar charges.

ANSWER: True TYPE: TF DIFFICULTY: Moderate KEYWORDS: Chi-square test for difference in proportions, conclusion

.

Simple Linear Regression 393

CHAPTER 12: SIMPLE LINEAR REGRESSION 1. The Y-intercept (b0) represents the a) predicted value of Y when X = 0. b) change in estimated average Y per unit change in X. c) predicted value of Y. d) variation around the sample regression line.

ANSWER: a TYPE: MC DIFFICULTY: Easy KEYWORDS: intercept, interpretation 2. The Y-intercept (b0) represents the a) estimated average Y when X = 0. b) change in estimated average Y per unit change in X. c) predicted value of Y. d) variation around the sample regression line.

ANSWER: a TYPE: MC DIFFICULTY: Easy KEYWORDS: intercept, interpretation 3. The slope (b1) represents a) predicted value of Y when X = 0. b) the estimated average change in Y per unit change in X. c) the predicted value of Y. d) variation around the line of regression.

ANSWER: b TYPE: MC DIFFICULTY: Easy KEYWORDS: slope, interpretation 4. The least squares method minimizes which of the following? a) SSR b) SSE c) SST d) All of the above ANSWER: b TYPE: MC DIFFICULTY: Difficult KEYWORDS: least squares, properties

.

394

Simple Linear Regression

TABLE 12-1 A large national bank charges local companies for using their services. A bank official reported the results of a regression analysis designed to predict the bank’s charges (Y) -- measured in dollars per month -- for services rendered to local companies. One independent variable used to predict service charge to a company is the company’s sales revenue (X) -- measured in millions of dollars. Data for 21 companies who use the bank’s services were used to fit the model:

E(Y ) = 0 + 1 X

The results of the simple linear regression are provided below.

Y = −2, 700 + 20 X , SYX = 65, two-tailed p value = 0.034 (for testing  ) 5. Referring to Table 12-1, interpret the estimate of  0 , the Y-intercept of the line. a) All companies will be charged at least $2,700 by the bank. b) There is no practical interpretation since a sales revenue of $0 is a nonsensical value. c) About 95% of the observed service charges fall within $2,700 of the least squares line. d) For every $1 million increase in sales revenue, we expect a service charge to decrease $2,700.

ANSWER: b TYPE: MC DIFFICULTY: Moderate KEYWORDS: intercept, interpretation 6. Referring to Table 12-1, interpret the estimate of  , the standard deviation of the random error term (standard error of the estimate) in the model. a) About 95% of the observed service charges fall within $65 of the least squares line. b) About 95% of the observed service charges equal their corresponding predicted values. c) About 95% of the observed service charges fall within $130 of the least squares line. d) For every $1 million increase in sales revenue, we expect a service charge to increase $65.

ANSWER: c TYPE: MC DIFFICULTY: Difficult KEYWORDS: standard error of the estimate, interpretation 7. Referring to Table 12-1, interpret the p-value for testing whether 1 exceeds 0. a) There is sufficient evidence (at the  = 0.05) to conclude that sales revenue (X) is a useful linear predictor of service charge (Y). b) There is insufficient evidence (at the  = 0.10) to conclude that sales revenue (X) is a useful linear predictor of service charge (Y). c) Sales revenue (X) is a poor predictor of service charge (Y). d) For every $1 million increase in sales revenue, we expect a service charge to increase $0.034.

ANSWER: a TYPE: MC DIFFICULTY: Moderate KEYWORDS: p-value, interpretation

.

Simple Linear Regression 395 8. Referring to Table 12-1, a 95% confidence interval for 1 is (15, 30). Interpret the interval. a) We are 95% confident that the mean service charge will fall between $15 and $30 per month. b) We are 95% confident that the sales revenue (X) will increase between $15 and $30 million for every $1 increase in service charge (Y). c) We are 95% confident that average service charge (Y) will increase between $15 and $30 for every $1 million increase in sales revenue (X). d) At the  = 0.05 level, there is no evidence of a linear relationship between service charge (Y) and sales revenue (X).

ANSWER: c TYPE: MC DIFFICULTY: Moderate KEYWORDS: confidence interval, interpretation TABLE 12-2 A candy bar manufacturer is interested in trying to estimate how sales are influenced by the price of their product. To do this, the company randomly chooses 6 small cities and offers the candy bar at different prices. Using candy bar sales as the dependent variable, the company will conduct a simple linear regression on the data below: City Price ($) Sales River Falls 1.30 100 Hudson 1.60 90 Ellsworth 1.80 90 Prescott 2.00 40 Rock Elm 2.40 38 Stillwater 2.90 32 9. Referring to Table 12-2, what is the estimated slope parameter for the candy bar price and sales data? a) 161.386 b) 0.784 c) – 3.810 d) – 48.193 ANSWER: d TYPE: MC DIFFICULTY: Moderate KEYWORDS: slope, estimation 10. Referring to Table 12-2, what is the estimated average change in the sales of the candy bar if price goes up by $1.00? a) 161.386 b) 0.784 c) – 3.810 d) – 48.193 ANSWER: d TYPE: MC DIFFICULTY: Moderate KEYWORDS: slope, interpretation .

396

Simple Linear Regression

11. Referring to Table 12-2, what is the coefficient of correlation for these data? a) – 0.8854 b) – 0.7839 c) 0.7839 d) 0.8854 ANSWER: a TYPE: MC DIFFICULTY: Moderate KEYWORDS: coefficient of correlation, estimation 12. Referring to Table 12-2, what is the percentage of the total variation in candy bar sales explained by the regression model? a) 100% b) 88.54% c) 78.39% d) 48.19% ANSWER: c TYPE: MC DIFFICULTY: Moderate KEYWORDS: coefficient of determination, interpretation 13. Referring to Table 12-2, what percentage of the total variation in candy bar sales is explained by prices? a) 100% b) 88.54% c) 78.39% d) 48.19% ANSWER: c TYPE: MC DIFFICULTY: Moderate KEYWORDS: coefficient of determination, interpretation 14. Referring to Table 12-2, what is the standard error of the estimate, SYX, for the data? a) 0.784 b) 0.885 c) 12.650 d) 16.299 ANSWER: d TYPE: MC DIFFICULTY: Moderate KEYWORDS: standard error of the estimate, estimation

.

Simple Linear Regression 397

15. Referring to Table 12-2, what is the standard error of the regression slope estimate, S b1 ? a) b) c) d)

0.784 0.885 12.650 16.299

ANSWER: c TYPE: MC DIFFICULTY: Moderate KEYWORDS: standard error, slope 16. Referring to Table 12-2, what is (X – X ) for these data? a) 0 b) 1.66 c) 2.54 d) 25.66 2

ANSWER: b TYPE: MC DIFFICULTY: Moderate KEYWORDS: sum of squares 17. Referring to Table 12-2, to test that the regression coefficient, 1 , is not equal to 0, what would be the critical values? Use  = 0.05. a)  2.5706 b)  2.7764 c)  3.1634 d)  3.4954 ANSWER: b TYPE: MC DIFFICULTY: Difficult KEYWORDS: t test on slope, critical value 18. Referring to Table 12-2, to test whether a change in price will have any impact on average sales, what would be the critical values? Use  = 0.05. a)  2.5706 b)  2.7765 c)  3.1634 d)  3.4954 ANSWER: b TYPE: MC DIFFICULTY: Difficult EXPLANATION: This is a test for 1 is not equal to 0.

KEYWORDS: t test on slope, critical value

.

398

Simple Linear Regression

19. Referring to Table 12-2, if the price of the candy bar is set at $2, the estimated average sales will be a) 30 b) 65 c) 90 d) 100 ANSWER: b TYPE: MC DIFFICULTY: Moderate KEYWORDS: estimation of mean values 20. Referring to Table 12-2, if the price of the candy bar is set at $2, the predicted sales will be a) 30 b) 65 c) 90 d) 100 ANSWER: b TYPE: MC DIFFICULTY: Moderate KEYWORDS: prediction of individual values 21. True of False: The Chancellor of a university has commissioned a team to collect data on students’ GPAs and the amount of time they spend bar hopping every week (measured in minutes). He wants to know if imposing much tougher regulations on all campus bars to make it more difficult for students to spend time in any campus bar will have a significant impact on general students' GPAs. His team should use a t test on the slope of the population regression.

ANSWER: True TYPE: TF DIFFICULTY: Moderate KEYWORDS: t test on slope 22. The residual represents the discrepancy between the observed dependent variable and its _______ value.

ANSWER: predicted or estimated average TYPE: FI DIFFICULTY: Moderate KEYWORDS: estimation of mean values, prediction of individual values

.

Simple Linear Regression 399

TABLE 12-3 The director of cooperative education at a state college wants to examine the effect of cooperative education job experience on marketability in the work place. She takes a random sample of 4 students. For these 4, she finds out how many times each had a cooperative education job and how many job offers they received upon graduation. These data are presented in the table below.

Student 1 2 3 4

CoopJobs 1 2 1 0

JobOffer 4 6 3 1

23. Referring to Table 12-3, set up a scatter diagram.

ANSWER: Scatter Diagram 6

Job Offers

5 4 3 2 1 0 0

1

2

Coop Jobs

TYPE: PR DIFFICULTY: Moderate KEYWORDS: scatter plot 24. Referring to Table 12-3, the least squares estimate of the slope is __________.

ANSWER: 2.50 TYPE: FI DIFFICULTY: 2 Moderate KEYWORDS: slope, estimation, least squares 25. Referring to Table 12-3, the least squares estimate of the Y-intercept is __________.

ANSWER: 1.00 TYPE: FI DIFFICULTY: Moderate KEYWORDS: intercept, estimation, least squares .

400

Simple Linear Regression

26. Referring to Table 12-3, the prediction for the number of job offers for a person with 2 coop jobs is __________.

ANSWER: 6 TYPE: FI DIFFICULTY: Moderate KEYWORDS: prediction of individual values 27. Referring to Table 12-3, the total sum of squares (SST) is __________.

ANSWER: 13.0 TYPE: FI DIFFICULTY: Moderate KEYWORDS: sum of squares 28. Referring to Table 12-3, the regression sum of squares (SSR) is __________.

ANSWER: 12.5 TYPE: FI DIFFICULTY: Moderate KEYWORDS: sum of squares 29. Referring to Table 12-3, the error or residual sum of squares (SSE) is __________.

ANSWER: 0.50 TYPE: FI DIFFICULTY: Moderate KEYWORDS: sum of squares 30. Referring to Table 12-3, the coefficient of determination is __________.

ANSWER: 0.962 TYPE: FI DIFFICULTY: Moderate KEYWORDS: coefficient of determination 31. Referring to Table 12-3, the standard error of estimate is __________.

ANSWER: 0.50 TYPE: FI DIFFICULTY: Moderate KEYWORDS: standard error of estimate 32. Referring to Table 12-3, the coefficient of correlation is __________.

ANSWER: 0.981 TYPE: FI DIFFICULTY: Moderate KEYWORDS: coefficient of correlation

.

Simple Linear Regression 401

33. Referring to Table 12-3, suppose the director of cooperative education wants to obtain a 95% confidence-interval estimate for the mean number of job offers received by people who have had exactly one cooperative education job. The t critical value she would use is ________.

ANSWER: 4.3027 TYPE: FI DIFFICULTY: Easy KEYWORDS: confidence interval, estimation of mean values, critical value 34. Referring to Table 12-3, suppose the director of cooperative education wants to obtain a 95% confidence interval estimate for the mean number of job offers received by people who have had exactly one cooperative education job. The confidence interval is from ________ to ________.

ANSWER: 2.424 to 4.576 TYPE: FI DIFFICULTY: Moderate KEYWORDS: confidence interval, estimation of mean values, critical value 35. Referring to Table 12-3, suppose the director of cooperative education wants to obtain a 95% prediction interval estimate for the number of job offers received by people who have had exactly one cooperative education job. The prediction interval is from ________ to ________.

ANSWER: 1.0947 to 5.9053 TYPE: FI DIFFICULTY: Moderate KEYWORDS: prediction interval, prediction of individual values 36. True or False: Referring to Table 12-3, suppose the director of cooperative education wants to obtain two 95% confidence interval estimates. One is for the mean number of job offers received by people who have had exactly one cooperative education job and one for people who have had two. The confidence interval for people who have had one cooperative education job would be the wider of the two intervals.

ANSWER: False TYPE: TF DIFFICULTY: Moderate KEYWORDS: confidence interval, estimation of mean values 37. Referring to Table 12-3, suppose the director of cooperative education wants to obtain a 95% prediction interval for the number of job offers received by a person who has had exactly two cooperative education jobs. The t critical value she would use is ________.

ANSWER: 4.3027 TYPE: FI DIFFICULTY: Moderate KEYWORDS: prediction interval, prediction of individual values, critical value

.

402

Simple Linear Regression

38. Referring to Table 12-3, suppose the director of cooperative education wants to obtain a 95% prediction interval for the number of job offers received by a person who has had exactly two cooperative education jobs. The prediction interval is from ________ to ________.

ANSWER: 3.154 to 8.846 TYPE: FI DIFFICULTY: Moderate KEYWORDS: prediction interval, prediction of individual values 39. True or False: Referring to Table 12-3, suppose the director of cooperative education wants to obtain both a 95% confidence interval estimate and a 95% prediction interval for X = 2. The confidence interval estimate would be the wider of the two intervals.

ANSWER: False TYPE: TF DIFFICULTY: Moderate KEYWORDS: confidence interval, prediction interval, estimation of mean values, prediction of individual values, properties 40. Referring to Table 12-3, the director of cooperative education wanted to test the hypothesis that the true slope was equal to 0. The denominator of the test statistic is s b1 . The value of s b1 in this sample is ________.

ANSWER: 0.3536 TYPE: FI DIFFICULTY: Moderate KEYWORDS: t test on slope, standard error, slope 41. Referring to Table 12-3, the director of cooperative education wanted to test the hypothesis that the true slope was equal to 0. The value of the test statistic is ________.

ANSWER: 7.07 TYPE: FI DIFFICULTY: Moderate KEYWORDS: t test on slope, test statistic, slope 42. Referring to Table 12-3, the director of cooperative education wanted to test the hypothesis that the true slope was equal to 3.0. The value of the test statistic is ________.

ANSWER: -1.4142 TYPE: FI DIFFICULTY: Moderate KEYWORDS: t test on slope, test statistic, slope

43. Referring to Table 12-3, the director of cooperative education wanted to test the hypothesis that the true slope was equal to 0. For a test with a level of significance of 0.05, the null hypothesis should be rejected if the value of the test statistic is ________. .

Simple Linear Regression 403

ANSWER: > 4.3027 or < -4.3027 TYPE: FI DIFFICULTY: Easy KEYWORDS: t test on slope, critical value, slope, decision 44. Referring to Table 12-3, the director of cooperative education wanted to test the hypothesis that the true slope was equal to 3.0. For a test with a level of significance of 0.05, the null hypothesis should be rejected if the value of the test statistic is ________.

ANSWER: > 4.3027 or < -4.3027 TYPE: FI DIFFICULTY: Easy KEYWORDS: t test on slope, critical value, slope, decision 45. Referring to Table 12-3, the director of cooperative education wanted to test the hypothesis that the true slope was equal to 0. The p-value of the test is between ________ and ________.

ANSWER: 0.01 and 0.02 using the Table, or 0.0194 using Excel TYPE: FI DIFFICULTY: Moderate KEYWORDS: t test on slope, p-value, slope 46. Referring to Table 12-3, the director of cooperative education wanted to test the hypothesis that the true slope was equal to 3.0. The p-value of the test is between ________ and ________.

ANSWER: 0.2 and 0.5 using the Table, or 0.2929 using Excel TYPE: FI DIFFICULTY: Difficult EXPLANATION: The t-test statistic is t =

( b1 − 1 ) = ( 2.5 − 3) = −1.4142

KEYWORDS: t test on slope, p-value, slope

TABLE 12-4

.

Sb1

0.3536

Simple Linear Regression

The managers of a brokerage firm are interested in finding out if the number of new clients a broker brings into the firm affects the sales generated by the broker. They sample 12 brokers and determine the number of new clients they have enrolled in the last year and their sales amounts in thousands of dollars. These data are presented in the table that follows.

Broker Clients Sales 1 27 52 2 11 37 3 42 64 4 33 55 5 15 29 6 15 34 7 25 58 8 36 59 9 28 44 10 30 48 11 17 31 12 22 38 47. Referring to Table 12-4, set up a scatter diagram.

ANSWER:

Scatter Diagram 70 60 50

Sales

404

40 30 20 10 0 0

5

10

15

20

25

30

35

40

Clients TYPE: PR DIFFICULTY: Moderate KEYWORDS: scatter plot

48. Referring to Table 12-4, the least squares estimate of the slope is __________. .

45

Simple Linear Regression 405

ANSWER: 1.12 TYPE: FI DIFFICULTY: Moderate KEYWORDS: slope, estimation, least squares 49. Referring to Table 12-4, the least squares estimate of the Y-intercept is __________.

ANSWER: 17.7 TYPE: FI DIFFICULTY: Moderate KEYWORDS: intercept, estimation, least squares 50. Referring to Table 12-4, the prediction for the amount of sales (in $1,000s) for a person who brings 25 new clients into the firm is ________.

ANSWER: 45.66 TYPE: FI DIFFICULTY: Moderate KEYWORDS: prediction of individual values 51. Referring to Table 12-4, the total sum of squares (SST) is __________.

ANSWER: 1,564.2 TYPE: FI DIFFICULTY: Moderate KEYWORDS: sum of squares 52. Referring to Table 12-4, the regression sum of squares (SSR) is __________.

ANSWER: 1,227.4 TYPE: FI DIFFICULTY: Moderate KEYWORDS: sum of squares 53. Referring to Table 12-4, the error or residual sum of squares (SSE) is __________.

ANSWER: 336.9 TYPE: FI DIFFICULTY: Moderate KEYWORDS: sum of squares 54. Referring to Table 12-4, the coefficient of determination is __________.

ANSWER: 0.785 TYPE: FI DIFFICULTY: Moderate KEYWORDS: coefficient of determination

55. Referring to Table 12-4, ______% of the total variation in sales generated can be explained by the number of new clients brought in.

.

406

Simple Linear Regression

ANSWER: 78.5 TYPE: FI DIFFICULTY: Moderate KEYWORDS: coefficient of determination, interpretation 56. Referring to Table 12-4, the standard error of the estimated slope coefficient is __________.

ANSWER: 0.185 TYPE: FI DIFFICULTY: Moderate KEYWORDS: slope, standard error 57. Referring to Table 12-4, the standard error of estimate is __________.

ANSWER: 5.804 TYPE: FI DIFFICULTY: Moderate KEYWORDS: standard error of estimate 58. Referring to Table 12-4, the coefficient of correlation is __________.

ANSWER: 0.886 TYPE: FI DIFFICULTY: Moderate KEYWORDS: coefficient of correlation 59. Referring to Table 12-4, suppose the managers of the brokerage firm want to obtain a 99% confidence interval estimate for the mean sales made by brokers who have brought into the firm 24 new clients. The t critical value they would use is ________.

ANSWER: 3.1693 TYPE: FI DIFFICULTY: Easy KEYWORDS: estimation of mean values, confidence interval, critical value 60. Referring to Table 12-4, suppose the managers of the brokerage firm want to obtain a 99% confidence interval estimate for the mean sales made by brokers who have brought into the firm 24 new clients. The confidence interval is from ________ to ________.

ANSWER: 39.19 to 49.89 TYPE: FI DIFFICULTY: Moderate KEYWORDS: estimation of mean values, confidence interval

61. Referring to Table 12-4, suppose the managers of the brokerage firm want to obtain a 99% prediction interval for the sales made by a broker who has brought into the firm 18 new clients.

The t critical value they would use is ________.

ANSWER: .

Simple Linear Regression 407

3.1693 TYPE: FI DIFFICULTY: Moderate KEYWORDS: prediction of individual values, prediction interval, critical value 62. Referring to Table 12-4, suppose the managers of the brokerage firm want to obtain a 99% prediction interval for the sales made by a broker who has brought into the firm 18 new clients.

The prediction interval is from ________ to ________.

ANSWER: 18.23 to 57.42 TYPE: FI DIFFICULTY: Moderate KEYWORDS: prediction of individual values, prediction interval 63. Referring to Table 12-4, suppose the managers of the brokerage firm want to obtain both a 99% confidence interval estimate and a 99% prediction interval for X = 24. The confidence interval estimate would be the __________ (wider or narrower) of the two intervals.

ANSWER: narrower TYPE: FI DIFFICULTY: Moderate KEYWORDS: prediction of individual values, prediction interval, estimation of mean values, confidence interval, properties 64. Referring to Table 12-4, the managers of the brokerage firm wanted to test the hypothesis that the true slope was equal to 0. The denominator of the test statistic is s b1 . The value of s b1 in this sample is ________.

ANSWER: 0.1853 TYPE: FI DIFFICULTY: Moderate KEYWORDS: t test on slope, slope, standard error 65. Referring to Table 12-4, the managers of the brokerage firm wanted to test the hypothesis that the true slope was equal to 0. The value of the test statistic is _______.

ANSWER: 6.04 TYPE: FI DIFFICULTY: Moderate KEYWORDS: t test on slope, slope, test statistic

66. Referring to Table 12-4, the managers of the brokerage firm wanted to test the hypothesis that the number of new clients brought in did not affect the amount of sales generated. The value of the test statistic is _______.

ANSWER: 6.04 TYPE: FI DIFFICULTY: Moderate KEYWORDS: t test on slope, slope, test statistic .

408

Simple Linear Regression

67. Referring to Table 12-4, the managers of the brokerage firm wanted to test the hypothesis that the true slope was equal to 0. For a test with a level of significance of 0.01, the null hypothesis should be rejected if the value of the test statistic is ________.

ANSWER: > 3.1693 or < -3.1693 TYPE: FI DIFFICULTY: Moderate KEYWORDS: t test on slope, slope, critical value, decision 68. Referring to Table 12-4, the managers of the brokerage firm wanted to test the hypothesis that the true slope was equal to 0. The p-value of the test is ________.

ANSWER: < 0.01 using the Table, or 0.000126 using Excel TYPE: FI DIFFICULTY: Moderate KEYWORDS: t test on slope, slope, p-value 69. Referring to Table 12-4, the managers of the brokerage firm wanted to test the hypothesis that the true slope was equal to 0. At a level of significance of 0.01, the null hypothesis should be _______ (accepted or rejected).

ANSWER: rejected TYPE: FI DIFFICULTY: Moderate KEYWORDS: t test on slope, slope, decision 70. Referring to Table 12-4, the managers of the brokerage firm wanted to test the hypothesis that the true slope was equal to 0. At a level of significance of 0.01, the decision that should be made implies that _____ (there is or there is no) linear dependent relation between the independent and dependent variables.

