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CHAPTER 1: INTRODUCTION AND DATA COLLECTION CHAPTER 2: PRESENTING DATA IN TABLES AND CHARTS CHAPTER 3: NUMERICAL DESCRIPTIVE MEASURES CHAPTER 4: BASIC PROBABILITY CHAPTER 5: SOME IMPORTANT DISCRETE PROBABILITY DISTRIBUTIONS CHAPTER 6: THE NORMAL DISTRIBUTION AND OTHER CONTINUOUS DISTRIBUTIONS CHAPTER 7: SAMPLING DISTRIBUTIONS CHAPTER 8: CONFIDENCE INTERVAL ESTIMATION CHAPTER 9: FUNDAMENTALS OF HYPOTHESIS TESTING: ONE-SAMPLE TESTS CHAPTER 10: TWO-SAMPLE TESTS WITH NUMERICAL DATA CHAPTER 11: ANALYSIS OF VARIANCE CHAPTER 12: TESTS FOR TWO OR MORE SAMPLES WITH CATEGORICAL DATA CHAPTER 13: SIMPLE LINEAR REGRESSION CHAPTER 14: INTRODUCTION TO MULTIPLE REGRESSION CHAPTER 15: MULTIPLE REGRESSION MODEL BUILDING CHAPTER 16: TIME-SERIES FORECASTING AND INDEX NUMBERS
Introduction and Data Collection
CHAPTER 1: INTRODUCTION AND DATA COLLECTION 1. The process of using sample statistics to draw conclusions about true population parameters is called a) statistical inference. b) the scientific method. c) sampling. d) descriptive statistics. ANSWER: a TYPE: MC DIFFICULTY: Easy KEYWORDS: inferential statistics 2. Those methods involving the collection, presentation, and characterization of a set of data in order to properly describe the various features of that set of data are called a) statistical inference. b) the scientific method. c) sampling. d) descriptive statistics. ANSWER: d TYPE: MC DIFFICULTY: Easy KEYWORDS: descriptive statistics 3. The collection and summarization of the socioeconomic and physical characteristics of the employees of a particular firm is an example of a) inferential statistics. b) descriptive statistics. c) a parameter. d) a statistic. ANSWER: b TYPE: MC DIFFICULTY: Easy KEYWORDS: descriptive statistics 4. The estimation of the population average family expenditure on food based on the sample average expenditure of 1,000 families is an example of a) inferential statistics. b) descriptive statistics. c) a parameter. d) a statistic. ANSWER: a TYPE: MC DIFFICULTY: Easy KEYWORDS: inferential statistics 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.
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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. 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 to investigate a sample than the entire population. ANSWER: d TYPE: MC DIFFICULTY: Moderate KEYWORDS: reasons for sampling
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13. 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 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 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 15. 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
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16. 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 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 costs of 250 first-year students and found that their average textbook costs were $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 18. 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 costs of 250 first-year students and found that their average textbook costs were $300 per semester. Identify the variable of interest to the researcher. a) the textbook costs 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
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19. 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 costs of 250 first-year students and found that their average textbook costs were $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 20. 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 21. 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
Introduction and Data Collection
22. 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 23. The British Airways Internet site provides a questionnaire 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 24. 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
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25. 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 salary in terms of real income 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 26. Which of the 4 methods of data collection is involved when a person records the use of the Los Angeles freeway system? a) published sources b) experimentation c) surveying d) observation ANSWER: d TYPE: MC DIFFICULTY: Moderate KEYWORDS: sources of data 27. 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 28. 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 29. 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
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b) a published source c) a random sample d) observational data ANSWER: c TYPE: MC DIFFICULTY: Easy KEYWORDS: sources of data 30. 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 31. 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 32. To monitor campus security, the campus police department is taking a survey of the number of students in a parking lot during 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 during 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 33. 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
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TYPE: MC DIFFICULTY: Moderate KEYWORDS: continuous random variable, types of data 34. 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 35. 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
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36. The chancellor of a major university was concerned about alcohol abuse on her campus and wanted to find out the portion of students at her university who visited campus bars on the weekend before the final exam week. Her advisor 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 37. The chancellor of a major university was concerned about alcohol abuse on her campus and wanted to find out the portion of students at her university who visited campus bars on the weekend before the final exam week. Her advisor 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 38. The chancellor of a major university was concerned about alcohol abuse on her campus and wanted to find out the portion of students at her university who visited campus bars on the weekend before the final exam week. Her advisor 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
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39. The chancellor of a major university was concerned about alcohol abuse on her campus and wanted to find out the portion of students at her university who visited campus bars on the weekend before the final exam week. Her advisor took a random sample of 250 students. Whether or not a particular student in the sample 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: a TYPE: MC DIFFICULTY: Difficult KEYWORDS: continuous random variable, types of data 40. Which of the following sampling methods is a probability sample? a) chunk b) quota sample c) stratified sample d) judgment sample ANSWER: c TYPE: MC DIFFICULTY: Easy KEYWORDS: probability sample 41. 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
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42. 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 43. Which of the following types of samples can you use if you want to make 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 44. The evening host of a dinner dance reached into a bowl, mixed all the tickets around, and selected the ticket to award the grand door prize. What sampling method was used? a) simple random sample b) systematic sample c) stratified sample d) cluster sample ANSWER: a TYPE: MC DIFFICULTY: Easy KEYWORDS: simple random sample, probability sample, sampling method 45. 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
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Introduction and Data Collection 46. 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 47. 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 48. A population frame for a survey contains a listing of 72,345 names. Using a table of random numbers, how many digits will the code numbers for each member of your population contain? a) 3 b) 4 c) 5 d) 6 ANSWER: c TYPE: MC DIFFICULTY: Easy KEYWORDS: random number 49. 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
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50. Which of the following can be reduced by proper interviewer training? a) sampling error b) measurement error c) both of the above d) none of the above ANSWER: b TYPE: MC DIFFICULTY: Difficult KEYWORDS: measurement error, survey worthiness TABLE 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. 51. Referring to Table 1-1, 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 52. Referring to Table 1-1, if there are 4 different brands of videocassette recorders made by the company, the best sampling strategy would be to use 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
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Introduction and Data Collection
53. Referring to Table 1-1, 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 54. 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 55. Referring to Table 1-1, if a customer survey questionnaire is included in all the videocassette recorders made and sold by the company over the next 12 months, this method of collecting data will most likely 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 56. 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 57. 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.
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ANSWER: c TYPE: MC DIFFICULTY: Easy KEYWORDS: categorical random variable, types of data 58. True or False: A population is the totality of items or things under consideration. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: population 59. True or False: A sample is the portion of the universe that is selected for analysis. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: sample 60. 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 61. 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 62. 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 63. 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
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Introduction and Data Collection 64. 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 65. True or False: In a simple random sample, 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 66. 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 67. 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
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68. 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 69. 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 70. 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 71. 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 72. 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 73. 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 74. 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 75. True or False: There can be only one sample drawn from a population. ANSWER: False
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TYPE: TF DIFFICULTY: Easy KEYWORDS: sample, sampling method 76. 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 77. True or False: Sampling error can be completely eliminated by taking larger sample sizes. ANSWER: False TYPE: TF DIFFICULTY: Difficult KEYWORDS: sampling error 78. True or False: Sampling error can be reduced by taking larger sample sizes. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: sampling error 79. 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 80. 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
Introduction and Data Collection 81. True or False: The answer to the question “How many hours on average do you spend watching TV every week?” is an example of a ratio scaled variable. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: ratio scale 82. True or False: The answer to the question “What is your sleeping bag temperature rating?” is an example of a ratio scaled variable. ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: interval scale 83. True or False: Chuck sample is a type of probability sample. ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: chunk sample 84. 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 85. 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 86. True or False: Systematic samples are less efficient than stratified samples. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: systematic sample, stratified sample
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87. 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 88. 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 89. 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. ANSWER: False TYPE: TF DIFFICULTY: Moderate KEYWORDS: systematic sample 90. 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 91. 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. ANSWER: True TYPE: TF DIFFICULTY: Moderate KEYWORDS: measurement error 92. 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
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93. 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 94. 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 95. 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 96. 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
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Introduction and Data Collection
97. 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 98. 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 99. 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 100. 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
Introduction and Data Collection
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101. 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 102. 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 103. 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
26
Introduction and Data Collection
104. 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 105. The Commissioner of Health in New York State wanted to study malpractice litigation in New York. A sample of 31,000 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 106. The Commissioner of Health in New York State wanted to study malpractice litigation in New York. A sample of 31,000 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 107. The Commissioner of Health in New York State wanted to study malpractice litigation in New York. A sample of 31,000 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
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108. The Commissioner of Health in New York State wanted to study malpractice litigation in New York. A sample of 31,000 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 109. An insurance company evaluates many numerical variables about a person before deciding on an appropriate rate for automobile insurance. The number of claims a person has made in the last 3 years is an example of a _______ variable. ANSWER: discrete TYPE: FI DIFFICULTY: Easy KEYWORDS: discrete random variable, types of data 110. 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 111. An insurance company evaluates many numerical variables about a person before deciding on an appropriate rate for automobile insurance. A person's age is an example of a _______ variable. ANSWER: continuous TYPE: FI DIFFICULTY: Easy KEYWORDS: continuous random variable, types of data 112. An insurance company evaluates many numerical variables about a person before deciding on an appropriate rate for automobile insurance. How long a person has been a licensed driver is an example of a _______ variable. ANSWER: continuous TYPE: FI DIFFICULTY: Moderate KEYWORDS: continuous random variable, types of data
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Introduction and Data Collection
113. An insurance company evaluates many numerical variables about a person before deciding on an appropriate rate for automobile insurance. The number of tickets a person has received in the last 3 years is an example of a _______ variable. ANSWER: discrete TYPE: FI DIFFICULTY: Easy KEYWORDS: discrete random variable, types of data 114. 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 115. 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 116. 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 117. 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 118. 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 119. 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: continuous TYPE: FI DIFFICULTY: Moderate
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KEYWORDS: continuous random variable, types of data 120. 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 121. 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 122. 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 123. The Dean of Students conducted a survey on campus. SAT score in mathematics is an example of a _______ numerical variable. ANSWER: continuous TYPE: FI DIFFICULTY: Easy KEYWORDS: continuous random variable, types of data 124. 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
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Introduction and Data Collection
125. 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 126. 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 127. 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 128. 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 129. 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 130. 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
Introduction and Data Collection
131. 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 132. 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 133. ________ results from the exclusion of certain groups of subjects from a population frame. ANSWER: Coverage error TYPE: FI DIFFICULTY: Difficult KEYWORDS: coverage error, survey worthiness 134. Coverage error results in a ________. ANSWER: selection bias TYPE: FI DIFFICULTY: Difficult KEYWORDS: selection bias, survey worthiness 135. ________ 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 136. The sampling process begins by locating appropriate data sources called ___________. ANSWER: frames TYPE: FI DIFFICULTY: Easy KEYWORDS: frames, sampling method
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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
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3. A type of vertical bar chart in which the categories are plotted in the descending rank order of the magnitude of their frequencies is called a a) contingency table. b) Pareto diagram. c) dot plot. d) pie chart. ANSWER: b TYPE: MC DIFFICULTY: Easy KEYWORDS: Pareto diagram 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
Introduction and Data Collection
6. The width of each bar in a histogram corresponds to the a) differences between the boundaries of the class. b) number of observations in each class. c) midpoint of each class. d) percentage of observations in each class. ANSWER: a TYPE: MC DIFFICULTY: Easy KEYWORDS: frequency distribution TABLE 2-3 Every spring semester, the School of Business coordinates with local business leaders a luncheon for graduating seniors, their families, and friends. Corporate sponsorship pays for the lunches of each of the seniors, but students have to purchase tickets to cover the cost of lunches served to the 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 85
Frequency
34
80 60 40 20
18
17
3
0
4
5
0 0
1
2 Guests per St udent
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
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8. Referring to the histogram from Table 2-3, if all the tickets purchased were used, how many guests attended the luncheon? a) 4 b) 152 c) 275 d) 388 ANSWER: d TYPE: MC DIFFICULTY: Difficult EXPLANATION: The total number of guests is
6 i 1
X i fi
KEYWORDS: interpretation, histogram 9. A professor of economics at a small Texas university wanted to determine which 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
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10. When polygons or histograms are constructed, which axis must show the true zero or "origin?" a) the horizontal axis b) the vertical axis c) both the horizontal and vertical axes d) neither the horizontal nor the vertical axis ANSWER: b TYPE: MC DIFFICULTY: Easy KEYWORDS: polygon, histogram 11. When constructing charts, the following is plotted at the class midpoints: a) frequency histograms. b) percentage polygons. c) cumulative relative frequency ogives. d) all of the above ANSWER: b TYPE: MC DIFFICULTY: Easy KEYWORDS: percentage polygon TABLE 2-4 A survey was conducted to determine how people rated the quality of programming available on television. Respondents were asked to rate the overall quality from 0 (no quality at all) to 100 (extremely good quality). The stem-and-leaf display of the data is shown below. Stem Leaves 3 24 4 03478999 5 0112345 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) 0.00 b) 0.04 c) 0.96 d) 1.00 ANSWER: b TYPE: MC DIFFICULTY: Easy KEYWORDS: stem-and-leaf display, interpretation
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13. Referring to Table 2-4, what percentage of the respondents rated overall television quality with a rating of 50 or below? a) 0.11 b) 0.40 c) 0.44 d) 0.56 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) 0.11 b) 0.40 c) 0.44 d) 0.56 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
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ANSWER: d TYPE: MC DIFFICULTY: Moderate KEYWORDS: relative frequency distribution, interpretation 17. Referring to Table 2-5, if 100 calls were randomly sampled, how many calls lasted 15 minutes or longer? a) 10 b) 14 c) 26 d) 74 ANSWER: c TYPE: MC DIFFICULTY: Moderate KEYWORDS: relative frequency distribution, interpretation 18. Referring to Table 2-5, if 10 calls lasted 30 minutes or more, how many calls lasted less than 5 minutes? a) 10 b) 185 c) 295 d) 500 ANSWER: b TYPE: MC DIFFICULTY: Moderate KEYWORDS: relative frequency distribution, interpretation 19. Referring to Table 2-5, what is the cumulative relative frequency for the percentage of calls that lasted under 20 minutes? a) 0.10 b) 0.59 c) 0.76 d) 0.84 ANSWER: d TYPE: MC DIFFICULTY: Easy KEYWORDS: cumulative relative frequency
Introduction and Data Collection
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 between 15 minutes but less than 20 minutes. a) 0.10 b) 0.16 c) 10 d) 16 ANSWER: c TYPE: MC DIFFICULTY: Easy KEYWORDS: relative frequency distribution, interpretation 22. Referring to Table 2-5, if 100 calls were sampled, _______ 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
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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. ANSWER: a TYPE: MC DIFFICULTY: Easy KEYWORDS: contingency table TABLE 2-6 A sample of 200 students at a Big Ten university was taken after the midterm to ask them whether they went bar hopping the weekend before the midterm or spent the weekend studying, and whether they did well or poorly on the midterm. The following table contains the result:
Studying for Exam Went Bar Hopping
Did Well on Midterm 80 30
Did Poorly on Midterm 20 70
26. Referring to Table 2-6, of those who went bar hopping the weekend before the midterm in the sample, _______ percent of them did well on the midterm. a) 15 b) 27.27 c) 30 d) 55 ANSWER: c TYPE: MC DIFFICULTY: Easy KEYWORDS: contingency table, interpretation
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27. 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 28. Referring to Table 2-6, _______ percent of the students in the sample went bar hopping the weekend before the midterm and did well on the midterm. a) 15 b) 27.27 c) 30 d) 50 ANSWER: a TYPE: MC DIFFICULTY: Easy KEYWORDS: contingency table, interpretation 29. Referring to Table 2-6, _______ percent of the students in the sample spent the weekend studying and did well on the midterm. a) 40 b) 50 c) 72.72 d) 80 ANSWER: a TYPE: MC DIFFICULTY: Easy KEYWORDS: contingency table, interpretation 30. Referring to Table 2-6, if the sample is a good representation of the population, we can expect _______ percent of the students in the population to spend the weekend studying and do poorly on the midterm. a) 10 b) 20 c) 45 d) 50 ANSWER: a TYPE: MC DIFFICULTY: Easy KEYWORDS: contingency table, interpretation 31. Referring to Table 2-6, if the sample is a good representation of the population, we can expect _______ percent of those who spent the weekend studying to do poorly on the midterm. a) 10 b) 20 c) 45 d) 50
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ANSWER: b TYPE: MC DIFFICULTY: Moderate KEYWORDS: contingency table, interpretation 32. Referring to Table 2-6, if the sample is a good representation of the population, we can expect _______ percent of those who did poorly on the midterm to have spent the weekend studying. a) 10 b) 22.22 c) 45 d) 50 ANSWER: b TYPE: MC DIFFICULTY: Moderate KEYWORDS: contingency table, interpretation 33. In a contingency table, the number of rows and columns a) must always be the same. b) must always be 2. c) must add to 100%. d) none of the above ANSWER: d TYPE: MC DIFFICULTY: Moderate KEYWORDS: contingency table
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34. Retailers are always interested in determining why a customer selected their store to make a purchase. A sporting goods retailer conducted a customer survey to determine why its customers shopped at the store. The results are shown in the bar chart below. What proportion of the customers responded that they shopped at the store because of the merchandise or the convenience?
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 1 234447899 2 22223455678889 3 0011135778 4 02345579 5 112466 6 158 35. Referring to Table 2-7, locate the first leaf, i.e., the smallest leaf with the smallest stem. This represents a wait of ________ months. ANSWER: 12 TYPE: FI DIFFICULTY: 1 Easy KEYWORDS: stem-and-leaf display, interpretation
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36. Referring to Table 2-7, the civil suit with the longest wait between when the suit was filed and when it was adjudicated had a wait of ________ months. ANSWER: 68 TYPE: FI DIFFICULTY: Easy KEYWORDS: stem-and-leaf display, interpretation 37. Referring to Table 2-7, the civil suit with the fourth shortest waiting time between when the suit was filed and when it was adjudicated had a wait of ________ months. ANSWER: 14 TYPE: FI DIFFICULTY: Moderate KEYWORDS: stem-and-leaf display, interpretation 38. Referring to Table 2-7, ________ percent of the cases were adjudicated within the first 2 years. ANSWER: 30 TYPE: FI DIFFICULTY: Moderate KEYWORDS: stem-and-leaf display, interpretation 39. Referring to Table 2-7, ________ percent of the cases were not adjudicated within the first 4 years. ANSWER: 20 TYPE: FI DIFFICULTY: Moderate KEYWORDS: stem-and-leaf display, interpretation 40. Referring to Table 2-7, if a frequency distribution with equal sized classes was made from this data, and the first class was "10 but less than 20," the frequency of that class would be ________. ANSWER: 9 TYPE: FI DIFFICULTY: Easy KEYWORDS: stem-and-leaf display, interpretation 41. Referring to Table 2-7, if a frequency distribution with equal sized classes was made from this data, and the first class was "10 but less than 20," the relative frequency of the third class would be ________. ANSWER: 0.20 or 20% or 10/50 TYPE: FI DIFFICULTY: Moderate KEYWORDS: stem-and-leaf display, relative frequency distribution 42. Referring to Table 2-7, if a frequency distribution with equal sized classes was made from this data, and the first class was "10 but less than 20," the cumulative percentage of the second class would be ________. ANSWER: 46% or 0.46 or 23/50 TYPE: FI DIFFICULTY: Moderate KEYWORDS: stem-and-leaf display, cumulative percentage distribution TABLE 2-8
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The stem-and-leaf display represents the number of times in a year that a random sample of 100 "lifetime" members of a health club actually visited the facility. Stem Leaves 0 012222233333344566666667789999 1 1111222234444455669999 2 00011223455556889 3 0000446799 4 011345567 5 0077 6 8 7 67 8 3 9 0247 43. Referring to Table 2-8, the person who has the largest leaf associated with the smallest stem visited the facility ________ times. ANSWER: 9 TYPE: FI DIFFICULTY: Moderate KEYWORDS: stem-and-leaf display, interpretation 44. Referring to Table 2-8, the person who visited the health club less than anyone else in the sample visited the facility ________ times. ANSWER: 0 or no TYPE: FI DIFFICULTY: Easy KEYWORDS: stem-and-leaf display, interpretation 45. Referring to Table 2-8, the person who visited the health club more than anyone else in the sample visited the facility ________ times. ANSWER: 97 TYPE: FI DIFFICULTY: Easy KEYWORDS: stem-and-leaf display, interpretation
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Introduction and Data Collection
46. 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 47. 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 48. Referring to Table 2-8, if a frequency distribution with equal sized classes was made from this data, and the first class was "0 but less than 10," the frequency of the fifth class would be ________. ANSWER: 9 TYPE: FI DIFFICULTY: Moderate KEYWORDS: stem-and-leaf display, frequency distribution 49. Referring to Table 2-8, if a frequency distribution with equal sized classes was made from this data, and the first class was "0 but less than 10," the relative frequency of the last class would be ________. ANSWER: 4% or 0.04 or 4/100 TYPE: FI DIFFICULTY: Moderate KEYWORDS: stem-and-leaf display, relative frequency distribution 50. Referring to Table 2-8, if a frequency distribution with equal sized classes was made from this data, and the first class was "0 but less than 10," the cumulative percentage of the next-to-last class would be ________. ANSWER: 96% or 0.96 or 96/100 TYPE: FI DIFFICULTY: Moderate KEYWORDS: stem-and-leaf display, cumulative percentage distribution
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51. Referring to Table 2-8, if a frequency distribution with equal sized classes was made from this data, and the first class was "0 but less than 10," the class midpoint of the third class would be ________. ANSWER: 25 or (20+30)/2 TYPE: FI DIFFICULTY: Moderate KEYWORDS: stem-and-leaf display, class midpoint TABLE 2-9 The frequency distribution below represents the rents of 250 randomly selected federally subsidized apartments in Minneapolis. Rents in $ Frequency 300 but less than 400 113 400 but less than 500 85 500 but less than 600 32 600 but less than 700 16 700 but less than 800 4 52. Referring to Table 2-9, ________ apartments rented for at least $400 but less than $600. ANSWER: 117 TYPE: FI DIFFICULTY: Easy KEYWORDS: frequency distribution 53. Referring to Table 2-9, ________ percent of the apartments rented for less than $600. ANSWER: 92% or 230/250 TYPE: FI DIFFICULTY: Easy KEYWORDS: frequency distribution, cumulative percentage distribution 54. Referring to Table 2-9, ________ percent of the apartments rented for at least $500. ANSWER: 20.8% or 52/250 TYPE: FI DIFFICULTY: Moderate KEYWORDS: frequency distribution, cumulative percentage distribution 55. Referring to Table 2-9, the class midpoint of the second class is ________. ANSWER: 450 TYPE: FI DIFFICULTY: Easy KEYWORDS: frequency distribution, class midpoint
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56. 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 57. Referring to Table 2-9, the percentage of apartments renting for less than $600 is ________. ANSWER: 230/250 or 23/25 or 92% or 0.92 TYPE: FI DIFFICULTY: Moderate KEYWORDS: frequency distribution, cumulative percentage distribution TABLE 2-10 The histogram below represents scores achieved by 200 job applicants on a personality profile. 0.30
Rel.Freq.
0.20
0.20
0.20
0.20
0.10
0.10
0.10
0.10
0.10
0.00 0
10
20
30
40
50
60
70
58. 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 59. 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
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60. Referring to the histogram from Table 2-10, the number of job applicants who scored between 30 and 60 is _______. ANSWER: 80 TYPE: FI DIFFICULTY: Moderate KEYWORDS: histogram 61. Referring to the histogram from Table 2-10, the number of job applicants who scored 50 or above is _______. ANSWER: 40 TYPE: FI DIFFICULTY: Moderate KEYWORDS: histogram 62. Referring to the histogram from Table 2-10, 90% of the job applicants scored above or equal to ________. ANSWER: 10 TYPE: FI DIFFICULTY: Moderate KEYWORDS: histogram, cumulative percentage distribution 63. Referring to the histogram from Table 2-10, half of the job applicants scored below ________. ANSWER: 30 TYPE: FI DIFFICULTY: Moderate KEYWORDS: histogram, cumulative percentage distribution 64. Referring to the histogram from Table 2-10, _______ percent of the applicants scored below 20 or at least 50. ANSWER: 50% TYPE: FI DIFFICULTY: Moderate KEYWORDS: histogram, cumulative percentage distribution 65. 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
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Introduction and Data Collection
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
66. Referring to Table 2-11, if a frequency distribution for the defects data is constructed, using "0 but less than 5" as the first class, the frequency of the “20 but less than 25” class would be ________. ANSWER: 4 TYPE: FI DIFFICULTY: Easy KEYWORDS: frequency distribution 67. Referring to Table 2-11, if a frequency distribution for the defects data is constructed, using "0 but less than 5" as the first class, the relative frequency of the “15 but less than 20” class would be ________. ANSWER: 0.08 or 8% or 2/25 TYPE: FI DIFFICULTY: Moderate KEYWORDS: relative frequency distribution 68. Referring to Table 2-11, construct a frequency distribution for the defects data, using "0 but less than 5" as the first class. ANSWER: Defects Frequency 0 but less than 5 4 5 but less than 10 6 10 but less than 15 2 15 but less than 20 2 20 but less than 25 4 25 but less than 30 7 TYPE: PR DIFFICULTY: Easy KEYWORDS: frequency distribution
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69. Referring to Table 2-11, construct a relative frequency or percentage distribution for the defects data, using "0 but less than 5" as the first class. ANSWER: Defects Percentage 0 but less than 5 16 5 but less than 10 24 10 but less than 15 8 15 but less than 20 8 20 but less than 25 16 25 but less than 30 28 TYPE: PR DIFFICULTY: Moderate KEYWORDS: relative frequency distribution, percentage distribution 70. Referring to Table 2-11, construct a cumulative percentage distribution for the defects data if the corresponding frequency distribution uses "0 but less than 5" as the first class. ANSWER: Defects CumPct 0 0 5 16 10 40 15 48 20 56 25 72 30 100 TYPE: PR DIFFICULTY: Moderate KEYWORDS: cumulative percentage distribution
Introduction and Data Collection
71. Referring to Table 2-11, construct a histogram for the defects data, using "0 but less than 5" as the first class. ANSWER: 7
Frequency
7
6 6 5 4
4
4 3 2
2
2 1 0
0
10
5
15 20 Number of Defect s
25
30
TYPE: PR DIFFICULTY: Easy KEYWORDS: histogram, frequency distribution 72. Referring to Table 2-11, construct a cumulative percentage polygon for the defects data if the corresponding frequency distribution uses "0 but less than 5" as the first class.
ANSWER: 100 90 80 70
Percentage of Chips
52
60 50 40 30 20 10 0 0
5
10
15 20 Number of Defect s
TYPE: PR DIFFICULTY: Moderate KEYWORDS: cumulative percentage polygon
25
30
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73. The point halfway between the boundaries of each class interval in a grouped frequency distribution is called the _______. ANSWER: class midpoint TYPE: FI DIFFICULTY: Easy KEYWORDS: cumulative percentage polygon, frequency distribution 74. A _______ is a vertical bar chart in which the rectangular bars are constructed at the boundaries of each class interval. ANSWER: histogram TYPE: FI DIFFICULTY: Easy KEYWORDS: histogram 75. It is essential that each class grouping or interval in a frequency distribution be ________. ANSWER: non-overlapping TYPE: FI DIFFICULTY: Easy KEYWORDS: frequency distribution, class interval 76. In order to compare one large batch of numerical data to another, a ________ distribution must be developed from the frequency distribution. ANSWER: relative frequency or percentage TYPE: FI DIFFICULTY: Easy KEYWORDS: relative frequency distribution, percentage distribution 77. When comparing two or more large batches of numerical data, the distributions being developed should use the same ________. ANSWER: class boundaries TYPE: FI DIFFICULTY: Easy KEYWORDS: class boundaries 78. It is desirable that the width of each class grouping or interval in a frequency distribution be ________. ANSWER: the same or equal TYPE: FI DIFFICULTY: Easy KEYWORDS: class interval, frequency distribution
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79. 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 80. A _______ is a summary table in which numerical data are tallied into class intervals or categories. ANSWER: frequency distribution TYPE: FI DIFFICULTY: Easy KEYWORDS: frequency distribution, class interval 81. True or False: In general, grouped frequency distributions should have between 5 and 15 class intervals. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: frequency distribution, number of classes 82. True or False: The sum of relative frequencies in a distribution always equals 1. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: relative frequency 83. True or False: The sum of cumulative frequencies in a distribution always equals 1. ANSWER: False TYPE: TF DIFFICULTY: Moderate KEYWORDS: cumulative frequency distribution 84. True or False: In graphing bivariate categorical data, the side-by-side bar chart is best suited when the primary interest is in demonstrating differences in magnitude rather than differences in percentages. ANSWER: True TYPE: TF DIFFICULTY: Moderate KEYWORDS: side-by-side chart
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85. True or False: When constructing a frequency distribution, classes should be selected in such a way that they are of equal width. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: frequency distribution 86. True or False: A research analyst was directed to arrange raw data collected on the yield of wheat, ranging from 40 to 93 bushels per acre, in a frequency distribution. He should choose 30 as the class interval width. ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: frequency distribution, class interval 87. True or False: If the values of the seventh and eighth classes in a cumulative frequency distribution are the same, we know that there are no observations in the eighth class. ANSWER: True TYPE: TF DIFFICULTY: Moderate KEYWORDS: cumulative frequency distribution 88. True or False: The bar chart is preferred to the pie chart, because the human eye can more accurately judge length comparisons against a fixed scale (as in a bar chart) than angular measures (as in a pie chart). ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: bar chart, pie chart 89. True or False: One of the advantages of a pie chart is that it clearly shows that the total of all the categories of the pie adds to 100%. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: pie chart 90. True or False: The larger the number of observations in a numerical data set, the larger the number of class intervals needed for a grouped frequency distribution. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: class interval, frequency distribution
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91. True or False: Determining the class boundaries of a frequency distribution is highly subjective. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: class boundaries, frequency distribution 92. True or False: The original data values cannot be assessed once they are grouped into a frequency distribution table. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: frequency distribution 93. True or False: The percentage distribution cannot be constructed from the frequency distribution directly. ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: percentage distribution, frequency distribution 94. True or False: The stem-and-leaf display is often superior to the frequency distribution in that it maintains the original values for further analysis. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: stem-and-leaf display, frequency distribution 95. True or False: The relative frequency is the frequency in each class divided by the total number of observations. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: relative frequency distribution 96. True or False: Ogives are plotted at the midpoints of the class groupings. ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: ogives, midpoint
Introduction and Data Collection
97. True or False: Percentage polygons are plotted at the boundaries of the class groupings. ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: percentage polygons 98. True or False: The main principle behind the Pareto diagram is the ability to track the "vital few" from the "trivial many." ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: Pareto diagram 99. True or False: A histogram can have gaps between the bars, whereas bar charts cannot have gaps. ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: histogram, bar chart 100. True or False: Histograms are used for numerical data, while bar charts are suitable for categorical data. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: histogram, bar chart 101. True or False: A Wal-Mart store in a small town monitors customer complaints and organizes these complaints into six distinct categories. Over the past year, the company has received 534 complaints. One possible graphical method for representing these data would be a Pareto chart. ANSWER: True TYPE: TF DIFFICULTY: Moderate KEYWORDS: Pareto diagram 102. True or False: Apple Computer, Inc. collected information on the age of their customers. The youngest customer was 12 and the oldest was 72. To study the distribution of age among its customers, it can use a Pareto diagram. ANSWER: False TYPE: TF DIFFICULTY: Moderate KEYWORDS: Pareto diagram
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103. True or False: Apple Computer, Inc. collected information on the age of their customers. The youngest customer was 12 and the oldest was 72. To study the distribution of age among its customers, it can use a pie chart. ANSWER: False TYPE: TF DIFFICULTY: Moderate KEYWORDS: pie chart 104. True or False: Apple Computer, Inc. collected information on the age of their customers. The youngest customer was 12 and the oldest was 72. To study the distribution of age among its customers, it can use a percentage polygon. ANSWER: True TYPE: TF DIFFICULTY: Moderate KEYWORDS: percentage polygon 105. True or False: Apple Computer, Inc. collected information on the age of their customers. The youngest customer was 12 and the oldest was 72. To study the percentage of their customers who are below a certain age, it can use an ogive. ANSWER: True TYPE: TF DIFFICULTY: Moderate KEYWORDS: ogive 106. True or False: If you wish to construct a graph of a relative frequency distribution, you would most likely construct an ogive first. ANSWER: False TYPE: TF DIFFICULTY: Moderate KEYWORDS: ogive 107. True or False: An ogive is a cumulative percentage polygon. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: Ogive, cumulative percentage polygon 108. True or False: A side-by-side chart is two histograms plotted along side each other. ANSWER: False TYPE: TF DIFFICULTY: Moderate KEYWORDS: side-by-side chart
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109. True or False: A good choice for the number of class groups to use in constructing frequency distribution is approximately
n.
ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: number of classes 110. True or False: In general, a frequency distribution should have at least 8 class groups but no more than 20. ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: number of classes 111. True of False: To determine the width of class interval, divide the number of class groups by the range of the data. ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: class interval 112. True or False: The percentage polygon is formed by having the lower boundary of each class represent the data in that class and then connecting the sequence of lower boundaries at their respective class percentages. ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: percentage polygon 113. True or False: A polygon can be constructed from a histogram. ANSWER: True TYPE: TF DIFFICULTY: Moderate KEYWORDS: polygon 114. To evaluate two categorical variables at the same time, a _______ should be developed. ANSWER: contingency or cross-classification table TYPE: FI DIFFICULTY: Easy KEYWORDS: contingency table, cross-classification table
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115. Relationships in a contingency table can be examined more fully if the frequencies are converted into _______ . ANSWER: percentages or proportions TYPE: FI DIFFICULTY: Easy KEYWORDS: contingency table TABLE 2-12 The table below contains the opinions of a sample of 200 people broken down by gender about the latest congressional plan to eliminate anti-trust exemptions for professional baseball.
Female Male Totals
For 38 12 50
Neutral Against 54 12 36 48 90 60
Totals 104 96 200
116. Referring to Table 2-12, construct a table of row percentages. ANSWER: For Neutral Against Totals Female 36.54 51.92 11.54 100.00 Male 12.50 37.50 50.00 100.00 Totals 25.00 45.00 30.00 100.00 TYPE: PR DIFFICULTY: Easy KEYWORDS: row percentages 117. Referring to Table 2-12, construct a table of column percentages. ANSWER: For Neutral Against Totals Female 76.00 60.00 20.00 52.00 Male 24.00 40.00 80.00 48.00 Totals 100.00 100.00 100.00 100.00 TYPE: PR DIFFICULTY: Easy KEYWORDS: column percentages 118. Referring to Table 2-12, construct a table of total percentages. ANSWER: For Neutral Against Totals Female 19.00 27.00 6.00 52.00 Male 6.00 18.00 24.00 48.00 Totals 25.00 45.00 30.00 100.00 TYPE: PR DIFFICULTY: Easy KEYWORDS: total percentages
Introduction and Data Collection
119. Referring to Table 2-12, of those for the plan in the sample, ________ percent were females. ANSWER: 76% TYPE: FI DIFFICULTY: Moderate KEYWORDS: contingency table, column percentages 120. Referring to Table 2-12, of those neutral in the sample, ________ percent were males. ANSWER: 40% TYPE: FI DIFFICULTY: Moderate KEYWORDS: contingency table, column percentages 121. Referring to Table 2-12, of the males in the sample, ________ percent were for the plan. ANSWER: 12.50% TYPE: FI DIFFICULTY: Moderate KEYWORDS: contingency table 122. Referring to Table 2-12, of the females in the sample, ________ percent were against the plan. ANSWER: 11.54% TYPE: FI DIFFICULTY: Moderate KEYWORDS: contingency table 123. Referring to Table 2-12, of the females in the sample, ________ percent were either neutral or against the plan. ANSWER: 63.46% or (51.92+11.54)% TYPE: FI DIFFICULTY: Moderate KEYWORDS: contingency table 124. Referring to Table 2-12, ________ percent of the 200 were females who were against the plan. ANSWER: 6% TYPE: FI DIFFICULTY: Moderate KEYWORDS: contingency table 125. Referring to Table 2-12, ________ percent of the 200 were males who were neutral. ANSWER: 18% TYPE: FI DIFFICULTY: Moderate KEYWORDS: contingency table 126. Referring to Table 2-12, ________ percent of the 200 were females who were either neutral or against the plan. ANSWER: 33%
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TYPE: FI DIFFICULTY: Difficult KEYWORDS: contingency table 127. Referring to Table 2-12, _______ percent of the 200 were males who were not against the plan. ANSWER: 24% TYPE: FI DIFFICULTY: Difficult KEYWORDS: contingency table 128. Referring to Table 2-12, _______ percent of the 200 were not neutral. ANSWER: 55% TYPE: FI DIFFICULTY: Difficult KEYWORDS: contingency table, row percentages 129. Referring to Table 2-12, _______ percent of the 200 were against the plan. ANSWER: 30% TYPE: FI DIFFICULTY: Moderate KEYWORDS: contingency table, row percentages 130. Referring to Table 2-12, ________ percent of the 200 were males. ANSWER: 48% TYPE: FI DIFFICULTY: Easy KEYWORDS: contingency table, column percentages 131. Referring to Table 2-12, if the sample is a good representation of the population, we can expect _______ percent of the population will be for the plant. ANSWER: 25% TYPE: FI DIFFICULTY: Moderate KEYWORDS: contingency table, row percentages 132. Referring to Table 2-12, if the sample is a good representation of the population, we can expect _______ percent of the population will be males. ANSWER: 48% TYPE: FI DIFFICULTY: Moderate KEYWORDS: contingency table, column percentages 133. Referring to Table 2-12, if the sample is a good representation of the population, we can expect _______ percent of those for the plan in the population will be males. ANSWER: 24% TYPE: FI DIFFICULTY: Moderate KEYWORDS: contingency table
Introduction and Data Collection
134. Referring to Table 2-12, if the sample is a good representation of the population, we can expect _______ percent of the males in the population will be against the plan. ANSWER: 50% TYPE: FI DIFFICULTY: Moderate KEYWORDS: contingency table 135. Referring to Table 2-12, if the sample is a good representation of the population, we can expect _______ percent of the females in the population will not be against the plan. ANSWER: 88.46% or (36.54+51.92) TYPE: FI DIFFICULTY: Moderate KEYWORDS: contingency table
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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) geometric mean ANSWER: c TYPE: MC DIFFICULTY: Easy KEYWORDS: measure of central tendency, mode, arithmetic mean, median, geometric mean 3. Which of the arithmetic mean, median, mode, and geometric mean are resistant measures of central tendency? a) the arithmetic mean and median only b) the median only c) the mode and geometric mean only d) the arithmetic mean and mode only ANSWER: b TYPE: MC DIFFICULTY: Moderate KEYWORDS: measure of central tendency, resistant to outliers, mean, median, mode 4. In a right-skewed distribution, a) the median equals the arithmetic mean. b) the median is less than the arithmetic mean. c) the median is larger than the arithmetic mean. d) none of the above ANSWER: b TYPE: MC DIFFICULTY: Easy KEYWORDS: shape
Introduction and Data Collection
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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 of 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
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9. In general, which of the following descriptive summary measures cannot be easily approximated from a boxand-whisker plot? a) the variance b) the range c) the interquartile range d) the median ANSWER: a TYPE: MC DIFFICULTY: Easy KEYWORDS: box-and-whisker plot, variance, range, interquartile range, median 10. The smaller the spread of scores around the arithmetic mean, a) the smaller the interquartile range. b) the smaller the standard deviation. c) the smaller the coefficient of variation. d) all of 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
Introduction and Data Collection
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
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16. According to the empirical rule, if the data form a "bell-shaped" normal distribution, _______ percent of the observations will be contained within 1 standard deviation around the arithmetic mean. a) 68.26 b) 75.00 c) 88.89 d) 93.75 ANSWER: a TYPE: MC DIFFICULTY: Moderate KEYWORDS: empirical rule, normal distribution 17. According to the empirical rule, if the data form a "bell-shaped" normal distribution, _______ percent of the observations will be contained within 3 standard deviations around the arithmetic mean. a) 68.26 b) 75.00 c) 95.0 d) 99.7 ANSWER: d TYPE: MC DIFFICULTY: Moderate KEYWORDS: empirical rule, normal distribution 18. Which of the following is NOT a measure of central tendency? a) the arithmetic mean b) the geometric mean c) the mode d) the interquartile range ANSWER: d TYPE: MC DIFFICULTY: Easy KEYWORDS: measure of central tendency, arithmetic mean, geometric mean, mode, interquartile range 19. Which of the following is NOT sensitive to extreme values? a) the range b) the standard deviation c) the interquartile range d) the coefficient of variation ANSWER: c TYPE: MC DIFFICULTY: Moderate KEYWORDS: resistant to outliers, range, standard deviation, interquartile range, coefficient of variation
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20. Which of the following is sensitive to extreme values? a) the median b) the interquartile range c) the arithmetic mean d) the 1st quartile ANSWER: c TYPE: MC DIFFICULTY: Moderate KEYWORDS: resistant to outliers, median, interquartile range, arithmetic mean, quartiles 21. Which of the following is the easiest to compute? a) the arithmetic mean b) the median c) the mode d) the geometric mean ANSWER: c TYPE: MC DIFFICULTY: Easy KEYWORDS: mode, arithmetic mean, median, geometric mean 22. According to the Bienayme-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: Bienayme-Chebyshev rule 23. According to the Bienayme-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: Easy KEYWORDS: Bienayme-Chebyshev rule
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24. According to the Bienayme-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: Bienayme-Chebyshev rule 25. According to the Bienayme-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: Bienayme-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
Introduction and Data Collection
28. Referring to Table 3-1, identify the first quartile of the ages of the uninsured senior citizens. ANSWER: 65 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 years TYPE: PR DIFFICULTY: Moderate KEYWORDS: quartiles 30. Referring to Table 3-1, identify the interquartile range of the ages of the uninsured senior citizens. ANSWER: 16 years TYPE: PR DIFFICULTY: Moderate KEYWORDS: interquartile range 31. Referring to Table 3-1, identify which of the following is the correct statement. a) One fourth of the senior citizens sampled are below 65.5 years of age. b) The middle 50% of the senior citizens sampled are between 65.5 and 73.0 years of age. c) The average age of senior citizens sampled is 73.5 years of age. d) All of the above are correct. ANSWER: a TYPE: MC DIFFICULTY: Moderate KEYWORDS: quartiles, arithmetic mean 32. Referring to Table 3-1, identify which of the following is the correct statement. a) One fourth of the senior citizens sampled are below 64 years of age. b) The middle 50% of the senior citizens sampled are between 65.5 and 73.0 years of age. c) 25% of the senior citizens sampled are older than 81 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
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TYPE: PR DIFFICULTY: Moderate KEYWORDS: variance 35. Referring to Table 3-1, calculate the standard deviation of the ages of the uninsured senior citizens correct to the nearest hundredth of a year. ANSWER: 9.74 years TYPE: PR DIFFICULTY: Moderate KEYWORDS: standard deviation 36. Referring to Table 3-1, calculate the coefficient of variation of the ages of the uninsured senior citizens. ANSWER: 13.16% TYPE: PR DIFFICULTY: Moderate KEYWORDS: coefficient of variation 37. True or False: The median of the values 3.4, 4.7, 1.9, 7.6, and 6.5 is 1.9. ANSWER: False TYPE: TF DIFFICULTY: Moderate KEYWORDS: median 38. True or False: The median of the values 3.4, 4.7, 1.9, 7.6, and 6.5 is 4.05. ANSWER: False TYPE: TF DIFFICULTY: Moderate KEYWORDS: median 39. True or False: In a set of numerical data, the value for Q3 can never be smaller than the value for Q1. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: quartiles
Introduction and Data Collection
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 box-and-whisker plot. ANSWER: True TYPE: TF DIFFICULTY: Moderate KEYWORDS: box-and-whisker plot, quartiles, shape 42. True or False: In right-skewed distributions, the distance from Q3 to the largest observation exceeds the distance from the smallest observation to Q1. ANSWER: True TYPE: TF DIFFICULTY: Moderate KEYWORDS: quartiles, shape 43. True or False: In left-skewed distributions, the distance from the smallest observation to Q1 exceeds the distance from Q3 to the largest observation. ANSWER: True TYPE: TF DIFFICULTY: Moderate KEYWORDS: quartiles, shape 44. True or False: A box-and-whisker plot is a graphical representation of a 5-number summary. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: box-and-whisker plot, five-number summary 45. True or False: The 5-number summary consists of the smallest observation, the first quartile, the median, the third quartile, and the largest observation. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: five-number summary
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46. True or False: In a box-and-whisker plot, the box portion represents the data between the first and third quartile values. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: box-and-whisker plot 47. True or False: The line drawn within the box of the box-and-whisker plot always represents the arithmetic mean. ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: box-and-whisker plot, arithmetic mean 48. True or False: The line drawn within the box of the box-and-whisker plot always represents the median. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: box-and-whisker plot, median 49. True or False: In a sample of size 40, the sample mean is 15. In this case, the sum of all observations in the sample is X i = 600. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: arithmetic mean 50. True or False: A population with 200 elements has an arithmetic mean of 10. From this information, it can be shown that the population standard deviation is 15. ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: arithmetic mean, standard deviation 51. True or False: In exploratory data analysis, a box-and-whisker plot can be used to illustrate the median, quartiles, and extreme values. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: box-and-whisker plot, five-number summary
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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 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: 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
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58. True or False: The interquartile range is a measure of central tendency in a set of data. ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: interquartile range, measure of variation 59. True or False: The geometric mean is a measure of variation or dispersion in a set of data. ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: geometric mean, measure of central tendency 60. True or False: The geometric mean is useful in measuring the rate of change of a variable over time. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: geometric mean 61. True or False: If a set of data is perfectly symmetrical, the arithmetic mean must be identical to the median. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: shape, arithmetic mean, median 62. True or False: The coefficient of variation is a measure of relative variation. ANSWER: True TYPE: TF DIFFICULTY: Moderate KEYWORDS: coefficient of variation 63. True or False: If the data set is approximately bell-shaped, the empirical rule will more accurately reflect the greater concentration of data close to the mean, as compared to the Bienayme-Chebyshev rule. ANSWER: True TYPE: TF DIFFICULTY: Moderate KEYWORDS: empirical rule, Bienayme-Chebyshev rule, normal distribution
Introduction and Data Collection
64. If the arithmetic mean of a numerical data set exceeds the median, the data are considered to be _______ skewed. ANSWER: positive or right TYPE: FI DIFFICULTY: Easy KEYWORDS: shape, arithmetic mean, median TABLE 3-2 The data below represent the amount of grams of carbohydrates in a serving of breakfast cereal. 11 15 23 29 19 22 21 20 15 25 17 65. Referring to Table 3-2, the arithmetic mean carbohydrates in this sample is ________ grams. ANSWER: 217/11 = 19.73 TYPE: FI DIFFICULTY: Moderate KEYWORDS: arithmetic mean 66. Referring to Table 3-2, the median carbohydrate amount in the cereal is ________ grams. ANSWER: 20 TYPE: FI DIFFICULTY: Moderate KEYWORDS: median 67. Referring to Table 3-2, the first quartile of the carbohydrate amounts is ________ grams. ANSWER: 15 TYPE: FI DIFFICULTY: Moderate KEYWORDS: quartiles 68. Referring to Table 3-2, the third quartile of the carbohydrate amounts is ________ grams. ANSWER: 23 TYPE: FI DIFFICULTY: Moderate KEYWORDS: quartiles 69. Referring to Table 3-2, the range in the carbohydrate amounts is ________ grams. ANSWER: 18 TYPE: FI DIFFICULTY: Easy KEYWORDS: range
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70. Referring to Table 3-2, the interquartile range in the carbohydrate amounts is ________ grams. ANSWER: 8 TYPE: FI DIFFICULTY: Moderate KEYWORDS: interquartile range 71. Referring to Table 3-2, the variance of the carbohydrate amounts is ________ (grams squared). ANSWER: 26.02 TYPE: FI DIFFICULTY: Moderate KEYWORDS: variance 72. Referring to Table 3-2, the standard deviation of the carbohydrate amounts is ________ grams. ANSWER: 5.10 TYPE: FI DIFFICULTY: Moderate KEYWORDS: standard deviation 73. Referring to Table 3-2, the coefficient of variation of the carbohydrate amounts is ________ percent. ANSWER: 25.86% TYPE: FI DIFFICULTY: Moderate KEYWORDS: coefficient of variation 74. Referring to Table 3-2, the five-number summary of the carbohydrate amounts consists of ________, ________, ________, ________, ________. ANSWER: 11, 15, 20, 23, 29 TYPE: FI DIFFICULTY: Moderate KEYWORDS: five-number summary
Introduction and Data Collection
75. Referring to Table 3-2, construct a box-and-whisker plot for the carbohydrate amounts. ANSWER: TYPE: PR DIFFICULTY: Moderate Box-and-whisker Plot
10
15
20
25
30
KEYWORDS: box-and-whisker plot 76. Referring to Table 3-2, what type of shape does the distribution of the sample appear to have? ANSWER: Slightly positive or right-skewed TYPE: PR DIFFICULTY: Moderate KEYWORDS: shape TABLE 3-3 The stem-and-leaf display below represents the number of vitamin supplements sold by a health food store in a sample of 16 days. Stem Leaves 1 99 2 0023567 3 034568 4 1 Note: For this sample, the sum of the observations is 448, the sum of the squares of the observations is 13,356, and the sum of the squared differences between each observation and the mean is 812. 77. Referring to Table 3-3, the arithmetic mean of the number of vitamin supplements sold in this sample is ________. ANSWER: 28 TYPE: FI DIFFICULTY: Easy KEYWORDS: arithmetic mean
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78. Referring to Table 3-3, the first quartile of the number of vitamin supplements sold in this sample is ________. ANSWER: 20 TYPE: FI DIFFICULTY: Moderate KEYWORDS: quartiles 79. Referring to Table 3-3, the third quartile of the number of vitamin supplements sold in this sample is ________. ANSWER: 35 TYPE: FI DIFFICULTY: Moderate KEYWORDS: quartiles 80. Referring to Table 3-3, the median number of vitamin supplements sold in this sample is ________. ANSWER: 26.5 TYPE: FI DIFFICULTY: Moderate KEYWORDS: median 81. Referring to Table 3-3, the range of the number of vitamin supplements sold in this sample is ________. ANSWER: 22 TYPE: FI DIFFICULTY: Easy KEYWORDS: range 82. Referring to Table 3-3, the interquartile range of the number of vitamin supplements sold in this sample is ________. ANSWER: 15 TYPE: FI DIFFICULTY: Moderate KEYWORDS: interquartile range 83. Referring to Table 3-3, the variance of the number of vitamin supplements sold in this sample is ________. ANSWER: 54.1 TYPE: FI DIFFICULTY: Easy KEYWORDS: variance
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84. Referring to Table 3-3, the standard deviation of the number of vitamin supplements sold in this sample is ________. ANSWER: 7.4 TYPE: FI DIFFICULTY: Easy KEYWORDS: standard deviation 85. Referring to Table 3-3, the coefficient of variation of the number of vitamin supplements sold in this sample is ________ percent. ANSWER: 26.3 TYPE: FI DIFFICULTY: Moderate KEYWORDS: coefficient of variation 86. Referring to Table 3-3, the five-number summary of the data in this sample consists of ________, ________, ________, ________, ________. ANSWER: 19, 20, 26.5, 35, 41 TYPE: FI DIFFICULTY: Moderate KEYWORDS: five-number summary 87. Referring to Table 3-3, construct a box-and-whisker plot for the data in this sample. ANSWER: Box-and-whisker Plot
10
20
30
TYPE: PR DIFFICULTY: Moderate KEYWORDS: box-and-whisker plot
40
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88. Referring to Table 3-3, what type of shape does the distribution of the sample appear to have? ANSWER: Slightly positive or right-skewed. TYPE: PR DIFFICULTY: Moderate KEYWORDS: shape TABLE 3-4 The stem-and-leaf display below represents the number of cargo manifests approved by customs inspectors of the Port of New York in a sample of 35 days. Stem Leaves 1 67889 2 00111222233334445566678899 3 1122 Note: For this sample, the sum of the observations is 838, the sum of the squares of the observations is 20,684, and the sum of the squared differences between each observation and the mean is 619.89. 89. Referring to Table 3-4, the arithmetic mean of the customs data is ________. ANSWER: 23.9 TYPE: FI DIFFICULTY: Easy KEYWORDS: arithmetic mean 90. Referring to Table 3-4, the median of the customs data is ________. ANSWER: 23 TYPE: FI DIFFICULTY: Moderate KEYWORDS: median 91. Referring to Table 3-4, the first quartile of the customs data is ________. ANSWER: 21 TYPE: FI DIFFICULTY: Moderate KEYWORDS: quartiles 92. Referring to Table 3-4, the third quartile of the customs data is ________. ANSWER: 27 TYPE: FI DIFFICULTY: Moderate KEYWORDS: quartiles
Introduction and Data Collection
93. Referring to Table 3-4, the range of the customs data is ________. ANSWER: 16 TYPE: FI DIFFICULTY: Easy KEYWORDS: range 94. Referring to Table 3-4, the interquartile range of the customs data is ________. ANSWER: 6 TYPE: FI DIFFICULTY: Moderate KEYWORDS: interquartile range 95. Referring to Table 3-4, the variance of the customs data is ________. ANSWER: 18.2 TYPE: FI DIFFICULTY: Easy KEYWORDS: variance 96. Referring to Table 3-4, the standard deviation of the customs data is ________. ANSWER: 4.3 TYPE: FI DIFFICULTY: Easy KEYWORDS: standard deviation 97. Referring to Table 3-4, the coefficient of variation of the customs data is ________ percent. ANSWER: 17.8 TYPE: FI DIFFICULTY: Moderate KEYWORDS: coefficient of variation 98. Referring to Table 3-4, the five-number summary for the data in the customs sample consists of ________, ________, ________, ________, ________. ANSWER: 16, 21, 23, 27, 32 TYPE: FI DIFFICULTY: Moderate KEYWORDS: five-number summary
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99. Referring to Table 3-4, construct a box-and-whisker plot of this sample. ANSWER: TYPE: PR DIFFICULTY: Moderate Box-and-whisker Plot
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25
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35
KEYWORDS: box-and-whisker plot TABLE 3-5 The rates of return of a Fortune 500 company over the last 5 years are: 3.17%, 4.43%, 5.93%, 5.43%, 7.29%, 8.21%, 6.23%, 5.23%, 4.34%, 6.68%, 7.14%, -5.56%, -5.23%, -5.73%, -10.34%. 100. Referring to Table 3-5, compute the arithmetic mean rate of return. ANSWER: 2.48% TYPE: PR DIFFICULTY: Easy KEYWORDS: arithmetic mean 101. Referring to Table 3-5, compute the geometric mean rate of return. ANSWER: 2.31% TYPE: PR DIFFICULTY: moderate KEYWORDS: geometric mean rate of return 102. Referring to Table 3-5, what is the range of the rate of return? ANSWER: 18.55% TYPE: PR DIFFICULTY: Easy KEYWORDS: range
Introduction and Data Collection
103. Referring to Table 3-5, construct a box-and-whisker plot for the rate of return. ANSWER: TYPE: PR DIFFICULTY: Easy Box-and-whisker Plot
-0.15
-0.1
-0.05
0
0.05
0.1
KEYWORDS: box-and-whisker plot 104. Referring to Table 3-5, what is the shape of the distribution for the rate of return? ANSWER: Left-skewed TYPE: PR DIFFICULTY: Easy KEYWORDS: shape TABLE 3-6 The rates of return of an Internet Service Provider over a 10 year period are: 10.25%, 12.64%, 8.37%, 9.29%, 6.23%, 42.53%, 29.23%, 15.25%, 21.52%, -2.35%. 105. Referring to Table 3-6, compute the arithmetic mean rate of return. ANSWER: 15.30% TYPE: PR DIFFICULTY: Easy KEYWORDS: arithmetic mean 106. Referring to Table 3-6, compute the geometric mean rate of return. ANSWER: 14.68% TYPE: PR DIFFICULTY: Moderate KEYWORDS: geometric mean rate of return
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107. Referring to Table 3-6, construct a box-and-whisker plot for the rate of return. ANSWER: Box-and-whisker Plot
-0.1
0
0.1
0.2
0.3
0.4
0.5
TYPE: PR DIFFICULTY: Moderate KEYWORDS: box-and-whisker plot 108. Referring to Table 3-6, what is the shape of the distribution for the rate of return? ANSWER: Right-skewed TYPE: PR DIFFICULTY: Moderate KEYWORDS: shape
Introduction and Data Collection
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
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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 a) 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 a) 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 a) Cannot be determined from the information given. ANSWER: b TYPE: MC DIFFICULTY: easy KEYWORDS: collectively exhaustive, mutually exclusive
Introduction and Data Collection
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 a) 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
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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
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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
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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
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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
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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
Introduction and Data Collection
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
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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
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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
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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
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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 that a randomly chosen widget was tested and found to be defective, what is the probability it was produced by the new machine?
a) b) c) d)
0.08 0.15 0.489 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
1 50 25 75
Number of Vehicles Involved 2 3 100 20 175 30 275 50
Totals 170 230 400
43. Referring to Table 4-1, what proportion of accidents involved more than one vehicle? a) 50/400 b) 75/400 c) 275/400 d) 325/400 ANSWER: d TYPE: MC DIFFICULTY: Easy KEYWORDS: contingency table, empirical classical probability, addition rule
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44. Referring to Table 4-1, what proportion of accidents involved alcohol and a single vehicle? a) 25/400 b) 50/400 c) 195/400 d) 245/400 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 b) 50/400 c) 195/400 d) 245/400 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 b) 50/170 c) 120/170 d) 120/400 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 b) 120/230 c) 120/325 d) 120/400 ANSWER: c TYPE: MC DIFFICULTY: Easy KEYWORDS: contingency table, empirical classical probability, conditional probability, addition rule
Introduction and Data Collection
48. Referring to Table 4-1, given that 3 vehicles were involved, what proportion of accidents involved alcohol? a) 20/30 b) 20/50 c) 20/170 d) 20/400 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 b) 25/230 c) 50/170 d) 25/75 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 b) 120/170 c) 205/230 d) 25/230 ANSWER: c TYPE: MC DIFFICULTY: Easy KEYWORDS: contingency table, empirical classical probability, conditional probability, complement
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TABLE 4-2 An alcohol awareness task force at a Big Ten university sampled 200 students after the midterm to ask them whether they went bar hopping the weekend before the midterm or spent the weekend studying, and whether they did well or poorly on the midterm. The following result was obtained.
