Chapter 1 Overview of Statistics 1.1
a. Statistics can be used to 1) determine what a typical commission is and then 2) use that value to identify commissions that appear to be unusually high. b. She could use statistics to show the average energy use compared to previous models. She could also use statistics to show how durable the monitor would be in the field. c. He could use statistics to calculate the average absenteeism at each plant and then compare across the three plants. d. He could calculate average number of defects in each shipment. He could determine variation in number of defects between the three shipments. Learning Objective: 01-1
1.2
a. He could calculate the job turnover for each gender for each restaurant. He could then look at the difference between the various restaurants as well as the difference between genders. b. He could calculate the average number of emails received and sent for employees in different job classifications and make comparisons. c. The portfolio manager could calculate both the average return and the variation on return for the six different investments and make comparisons. d. By studying the busiest times of day for surgery, the administrator could work with surgeons to spread their surgeries out to better use the facilities. He might also look at which surgeries take the longest and which are shorter to help with scheduling. Learning Objective: 01-1
1.3
a. The average business school graduate should expect to use computers to manipulate the data. b. Answers will vary. Weak quantitative skills lead to poor decision making because databased decision making is a hallmark of successful businesses. If one cannot analyze data or understand summary analyses, one will be making decisions without full information. Learning Objective: 01-2
1.4
a. Answers will vary. Why Not Study: It is difficult to become a statistical “expert” after taking one introductory college course. A business person should hire statistical experts and have faith that those who are using statistics are doing it correctly. Why Study: In fact, most college graduates will use statistics every day. Relying on a consultant to perform simple or even complex statistical analyses means turning over part of the business decision-making to someone who doesn’t know your business as well as you do. b. Answers will vary. Answers provided in part a will be similar for the subjects of accounting. Foreign languages are essential in this global business environment of today. While learning a foreign language can take considerably more time as an adult, the investment is worth it. Businesses are looking for college graduates that have 1
quantitative skills and speak a foreign language. Chinese and Spanish are popular choices. c. To arrive at an absurd result, and then conclude the original assumption must have been wrong, since it gave us this absurd result. This is also known as proof by contradiction. It makes use of the law of excluded middle — a statement which cannot be false, must then be true. If you state that you will never use statistics in your business profession then you might conclude that you shouldn’t study statistics. However, the original assumption of never using statistics is wrong; therefore the conclusion of not needing to study statistics is also wrong. Learning Objective: 01-2 1.5
a. Answers will vary. b. An hour with an expert at the beginning of a project could be the smartest move a manager can make. A consultant is helpful when your team lacks certain critical skills, or when an unbiased or informed view cannot be found inside your organization. Expert consultants can handle domineering or indecisive team members, personality clashes, fears about adverse findings, and local politics. As in any business decision, the costs of paying for statistical assistance must be weighed against the benefits. Costs are: statistician’s time, more time invested in the beginning of a project which may mean results are not immediate. Benefits include: better sampling strategies which can result in more useful data, a better understanding of what information can be extracted from the data, greater confidence in the results. Learning Objective: 01-2
1.6
a. The ethical issue is that credit card companies are using unfair marketing practices to entice students to use credit cards. Students are a vulnerable group that has not been educated about personal finance. Credit card companies are also purchasing student lists from universities and various student groups. This is also an unethical practice by the universities. b. They used an in-person survey given to 1500 students. These students were randomly solicited at popular places on campus. The sampling technique was a convenience sample. The report did not attempt to make inferences about the population of college students. The report simply provided statistics collected from their sample. The naïve reader would most likely make the inference that the numbers from the sample apply to the population as a whole. This should be made clearer in the report. c. The subjects surveyed include 1) how students pay for their education, 2) how they use credit cards, 3) how many of them use credit cards, and 4) attitudes toward credit card marketing on campus. It would be interesting to see what the questions were and how they were worded. d. Because the survey focused on students’ opinions and did not provide information about credit card use by the general population it is difficult to conclude that the marketing practices companies use on campuses are different from those used to market cards to the general public. Furthermore, it was not obvious that those students who had credit cards obtained those cards as a direct result of the campus marketing efforts. What was impressive was the amount of research the group did to support their claims of unethical practices. The references to other reports and court cases were a better support for their claims in question 1. 2