ANSWER: there is TYPE: FI DIFFICULTY: Moderate KEYWORDS: t test on slope, slope, conclusion

71. Referring to Table 12-4, the managers of the brokerage firm wanted to test the hypothesis that the number of new clients brought in had a positive impact on the amount of sales generated. The value of the test statistic is _______.

ANSWER: 6.04 TYPE: FI DIFFICULTY: Moderate KEYWORDS: t test on slope, slope, test statistic

.

Simple Linear Regression 409

72. Referring to Table 12-4, the managers of the brokerage firm wanted to test the hypothesis that the number of new clients brought in had a positive impact on the amount of sales generated. For a test with a level of significance of 0.01, the null hypothesis should be rejected if the value of the test statistic is ________.

ANSWER: > 2.7638 TYPE: FI DIFFICULTY: Moderate KEYWORDS: t test on slope, slope, critical value, decision 73. Referring to Table 12-4, the managers of the brokerage firm wanted to test the hypothesis that the number of new clients brought in had a positive impact on the amount of sales generated. The pvalue of the test is ________.

ANSWER: < 0.005 using the Table, or 0.000126/2 using Excel TYPE: FI DIFFICULTY: Moderate KEYWORDS: t test on slope, slope, p-value 74. Referring to Table 12-4, the managers of the brokerage firm wanted to test the hypothesis that the number of new clients brought in had a positive impact on the amount of sales generated. At a level of significance of 0.01, the null hypothesis should be _______ (accepted or rejected).

ANSWER: rejected TYPE: FI DIFFICULTY: Moderate KEYWORDS: t test on slope, slope, decision 75. Referring to Table 12-4, the managers of the brokerage firm wanted to test the hypothesis that the number of new clients brought in had a positive impact on the amount of sales generated. At a level of significance of 0.01, the decision that should be made implies that the number of new clients brought in _____ (had or did not have) a positive impact on the amount of sales generated.

ANSWER: had TYPE: FI DIFFICULTY: Moderate KEYWORDS: t test on slope, slope, conclusion

TABLE 12-5 The managing partner of an advertising agency believes that his company's sales are related to the industry sales. He uses Microsoft Excel’s Data Analysis tool to analyze the last 4 years of quarterly data (i.e., n = 16) with the following results: Regression Statistics Multiple R 0.802 R Square 0.643 .

410

Simple Linear Regression

Adjusted R Square 0.618 Standard Error SYX0.9224 Observations 16 ANOVA df Regression 1 Error 14 Total 15 Predictor Coef Intercept 3.962 Industry 0.040451

SS 21.497 11.912 33.409 StdError 1.440 0.008048

MS 21.497 0.851

F 25.27

t Stat 2.75 5.03

Durbin-Watson Statistic

Sig.F 0.000

P-value 0.016 0.000

1.59

76. Referring to Table 12-5, the value of the quantity that the least squares regression line minimizes is ________.

ANSWER: 11.912 TYPE: FI DIFFICULTY: Moderate KEYWORDS: least squares, properties, sum of squares 77. Referring to Table 12-5, the estimates of the Y-intercept and slope are ________ and ________, respectively.

ANSWER: 3.962 and 0.040451 TYPE: FI DIFFICULTY: Easy KEYWORDS: intercept, slope, estimation 78. Referring to Table 12-5, the prediction for a quarter in which X = 120 is Y = ________.

ANSWER: 8.816 TYPE: FI DIFFICULTY: Moderate KEYWORDS: prediction of individual values

79. Referring to Table 12-5, the standard error of the estimate is ________.

ANSWER: 0.9224 TYPE: FI DIFFICULTY: Easy KEYWORDS: standard error of estimate 80. Referring to Table 12-5, the coefficient of determination is ________.

ANSWER: .

Simple Linear Regression 411

0.643 TYPE: FI DIFFICULTY: Easy KEYWORDS: coefficient of determination 81. Referring to Table 12-5, the standard error of the estimated slope coefficient is ________.

ANSWER: 0.008 TYPE: FI DIFFICULTY: Easy KEYWORDS: slope, standard error 82. Referring to Table 12-5, the correlation coefficient is ________.

ANSWER: 0.802 TYPE: FI DIFFICULTY: Easy KEYWORDS: coefficient of correlation 83. Referring to Table 12-5, the partner wants to test for autocorrelation using the Durbin-Watson statistic. Using a level of significance of 0.05, the critical values of the test are dL = ________, and dU = ________.

ANSWER: 1.10; 1.37 TYPE: FI DIFFICULTY: Easy KEYWORDS: Durbin-Watson statistic, critical value, autocorrelation 84. Referring to Table 12-5, the partner wants to test for autocorrelation using the Durbin-Watson statistic. Using a level of significance of 0.05, the decision he should make is: a) there is evidence of autocorrelation. b) the test is unable to make a definite conclusion. c) there is no evidence of autocorrelation. d) there is not enough information to perform the test.

ANSWER: c TYPE: MC DIFFICULTY: Moderate KEYWORDS: Durbin-Watson statistic, decision, autocorrelation

85. If the Durbin-Watson statistic has a value close to 0, which assumption is violated? a) Normality of the errors. b) Independence of errors. c) Homoscedasticity. d) None of the above.

ANSWER: b TYPE: MC Difficulty: Easy KEYWORDS: Durbin-Watson statistic, autocorrelation, assumption .

412

Simple Linear Regression

86. If the Durbin-Watson statistic has a value close to 4, which assumption is violated? a) Normality of the errors. b) Independence of errors. c) Homoscedasticity. d) None of the above.

ANSWER: b TYPE: MC Difficulty: Easy KEYWORDS: Durbin-Watson statistic, assumption, autocorrelation 87. The standard error of the estimate is a measure of a) total variation of the Y variable. b) the variation around the sample regression line. c) explained variation. d) the variation of the X variable.

ANSWER: b TYPE: MC DIFFICULTY: Easy KEYWORDS: standard error of estimate, interpretation 88. The coefficient of determination (r2) tells us a) that the coefficient of correlation (r) is larger than 1. b) whether r has any significance. c) that we should not partition the total variation. d) the proportion of total variation that is explained.

ANSWER: d TYPE: MC DIFFICULTY: Easy KEYWORDS: coefficient of determination, interpretation

TABLE 12-6 The following EXCEL tables are obtained when "Score received on an exam (measured in percentage points)" (Y) is regressed on "percentage attendance" (X) for 22 students in a Statistics for Business and Economics course.

Regression Statistics Multiple R 0.142620229 R Square 0.02034053 Adjusted R Square -0.028642444 .

Simple Linear Regression 413

Standard Error Observations

20.25979924 22 Coefficients Standard Error 39.39027309 37.24347659 0.340583573 0.52852452

Intercept Attendance

T Stat 1.057642216 0.644404489

P-value 0.302826622 0.526635689

89. Referring to Table 12-6, which of the following statements is true? a) -2.86% of the total variability in score received can be explained by percentage attendance. b) -2.86% of the total variability in percentage attendance can be explained by score received. c) 2% of the total variability in score received can be explained by percentage attendance. d) 2% of the total variability in percentage attendance can be explained by score received.

ANSWER: c TYPE: MC DIFFICULTY: Easy KEYWORDS: coefficient of determination, interpretation 90. Referring to Table 12-6, which of the following statements is true? a) If attendance increases by 0.341%, the estimated average score received will increase by 1 percentage point. b) If attendance increases by 1%, the estimated average score received will increase by 39.39 percentage points. c) If attendance increases by 1%, the estimated average score received will increase by 0.341 percentage points. d) If the score received increases by 39.39%, the estimated average attendance will go up by 1%.

ANSWER: c TYPE: MC DIFFICULTY: Easy KEYWORDS: slope, interpretation

91. True or False: The Regression Sum of Squares (SSR) can never be greater than the Total Sum of Squares (SST).

ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: sum of squares, properties 92. True or False: The coefficient of determination represents the ratio of SSR to SST. ANSWER: .

414

Simple Linear Regression

True TYPE: TF DIFFICULTY: Easy KEYWORDS: coefficient of determination, properties 93. True or False: Regression analysis is used for prediction, while correlation analysis is used to measure the strength of the association between two numerical variables.

ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: coefficient of correlation, slope, interpretation 94. True or False: The value of r is always positive.

ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: coefficient of correlation, properties 95. In performing a regression analysis involving two numerical variables, we are assuming a) the variances of X and Y are equal. b) the variation around the line of regression is the same for each X value. c) that X and Y are independent. d) All of the above.

ANSWER: b TYPE: MC DIFFICULTY: Moderate KEYWORDS: assumption

96. Which of the following assumptions concerning the probability distribution of the random error term is stated incorrectly? a) The distribution is normal. b) The mean of the distribution is 0. c) The variance of the distribution increases as X increases. d) The errors are independent.

ANSWER: c TYPE: MC DIFFICULTY: Moderate KEYWORDS: assumption

.

Simple Linear Regression 415

97. Based on the residual plot below, you will conclude that there might be a violation of which of the following assumptions.

Footage Residual Plot 6000

Residuals

4000 2000 0 0

1,000 2,000 3,000 4,000 5,000 6,000

-2000 -4000 Footage

a) b) c) d)

Linearity of the relationship Normality of errors Homoscedasticity Independence of errors

ANSWER: c TYPE: MC DIFFICULTY: Moderate KEYWORDS: assumption, homoscedasticity 98. True or False: Data that exhibit an autocorrelation effect violate the regression assumption of independence.

ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: assumption, autocorrelation

99. True or False: The Durbin-Watson D statistic is used to check the assumption of normality.

ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: Durbin-Watson statistic, assumption, autocorrelation 100. If the residuals in a regression analysis of time ordered data are not correlated, the value of the Durbin-Watson D statistic should be near __________.

ANSWER: 2 .

416

Simple Linear Regression

TYPE: FI DIFFICULTY: Moderate KEYWORDS: Durbin-Watson statistic, autocorrelation, properties 101. The residuals represent a) the difference between the actual Y values and the mean of Y. b) the difference between the actual Y values and the predicted Y values. c) the square root of the slope. d) the predicted value of Y for the average X value.

ANSWER: b TYPE: MC DIFFICULTY: Easy KEYWORDS: residual, properties 102. If the plot of the residuals is fan shaped, which assumption is violated? a) Normality. b) Homoscedasticity. c) Independence of errors. d) No assumptions are violated, the graph should resemble a fan.

ANSWER: b TYPE: MC DIFFICULTY: Moderate KEYWORDS: residual plot, homoscedasticity 103. What do we mean when we say that a simple linear regression model is “statistically” useful? a) All the statistics computed from the sample make sense. b) The model is an excellent predictor of Y. c) The model is “practically” useful for predicting Y. d) The model is a better predictor of Y than the sample mean, Y .

ANSWER: d TYPE: MC DIFFICULTY: Difficult KEYWORDS: slope, interpretation

104. If the correlation coefficient (r) = 1.00, then a) the Y-intercept (b0) must equal 0. b) the explained variation equals the unexplained variation. c) there is no unexplained variation. d) there is no explained variation.

ANSWER: c TYPE: MC DIFFICULTY: Moderate KEYWORDS: coefficient of correlation, properties, interpretation 105. If the correlation coefficient (r) = 1.00, then .

Simple Linear Regression 417

a) b) c) d)

all the data points must fall exactly on a straight line with a slope that equals 1.00. all the data points must fall exactly on a straight line with a negative slope. all the data points must fall exactly on a straight line with a positive slope. all the data points must fall exactly on a horizontal straight line with a zero slope.

ANSWER: c TYPE: MC DIFFICULTY: - Moderate KEYWORDS: coefficient of correlation, properties 106. Assuming a linear relationship between X and Y, if the coefficient of correlation (r) equals – 0.30, a) there is no correlation. b) the slope (b1) is negative. c) variable X is larger than variable Y. d) the variance of X is negative.

ANSWER: b TYPE: MC DIFFICULTY: Moderate KEYWORDS: coefficient of correlation, properties, interpretation 107. Testing for the existence of correlation is equivalent to a) testing for the existence of the slope (  ).

b) testing for the existence of the Y-intercept (   ). c) the confidence interval estimate for predicting Y. d) None of the above.

ANSWER: a TYPE: MC DIFFICULTY: Moderate KEYWORDS: coefficient of correlation, t test for correlation coefficient, t test on slope

108. The strength of the linear relationship between two numerical variables may be measured by the a) scatter diagram. b) coefficient of correlation. c) slope. d) Y-intercept.

ANSWER: b TYPE: MC DIFFICULTY: Easy KEYWORDS: coefficient of correlation, interpretation 109. In a simple linear regression problem, r and b1 .

418

Simple Linear Regression

a) b) c) d)

may have opposite signs. must have the same sign. must have opposite signs. are equal.

ANSWER: b TYPE: MC DIFFICULTY: Easy KEYWORDS: coefficient of correlation, slope, properties 110. The sample correlation coefficient between X and Y is 0.375. It has been found out that the pvalue is 0.256 when testing H 0 :  = 0 against the two-sided alternative H1 :   0 . To test

H0 :  = 0 against the one-sided alternative H1 :   0 at a significance level of 0.2, the p-value is a) 0.256 / 2 b) ( 0.256 ) 2 c) 1 − 0.256 d) 1 − 0.256 / 2 ANSWER: d TYPE: MC DIFFICULTY: Difficult KEYWORDS: coefficient of correlation, t test for correlation coefficient, p-value

111. The sample correlation coefficient between X and Y is 0.375. It has been found out that the pvalue is 0.256 when testing H 0 :  = 0 against the two-sided alternative H1 :   0 . To test

H0 :  = 0 against the one-sided alternative H1 :   0 at a significance level of 0.2, the p-value is a) 0.256 / 2 b) ( 0.256 ) 2 c) 1 − 0.256 d) 1 − 0.256 / 2 ANSWER: a TYPE: MC DIFFICULTY: Difficult KEYWORDS: coefficient of correlation, t test for correlation coefficient, p-value .

Simple Linear Regression 419

112. The sample correlation coefficient between X and Y is 0.375. It has been found out that the pvalue is 0.256 when testing H 0 :  = 0 against the one-sided alternative H1 :   0 . To test

H0 :  = 0 against the two-sided alternative H1 :   0 at a significance level of 0.2, the p-value is a) 0.256 / 2 b) (0.256)(2) c) 1 − 0.256 d) 1 − 0.256 / 2 ANSWER: b TYPE: MC DIFFICULTY: Difficult KEYWORDS: coefficient of correlation, t test for correlation coefficient, p-value 113. The sample correlation coefficient between X and Y is 0.375. It has been found out that the pvalue is 0.744 when testing H 0 :  = 0 against the one-sided alternative H1 :   0 . To test

H0 :  = 0 against the two-sided alternative H1 :   0 at a significance level of 0.2, the p-value is a) 0.744 / 2 b) (0.744)(2) c) 1 − 0.744 d) (1 − 0.744)(2) ANSWER: d TYPE: MC DIFFICULTY: Difficult KEYWORDS: coefficient of correlation, t test for correlation coefficient, p-value

114. If you wanted to find out if alcohol consumption (measured in fluid oz.) and grade point average on a 4-point scale are linearly related, you would perform a a.  2 test for the difference in two proportions. b.  2 test for independence. c. a Z test for the difference in two proportions. d. a t test for a correlation coefficient.

ANSWER: d TYPE: MC DIFFICULTY: Difficult KEYWORDS: coefficient of correlation, t test for correlation coefficient 115. True or False: When r = – 1, it indicates a perfect relationship between X and Y. ANSWER: .

420

Simple Linear Regression

True TYPE: TF DIFFICULTY: Easy KEYWORDS: coefficient of correlation, properties

TABLE 12-7 An investment specialist claims that if one holds a portfolio that moves in the opposite direction to the market index like the S&P 500, then it is possible to reduce the variability of the portfolio's return. In other words, one can create a portfolio with positive returns but less exposure to risk.

A sample of 26 years of S&P 500 index and a portfolio consisting of stocks of private prisons, which are believed to be negatively related to the S&P 500 index, is collected. A regression analysis was performed by regressing the returns of the prison stocks portfolio (Y) on the returns of S&P 500 index (X) to prove that the prison stocks portfolio is negatively related to the S&P 500 index at a 5% level of significance. The results are given in the following EXCEL output.

Coefficients Standard Error 4.866004258 0.35743609 -0.502513506 0.071597152

Intercept S&P

T Stat 13.61363441 -7.01862425

P-value 8.7932E-13 2.94942E-07

116. Referring to Table 12-7, to test whether the prison stocks portfolio is negatively related to the S&P 500 index, the appropriate null and alternative hypotheses are, respectively, a) H 0 :   0 vs. H1 :   0 b) H 0 :   0 vs. H1 :   0

H 0 : r  0 vs. H1 : r  0 d) H 0 : r  0 vs. H1 : r  0 c)

ANSWER: a TYPE: MC DIFFICULTY: Easy KEYWORDS: coefficient of correlation, t test for correlation coefficient, form of hypothesis

117. Referring to Table 12-7, to test whether the prison stocks portfolio is negatively related to the S&P 500 index, the measured value of the test statistic is a) -7.019 b) -0.503 c) 0.072 d) 0.357 ANSWER: a TYPE: MC DIFFICULTY: Easy KEYWORDS: coefficient of correlation, t test for correlation coefficient, test statistic 118. Referring to Table 12-7, to test whether the prison stocks portfolio is negatively related to the S&P 500 index, the p-value of the associated test statistic is a) 2.94942E-07 b) 2.94942E-07 / 2 .

Simple Linear Regression 421

c)

( 2.94942E-07 ) 2

d) 8.7932E-13 ANSWER: b TYPE: MC DIFFICULTY: Easy KEYWORDS: coefficient of correlation, t test for correlation coefficient, p-value 119. Referring to Table 12-7, which of the following will be a correct conclusion? a) We cannot reject the null hypothesis and, therefore, conclude that there is sufficient evidence to show that the prisons stock portfolio and S&P 500 index are negatively related. b) We can reject the null hypothesis and, therefore, conclude that there is sufficient evidence to show that the prisons stock portfolio and S&P 500 index are negatively related. c) We cannot reject the null hypothesis and, therefore, conclude that there is not sufficient evidence to show that the prisons stock portfolio and S&P 500 index are negatively related. d) We can reject the null hypothesis and conclude that there is not sufficient evidence to show that the prisons stock portfolio and S&P 500 index are negatively related.

ANSWER: b TYPE: MC DIFFICULTY: Easy KEYWORDS: coefficient of correlation, t test for correlation coefficient, decision, conclusion

TABLE 12-8 It is believed that GPA (grade point average, based on a four point scale) should have a positive linear relationship with ACT scores. Given below is the Excel output from regressing GPA on ACT scores using a data set of 8 randomly chosen students from a Big-Ten university.

Regressing GPA on ACT Regression Statistics Multiple R 0.7598 R Square 0.5774 Adjusted R Square 0.5069 Standard Error 0.2691 Observations 8 ANOVA .

422

Simple Linear Regression

df Regression Residual Total

SS 1 6 7

0.5940 0.4347 1.0287

Coefficients Standard Error 0.5681 0.9284 0.1021 0.0356

Intercept ACT

MS 0.5940 0.0724

F Significance F 8.1986 0.0286

t Stat P-value 0.6119 0.5630 2.8633 0.0286

Lower 95% Upper 95% -1.7036 2.8398 0.0148 0.1895

120. Referring to Table 12-8, the interpretation of the coefficient of determination in this regression is a) 57.74% of the total variation of ACT scores can be explained by GPA. b) ACT scores account for 57.74% of the total fluctuation in GPA. c) GPA accounts for 57.74% of the variability of ACT scores. d) None of the above.

ANSWER: b TYPE: MC DIFFICULTY: Easy KEYWORDS: coefficient of determination, interpretation 121. Referring to Table 12-8, the value of the measured test statistic to test whether there is any linear relationship between GPA and ACT is a) 0.0356 b) 0.1021 c) 0.7598 d) 2.8633 ANSWER: d TYPE: MC DIFFICULTY: Easy KEYWORDS: coefficient of correlation, t test for correlation coefficient, test statistic

122. Referring to Table 12-8, what is the predicted average value of GPA when ACT = 20? a. 2.61 b. 2.66 c. 2.80 d. 3.12 ANSWER: a TYPE: MC DIFFICULTY: Easy KEYWORDS: prediction of individual values 123. Referring to Table 12-8, what are the decision and conclusion on testing whether there is any linear relationship at 1% level of significance between GPA and ACT scores? a) Do not reject the null hypothesis; hence there is not sufficient evidence to show that ACT scores and GPA are linearly related.

.

Simple Linear Regression 423

b) Reject the null hypothesis; hence there is not sufficient evidence to show that ACT scores and GPA are linearly related. c) Do not reject the null hypothesis; hence there is sufficient evidence to show that ACT scores and GPA are linearly related. d) Reject the null hypothesis; hence there is sufficient evidence to show that ACT scores and GPA are linearly related.

ANSWER: a TYPE: MC DIFFICULTY: Easy KEYWORDS: coefficient of correlation, t test for correlation coefficient, decision, conclusion 124. Referring to Table 12-8, the value of the measured (observed) test statistic of the F-test for

H0 :  = 0 vs. H1 :   0 a) b) c) d)

may be negative. is always positive. is always negative. has the same sign as the corresponding t test statistic.

ANSWER: b TYPE: MC DIFFICULTY: Easy KEYWORDS: F test on slope, test statistic

TABLE 12-9 It is believed that, the average numbers of hours spent studying per day (HOURS) during undergraduate education should have a positive linear relationship with the starting salary (SALARY, measured in thousands of dollars per month) after graduation. Given below is the Excel output from regressing starting salary on number of hours spent studying per day for a sample of 51 students.

NOTE: Some of the numbers in the output are purposely erased.

Regression Statistics Multiple R R Square Adjusted R Square Standard Error Observations

.

0.8857 0.7845 0.7801 1.3704 51

424

Simple Linear Regression

ANOVA df Regression Residual Total

Intercept Hours

1

SS 335.0472

50

427.0798

Coefficients -1.8940 0.9795

Standard Error 0.4018 0.0733

MS 335.0473 1.8782

F Significance F 178.3859

t Stat -4.7134 13.3561

P-value 2.051E-05 5.944E-18

Lower 95% Upper 95% -2.7015 -1.0865 0.8321 1.1269

125. Referring to Table 12-9, the estimated average change in salary (in thousands of dollars) as a result of spending an extra hour per day studying is a. -1.8940 b. 0.7845 c. 0.9795 d. 335.0473 ANSWER: c TYPE: MC DIFFICULTY: Moderate KEYWORDS: slope, interpretation 126. Referring to Table 12-9, the value of the measured t-test statistic to test whether average SALARY depends linearly on HOURS is a) -4.7134 b) -1.8940 c) 0.9795 d) 13.3561 ANSWER: d TYPE: MC DIFFICULTY: Moderate KEYWORDS: t test on slope, slope, test statistic 127. Referring to Table 12-9, the p-value of the measured F-test statistic to test whether HOURS affects SALARY is a) (5.944E-18)/2 b) 5.944E-18 c) (2.051E-05)/2 d) 2.051E-05 ANSWER: b TYPE: MC DIFFICULTY: Moderate KEYWORDS: F test on slope, slope, test statistic 128. Referring to Table 12-9, the degrees of freedom for the F test on whether HOURS affects SALARY are a) 1, 49 b) 1, 50 .