Studying for Exam Went Bar Hopping
Did Well on Midterm 80 30
Did Poorly on Midterm 20 70
51. Referring to Table 4-2, what is the probability that a randomly selected student who went bar hopping will do well on the midterm? a. 30/100 b. 30/110 c. 30/200 d. (100/200)*(110/200) 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 b) (80+30)/200 c) (30+70)/200 d) (80+30+70)/200 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 b) (80+30)/200 c) (30+70)/200 d) (80+30+70)/200 ANSWER: a TYPE: MC DIFFICULTY: Easy KEYWORDS: contingency table, empirical classical probability, joint probability
Introduction and Data Collection
54. Referring to Table 4-2, the events "Did Well on Midterm" and "Studying for Exam" are a) statistically dependent. b) mutually exclusive. c) collective exhaustive. d) none of the above ANSWER: a TYPE: MC DIFFICULTY: Easy KEYWORDS: contingency table, empirical classical probability, statistical independence, joint probability 55. Referring to Table 4-2, the events "Did Well on Midterm" and "Studying for Exam" are a) not statistically dependent. b) not mutually exclusive. c) collective exhaustive. d) none of the above ANSWER: b TYPE: MC DIFFICULTY: Easy KEYWORDS: contingency table, empirical classical probability, mutually exclusive, joint probability 56. Referring to Table 4-2, the events "Did Well on Midterm" and "Did Poorly on Midterm" are a) statistically dependent. b) mutually exclusive. c) collective exhaustive. d) all of the above ANSWER: d TYPE: MC DIFFICULTY: Easy KEYWORDS: contingency table, empirical classical probability, statistical independence, mutually exclusive, collective exhaustive, joint probability 57. True or False: When A and B are mutually exclusive, P(A or B) can be found by adding P(A) and P(B). ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: mutually exclusive, addition rule 58. True or False: The collection of all the possible events is called a sample space. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: sample space 59. True or False: If A and B cannot occur at the same time, they are called mutually exclusive. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: mutually exclusive
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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
Introduction and Data Collection
66. True or False: If P(A and B) = 0, then A and B must be mutually exclusive. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: mutually exclusive 67. True or False: If P(A and B) = 0, then A and B must be collectively exhaustive. ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: collectively exhaustive, mutually exclusive 68. True or False: If P(A and B) = 1, then A and B must be collectively exhaustive. ANSWER: True TYPE: TF DIFFICULTY: Difficult KEYWORDS: collective exhaustive 69. True or False: If P(A and B) = 1, then A and B must be mutually exclusive. ANSWER: False TYPE: TF DIFFICULTY: Difficult KEYWORDS: mutually exclusive, collective exhaustive 70. Suppose A and B are independent events where P(A) = 0.4 and P(B) = 0.5. Then P(A and B) = __________. ANSWER: 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
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72. Suppose A and B are mutually exclusive events where P(A) = 0.4 and P(B) = 0.5. Then P(A or B) = __________. ANSWER: 0.9 TYPE: FI DIFFICULTY: Easy KEYWORDS: mutually exclusive, addition rule 73. Suppose A and B are independent events where P(A) = 0.4 and P(B) = 0.5. Then P(A or B) = __________. ANSWER: 0.7 TYPE: FI DIFFICULTY: Easy KEYWORDS: statistical independence, addition rule 74. Suppose A and B are events where P(A) = 0.4, P(B) = 0.5, and P(A and B) = 0.1. Then P(A or B) = __________. ANSWER: 0.8 TYPE: FI DIFFICULTY: Easy KEYWORDS: addition rule 75. Suppose A and B are events where P(A) = 0.4, P(B) = 0.5, and P(A and B) = 0.1. Then P(A|B) = __________. ANSWER: 0.2 TYPE: FI DIFFICULTY: Moderate KEYWORDS: conditional probability 76. Suppose A and B are events where P(A) = 0.4, P(B) = 0.5, and P(A and B) = 0.1. Then P(B|A) = __________. ANSWER: 0.25 TYPE: FI DIFFICULTY: Moderate KEYWORDS: conditional probability
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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 TYPE: FI DIFFICULTY: Moderate KEYWORDS: empirical classical probability, conditional probability, marginal probability 78. Referring to Table 4-3, the probability that a randomly selected individual is an adult or a child is __________. ANSWER: 200/200 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 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 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 TYPE: FI DIFFICULTY: Moderate KEYWORDS: empirical classical probability, addition rule
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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 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 TYPE: FI DIFFICULTY: Moderate KEYWORDS: empirical classical probability, conditional probability, Bayes‟ theorem 84. Referring to Table 4-3, assume we know that a person has ordered hamburger. The probability that this individual is a child is __________. ANSWER: 55/120 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 that 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 that a randomly selected patron is a female is __________. ANSWER: 0.40 TYPE: FI DIFFICULTY: Moderate KEYWORDS: addition rule, multiplication rule, marginal probability
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87. Referring to Table 4-4, the probability that 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 that 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. The probability that 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. The probability that the patron is a male is __________. ANSWER: 0.69 TYPE: FI DIFFICULTY: Difficult KEYWORDS: addition rule, multiplication rule, conditional probability 91. Referring to Table 4-4, suppose a randomly selected patron is a female. The probability that 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 "liking" beer and "liking" wine are statistically independent. ANSWER: False TYPE: TF DIFFICULTY: Moderate KEYWORDS: addition rule, multiplication rule, statistical independence
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93. True or False: Referring to Table 4-4, the two events "liking" 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 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
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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-ofstaters have passed the ELM. 99. Referring to Table 4-6, the probability that a randomly selected student is someone from northern Texas who has not passed the ELM is ________. ANSWER: 0.09 TYPE: FI DIFFICULTY: Difficult KEYWORDS: joint probability, multiplication rule 100. Referring to Table 4-6, the probability that a randomly selected student has passed the ELM is ________. ANSWER: 0.66 TYPE: FI DIFFICULTY: Difficult KEYWORDS: marginal probability, multiplication rule, addition rule 101. Referring to Table 4-6, if a randomly selected student has passed the ELM, the probability that 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 that the student is from southern Texas is ________. ANSWER: 0.706 TYPE: FI DIFFICULTY: Difficult KEYWORDS: conditional probability, complement, Bayes‟ theorem
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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 that 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 that 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 that the student has passed the ELM is ________. ANSWER: 0.633 TYPE: FI DIFFICULTY: Difficult KEYWORDS: conditional probability, complement, addition rule, multiplication rule, Bayes‟ theorem
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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: 0.1 TYPE: FI DIFFICULTY: Moderate KEYWORDS: conditional probability, addition rule 110. A new model car from Ford Motor Company offers a keyless entry system that utilizes a four-letter code. How many different possible combinations are there for the code? ANSWER: 456976 TYPE: PR DIFFICULTY: Easy KEYWORDS: counting rule 111. At the International Pancakes Hut, there are 4 different ways to have an egg cooked, 7 different choices of pancakes, 5 different types of syrups and 8 different beverages. How many different ways are there to order an egg, a pancake with a choice of syrup, and a beverage? ANSWER: 1120 TYPE: PR DIFFICULTY: Easy KEYWORDS: counting rule 112. There are 10 finalists at a national dog show. How many different orders of finishing can there be for all the 10 finalists? ANSWER:
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3628800 TYPE: PR DIFFICULTY: Easy KEYWORDS: counting rule, permutation 113. Eleven freshmen are to be assigned to eleven empty rooms in a student dormitory. Each room is considered unique, so that it matters who is being assigned to which room. How many different ways can those eleven freshmen be allocated? ANSWER: 39916800 TYPE: PR DIFFICULTY: Easy KEYWORDS: counting rule, permutation 114. There are only 4 empty rooms available in a student dormitory for eleven new freshmen. Each room is considered unique, so that it matters who is being assigned to which room. How many different ways can those 4 empty rooms be filled? ANSWER: 7920 TYPE: PR DIFFICULTY: Easy KEYWORDS: counting rule, permutation 115. There are only 4 empty rooms available in a student dormitory for eleven new freshmen. All the rooms are considered homogenous, so that it does not matter who is being assigned to which room. How many different ways can those 4 empty rooms be filled? ANSWER: 330 TYPE: PR DIFFICULTY: Easy KEYWORDS: counting rule, combination 116. Four freshmen are to be assigned to eleven empty rooms in a student dormitory. All the rooms are considered homogenous, so that it does not matter who is being assigned to which room. How many different ways can those 4 freshmen be assigned? ANSWER: 330 TYPE: PR DIFFICULTY: Easy KEYWORDS: counting rule, combination
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117. There are 47 contestants at a national dog show. How many different top three finishes can there be? ANSWER: 97290 TYPE: PR DIFFICULTY: Easy KEYWORDS: counting rule, permutation 118. Seven passengers are on a waiting list for an overbooked flight. As a result of cancellations, 3 seats become available. How many different ways can those 3 available seats be filled, regardless of the order? ANSWER: 35 TYPE: PR DIFFICULTY: Easy KEYWORDS: counting rule, combination 119. A high school debate team of 4 is to be chosen from a class of 35. How many possible ways can the team be formed? ANSWER: 52360 TYPE: PR DIFFICULTY: Easy KEYWORDS: counting rule, combination 120. A debate team of 4 is to be chosen from a class of 35. There are two twin brothers in the class. How many possible ways can the team be formed which will include only one of the twin brothers? ANSWER: 10912 TYPE: PR DIFFICULTY: Moderate KEYWORDS: counting rule, combination 121. A debate team of 4 is to be chosen from a class of 35. There are two twin brothers in the class. How many possible ways can the team be formed which will not include either of the twin brothers? ANSWER: 40920 TYPE: PR DIFFICULTY: Moderate KEYWORDS: counting rule, combination 122. A debate team of 4 is to be chosen from a class of 35. There are two twin brothers in the class. How many possible ways can the team be formed which will include both of the twin brothers? ANSWER: 528 TYPE: PR DIFFICULTY: Moderate KEYWORDS: counting rule, combination 123. An exploration team of 2 women and 3 men is to be chosen from a candidate pool of 6 women and 7 men. How many different ways can this team of 5 be formed? ANSWER: 525 TYPE: PR DIFFICULTY: Moderate KEYWORDS: counting rule, combination
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124. Twelve students in a Business Statistics class are to be formed into three teams of four. How many different ways can this be done? ANSWER: 5775 TYPE: PR DIFFICULTY: Difficult EXPLANATION: 12!/ 4!8! 8!/ 4!4! / 3!
KEYWORDS: counting rule, combination
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CHAPTER 5: SOME IMPORTANT DISCRETE PROBABILITY DISTRIBUTIONS 125. Thirty-six of the staff of 80 teachers at a local intermediate school are certified in Cardio-Pulmonary 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 126. 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 127. a) b) c) d)
A probability distribution is an equation that associates a particular probability of occurrence with each outcome in the sample space. measures outcomes and assigns values of X to the simple events. assigns a value to the variability in the sample space. assigns a value to the center of the sample space.
ANSWER: a TYPE: MC DIFFICULTY: Moderate KEYWORDS: probability distribution
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Introduction and Data Collection 128. a) b) c) d)
The connotation "expected value" or "expected gain" from playing roulette at a casino means the amount you expect to "gain" on a single play. the amount you expect to "gain" in the long run over many plays. the amount you need to "break even" over many plays. the amount you should expect to gain if you are lucky.
ANSWER: b TYPE: MC DIFFICULTY: Easy KEYWORDS: expected value 129. a) b) c) d)
Which of the following about the binomial distribution is not a true statement? The probability of success must be constant from trial to trial. Each outcome is independent of the other. Each outcome may be classified as either "success" or "failure." The random variable of interest is continuous.
ANSWER: d TYPE: MC DIFFICULTY: Easy KEYWORDS: binomial distribution, properties 130. a) b) c) d)
In a binomial distribution the random variable X is continuous. the probability of success p is stable from trial to trial. the number of trials n must be at least 30. the results of one trial are dependent on the results of the other trials.
ANSWER: b TYPE: MC DIFFICULTY: Easy KEYWORDS: binomial distribution, properties 131. a) b) c) d)
Whenever p = 0.5, the binomial distribution will always be symmetric. be symmetric only if n is large. be right-skewed. be left-skewed.
ANSWER: a TYPE: MC DIFFICULTY: Moderate KEYWORDS: binomial distribution, properties
Introduction and Data Collection 132. a) b) c) d)
Whenever p = 0.1 and n is small, the binomial distribution will be symmetric. right-skewed. left-skewed. None of the above.
ANSWER: b TYPE: MC DIFFICULTY: Moderate KEYWORDS: binomial distribution, properties 133. a) b) c) d)
If n = 10 and p = 0.70, then the mean of the binomial distribution is 0.07 1.45. 7.00 14.29
ANSWER: c TYPE: MC DIFFICULTY: Easy KEYWORDS: binomial distribution, mean 134. a) b) c) d)
If n = 10 and p = 0.70, then the standard deviation of the binomial distribution is 0.07 1.45 7.00 14.29
ANSWER: b TYPE: MC DIFFICULTY: Easy KEYWORDS: binomial distribution, standard deviation 135. a) b) c) d)
If the outcomes of a random variable follow a Poisson distribution, then their mean equals the standard deviation. median equals the standard deviation. mean equals the variance. median equals the variance.
ANSWER: c TYPE: MC DIFFICULTY: Easy KEYWORDS: Poisson distribution, mean, standard deviation, properties
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Introduction and Data Collection 136. What type of probability distribution will the consulting firm most likely employ to analyze the insurance claims in the following problem? An insurance company has called a consulting firm to determine if the company has an unusually high number of false insurance claims. It is known that the industry proportion for false claims is 3%. The consulting firm has decided to randomly and independently sample 100 of the company’s insurance claims. They believe the number of these 100 that are false will yield the information the company desires. a) binomial distribution. b) Poisson distribution. c) hypergeometric distribution. d) none of the above. ANSWER: a TYPE: MC DIFFICULTY: Difficult KEYWORDS: binomial distribution, properties 137. The covariance a) must be between -1 and +1. b) must be positive. c) can be positive or negative. d) must be less than +1. ANSWER: c TYPE: MC DIFFICULTY: Easy KEYWORDS: covariance, properties 138. What type of probability distribution will most likely be used to analyze warranty repair needs on new cars in the following problem? The service manager for a new automobile dealership reviewed dealership records of the past 20 sales of new cars to determine the number of warranty repairs he will be called on to perform in the next 90 days. Corporate reports indicate that the probability any one of their new cars needs a warranty repair in the first 90 days is 0.05. The manager assumes that calls for warranty repair are independent of one another and is interested in predicting the number of warranty repairs he will be called on to perform in the next 90 days for this batch of 20 new cars sold. a) binomial distribution. b) Poisson distribution. c) hypergeometric distribution. d) none of the above. ANSWER: a TYPE: MC DIFFICULTY: Difficult KEYWORDS: binomial distribution, properties
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139. What type of probability distribution will most likely be used to analyze the number of chocolate chip parts per cookie in the following problem? The quality control manager of Marilyn’s Cookies is inspecting a batch of chocolate chip cookies. When the production process is in control, the average number of chocolate chip parts per cookie is 6.0. The manager is interested in analyzing the probability that any particular cookie being inspected has fewer than 5.0 chip parts. a) binomial distribution. b) Poisson distribution. c) hypergeometric distribution. d) none of the above. ANSWER: b TYPE: MC DIFFICULTY: Moderate KEYWORDS: Poisson distribution, properties 140. What type of probability distribution will most likely be used to analyze the number of cars with defective radios in the following problem? From an inventory of 48 new cars being shipped to local dealerships, corporate reports indicate that 12 have defective radios installed. The sales manager of one dealership wants to predict the probability out of the 8 new cars it just received that, when each is tested, no more than 2 of the cars have defective radios. a) binomial distribution. b) Poisson distribution. c) hypergeometric distribution. d) none of the above. ANSWER: c TYPE: MC DIFFICULTY: Moderate KEYWORDS: hypergeometric distribution, properties 141. A stock analyst was provided with a list of 25 stocks. He was expected to pick 3 stocks from the list whose prices are expected to rise by more than 20% after 30 days. In reality, the prices of only 5 stocks would rise by more than 20% after 30 days. If he randomly selected 3 stocks from the list, he would use what type of probability distribution to compute the probability that all of the chosen stocks would appreciate more than 20% after 30 days? a) binomial distribution. b) Poisson distribution. c) hypergeometric distribution. d) none of the above. ANSWER: c TYPE: MC DIFFICULTY: Moderate KEYWORDS: hypergeometric distribution, properties
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Introduction and Data Collection 142. A professor receives, on average, 24.7 e-mails from students the day before the midterm exam. To compute the probability of receiving at least 10 e-mails on such a day, he will use what type of probability distribution? a) binomial distribution. b) Poisson distribution. c) hypergeometric distribution. d) none of the above. ANSWER: b TYPE: MC DIFFICULTY: Moderate KEYWORDS: Poisson distribution, properties 143. A company has 125 personal computers. The probability that any one of them will require repair on a given day is 0.025. To find the probability that exactly 20 of the computers will require repair on a given day, one will use what type of probability distribution? a) binomial distribution. b) Poisson distribution. c) hypergeometric distribution. d) none of the above. ANSWER: a TYPE: MC DIFFICULTY: Moderate KEYWORDS: binomial distribution, properties 144. The portfolio expected return of two investments a) will be higher when the covariance is zero. b) will be higher when the covariance is negative. c) will be higher when the covariance is positive. d) does not depend on the covariance. ANSWER: d TYPE: MC DIFFICULTY: Easy KEYWORDS: portfolio, mean 145. A financial analysis is presented with information on the past records of 60 start-up companies and told that in fact only 3 of them have managed to become highly successful. He selected 3 companies from this group as the candidates for success. To analyze his ability to spot the companies that will eventually become highly successful, he will use what type of probability distribution? a) binomial distribution. b) Poisson distribution. c) hypergeometric distribution. d) none of the above. ANSWER: c TYPE: MC DIFFICULTY: Moderate KEYWORDS: hypergeometric distribution, properties 146. 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?
Introduction and Data Collection a) b) c) d)
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binomial distribution. Poisson distribution. hypergeometric distribution. none of the above.
ANSWER: b TYPE: MC DIFFICULTY: Moderate KEYWORDS: Poisson distribution, properties 147. A multiple-choice test has 30 questions. There are 4 choices for each question. A student who has not studied for the test decides to answer all questions randomly. What type of probability distribution can be used to figure out his chance of getting at least 20 questions right? a) binomial distribution. b) Poisson distribution. c) hypergeometric distribution. d) none of the above. ANSWER: a TYPE: MC DIFFICULTY: Moderate KEYWORDS: binomial distribution, properties 148. 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
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Introduction and Data Collection 149. A lab orders 100 rats a week for each of the 52 weeks in the year for experiments that the lab conducts. Prices for 100 rats follow the following distribution: Price: $10.00 $12.50 $15.00 Probability: 0.35 0.40 0.25 How much should the lab budget for next year’s rat orders be, assuming this distribution does not change? a) $520 b) $637 c) $650 d) $780 ANSWER: b TYPE: MC DIFFICULTY: Moderate KEYWORDS: mean 150. 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 151.
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 152.
True or False: Another name for the mean of a probability distribution is its expected value.
ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: mean
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153. 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 154. 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 155. True or False: The number of males selected in a sample of 5 students taken without replacement from a class of 9 females and 18 males has a binomial distribution. ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: hypergeometric distribution, properties 156. True or False: The number of males selected in a sample of 5 students taken without replacement from a class of 9 females and 18 males has a hypergeometric distribution. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: hypergeometric distribution, properties 157.
True or False: The diameters of 10 randomly selected bolts have a binomial distribution.
ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: binomial distribution, properties 158.
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
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Introduction and Data Collection 159.
True or False: In a Poisson distribution, the mean and standard deviation are equal.
ANSWER: False TYPE: TF DIFFICULTY: Moderate KEYWORDS: Poisson distribution, properties 160.
True or False: In a Poisson distribution, the mean and variance are equal.
ANSWER: True TYPE: TF DIFFICULTY: Moderate KEYWORDS: Poisson distribution, properties 161. True or False: If p remains constant in a binomial distribution, an increase in n will increase the variance. ANSWER: True TYPE: TF DIFFICULTY: Moderate KEYWORDS: binomial distribution, properties 162. 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 163. 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 164. True or False: Suppose that the number of airplanes arriving at an airport per minute is a Poisson process. The average number of airplanes arriving per minute is 3. The probability that exactly 6 planes arrive in the next minute is 0.0504. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: Poisson distribution, probability
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165. True or False: The covariance between two investments is equal to the sum of the variances of the investments. ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: covariance, variance 166. True or False: If the covariance between two investments is zero, the variance of the sum of the two investments will be equal to the sum of the variances of the investments. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: covariance, variance 167. True or False: The expected return of the sum of two investments will be equal to the sum of the expected returns of the two investments plus twice the covariance between the investments. ANSWER: False TYPE: TF DIFFICULTY: Moderate KEYWORDS: mean, portfolio 168. True or False: The variance of the sum of two investments will be equal to the sum of the variances of the two investments plus twice the covariance between the investments. ANSWER: True TYPE: TF DIFFICULTY: Moderate KEYWORDS: covariance, variance 169. True or False: The variance of the sum of two investments will be equal to the sum of the variances of the two investments when the covariance between the investments is zero. ANSWER: True TYPE: TF DIFFICULTY: Moderate KEYWORDS: covariance, variance 170. True or False: The expected return of a two-asset portfolio is equal to the product of the weight assigned to the first asset and the expected return of the first asset plus the product of the weight assigned to the second asset and the expected return of the second asset. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: mean, portfolio
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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. 171.
Referring to Table 5-1, the probability that both sound an alarm in the presence of smoke is ________.
ANSWER: 0.64 TYPE: FI DIFFICULTY: Moderate KEYWORDS: binomial distribution 172. 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 173. 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). 174.
Referring to Table 5-2, the probability that all 3 businesses succeed is ________.
ANSWER: 0.216 TYPE: FI DIFFICULTY: Moderate KEYWORDS: binomial distribution 175.
Referring to Table 5-2, the probability that all 3 businesses fail is ________.
ANSWER: 0.064 TYPE: FI DIFFICULTY: Moderate KEYWORDS: binomial distribution
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176.
Referring to Table 5-2, the probability that at least 1 business succeeds is ________.
ANSWER: 0.936 TYPE: FI DIFFICULTY: Moderate KEYWORDS: binomial distribution 177.
Referring to Table 5-2, the probability that exactly 1 business succeeds is ________.
ANSWER: 0.288 TYPE: FI DIFFICULTY: Moderate KEYWORDS: binomial distribution 178.
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 179.
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 180.
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 181. 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 182. 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 183. Suppose that past history shows that 60% of college students prefer Brand C cola. A sample of 5 students is to be selected. The probability that exactly 3 prefer brand C is ________.
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Introduction and Data Collection ANSWER: 0.3456 TYPE: FI DIFFICULTY: Easy KEYWORDS: binomial distribution 184. Suppose that past history shows that 60% of college students prefer Brand C cola. A sample of 5 students is to be selected. The probability that exactly 4 prefer brand C is ________. ANSWER: 0.2592 TYPE: FI DIFFICULTY: Easy KEYWORDS: binomial distribution 185. 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 186. 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 187. 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 188. 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
Introduction and Data Collection 189. 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. X 0 1 2 3 P(X 0.35 0.35 0.25 0.05 ) 190.
Referring to Table 5-3, the probability of no retransmissions is ________.
ANSWER: 0.35 TYPE: FI DIFFICULTY: Easy KEYWORDS: probability distribution 191.
Referring to Table 5-3, the probability of at least one retransmission is ________.
ANSWER: 0.65 TYPE: FI DIFFICULTY: Easy KEYWORDS: probability distribution 192.
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 193.
Referring to Table 5-3, the variance for the number of retransmissions is ________.
ANSWER: 0.80 TYPE: FI DIFFICULTY: Easy KEYWORDS: probability distribution, variance
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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 195. 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 TYPE: FI DIFFICULTY: Moderate KEYWORDS: binomial distribution 196. 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 197. 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 198. 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 199. 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 200. 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 ________.
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ANSWER: 0.9375 TYPE: FI DIFFICULTY: Moderate KEYWORDS: binomial distribution, variance 201. 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 0.05 202.
Referring to Table 5-4, the probability of 3 accidents is ________.
ANSWER: 0.15 TYPE: FI DIFFICULTY: Easy KEYWORDS: probability distribution 203.
Referring to Table 5-4, the probability of at least 1 accident is ________.
ANSWER: 0.90 TYPE: FI DIFFICULTY: Easy KEYWORDS: probability distribution 204.
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,
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Referring to Table 5-4, the variance of the number of accidents is ________.
ANSWER: 1.4 TYPE: FI DIFFICULTY: Easy KEYWORDS: probability distribution, variance 206.
Referring to Table 5-4, the standard deviation of the number of accidents is ________.
ANSWER: 1.18 TYPE: FI DIFFICULTY: Easy KEYWORDS: probability distribution, standard deviation 207. 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 208. 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 209. 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 210. 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 211. 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 power outages in a year is ____________. ANSWER: 0.1487 TYPE: FI DIFFICULTY: Difficult KEYWORDS: Poisson distribution
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212. 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 213. 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 214. 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 215. 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 216. 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
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Introduction and Data Collection 217. An Undergraduate Study Committee of 6 members at a major university is to be formed from a pool of faculty of 18 men and 6 women. If the committee members are chosen randomly, what is the probability that precisely half of the members will be women? ANSWER: 0.1213 TYPE: FI DIFFICULTY: Easy KEYWORDS: hypergeometric distribution 218. An Undergraduate Study Committee of 6 members at a major university is to be formed from a pool of faculty of 18 men and 6 women. If the committee members are chosen randomly, what is the probability that all of the members will be men? ANSWER: 0.1379 TYPE: FI DIFFICULTY: Easy KEYWORDS: hypergeometric distribution 219. A debate team of 4 members for a high school will be chosen randomly from a potential group of 15 students. Ten of the 15 students have no prior competition experience while the others have some degree of experience. What is the probability that none of the members chosen for the team have any competition experience? ANSWER: 0.1538 TYPE: FI DIFFICULTY: Easy KEYWORDS: hypergeometric distribution 220. A debate team of 4 members for a high school will be chosen randomly from a potential group of 15 students. Ten of the 15 students have no prior competition experience while the others have some degree of experience. What is the probability that at least 1 of the members chosen for the team have some prior competition experience? ANSWER: 0.8462 TYPE: FI DIFFICULTY: Moderate KEYWORDS: hypergeometric distribution 221. A debate team of 4 members for a high school will be chosen randomly from a potential group of 15 students. Ten of the 15 students have no prior competition experience while the others have some degree of experience. What is the probability that at most 1 of the members chosen for the team have some prior competition experience? ANSWER: 0.5934 TYPE: FI DIFFICULTY: Moderate KEYWORDS: hypergeometric distribution
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222. A debate team of 4 members for a high school will be chosen randomly from a potential group of 15 students. Ten of the 15 students have no prior competition experience while the others have some degree of experience. What is the probability that exactly half of the members chosen for the team have some prior competition experience? ANSWER: 0.3297 TYPE: FI DIFFICULTY: Easy KEYWORDS: hypergeometric distribution 223. 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 224. 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 225. The on-line access computer service industry is growing at an extraordinary rate. Current estimates suggest that only 20% of the home-based computers have access to on-line services. This number is expected to grow quickly over the next 5 years. Suppose 25 people with home-based computers were randomly and independently sampled. Find the probability that fewer than 10 of those sampled currently have access to on-line services. ANSWER: 0.983 TYPE: PR DIFFICULTY: Moderate KEYWORDS: binomial distribution 226. The on-line access computer service industry is growing at an extraordinary rate. Current estimates suggest that only 20% of the home-based computers have access to on-line services. This number is expected to grow quickly over the next 5 years. Suppose 25 people with home-based computers were randomly and independently sampled. Find the probability that more than 20 of those sampled currently do not have access to on-line services. ANSWER: 0.421 TYPE: PR DIFFICULTY: Difficult KEYWORDS: binomial distribution
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Introduction and Data Collection 227. 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 228. 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 TYPE: PR DIFFICULTY: Moderate KEYWORDS: binomial distribution 229. 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 230. 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
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231. The local police department must write, on average, 5 tickets a day to keep department revenues at budgeted levels. Suppose the number of tickets written per day follows a Poisson distribution with a mean of 6.4 tickets per day. Find the probability that exactly 6 tickets are written on a randomly selected day from this population. ANSWER: 0.159 TYPE: PR DIFFICULTY: Easy KEYWORDS: Poisson distribution TABLE 5-5 From an inventory of 48 new cars being shipped to local dealerships, corporate reports indicate that 12 have defective radios installed. 232. Referring to Table 5-5, what is the probability out of the 8 new cars it just received that, when each is tested, no more than 2 of the cars have defective radios? ANSWER: 0.6863 TYPE: PR DIFFICULTY: Moderate KEYWORDS: hypergeometric distribution 233. Referring to Table 5-5, what is the probability out of the 8 new cars it just received that, when each is tested, exactly half of the cars have defective radios? ANSWER: 0.0773 TYPE: PR DIFFICULTY: Easy KEYWORDS: hypergeometric distribution 234. Referring to Table 5-5, what is the probability out of the 8 new cars it just received that, when each is tested, none of the cars have defective radios? ANSWER: 0.08019 TYPE: PR DIFFICULTY: Easy KEYWORDS: hypergeometric distribution 235. Referring to Table 5-5, what is the probability out of the 8 new cars it just received that, when each is tested, at least half of the cars have defective radios? ANSWER: 0.09388 TYPE: PR DIFFICULTY: Moderate KEYWORDS: hypergeometric distribution
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236. Referring to Table 5-5, what is the probability out of the 8 new cars it just received that, when each is tested, no more than half of the cars have defective radios? ANSWER: 0.9834 TYPE: PR DIFFICULTY: Moderate KEYWORDS: hypergeometric distribution 237. Referring to Table 5-5, what is the probability out of the 8 new cars it just received that, when each is tested, at most 2 of the cars have defective radios? ANSWER: 0.6863 TYPE: PR DIFFICULTY: Moderate KEYWORDS: hypergeometric distribution 238. Referring to Table 5-5, what is the probability out of the 8 new cars it just received that, when each is tested, exactly two of the cars have non-defective radios? ANSWER: 0.001543 TYPE: PR DIFFICULTY: Difficult KEYWORDS: hypergeometric distribution 239. Referring to Table 5-5, what is the probability out of the 8 new cars it just received that, when each is tested, at most three of the cars have non-defective radios? ANSWER: 0.01661 TYPE: PR DIFFICULTY: Difficult KEYWORDS: hypergeometric distribution 240. Referring to Table 5-5, what is the probability out of the 8 new cars it just received that, when each is tested, no more than half of the cars have non-defective radios? ANSWER: 0.09388 TYPE: PR DIFFICULTY: Difficult KEYWORDS: hypergeometric distribution
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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. 241. 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 242. 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 243. Referring to Table 5-6, what is the probability that any particular cookie being inspected has at least 6.0 chip parts. ANSWER: 0.5543 TYPE: PR DIFFICULTY: Moderate KEYWORDS: Poisson distribution 244. Referring to Table 5-6, what is the probability that any particular cookie being inspected has between 5.0 and 8.0 chip parts. ANSWER: 0.5622 TYPE: PR DIFFICULTY: Moderate KEYWORDS: Poisson distribution 245. 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
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CHAPTER 6: THE NORMAL DISTRIBUTION AND OTHER CONTINUOUS DISTRIBUTIONS 1. In its standardized form, the normal distribution a) has a mean of 0 and a standard deviation of 1. b) has a mean of 1 and a variance of 0. c) has an area equal to 0.5. d) cannot be used to approximate discrete probability distributions. ANSWER: a TYPE: MC DIFFICULTY: Easy KEYWORDS: standardized normal distribution, properties 2. Which of the following about the normal distribution is NOT true? a) Theoretically, the mean, median, and mode are the same. b) About 2/3 of the observations fall within 1 standard deviation from the mean. c) It is a discrete probability distribution. d) Its parameters are the mean, , and standard deviation, . ANSWER: c TYPE: MC DIFFICULTY: Easy KEYWORDS: normal distribution, properties 3. If a particular batch of data is approximately normally distributed, we would find that approximately a) 2 of every 3 observations would fall between 1 standard deviation around the mean. b) 4 of every 5 observations would fall between 1.28 standard deviations around the mean. c) 19 of every 20 observations would fall between 2 standard deviations around the mean. d) all of the above ANSWER: d TYPE: MC DIFFICULTY: Easy KEYWORDS: normal distribution, properties
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4. For some positive value of Z, the probability that a standardized 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 5. For some value of Z, the probability that a standardized 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 standardized 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 standardized 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
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8. For some positive value of X, the probability that a standardized 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 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 TYPE: PR DIFFICULTY: Moderate KEYWORDS: normal distribution, probability
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12. A company that sells annuities must base the annual payout on the probability distribution of the length of life of the participants in the plan. Suppose the probability distribution of the lifetimes of the participants is approximately a normal distribution with a mean of 68 years and a standard deviation of 3.5 years. Find the age at which payments have ceased for approximately 86% of the plan participants. ANSWER: 71.78 years old TYPE: PR DIFFICULTY: Difficult KEYWORDS: normal distribution, value 13. A catalog company that receives the majority of its orders by telephone conducted a study to determine how long customers were willing to wait on hold before ordering a product. The length of time was found to be a random variable best approximated by an exponential distribution with a mean equal to 3 minutes. What proportion of customers having to hold more than 4.5 minutes will hang up before placing an order? a) 0.22313 b) 0.48658 c) 0.51342 d) 0.77687 ANSWER: a TYPE: MC DIFFICULTY: Moderate KEYWORDS: exponential distribution, probability 14. A catalog company that receives the majority of its orders by telephone conducted a study to determine how long customers were willing to wait on hold before ordering a product. The length of time was found to be a random variable best approximated by an exponential distribution with a mean equal to 3 minutes. What proportion of customers having to hold more than 1.5 minutes will hang up before placing an order? a) 0.86466 b) 0.60653 c) 0.39347 d) 0.13534 ANSWER: b TYPE: MC DIFFICULTY: Moderate KEYWORDS: exponential distribution, probability
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15. A catalog company that receives the majority of its orders by telephone conducted a study to determine how long customers were willing to wait on hold before ordering a product. The length of time was found to be a random variable best approximated by an exponential distribution with a mean equal to 3 minutes. Find the waiting time at which only 10% of the customers will continue to hold. a) 2.3 minutes b) 3.3 minutes c) 6.9 minutes d) 13.8 minutes ANSWER: c TYPE: MC DIFFICULTY: Moderate KEYWORDS: exponential distribution, value 16. A catalog company that receives the majority of its orders by telephone conducted a study to determine how long customers were willing to wait on hold before ordering a product. The length of time was found to be a random variable best approximated by an exponential distribution with a mean equal to 2.8 minutes. What proportion of callers is put on hold longer than 2.8 minutes? a) 0.60810 b) 0.367879 c) 0.50 d) 0.632121 ANSWER: b TYPE: MC DIFFICULTY: Moderate KEYWORDS: exponential distribution, probability 17. A catalog company that receives the majority of its orders by telephone conducted a study to determine how long customers were willing to wait on hold before ordering a product. The length of time was found to be a random variable best approximated by an exponential distribution with a mean equal to 2.8 minutes. What is the probability that a randomly selected caller is placed on hold fewer than 7 minutes? a) 0.0009119 b) 0.082085 c) 0.917915 d) 0.9990881 ANSWER: c TYPE: MC DIFFICULTY: Moderate KEYWORDS: exponential distribution, probability
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18. If we know that the length of time it takes a college student to find a parking spot in the library parking lot follows a normal distribution with a mean of 3.5 minutes and a standard deviation of 1 minute, find the probability that a randomly selected college student will find a parking spot in the library parking lot in less than 3 minutes. a) 0.3551 b) 0.3085 c) 0.2674 d) 0.1915 ANSWER: b TYPE: MC DIFFICULTY: Easy KEYWORDS: normal distribution, probability 19. If we know that the length of time it takes a college student to find a parking spot in the library parking lot follows a normal distribution with a mean of 3.5 minutes and a standard deviation of 1 minute, find the probability that a randomly selected college student will take between 2 and 4.5 minutes to find a parking spot in the library parking lot. a) 0.0919 b) 0.2255 c) 0.4938 d) 0.7745 ANSWER: d TYPE: MC DIFFICULTY: Easy KEYWORDS: normal distribution, probability 20. If we know that the length of time it takes a college student to find a parking spot in the library parking lot follows a normal distribution with a mean of 3.5 minutes and a standard deviation of 1 minute, find the point in the distribution that 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
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21. Let X represent the amount of time it takes a student to park in the library parking lot at the university. If we know that the distribution of parking times can be modeled using an exponential distribution with a mean of 4 minutes, find the probability that it will take a randomly selected student more than 10 minutes to park in the library lot. a) 0.917915 b) 0.670320 c) 0.329680 d) 0.082085 ANSWER: d TYPE: MC DIFFICULTY: Moderate KEYWORDS: exponential distribution, probability 22. Let X represent the amount of time it takes a student to park in the library parking lot at the university. If we know that the distribution of parking times can be modeled using an exponential distribution with a mean of 4 minutes, find the probability that it will take a randomly selected student between 2 and 12 minutes to park in the library lot. a) 0.049787 b) 0.556744 c) 0.606531 d) 0.656318 ANSWER: b TYPE: MC DIFFICULTY: Moderate KEYWORDS: exponential distribution, probability 23. The owner of a fish market determined that the average weight for a catfish is 3.2 pounds with a standard deviation of 0.8 pounds. Assuming the weights of catfish are normally distributed, the probability that a randomly selected catfish will weigh more than 4.4 pounds is _______. ANSWER: 0.0668 TYPE: FI DIFFICULTY: Easy KEYWORDS: normal distribution, probability 24. The owner of a fish market determined that the average weight for a catfish is 3.2 pounds with a standard deviation of 0.8 pounds. 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
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25. The owner of a fish market determined that the average weight for a catfish is 3.2 pounds with a standard deviation of 0.8 pounds. A citation catfish should be one of the top 2% in weight. Assuming the weights of catfish are normally distributed, at what weight (in pounds) should the citation designation be established? a) 1.56 pounds b) 4.84 pounds c) 5.20 pounds d) 7.36 pounds ANSWER: b TYPE: MC DIFFICULTY: Moderate KEYWORDS: normal distribution, value 26. The owner of a fish market determined that the average weight for a catfish is 3.2 pounds with a standard deviation of 0.8 pounds. Assuming the weights of catfish are normally distributed, above what weight (in pounds) do 89.80% of the weights occur? ANSWER: 2.184 pounds TYPE: PR DIFFICULTY: Moderate KEYWORDS: normal distribution, value 27. The owner of a fish market determined that the average weight for a catfish is 3.2 pounds with a standard deviation of 0.8 pounds. Assuming the weights of catfish are normally distributed, the probability that a randomly selected catfish will weigh less than 2.2 pounds is _______. ANSWER: 0.1056 TYPE: FI DIFFICULTY: Easy KEYWORDS: normal distribution, probability 28. The Tampa International Airport (TIA) has been criticized for the waiting times associated with departing flights. While the critics acknowledge that many flights have little or no waiting times, their complaints deal more specifically with the longer waits attributed to some flights. The critics are interested in showing, mathematically, exactly what the problems are. Which type of distribution would best model the waiting times of the departing flights at TIA? a) uniform distribution b) binomial distribution c) normal distribution d) exponential distribution ANSWER: d TYPE: MC DIFFICULTY: Moderate KEYWORDS: exponential distribution, properties
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29. Scientists in the Amazon are trying to find a cure for a deadly disease that is attacking the rain forests there. One of the variables that the scientists have been measuring involves the diameter of the trunk of the trees that have been affected by the disease. Scientists have calculated that the average diameter of the diseased trees is 42 centimeters. They also know that approximately 95% of the diameters fall between 32 and 52 centimeters, and almost all of the diseased trees have diameters between 27 and 57 centimeters. When modeling the diameters of diseased trees, which distribution should the scientists use? a) uniform distribution b) binomial distribution c) normal distribution d) exponential distribution ANSWER: c TYPE: MC DIFFICULTY: Moderate KEYWORDS: normal distribution, properties 30. In the game Wheel of Fortune, which of the following distributions can best be used to compute the probability of winning the special vacation package in a single spin? a) uniform distribution b) binomial distribution c) normal distribution d) exponential distribution ANSWER: a TYPE: MC DIFFICULTY: Moderate KEYWORDS: uniform distribution, properties 31. A food processor packages orange juice in small jars. The weights of the filled jars are approximately normally distributed with a mean of 10.5 ounces and a standard deviation of 0.3 ounces. Find the proportion of all jars packaged by this process that have weights that fall below 10.875 ounces. ANSWER: 0.8944 TYPE: PR DIFFICULTY: Easy KEYWORDS: normal distribution, probability 32. A food processor packages orange juice in small jars. The weights of the filled jars are approximately normally distributed with a mean of 10.5 ounces and a standard deviation of 0.3 ounces. 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 33. True or False: The probability that a standardized 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
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34. True or False: The probability that a standardized normal random variable, Z, falls between 1.50 and 2.10 is the same as the probability that Z is between – 2.10 and – 1.50. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: standardized normal distribution, probability 35. True or False: The probability that a standardized normal random variable, Z, is below 1.96 is 0.4750. ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: standardized normal distribution, probability 36. True or False: The probability that a standardized normal random variable, Z, is between 1.00 and 3.00 is 0.1574. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: standardized normal distribution, probability 37. True or False: The probability that a standardized normal random variable, Z, falls between –2.00 and –0.44 is 0.6472. ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: standardized normal distribution, probability 38. True or False: The probability that a standardized normal random variable, Z, is less than 50 is approximately 0. ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: standardized normal distribution, probability
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39. True or False: A worker earns $15 per hour at a plant and is told that only 2.5% of all workers make a higher wage. If the wage is assumed to be normally distributed and the standard deviation of wage rates is $5 per hour, the average wage for the plant is $7.50 per hour. ANSWER: False TYPE: TF DIFFICULTY: Moderate KEYWORDS: normal distribution, mean 40. True or False: Theoretically, the mean, median, and the mode are all equal for a normal distribution. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: normal distribution, properties 41. True or False: Any set of normally distributed data can be transformed to its standardized form. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: normal distribution, properties 42. True or False: The "middle spread," that is, the middle 50% of the normal distribution, is equal to one standard deviation. ANSWER: False TYPE: TF DIFFICULTY: Moderate KEYWORDS: normal distribution, probability, value 43. True or False: A normal probability plot may be used to assess the assumption of normality for a particular batch of data. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: normal probability plot 44. True or False: If a data batch is approximately normally distributed, its normal probability plot would be Sshaped. ANSWER: False TYPE: TF DIFFICULTY: Moderate KEYWORDS: normal probability plot
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45. The probability that a standardized normal variable Z is positive is ________. ANSWER: 0.50 TYPE: FI DIFFICULTY: Easy KEYWORDS: standardized normal distribution 46. The amount of pyridoxine (in grams) in a multiple vitamin is normally distributed with = 110 grams and = 25 grams. What is the probability that a randomly selected vitamin will contain between 100 and 110 grams of pyridoxine? ANSWER: 0.1554 TYPE: PR DIFFICULTY: Easy KEYWORDS: normal distribution, probability 47. The amount of pyridoxine (in grams) in a multiple vitamin is normally distributed with = 110 grams and = 25 grams. What is the probability that a randomly selected vitamin will contain between 82 and 100 grams of pyridoxine? ANSWER: 0.2132 TYPE: PR DIFFICULTY: Easy KEYWORDS: normal distribution, probability 48. The amount of pyridoxine (in grams) in a multiple vitamin is normally distributed with = 110 grams and = 25 grams. What is the probability that a randomly selected vitamin will contain at least 100 grams of pyridoxine? ANSWER: 0.6554 TYPE: PR DIFFICULTY: Easy KEYWORDS: normal distribution, probability 49. The amount of pyridoxine (in grams) in a multiple vitamin is normally distributed with = 110 grams and = 25 grams. What is the probability that a randomly selected vitamin will contain between 100 and 120 grams of pyridoxine? ANSWER: 0.3108 TYPE: PR DIFFICULTY: Moderate KEYWORDS: normal distribution, probability
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55. The true length of boards cut at a mill with a listed length of 10 feet is normally distributed with a mean of 123 inches and a standard deviation of 1 inch. What proportion of the boards will be less than 124 inches? ANSWER: 0.8413 TYPE: PR DIFFICULTY: Easy KEYWORDS: normal distribution, probability 56. Suppose Z has a standardized normal distribution with a mean of 0 and a 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 57. Suppose Z has a standardized normal distribution with a mean of 0 and a standard deviation of 1. The probability that Z is more than 0.77 is __________. ANSWER: 0.2207 TYPE: FI DIFFICULTY: Easy KEYWORDS: standardized normal distribution, probability 58. Suppose Z has a standardized normal distribution with a mean of 0 and a standard deviation of 1. The probability that Z is less than -2.20 is __________. ANSWER: 0.0139 TYPE: FI DIFFICULTY: Easy KEYWORDS: standardized normal distribution, probability 59. Suppose Z has a standardized normal distribution with a mean of 0 and a standard deviation of 1. The probability that Z is more than -0.98 is __________. ANSWER: 0.8365 TYPE: FI DIFFICULTY: Easy KEYWORDS: standardized normal distribution, probability 60. Suppose Z has a standardized normal distribution with a mean of 0 and a standard deviation of 1. The probability that Z is between -2.33 and 2.33 is __________. ANSWER: 0.9802 TYPE: FI DIFFICULTY: Easy KEYWORDS: standardized normal distribution, probability 61. Suppose Z has a standardized normal distribution with a mean of 0 and a standard deviation of 1. The probability that Z is between -2.89 and -1.03 is __________. ANSWER: 0.1496 TYPE: FI DIFFICULTY: Easy
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KEYWORDS: standardized normal distribution, probability 62. Suppose Z has a standardized normal distribution with a mean of 0 and a 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 63. Suppose Z has a standardized normal distribution with a mean of 0 and a standard deviation of 1. The probability that Z values are larger than __________ is 0.3483. ANSWER: 0.39 TYPE: FI DIFFICULTY: Moderate KEYWORDS: standardized normal distribution, probability 64. Suppose Z has a standardized normal distribution with a mean of 0 and a standard deviation of 1. The probability that Z values are larger than __________ is 0.6985. ANSWER: -0.52 TYPE: FI DIFFICULTY: Moderate KEYWORDS: standardized normal distribution, probability 65. Suppose Z has a standardized normal distribution with a mean of 0 and a standard deviation of 1. So, 27% of the possible Z values are smaller than __________. ANSWER: -0.61 TYPE: FI DIFFICULTY: Moderate KEYWORDS: standardized normal distribution, value 66. Suppose Z has a standardized normal distribution with a mean of 0 and a 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
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67. Suppose Z has a standardized normal distribution with a mean of 0 and a 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 TYPE: FI DIFFICULTY: Moderate KEYWORDS: standardized normal distribution, value 68. Suppose Z has a standardized normal distribution with a mean of 0 and a 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 TYPE: FI DIFFICULTY: Moderate KEYWORDS: standardized normal distribution, value 69. The amount of time necessary for assembly line workers to complete a product is a normal random variable with a mean of 15 minutes and a standard deviation of 2 minutes. The probability is __________ that a product is assembled in less than 12 minutes. ANSWER: 0.0668 TYPE: FI DIFFICULTY: Easy KEYWORDS: normal distribution, probability 70. The amount of time necessary for assembly line workers to complete a product is a normal random variable with a mean of 15 minutes and a standard deviation of 2 minutes. The probability is __________ that a product is assembled in between 14 and 16 minutes. ANSWER: 0.3829 TYPE: FI DIFFICULTY: Easy KEYWORDS: normal distribution, probability 71. The amount of time necessary for assembly line workers to complete a product is a normal random variable with a mean of 15 minutes and a standard deviation of 2 minutes. The probability is __________ that a product is assembled in between 10 and 12 minutes. ANSWER: 0.0606 TYPE: FI DIFFICULTY: Easy KEYWORDS: normal distribution, probability
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72. The amount of time necessary for assembly line workers to complete a product is a normal random variable with a mean of 15 minutes and a standard deviation of 2 minutes. The probability is __________ that a product is assembled in between 15 and 21 minutes. ANSWER: 0.49865 TYPE: FI DIFFICULTY: Easy KEYWORDS: normal distribution, probability 73. The amount of time necessary for assembly line workers to complete a product is a normal random variable with a mean of 15 minutes and a standard deviation of 2 minutes. The probability is __________ that a product is assembled in between 16 and 21 minutes. ANSWER: 0.30719 TYPE: FI DIFFICULTY: Easy KEYWORDS: normal distribution, probability 74. The amount of time necessary for assembly line workers to complete a product is a normal random variable with a mean of 15 minutes and a standard deviation of 2 minutes. The probability is __________ that a product is assembled in more than 11 minutes. ANSWER: 0.97725 TYPE: FI DIFFICULTY: Easy KEYWORDS: normal distribution, probability 75. The amount of time necessary for assembly line workers to complete a product is a normal random variable with a mean of 15 minutes and a standard deviation of 2 minutes. The probability is __________ that a product is assembled in more than 19 minutes. ANSWER: 0.0228 TYPE: FI DIFFICULTY: Easy KEYWORDS: normal distribution, probability 76. The amount of time necessary for assembly line workers to complete a product is a normal random variable with a mean of 15 minutes and a standard deviation of 2 minutes. The probability is __________ that a product is assembled in less than 20 minutes. ANSWER: 0.9938 TYPE: FI DIFFICULTY: Easy KEYWORDS: normal distribution, probability
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77. The amount of time necessary for assembly line workers to complete a product is a normal random variable with a mean of 15 minutes and a standard deviation of 2 minutes. So, 15% of the products require more than __________ minutes for assembly. ANSWER: 17.0729 TYPE: FI DIFFICULTY: Moderate KEYWORDS: normal distribution, value 78. The amount of time necessary for assembly line workers to complete a product is a normal random variable with a mean of 15 minutes and a standard deviation of 2 minutes. So, 90% of the products require more than __________ minutes for assembly. ANSWER: 12.44 TYPE: FI DIFFICULTY: Moderate KEYWORDS: normal distribution, value 79. The amount of time necessary for assembly line workers to complete a product is a normal random variable with a mean of 15 minutes and a standard deviation of 2 minutes. So, 60% of the products would be assembled within __________ and __________ minutes (symmetrically distributed about the mean). ANSWER: 13.32 and 16.68 TYPE: FI DIFFICULTY: Difficult KEYWORDS: normal distribution, value 80. The amount of time necessary for assembly line workers to complete a product is a normal random variable with a mean of 15 minutes and a standard deviation of 2 minutes. So, 17% of the products would be assembled within __________ minutes. ANSWER: 13.1 TYPE: FI DIFFICULTY: Moderate KEYWORDS: normal distribution, value 81. The amount of time necessary for assembly line workers to complete a product is a normal random variable with a mean of 15 minutes and a standard deviation of 2 minutes. So, 70% of the products would be assembled within __________ minutes. ANSWER: 16.0488 TYPE: FI DIFFICULTY: Moderate KEYWORDS: normal distribution, value
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TABLE 6-1 The manager of a surveying company believes that the average number of phone surveys completed per hour by her employees has a normal distribution. She takes a sample of 15 days output from her employees and determines the average number of surveys per hour on these days. The ordered array for this data is: 10.0, 10.1, 10.3, 10.5, 10.7, 11.2, 11.4, 11.5, 11.7, 11.8, 11.8, 12.0, 12.2, 12.2, 12.5. 82. Referring to Table 6-1, the first standardized normal quantile is ________. ANSWER: -1.53 TYPE: FI DIFFICULTY: Moderate KEYWORDS: standardized normal quantile, normal probability plot 83. Referring to Table 6-1, the fourth standardized normal quantile is ________. ANSWER: -0.67 TYPE: FI DIFFICULTY: Moderate KEYWORDS: standardized normal quantile, normal probability plot 84. Referring to Table 6-1, the ninth standardized normal quantile is ________. ANSWER: +0.16 TYPE: FI DIFFICULTY: Moderate KEYWORDS: standardized normal quantile, normal probability plot 85. Referring to Table 6-1, the fourteenth standardized normal quantile is ________. ANSWER: +1.15 TYPE: FI DIFFICULTY: Moderate KEYWORDS: standardized normal quantile, normal probability plot 86. Referring to Table 6-1, the last standardized normal quantile is ________. ANSWER: +1.53 TYPE: FI DIFFICULTY: Moderate KEYWORDS: standardized normal quantile, normal probability plot
Introduction and Data Collection
87. Referring to Table 6-1, construct a normal probability plot for the data. ANSWER: 13
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TYPE: PR DIFFICULTY: Moderate KEYWORDS: normal probability plot 88. True or False: Referring to Table 6-1, the data appear reasonably normal but not perfectly normal. ANSWER: True TYPE: TF DIFFICULTY: Moderate KEYWORDS: normal probability plot TABLE 6-2 The city manager of a large city believes that the number of reported accidents on any weekend has a normal distribution. She takes a sample of nine weekends and determines the number of reported accidents during each. The ordered array for this data is: 15, 46, 53, 54, 55, 76, 82, 256, 407. 89. Referring to Table 6-2, the first standardized normal quantile is ________. ANSWER: -1.28 TYPE: FI DIFFICULTY: Moderate KEYWORDS: standardized normal quantile, normal probability plot
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90. Referring to Table 6-2, the fifth standardized normal quantile is ________. ANSWER: 0 TYPE: FI DIFFICULTY: Moderate KEYWORDS: standardized normal quantile, normal probability plot 91. Referring to Table 6-2, the sixth standardized normal quantile is ________. ANSWER: +0.25 TYPE: FI DIFFICULTY: Moderate KEYWORDS: standardized normal quantile, normal probability plot 92. Referring to Table 6-2, the second standardized normal quantile is ________. ANSWER: -0.84 TYPE: FI DIFFICULTY: Moderate KEYWORDS: standardized normal quantile, normal probability plot 93. Referring to Table 6-2, the seventh standardized normal quantile is ________. ANSWER: +0.52 TYPE: FI DIFFICULTY: Moderate KEYWORDS: standardized normal quantile, normal probability plot
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94. Referring to Table 6-2, construct a normal probability plot. ANSWER: 450
Accide n ts
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TYPE: PR DIFFICULTY: Moderate KEYWORDS: normal probability plot 95. True or False: Referring to Table 6-2, the data appear normal. ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: normal probability plot 96. Times spent watching TV every week by first graders follow an exponential distribution with mean 10 hours. The probability that a given first grader spends less than 20 hours watching TV is ______. ANSWER: 0.8647 TYPE: FI DIFFICULTY: Easy KEYWORDS: exponential distribution, probability 97. Times spent watching TV every week by first graders follow an exponential distribution with a mean of 10 hours. The probability that a given first grader spends more than 5 hours watching TV is ______. ANSWER: 0.6065 TYPE: FI DIFFICULTY: Easy KEYWORDS: exponential distribution, probability 98. Times spent watching TV every week by first graders follow an exponential distribution with a mean of 10 hours. The probability that a given first grader spends between 10 and 15 hours watching TV is ______. ANSWER: 0.1447 TYPE: FI DIFFICULTY: Moderate KEYWORDS: exponential distribution, probability
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99. Patients arriving at an outpatient clinic follow an exponential distribution with a mean of 15 minutes. What is the average number of arrivals per minute? ANSWER: 0.06667 TYPE: PR DIFFICULTY: Easy KEYWORDS: exponential distribution, mean 100. Patients arriving at an outpatient clinic follow an exponential distribution with a mean of 15 minutes. What is the probability that a randomly chosen arrival would be more than 18 minutes? ANSWER: 0.3012 TYPE: PR DIFFICULTY: Easy KEYWORDS: exponential distribution, probability 101. Patients arriving at an outpatient clinic follow an exponential distribution with a mean of 15 minutes. What is the probability that a randomly chosen arrival would be less than 15 minutes? ANSWER: 0.6321 TYPE: PR DIFFICULTY: Easy KEYWORDS: exponential distribution, probability 102. Patients arriving at an outpatient clinic follow an exponential distribution at a rate of 15 patients per hour. What is the probability that a randomly chosen arrival would be less than 15 minutes? ANSWER: 0.9765 TYPE: PR DIFFICULTY: Easy KEYWORDS: exponential distribution, probability 103. Patients arriving at an outpatient clinic follow an exponential distribution at a rate of 15 patients per hour. What is the probability that a randomly chosen arrival would be more than 5 minutes? ANSWER: 0.2865 TYPE: PR DIFFICULTY: Easy KEYWORDS: exponential distribution, probability 104. Patients arriving at an outpatient clinic follow an exponential distribution at a rate of 15 patients per hour. What is the probability that a randomly chosen arrival would be between 5 minutes and 15 minutes? ANSWER: 0.2630 TYPE: PR DIFFICULTY: Moderate KEYWORDS: exponential distribution, probability 105. Patients arriving at an outpatient clinic follow an exponential distribution at a rate of 1 patient per hour. What is the probability that a randomly chosen arrival would be more than 1 hour? ANSWER: 0.3679
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TYPE: PR DIFFICULTY: Easy KEYWORDS: exponential distribution, probability 106. Patients arriving at an outpatient clinic follow an exponential distribution at a rate of 1 patient per hour. What is the probability that a randomly chosen arrival would be more than 2.5 hours? ANSWER: 0.0821 TYPE: PR DIFFICULTY: Easy KEYWORDS: exponential distribution, probability 107. Patients arriving at an outpatient clinic follow an exponential distribution at a rate of 1 patient per hour. What is the probability that a randomly chosen arrival would be less than 20 minutes? ANSWER: 0.2835 TYPE: PR DIFFICULTY: Easy KEYWORDS: exponential distribution, probability 108. Patients arriving at an outpatient clinic follow an exponential distribution at a rate of 1.5 patients per hour. What is the probability that a randomly chosen arrival would be less than 10 minutes? ANSWER: 0.2212 TYPE: PR DIFFICULTY: Easy KEYWORDS: exponential distribution, probability
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109. Patients arriving at an outpatient clinic follow an exponential distribution at a rate of 1.5 patients per hour. What is the probability that a randomly chosen arrival would be between 10 and 15 minutes? ANSWER: 0.0915 TYPE: PR DIFFICULTY: Moderate KEYWORDS: exponential distribution, probability TABLE 6-3 The number of column inches of classified advertisements appearing on Mondays in a certain daily newspaper is normally distributed with a population mean of 320 and a population standard deviation of 20 inches. 110. Referring to Table 6-3, for a randomly chosen Monday, what is the probability there will be less than 340 column inches of classified advertisements? ANSWER: 0.8413 TYPE: PR DIFFICULTY: Easy KEYWORDS: normal distribution, probability 111. Referring to Table 6-3, for a randomly chosen Monday, what is the probability there will be between 280 and 360 column inches of classified advertisements? ANSWER: 0.9545 TYPE: PR DIFFICULTY: Moderate KEYWORDS: normal distribution, probability 112. Referring to Table 6-3, for a randomly chosen Monday, the probability is 0.1 that there will be less than how many column inches of classified advertisements? ANSWER: 294.4 TYPE: PR DIFFICULTY: Moderate KEYWORDS: normal distribution, value 113. Referring to Table 6-3, a single Monday is chosen at random. State in which of the following ranges the number of column inches of classified advertisements 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 a mean of $600 and a standard deviation of $40. At night he
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works as a waiter, for which his monthly income is normally distributed with a mean of $100 and a standard deviation of $30. John's income levels from these two sources are independent of each other. 114. Referring to Table 6-4, for a given month, what is the probability that John's commission from the jewelry store is less than $640? ANSWER: 0.8413 TYPE: PR DIFFICULTY: Easy KEYWORDS: normal distribution, probability 115. Referring to Table 6-4, for a given month, what is the probability that John's income as a waiter is between $70 and $160? ANSWER: 0.8186 TYPE: PR DIFFICULTY: Easy KEYWORDS: normal distribution, probability 116. Referring to Table 6-4, the probability is 0.9 that John's income as a waiter is less than how much in a given month? ANSWER: $138.45 TYPE: PR DIFFICULTY: Moderate KEYWORDS: normal distribution, value 117. Referring to Table 6-4, find the mean and standard deviation of John's total income from these two jobs for a given month. ANSWER: $900; $50 TYPE: PR DIFFICULTY: Difficult EXPLANATION: Total mean = $200 + $600 + $100 = $900; Total variance = 40 2 + 302 = 2,500 KEYWORDS: normal distribution, mean, standard deviation 118. Referring to Table 6-4, for a given month, what is the probability that John's total income from these two jobs is less than $825? ANSWER: 0.0668 TYPE: PR DIFFICULTY: Difficult EXPLANATION: Total mean = $900, Total standard deviation = $50 KEYWORDS: normal distribution, probability 119. Referring to Table 6-4, the probability is 0.2 that John's total income from these two jobs in a given month is less than how much? ANSWER: $857.92 TYPE: PR DIFFICULTY: Difficult KEYWORDS: normal distribution, value TABLE 6-5
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Suppose the time interval between two consecutive defective light bulbs from a production line has a uniform distribution over an interval from 0 to 90 minutes. 120. Referring to Table 6-5, what is the mean of the time interval? ANSWER: 45 TYPE: PR DIFFICULTY: Easy KEYWORDS: uniform distribution, mean 121. Referring to Table 6-5, what is the variance of the time interval? ANSWER: 675 TYPE: PR DIFFICULTY: Easy KEYWORDS: uniform distribution, variance 122. Referring to Table 6-5, what is the standard deviation of the time interval? ANSWER: 25.9808 TYPE: PR DIFFICULTY: Easy KEYWORDS: uniform distribution, standard deviation 123. Referring to Table 6-5, what is the probability that the time interval between two consecutive defective light bulbs will be exactly 10 minutes? ANSWER: 0.0 TYPE: PR DIFFICULTY: Moderate KEYWORDS: uniform distribution, probability 124. Referring to Table 6-5, what is the probability that the time interval between two consecutive defective light bulbs will be less than 10 minutes? ANSWER: 0.1111 TYPE: PR DIFFICULTY: Easy KEYWORDS: uniform distribution, probability 125. Referring to Table 6-5, what is the probability that the time interval between two consecutive defective light bulbs will be between 10 and 20 minutes? ANSWER: 0.1111 TYPE: PR DIFFICULTY: Easy KEYWORDS: uniform distribution, probability 126. Referring to Table 6-5, what is the probability that the time interval between two consecutive defective light bulbs will be between 10 and 35 minutes? ANSWER: 0.2778
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TYPE: PR DIFFICULTY: Easy KEYWORDS: uniform distribution, probability 127. Referring to Table 6-5, what is the probability that the time interval between two consecutive defective light bulbs will be at least 50 minutes? ANSWER: 0.4444 TYPE: PR DIFFICULTY: Easy KEYWORDS: uniform distribution, probability 128. Referring to Table 6-5, what is the probability that the time interval between two consecutive defective light bulbs will be at least 80 minutes? ANSWER: 0.1111 TYPE: PR DIFFICULTY: Easy KEYWORDS: uniform distribution, probability 129. Referring to Table 6-5, what is the probability that the time interval between two consecutive defective light bulbs will be at least 90 minutes? ANSWER: 0.0 TYPE: PR DIFFICULTY: Moderate KEYWORDS: uniform distribution, probability 130. Referring to Table 6-5, the probability is 50% that the time interval between two consecutive defective light bulbs will fall between which two values that are the same distance from the mean? ANSWER: 22.5 and 67.5 TYPE: PR DIFFICULTY: Difficult KEYWORDS: uniform distribution, value
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131. Referring to Table 6-5, the probability is 75% that the time interval between two consecutive defective light bulbs will fall between which two values that are the same distance from the mean? ANSWER: 11.25 and 78.75 TYPE: PR DIFFICULTY: Difficult KEYWORDS: uniform distribution, value 132. Referring to Table 6-5, the probability is 90% that the time interval between two consecutive defective light bulbs will fall between which two values that are the same distance from the mean? ANSWER: 4.5 and 85.5 TYPE: PR DIFFICULTY: Difficult KEYWORDS: uniform distribution, value TABLE 6-6 The interval between consecutive hits at a web site is assumed to follow an exponential distribution with an average of 40 hits per minute. 133. Referring to Table 6-6, what is the average time between consecutive hits? ANSWER: 0.025 minutes TYPE: PR DIFFICULTY: Easy KEYWORDS: exponential distribution, mean 134. Referring to Table 6-6, what is the probability that the next hit at the web site will occur within 10 seconds after just being hit by a visitor? ANSWER: 0.9987 TYPE: PR DIFFICULTY: Easy KEYWORDS: exponential distribution, probability 135. Referring to Table 6-6, what is the probability that the next hit at the web site will occur within no sooner than 5 seconds after just being hit by a visitor? ANSWER: 0.0357 TYPE: PR DIFFICULTY: Moderate KEYWORDS: exponential distribution, probability
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136. Referring to Table 6-6, what is the probability that the next hit at the web site will occur between the next 1.2 and 1.5 seconds after just being hit by a visitor? ANSWER: 0.08145 TYPE: PR DIFFICULTY: Moderate KEYWORDS: exponential distribution, probability
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CHAPTER 7: SAMPLING DISTRIBUTIONS 1. Sampling distributions describe the distribution of a) parameters. b) statistics. c) both parameters and statistics. d) neither parameters nor statistics. ANSWER: b TYPE: MC DIFFICULTY: Easy KEYWORDS: statistics, sampling distribution 2. The standard error of the mean a) is never larger than the standard deviation of the population. b) decreases as the sample size increases. c) measures the variability of the mean from sample to sample. d) all of the above ANSWER: d TYPE: MC DIFFICULTY: Easy KEYWORDS: standard error, mean 3. The Central Limit Theorem is important in statistics because a) for a large n, it says the population is approximately normal. b) for any population, it says the sampling distribution of the sample mean is approximately normal, regardless of the sample size. c) for a large n, it says the sampling distribution of the sample mean is approximately normal, regardless of the shape of the population. d) for any sized sample, it says the sampling distribution of the sample mean is approximately normal. ANSWER: c TYPE: MC DIFFICULTY: Difficult KEYWORDS: central limit theorem 4. If the expectation of a sampling distribution is located at the parameter it is estimating, then we call that statistic a) unbiased. b) minimum variance. c) biased. d) random. ANSWER: a TYPE: MC DIFFICULTY: Moderate KEYWORDS: unbiased
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5. For air travelers, one of the biggest complaints involves the waiting time between when the airplane taxis away from the terminal until the flight takes off. This waiting time is known to have a skewed-right distribution with a mean of 10 minutes and a standard deviation of 8 minutes. Suppose 100 flights have been randomly sampled. Describe the sampling distribution of the mean waiting time between when the airplane taxis away from the terminal until the flight takes off for these 100 flights. a) Distribution is skewed-right with mean = 10 minutes and standard error = 0.8 minutes. b) Distribution is skewed-right with mean = 10 minutes and standard error = 8 minutes. c) Distribution is approximately normal with mean = 10 minutes and standard error = 0.8 minutes. d) Distribution is approximately normal with mean = 10 minutes and standard error = 8 minutes. ANSWER: c TYPE: MC DIFFICULTY: Moderate KEYWORDS: central limit theorem 6. Which of the following statements about the sampling distribution of the sample mean is incorrect? a) The sampling distribution of the sample mean is approximately normal whenever the sample size is sufficiently large ( n 30 ). b) The sampling distribution of the sample mean is generated by repeatedly taking samples of size n and computing the sample means. c) The mean of the sampling distribution of the sample mean is equal to . d) The standard deviation of the sampling distribution of the sample mean is equal to . ANSWER: d TYPE: MC DIFFICULTY: Easy KEYWORDS: sampling distribution, properties 7. Which of the following is true about the sampling distribution of the sample mean? a) The mean of the sampling distribution is always . b) The standard deviation of the sampling distribution is always . c) The shape of the sampling distribution is always approximately normal. d) All of the above are true. ANSWER: a TYPE: MC DIFFICULTY: Moderate KEYWORDS: sampling distribution, properties
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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 sample mean is equal to 23 years. b) The standard deviation of the sample mean is equal to 3 years. c) The shape of the sampling distribution is approximately normal. d) The standard error of the sample mean is equal to 0.3 years. ANSWER: b TYPE: MC DIFFICULTY: Easy KEYWORDS: sampling distribution, central limit theorem 10. Why is the Central Limit Theorem so important to the study of sampling distributions? a) It allows us to disregard the size of the sample selected when the population is not normal. b) It allows us to disregard the shape of the sampling distribution when the size of the population is large. c) It allows us to disregard the size of the population we are sampling from. d) It allows us to disregard the shape of the population when n is large. ANSWER: d TYPE: MC DIFFICULTY: Moderate KEYWORDS: central limit theorem 11. A sample that does not provide a good representation of the population from which it was collected is referred to as a(n) sample. ANSWER: biased TYPE: FI DIFFICULTY: Moderate KEYWORDS: unbiased 12. True or False: The Central Limit Theorem is considered powerful in statistics because it works for any population distribution, provided the sample size is sufficiently large and the population mean and standard deviation are known. ANSWER: True TYPE: TF DIFFICULTY: Moderate KEYWORDS: central limit theorem 13. Suppose a sample of n = 50 items is drawn from a population of manufactured products and the weight, X, of each item is recorded. Prior experience has shown that the weight has a probability distribution with = 6 ounces and = 2.5 ounces. Which of the following is true about the sampling distribution of the sample mean if a sample of size 15 is selected? a) The mean of the sampling distribution is 6 ounces.