e. Answers vary. In general, because the study was based on a convenience sample the results cannot be assumed to hold for the general population of students. Also, it would be good to know how students’ use of credit cards compares to the general public. f. The list of schools included in the survey is focused in a few states such as Massachusetts, California, and Colorado. There were very few schools from the southeast and virtually no private colleges or universities. Because it was a convenience sample it is not appropriate to extend the results from the survey to the larger population of college students. However, many of the references cited did suggest that the problem was widespread. g. First, they suggest eliminating "freebies" or gifts that would entice students to sign up for a credit card. Second, they suggest limiting posted marketing materials on campus. The third solution is to disallow acquisition of students’ lists. Fourth, they suggest that sponsorship should be discontinued. In other words, credit card companies cannot pay for student groups to get them new customers. The fifth recommendation is to enhance student awareness about credit card problems. Finally, they suggest that certain terms that take advantage of students should be discouraged. For example, hidden fees, changing contracts, and universal default should be eliminated. As a follow up to this report note that the Credit Card Accountability Responsibility and Disclosure Act of 2009 or Credit CARD Act of 2009 is a federal law passed by the United States Congress and signed by President Barack Obama on May 22, 2009. It is comprehensive credit card reform legislation that aims "...to establish fair and transparent practices relating to the extension of credit under an open end consumer credit plan, and for other purposes."[1] Learning Objective: 01-3 1.7
Answers will vary. Examples include: linking expectations on conduct to the company’s mission, explaining what is considered acceptable and unacceptable behavior, providing courses of action for employees when they have questions or believe an ethical guideline has been violated, documentation so that employees will know the code, addressing nepotism, addressing romantic relationships, addressing customer relationships and vendor relationships. Learning Objective: 01-3
1.8
Mary is falling for Pitfall 3: Conclusions from Rare Events. Just because an event is unlikely doesn’t mean it will never happen. The probability may be small but there is a chance of observing the same five winning numbers in the same sequence on two different days. Learning Objective: 01-4
1.9
Bob is falling for Pitfall 5: Assuming a Causal Link. Statistical association does not prove causation. Bob also appears to be falling for Pitfall 6: Generalization to Individuals. Even if there were a link between cell phones and binge drinking, the link would be between groups of cell phone users and groups of binge drinkers, not necessarily a one to one link. Learning Objective: 01-4
3
1.10
a. It is not obvious that there is a direct cause and effect relationship between an individual choosing to use a radar detector and that individual choosing to vote and wear a seatbelt. b. Increasing the use of radar detectors may not influence those who obey laws and are less concerned with government limitations. Learning Objective: 01-4
1.11
a. No, the method did not “work” in the sense that he increased his chances of winning by picking the numbers the way he did. Every combination of six numbers has the same chance of winning. The fact that this winner chose his numbers based on his families’ birthdays and school grade does not increase the chance of him winning. b. Someone who picks 1-2-3-4-5-6 has just as much chance of winning as anyone else (see (a)). Learning Objective: 01-4
1.12
a. The phrase “much more” is not quantified. The study report is not mentioned. There is no way to determine the veracity of this statement. Six causes of car accidents could be poor weather, road construction, heavy traffic, inexperienced driver, engine failure, and drinking. Smoking is not on this list. b. Smokers might pay less attention to driving when lighting a cigarette. Learning Objective: 01-4
1.13
The percent reduction might not seem important because it appears so small but when multiplied by millions of calls each day it does translate into a significant improvement. Learning Objective: 01-4
1.14
a. The conclusion was a broad generalization. This can cause ethical issues if action was taken against employees that had legitimate complaints about management decisions. b. The sample is too small, and non-random, creating bias. c. This test is only going to be performed in the laboratory which limits the ability to see how it works out in the world. Additionally, there is no informed consent. d. The sample is once again way too small and is non random. This again creates bias which creates no causal link. Learning Objective: 01-4
1.15
Sarah’s statement is not correct. Most people would feel the difference is practically important. 0.9% of 231,164 is 2,080 patients. Learning Objective: 01-4
1.16
The headline implies causation but we cannot assume causation just because the repeal came before the increase in deaths. The deaths could have been going up each year before this study or there could be other variables contributing to the increase in deaths. Although the research accounts for some third variables such as increased registrations, it does not prove causality. Learning Objective: 01-4