Simple Linear Regression 425

c) 49, 1 d) 50, 1 ANSWER: a TYPE: MC DIFFICULTY: Moderate KEYWORDS: F test on slope, slope, degrees of freedom 129. Referring to Table 12-9, the error sum of squares (SSE) of the above regression is a) 1.878215 b) 92.0325465 c) 335.047257 d) 427.079804 ANSWER: b TYPE: MC DIFFICULTY: Moderate KEYWORDS: sum of squares 130. Referring to Table 12-9, the 90% confidence interval for the average change in SALARY (in thousands of dollars) as a result of spending an extra hour per day studying is a) wider than [-2.70159, -1.08654]. b) narrower than [-2.70159, -1.08654]. c) wider than [0.8321927, 1.12697]. d) narrower than [0.8321927, 1.12697].

ANSWER: d TYPE: MC DIFFICULTY: Moderate KEYWORDS: slope, interpretation, confidence interval

131. Referring to Table 12-9, to test the claim that average SALARY depends positively on HOURS against the null hypothesis that average SALARY does not depend linearly on HOURS, the pvalue of the test statistic is a) (5.944E-18)/2 b) 5.944E-18 c) (2.051E-05)/2 d) 2.051E-05 ANSWER: a TYPE: MC DIFFICULTY: Moderate KEYWORDS: t test on slope, slope, p-value 132. True or False: A zero population correlation coefficient between a pair of random variables means that there is no linear relationship between the random variables.

ANSWER: True .

426

Simple Linear Regression

TYPE: TF DIFFICULTY: Easy KEYWORDS: coefficient of correlation, interpretation 133. True or False: You give a pre-employment examination to your applicants. The test is scored from 1 to 100. You have data on their sales at the end of one year measured in dollars. You want to know if there is any linear relationship between pre-employment examination score and sales.

An appropriate test to use is the t test on the population correlation coefficient.

ANSWER: True TYPE: MC DIFFICULTY: Moderate KEYWORDS: coefficient of correlation, t test for correlation coefficient 134. The width of the prediction interval for the predicted value of Y is dependent on a) the standard error of the estimate. b) the value of X for which the prediction is being made. c) the sample size. d) All of the above.

ANSWER: d TYPE: MC DIFFICULTY: Moderate KEYWORDS: prediction of individual values, prediction interval, properties 135. True or False: The confidence interval for the mean of Y is always narrower than the prediction interval for an individual response Y given the same data set, X value, and confidence level.

ANSWER: True TYPE: TF DIFFICULTY: Moderate KEYWORDS: estimation of mean values, confidence interval, prediction of individual values, prediction interval, properties

TABLE 12-10 The management of a chain electronic store would like to develop a model for predicting the weekly sales (in thousand of dollars) for individual stores based on the number of customers who made purchases. A random sample of 12 stores yields the following results:

Customers Sales (Thousands of Dollars) 907 11.20 926 11.05 713 8.21 741 9.21 780 9.42 898 10.08 510 6.73 529 7.02 460 6.12 872 9.52 .

Simple Linear Regression 427

650 603

7.53 7.25

136. Referring to Table 12-10, generate the scatter plot.

ANSWER: Scatter Diagram

Salses (Thousands of Dollars)

12 10 8 6 4 2 0 0

200

400

600

800

1000

Customers

TYPE: PR DIFFICULTY: Moderate KEYWORDS: scatter plot 137. Referring to Table 12-10, what are the values of the estimated intercept and slope?

ANSWER: 1.4464, 0.0100 TYPE: PR DIFFICULTY: Moderate KEYWORDS: intercept, slope, estimation

138. Referring to Table 12-10, what is the value of the coefficient of determination?

ANSWER: 0.9453 TYPE: PR DIFFICULTY: Moderate KEYWORDS: coefficient of determination 139. Referring to Table 12-10, what is the value of the coefficient of correlation?

ANSWER: 0.9723 TYPE: PR DIFFICULTY: Moderate KEYWORDS: coefficient of correlation 140. Referring to Table 12-10, what is the value of the standard error of the estimate?

ANSWER: 0.4191 TYPE: PR DIFFICULTY: Moderate .

428

Simple Linear Regression

KEYWORDS: standard error of estimate 141. Referring to Table 12-10, which is the correct null hypothesis for testing whether the number of customers who make purchase affects weekly sales? a) H 0 : 0 = 0 b) H 0 : 1 = 0

H0 :  = 0 d) H 0 :  = 0 c)

ANSWER: b TYPE: MC DIFFICULTY: Easy KEYWORDS: t test on slope, slope, form of hypothesis 142. Referring to Table 12-10, what is the value of the t test statistic when testing whether the number of customers who make purchase affects weekly sales?

ANSWER: 13.1464 TYPE: PR DIFFICULTY: Moderate KEYWORDS: t test on slope, slope, test statistic 143. Referring to Table 12-10, what are the degrees of freedom of the t test statistic when testing whether the number of customers who make purchase affects weekly sales?

ANSWER: 10 TYPE: PR DIFFICULTY: Easy KEYWORDS: t test on slope, slope, degrees of freedom

144. Referring to Table 12-10, what is the p-value of the t test statistic when testing whether the number of customers who make purchase affects weekly sales?

ANSWER: 1.23323E-07 or smaller than 0.01 TYPE: PR DIFFICULTY: Moderate KEYWORDS: t test on slope, slope, p-value 145. True or False: Referring to Table 12-10, the null hypothesis for testing whether the number of customers who make purchase effects weekly sales cannot be rejected if 1% probability of committing a type I error is desired.

ANSWER: False TYPE: TF DIFFICULTY: Moderate KEYWORDS: t test on slope, slope, decision 146. True or False: Referring to Table 12-10, the average weekly sales will increase by an estimated $0.01 for each additional purchasing customer. .

Simple Linear Regression 429

ANSWER: False TYPE: TF DIFFICULTY: Moderate KEYWORDS: t test on slope, slope, interpretation 147. True or False: Referring to Table 12-10, the average weekly sales will increase by an estimated $10 for each additional purchasing customer.

ANSWER: True TYPE: TF DIFFICULTY: Moderate KEYWORDS: t test on slope, slope, interpretation 148. True or False: Referring to Table 12-10, 93.98% of the total variation in weekly sales can be explained by the variation in the number of customers who make purchases.

ANSWER: False TYPE: TF DIFFICULTY: Moderate KEYWORDS: coefficient of determination, interpretation 149. Referring to Table 12-10, what are the degrees of freedom of the F test statistic when testing whether the number of customers who make purchases is a good predictor for weekly sales?

ANSWER: 1 in the numerator and 10 in the denominator TYPE: PR DIFFICULTY: Easy KEYWORDS: F test on slope, slope, test statistic

150. Referring to Table 12-10, what is the value of the F test statistic when testing whether the number of customers who make purchases is a good predictor for weekly sales?

ANSWER: 172.8265 TYPE: PR DIFFICULTY: Moderate KEYWORDS: F test on slope, slope, test statistic 151. Referring to Table 12-10, what is the p-value of the F test statistic when testing whether the number of customers who make purchases is a good predictor for weekly sales?

ANSWER: 1.23323E-07 or smaller than 0.01 TYPE: PR DIFFICULTY: Moderate KEYWORDS: F test on slope, slope, p-value 152. True or False: Referring to Table 12-10, the p-value of the t test and F test should be the same when testing whether the number of customers who make purchases is a good predictor for weekly sales.

.

430

Simple Linear Regression

ANSWER: True TYPE: TF DIFFICULTY: Moderate KEYWORDS: t test on slope, F test on slope, p-value 153. True or False: Referring to Table 12-10, the value of the t test statistic and F test statistic should be the same when testing whether the number of customers who make purchases is a good predictor for weekly sales.

ANSWER: False TYPE: TF DIFFICULTY: Moderate KEYWORDS: t test on slope, F test on slope, test statistic 154. True or False: Referring to Table 12-10, the value of the F test statistic equals the square of the t test statistic when testing whether the number of customers who make purchases is a good predictor for weekly sales.

ANSWER: True TYPE: TF DIFFICULTY: Moderate KEYWORDS: t test on slope, F test on slope, test statistic

155. Referring to Table 12-10, generate the residual plot.

ANSWER: Customers Residual Plot 0.8 0.6

Residuals

0.4 0.2 0 -0.2 -0.4 -0.6 -0.8 0

200

400

600

Customers

TYPE: PR DIFFICULTY: Moderate .

800

1000

Simple Linear Regression 431

KEYWORDS: residual plot 156. Referring to Table 12-10, the residual plot indicates possible violation of which assumptions? a) Linearity of the relationship b) Homoscedasticity c) Autocorrelation d) Normality ANSWER: b TYPE: MC DIFFICULTY: Moderate KEYWORDS: homoscedasticity, residual plot, assumption 157. True or False: Referring to Table 12-10, it is inappropriate to compute the Durbin-Watson statistic and test for autocorrelation in this case.

ANSWER: True TYPE: TF DIFFICULTY: Moderate KEYWORDS: Durbin-Watson statistic, autocorrelation 158. Referring to Table 12-10, construct a 95% confidence interval for the change in average weekly sales when the number of customers who make purchases increases by one.

ANSWER: 0.0083 thousands of dollars to 0.0117 thousands of dollars TYPE: PR DIFFICULTY: Difficult KEYWORDS: slope, confidence interval, interpretation

159. Referring to Table 12-10, construct a 95% confidence interval for the average weekly sales when the number of customers who make purchases is 600.

ANSWER: 7.1194 to 7.7864 thousands of dollars TYPE: PR DIFFICULTY: Difficult KEYWORDS: estimate of mean values, confidence interval 160. Referring to Table 12-10, construct a 95% prediction interval for the weekly sales of a store that has 600 purchasing customers.

ANSWER: 6.4614 to 8.4444 thousands of dollars TYPE: PR DIFFICULTY: Difficult KEYWORDS: prediction of individual values, prediction interval TABLE 12-11 A company that has the distribution rights to home video sales of previously released movies would like to use the box office gross (in millions of dollars) to estimate the number of units (in thousands of units) that it can expect to sell. Following is the output from a simple linear regression along with the residual plot and normal probability plot obtained from a data set of 30 different movie titles: .

Simple Linear Regression

Regression Statistics Multiple R 0.8531 R Square 0.7278 Adjusted R 0.7180 Square Standard 47.8668 Error Observations 30 ANOVA df

SS

MS

1 28 29

171499.78 64154.42 235654.20

171499.78 2291.23

Coefficients

Standard Error 11.8318 0.5008

Regression Residual Total

Intercept Gross

76.5351 4.3331

F 74.8505

Significance F 2.1259E-09

t Stat

P-value

Lower 95%

6.4686 8.6516

5.24E-07 2.13E-09

Gross Residual Plot 150

Residuals

100 50 0 -50 -100 0

10

20

30

40

50

60

70

Gross

Normal Probability Plot

Residuals

432

120 100 80 60 40 20 0 -20-2.5 -40 -60 -80 -100

-2

-1.5

-1

-0.5

0

Z Value

.

0.5

1

1.5

2

2.5

52.2987 3.3072

Upper 95% 100.7716 5.3590

Simple Linear Regression 433

161. Referring to Table 12-11, which of the following is the correct interpretation for the slope coefficient? a) For each increase of 1 million dollars in box office gross, expected home video units sold are estimated to increase by 4.3331 units. b) For each increase of 1 million dollars in box office gross, expected home video units sold are estimated to increase by 4.3331 thousand units. c) For each increase of 1 dollar in box office gross, expected home video units sold are estimated to increase by 4.3331 units. d) For each increase of 1 dollar in box office gross, expected home video units sold are estimated to increase by 4.3331 thousand units.

ANSWER: b TYPE: MC DIFFICULTY: Easy KEYWORDS: slope, interpretation 162. Referring to Table 12-11, predict the video unit sales for a movie that had a box office gross of $30 millions.

ANSWER: 206.53 thousand units TYPE: PR DIFFICULTY: Easy KEYWORDS: prediction of individual values

.

434

Simple Linear Regression

163. Referring to Table 12-11, which of the following is the correct interpretation for the coefficient of determination? a) 71.8% of the variation in the video unit sales can be explained by the variation in the box office gross. b) 72.8% of the variation in the video unit sales can be explained by the variation in the box office gross. c) 71.8% of the variation in the box office gross can be explained by the variation in the video unit sales. d) 72.8% of the variation in the box office gross can be explained by the variation in the video unit sales.

ANSWER: b TYPE: MC DIFFICULTY: Easy KEYWORDS: interpretation, coefficient of determination 164. Referring to Table 12-11, what is the standard error of estimate?

ANSWER: 47.8668 thousand units TYPE: PR DIFFICULTY: Moderate KEYWORDS: standard error of estimate 165. Referring to Table 12-11, what is the standard deviation around the regression line?

ANSWER: 47.8668 thousand units TYPE: PR DIFFICULTY: Difficult KEYWORDS: interpretation, standard error of estimate 166. Referring to Table 12-11, which of the following assumptions appears to have been violated? a) Normality of error b) Homoscedasticity c) Independence of errors d) None of the above ANSWER: d TYPE: MC DIFFICULTY: Moderate KEYWORDS: interpretation, assumption, residual plot, homoscedasticity 167. True or False: Referring to Table 12-11, the normality of error assumption appears to have been violated.

ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: interpretation, assumption

.

Simple Linear Regression 435

168. True or False: Referring to Table 12-11, the homoscedasticity of error assumption appears to have been violated.

ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: interpretation, assumption, homoscedasticity 169. True or False: Referring to Table 12-11, there appears to be autocorrelation in the residuals.

ANSWER: False TYPE: TF DIFFICULTY: Moderate KEYWORDS: interpretation, autocorrelation 170. True or False: Referring to Table 12-11, the Durbin-Watson statistic is inappropriate for this data set.

ANSWER: True TYPE: TF DIFFICULTY: Moderate KEYWORDS: Durbin-Watson statistic, assumption 171. True or False: Referring to Table 12-11, the null hypothesis for testing whether there is a linear relationship between box office gross and home video unit sales is “There is no linear relationship between box office gross and home video unit sales”.

ANSWER: True TYPE: TF DIFFICULTY: Moderate KEYWORDS: slope, t test for slope, form of hypothesis 172. Referring to Table 12-11, which of the following is the correct null hypothesis for testing whether there is a linear relationship between box office gross and home video unit sales? a) H 0 : b1 = 0

H 0 : b1  0 c) H 0 : 1 = 0 d) H 0 : 1  0 b)

ANSWER: c TYPE: MC DIFFICULTY: Easy KEYWORDS: slope, t test for slope, form of hypothesis

.

436

Simple Linear Regression

173. Referring to Table 12-11, which of the following is the correct alternative hypothesis for testing whether there is a linear relationship between box office gross and home video unit sales? a) H1 : b1 = 0 b) H1 : b1  0

H1 : 1 = 0 d) H1 : 1  0 c)

ANSWER: d TYPE: MC DIFFICULTY: Easy KEYWORDS: slope, t test for slope, form of hypothesis 174. Referring to Table 12-11, what is the value of the test statistic for testing whether there is a linear relationship between box office gross and home video unit sales?

ANSWER: 8.65 TYPE: PR DIFFICULTY: Easy KEYWORDS: test statistic, slope, t test for slope 175. Referring to Table 12-11, what is the critical value for testing whether there is a linear relationship between box office gross and home video unit sales at a 5% level of significance?

ANSWER: 2.0484 TYPE: PR DIFFICULTY: Easy KEYWORDS: critical value, slope, t test for slope 176. Referring to Table 12-11, what is the p-value for testing whether there is a linear relationship between box office gross and home video unit sales at a 5% level of significance?

ANSWER: virtually zero or 2.13E-09 TYPE: PR DIFFICULTY: Easy KEYWORDS: p-value , slope, t test for slope 177. True or False: Referring to Table 12-11, the null hypothesis that there is no linear relationship between box office gross and home video unit sales should be reject at a 5% level of significance.

ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: decision, slope, t test for slope

178. True or False: Referring to Table 12-11, there is sufficient evidence that box office gross and home video unit sales are linearly related at a 5% level of significance. .

Simple Linear Regression 437

ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: conclusion, slope, t test for slope 179. Referring to Table 12-11, what are, respectively, the lower and upper limits of the 95% confidence interval estimate for population slope?

ANSWER: 3.3072 and 5.3590 TYPE: PR DIFFICULTY: Easy KEYWORDS: slope, confidence interval 180. Referring to Table 12-11, what are, respectively, the lower and upper limits of the 95% confidence interval estimate for the average change in video unit sales as a result of a one million dollars increase in box office?

ANSWER: 3.3072 and 5.3590 thousand units TYPE: PR DIFFICULTY: Difficult KEYWORDS: slope, confidence interval

TABLE 12-12

.

Simple Linear Regression

The manager of the purchasing department of a large banking organization would like to develop a model to predict the amount of time (measured in hours) it takes to process invoices. Data are collected from a sample of 30 days, and the number of invoices processed and completion time in hours is recorded. Below is the regression output: Regression Statistics Multiple R 0.9447 R Square 0.8924 Adjusted R 0.8886 Square Standard 0.3342 Error Observations 30 ANOVA df

SS

MS

F

1 28 29

25.9438 3.1282 29.072

25.9438 0.1117

232.2200

Significance F 4.3946E-15

Coefficients

Standard Error 0.1236 0.0008

t Stat

P-value

Lower 95%

Upper 95%

3.2559 15.2388

0.0030 4.3946E-15

0.1492 0.0109

0.6555 0.0143

Regression Residual Total

Intercept Invoices Processed

0.4024 0.0126

Invoices Processed Residual Plot 0.8 0.6 0.4

Residuals

438

0.2 0 -0.2 -0.4 -0.6 -0.8 0

50

100

150

200

Invoices Processed

.

250

300

350

Simple Linear Regression 439

181. Referring to Table 12-12, the estimated average amount of time it takes to process one additional invoice is a) 0.4024 fewer hours b) 0.4024 more hours c) 0.0126 fewer hours d) 0.0126 more hours ANSWER: d TYPE: MC DIFFICULTY: Easy KEYWORDS: slope, interpretation 182. Referring to Table 12-12, the value of the measured t-test statistic to test whether the amount of time depends linearly on the number of invoices processed is a) 0.8924 b) 3.2559 c) 15.2388 d) 232.2200 ANSWER: c TYPE: MC DIFFICULTY: Easy KEYWORDS: t test on slope, slope, test statistic 183. Referring to Table 12-12, the p-value of the measured F-test statistic to test whether the number of invoices proceesed affects the amount of time is a) (4.3946E-15)/2 b) 4.3946E-15 c) (0.0030)/2 d) 0.0030 ANSWER: b TYPE: MC DIFFICULTY: Moderate KEYWORDS: F test on slope, slope, test statistic

.

440

Simple Linear Regression

184. Referring to Table 12-12, the p-value of the measured t-test statistic to test whether the number of invoices processed affects the amount of time is a) (4.3946E-15)/2 b) 4.3946E-15 c) (0.0030)/2 d) 0.0030 ANSWER: b TYPE: MC DIFFICULTY: Moderate KEYWORDS: t test on slope, slope, test statistic 185. Referring to Table 12-12, the degrees of freedom for the F test on whether the number of invoices processed affects the amount of time are a) 1, 28 b) 1, 29 c) 28, 1 d) 29, 1 ANSWER: a TYPE: MC DIFFICULTY: Moderate KEYWORDS: F test on slope, slope, degrees of freedom 186. Referring to Table 12-12, the degrees of freedom for the t test on whether the number of invoices processed affects the amount of time are a) 1 b) 28 c) 29 d) 30 ANSWER: b TYPE: MC DIFFICULTY: Moderate KEYWORDS: t test on slope, slope, degrees of freedom 187. Referring to Table 12-12, the error sum of squares (SSE) of the above regression is a) 0.1117 b) 3.1282 c) 25.9438 d) 29.0720 ANSWER: b TYPE: MC DIFFICULTY: Moderate KEYWORDS: sum of squares

.

Simple Linear Regression 441

188. Referring to Table 12-12, the 90% confidence interval for the average change in the amount of time needed as a result of processing one additional invoice is a) wider than [0.1492, 0.6555]. b) narrower than [0.1492, 0.6555]. c) wider than [0.0109, 0.0143]. d) narrower than [0.0109, 0.0143].

ANSWER: d TYPE: MC DIFFICULTY: Moderate KEYWORDS: slope, interpretation, confidence interval 189. True or False: Referring to Table 12-12, you can be 95% confident that the average amount of time needed to process one additional invoice is somewhere between 0.0109 and 0.0143 hours.

ANSWER: True TYPE: TF DIFFICULTY: Moderate KEYWORDS: confidence interval, slope, interpretation 190. True or False: Referring to Table 12-12, there is a 95% probability that the average amount of time needed to process one additional invoice is somewhere between 0.0109 and 0.0143 hours.

ANSWER: False TYPE: TF DIFFICULTY: Difficult KEYWORDS: confidence interval, slope, interpretation 191. True or False: Referring to Table 12-12, there is sufficient evidence that the amount of time needed linearly depends on the number of invoices processed at a 5% level of significance.

ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: conclusion, slope, t test for slope 192. True or False: Referring to Table 12-12, there is sufficient evidence that the amount of time needed linearly depends on the number of invoices processed at a 1% level of significance.

ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: conclusion, slope, t test for slope

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Simple Linear Regression

193. Referring to Table 12-12, to test the claim that the average amount of time depends positively on the number of invoices processed against the null hypothesis that the average amount of time does not depend linearly on the number of invoices processed, the p-value of the test statistic is a) (4.3946E-15)/2 b) 4.3946E-15 c) (4.3946E-15)*2 d) 0.0030 ANSWER: a TYPE: MC DIFFICULTY: Difficult KEYWORDS: t test on slope, slope, p-value 194. Referring to Table 12-12, predict the amount of time it would take to process 150 invoices.

ANSWER: 2.2924 hours TYPE: PR DIFFICULTY: Easy KEYWORDS: prediction of individual values 195. Referring to Table 12-12, what percentage of the variation in the amount of time needed can be explained by the variation in the number of invoices processed?

ANSWER: 89.24% TYPE: PR DIFFICULTY: Moderate KEYWORDS: coefficient of determination, interpretation 196. True or False: Referring to Table 12-12, the model appears to be adequate based on the residual analyses.