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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 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
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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 centimeters. 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 centimeters. 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 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 centimeters. A random sample of 12 computer chips is taken. What is the probability that the sample mean will be below 0.95 centimeters? ANSWER: 0.0416 TYPE: PR DIFFICULTY: Moderate KEYWORDS: sampling distribution, mean, probability
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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 centimeters. 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 a standard deviation of 0.8 pounds. 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 a standard deviation of 0.8 pounds. 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 a standard deviation of 0.8 pounds. If a sample of 64 fish yields a mean of 3.4 pounds, what is probability of obtaining a sample mean this large or larger? a) 0.0001 b) 0.0013 c) 0.0228 d) 0.4987 ANSWER: c TYPE: MC DIFFICULTY: Moderate KEYWORDS: sampling distribution, mean, probability 25. The owner of a fish market has an assistant who has determined that the weights of catfish are normally distributed, with mean of 3.2 pounds and a standard deviation of 0.8 pounds. What percentage of samples of 4 fish will have sample means between 3.0 and 4.0 pounds? a) 84% b) 67%
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c) 29% d) 16% ANSWER: b TYPE: MC DIFFICULTY: Moderate KEYWORDS: sampling distribution, mean, probability 26. The use of the finite population correction factor, when sampling without replacement from finite populations, will a) increase the standard error of the mean. b) not affect the standard error of the mean. c) reduce the standard error of the mean. d) only affect the proportion, not the mean. ANSWER: c TYPE: MC DIFFICULTY: Easy KEYWORDS: finite population correction 27. For sample size 16, the sampling distribution of the mean will be approximately normally distributed a) regardless of the shape of the population. b) if the shape of the population is symmetrical. c) if the sample standard deviation is known. d) if the sample is normally distributed. ANSWER: b TYPE: MC DIFFICULTY: Moderate KEYWORDS: sampling distribution, mean, central limit theorem 28. The standard error of the mean for a sample of 100 is 30. In order to cut the standard error of the mean to 15, we would a) increase the sample size to 200. b) increase the sample size to 400. c) decrease the sample size to 50. d) decrease the sample to 25. ANSWER: b TYPE: MC DIFFICULTY: Moderate KEYWORDS: standard error, mean
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29. Which of the following is true regarding the sampling distribution of the mean for a large sample size? a) It has the same shape, mean, and standard deviation as the population. b) It has a normal distribution with the same mean and standard deviation as the population. c) It has the same shape and mean as the population, but has a smaller standard deviation. d) It has a normal distribution with the same mean as the population but with a smaller standard deviation. ANSWER: d TYPE: MC DIFFICULTY: Moderate KEYWORDS: sampling distribution, mean, central limit theorem 30. For sample sizes greater than 30, the sampling distribution of the mean will be approximately normally distributed 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 31. For sample size 1, the sampling distribution of the mean will be normally distributed a) regardless of the shape of the population. b) only if the shape of the population is symmetrical. c) only if the population values are positive. d) only if the population is normally distributed. ANSWER: d TYPE: MC DIFFICULTY: Moderate KEYWORDS: sampling distribution, mean, central limit theorem 32. The standard error of the proportion will become larger a) as p approaches 0. b) as p approaches 0.50. c) as p approaches 1.00. d) as n increases. ANSWER: b TYPE: MC DIFFICULTY: Moderate KEYWORDS: standard error, proportion
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33. True or False: As the sample size increases, the standard error of the mean increases. ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: standard error, mean 34. True or False: If the population distribution is symmetric, the sampling distribution of the mean can be approximated by the normal distribution if the samples contain at least 15 observations. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: population distribution, sampling distribution, mean, central limit theorem 35. True or False: If the population distribution is unknown, in most cases the sampling distribution of the mean can be approximated by the normal distribution if the samples contain at least 30 observations. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: sampling distribution, mean, central limit theorem 36. True or False: If the amount of gasoline purchased per car at a large service station has a population mean of $15 and a population standard deviation of $4, then 99.73% of all cars will purchase between $3 and $27. ANSWER: False TYPE: TF DIFFICULTY: Moderate KEYWORDS: sampling distribution, mean, probability 37. True or False: If the amount of gasoline purchased per car at a large service station has a population mean of $15 and a population standard deviation of $4, and a random sample of 4 cars is selected, there is approximately a 68.26% chance that the sample mean will be between $13 and $17. ANSWER: False TYPE: TF DIFFICULTY: Moderate EXPLANATION: The sample is too small for the normal approximation. KEYWORDS: sampling distribution, mean, probability
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38. True or False: If the amount of gasoline purchased per car at a large service station has a population mean of $15 and a population standard deviation of $4, and it is assumed that the amount of gasoline purchased per car is symmetric, there is approximately a 68.26% chance that a random sample of 16 cars will have a sample mean between $14 and $16. ANSWER: True TYPE: TF DIFFICULTY: Moderate KEYWORDS: sampling distribution, mean, probability 39. True or False: If the amount of gasoline purchased per car at a large service station has a population mean of $15 and a population standard deviation of $4, and a random sample of 64 cars is selected, there is approximately a 95.44% chance that the sample mean will be between $14 and $16. ANSWER: True TYPE: TF DIFFICULTY: Difficult KEYWORDS: sampling distribution, mean, probability 40. True or False: As the sample size increases, the effect of an extreme value on the sample mean becomes smaller. ANSWER: True TYPE: TF DIFFICULTY: Moderate KEYWORDS: sampling distribution, law of large numbers 41. True or False: If the population distribution is skewed, in most cases the sampling distribution of the mean can be approximated by the normal distribution if the samples contain at least 30 observations. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: sampling distribution, mean, central limit theorem 42. True or False: A sampling distribution is a probability distribution for a statistic. ANSWER: True TYPE: TF DIFFICULTY: Moderate KEYWORDS: sampling distribution
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ANSWER: True TYPE: TF DIFFICULTY: Moderate KEYWORDS: sampling distribution, mean, probability 44. 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 45. 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 46. True or False: The Central Limit Theorem ensures that the sampling distribution of the sample mean approaches normal as the sample size increases. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: central limit theorem 47. True or False: The standard error of the mean is also known as the standard deviation of the sampling distribution of the sample mean. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: standard error, mean 48. True or False: A sampling distribution is defined as the probability distribution of possible sample sizes that can be observed from a given population. ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: sampling distribution
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49. True or False: As the size of the sample is increased, the standard deviation of the sampling distribution of the sample mean for a normally distributed population will stay the same. ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: standard error, properties 50. True or False: For distributions such as the normal distribution, the arithmetic mean is considered more stable from sample to sample than other measures of central tendency. ANSWER: True TYPE: TF DIFFICULTY: Moderate KEYWORDS: sampling distribution, mean 51. True or False: The fact that the sample means are less variable than the population data can be observed from the standard error of the mean. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: sampling distribution, mean, standard error 52. The amount of pyridoxine (in grams) per multiple vitamin is normally distributed with = 110 grams and = 25 grams. A sample of 25 vitamins is to be selected. What is the probability that the sample mean will be between 100 and 120 grams? ANSWER: 0.9545 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 less than 100 grams? ANSWER: 0.0228 TYPE: PR DIFFICULTY: Moderate KEYWORDS: sampling distribution, mean, probability 54. The amount of pyridoxine (in grams) per multiple vitamin is normally distributed with = 110 grams and = 25 grams. A sample of 25 vitamins is to be selected. What is the probability that the sample mean will be greater than 100 grams? ANSWER: 0.9772 TYPE: PR DIFFICULTY: Moderate KEYWORDS: sampling distribution, mean, probability 55. The amount of pyridoxine (in grams) per multiple vitamin is normally distributed with = 110 grams and = 25 grams. A sample of 25 vitamins is to be selected. So, 95% of all sample means will be greater than how many grams?
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ANSWER: 101.7757 TYPE: PR DIFFICULTY: Difficult KEYWORDS: sampling distribution, mean, value 56. The amount of pyridoxine (in grams) per multiple vitamin is normally distributed with = 110 grams and = 25 grams. A sample of 25 vitamins is to be selected. So, the middle 70% of all sample means will fall between what two values? ANSWER: 104.8 and 115.2 TYPE: PR DIFFICULTY: Difficult KEYWORDS: sampling distribution, mean, value 57. The amount of time required for an oil and filter change on an automobile is normally distributed with a mean of 45 minutes and a standard deviation of 10 minutes. A random sample of 16 cars is selected. What would you expect the standard error of the mean to be? ANSWER: 2.5 minutes TYPE: PR DIFFICULTY: Easy KEYWORDS: standard error, mean 58. The amount of time required for an oil and filter change on an automobile is normally distributed with a mean of 45 minutes and a standard deviation of 10 minutes. A random sample of 16 cars is selected. What is the probability that the sample mean is between 45 and 52 minutes? ANSWER: 0.4974 TYPE: PR DIFFICULTY: Moderate KEYWORDS: sampling distribution, mean, probability 59. The amount of time required for an oil and filter change on an automobile is normally distributed with a mean of 45 minutes and a standard deviation of 10 minutes. A random sample of 16 cars is selected. What is the probability that the sample mean will be between 39 and 48 minutes? ANSWER: 0.8767 TYPE: PR DIFFICULTY: Moderate KEYWORDS: sampling distribution, mean, probability
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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, 95% of all sample means will fall between what two values? ANSWER: 40.1 and 49.9 minutes TYPE: PR DIFFICULTY: Difficult KEYWORDS: sampling distribution, mean, probability 61. The amount of time required for an oil and filter change on an automobile is normally distributed with a mean of 45 minutes and a standard deviation of 10 minutes. A random sample of 16 cars is selected. So, 90% of the sample means will be greater than what value? ANSWER: 41.8 minutes TYPE: PR DIFFICULTY: Difficult KEYWORDS: sampling distribution, mean, probability 62. True or False: The amount of bleach a machine pours into bottles has a mean of 36 oz. with a standard deviation of 0.15 oz. Suppose we take a random sample of 36 bottles filled by this machine. The sampling distribution of the sample mean has a mean of 36. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: sampling distribution, mean, unbiased 63. True or False: The amount of bleach a machine pours into bottles has a mean of 36 oz. with a standard deviation of 0.15 oz. Suppose we take a random sample of 36 bottles filled by this machine. The sampling distribution of the sample mean has a standard error of 0.15. ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: sampling distribution, mean, standard error 64. True or False: The amount of bleach a machine pours into bottles has a mean of 36 oz. with a standard deviation of 0.15 oz. Suppose we take a random sample of 36 bottles filled by this 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
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65. The amount of bleach a machine pours into bottles has a mean of 36 oz. with a standard deviation of 0.15 oz. Suppose we take a random sample of 36 bottles filled by this machine. The probability that the mean of the sample exceeds 36.01 oz. is __________. ANSWER: 0.3446 TYPE: FI DIFFICULTY: Moderate KEYWORDS: sampling distribution, mean, probability, central limit theorem 66. The amount of bleach a machine pours into bottles has a mean of 36 oz. with a standard deviation of 0.15 oz. Suppose we take a random sample of 36 bottles filled by this machine. The probability that the mean of the sample is less than 36.03 is __________. ANSWER: 0.8849 TYPE: FI DIFFICULTY: Moderate KEYWORDS: sampling distribution, mean, probability, central limit theorem 67. The amount of bleach a machine pours into bottles has a mean of 36 oz. with a standard deviation of 0.15 oz. Suppose we take a random sample of 36 bottles filled by this machine. The probability that the mean of the sample is between 35.94 and 36.06 oz. is __________. ANSWER: 0.9836 TYPE: FI DIFFICULTY: Moderate KEYWORDS: sampling distribution, mean, probability, central limit theorem 68. The amount of bleach a machine pours into bottles has a mean of 36 oz. with a standard deviation of 0.15 oz. Suppose we take a random sample of 36 bottles filled by this machine. The probability that the mean of the sample is between 35.95 and 35.98 oz. is __________. ANSWER: 0.1891 TYPE: FI DIFFICULTY: Moderate KEYWORDS: sampling distribution, mean, probability, central limit theorem 69. The amount of bleach a machine pours into bottles has a mean of 36 oz. with a standard deviation of 0.15 oz. Suppose we take a random sample of 36 bottles filled by this machine. So, 95% of the sample means based on samples of size 36 will be between __________ and __________. ANSWER: 35.951 ; 36.049 ounces TYPE: FI DIFFICULTY: Difficult KEYWORDS: sampling distribution, mean, value, central limit theorem
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70. A manufacturer of power tools claims that the average amount of time required to assemble their top-of-the-line table saw is 80 minutes with a standard deviation of 40 minutes. Suppose a random sample of 64 purchasers of this table saw is taken. The mean of the sampling distribution of the sample mean is __________ minutes. ANSWER: 80 TYPE: FI DIFFICULTY: Easy KEYWORDS: sampling distribution, mean, unbiased 71. A manufacturer of power tools claims that the average amount of time required to assemble their top-of-the-line table saw is 80 minutes with a standard deviation of 40 minutes. Suppose a random sample of 64 purchasers of this table saw is taken. The standard deviation of the sampling distribution of the sample mean is __________ minutes. ANSWER: 5 TYPE: FI DIFFICULTY: Easy KEYWORDS: sampling distribution, mean, standard error 72. A manufacturer of power tools claims that the average amount of time required to assemble their top-of-the-line table saw is 80 minutes with a standard deviation of 40 minutes. Suppose a random sample of 64 purchasers of this table saw is taken. The probability that the sample mean will be less than 82 minutes is __________. ANSWER: 0.6554 TYPE: FI DIFFICULTY: Moderate KEYWORDS: sampling distribution, mean, probability, central limit theorem 73. A manufacturer of power tools claims that the average amount of time required to assemble their top-of-the-line table saw is 80 minutes with a standard deviation of 40 minutes. Suppose a random sample of 64 purchasers of this table saw is taken. The probability that the sample mean will be between 77 and 89 minutes is __________. ANSWER: 0.6898 TYPE: FI DIFFICULTY: Moderate KEYWORDS: sampling distribution, mean, probability, central limit theorem 74. A manufacturer of power tools claims that the average amount of time required to assemble their top-of-the-line table saw is 80 minutes with a standard deviation of 40 minutes. Suppose a random sample of 64 purchasers of this table saw is taken. The probability that the sample mean will be greater than 88 minutes is __________. ANSWER: 0.0548 TYPE: FI DIFFICULTY: Moderate KEYWORDS: sampling distribution, mean, probability, central limit theorem 75. A manufacturer of power tools claims that the average amount of time required to assemble their top-of-the-line table saw is 80 minutes with a standard deviation of 40 minutes. Suppose a random sample of 64 purchasers of this table saw is taken. So, 95% of the sample means based on samples of size 64 will be between __________ and __________. ANSWER: 70.2 ; 89.8 minutes TYPE: FI DIFFICULTY: Difficult
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KEYWORDS: sampling distribution, mean, value, central limit theorem 76. To use the normal distribution to approximate the binomial distribution, we need ______ and ______ to be at least 5. ANSWER: np ; n(1-p) TYPE: FI DIFFICULTY: Easy KEYWORDS: sampling distribution, proportion, central limit theorem 77. True or False: The sample mean is an unbiased estimate of the population mean. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: mean, unbiased 78. True or False: The sample proportion is an unbiased estimate of the population proportion. ANSWER: True TYPE: TF DIFFICULTY: Difficult KEYWORDS: proportion, unbiased 79. True or False: The mean of the sampling distribution of a sample proportion is the population proportion, p. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: sampling distribution, proportion
80. True or False: The standard error of the sampling distribution of a sample proportion is
pS 1 pS where n
pS is the sample proportion. ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: sampling distribution, proportion
81. True or False: The standard deviation of the sampling distribution of a sample proportion is
p 1 p where n
p is the population proportion. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: sampling distribution, proportion 82. True or False: A sample of size 25 provides a sample variance of 400. The standard error, in this case equal to 4, is best described as the estimate of the standard deviation of means calculated from samples of size 25.
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ANSWER: True TYPE: TF DIFFICULTY: Moderate KEYWORDS: standard error 83. True or False: An unbiased estimator will have a value, on average across samples, equal to the population parameter value. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: unbiased 84. True or False: In inferential statistics, the standard error of the sample mean assesses the uncertainty or error of estimation. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: mean, standard error 85. Assume that house prices in a neighborhood are normally distributed with a standard deviation of $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 TYPE: PR DIFFICULTY: Moderate KEYWORDS: sampling distribution, mean, probability
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TABLE 7-1 Times spent studying by students in the week before final exams follow a normal distribution with a standard deviation of 8 hours. A random sample of 4 students was taken in order to estimate the mean study time for the population of all students. 86. Referring to Table 7-1, what is the probability that the sample mean exceeds the population mean by more than 2 hours? ANSWER: 0.3085 TYPE: PR DIFFICULTY: Moderate KEYWORDS: sampling distribution, mean, probability 87. Referring to Table 7-1, what is the probability that the sample mean is more than 3 hours below the population mean? ANSWER: 0.2266 TYPE: PR DIFFICULTY: Moderate KEYWORDS: sampling distribution, mean, probability 88. Referring to Table 7-1, what is the probability that the sample mean differs from the population mean by less than 2 hours? ANSWER: 0.3829 TYPE: PR DIFFICULTY: Moderate KEYWORDS: sampling distribution, mean, probability 89. Referring to Table 7-1, what is the probability that the sample mean differs from the population mean by more than 3 hours? ANSWER: 0.4533 TYPE: PR DIFFICULTY: Moderate KEYWORDS: sampling distribution, mean, probability TABLE 7-2 The mean selling price of new homes in a city over a 1-year period was $115,000. The population standard deviation was $25,000. A random sample of 100 new home sales from this city was taken. 90. Referring to Table 7-2, what is the probability that the sample mean selling price was more than $110,000? ANSWER: 0.9772 TYPE: PR DIFFICULTY: Easy KEYWORDS: sampling distribution, mean, probability, central limit theorem 91. Referring to Table 7-2, what is the probability that the sample mean selling price was between $113,000 and $117,000? ANSWER:
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0.5763 TYPE: PR DIFFICULTY: Easy KEYWORDS: sampling distribution, mean, probability, central limit theorem 92. Referring to Table 7-2, what is the probability that the sample mean selling price was between $114,000 and $116,000? ANSWER: 0.3108 TYPE: PR DIFFICULTY: Easy KEYWORDS: sampling distribution, mean, probability, central limit theorem 93. Referring to Table 7-2, without doing the calculations, state in which of the following ranges the sample mean selling price is most likely to lie. a) $113,000-$115,000 b) $114,000-$116,000 c) $115,000-$117,000 d) $116,000-$118,000 ANSWER: b TYPE: MC DIFFICULTY: Easy KEYWORDS: sampling distribution, mean, probability, central limit theorem TABLE 7-3 The lifetimes of a certain brand of light bulbs are known to be normally distributed with a mean of 1,600 hours and a standard deviation of 400 hours. A random sample of 64 of these light bulbs is taken. 94. Referring to Table 7-3, what is the probability that the sample mean lifetime is more than 1,550 hours? ANSWER: 0.8413 TYPE: PR DIFFICULTY: Easy KEYWORDS: sampling distribution, mean, probability 95. Referring to Table 7-3, the probability is 0.15 that the sample mean lifetime is more than how many hours? ANSWER: 1,651.82 hours TYPE: PR DIFFICULTY: Moderate KEYWORDS: sampling distribution, mean, probability 96. Referring to Table 7-3, the probability is 0.20 that the sample mean lifetime differs from the population mean lifetime by at least how many hours? ANSWER: 64.08 hours TYPE: PR DIFFICULTY: Difficult KEYWORDS: sampling distribution, mean, value TABLE 7-4
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According to a survey, only 15% of customers who visited the web site of a major retail store made a purchase. Random samples of size 50 are selected. 97. Referring to Table 7-4, the average of all the sample proportions of customers who will make a purchase after visiting the web site is _______. ANSWER: 0.15 or 15% TYPE: FI DIFFICULTY: Moderate KEYWORDS: sampling distribution, proportion, mean 98. Referring to Table 7-4, the standard deviation of all the sample proportions of customers who will make a purchase after visiting the web site is ________. ANSWER: 0.05050 TYPE: FI DIFFICULTY: Moderate KEYWORDS: sampling distribution, proportion, standard error 99. True of False: Referring to Table 7-4, the requirements for using a normal distribution to approximate a binomial distribution are fulfilled. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: sampling distribution, proportion, central limit theorem 100. Referring to Table 7-4, what proportion of the samples will have between 20% and 30% of customers who will make a purchase after visiting the web site? ANSWER: 0.1596 TYPE: PR DIFFICULTY: Easy KEYWORDS: sampling distribution, proportion, probability 101. Referring to Table 7-4, what proportion of the samples will have less than 15% of customers who will make a purchase after visiting the web site? ANSWER: 0.5 TYPE: PR DIFFICULTY: Easy KEYWORDS: sampling distribution, proportion, probability 102. Referring to Table 7-4, what is the probability that a random sample of 50 will have at least 30% of customers who will make a purchase after visiting the web site? ANSWER: 0.0015 TYPE: PR DIFFICULTY: Easy KEYWORDS: sampling distribution, proportion, probability 103. Referring to Table 7-4, 90% of the samples will have less than what percentage of customers who will make a purchase after visiting the web site? ANSWER:
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21.47 TYPE: PR DIFFICULTY: Difficult KEYWORDS: sampling distribution, proportion, value 104. Referring to Table 7-4, 90% of the samples will have more than what percentage of customers who will make a purchase after visiting the web site? ANSWER: 8.528 TYPE: PR DIFFICULTY: Difficult KEYWORDS: sampling distribution, proportion, value 105. A study at a college on 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 DIFFICULTY: Moderate KEYWORDS: sampling distribution, proportion, mean 106. A study at a college on 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 DIFFICULTY: Moderate KEYWORDS: sampling distribution, proportion, standard error 107. A study at a college on 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 TYPE: FI DIFFICULTY: Easy KEYWORDS: sampling distribution, proportion, probability 108. A study at a college on 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 TYPE: FI DIFFICULTY: Easy KEYWORDS: sampling distribution, proportion, probability 109. A study at a college on 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 DIFFICULTY: Difficult KEYWORDS: sampling distribution, proportion, value
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110. A study at a college on 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 DIFFICULTY: Difficult KEYWORDS: sampling distribution, proportion, value
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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 p is estimated to be, the __________ the sample size required. a) smaller b) larger c) The sample size is not affected. d) The effect cannot be determined from the information given. 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) 2.7970. b) 2.7874. c) 2.4922. d) 2.4851. ANSWER: a TYPE: MC DIFFICULTY: Easy KEYWORDS: critical value, t distribution 5. Which of the following is NOT true about the Student‟s t distribution? a) It has more area in the tails and less in the center than does the normal distribution. b) It is used to construct confidence intervals for the population mean when the population standard deviation is known.
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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 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” in 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) Decrease the confidence level. c) Increase the sample size and decrease the confidence level. d) Increase the confidence level and decrease the sample size. ANSWER: c TYPE: MC DIFFICULTY: Moderate KEYWORDS: confidence interval, properties, width
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15. Suppose a 95% confidence interval for has been constructed. If it is decided to take a larger sample and to decrease the confidence level of the interval, then the resulting interval width would . (Assume that the sample statistics gathered would not change very much for the new sample.) a) be larger than the current interval width b) be narrower than the current interval width c) be the same as the current interval width d) be unknown until actual sample sizes and reliability levels were determined ANSWER: b TYPE: MC DIFFICULTY: Moderate KEYWORDS: confidence interval, properties, width 16. In the construction of confidence intervals, if all other quantities are unchanged, an increase in the sample size will lead to a 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 2 analyzed with the following results: X $50.50 and s 4 00 . 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
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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 2 analyzed with the following results: X $50.50 and s 4 00 . 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 $119.605 ANSWER: d TYPE: MC DIFFICULTY: Moderate KEYWORDS: confidence interval, mean, t distribution
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21. A university system enrolling hundreds of thousands of students is considering a change in the way students pay for their education. Presently the students pay $55 per credit hour. The university system administrators are contemplating charging each student a set fee of $750 per quarter, regardless of how many credit hours each takes. To see if this proposal would be economically feasible, the administrators would like to know how many credit hours, on the average, each student takes per quarter. A random sample of 250 students yields a mean of 14.1 credit hours per quarter and a standard deviation of 2.3 credit hours per quarter. Suppose the administration wanted to estimate the mean to within 0.1 hours at 95% reliability and assumed that the sample standard deviation provided a good estimate for the population standard deviation. How large a 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 1 9.8 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 1 9.8 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
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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 1 9.8 and s2 = 25. If the director wishes to estimate the mean number of admissions per 24-hour period to within 1 admission with 99% reliability, what size sample should she choose? ANSWER: n = 166 TYPE: PR DIFFICULTY: Moderate KEYWORDS: mean, sample size determination 25. A university dean is interested in determining the proportion of students who receive some sort of financial aid. Rather than examine the records for all students, the dean randomly selects 200 students and finds that 118 of them are receiving financial aid. Use a 90% confidence interval to estimate the true proportion of students who receive financial aid. ANSWER: 0.59 0.057 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 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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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, 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, 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 error is not known. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: confidence interval, mean, t distribution 43. True or False: The t distribution allows the calculation of confidence intervals for means for small samples when the population variance is not known, regardless of the shape of the distribution in the population. ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: confidence interval, mean, t distribution
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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 0.95 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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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: 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 58. True or False: The difference between the sample mean and the population mean is called the sampling error. ANSWER: True TYPE: TF DIFFICULTY: Moderate KEYWORDS: sampling error 59. True or False: The difference between the sample proportion and the population proportion is called the sampling error. ANSWER: True TYPE: TF DIFFICULTY: Moderate KEYWORDS: sampling error 60. 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
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61. 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 62. 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 63. True or False: Sampling error equals to half the width of a confidence interval. ANSWER: True TYPE: TF DIFFICULTY: Moderate KEYWORDS: sampling error 64. True or False: The width of a confidence interval equals twice the sampling error. ANSWER: True TYPE: TF DIFFICULTY: Moderate KEYWORDS: sampling error 65. True or False: The sampling error can either be positive or negative. ANSWER: True TYPE: TF DIFFICULTY: Difficult KEYWORDS: sampling error 66. 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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67. True or False: For a given data set and confidence level, the confidence interval will be wider for 95% confidence than for 90% confidence. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: confidence interval, properties, width TABLE 8-1 A random sample of 100 stores from a large chain of 1,000 garden supply stores was selected to determine the average number of lawnmowers sold at an end-of-season clearance sale. The sample results indicated an average of 6 and a standard deviation of 2 lawnmowers sold. A 95% confidence interval (5.623 to 6.377) was established based on these results. 68. 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 69. 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 70. 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 71. 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 72. 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
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73. True or False: Referring to Table 8-1, we do not know for sure whether the true population mean is between 5.623 and 6.377 lawnmowers sold. ANSWER: True TYPE: TF DIFFICULTY: Moderate KEYWORDS: confidence interval, mean, interpretation TABLE 8-2 The managers of a company are worried about the morale of their employees. In order to determine if a problem in this area exists, they decide to evaluate the attitudes of their employees with a standardized test. They select the Fortunato test of job satisfaction, which has a known standard deviation of 24 points. 74. 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 75. 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 76. 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
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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. 77. 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 78. Referring to Table 8-3, the confidence interval goes from ________ to ________. ANSWER: 12.09; 12.19 TYPE: FI DIFFICULTY: Moderate KEYWORDS: confidence interval, mean, standardized normal distribution 79. 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 80. 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 81. 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: 166 TYPE: FI DIFFICULTY: Moderate KEYWORDS: mean, sample size determination
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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. 82. 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 83. 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 84. 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 85. 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; 7.34 TYPE: FI DIFFICULTY: Moderate KEYWORDS: confidence interval, mean, t distribution 86. 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
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87. 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 88. 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 89. 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 0.95. 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. 90. 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 91. 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
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92. Referring to Table 8-5, a 99% confidence interval for the mean voltage of the power packs is from __________ to __________. ANSWER: 11.6367; 12.0891 TYPE: FI DIFFICULTY: Moderate KEYWORDS: confidence interval, mean, t distribution 93. 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 94. 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 95. 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 96. 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 97. 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
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98. True or False: Referring to Table 8-5, it is possible that the 99% confidence interval calculated from the data will not contain the mean voltage for the entire population. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: confidence interval, mean, interpretation TABLE 8-6 A sample of salary offers (in thousands of dollars) given to management majors is: 28, 31, 26, 32, 27, 28, 27, 30, 31, and 29. Using this data to obtain a 95% confidence interval resulted in an interval from 27.5 to 30.3. 99. 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 100. 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 101. 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 102. 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
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103. 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 104. True or False: Referring to Table 8-6, 95% of all confidence intervals constructed similarly to this one will contain the mean of the population. ANSWER: True TYPE: TF DIFFICULTY: Moderate KEYWORDS: confidence interval, mean, interpretation 105. 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 106. 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 107. 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
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108. A prison official wants to estimate the proportion of cases of recidivism. Examining the records of 250 convicts, the official determines that there are 65 cases of recidivism. A 99% confidence interval for the proportion of cases of recidivism would go from __________ to __________. ANSWER: 0.189; 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. 109. 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 110. 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 111. 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 112. 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
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113. 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 114. 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 115. 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 116. 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 117. 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
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118. 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 119. The head of a computer science department is interested in estimating the proportion of students entering the department who will choose the new computer engineering option. A preliminary sample indicates that the proportion will be around 0.25. Therefore, what size sample should the department head take if she wants to be 95% confident that the estimate is within 0.10 of the true proportion? ANSWER: 73 TYPE: PR DIFFICULTY: Easy KEYWORDS: proportion, sample size determination TABLE 8-8 A hotel chain wants to estimate the average number of rooms rented daily in each month. The population of rooms rented daily is assumed to be normally distributed for each month with a standard deviation of 24 rooms. 120. 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: Moderate KEYWORDS: confidence interval, mean 121. 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
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122. 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 123. The county clerk wants to estimate the proportion of retired voters who will need special election facilities. Suppose a sample of 400 retired voters was taken. If 150 need special election facilities, calculate an 80% confidence interval for the population proportion. ANSWER: 0.344 to 0.406 TYPE: PR DIFFICULTY: Easy KEYWORDS: confidence interval, proportion TABLE 8-9 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. From past experience, the standard deviation of market housing prices is estimated to be $40,500. 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 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 were appraised for higher than the market prices was 0.24. 124. Referring to Table 8-9, if he wants a 95% confidence on estimating the true population average market price of the houses in the suburban county to be within $10,000, how large a sample will he need? ANSWER: 62 TYPE: PR DIFFICULTY: Moderate EXPLANATION: This is a sample size with a finite population correction factor. KEYWORDS: mean, sample size determination, finite population correction 125. Referring to Table 8-9, what will be the 90% confidence interval for the average market price of the houses in the suburban county constructed by the consulting group? ANSWER: $220,604.42 to $229,395.58 TYPE: PR DIFFICULTY: Moderate EXPLANATION: This is a confidence interval estimate for the mean with an unknown standard deviation and a finite population correction factor. KEYWORDS: confidence interval, mean, finite population correction 126. Referring to Table 8-9, what will be the 90% confidence interval for the total market price of the houses in the suburban county constructed by the consulting group? ANSWER: $795,278,939.53 to $826,971,060.47 TYPE: PR DIFFICULTY: Moderate KEYWORDS: confidence interval, total amount, finite population correction
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127. Referring to Table 8-9, what will be the 90% confidence interval for the total difference between the market prices and appraised prices of the houses in the suburban county constructed by the consulting group? ANSWER: $449,232,839.65 to $452,017,160.35 TYPE: PR DIFFICULTY: Moderate KEYWORDS: confidence interval, total difference, finite population correction 128. Referring to Table 8-9, what will be the 90% confidence interval for the population proportion of houses that will be appraised for higher than the market prices? ANSWER: 0.1917 to 0.2883 TYPE: PR DIFFICULTY: Moderate EXPLANATION: This is a confidence interval estimate for the proportion with a finite population correction factor. KEYWORDS: confidence interval, proportion, finite population correction 129. 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; 0.117 TYPE: FI DIFFICULTY: Moderate KEYWORDS: confidence interval, proportion TABLE 8-10 The president of a university would like to estimate the proportion of the student population who owns a personal computer. In a sample of 500 students, 417 own a personal computer. 130. 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 131. True or False: Referring to Table 8-10, the parameter of interest is the proportion of the student population who own a personal computer. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: confidence interval, proportion, parameter 132. 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
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133. Referring to Table 8-10, a 99% confidence interval for the proportion of the student population who own a personal computer is from __________ to __________. ANSWER: 0.7911; 0.8769 TYPE: FI DIFFICULTY: Moderate KEYWORDS: confidence interval, proportion 134. Referring to Table 8-10, the sampling error of a 99% confidence interval for the proportion of the student population who own a personal computer is __________. ANSWER: 0.04286 TYPE: FI DIFFICULTY: Moderate KEYWORDS: confidence interval, proportion, sampling error 135. True or False: Referring to Table 8-10, a 95% confidence interval for the proportion of the 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 136. 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
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137. 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 138. 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 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 sample proportion of students who own a personal computer. ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: confidence interval, proportion, interpretation 140. True or False: Referring to Table 8-10, it is possible that the 99% confidence interval calculated from the data will not contain the proportion of the student population who own a personal computer. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: confidence interval, proportion, interpretation 141. True or False: Referring to Table 8-10, we are 99% confident that the average numbers of the 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 142. 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 showed that 12 students skipped more than 10 days of school last year.
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143. Referring to Table 8-11, what is the critical value for the 95% one-sided confidence interval for the proportion of students who skipped more than 10 days of school last year? ANSWER: 1.6449 TYPE: PR DIFFICULTY: Easy KEYWORDS: confidence interval, proportion, one-sided, critical value 144. Referring to Table 8-11, what is the upper bound of the 95% one-sided confidence interval for the proportion of students who skipped more than 10 days of school last year? ANSWER: 0.1204 TYPE: PR DIFFICULTY: Easy KEYWORDS: confidence interval, proportion, one-sided 145. True or False: Referring to Table 8-11, the superintendent can conclude with a 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 1% acceptable level. A random sample of 250 students revealed that 7 of them had used illicit drugs during the last 12 months. 146. 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 drugs during the last 12 months? ANSWER: 1.2816 TYPE: PR DIFFICULTY: Easy KEYWORDS: confidence interval, proportion, one-sided, critical value
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147. 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 drugs during the last 12 months? ANSWER: 0.04137 TYPE: PR DIFFICULTY: Easy KEYWORDS: confidence interval, proportion, one-sided 148. True or False: Referring to Table 8-12, the superintendent can be 90% confident that no more than 5% of the students at the university had used illicit drugs during the last 12 months. ANSWER: True TYPE: TF DIFFICULTY: Moderate KEYWORDS: confidence interval, proportion, one-sided, interpretation 149. True or False: Referring to Table 8-12, using the 90% one-sided confidence interval, the superintendent can be 95% confident that no more than 5% of the students at the university had used illicit drugs during the last 12 months. ANSWER: False TYPE: TF DIFFICULTY: Moderate KEYWORDS: confidence interval, proportion, one-sided, interpretation 150. True or False: Referring to Table 8-12, using the 90% one-sided confidence interval, the superintendent can be 85% confident that no more than 5% of the students at the university had used illicit drugs during the last 12 months. ANSWER: True TYPE: TF DIFFICULTY: Moderate KEYWORDS: confidence interval, proportion, one-sided, interpretation
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CHAPTER 9: FUNDAMENTALS OF HYPOTHESIS TESTING: ONESAMPLE 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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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. If we are performing a two-tailed test of whether = 100, the probability of detecting a shift of the mean to 105 will be ________ the probability of detecting a shift of the mean to 110. a) less than b) greater than c) equal to d) not comparable to ANSWER: a TYPE: MC DIFFICULTY: Moderate KEYWORDS: power
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Introduction and Data Collection 9. True or False: For a given level of significance, if the sample size is increased, the power of the test will increase. ANSWER: True TYPE: TF DIFFICULTY: Moderate KEYWORDS: power, level of significance, sample size 10. True or False: For a given level of significance, if the sample size is increased, the probability of committing a Type I error will increase. ANSWER: False TYPE: TF DIFFICULTY: Moderate KEYWORDS: level of significance, sample size, type I error 11. True or False: For a given level of significance, if the sample size is increased, the probability of committing a Type II error will increase. ANSWER: False TYPE: TF DIFFICULTY: Moderate KEYWORDS: level of significance, sample size, type II error 12. True or False: For a given sample size, the probability of committing a Type II error will increase when the probability of committing a Type I error is reduced. ANSWER: True TYPE: TF DIFFICULTY: Moderate KEYWORDS: sample size, type I error, type II error 13. If an economist wishes to determine whether there is evidence that average family income in a community exceeds $25,000 a) either a one-tailed or two-tailed test could be used with equivalent results. b) a one-tailed test should be utilized. c) a two-tailed test should be utilized. d) none of the above ANSWER: b TYPE: MC DIFFICULTY: Easy KEYWORDS: one-tailed test
Introduction and Data Collection 14. If an economist wishes to determine whether there is evidence that average family income in a community equals $25,000 a) either a one-tailed or two-tailed test could be used with equivalent results. b) a one-tailed test should be utilized. c) a two-tailed test should be utilized. d) none of the above ANSWER: c TYPE: MC DIFFICULTY: Easy KEYWORDS: two-tailed test 15. If the p-value is less than in a two-tailed test, a) 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 16. 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 17. If the Type I error ( ) for a given test is to be decreased, then for a fixed sample size n a) the Type II error ( ) will also decrease. b) the Type II error ( ) will increase. c) the power of the test will increase. d) a one-tailed test must be utilized. ANSWER: b TYPE: MC DIFFICULTY: Moderate KEYWORDS: type I error, type II error, sample size
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Introduction and Data Collection 18. 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 19. For a given level of significance ( ), if the sample size n is increased, the probability of a Type II error ( ) a) will decrease. b) will increase. c) will remain the same. d) cannot be determined. ANSWER: a TYPE: MC DIFFICULTY: Moderate KEYWORDS: level of significance, beta risk, sample size 20. If a researcher rejects a true null hypothesis, she has made a _______error. ANSWER: Type I TYPE: FI DIFFICULTY: Easy KEYWORDS: type I error 21. If a researcher accepts a true null hypothesis, she has made a _______decision. ANSWER: correct TYPE: FI DIFFICULTY: Easy KEYWORDS: decision 22. If a researcher rejects a false null hypothesis, she has made a _______decision. ANSWER: correct TYPE: FI DIFFICULTY: Easy KEYWORDS: decision 23. If a researcher accepts a false null hypothesis, she has made a _______error. ANSWER: Type II TYPE: FI DIFFICULTY: Easy KEYWORDS: type II error 24. It is possible to directly compare the results of a confidence interval estimate to the results obtained by testing a null hypothesis if a) a two-tailed test for is used. b) a one-tailed test for is used. c) Both of the previous statements are true.
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d) None of the previous statements is true. ANSWER: a TYPE: MC DIFFICULTY: Moderate KEYWORDS: confidence interval, two-tailed test 25. The power of a statistical test is a) the probability of not rejecting H0 when it is false. b) the probability of rejecting H0 when it is true. c) the probability of not rejecting H0 when it is true. d) the probability of rejecting H0 when it is false. ANSWER: d TYPE: MC DIFFICULTY: Moderate KEYWORDS: power 26. The symbol for the power of a statistical test is a) . b) 1 – . c) . d) 1 – . ANSWER: d TYPE: MC DIFFICULTY: Easy KEYWORDS: power 27. Suppose we wish to test H0: 47 versus H1: > 47. What will result if we conclude that the mean is greater than 47 when its true value is really 52? a) We have made a Type I error. b) We have made a Type II error. c) We have made a correct decision. d) None of the above are correct. ANSWER: c TYPE: MC DIFFICULTY: Moderate KEYWORDS: one-tailed test, conclusion
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Introduction and Data Collection 28. How many Kleenex should the Kimberly Clark Corporation package of tissues contain? Researchers determined that 60 tissues is the average number of tissues used during a cold. Suppose a random sample of 100 Kleenex users yielded the following data on the number of tissues used during a cold: X = 52, s = 22. Give the null and alternative hypotheses to determine if the number of tissues used during a cold is less than 60. a) H0 : 60 and H1 : 60. b) H0 : 60 and H1 : 60.
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, mean, t test 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 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 30. How many Kleenex should the Kimberly Clark Corporation package of tissues contain? Researchers determined that 60 tissues is the average number of tissues used during a cold. Suppose a random sample of 100 Kleenex users yielded the following data on the number of tissues used during a cold: X = 52, s = 22. Suppose the alternative we wanted to test was H1 : 60 . State the correct rejection region for = 0.05. a) Reject H0 if t > 1.6604. b) Reject H0 if t < – 1.6604. c) Reject H0 if t > 1.9842 or Z < – 1.9842. d) Reject H0 if t < – 1.9842. ANSWER: b TYPE: MC DIFFICULTY: Moderate KEYWORDS: one-tailed test, mean, t test, rejection region
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31. How many Kleenex should the Kimberly Clark Corporation package of tissues contain? Researchers determined that 60 tissues is the average number of tissues used during a cold. Suppose a random sample of 100 Kleenex users yielded the following data on the number of tissues used during a cold: X = 52, s = 22. Suppose the test statistic does fall in the rejection region at = 0.05. Which of the following decisions 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 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 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 conclusions is correct? a) At = 0.05, there is not sufficient evidence to conclude that the average number of tissues used during a cold is 60 tissues. b) At = 0.05, there is sufficient evidence to conclude that the average number of tissues used during a cold is 60 tissues. c) At = 0.05, there is not sufficient evidence to conclude that the average number of tissues used during a cold is not 60 tissues. d) At = 0.10, there is sufficient evidence to conclude that the average number of tissues used during a cold is not 60 tissues. ANSWER: d TYPE: MC DIFFICULTY: Moderate KEYWORDS: one-tailed test, mean, t test, conclusion 33. We have created a 95% confidence interval for with the result (10, 15). What decision will we make if we test H0 : 16 versus H1 : 16 at = 0.05? a) Reject H0 in favor of H1. b) Accept H0 in favor of H1. c) Fail to reject H0 in favor of H1. d) We cannot tell what our decision will be from the information given. ANSWER: a TYPE: MC DIFFICULTY: Easy KEYWORDS: two-tailed test, confidence interval, mean, t test, decision
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Introduction and Data Collection 34. We have created a 95% confidence interval for with the result (10, 15). What decision will we make if we test H0 : 16 versus H1 : 16 at = 0.10? a) Reject H0 in favor of H1. b) Accept H0 in favor of H1. c) Fail to reject H0 in favor of H1. d) We cannot tell what our decision will be from the information given. ANSWER: a TYPE: MC DIFFICULTY: Difficult EXPLANATION: The 90% confidence interval is narrower than (10, 15), which still does not contain 16. KEYWORDS: two-tailed test, confidence interval, mean, t test, decision 35. We have created a 95% confidence interval for with the result (10, 15). What decision will we make if we test H0 : 16 versus H1 : 16 at = 0.025? a) Reject H0 in favor of H1. b) Accept H0 in favor of H1. c) Fail to reject H0 in favor of H1. d) We cannot tell what our decision will be from the information given. ANSWER: d TYPE: MC DIFFICULTY: Difficult EXPLANATION: The 97.5% confidence interval is wider than (10, 15), which could have contained 16 or not have contained 16. KEYWORDS: two-tailed test, confidence interval, mean, t test, decision 36. Suppose we want to test H0 : 30 versus H1 : 30. Which of the following possible sample results based on a sample of size 36 gives the strongest evidence to reject H0 in favor of H1? a) X = 28, s = 6 b) X = 27, s = 4 c) X = 32, s = 2 d) X = 26, s = 9 ANSWER: b TYPE: MC DIFFICULTY: Moderate KEYWORDS: one-tailed test, rejection region
Introduction and Data Collection 37. Which of the following statements is NOT true about the level of significance in a hypothesis test? a) The larger the level of significance, the more likely you are to reject the null hypothesis. b) The level of significance is the maximum risk we are willing to accept in making a Type I error. c) The significance level is also called the level. d) The significance level is another name for Type II error. ANSWER: d TYPE: MC DIFFICULTY: Moderate KEYWORDS: level of significance 38. If, as a result of a hypothesis test, we reject the null hypothesis when it is false, then we have committed a) a Type II error. b) a Type I error. c) no error. d) an acceptance error. ANSWER: c TYPE: MC DIFFICULTY: Moderate KEYWORDS: decision, type I error, type II error 39. The value that separates a rejection region from a non-rejection region is called the _______. ANSWER: critical value TYPE: FI DIFFICULTY: Easy KEYWORDS: critical value, rejection region 40. A is a numerical quantity computed from the data of a sample and is used to reach 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
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Introduction and Data Collection 41. 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.