4
1.17
a. Many people have math “phobia”, and because statistics involves math the subject can sound scary. The subject of statistics has a reputation for being difficult and this can cause fear of the unknown. b. There is usually not the same fear towards an ethics class. This is because there is much more emphasis on unethical behavior in the media, to the extent that ethical behavior and an understanding of how to be ethical is widely accepted as a requirement to graduate and then succeed. Learning Objective: 01-2
1.18
Random sampling of cans of sauce for a specific manufacturer can be used to assess quality control. Learning Objective: 01-1
1.19
a. The consultant can analyze the responses from the 80 purchasing managers noting that the linen supplier should not make any conclusions about the managers who did not respond. The consultant should not use the responses to ambiguous questions. She should suggest that the supplier redesign both the survey questions and the survey methods to increase the response rates. b. An imperfect analysis would be a mistake because the supplier may make changes to their business that upset those customers not responding to the survey or those customers not sent a survey. Learning Objective: 01-1
1.20
All of these involve taking samples from the population of interest and estimating the value of the variable of interest (e.g., average height, average width of a car, etc.) Learning Objective: 01-1
1.21
Hannah’s statement is correct in the sense that we should not use the percentage to describe the population of all musicians. This was not a random sample. It is merely a statistic that describes this group of deceased musicians. She correctly recognized Pitfall 2: Making conclusions from nonrandom samples. Learning Objective: 01-4
1.22
Sarah is falling for Pitfalls 5 and 6. A statistical association does not mean causation. And even if there were evidence to support cause and effect, extending the result about groups to an individual is not appropriate. Learning Objective: 01-4
1.23
Tom is falling for Pitfall 2: Conclusions from Nonrandom Samples. Tom’s team is a nonrandom sample. In order to conclude that helmets are not needed, Tom would need to take a random sample of lacrosse players from several age groups, ability levels, and geographies. Learning Objective: 01-4
5
1.24
Just because it is not statistically significant does not mean it is not important. Reducing the risk of death from prostate cancer is extremely important even if it is a small reduction. Something can also be statistically significant without being important. Learning Objective: 01-4
1.25
a. Class attendance, time spent studying, natural ability of student, interest level in subject, instructor’s ability, or performance in course prerequisites. Smoking is not on the list. b. Most likely students who earn A’s are also making good decisions about their health. Students who smoke might also be making poor choices surrounding their study habits. c. Giving up smoking alone may not stop a student from using poor study habits nor is it likely to increase their interest in a topic. Learning Objective: 01-4
1.26
Curiosity, parents’ who smoke, friends who smoke, seeing teenagers smoke in movies and TV, boredom, wanting to look cool. Yes, seeing movie and TV stars smoking was on the list. Learning Objective: 01-4
1.27
a. We need to know the total number of philosophy majors to evaluate this. b. We don’t know the number of students in each major. c. This statement suffers from self-selection bias. There are likely many more marketing majors who choose to take the GMAT and therefore a wider range of abilities than the abilities of physics majors who choose to take the GMAT. d. The GMAT is just one indicator of managerial skill and ability. It is not the only predictor of success in management. Learning Objective: 01-4
1.28
a. The graph is much more useful. We can clearly see that as square feet in the restaurant increases, so does interior seats. It is a fairly strong linear relationship. b. The last three data points on the far right show restaurants that have large square feet as well as a lot of interior seats. The point all the way to the left shows the opposite, small square footage with hardly any interior seats. Learning Objective: 01-1
1.29
a. The graph is more helpful. The visual illustration allows you to quickly see the results and the obvious decline in salad sales. b. We can quickly see that the month of May saw most salads sold, while the month of December saw the least salads sold. Learning Objective: 01-1
1.30
Answers will vary. Learning Objective: 01-4
6
ASBE 6e Solutions for Instructors
Chapter 2 Data Collection 2.1
a. Categorical b. Categorical c. Discrete numerical Learning Objective: 02-2
2.2
a. Continuous numerical b. Discrete numerical c. Categorical d. Continuous numerical Learning Objective: 02-2
2.3
a. Continuous numerical b. Continuous numerical (often reported as an integer) c. Categorical d. Categorical Learning Objective: 02-2
2.4
Answers will vary. Learning Objective: 02-2
2.5
a. Cross-sectional b. Time series c. Time series d. Cross-sectional. Learning Objective: 02-3
2.6
a. Time series b. Cross-sectional c. Time series d. Cross-sectional Learning Objective: 02-3
2.7
a. Time series b. Cross-sectional. c. Time series. d. Cross-sectional. Learning Objective: 02-3
2.8
Answers will vary. Learning Objective: 02-3 7
ASBE 6e Solutions for Instructors
2.9