ANSWER: True TYPE: TF DIFFICULTY: Moderate KEYWORDS: residual plot, homoscedasticity, autocorrelation 197. Referring to Table 12-12, what are the critical values of the Durbin-Watson test statistic using the 5% level of significance to test for evidence of positive autocorrelation?

ANSWER: 1.35 and 1.49 TYPE: PR DIFFICULTY: Easy KEYWORDS: autocorrelation, Durbin-Watson statistic, critical value 198. True or False: Referring to Table 12-12, there is no evidence of positive autocorrelation if the Durbin-Watson test statistic is found to be 1.78.

ANSWER: True TYPE: TF DIFFICULTY: Moderate KEYWORDS: autocorrelation, Durbin-Watson statistic, decision

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Multiple Regression

443

CHAPTER 13: MULTIPLE REGRESSION 1. In a multiple regression problem involving two independent variables, if b1 is computed to be +2.0, it means that a) the relationship between X1 and Y is significant. b) the estimated average of Y increases by 2 units for each increase of 1 unit of X1, holding X2 constant. c) the estimated average of Y increases by 2 units for each increase of 1 unit of X1, without regard to X2. d) the estimated average of Y is 2 when X1 equals zero.

ANSWER: b TYPE: MC DIFFICULTY: Easy KEYWORDS: slope, interpretation 2. The coefficient of multiple determination r2Y.12 a) measures the variation around the predicted regression equation. b) measures the proportion of variation in Y that is explained by X1 and X2. c) measures the proportion of variation in Y that is explained by X1 holding X2 constant. d) will have the same sign as b1.

ANSWER: b TYPE: MC DIFFICULTY: Easy KEYWORDS: coefficient of multiple determination, interpretation 3. In a multiple regression model, the value of the coefficient of multiple determination a) has to fall between -1 and +1. b) has to fall between 0 and +1. c) has to fall between -1 and 0. d) can fall between any pair of real numbers.

ANSWER: b TYPE: MC DIFFICULTY: Easy KEYWORDS: coefficient of multiple determination, properties 4. In a multiple regression model, which of the following is correct regarding the value of the adjusted r 2 ? a) It can be negative. b) It has to be positive. c) It has to be larger than the coefficient of multiple determination. d) It can be larger than 1.

ANSWER: a TYPE: MC DIFFICULTY: Moderate KEYWORDS: adjusted r-square, properties .

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Multiple Regression

TABLE 13-1 A manager of a product sales group believes the number of sales made by an employee (Y) depends on how many years that employee has been with the company (X1) and how he/she scored on a business aptitude test (X2). A random sample of 8 employees provides the following: Employee Y X1 X2 1 100 10 7 2 90 3 10 3 80 8 9 4 70 5 4 5 60 5 8 6 50 7 5 7 40 1 4 8 30 1 1 5. Referring to Table 13-1, for these data, what is the value for the regression constant, b0? a) 0.998 b) 3.103 c) 4.698 d) 21.293 ANSWER: d TYPE: MC DIFFICULTY: Moderate KEYWORDS: intercept, estimation 6. Referring to Table 13-1, for these data, what is the estimated coefficient for the variable representing years an employee has been with the company, b1? a) 0.998 b) 3.103 c) 4.698 d) 21.293 ANSWER: b TYPE: MC DIFFICULTY: Moderate KEYWORDS: slope, estimation 7. Referring to Table 13-1, for these data, what is the estimated coefficient for the variable representing scores on the aptitude test, b2? a) 0.998 b) 3.103 c) 4.698 d) 21.293 ANSWER: c TYPE: MC DIFFICULTY: Moderate KEYWORDS: slope, estimation

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Multiple Regression

445

8. Referring to Table 13-1, if an employee who had been with the company 5 years scored a 9 on the aptitude test, what would his estimated expected sales be? a) 79.09 b) 60.88 c) 55.62 d) 17.98 ANSWER: a TYPE: MC DIFFICULTY: Moderate KEYWORDS: estimation of mean values

TABLE 13-2 A professor of industrial relations believes that an individual’s wage rate at a factory (Y) depends on his performance rating (X1) and the number of economics courses the employee successfully completed in college (X2). The professor randomly selects 6 workers and collects the following information: Employee Y ($) X1 X2 1 10 3 0 2 12 1 5 3 15 8 1 4 17 5 8 5 20 7 12 6 25 10 9 9. Referring to Table 13-2, for these data, what is the value for the regression constant, b0? a) 0.616 b) 1.054 c) 6.932 d) 9.103 ANSWER: c TYPE: MC DIFFICULTY: Moderate KEYWORDS: intercept, estimation 10. Referring to Table 13-2, for these data, what is the estimated coefficient for performance rating, b1? a) 0.616 b) 1.054 c) 6.932 d) 9.103 ANSWER: b TYPE: MC DIFFICULTY: Moderate KEYWORDS: slope, estimation

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446

Multiple Regression

11. Referring to Table 13-2, for these data, what is the estimated coefficient for the number of economics courses taken, b2? a) 0.616 b) 1.054 c) 6.932 d) 9.103 ANSWER: a TYPE: MC DIFFICULTY: Moderate KEYWORDS: slope, estimation 12. Referring to Table 13-2, suppose an employee had never taken an economics course and managed to score a 5 on his performance rating. What is his estimated expected wage rate? a) 10.90 b) 12.20 c) 17.23 d) 25.11 ANSWER: b TYPE: MC DIFFICULTY: Moderate KEYWORDS: estimation of mean values 13. Referring to Table 13-2, an employee who took 12 economics courses scores 10 on the performance rating. What is her estimated expected wage rate? a) 10.90 b) 12.20 c) 24.87 d) 25.70 ANSWER: c TYPE: MC DIFFICULTY: Moderate KEYWORDS: estimation of mean values 14. The variation attributable to factors other than the relationship between the independent variables and the explained variable in a regression analysis is represented by a) regression sum of squares. b) error sum of squares. c) total sum of squares. d) regression mean squares.

ANSWER: b TYPE: MC DIFFICULTY: Moderate KEYWORDS: sum of squares

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Multiple Regression

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TABLE 13-3 An economist is interested to see how consumption for an economy (in $ billions) is influenced by gross domestic product ($ billions) and aggregate price (consumer price index). The Microsoft Excel output of this regression is partially reproduced below.

SUMMARY OUTPUT Regression Statistics Multiple R 0.991 R Square 0.982 Adjusted R Square 0.976 Standard Error 0.299 Observations 10 ANOVA Regression Residual Total

Intercept GDP Price

df 2 7 9 Coeff – 0.0861 0.7654 – 0.0006

SS 33.4163 0.6277 34.0440

MS 16.7082 0.0897

StdError 0.5674 0.0574 0.0028

t Stat – 0.152 13.340 – 0.219

F 186.325

Signif F 0.0001

P-value 0.8837 0.0001 0.8330

15. Referring to Table 13-3, when the economist used a simple linear regression model with consumption as the dependent variable and GDP as the independent variable, he obtained an r2 value of 0.971. What additional percentage of the total variation of consumption has been explained by including aggregate prices in the multiple regression? a) 98.2 b) 11.1 c) 2.8 d) 1.1 ANSWER: d TYPE: MC DIFFICULTY: Moderate KEYWORDS: coefficient of multiple determination 16. Referring to Table 13-3, the p-value for GDP is a) 0.05 b) 0.01 c) 0.001 d) None of the above.

ANSWER: d TYPE: MC DIFFICULTY: Easy KEYWORDS: t test on slope, p-value 17. Referring to Table 13-3, the p-value for the aggregated price index is .

448

Multiple Regression

a) b) c) d)

0.05 0.01 0.001 None of the above.

ANSWER: d TYPE: MC DIFFICULTY: Easy KEYWORDS: t test on slope, p-value 18. Referring to Table 13-3, the p-value for the regression model as a whole is a) 0.05 b) 0.01 c) 0.001 d) None of the above.

ANSWER: d TYPE: MC DIFFICULTY: Easy KEYWORDS: F test for the entire regression, p-value 19. Referring to Table 13-3, what is the predicted consumption level for an economy with GDP equal to $4 billion and an aggregate price index of 150? a) $1.39 billion b) $2.89 billion c) $4.75 billion d) $9.45 billion ANSWER: b TYPE: MC DIFFICULTY: Moderate KEYWORDS: prediction of individual values 20. Referring to Table 13-3, what is the estimated average consumption level for an economy with GDP equal to $4 billion and an aggregate price index of 150? a) $1.39 billion b) $2.89 billion c) $4.75 billion d) $9.45 billion ANSWER: b TYPE: MC DIFFICULTY: Moderate KEYWORDS: estimation of mean values

21. Referring to Table 13-3, what is the estimated average consumption level for an economy with GDP equal to $2 billion and an aggregate price index of 90? .

Multiple Regression

a) b) c) d)

449

$1.39 billion $2.89 billion $4.75 billion $9.45 billion

ANSWER: a TYPE: MC DIFFICULTY: Moderate KEYWORDS: estimation of mean values 22. Referring to Table 13-3, one economy in the sample had an aggregate consumption level of $3 billion, a GDP of $3.5 billion, and an aggregate price level of 125. What is the residual for this data point? a) $2.52 billion b) $0.48 billion c) – $1.33 billion d) – $2.52 billion ANSWER: b TYPE: MC DIFFICULTY: Moderate KEYWORDS: residual 23. Referring to Table 13-3, one economy in the sample had an aggregate consumption level of $4 billion, a GDP of $6 billion, and an aggregate price level of 200. What is the residual for this data point? a) $4.39 billion b) $0.39 billion c) – $0.39 billion d) – $1.33 billion ANSWER: c TYPE: MC DIFFICULTY: Moderate KEYWORDS: residual 24. Referring to Table 13-3, to test for the significance of the coefficient on aggregate price index, the value of the relevant t-statistic is a) 2.365 b) 0.143 c) – 0.219 d) – 1.960 ANSWER: c TYPE: MC DIFFICULTY: Easy KEYWORDS: t test on slope, test statistic

25. Referring to Table 13-3, to test for the significance of the coefficient on aggregate price index, the p-value is a) 0.0001 .

450

Multiple Regression

b) 0.8330 c) 0.8837 d) 0.9999 ANSWER: b TYPE: MC DIFFICULTY: Easy KEYWORDS: t test on slope, p-value 26. Referring to Table 13-3, to test for the significance of the coefficient on gross domestic product, the p-value is a) 0.0001 b) 0.8330 c) 0.8837 d) 0.9999 ANSWER: a TYPE: MC DIFFICULTY: Easy KEYWORDS: t test on slope, p-value 27. Referring to Table 13-3, to test whether aggregate price index has a negative impact on consumption, the p-value is _______? a) 0.0001 b) 0.4165 c) 0.8330 d) 0.8837 ANSWER: b TYPE: MC DIFFICULTY: Moderate KEYWORDS: t test on slope, p-value 28. Referring to Table 13-3, to test whether aggregate price index has a positive impact on consumption, the p-value is a) 0.0001 b) 0.4165 c) 0.5835 d) 0.8330 ANSWER: c TYPE: MC DIFFICULTY: Moderate KEYWORDS: t test on slope, p-value

29. Referring to Table 13-3, to test whether gross domestic product has a positive impact on consumption, the p-value is a) 0.00005 b) 0.0001 .

Multiple Regression

451

c) 0.9999 d) 0.99995 ANSWER: a TYPE: MC DIFFICULTY: Easy KEYWORDS: t test on slope, p-value

TABLE 13-4 A real estate builder wishes to determine how house size (House) is influenced by family income (Income), family size (Size), and education of the head of household (School). House size is measured in hundreds of square feet, income is measured in thousands of dollars, and education is in years. The builder randomly selected 50 families and ran the multiple regression. Microsoft Excel output is provided below: SUMMARY OUTPUT Regression Statistics Multiple R 0.865 R Square 0.748 Adjusted R Square 0.726 Standard Error 5.195 Observations 50 ANOVA df Regression Residual Total

49

Intercept Income Size School

Coeff – 1.6335 0.4485 4.2615 – 0.6517

SS 3605.7736 1214.2264 4820.0000 StdError 5.8078 0.1137 0.8062 0.4319

MS 1201.9245 26.3962

t Stat – 0.281 3.9545 5.286 – 1.509

F

Signif F 0.0000

P-value 0.7798 0.0003 0.0001 0.1383

30. Referring to Table 13-4, what fraction of the variability in house size is explained by income, size of family, and education? a) 27.0% b) 33.4% c) 74.8% d) 86.5% .

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ANSWER: c TYPE: MC DIFFICULTY: Easy KEYWORDS: coefficient of multiple determination, interpretation 31. Referring to Table 13-4, which of the independent variables in the model are significant at the 2% level? a) Income, Size, School b) Income, Size c) Size, School d) Income, School ANSWER: b TYPE: MC DIFFICULTY: Easy KEYWORDS: t test on slope, critical value, p-value, decision, conclusion 32. Referring to Table 13-4, when the builder used a simple linear regression model with house size (House) as the dependent variable and education (School) as the independent variable, he obtained an r2 value of 23.0%. What additional percentage of the total variation in house size has been explained by including family size and income in the multiple regression? a) 2.8% b) 51.8% c) 72.6% d) 74.8% ANSWER: b TYPE: MC DIFFICULTY: Moderate KEYWORDS: coefficient of multiple determination 33. Referring to Table 13-4, which of the following values for the level of significance is the smallest for which every explanatory variable is significant individually? a) 0.01 b) 0.025 c) 0.05 d) 0.15 ANSWER: d TYPE: MC DIFFICULTY: Moderate KEYWORDS: p-value, interpretation

34. Referring to Table 13-4, which of the following values for the level of significance is the smallest for which at least two explanatory variables are significant individually? a) 0.01 b) 0.025 c) 0.05 d) 0.15

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Multiple Regression

453

ANSWER: a TYPE: MC DIFFICULTY: Moderate KEYWORDS: p-value, interpretation 35. Referring to Table 13-4, which of the following values for the level of significance is the smallest for which the regression model as a whole is significant? a) 0.0005 b) 0.001 c) 0.01 d) 0.05 ANSWER: a TYPE: MC DIFFICULTY: Moderate KEYWORDS: p-value, interpretation 36. Referring to Table 13-4, what is the predicted house size (in hundreds of square feet) for an individual earning an annual income of $40,000, having a family size of 4, and going to school a total of 13 years? a) 11.43 b) 15.15 c) 24.88 d) 53.87 ANSWER: c TYPE: MC DIFFICULTY: Moderate KEYWORDS: prediction of individual values 37. Referring to Table 13-4, what minimum annual income would an individual with a family size of 4 and 16 years of education need to attain a predicted 10,000 square foot home (House = 100)? a) $44.14 thousand b) $56.75 thousand c) $178.33 thousand d) $211.85 thousand ANSWER: d TYPE: MC DIFFICULTY: Moderate KEYWORDS: prediction of individual values

38. Referring to Table 13-4, what minimum annual income would an individual with a family size of 9 and 10 years of education need to attain a predicted 5,000 square foot home (House = 50)? a) $44.14 thousand b) $56.75 thousand c) $178.33 thousand d) $211.85 thousand ANSWER: .

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a TYPE: MC DIFFICULTY: Moderate KEYWORDS: prediction of individual values 39. Referring to Table 13-4, one individual in the sample had an annual income of $100,000, a family size of 10, and an education of 16 years. This individual owned a home with an area of 7,000 square feet (House = 70.00). What is the residual (in hundreds of square feet) for this data point? a) 7.40 b) 2.52 c) – 2.52 d) – 5.40 ANSWER: d TYPE: MC DIFFICULTY: Moderate KEYWORDS: residual 40. Referring to Table 13-4, one individual in the sample had an annual income of $10,000, a family size of 1, and an education of 8 years. This individual owned a home with an area of 1,000 square feet (House = 10.00). What is the residual (in hundreds of square feet) for this data point? a) 8.10 b) 5.40 c) – 5.40 d) – 8.10 ANSWER: a TYPE: MC DIFFICULTY: Moderate KEYWORDS: residual 41. Referring to Table 13-4, suppose the builder wants to test whether the coefficient on Income is significantly different from 0. What is the value of the relevant t-statistic? a) 5.286 b) 5.195 c) 3.945 d) – 1.509 ANSWER: c TYPE: MC DIFFICULTY: Easy KEYWORDS: t test on slope, test statistic

42. Referring to Table 13-4, at the 0.01 level of significance, what conclusion should the builder draw regarding the inclusion of Income in the regression model? a) Income is significant in explaining house size and should be included in the model because its p-value is less than 0.01. b) Income is significant in explaining house size and should be included in the model because its p-value is more than 0.01. c) Income is not significant in explaining house size and should not be included in the model because its p-value is less than 0.01.

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d) Income is not significant in explaining house size and should not be included in the model because its p-value is more than 0.01.

ANSWER: a TYPE: MC DIFFICULTY: Moderate KEYWORDS: t test on slope, decision, conclusion 43. Referring to Table 13-4, suppose the builder wants to test whether the coefficient on School is significantly different from 0. What is the value of the relevant t-statistic? a) 5.286 b) 5.195 c) 3.945 d) – 1.509 ANSWER: d TYPE: MC DIFFICULTY: Easy KEYWORDS: t test on slope, test statistic 44. Referring to Table 13-4, what is the value of the calculated F test statistic that is missing from the output for testing whether the whole regression model is significant? a) 0.0001 b) 0.0299 c) 0.726 d) 45.5340 ANSWER: d TYPE: MC DIFFICULTY: Moderate KEYWORDS: F test on the entire regression, test statistic 45. Referring to Table 13-4, the observed value of the F-statistic is missing from the printout. What are the degrees of freedom for this F-statistic? a) 46 for the numerator, 3 for the denominator b) 3 for the numerator, 49 for the denominator c) 46 for the numerator, 49 for the denominator d) 3 for the numerator, 46 for the denominator ANSWER: d TYPE: MC DIFFICULTY: Easy KEYWORDS: F test on the entire regression, degrees of freedom 46. Referring to Table 13-4, at the 0.01 level of significance, what conclusion should the builder draw regarding the inclusion of School in the regression model? a) School is significant in explaining house size and should be included in the model because its p-value is less than 0.01. b) School is significant in explaining house size and should be included in the model because its p-value is more than 0.01. c) School is not significant in explaining house size and should not be included in the model because its p-value is less than 0.01. .

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Multiple Regression

d) School is not significant in explaining house size and should not be included in the model because its p-value is more than 0.01.

ANSWER: d TYPE: MC DIFFICULTY: Moderate KEYWORDS: t test on slope, critical value, p-value, decision, conclusion 47. Referring to Table 13-4, what are the regression degrees of freedom that are missing from the output? a) 3 b) 46 c) 49 d) 50 ANSWER: a TYPE: MC DIFFICULTY: Easy KEYWORDS: degrees of freedom 48. Referring to Table 13-4, what are the residual degrees of freedom that are missing from the output? a) 3 b) 46 c) 49 d) 50 ANSWER: b TYPE: MC DIFFICULTY: Easy KEYWORDS: degrees of freedom

TABLE 13-5 A microeconomist wants to determine how corporate sales are influenced by capital and wage spending by companies. She proceeds to randomly select 26 large corporations and record information in millions of dollars. The Microsoft Excel output below shows results of this multiple regression.

SUMMARY OUTPUT .

Multiple Regression

457

Regression Statistics Multiple R 0.830 R Square 0.689 Adjusted R Square 0.662 Standard Error 17501.643 Observations 26 ANOVA Regression Residual Total

df 2 23 25

SS 15579777040 7045072780 22624849820

Intercept Capital Wages

Coeff 15800.0000 0.1245 7.0762

StdError 6038.2999 0.2045 1.4729

MS 7789888520 306307512

t Stat 2.617 0.609 4.804

F 25.432

Signif F 0.0001

P-value 0.0154 0.5485 0.0001

49. Referring to Table 13-5, what fraction of the variability in sales is explained by spending on capital and wages? a) 27.0% b) 50.9% c) 68.9% d) 83.0% ANSWER: c TYPE: MC DIFFICULTY: Easy KEYWORDS: coefficient of multiple determination, interpretation

50. Referring to Table 13-5, which of the independent variables in the model are significant at the 5% level? a) Capital, Wages b) Capital c) Wages d) None of the above ANSWER: c TYPE: MC DIFFICULTY: Easy .

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Multiple Regression

KEYWORDS: t test on slope, critical value, p-value, decision, conclusion 51. Referring to Table 13-5, when the microeconomist used a simple linear regression model with sales as the dependent variable and wages as the independent variable, she obtained an r2 value of 0.601. What additional percentage of the total variation of sales has been explained by including capital spending in the multiple regression? a) 60.1% b) 31.1% c) 22.9% d) 8.8% ANSWER: d TYPE: MC DIFFICULTY: Moderate KEYWORDS: coefficient of multiple determination 52. Referring to Table 13-5, what is the p-value for Wages? a) 0.01 b) 0.05 c) 0.0001 d) None of the above ANSWER: c TYPE: MC DIFFICULTY: Easy KEYWORDS: t test on slope, p-value 53. Referring to Table 13-5, what is the p-value for testing whether Wages have a positive impact on corporate sales? a) 0.01 b) 0.05 c) 0.0001 d) 0.00005 ANSWER: d TYPE: MC DIFFICULTY: Easy KEYWORDS: t test on slope, p-value

54. Referring to Table 13-5, what is the p-value for testing whether Wages have a negative impact on corporate sales? a) 0.05 b) 0.0001 c) 0.00005 d) 0.99995 ANSWER: d TYPE: MC DIFFICULTY: Moderate KEYWORDS: t test on slope, p-value .

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55. Referring to Table 13-5, what is the p-value for Capital? a) 0.01 b) 0.025 c) 0.05 d) None of the above ANSWER: d TYPE: MC DIFFICULTY: Easy KEYWORDS: t test on slope, p-value 56. Referring to Table 13-5, what is the p-value for testing whether Capital has a positive influence on corporate sales? a) 0.025 b) 0.05 c) 0.2743 d) 0.5485 ANSWER: c TYPE: MC DIFFICULTY: Easy KEYWORDS: t test on slope, p-value 57. Referring to Table 13-5, what is the p-value for testing whether Capital has a negative influence on corporate sales? a) 0.05 b) 0.2743 c) 0.5485 d) 0.7258 ANSWER: d TYPE: MC DIFFICULTY: Moderate KEYWORDS: t test on slope, p-value

58. Referring to Table 13-5, which of the following values for  is the smallest for which the regression model as a whole is significant? a) 0.00005 b) 0.001 c) 0.01 d) 0.05 ANSWER: b TYPE: MC DIFFICULTY: Moderate KEYWORDS: F test on the entire regression, p-value, interpretation

.