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 42. The owner of a local nightclub has recently surveyed a random sample of n = 250 customers of the club. She would now like to determine whether or not the mean age of her customers is over 30. If so, she plans to alter the entertainment to appeal to an older crowd. If not, no entertainment changes will be made. If she wants to be 99% confident in her decision, what rejection region should she use? a) Reject H0 if t < – 2.34. b) Reject H0 if t < – 2.55. c) Reject H0 if t > 2.34. d) Reject H0 if t > 2.58. ANSWER: c TYPE: MC DIFFICULTY: Moderate KEYWORDS: one-tailed test, mean, Z test, t test, rejection region 43. The owner of a local nightclub has recently surveyed a random sample of n = 250 customers of the club. She would now like to determine whether or not the mean age of her customers is over 30. If so, she plans to alter the entertainment to appeal to an older crowd. If not, no entertainment changes will be made. Suppose she found that the sample mean was 30.45 years and the sample standard deviation was 5 years. If she wants to be 99% confident in her decision, what decision should she make? a) Reject H0. b) Accept H0. c) Fail to reject H0. d) We cannot tell what her decision should be from the information given. ANSWER: c TYPE: MC DIFFICULTY: Moderate KEYWORDS: one-tailed test, mean, t test, decision
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44. The owner of a local nightclub has recently surveyed a random sample of n = 250 customers of the club. She would now like to determine whether or not the mean age of her customers is over 30. If so, she plans to alter the entertainment to appeal to an older crowd. If not, no entertainment changes will be made. Suppose she found that the sample mean was 30.45 years and the sample standard deviation was 5 years. If she wants to be 99% confident in her decision, what conclusion can she make? a) There is not sufficient evidence that the mean age of her customers is over 30. b) There is sufficient evidence that the mean age of her customers is over 30. c) There is not sufficient evidence that the mean age of her customers is not over 30. d) There is sufficient evidence that the mean age of her customers is not over 30. ANSWER: a TYPE: MC DIFFICULTY: Moderate KEYWORDS: one-tailed test, mean, t test, conclusion 45. The owner of a local nightclub has recently surveyed a random sample of n = 250 customers of 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 46. A survey claims that 9 out of 10 doctors recommend aspirin for their patients with headaches. To test this claim against the alternative that the actual proportion of doctors who recommend aspirin is less than 0.90, a random sample of 100 doctors results in 83 who indicate that they recommend aspirin. The value of the test statistic in this problem is approximately equal to: a) – 4.12. b) – 2.33. c) – 1.86. d) – 0.07. ANSWER: b TYPE: MC DIFFICULTY: Easy KEYWORDS: one-tailed test, proportion, test statistic
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Introduction and Data Collection 47. A survey claims that 9 out of 10 doctors recommend aspirin for their patients with headaches. To test this claim against the alternative that the actual proportion of doctors who recommend aspirin is less than 0.90, a random sample of 100 doctors was selected. Suppose that the test statistic is – 2.20. Can we conclude that H0 should be rejected at the (a) = 0.10, (b) = 0.05, and (c) = 0.01 level of Type I error? a) (a) yes; (b) yes; (c) yes b) (a) no; (b) no; (c) no c) (a) no; (b) no; (c) yes d) (a) yes; (b) yes; (c) no ANSWER: d TYPE: MC DIFFICULTY: Moderate KEYWORDS: one-tailed test, proportion, decision 48. A survey claims that 9 out of 10 doctors recommend aspirin for their patients with headaches. To test this claim against the alternative that the actual proportion of doctors who recommend aspirin is less than 0.90, a random sample of 100 doctors was selected. Suppose you reject the null hypothesis. What conclusion can you draw? a) There is not sufficient evidence that the proportion of doctors who recommend aspirin is not less than 0.90. b) There is sufficient evidence that the proportion of doctors who recommend aspirin is not less than 0.90. c) There is not sufficient evidence that the proportion of doctors who recommend aspirin is less than 0.90. d) There is sufficient evidence that the proportion of doctors who recommend aspirin is less than 0.90. ANSWER: d TYPE: MC DIFFICULTY: Moderate KEYWORDS: one-tailed test, proportion, conclusion 49. A major videocassette rental chain is considering opening a new store in an area that currently does not have any such stores. The chain will open if there is evidence that more than 5,000 of the 20,000 households in the area are equipped with videocassette recorders (VCRs). It conducts a telephone poll of 300 randomly selected households in the area and finds that 96 have VCRs. State the test of interest to the rental chain. a) H0 : p 0.32 versus H1 : p 0.32 b) H0 : p 0.25 versus H1 : p 0.25 c) H0 : p 5,000 versus H1 : p 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
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50. A major videocassette rental chain is considering opening a new store in an area that currently does not have any such stores. The chain will open if there is evidence that more than 5,000 of the 20,000 households in the area are equipped with videocassette recorders (VCRs). It conducts a telephone poll of 300 randomly selected households in the area and finds that 96 have VCRs. The value of the test statistic in this problem is approximately equal to: a) 2.80. b) 2.60. c) 1.94. d) 1.30. ANSWER: a TYPE: MC DIFFICULTY: Easy KEYWORDS: one-tailed test, proportion, Z test, test statistic 51. A major videocassette rental chain is considering opening a new store in an area that currently does not have any such stores. The chain will open if there is evidence that more than 5,000 of the 20,000 households in the area are equipped with videocassette recorders (VCRs). It conducts a telephone poll of 300 randomly selected households in the area and finds that 96 have VCRs. The p-value associated with the test statistic in this problem is approximately equal to: a) 0.0100. b) 0.0051. c) 0.0026. d) 0.0013. ANSWER: c TYPE: MC DIFFICULTY: Easy KEYWORDS: one-tailed test, proportion, Z test, p-value 52. A major videocassette rental chain is considering opening a new store in an area that currently does not have any such stores. The chain will open if there is evidence that more than 5,000 of the 20,000 households in the area are equipped with videocassette recorders (VCRs). It conducts a telephone poll of 300 randomly selected households in the area and finds that 96 have VCRs. The decision on the hypothesis test using a 3% level of significance is: a) to reject H0 in favor of H1. b) to accept H0 in favor of H1. c) to fail to reject H0 in favor of H1. d) We cannot tell what the decision should be from the information given. ANSWER: a TYPE: MC DIFFICULTY: Easy KEYWORDS: one-tailed test, proportion, Z test, decision
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Introduction and Data Collection 53. A major videocassette rental chain is considering opening a new store in an area that currently does not have any such stores. The chain will open if there is evidence that more than 5,000 of the 20,000 households in the area are equipped with videocassette recorders (VCRs). It conducts a telephone poll of 300 randomly selected households in the area and finds that 96 have VCRs. The rental chain's conclusion from the hypothesis test using a 3% level of significance is: a) to open a new store. b) not to open a new store. c) to delay opening a new store until additional evidence is collected. d) We cannot tell what the decision should be from the information given. ANSWER: a TYPE: MC DIFFICULTY: Easy KEYWORDS: one-tailed test, proportion, Z test, conclusion 54. An entrepreneur is considering the purchase of a coin-operated laundry. The present owner claims that over the past 5 years, the average daily revenue was $675 with a standard deviation of $75. A sample of 30 days reveals a daily average revenue of $625. If you were to test the null hypothesis that the daily average revenue was $675, which test would you use? a) Z-test of a population mean b) Z-test of a population proportion c) t test of a population mean d) 2 -test of population variance ANSWER: a TYPE: MC DIFFICULTY: Moderate KEYWORDS: two-tailed test, mean, Z test 55. An entrepreneur is considering the purchase of a coin-operated laundry. The present 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
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56. A manager of the credit department for an oil company would like to determine whether the average monthly balance of credit card holders is equal to $75. An auditor selects a random sample of 100 accounts and finds that the average owed is $83.40 with a sample standard deviation of $23.65. If you wanted to test whether the auditor should conclude that there is evidence that the average balance is different from $75, which test would you use? a) Z-test of a population mean b) Z-test of a population proportion c) t test of population mean 2 d) -test of population variance ANSWER: c TYPE: MC DIFFICULTY: Easy KEYWORDS: two-tailed test, mean, t test 57. A manager of the credit department for an oil company would like to determine whether the average monthly balance of credit card holders is equal to $75. An auditor selects a random sample of 100 accounts and finds that the average owed is $83.40 with a sample standard deviation of $23.65. If you 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 58. The marketing manager for an automobile manufacturer is interested in determining the proportion of new compact car owners who would have purchased a passenger-side inflatable air bag if it had been available for an additional cost of $300. The manager believes from previous information that the proportion is 0.30. Suppose that a survey of 200 new compact car owners is selected and 79 indicate that they would have purchased the inflatable air bag. 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 2 d) -test of population variance ANSWER: b TYPE: MC DIFFICULTY: Easy KEYWORDS: two-tailed test, proportion
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Introduction and Data Collection 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 bag. If you were to conduct a test to determine whether there is evidence that the proportion is different from 0.30 and decided not to reject the null hypothesis, what conclusion could you draw? a) There is sufficient evidence that the proportion is 0.30. b) There is not sufficient evidence that the proportion is 0.30. c) There is sufficient evidence that the proportion is 0.30. d) There is not sufficient evidence that the proportion is not 0.30. ANSWER: d TYPE: MC DIFFICULTY: Easy KEYWORDS: two-tailed test, proportion, conclusion 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 0.10. The following information was extracted from the Microsoft Excel output for the sample of 46 Monarch butterflies: n = 46; Arithmetic Mean = 28.00; Standard Deviation = 25.92; Standard Error = 3.82; Null Hypothesis: H0 : 20.000; = 0.10; df = 45; T Test Statistic = 2.09; One-Tailed Test Upper Critical Value = 1.3006; p-value = 0.021; Decision = Reject. 60. Referring to Table 9-1, the parameter the biologist is interested in is: a) the mean number of butterflies in Pismo Beach State Park. b) the mean number of parasites on these 46 butterflies. c) the mean number of parasites on Monarch butterflies in Pismo Beach State Park. d) the proportion of butterflies with parasites. ANSWER: c TYPE: MC DIFFICULTY: Easy KEYWORDS: mean, t test, parameter 61. Referring to Table 9-1, state the alternative hypothesis for this study. ANSWER:
H1 : 20.000
TYPE: PR DIFFICULTY: Easy KEYWORDS: one-tailed test, mean, t test, form of hypothesis
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62. Referring to Table 9-1, what critical value should the biologist use to determine the rejection region? a) 1.6794 b) 1.3006 c) 1.3002 d) 1.28 ANSWER: b TYPE: MC DIFFICULTY: Easy KEYWORDS: one-tailed test, mean, t test, critical value 63. True or False: Referring to Table 9-1, the null hypothesis would be rejected. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: one-tailed test, mean, t test, decision 64. True or False: Referring to Table 9-1, the null hypothesis would be rejected if a 4% probability of committing a Type I error is allowed. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: one-tailed test, mean, t test, decision 65. True or False: Referring to Table 9-1, the null hypothesis would be rejected if a 1% probability of committing a Type I error is allowed. ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: one-tailed test, mean, t test, decision 66. Referring to Table 9-1, the lowest probability at which the null hypothesis can be rejected is ______. ANSWER: 0.021 TYPE: FI DIFFICULTY: Easy KEYWORDS: one-tailed test, mean, t test, p-value 67. True or False: Referring to Table 9-1, this result proves beyond a doubt that the mean number of parasites on butterflies in Pismo Beach State Park is over 20. ANSWER: False TYPE: TF DIFFICULTY: Moderate KEYWORDS: one-tailed test, mean, t test, conclusion 68. True of False: Referring to Table 9-1, the biologist can conclude that there is sufficient evidence to show that the average number of parasites per butterfly is over 20 using a level of significance of 0.10. ANSWER: True TYPE: TF DIFFICULTY: Moderate
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Introduction and Data Collection KEYWORDS: one-tailed test, mean, t test, conclusion 69. True or False: Referring to Table 9-1, the same decision would have been reached if the biologist had selected a level of significance of 0.05. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: one-tailed test, mean, t test, decision 70. True or False: Referring to Table 9-1, the same decision would have been reached if the biologist had selected a level of significance of 0.01. ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: one-tailed test, mean, t test, decision 71. True or False: Referring to Table 9-1, the value of is 0.90. ANSWER: False TYPE: TF DIFFICULTY: Moderate KEYWORDS: one-tailed test, mean, t test, beta risk 72. True or False: Referring to Table 9-1, if these data were used to perform a two-tailed test, the p-value would be 0.042. ANSWER: True TYPE: TF DIFFICULTY: Moderate KEYWORDS: one-tailed test, mean, t test, p-value 73. 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
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74. 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 75. True or False: Suppose, in testing a hypothesis about a proportion, the Z test statistic is computed to be 2.04. The null hypothesis should be rejected if the chosen level of significance is 0.01 and a two-tailed test is used. ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: proportion, Z test, test statistic, critical value, decision 76. 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 77. 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 78. True or False: The statement of the null hypothesis always contains an equality. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: form of null hypothesis 79. True or False: The larger the p-value, the more likely one is to reject the null hypothesis. ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: p-value
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Introduction and Data Collection 80. True or False: The smaller the p-value, the stronger the evidence against the null hypothesis. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: p-value 81. 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 82. 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 83. 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: Difficult KEYWORDS: confidence interval, two-tailed test, decision
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84. 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: Difficult EXPLANATION: We are not sure if 20 will be in the wider confidence interval. KEYWORDS: confidence interval, two-tailed test, decision 85. 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: Difficult 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 he can correctly identify a business major 87% of the time, while 13% of the time, he mistakenly identifies an agriculture major as a business major. 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 being that the person is an agriculture major. 86. 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
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Introduction and Data Collection 87. 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 88. 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 89. Referring to Table 9-2, what is the “actual confidence coefficient?” a) 0.13 b) 0.16 c) 0.84 d) 0.87 ANSWER: d TYPE: MC DIFFICULTY: Difficult KEYWORDS: confidence coefficient 90. 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
?
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91. 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 92. Referring to Table 9-2, what is the power of the test? a) 0.13 b) 0.16 c) 0.84 d) 0.87 ANSWER: c TYPE: MC DIFFICULTY: Difficult KEYWORDS: power TABLE 9-3 An appliance manufacturer claims to have developed a compact microwave oven that consumes an average of no more than 250 W. From previous studies, it is believed that power consumption for microwave ovens is normally distributed with a standard deviation of 15 W. A consumer group has decided to try to discover if the claim appears true. They take a sample of 20 microwave ovens and find that they consume an average of 257.3 W. 93. Referring to Table 9-3, the population of interest is a) the power consumption in the 20 microwave ovens. b) the power consumption in all such microwave ovens. c) the mean power consumption in the 20 microwave ovens. d) the mean power consumption in all such microwave ovens. ANSWER: b TYPE: MC DIFFICULTY: Moderate KEYWORDS: one-tailed test, mean, Z test
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Introduction and Data Collection 94. 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 95. Referring to Table 9-3, the appropriate hypotheses to determine if the manufacturer's claim appears reasonable are: a) H 0 : 250 versus H1 : 250 b) H 0 : 250 versus H1 : 250 c)
H 0 : 250 versus H1 : 250
d) H 0 : 257.3 versus H1 : 257.3 ANSWER: c TYPE: MC DIFFICULTY: Moderate KEYWORDS: one-tailed test, mean, Z test, form of hypothesis 96. 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 97. 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 98. Referring to Table 9-3, the p-value of the test is ________. ANSWER: 0.0148 TYPE: FI DIFFICULTY: Moderate KEYWORDS: one-tailed test, mean, Z test, p-value
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99. True or False: Referring to Table 9-3, for this test to be valid, it is necessary that the power consumption for microwave ovens has a normal distribution. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: one-tailed test, mean, Z test, assumptions 100. True or False: Referring to Table 9-3, the null hypothesis will be rejected at a 5% level of significance. ANSWER: True TYPE: TF DIFFICULTY: Moderate KEYWORDS: one-tailed test, mean, Z test, decision 101. True or False: Referring to Table 9-3, the null hypothesis will be rejected at a 1% level of significance. ANSWER: False TYPE: TF DIFFICULTY: Moderate KEYWORDS: one-tailed test, mean, Z test, decision 102. 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: False TYPE: TF DIFFICULTY: Difficult KEYWORDS: one-tailed test, mean, Z test, conclusion
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Introduction and Data Collection 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 hypothesis test will be done to help make the decision. 103. 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 c)
H 0 : 7.4 versus H1 : 7.4
d) H 0 : 7.4 versus H1 : 7.4 ANSWER: c TYPE: MC DIFFICULTY: Easy KEYWORDS: one-tailed test, mean, Z test, form of hypothesis 104. 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 105. Referring to Table 9-4, a sample of size 36 results in a sample mean of 7.1. The value of the test statistic is ________. ANSWER: -1.50 TYPE: FI DIFFICULTY: Easy KEYWORDS: one-tailed test, mean, Z test, test statistic 106. Referring to Table 9-4, a sample of size 36 results in a sample mean of 7.1. The p-value of the test is ________. ANSWER: 0.0668 TYPE: FI DIFFICULTY: Moderate KEYWORDS: one-tailed test, mean, Z test, p-value
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107. True or False: Referring to Table 9-4, a sample of size 36 results in a sample mean of 7.1. 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 108. True or False: Referring to Table 9-4, a sample of size 36 results in a sample mean of 7.1. 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 109. True or False: Referring to Table 9-4, a sample of size 36 results in a sample mean of 7.1. 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 TABLE 9-5 A bank tests the null hypothesis that the mean age of the bank's mortgage holders is less than or equal to 45, versus an alternative that the mean age is greater than 45. They take a sample and calculate a p-value of 0.0202. 110. 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 111. 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
=
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Introduction and Data Collection 112. 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 113. Referring to Table 9-5, if the same sample was used to test the opposite one-tailed test, what would be this test's p-value? a) 0.0202 b) 0.0404 c) 0.9596 d) 0.9798 ANSWER: d TYPE: MC DIFFICULTY: Moderate KEYWORDS: one-tailed test, mean, p-value TABLE 9-6 The quality control engineer for a furniture manufacturer is interested in the mean amount of force necessary to produce cracks in stressed oak furniture. She performs a two-tailed test of the null hypothesis that the mean for the stressed oak furniture is 650. The calculated value of the Z test statistic is a positive number that leads to a p-value of 0.080 for the test. 114. True or False: Referring to Table 9-6, if the test is performed with a level of significance of 0.10, the null hypothesis would be rejected. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: one-tailed test, mean, decision 115. True or False: Referring to Table 9-6, if the test is performed with a level of significance of 0.10, the engineer can conclude that the mean amount of force necessary to produce cracks in stressed oak furniture is 650. ANSWER: False TYPE: TF DIFFICULTY: Moderate KEYWORDS: one-tailed test, mean, conclusion
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116. True or False: Referring to Table 9-6, if the test is performed with a level of significance of 0.05, the null hypothesis would be rejected. ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: one-tailed test, mean, decision 117. True or False: Referring to Table 9-6, if the test is performed with a level of significance of 0.05, the engineer can conclude that the mean amount of force necessary to produce cracks in stressed oak furniture is 650. ANSWER: False TYPE: TF DIFFICULTY: Moderate EXPLANATION: The engineer can conclude that there is insufficient evidence to show that the mean amount of force needed is not 650, but cannot conclude that there is evidence to show that the force needed is 650. KEYWORDS: one-tailed test, mean, conclusion 118. True or False: Referring to Table 9-6, suppose the engineer had decided that the alternative hypothesis to test was that the mean was greater than 650. Then, if the test is performed with a level of significance of 0.10, the null hypothesis would be rejected. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: one-tailed test, mean, decision 119. Referring to Table 9-6, suppose the engineer had decided that the alternative hypothesis to test was that the mean was greater than 650. What would be the p-value of this one-tailed test? a) 0.040 b) 0.160 c) 0.840 d) 0.960 ANSWER: a TYPE: MC DIFFICULTY: Moderate KEYWORDS: one-tailed test, mean, p-value
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Introduction and Data Collection 120. Referring to Table 9-6, suppose the engineer had decided that the alternative hypothesis to test was that the mean was less than 650. What would be the p-value of this one-tailed test? a) 0.040 b) 0.160 c) 0.840 d) 0.960 ANSWER: d TYPE: MC DIFFICULTY: Moderate KEYWORDS: one-tailed test, mean, p-value 121. True or False: Referring to Table 9-6, suppose the engineer had decided that the alternative hypothesis to test was that the mean was less than 650. Then, if the test is performed with a level of significance of 0.10, the null hypothesis would be rejected. ANSWER: False TYPE: TF DIFFICULTY: Moderate KEYWORDS: one-tailed test, mean, decision TABLE 9-7 A filling machine at a local soft drinks company is calibrated to fill the cans at an average amount of 12 fluid ounces and a standard deviation of 0.5 ounces. The company wants to test whether the standard deviation of the amount filled by the machine is indeed 0.5 ounces. A random sample of 15 cans filled by the machine reveals a standard deviation of 0.67 ounces. 122. Referring to Table 9-7, the parameter of interest in the test is ________. ANSWER: Population standard deviation or population variance TYPE: FI DIFFICULTY: Easy KEYWORDS: two-tailed test, variance, parameter 123. Referring to Table 9-7, which is the appropriate test to use? a) Z-test of a population mean b) Z-test of a population proportion c) t test of a population mean d) 2 -test of population variance ANSWER: d TYPE: MC DIFFICULTY: Moderate KEYWORDS: two-tailed test, variance
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124. True or False: Referring to Table 9-7, in order to perform the test, we need to assume that the amount filled by the machine has a normal distribution. ANSWER: True TYPE: TF DIFFICULTY: Moderate KEYWORDS: two-tailed test, variance, assumption 125. Referring to Table 9-7, what type of test should be performed? a) lower-tailed test b) upper-tailed test c) two-tailed test d) none of the above ANSWER: c TYPE: MC DIFFICULTY: Easy KEYWORDS: two-tailed test, variance, form of hypothesis 126. Referring to Table 9-7, what are the lower and upper critical values of the test when allowing for 5% probability of committing a Type I error? ANSWER: 5.6287 and 26.1189 TYPE: PR DIFFICULTY: Easy KEYWORDS: two-tailed test, variance, critical value 127. Referring to Table 9-7, what is the value of the test statistic? ANSWER: 25.1384 TYPE: PR DIFFICULTY: Easy KEYWORDS: two-tailed test, variance, test statistic 128. True or False: Referring to Table 9-7, the decision is to reject the null hypothesis when using a 5% level of significance. ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: two-tailed test, variance, decision 129. True or False: Referring to Table 9-7, there is sufficient evidence to conclude that the standard deviation of the amount filled by the machine is not exactly 0.5 ounces when using a 5% level of significance. ANSWER: False TYPE: TF DIFFICULTY: Moderate KEYWORDS: two-tailed test, variance, conclusion 130. True or False: Referring to Table 9-7, the decision is to reject the null hypothesis when using a 10% level of significance. ANSWER: True
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Introduction and Data Collection TYPE: TF DIFFICULTY: Easy KEYWORDS: two-tailed test, variance, decision 131. True or False: Referring to Table 9-7, there is sufficient evidence to conclude that the standard deviation of the amount filled by the machine is not exactly 0.5 ounces when using a 10% level of significance. ANSWER: True TYPE: TF DIFFICULTY: Moderate KEYWORDS: two-tailed test, variance, conclusion 132. True or False: Referring to Table 9-7, the p-value of the test is somewhere between 5% and 10%. ANSWER: True TYPE: TF DIFFICULTY: Moderate KEYWORDS: two-tailed test, variance, p-value
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CHAPTER 10: TWO-SAMPLE TESTS WITH NUMERICAL DATA 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 respective a) population sizes are equal. b) sample variances are equal. c) populations are approximately normal or sample sizes are large enough. d) all of the above ANSWER: c TYPE: MC DIFFICULTY: Moderate KEYWORDS: t test for 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 for mean difference, degrees of freedom
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4. If we are testing for the difference between the means of 2 independent populations with samples of n1 = 20 and n2 = 20, the number of degrees of freedom is equal to a) 39. b) 38. c) 19. d) 18. ANSWER: b 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 for 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 for mean difference, degrees of freedom
Introduction and Data Collection
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 .
H0 : D 0 . d) H 0 : D 0 . c)
ANSWER: b TYPE: MC DIFFICULTY: Easy KEYWORDS: t test for mean difference, form of hypothesis 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. In testing for differences between the median of two independent populations, the null hypothesis is a) H 0 : M D 0 . b) H 0 : M D 0 .
H 0 : M1 M 2 0 . d) H 0 : M1 M 2 0 . c)
ANSWER: c TYPE: MC DIFFICULTY: Easy KEYWORDS: Wilcoxon rank sum test, form of hypothesis
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11. In testing for whether the median difference of two related populations is zero, the null hypothesis is a) H 0 : M D 0 . b) H 0 : M D 0 .
H 0 : M1 M 2 0 . d) H 0 : M1 M 2 0 . c)
ANSWER: a TYPE: MC DIFFICULTY: Easy KEYWORDS: Wilcoxon rank sum test, form of hypothesis 12. 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 for equality of variances, degrees of freedom 13. 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 for equality of variances, degrees of freedom 14. 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 for equality of variances
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15. To use the Wilcoxon rank sum test as a test for location, we must assume that a) the obtained data are either ranks or numerical measurements both within and between the 2 samples. b) both samples are randomly and independently drawn from their respective populations. c) both underlying populations from which the samples were drawn are equivalent in shape and dispersion. d) all of the above ANSWER: d TYPE: MC DIFFICULTY: Moderate KEYWORDS: Wilcoxon rank sum test, assumption 16. 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 for equality of variances, rejection region 17. 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 for mean difference, form of hypothesis 18. 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 for mean difference, assumption 19. 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
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20. A "robust" test procedure is one which a) requires a sophisticated level of measurement. b) requires a semi-circular shaped population. c) is sensitive to slight violations in its assumptions. d) is insensitive to slight violations in its assumptions. ANSWER: d TYPE: MC DIFFICULTY: Moderate KEYWORDS: robust test 21. Which of the following is NOT a “robust” test procedure against the violation of distribution assumptions? a) pooled-variance t test for difference in means b) separate variance t test for difference in means c) F test for difference in variances d) Wilcoxon rank sum test for difference in medians ANSWER: c TYPE: MC DIFFICULTY: Easy KEYWORDS: robust test 22. Given the following information, calculate the degrees of freedom that should be used in the pooled-variance t test. s12 = 4 s22 = 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 23. Given the following information, calculate sp2, the pooled sample variance that should be used in the pooledvariance t test. s12 = 4 s22 = 6 n1 = 16 n2 = 25 a) sp2 = 6 b) sp2 = 5 c) sp2 = 5.23 d) sp2 = 4 ANSWER: c TYPE: MC DIFFICULTY: Easy KEYWORDS: pooled-variance t test TABLE 10-1
Introduction and Data Collection
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
24. Referring to Table 10-1, judging from the way the data were collected, which test would likely be most appropriate to employ? a) related samples t test for mean difference b) pooled-variance t test for the difference in means c) independent samples Z test for the difference in means d) related samples Z test for mean difference ANSWER: c TYPE: MC DIFFICULTY: Easy KEYWORDS: Z test for difference in means 25. 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 for mean differences, form of hypothesis
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26. 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 2 2 9.82 9.82 211 100 65.75– 79.83 d) Z 2 2 11.07 6.41 211 100 a)
Z
ANSWER: d TYPE: MC DIFFICULTY: Easy KEYWORDS: Z test for mean differences, test statistic 27. 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 for mean differences, p-value
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TABLE 10-2 A researcher randomly sampled 30 graduates of an MBA program and recorded data concerning their starting salaries. Of primary interest to the researcher was the effect of gender on starting salaries. Analysis of the mean salaries of the females and males in the sample is given below. Size Mean Std Dev 18 48,266.7 13,577.63 12 55,000 11,741.29 Std Error = 4,764.82 Means Diff = – 6,733.3 Z = – 1.4528 2-tailed p-value = 0.1463 t = – 1.4221 2-tailed p-value = 0.1574 Females Males
28. 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
c)
H1 : females males
d)
H1 : females males
ANSWER: b TYPE: MC DIFFICULTY: Moderate KEYWORDS: pooled-variance t test, form of hypothesis 29. 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) 4,634.72. b) – 1.4221. c) – 1.4528. d) – 6,733.33. ANSWER: b TYPE: MC DIFFICULTY: Moderate KEYWORDS: pooled-variance t test, test statistic
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30. 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 31. 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
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TABLE 10-3 The use of preservatives by food processors has become a controversial issue. Suppose 2 preservatives are extensively tested and determined safe for use in meats. A processor wants to compare the preservatives for their effects on retarding spoilage. Suppose 15 cuts of fresh meat are treated with preservative A and 15 are treated with preservative B, and the number of hours until spoilage begins is recorded for each of the 30 cuts of meat. The results are summarized in the table below. Preservative A Preservative B = 106.4 hours XA X B = 96.54 hours S A = 10.3 hours S B = 13.4 hours 32. Referring to Table 10-3, state the test statistic for determining if the population variances differ for preservatives A and B. a) F = – 3.10 b) F = 0.5908 c) F = 0.7687 d) F = 0.8250 ANSWER: b TYPE: MC DIFFICULTY: Moderate KEYWORDS: F test for equality of variances, test statistic 33. Referring to Table 10-3, what assumptions are necessary for a comparison of the population variances to be valid? a) Both sampled populations are normally distributed. b) Both samples are random and independent. c) Neither (a) nor (b) is necessary. d) Both (a) and (b) are necessary. ANSWER: d TYPE: MC DIFFICULTY: Easy KEYWORDS: F test for equality of variances, assumption
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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 the families in Metropolis have. 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 34. 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 35. 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 36. 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
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37. 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 38. 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 39. 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 40. Referring to Table 10-4, suppose = 0.10. Which of the following represents the result of the relevant hypothesis test? a) The alternative hypothesis is rejected. b) The null hypothesis is rejected. c) The null hypothesis is not rejected. d) Insufficient information exists on which to make a decision. ANSWER: b TYPE: MC DIFFICULTY: Easy KEYWORDS: pooled-variance t test, decision
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44. Referring to Table 10-4, suppose = 0.05. Which of the following represents the correction 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 45. Referring to Table 10-4, suppose = 0.01. Which of the following represents the correction 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
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TABLE 10-5 To test the effects of a business school preparation course, 8 students took a general business test before and after the course. The results are given below. Exam Score Exam Score Student Before Course (1) After Course (2) 1 530 670 2 690 770 3 910 1,000 4 700 710 5 450 550 6 820 870 7 820 770 8 630 610 46. Referring to Table 10-5, the number of degrees of freedom is a) 14. b) 13. c) 8. d) 7. ANSWER: d TYPE: MC DIFFICULTY: Easy KEYWORDS: t test for mean differences, degrees of freedom 47. Referring to Table 10-5, the value of the sample mean difference is _______ if the difference scores reflect the results of the exam after the course minus the results of the exam before the course. a) 0 b) 50 c) 68 d) 400 ANSWER: b TYPE: MC DIFFICULTY: Easy KEYWORDS: t test for mean differences, test statistic 48. Referring to Table 10-5, the value of the standard error of the difference scores is a) 65.027. b) 60.828. c) 22.991. d) 14.696. ANSWER: c TYPE: MC DIFFICULTY: Easy KEYWORDS: t test for mean differences, standard error
Introduction and Data Collection
49. Referring to Table 10-5, what is the critical value for testing the hypothesis at the 5% level if the difference scores reflect the results of the exam after the course minus the results of the exam before the course? a) 2.365 b) 2.145 c) 1.761 d) 1.895 ANSWER: d TYPE: MC DIFFICULTY: Easy KEYWORDS: t test for mean differences, critical value 50. Referring to Table 10-5, at the 0.05 level of significance, the decision for this hypothesis test would be: a) reject the null hypothesis. b) do not reject the null hypothesis. c) reject the alternative hypothesis. d) It cannot be determined from the information given. ANSWER: a TYPE: MC DIFFICULTY: Easy KEYWORDS: t test for mean differences, decision 51. Referring to Table 10-5, at the 0.05 level of significance, the conclusion for this hypothesis test would be: a) the business school preparation course does improve exam scores. b) the business school preparation course does not improve exam scores. c) the business school preparation course has no impact on exam scores. d) It cannot be drawn from the information given. ANSWER: a TYPE: MC DIFFICULTY: Easy KEYWORDS: t test for mean differences, conclusion 52. True or False: Referring to Table 10-5, one must assume that the population of difference scores is normally distributed. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: t test for mean differences, assumption 53. Referring to Table 10-5, the calculated value of the test statistic is ________. ANSWER: 2.175 TYPE: FI DIFFICULTY: Easy KEYWORDS: t test for mean differences, test statistic
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54. Referring to Table 10-5, the p-value of the test statistic is ________. ANSWER: 0.0331 TYPE: FI DIFFICULTY: Moderate EXPLANATION: p-value obtained from Excel KEYWORDS: t test for mean differences, p-value 55. True or False: Referring to Table 10-5, in examining the differences between related samples we are essentially sampling from an underlying population of difference "scores." ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: t test for mean differences, sampling distribution 56. A local real estate appraiser analyzed the sales prices of homes in 2 neighborhoods to the corresponding appraised values of the homes. The goal of the analysis was to compare the distribution of sale-to-appraised ratios from homes in the 2 neighborhoods. Random and independent samples were selected from the 2 neighborhoods from last year‟s homes sales, 8 from each of the 2 neighborhoods. Identify the nonparametric method that would be used to analyze the data. a) the Wilcoxon signed ranks test, using the test statistic Z b) the Wilcoxon signed ranks test, using the test statistic W c) the Wilcoxon rank sum test, using the test statistic T1 d) the Wilcoxon rank sum test, using the test statistic Z ANSWER: c TYPE: MC DIFFICULTY: Moderate KEYWORDS: Wilcoxon rank sum test, test statistic 57. 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 58. 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
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59. True or False: When testing for differences between the means of 2 related populations, we can use either a onetailed or two-tailed test. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: t test for mean difference, rejection region 60. 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 for mean difference 61. True or False: The test for the equality of 2 population variances assumes that each of the 2 populations is normally distributed. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: F test for equality of variances, assumption 62. 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 63. 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 64. True or False: The F distribution is symmetric. ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: F distribution, properties
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65. True or False: The F distribution can only have positive values. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: F distribution, properties 66. True or False: All F tests are one-tailed tests. ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: F distribution, properties 67. True or False: The procedure for the Wilcoxon rank sum test requires that we rank each group separately rather than together. ANSWER: False TYPE: TF DIFFICULTY: Moderate KEYWORDS: Wilcoxon rank sum test 68. True of False: When performing a two-tailed test, the lower-tailed critical value of the F distribution with n1 1 degrees of freedom in the numerator and n2 1 degrees of freedom in the denominator is exactly equivalent to the reciprocal of the upper-tailed critical value of the F distribution with n2 1 degrees of freedom in the numerator and n1 1 degrees of freedom in the denominator. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: F distribution, properties 69. Given the upper-tailed critical value of an F test, with 3 degrees of freedom in the numerator and 8 degrees of freedom in the denominator being 4.07, the lower-tailed critical value of an F test with 8 degrees of freedom in the numerator and 3 degrees of freedom in the denominator for the same level of significance will be _________. ANSWER: 0.2457 TYPE: FI DIFFICULTY: Easy KEYWORDS: F distribution, critical value, properties 70. True or False: A researcher is curious about the effect of sleep on students‟ test performances. He chooses 60 students and gives each 2 tests: one given after 2 hours‟ sleep and one after 8 hours‟ sleep. The test the researcher should use would be a related samples test. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: t test for mean difference, Wilcoxon signed ranks test 71. When testing H0 : 1 – 2 0 versus H1 : 1 – 2 0 , the observed value of the Z-score was found to be – 2.13. The p-value for this test would be a) 0.0166.
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b) 0.0332. c) 0.9668. d) 0.9834. ANSWER: b TYPE: MC DIFFICULTY: Easy KEYWORDS: Z test for difference in means, p-value 72. 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 for difference in 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: a TYPE: MC DIFFICULTY: Easy KEYWORDS: Z test for difference in means, p-value 74. True or False: A statistics professor wanted to test whether the grades on a statistics test were the same for upper and lower classmen. The professor took a random sample of size 10 from each, conducted a test, and found out that the variances were equal. For this situation, the professor should use a t test with related samples. ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: pooled-variance t test
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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 independent samples. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: pooled-variance t test 76. 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 for mean difference, degrees of freedom 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 10 degrees of freedom. ANSWER: True TYPE: TF DIFFICULTY: Moderate KEYWORDS: t test for 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 a standard deviation of 3.0, while the second sample has a mean of 33.0 and a standard deviation of 4.0. 78. Referring to Table 10-6, the pooled (i.e., combined) variance is _______. ANSWER: 12.5 TYPE: FI DIFFICULTY: Easy KEYWORDS: pooled-variance t test 79. 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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80. 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 81. 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 82. 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 83. 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; 0.02 TYPE: FI DIFFICULTY: Moderate KEYWORDS: pooled-variance t test, p-value 84. 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 (be rejected/not be rejected) _______ .
ANSWER: be rejected TYPE: FI DIFFICULTY: Easy KEYWORDS: pooled-variance t test, decision 85. 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 _______ and _______ . ANSWER: 0.005; 0.01 TYPE: FI DIFFICULTY: Moderate KEYWORDS: pooled-variance t test, p-value
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TABLE 10-7 A perfume manufacturer is trying to choose between 2 magazine advertising layouts. An expensive layout would include a small package of the perfume. A cheaper layout would include a "scratch-and-sniff" sample of the product. The manufacturer would use the more expensive layout only if there is evidence that it would lead to a higher approval rating. The manufacturer presents the more expensive layout to 4 groups and determines the approval rating for each group. He presents the "scratch-and-sniff" layout to 5 groups and again determines the approval rating of the perfume for each group. The data are given below. Use this to test the appropriate hypotheses with the Wilcoxon rank sum test with a level of significance of 0.05. Package 52 68 43 48
Scratch 37 43 53 39 47
86. Referring to Table 10-7, the hypotheses that should be used are: a) H0 : 1 2 versus H1 : 1 2 b) H0 : 1 2 versus H1 : 1 2 c) H0 : M1 M2 versus H1 : M1 M 2 d) H0 : M1 M2 versus H1 : M1 M2 ANSWER: d TYPE: MC DIFFICULTY: Easy KEYWORDS: Wilcoxon rank sum test, form of hypothesis 87. Referring to Table 10-7, the rank given to the last observation in the "scratch-and-sniff" group is ________. ANSWER: 5 TYPE: FI DIFFICULTY: Easy KEYWORDS: Wilcoxon rank sum test, test statistic 88. Referring to Table 10-7, the rank given to the second observation in the "scratch-and-sniff" group is ________. ANSWER: 3.5 TYPE: FI DIFFICULTY: Easy KEYWORDS: Wilcoxon rank sum test, test statistic 89. Referring to Table 10-7, the calculated value of the test statistic is ________. ANSWER: 25.5 TYPE: FI DIFFICULTY: Easy KEYWORDS: Wilcoxon rank sum test, test statistic 90. Referring to Table 10-7, the critical value of the test is ________. ANSWER: 28 TYPE: FI DIFFICULTY: Easy
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KEYWORDS: Wilcoxon rank sum test, critical value 91. True or False: Referring to Table 10-7, the null hypothesis should be rejected. ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: Wilcoxon rank sum test, decision 92. Referring to Table 10-7, the perfume manufacturer will a) use the "scratch-and-sniff" layout because there is insufficient evidence to do otherwise. b) use the package layout because there is insufficient evidence to do otherwise. c) use the "scratch-and-sniff" layout because there is sufficient evidence to conclude that this is the best course of action. d) use the package layout because there is sufficient evidence to conclude that this is the best course of action. ANSWER: a TYPE: MC DIFFICULTY: Moderate KEYWORDS: Wilcoxon rank sum test, conclusion TABLE 10-8 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 mean weight during this period. Suppose the mean decrease in weight over all 16 subjects in the study is 3.0 pounds with the standard deviation of differences computed as 6.0 pounds. 93. Referring to Table 10-8, the t test should be _______-tailed. ANSWER: one TYPE: FI DIFFICULTY: Easy KEYWORDS: t test for mean difference, rejection region 94. Referring to Table 10-8, the computed t statistic is _______. ANSWER: 2.00 TYPE: FI DIFFICULTY: Easy KEYWORDS: t test for mean difference, test statistic 95. Referring to Table 10-8, there are _______ degrees of freedom for this test. ANSWER: 15 TYPE: FI DIFFICULTY: Easy KEYWORDS: t test for mean difference, degrees of freedom 96. Referring to Table 10-8, the critical value for a one-tailed test of the null hypothesis of no difference at the 0.05 level of significance is _______.
=
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ANSWER: 1.7531 TYPE: FI DIFFICULTY: Easy KEYWORDS: t test for mean difference, critical value 97. Referring to Table 10-8, 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 for mean difference, decision 98. Referring to Table 10-8, the p-value for a one-tailed test whose computed t statistic is 2.00 is between _______ and _______ . ANSWER: 0.025; 0.05 TYPE: FI DIFFICULTY: Moderate KEYWORDS: t test for mean difference, p-value 99. Referring to Table 10-8, 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 for mean difference, decision 100. Referring to Table 10-8, the p-value for a two-tailed test whose computed statistic is 2.00 is between ________ and _______ . ANSWER: 0.05; 0.10 TYPE: FI DIFFICULTY: Moderate KEYWORDS: t test for mean difference, p-value
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TABLE 10-9 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 prices on various materials. He wants to compare these prices with those of his primary supplier. The data collected is presented in the table below, with some summary statistics presented (all of these might not be necessary to answer the questions which follow). The buyer believes that the differences are normally distributed and will use this sample to perform an appropriate test at a level of significance of 0.01. Primary Secondary Material Supplier Supplier Difference 1 $55 $45 $10 2 $48 $47 $1 3 $31 $32 – $1 4 $83 $77 $6 5 $37 $37 $0 6 $55 $54 $1 Sum: $309 $292 $17 Sum of Squares: $17,573 $15,472 $139 101. Referring to Table 10-9, 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 for mean difference, form of hypothesis 102. Referring to Table 10-9, the test to perform is a a) pooled-variance t test for differences in 2 means. b) separate-variance t test for differences in 2 means. c) Wilcoxon rank sum test for differences in 2 medians. d) t test for mean difference. ANSWER: d TYPE: MC DIFFICULTY: Easy KEYWORDS: t test for mean difference 103. Referring to Table 10-9, the decision rule is to reject the null hypothesis if ________. ANSWER: t > 3.3649 TYPE: FI DIFFICULTY: Easy KEYWORDS: t test for mean difference, decision
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104. Referring to Table 10-9, the calculated value of the test statistic is ________. ANSWER: 1.628 TYPE: FI DIFFICULTY: Easy KEYWORDS: t test for mean difference, test statistic 105. Referring to Table 10-9, the p-value of the test is between ________ and ________. ANSWER: 0.05; 0.1 TYPE: FI DIFFICULTY: Moderate KEYWORDS: t test for mean difference, p-value 106. True or False: Referring to Table 10-9, the null hypothesis should be rejected. ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: t test for mean difference, decision 107. Referring to Table 10-9, 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 for mean difference, conclusion 108. Referring to Table 10-9, if the buyer had decided to perform a two-tailed test, the p-value would have been between ________ and ________. ANSWER: 0.01; 0.2 TYPE: FI DIFFICULTY: Moderate KEYWORDS: t test for mean difference, p-value
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TABLE 10-10 A perfume manufacturer is trying to choose between 2 magazine advertising layouts. An expensive layout would include a small package of the perfume. A cheaper layout would include a "scratch-and-sniff" sample of the product. The manufacturer would use the more expensive layout only if there is evidence that it would lead to a higher approval rating. The manufacturer presents both layouts to 5 groups and determines the approval rating from each group on both layouts. The data are given below. Use this to test whether the median difference in approval rating is different from zero, in favor of the more expensive layout, with a level of significance of 0.05. Package 52 68 43 48 56
Scratch 37 43 53 39 47
109. Referring to Table 10-10, what is the right test to use? a) Wilcoxon rank sum test for difference in median b) Wilcoxon rank sum test for median difference c) Wilcoxon signed ranks test for difference in median d) Wilcoxon signed ranks test for median difference ANSWER: d TYPE: MC DIFFICULTY: Moderate KEYWORDS: Wilcoxon signed ranks test 110. Referring to Table 10-10, the hypotheses that should be used are: a) H 0 : M D 0 versus H1 : M D 0 b) H 0 : M D 0 versus H1 : M D 0 c) H0 : M1 M2 versus H1 : M1 M 2 d) H0 : M1 M2 versus H1 : M1 M2 ANSWER: b TYPE: MC DIFFICULTY: Moderate KEYWORDS: Wilcoxon signed ranks test, form of hypothesis 111. Referring to Table 10-10, what are the lower and upper critical values of the test? ANSWER: 0, 15 TYPE: PR DIFFICULTY: Easy KEYWORDS: Wilcoxon signed ranks test, critical value
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112. Referring to Table 10-10, what is the rank of the absolute difference for the last pair of observations? ANSWER: 1.5 TYPE: PR DIFFICULTY: Easy KEYWORDS: Wilcoxon signed ranks test, test statistic 113. Referring to Table 10-10, which pair(s) of observations has a negative signed rank? ANSWER: Third, or (43, 53) TYPE: FI DIFFICULTY: Easy KEYWORDS: Wilcoxon signed ranks test, test statistic 114. Referring to Table 10-10, what is the value of the test statistic? ANSWER: 12 TYPE: FI DIFFICULTY: Easy KEYWORDS: Wilcoxon signed ranks test, test statistic 115. True or False: Referring to Table 10-10, the null hypothesis should be rejected. ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: Wilcoxon signed ranks test, decision 116. Referring to Table 10-10, the perfume manufacturer will a) use the "scratch-and-sniff" layout because there is insufficient evidence to do otherwise. b) use the package layout because there is insufficient evidence to do otherwise. c) use the "scratch-and-sniff" layout because there is sufficient evidence to conclude that this is the best course of action. d) use the package layout because there is sufficient evidence to conclude that this is the best course of action. ANSWER: a TYPE: MC DIFFICULTY: Moderate KEYWORDS: Wilcoxon signed ranks test, conclusion
Introduction and Data Collection
CHAPTER 11: ANALYSIS OF VARIANCE 117. a) b) c) d)
In a one-way ANOVA, if the computed F statistic exceeds the critical F value, we may reject H0 since there is evidence all the means differ. reject H0 since there is evidence of a treatment effect. not reject H0 since there is no evidence of a difference. not reject H0 because a mistake has been made.
ANSWER: b TYPE: MC DIFFICULTY: Easy KEYWORDS: one-way analysis of variance, decision 118. a) b) c) d)
Which of the following components in an ANOVA table are not additive? sum of squares degrees of freedom mean squares It is not possible to tell.
ANSWER: c TYPE: MC DIFFICULTY: Easy KEYWORDS: one-way analysis of variance, sum of squares, properties 119. a) b) c) d)
Why would you use the Tukey-Kramer procedure? to test for normality to test for homogeneity of variance to test independence of errors to test for differences in pairwise means
ANSWER: d TYPE: MC DIFFICULTY: Easy KEYWORDS: Tukey-Kramer procedure 120. a) b) c) d)
A completely randomized design has only one factor, with several treatment groups. can have more than one factor, each with several treatment groups. has one factor and one block. has one factor, one block, and multiple values.
ANSWER: a TYPE: MC DIFFICULTY: Moderate KEYWORDS: completely randomized design
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121. a) b) c) d)
The F test statistic in a one-way ANOVA is MSW/MSA. SSW/SSA. MSA/MSW. SSA/SSW.
ANSWER: c TYPE: MC DIFFICULTY: Easy KEYWORDS: one-way analysis of variance, F test for factor 122. a) b) c) d)
The degrees of freedom for the F test in a one-way ANOVA are (n – c) and (c – 1). (c – 1) and (n – c). (c – n) and (n – 1). (n – 1) and (c – n).
ANSWER: b TYPE: MC DIFFICULTY: Easy KEYWORDS: degrees of freedom, F test for factor 123. a) b) c) d)
In a one-way ANOVA, the null hypothesis is always there is no treatment effect. there is some treatment effect. all the population means are different. some of the population means are different.
ANSWER: a TYPE: MC DIFFICULTY: Easy KEYWORDS: one-way analysis of variance, form of hypothesis 124. a) b) c) d)
In a one-way ANOVA an interaction term is present. an interaction effect can be tested. there is no interaction term. 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
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125. a) b) c) d)
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Interaction in an experimental design can be tested in a completely randomized model. a randomized block model. a two-factor model. all ANOVA models.
ANSWER: c TYPE: MC DIFFICULTY: Easy KEYWORDS: two-factor analysis of variance, interaction, properties 126. a) b) c) d)
In a two-way ANOVA, the degrees of freedom for the interaction term are (r – 1)(c – 1). rc(n – 1). (r – 1). rcn + 1.
ANSWER: a TYPE: MC DIFFICULTY: Easy KEYWORDS: two-factor analysis of variance, interaction, degrees of freedom 127. a) b) c) d)
In a two-way ANOVA, the degrees of freedom for the "error" term are (r – 1)(c – 1). rc(n – 1). (r – 1). rcn + 1.
ANSWER: b TYPE: MC DIFFICULTY: Easy KEYWORDS: two-factor analysis of variance, degrees of freedom 128. Suppose there is interest in comparing the median response time for three independent groups learning a specific task. The appropriate nonparametric procedure is the a) Wilcoxon rank sums test. b) Wilcoxon signed ranks test. c) Kruskal-Wallis rank test for differences in medians. d) none of the above ANSWER: c TYPE: MC DIFFICULTY: Easy KEYWORDS: Kruskal-Wallis rank test
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129. a) b) c) d)
The Kruskal-Wallis rank test for differences in more than two medians is a nonparametric alternative to Fisher's ANOVA F test for completely randomized experiments. Student's t test for related samples. Student's t test for independent samples. Wilcoxon's rank sum test for differences in two medians.
ANSWER: a TYPE: MC DIFFICULTY: Easy KEYWORDS: Kruskal-Wallis rank test TABLE 11-1 Psychologists have found that people are generally reluctant to transmit bad news to their peers. This phenomenon has been termed the “MUM effect.” To investigate the cause of the MUM effect, 40 undergraduates at Duke University participated in an experiment. Each subject was asked to administer an IQ test to another student and then provide the test taker with his or her percentile score. Unknown to the subject, the test taker was a bogus student who was working with the researchers. The experimenters manipulated two factors: subject visibility and success of test taker, each at two levels. Subject visibility was either visible or not visible to the test taker. Success of the test taker was either visible or not visible to the test taker. Success of the test taker was either top 20% or bottom 20%. Ten subjects were randomly assigned to each of the 2 x 2 = 4 experimental conditions, then the time (in seconds) between the end of the test and the delivery of the percentile score from the subject to the test taker was measured. (This variable is called the latency to feedback.) The data were subjected to appropriate analyses with the following results: Source df Subject visibility 1 Test taker success 1 Interaction 1 Error 36 Total 39 130. a) b) c) d)
SS 1380.24 1325.16 3385.80 11,664.00 17,755.20
MS 1380.24 1325.16 3385.80 324.00
F 4.26 4.09 10.45
PR > F 0.043 0.050 0.002
Referring to Table 11-1, what type of experimental design was employed in this study? completely randomized design with 4 treatments randomized block design with four treatments and 10 blocks 2 x 2 factorial design with 10 replications none of the above
ANSWER: c TYPE: MC DIFFICULTY: Moderate KEYWORDS: two-factor analysis of variance, two-factor factorial design
Introduction and Data Collection
131. a) b) c) d)
295
Referring to Table 11-1, at the 0.01 level, what conclusions can you draw from the analyses? At the 0.01 level, subject visibility and test taker success are significant predictors of latency feedback. At the 0.01 level, the model is not useful for predicting latency to feedback. At the 0.01 level, there is evidence to indicate that subject visibility and test taker success interact. At the 0.01 level, there is no evidence of interaction between subject visibility and test taker success.
ANSWER: c TYPE: MC DIFFICULTY: Moderate KEYWORDS: two-factor analysis of variance, F test for interaction, decision, conclusion, interaction 132. Referring to Table 11-1, in the context of this study, interpret the statement: “Subject visibility and test taker success interact.” a) The difference between the mean feedback time for visible and nonvisible subjects depends on the success of the test taker. b) The difference between the mean feedback time for test takers scoring in the top 20% and bottom 20% depends on the visibility of the subject. c) The relationship between feedback time and subject visibility depends on the success of the test taker. d) All of the above are correct interpretations. ANSWER: d TYPE: MC DIFFICULTY: Difficult KEYWORDS: two-factor analysis of variance, interaction, conclusion 133. 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 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
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TABLE 11-2 An airline wants to select a computer software package for its reservation system. Four software packages (1, 2, 3, and 4) are commercially available. The airline will choose the package that bumps as few passengers, on average, as possible during a month. An experiment is set up in which each package is used to make reservations for 5 randomly selected weeks. (A total of 20 weeks was included in the experiment.) The number of passengers bumped each week is obtained, which gives rise to the following Excel output: ANOVA Source of Variation Between Groups Within Groups
SS 212.4 136.4
Total
348.8
134. a) b) c) d)
df
MS 3
F 8.304985
8.525
Referring to Table 11-2, the within groups degrees of freedom is 3. 4. 16. 19.
ANSWER: c TYPE: MC DIFFICULTY: Easy KEYWORDS: one-way analysis of variance, degrees of freedom 135. a) b) c) d)
Referring to Table 11-2, the total degrees of freedom is 3. 4. 16. 19.
ANSWER: d TYPE: MC DIFFICULTY: Easy KEYWORDS: one-way analysis of variance, degrees of freedom 136. a) b) c) d)
p-value 0.001474
Referring to Table 11-2, the among group mean squares is 8.525. 70.8. 212.4. 637.2.
ANSWER: b TYPE: MC DIFFICULTY: Easy KEYWORDS: one-way analysis of variance, mean squares
F crit 3.238867
Introduction and Data Collection
137.
297
Referring to Table 11-2, at a significance level of 1%, a) there is insufficient evidence to conclude that the average numbers of customers bumped by the 4 packages are not all the same. b) there is insufficient evidence to conclude that the average numbers of customers bumped by the 4 packages are all the same. c) there is sufficient evidence to conclude that the average numbers of customers bumped by the 4 packages are not all the same. d) there is sufficient evidence to conclude that the average numbers of customers bumped by the 4 packages are all the same.
ANSWER: c TYPE: MC DIFFICULTY: Easy KEYWORDS: one-way analysis of variance, F test for factor, decision, conclusion TABLE 11-3 A realtor wants to compare the average sales-to-appraisal ratios of residential properties sold in four neighborhoods (A, B, C, and D). Four properties are randomly selected from each neighborhood and the ratios are 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 138. a) b) c) d)
df
SS 2.97
MS 0.990
F 8.31
PR > F 0.0260
12 4.40
Referring to Table 11-3, the among group degrees of freedom is 3. 4. 12. 16.
ANSWER: a TYPE: MC DIFFICULTY: Easy KEYWORDS: one-way analysis of variance, degrees of freedom 139. a) b) c) d)
Referring to Table 11-3, the within group sum of squares is 0.119. 1.43. 2.97. 4.40.
ANSWER: b TYPE: MC DIFFICULTY: Easy KEYWORDS: one-way analysis of variance, sum of squares
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140. a) b) c) d)
Referring to Table 11-3, the within group mean squares is 0.119. 0.990. 1.109. 8.31.
ANSWER: a TYPE: MC DIFFICULTY: Easy KEYWORDS: one-way analysis of variance, mean squares 141. a) b) c) d)
Referring to Table 11-3, at the 0.05 level of significance, the mean ratios for the 4 neighborhoods are not all the same. at the 0.01 level of significance, the mean ratios for the 4 neighborhoods are not all the same. at the 0.10 level of significance, the mean ratios for the 4 neighborhoods are not significantly different. 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 142. A campus researcher wanted to investigate the factors that affect visitor travel time in a complex, multilevel building on campus. Specifically, he wanted to determine whether different building signs (building maps versus wall signage) affect the total amount of time visitors require to reach their destination, and whether that time depends on whether the starting location is inside or outside the building. Three subjects were assigned to each of the combinations of signs and starting locations, and travel times in seconds from beginning to destination were recorded. How should the data be analyzed? Starting Room Interior Exterior Wall Signs 141, 119, 238 224, 339, 139 Map 85, 94, 126 226, 129, 130 a) b) c) d)
completely randomized design randomized block design 2 x 2 factorial design Kruskal-Wallis rank test
ANSWER: c TYPE: MC DIFFICULTY: Moderate KEYWORDS: two-factor factorial design
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TABLE 11-4 A campus researcher wanted to investigate the factors that affect visitor travel time in a complex, multilevel building on campus. Specifically, he wanted to determine whether different building signs (building maps versus wall signage) affect the total amount of time visitors require to reach their destination, and whether that time depends on whether the starting location is inside or outside the building. Three subjects were assigned to each of the combinations of signs and starting locations, and travel times in seconds from beginning to destination were recorded. An Excel output of the appropriate analysis is given below: ANOVA Source of Variation Signs Starting Location Interaction Within
SS 14008.33 12288 48 35305.33
Total
61649.67
143. a) b) c) d)
df
MS 14008.33 48 4413.167
F
p-value 0.11267 2.784395 0.13374 0.919506
11
Referring to Table 11-4, the degrees of freedom for the different building signs (factor A) is 1. 2. 3. 8.
ANSWER: a TYPE: MC DIFFICULTY: Easy KEYWORDS: two-factor analysis of variance, degrees of freedom 144. a) b) c) d)
F crit 5.317645 5.317645 5.317645
Referring to Table 11-4, the within (error) degrees of freedom is 1. 4. 8. 11.
ANSWER: c TYPE: MC DIFFICULTY: Easy KEYWORDS: two-factor analysis of variance, degrees of freedom
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145. a) b) c) d)
Referring to Table 11-4, the mean squares for starting location (factor B) is 48. 4,413.17. 12,288. 14,008.3.
ANSWER: c TYPE: MC DIFFICULTY: Easy KEYWORDS: two-factor analysis of variance, mean squares 146. a) b) c) d)
Referring to Table 11-4, the F test statistic for testing the main effect of types of signs is 0.0109. 2.7844. 3.1742. 5.3176.
ANSWER: c TYPE: MC DIFFICULTY: Easy KEYWORDS: two-factor analysis of variance, F test for factor 147. Referring to Table 11-4, the F test statistic for testing the interaction effect between the types of signs and the starting location is a) 0.0109. b) 2.7844. c) 3.1742. d) 5.3176. ANSWER: a TYPE: MC DIFFICULTY: Easy KEYWORDS: two-factor analysis of variance, F test for interaction, interaction 148.
Referring to Table 11-4, at 1% level of significance, a) there is insufficient evidence to conclude that the difference between the average traveling times for the different starting locations depends on the types of signs. b) there is insufficient evidence to conclude that the difference between the average traveling times for the different types of signs depends on the starting locations. c) there is insufficient evidence to conclude that the relationship between traveling times and the types of signs depends on the starting locations. d) all of the above
ANSWER: d TYPE: MC DIFFICULTY: Easy KEYWORDS: two-factor analysis of variance, F test for factor, F test for interaction, decision, conclusion
Introduction and Data Collection
149.
301
Referring to Table 11-4, at 10% level of significance, a) there is sufficient evidence to conclude that the difference between the average traveling times for the different starting locations depends on the types of signs. b) there is insufficient evidence to conclude that the difference between the average traveling times for the different types of signs depends on the starting locations. c) there is sufficient evidence to conclude that the difference between the average traveling times for the different starting locations does not depend on the types of signs. d) none of the above
ANSWER: b TYPE: MC DIFFICULTY: Easy KEYWORDS: two-factor analysis of variance, F test for factor, decision, conclusion 150. The Journal of Business Venturing reported on the activities of entrepreneurs during the organization creation process. As part of a designed study, a total of 71 entrepreneurs were interviewed and divided into 3 groups: those that were successful in founding a new firm (n1 = 34), those still actively trying to establish a firm (n2 = 21), and those who tried to start a new firm but eventually gave up (n3 = 16). The total number of activities undertaken (e.g., developed a business plan, sought funding, looked for facilities) by each group over a specified time period during organization creation was measured. The objective is to compare the mean number of activities of the 3 groups of entrepreneurs. Because of concerns over necessary assumption of the parametric analysis, it was decided to use a nonparametric analysis. Identify the nonparametric method that would be used to analyze the data. a) Wilcoxon rank sums test b) Wilcoxon signed ranks test c) Kruskal-Wallis rank test for differences in medians d) one-way ANOVA F test ANSWER: c TYPE: MC DIFFICULTY: Easy KEYWORDS: Kruskal-Wallis rank test
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TABLE 11-5 A physician and president of a Tampa Health Maintenance Organization (HMO) are attempting to show the benefits of managed health care to an insurance company. The physician believes that certain types of doctors are more costeffective than others. One theory is that Primary Specialty is an important factor in measuring the cost-effectiveness of physicians. To investigate this, the president obtained independent random samples of 20 HMO physicians from each of 4 primary specialties - General Practice (GP), Internal Medicine (IM), Pediatrics (PED), and Family Physicians (FP) - and recorded the total charges per member per month for each. A second factor which the president believes influences total charges per member per month is whether the doctor is a foreign or USA medical school graduate. The president theorizes that foreign graduates will have higher mean charges than USA graduates. To investigate this, the president also collected data on 20 foreign medical school graduates in each of the 4 primary specialty types described above. So information on charges for 40 doctors (20 foreign and 20 USA medical school graduates) was obtained for each of the 4 specialties. The results for the ANOVA are summarized in the following table. Source Specialty Med school Interaction Error Total 151. a) b) c) d)
df 3 1 3 152 159
SS 22,855 105 890 18,950 42,800
MS 7,618 105 297
F 60.94 0.84 2.38
PR > F 0.0001 0.6744 0.1348
Referring to Table 11-5, what was the total number of doctors included in the study? 20 40 159 160
ANSWER: d TYPE: MC DIFFICULTY: Easy KEYWORDS: two-factor analysis of variance, properties 152. Referring to Table 11-5, what degrees of freedom should be used to determine the critical value of the F ratio against which to test for interaction between the two factors? a) numerator df = 1, denominator df = 159 b) numerator df = 3, denominator df = 159 c) numerator df = 1, denominator df = 152 d) numerator df = 3, denominator df = 152 ANSWER: d TYPE: MC DIFFICULTY: Easy KEYWORDS: two-factor analysis of variance, F test for interaction, degrees of freedom
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Referring to Table 11-5, interpret the test for interaction. a) There is insufficient evidence to say at the 0.10 level of significance that the difference between the mean charges for foreign and USA graduates depends on primary specialty. b) There is sufficient evidence to say at the 0.10 level of significance that the difference between the mean charges for foreign and USA graduates depends on primary specialty. c) There is sufficient evidence at the 0.10 level of significance of a difference between the mean charges for foreign and USA medical graduates. d) There is sufficient evidence to say at the 0.10 level of significance that mean charges depend on both primary specialty and medical school.
ANSWER: a TYPE: MC DIFFICULTY: Difficult KEYWORDS: two-factor analysis of variance, interaction, interpretation 154. Referring to Table 11-5, what degrees of freedom should be used to determine the critical value of the F ratio against which to test for differences in the mean charges for doctors among the four primary specialty areas? a) numerator df = 1, denominator df = 159 b) numerator df = 3, denominator df = 159 c) numerator df = 1, denominator df = 152 d) numerator df = 3, denominator df = 152 ANSWER: d TYPE: MC DIFFICULTY: Easy KEYWORDS: two-factor analysis of variance, F test for factor, degrees of freedom 155. Referring to Table 11-5, what degrees of freedom should be used to determine the critical value of the F ratio against which to test for differences between the mean charges of foreign and USA medical school graduates? a) numerator df = 1, denominator df = 159 b) numerator df = 3, denominator df = 159 c) numerator df = 1, denominator df = 152 d) numerator df = 3, denominator df = 152 ANSWER: c TYPE: MC DIFFICULTY: Easy KEYWORDS: two-factor analysis of variance, F test for factor, degrees of freedom
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156. Referring to Table 11-5, is there evidence of a difference between the mean charges of foreign and USA medical school graduates? a) Yes, the test for the main effect for primary specialty is significant at = 0.10. b) No, the test for the main effect for medical school is not significant at = 0.10. c) No, the test for the interaction is not significant at = 0.10. d) Maybe, but we need information on the -estimates to fully answer the question. ANSWER: b TYPE: MC DIFFICULTY: Easy KEYWORDS: two-factor analysis of variance, F test for factor, decision, conclusion 157. Referring to Table 11-5, what assumption(s) need(s) to be made in order to conduct the test for differences between the mean charges of foreign and USA medical school graduates? a) There is no significant interaction effect between the area of primary specialty and the medical school on the doctors‟ mean charges. b) The charges in each group of doctors sampled are drawn from normally distributed populations. c) The charges in each group of doctors sampled are drawn from populations with equal variances. d) All of the above are necessary assumptions. ANSWER: d TYPE: MC DIFFICULTY: Moderate KEYWORDS: two-factor analysis of variance, assumptions 158.
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 159. True or False: The F test, in a completely randomized model, is just an expansion of the t test for independent samples. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: completely randomized design, F test for factor 160.