a. Ratio. The number of hits is an integer with zero a possibility. b. Ordinal. Ranking but difference in ranks is not meaningful. c. Nominal. Positions on the field have no ranking implied. d. Interval. Celsius is an interval measure because the zero is not meaningful. e. Ratio. Salary has a meaningful zero. f. Ordinal. Ranking but differences are not meaningful. Learning Objective: 02-4
2.10
a. Ratio. The number of employees is a count and you can have zero employees. b. Ratio. The number of returns is a count and you can have zero returns. c. Interval. The temperature difference from 70 degrees to 80 degrees is the same increase as 80 degrees to 90 degrees. However, zero temperature does not mean no temperature exists, therefore it is interval. d. Nominal. It is not a number and you could not rank order this cashier with others. e. Ordinal. Ratings of employees generally fall into categories such as "exceeds standards", etc. Therefore, we know it is either nominal or ordinal and since we can rank order this employee with others given their rankings, we can say it is ordinal. f. Nominal. There is no meaningful zero and distance between social security numbers has no meaning. We also would not rank order based on social security number so this is nominal even though it is a number. Learning Objective: 02-4
2.11
a. Ratio. The number of passengers is a count and the zero point means absence of passengers. b. Ratio. The waiting time is a continuous variable and you can have a waiting time equal to zero. c. Nominal. Brand names are categories with no ranking. d. Ordinal. Ticket class is a nonnumerical category and there is a ranking implied. e. Interval. The temperature difference from 70 degrees to 80 degrees is the same increase as 80 degrees to 90 degrees. However, zero temperature does not mean no temperature exists, therefore it is interval. f. Interval most common answer. Likert scales are typically assumed to be interval. Ordinal is a possible answer if the assumption is that the differences between ratings are not equal. Learning Objective: 02-4
2.12
a. Ordinal (possibly interval). There is no meaningful zero so we can eliminate ratio. There is a rank order to the "categories" so we can eliminate nominal. With only three responses on the scale most statisticians would call this ordinal meaning the intervals between responses are not equal. b. Ordinal. There is no meaningful zero so we can eliminate ratio. There is a rank order to the "categories" so we can eliminate nominal. But we cannot assume the difference between Rarely and Often is the same as the difference between Often and Very Often. c. Nominal. There is no meaningful zero or distance or ranking. d. Ratio. This is a number not a category and zero has meaning. 8
ASBE 6e Solutions for Instructors Learning Objective: 02-4 Learning Objective: 02-5 2.13
a. Interval, assuming intervals are equal, otherwise ordinal. b. Yes (assuming interval data) c. 10 point scale might give too many points and make it hard for guests to choose between. Learning Objective: 02-4 Learning Objective: 02-5
2.14
a. Interval because it is a ranking with meaningful intervals between scale points. b. No, we can only say that the difference between 3 and 4 is the same as the difference between 4 and 5. c. Yes, a 5 point Likert scale would work just as well. In fact, a 5 point scale might be preferred. It might be difficult for customers to differentiate between a 1 and a 2 on 10 point scale whereas a 5 point scale would make it easier for the customer to answer. Learning Objective: 02-4 Learning Objective: 02-5
2.15
a. Census. You can easily ask each of your friends this question. b. Census or Sample. If your class is large you might take a sample. c. Sample. The number of students at a university is too large to take a census. d. Census. You most likely have fewer than 7 classes so fewer than 7 professors. Learning Objective: 02-6
2.16
a. Sample. Over the lifetime of your computer you will recharge your battery a very high number of times. A sample makes sense in this case. b. Census or sample. If your class is large you might take a sample. c. Sample. The number of students at a university is too large to take a census. d. Census. You can easily ask each of your friends this question. Learning Objective: 02-6
2.17
a. Parameter. The S&P is the population. b. Parameter. Same as above. The S&P is the population. c. Statistic. We clearly stated a random sample. d. Statistic. This isn’t random but it could be considered a sample. Learning Objective: 02-6
2.18
Use the formula: N = 20×n a. N = 20×10 = 200 b. N = 20×50 = 1000 c. N = 20×100 = 2000 Learning Objective: 02-7
9
ASBE 6e Solutions for Instructors 2.19
a. Convenience. b. Systematic. c. Judgment or biased. Learning Objective: 02-7
2.20
a. Random b. Convenience c. Systematic Learning Objective: 02-7
2.21
Answers will vary. Learning Objective: 02-7
2.22
a. There were 24 ages under 30. The proportion is 24/48 = 0.50. b. Answers will vary. c. Answers will vary. Learning Objective: 02-7
2.23
Answers will vary. Learning Objective: 02-7
2.24
a. Response bias. The students might exaggerate the number of dates they’ve had. b. Self-selection bias, coverage error. By only asking folks outside of a church you might get a number that is higher than the number from the general public. c. Coverage error, self-selection bias. Same reasons as in part b. Learning Objective: 02-9
2.25
a. Telephone or web. A web-based survey might overestimate the numbers who prefer a web-based course. b. Direct observation of students on campus. c. Interview, web, or mail. Response rates would most likely differ with the three methods. Mail surveys tend to have lower response rates. d. Interview or web. Learning Objective: 02-9
2.26