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Multiple Regression

59. Referring to Table 13-5, what are the predicted sales (in millions of dollars) for a company spending $100 million on capital and $100 million on wages? a) 15,800.00 b) 16,520.07 c) 17,277.49 d) 20,455.98 ANSWER: b TYPE: MC DIFFICULTY: Moderate KEYWORDS: prediction of individual values 60. Referring to Table 13-5, what are the predicted sales (in millions of dollars) for a company spending $500 million on capital and $200 million on wages? a) 15,800.00 b) 16,520.07 c) 17,277.49 d) 20,455.98 ANSWER: c TYPE: MC DIFFICULTY: Moderate KEYWORDS: prediction of individual values 61. Referring to Table 13-5, one company in the sample had sales of $20 billion (Sales = 20,000).

This company spent $300 million on capital and $700 million on wages. What is the residual (in millions of dollars) for this data point? a) 874.55 b) 622.87 c) –790.69 d) –983.56 ANSWER: c TYPE: MC DIFFICULTY: Moderate KEYWORDS: residual

62. Referring to Table 13-5, one company in the sample had sales of $21.439 billion (Sales = 21,439). This company spent $300 million on capital and $700 million on wages. What is the residual (in millions of dollars) for this data point? a) 790.69 b) 648.31 c) – 648.31 d) – 790.69 ANSWER: b TYPE: MC DIFFICULTY: Moderate KEYWORDS: residual

.

Multiple Regression

461

63. Referring to Table 13-5, suppose the microeconomist wants to test whether the coefficient on Capital is significantly different from 0. What is the value of the relevant t-statistic? a) 0.609 b) 2.617 c) 4.804 d) 25.432 ANSWER: a TYPE: MC DIFFICULTY: Easy KEYWORDS: t test on slope, test statistic 64. Referring to Table 13-5, at the 0.01 level of significance, what conclusion should the microeconomist draw regarding the inclusion of Capital in the regression model? a) Capital is significant in explaining corporate sales and should be included in the model because its p-value is less than 0.01. b) Capital is significant in explaining corporate sales and should be included in the model because its p-value is more than 0.01. c) Capital is not significant in explaining corporate sales and should not be included in the model because its p-value is less than 0.01. d) Capital is not significant in explaining corporate sales and should not be included in the model because its p-value is more than 0.01.

ANSWER: d TYPE: MC DIFFICULTY: Moderate KEYWORDS: t test on the slope, critical value, p-value, decision, conclusion

65. Referring to Table 13-5, the observed value of the F-statistic is given on the printout as 25.432.

What are the degrees of freedom for this F-statistic? a) 25 for the numerator, 2 for the denominator b) 2 for the numerator, 23 for the denominator c) 23 for the numerator, 25 for the denominator d) 2 for the numerator, 25 for the denominator ANSWER: b TYPE: MC DIFFICULTY: Easy KEYWORDS: F test on the entire regression, degrees of freedom

TABLE 13-6 .

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Multiple Regression

One of the most common questions of prospective house buyers pertains to the average cost of heating in dollars (Y). To provide its customers with information on that matter, a large real estate firm used the following 4 variables to predict heating costs: the daily minimum outside temperature in degrees of Fahrenheit ( X 1 ), the amount of insulation in inches ( X 2 ), the number of windows in the house ( X 3 ), and the age of the furnace in years ( X 4 ). Given below is the EXCEL outputs.

Regression Statistics R Square 0.8080 Adjusted R Square 0.7568 Observations 20 ANOVA df Regression 4 Residual 15 Total 19

SS 169503.4241 40262.3259 209765.75

Coefficients Standard Error 421.4277 77.8614 -4.5098 0.8129 -14.9029 5.0508 0.2151 4.8675 6.3780 4.1026

Intercept X1 (Temperature) X2 (Insulation) X3 (Windows) X4 (Furnace Age)

MS 42375.86 2684.155

F Significance F 15.7874 2.96869E-05

t Stat P-value Lower 90.0% Upper 90.0% 5.4125 7.2E-05 284.9327 557.9227 -5.5476 5.58E-05 -5.9349 -3.0847 -2.9505 0.0099 -23.7573 -6.0485 0.0442 0.9653 -8.3181 8.7484 1.5546 0.1408 -0.8140 13.5702

66. Referring to Table 13-6, the estimated value of the partial regression parameter 1 means that a) all else equal, an estimated expected $1 increase in average heating costs is associated with a decrease in the daily minimum outside temperature by 4.51 degrees. b) all else equal, a 1 degree increase in the daily minimum outside temperature results in a decrease in average heating costs by $4.51. c) all else equal, a 1 degree increase in the daily minimum outside temperature results in an estimated expected decrease in average heating costs by $4.51. d) all else equal, a 1% increase in the daily minimum outside temperature results in an estimated expected decrease in average heating costs by 4.51%.

ANSWER: c TYPE: MC DIFFICULTY: Easy KEYWORDS: slope, interpretation .

Multiple Regression

463

67. Referring to Table 13-6, what can we say about the regression model? a) The model explains 77.7% of the sample variability of average heating costs; after correcting for the degrees of freedom, the model explains 75.1% of the sample variability of average heating costs. b) The model explains 75.1% of the sample variability of average heating costs; after correcting for the degrees of freedom, the model explains 77.7% of the sample variability of average heating costs. c) The model explains 80.8% of the sample variability of average heating costs; after correcting for the degrees of freedom, the model explains 75.7% of the sample variability of average heating costs. d) The model explains 75.7% of the sample variability of average heating costs; after correcting for the degrees of freedom, the model explains 80.8% of the sample variability of average heating costs.

ANSWER: c TYPE: MC DIFFICULTY: Moderate KEYWORDS: adjusted r-square 68. Referring to Table 13-6, what is your decision and conclusion for the test H0 :  2 = 0 vs. H1 :  2  0 at the  = 0.01 level of significance? a) Do not reject H0 and conclude that the amount of insulation has a linear effect on average heating cots. b) Reject H0 and conclude that the amount of insulation does not have a linear effect on average heating costs. c) Reject H0 and conclude that the amount of insulation has a negative linear effect on average heating costs. d) Do not reject H0 and conclude that the amount of insulation has a negative linear effect on average heating costs.

ANSWER: c TYPE: MC DIFFICULTY: Easy KEYWORDS: t test on slope, critical value, p-value, decision, conclusion

69. Referring to Table 13-6, what is the 90% confidence interval for the expected change in average heating costs as a result of a 1 degree Fahrenheit change in the daily minimum outside temperature? a) [−6.58, −3.65] b) [−6.24, −2.78] c) [−5.94, −3.08] d) [−2.37, 15.12] ANSWER: c TYPE: MC DIFFICULTY: Moderate KEYWORDS: confidence interval, slope, interpretation

.

464

Multiple Regression

70. Referring to Table 13-6 and allowing for a 1% probability of committing a type I error, what is the decision and conclusion for the test H 0 : 1 =  2 = 3 =  4 = 0 vs. H1 : At least one  j  0, j = 1, 2, , 4 ? a) Do not reject H0 and conclude that the 4 independent variables have significant individual linear effects on average heating costs. b) Reject H0 and conclude that the 4 independent variables taken as a group have significant linear effects on average heating costs. c) Do not reject H0 and conclude that the 4 independent variables taken as a group do not have significant linear effects on average heating costs. d) Reject H0 and conclude that the 4 independent variables taken as a group do not have significant linear effects on average heating costs.

ANSWER: b TYPE: MC DIFFICULTY: Easy KEYWORDS: F test on the entire regression, critical value, p-value, decision, conclusion 71. True or False: The interpretation of the slope is different in a multiple linear regression model as compared to a simple linear regression model.

ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: slope, interpretation 72. True or False: The coefficient of multiple determination r2Y.12 measures the proportion of variation in Y that is explained by X1 and X2.

ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: coefficient of multiple determination

73. True or False: When an additional explanatory variable is introduced into a multiple regression model, the coefficient of multiple determination will never decrease.

ANSWER: True TYPE: TF DIFFICULTY: Moderate KEYWORDS: coefficient of multiple determination, properties 74. True or False: When an additional explanatory variable is introduced into a multiple regression model, the adjusted r 2 can never decrease.

ANSWER: False TYPE: TF DIFFICULTY: Moderate .

Multiple Regression

465

KEYWORDS: adjusted r-square, properties 75. True or False: When an explanatory variable is dropped from a multiple regression model, the coefficient of multiple determination can increase.

ANSWER: False TYPE: TF DIFFICULTY: Moderate KEYWORDS: coefficient of multiple determination, properties 76. True or False: When an explanatory variable is dropped from a multiple regression model, the adjusted r 2 can increase.

ANSWER: True TYPE: TF DIFFICULTY: Moderate KEYWORDS: adjusted r-square, properties 77. True or False: The slopes in a multiple regression model are called net regression coefficients.

ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: slope 78. True or False: In calculating the standard error of the estimate, SYX = MSE , there are n – k – 1 degrees of freedom, where n is the sample size and k represents the number of independent variables in the model.

ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: standard error of estimate

79. True or False: The total sum of squares (SST) in a regression model will never exceed the regression sum of squares (SSR).

ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: sum of squares, properties 80. True or False: The coefficient of multiple determination measures the fraction of the total variation in the dependent variable that is explained by the regression plane.

ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: coefficient of multiple determination, interpretation .

466

Multiple Regression

81. True or False: The coefficient of multiple determination is calculated by taking the ratio of the regression sum of squares over the total sum of squares (SSR/SST) and subtracting that value from 1.

ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: coefficient of multiple determination, properties 82. True or False: In a particular model, the sum of the squared residuals was 847. If the model had 5 independent variables, and the data set contained 40 points, the value of the standard error of the estimate is 24.911.

ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: sum of squares, standard error of estimate, properties 83. True or False: A multiple regression is called “multiple” because it has several data points.

ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: properties, interpretation 84. True or False: A multiple regression is called “multiple” because it has several explanatory variables.

ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: properties, interpretation

85. True or False: If we have taken into account all relevant explanatory factors, the residuals from a multiple regression should be random.

ANSWER: True TYPE: TF DIFFICULTY: Moderate KEYWORDS: residual, properties 86. True or False: Multiple regression is the process of using several independent variables to predict a number of dependent variables.

ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: properties

.

Multiple Regression

467

87. True or False: You have just run a regression in which the value of coefficient of multiple determination is 0.57. To determine if this indicates that the independent variables explain a significant portion of the variation in the dependent variable, you would perform an F-test.

ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: F test on the entire regression 88. True or False: From the coefficient of multiple determination, we cannot detect the strength of the relationship between Y and any individual independent variable.

ANSWER: True TYPE: TF DIFFICULTY: Moderate KEYWORDS: coefficient of multiple determination, interpretation 89. True or False: Consider a regression in which b2 = – 1.5 and the standard error of this coefficient equals 0.3. To determine whether X2 is a significant explanatory variable, you would compute an observed t-value of – 5.0.

ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: t test on slope, test statistic 90. True or False: A regression had the following results: SST = 82.55, SSE = 29.85. It can be said that 73.4% of the variation in the dependent variable is explained by the independent variables in the regression.

ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: coefficient multiple determination, sum of squares, properties 91. True or False: A regression had the following results: SST = 82.55, SSE = 29.85. It can be said that 63.84% of the variation in the dependent variable is explained by the independent variables in the regression.

ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: coefficient multiple determination, sum of squares, properties 92. True or False: A regression had the following results: SST = 102.55, SSE = 82.04. It can be said that 90.0% of the variation in the dependent variable is explained by the independent variables in the regression.

ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: coefficient multiple determination, sum of squares, properties .

468

Multiple Regression

93. True or False: A regression had the following results: SST = 102.55, SSE = 82.04. It can be said that 20.0% of the variation in the dependent variable is explained by the independent variables in the regression.

ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: coefficient multiple determination, sum of squares, properties

TABLE 13-7 The department head of the accounting department wanted to see if she could predict the GPA of students using the number of course units (credits) and total SAT scores of each. She takes a sample of students and generates the following Microsoft Excel output: SUMMARY OUTPUT Regression Statistics Multiple R 0.916 R Square 0.839 Adjusted R Square 0.732 Standard Error 0.24685 Observations 6 ANOVA df .

SS

MS

F

Signif F

Multiple Regression

Regression Residual Total

2 3 5

Coeff Intercept 4.593897 Units – 0.247270 SAT Total 0.001443

0.95219 0.18281 1.13500 StdError 1.13374542 0. 06268485 0.00101241

0.47610 0.06094

t Stat 4.052 – 3.945 1.425

7.813

469

0.0646

P-value 0.0271 0.0290 0.2494

94. Referring to Table 13-7, the estimate of the unit change in the mean of Y per unit change in X1, holding X2 constant, is ________.

ANSWER: -0.247 TYPE: FI DIFFICULTY: Easy KEYWORDS: slope, interpretation 95. Referring to Table 13-7, the net regression coefficient of X2 is ________.

ANSWER: 0.001443 TYPE: FI DIFFICULTY: Easy KEYWORDS: slope, interpretation 96. Referring to Table 13-7, the predicted GPA for a student carrying 15 course units and who has a total SAT of 1,100 is ________.

ANSWER: 2.472 TYPE: FI DIFFICULTY: Moderate KEYWORDS: prediction of individual values 97. Referring to Table 13-7, the value of the coefficient of multiple determination, r2Y.12, is ________.

ANSWER: 0.839 TYPE: FI DIFFICULTY: Easy KEYWORDS: coefficient of multiple determination 98. Referring to Table 13-7, the value of the adjusted coefficient of multiple determination, r2adj, is ________.

ANSWER: 0.732 TYPE: FI DIFFICULTY: Easy KEYWORDS: adjusted r-square 99. Referring to Table 13-7, the department head wants to test H0: 1 = 2 = 0 . The appropriate alternative hypothesis is ________. .

470

Multiple Regression

ANSWER: at least one  i  0 .

TYPE: FI DIFFICULTY: Easy KEYWORDS: F test on the entire regression, form of hypothesis 100. Referring to Table 13-7, the department head wants to test H0: 1 = 2 = 0 . The critical value of the F test for a level of significance of 0.05 is ________.

ANSWER: 9.55 TYPE: FI DIFFICULTY: Easy KEYWORDS: F test on the entire regression, critical value 101. Referring to Table 13-7, the department head wants to test H0: 1 = 2 = 0 . The value of the F-test statistic is ________.

ANSWER: 7.813 TYPE: FI DIFFICULTY: Easy KEYWORDS: F test on the entire regression, test statistic 102. Referring to Table 13-7, the department head wants to test H0: 1 = 2 = 0 . The p-value of the test is ________.

ANSWER: 0.0646 TYPE: FI DIFFICULTY: Easy KEYWORDS: F test on the entire regression, p-value

103. True or False: Referring to Table 13-7, the department head wants to test H0: 1 = 2 = 0 . At a level of significance of 0.05, the null hypothesis is rejected.

ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: F test on the entire regression, decision 104. Referring to Table 13-7, the department head wants to use a t test to test for the significance of the coefficient of X1. For a level of significance of 0.05, the critical values of the test are ________.

ANSWER:  3.1824 TYPE: FI DIFFICULTY: Easy KEYWORDS: t test on slope, critical value 105. Referring to Table 13-7, the department head wants to use a t test to test for the significance of the coefficient of X1. The value of the test statistic is ________. .

Multiple Regression

471

ANSWER: -3.945 TYPE: FI DIFFICULTY: Easy KEYWORDS: t test on slope, test statistic 106. Referring to Table 13-7, the department head wants to use a t test to test for the significance of the coefficient of X1. The p-value of the test is ________.

ANSWER: 0.0290 TYPE: FI DIFFICULTY: Easy KEYWORDS: t test on slope, p-value 107. True or False: Referring to Table 13-7, the department head wants to use a t test to test for the significance of the coefficient of X1. At a level of significance of 0.05, the department head would decide that 1  0 .

ANSWER: True TYPE: TF DIFFICULTY: Moderate KEYWORDS: t test on slope, decision, conclusion 108. Referring to Table 13-7, the department head decided to obtain a 95% confidence interval for 1 . The confidence interval is from ________ to ________.

ANSWER: -0.4468 to -0.0478 TYPE: FI DIFFICULTY: Moderate KEYWORDS: slope, confidence interval

TABLE 13-8 A financial analyst wanted to examine the relationship between salary (in $1,000) and 4 variables: age (X1 = Age), experience in the field (X2 = Exper), number of degrees (X3 = Degrees), and number of previous jobs in the field (X4 = Prevjobs). He took a sample of 20 employees and obtained the following Microsoft Excel output: SUMMARY OUTPUT Regression Statistics Multiple R 0.992 R Square 0.984 Adjusted R Square 0.979 Standard Error 2.26743 Observations 20 ANOVA Regression Residual .

df 4 15

SS 4609.83164 77.11836

MS 1152.45791 5.14122

F 224.160

Signif F 0.0001

472

Multiple Regression

Total

19

4686.95000

Intercept Age Exper Degrees Prevjobs

Coeff – 9.611198 1.327695 – 0.106705 7.311332 – 0.504168

StdError t Stat 2.77988638 – 3.457 0.11491930 11.553 0.14265559 – 0.748 0.80324187 9.102 0.44771573 – 1.126

P-value 0.0035 0.0001 0.4660 0.0001 0.2778

109. Referring to Table 13-8, the estimate of the unit change in the mean of Y per unit change in X4, taking into account the effects of the other 3 variables, is ________.

ANSWER: -0.504 TYPE: FI DIFFICULTY: Easy KEYWORDS: slope, interpretation 110. Referring to Table 13-8, the net regression coefficient of X2 is ________.

ANSWER: -0.107 TYPE: FI DIFFICULTY: Easy KEYWORDS: slope, interpretation 111. Referring to Table 13-8, the predicted salary for a 35-year-old person with 10 years of experience, 3 degrees, and 1 previous job is ________.

ANSWER: 57.2 TYPE: FI DIFFICULTY: Moderate KEYWORDS: prediction of individual values 112. Referring to Table 13-8, the value of the coefficient of multiple determination, r2Y.1234, is ________.

ANSWER: 0.984 TYPE: FI DIFFICULTY: Easy KEYWORDS: coefficient of multiple determination 113. Referring to Table 13-8, the value of the adjusted coefficient of multiple determination, adj r2, is ________.

ANSWER: 0.979 TYPE: FI DIFFICULTY: Easy KEYWORDS: adjusted r-square 114. Referring to Table 13-8, the analyst wants to use an F-test to test H0 : 1 = 2 = 3 =  4 = 0 . The appropriate alternative hypothesis is ________.

ANSWER: .

Multiple Regression

473

at least one  i  0 .

TYPE: FI DIFFICULTY: Easy KEYWORDS: F test on the entire regression, form of hypothesis 115. Referring to Table 13-8, the critical value of an F test on the entire regression for a level of significance of 0.01 is ________.

ANSWER: 4.89 TYPE: FI DIFFICULTY: Easy KEYWORDS: F test on the entire regression, critical value 116. Referring to Table 13-8, the value of the F-statistic for testing the significance of the entire regression is ________.

ANSWER: 224.160 TYPE: FI DIFFICULTY: Easy KEYWORDS: F test on the entire regression, test statistic 117. Referring to Table 13-8, the p-value of the F test for the significance of the entire regression is ________.

ANSWER: 0.0001 TYPE: FI DIFFICULTY: Easy KEYWORDS: F test on the entire regression, p-value

118. True or False: Referring to Table 13-8, the F test for the significance of the entire regression performed at a level of significance of 0.01 leads to a rejection of the null hypothesis.

ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: F test on the entire regression, decision 119. Referring to Table 13-8, the analyst wants to use a t test to test for the significance of the coefficient of X3. For a level of significance of 0.01, the critical values of the test are ________.

ANSWER:  2.9467 TYPE: FI DIFFICULTY: Easy KEYWORDS: t test on the slope, critical value 120. Referring to Table 13-8, the analyst wants to use a t test to test for the significance of the coefficient of X3. The value of the test statistic is ________.

ANSWER: 9.102 TYPE: FI DIFFICULTY: Easy .

474

Multiple Regression

KEYWORDS: t test on the slope, test statistic 121. Referring to Table 13-8, the analyst wants to use a t test to test for the significance of the coefficient of X3. The p-value of the test is ________.

ANSWER: 0.0001 TYPE: FI DIFFICULTY: Easy KEYWORDS: t test on the slope, p-value 122. True or False: Referring to Table 13-8, the analyst wants to use a t test to test for the significance of the coefficient of X3. At a level of significance of 0.01, the department head would decide that  3  0 .

ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: t test on the slope, decision, conclusion 123. Referring to Table 13-8, the analyst decided to obtain a 99% confidence interval for  3 . The confidence interval is from ________ to ________.

ANSWER: 4.944 to 9.678 TYPE: FI DIFFICULTY: Moderate KEYWORDS: slope, confidence interval

TABLE 13-9 You decide to predict gasoline prices in different cities and towns in the United States for your term project. Your dependent variable is price of gasoline per gallon and your explanatory variables are per capita income, the number of firms that manufacture automobile parts in and around the city, the number of new business starts in the last year, population density of the city, percentage of local taxes on gasoline, and the number of people using public transportation. You collected data of 32 cities and obtained a regression sum of squares SSR= 122.8821. Your computed value of standard error of the estimate is 1.9549.

124. Referring to Table 13-9, what is the value of the coefficient of multiple determination? a) 0.2225 b) 0.4576 c) 0.5626 d) 0.6472 ANSWER: c TYPE: MC DIFFICULTY: Difficult KEYWORDS: coefficient of multiple determination 125. Referring to Table 13-9, the value of adjusted r 2 is .

Multiple Regression

a) b) c) d)

475

0.4576 0.5626 0.6472 95.5414

ANSWER: a TYPE: MC DIFFICULTY: Moderate KEYWORDS: adjusted r-square 126. Referring to Table 13-9, if variables that measure the number of new business starts in the last year and population density of the city were removed from the multiple regression model, which of the following would be true? a) The adjusted r 2 will definitely increase. b) The adjusted r 2 cannot increase. c) The coefficient of multiple determination will not increase. d) The coefficient of multiple determination will definitely increase.

ANSWER: c TYPE: MC DIFFICULTY: Moderate KEYWORDS: coefficient of multiple determination, adjusted r-square, properties

TABLE 13-10 You worked as an intern at We Always Win Car Insurance Company last summer. You notice that individual car insurance premiums depend very much on the age of the individual, the number of traffic tickets received by the individual, and the population density of the city in which the individual lives. You performed a regression analysis in EXCEL and obtained the following information: Regression Analysis Regression Statistics Multiple R 0.63 R Square 0.40 Adjusted R Square 0.23 Standard Error 50.00 Observations 15.00 ANOVA df Regression Residual Total

SS 3 11

.

27496.82 45479.54

MS 5994.24

F 2.40

Significance F 0.12

476

Multiple Regression

Coefficients Standard Error 123.80 48.71 -0.82 0.87 21.25 10.66 -3.14 6.46

Intercept AGE TICKETS DENSITY

t Stat

P-value

Lower 99.0%

Upper 99.0%

0.03 0.36 0.07 0.64

-27.47 -3.51 -11.86 -23.19

275.07 1.87 54.37 16.91

2.54 -0.95 1.99 -0.49

127. Referring to Table 13-10, the proportion of the total variability in insurance premiums that can be explained by AGE, TICKETS, and DENSITY is _________.