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
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161. True or False: A completely randomized design with 4 groups would have 6 possible pairwise comparisons. ANSWER: True TYPE: TF DIFFICULTY: Moderate KEYWORDS: completely randomized design, properties 162. True or False: If you are comparing the average sales among 3 different brands, you are dealing with a three-way ANOVA design. ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: one-way analysis of variance, properties 163.
True or False: The MSE must always be positive.
ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: mean squares, properties 164. True or False: In a two-way ANOVA, it is easier to interpret main effects when the interaction component is not significant. ANSWER: True TYPE: TF DIFFICULTY: Moderate KEYWORDS: two-factor analysis of variance, interpretation 165. True or False: In a one-factor ANOVA analysis, the among sum of squares and within sum of squares must add up to the total sum of squares. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: one-way analysis of variance, sum of squares, properties 166. True or False: In a two-factor ANOVA analysis, the sum of squares due to both factors, the interaction sum of squares, and the within sum of squares must add up to the total sum of squares. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: two-factor analysis of variance, sum of squares, properties
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TABLE 11-6 As part of an evaluation program, a sporting goods retailer wanted to compare the downhill coasting speeds of 4 brands of bicycles. She took 3 of each brand and determined their maximum downhill speeds. The results are presented in miles per hour in the table below. Trial Barth Tornado Reiser Shaw 1 43 37 41 43 2 46 38 45 45 3 43 39 42 46 167. Referring to Table 11-6, the sporting goods retailer decided to perform an ANOVA F test. The amount of total variation or SST is __________. ANSWER: 102.67 TYPE: FI DIFFICULTY: Moderate KEYWORDS: one-way analysis of variance, sum of squares, interpretation 168.
Referring to Table 11-6, the among group variation or SSA is __________.
ANSWER: 81.33 TYPE: FI DIFFICULTY: Moderate KEYWORDS: one-way analysis of variance, sum of squares 169.
Referring to Table 11-6, the within group variation or SSW is __________.
ANSWER: 21.33 TYPE: FI DIFFICULTY: Moderate KEYWORDS: one-way analysis of variance, sum of squares 170.
Referring to Table 11-6, the value of MSA is __________, while MSW is __________.
ANSWER: 27.11; 2.67 TYPE: FI DIFFICULTY: Moderate KEYWORDS: one-way analysis of variance, mean squares 171. Referring to Table 11-6, the null hypothesis that the average downhill coasting speeds of the 4 brands of bicycles are equal, will be rejected at a level of significance of 0.05 if the value of the test statistic is greater than __________. ANSWER: 4.07 TYPE: FI DIFFICULTY: Easy KEYWORDS: one-way analysis of variance, F test for factor, decision
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172. Referring to Table 11-6, in testing the null hypothesis that the average downhill coasting speeds of the 4 brands of bicycles are equal, the value of the test statistic is __________. ANSWER: 10.17 TYPE: FI DIFFICULTY: Moderate KEYWORDS: one-way analysis of variance, F test for factor, test statistic 173.
Referring to Table 11-6, 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*
174. True or False: Referring to Table 11-6, the null hypothesis should be rejected at a 5% level of significance. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: one-way analysis of variance, decision 175. True or False: Referring to Table 11-6, the decision made implies that all 4 means are significantly different. ANSWER: False TYPE: TF DIFFICULTY: Moderate KEYWORDS: one-way analysis of variance, conclusion, interpretation 176. True or False: Referring to Table 11-6, the test is valid only if the population of speeds has the same variance for the 4 brands. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: one-way analysis of variance, assumption
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177. True or False: Referring to Table 11-6, the test is less sensitive to the assumption that the population of speeds has the same variance for the 4 brands if the sample sizes of the 4 brands are equal. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: one-way analysis of variance, assumption 178. True or False: Referring to Table 11-6, the test is valid only if the population of speeds is normally distributed. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: one-way analysis of variance, assumption 179. True or False: Referring to Table 11-6, the test is robust to the violation of the assumption that the population of speeds is normally distributed. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: one-way analysis of variance, assumption 180. Referring to Table 11-6, the sporting goods retailer decided to compare the 4 treatment means by using the Tukey-Kramer procedure with an overall level of significance of 0.05. There are ________ pairwise comparisons that can be made. ANSWER: 6 TYPE: FI DIFFICULTY: Moderate KEYWORDS: one-way analysis of variance, Tukey-Kramer procedure 181. Referring to Table 11-6, using an overall level of significance of 0.05, the critical value of the Studentized range Q used in calculating the critical range for the Tukey-Kramer procedure is ________. ANSWER: 4.53 TYPE: FI DIFFICULTY: Moderate KEYWORDS: one-way analysis of variance, Tukey-Kramer procedure, critical value 182. Referring to Table 11-6, using an overall level of significance of 0.05, the critical range for the TukeyKramer procedure is ________. ANSWER: 4.27 TYPE: FI DIFFICULTY: Easy KEYWORDS: one-way analysis of variance, Tukey-Kramer procedure, critical value 183. True or False: Referring to Table 11-6, based on the Tukey-Kramer procedure with an overall level of significance of 0.05, the retailer would decide that there is a significant difference between all pairs of mean speeds. ANSWER:
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False TYPE: TF DIFFICULTY: Moderate KEYWORDS: one-way analysis of variance, Tukey-Kramer procedure, decision, conclusion 184. True or False: Referring to Table 11-6, based on the Tukey-Kramer procedure with an overall level of significance of 0.05, the retailer would decide that there is no significant difference between any pair of mean speeds. ANSWER: False TYPE: TF DIFFICULTY: Moderate KEYWORDS: one-way analysis of variance, Tukey-Kramer procedure, decision, conclusion 185. True or False: Referring to Table 11-6, based on the Tukey-Kramer procedure with an overall level of significance of 0.05, the retailer would decide that the mean speed for the Tornado brand is significantly different from each of the mean speeds for other brands. ANSWER: True TYPE: TF DIFFICULTY: Moderate KEYWORDS: one-way analysis of variance, Tukey-Kramer procedure, decision, conclusion 186. True or False: Referring to Table 11-6, based on the Tukey-Kramer procedure with an overall level of significance of 0.05, the retailer would decide that the 3 means other than the mean for Tornado are not significantly different from each other. ANSWER: True TYPE: TF DIFFICULTY: Moderate KEYWORDS: one-way analysis of variance, Tukey-Kramer procedure, decision, conclusion 187. Referring to Table 11-6, the sporting goods retailer decided to perform a Kruskal-Wallis test. The null hypothesis of the test is ________. ANSWER: H0: M1 = M2 = M3 = M4 TYPE: FI DIFFICULTY: Easy KEYWORDS: one-way analysis of variance, Kruskal-Wallis rank test, form of hypothesis 188.
Referring to Table 11-6, the alternative hypothesis of the Kruskal-Wallis test is that ________.
ANSWER: not all the medians are equal TYPE: FI DIFFICULTY: Easy KEYWORDS: one-way analysis of variance, Kruskal-Wallis rank test, form of hypothesis 189. Referring to Table 11-6, the decision rule for a level of significance of 0.05 using the Kruskal-Wallis test is to reject the null hypothesis if the test statistic H is ________. ANSWER: greater than 7.815 TYPE: FI DIFFICULTY: Easy KEYWORDS: one-way analysis of variance, Kruskal-Wallis rank test, critical value, decision
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190. Referring to Table 11-6, the calculation of the Kruskal-Wallis test statistic H involves ranking the observations. Construct a table containing these ranks. ANSWER: Barth Tornado Reiser Shaw 7.0 1.0 4.0 7.0 11.5 2.0 9.5 9.5 7.0 3.0 5.0 11.5 TYPE: PR DIFFICULTY: Moderate KEYWORDS: one-way analysis of variance, Kruskal-Wallis test, test statistic 191. Referring to Table 11-6, the calculation of the Kruskal-Wallis test statistic H involves obtaining the total of the ranks for each sample. These totals are ________, ________, ________, and ________. ANSWER: 25.5, 6.0, 18.5, 28.0 TYPE: FI DIFFICULTY: Moderate KEYWORDS: one-way analysis of variance, Kruskal-Wallis rank test, test statistic 192.
Referring to Table 11-6, the calculated value of the Kruskal-Wallis test statistic H is ________.
ANSWER: 7.47 TYPE: FI DIFFICULTY: Moderate KEYWORDS: one-way analysis of variance, Kruskal-Wallis rank test, test statistic
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TABLE 11-7 An agronomist wants to compare the crop yield of 3 varieties of chickpea seeds. She plants 15 fields, 5 with each variety. She then measures the crop yield in bushels per acre. Treating this as a completely randomized design, the results are presented in the table that follows. Trial Smith Walsh Trevor 1 11.1 19.0 14.6 2 13.5 18.0 15.7 3 15.3 19.8 16.8 4 14.6 19.6 16.7 5 9.8 16.6 15.2 193. Referring to Table 11-7, the agronomist decided to perform an ANOVA F test. The amount of total variation or SST is __________. ANSWER: 114.82 TYPE: FI DIFFICULTY: Moderate KEYWORDS: one-way analysis of variance, sum of squares 194.
Referring to Table 11-7, the among group variation or SSA is __________.
ANSWER: 82.39 TYPE: FI DIFFICULTY: Moderate KEYWORDS: one-way analysis of variance, sum of squares 195.
Referring to Table 11-7, the within group variation or SSW is __________.
ANSWER: 32.43 TYPE: FI DIFFICULTY: Moderate KEYWORDS: one-way analysis of variance, sum of squares 196.
Referring to Table 11-7, the value of MSA is __________, while MSW is __________.
ANSWER: 41.19; 2.70 TYPE: FI DIFFICULTY: Moderate KEYWORDS: one-way analysis of variance, mean squares 197. Referring to Table 11-7, the null hypothesis will be rejected at a level of significance of 0.01 if the value of the test statistic is greater than __________. ANSWER: 6.93 TYPE: FI DIFFICULTY: Easy KEYWORDS: one-way analysis of variance, F test for factor, critical value
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198.
Referring to Table 11-7, the value of the test statistic is __________.
ANSWER: 15.24 TYPE: FI DIFFICULTY: Moderate KEYWORDS: one-way analysis of variance, F test for factor, test statistic 199.
Referring to Table 11-7, construct the ANOVA table from the sample data.
ANSWER: Analysis of Variance Source df SS MS F p Seed Varieties 2 82.39 41.19 15.24 0.000508* Error 12 32.43 2.70 Total 14 114.82 * or p < 0.005, tabular value TYPE: PR DIFFICULTY: Moderate KEYWORDS: one-way analysis of variance, F test for factor, properties 200.
Referring to Table 11-7, state the null hypothesis that can be tested.
ANSWER: H0: Smith Walsh Trevor TYPE: PR DIFFICULTY: Easy KEYWORDS: one-way analysis of variance, F test for factor, form of hypothesis 201. True or False: Referring to Table 11-7, the null hypothesis should be rejected at 0.005 level of significance. ANSWER: True TYPE: TF DIFFICULTY: Moderate KEYWORDS: one-way analysis of variance, F test for factor, decision 202. True or False: Referring to Table 11-7, the decision made at 0.005 level of significance implies that all 3 means are significantly different. ANSWER: False TYPE: TF DIFFICULTY: Moderate KEYWORDS: one-way analysis of variance, F test for factor, conclusion 203. True or False: Referring to Table 11-7, the test is valid only if the population of crop yields has the same variance for the 3 varieties. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: one-way analysis of variance, F test for factor, assumption
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204. True or False: Referring to Table 11-7, the test is valid only if the population of crop yields is normally distributed for the 3 varieties. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: one-way analysis of variance, F test for factor, assumption 205. Referring to Table 11-7, the agronomist decided to compare the 3 treatment means by using the TukeyKramer 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 206. Referring to Table 11-7, using an overall level of significance of 0.01, the critical value of the Studentized range Q used in calculating the critical range for the Tukey-Kramer procedure is ________. ANSWER: 5.04 TYPE: FI DIFFICULTY: Easy KEYWORDS: one-way analysis of variance, Tukey-Kramer procedure, critical value 207. Referring to Table 11-7, using an overall level of significance of 0.01, the critical range for the TukeyKramer procedure is ________. ANSWER: 3.70 TYPE: FI DIFFICULTY: Moderate KEYWORDS: one-way analysis of variance, Tukey-Kramer procedure, critical value 208. True or False: Referring to Table 11-7, based on the Tukey-Kramer procedure with an overall level of significance of 0.01, the agronomist would decide that there is a significant difference between the crop yield of Smith and Walsh seeds. ANSWER: True TYPE: TF DIFFICULTY: Moderate KEYWORDS: one-way analysis of variance, Tukey-Kramer procedure, decision, conclusion 209. True or False: Referring to Table 11-7, based on the Tukey-Kramer procedure with an overall level of significance of 0.01, the agronomist would decide that there is a significant difference between the crop yield of Smith and Trevor seeds. ANSWER: False TYPE: TF DIFFICULTY: Moderate KEYWORDS: one-way analysis of variance, Tukey-Kramer procedure, decision, conclusion 210. True or False: Referring to Table 11-7, based on the Tukey-Kramer procedure with an overall level of significance of 0.01, the agronomist would decide that there is a significant difference between the crop yield of Walsh and Trevor seeds.
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ANSWER: False TYPE: TF DIFFICULTY: Moderate KEYWORDS: one-way analysis of variance, Tukey-Kramer procedure, decision, conclusion TABLE 11-8 A hotel chain has identically sized resorts in 5 locations. The data that follow resulted from analyzing the hotel occupancies on randomly selected days in the 5 locations. ROW Caymen Pennkamp California Mayaguez Maui 1 28 40 21 37 22 2 33 35 21 47 19 3 41 33 27 45 25 Analysis of Variance Source df SS Location 4 963.6 Error 10 210.0 Total
MS
F 11.47
p 0.001
211. Referring to Table 11-8, the value of the element in the ANOVA table that always provides an estimate of the population variance is ________. ANSWER: 21.0 TYPE: FI DIFFICULTY: Difficult KEYWORDS: one-way analysis of variance, properties 212. Referring to Table 11-8, the test ratio involves the ratio of 2 elements of the ANOVA table. Of these elements, the value of the one that provides an estimate of the population variance only when the null hypothesis is true is ________. ANSWER: 240.9 TYPE: FI DIFFICULTY: Difficult KEYWORDS: one-way analysis of variance, properties 213. True or False: Referring to Table 11-8, if a level of significance of 0.05 is chosen, the null hypothesis should be rejected. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: one-way analysis of variance, F test for factor, decision 214. True or False: Referring to Table 11-8, if a level of significance of 0.05 is chosen, the decision made indicates that all 5 locations have different mean occupancy rates. ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: one-way analysis of variance, F test for factor, conclusion
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215. True or False: Referring to Table 11-8, if a level of significance of 0.05 is chosen, the decision made indicates that at least 2 of the 5 locations have different mean occupancy rates. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: one-way analysis of variance, F test for factor, conclusion 216. Referring to Table 11-8, the among group variation or SSA is _________. ANSWER: 963.6 TYPE: FI DIFFICULTY: Easy KEYWORDS: one-way analysis of variance, sum of squares 217. Referring to Table 11-8, the within group variation or SSW is _________. ANSWER: 210.0 TYPE: FI DIFFICULTY: Easy KEYWORDS: one-way analysis of variance, sum of squares 218. Referring to Table 11-8, the total variation or SST is ________. ANSWER: 1,173.6 TYPE: FI DIFFICULTY: Easy KEYWORDS: one-way analysis of variance, sum of squares 219. Referring to Table 11-8, the value of MSA is ______ while MSW is _______. ANSWER: 240.9; 21.0 TYPE: FI DIFFICULTY: Easy KEYWORDS: one-way analysis of variance, mean squares 220. Referring to Table 11-8, the numerator and denominator degrees of freedom of the test ratio are ________ and ________, respectively. ANSWER: 4; 10 TYPE: FI DIFFICULTY: Easy KEYWORDS: one-way analysis of variance, degrees of freedom 221. True or False: Referring to Table 11-8, the total mean squares is 261.90. ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: one-way analysis of variance, sum of squares TABLE 11-9 The marketing manager of a company producing a new cereal aimed for children wants to examine the effect of the color and shape of the box's logo on the approval rating of the cereal. He combined 4 colors and 3 shapes to produce
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a total of 12 designs. Each logo was presented to 2 different groups (a total of 24 groups) and the approval rating for each was recorded and is shown below. The manager analyzed these data using the = 0.05 level of significance for all inferences.
SHAPES Circle Square Diamond
Red 54 44 34 36 46 48
Analysis of Variance Source df SS Colors 3 2711.17 Shapes 2 579.00 Interaction 6 150.33 Error 12 150.00 Total 23 3590.50
COLORS Green Blue 67 36 61 44 56 36 58 30 60 34 60 38
MS 903.72 289.50 25.06 12.50
F 72.30 23.16 2.00
Yellow 45 41 21 25 31 33
p 0.000 0.000 0.144
222. Referring to Table 11-9, the mean square for the factor color is ________. ANSWER: 903.72 TYPE: FI DIFFICULTY: Easy KEYWORDS: two-factor analysis of variance, mean squares 223. Referring to Table 11-9, the mean square for the factor shape is ________. ANSWER: 289.50 TYPE: FI DIFFICULTY: Easy KEYWORDS: two-factor analysis of variance, mean squares
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224. Referring to Table 11-9, the mean square for the interaction of color and shape is ________. ANSWER: 25.06 TYPE: FI DIFFICULTY: Easy KEYWORDS: two-factor analysis of variance, mean squares 225. Referring to Table 11-9, the mean square for error is ________. ANSWER: 12.50 TYPE: FI DIFFICULTY: Easy KEYWORDS: two-factor analysis of variance, mean squares 226. Referring to Table 11-9, the critical value of the test for significant differences between colors is ________. ANSWER: 3.49 TYPE: FI DIFFICULTY: Easy KEYWORDS: two-factor analysis of variance, F test for factor, critical value 227. Referring to Table 11-9, the value of the statistic used to test for significant differences between colors is ________. ANSWER: 72.30 TYPE: FI DIFFICULTY: Easy KEYWORDS: two-factor analysis of variance, F test for factor, test statistic 228. True or False: Referring to Table 11-9, based on the results of the hypothesis test, it appears that there is a significant effect on the approval rating associated with the color of the logo. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: two-factor analysis of variance, F test for factor, decision, conclusion 229. Referring to Table 11-9, the critical value in the test for significant differences between shapes is ________. ANSWER: 3.89 TYPE: FI DIFFICULTY: Easy KEYWORDS: two-factor analysis of variance, F test for factor, critical value
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230. Referring to Table 11-9, the value of the statistic used to test for significant differences between shapes is ________. ANSWER: 23.16 TYPE: FI DIFFICULTY: Easy KEYWORDS: two-factor analysis of variance, F test for factor, test statistic 231. True or False: Referring to Table 11-9, based on the results of the hypothesis test, it appears that there is a significant effect associated with the shape of the logo. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: two-factor analysis of variance, F test for factor, decision, conclusion 232. Referring to Table 11-9, the critical value in the test for a significant interaction is ________. ANSWER: 3.00 TYPE: FI DIFFICULTY: Easy KEYWORDS: two-factor analysis of variance, F test for interaction, critical value 233. Referring to Table 11-9, the value of the statistic used to test for an interaction is ________. ANSWER: 2.00 TYPE: FI DIFFICULTY: Easy KEYWORDS: two-factor analysis of variance, F test for interaction, test statistic 234. True or False: Referring to Table 11-9, based on the results of the hypothesis test, it appears that there is a significant interaction. ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: two-factor analysis of variance, F test for interaction, decision, conclusion
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TABLE 11-10 An agronomist wants to compare the crop yield of 3 varieties of chickpea seeds. She plants all 3 varieties of the seeds on each of 5 different patches of fields. She then measures the crop yield in bushels per acre. Treating this as a randomized block design, the results are presented in the table that follows: Fields Smith Walsh Trevor 1 11.1 19.0 14.6 2 13.5 18.0 15.7 3 15.3 19.8 16.8 4 14.6 19.6 16.7 5 9.8 16.6 15.2 235. Referring to Table 11-10, the agronomist decided to perform a randomized block F test for the difference in the means. The amount of total variation or SST is __________. ANSWER: 114.82 TYPE: FI DIFFICULTY: Moderate KEYWORDS: randomized block design, sum of squares 236. Referring to Table 11-10, the among group variation or SSA is __________. ANSWER: 82.39 TYPE: FI DIFFICULTY: Moderate KEYWORDS: randomized block design, sum of squares 237. Referring to Table 11-10, the among block variation or SSBL is __________. ANSWER: 24.46 TYPE: FI DIFFICULTY: Moderate KEYWORDS: randomized block design, sum of squares 238. Referring to Table 11-10, the value of MSA is __________, while MSBL is __________. ANSWER: 41.19; 6.11 TYPE: FI DIFFICULTY: Moderate KEYWORDS: randomized block design, mean squares
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239. Referring to Table 11-10, the null hypothesis for the randomized block F test for the difference in the means is a) H 0 : Field 1 Field 2 Field 3 Field 4 Field 5 b) H 0 : Smith Walsh Trevor c)
H 0 : M Field 1 M Field 2 M Field 3 M Field 4 M Field 5
d) H 0 : M Smith M Walsh M Trevor ANSWER: b TYPE: MC DIFFICULTY: Easy KEYWORDS: randomized block design, F test for factor, form of hypothesis 240. Referring to Table 11-10, what are the degrees of freedom of the randomized block F test for the difference in the means at a level of significance of 0.01? ANSWER: 2 numerator and 8 denominator degrees of freedom TYPE: PR DIFFICULTY: Easy KEYWORDS: randomized block design, F test for factor, degrees of freedom 241. Referring to Table 11-10, what is the critical value of the randomized block F test for the difference in the means at a level of significance of 0.01? ANSWER: 8.65 TYPE: PR DIFFICULTY: Easy KEYWORDS: randomized block design, F test for factor, critical value 242. Referring to Table 11-10, what is the value of the test statistic for the randomized block F test for the difference in the means? ANSWER: 41.32 TYPE: PR DIFFICULTY: Moderate KEYWORDS: randomized block design, F test for factor, test statistic 243. Referring to Table 11-10, what is the p-value of the test statistic for the randomized block F test for the difference in the means? ANSWER: 6.07E-05 TYPE: PR DIFFICULTY: Moderate KEYWORDS: randomized block design, F test for factor, p-value
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244. True or False: Referring to Table 11-10, the null hypothesis for the randomized block F test for the difference in the means should be rejected at a 0.01 level of significance. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: randomized block design, F test for factor, decision 245. True or False: Referring to Table 11-10, the decision made at a 0.01 level of significance on the randomized block F test for the difference in means implies that all 3 means are significantly different. ANSWER: False TYPE: TF DIFFICULTY: Moderate KEYWORDS: randomized block design, F test for factor, conclusion 246. True or False: Referring to Table 11-10, the randomized block F test is valid only if the population of crop yields has the same variance for the 3 varieties. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: randomized block design, F test for factor, assumption 247. True or False: Referring to Table 11-10, the randomized block F test is valid only if the population of crop yields is normally distributed for the 3 varieties. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: randomized block design, F test for factor, assumption 248. True or False: Referring to Table 11-10, the randomized block F test is valid only if there is no interaction between the variety of seeds and the patches of fields. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: randomized block design, F test for factor, assumption 249. Referring to Table 11-10, the agronomist decided to compare the 3 treatment means by using the Tukey multiple comparison procedure with an overall level of significance of 0.01. How many pairwise comparisons can be made? ANSWER: 3 TYPE: PR DIFFICULTY: Moderate KEYWORDS: randomized block design, Tukey procedure, properties 250. Referring to Table 11-10, using an overall level of significance of 0.01, what is the critical value of the Studentized range Q used in calculating the critical range for the Tukey multiple comparison procedure? ANSWER: 5.63
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TYPE: PR DIFFICULTY: Easy KEYWORDS: randomized block design, Tukey procedure, critical value 251. Referring to Table 11-10, using an overall level of significance of 0.01, what is the critical range for the Tukey multiple comparison procedure? ANSWER: 2.51 TYPE: PR DIFFICULTY: Moderate KEYWORDS: randomized block design, Tukey procedure, critical value 252. True or False: Referring to Table 11-10, based on the Tukey multiple comparison procedure with an overall level of significance of 0.01, the agronomist would decide that there is a significant difference between the crop yield of Smith and Walsh seeds. ANSWER: True TYPE: TF DIFFICULTY: Moderate KEYWORDS: randomized block design, Tukey procedure, decision, conclusion 253. True or False: Referring to Table 11-10, based on the Tukey-Kramer procedure with an overall level of significance of 0.01, the agronomist would decide that there is a significant difference between the crop yield of Smith and Trevor seeds. ANSWER: True TYPE: TF DIFFICULTY: Moderate KEYWORDS: randomized block design, Tukey procedure, decision, conclusion 254. True or False: Referring to Table 11-10, based on the Tukey multiple comparison procedure with an overall level of significance of 0.01, the agronomist would decide that there is a significant difference between the crop yield of Walsh and Trevor seeds. ANSWER: True TYPE: TF DIFFICULTY: Moderate KEYWORDS: randomized block design, Tukey procedure, decision, conclusion
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255. Referring to Table 11-10, what is the null hypothesis for testing the block effects? a) H 0 : Field 1 Field 2 Field 3 Field 4 Field 5 b) H 0 : Smith Walsh Trevor c)
H 0 : M Field 1 M Field 2 M Field 3 M Field 4 M Field 5
d) H 0 : M Smith M Walsh M Trevor ANSWER: a TYPE: MC DIFFICULTY: Easy KEYWORDS: randomized block design, F test for block effects, form of hypothesis 256. Referring to Table 11-10, what are the degrees of freedom of the F test statistic for testing the block effects? ANSWER: 4 numerator and 8 denominator degrees of freedom TYPE: PR DIFFICULTY: Easy KEYWORDS: randomized block design, F test for block effects, degrees of freedom 257. Referring to Table 11-10, what is the value of the F test statistic for testing the block effects? ANSWER: 6.13 TYPE: PR DIFFICULTY: Moderate KEYWORDS: randomized block design, F test for block effects, test statistic 258. Referring to Table 11-10, what is the critical value for testing the block effects at a 0.01 level of significance? ANSWER: 7.01 TYPE: PR DIFFICULTY: Moderate KEYWORDS: randomized block design, F test for block effects, critical value 259. Referring to Table 11-10, what is the p-value of the F test statistic for testing the block effects? ANSWER: 0.015 or between 0.01 and 0.025 TYPE: PR DIFFICULTY: Moderate KEYWORDS: randomized block design, F test for block effects, p-value 260. True or False: Referring to Table 11-10, the null hypothesis for the F test for the block effects should be rejected at a 0.01 level of significance. ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: randomized block design, F test for block effects, decision 261. True or False: Referring to Table 11-10, the decision made at a 0.01 level of significance on the F test for the block effects implies that the blocking has been advantageous in reducing the experiment error. ANSWER: False TYPE: TF DIFFICULTY: Moderate
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KEYWORDS: randomized block design, F test for block effects, conclusion 262. Referring to Table 11-10, what is the estimated relative efficiency? ANSWER: 2.47 TYPE: PR DIFFICULTY: Moderate KEYWORDS: randomized block design, F test for block effects, relative efficiency 263. True or False: Referring to Table 11-10, the relative efficiency means that 2.47 times as many observations in each variety group would be needed in a one-way ANOVA design, as compared to the randomized block design in order to obtain the same precision for comparison of the variety means. ANSWER: True TYPE: TF DIFFICULTY: Moderate KEYWORDS: randomized block design, F test for block effects, relative efficiency, interpretation
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CHAPTER 12: TESTS FOR TWO OR MORE SAMPLES WITH CATEGORICAL DATA 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 successes is higher in group 1 than in group 2, the appropriate test to use is a) the Z test. 2 b) the test. c) both of the above d) none of the above ANSWER: a TYPE: MC DIFFICULTY: Easy KEYWORDS: chi-square test for difference in proportions
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4. If we wish to determine whether there is evidence that the proportion of successes is the same in group 1 as in group 2, the appropriate test to use is a) the Z test. 2 b) the test. c) both of the above d) none of the above ANSWER: c TYPE: MC DIFFICULTY: Easy KEYWORDS: chi-square test for difference in proportions 5. 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, properties TABLE 12-1 A study published in the American Journal of Public Health was conducted to determine whether the use of seat belts in motor vehicles depends on ethnic status in San Diego County. A sample of 792 children treated for injuries sustained from motor vehicle accidents was obtained, and each child was classified according to (1) ethnic status (Hispanic or non-Hispanic) and (2) seat belt usage (worn or not worn) during the accident. The number of children in each category is given in the table below.
Seat belts worn Seat belts not worn
Hispanic 31 283
Non-Hispanic 148 330
6. Referring to Table 12-1, which test would be used to properly analyze the data in this experiment? a) test for independence. b) test for difference between proportions. c) ANOVA F test for interaction in a 2 x 2 factorial design. d) test for goodness of fit. ANSWER: a TYPE: MC DIFFICULTY: Moderate KEYWORDS: chi-square test of independence
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7. Referring to Table 12-1, the calculated test statistic is a) -0.9991. b) -0.1368. c) 48.1849. d) 72.8063. ANSWER: c TYPE: MC DIFFICULTY: Easy KEYWORDS: chi-square test of independence, test statistic 8. Referring to Table 12-1, at 5% level of significance, the critical value of the test statistic is a) 3.8415. b) 5.9914. c) 9.4877. d) 13.2767. ANSWER: a TYPE: MC DIFFICULTY: Easy KEYWORDS: chi-square test of independence, critical value 9. Referring to Table 12-1, at 5% level of significance, there is sufficient evidence to conclude that the 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
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TABLE 12-2 Many companies use well-known celebrities as spokespersons in their TV advertisements. A study was conducted to determine whether brand awareness of female TV viewers and the gender of the spokesperson are independent. Each in a sample of 300 female TV viewers was asked to identify a product advertised by a celebrity spokesperson. The gender of the spokesperson and whether or not the viewer could identify the product was recorded. The numbers in each category are given below.
Identified product Could not identify
Male Celebrity 41 109
Female Celebrity 61 89
10. Referring to Table 12-2, which test would be used to properly analyze the data in this experiment? a) test for independence. b) test for difference between proportions. c) ANOVA F test for main treatment effect. d) test for goodness of fit. ANSWER: a TYPE: MC DIFFICULTY: Moderate KEYWORDS: chi-square test of independence 11. Referring to Table 12-2, the calculated test statistic is a) -0.1006. b) 0.00. c) 5.9418. d) 6.1194. ANSWER: c TYPE: MC DIFFICULTY: Easy KEYWORDS: chi-square test of independence, test statistic 12. Referring to Table 12-2, at 5% level of significance, the critical value of the test statistic is a) 3.8415. b) 5.9914. c) 9.4877. d) 13.2767. ANSWER: a TYPE: MC DIFFICULTY: Easy KEYWORDS: chi-square test of independence, critical value
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13. Referring to Table 12-2, the degrees of freedom of the test statistic are a) 1. b) 2. c) 4. d) 299. ANSWER: a TYPE: MC DIFFICULTY: Easy KEYWORDS: chi-square test of independence, degrees of freedom 14. Referring to Table 12-2, at 5% level of significance, the conclusion is that a) brand awareness of female TV viewers and the gender of the spokesperson are independent. b) brand awareness of female TV viewers and the gender of the spokesperson are not independent. c) brand awareness of female TV viewers and the gender of the spokesperson are related. d) both (b) and (c) ANSWER: d TYPE: MC DIFFICULTY: Moderate KEYWORDS: chi-square test of independence, decision, conclusion TABLE 12-3 A computer used by a 24-hour banking service is supposed to randomly assign each transaction to one of 5 memory locations. A check at the end of a day‟s transactions gave the counts shown in the table to each of the 5 memory locations, along with the number of reported errors. Memory Location:
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. 15. Referring to Table 12-3, which test would be used to properly analyze the data in this experiment? a) test for independence b) test for difference between proportions c) ANOVA F test for main treatment effect d) test for goodness of fit ANSWER: b TYPE: MC DIFFICULTY: Moderate KEYWORDS: chi-square test for difference in proportions
16. Referring to Table 12-3, the degrees of freedom of the test statistic is a) 4. b) 8. c) 10. d) 448.
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ANSWER: a TYPE: MC DIFFICULTY: Easy KEYWORDS: chi-square test for difference in proportions, degrees of freedom 17. Referring to Table 12-3, the critical value of the test statistic at 1% level of significance is a) 7.7794. b) 13.2767. c) 20.0902. d) 23.2093. ANSWER: b TYPE: MC DIFFICULTY: Easy KEYWORDS: chi-square test for difference in proportions, critical value 18. Referring to Table 12-3, the calculated value of the test statistic is a) -0.1777. b) -0.0185. c) 1.4999. d) 1.5190. ANSWER: c TYPE: MC DIFFICULTY: Moderate KEYWORDS: chi-square test for difference in proportions, test statistic 19. Referring to Table 12-3, at 1% level of significance a) there is sufficient evidence to conclude that the proportions of errors in transactions assigned to each of the 5 memory locations are all different. b) there is insufficient evidence to conclude that the proportions of errors in transactions assigned to each of the 5 memory locations are all different. c) there is sufficient evidence to conclude that the proportion of errors in transactions assigned to each of the 5 memory locations are not all the same. d) there is insufficient evidence to conclude that the proportion of errors in transactions assigned to each of the 5 memory locations are not all the same. ANSWER: d TYPE: MC DIFFICULTY: Moderate KEYWORDS: chi-square test for difference in proportions, decision, conclusion
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20. 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) test for independence b) test for goodness of fit c) ANOVA F test for main treatment effect d) test for the difference between proportions ANSWER: d TYPE: MC DIFFICULTY: Moderate KEYWORDS: Z test for difference in two proportions 21. The Wall Street Journal recently ran an article indicating differences in the 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.” 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: pW – pM 0 versus H1: pW – pM < 0 b) H0: pW – pM 0 versus H1: pW – pM > 0 c) H0: pW – pM = 0 versus H1: pW – pM 0 d) H0: W – M 0 versus H1: W – M > 0 ANSWER: b TYPE: MC DIFFICULTY: Easy KEYWORDS: Z test for difference in two proportions, form of hypothesis
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22. 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 for difference in two proportions, critical value 23. 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 for difference in two proportions, test statistic
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24. 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 being 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 being much more prevalent than do men. c) There is no evidence of a significant difference between the men and women in their perception. d) More information is needed to draw any conclusions from the data set. ANSWER: a TYPE: MC DIFFICULTY: Moderate KEYWORDS: Z test for difference in two proportions, decision, conclusion 25. 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. Which of the following are the appropriate null and alternative hypotheses to test the group‟s claim? a) H0: pW – pM 0 versus H1: pW – pM < 0 b) H0: pW – pM 0 versus H1: pW – pM > 0 c) H0: pW – pM = 0 versus H1: pW – pM 0 d) H0: W – M 0 versus H1: W – M > 0 ANSWER: c TYPE: MC DIFFICULTY: Moderate KEYWORDS: Z test for difference in two proportions, form of hypothesis 26. A powerful women‟s group has claimed that men and women differ in attitudes about sexual discrimination. A group of 50 men (group 1) and 40 women (group 2) were asked if they thought sexual discrimination is a problem in the United States. Of those sampled, 11 of the men and 19 of the women did believe that sexual discrimination is a problem. Find the value of the test statistic. a) Z = – 2.55 b) Z = – 0.85 c) Z = – 1.05 d) Z = – 1.20 ANSWER: a TYPE: MC DIFFICULTY: Moderate KEYWORDS: Z test for difference in two proportions, test statistic
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Introduction and Data Collection 27. 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 for difference in two proportions, decision, conclusion TABLE 12-4 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, pSM = 0.422018 pSM – pSF = 0.172018
Females: n = 52, pSF = 0.25 Z = 2.11825
28. Referring to Table 12-4, 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: pM – pF 0 d) H1: pM – pF = 0 ANSWER: c TYPE: MC DIFFICULTY: Easy KEYWORDS: Z test for difference in two proportions, form of hypothesis 29. Referring to Table 12-4, 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 for difference in two proportions, p-value
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30. Referring to Table 12-4, suppose Pepsi wanted to test to determine if the males preferred Pepsi less than the females. Using the test statistic given, compute the appropriate p-value for the test. a) 0.0170 b) 0.0340 c) 0.9660 d) 0.9830 ANSWER: d TYPE: MC DIFFICULTY: Moderate KEYWORDS: Z test for difference in two proportions, p-value 31. Referring to Table 12-4, 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 for difference in two proportions, decision, conclusion
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TABLE 12-5 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
163 227 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
154 262 0.718061674 0.58778626 0.130275414 0.648261759 3.00875353 -1.959961082 1.959961082 0.002623357
32. Referring to Table 12-5, 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? e. 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. f. 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. g. 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. h. 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 for difference in two proportions, decision, conclusion
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33. Referring to Table 12-5, if you want to test the claim that "Resort 1 (Group 1) has a higher proportion of satisfied guests than 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 for difference in two proportions, p-value 34. Referring to Table 12-5, if you want to test the claim that "Resort 1 (Group 1) has a lower proportion of satisfied guests than Resort 2 (Group 2)," you will use a) a t test for the difference in two proportions. b) a Z test for the difference in two proportions. c) a 2 test for the difference in two proportions. d) a 2 test for independence. ANSWER: b TYPE: MC DIFFICULTY: Moderate KEYWORDS: Z test for difference in two proportions TABLE 12-6 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.
Defect Rate:
High Average Low
< 1 Year 6 9 7
Years Since Training Period 1 – 4 Years 5 – 9 Years 9 9 19 23 8 10
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35. Referring to Table 12-6, which test would be used to properly analyze the data in this experiment to determine whether there is a relationship between defect rate and years of experience? 2 a) test for independence in a two-way contingency table 2 b) test for equal proportions in a one-way table c) ANOVA F test for main treatment effect d) Z test for the difference in two proportions ANSWER: a TYPE: MC DIFFICULTY: Moderate KEYWORDS: chi-square test of independence 36. Referring to Table 12-6, 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 37. Referring to Table 12-6, 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 38. Referring to Table 12-6, 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 39. Referring to Table 12-6, of the cell for 1 to 4 years of training time and a high defect rate, what is the 2 contribution to the overall statistic for the independence test? a) 0.36 b) 0.1296
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c) 0.015 d) 0.0144
ANSWER: c TYPE: MC DIFFICULTY: Moderate KEYWORDS: chi-square test of independence, contingency table, properties 40. Referring to Table 12-6, a test was conducted to determine if a relationship exists between defect rate and years of experience. Which of the following p-values would indicate that defect rate and years of experience are dependent? Assume you are testing at = 0.05. a) 0.045 b) 0.055 c) 0.074 d) 0.080 ANSWER: a TYPE: MC DIFFICULTY: Easy KEYWORDS: chi-square test of independence, p-value, decision TABLE 12-7 A corporation randomly selects 150 salespeople and finds that 66% who have never taken a self-improvement course would like to take such a course. The firm did a similar study 10 years ago in which 60% of a random sample of 160 salespeople wanted to take a self-improvement course. The groups are assumed to be independent random samples. Let p1 and p2 represent the true proportion of workers who would like to attend a self-improvement course in the recent study and the past study, respectively. 41. Referring to Table 12-7, if the firm wanted to test whether this proportion has changed from the previous study, which represents the relevant hypotheses? a) H0: p1 – p2 = 0 versus H1: p1 – p2 0 b) H0: p1 – p2 0 versus H1: p1 – p2 = 0 c) H0: p1 – p2 0 versus H1: p1 – p2 > 0 d) H0: p1 – p2 0 versus H1: p1 – p2 < 0 ANSWER: a TYPE: MC DIFFICULTY: Easy KEYWORDS: chi-square test for difference in proportions, Z test for difference in two proportions
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42. Referring to Table 12-7, 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: p1 – p2 = 0 versus H1: p1 – p2 0 b) H0: p1 – p2 0 versus H1: p1 – p2 = 0 c) H0: p1 – p2 0 versus H1: p1 – p2 > 0 d) H0: p1 – p2 0 versus H1: p1 – p2 < 0 ANSWER: c TYPE: MC DIFFICULTY: Easy KEYWORDS: Z test for difference in two proportions, form of hypothesis 43. Referring to Table 12-7, what is the unbiased point estimate for the difference between the two population proportions? a) 0.06 b) 0.10 c) 0.15 d) 0.22 ANSWER: a TYPE: MC DIFFICULTY: Easy KEYWORDS: point estimate 44. Referring to Table 12-7, 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 for difference in two proportions, critical value 45. Referring to Table 12-7, 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 for difference in two proportions, critical value 46. Referring to Table 12-7, 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
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ANSWER: b TYPE: MC DIFFICULTY: Easy KEYWORDS: chi-square test for difference in proportions, critical value 47. Referring to Table 12-7, what is the estimated standard error of the difference between the two sample proportions? a) 0.629 b) 0.500 c) 0.055 d) 0 ANSWER: c TYPE: MC DIFFICULTY: Moderate KEYWORDS: Z test for difference in two proportions, standard error 48. Referring to Table 12-7, 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 for difference in two proportions, chi-square test for difference in proportions, test statistic
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49. Referring to Table 12-7, the company tests at the 0.05 level to determine 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 selfimprovement 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 selfimprovement 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 for difference in two proportions, chi-square test for difference in proportions, decision, conclusion 50. True or False: In testing the difference between two proportions, 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 51. 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 52. True or False: A test for the difference between two proportions can be performed using the chi-square distribution. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: chi-square test for difference in proportions 53. 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 54. 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:
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True TYPE: TF DIFFICULTY: Difficult KEYWORDS: chi-square test of independence, properties, assumption 55. 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 56. If we wish to determine whether there is evidence that the proportion of successes is higher in Group 1 than in Group 2, and the test statistic for Z = +2.07, the p-value is equal to ______. ANSWER: 0.0192 TYPE: FI DIFFICULTY: Moderate KEYWORDS: Z test for difference in two proportions, p-value TABLE 12-8 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. 57. Referring to Table 12-8, the hypotheses the dean should use are: a) H0: p1 – p2 = 0 versus H1: p1 – p2 0. b) H0: p1 – p2 0 versus H1: p1 – p2 = 0. c) H0: p1 – p2 0 versus H1: p1 – p2 > 0. d) H0: p1 – p2 0 versus H1: p1 – p2 < 0. ANSWER: a TYPE: MC DIFFICULTY: Easy KEYWORDS: chi-square test for difference in proportions, Z test for difference in two proportions, form of hypothesis
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58. Referring to Table 12-8, the null hypothesis will be rejected if the test statistic is ________. ANSWER: 2 Z > 1.96 or < -1.96 or > 3.841 TYPE: FI DIFFICULTY: Easy KEYWORDS: chi-square test for difference in proportions, Z test for difference in two proportions, critical value 59. Referring to Table 12-8, the value of the test statistic is ________. ANSWER: 2 Z = 1.866 or = 3.4806 TYPE: FI DIFFICULTY: Moderate KEYWORDS: chi-square test for difference in proportions, Z test for difference in two proportions, test statistic 60. Referring to Table 12-8, the p-value of the test is ________. ANSWER: 0.0621 TYPE: FI DIFFICULTY: Moderate KEYWORDS: chi-square test for difference in proportions, Z test for difference in two proportions, p-value 61. True or False: Referring to Table 12-8, the null hypothesis should be rejected. ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: chi-square test for difference in proportions, Z test for difference in two proportions, decision 62. True or False: Referring to Table 12-8, 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, Z test for difference in two proportions, decision 63. True or False: Referring to Table 12-8, 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, Z test for difference in two proportions, decision TABLE 12-9 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. 64. Referring to Table 12-9, the hypotheses that should be tested are: a) H0: p1 – p2 = 0 versus H1: p1 – p2 0.
Introduction and Data Collection b) H0: p1 – p2 0 versus H1: p1 – p2 = 0. c) H0: p1 – p2 0 versus H1: p1 – p2 > 0. d) H0: p1 – p2 0 versus H1: p1 – p2 < 0. ANSWER: c TYPE: MC DIFFICULTY: Easy KEYWORDS: Z test for difference in two proportions, form of hypothesis 65. Referring to Table 12-9, the null hypothesis will be rejected if the test statistic is ________. ANSWER: Z > 2.33 TYPE: FI DIFFICULTY: Easy KEYWORDS: Z test for difference in two proportions, critical value 66. Referring to Table 12-9, the value of the test statistic is ________. ANSWER: 2.49 TYPE: FI DIFFICULTY: Moderate KEYWORDS: Z test for difference in two proportions, test statistic 67. Referring to Table 12-9, the p-value of the test is ________. ANSWER: 0.0064 TYPE: FI DIFFICULTY: Moderate KEYWORDS: Z test for difference in two proportions, p-value 68. True or False: Referring to Table 12-9, the null hypothesis should be rejected. ANSWER: True TYPE: TF DIFFICULTY: Moderate KEYWORDS: Z test for difference in two proportions, decision
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69. True or False: Referring to Table 12-9, the same decision would be made with this test if the level of significance had been 0.05 rather than 0.01. ANSWER: True TYPE: TF DIFFICULTY: Moderate KEYWORDS: Z test for difference in two proportions, decision 70. True or False: Referring to Table 12-9, 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 for difference in two proportions, decision TABLE 12-10 The director of transportation of a large company is interested in the usage of her van pool. She considers her routes to be divided into local and non-local. She is particularly interested in learning if there is a difference in the proportion of males and females who use the local routes. She takes a sample of a day's riders and finds the following:
Local Non-Local Total
Male 27 33 60
Female 44 25 69
Total 71 58 129
She will use this information to perform a chi-square hypothesis test using a level of significance of 0.05. 71. Referring to Table 12-10, 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 72. Referring to Table 12-10, 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
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73. Referring to Table 12-10, 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 74. Referring to Table 12-10, 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 75. Referring to Table 12-10, the critical value of the test is _________. ANSWER: 3.841 TYPE: FI DIFFICULTY: Easy KEYWORDS: chi-square test for difference in proportions, critical value 76. Referring to Table 12-10, the value of the test statistic is _________. ANSWER: 4.568 TYPE: FI DIFFICULTY: Moderate KEYWORDS: chi-square test for difference in proportions, test statistic 77. True or False: Referring to Table 12-10, the null hypothesis will be rejected. ANSWER: True TYPE: TF DIFFICULTY: Moderate KEYWORDS: chi-square test for difference in proportions, decision 78. True or False: Referring to Table 12-10, the decision made suggests that there is a difference between the proportion of males and females who ride local versus non-local routes. ANSWER: True TYPE: TF DIFFICULTY: Median KEYWORDS: chi-square test for difference in proportions, conclusion
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TABLE 12-11 Four surgical procedures are currently 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. 79. Referring to Table 12-11, the test will involve _________ degrees of freedom. ANSWER: 3 TYPE: FI DIFFICULTY: Easy KEYWORDS: chi-square test for difference in proportions, degrees of freedom 80. Referring to Table 12-11, 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 81. Referring to Table 12-11, 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 82. Referring to Table 12-11, the expected cell frequency for the Procedure D/Return cell is __________. ANSWER: 5.83 TYPE: FI DIFFICULTY: Moderate KEYWORDS: chi-square test for difference in proportions, contingency table, properties
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83. Referring to Table 12-11, the critical value of the test is ________. ANSWER: 7.815 TYPE: FI DIFFICULTY: Easy KEYWORDS: chi-square test for difference in proportions, critical value 84. Referring to Table 12-11, the value of the test statistic is _________. ANSWER: 7.867 TYPE: FI DIFFICULTY: Moderate KEYWORDS: chi-square test for difference in proportions, test statistic 85. True or False: Referring to Table 12-11, the null hypothesis will be rejected. ANSWER: True TYPE: TF DIFFICULTY: Moderate KEYWORDS: chi-square test for difference in proportions, decision 86. True or False: Referring to Table 12-11, 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 87. True or False: Referring to Table 12-11, 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 12-12 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. 88. Referring to Table 12-12, the test will involve _________ degrees of freedom. ANSWER:
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4 TYPE: FI DIFFICULTY: Easy KEYWORDS: chi-square test of independence, degrees of freedom 89. Referring to Table 12-12, the critical value of the test is _________. ANSWER: 13.277 TYPE: FI DIFFICULTY: Easy KEYWORDS: chi-square test of independence, critical value 90. Referring to Table 12-12, 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 91. Referring to Table 12-12, the value of the test statistic is _________. ANSWER: 12.624 TYPE: FI DIFFICULTY: Moderate KEYWORDS: chi-square test of independence, test statistic 92. True or False: Referring to Table 12-12, the null hypothesis will be rejected. ANSWER: False TYPE: TF DIFFICULTY: Moderate KEYWORDS: chi-square test of independence, decision 93. True or False: Referring to Table 12-12, the p-value of this test is greater than 0.01. ANSWER: True TYPE: TF DIFFICULTY: Moderate KEYWORDS: chi-square test of independence, p-value 94. True or False: Referring to Table 12-12, 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 95. True or False: Referring to Table 12-12, 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
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96. True or False: Referring to Table 12-12, 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 97. True or False: Referring to Table 12-12, 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 98. True or False: Referring to Table 12-12, 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 99. True or False: The chi-square test of independence requires that the number of expected frequency in each cell to be at least 5. ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: chi-square test of independence, assumption 100. True or False: The chi-square test of independence requires that the number of expected frequency in each cell to be at least 1. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: chi-square test of independence, assumption
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TABLE 12-13 Parents complain that children read too few story books 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 story books:
Average time spent watching TV Less than 2 hours More than 2 hours
Average time spent reading story books Less than 1 hour Between 1 and 2 hours
More than 2 hours
90 655
130 8
85 32
101. Referring to Table 12-13, 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 102. Referring to Table 12-13, how many children in the survey spent less than 2 hours watching TV and no more than 2 hours reading story books, on average? a) 8 b) 130 c) 175 d) 687 ANSWER: c TYPE: MC DIFFICULTY: Easy KEYWORDS: contingency table, properties 103. Referring to Table 12-13, 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 104. Referring to Table 12-13, 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?
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a) b) c) d)
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42.09 155.25 262.92 987.75
ANSWER: a TYPE: MC DIFFICULTY: Moderate KEYWORDS: chi-square test of independence, contingency table, properties 105. Referring to Table 12-13, 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 106. Referring to Table 12-13, suppose we want to constrain the probability of committing a Type I error to 5% when testing whether there is any relationship between average time spent watching TV and average time spent reading story books. The critical value will be a) 5.991. b) 7.378. c) 12.592. d) 14.449. ANSWER: a TYPE: MC DIFFICULTY: Easy KEYWORDS: chi-square test of independence, critical value
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107. Referring to Table 12-13, we want to test whether there is any relationship between average time spent watching TV and average time spent reading story books. 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 a connection between time spent reading story books and time spent watching TV. b) there is no connection between time spent reading story books and time spent watching TV. c) more time spent reading story books leads to less time spent watching TV. d) more time spent watching TV leads to less time spent reading story books. ANSWER: a TYPE: MC DIFFICULTY: Moderate KEYWORDS: chi-square test of independence, decision, conclusion TABLE 12-14 Recent studies have found that American children are more obese than in the past. The amount of time children spend watching television has received much of the blame. A survey of 100 ten-year-olds revealed the following with regards to weights and average number of hours a day spent watching television. We are interested in testing whether the average number of hours spent watching TV and weights are independent at 1% level of significance. Weights More than 10 lbs. overweight Within 10 lbs. of normal weight More than 10 lbs. underweight Total
0-3 1 20 10 31
TV Hours 3-6 9 15 5 29
6+ 20 15 5 40
Total 30 50 20 100
108. Referring to Table 12-14, 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
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109. Referring to Table 12-14, 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 110. Referring to Table 12-14, 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 111. Referring to Table 12-14, how many children in the survey spend no more than 6 hours watching TV and are more than 10 lbs. underweight? a) 5 b) 10 c) 15 d) 20 ANSWER: c TYPE: MC DIFFICULTY: Easy KEYWORDS: chi-square test of independence, contingency table, properties 112. Referring to Table 12-14, 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
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113. Referring to Table 12-14, 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 114. Referring to Table 12-14, 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 the average number of hours spent watching TV and weights are independent. b) We will reject the null and conclude that the average number of hours spent watching TV and weights are independent. c) We will accept the null and conclude that the average number of hours spent watching TV and weights are not independent. d) We will reject the null and conclude that the 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 115. Referring to Table 12-14, 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 116. Referring to Table 12-14, 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 117. True or False: Referring to Table 12-14, the test is always a one-tailed test. ANSWER: True TYPE: TF DIFFICULTY: Easy
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KEYWORDS: chi-square test of independence, properties Table 12-15 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 paid by check over three consecutive years at a 0.05 level of significance, the results of a survey of 500 shoppers during three consecutive years are obtained and presented below.
Check Written Yes No
Year 1 225 275
Year Year 2 175 325
Year 3 125 375
118. Referring to Table 12-15, what is the expected number of shoppers who paid by check in year 1 if there was no difference in the proportion of shoppers who paid by check over the three years? ANSWER: 175 TYPE: PR DIFFICULTY: Easy KEYWORDS: chi-square test for difference in proportions, properties 119. Referring to Table 12-15, what is the expected number of shoppers who did not pay by check in year 3 if there was no difference in the proportion of shoppers who paid by check over the three years? ANSWER: 325 TYPE: PR DIFFICULTY: Easy KEYWORDS: chi-square test for difference in proportions, properties 120. Referring to Table 12-15, what is the form of the null hypothesis? a) H 0 : p1 p2 p3 b) H 0 : p1 p2 p3 c)
H 0 : p1 p2 p3
d) H 0 : p1 p2 p3 ANSWER: a TYPE: MC DIFFICULTY: Easy KEYWORDS: chi-square test for difference in proportions, forms of hypothesis
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121. Referring to Table 12-15, what is the form of the alternative hypothesis? a) H1 : p1 p2 p3 b) H1 : p1 p2 p3 c)
H1 : p1 p2 p3
d) H1 : not all p j are the same ANSWER: d TYPE: MC DIFFICULTY: Easy KEYWORDS: chi-square test for difference in proportions, forms of hypothesis 122. True or False: Referring to Table 12-15, the assumptions needed to perform the test are satisfied. ANSWER: True TYPE: TF DIFFICULTY: Moderate KEYWORDS: chi-square test for difference in proportions, assumption 123. Referring to Table 12-15, 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 124. Referring to Table 12-15, 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 125. Referring to Table 12-15, what is the critical value? ANSWER: 5.99 TYPE: PR DIFFICULTY: Easy KEYWORDS: chi-square test for difference in proportions, critical value 126. Referring to Table 12-15, 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
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127. True or False: Referring to Table 12-15, the null hypothesis cannot be rejected. ANSWER: False TYPE: TF DIFFICULTY: Moderate KEYWORDS: chi-square test for difference in proportions, decision 128. Referring to Table 12-15, 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 129. Referring to Table 12-15, what is the value of the critical range for the Marascuilo procedure to test for the difference in proportions between year 1 and year 2 using a 0.05 level of significance? ANSWER: 0.0754 TYPE: PR DIFFICULTY: Difficult KEYWORDS: chi-square test for difference in proportions, Marascuilo procedure, critical value 130. Referring to Table 12-15, what is the value of the critical range for the Marascuilo procedure to test for the difference in proportions between year 1 and year 3 using a 0.05 level of significance? ANSWER: 0.0722 TYPE: PR DIFFICULTY: Difficult KEYWORDS: chi-square test for difference in proportions, Marascuilo procedure, critical range 131. Referring to Table 12-15, what is the value of the critical range for the Marascuilo procedure to test for the difference in proportions between year 2 and year 3 using a 0.05 level of significance? ANSWER: 0.0705 TYPE: PR DIFFICULTY: Difficult KEYWORDS: chi-square test for difference in proportions, Marascuilo procedure, critical value
132. True or False: Referring to Table 12-15, there is sufficient evidence to conclude that the proportions between year 1 and year 2 are different at a 0.05 level of significance. ANSWER: True TYPE: TF DIFFICULTY: Moderate KEYWORDS: chi-square test for difference in proportions, Marascuilo procedure, decision, conclusion
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133. True or False: Referring to Table 12-15, there is not sufficient evidence to conclude that the proportions between year 1 and year 3 are different at a 0.05 level of significance. ANSWER: False TYPE: TF DIFFICULTY: Moderate KEYWORDS: chi-square test for difference in proportions, Marascuilo procedure, decision, conclusion 134. True or False: Referring to Table 12-15, there is sufficient evidence to conclude that the proportions between year 2 and year 3 are different at a 0.05 level of significance. ANSWER: True TYPE: TF DIFFICULTY: Moderate KEYWORDS: chi-square test for difference in proportions, Marascuilo procedure, decision conclusion
Introduction and Data Collection
CHAPTER 13: SIMPLE LINEAR REGRESSION 1. The Y intercept (b0) represents the a) predicted value of Y when X = 0. b) change in estimated average Y per unit change in X. c) predicted value of Y. d) variation around the sample regression line. ANSWER: a TYPE: MC DIFFICULTY: Easy KEYWORDS: intercept, interpretation 2. The Y intercept (b0) represents the a) estimated average Y when X = 0. b) change in estimated average Y per unit change in X. c) predicted value of Y. d) variation around the sample regression line. ANSWER: a TYPE: MC DIFFICULTY: Easy KEYWORDS: intercept, interpretation 3. The slope (b1) represents a) predicted value of Y when X = 0. b) the estimated average change in Y per unit change in X. c) the predicted value of Y. d) variation around the line of regression. ANSWER: b TYPE: MC DIFFICULTY: Easy KEYWORDS: slope, interpretation 4. The least squares method minimizes which of the following? a) SSR b) SSE c) SST d) all of the above ANSWER: b TYPE: MC DIFFICULTY: Difficult KEYWORDS: least squares, properties
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TABLE 13-1 A large national bank charges local companies for using their services. A bank official reported the results of a regression analysis designed to predict the bank‟s charges (Y) -- measured in dollars per 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.
Yˆ 2,700 20 X , SYX 65, two-tailed p value 0.034 (for testing ) 5. Referring to Table 13-1, interpret the estimate of 0 , the Y intercept of the line. a) 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 13-1, interpret the standard error of the estimate in the model. a) About 95% of the observed service charges fall within $65 of the least squares line. b) About 95% of the observed service charges equal their corresponding predicted values. c) About 95% of the observed service charges fall within $130 of the least squares line. d) For every $1 million increase in sales revenue, we expect a service charge to increase $65. ANSWER: c TYPE: MC DIFFICULTY: Difficult KEYWORDS: standard error of the estimate, interpretation 7. Referring to Table 13-1, interpret the p-value for testing whether 1 exceeds 0. a) There is sufficient evidence (at the = 0.05) to conclude that sales revenue (X) is a useful linear predictor of service charge (Y). b) There is insufficient evidence (at the = 0.10) to conclude that sales revenue (X) is a useful linear predictor of service charge (Y). c) Sales revenue (X) is a poor predictor of service charge (Y). d) For every $1 million increase in sales revenue, we expect a service charge to increase $0.034. ANSWER: a TYPE: MC DIFFICULTY: Moderate KEYWORDS: p-value, interpretation 8. Referring to Table 13-1, a 95% confidence interval for 1 is (15, 30). Interpret the interval. a) We are 95% confident that the mean service charge will fall between $15 and $30 per month. 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 the average service charge (Y) will increase between $15 and $30 for every $1 million increase in sales revenue (X). d) At the = 0.05 level, there is no evidence of a linear relationship between service charge (Y) and sales revenue (X).