a. Mail or interview. You would most likely have a list of customer addresses but you could also just ask customers that come in. A mail survey might have a lower response rate. b. Direct observation, through customer invoices/receipts. c. If you track zip codes as well as invoices/receipts this could be done via direct observation of your records. However, mail would be another option if that data is unavailable. d. Interview since you only have to ask seven employees. Learning Objective: 02-9
2.27
Version 1: Most would say yes. Version 2: More varied responses. 10
ASBE 6e Solutions for Instructors Learning Objective: 02-9 2.28
Does not include all possible responses or allow for the responder to pick something other than those presented. Learning Objective: 02-9
2.29
a. Continuous numerical. Age can be measured with fractions. b. Categorical. Nationality is not a numerical measure. c. Discrete numerical. We can count the double-faults using integers. Learning Objective: 02-2
2.30
a. Discrete numerical. We can count the number of spectators using integers. b. Continuous numerical. The amount of water can be fractions of liters. c. Categorical. Gender is a category, not a number. Learning Objective: 02-2
2.31
a. Ordinal. We do have a ranking but the differences between rankings would not be equal. b. Interval measure if using a noise meter to measures decibels (20dB is not twice as much as 10dB; 0dB does not mean no sound). But if the noise level is based on a word description such as "noisy" or "quiet" then the measurement scale would be ordinal. c. Ratio because this is a count. Learning Objective: 02-4
2.32
a. Ratio because this is a count. b. Ratio if there were a way to measure the actual amount. Most likely this would be an ordinal measure because one would characterize the consumption as high, medium or low. c. Categorical. Type of vehicle is not a numerical measure and no ranking is implied. Learning Objective: 02-4
2.33
Q1 Categorical, nominal. Not numerical and no ranking. Q2 Continuous, ratio. Can take on decimal values and has a clearly defined zero value. Q3 Continuous, ratio. Can take on decimal values and has a clearly defined zero value. Q4 Discrete, ratio. Integer, clearly defined zero. Q5 Categorical, ordinal or interval. Interval if differences are equal. Learning Objective: 02-4 Learning Objective: 02-5
2.34
Q1 Categorical, ordinal or interval. Interval if differences are equal. Q2 Discrete, ratio. Integer, clearly defined zero. Q3 Continuous, ratio. Can take on decimal values and has a clearly defined zero value. Q4 Discrete, ratio. Integer, clearly defined zero. Q5 Categorical, ordinal. Ranking but not numerical so no calculations possible. Learning Objective: 02-4 Learning Objective: 02-5
11
ASBE 6e Solutions for Instructors 2.35
Q1 Continuous, ratio. Can take on decimal values and has a clearly defined zero value. Q2 Discrete, ratio. Integer, clearly defined zero. Q3 Categorical, ordinal or interval. Interval if differences are equal. Q4 Categorical, nominal. Binary, no ranking. Q5 Categorical, ordinal or interval. Interval if differences are equal. Learning Objective: 02-4 Learning Objective: 02-5
2.36
a. Cross-sectional. A single point in time: end of 2017. b. Time series. Data is collected over an 8 year time period. c. Time series. Data collected over 52 weeks. d. Cross-sectional. Single point in time: Christmas Day 2017. Learning Objective: 02-3
2.37
a. Time series. Data collected over 31 days in January. b. Cross-sectional. Single point in time: start of a particular semester. c. Cross-sectional. Single point in time: summary for a particular week. d. Time series. Data collected for the past 10 years. Learning Objective: 02-3
2.38
a. Census. It would be easy enough to count all of them. b. Sample. It would be too costly to track each can. c. Census. You can count them all quickly and cheaply. Learning Objective: 02-6
2.39
a. Census. This is assuming the company can easily generate the value from its human resource center. b. Sample. Impossible to observe prices of all cans in grocery stores. c. Census. This should be in Campbell Soup’s data base. Learning Objective: 02-6
2.40
a. Statistic. The data collected at your local supermarket would be a sample for the population of all soup sold by the company. b. Parameter. The population is all soup sold last year. c. Statistic. The sample consists of 10 students. Learning Objective: 02-6
2.41
a. Statistic. The week of visits is the sample. b. Parameter. The population is all books sold to date. c. Parameter. The population is all books sold. Learning Objective: 02-6
2.42
No, a census would be too difficult since this is an infinite population (people can continue to send e-mails). Learning Objective: 02-6
12
ASBE 6e Solutions for Instructors 2.43
a. The patient’s complaint. b. The number of patient visits is discrete numerical. The waiting time is continuous. Learning Objective: 02-3
2.44
a. Simple Random Sample. It is easy enough to use a computerized random number generator to choose 15 ports of entry. Learning Objective: 02-7
2.45
No a census could not be used. It would be impossible to ask each taxpayer how much time they spent in preparation. A sample is more appropriate. Learning Objective: 02-6
2.46
b. Cluster sampling. Easier to define geographic areas within a state where gasoline is sold. Gasoline stations are not everywhere, thus simple random sample or stratified sampling doesn’t make sense. Learning Objective: 02-7
2.47
a. Cluster sampling. It makes sense to take samples from geographic regions. b. No, population is effectively infinite. Learning Objective: 02-7
2.48