ANSWER: 0.40 TYPE: FI DIFFICULTY: Easy KEYWORDS: coefficient of multiple determination, interpretation 128. Referring to Table 13-10, the adjusted r 2 is _________.

ANSWER: 0.23 TYPE: FI DIFFICULTY: Easy KEYWORDS: adjusted r-square

129. Referring to Table 13-10, the standard error of the estimate is _________.

ANSWER: 50.0 TYPE: FI DIFFICULTY: Easy KEYWORDS: standard error of estimate 130. Referring to Table 13-10, the estimated average change in insurance premiums for every 10 additional tickets received is _____.

ANSWER: 212.5 TYPE: FI DIFFICULTY: Difficult KEYWORDS: slope, interpretation 131. Referring to Table 13-10, the 99% confidence interval for the change in average insurance premiums of a person who has become 1 year older (i.e., the slope coefficient for AGE) is _________.

ANSWER:

-0.82  2.69

TYPE: FI DIFFICULTY: Moderate KEYWORDS: slope, interpretation, confidence interval .

Multiple Regression

477

132. Referring to Table 13-10, the total degrees of freedom that are missing in the ANOVA table should be ______.

ANSWER: 14 TYPE: FI DIFFICULTY: Easy KEYWORDS: degrees of freedom 133. Referring to Table 13-10, the regression sum of squares that is missing in the ANOVA table should be ______.

ANSWER: 17,982.72 TYPE: FI DIFFICULTY: Easy KEYWORDS: sum of squares 134. Referring to Table 13-10, the residual mean squares (MSE) that are missing in the ANOVA table should be _____.

ANSWER: 2,499.71 TYPE: FI DIFFICULTY: Easy KEYWORDS: mean squares

135. Referring to Table 13-10, to test the significance of the multiple regression model, what is the form of the null hypothesis? a) H 0 : 0 b) H 0 : 1

H0 : 1 = 2 = 3 d) H 0 : 0 = 1 = 2 = 3 c)

ANSWER: c TYPE: MC DIFFICULTY: Easy KEYWORDS: F test on the entire regression, form of hypothesis 136. Referring to Table 13-10, to test the significance of the multiple regression model, the value of the test statistic is ______.

ANSWER: 2.40 TYPE: FI DIFFICULTY: Easy KEYWORDS: F test on the entire regression, test statistic 137. Referring to Table 13-10, to test the significance of the multiple regression model, the p-value of the test statistic in the sample is ______. .

478

Multiple Regression

ANSWER: 0.12 TYPE: FI DIFFICULY: Easy KEYWORDS: F test on the entire regression, p-value 138. Referring to Table 13-10, to test the significance of the multiple regression model, what are the degrees of freedom?

ANSWER: 3 for the numerator, 11 for the denominator TYPE: PR DIFFICULY: Easy KEYWORDS: F test on the entire regression, degrees of freedom 139. True or False: Referring to Table 13-10, to test the significance of the multiple regression model, the null hypothesis should be rejected while allowing for 1% probability of committing a type I error.

ANSWER: False TYPE: TF DIFFICULY: Easy KEYWORDS: F test on the entire regression, decision

140. True or False: Referring to Table 13-10, the multiple regression model is significant at a 10% level of significance.

ANSWER: False TYPE: TF DIFFICULY: Easy KEYWORDS: F test on the entire regression, decision, conclusion

141. If a categorical independent variable contains 2 categories, then _________ dummy variable(s) will be needed to uniquely represent these categories. a) 1 b) 2 c) 3 d) 4 ANSWER: a TYPE: MC DIFFICULTY: Easy KEYWORDS: dummy variable, properties 142. If a categorical independent variable contains 4 categories, then _________ dummy variable(s) will be needed to uniquely represent these categories. a) 1 b) 2 .

Multiple Regression

479

c) 3 d) 4 ANSWER: c TYPE: MC DIFFICULTY: Easy KEYWORDS: dummy variable, properties 143. A dummy variable is used as an independent variable in a regression model when a) the variable involved is numerical. b) the variable involved is categorical. c) a curvilinear relationship is suspected. d) when 2 independent variables interact.

ANSWER: b TYPE: MC DIFFICULTY: Easy KEYWORDS: dummy variable, properties

144. An interaction term in a multiple regression model may be used when a) the coefficient of determination is small. b) there is a curvilinear relationship between the dependent and independent variables. c) neither one of 2 independent variables contribute significantly to the regression model. d) the relationship between X1 and Y changes for differing values of X2.

ANSWER: d TYPE: MC DIFFICULTY: Moderate KEYWORDS: interaction, properties 145. To explain personal consumption (CONS) measured in dollars, data is collected for INC: CRDTLIM: APR: ADVT: SEX:

personal income in dollars $1 plus the credit limit in dollars available to the individual average annualized percentage interest rate for borrowing for the individual per person advertising expenditure in dollars by manufacturers in the city where the individual lives gender of the individual; 1 if female, 0 if male

A regression analysis was performed with CONS as the dependent variable and ln(CRDTLIM), ln(APR), ln(ADVT), and SEX as the independent variables. The estimated model was y = 2.28 - 0.29 ln(CRDTLIM) + 5.77 ln(APR) + 2.35 ln(ADVT) + 0.39 SEX .

480

Multiple Regression

What is the correct interpretation for the estimated coefficient for SEX? a) Holding everything else fixed, personal consumption for females is estimated to be $0.39 higher than males on the average. b) Holding everything else fixed, personal consumption for males is estimated to be $0.39 higher than females on the average. c) Holding everything else fixed, personal consumption for females is estimated to be 0.39% higher than males on the average. d) Holding everything else fixed, personal consumption for males is estimated to be 0.39% higher than females on the average.

ANSWER: a TYPE: MC DIFFICULTY: Easy KEYWORDS: dummy variable, slope, interpretation

TABLE 13-11 A weight-loss clinic wants to use regression analysis to build a model for weight-loss of a client (measured in pounds). Two variables thought to affect weight-loss are client’s length of time on the weight loss program and time of session. These variables are described below: Y = Weight-loss (in pounds) X1 = Length of time in weight-loss program (in months) X2 = 1 if morning session, 0 if not X3 = 1 if afternoon session, 0 if not (Base level = evening session) Data for 12 clients on a weight-loss program at the clinic were collected and used to fit the interaction Y = 0 + 1 X1 + 2 X2 + 3 X 3 + 4 X1 X2 + 5 X1 X3 +  model: Partial output from Microsoft Excel follows: Regression Statistics Multiple R 0.73514 R Square 0.540438 Adjusted R Square 0.157469 Standard Error 12.4147 Observations 12 ANOVA F = 5.41118

Intercept .

Significance F = 0.040201 Coeff 0.089744

StdError 14.127

t Stat 0.0060

P-value 0.9951

Multiple Regression

Length (X1) Morn Ses (X2) Aft Ses (X3) Length*Morn Ses Length*Aft Ses

6.22538 2.217272 11.8233 0.77058 – 0.54147

2.43473 22.1416 3.1545 3.562 3.35988

2.54956 0.100141 3.558901 0.216334 – 0.161158

481

0.0479 0.9235 0.0165 0.8359 0.8773

146. Referring to Table 13-11, what is the experimental unit for this analysis? a) A clinic b) A client on a weight-loss program c) A month d) A morning, afternoon, or evening session ANSWER: b TYPE: MC DIFFICULTY: Moderate KEYWORDS: model building

147. Referring to Table 13-11, what null hypothesis would you test to determine whether the slope of the linear relationship between weight-loss (Y) and time in the program (X1) varies according to time of session? a) H0 : 1 = 2 = 3 = 4 = 5 = 0 b) H0 : 2 = 3 = 4 =  5 = 0 c) H0 : 4 = 5 = 0 d) H0 : 2 = 3 = 0 ANSWER: c TYPE: MC DIFFICULTY: Moderate KEYWORDS: interaction, form of hypothesis 148. Referring to Table 13-11, in terms of the  's in the model, give the average change in weightloss (Y) for every 1 month increase in time in the program (X1) when attending the evening session. a) 1 + 4 b)  1 +  5 c) 1 d)  4 +  5 ANSWER: c TYPE: MC DIFFICULTY: Moderate KEYWORDS: interaction, slope, interpretation

.

482

Multiple Regression 149. Referring to Table 13-11, in terms of the  's in the model, give the average change in weightloss (Y) for every 1 month increase in time in the program (X1) when attending the morning session. a) 1 + 4 b)  1 +  5 c) 1 d)  4 +  5 ANSWER: a TYPE: MC DIFFICULTY: Moderate KEYWORDS: interaction, slope, interpretation

150. Referring to Table 13-11, in terms of the  's in the model, give the average change in weightloss (Y) for every 1 month increase in time in the program (X1) when attending the afternoon session. a) 1 + 4 b)  1 +  5 c) 1 d)  4 +  5 ANSWER: b TYPE: MC DIFFICULTY: Moderate KEYWORDS: interaction, slope, interpretation 151. True or False: Referring to Table 13-11, the overall model for predicting weight-loss (Y) is statistically significant at the 0.05 level.

ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: interaction, F test on the entire regression, decision, conclusion 152. Referring to Table 13-11, which of the following statements is supported by the analysis shown? a) There is sufficient evidence (at  = 0.05) of curvature in the relationship between weight-loss (Y) and months in program(X1). b) There is sufficient evidence (at  = 0.05) to indicate that the relationship between weight-loss (Y) and months in program (X1) depends on session time. c) There is sufficient evidence (at  = 0.10) to indicate that the session time (morning, afternoon, evening) affects weight-loss (Y). .

Multiple Regression

483

d) There is insufficient evidence (at  = 0.10) to indicate that the relationship between weight-loss (Y) and months in program(X1) depends on session time.

ANSWER: d TYPE: MC DIFFICULTY: Difficult KEYWORDS: interaction, t test on slope, decision, conclusion 153. In a multiple regression model, the adjusted r 2 a) cannot be negative. b) can sometimes be negative. c) can sometimes be greater than +1. d) has to fall between 0 and +1.

ANSWER: b TYPE: MC DIFFICULTY: Difficult KEYWORDS: adjusted r-square, properties

TABLE 13-12 As a project for his business statistics class, a student examined the factors that determined parking meter rates throughout the campus area. Data were collected for the price per hour of parking, blocks to the quadrangle, and one of the three jurisdictions: on campus, in downtown and off campus, or outside of downtown and off campus. The population regression model hypothesized is

Yi =  + 1 X 1i +  2 X 2i +  3 X 3i + 

where Y is the meter price X1 is the number of blocks to the quad X2 is a dummy variable that takes the value 1 if the meter is located in downtown and off campus and the value 0 otherwise X3 is a dummy variable that takes the value 1 if the meter is located outside of downtown and off campus, and the value 0 otherwise The following Excel results are obtained.

Regression Statistics Multiple R 0.9659 R Square 0.9331 Adjusted R Square 0.9294 Standard Error 0.0327 Observations 58 ANOVA Df Regression 3 Residual 54 Total 57 .

SS 0.8094 0.0580 0.8675

MS 0.2698 0.0010

F Significance F 251.1995 1.0964E-31

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Multiple Regression

Coefficient s Standard Error 0.5118 0.0136 -0.0045 0.0034 -0.2392 0.0123 -0.0002 0.0123

Intercept X1 X2 X3

t Stat 37.4675 -1.3276 -19.3942 -0.0214

P-value 2.4904 0.1898 5.3581E-26 0.9829

154. Referring to Table 13-12, what is the correct interpretation for the estimated coefficient for X2? a) All else equal, the estimated average difference in costs between parking on campus, and parking outside of downtown and off campus is −$0.24 per hour. b) All else equal, the estimated average difference in costs between parking in downtown and off campus, and parking on campus is −$0.24 per hour. c) All else equal, the estimated average difference in costs between parking in downtown and off campus, and parking outside of downtown and off campus is −$0.24 per hour. d) All else equal, the estimated average difference in costs between parking in downtown and off campus, and parking either outside of downtown and off campus or on campus is −$0.24 per hour.

ANSWER: b TYPE: MC DIFFICULTY: Moderate KEYWORDS: dummy variable, slope, interpretation 155. Referring to Table 13-12, predict the meter rate per hour if one parks outside of downtown and off campus 3 blocks from the quad. a) $−0.0139 b) $0.2589 c) $0.2604 d) $0.4981 ANSWER: d TYPE: MC DIFFICULTY: Easy KEYWORDS: prediction of individual values

.

Multiple Regression

485

156. Referring to Table 13-12, if one is already outside of downtown and off campus but decides to park 3 more blocks from the quad, the estimated average parking meter rate will a) decrease by 0.0045. b) decrease by 0.0135 . c) decrease by 0.0139. d) decrease by 0.4979.

ANSWER b TYPE: MC DIFFICULTY: Moderate KEYWORDS: dummy variable, slope, interpretation 157. True or False: An interaction term in a multiple regression model may be used when the relationship between X1 and Y changes for differing values of X2.

ANSWER: True TYPE: TF DIFFICULTY: Moderate KEYWORDS: interaction, properties

158. True or False: When a dummy variable is included in a multiple regression model, the interpretation of the estimated slope coefficient does not make any sense anymore.

ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: dummy variable, slope, interpretation 159. True or False: In trying to obtain a model to estimate grades on a statistics test, a professor wanted to include, among other factors, whether the person had taken the course previously. To do this, the professor included a dummy variable in her regression that was equal to 1 if the person had previously taken the course, and 0 otherwise. The interpretation of the coefficient associated with this dummy variable would be the average amount the repeat students tended to be above or below non-repeaters, with all other factors the same.

ANSWER: True TYPE: TF DIFFICULTY: Moderate KEYWORDS: dummy variable, slope, interpretation TABLE 13-13 An econometrician is interested in evaluating the relation of demand for building materials to mortgage rates in Los Angeles and San Francisco. He believes that the appropriate model is Y = 10 + 5X1 + 8X2 where

X1 = mortgage rate in % X2 = 1 if SF, 0 if LA .

486

Multiple Regression

Y = demand in $100 per capita 160. Referring to Table 13-13, holding constant the effect of city, each additional increase of 1% in the mortgage rate would lead to an estimated increase of ________ in the mean demand.

ANSWER: $500 per capita TYPE: FI DIFFICULTY: Moderate KEYWORDS: slope, interpretation 161. Referring to Table 13-13, the effect of living in San Francisco rather than Los Angeles is to increase the mean demand by an estimated ________.

ANSWER: $800 per capita TYPE: FI DIFFICULTY: Moderate KEYWORDS: dummy variable, slope, interpretation

162. Referring to Table 13-13, the predicted demand in Los Angeles when the mortgage rate is 8% is ________.

ANSWER: $5,000 per capita TYPE: FI DIFFICULTY: Moderate KEYWORDS: prediction of individual values, dummy variable 163. Referring to Table 13-13, the predicted demand in San Francisco when the mortgage rate is 10% is ________.

ANSWER: $6,800 per capita TYPE: FI DIFFICULTY: Moderate KEYWORDS: prediction of individual values, dummy variable 164. Referring to Table 13-13, the fitted model for predicting demand in Los Angeles is ________. a) 10 + 5X1 b) 10 + 13X1 c) 15 + 8X2 d) 18 + 5X2 ANSWER: a TYPE: MC DIFFICULTY: Moderate KEYWORDS: prediction of individual values, dummy variable 165. Referring to Table 13-13, the fitted model for predicting demand in San Francisco is ________. a) 10 + 5X1 b) 10 + 13X1 .

Multiple Regression

487

c) 15 + 8X2 d) 18 + 5X1 ANSWER: d TYPE: MC DIFFICULTY: Moderate KEYWORDS: prediction of individual values, dummy variable

TABLE 13-14 An automotive engineer would like to be able to predict automobile mileages. She believes that the two most important characteristics that affect mileage are horsepower and the number of cylinders (4 or 6) of a car. She believes that the appropriate model is Y = 40 – 0.05X1 + 20X2 – 0.1X1X2 where X1 = horsepower X2 = 1 if 4 cylinders, 0 if 6 cylinders Y = mileage.

166. Referring to Table 13-14, the predicted mileage for a 300 horsepower, 6-cylinder car is ________.

ANSWER: 25 TYPE: FI DIFFICULTY: Moderate KEYWORDS: prediction of individual values, dummy variable 167. Referring to Table 13-14, the predicted mileage for a 200 horsepower, 4-cylinder car is ________.

ANSWER: 30 TYPE: FI DIFFICULTY: Moderate KEYWORDS: prediction of individual values, dummy variable 168. Referring to Table 13-14, the fitted model for predicting mileages for 6-cylinder cars is ________. a) 40 – 0.05X1 b) 40 – 0.10X1 .

488

Multiple Regression c) 60 – 0.10X1 d) 60 – 0.15X1 ANSWER: a TYPE: MC DIFFICULTY: Moderate KEYWORDS: prediction of individual values, dummy variable 169. Referring to Table 13-14, the fitted model for predicting mileages for 4-cylinder cars is ________. a) 40 – 0.05X1 b) 40 – 0.10X1 c) 60 – 0.10X1 d) 60 – 0.15X1 ANSWER: d TYPE: MC DIFFICULTY: Moderate KEYWORDS: prediction of individual values, dummy variable

TABLE 13-15 The superintendent of a school district wanted to predict the percentage of students passing a sixthgrade proficiency test. She obtained the data on percentage of students passing the proficiency test (% Passing), daily average of the percentage of students attending class (% Attendance), average teacher salary in dollars (Salaries), and instructional spending per pupil in dollars (Spending) of 47 schools in the state.

Following is the multiple regression output with Y = % Passing as the dependent variable, X 1 =:% Attendance, X 2 = Salaries and X 3 = Spending: Regression Statistics Multiple R 0.7930 R Square 0.6288 Adjusted R 0.6029 Square Standard 10.4570 Error Observations 47 ANOVA df Regression Residual Total

3 43 46 Coefficients

Intercept

-753.4225

SS 7965.08 4702.02 12667.11

MS 2655.03 109.35

F 24.2802

Significance F 2.3853E-09

Standard Error 101.1149

t Stat

P-value

Lower 95%

-7.4511

2.88E-09

-957.3401

Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall

Upper 95% -549.5050

Multiple Regression % Attendance Salary Spending

8.5014 6.85E-07 0.0060

1.0771 0.0006 0.0046

7.8929 0.0011 1.2879

6.73E-10 0.9991 0.2047

6.3292 -0.0013 -0.0034

489

10.6735 0.0013 0.0153

170. Referring to Table 13-15, which of the following is a correct statement? a) The average percentage of students passing the proficiency test is estimated to go up by 8.50% when daily average of percentage of students attending class increases by 1%. b) The daily average of the percentage of students attending class is expected to go up by an estimated 8.50% when the percentage of students passing the proficiency test increases by 1%. c) The average percentage of students passing the proficiency test is estimated to go up by 8.50% when daily average of the percentage of students attending class increases by 1% holding constant the effects of all the remaining independent variables. d) The daily average of the percentage of students attending class is expected to go up by an estimated 8.50% when the percentage of students passing the proficiency test increases by 1% holding constant the effects of all the remaining independent variables.

ANSWER: c TYPE: MC DIFFICULTY: Moderate KEYWORDS: interpretation, slope 171. Referring to Table 13-15, which of the following is a correct statement? a) 62.88% of the total variation in the percentage of students passing the proficiency test can be explained by daily average of the percentage of students attending class, average teacher salary, and instructional spending per pupil. b) 62.88% of the total variation in the percentage of students passing the proficiency test can be explained by daily average of the percentage of students attending class, average teacher salary, and instructional spending per pupil after adjusting for the number of predictors and sample size. c) 62.88% of the total variation in the percentage of students passing the proficiency test can be explained by daily average of the percentage of students attending class holding constant the effect of average teacher salary, and instructional spending per pupil. d) 62.88% of the total variation in the percentage of students passing the proficiency test can be explained by daily average of the percentage of students attending class after adjusting for the effect of average teacher salary, and instructional spending per pupil.

ANSWER: a TYPE: MC DIFFICULTY: Moderate KEYWORDS: interpretation, coefficient of multiple determination 172. Referring to Table 13-15, which of the following is a correct statement? a) 60.29% of the total variation in the percentage of students passing the proficiency test can be explained by daily average of the percentage of students attending class, average teacher salary, and instructional spending per pupil. b) 60.29% of the total variation in the percentage of students passing the proficiency test can be explained by daily average of the percentage of students attending class, average teacher salary, and instructional spending per pupil after adjusting for the number of predictors and sample size. c) 60.29% of the total variation in the percentage of students passing the proficiency test can be explained by daily average of the percentage of students attending class holding constant the effect of average teacher salary, and instructional spending per pupil. .

490

Multiple Regression

d) 60.29% of the total variation in the percentage of students passing the proficiency test can be explained by daily average of the percentage of students attending class after adjusting for the effect of average teacher salary, and instructional spending per pupil.

ANSWER: b TYPE: MC DIFFICULTY: Moderate KEYWORDS: interpretation, adjusted r-square 173. Referring to Table 13-15, what is the standard error of estimate?

ANSWER: 10.46 (%) TYPE: PR DIFFICULTY: Easy KEYWORDS: standard error of estimate

174. Referring to Table 13-15, predict the percentage of students passing the proficiency test for a school which has a daily average of 95% of students attending class, an average teacher salary of 40,000 dollars, and an instructional spending per pupil of 2000 dollars.

ANSWER: 66.20 (%) TYPE: PR DIFFICULTY: Easy KEYWORDS: prediction of individual values 175. Referring to Table 13-15, estimate the average percentage of students passing the proficiency test for all the schools that have a daily average of 95% of students attending class, an average teacher salary of 40,000 dollars, and an instructional spending per pupil of 2000 dollars.

ANSWER: 66.20 (%) TYPE: PR DIFFICULTY: Easy KEYWORDS: estimation of mean values 176. Referring to Table 13-15, which of the following is the correct null hypothesis to test whether instructional spending per pupil has any effect on percentage of students passing the proficiency test? a) H 0 : 0 = 0 b) H 0 : 1 = 0

H 0 : 2 = 0 d) H 0 : 3 = 0 c)

ANSWER: d TYPE: MC DIFFICULTY: Easy KEYWORDS: t test on slope, form of hypothesis

.

Multiple Regression

491

177. Referring to Table 13-15, which of the following is the correct alternative hypothesis to test whether instructional spending per pupil has any effect on percentage of students passing the proficiency test? a) H1 : 0  0 b) H1 : 1  0

H1 :  2  0 d) H1 : 3  0 c)

ANSWER: d TYPE: MC DIFFICULTY: Easy KEYWORDS: t test on slope, form of hypothesis

178. Referring to Table 13-15, what is the value of the test statistic when testing whether instructional spending per pupil has any effect on percentage of students passing the proficiency test?