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ANSWER: c TYPE: MC DIFFICULTY: Moderate KEYWORDS: confidence interval, interpretation TABLE 13-2 A candy bar manufacturer is interested in trying to estimate how sales are influenced by the price of their product. To do this, the company randomly chooses 6 small cities and offers the candy bar at different prices. Using candy bar sales as the dependent variable, the company will conduct a simple linear regression on the data below: City Price ($) Sales River Falls 1.30 100 Hudson 1.60 90 Ellsworth 1.80 90 Prescott 2.00 40 Rock Elm 2.40 38 Stillwater 2.90 32 9. Referring to Table 13-2, what is the estimated slope parameter for the candy bar price and sales data? a) 161.386 b) 0.784 c) – 3.810 d) – 48.193 ANSWER: d TYPE: MC DIFFICULTY: Moderate KEYWORDS: slope, estimation
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10. Referring to Table 13-2, what is the estimated average change in the sales of the candy bar if price goes up by $1.00? a) 161.386 b) 0.784 c) – 3.810 d) – 48.193 ANSWER: d TYPE: MC DIFFICULTY: Moderate KEYWORDS: slope, interpretation 11. Referring to Table 13-2, what is the coefficient of correlation for these data? a) – 0.8854 b) – 0.7839 c) 0.7839 d) 0.8854 ANSWER: a TYPE: MC DIFFICULTY: Moderate KEYWORDS: coefficient of correlation, estimation 12. Referring to Table 13-2, what is the percentage of the total variation in candy bar sales explained by the regression model? a) 100% b) 88.54% c) 78.39% d) 48.19% ANSWER: c TYPE: MC DIFFICULTY: Moderate KEYWORDS: coefficient of determination, interpretation 13. Referring to Table 13-2, what percentage of the total variation in candy bar sales is explained by prices? a) 100% b) 88.54% c) 78.39% d) 48.19% ANSWER: c TYPE: MC DIFFICULTY: Moderate KEYWORDS: coefficient of determination, interpretation
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14. Referring to Table 13-2, what is the standard error of the estimate, SYX, for the data? a) 0.784 b) 0.885 c) 12.650 d) 16.299 ANSWER: d TYPE: MC DIFFICULTY: Moderate KEYWORDS: standard error of the estimate, estimation 15. Referring to Table 13-2, what is the standard error of the regression slope estimate, sb1 ? a) 0.784 b) 0.885 c) 12.650 d) 16.299 ANSWER: c TYPE: MC DIFFICULTY: Moderate KEYWORDS: standard error, slope 16. Referring to Table 13-2, 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 13-2, to test that the regression coefficient, 1 , is not equal to 0, what would be the limits of the rejection region for b1 ? Use = 0.05. a) – 48.193 45.245 b) – 48.193 35.123 c) 0 45.245 d) 0 35.123 ANSWER: d TYPE: MC DIFFICULTY: Difficult KEYWORDS: t test on slope, critical value
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18. Referring to Table 13-2, to test whether a change in price will have any impact on average sales, what would be the limits of the rejection region for b1 ? Use = 0.05. a) – 48.193 45.245 b) – 48.193 35.117 c) – 48.193 2.776 d) 0 35.117 ANSWER: d TYPE: MC DIFFICULTY: Difficult EXPLANATION: This is a test for 1 that is not equal to 0. KEYWORDS: t test on slope, critical value 19. Referring to Table 13-2, if the price of the candy bar is set at $2, the estimated average sales will be a) 30. b) 65. c) 90. d) 100. ANSWER: b TYPE: MC DIFFICULTY: Moderate KEYWORDS: estimation of mean values 20. Referring to Table 13-2, if the price of the candy bar is set at $2, the predicted sales will be a) 30. b) 65. 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
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22. The residual represents the discrepancy between the observed dependent variable and its _______ value. ANSWER: predicted or estimated average TYPE: FI DIFFICULTY: Moderate KEYWORDS: estimation of mean values, prediction of individual values TABLE 13-3 The director of cooperative education at a state college wants to examine the effect of cooperative education job experience on marketability in the workplace. She takes a random sample of 4 students. For these 4, she finds out how many times each had a cooperative education job and how many job offers they received upon graduation. These data are presented in the table below. Student 1 2 3 4
CoopJobs 1 2 1 0
JobOffer 4 6 3 1
23. Referring to Table 13-3, set up a scatter diagram. ANSWER: Scatte r Diagram 6
Job Offers
5 4 3 2 1 0 0
1 C oop Jobs
TYPE: PR DIFFICULTY: Moderate KEYWORDS: scatter plot
2
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24. Referring to Table 13-3, the least squares estimate of the slope is __________. ANSWER: 2.50 TYPE: FI DIFFICULTY: Moderate KEYWORDS: slope, estimation, least squares 25. Referring to Table 13-3, the least squares estimate of the Y intercept is __________. ANSWER: 1.00 TYPE: FI DIFFICULTY: Moderate KEYWORDS: intercept, estimation, least squares 26. Referring to Table 13-3, the prediction for the number of job offers for a person with 2 cooperative education jobs is __________. ANSWER: 6 TYPE: FI DIFFICULTY: Moderate KEYWORDS: prediction of individual values 27. Referring to Table 13-3, the total sum of squares (SST) is __________. ANSWER: 13.0 TYPE: FI DIFFICULTY: Moderate KEYWORDS: sum of squares 28. Referring to Table 13-3, the regression sum of squares (SSR) is __________. ANSWER: 12.5 TYPE: FI DIFFICULTY: Moderate KEYWORDS: sum of squares 29. Referring to Table 13-3, the error or residual sum of squares (SSE) is __________. ANSWER: 0.50 TYPE: FI DIFFICULTY: Moderate KEYWORDS: sum of squares 30. Referring to Table 13-3, the coefficient of determination is __________. ANSWER: 0.962 TYPE: FI DIFFICULTY: Moderate KEYWORDS: coefficient of determination
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31. Referring to Table 13-3, the standard error of estimate is __________. ANSWER: 0.50 TYPE: FI DIFFICULTY: Moderate KEYWORDS: standard error of estimate 32. Referring to Table 13-3, the coefficient of correlation is __________. ANSWER: 0.981 TYPE: FI DIFFICULTY: Moderate KEYWORDS: coefficient of correlation 33. Referring to Table 13-3, suppose the director of cooperative education wants to obtain a 95% confidence interval estimate for the mean number of job offers received by people who have had exactly one cooperative education job. The t critical value she would use is ________. ANSWER: 4.3027 TYPE: FI DIFFICULTY: Easy KEYWORDS: confidence interval, estimation of mean values, critical value 34. Referring to Table 13-3, suppose the director of cooperative education wants to obtain a 95% confidence interval estimate for the mean number of job offers received by people who have had exactly one cooperative education job. The confidence interval is from ________ to ________. ANSWER: 2.424; 4.576 TYPE: FI DIFFICULTY: Moderate KEYWORDS: confidence interval, estimation of mean values, critical value 35. Referring to Table 13-3, suppose the director of cooperative education wants to obtain a 95% prediction interval estimate for the mean number of job offers received by people who have had exactly one cooperative education job. The prediction interval is from ________ to ________. ANSWER: 1.0947; 5.9053 TYPE: FI DIFFICULTY: Moderate KEYWORDS: prediction interval, prediction of individual values 36. True or False: Referring to Table 13-3, suppose the director of cooperative education wants to obtain two 95% confidence interval estimates. One is for the mean number of job offers received by people who have had exactly one cooperative education job and one for people who have had two. The confidence interval for people who have had one cooperative education job would be the wider of the two intervals. ANSWER: False TYPE: TF DIFFICULTY: Moderate KEYWORDS: confidence interval, estimation of mean values 37. Referring to Table 13-3, suppose the director of cooperative education wants to obtain a 95% prediction interval for the number of job offers received by a person who has had exactly two cooperative education jobs. The t critical value she would use is ________.
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ANSWER: 4.3027 TYPE: FI DIFFICULTY: Moderate KEYWORDS: prediction interval, prediction of individual values, critical value 38. Referring to Table 13-3, suppose the director of cooperative education wants to obtain a 95% prediction interval for the number of job offers received by a person who has had exactly two cooperative education jobs. The prediction interval is from ________ to ________. ANSWER: 3.154; 8.846 TYPE: FI DIFFICULTY: Moderate KEYWORDS: prediction interval, prediction of individual values 39. True or False: Referring to Table 13-3, suppose the director of cooperative education wants to obtain both a 95% confidence interval estimate and a 95% prediction interval for X = 2. The confidence interval estimate would be the wider of the two intervals. ANSWER: False TYPE: TF DIFFICULTY: Moderate KEYWORDS: confidence interval, prediction interval, estimation of mean values, prediction of individual values, properties 40. Referring to Table 13-3, the director of cooperative education wanted to test the hypothesis that the true slope was equal to 0. The denominator of the test statistic is sb1 . The value of sb1 in this sample is ________. ANSWER: 0.3536 TYPE: FI DIFFICULTY: Moderate KEYWORDS: t test on slope, standard error, slope 41. Referring to Table 13-3, the director of cooperative education wanted to test the hypothesis that the true slope was equal to 0. The value of the test statistic is ________. ANSWER: 7.07 TYPE: FI DIFFICULTY: Moderate KEYWORDS: t test on slope, test statistic, slope
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42. Referring to Table 13-3, the director of cooperative education wanted to test the hypothesis that the true slope was equal to 3.0. The value of the test statistic is ________. ANSWER: -1.4142 TYPE: FI DIFFICULTY: Moderate KEYWORDS: t test on slope, test statistic, slope 43. Referring to Table 13-3, the director of cooperative education wanted to test the hypothesis that the true slope was equal to 0. For a test with a level of significance of 0.05, the null hypothesis should be rejected if the value of the test statistic is ________. ANSWER: > 4.3027 or < -4.3027 TYPE: FI DIFFICULTY: Easy KEYWORDS: t test on slope, critical value, slope, decision 44. Referring to Table 13-3, the director of cooperative education wanted to test the hypothesis that the true slope was equal to 3.0. For a test with a level of significance of 0.05, the null hypothesis should be rejected if the value of the test statistic is ________. ANSWER: > 4.3027 or < -4.3027 TYPE: FI DIFFICULTY: Easy KEYWORDS: t test on slope, critical value, slope, decision 45. Referring to Table 13-3, the director of cooperative education wanted to test the hypothesis that the true slope was equal to 0. The p-value of the test is between ________ and ________. ANSWER: 0.01 and 0.02 using the Table, or 0.0194 using Excel TYPE: FI DIFFICULTY: Moderate KEYWORDS: t test on slope, p-value, slope 46. Referring to Table 13-3, the director of cooperative education wanted to test the hypothesis that the true slope was equal to 3.0. The p-value of the test is between ________ 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
Sb1
0.3536
Introduction and Data Collection
TABLE 13-4 The managers of a brokerage firm are interested in finding out if the number of new clients a broker brings into the firm affects the sales generated by the broker. They sample 12 brokers and determine the number of new clients they have enrolled in the last year and their sales amounts in thousands of dollars. These data are presented in the table that follows. Broker Clients Sales 1 27 52 2 11 37 3 42 64 4 33 55 5 15 29 6 15 34 7 25 58 8 36 59 9 28 44 10 30 48 11 17 31 12 22 38 47. Referring to Table 13-4, set up a scatter diagram. ANSWER:
Scatte r Diagram 70 60 50
Sales
372
40 30 20 10 0 0
5
10
15
20
25
C li e nts TYPE: PR DIFFICULTY: Moderate KEYWORDS: scatter plot
30
35
40
45
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48. Referring to Table 13-4, the least squares estimate of the slope is __________. ANSWER: 1.12 TYPE: FI DIFFICULTY: Moderate KEYWORDS: slope, estimation, least squares 49. Referring to Table 13-4, the least squares estimate of the Y intercept is __________. ANSWER: 17.7 TYPE: FI DIFFICULTY: Moderate KEYWORDS: intercept, estimation, least squares 50. Referring to Table 13-4, the prediction for the amount of sales (in $1,000s) for a person who brings 25 new clients into the firm is ________. ANSWER: 45.66 TYPE: FI DIFFICULTY: Moderate KEYWORDS: prediction of individual values 51. Referring to Table 13-4, the total sum of squares (SST) is __________. ANSWER: 1,564.2 TYPE: FI DIFFICULTY: Moderate KEYWORDS: sum of squares 52. Referring to Table 13-4, the regression sum of squares (SSR) is __________. ANSWER: 1,227.4 TYPE: FI DIFFICULTY: Moderate KEYWORDS: sum of squares 53. Referring to Table 13-4, the error or residual sum of squares (SSE) is __________. ANSWER: 336.9 TYPE: FI DIFFICULTY: Moderate KEYWORDS: sum of squares 54. Referring to Table 13-4, the coefficient of determination is __________. ANSWER: 0.785 TYPE: FI DIFFICULTY: Moderate KEYWORDS: coefficient of determination 55. Referring to Table 13-4, ______% of the total variation in sales generated can be explained by the number of new clients brought in. ANSWER: 78.5
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TYPE: FI DIFFICULTY: Moderate KEYWORDS: coefficient of determination, interpretation 56. Referring to Table 13-4, the standard error of the estimated slope coefficient is __________. ANSWER: 0.185 TYPE: FI DIFFICULTY: Moderate KEYWORDS: slope, standard error 57. Referring to Table 13-4, the standard error of estimate is __________. ANSWER: 5.804 TYPE: FI DIFFICULTY: Moderate KEYWORDS: standard error of estimate 58. Referring to Table 13-4, the coefficient of correlation is __________. ANSWER: 0.886 TYPE: FI DIFFICULTY: Moderate KEYWORDS: coefficient of correlation 59. Referring to Table 13-4, suppose the managers of the brokerage firm want to obtain a 99% confidence interval estimate for the mean sales made by brokers who have brought into the firm 24 new clients. The t critical value they would use is ________. ANSWER: 3.1693 TYPE: FI DIFFICULTY: Easy KEYWORDS: estimation of mean values, confidence interval, critical value 60. Referring to Table 13-4, suppose the managers of the brokerage firm want to obtain a 99% confidence interval estimate for the mean sales made by brokers who have brought into the firm 24 new clients. The confidence interval is from ________ to ________. ANSWER: 39.19; 49.89 TYPE: FI DIFFICULTY: Moderate KEYWORDS: estimation of mean values, confidence interval
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61. Referring to Table 13-4, suppose the managers of the brokerage firm want to obtain a 99% prediction interval for the sales made by a broker who has brought into the firm 18 new clients. The t critical value they would use is ________. ANSWER: 3.1693 TYPE: FI DIFFICULTY: Moderate KEYWORDS: prediction of individual values, prediction interval, critical value 62. Referring to Table 13-4, suppose the managers of the brokerage firm want to obtain a 99% prediction interval for the sales made by a broker who has brought into the firm 18 new clients. The prediction interval is from ________ to ________. ANSWER: 18.23; 57.42 TYPE: FI DIFFICULTY: Moderate KEYWORDS: prediction of individual values, prediction interval 63. Referring to Table 13-4, suppose the managers of the brokerage firm want to obtain both a 99% confidence interval estimate and a 99% prediction interval for X = 24. The confidence interval estimate would be the __________ (wider or narrower) of the two intervals. ANSWER: narrower TYPE: FI DIFFICULTY: Moderate KEYWORDS: prediction of individual values, prediction interval, estimation of mean values, confidence interval, properties 64. Referring to Table 13-4, the managers of the brokerage firm wanted to test the hypothesis that the true slope was equal to 0. The denominator of the test statistic is sb1 . The value of sb1 in this sample is ________. ANSWER: 0.1853 TYPE: FI DIFFICULTY: Moderate KEYWORDS: t test on slope, slope, standard error 65. Referring to Table 13-4, the managers of the brokerage firm wanted to test the hypothesis that the true slope was equal to 0. The value of the test statistic is _______. ANSWER: 6.04 TYPE: FI DIFFICULTY: Moderate KEYWORDS: t test on slope, slope, test statistic
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66. Referring to Table 13-4, the managers of the brokerage firm wanted to test the hypothesis that the number of new clients brought in did not affect the amount of sales generated. The value of the test statistic is _______. ANSWER: 6.04 TYPE: FI DIFFICULTY: Moderate KEYWORDS: t test on slope, slope, test statistic 67. Referring to Table 13-4, the managers of the brokerage firm wanted to test the hypothesis that the true slope was equal to 0. For a test with a level of significance of 0.01, the null hypothesis should be rejected if the value of the test statistic is ________. ANSWER: > 3.1693 or < -3.1693 TYPE: FI DIFFICULTY: Moderate KEYWORDS: t test on slope, slope, critical value, decision 68. Referring to Table 13-4, the managers of the brokerage firm wanted to test the hypothesis that the true slope was equal to 0. The p-value of the test is ________. ANSWER: < 0.01 using the Table, or 0.000126 using Excel TYPE: FI DIFFICULTY: Moderate KEYWORDS: t test on slope, slope, p-value 69. Referring to Table 13-4, the managers of the brokerage firm wanted to test the hypothesis that the true slope was equal to 0. At a level of significance of 0.01, the null hypothesis should be _______ (accepted or rejected). ANSWER: rejected TYPE: FI DIFFICULTY: Moderate KEYWORDS: t test on slope, slope, decision 70. Referring to Table 13-4, the managers of the brokerage firm wanted to test the hypothesis that the true slope was equal to 0. At a level of significance of 0.01, the decision that should be made implies that _____ (there is or there is no) linear dependent relation between the independent and dependent variables. ANSWER: there is TYPE: FI DIFFICULTY: Moderate KEYWORDS: t test on slope, slope, conclusion
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71. Referring to Table 13-4, the managers of the brokerage firm wanted to test the hypothesis that the number of new clients brought in had a positive impact on the amount of sales generated. The value of the test statistic is _______. ANSWER: 6.04 TYPE: FI DIFFICULTY: Moderate KEYWORDS: t test on slope, slope, test statistic 72. Referring to Table 13-4, the managers of the brokerage firm wanted to test the hypothesis that the number of new clients brought in had a positive impact on the amount of sales generated. For a test with a level of significance of 0.01, the null hypothesis should be rejected if the value of the test statistic is ________. ANSWER: > 2.7638 TYPE: FI DIFFICULTY: Moderate KEYWORDS: t test on slope, slope, critical value, decision 73. Referring to Table 13-4, the managers of the brokerage firm wanted to test the hypothesis that the number of new clients brought in had a positive impact on the amount of sales generated. The p-value of the test is ________. ANSWER: < 0.005 using the Table, or 0.000126/2 using Excel TYPE: FI DIFFICULTY: Moderate KEYWORDS: t test on slope, slope, p-value 74. Referring to Table 13-4, the managers of the brokerage firm wanted to test the hypothesis that the number of new clients brought in had a positive impact on the amount of sales generated. At a level of significance of 0.01, the null hypothesis should be _______ (accepted or rejected). ANSWER: rejected TYPE: FI DIFFICULTY: Moderate KEYWORDS: t test on slope, slope 75. Referring to Table 13-4, the managers of the brokerage firm wanted to test the hypothesis that the number of new clients brought in had a positive impact on the amount of sales generated. At a level of significance of 0.01, the decision that should be made implies that the number of new clients brought in _____ (had or did not have) a positive impact on the amount of sales generated. ANSWER: had TYPE: FI DIFFICULTY: Moderate KEYWORDS: t test on slope, slope, conclusion
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TABLE 13-5 The managing partner of an advertising agency believes that his company's sales are related to the industry‟s 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 R Square Adjusted R Square Standard Error SYX Observations
0.802 0.643 0.618 0.9224 16
ANOVA Regression Error Total Predictor Intercept Industry
df 1 14 15 Coef 3.962 0.040451
Durbin-Watson Statistic
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
Sig.F 0.000
p-value 0.016 0.000
1.59
76. Referring to Table 13-5, the value of the quantity that the least squares regression line minimizes is ________. ANSWER: 11.912 TYPE: FI DIFFICULTY: Moderate KEYWORDS: least squares, properties, sum of squares 77. Referring to Table 13-5, the estimates of the Y intercept and slope are ________ and ________, respectively. ANSWER: 3.962; 0.040451 TYPE: FI DIFFICULTY: Easy KEYWORDS: intercept, slope, estimation 78. Referring to Table 13-5, the prediction for a quarter in which X = 120 is Y = ________. ANSWER: 8.816 TYPE: FI DIFFICULTY: Moderate KEYWORDS: prediction of individual values
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79. Referring to Table 13-5, the standard error of the estimate is ________. ANSWER: 0.9224 TYPE: FI DIFFICULTY: Easy KEYWORDS: standard error of estimate 80. Referring to Table 13-5, the coefficient of determination is ________. ANSWER: 0.643 TYPE: FI DIFFICULTY: Easy KEYWORDS: coefficient of determination 81. Referring to Table 13-5, the standard error of the estimated slope coefficient is ________. ANSWER: 0.008 TYPE: FI DIFFICULTY: Easy KEYWORDS: slope, standard error 82. Referring to Table 13-5, the correlation coefficient is ________. ANSWER: 0.802 TYPE: FI DIFFICULTY: Easy KEYWORDS: coefficient of correlation 83. Referring to Table 13-5, the partner wants to test for autocorrelation using the Durbin-Watson statistic. Using a level of significance of 0.05, the critical values of the test are dL = ________, and dU = ________. ANSWER: 1.10; 1.37 TYPE: FI DIFFICULTY: Easy KEYWORDS: Durbin-Watson statistic, critical value, autocorrelation 84. Referring to Table 13-5, the partner wants to test for autocorrelation using the Durbin-Watson statistic. Using a level of significance of 0.05, the decision he should make is: a) there is evidence of autocorrelation. b) the test is unable to make a definite conclusion. c) there is no evidence of autocorrelation. d) there is not enough information to perform the test. ANSWER: c TYPE: MC DIFFICULTY: Moderate KEYWORDS: Durbin-Watson statistic, decision, autocorrelation
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85. If the Durbin-Watson statistic has a value close to 0, which assumption is violated? a) normality of the errors b) independence of errors c) homoscedasticity d) none of the above ANSWER: b TYPE: MC DIFFICULTY: Easy KEYWORDS: Durbin-Watson statistic, autocorrelation, assumption 86. If the Durbin-Watson statistic has a value close to 4, which assumption is violated? a) normality of the errors b) independence of errors c) homoscedasticity d) none of the above ANSWER: b TYPE: MC DIFFICULTY: Easy KEYWORDS: Durbin-Watson statistic, assumption, autocorrelation 87. The standard error of the estimate is a measure of a) total variation of the Y variable. b) the variation around the sample regression line. c) explained variation. d) the variation of the X variable. ANSWER: b TYPE: MC DIFFICULTY: Easy KEYWORDS: standard error of estimate, interpretation 88. The coefficient of determination (r2) tells us a) that the coefficient of correlation (r) is larger than 1. b) whether r has any significance. c) that we should not partition the total variation. d) the proportion of total variation that is explained. ANSWER: d TYPE: MC DIFFICULTY: Easy KEYWORDS: coefficient of determination, interpretation
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TABLE 13-6 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 Standard Error 20.25979924 Observations 22
Intercept Attendance
Coefficients Standard Error 39.39027309 37.24347659 0.340583573 0.52852452
t Stat 1.057642216 0.644404489
p-value 0.302826622 0.526635689
89. Referring to Table 13-6, which of the following statements is true? a) -2.86% of the total variability in score received can be explained by percentage attendance. b) -2.86% of the total variability in percentage attendance can be explained by score received. c) 2% of the total variability in score received can be explained by percentage attendance. d) 2% of the total variability in percentage attendance can be explained by score received. ANSWER: c TYPE: MC DIFFICULTY: Easy KEYWORDS: coefficient of determination, interpretation 90. Referring to Table 13-6, which of the following statements is true? a) If attendance increases by 0.341%, the estimated average score received will increase by 1 percentage point. b) If attendance increases by 1%, the estimated average score received will increase by 39.39 percentage points. c) If attendance increases by 1%, the estimated average score received will increase by 0.341 percentage points. d) If the score received increases by 39.39%, the estimated average attendance will go up by 1%. ANSWER: c TYPE: MC DIFFICULTY: Easy KEYWORDS: slope, interpretation
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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: 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
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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 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
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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 DurbinWatson D statistic should be near __________. ANSWER: 2 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
Introduction and Data Collection
104. If the correlation coefficient (r) = 1.00, then a) the Y-intercept (b0) must equal 0. b) the explained variation equals the unexplained variation. c) there is no unexplained variation. d) there is no explained variation. ANSWER: c TYPE: MC DIFFICULTY: Moderate KEYWORDS: coefficient of correlation, properties, interpretation 105. If the correlation coefficient (r) = 1.00, then a) all the data points must fall exactly on a straight line with a slope that equals 1.00. b) all the data points must fall exactly on a straight line with a negative slope. c) all the data points must fall exactly on a straight line with a positive slope. d) 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
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108. The strength of the linear relationship between two numerical variables may be measured by the a) scatter diagram. b) coefficient of correlation. c) slope. d) Y intercept. ANSWER: b TYPE: MC DIFFICULTY: Easy KEYWORDS: coefficient of correlation, interpretation 109. In a simple linear regression problem, r and b1 a) may have opposite signs. b) must have the same sign. c) must have opposite signs. d) 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 p-value is 0.256 when testing H 0 : 0 against the two-sided alternative H1 : 0 . To test H 0 : 0 against the one-sided alternative H1 : 0 at a significance level of 0.193, 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
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111. The sample correlation coefficient between X and Y is 0.375. It has been found out that the p-value is 0.256 when testing H 0 : 0 against the two-sided alternative H1 : 0 . To test H 0 : 0 against the one-sided alternative H1 : 0 at a significance level of 0.193, the p-value is a) 0.256 / 2 b) 0.256 2 c) 1 0.256 d) 1 0.256 / 2 ANSWER: a TYPE: MC DIFFICULTY: Difficult KEYWORDS: coefficient of correlation, t test for correlation coefficient, p-value 112. The sample correlation coefficient between X and Y is 0.375. It has been found out that the p-value is 0.256 when testing H 0 : 0 against the one-sided alternative H1 : 0 . To test H 0 : 0 against the two-sided alternative H1 : 0 at a significance level of 0.193, 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 p-value is 0.744 when testing H 0 : 0 against the one-sided alternative H1 : 0 . To test H 0 : 0 against the two-sided alternative H1 : 0 at a significance level of 0.193, 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
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114. If you wanted to find out if alcohol consumptions (measured in fluid oz.) and grade point average on a 4-point scale are linearly related, you would perform a i. 2 test for the difference in two proportions. j. 2 test for independence. k. a Z test for the difference in two proportions. l. a t test for independence. ANSWER: d TYPE: MC DIFFICULTY: Difficult KEYWORDS: coefficient of correlation, t test for correlation coefficient 115. True or False: When r = – 1, it indicates a perfect relationship between X and Y. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: coefficient of correlation, properties TABLE 13-7 An investment specialist claims that if one holds a portfolio that moves in the opposite direction to the market index like the S&P 500, then it is possible to reduce the variability of the portfolio's return. In other words, one can create a portfolio with positive returns but less exposure to risk. A sample of 26 years of S&P 500 index data and a portfolio consisting of stocks of private prisons, which are believed to be negatively related to the S&P 500 index, is collected. A regression analysis was performed by regressing the returns of the prison stocks portfolio (Y) on the returns of S&P 500 index (X) to prove that the prison stocks portfolio is negatively related to the S&P 500 index at a 5% level of significance. The results are given in the following Excel output.
Intercept S&P
Coefficients Standard Error 4.866004258 0.35743609 -0.502513506 0.071597152
t Stat 13.61363441 -7.01862425
p-value 8.7932E-13 2.94942E-07
116. Referring to Table 13-7, to test whether the prison stocks portfolio is negatively related to the S&P 500 index, the appropriate null and alternative hypotheses are, respectively, a) H 0 : 0 vs. H1 : 0 b)
H 0 : 0 vs. H1 : 0
c)
H 0 : r 0 vs. H1 : r 0
d)
H 0 : r 0 vs. H1 : r 0
ANSWER: a TYPE: MC DIFFICULTY: Easy KEYWORDS: coefficient of correlation, t test for correlation coefficient, form of hypothesis 117. Referring to Table 13-7, to test whether the prison stocks portfolio is negatively related to the S&P 500 index, the measured value of the test statistic is a) -7.019. b) -0.503. c) 0.072. d) 0.357.
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ANSWER: a TYPE: MC DIFFICULTY: Easy KEYWORDS: coefficient of correlation, t test for correlation coefficient, test statistic 118. Referring to Table 13-7, to test whether the prison stocks portfolio is negatively related to the S&P 500 index, the p-value of the associated test statistic is a) 2.94942E-07 b) 2.94942E-07 / 2 c) 2.94942E-07 2 d) 8.7932E-13 ANSWER: b TYPE: MC DIFFICULTY: Easy KEYWORDS: coefficient of correlation, t test for correlation coefficient, p-value 119. Referring to Table 13-7, which of the following will be a correct conclusion? a) We cannot reject the null hypothesis and, therefore, conclude that there is sufficient evidence to show that the prisons stock portfolio and S&P 500 index are negatively related. b) We can reject the null hypothesis and, therefore, conclude that there is sufficient evidence to show that the prisons stock portfolio and S&P 500 index are negatively related. c) We cannot reject the null hypothesis and, therefore, conclude that there is not sufficient evidence to show that the prisons stock portfolio and S&P 500 index are negatively related. d) We can reject the null hypothesis and conclude that there is not sufficient evidence to show that the prisons stock portfolio and S&P 500 index are negatively related. ANSWER: b TYPE: MC DIFFICULTY: Easy KEYWORDS: coefficient of correlation, t test for correlation coefficient, decision, conclusion
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TABLE 13-8 It is believed that GPA (grade point average, based on a four point scale) should have a positive linear relationship with ACT scores. Given below is the Excel output from regressing GPA on ACT scores using a data set of 8 randomly chosen students from a Big Ten university. Regressing GPA on ACT Regression Statistics Multiple R 0.7598 R Square 0.5774 Adjusted R Square 0.5069 Standard Error 0.2691 Observations 8 ANOVA df Regression Residual Total
Intercept ACT
SS 1 6 7
0.5940 0.4347 1.0287
Coefficients Standard Error 0.5681 0.9284 0.1021 0.0356
MS 0.5940 0.0724
F Significance F 8.1986 0.0286
t Stat p-value 0.6119 0.5630 2.8633 0.0286
Lower 95% Upper 95% -1.7036 2.8398 0.0148 0.1895
120. Referring to Table 13-8, the interpretation of the coefficient of determination in this regression is that a) 57.74% of the total variation of ACT scores can be explained by GPA. b) ACT scores account for 57.74% of the total fluctuation in GPA. c) GPA accounts for 57.74% of the variability of ACT scores. d) none of the above ANSWER: b TYPE: MC DIFFICULTY: Easy KEYWORDS: coefficient of determination, interpretation 121. Referring to Table 13-8, the value of the measured test statistic to test whether there is any linear relationship between GPA and ACT is a) 0.0356. b) 0.1021. c) 0.7598. d) 2.8633. ANSWER: d TYPE: MC DIFFICULTY: Easy KEYWORDS: coefficient of correlation, t test for correlation coefficient, test statistic
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122. Referring to Table 13-8, what is the predicted average value of GPA when ACT = 20? a) 2.61 b) 2.66 c) 2.80 d) 3.12 ANSWER: a TYPE: MC DIFFICULTY: Easy KEYWORDS: prediction of individual values 123. Referring to Table 13-8, what are the decision and conclusion on testing whether there is any linear relationship at the 1% level of significance between GPA and ACT scores? a) Do not reject the null hypothesis; hence, there is not sufficient evidence to show that ACT scores and GPA are linearly related. b) Reject the null hypothesis; hence, there is not sufficient evidence to show that ACT scores and GPA are linearly related. c) Do not reject the null hypothesis; hence, there is sufficient evidence to show that ACT scores and GPA are linearly related. d) Reject the null hypothesis; hence, there is sufficient evidence to show that ACT scores and GPA are linearly related. ANSWER: a TYPE: MC DIFFICULTY: Easy KEYWORDS: coefficient of correlation, t test for correlation coefficient, decision, conclusion 124. Referring to Table 13-8, the value of the measured (observed) test statistic of the F test for
H 0 : 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
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TABLE 13-9 It is believed that the average number of hours spent studying per day (HOURS) during undergraduate education should have a positive linear relationship with starting salary (SALARY, measured in thousands of dollars per month) after graduation. Given below is the Excel output from regressing starting salary on number of hours spent studying per day for a sample of 51 students. NOTE: Some of the numbers in the output are purposely erased. Regression Statistics Multiple R R Square Adjusted R Square Standard Error Observations
0.8857 0.7845 0.7801 1.3704 51
ANOVA df Regression Residual Total
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
t Stat -4.7134 13.3561
F Significance F 178.3859
p-value 2.051E-05 5.944E-18
Lower 95% Upper 95% -2.7015 -1.0865 0.8321 1.1269
125. Referring to Table 13-9, the estimated average change in salary (in thousands of dollars) as a result of spending an extra hour per day studying is a) -1.8940. b) 0.7845. c) 0.9795. d) 335.0473. ANSWER: c TYPE: MC DIFFICULTY: Moderate KEYWORDS: slope, interpretation 126. Referring to Table 13-9, the value of the measured t test statistic to test whether average SALARY depends linearly on HOURS is a) -4.7134. b) -1.8940. c) 0.9795. d) 13.3561. ANSWER: d TYPE: MC DIFFICULTY: Moderate KEYWORDS: t test on slope, slope, test statistic 127. Referring to Table 13-9, the p-value of the measured F test statistic to test whether HOURS affect SALARY is a) (5.944E-18)/2. b) 5.944E-18. c) (2.051E-05)/2.
Introduction and Data Collection
d) 2.051E-05. ANSWER: b TYPE: MC DIFFICULTY: Moderate KEYWORDS: F test on slope, slope, test statistic 128. Referring to Table 13-9, the degrees of freedom for testing whether HOURS affect SALARY are a) 1, 49. b) 1, 50. c) 49, 1. d) 50, 1. ANSWER: a TYPE: MC DIFFICULTY: Moderate KEYWORDS: F test on slope, slope, degrees of freedom 129. Referring to Table 13-9, the error sum of squares (SSE) of the above regression is a) 1.878215. b) 92.0325465. c) 335.047257. d) 427.079804. ANSWER: b TYPE: MC DIFFICULTY: Moderate KEYWORDS: sum of squares 130. Referring to Table 13-9, the 90% confidence interval for the average change in SALARY (in thousands of dollars) as a result of spending an extra hour per day studying is a) wider than [-2.70159, -1.08654]. b) narrower than [-2.70159, -1.08654]. c) wider than [0.8321927, 1.12697]. d) narrower than [0.8321927, 1.12697]. ANSWER: d TYPE: MC DIFFICULTY: Moderate KEYWORDS: slope, interpretation, confidence interval
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131. Referring to Table 13-9, to test the claim that average SALARY depends positively on HOURS, against the null hypothesis that average SALARY does not depend linearly on HOURS, the p-value of the test statistic is a) (5.944E-18)/2. b) 5.944E-18. c) (2.051E-05)/2. d) 2.051E-05. ANSWER: a TYPE: MC DIFFICULTY: Moderate KEYWORDS: t test on slope, slope, p-value 132. True or False: A zero population correlation coefficient between a pair of random variables means that there is no linear relationship between the random variables. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: coefficient of correlation, interpretation 133. 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
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135. True or False: The confidence interval for the mean of Y is always narrower than the prediction interval for an individual response Y given the same data set, X value, and confidence level. ANSWER: True TYPE: TF DIFFICULTY: Moderate KEYWORDS: estimation of mean values, confidence interval, prediction of individual values, prediction interval, properties Table 13-10 The management of a chain electronic store would like to develop a model for predicting the weekly sales (in thousand of dollars) for individual stores based on the number of customers who made purchases. A random sample of 12 stores yields the following results:
Customers Sales (Thousands of Dollars) 907 11.20 926 11.05 713 8.21 741 9.21 780 9.42 898 10.08 510 6.73 529 7.02 460 6.12 872 9.52 650 7.53 603 7.25
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136. Referring to Table 13-10, generate the scatter plot. ANSWER: TYPE: PR DIFFICULTY: Moderate Scatter Diagram
Salses (Thousands of Dollars)
12 10 8 6 4 2 0 0
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Customers
KEYWORDS: scatter plot 137. Referring to Table 13-10, what are the values of the estimated intercept and slope? ANSWER: 1.4464, 0.0100 TYPE: PR DIFFICULTY: Moderate KEYWORDS: intercept, slope, estimation 138. Referring to Table 13-10, what is the value of the coefficient of determination? ANSWER: 0.9453 TYPE: PR DIFFICULTY: Moderate KEYWORDS: coefficient of determination 139. Referring to Table 13-10, what is the value of the coefficient of correlation? ANSWER: 0.9723 TYPE: PR DIFFICULTY: Moderate KEYWORDS: coefficient of correlation 140. Referring to Table 13-10, what is the value of the standard error of the estimate? ANSWER: 0.4191 TYPE: PR DIFFICULTY: Moderate KEYWORDS: standard error of estimate 141. Referring to Table 13-10, which is the correct null hypothesis for testing whether the number of customers who make purchases affects weekly sales? a) H 0 : 0 0 b) H 0 : 1 0 c)
H0 : 0
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d) H 0 : 0 ANSWER: b TYPE: MC DIFFICULTY: Easy KEYWORDS: t test on slope, slope, form of hypothesis 142. Referring to Table 13-10, what is the value of the t test statistic when testing whether the number of customers who make purchases affects weekly sales? ANSWER: 13.1464 TYPE: PR DIFFICULTY: Moderate KEYWORDS: t test on slope, slope, test statistic 143. Referring to Table 13-10, what are the degrees of freedom of the t test statistic when testing whether the number of customers who make purchases affects weekly sales? ANSWER: 10 TYPE: PR DIFFICULTY: Easy KEYWORDS: t test on slope, slope, degrees of freedom 144. Referring to Table 13-10, what is the p-value of the t test statistic when testing whether the number of customers who make purchases affects weekly sales? ANSWER: 1.23323E-07 or smaller than 0.01 TYPE: PR DIFFICULTY: Moderate KEYWORDS: t test on slope, slope, p-value 145. True or False: Referring to Table 13-10, the null hypothesis for testing whether the number of customers who make purchases affects weekly sales cannot be rejected if a 1% probability of committing a Type I error is desired. ANSWER: False TYPE: TF DIFFICULTY: Moderate KEYWORDS: t test on slope, slope, decision
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146. True or False: Referring to Table 13-10, the average weekly sales will increase by an estimated $0.01 for each additional purchasing customer. ANSWER: False TYPE: TF DIFFICULTY: Moderate KEYWORDS: t test on slope, slope, interpretation 147. True or False: Referring to Table 13-10, the average weekly sales will increase by an estimated $10 for each additional purchasing customer. ANSWER: True TYPE: TF DIFFICULTY: Moderate KEYWORDS: t test on slope, slope, interpretation 148. True or False: Referring to Table 13-10, 93.98% of the total variation in weekly sales can be explained by the variation in the number of customers who make purchases. ANSWER: False TYPE: TF DIFFICULTY: Moderate KEYWORDS: coefficient of determination, interpretation 149. Referring to Table 13-10, what are the degrees of freedom of the F test statistic when testing whether the number of customers who make purchases is a good predictor of weekly sales? ANSWER: 1 in the numerator and 10 in the denominator TYPE: PR DIFFICULTY: Easy KEYWORDS: F test on slope, slope, test statistic 150. Referring to Table 13-10, what is the value of the F test statistic when testing whether the number of customers who make purchases is a good predictor of weekly sales? ANSWER: 172.8265 TYPE: PR DIFFICULTY: Moderate KEYWORDS: F test on slope, slope, test statistic 151. Referring to Table 13-10, what is the p-value of the F test statistic when testing whether the number of customers who make purchases is a good predictor of weekly sales? ANSWER: 1.23323E-07 or smaller than 0.01 TYPE: PR DIFFICULTY: Moderate KEYWORDS: F test on slope, slope, p-value
152. True or False: Referring to Table 13-10, the p-value of the t test and F test should be the same when testing whether the number of customers who make purchases is a good predictor of weekly sales. ANSWER: True TYPE: TF DIFFICULTY: Moderate
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KEYWORDS: t test on slope, F test on slope, p-value 153. True or False: Referring to Table 13-10, the value of the t test statistic and F test statistic should be the same when testing whether the number of customers who make purchases is a good predictor of 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 13-10, the value of the F test statistic equals the square of the t test statistic when testing whether the number of customers who make purchases is a good predictor of weekly sales. ANSWER: True TYPE: TF DIFFICULTY: Moderate KEYWORDS: t test on slope, F test on slope, test statistic 155. Referring to Table 13-10, generate the residual plot. ANSWER: Customers Residual Plot 0.8 0.6
Residuals
0.4 0.2 0 -0.2 -0.4 -0.6 -0.8 0
200
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Customers
TYPE: PR DIFFICULTY: Moderate KEYWORDS: residual plot
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156. Referring to Table 13-10, the residual plot indicates possible violation of which assumptions? a) linearity of the relationship b) homoscedasticity c) autocorrelation d) normality ANSWER: b TYPE: MC DIFFICULTY: Moderate KEYWORDS: homoscedasticity, residual plot, assumption 157. True or False: Referring to Table 13-10, it is inappropriate to compute the Durbin-Watson statistic and test for autocorrelation in this case. ANSWER: True TYPE: TF DIFFICULTY: Moderate KEYWORDS: Durbin-Watson statistic, autocorrelation 158. Referring to Table 13-10, construct a 95% confidence interval for the change in average weekly sales when the number of customers who make purchases increases by one. ANSWER: 0.0083 to 0.0117 TYPE: PR DIFFICULTY: Difficult KEYWORDS: slope, confidence interval, interpretation 159. Referring to Table 13-10, construct a 95% confidence interval for the average weekly sales when the number of customers who make purchases is 600. ANSWER: 7.1194 to 7.7864 thousands of dollars TYPE: PR DIFFICULTY: Difficult KEYWORDS: estimate of mean values, confidence interval 160. Referring to Table 13-10, construct a 95% prediction interval for the weekly sales of a store that has 600 purchasing customers. ANSWER: 6.4614 to 8.4444 thousands of dollars TYPE: PR DIFFICULTY: Difficult KEYWORDS: prediction of individual values, prediction interval
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CHAPTER 14: INTRODUCTION TO MULTIPLE REGRESSION 1. In a multiple regression problem involving two independent variables, if b1 is computed to be +2.0, it means that a) the relationship between X1 and Y is significant. b) the estimated average of Y increases by 2 units for each increase of 1 unit of X1, holding X2 constant. c) the estimated average of Y increases by 2 units for each increase of 1 unit of X1, without regard to X2. d) the estimated average of Y is 2 when X1 equals zero. ANSWER: b TYPE: MC DIFFICULTY: Easy KEYWORDS: slope, interpretation 2. The coefficient of multiple determination r2Y.12 a) measures the variation around the predicted regression equation. b) measures the proportion of variation in Y that is explained by X1 and X2. c) measures the proportion of variation in Y that is explained by X1 holding X2 constant. d) will have the same sign as b1. ANSWER: b TYPE: MC DIFFICULTY: Easy KEYWORDS: coefficient of multiple determination, interpretation 3. In a multiple regression model, the value of the coefficient of multiple determination a) has to fall between -1 and +1. b) has to fall between 0 and +1. c) has to fall between -1 and 0. d) can fall between any pair of real numbers. ANSWER: b TYPE: MC DIFFICULTY: Easy KEYWORDS: coefficient of multiple determination, properties
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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 TABLE 14-1 A manager of a product sales group believes the number of sales made by an employee (Y) depends on how many years that employee has been with the company (X1) and how he/she scored on a business aptitude test (X2). A random sample of 8 employees provides the following: Employee Y X1 X2 1 100 10 7 2 90 3 10 3 80 8 9 4 70 5 4 5 60 5 8 6 50 7 5 7 40 1 4 8 30 1 1 5. Referring to Table 14-1, for these data, what is the value for the regression constant, b0? a) 0.998 b) 3.103 c) 4.698 d) 21.293 ANSWER: d TYPE: MC DIFFICULTY: Moderate KEYWORDS: intercept, estimation 6. Referring to Table 14-1, for these data, what is the estimated coefficient for the variable representing the years an employee has been with the company, b1? a) 0.998 b) 3.103 c) 4.698 d) 21.293 ANSWER: b TYPE: MC DIFFICULTY: Moderate KEYWORDS: slope, estimation 7. Referring to Table 14-1, for these data, what is the estimated coefficient for the variable representing scores on the aptitude test, b2? a) 0.998 b) 3.103 c) 4.698 d) 21.293
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ANSWER: c TYPE: MC DIFFICULTY: Moderate KEYWORDS: slope, estimation 8. Referring to Table 14-1, if an employee who had been with the company 5 years scored a 9 on the aptitude test, what would his estimated expected sales be? a) 79.09 b) 60.88 c) 55.62 d) 17.98 ANSWER: a TYPE: MC DIFFICULTY: Moderate KEYWORDS: estimation of mean values TABLE 14-2 A professor of industrial relations believes that an individual‟s wage rate at a factory (Y) depends on his performance rating (X1) and the number of economics courses the employee successfully completed in college (X2). The professor randomly selects 6 workers and collects the following information: Employee Y ($) X1 X2 1 10 3 0 2 12 1 5 3 15 8 1 4 17 5 8 5 20 7 12 6 25 10 9 9. Referring to Table 14-2, for these data, what is the value for the regression constant, b0? a) 0.616 b) 1.054 c) 6.932 d) 9.103 ANSWER: c TYPE: MC DIFFICULTY: Moderate KEYWORDS: intercept, estimation
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10. Referring to Table 14-2, for these data, what is the estimated coefficient for performance rating, b1? a) 0.616 b) 1.054 c) 6.932 d) 9.103 ANSWER: b TYPE: MC DIFFICULTY: Moderate KEYWORDS: slope, estimation 11. Referring to Table 14-2, for these data, what is the estimated coefficient for the number of economics courses taken, b2? a) 0.616 b) 1.054 c) 6.932 d) 9.103 ANSWER: a TYPE: MC DIFFICULTY: Moderate KEYWORDS: slope, estimation 12. Referring to Table 14-2, suppose an employee had never taken an economics course and managed to score a 5 on his performance rating. What is his estimated expected wage rate? a) 10.90 b) 12.20 c) 17.23 d) 25.11 ANSWER: b TYPE: MC DIFFICULTY: Moderate KEYWORDS: estimation of mean values 13. Referring to Table 14-2, an employee who took 12 economics courses scores 10 on the performance rating. What is her estimated expected wage rate? a) 10.90 b) 12.20 c) 24.87 d) 25.70 ANSWER: c TYPE: MC DIFFICULTY: Moderate KEYWORDS: estimation of mean values
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14. The variation attributable to factors other than the relationship between the independent variables and the explained variable in a regression analysis is represented by a) regression sum of squares. b) error sum of squares. c) total sum of squares. d) regression mean squares. ANSWER: b TYPE: MC DIFFICULTY: Moderate KEYWORDS: sum of squares TABLE 14-3 An economist is interested to see how consumption for an economy (in $ billions) is influenced by gross domestic product ($ billions) and aggregate price (consumer price index). The Microsoft Excel output of this regression is partially reproduced below: SUMMARY OUTPUT Regression Statistics Multiple R 0.991 R Square 0.982 Adjusted R Square 0.976 Standard Error 0.299 Observations 10 ANOVA Regression Residual Total
df 2 7 9
SS 33.4163 0.6277 34.0440
MS 16.7082 0.0897
F 186.325
Signif F 0.0001
Coeff StdError t Stat p-value Intercept – 0.0861 0.5674 – 0.152 0.8837 GDP 0.7654 0.0574 13.340 0.0001 Price – 0.0006 0.0028 – 0.219 0.8330 15. Referring to Table 14-3, when the economist used a simple linear regression model with consumption as the dependent variable and GDP as the independent variable, he obtained an r2 value of 0.971. What additional percentage of the total variation of consumption has been explained by including aggregate prices in the multiple regression? a) 98.2 b) 11.1 c) 2.8 d) 1.1 ANSWER: d TYPE: MC DIFFICULTY: Moderate KEYWORDS: coefficient of multiple determination 16. Referring to Table 14-3, the p-value for GDP is a) 0.05. b) 0.01. c) 0.001. d) none of the above
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ANSWER: d TYPE: MC DIFFICULTY: Easy KEYWORDS: t test on slope, p-value 17. Referring to Table 14-3, the p-value for the aggregated price index is a) 0.05. b) 0.01. c) 0.001. d) none of the above ANSWER: d TYPE: MC DIFFICULTY: Easy KEYWORDS: t test on slope, p-value 18. Referring to Table 14-3, the p-value for the regression model as a whole is a) 0.05. b) 0.01. c) 0.001. d) none of the above ANSWER: d TYPE: MC DIFFICULTY: Easy KEYWORDS: F test for the entire regression, p-value 19. Referring to Table 14-3, what is the predicted consumption level for an economy with a 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
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20. Referring to Table 14-3, what is the estimated average consumption level for an economy with a GDP equal to $4 billion and an aggregate price index of 150? a) $1.39 billion b) $2.89 billion c) $4.75 billion d) $9.45 billion ANSWER: b TYPE: MC DIFFICULTY: Moderate KEYWORDS: estimation of mean values 21. Referring to Table 14-3, what is the estimated average consumption level for an economy with a GDP equal to $2 billion and an aggregate price index of 90? a) $1.39 billion b) $2.89 billion c) $4.75 billion d) $9.45 billion ANSWER: a TYPE: MC DIFFICULTY: Moderate KEYWORDS: estimation of mean values 22. Referring to Table 14-3, one economy in the sample had an aggregate consumption level of $3 billion, a GDP of $3.5 billion, and an aggregate price level of 125. What is the residual for this data point? a) $2.52 billion b) $0.48 billion c) – $1.33 billion d) – $2.52 billion ANSWER: b TYPE: MC DIFFICULTY: Moderate KEYWORDS: residual 23. Referring to Table 14-3, one economy in the sample had an aggregate consumption level of $4 billion, a GDP of $6 billion, and an aggregate price level of 200. What is the residual for this data point? a) $4.39 billion b) $0.39 billion c) – $0.39 billion d) – $1.33 billion ANSWER: c TYPE: MC DIFFICULTY: Moderate KEYWORDS: residual 24. Referring to Table 14-3, to test for the significance of the coefficient on aggregate price index, the value of the relevant t-statistic is a) 2.365. b) 0.143. c) – 0.219. d) – 1.960.
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ANSWER: c TYPE: MC DIFFICULTY: Easy KEYWORDS: t test on slope, test statistic 25. Referring to Table 14-3, to test for the significance of the coefficient on aggregate price index, the p-value is a) 0.0001. b) 0.8330. c) 0.8837. d) 0.9999. ANSWER: b TYPE: MC DIFFICULTY: Easy KEYWORDS: t test on slope, p-value 26. Referring to Table 14-3, to test for the significance of the coefficient on gross domestic product, the p-value is a) 0.0001. b) 0.8330. c) 0.8837. d) 0.9999. ANSWER: a TYPE: MC DIFFICULTY: Easy KEYWORDS: t test on slope, p-value 27. Referring to Table 14-3, to test whether aggregate price index has a negative impact on consumption, the pvalue 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
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28. Referring to Table 14-3, to test whether aggregate price index has a positive impact on consumption, the p-value is a) 0.0001. b) 0.4165. c) 0.5835. d) 0.8330. ANSWER: c TYPE: MC DIFFICULTY: Moderate KEYWORDS: t test on slope, p-value 29. Referring to Table 14-3, to test whether gross domestic product has a positive impact on consumption, the pvalue is a) 0.00005. b) 0.0001. c) 0.9999. d) 0.99995. ANSWER: a TYPE: MC DIFFICULTY: Easy KEYWORDS: t test on slope, p-value
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TABLE 14-4 A real estate builder wishes to determine how house size (House) is influenced by family income (Income), family size (Size), and education of the head of household (School). House size is measured in hundreds of square feet, income is measured in thousands of dollars, and education is measured 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
Intercept Income Size School
49 Coeff – 1.6335 0.4485 4.2615 – 0.6517
SS 3605.7736 1214.2264 4820.0000 StdError 5.8078 0.1137 0.8062 0.4319
MS 901.4434 26.9828
t Stat – 0.281 3.9545 5.286 – 1.509
F
Signif F 0.0001
p-value 0.7798 0.0003 0.0001 0.1383
30. Referring to Table 14-4, what fraction of the variability in house size is explained by income, size of family, and education? a) 27.0% b) 33.4% c) 74.8% d) 86.5% ANSWER: c TYPE: MC DIFFICULTY: Easy KEYWORDS: coefficient of multiple determination, interpretation
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31. Referring to Table 14-4, which of the independent variables in the model are significant at the 2% level? a) Income, Size, School b) Income, Size c) Size, School d) Income, School ANSWER: b TYPE: MC DIFFICULTY: Easy KEYWORDS: t test on slope, critical value, p-value, decision, conclusion 32. Referring to Table 14-4, when the builder used a simple linear regression model with house size (House) as the dependent variable and education (School) as the independent variable, he obtained an r2 value of 23.0%. What additional percentage of the total variation in house size has been explained by including family size and income in the multiple regression? a) 2.8% b) 51.8% c) 72.6% d) 74.8% ANSWER: b TYPE: MC DIFFICULTY: Moderate KEYWORDS: coefficient of multiple determination 33. Referring to Table 14-4, which of the following values for the level of significance is the smallest, for which all explanatory variables are significant individually? a) 0.01 b) 0.025 c) 0.05 d) 0.15 ANSWER: d TYPE: MC DIFFICULTY: Moderate KEYWORDS: p-value, interpretation 34. Referring to Table 14-4, which of the following values for the level of significance is the smallest, for which at least two explanatory variables are significant individually? a) 0.01 b) 0.025 c) 0.05 d) 0.15 ANSWER: a TYPE: MC DIFFICULTY: Moderate KEYWORDS: p-value, interpretation 35. Referring to Table 14-4, which of the following values for the level of significance is the smallest, for which the regression model as a whole is significant? a) 0.00005 b) 0.001 c) 0.01 d) 0.05
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ANSWER: b TYPE: MC DIFFICULTY: Moderate KEYWORDS: p-value, interpretation 36. Referring to Table 14-4, what is the predicted house size (in hundreds of square feet) for an individual earning an annual income of $40,000, having a family size of 4, and going to school a total of 13 years? a) 11.43 b) 15.15 c) 24.88 d) 53.87 ANSWER: c TYPE: MC DIFFICULTY: Moderate KEYWORDS: prediction of individual values 37. Referring to Table 14-4, what minimum annual income would an individual with a family size of 4 and 16 years of education need to attain a predicted 10,000 square foot home (House = 100)? a) $44.14 thousand b) $56.75 thousand c) $178.33 thousand d) $211.85 thousand ANSWER: d TYPE: MC DIFFICULTY: Moderate KEYWORDS: prediction of individual values 38. Referring to Table 14-4, what minimum annual income would an individual with a family size of 9 and 10 years of education need to attain a predicted 5,000 square foot home (House = 50)? a) $44.14 thousand b) $56.75 thousand c) $178.33 thousand d) $211.85 thousand ANSWER: a TYPE: MC DIFFICULTY: Moderate KEYWORDS: prediction of individual values
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39. Referring to Table 14-4, one individual in the sample had an annual income of $100,000, a family size of 10, and an education of 16 years. This individual owned a home with an area of 7,000 square feet (House = 70.00). What is the residual (in hundreds of square feet) for this data point? a) 7.40 b) 2.52 c) – 2.52 d) – 5.40 ANSWER: d TYPE: MC DIFFICULTY: Moderate KEYWORDS: residual 40. Referring to Table 14-4, one individual in the sample had an annual income of $10,000, a family size of 1, and an education of 8 years. This individual owned a home with an area of 1,000 square feet (House = 10.00). What is the residual (in hundreds of square feet) for this data point? a) 8.10 b) 5.40 c) – 5.40 d) – 8.10 ANSWER: a TYPE: MC DIFFICULTY: Moderate KEYWORDS: residual 41. Referring to Table 14-4, suppose the builder wants to test whether the coefficient on Income is significantly different from 0. What is the value of the relevant t statistic? a) 5.286 b) 5.195 c) 3.945 d) – 1.509 ANSWER: c TYPE: MC DIFFICULTY: Easy KEYWORDS: t test on slope, test statistic
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42. Referring to Table 14-4, at the 0.01 level of significance, what conclusion should the builder draw regarding the inclusion of Income in the regression model? a) Income is significant in explaining house size and should be included in the model because its p-value is less than 0.01. b) Income is significant in explaining house size and should be included in the model because its p-value is more than 0.01. c) Income is not significant in explaining house size and should not be included in the model because its pvalue is less than 0.01. d) Income is not significant in explaining house size and should not be included in the model because its pvalue is more than 0.01. ANSWER: a TYPE: MC DIFFICULTY: Moderate KEYWORDS: t test on slope, decision, conclusion 43. Referring to Table 14-4, suppose the builder wants to test whether the coefficient on School is significantly different from 0. What is the value of the relevant t statistic? a) 5.286 b) 5.195 c) 3.945 d) – 1.509 ANSWER: d TYPE: MC DIFFICULTY: Easy KEYWORDS: t test on slope, test statistic 44. Referring to Table 14-4, what is the value of the calculated F test statistic that is missing from the output for testing whether the whole regression model is significant? a) 0.0001 b) 0.0299 c) 0.726 d) 33.408 ANSWER: d TYPE: MC DIFFICULTY: Moderate KEYWORDS: F test on the entire regression, test statistic 45. Referring to Table 14-4, the observed value of the F statistic is missing from the printout. What are the degrees of freedom for this F statistic? a) 45 for the numerator, 4 for the denominator b) 4 for the numerator, 49 for the denominator c) 45 for the numerator, 49 for the denominator d) 4 for the numerator, 45 for the denominator ANSWER: d TYPE: MC DIFFICULTY: Easy KEYWORDS: F test on the entire regression, degrees of freedom 46. Referring to Table 14-4, at the 0.01 level of significance, what conclusion should the builder draw regarding the inclusion of School in the regression model?