a. Sample – this information is most likely collected by a survey of a sample of customers. b. Census – this information can be collected from the point-of-sale system. c. Sample – this is most likely monitored by sampling coffee served. d. Census – this can be tracked on the point-of-sale system and will be population data. Learning Objective: 02-6
2.49
a. Census – this information is collected for all restaurants. b. Sample – this cannot be tracked for all customers, must be taken from a sample. c. Sample – this cannot be tracked for all customers, must be taken from a sample. d. Census – this can be tracked on the point-of-sale system and will be population data. Learning Objective: 02-6
2.50
Simple random sample or systematic sampling. A simple random sample is always best because it reduces bias. If it is truly random, every major stock fund was equally likely to be chosen. One way to do that is to: Create an excel spreadsheet with the funds listed and numbered Click on a separate cell and use the excel function =RANDBETWEEN(1,1699). This will give you one random number between 1 and 1,699. Whatever that number is can represent the first randomly chosen fund. For example, if the random number is 42, you would select the fund that you had listed under the #42. To get the other 20 randomly chosen funds, you would simply drag the bottom right corner of the cell that has the first number, to the next 19 cells below it. Another way to get a random sample is to use systematic sampling. For example, I might decide to take every 5th fund until I have 20. Pick a random starting point and then take every 20th fund from the starting point. 13
ASBE 6e Solutions for Instructors Learning Objective: 02-7 2.51
a. Cluster sample. Most likely choose businesses within a geographic region then take a random sample within the region. b. Cluster sample. Most likely choose practices within a geographic region then take a random sample within the region. c. Simple Random Sample (SRS), fairly accurate. d. The statistic is most likely based on sales data reported by cigarette companies. While the data does not come from a random sample, this information is available for almost all companies and therefore fairly accurate. Learning Objective: 02-7
2.52
a. This is a simple random sample. This population is effectively infinite because n = 780 and 780×20 = 15,600. This value is much less than N = 999,645. Learning Objective: 02-7
2.53
a. Cluster sampling, neighborhoods are natural clusters. b. Picking a day near a holiday with light trash. Learning Objective: 02-7 Learning Objective: 02-9
2.54
Yes, the population is effectively infinite. The population size, 30000, is at least 20 times greater than the sample size: 20×600 = 12000 < 30000. Learning Objective: 02-6
2.55
a. Yes, the population is effectively infinite because 18×20 < 11,000. b. 1/39 is the value from the sample therefore it is the statistic. Learning Objective: 02-6
2.56
Because 1200×20 = 24,000 and this value is less than the population we can consider the population effectively infinite. Learning Objective: 02-6
2.57
Education and income could affect who uses the no-call list. a. They won’t reach those who purchase such services. Same response for b and c. Learning Objective: 02-9
2.58
For each question, the difficulty is deciding what the possible responses should be and giving a realistic range of responses. Learning Objective: 02-9
2.59
a. Rate the effectiveness of this professor. 1 – Excellent to 5 – Poor. b. Rate your satisfaction with the President’s economic policy. 1 – Very Satisfied to 5 – Very dissatisfied. 14
ASBE 6e Solutions for Instructors c. How long did you wait to see your doctor? Less than 15 minutes, between 15 and 30 minutes, between 30 minutes and 1 hour, more than 1 hour. Learning Objective: 02-4 Learning Objective: 02-9 2.60
Ordinal measure. There is no numerical scale and the intervals are not considered equal. Learning Objective: 02-4 Learning Objective: 02-9
2.61
a. Ordinal. b. That the intervals are equal. Learning Objective: 02-4 Learning Objective: 02-9
2.62
a. A binary response scale. b. A Likert scale would be better. c. Self-selection bias. People with very bad experiences might respond more often than people with acceptable experiences. Learning Objective: 02-5 Learning Objective: 02-9
2.63
Answers will vary. Learning Objective: 02-2
2.64
Answers will vary. Learning Objective: 02-3
2.65
Answers will vary. Learning Objective: 02-7
2.66
Answers will vary. Learning Objective: 02-7
2.67
Answers vary for a-c; most appropriate method is simple random sampling (or stratified based on department). Learning Objective: 02-7
2.68
We can use the =RANDBETWEEN(1,52) function in excel to get random numbers which will allow us to choose 5 random cards. A stratified sample would not work since the cards are listed in order and spades and hearts are listed first. Therefore, if I chose every 5th card and stopped after 5 cards, I would not have any clubs or diamonds represented. Stratified sampling doesn't really make sense since there is already an equal number of each suit and the same numbers within each suit. Cluster sampling doesn't make sense since we are not concerned with geographic region. Judgment is not necessary when choosing playing cards, and convenience would be possible but not necessary and we want to avoid it if possible. 15
ASBE 6e Solutions for Instructors Learning Objective: 02-7 2.69
Answers will vary. Learning Objective: 02-7
2.70
Answers will vary. Learning Objective: 02-7
2.71
Answers will vary. Learning Objective: 02-7
16
ASBE 6e Solutions for Instructors
Chapter 3 Describing Data Visually 3.1
a. Stem and Leaf plot for stem unit = leaf unit =
#1 10 1
Frequency 1 5 7 10
Stem 0 1 2 3
1
4
Leaf 9 56889 1145667 1222334559 2
24
b.