ANSWER: 1.2879 TYPE: PR DIFFICULTY: Easy KEYWORDS: t test on slope, test statistic 179. Referring to Table 13-15, what is the p-value of the test statistic when testing whether instructional spending per pupil has any effect on percentage of students passing the proficiency test?

ANSWER: 0.2047 TYPE: PR DIFFICULTY: Easy KEYWORDS: t test on slope, p-value 180. True or False: Referring to Table 13-15, the null hypothesis should be rejected at a 5% level of significance when testing whether instructional spending per pupil has any effect on percentage of students passing the proficiency test.

ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: t test on slope, decision 181. True or False: Referring to Table 13-15, there is sufficient evidence that instructional spending per pupil has an effect on percentage of students passing the proficiency test while holding constant the effect of all the other independent variables at a 5% level of significance.

ANSWER: .

492

Multiple Regression

False TYPE: TF DIFFICULTY: Easy KEYWORDS: t test on slope, conclusion 182. Referring to Table 13-15, which of the following is the correct null hypothesis to test whether daily average of the percentage of students attending class has any effect on percentage of students passing the proficiency test? a) H 0 : 0 = 0 b) H 0 : 1 = 0

H 0 : 2 = 0 d) H 0 : 3 = 0 c)

ANSWER: b TYPE: MC DIFFICULTY: Easy KEYWORDS: t test on slope, form of hypothesis

183. Referring to Table 13-15, which of the following is the correct alternative hypothesis to test whether daily average of the percentage of students attending class has any effect on percentage of students passing the proficiency test? a) H1 : 0  0 b) H1 : 1  0

H1 :  2  0 d) H1 : 3  0 c)

ANSWER: b TYPE: MC DIFFICULTY: Easy KEYWORDS: t test on slope, form of hypothesis 184. Referring to Table 13-15, what is the value of the test statistic when testing whether daily average of the percentage of students attending class has any effect on percentage of students passing the proficiency test?

ANSWER: 7.8929 TYPE: PR DIFFICULTY: Easy KEYWORDS: t test on slope, test statistic 185. Referring to Table 13-15, what is the p-value of the test statistic when testing whether daily average of the percentage of students attending class has any effect on percentage of students passing the proficiency test?

ANSWER: 6.73E−10 or virtually zero TYPE: PR DIFFICULTY: Easy KEYWORDS: t test on slope, p-value .

Multiple Regression

493

186. True or False: Referring to Table 13-15, the null hypothesis should be rejected at a 5% level of significance when testing whether daily average of the percentage of students attending class has any effect on percentage of students passing the proficiency test.

ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: t test on slope, decision 187. True or False: Referring to Table 13-15, there is sufficient evidence that daily average of the percentage of students attending class has an effect on percentage of students passing the proficiency test while holding constant the effect of all the other independent variables at a 5% level of significance.

ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: t test on slope, conclusion 188. Referring to Table 13-15, which of the following is the correct null hypothesis to determine whether there is a significant relationship between percentage of students passing the proficiency test and the entire set of explanatory variables? a) H 0 : 0 = 1 = 2 = 3 = 0 b) H 0 : 1 = 2 = 3 = 0

H0 : 0 = 1 = 2 = 3  0 d) H 0 : 1 = 2 = 3  0 c)

ANSWER: b TYPE: MC DIFFICULTY: Easy KEYWORDS: F test on the entire regression, form of hypothesis 189. Referring to Table 13-15, which of the following is the correct alternative hypothesis to determine whether there is a significant relationship between percentage of students passing the proficiency test and the entire set of explanatory variables? a) H1 : 0 = 1 = 2 = 3  0 b) H1 : 1 = 2 = 3  0 c)

H1 : At least one of  j  0 for j = 0, 1, 2, 3

d) H1 : At least one of  j  0 for j = 1, 2, 3 ANSWER: d TYPE: MC DIFFICULTY: Easy KEYWORDS: F test on the entire regression, form of hypothesis 190. Referring to Table 13-15, the null hypothesis H 0 : 1 = 2 = 3 = 0 implies that percentage of students passing the proficiency test is not affected by any of the explanatory variables.

.

494

Multiple Regression

ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: F test on the entire regression, interpretation, form of hypothesis 191. Referring to Table 13-15, the null hypothesis H 0 : 1 = 2 = 3 = 0 implies that percentage of students passing the proficiency test is not affected by some of the explanatory variables.

ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: F test on the entire regression, interpretation, form of hypothesis

192. Referring to Table 13-15, the null hypothesis H 0 : 1 = 2 = 3 = 0 implies that percentage of students passing the proficiency test is not related to any of the explanatory variables.

ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: F test on the entire regression, interpretation, form of hypothesis 193. Referring to Table 13-15, the null hypothesis H 0 : 1 = 2 = 3 = 0 implies that percentage of students passing the proficiency test is not related to one of the explanatory variables.

ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: F test on the entire regression, interpretation, form of hypothesis 194. Referring to Table 13-15, the alternative hypothesis H1 : At least one of  j  0 for j = 1, 2, 3 implies that percentage of students passing the proficiency test is related to all of the explanatory variables.

ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: F test on the entire regression, interpretation, form of hypothesis 195. Referring to Table 13-15, the alternative hypothesis H1 : At least one of  j  0 for j = 1, 2, 3 implies that percentage of students passing the proficiency test is related to at least one of the explanatory variables.

ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: F test on the entire regression, interpretation, form of hypothesis .

Multiple Regression

495

196. Referring to Table 13-15, the alternative hypothesis H1 : At least one of  j  0 for j = 1, 2, 3 implies that percentage of students passing the proficiency test is affected by all of the explanatory variables.

ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: F test on the entire regression, interpretation, form of hypothesis

197. Referring to Table 13-15, the alternative hypothesis H1 : At least one of  j  0 for j = 1, 2, 3 implies that percentage of students passing the proficiency test is affected by at least one of the explanatory variables.

ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: F test on the entire regression, interpretation, form of hypothesis 198. Referring to Table 13-15, what are the numerator and denominator degrees of freedom, respectively, for the test statistic to determine whether there is a significant relationship between percentage of students passing the proficiency test and the entire set of explanatory variables?

ANSWER: 3 and 43 TYPE: PR DIFFICULTY: Easy KEYWORDS: F test on the entire regression, degrees of freedom 199. Referring to Table 13-15, what is the value of the test statistic to determine whether there is a significant relationship between percentage of students passing the proficiency test and the entire set of explanatory variables?

ANSWER: 24.2802 TYPE: PR DIFFICULTY: Easy KEYWORDS: F test on the entire regression, test statistic 200. Referring to Table 13-15, what is the p-value of the test statistic to determine whether there is a significant relationship between percentage of students passing the proficiency test and the entire set of explanatory variables?

ANSWER: 2.3853E−09 or virtually zero TYPE: PR DIFFICULTY: Easy KEYWORDS: F test on the entire regression, p-value

.

496

Multiple Regression

201. True or False: Referring to Table 13-15, the null hypothesis should be rejected at a 5% level of significance when testing whether there is a significant relationship between percentage of students passing the proficiency test and the entire set of explanatory variables.

ANSWER: True TYPE: PR DIFFICULTY: Easy KEYWORDS: F test on the entire regression, decision

202. True or False: Referring to Table 13-15, there is sufficient evidence that at least one of the explanatory variables is related to the percentage of students passing the proficiency test.

ANSWER: True TYPE: PR DIFFICULTY: Easy KEYWORDS: F test on the entire regression, conclusion 203. True or False: Referring to Table 13-15, there is sufficient evidence that the percentage of students passing the proficiency test depends on at least one of the explanatory variables.

ANSWER: True TYPE: PR DIFFICULTY: Easy KEYWORDS: F test on the entire regression, conclusion 204. True or False: Referring to Table 13-15, there is sufficient evidence that all of the explanatory variables is related to the percentage of students passing the proficiency test.

ANSWER: False TYPE: PR DIFFICULTY: Moderate KEYWORDS: F test on the entire regression, conclusion 205. True or False: Referring to Table 13-15, there is sufficient evidence that the percentage of students passing the proficiency test depends on all of the explanatory variables.

ANSWER: False TYPE: PR DIFFICULTY: Moderate KEYWORDS: F test on the entire regression, conclusion 206. Referring to Table 13-15, what are the lower and upper limits of the 95% confidence interval estimate for the effect of a one dollar increase in instructional spending per pupil on average percentage of students passing the proficiency test?

ANSWER: −0.0034 (%) to 0.0153 (%) .

Multiple Regression

497

TYPE: PR DIFFICULTY: Moderate KEYWORDS: F test on the entire regression, confidence interval, slope, interpretation 207. Referring to Table 13-15, what are the lower and upper limits of the 95% confidence interval estimate for the effect of a one dollar increase in average teacher salary on average percentage of students passing the proficiency test?

ANSWER: −0.0013 (%) to 0.0013 (%) TYPE: PR DIFFICULTY: Moderate KEYWORDS: F test on the entire regression, confidence interval, slope, interpretation

208. True or False: Referring to Table 13-15, we can conclude that average teacher salary has no impact on average percentage of students passing the proficiency test at a 5% level of significance using the 95% confidence interval estimate for  2 .

ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: F test on the entire regression, confidence interval, t test on slope, decision 209. True or False: Referring to Table 13-15, we can conclude that instructional spending per pupil has no impact on average percentage of students passing the proficiency test at a 5% level of significance using the 95% confidence interval estimate for  3 .

ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: F test on the entire regression, confidence interval, t test on slope, decision 210. True or False: Referring to Table 13-15, we can conclude that average teacher salary has no impact on average percentage of students passing the proficiency test at a 1% level of significance based solely on the 95% confidence interval estimate for  2 .

ANSWER: True TYPE: TF DIFFICULTY: Moderate KEYWORDS: F test on the entire regression, confidence interval, t test on slope, decision 211. True or False: Referring to Table 13-15, we can conclude that instructional spending per pupil has no impact on average percentage of students passing the proficiency test at a 1% level of significance based solely on the 95% confidence interval estimate for  3 .

ANSWER: True TYPE: TF DIFFICULTY: Moderate KEYWORDS: F test on the entire regression, confidence interval, t test on slope, decision

.

498

Multiple Regression

212. True or False: Referring to Table 13-15, we can conclude that average teacher salary has no impact on average percentage of students passing the proficiency test at a 10% level of significance based solely on the 95% confidence interval estimate for  2 .

ANSWER: False TYPE: TF DIFFICULTY: Moderate KEYWORDS: F test on the entire regression, confidence interval, t test on slope, decision

213. True or False: Referring to Table 13-15, we can conclude that instructional spending per pupil has no impact on average percentage of students passing the proficiency test at a 10% level of significance based solely on the 95% confidence interval estimate for  3 .

ANSWER: False TYPE: TF DIFFICULTY: Moderate KEYWORDS: F test on the entire regression, confidence interval, t test on slope, decision

.

500

Statistical Applications in Quality Management

CHAPTER 14: STATISTICAL APPLICATIONS IN QUALITY MANAGEMENT 1. The control chart a) focuses on the time dimension of a system. b) captures the natural variability in the system. c) can be used for categorical, discrete, or continuous variables. d) All of the above.

ANSWER: d TYPE: MC DIFFICULTY: Moderate KEYWORDS: control chart 2. Variation signaled by individual fluctuations or patterns in the data is called a) special or assignable causes. b) common or chance causes. c) explained variation. d) the standard deviation.

ANSWER: a TYPE: MC DIFFICULTY: Easy KEYWORDS: special causes of variation 3. Variation due to the inherent variability in a system of operation is called a) special or assignable causes. b) common or chance causes. c) explained variation. d) the standard deviation.

ANSWER: b TYPE: MC DIFFICULTY: Easy KEYWORDS: common causes of variation 4. Which of the following is not one of Deming's 14 points? a) Belief in mass inspection. b) Create constancy of purpose for improvement of product or service. c) Adopt and institute leadership. d) Drive out fear.

ANSWER: a TYPE: MC DIFFICULTY: Moderate KEYWORDS: Deming’s 14 points

.

Statistical Applications in Quality Management

501

5. Which of the following is not one of Deming's 14 points? a) Create constancy of purpose for improvement of product or service. b) Award business on the basis of price tag alone. c) Break down barriers between staff areas. d) Drive out fear.

ANSWER: b TYPE: MC DIFFICULTY: Moderate KEYWORDS: Deming’s 14 points 6. The principal focus of the control chart is the attempt to separate special or assignable causes of variation from common causes of variation. What cause of variation can be reduced only by changing the system? a) Special or assignable causes b) Common causes c) Total causes d) None of the above ANSWER: b TYPE: MC DIFFICULTY: Easy KEYWORDS: common causes of variation 7. Once the control limits are set for a control chart, one attempts to a) discern patterns that might exist in values over time. b) determine whether any points fall outside the control limits. c) Both of the above. d) None of the above.

ANSWER: c TYPE: MC DIFFICULTY: Easy KEYWORDS: control chart 8. Which of the following is not part of the Shewhart-Deming cycle? a) Plan b) Do c) React d) Act ANSWER: c TYPE: MC DIFFICULTY: Moderate KEYWORDS: Shewhart-Deming cycle

.

502

Statistical Applications in Quality Management 9. The Shewhart-Deming cycle plays an important role in which of the following Deming’s 14 points for management? a) Adopt the new philosophy. b) Break down barriers between staff areas. c) Create constancy of purpose for improvement of product and services. d) Eliminate slogans, exhortation, and targets for the workforce.

ANSWER: c TYPE: MC DIFFICULTY: Moderate KEYWORDS: Shewhart-Deming cycle 10. Which of the following situations suggests a process that appears to be operating in a state of statistical control? a) A control chart with a series of consecutive points that are above the center line and a series of consecutive points that are below the center line. b) A control chart in which no points fall outside either the upper control limit or the lower control limit and no patterns are present. c) A control chart in which several points fall outside the upper control limit. d) All of the above.

ANSWER: b TYPE: MC DIFFICULTY: Easy KEYWORDS: control chart, statistical control 11. Which of the following situations suggests a process that appears to be operating out of statistical control? a) A control chart with a series of consecutive points that are above the center line and a series of consecutive points that are below the center line. b) A control chart in which points fall outside the lower control limit. c) A control chart in which several points fall outside the upper control limit. d) All of the above.

ANSWER: d TYPE: MC DIFFICULTY: Moderate KEYWORDS: control chart, statistical control 12. A process is said to be out of control if a) a point falls above the upper or below the lower control lines. b) a run of 8 of more points is observed. c) Either of the above. d) None of the above.

ANSWER: c TYPE: MC DIFFICULTY: Moderate KEYWORDS: statistical control

.

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503

13. One of the morals of the red bead experiment is: a) variation is part of the process. b) only management can change the system. c) it is the system that primarily determines performance. d) All of the above.

ANSWER: d TYPE: MC DIFFICULTY: Moderate KEYWORDS: red bead experiment 14. Which famous statistician developed the 14 Points of Quality? a) Shewhart b) Deming c) Chebyshev d) Taguchi ANSWER: b TYPE: MC DIFFICULTY: Easy KEYWORDS: Deming’s 14 points TABLE 14-1 A local newspaper has 10 delivery boys who each deliver the morning paper to 50 customers every day. The owner decides to record the number of papers delivered on time for a 10-day period and construct a p chart to see whether the percentage is too erratic.

Number of Papers Day Delivered on Time 1 46 2 45 3 46 4 45 5 43 6 48 7 46 8 49 9 48 10 47 15. Referring to Table 14-1, what is the numerical value of the center line for the p chart? a) 0.926 b) 0.911 c) 0.885 d) 0.500 ANSWER: a TYPE: MC DIFFICULTY: Moderate KEYWORDS: p chart, center line

.

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Statistical Applications in Quality Management

16. Referring to Table 14-1, what is the numerical value of the lower control limit for the p chart? a) 0.920 b) 0.911 c) 0.815 d) 0.798 ANSWER: c TYPE: MC DIFFICULTY: Moderate KEYWORDS: p chart, control limit 17. Referring to Table 14-1, what is the numerical value of the upper control limit for the p chart? a) 0.926 b) 0.961 c) 0.979 d) 1.000 ANSWER: d TYPE: MC DIFFICULTY: Moderate KEYWORDS: p chart, control limit 18. Referring to Table 14-1, which expression best characterizes the p chart? a) Cycles b) Increasing trend c) In-control d) Individual outliers ANSWER: c TYPE: MC DIFFICULTY: Moderate KEYWORDS: p chart TABLE 14-2 A political pollster randomly selects a sample of 100 voters each day for 8 successive days and asks how many will vote for the incumbent. The pollster wishes to construct a p chart to see if the percentage favoring the incumbent candidate is too erratic.

Sample Number Favoring (Day) Incumbent Candidate 1 57 2 57 3 53 4 51 5 55 6 60 7 56 8 59

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Statistical Applications in Quality Management

19. Referring to Table 14-2, what is the numerical value of the center line for the p chart? a) 0.53 b) 0.56 c) 0.63 d) 0.66 ANSWER: b TYPE: MC DIFFICULTY: Moderate KEYWORDS: p chart, center line 20. Referring to Table 14-2, what is the numerical value of the lower control limit for the p chart? a) 0.37 b) 0.41 c) 0.50 d) 0.71 ANSWER: b TYPE: MC DIFFICULTY: Moderate KEYWORDS: p chart, control limit 21. Referring to Table 14-2, what is the numerical value of the upper control limit for the p chart? a) 0.92 b) 0.89 c) 0.71 d) 0.62 ANSWER: c TYPE: MC DIFFICULTY: Moderate KEYWORDS: p chart, control limit 22. Referring to Table 14-2, which expression best characterizes the p chart? a) Hugging the center line b) Cycles c) Hugging the control line d) Individual outliers ANSWER: a TYPE: MC DIFFICULTY: Moderate KEYWORDS: p chart

.

505

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Statistical Applications in Quality Management

TABLE 14-3 A quality control analyst for a light bulb manufacturer is concerned that the time it takes to produce a batch of light bulbs is too erratic. Accordingly, the analyst randomly surveys 10 production periods each day for 14 days and records the sample mean and range for each day.

Day R X (in minutes) 1 58.5 5.1 2 47.6 7.8 3 64.3 6.1 4 60.6 5.7 5 63.7 6.2 6 57.5 6.0 7 55.0 5.4 8 54.9 6.1 9 55.0 5.9 10 62.7 5.0 11 61.9 7.1 12 60.0 6.5 13 58.3 5.9 14 52.0 5.2 23. Referring to Table 14-3, suppose the analyst constructs an R chart to see if the variability in production times is in-control. What is the center line of this R chart? a) 4.8 b) 6.0 c) 6.9 d) 7.1 ANSWER: b TYPE: MC DIFFICULTY: Moderate KEYWORDS: R chart, center line 24. Referring to Table 14-3, suppose the analyst constructs an R chart to see if the variability in production times is in-control. What is the lower control limit for this R chart? a) 4.84 b) 3.37 c) 2.98 d) 1.34 ANSWER: d TYPE: MC DIFFICULTY: Moderate KEYWORDS: R chart, control limit

25. Referring to Table 14-3, suppose the analyst constructs an R chart to see if the variability in production times is in-control. What is the upper control limit for this R chart? .

Statistical Applications in Quality Management

a) b) c) d)

507

10.66 9.37 7.98 6.34

ANSWER: a TYPE: MC DIFFICULTY: Moderate KEYWORDS: R chart, control limit 26. Referring to Table 14-3, suppose the analyst constructs an R chart to see if the variability in production times is in-control. The R chart is characterized by which of the following? a) Increasing trend b) Jump in the level around which the observations vary c) Hugging the center line d) Hugging the control limits ANSWER: c TYPE: MC DIFFICULTY: Moderate KEYWORDS: R chart 27. Referring to Table 14-3, suppose the sample mean and range data were based on 11 observations per day instead of 10. How would this change affect the lower and upper control limits of the R chart? a) LCL would increase; UCL would decrease. b) LCL would remain the same; UCL would decrease. c) Both LCL and UCL would remain the same. d) LCL would decrease; UCL would increase.

ANSWER: a TYPE: MC DIFFICULTY: Moderate KEYWORDS: control limit 28. Referring to Table 14-3, suppose the analyst constructs an X chart to see if the production process is in-control. What is the center line for this chart? a) 64.3 b) 59.5 c) 58.0 d) 57.1 ANSWER: c TYPE: MC DIFFICULTY: Moderate KEYWORDS: XBar chart, center line

29. Referring to Table 14-3, suppose the analyst constructs an X chart to see if the production process is in-control. What is the lower control limit (LCL) for this chart? a) 47.60 b) 56.15 .

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Statistical Applications in Quality Management

c) 57.15 d) 58.05 ANSWER: b TYPE: MC DIFFICULTY: Moderate KEYWORDS: XBar chart, control limit 30. Referring to Table 14-3, suppose the analyst constructs an X chart to see if the production process is in-control. What is the upper control limit (UCL) for this chart? a) 62.15 b) 60.95 c) 59.85 d) 58.75 ANSWER: c TYPE: MC DIFFICULTY: Moderate KEYWORDS: XBar chart, control limit 31. Referring to Table 14-3, suppose the analyst constructs an X chart to see if the production process is in-control. Which expression best describes this chart? a) In-control b) Increasing trend c) Hugging the center line d) Individual outliers ANSWER: d TYPE: MC DIFFICULTY: Moderate KEYWORDS: XBar chart

TABLE 14-4 A factory supervisor is concerned that the time it takes workers to complete an important production task (measured in seconds) is too erratic and adversely affects expected profits. The supervisor .

Statistical Applications in Quality Management

509

proceeds by randomly sampling 5 individuals per hour for a period of 10 hours. The sample mean and range for each hour are listed below.

Hour R X 1 18.4 25 2 16.9 27 3 23.0 30 4 21.2 23 5 21.0 24 6 24.0 25 7 19.3 12 8 15.8 14 9 20.0 13 10 23.0 11 She also decides that lower and upper specification limit for the critical-to-quality variable should be 10 and 30 seconds, respectively.

32. Referring to Table 14-4, suppose the supervisor constructs an R chart to see if the variability in collection times is in-control. What is the center line of this R chart? a) 20.00 b) 20.56 c) 20.40 d) 24.00 ANSWER: c TYPE: MC DIFFICULTY: Moderate KEYWORDS: R chart, center line 33. Referring to Table 14-4, suppose the supervisor constructs an R chart to see if the variability in collection times is in-control. What are the lower and upper control limits for this R chart? a) – 2.33, 43.13 b) – 2.28, 42.28 c) 0, 42.28 d) 0, 43.13 ANSWER: d TYPE: MC DIFFICULTY: Moderate KEYWORDS: R chart, control limit

34. Referring to Table 14-4, suppose the supervisor constructs an R chart to see if the variability in collection times is in-control. This R chart is characterized by which of the following? a) Jump in the level around which the observations vary b) Increasing trend c) Cycles d) Hugging the control limits .