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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 pvalue is less than 0.01. d) School is not significant in explaining house size and should not be included in the model because its pvalue is more than 0.01. ANSWER: d TYPE: MC DIFFICULTY: Moderate KEYWORDS: t test on slope, critical value, p-value, decision, conclusion 47. Referring to Table 14-4, what are the regression degrees of freedom that are missing from the output? a) 4 b) 45 c) 49 d) 50 ANSWER: a TYPE: MC DIFFICULTY: Easy KEYWORDS: degrees of freedom 48. Referring to Table 14-4, what are the residual degrees of freedom that are missing from the output? a) 4 b) 45 c) 49 d) 50 ANSWER: b TYPE: MC DIFFICULTY: Easy KEYWORDS: degrees of freedom
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TABLE 14-5 A microeconomist wants to determine how corporate sales are influenced by capital and wage spending by companies. She proceeds to randomly select 26 large corporations and record information in millions of dollars. The Microsoft Excel output below shows the results of this multiple regression. SUMMARY OUTPUT 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
Intercept Capital Wages
df 2 23 25
SS 15579777040 7045072780 22624849820
Coeff 15800.0000 0.1245 7.0762
StdError 6038.2999 0.2045 1.4729
MS 7789888520 306307512
t Stat 2.617 0.609 4.804
F 25.432
Signif F 0.0001
p-value 0.0154 0.5485 0.0001
49. Referring to Table 14-5, what fraction of the variability in sales is explained by spending on capital and wages? a) 27.0% b) 50.9% c) 68.9% d) 83.0% ANSWER: c TYPE: MC DIFFICULTY: Easy KEYWORDS: coefficient of multiple determination, interpretation 50. Referring to Table 14-5, which of the independent variables in the model are significant at the 5% level? a) capital, wages b) capital c) wages d) none of the above ANSWER: c TYPE: MC DIFFICULTY: Easy KEYWORDS: t test on slope, critical value, p-value, decision, conclusion 51. Referring to Table 14-5, when the microeconomist used a simple linear regression model, with sales as the dependent variable and wages as the independent variable, he obtained an r2 value of 0.601. What additional percentage of the total variation of sales has been explained by including capital spending in the multiple regression? a) 60.1% b) 31.1% c) 22.9%
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d) 8.8% ANSWER: d TYPE: MC DIFFICULTY: Moderate KEYWORDS: coefficient of multiple determination 52. Referring to Table 14-5, what is the p-value for Wages? a) 0.01 b) 0.05 c) 0.0001 d) none of the above ANSWER: c TYPE: MC DIFFICULTY: Easy KEYWORDS: t test on slope, p-value 53. Referring to Table 14-5, what is the p-value for testing whether Wages have a positive impact on corporate sales? a) 0.01 b) 0.05 c) 0.0001 d) 0.00005 ANSWER: d TYPE: MC DIFFICULTY: Easy KEYWORDS: t test on slope, p-value 54. Referring to Table 14-5, what is the p-value for testing whether Wages have a negative impact on corporate sales? a) 0.05 b) 0.0001 c) 0.00005 d) 0.99995 ANSWER: d TYPE: MC DIFFICULTY: Moderate KEYWORDS: t test on slope, p-value
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55. Referring to Table 14-5, what is the p-value for Capital? a) 0.01 b) 0.025 c) 0.05 d) none of the above ANSWER: d TYPE: MC DIFFICULTY: Easy KEYWORDS: t test on slope, p-value 56. Referring to Table 14-5, what is the p-value for testing whether Capital has a positive influence on corporate sales? a) 0.025 b) 0.05 c) 0.2743 d) 0.5485 ANSWER: c TYPE: MC DIFFICULTY: Easy KEYWORDS: t test on slope, p-value 57. Referring to Table 14-5, what is the p-value for testing whether Capital has a negative influence on corporate sales? a) 0.05 b) 0.2743 c) 0.5485 d) 0.7258 ANSWER: d TYPE: MC DIFFICULTY: Moderate KEYWORDS: t test on slope, p-value 58. Referring to Table 14-5, which of the following values for whole is significant? a) 0.00005 b) 0.001 c) 0.01 d) 0.05
is the smallest for which the regression model as a
ANSWER: b TYPE: MC DIFFICULTY: Moderate KEYWORDS: F test on the entire regression, p-value, interpretation
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59. Referring to Table 14-5, what are the predicted sales (in millions of dollars) for a company spending $100 million on capital and $100 million on wages? a) 15,800.00 b) 16,520.07 c) 17,277.49 d) 20,455.98 ANSWER: b TYPE: MC DIFFICULTY: Moderate KEYWORDS: prediction of individual values 60. Referring to Table 14-5, what are the predicted sales (in millions of dollars) for a company spending $500 million on capital and $200 million on wages? a) 15,800.00 b) 16,520.07 c) 17,277.49 d) 20,455.98 ANSWER: c TYPE: MC DIFFICULTY: Moderate KEYWORDS: prediction of individual values 61. Referring to Table 14-5, one company in the sample had sales of $20 billion (Sales = 20,000). This company spent $300 million on capital and $700 million on wages. What is the residual (in millions of dollars) for this data point? a) 874.55 b) 622.87 c) –790.69 d) –983.56 ANSWER: c TYPE: MC DIFFICULTY: Moderate KEYWORDS: residual 62. Referring to Table 14-5, one company in the sample had sales of $21.439 billion (Sales = 21,439). This company spent $300 million on capital and $700 million on wages. What is the residual (in millions of dollars) for this data point? a) 790.69 b) 648.31 c) – 648.31 d) – 790.69 ANSWER: b TYPE: MC DIFFICULTY: Moderate KEYWORDS: residual 63. Referring to Table 14-5, suppose the microeconomist wants to test whether the coefficient on Capital is significantly different from 0. What is the value of the relevant t statistic? a) 0.609 b) 2.617 c) 4.804
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d) 25.432 ANSWER: a TYPE: MC DIFFICULTY: Easy KEYWORDS: t test on slope, test statistic 64. Referring to Table 14-5, at the 0.01 level of significance, what conclusion should the microeconomist draw regarding the inclusion of Capital in the regression model? a) Capital is significant in explaining corporate sales and should be included in the model because its pvalue is less than 0.01. b) Capital is significant in explaining corporate sales and should be included in the model because its pvalue 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 14-5, the observed value of the F statistic is given on the printout as 25.432. What are the degrees of freedom for this F statistic? a) 25 for the numerator, 2 for the denominator b) 2 for the numerator, 23 for the denominator c) 23 for the numerator, 25 for the denominator d) 2 for the numerator, 25 for the denominator ANSWER: b TYPE: MC DIFFICULTY: Easy KEYWORDS: F test on the entire regression, degrees of freedom
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TABLE 14-6 One of the most common questions of prospective house buyers pertains to the average cost of heating in dollars (Y). To provide its customers with information on that matter, a large real estate firm used the following 4 variables to predict heating costs: the daily minimum outside temperature in degrees of Fahrenheit ( 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 are the Excel outputs of two regression models. Model 1 Regression Statistics R Square 0.8080 Adjusted R Square 0.7568 Observations 20 ANOVA df Regression 4 Residual 15 Total 19
Intercept X1 (Temperature) X2 (Insulation) X3 (Windows) X4 (Furnace Age)
SS MS 169503.4241 42375.86 40262.3259 2684.155 209765.75
Coefficients Standard Error 421.4277 77.8614 -4.5098 0.8129 -14.9029 5.0508 0.2151 4.8675 6.3780 4.1026
F Significance F 15.7874 2.96869E-05
t Stat p-value Lower 90.0% Upper 90.0% 5.4125 7.2E-05 284.9327 557.9227 -5.5476 5.58E-05 -5.9349 -3.0847 -2.9505 0.0099 -23.7573 -6.0485 0.0442 0.9653 -8.3181 8.7484 1.5546 0.1408 -0.8140 13.5702
Model 2 Regression Statistics R Square 0.7768 Adjusted R Square 0.7506 Observations 20 ANOVA df Regression Residual Total
Intercept X1 (Temperature) X2 (Insulation)
SS MS F 2 162958.2277 81479.11 29.5923 17 46807.5222 2753.384 19 209765.75
Significance F 2.9036E-06
Coefficients Standard Error t Stat p-value Lower 95% Upper 95% 489.3227 43.9826 11.1253 3.17E-09 396.5273 582.1180 -5.1103 0.6951 -7.3515 1.13E-06 -6.5769 -3.6437 -14.7195 4.8864 -3.0123 0.0078 -25.0290 -4.4099
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= 0.01 level of significance using Model 1? a) Do not reject H0 and conclude that the amount of insulation has a linear effect on average heating costs. b) Reject H0 and conclude that the amount of insulation does not have a linear effect on average heating costs. c) Reject H0 and conclude that the amount of insulation has a negative linear effect on average heating costs. d) Do not reject H0 and conclude that the amount of insulation has a negative linear effect on average heating costs. ANSWER: c TYPE: MC DIFFICULTY: Easy KEYWORDS: t test on slope, critical value, p-value, decision, conclusion 69. Referring to Table 14-6, what is the 90% confidence interval for the expected change in average heating costs as a result of a 1 degree Fahrenheit change in the daily minimum outside temperature using Model 1? a) [6.58, 3.65] b) [6.24, 2.78] c) [5.94, 3.08] d) [2.37, 15.12] ANSWER: c
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TYPE: MC DIFFICULTY: Moderate KEYWORDS: confidence interval, slope, interpretation 70. Referring to Table 14-6 and allowing for a 1% probability of committing a Type I error, what is the decision and conclusion for the test H 0 : 1 2 3 4 0 vs. H1 : At least one j 0, j 1, 2, , 4 using Model 1? a) Do not reject H0 and conclude that the 4 independent variables have significant individual linear effects on average heating costs. b) Reject H0 and conclude that the 4 independent variables, taken as a group, have significant linear effects on average heating costs. c) Do not reject H0 and conclude that the 4 independent variables, taken as a group, do not have significant linear effects on average heating costs. d) Reject H0 and conclude that the 4 independent variables, taken as a group, do not have significant linear effects on average heating costs. ANSWER: b TYPE: MC DIFFICULTY: Easy KEYWORDS: F test on the entire regression, critical value, p-value, decision, conclusion 71. Referring to Table 14-6, what is the value of the partial F test statistic for H 0 : 3 4 0 vs. H1 : At least one j 0, j 3, 4 ? a) b) c) d)
0.820 1.219 1.382 15.787
ANSWER: b TYPE: MC DIFFICULTY: Difficult KEYWORDS: partial F test, test statistic
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72. Referring to Table 14-6, what are the degrees of freedom of the partial F test for H 0 : 3 4 0 vs. H1 : At least one j 0, j 3, 4 ? a) b) c) d)
2 numerator degrees of freedom and 15 denominator degrees of freedom 15 numerator degrees of freedom and 2 denominator degrees of freedom 2 numerator degrees of freedom and 17 denominator degrees of freedom 17 numerator degrees of freedom and 2 denominator degrees of freedom
ANSWER: a TYPE: MC DIFFICULTY: Moderate KEYWORDS: partial F test, degrees of freedom 73. True or False: The interpretation of the slope is different in a multiple linear regression model as compared to a simple linear regression model. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: slope, interpretation 74. True or False: The coefficient of multiple determination 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 75. True or False: When an additional explanatory variable is introduced into a multiple regression model, the coefficient of multiple determination will never decrease. ANSWER: True TYPE: TF DIFFICULTY: Moderate KEYWORDS: coefficient of multiple determination, properties 76. True or False: When an additional explanatory variable is introduced into a multiple regression model, the adjusted r 2 can never decrease. ANSWER: False TYPE: TF DIFFICULTY: Moderate KEYWORDS: adjusted r-square, properties
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77. True or False: When an explanatory variable is dropped from a multiple regression model, the coefficient of multiple determination can increase. ANSWER: False TYPE: TF DIFFICULTY: Moderate KEYWORDS: coefficient of multiple determination, properties 78. True or False: When an explanatory variable is dropped from a multiple regression model, the adjusted r 2 can increase. ANSWER: True TYPE: TF DIFFICULTY: Moderate KEYWORDS: adjusted r-square, properties 79. True or False: The slopes in a multiple regression model are called net regression coefficients. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: slope 80. True or False: In calculating the standard error of the estimate, SYX MSE , there are n – p – 1 degrees of freedom, where n is the sample size and p represents the number of independent variables in the model. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: standard error of estimate 81. True or False: The total sum of squares (SST) in a regression model will never exceed the regression sum of squares (SSR). ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: sum of squares, properties 82. True or False: The coefficient of multiple determination measures the fraction of the total variation in the dependent variable that is explained by the regression plane. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: coefficient of multiple determination, interpretation 83. True or False: The coefficient of multiple determination is calculated by taking the ratio of the regression sum of squares over the total sum of squares (SSR/SST) and subtracting that value from 1. ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: coefficient of multiple determination, properties
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84. True or False: In a particular model, the sum of the squared residuals was 847. If the model had 5 independent variables, and the data set contained 40 points, the value of the standard error of the estimate is 24.911. ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: sum of squares, standard error of estimate, properties 85. True or False: A multiple regression is called “multiple” because it has several data points. ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: properties, interpretation 86. True or False: A multiple regression is called “multiple” because it has several explanatory variables. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: properties, interpretation 87. True or False: If we have taken into account all relevant explanatory factors, the residuals from a multiple regression should be random. ANSWER: True TYPE: TF DIFFICULTY: Moderate KEYWORDS: residual, properties 88. True or False: Multiple regression is the process of using several independent variables to predict a number of dependent variables. ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: properties 89. True or False: You have just run a regression in which the value of coefficient of multiple determination is 0.57. To determine if this indicates that the independent variables explain a significant portion of the variation in the dependent variable, you would perform an F test. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: F test on the entire regression 90. True or False: From the coefficient of multiple determination, we cannot detect the strength of the relationship between Y and any individual independent variable. ANSWER: True TYPE: TF DIFFICULTY: Moderate KEYWORDS: coefficient of multiple determination, interpretation
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91. True or False: Consider a regression in which b2 = – 1.5 and the standard error of this coefficient equals 0.3. To determine whether X2 is a significant explanatory variable, you would compute an observed t-value of – 5.0. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: t test on slope, test statistic 92. True or False: A regression had the following results: SST = 82.55, SSE = 29.85. It can be said that 73.4% of the variation in the dependent variable is explained by the independent variables in the regression. ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: coefficient multiple determination, sum of squares, properties 93. True or False: A regression had the following results: SST = 82.55, SSE = 29.85. It can be said that 63.84% of the variation in the dependent variable is explained by the independent variables in the regression. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: coefficient multiple determination, sum of squares, properties
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94. True or False: A regression had the following results: SST = 102.55, SSE = 82.04. It can be said that 90.0% of the variation in the dependent variable is explained by the independent variables in the regression. ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: coefficient multiple determination, sum of squares, properties 95. True or False: A regression had the following results: SST = 102.55, SSE = 82.04. It can be said that 20.0% of the variation in the dependent variable is explained by the independent variables in the regression. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: coefficient multiple determination, sum of squares, properties TABLE 14-7 The department head of the accounting department wanted to see if she could predict the GPA of students using the number of course units (credits) and total SAT scores of each. She takes a sample of students and generates the following Microsoft Excel output: SUMMARY OUTPUT Regression Statistics Multiple R 0.916 R Square 0.839 Adjusted R Square 0.732 Standard Error 0.24685 Observations 6 ANOVA Regression Residual Total
df 2 3 5
Coeff Intercept 4.593897 Units – 0.247270 SAT Total 0.001443
SS 0.95219 0.18281 1.13500 StdError 1.13374542 0. 06268485 0.00101241
MS 0.47610 0.06094
t Stat 4.052 – 3.945 1.425
F 7.813
p-value 0.0271 0.0290 0.2494
Signif F 0.0646
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96. Referring to Table 14-7, the estimate of the unit change in the mean of Y per unit change in X1, taking into account the effect of X2, is ________. ANSWER: -0.247 TYPE: FI DIFFICULTY: Easy KEYWORDS: slope, interpretation 97. Referring to Table 14-7, the net regression coefficient of X2 is ________. ANSWER: 0.001443 TYPE: FI DIFFICULTY: Easy KEYWORDS: slope, interpretation 98. Referring to Table 14-7, the predicted GPA for a student carrying 15 course units and who has a total SAT score of 1,100 is ________. ANSWER: 2.472 TYPE: FI DIFFICULTY: Moderate KEYWORDS: prediction of individual values 99. Referring to Table 14-7, the value of the coefficient of multiple determination, r2Y.12, is ________. ANSWER: 0.839 TYPE: FI DIFFICULTY: Easy KEYWORDS: coefficient of multiple determination 100. Referring to Table 14-7, the value of the adjusted coefficient of multiple determination, r2adj, is ________. ANSWER: 0.732 TYPE: FI DIFFICULTY: Easy KEYWORDS: adjusted r-square 101. Referring to Table 14-7, the department head wants to test H0: 1 2 0 . The appropriate alternative hypothesis is ________. ANSWER: at least one i 0 . TYPE: FI DIFFICULTY: Easy KEYWORDS: F test on the entire regression, form of hypothesis 102. Referring to Table 14-7, the department head wants to test H0: 1 2 0 . The critical value of the F test for a level of significance of 0.05 is ________. ANSWER: 9.55 TYPE: FI DIFFICULTY: Easy KEYWORDS: F test on the entire regression, critical value
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Introduction and Data Collection 103. Referring to Table 14-7, the department head wants to test H0: 1 2 0 . The value of the F test statistic is ________. ANSWER: 7.813 TYPE: FI DIFFICULTY: Easy KEYWORDS: F test on the entire regression, test statistic 104. Referring to Table 14-7, the department head wants to test H0: 1 2 0 . The p-value of the test is ________. ANSWER: 0.0646 TYPE: FI DIFFICULTY: Easy KEYWORDS: F test on the entire regression, p-value 105. True or False: Referring to Table 14-7, the department head wants to test H0: 1 2 0 . At a level of significance of 0.05, the null hypothesis is rejected. ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: F test on the entire regression, decision 106. Referring to Table 14-7, the department head wants to use a t test to test for the significance of the coefficient of X1. For a level of significance of 0.05, the critical values of the test are ________. ANSWER: 3.1824 TYPE: FI DIFFICULTY: Easy KEYWORDS: t test on slope, critical value 107. Referring to Table 14-7, the department head wants to use a t test to test for the significance of the coefficient of X1. The value of the test statistic is ________. ANSWER: -3.945 TYPE: FI DIFFICULTY: Easy KEYWORDS: t test on slope, test statistic 108. Referring to Table 14-7, the department head wants to use a t test to test for the significance of the coefficient of X1. The p-value of the test is ________. ANSWER: 0.0290 TYPE: FI DIFFICULTY: Easy KEYWORDS: t test on slope, p-value 109. True or False: Referring to Table 14-7, the department head wants to use a t test to test for the significance of the coefficient of X1. At a level of significance of 0.05, the department head would decide that 1 0 . ANSWER: True TYPE: TF DIFFICULTY: Moderate KEYWORDS: t test on slope, decision, conclusion
Introduction and Data Collection
110. Referring to Table 14-7, the department head decided to obtain a 95% confidence interval for 1 . The confidence interval is from ________ to ________. ANSWER: -0.4468; -0.0478 TYPE: FI DIFFICULTY: Moderate KEYWORDS: slope, confidence interval
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TABLE 14-8 A financial analyst wanted to examine the relationship between salary (in $1,000) and 4 variables: age (X1 = Age), experience in the field (X2 = Exper), number of degrees (X3 = Degrees), and number of previous jobs in the field (X4 = Prevjobs). He took a sample of 20 employees and obtained the following Microsoft Excel output: SUMMARY OUTPUT Regression Statistics Multiple R 0.992 R Square 0.984 Adjusted R Square 0.979 Standard Error 2.26743 Observations 20 ANOVA Regression Residual Total
Intercept Age Exper Degrees Prevjobs
df 4 15 19
SS 4609.83164 77.11836 4686.95000
Coeff – 9.611198 1.327695 – 0.106705 7.311332 – 0.504168
MS 1152.45791 5.14122
StdError 2.77988638 0.11491930 0.14265559 0.80324187 0.44771573
t Stat – 3.457 11.553 – 0.748 9.102 – 1.126
F 224.160
Signif F 0.0001
p-value 0.0035 0.0001 0.4660 0.0001 0.2778
111. Referring to Table 14-8, the estimate of the unit change in the mean of Y per unit change in X4, taking into account the effects of the other 3 variables, is ________. ANSWER: -0.504 TYPE: FI DIFFICULTY: Easy KEYWORDS: slope, interpretation 112. Referring to Table 14-8, the net regression coefficient of X2 is ________. ANSWER: -0.107 TYPE: FI DIFFICULTY: Easy KEYWORDS: slope, interpretation 113. Referring to Table 14-8, the predicted salary for a 35-year-old person with 10 years of experience, 3 degrees, and 1 previous job is ________. ANSWER: 57.2 TYPE: FI DIFFICULTY: Moderate KEYWORDS: prediction of individual values 114. Referring to Table 14-8, the value of the coefficient of multiple determination, r2Y.1234, is ________. ANSWER: 0.984
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TYPE: FI DIFFICULTY: Easy KEYWORDS: coefficient of multiple determination 115. Referring to Table 14-8, the value of the adjusted coefficient of multiple determination, adj r2, is ________. ANSWER: 0.979 TYPE: FI DIFFICULTY: Easy KEYWORDS: adjusted r-square 116. Referring to Table 14-8, the analyst wants to use an F test to test H0 : 1 2 3 4 0 . The appropriate alternative hypothesis is ________. ANSWER: at least one i 0 . TYPE: FI DIFFICULTY: Easy KEYWORDS: F test on the entire regression, form of hypothesis 117. Referring to Table 14-8, the critical value of an F test on the entire regression for a level of significance of 0.01 is ________. ANSWER: 4.89 TYPE: FI DIFFICULTY: Easy KEYWORDS: F test on the entire regression, critical value 118. Referring to Table 14-8, the value of the F statistic for testing the significance of the entire regression is ________. ANSWER: 224.160 TYPE: FI DIFFICULTY: Easy KEYWORDS: F test on the entire regression, test statistic
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119. Referring to Table 14-8, the p-value of the F test for the significance of the entire regression is ________. ANSWER: 0.0001 TYPE: FI DIFFICULTY: Easy KEYWORDS: F test on the entire regression, p-value 120. True or False: Referring to Table 14-8, the F test for the significance of the entire regression performed at a level of significance of 0.01 leads to a rejection of the null hypothesis. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: F test on the entire regression, decision 121. Referring to Table 14-8, the analyst wants to use a t test to test for the significance of the coefficient of X3. For a level of significance of 0.01, the critical values of the test are ________. ANSWER: 2.9467 TYPE: FI DIFFICULTY: Easy KEYWORDS: t test on the slope, critical value 122. Referring to Table 14-8, the analyst wants to use a t test to test for the significance of the coefficient of X3. The value of the test statistic is ________. ANSWER: 9.102 TYPE: FI DIFFICULTY: Easy KEYWORDS: t test on the slope, test statistic 123. Referring to Table 14-8, the analyst wants to use a t test to test for the significance of the coefficient of X3. The p-value of the test is ________. ANSWER: 0.0001 TYPE: FI DIFFICULTY: Easy KEYWORDS: t test on the slope, p-value 124. True or False: Referring to Table 14-8, the analyst wants to use a t test to test for the significance of the coefficient of X3. At a level of significance of 0.01, the department head would decide that 3 0 . ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: t test on the slope, decision, conclusion
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125. Referring to Table 14-8, the analyst decided to obtain a 99% confidence interval for 3 . The confidence interval is from ________ to ________. ANSWER: 4.944; 9.678 TYPE: FI DIFFICULTY: Moderate KEYWORDS: slope, confidence interval TABLE 14-9 You decide to predict gasoline prices in different cities and towns in the United States for your term project. Your dependent variable is price of gasoline per gallon and your explanatory variables are per capita income, the number of firms that manufacture automobile parts in and around the city, the number of new business starts in the last year, population density of the city, percentage of local taxes on gasoline, and the number of people using public transportation. You collect data of 32 cities and obtain a regression sum of squares SSR= 122.8821. Your computed value of standard error of the estimate is 1.9549. 126. Referring to Table 14-9, what is the value of the coefficient of multiple determination? a) 0.2225 b) 0.4576 c) 0.5626 d) 0.6472 ANSWER: c TYPE: MC DIFFICULTY: Easy KEYWORDS: coefficient of multiple determination 127. Referring to Table 14-9, the value of adjusted r 2 is a) 0.4576. b) 0.5626. c) 0.6472. d) 95.5414. ANSWER: a TYPE: MC DIFFICULTY: Moderate KEYWORDS: adjusted r-square
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128. Referring to Table 14-9, if variables that measure the number of new business starts in the last year and population density of the city are removed from the multiple regression model, which of the following is true? a) The adjusted r 2 will definitely increase. b) The adjusted r 2 can never increase. c) The coefficient of multiple determination can never increase. d) The coefficient of multiple determination will definitely increase. ANSWER: c TYPE: MC DIFFICULTY: Moderate KEYWORDS: coefficient of multiple determination, adjusted r-square, properties TABLE 14-10 You worked as an intern at We Always Win Car Insurance Company last summer. You notice that individual car insurance premiums depend very much on the age of the individual, the number of traffic tickets received by the individual, and the population density of the city in which the individual lives. You performed a regression analysis in Excel and obtained the following information: Regression Analysis Regression Statistics Multiple R 0.63 R Square 0.40 Adjusted R Square 0.23 Standard Error 50.00 Observations 15.00 ANOVA df Regression Residual Total
Intercept AGE TICKETS DENSITY
SS 3 11
MS 5994.24
F 2.40
Significance F 0.12
27496.82 45479.54
Coefficients Standard Error 123.80 48.71 -0.82 0.87 21.25 10.66 -3.14 6.46
t Stat 2.54 -0.95 1.99 -0.49
p-value 0.03 0.36 0.07 0.64
Lower 99.0% -27.47 -3.51 -11.86 -23.19
Upper 99.0% 275.07 1.87 54.37 16.91
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129. Referring to Table 14-10, the portion 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 130. Referring to Table 14-10, the adjusted r 2 is _________. ANSWER: 0.23 TYPE: FI DIFFICULTY: Easy KEYWORDS: adjusted r-square 131. Referring to Table 14-10, the standard error of the estimate is _________. ANSWER: 50.0 TYPE: FI DIFFICULTY: Easy KEYWORDS: standard error of estimate 132. Referring to Table 14-10, the estimated average change in insurance premium for every 10 additional tickets received is _____. ANSWER: 212.5 TYPE: FI DIFFICULTY: Difficult KEYWORDS: slope, interpretation 133. Referring to Table 14-10, the 99% confidence interval for the change in average insurance premium of a person who has become 1 year older (i.e., the slope coefficient for AGE) is _________. ANSWER:
-0.82 2.69
TYPE: FI DIFFICULTY: Moderate KEYWORDS: slope, interpretation, confidence interval 134. Referring to Table 14-10, the total degrees of freedom that are missing in the ANOVA table should be ______. ANSWER: 14 TYPE: FI DIFFICULTY: Easy KEYWORDS: degrees of freedom
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135. Referring to Table 14-10, the regression sum of squares that is missing in the ANOVA table should be ______. ANSWER: 17,982.72 TYPE: FI DIFFICULTY: Easy KEYWORDS: sum of squares 136. Referring to Table 14-10, the residual mean squares (MSE) that are missing in the ANOVA table should be _____. ANSWER: 2,499.71 TYPE: FI DIFFICULTY: Easy KEYWORDS: mean squares 137. Referring to Table 14-10, to test the significance of the multiple regression model, what is the form of the null hypothesis? a) H 0 : 0 b) H 0 : 1 c)
H 0 : 1 2 3
d) H 0 : 0 1 2 3 ANSWER: c TYPE: MC DIFFICULTY: Easy KEYWORDS: F test on the entire regression, form of hypothesis 138. Referring to Table 14-10, to test the significance of the multiple regression model, the value of the test statistic is ______. ANSWER: 2.40 TYPE: FI DIFFICULTY: Easy KEYWORDS: F test on the entire regression, test statistic 139. Referring to Table 14-10, to test the significance of the multiple regression model, the p-value of the test statistic in the sample is ______. ANSWER: 0.12 TYPE: FI DIFFICULTY: Easy KEYWORDS: F test on the entire regression, p-value
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140. Referring to Table 14-10, to test the significance of the multiple regression model, what are the degrees of freedom? ANSWER: 3 for the numerator, 11 for the denominator TYPE: PR DIFFICULTY: Easy KEYWORDS: F test on the entire regression, degrees of freedom 141. True or False: Referring to Table 14-10, to test the significance of the multiple regression model, the null hypothesis should be rejected, while allowing for a 1% probability of committing a Type I error. ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: F test on the entire regression, decision 142. True or False: Referring to Table 14-10, the multiple regression model is significant at a 10% level of significance. ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: F test on the entire regression, decision, conclusion TABLE 14-11 A logistic regression model was estimated in order to predict the probability that a randomly chosen university or college would be a private university using information on average total Scholastic Aptitude Test score (SAT) at the university or college, the room and board expense measured in thousands of dollars (Room/Brd), and whether the TOEFL criterion is at least 550 (Toefl550 = 1 if yes, 0 otherwise.) The dependent variable, Y, is school type (Type = 1 if private and 0 otherwise). The Minitab output is given below: Logistic Regression Table Predictor Constant SAT Toefl550 Room/Brd
Coef -27.118 0.015 -0.390 2.078
SE Coef 6.696 0.004666 0.9538 0.5076
Z -4.05 3.17 -0.41 4.09
P 0.000 0.002 0.682 0.000
Odds Ratio 1.01 0.68 7.99
95% CI Lower Upper 1.01 0.10 2.95
Log-Likelihood = -21.883 Test that all slopes are zero: G = 62.083, DF = 3, P-Value = 0.000 Goodness-of-Fit Tests Method Pearson Deviance Hosmer-Lemeshow
Chi-Square 143.551 43.767 15.731
DF 76 76 8
P 0.000 0.999 0.046
1.02 4.39 21.60
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143. Referring to Table 14-11, which of the following is the correct expression for the estimated model? a) Y 27.118 0.015 SAT 0.390 Toefl 550 2.078 Room / Brd b) Yˆ 27.118 0.015 SAT 0.390 Toefl 550 2.078 Room / Brd c) ln odds ratio 27.118 0.015 SAT 0.390 Toefl 550 2.078 Room / Brd d) ln estimated odds ratio 27.118 0.015 SAT 0.390 Toefl 550 2.078 Room / Brd ANSWER: d TYPE: MC DIFFICULTY: Moderate KEYWORDS: logistic regression, estimation 144. Referring to Table 14-11, what is the estimated odds ratio for a school with an average SAT score of 1250, a TOEFL criterion that is at least 550, and a room and board expense of 5 thousand dollars? ANSWER: 5.114 TYPE: PR DIFFICULTY: Moderate KEYWORDS: logistic regression, odds ratio, estimation 145. Referring to Table 14-11, what is the estimated probability that a school with an average SAT score of 1250, a TOEFL criterion that is at least 550, and a room and board expense of 5 thousand dollars will be a private school? ANSWER: 0.836 TYPE: PR DIFFICULTY: Moderate KEYWORDS: logistic regression, odds ratio, estimation 146. Referring to Table 14-11, what is the estimated odds ratio for a school with an average SAT score of 1100, a TOEFL criterion that is not at least 550, and a room and board expense of 7 thousand dollars? ANSWER: 50.805 TYPE: PR DIFFICULTY: Moderate KEYWORDS: logistic regression, odds ratio, estimation 147. Referring to Table 14-11, what is the estimated probability that a school with an average SAT score of 1100, a TOEFL criterion that is not at least 550, and a room and board expense of 7 thousand dollars will be a private school? ANSWER: 0.981 TYPE: PR DIFFICULTY: Moderate KEYWORDS: logistic regression, odds ratio, estimation
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148. Referring to Table 14-11, which of the following is the correct interpretation for the SAT slope coefficient? a) Holding constant the effect of the other variables, the estimated average value of school type increases by 0.015 for each increase of one point in average SAT score. b) Holding constant the effect of the other variables, the estimated school type increases by 0.015 for each increase of one point in average SAT score. c) Holding constant the effect of the other variables, the estimated probability of the school being a private school increases by 0.015 for each increase of one point in average SAT score. d) Holding constant the effect of the other variables, the estimated natural logarithm of the odds ratio of the school being a private school increases by 0.015 for each increase of one point in average SAT score. ANSWER: d TYPE: MC DIFFICULTY: Moderate KEYWORDS: logistic regression, slope, interpretation 149. Referring to Table 14-11, which of the following is the correct interpretation for the Toefl500 slope coefficient? a) Holding constant the effect of the other variables, the estimated average value of school type is 0.39 lower when the school has a TOEFL criterion that is at least 550. b) Holding constant the effect of the other variables, the estimated school type decreases by 0.39 when the school has a TOEFL criterion that is at least 550. c) Holding constant the effect of the other variables, the estimated natural logarithm of the odds ratio of the school being a private school is 0.39 lower for a school that has a TOEFL criterion that is at least 550 than one that does not. d) Holding constant the effect of the other variables, the estimated probability of the school being a private school is 0.39 lower for a school that has a TOEFL criterion that is at least 550 than one that does not. ANSWER: c TYPE: MC DIFFICULTY: Moderate KEYWORDS: logistic regression, slope, interpretation 150. Referring to Table 14-11, what are the degrees of freedom for the chi-square distribution when testing whether the model is a good-fitting model? ANSWER: 76 TYPE: PR DIFFICULTY: Moderate KEYWORDS: logistic regression, deviance statistic, degrees of freedom 151. Referring to Table 14-11, what is the p-value of the test statistic when testing whether the model is a goodfitting model? ANSWER: 0.999 TYPE: PR DIFFICULTY: Moderate KEYWORDS: logistic regression, deviance statistic, p-value 152. True or False: Referring to Table 14-11, the null hypothesis that the model is a good-fitting model cannot be rejected when allowing for a 5% probability of making a Type I error. ANSWER: True TYPE: TF DIFFICULTY: Moderate
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KEYWORDS: logistic regression, deviance statistic, decision 153. True or False: Referring to Table 14-11, there is not enough evidence to conclude that the model is not a good-fitting model at a 0.05 level of significance. ANSWER: True TYPE: TF DIFFICULTY: Moderate KEYWORDS: logistic regression, deviance statistic, conclusion 154. Referring to Table 14-11, what is the p-value of the test statistic when testing whether SAT makes a significant contribution to the model in the presence of the other independent variables? ANSWER: 0.002 TYPE: PR DIFFICULTY: Easy KEYWORDS: logistic regression, Wald statistic, p-value 155. Referring to Table 14-11, what should be the decision („reject‟ or „do not reject‟) on the null hypothesis when testing whether SAT makes a significant contribution to the model in the presence of the other independent variables at a 0.05 level of significance? ANSWER: Reject TYPE: PR DIFFICULTY: Easy KEYWORDS: logistic regression, Wald statistic, decision 156. True or False: Referring to Table 14-11, there is not enough evidence to conclude that SAT makes a significant contribution to the model in the presence of the other independent variables at a 0.05 level of significance. ANSWER: False TYPE: TF DIFFICULTY: moderate KEYWORDS: logistic regression, Wald statistic, conclusion
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157. Referring to Table 14-11, what is the p-value of the test statistic when testing whether Toefl500 makes a significant contribution to the model in the presence of the other independent variables? ANSWER: 0.682 TYPE: PR DIFFICULTY: Easy KEYWORDS: logistic regression, Wald statistic, p-value 158. Referring to Table 14-11, what should be the decision („reject‟ or „do not reject‟) on the null hypothesis when testing whether Toefl500 makes a significant contribution to the model in the presence of the other independent variables at a 0.05 level of significance? ANSWER: Do not reject TYPE: PR DIFFICULTY: Easy KEYWORDS: logistic regression, Wald statistic, decision 159. True or False: Referring to Table 14-11, there is not enough evidence to conclude that Toefl500 makes a significant contribution to the model in the presence of the other independent variables at a 0.05 level of significance. ANSWER: True TYPE: TF DIFFICULTY: moderate KEYWORDS: logistic regression, Wald statistic, conclusion 160. 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 161. If a categorical independent variable contains 4 categories, then _________ dummy variable(s) will be needed to uniquely represent these categories. a) 1 b) 2 c) 3 d) 4 ANSWER: c TYPE: MC DIFFICULTY: Easy KEYWORDS: dummy variable, properties
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162. 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 163. 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 164. 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
What is the correct interpretation for the estimated coefficient for SEX? a) Holding everything else fixed, personal consumption for females is estimated to be $0.39 higher than males, on the average. b) Holding everything else fixed, personal consumption for males is estimated to be $0.39 higher than females, on the average. c) Holding everything else fixed, personal consumption for females is estimated to be 0.39% higher than males, on the average. d) Holding everything else fixed, personal consumption for males is estimated to be 0.39% higher than females, on the average. ANSWER: a TYPE: MC DIFFICULTY: Easy KEYWORDS: dummy variable, slope, interpretation TABLE 14-12
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A weight loss clinic wants to use regression analysis to build a model for the weight loss of a client (measured in pounds). Two variables thought to affect weight loss are client‟s length of time on the weight loss program and time of session. These variables are described below: Y = Weight loss (in pounds) X1 = Length of time in weight loss program (in months) X2 = 1 if morning session, 0 if not X3 = 1 if afternoon session, 0 if not (Base level = evening session) Data for 12 clients on a weight loss program at the clinic were collected and used to fit the interaction model:
Y 0 1 X1 2 X2 3 X 3 4 X1 X2 5 X1 X3
Partial output from Microsoft Excel follows: Regression Statistics Multiple R 0.73514 R Square 0.540438 Adjusted R Square 0.157469 Standard Error 12.4147 Observations 12 ANOVA F = 5.41118
Intercept Length (X1) Morn Ses (X2) Aft Ses (X3) Length*Morn Ses Length*Aft Ses
Significance F = 0.040201 Coeff 0.089744 6.22538 2.217272 11.8233 0.77058 – 0.54147
StdError 14.127 2.43473 22.1416 3.1545 3.562 3.35988
t Stat 0.0060 2.54956 0.100141 3.558901 0.216334 – 0.161158
p-value 0.9951 0.0479 0.9235 0.0165 0.8359 0.8773
165. Referring to Table 14-12, 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
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166. Referring to Table 14-12, 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 167. Referring to Table 14-12, in terms of the 's in the model, give the average change in weight loss (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 168. Referring to Table 14-12, in terms of the 's in the model, give the average change in weight loss (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
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169. Referring to Table 14-12, in terms of the 's in the model, give the average change in weight loss (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 170. True or False: Referring to Table 14-12, 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 171. Referring to Table 14-12, which of the following statements is supported by the analysis shown? a) There is sufficient evidence (at = 0.05) of curvature in the relationship between weight loss (Y) and months in program(X1). b) There is sufficient evidence (at = 0.05) to indicate that the relationship between weight loss (Y) and months in program (X1) depends on session time. c) There is sufficient evidence (at = 0.10) to indicate that the session time (morning, afternoon, evening) affects weight loss (Y). d) There is insufficient evidence (at = 0.10) to indicate that the relationship between weight loss (Y) and months in program(X1) depends on session time. ANSWER: d TYPE: MC DIFFICULTY: Difficult KEYWORDS: interaction, t test on slope, decision, conclusion 172. 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
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TABLE 14-13 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 x1i 2 x2i 3 x3i i where Y is the meter price x1 is the number of blocks to the quad x2 is a dummy variable that takes the value 1 if the meter is located in downtown and off campus and the value 0 otherwise x3 is a dummy variable that takes the value 1 if the meter is located outside of downtown and off campus, and the value 0 otherwise The following Excel results are obtained: Regression Statistics Multiple R 0.9659 R Square 0.9331 Adjusted R Square 0.9294 Standard Error 0.0327 Observations 58 ANOVA Df Regression 3 Residual 54 Total 57
Intercept X1 X2 X3
SS 0.8094 0.0580 0.8675
Coefficient s Standard Error 0.5118 0.0136 -0.0045 0.0034 -0.2392 0.0123 -0.0002 0.0123
MS 0.2698 0.0010
t Stat 37.4675 -1.3276 -19.3942 -0.0214
F Significance F 251.1995 1.0964E-31
p-value 2.4904 0.1898 5.3581E-26 0.9829
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173. Referring to Table 14-13, 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 174. Referring to Table 14-13, 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 175. Referring to Table 14-13, 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 176. 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 177. 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
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178. True or False: In trying to obtain a model to estimate grades on a statistics test, a professor wanted to include, among other factors, whether the person had taken the course previously. To do this, the professor included a dummy variable in her regression that was equal to 1 if the person had previously taken the course, and 0 otherwise. The interpretation of the coefficient associated with this dummy variable would be the average amount the repeat students tended to be above or below non-repeaters, with all other factors the same. ANSWER: True TYPE: TF DIFFICULTY: Moderate KEYWORDS: dummy variable, slope, interpretation TABLE 14-14 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 Y = demand in $100 per capita
179. Referring to Table 14-14, 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 180. Referring to Table 14-14, 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
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181. Referring to Table 14-14, 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 182. Referring to Table 14-14, 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 183. Referring to Table 14-14, 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 184. Referring to Table 14-14, the fitted model for predicting demand in San Francisco is ________. a) 10 + 5X1 b) 10 + 13X1 c) 15 + 8X2 d) 18 + 5X1 ANSWER: d TYPE: MC DIFFICULTY: Moderate KEYWORDS: prediction of individual values, dummy variable
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TABLE 14-15 An automotive engineer would like to be able to predict automobile mileages. She believes that the 2 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 185. Referring to Table 14-15, the predicted mileage for a 300 horsepower, 6-cylinder car is ________. ANSWER: 25 TYPE: FI DIFFICULTY: Moderate KEYWORDS: prediction of individual values, dummy variable 186. Referring to Table 14-15, the predicted mileage for a 200 horsepower, 4-cylinder car is ________. ANSWER: 30 TYPE: FI DIFFICULTY: Moderate KEYWORDS: prediction of individual values, dummy variable 187. Referring to Table 14-15, the fitted model for predicting mileages for 6-cylinder cars is ________. a) 40 – 0.05X1 b) 40 – 0.10X1 c) 60 – 0.10X1 d) 60 – 0.15X1 ANSWER: a TYPE: MC DIFFICULTY: Moderate KEYWORDS: prediction of individual values, dummy variable 188. Referring to Table 14-15, 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
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CHAPTER 15: MULTIPLE REGRESSION MODEL BUILDING 189. 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 measured in years. The builder randomly selects 50 families and runs the multiple regression. The business literature involving human capital shows that education influences an individual‟s annual income. Combined, these may influence family size. With this in mind, what should the real estate builder be particularly concerned with when analyzing the multiple regression model? a) randomness of error terms b) collinearity c) normality of residuals d) missing observations ANSWER: b TYPE: MC DIFFICULTY: Moderate KEYWORDS: collinearity, assumption 190. 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. A statistical analyst discovers that capital spending by corporations has a significant inverse relationship with wage spending. What should the microeconomist who developed this multiple regression model be particularly concerned with? a) randomness of error terms b) collinearity c) normality of residuals d) missing observations ANSWER: b TYPE: MC DIFFICULTY: Moderate KEYWORDS: collinearity, assumption 191. In multiple regression, the __________ procedure permits variables to enter and leave the model at different stages of its development. a) forward selection b) residual analysis c) backward elimination d) stepwise regression ANSWER: d TYPE: MC DIFFICULTY: Easy KEYWORDS: stepwise regression, model building
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192. a) b) c) d)
A regression diagnostic tool used to study the possible effects of collinearity is the slope. the Y intercept. the VIF. the standard error of the estimate.
ANSWER: c TYPE: MC DIFFICULTY: Easy KEYWORDS: variance inflationary factor, collinearity 193. a) b) c) d)
Which of the following is not used to find a "best" model? adjusted r2 Mallow's Cp odds ratio all of the above
ANSWER: c TYPE: MC DIFFICULTY: Moderate KEYWORDS: model building 194. a) b) c) d)
The Variance Inflationary Factor (VIF) measures the correlation of the X variables with the Y variable. correlation of the X variables with each other. contribution of each X variable with the Y variable after all other X variables are included in the model. standard deviation of the slope.
ANSWER: b TYPE: MC DIFFICULTY: Easy KEYWORDS: variance inflationary factor, collinearity 195.
The C p statistic is used a. b. c. d.
to determine if there is a problem of collinearity. if the variances of the error terms are all the same in a regression model. to choose the best model. to determine if there is an irregular component in a time series.
ANSWER: c TYPE: MC DIFFICULTY: Easy KEYWORDS: C-p statistic, model building
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TABLE 15-1 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
196.
Referring to Table 15-1, what is the correct interpretation for the estimated coefficient for ADVT? a) A $1 increase in per person advertising expenditure by the manufacturer will result in an estimated average increase of $2.35 on personal consumption, holding other variables constant. b) A 100% increase in per person advertising expenditure by the manufacturer will result in an estimated average increase of $2.35 on personal consumption, holding other variables constant. c) A 100% increase in per person advertising expenditure by the manufacturer will result in an estimated average increase of 2.35% on personal consumption, holding other variables constant. d) A 1% increase in per person advertising expenditure by the manufacturer will result in an estimated average increase of 2.35% on personal consumption, holding other variables constant.
ANSWER: b TYPE: MC DIFFICULTY: Difficult KEYWORDS: transformation, slope, interpretation
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197.
Referring to Table 15-1, what is the correct interpretation for the estimated coefficient for APR? a) A one percentage point increase in average annualized percentage interest rate will result in an estimated average increase of $5.77 on personal consumption, holding other variables constant. b) A 100% increase in average annualized percentage interest rate will result in an estimated average increase of 5.77% on personal consumption, holding other variables constant. c) A 100% increase in average annualized percentage interest rate will result in an estimated average increase of $5.77 on personal consumption, holding other variables constant. d) A 1% increase in average annualized percentage interest rate will result in an estimated average increase of 5.77% on personal consumption, holding other variables constant.
ANSWER: c TYPE: MC DIFFICULTY: Moderate KEYWORDS: transformation, slope, interpretation TABLE 15-2 A certain type of rare gem serves as a status symbol for many of its owners. In theory, for low prices, the demand decreases as the price of the gem increases. However, experts hypothesize that when the gem is valued at very high prices, the demand increases with price due to the status owners believe they will gain by obtaining the gem. Thus, the model proposed to best explain the demand for the gem by its price is the quadratic model:
Y 0 1 X 2 X 2
where Y = demand (in thousands) and X = retail price per carat. This model was fit to data collected for a sample of 12 rare gems of this type. A portion of the computer analysis obtained from Microsoft Excel is shown below: SUMMARY OUTPUT Regression Statistics Multiple R 0.994 R Square 0.988 Standard Error 12.42 Observations 12 ANOVA Regression Residual Total
Intercept Price Price Sq
df 2 9 11 Coeff 286.42 – 0.31 0.000067
SS 115145 1388 116533 StdError 9.66 0.06 0.00007
MS 57573 154
t Stat 29.64 – 5.14 0.95
F 373
Signif F 0.0001
p-value 0.0001 0.0006 0.3647
198. Referring to Table 15-2, what is the value of the test statistic for testing whether there is an upward curvature in the response curve relating the demand (Y) and the price (X)? a) -5.14 b) 0.95 c) 373 d) none of the above ANSWER: b
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TYPE: MC DIFFICULTY: Easy KEYWORDS: quadratic regression, t test on slope, test statistic 199. Referring to Table 15-2, what is the p-value associated with the test statistic for testing whether there is an upward curvature in the response curve relating the demand (Y) and the price (X)? a) 0.0001 b) 0.0006 c) 0.3647 d) none of the above ANSWER: c TYPE: MC DIFFICULTY: Easy KEYWORDS: quadratic regression, t test on slope, p-value 200. a) b) c) d)
Referring to Table 15-2, what is the correct interpretation of the coefficient of multiple determination? 98.8% of the total variation in demand can be explained by the linear relationship between demand and price. 98.8% of the total variation in demand can be explained by the quadratic relationship between demand and price. 98.8% of the total variation in demand can be explained by the addition of the square term in price. 98.8% of the total variation in demand can be explained by just the square term in price.
ANSWER: b TYPE: MC DIFFICULTY: Moderate KEYWORDS: coefficient of multiple determination, interpretation
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201. Referring to Table 15-2, does there appear to be significant upward curvature in the response curve relating the demand (Y) and the price (X) at 10% level of significance? a) Yes, since the p-value for the test is less than 0.10. b) No, since the value of 2 is near 0. c) No, since the p-value for the test is greater than 0.10. d) Yes, since the value of 2 is positive. ANSWER: c TYPE: MC DIFFICULTY: Moderate KEYWORDS: quadratic regression, t test on slope, decision, conclusion 202. True or False: Referring to Table 15-2, a more parsimonious simple linear model is likely to be statistically superior to the fitted curvilinear for predicting sale price (Y). ANSWER: True TYPE: TF DIFFICULTY: Moderate KEYWORDS: quadratic regression, t test on slope, interpretation
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TABLE 15-3 In Hawaii, condemnation proceedings are underway to enable private citizens to own the property that their homes are built on. Until recently, only estates were permitted to own land, and homeowners leased the land from the estate. In order to comply with the new law, a large Hawaiian estate wants to use regression analysis to estimate the fair market value of the land. Each of the following 3 models were fit to data collected for n = 20 properties, 10 of which are located near a cove. 2 2 Model 1: Y 0 1 X1 2 X2 3 X1 X2 4 X1 5 X1 X2 where Y = Sale price of property in thousands of dollars X1 = Size of property in thousands of square feet X2 = 1 if property located near cove, 0 if not Using the data collected for the 20 properties, the following partial output obtained from Microsoft Excel is shown: SUMMARY OUTPUT Regression Statistics Multiple R 0.985 R Square 0.970 Standard Error 9.5 Observations 20 ANOVA Regression Residual Total
Intercept Size Cove Size*Cove SizeSq SizeSq*Cove
df 5 14 19
SS 28324 1279 29063
MS 5664 91
Coeff StdError – 32.1 35.7 12.2 5.9 – 104.3 53.5 17.0 8.5 – 0.3 0.2 – 0.3 0.3
F 62.2
t Stat – 0.90 2.05 – 1.95 1.99 – 1.28 – 1.13
Signif F 0.0001
p-value 0.3834 0.0594 0.0715 0.0661 0.2204 0.2749
203. Referring to Table 15-3, is the overall model statistically adequate at a 0.05 level of significance for predicting sale price (Y)? a) No, since some of the t tests for the individual variables are not significant. b) No, since the standard deviation of the model is fairly large. c) Yes, since none of the -estimates are equal to 0. d) Yes, since the p-value for the test is smaller than 0.05. ANSWER: d TYPE: MC DIFFICULTY: Easy KEYWORDS: F test on the entire regression, p-value, decision, conclusion
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204. Referring to Table 15-3, given a quadratic relationship between sale price (Y) and property size (X1), what null hypothesis would you test to determine whether the curves differ from cove and non-cove properties? a) H0 : 2 3 5 0 b) H0 : 4 5 0 c) H0 : 3 5 0 d) H0 : 2 0 ANSWER: c TYPE: MC DIFFICULTY: Difficult KEYWORDS: interaction, partial F test, form of hypothesis 205. Referring to Table 15-3, given a quadratic relationship between sale price (Y) and property size (X1), what test should be used to test whether the curves differ from cove and non-cove properties? a) F test for the entire regression model. b) t test on each of the coefficients in the entire regression model. c) Partial F test on the subset of the appropriate coefficients. d) t test on each of the subsets of the appropriate coefficients. ANSWER: c TYPE: MC DIFFICULTY: Difficult KEYWORDS: interaction, partial F test, interpretation 206. If a group of independent variables are not significant individually, but are significant as a group at a specified level of significance, this is most likely due to a) autocorrelation. b) the presence of dummy variables. c) the absence of dummy variables. d) collinearity. ANSWER: d TYPE: MC DIFFICULTY: Easy KEYWORDS: collinearity, assumption, properties
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207. 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 0 1 x1i 2 x2i 3 x3i i where Y is the meter price x1 is the number of blocks to the quad x2 is a dummy variable that takes the value 1 if the meter is located in downtown and off campus and the value 0 otherwise x3 is a dummy variable that takes the value 1 if the meter is located outside of downtown and off campus, and the value 0 otherwise Suppose that whether the meter is located on campus is an important explanatory factor. Why should the variable that depicts this attribute not be included in the model? a) Its inclusion will introduce autocorrelation. b) Its inclusion will introduce collinearity. c) Its inclusion will inflate the standard errors of the estimated coefficients. d) both (b) and (c) ANSWER: d TYPE: MC DIFFICULTY: Moderate KEYWORDS: dummy variable, collinearity, properties 208. True or False: The Variance Inflationary Factor (VIF) measures the correlation of the X variables with the Y variable. ANSWER: False TYPE: TF DIFFICULTY: Moderate KEYWORDS: variance inflationary factor, collinearity 209. True or False: Collinearity is present when there is a high degree of correlation between independent variables. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: collinearity 210. True or False: Collinearity is present when there is a high degree of correlation between the dependent variable and any of the independent variables. ANSWER: False TYPE: TF DIFFICULTY: Moderate KEYWORDS: collinearity, properties
462
Introduction and Data Collection 211. True or False: A high value of R2 significantly above 0 in multiple regression, accompanied by insignificant t-values on all parameter estimates, very often indicates a high correlation between independent variables in the model. ANSWER: True TYPE: TF DIFFICULTY: Difficult KEYWORDS: collinearity, properties 212. True or False: One of the consequences of collinearity in multiple regression is inflated standard errors in some or all of the estimated slope coefficients. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: collinearity, properties 213. True or False: One of the consequences of collinearity in multiple regression is biased estimates on the slope coefficients. ANSWER: False TYPE: TF DIFFICULTY: Difficult KEYWORDS: collinearity, properties 214. True or False: Collinearity is present if the dependent variable is linearly related to one of the explanatory variables. ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: collinearity, properties 215. True or False: Collinearity will result in excessively low standard errors of the parameter estimates reported in the regression output. ANSWER: False TYPE: TF DIFFICULTY: Difficult KEYWORDS: collinearity, properties 216. True or False: The parameter estimates are biased when collinearity is present in a multiple regression equation. ANSWER: False TYPE: TF DIFFICULTY: Difficult KEYWORDS: collinearity, properties
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217. True or False: Two simple regression models were used to predict a single dependent variable. Both models were highly significant, but when the two independent variables were placed in the same multiple regression model for the dependent variable, R2 did not increase substantially and the parameter estimates for the model were not significantly different from 0. This is probably an example of collinearity. ANSWER: True TYPE: TF DIFFICULTY: Moderate KEYWORDS: collinearity, properties 218. True or False: So that we can fit curves as well as lines by regression, we often use mathematical manipulations for converting one variable into a different form. These manipulations are called dummy variables. ANSWER: False TYPE: TF DIFFICULTY: Moderate KEYWORDS: quadratic regression, transformation TABLE 15-4 A chemist employed by a pharmaceutical firm has developed a muscle relaxant. She took a sample of 14 people suffering from extreme muscle constriction. She gave each a vial containing a dose (X) of the drug and recorded the time to relief (Y) measured in seconds for each. She fit a “centered” curvilinear model to this data. The results obtained by Microsoft Excel follow, where the dose (X) given has been “centered.” SUMMARY OUTPUT Regression Statistics Multiple R 0.747 R Square 0.558 Adjusted R Square 0.478 Standard Error 863.1 Observations 14 ANOVA Regression Residual Total
Intercept CenDose CenDoseSq
df 2 11 13
SS 10344797 8193929 18538726
MS 5172399 744903
F
Coeff StdError t Stat 1283.0 352.0 3.65 25.228 8.631 2.92 0.8604 0.3722 2.31
6.94
Signif F 0.0110
p-value 0.0040 0.0140 0.0410
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219. Referring to Table 15-4, the prediction of time to relief for a person receiving a dose of the drug 10 units above the average dose (i.e., the prediction of Y for X = 10), is ________. ANSWER: 1,621 TYPE: FI DIFFICULTY: Moderate KEYWORDS: quadratic regression, prediction of individual values 220. Referring to Table 15-4, suppose the chemist decides to use an F test to determine if there is a significant curvilinear relationship between time and dose. The p-value of the test is ________. ANSWER: 0.041 TYPE: FI DIFFICULTY: Difficult KEYWORDS: quadratic regression, partial F test, p-value 221. Referring to Table 15-4, suppose the chemist decides to use an F test to determine if there is a significant curvilinear relationship between time and dose. The value of the test statistic is ________. ANSWER: 2.31**2 or 5.3361 TYPE: FI DIFFICULTY: Moderate KEYWORDS: quadratic regression, partial F test, test statistic 222. True or False: Referring to Table 15-4, suppose the chemist decides to use an F test to determine if there is a significant curvilinear relationship between time and dose. If she chooses to use a level of significance of 0.05, she would decide that there is a significant curvilinear relationship. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: quadratic regression, partial F test, decision, conclusion 223. True or False: Referring to Table 15-4, suppose the chemist decides to use an F test to determine if there is a significant curvilinear relationship between time and dose. If she chooses to use a level of significance of 0.01, she would decide that there is a significant curvilinear relationship. ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: quadratic regression, partial F test, conclusion
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224. Referring to Table 15-4, suppose the chemist decides to use a t test to determine if there is a significant difference between a linear model and a curvilinear model that includes a linear term. The p-value of the test statistic for the contribution of the curvilinear term is ________. ANSWER: 0.041 TYPE: FI DIFFICULTY: Moderate KEYWORDS: quadratic regression, t test on slope, p-value 225. Referring to Table 15-4, suppose the chemist decides to use a t test to determine if there is a significant difference between a curvilinear model without a linear term and a curvilinear model that includes a linear term. The value of the test statistic is ______. ANSWER: 2.92 TYPE: FI DIFFICULTY: Moderate KEYWORDS: quadratic regression, t test on slope, test statistic 226. Referring to Table 15-4, suppose the chemist decides to use a t test to determine if there is a significant difference between a curvilinear model without a linear term and a curvilinear model that includes a linear term. The p-value of the test is ______. ANSWER: 0.0140 TYPE: FI DIFFICULTY: Moderate KEYWORDS: quadratic regression, t test on slope, p-value 227. True or False: Referring to Table 15-4, suppose the chemist decides to use a t test to determine if there is a significant difference between a linear model and a curvilinear model that includes a linear term. If she used a level of significance of 0.05, she would decide that the linear model is sufficient. ANSWER: False TYPE: TF DIFFICULTY: Moderate KEYWORDS: quadratic regression, t test on slope, decision 228. True or False: Referring to Table 15-4, suppose the chemist decides to use a t test to determine if there is a significant difference between a linear model and a curvilinear model that includes a linear term. If she used a level of significance of 0.02, she would decide that the linear model is sufficient. ANSWER: True TYPE: TF DIFFICULTY: Moderate KEYWORDS: quadratic regression, t test on slope, decision
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229. True or False: Referring to Table 15-4, suppose the chemist decides to use a t test to determine if there is a significant difference between a curvilinear model without a linear term and a curvilinear model that includes a linear term. Using a level of significance of 0.05, she would decide that the curvilinear model should include a linear term.
ANSWER: True TYPE: TF DIFFICULTY: Moderate KEYWORDS: quadratic regression, t test on slope, decision 230. In multiple regression, the __________ procedure permits variables to enter and leave the model at different stages of its development. ANSWER: stepwise regression TYPE: FI DIFFICULTY: Easy KEYWORDS: stepwise regression 231.
A regression diagnostic tool used to study the possible effects of collinearity is ______.
ANSWER: VIF TYPE: FI DIFFICULTY: Moderate KEYWORDS: collinearity, variance inflationary factor 232.
The _______ the value of the Variance Inflationary Factor, the higher the collinearity of the X variables.
ANSWER: larger TYPE: FI DIFFICULTY: Moderate KEYWORDS: variance inflationary factor, collinearity, properties 233. a) b) c) d)
The logarithm transformation can be used to overcome violations of the autocorrelation assumption. to test for possible violations of the autocorrelation assumption. to overcome violations of the homoscedasticity assumption. to test for possible violations of the homoscedasticity assumption.
ANSWER: c TYPE: MC DIFFICULTY: Moderate KEYWORDS: transformation, autocorrelation, homoscedasticity, assumption
Introduction and Data Collection
234. a) b) c) d)
The logarithm transformation can be used to overcome violations of the autocorrelation assumption. to test for possible violations of the autocorrelation assumption. to change a nonlinear model into a linear model. to change a linear independent variable into a nonlinear independent variable.