Object 3
c. The stem and leaf plot appears to be left skewed with central tendency near 30 and variation between 9 and 42. The dot plot appears to be symmetric but with no clear central tendency and variation between 9 and 42. Learning Objective: 03-1 3.2
a. Stem and Leaf plot for stem unit = leaf unit = Frequency 4 2 4 11 5 4 1 0 0 1 32
Defect s 10 1 Stem 8 9 10 11 12 13 14 15 16 17
17
Leaf 3678 35 2679 01111334447 12669 0035 6
0
ASBE 6e Solutions for Instructors b.
Object 5
c. The stem and leaf plot appears to be slightly left skewed with central tendency near 110 and variation between 83 and 146. There is one outlier at 170. The dot plot shows a similar pattern to the stem and leaf plot. Learning Objective: 03-1 3.3
a. Sarah’s Calls:
Bob’s Calls:
b. Sarah makes more calls than Bob and her calls are shorter in duration. Learning Objective: 03-1 3.4
a. We would use 9 bins, each with a width of 20. The frequency distribution and histogram follow.
18
ASBE 6e Solutions for Instructors
Object 7
b. The distribution has central tendency between 40 and 60 days with most values falling between 0 and 100 days. The shape is somewhat uniform in this range. There are several outliers over 100 days: 121, 176 and 179 Days on Market. Learning Objective: 03-2 Learning Objective: 03-3 3.5
a. We would use 9 bins, each with a width of 10. The frequency distribution and histogram follow.
Object 9
19
ASBE 6e Solutions for Instructors b. The distribution has central tendency at approximately 80 with most values falling between 50 and 100. The shape is clearly skewed left. There are two outliers at 18 and 27. Learning Objective: 03-2 Learning Objective: 03-3 3.6
a. We would use 7 bins, each with a width of 20. The frequency distribution and histogram follow.
Object 11
b. Answers will vary. The histogram above shows a fairly symmetrical distribution with clear central tendency and a few high values. Using fewer bins will smooth the histogram and disguise the shape making it difficult to identify central tendency. Using too many bins (more than 15) will make the histogram appear “choppy” and also make it difficult to identify central tendency. Learning Objective: 03-2 Learning Objective: 03-3 20
ASBE 6e Solutions for Instructors
3.7
a. We would use 6 bins, each with a width of 100. The frequency distribution and histogram follow.
Object 14
b. Answers will vary. The histogram above shows a slightly skewed right distribution with clear central tendency. Using fewer bins will smooth the histogram and disguise the shape making it difficult to identify skewness. Using too many bins (more than 15) will make the histogram appear “choppy” and also make it difficult to identify central tendency and skewness. Learning Objective: 03-2 Learning Objective: 03-3
21
ASBE 6e Solutions for Instructors
3.8
a. The histogram below has bin widths of 5 starting at 0 and ending at 195 resulting in 39 bins. The frequency distribution table is too large to show. Notice that many of the bins have a frequency of 0.