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Statistical Applications in Quality Management

ANSWER: a TYPE: MC DIFFICULTY: Moderate KEYWORDS: R chart 35. Referring to Table 14-4, suppose the sample mean and range data were based on 6 observations per hour instead of 5. How would this change affect the lower and upper control limits of an R chart? a) LCL would increase; UCL would decrease. b) LCL would remain the same; UCL would decrease. c) Both LCL and UCL would remain the same. d) LCL would decrease; UCL would increase.

ANSWER: b TYPE: MC DIFFICULTY: Moderate KEYWORDS: R chart, control limit 36. Referring to Table 14-4, suppose the supervisor constructs an X chart to see if the process is incontrol. What is the center line of the chart? a) 20.00 b) 20.26 c) 21.00 d) 24.26 ANSWER: b TYPE: MC DIFFICULTY: Moderate KEYWORDS: XBar chart, center line 37. Referring to Table 14-4, suppose the supervisor constructs an X chart to see if the process is incontrol. What are the lower and upper control limits of this chart? a) 10.00, 30.00 b) 8.49, 32.03 c) 5.39, 35.13 d) 4.96, 35.56 ANSWER: b TYPE: MC DIFFICULTY: Moderate KEYWORDS: XBar chart, control limit

38. Referring to Table 14-4, suppose the supervisor constructs an X chart to see if the process is incontrol. Which expression best describes this chart? a) Decreasing trend b) Hugging the center line c) Increasing trend d) Cycles ANSWER: .

Statistical Applications in Quality Management

511

b TYPE: MC DIFFICULTY: Moderate KEYWORDS: XBar chart 39. Referring to Table 14-4, what is the value of d 2 factor?

ANSWER: 2.326 TYPE: PR DIFFICULTY: Easy KEYWORDS: d2 factor

40. True or False: The control limits are based on the standard deviation of the process.

ANSWER: True TYPE: TF DIFFICULTY: Moderate KEYWORDS: control limit 41. True or False: The purpose of a control chart is to eliminate common cause variation.

ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: control chart, common causes of variation 42. True or False: Special or assignable causes of variation are signaled by individual fluctuations or patterns in the data.

ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: special causes of variation 43. True or False: One of the focuses of management by process is continuous improvement of a product or service.

ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: Deming’s 14 points 44. True or False: Common causes of variation represent variation due to the inherent variability in the system.

ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: common causes of variation 45. True or False: Common causes of variation are correctable without modifying the system. .

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Statistical Applications in Quality Management

ANSWER: False TYPE: TF DIFFICULTY: Moderate KEYWORDS: common causes of variation 46. True or False: Changes in the system to reduce common cause variation are the responsibility of management.

ANSWER: True TYPE: TF DIFFICULTY: Moderate KEYWORDS: common causes of variation 47. True or False: One of the morals of the red bead experiment is that variation is part of any process.

ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: red bead experiment 48. True or False: In the United States, the control limits on a control chart are placed so that they are 3 standard deviations above and below a central line.

ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: control chart, control limit 49. True or False: The p chart is a control chart used for monitoring the proportion of items in a batch that meet given specifications.

ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: p chart 50. True or False: The R chart is a control chart used to monitor a process mean.

ANSWER: False TYPE: TF DIFFICULTY: Moderate KEYWORDS: R chart 51. True or False: It is not possible for the X chart to be out-of-control when the R chart is in control.

ANSWER: False TYPE: TF DIFFICULTY: Moderate KEYWORDS: XBar chart, R chart, statistical control 52. The cause of variation that can be reduced only by changing the system is ______ cause variation. .

Statistical Applications in Quality Management

513

ANSWER: common TYPE: FI DIFFICULTY: Easy KEYWORDS: common causes of variation 53. _______ causes of variation are correctable without modifying the system.

ANSWER: Special TYPE: FI DIFFICULTY: Easy KEYWORDS: special causes of variation TABLE 14-5 A manufacturer of computer disks took samples of 240 disks on 15 consecutive days. The number of disks with bad sectors was determined for each of these samples. The results are in the table that follows.

Day 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

Bad 9 7 4 6 8 3 6 10 16 24 15 9 4 8 6

% Bad 0.037500 0.029167 0.016667 0.025000 0.033333 0.012500 0.025000 0.041667 0.066667 0.100000 0.062500 0.037500 0.016667 0.033333 0.025000

54. Referring to Table 14-5, the best estimate of the average proportion of disks with bad sectors is __________.

ANSWER: 0.0375 TYPE: FI DIFFICULTY: Moderate KEYWORDS: p chart, center line 55. Referring to Table 14-5, a p control chart is to be made for these data. The center line of the control chart is __________.

.

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Statistical Applications in Quality Management

ANSWER: 0.0375 TYPE: FI DIFFICULTY: Moderate KEYWORDS: p chart, center line 56. Referring to Table 14-5, a p control chart is to be made for these data. The estimate of the standard error of the proportion of disks with bad sectors is __________.

ANSWER: 0.03679/3 or 0.01226 TYPE: FI DIFFICULTY: Moderate KEYWORDS: p chart, standard error 57. Referring to Table 14-5, a p control chart is to be made for these data. The upper control limit is __________, and the lower control limit is __________.

ANSWER: 0.0743; 0.00071 TYPE: FI DIFFICULTY: Moderate KEYWORDS: p chart, control limit

58. Referring to Table 14-5, construct a p control chart for these data.

ANSWER: Proportion of Bad Disks in Samples of Size 240

.

Statistical Applications in Quality Management

515

p Chart 0.1 0.09 0.08

Proportion

0.07 0.06 0.05 0.04 0.03 0.02 0.01 0 0

2

4

6

8

10

12

14

16

Sample Number

TYPE: PR DIFFICULTY: Difficult KEYWORDS: p chart 59. True or False: Referring to Table 14-5, the process seems to be in control.

ANSWER: False TYPE: TF DIFFICULTY: Moderate KEYWORDS: p chart, statistical control

TABLE 14-6 The maker of a packaged candy wants to evaluate the quality of her production process. On each of 16 consecutive days, she samples 600 bags of candy and determines the number in each day's sample that she considers to be of poor quality. The data that she developed follow. .

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Statistical Applications in Quality Management

Number Poor 33 29 31 32 43 45 46 48 48 46 28 32 28 32 31 24

Day 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16

Proportion Poor 0.0550000 0.0483333 0.0516667 0.0533333 0.0716667 0.0750000 0.0766667 0.0800000 0.0800000 0.0766667 0.0466667 0.0533333 0.0466667 0.0533333 0.0516667 0.0400000

60. Referring to Table 14-6, the estimate of the proportion of poor quality bags of candy is __________.

ANSWER: 0.06 TYPE: FI DIFFICULTY: Moderate KEYWORDS: p chart, center line 61. Referring to Table 14-6, a p control chart is to be constructed for these data. The center line for the chart should be located at __________.

ANSWER: 0.06 TYPE: FI DIFFICULTY: Moderate KEYWORDS: p chart, center line 62. Referring to Table 14-6, a p control chart is to be constructed for these data. The estimate of the standard error of the sample proportion is __________.

ANSWER: 0.009695 TYPE: FI DIFFICULTY: Moderate KEYWORDS: p chart, standard error

63. Referring to Table 14-6, a p control chart is to be constructed for these data. The lower control limit is __________, while the upper control limit is __________.

ANSWER: 0.0309139; 0.0890861 TYPE: FI DIFFICULTY: Moderate .

Statistical Applications in Quality Management

517

KEYWORDS: p chart, control limit 64. Referring to Table 14-6, construct a p control chart for these data.

ANSWER: P Chart: Sample Sizes = 600

p Chart 0.09 0.08

Proportion

0.07 0.06 0.05 0.04 0.03 0.02 0.01 0 0

2

4

6

8

10

12

14

16

Sample Number TYPE: PR DIFFICULTY: Difficult KEYWORDS: p chart

TABLE 14-7 A supplier of silicone sheets for producers of computer chips wants to evaluate her manufacturing process. She takes samples of size 5 from each day's output and counts the number of blemishes on each silicone sheet. The results from 20 days of such evaluations are presented below.

Sheet .

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Statistical Applications in Quality Management

Day 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20

1 8 8 10 5 8 9 10 10 6 6 8 6 7 5 7 7 8 11 6 6

2 10 13 12 9 3 7 10 9 10 9 5 4 5 8 12 11 4 4 10 12

3 14 6 7 12 8 9 5 10 6 8 6 7 7 8 10 4 5 11 6 12

4 6 6 7 7 9 6 7 6 9 6 10 7 6 7 6 7 4 11 10 6

5 5 10 9 10 10 9 6 5 9 8 10 12 9 6 10 8 7 10 10 8

Mean 8.6 8.6 9.0 8.6 7.6 8.0 7.6 8.0 8.0 7.4 7.8 7.2 6.8 6.8 9.0 7.4 5.6 9.4 8.4 8.8

Range 9 7 5 7 7 3 5 5 4 3 5 8 4 3 6 7 4 7 4 6

She also decides that the upper specification limit is 10 blemishes.

65. Referring to Table 14-7, an R chart is to be constructed for the number of blemishes. The center line of this R chart is located at __________.

ANSWER: 5.45 TYPE: FI DIFFICULTY: Moderate KEYWORDS: R chart, center line 66. Referring to Table 14-7, an R chart is to be constructed for the number of blemishes. One way to create the lower control limit involves multiplying the average of the sample ranges by D3. For this data set, the value of D3 is ________.

ANSWER: 0 TYPE: FI DIFFICULTY: Easy KEYWORDS: R chart, D3 factor

67. Referring to Table 14-7, an R chart is to be constructed for the number of blemishes. One way to create the upper control limit involves multiplying the average of the sample ranges by D4. For this data set, the value of D4 is ________.

ANSWER: 2.114 TYPE: FI DIFFICULTY: Easy KEYWORDS: R chart, D4 factor

.

Statistical Applications in Quality Management

519

68. Referring to Table 14-7, an R chart is to be constructed for the number of blemishes. The lower control limit for this data set is ________.

ANSWER: 0 or has no limit TYPE: FI DIFFICULTY: Moderate KEYWORDS: R chart, control limit 69. Referring to Table 14-7, an R chart is to be constructed for the number of blemishes. The upper control limit for this data set is ________.

ANSWER: 11.52 TYPE: FI DIFFICULTY: Moderate KEYWORDS: R chart, control limit 70. Referring to Table 14-7, construct an R chart for the number of blemishes.

ANSWER: R Chart 12 10

Range

8 6 4 2 0 0

5

10

15

20

Sample Number

TYPE: PR DIFFICULTY: Difficult KEYWORDS: R chart 71. True or False: Referring to Table 14-7, based on the R chart, it appears that the process is out of control.

ANSWER: False TYPE: TF DIFFICULTY: Moderate KEYWORDS: R chart, statistical control 72. Referring to Table 14-7, an X chart is to be used for the number of blemishes. The center line of this chart is located at __________. .

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Statistical Applications in Quality Management

ANSWER: 7.930 TYPE: FI DIFFICULTY: Moderate KEYWORDS: XBar chart, center line 73. Referring to Table 14-7, an X chart is to be used for the number of blemishes. One way to obtain the control limits is to take the grand mean and add and subtract the product of A2 times the average of the sample ranges. For this data set, the value of A2 is ________.

ANSWER: 0.577 TYPE: FI DIFFICULTY: Easy KEYWORDS: XBar chart, A2 factor 74. Referring to Table 14-7, an X chart is to be used for the number of blemishes. The lower control limit for this data set is ________, while the upper control limit is ________.

ANSWER: 4.785; 11.075 TYPE: FI DIFFICULTY: Moderate KEYWORDS: XBar chart, control limit

75. Referring to Table 14-7, construct an X chart for the number of blemishes.

ANSWER:

.

Statistical Applications in Quality Management

521

Mean Chart 12 10

Mean

8 6 4 2 0 0

5

10

15

20

Sample Number TYPE: PR DIFFICULTY: Difficult KEYWORDS: XBar chart 76. True or False: Referring to Table 14-7, based on the X chart for the number of blemishes, it appears that the process is out of control.

ANSWER: False TYPE: TF DIFFICULTY: Moderate KEYWORDS: XBar chart, statistical control 77. Referring to Table 14-7, what is the value of d 2 factor?

ANSWER: 2.326 TYPE: PR DIFFICULTY: Easy KEYWORDS: d2 factor

TABLE 14-8 Recently, a university switched to a new type of computer-based registration. The registrar is concerned with the amount of time students are spending on the computer registering under the new .

522

Statistical Applications in Quality Management

system. She decides to randomly select 8 students on each of the 12 days of the registration and determine the time each spends on the computer registering. The range, mean, and standard deviation of the times required to register are in the table that follows.

Day Range Mean Std.Dev. 1 10 5.250 3.4949 2 31 15.250 10.3060 3 13 20.375 4.9262 4 21 22.875 8.3911 5 35 8.500 11.3767 6 18 7.875 6.9372 7 25 11.250 8.5815 8 30 7.875 9.5235 9 17 10.250 6.3640 10 22 9.500 7.8740 11 27 7.875 8.7086 12 26 12.875 9.3723 78. Referring to Table 14-8, an R chart is to be constructed for the time required to register. The center line of this R chart is located at __________.

ANSWER: 22.92 TYPE: FI DIFFICULTY: Moderate KEYWORDS: R chart, center line 79. Referring to Table 14-8, an R chart is to be constructed for the time required to register. One way to create the lower control limit involves multiplying the average of the sample ranges by D3. For this data set, the value of D3 is ________.

ANSWER: 0.136 TYPE: FI DIFFICULTY: Easy KEYWORDS: R chart, D3 factor 80. Referring to Table 14-8, an R chart is to be constructed for the time required to register. One way to create the upper control limit involves multiplying the average of the sample ranges by D4. For this data set, the value of D4 is ________.

ANSWER: 1.864 TYPE: FI DIFFICULTY: Easy KEYWORDS: R chart, D4 factor

81. Referring to Table 14-8, an R chart is to be constructed for the time required to register. The lower control limit for this data set is ________.

ANSWER: 3.117 .

Statistical Applications in Quality Management

523

TYPE: FI DIFFICULTY: Moderate KEYWORDS: R chart, control limit 82. Referring to Table 14-8, an R chart is to be constructed for the time required to register. The upper control limit for this data set is ________.

ANSWER: 42.72 TYPE: FI DIFFICULTY: Moderate KEYWORDS: R chart, control limit 83. Referring to Table 14-8, construct an R chart for the time required to register.

ANSWER: R Chart 45 40 35

Range

30 25 20 15 10 5 0 0

2

4

6

8

10

12

Sample Number

TYPE: PR DIFFICULTY: Difficult KEYWORDS: R chart 84. True or False: Referring to Table 14-8, based on the R chart, it appears that the process is out of control.

ANSWER: False TYPE: TF DIFFICULTY: Moderate KEYWORDS: R chart, statistical control 85. Referring to Table 14-8, an X chart is to be used for the time required to register. The center line of this chart is located at __________.

ANSWER: 11.65 TYPE: FI DIFFICULTY: Moderate .

Statistical Applications in Quality Management

KEYWORDS: XBar chart, center line 86. Referring to Table 14-8, an X chart is to be used for the time required to register. One way to obtain the control limits is to take the grand mean and add and subtract the product of A2 times the average of the sample ranges. For this data set, the value of A2 is ________.

ANSWER: 0.373 TYPE: FI DIFFICULTY: Easy KEYWORDS: XBar chart, A2 factor 87. Referring to Table 14-8, an X chart is to be used for the time required to register. The lower control limit for this data set is ________, while the upper control limit is ________.

ANSWER: 3.098; 20.194 TYPE: FI DIFFICULTY: Moderate KEYWORDS: XBar chart, control limit 88. Referring to Table 14-8, construct an X chart for the time required to register.

ANSWER: Mean Chart 25 20

Mean

524

15 10 5 0 0

2

4

6

8

10

12

Sample Number

TYPE: PR DIFFICULTY: Difficulty KEYWORDS: XBar char

89. True or False: Referring to Table 14-8, based on the X chart, it appears that the process is in control.

ANSWER: False TYPE: TF DIFFICULTY: Moderate KEYWORDS: XBar chart, statistical control .

Statistical Applications in Quality Management

525

90. Which of the following is a method for breaking process into a series of steps in order to

eliminate defects and produce near perfect results? a) Deming’s 14 points management. b) Six Sigma management. c) Shewhart-Deming cycle. d) R chart and X chart.

ANSWER: b TYPE: MC DIFFICULTY: Easy KEYWORDS: six sigma management

91. Which of the following is not part of the DMAIC process in Six Sigma management? a) Define b) Do c) Analyze d) Control

ANSWER: b TYPE: MC DIFFICULTY: Easy KEYWORDS: six sigma management 92. Developing operational definitions for each critical-to-quality characteristic involves which part of the DMAIC process? a) Define b) Measure c) Analyze d) Improve e) Control ANSWER: b TYPE: MC DIFFICULTY: Easy KEYWORDS: six sigma management

93.

Determining the root causes of why defects can occur along with the variables in the process that cause these defects to occur involves which part of the DMAIC process? a) Define b) Measure c) Analyze d) Improve e) Control

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Statistical Applications in Quality Management

ANSWER: c TYPE: MC DIFFICULTY: Easy KEYWORDS: six sigma management 94. Maintaining the gains that have been made with a revised process in the long term by avoiding potential problems that can occur when a process is changed involves which part of the DMAIC process? a) Define b) Measure c) Analyze d) Improve e) Control ANSWER: e TYPE: MC DIFFICULTY: Easy KEYWORDS: six sigma management 95. Which of the following best measures the ability of a process to consistently meet specified customer-driven requirement? a) Process capability. b) Specification limits c) Upper control limit d) Lower control limit ANSWER: a TYPE: MC DIFFICULTY: Easy KEYWORDS: process capability 96. True or False: An in-control process must be achieved before being able to estimate process capability.

ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: process capability

97. True or False: A process capability is estimated by the percentage of product or service that fall outside the specification limits.

ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: process capability

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Statistical Applications in Quality Management

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98. True or False: The smaller the capability index, the more capable a process is of meeting customer requirement.

ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: capability index 99. True or False: The Cp index measures the potential of a process, not its actual performance.

ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: capability index 100. True or False: C p >1 indicates that if the process average can be centered, then more than 99.73% of the observations will fall inside the specification limits.

ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: capability index

TABLE 14-9 The manufacturer of cat food constructed control charts and analyzed several quality characteristics.

One characteristic of interest is the weight of the filled cans. The lower specification limit for weight is 2.95 pounds. The table below provides the range and mean of the weights of five cans tested every fifteen minutes during a day’s production.

Number

XBar

Range

Number

XBar .

Range

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Statistical Applications in Quality Management

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

3.05072 3.01228 3.03558 3.01014 3.00858 3.02704 3.04268 3.01052 3.03464 3.02034 3.03764 3.0409 3.0519 3.03994 3.03788

0.0215 0.087 0.0851 0.0674 0.0983 0.0527 0.0508 0.0791 0.0663 0.0538 0.0584 0.0434 0.0762 0.0833 0.0601

16 17 18 19 20 21 22 23 24 25 26 27 28 29

3.04646 3.03868 3.03338 3.03322 3.05078 3.04808 3.05348 3.05516 3.03426 3.0516 3.05562 3.04402 3.06458 3.0544

0.073 0.0696 0.0329 0.0455 0.0215 0.0396 0.0581 0.0682 0.0325 0.0641 0.0809 0.0374 0.0284 0.0738

101. Referring to Table 14-9, an R chart is to be constructed for the weight. The center line of this R chart is located at __________.

ANSWER: 0.0589 TYPE: FI DIFFICULTY: Moderate KEYWORDS: R chart, center line 102. Referring to Table 14-9, an R chart is to be constructed for the weight. One way to create the lower control limit involves multiplying the average of the sample ranges by D3. For this data set, the value of D3 is ________.

ANSWER: 0 TYPE: FI DIFFICULTY: Easy KEYWORDS: R chart, D3 factor 103. Referring to Table 14-9, an R chart is to be constructed for the weight. One way to create the upper control limit involves multiplying the average of the sample ranges by D4. For this data set, the value of D4 is ________.

ANSWER: 2.114 TYPE: FI DIFFICULTY: Easy KEYWORDS: R chart, D4 factor 104. Referring to Table 14-9, an R chart is to be constructed for the weight. The lower control limit for this data set is ________.

ANSWER: 0 TYPE: FI DIFFICULTY: Moderate KEYWORDS: R chart, control limit

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Statistical Applications in Quality Management

529

105. Referring to Table 14-9, an R chart is to be constructed for the weight. The upper control limit for this data set is ________.

ANSWER: 0.1246 TYPE: FI DIFFICULTY: Moderate KEYWORDS: R chart, control limit 106. Referring to Table 14-9, construct an R chart for the time required to register.

ANSWER:

Control Chart Calculations 0.14 UCL

0.12

0.1 0.08

0.06

RBar

0.04

0.02 0 0

5

10

15

20

25

LCL 30

35

TYPE: PR DIFFICULTY: Difficult KEYWORDS: R chart 107. True or False: Referring to Table 14-9, based on the R chart, it appears that the process is out of control.

ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: R chart, statistical control 108. Referring to Table 14-9, an X chart is to be used for the weight. The center line of this chart is located at __________.

ANSWER: 3.0388 TYPE: FI DIFFICULTY: Moderate KEYWORDS: XBar chart, center line .

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Statistical Applications in Quality Management

109. Referring to Table 14-9, an X chart is to be used for the weight. One way to obtain the control limits is to take the grand mean and add and subtract the product of A2 times the average of the sample ranges. For this data set, the value of A2 is ________.

ANSWER: 0.577 TYPE: FI DIFFICULTY: Easy KEYWORDS: XBar chart, A2 factor 110. Referring to Table 14-9, an X chart is to be used for the weight. The lower control limit for this data set is ________, while the upper control limit is ________.

ANSWER: 3.0048; 3.0728 TYPE: FI DIFFICULTY: Moderate KEYWORDS: XBar chart, control limit 111. Referring to Table 14-9, construct an X chart for the weight.

ANSWER: Control Chart Calculations 3.08 UCL

3.07 3.06 3.05 3.04

XBar

3.03 3.02 3.01 LCL 3 0

5

10

15

20

25

30

35

TYPE: PR DIFFICULTY: Difficulty KEYWORDS: XBar char 112. True or False: Referring to Table 14-9, based on the X chart, it appears that the process is in control.

ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: XBar chart, statistical control

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Statistical Applications in Quality Management

113. Referring to Table 14-9, estimate the percentage of the cans whose weight is inside the specification limits.

ANSWER: 0.9998 or 99.98% TYPE: FI DIFFICULTY: Difficult KEYWORDS: XBar chart, control limit

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