ANSWER: c TYPE: MC DIFFICULTY: Moderate KEYWORDS: transformation, autocorrelation, homoscedasticity, assumption 235. a) b) c) d)
Which of the following will NOT change a nonlinear model into a linear model? quadratic regression model logarithmic transformation square-root transformation variance inflationary factor
ANSWER: d TYPE: MC DIFFICULTY: Easy KEYWORDS: transformation 236. model? a) b) c) d)
Which of the following is used to determine observations that have an influential effect on the fitted Durbin-Watson statistic variance inflationary factor the Cp statistic Cook‟s distance statistic
ANSWER: d TYPE: MC DIFFICULTY: Easy KEYWORDS: influence analysis 237. Which of the following is NOT used to determine observations that have an influential effect on the fitted model? a) the hat matrix elements hi b) the Studentized deleted residuals ti c) the Cp statistic d) Cook‟s distance statistic ANSWER: c TYPE: MC DIFFICULTY: Easy KEYWORDS: influence analysis
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238. Using the hat matrix elements hi to determine influential points in a multiple regression model with k independent variable and n observations, Xi is an influential point if a) hi 2 k 1 / n b) hi 2 k 1 / n c)
hi n k 1 / 2
d) hi n k 1 / 2 ANSWER: b TYPE: MC DIFFICULTY: Easy KEYWORDS: influence analysis 239. Using the Studentized residuals ti to determine influential points in a multiple regression model with k independent variable and n observations, and letting tn-k-2 denote the upper critical value of a two-tail t test with a 0.10 level of significance, Xi is an influential point if a) ti tn k 1 b)
ti tn k 1
c)
ti tn k 2
d)
ti tn k 2
ANSWER: d TYPE: MC DIFFICULTY: Easy KEYWORDS: influence analysis 240. Using the Cook‟s distance statistic Di to determine influential points in a multiple regression model with k independent variable and n observations and letting F1 , 2 denote the critical value of an F distribution with 1 and 2 degrees of freedom at a 0.50 level of significance, Xi is an influential point if a)
Di Fk 1,nk 1
b) Di Fk 1,nk 1 c)
Di Fnk 1,k 1
d) Di Fnk 1,k 1 ANSWER: b TYPE: MC DIFFICULTY: Easy KEYWORDS: influence analysis
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241. True or False: Only when all three of the hat matrix elements hi, the Studentized deleted residuals ti, and the Cook‟s distance statistic Di, reveal a consistent result, should an observation be removed from the regression analysis. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: influence analysis 242.
An independent variable Xj is considered highly correlated with the other independent variables if a) VIFj 5 b) VIFj 5 c) VIFj VIFi for i j d) VIFj VIFi for i j
ANSWER: b TYPE: MC DIFFICULTY: Easy KEYWORDS: variance inflationary factor, collinearity 243. True or False: The goals of model building are to find a good model with the fewest independent variables that is easier to interpret and has a lower probability of collinearity. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: model building, collinearity 244.
Using the best-subsets approach to model building, models are being considered when their a)
Cp k
b) C p k c)
C p k 1
d) C p k 1 ANSWER: d TYPE: MC DIFFICULTY: Easy KEYWORDS: model building, C-p statistic
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245. True or False: In data mining, where huge data sets are being explored to discover relationships among a large number of variables, the best-subsets approach is more practical than the stepwise regression approach. ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: model building, stepwise regression, C-p 246.
True or False: The stepwise regression approach takes into consideration all possible models.
ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: model building, stepwise regression 247. True or False: In stepwise regression, an independent variable is not allowed to be removed from the model once it has been entered into the model. ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: model building, stepwise regression 248.
True or False: Using the Cp statistic in model building, all models with C p k 1 are equally good.
ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: model building, C-p
Introduction and Data Collection
CHAPTER 16: TIME-SERIES FORECASTING AND INDEX NUMBERS
1. The effect of an unpredictable, rare event will be contained in the ___________ component. a) trend b) cyclical c) irregular d) seasonal ANSWER: c TYPE: MC DIFFICULTY: Easy KEYWORDS: component factors, properties 2. The overall upward or downward pattern of the data in an annual time series will be contained in the ____________ component. a) trend b) cyclical c) irregular d) seasonal ANSWER: a TYPE: MC DIFFICULTY: Easy KEYWORDS: component factors, properties 3. The fairly regular fluctuations that occur within each year would be contained in the ____________ component. a) trend b) cyclical c) irregular d) seasonal ANSWER: d TYPE: MC DIFFICULTY: Easy KEYWORDS: component factors, properties 4. The annual multiplicative time-series model does not possess _______ component. a) a trend b) a cyclical c) an irregular d) a seasonal ANSWER: d TYPE: MC DIFFICULTY: Easy KEYWORDS: component factors, properties
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5. Based on the following scatter plot, which of the time-series components is not present in this quarterly time series? 350 300 Stock Returns
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250 200 150 100 50 0 0
10
20
30
40
50
60
Quarters
e. f. g. h.
trend seasonal cyclical irregular
ANSWER: c TYPE: MC DIFFICULTY: Easy KEYWORDS: component factors, properties 6. The method of moving averages is used a) to plot a series. b) to exponentiate a series. c) to smooth a series. d) in regression analysis. ANSWER: c TYPE: MC DIFFICULTY: Easy KEYWORDS: moving averages 7. When using the exponentially weighted moving average for purposes of forecasting rather than smoothing, a) the previous smoothed value becomes the forecast. b) the current smoothed value becomes the forecast. c) the next smoothed value becomes the forecast. d) none of the above ANSWER: b TYPE: MC DIFFICULTY: Moderate KEYWORDS: exponential smoothing, properties
Introduction and Data Collection
8. In selecting an appropriate forecasting model, the following approaches are suggested: a) Perform a residual analysis. b) Measure the size of the forecasting error. c) Use the principle of parsimony. d) all of the above ANSWER: d TYPE: MC DIFFICULTY: Moderate KEYWORDS: model selection 9. To assess the adequacy of a forecasting model, one measure that is often used is a) quadratic trend analysis. b) the MAD. c) exponential smoothing. d) moving averages. ANSWER: b TYPE: MC DIFFICULTY: Easy KEYWORDS: model selection 10. Which of the following methods should not be used for short-term forecasts into the future? a) exponential smoothing b) moving averages c) linear trend model d) autoregressive modeling ANSWER: b TYPE: MC DIFFICULTY: Moderate KEYWORDS: moving averages, properties 11. A model that can be used to make predictions about long-term future values of a time series is a) linear trend. b) quadratic trend. c) exponential trend. d) all of the above ANSWER: d TYPE: MC DIFFICULTY: Easy KEYWORDS: least-squares trend fitting, properties
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12. You need to decide whether you should invest in a particular stock. You would like to invest if the price is likely to rise in the long run. You have data on the daily average price of this stock over the past 12 months. Your best action is to a) compute moving averages. b) perform exponential smoothing. c) estimate a least square trend model. d) compute the MAD statistic. ANSWER: c TYPE: MC DIFFICULTY: Moderate KEYWORDS: least-squares trend fitting, properties 13. When a time series appears to be increasing at an increasing rate, such that the percentage difference from observation to observation is constant, the appropriate model to fit is the a. linear trend. b. quadratic trend. c. exponential trend. d. none of the above ANSWER: c TYPE: MC DIFFICULTY: Easy KEYWORDS: least-squares trend fitting, model selection 14. The method of least squares is used on time-series data for a) eliminating irregular movements. b) deseasonalizing the data. c) obtaining the trend equation. d) exponentially smoothing a series. ANSWER: c TYPE: MC DIFFICULTY: Moderate KEYWORDS: least-squares trend fitting, properties 15. Which of the following statements about the method of exponential smoothing is NOT true? a) It gives greater weight to more recent data. b) It can be used for forecasting. c) It uses all earlier observations in each smoothing calculation. d) It gives greater weight to the earlier observations in the series. ANSWER: d TYPE: MC DIFFICULTY: Moderate KEYWORDS: exponential smoothing, properties
Introduction and Data Collection
16. Which of the following is NOT an advantage of exponential smoothing? a) It enables us to perform one-period ahead forecasting. b) It enables us to perform more than one-period ahead forecasting. c) It enables us to smooth out seasonal components. d) It enables us to smooth out cyclical components. ANSWER: b TYPE: MC DIFFICULTY: Moderate KEYWORDS: exponential smoothing, properties 17. Which of the following statements about moving averages is NOT true? a) They can be used to smooth a series. b) They give equal weight to all values in the computation. c) They are simpler than the method of exponential smoothing. d) They give greater weight to more recent data. ANSWER: d TYPE: MC DIFFICULTY: Moderate KEYWORDS: moving averages, properties 18. After estimating a trend model for annual time-series data, you obtain the following residual plot against time; the problem with your model is that: a) the cyclical component has not been accounted for. b) the seasonal component has not been accounted for. c) the trend component has not been accounted for. d) the irregular component has not been accounted for. 1 0.8
Residuals
0.6 0.4 0.2 0 -0.2
0
2
4
6
-0.4 -0.6 -0.8 Time (Year)
ANSWER: a TYPE: MC DIFFICULTY: Moderate KEYWORDS: component factors
8
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TABLE 16-1 The number of cases of chardonnay wine sold by a Paso Robles winery in an 8-year period follows: Year Cases of Wine 1991 270 1992 356 1993 398 1994 456 1995 438 1996 478 1997 460 1998 480 19. Referring to Table 16-1, set up a scatter diagram (i.e., a time-series plot) with year on the horizontal X-axis. ANSWER: 500
Number of Cases of W ine
450 400 350 300 250 200 150 100 50 0 1991
1992
1993
1994
TYPE: PR DIFFICULTY: Easy KEYWORDS: scatter plot
Year
1995
1996
1997
1998
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20. Referring to Table 16-1, does there appear to be a relationship between the year and the number of cases of wine sold? a) No, there appears to be no relationship between the year and the number of cases of wine sold by the vintner. b) Yes, there appears to be a slight negative linear relationship between the year and the number of cases of wine sold by the vintner. c) Yes, there appears to be a slight positive relationship between the year and the number of cases of wine sold by the vintner. d) Yes, there appears to be a negative nonlinear relationship between the year and the number of cases of wine sold by the vintner. ANSWER: c TYPE: MC DIFFICULTY: Easy KEYWORDS: scatter plot, component factor
Residuals
21. After estimating a trend model for annual time-series data, you obtain the following residual plot against time; the problem with your model is that a) the cyclical component has not been accounted for. b) the seasonal component has not been accounted for. c) the trend component has not been accounted for. d) the irregular component has not been accounted for. 1 0.8 0.6 0.4 0.2 0 -0.2 0 -0.4 -0.6 -0.8 -1
2
4
6
8
10
12
Time (Year)
ANSWER: c TYPE: MC DIFFICULTY: Moderate KEYWORDS: component factor 22. The cyclical component of a time series a) represents periodic fluctuations which reoccur within 1 year. b) represents periodic fluctuations which usually occur in 2 or more years. c) is obtained by adding up the seasonal indexes. d) is obtained by adjusting for calendar variation. ANSWER: b TYPE: MC DIFFICULTY: Easy KEYWORDS: component factor
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23. Which of the following terms describes the overall long-term tendency of a time series? a) trend b) cyclical component c) irregular component d) seasonal component ANSWER: a TYPE: MC DIFFICULTY: Easy KEYWORDS: component factor 24. Which of the following terms describes the up and down movements of a time series that vary both in length and intensity? a) trend b) cyclical component c) irregular component d) seasonal component ANSWER: b TYPE: MC DIFFICULTY: Easy KEYWORDS: component factor 25. The following is the list of MAD statistics for each of the models you have estimated from time-series data: Model MAD Linear Trend 1.38 Quadratic Trend 1.22 Exponential Trend 1.39 AR(2) 0.71 Based on the MAD criterion, the most appropriate model is a) linear trend. b) quadratic trend. c) exponential trend. d) second order autoregressive model. ANSWER: d TYPE: MC DIFFICULTY: Easy KEYWORDS: model selection
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TABLE 16-2 The monthly advertising expenditures of a department store chain (in $1,000,000s) were collected over the last decade. The last 14 months of this time series follows: Month Expenditures ($) 1 1.4 2 1.8 3 1.6 4 1.5 5 1.8 6 1.7 7 1.9 8 2.2 9 1.9 10 1.9 11 2.1 12 2.4 13 2.8 14 3.1 26. Referring to Table 16-2, set up a scatter diagram (i.e., time-series plot) with months on the horizontal Xaxis. ANSWER: 3.5
Advertising Expenditures
3 2.5 2 1.5 1 0.5 0 0
2
4
6 8 Number of Months
10
12
14
TYPE: PR DIFFICULTY: Easy KEYWORDS: scatter plot 27. True or False: Referring to Table 16-2, advertising expenditures appear to be increasing in a linear rather than curvilinear manner over time. ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: least-squares trend fitting
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TABLE 16-3 The following table contains the number of complaints received in a department store for the first 6 months of last year. Month Complaints January 36 February 45 March 81 April 90 May 108 June 144 28. Referring to Table 16-3, if a three-term moving average is used to smooth this series, what would be the second calculated term? a) 36 b) 40.5 c) 54 d) 72 ANSWER: d TYPE: MC DIFFICULTY: Easy KEYWORDS: moving averages 29. Referring to Table 16-3, if a three-term moving average is used to smooth this series, what would be the last calculated term? a) 72 b) 93 c) 114 d) 126 ANSWER: c TYPE: MC DIFFICULTY: Easy KEYWORDS: moving averages 30. Referring to Table 16-3, if a three-term moving average is used to smooth this series, how many terms would it have? a) 2 b) 3 c) 4 d) 5 ANSWER: c TYPE: MC DIFFICULTY: Easy KEYWORDS: moving averages, properties
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31. Referring to Table 16-3, if this series is smoothed using exponential smoothing with a smoothing constant of 1/3, how many terms would it have? a) 3 b) 4 c) 5 d) 6 ANSWER: d TYPE: MC DIFFICULTY: Easy KEYWORDS: exponential smoothing, properties 32. Referring to Table 16-3, if this series is smoothed using exponential smoothing with a smoothing constant of 1/3, what would be the first term? a) 36 b) 39 c) 42 d) 45 ANSWER: a TYPE: MC DIFFICULTY: Moderate KEYWORDS: exponential smoothing 33. Referring to Table 16-3, if this series is smoothed using exponential smoothing with a smoothing constant of 1/3, what would be the second term? a) 39 b) 42 c) 45 d) 53 ANSWER: a TYPE: MC DIFFICULTY: Moderate KEYWORDS: exponential smoothing 34. Referring to Table 16-3, if this series is smoothed using exponential smoothing with a smoothing constant of 1/3, what would be the third term? a) 53 b) 65.33 c) 68 d) 81 ANSWER: a TYPE: MC DIFFICULTY: Moderate KEYWORDS: exponential smoothing
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35. Referring to Table 16-3, suppose the last two smoothed values are 81 and 96. (Note: they are not.) What would you forecast as the value of the time series for July? a) 81 b) 86 c) 91 d) 96 ANSWER: d TYPE: MC DIFFICULTY: Moderate KEYWORDS: exponential smoothing, forecasting 36. Referring to Table 16-3, suppose the last two smoothed values are 81 and 96. (Note: they are not.) What would you forecast as the value of the time series for September? a) 81 b) 86 c) 91 d) 96 ANSWER: d TYPE: MC DIFFICULTY: Moderate KEYWORDS: exponential smoothing, forecasting 37. If you want to recover the trend using exponential smoothing, you will choose a weight (W) that falls in the range a) 0, 0.2 b) c) d)
0.2, 0.4 0.6, 0.8 0.8,1.0
ANSWER: a TYPE: MC DIFFICULTY: Easy KEYWORDS: exponential smoothing, forecasting, properties
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TABLE 16-4 Given below are Excel outputs for various estimated autoregressive models for Coca-Cola's real operating revenues (in billions of dollars) from 1975 to 1998. From the data, we also know that the real operating revenues for 1996, 1997, and 1998 are 11.7909, 11.7757 and 11.5537, respectively. First Order Autoregressive Model: Coefficients Standard Error t Stat p-value Intercept 0.1802077 0.39797154 0.452815546 0.655325119 XLag1 1.011222533 0.049685158 20.35260757 2.64373E-15 Second Order Autoregressive Model: Coefficients Standard Error t Stat p-value Intercept 0.30047473 0.4407641 0.681713257 0.503646149 X Lag 1 1.17322186 0.234737881 4.998008229 7.98541E-05 X Lag 2 -0.183028189 0.250716669 -0.730020026 0.474283347 Third Order Autoregressive Model: Coefficients Standard Error Intercept 0.313043288 0.514437257 XLag1 1.173719587 0.246490594 XLag2 -0.069378567 0.373086508 XLag3 -0.122123515 0.282031297
t Stat 0.608515972 4.761721601 -0.185958391 -0.433014053
p-value 0.550890271 0.000180926 0.854678245 0.670448392
38. Referring to Table 16-4 and using a 5% level of significance, what is the appropriate AR model for CocaCola's real operating revenue? a) AR(1) b) AR(2) c) AR(3) d) any of the above ANSWER: a TYPE: MC DIFFICULTY: Moderate KEYWORDS: autoregressive model, t test on slope 39. Referring to Table 16-4, if one decides to use AR(3), what will the predicted real operating revenue for Coca-Cola be in 2001? a) $11.59 billion b) $11.68 billion c) $11.84 billion d) $12.47 billion ANSWER: c TYPE: MC DIFFICULTY: Easy KEYWORDS: autoregressive model, forecasting
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TABLE 16-5 A contractor developed a multiplicative time-series model to forecast the number of contracts in future quarters, using quarterly data on number of contracts during the 3-year period from 1996 to 1998. The following is the resulting regression equation: ln Yˆ = 3.37 + 0.117 X – 0.083 Q1 + 1.28 Q2 + 0.617 Q3 where Yˆ is the estimated number of contracts in a quarter X is the coded quarterly value with X = 0 in the first quarter of 1996. Q1 is a dummy variable equal to 1 in the first quarter of a year and 0 otherwise. Q2 is a dummy variable equal to 1 in the second quarter of a year and 0 otherwise. Q3 is a dummy variable equal to 1 in the third quarter of a year and 0 otherwise. 40. Referring to Table 16-5 , the best interpretation of the constant 3.37 in the regression equation is: a) the fitted value for the first quarter of 1996, prior to seasonal adjustment, is ln 3.37. b) the fitted value for the first quarter of 1996, after seasonal adjustment, is ln 3.37. c) the fitted value for the first quarter of 1996, prior to seasonal adjustment, is e 3.37. d) the fitted value for the first quarter of 1996, after seasonal adjustment, is e 3.37. ANSWER: c TYPE: MC DIFFICULTY: Moderate KEYWORDS: exponential model, intercept, interpretation 41. Referring to Table 16-5, the best interpretation of the coefficient of X (0.117) in the regression equation is: a) the quarterly growth rate in contracts is around 11.7%. b) the annual growth rate in contracts is around 11.7%. c) the quarterly growth rate in contracts is around 17%. d) the annual growth rate in contracts is around 17%. ANSWER: a TYPE: MC DIFFICULTY: Moderate KEYWORDS: exponential model, slope, interpretation
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42. Referring to Table 16-5, the best interpretation of the coefficient of Q3 (0.617) in the regression equation is: a) the number of contracts in the third quarter of a year is approximately 62% higher than the average over all 4 quarters. b) the number of contracts in the third quarter of a year is approximately 62% higher than it would be during the fourth quarter. c) the number of contracts in the third quarter of a year is approximately 85% higher than the average over all 4 quarters. d) the number of contracts in the third quarter of a year is approximately 85% higher than it would be during the fourth quarter. ANSWER: d TYPE: MC DIFFICULTY: Moderate KEYWORDS: exponential model, slope, interpretation 43. Referring to Table 16-5, to obtain a forecast for the first quarter of 1999 using the model, which of the following sets of values should be used in the regression equation? a) X = 12, Q1 = 0, Q2 = 0, Q3 = 0 b) X = 12, Q1 = 1, Q2 = 0, Q3 = 0 c) X = 13, Q1 = 0, Q2 = 0, Q3 = 0 d) X = 13, Q1 = 1, Q2 = 0, Q3 = 0 ANSWER: b TYPE: MC DIFFICULTY: Moderate KEYWORDS: exponential model, forecasting 44. Referring to Table 16-5, to obtain a forecast for the fourth quarter of 1999 using the model, which of the following sets of values should be used in the regression equation? a) X = 15, Q1 = 0, Q2 = 0, Q3 = 0 b) X = 15, Q1 = 1, Q2 = 0, Q3 = 0 c) X = 16, Q1 = 0, Q2 = 0, Q3 = 0 d) X = 16, Q1 = 1, Q2 = 0, Q3 = 0 ANSWER: a TYPE: MC DIFFICULTY: Moderate KEYWORDS: exponential model, forecasting
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45. Referring to Table 16-5, using the regression equation, which of the following values is the best forecast for the number of contracts in the third quarter of 1999? a) 228 b) 252 c) 277 d) 311 ANSWER: c TYPE: MC DIFFICULTY: Moderate KEYWORDS: exponential model, forecasting 46. Referring to Table 16-5, using the regression equation, which of the following values is the best forecast for the number of contracts in the second quarter of 2000? a) 212 b) 272 c) 592 d) 764 ANSWER: d TYPE: MC DIFFICULTY: Moderate KEYWORDS: exponential model, forecasting 47. Referring to Table 16-5, in testing the coefficient of X in the regression equation (0.117), the results were a t-statistic of 9.08 and an associated p-value of 0.0000. Which of the following is the best interpretation of this result? a) The quarterly growth rate in the number of contracts is significantly different than 0% ( = 0.05). b) The quarterly growth rate in the number of contracts is not significantly different than 0% ( = 0.05). c) The quarterly growth rate in the number of contracts is significantly different than 100% ( = 0.05). d) The quarterly growth rate in the number of contracts is not significantly different than 100% ( = 0.05). ANSWER: a TYPE: MC DIFFICULTY: Moderate KEYWORDS: exponential model, t test on slope, decision, conclusion, interpretation
Introduction and Data Collection
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48. Referring to Table 16-5, in testing the coefficient for Q1 in the regression equation (–0.083), the results were a t-statistic of –0.66 and an associated p-value of 0.530. Which of the following is the best interpretation of this result? a) The number of contracts in the first quarter of the year is significantly different than the number of contracts in an average quarter ( = 0.05). b) The number of contracts in the first quarter of the year is not significantly different than the number of contracts in an average quarter ( = 0.05). c) The number of contracts in the first quarter of the year is significantly different than the number of contracts in the fourth quarter for a given coded quarterly value of X ( = 0.05). d) The number of contracts in the first quarter of the year is not significantly different than the number of contracts in the fourth quarter for a given coded quarterly value of X ( = 0.05). ANSWER: d TYPE: MC DIFFICULTY: Moderate KEYWORDS: exponential model, t test on slope, decision, conclusion, interpretation 49. True or False: A trend is a persistent pattern in annual time-series data that has to be followed for several years. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: component factor 50. True or False: Given a data set with 15 yearly observations, a 3-year moving average will have fewer observations than a 5-year moving average. ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: moving averages, properties 51. True or False: Given a data set with 15 yearly observations, there are only thirteen 3-year moving averages. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: moving averages, properties 52. True or False: Given a data set with 15 yearly observations, there are only seven 9-year moving averages. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: moving averages, properties 53. True or False: MAD is the summation of the residuals divided by the sample size. ANSWER: False TYPE: TF DIFFICULTY: Moderate KEYWORDS: model selection 54. True or False: A least squares linear trend line is just a simple regression line with the years recoded.
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ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: least-squares trend fitting 55. True or False: If a time series does not exhibit a long-term trend, the method of exponential smoothing may be used to obtain short-term predictions about the future. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: exponential smoothing 56. True or False: The method of least squares may be used to estimate both linear and curvilinear trends. ANSWER: True TYPE: TF DIFFICULTY: Moderate KEYWORDS: least-squares trend fitting 57. True or False: The principle of parsimony indicates that the simplest model that gets the job done adequately should be used. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: model selection 58. True or False: In selecting a forecasting model, we should perform a residual analysis. ANSWER: True TYPE: TF DIFFICULTY: Moderate KEYWORDS: model selection
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59. True or False: Each forecast using the method of exponential smoothing depends on all the previous observations in the time series. ANSWER: True TYPE: TF DIFFICULTY: Moderate KEYWORDS: exponential smoothing, properties 60. True or False: The MAD is a measure of the average of the absolute discrepancies between the actual and fitted values in a given time series. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: model selection TABLE 16-6 The number of cases of merlot wine sold by a Paso Robles winery in an 8-year period follows: Year Cases of Wine 1991 270 1992 356 1993 398 1994 456 1995 358 1996 500 1997 410 1998 376 61. Referring to Table 16-6, a centered 3-year moving average is to be constructed for the wine sales. The result of this process will lead to a total of __________ moving averages. ANSWER: 6 TYPE: FI DIFFICULTY: Easy KEYWORDS: moving averages, properties 62. Referring to Table 16-6, a centered 3-year moving average is to be constructed for the wine sales. The moving average for 1992 is __________. ANSWER: 341.33 TYPE: FI DIFFICULTY: Easy KEYWORDS: moving averages, properties
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63. Referring to Table 16-6, a centered 3-year moving average is to be constructed for the wine sales. The moving average for 1995 is __________. ANSWER: 438 TYPE: FI DIFFICULTY: Easy KEYWORDS: moving averages, properties 64. Referring to Table 16-6, construct a centered 3-year moving average for the wine sales. ANSWER: Period Cases MA 1 270 * 2 356 341.333 3 398 403.333 4 456 404.000 5 358 438.000 6 500 422.667 7 410 428.667 8 376 * TYPE: PR DIFFICULTY: Moderate KEYWORDS: moving averages 65. Referring to Table 16-6, a centered 5-year moving average is to be constructed for the wine sales. The number of moving averages that will be calculated is __________. ANSWER: 4 TYPE: FI DIFFICULTY: Easy KEYWORDS: moving averages 66. Referring to Table 16-6, a centered 5-year moving average is to be constructed for the wine sales. The moving average for 1993 is __________. ANSWER: 367.6 TYPE: FI DIFFICULTY: Moderate KEYWORDS: moving averages 67. Referring to Table 16-6, a centered 5-year moving average is to be constructed for the wine sales. The moving average for 1996 is __________. ANSWER: 420.0 TYPE: FI DIFFICULTY: Moderate KEYWORDS: moving averages
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68. Referring to Table 16-6, construct a centered 5-year moving average for the wine sales. ANSWER: Period Cases MA 1 270 * 2 356 * 3 398 367.6 4 456 413.6 5 358 424.4 6 500 420.0 7 410 * 8 376 * TYPE: PR DIFFICULTY: Moderate KEYWORDS: moving averages 69. Referring to Table 16-6, exponential smoothing with a weight or smoothing constant of 0.2 will be used to smooth the wine sales. The value of E2, the smoothed value for 1992, is __________. ANSWER: 287.2 TYPE: FI DIFFICULTY: Moderate KEYWORDS: exponential smoothing 70. Referring to Table 16-6, exponential smoothing with a weight or smoothing constant of 0.2 will be used to smooth the wine sales. The value of E4, the smoothed value for 1994, is __________. ANSWER: 338.7 TYPE: FI DIFFICULTY: Moderate KEYWORDS: exponential smoothing 71. Referring to Table 16-6, exponential smoothing with a weight or smoothing constant of 0.2 will be used to forecast wine sales. The forecast for 1999 is __________. ANSWER: 380.2 TYPE: FI DIFFICULTY: Moderate KEYWORDS: exponential smoothing, forecasting
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72. Referring to Table 16-6, exponentially smooth the wine sales with a weight or smoothing constant of 0.2. ANSWER: Time CaseWine Smooth 1 270 270.000 2 356 287.200 3 398 309.360 4 456 338.688 5 358 342.550 6 500 374.040 7 410 381.232 8 376 380.186 TYPE: PR DIFFICULTY: Difficult KEYWORDS: exponential smoothing 73. Referring to Table 16-6, exponential smoothing with a weight or smoothing constant of 0.4 will be used to smooth the wine sales. The value of E2, the smoothed value for 1992, is __________. ANSWER: 304.4 TYPE: FI DIFFICULTY: Moderate KEYWORDS: exponential smoothing 74. Referring to Table 16-6, exponential smoothing with a weight or smoothing constant of 0.4 will be used to smooth the wine sales. The value of E5, the smoothed value for 1995, is __________. ANSWER: 375.7 TYPE: FI DIFFICULTY: Moderate KEYWORDS: exponential smoothing 75. Referring to Table 16-6, exponential smoothing with a weight or smoothing constant of 0.4 will be used to forecast wine sales. The forecast for 1999 is __________. ANSWER: 401.95 TYPE: FI DIFFICULTY: Moderate KEYWORDS: exponential smoothing, forecasting
Introduction and Data Collection
493
76. Referring to Table 16-6, exponentially smooth the wine sales with a weight or smoothing constant of 0.4. ANSWER: Time CaseWine Smooth 1 270 270.000 2 356 304.400 3 398 341.840 4 456 387.504 5 358 375.702 6 500 425.421 7 410 419.253 8 376 401.952 TYPE: PR DIFFICULTY: Difficult KEYWORDS: exponential smoothing 77. Referring to Table 16-6, the Holt-Winters method for forecasting with a smoothing constant of 0.2 for both level and trend will be used to smooth the wine sales. The smoothed values of the level and trend for 1992 are ______ and ______, respectively. ANSWER: 356; 86 TYPE: FI DIFFICULTY: Easy KEYWORDS: Holt-Winters method 78. Referring to Table 16-6, the Holt-Winters method for forecasting with a smoothing constant of 0.2 for both level and trend will be used to smooth the wine sales. The smoothed values of the level and trend for 1993 are ______ and ______, respectively. ANSWER: 406.8; 57.84 TYPE: FI DIFFICULTY: Moderate KEYWORDS: Holt-Winters method 79. Referring to Table 16-6, the Holt-Winters method for forecasting with a smoothing constant of 0.2 for both level and trend will be used to smooth the wine sales. The smoothed values of the level and trend for 1998 are ______ and ______, respectively. ANSWER: 383.82; -42.70 TYPE: FI DIFFICULTY: Difficult KEYWORDS: Holt-Winters method
494
Introduction and Data Collection
80. Referring to Table 16-6, the Holt-Winters method for forecasting with a smoothing constant of 0.2 for both level and trend will be used to forecast the wine sales. The forecast for 1999 is _____. ANSWER: 341.12 TYPE: FI DIFFICULTY: Moderate KEYWORDS: Holt-Winters method, forecasting 81. Referring to Table 16-6, the Holt-Winters method for forecasting with a smoothing constant of 0.2 for both level and trend will be used to forecast the wine sales. The forecast for 2002 is _____. ANSWER: 213.01 TYPE: FI DIFFICULTY: Moderate KEYWORDS: Holt-Winters method, forecasting 82. Referring to Table 16-6, use the Holt-Winters method of fitting wine sales to compute the smoothed level and trend with a smoothing constant of 0.2 for both level and trend. ANSWER: Year Series (Y) Level (E) Trend (T) 1991 270 1992 356 356 86 1993 398 406.8 57.84 1994 456 457.728 52.3104 1995 358 388.4077 -44.9942 1996 500 468.6827 55.22118 1997 410 432.7808 -17.6773 1998 376 383.8207 -42.7035
TYPE: PR DIFFICULTY: Difficult KEYWORDS: Holt-Winters method
Introduction and Data Collection
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83. Referring to Table 16-6, plot both the wine sales series and the series of Holt-Winters forecasts for 1999 to 2002 using a smoothing constant of 0.2 for both level and trend. ANSWER: 600 500 400 Prediction
300
Sales (Y)
200 100 0 1
2
3
4
5
6
7
8
9
10 11 12
TYPE: PR DIFFICULTY: Difficult KEYWORDS: Holt-Winters method, forecasting, scatter plot 84. Referring to Table 16-6, the Holt-Winters method for forecasting with a smoothing constant of 0.8 for both level and trend will be used to smooth the wine sales. The smoothed values of the level and trend for 1992 are ______ and ______, respectively. ANSWER: 356; 86 TYPE: FI DIFFICULTY: Easy KEYWORDS: Holt-Winters method 85. Referring to Table 16-6, the Holt-Winters method for forecasting with a smoothing constant of 0.8 for both level and trend will be used to smooth the wine sales. The smoothed values of the level and trend for 1993 are ______ and ______, respectively. ANSWER: 433.2; 84.24 TYPE: FI DIFFICULTY: Moderate KEYWORDS: Holt-Winters method 86. Referring to Table 16-6, the Holt-Winters method for forecasting with a smoothing constant of 0.8 for both level and trend will be used to smooth the wine sales. The smoothed values of the level and trend for 1998 are ______ and ______, respectively. ANSWER: 609.11; 46.46 TYPE: FI DIFFICULTY: Difficult KEYWORDS: Holt-Winters method 87. Referring to Table 16-6, the Holt-Winters method for forecasting with a smoothing constant of 0.8 for both level and trend will be used to forecast the wine sales. The forecast for 1999 is _____. ANSWER: 655.56
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TYPE: FI DIFFICULTY: Moderate KEYWORDS: Holt-Winters method, forecasting 88. Referring to Table 16-6, the Holt-Winters method for forecasting with a smoothing constant of 0.8 for both level and trend will be used to forecast the wine sales. The forecast for 2002 is _____. ANSWER: 794.93 TYPE: FI DIFFICULTY: Moderate KEYWORDS: Holt-Winters method, forecasting 89. Referring to Table 16-6, use the Holt-Winters method of fitting wine sales to compute the smoothed level and trend with a smoothing constant of 0.8 for both level and trend. ANSWER: Year Series (Y) Level (E) Trend (T) 1991 270 1992 356 356 86 1993 398 433.2 84.24 1994 456 505.152 81.7824 1995 358 541.1475 72.62502 1996 500 591.018 68.07412 1997 410 609.2737 58.11044 1998 376 609.1073 46.45507
TYPE: PR DIFFICULTY: Difficult KEYWORDS: Holt-Winters method
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90. Referring to Table 16-6, plot both the wine sales series and the series of Holt-Winters forecasts for 1999 to 2002 using a smoothing constant of 0.8 for both level and trend. ANSWER: 900 800 700 600 500
Prediction
400
Sales (Y)
300 200 100 0 1
2
3
4
5
6
7
8
9 10 11 12
TYPE: PR DIFFICULTY: Difficult KEYWORDS: Holt-Winters method, forecasting, scatter plot TABLE 16-7 The number of passengers arriving at San Francisco on the Amtrak cross country express on 6 successive Mondays were: 60, 72, 96, 84, 36, and 48. 91. Referring to Table 16-7, the number of arrivals will be smoothed with a 3-term moving average. There will be a total of __________ smoothed values. ANSWER: 4 TYPE: FI DIFFICULTY: Easy KEYWORDS: moving averages, properties 92. Referring to Table 16-7, the number of arrivals will be smoothed with a 3-term moving average. The first smoothed value will be __________. ANSWER: 76 TYPE: FI DIFFICULTY: Easy KEYWORDS: moving averages
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93. Referring to Table 16-7, the number of arrivals will be smoothed with a 3-term moving average. The last smoothed value will be __________. ANSWER: 56 TYPE: FI DIFFICULTY: Easy KEYWORDS: moving averages 94. Referring to Table 16-7, the number of arrivals will be smoothed with a 5-term moving average. The first smoothed value will be __________. ANSWER: 69.6 TYPE: FI DIFFICULTY: Moderate KEYWORDS: moving averages 95. Referring to Table 16-7, the number of arrivals will be smoothed with a 5-term moving average. The last smoothed value will be __________. ANSWER: 67.2 TYPE: FI DIFFICULTY: Moderate KEYWORDS: moving averages 96. Referring to Table 16-7, the number of arrivals will be exponentially smoothed with a smoothing constant of 0.1. The smoothed value for the second Monday will be __________. ANSWER: 61.2 TYPE: FI DIFFICULTY: Moderate KEYWORDS: exponential smoothing 97. Referring to Table 16-7, the number of arrivals will be exponentially smoothed with a smoothing constant of 0.1. The smoothed value for the sixth Monday will be __________. ANSWER: 62.0 TYPE: FI DIFFICULTY: Moderate KEYWORDS: exponential smoothing 98. Referring to Table 16-7, the number of arrivals will be exponentially smoothed with a smoothing constant of 0.1. The forecast for the seventh Monday will be __________. ANSWER: 62.0 TYPE: FI DIFFICULTY: Moderate KEYWORDS: exponential smoothing, forecasting
99. Referring to Table 11.7, exponentially smooth the number of arrivals using a smoothing constant of 0.1. ANSWER: Time
Arrivals Smooth
Introduction and Data Collection
499
1 60 60.0000 2 72 61.2000 3 96 64.6800 4 84 66.6120 5 36 63.5508 6 48 61.9957 TYPE: PR DIFFICULTY: Difficult KEYWORDS: exponential smoothing 100. Referring to Table 16-7, the number of arrivals will be exponentially smoothed with a smoothing constant of 0.25. The smoothed value for the second Monday will be __________. ANSWER: 63.0 TYPE: FI DIFFICULTY: Moderate KEYWORDS: exponential smoothing 101. Referring to Table 16-7, the number of arrivals will be exponentially smoothed with a smoothing constant of 0.25. The smoothed value for the third Monday will be __________. ANSWER: 71.25 TYPE: FI DIFFICULTY: Moderate KEYWORDS: exponential smoothing 102. Referring to Table 16-7, the number of arrivals will be exponentially smoothed with a smoothing constant of 0.25. The forecast of the number of arrivals on the seventh Monday will be __________. ANSWER: 60.6 TYPE: FI DIFFICULTY: Moderate KEYWORDS: exponential smoothing, forecasting
500
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103. Referring to Table 16-7, exponentially smooth the number of arrivals using a smoothing constant of 0.25. ANSWER: Time Arrivals Smooth 1 60 60.0000 2 72 63.0000 3 96 71.2500 4 84 74.4375 5 36 64.8281 6 48 60.6211 TYPE: PR DIFFICULTY: Difficult KEYWORDS: exponential smoothing 104. Referring to Table 16-7, the Holt-Winters method for forecasting with a smoothing constant of 0.3 for both level and trend will be used to smooth the number of arrivals. The smoothed values of the level and trend for the second Monday are ______ and ______, respectively. ANSWER: 72; 12 TYPE: FI DIFFICULTY: Easy KEYWORDS: Holt-Winters 105. Referring to Table 16-7, the Holt-Winters method for forecasting with a smoothing constant of 0.3 for both level and trend will be used to smooth the number of arrivals. The smoothed values of the level and trend for the sixth Monday are ______ and ______, respectively. ANSWER: 42.43; -15.74 TYPE: FI DIFFICULTY: Easy KEYWORDS: Holt-Winters 106. Referring to Table 16-7, the Holt-Winters method for forecasting with a smoothing constant of 0.3 for both level and trend will be used to forecast the number of arrivals. The forecast for the seventh Monday is _____. ANSWER: 26.70 TYPE: FI DIFFICULTY: Moderate KEYWORDS: Holt-Winters, forecasting 107. Referring to Table 16-7, the Holt-Winters method for forecasting with a smoothing constant of 0.3 for both level and trend will be used to forecast the number of arrivals. The forecast for the twelfth Monday is _____. ANSWER: -51.98 TYPE: FI DIFFICULTY: Moderate KEYWORDS: Holt-Winters, forecasting 108. Referring to Table 16-7, use the Holt-Winters method of fitting number of arrivals to compute the smoothed level and trend with a smoothing constant of 0.3 for both level and trend. ANSWER: Mondays Arrivals Level (E) Trend (T) 1 60 2 72 72 12
Introduction and Data Collection
3 4 5 6
96 92.4 84 91.884 36 54.26604 48 42.43029
501
17.88 5.0028 -24.8317 -15.7345
TYPE: PR DIFFICULTY: Difficult KEYWORDS: Holt-Winters 109. Referring to Table 16-7, plot both the number of arrivals series and the series of Holt-Winters forecasts for the seventh through twelfth Mondays using a smoothing constant of 0.3 for both level and trend. ANSWER: 120
100 80 60 40
Prediction
20
Arrivals
0 -20
1
2
3
4
5
6
7
8
9
10 11 12
-40
-60
TYPE: PR DIFFICULTY: Difficult KEYWORDS: Holt-Winters, forecasting, scatter plot 110. Referring to Table 16-7, the Holt-Winters method for forecasting with a smoothing constant of 0.9 for both level and trend will be used to smooth the number of arrivals. The smoothed values of the level and trend for the second Monday are ______ and ______, respectively. ANSWER: 72; 12 TYPE: FI DIFFICULTY: Easy KEYWORDS: Holt-Winters
502
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111. Referring to Table 16-7, the Holt-Winters method for forecasting with a smoothing constant of 0.9 for both level and trend will be used to smooth the number of arrivals. The smoothed values of the level and trend for the sixth Monday are ______ and ______, respectively. ANSWER: 105.64; 10.63 TYPE: FI DIFFICULTY: Easy KEYWORDS: Holt-Winters 112. Referring to Table 16-7, the Holt-Winters method for forecasting with a smoothing constant of 0.9 for both level and trend will be used to forecast the number of arrivals. The forecast for the seventh Monday is _____. ANSWER: 116.27 TYPE: FI DIFFICULTY: Moderate KEYWORDS: Holt-Winters, forecasting 113. Referring to Table 16-7, the Holt-Winters method for forecasting with a smoothing constant of 0.9 for both level and trend will be used to forecast the number of arrivals. The forecast for the twelfth Monday is _____. ANSWER: 169.40 TYPE: FI DIFFICULTY: Moderate KEYWORDS: Holt-Winters, forecasting 114. Referring to Table 16-7, use the Holt-Winters method of fitting the number of arrivals to compute the smoothed level and trend with a smoothing constant of 0.9 for both level and trend. ANSWER: Mondays Arrivals Level (E) Trend (T) 1 60 2 72 72 12 3 96 85.2 12.12 4 84 95.988 11.9868 5 36 100.7773 11.26705 6 48 105.6399 10.62661
TYPE: PR DIFFICULTY: Difficult KEYWORDS: Holt-Winters
Introduction and Data Collection
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115. Referring to Table 16-7, plot both the number of arrivals series and the series of Holt-Winters forecasts for the seventh through twelfth Mondays using a smoothing constant of 0.9 for both level and trend. ANSWER: 180 160 140 120 Prediction
100
Arrivals
80 60 40 20 0 1
2
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4
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10 11 12
TYPE: PR DIFFICULTY: Difficult KEYWORDS: Holt-Winters, forecasting, scatter plot TABLE 16-8 The president of a chain of department stores believes that her stores' total sales have been showing a linear trend since 1980. She uses Microsoft Excel to obtain the partial output below. The dependent variable is sales (in millions of dollars), while the independent variable is coded years, where 1980 is coded as 0, 1981 is coded as 1, etc. SUMMARY OUTPUT Regression Statistics Multiple R R Square Adjusted R Square Standard Error Observations
Intercept Coded Year
0.604 0.365 0.316 4.800 17
Coefficients 31.2 0.78
116. Referring to Table 16-8, the fitted trend value for 1980 is __________. ANSWER: 31.2 TYPE: FI DIFFICULTY: Moderate KEYWORDS: least-squares trend fitting, fitted value
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Introduction and Data Collection
117. Referring to Table 16-8, the fitted trend value for 1985 is __________. ANSWER: 35.1 TYPE: FI DIFFICULTY: Moderate KEYWORDS: least-squares trend fitting, fitted value 118. Referring to Table 16-8, the estimate of the rate at which sales are increasing each year is __________. ANSWER: 0.78 TYPE: FI DIFFICULTY: Moderate KEYWORDS: least-squares trend fitting, slope, interpretation 119. Referring to Table 16-8, the forecast for sales in 2000 is __________. ANSWER: 46.8 TYPE: FI DIFFICULTY: Moderate KEYWORDS: least-squares trend fitting, forecasting 120. Referring to Table 16-8, the forecast for sales in 2005 is __________. ANSWER: 50.7 TYPE: FI DIFFICULTY: Moderate KEYWORDS: least-squares trend fitting, forecasting
Introduction and Data Collection
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TABLE 16-9 The executive vice-president of a drug manufacturing firm believes that the demand for the firm's most popular drug has been evidencing an exponential trend since 1985. She uses Microsoft Excel to obtain the partial output below. The dependent variable is the log base 10 of the demand for the drug, while the independent variable is years, where 1985 is coded as 0, 1986 is coded as 1, etc. SUMMARY OUTPUT Regression Statistics Multiple R R Square Adjusted R Square Standard Error Observations
Intercept Coded Year
0.996 0.992 0.991 0.02831 12
Coefficients 1.44 0.068
121. Referring to Table 16-9, the fitted trend value for 1985 is __________. ANSWER: 27.54 TYPE: FI DIFFICULTY: Moderate KEYWORDS: least-squares trend fitting, fitted value 122. Referring to Table 16-9, the fitted trend value for 1990 is __________. ANSWER: 60.26 TYPE: FI DIFFICULTY: Moderate KEYWORDS: least-squares trend fitting, fitted value 123. Referring to Table 16-9, the fitted exponential trend equation to predict Y is __________. ANSWER: 27.54(1.17)X TYPE: FI DIFFICULTY: Difficult KEYWORDS: least-squares trend fitting, fitted value 124. Referring to Table 16-9, the forecast for the demand in 1999 is __________. ANSWER: 246.6 TYPE: FI DIFFICULTY: Moderate KEYWORDS: least-squares trend fitting, forecasting
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125. Referring to Table 16-9, the forecast for the demand in 2002 is __________. ANSWER: 394.46 TYPE: FI DIFFICULTY: Moderate KEYWORDS: least-squares trend fitting, forecasting TABLE 16-10 The manager of a marketing consulting firm has been examining his company's yearly profits. He believes that these profits have been showing a quadratic trend since 1980. He uses Microsoft Excel to obtain the partial output below. The dependent variable is profit (in thousands of dollars), while the independent variables are coded years and squares of coded years, where 1980 is coded as 0, 1981 is coded as 1, etc. SUMMARY OUTPUT Regression Statistics Multiple R R Square Adjusted R Square Standard Error Observations
Intercept Coded Year Year Squared
0.998 0.996 0.996 4.996 17
Coefficients 35.5 0.45 1.00
126. Referring to Table 16-10, the fitted value for 1980 is __________. ANSWER: 35.5 TYPE: FI DIFFICULTY: Moderate KEYWORDS: least-squares trend fitting, fitted value 127. Referring to Table 16-10, the fitted value for 1985 is __________. ANSWER: 62.75 TYPE: FI DIFFICULTY: Moderate KEYWORDS: least-squares trend fitting, fitted value 128. Referring to Table 16-10, the forecast for profits in 2000 is __________. ANSWER: 444.5 TYPE: FI DIFFICULTY: Moderate KEYWORDS: least-squares trend fitting, forecasting 129. Referring to Table 16-10, the forecast for profits in 2005 is __________. ANSWER: 671.75 TYPE: FI DIFFICULTY: Moderate
Introduction and Data Collection
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KEYWORDS: least-squares trend fitting, forecasting 130. Microsoft Excel was used to obtain the following quadratic trend equation: Sales = 100 – 10X + 15X2. The data used was from 1989 through 1998, coded 0 to 9. The forecast for 1999 is __________. ANSWER: 1,500 TYPE: FI DIFFICULTY: Moderate KEYWORDS: least-squares trend fitting, forecasting 131. The manager of a company believed that her company's profits were following an exponential trend. She used Microsoft Excel to obtain a prediction equation for the logarithm (base 10) of profits: log10(Profits) = 2 + 0.3X. The data she used were from 1993 through 1998, coded 0 to 5. The forecast for 1999 profits is __________. ANSWER: 6,309.57 TYPE: FI DIFFICULTY: Moderate KEYWORDS: least-squares trend fitting, forecasting 132. A first-order autoregressive model for stock sales is: Salesi = 800 + 1.2(Sales)i-1. If sales in 1998 are 6000, the forecast of sales for 1999 is __________. ANSWER: 8,000 TYPE: FI DIFFICULTY: Moderate KEYWORDS: autoregressive model, forecasting 133. A second-order autoregressive model for average mortgage rate is: Ratei = – 2.0 + 1.8(Rate)i-1 – 0.5 (Rate)i-2. If the average mortgage rate in 1998 was 7.0, and in 1997 was 6.4, the forecast for 1999 is __________. ANSWER: 7.4 TYPE: FI DIFFICULTY: Moderate KEYWORDS: autoregressive model, forecasting
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134. A second-order autoregressive model for average mortgage rate is: Ratei = – 2.0 + 1.8(Rate)i-1 – 0.5 (Rate)i-2. If the average mortgage rate in 1998 was 7.0, and in 1997 was 6.4, the forecast for 2000 is __________. ANSWER: 7.82 TYPE: FI DIFFICULTY: Difficult KEYWORDS: autoregressive model, forecasting TABLE 16-11 Business closures in Laramie, Wyoming from 1989 to 1994 were: 1993 10 1994 11 1995 13 1996 19 1997 24 1998 35 Microsoft Excel was used to fit both first-order and second-order autoregressive models, resulting in the following partial outputs: SUMMARY OUTPUT – 2nd Order Model
Intercept X Variable 1 X Variable 2
Coefficients -5.77 0.80 1.14
SUMMARY OUTPUT – 1st Order Model
Intercept X Variable 1
Coefficients -4.16 1.59
135. Referring to Table 16-11, the fitted values for the first-order autoregressive model are ________, ________, ________, ________, and ________. ANSWER: 11.74, 13.33, 16.51, 26.05, 34.00 TYPE: FI DIFFICULTY: Moderate KEYWORDS: autoregressive model, fitted value
Introduction and Data Collection
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136. Referring to Table 16-11, the residuals for the first-order autoregressive model are ________, ________, ________, ________, and ________. ANSWER: -.74, -.33, 2.49, -2.05, 1.00 TYPE: FI DIFFICULTY: Moderate KEYWORDS: autoregressive model, residual 137. Referring to Table 16-11, the value of the MAD for the first-order autoregressive model is ________. ANSWER: 1.322 TYPE: FI DIFFICULTY: Moderate KEYWORDS: autoregressive model, model selection 138. Referring to Table 16-11, the fitted values for the second-order autoregressive model are ________, ________, ________, and ________. ANSWER: 14.43, 17.17, 24.25, 35.09 TYPE: FI DIFFICULTY: Moderate KEYWORDS: autoregressive model, fitted value 139. Referring to Table 16-11, the residuals for the second-order autoregressive model are ________, ________, ________, and ________. ANSWER: – 1.43, 1.83, – 0.25, – 0.09 TYPE: FI DIFFICULTY: Moderate KEYWORDS: autoregressive model, residual 140. Referring to Table 16-11, the value of the MAD for the second-order autoregressive model is ________. ANSWER: 0.90 TYPE: FI DIFFICULTY: Moderate KEYWORDS: autoregressive model, model selection 141. True or False: Referring to Table 16-11, the values of the MAD for the two models indicate that the firstorder model should be used for forecasting. ANSWER: False TYPE: TF DIFFICULTY: Moderate KEYWORDS: autoregressive model, model selection
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Introduction and Data Collection
TABLE 16-12 The manager of a health club has recorded average attendance in the club‟s newly introduced step classes over the last 15 months: 32.1, 39.5, 40.3, 46.0, 65.2, 73.1, 83.7, 106.8, 118.0, 133.1, 163.3, 182.8, 205.6, 249.1, and 263.5. She then used Microsoft Excel to obtain the following partial output for both a first- and second-order autoregressive model. SUMMARY OUTPUT – 2nd Order Model Regression Statistics Multiple R R Square Adjusted R Square Standard Error Observations
Intercept X Variable 1 X Variable 2
0.993 0.987 0.985 9.276 15
Coefficients 5.86 0.37 0.85
SUMMARY OUTPUT – 1st Order Model Regression Statistics Multiple R R Square Adjusted R Square Standard Error Observations
Intercept X Variable 1
0.993 0.987 0.985 9.150 15
Coefficients 5.66 1.10
142. Referring to Table 16-12, using the first-order model, the forecast of average attendance for month 16 is __________. ANSWER: 295.51 TYPE: FI DIFFICULTY: Moderate KEYWORDS: autoregressive model, forecasting
Introduction and Data Collection
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143. Referring to Table 16-12, using the first-order model, the forecast of average attendance for month 17 is __________. ANSWER: 330.72 TYPE: FI DIFFICULTY: Moderate KEYWORDS: autoregressive model, forecasting 144. Referring to Table 16-12, using the second-order model, the forecast of average attendance for month 16 is __________. ANSWER: 315.09 TYPE: FI DIFFICULTY: Moderate KEYWORDS: autoregressive model, forecasting 145. Referring to Table 16-12, using the second-order model, the forecast of average attendance for month 17 is __________. ANSWER: 346.42 TYPE: FI DIFFICULTY: Moderate KEYWORDS: autoregressive model, forecasting 146. True or False: Referring to Table 16-12, based on the parsimony principle, the second-order model is the better model for making forecasts. ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: autoregressive model, model selection
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Introduction and Data Collection
TABLE 16-13 A local store developed a multiplicative time-series model to forecast its revenues in future quarters, using quarterly data on its revenues during the 4-year period from 1998 to 2002. The following is the resulting regression equation:
log10 Yˆ = 6.102 + 0.012 X – 0.129 Q1 – 0.054 Q2 + 0.098 Q3 where Yˆ is the estimated number of contracts in a quarter X is the coded quarterly value with X = 0 in the first quarter of 1998. Q1 is a dummy variable equal to 1 in the first quarter of a year and 0 otherwise. Q2 is a dummy variable equal to 1 in the second quarter of a year and 0 otherwise. Q3 is a dummy variable equal to 1 in the third quarter of a year and 0 otherwise. 147. Referring to Table 16-13, the best interpretation of the constant 6.102 in the regression equation is: a) the fitted value for the first quarter of 1998, prior to seasonal adjustment, is log 10(6.102). b) the fitted value for the first quarter of 1998, after seasonal adjustment, is log10(6.102). c) the fitted value for the first quarter of 1998, prior to seasonal adjustment, is 10 6.102. d) the fitted value for the first quarter of 1998, after seasonal adjustment, is 10 6.102. ANSWER: c TYPE: MC DIFFICULTY: Moderate KEYWORDS: exponential model, intercept, interpretation 148. Referring to Table 16-13, the best interpretation of the coefficient of X (0.012) in the regression equation is: a) the quarterly growth rate in revenues is around 1.2%. b) the annual growth rate in revenues is around 1.2%. c) the quarterly growth rate in revenues is around 12%. d) the annual growth rate in revenues is around 12%. ANSWER: a TYPE: MC DIFFICULTY: Moderate KEYWORDS: exponential model, slope, interpretation 149. Referring to Table 16-13, the estimated quarterly compound growth rate in revenues is: a) the quarterly growth rate in revenues is around 1.2%. b) the annual growth rate in revenues is around 2.8%. c) the quarterly growth rate in revenues is around 12%. d) the annual growth rate in revenues is around 28%. ANSWER: b TYPE: MC DIFFICULTY: Moderate KEYWORDS: exponential model, slope, interpretation 150. Referring to Table 16-13, the best interpretation of the coefficient of Q2 (–0.054) in the regression equation is: a) the revenues in the second quarter of a year are approximately 5.4% lower than the average over all 4 quarters. b) the revenues in the second quarter of a year are approximately 5.4% lower than they would be during the fourth quarter. c) the revenues in the second quarter of a year are approximately 11.69% lower than the average over all 4 quarters.
Introduction and Data Collection
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d) the revenues in the second quarter of a year are approximately 11.69% lower than they would be during the fourth quarter. ANSWER: d TYPE: MC DIFFICULTY: Moderate KEYWORDS: exponential model, slope, interpretation 151. Referring to Table 16-13, the best interpretation of the coefficient of Q3 (0.098) in the regression equation is: a) the revenues in the third quarter of a year are approximately 9.8% higher than the average over all 4 quarters. b) the revenues in the third quarter of a year are approximately 9.8% higher than they would be during the fourth quarter. c) the revenues in the third quarter of a year are approximately 25.31% higher than the average over all 4 quarters. d) the revenues in the third quarter of a year are approximately 25.31% higher than they would be during the fourth quarter. ANSWER: d TYPE: MC DIFFICULTY: Moderate KEYWORDS: exponential model, slope, interpretation 152. Referring to Table 16-13, to obtain a forecast for the first quarter of 2002 using the model, which of the following sets of values should be used in the regression equation? a) X = 16, Q1 = 1, Q2 = 0, Q3 = 0 b) X = 16, Q1 = 0, Q2 = 1, Q3 = 0 c) X = 17, Q1 = 1, Q2 = 0, Q3 = 0 d) X = 17, Q1 = 0, Q2 = 1, Q3 = 0 ANSWER: a TYPE: MC DIFFICULTY: Moderate KEYWORDS: exponential model, forecasting
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Introduction and Data Collection
153. Referring to Table 16-13, to obtain a forecast for the fourth quarter of 1999 using the model, which of the following sets of values should be used in the regression equation? a) X = 7, Q1 = 0, Q2 = 0, Q3 = 0 b) X = 7, Q1 = 1, Q2 = 0, Q3 = 0 c) X = 8, Q1 = 0, Q2 = 0, Q3 = 0 d) X = 8, Q1 = 1, Q2 = 0, Q3 = 0 ANSWER: a TYPE: MC DIFFICULTY: Moderate KEYWORDS: exponential model, forecasting 154. Referring to Table 16-13, to obtain a forecast for the third quarter of 2003 using the model, which of the following sets of values should be used in the regression equation? a) X = 22, Q1 = 0, Q2 = 0, Q3 = 0 b) X = 22, Q1 = 0, Q2 = 0, Q3 = 1 c) X = 23, Q1 = 0, Q2 = 0, Q3 = 0 d) X = 23, Q1 = 0, Q2 = 0, Q3 = 1 ANSWER: b TYPE: MC DIFFICULTY: Moderate KEYWORDS: exponential model, forecasting 155. Referring to Table 16-13, using the regression equation, what is the forecast for revenues in the third quarter of 2003? ANSWER: 2910717.12 TYPE: PR DIFFICULTY: Moderate KEYWORDS: exponential model, forecasting 156. Referring to Table 16-13, using the regression equation, what is the forecast for revenues in the first quarter of 2005? ANSWER: 2037042.08 TYPE: PR DIFFICULTY: Moderate KEYWORDS: exponential model, forecasting 157. Referring to Table 16-13, using the regression equation, what is the forecast for revenues in the fourth quarter of 2004? ANSWER: 2666858.67 TYPE: PR DIFFICULTY: Moderate KEYWORDS: exponential model, forecasting
Introduction and Data Collection
515
158. Referring to Table 16-13, in testing the significance of the coefficient of X in the regression equation (0.012), which has a p-value of 0.0000, which of the following is the best interpretation of this result? a) The quarterly growth rate in revenues is significantly different than 0% ( = 0.05). b) The quarterly growth rate in revenues is not significantly different than 0% ( = 0.05). c) The quarterly growth rate in revenues is significantly different than 1.2% ( = 0.05). d) The quarterly growth rate in revenues is not significantly different than 1.2% ( = 0.05). ANSWER: a TYPE: MC DIFFICULTY: Moderate KEYWORDS: exponential model, t test on slope, decision, conclusion, interpretation 159. Referring to Table 16-13, in testing the significance of the coefficient for Q1 in the regression equation (– 0.129), which has a p-value of 0.492, which of the following is the best interpretation of this result? a) The revenues in the first quarter of the year are significantly different than the revenues in an average quarter ( = 0.05). b) The revenues in the first quarter of the year are not significantly different than the revenues in an average quarter ( = 0.05). c) The revenues in the first quarter of the year are significantly different than the revenues in the fourth quarter ( = 0.05). d) The revenues in the first quarter of the year are not significantly different than the revenues in the fourth quarter ( = 0.05). ANSWER: d TYPE: MC DIFFICULTY: Moderate KEYWORDS: exponential model, t test on slope, decision, conclusion, interpretation
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