Object 16
b. The histogram is strongly skewed right. c. There are two unusual values: 66.08 million and 192.92 million. Learning Objective: 03-2 Learning Objective: 03-3 Learning Objective: 03-4 3. 9
a. Suggest 7 bins with width 5. Sturges’ Rule shows the number of bins to be 6 with bin width (38.7-9.4)/6 = 4.88 so rounding up, 5. Sturges’ Rule is close. b. Suggest 8 bins with width 10. Sturges’ Rule shows the number of bins to be between 6 and 7 with bin width (85-12)/7 = 10.43 so 10. Sturges’ rule suggests fewer bins. c. Suggest 10 bins with width .15. Sturges’ Rule shows the number of bins to be 9 with bin width (3.71-2.25)/9 = 0.162. Sturges’ Rule is close. d. Suggest 8 bins with width .01. Sturges’ Rule shows the number of bins to be 8 with bin width (.097-.023)/8 = .00925 so rounding up, .01. Yes, Sturges’ rule does agree. Learning Objective: 03-3
3.10
a. The frequency distribution and histogram shown below are from MegaStat. We did not choose a bin # or bin width so it defaulted to 8 bins with a width of .20. Sturges' Rule results are below.
k = 1 + 3.3log(n) k = 1 + 3.3log(n) = 1 + 3.3(74) = 1 + 3.3(1.8692) = 7.1685 Approximately 7 bins. Using the bin width formula:
22
ASBE 6e Solutions for Instructors
7.97 - 6.54 = .20 7
43. These are similar to what MegaStat gives us.
b. The distribution appears to be slightly skewed right.
Object 18
c. This histogram has 11 bins with a width of .15. The shape is still skewed right.
Object 20
d. The histograms show the same distribution shape so increasing the number of bins did not change the perception. 23
ASBE 6e Solutions for Instructors Learning Objective: 03-2 Learning Objective: 03-3 Learning Objective: 03-4 3.11
a.
Object 22
b. Increasing trend until 2005, then sharp decrease. Learning Objective: 03-5 3.12
a.
Object 25
b. In spite of variation from year to year the number of visits showed a steady increase until a sharp drop in 2010-2011 season. Learning Objective: 03-5
24
ASBE 6e Solutions for Instructors
3.13
a. Sample default graph given. Answers will vary as to modification.
Object 27
b. The pattern shows a strong decrease from 1940 to 2015. Learning Objective: 03-5 3.14
a. Default graph presented, answers will vary with respect to modifications made.
Object 29
b. The number of kidney and liver transplants increased up to 2006 and since then all three organ transplants have stayed fairly constant. Learning Objective: 03-5
25
ASBE 6e Solutions for Instructors 3.15 Default graphs presented, answers will vary with respect to modifications made. a.
Object 31
b.
Object 33
c. The line chart is preferred by more people because the increasing trend is more evident in the line than in the columns. Learning Objective: 03-5 Learning Objective: 03-6
3.16
a. 26
ASBE 6e Solutions for Instructors
Object 35
b.
Object 37
Many individuals will see the 3-D as more distracting. When there are several graphs in a report it might be nice to add variety in the graph type. Consider using the easiest graph to read first when including many graphs in a report.
27
ASBE 6e Solutions for Instructors c.
Object 40
Stacked column charts can be useful to see how much a category makes up of a total. For each screen size the sales in the three years are similar while sales vary for different screen sizes. d.
Object 42
The labels are distracting because they overlap in some places. Learning Objective: 03-6
3.17
a. 28
ASBE 6e Solutions for Instructors
Object 44
b. Wait too long, Rude service, No real person on line c. Wait too long Learning Objective: 03-6 3.18
Sample default graphs presented for a, b, and c. a.
Object 47
b. 29
ASBE 6e Solutions for Instructors
Object 49
c.
Object 51
Answers will vary. The exploded pie chart makes the sizes and colors of the "pieces" stand out but many statisticians prefer the bar chart. It can be difficult for people to accurately compare areas of pie slices. The differences in lengths of the bars are easier for people to judge. Learning Objective: 03-7
30
ASBE 6e Solutions for Instructors 3.19
Sample default graphs presented for a, b, and c. Most statisticians prefer the bar chart. a. 2-D pie chart
Object 54
b. 3-D pie chart
Object 56
c. Bar chart 31
ASBE 6e Solutions for Instructors
Object 58
Learning Objective: 03-7 3.20
Sample default graphs presented for a and b.
Object 60
Object 62
32
ASBE 6e Solutions for Instructors Pie charts can get cluttered with this many pieces. The ideal number of slices is between 2 and 3 even though you often see more than that in the media. A bar chart may be more clear. Learning Objective: 03-7 3.21
a. Sample default graph presented.
Object 64
b. There is moderate, negative linear relationship. Learning Objective: 03-8 3.22
a. Sample default graph presented.
Object 66
b. There is a strong, negative linear relationship between vehicle weight and city MPG. Learning Objective: 03-8
3.23
a. Sample default graph presented. 33
ASBE 6e Solutions for Instructors
Object 68
b. The relationship shows a moderate, positive linear trend between coupon redemption and revenue. Learning Objective: 03-8 3.24
a. Sample default graph presented.
Object 70
b. There is a weak, positive linear relationship. Learning Objective: 03-8
3.25
a. Stem and Leaf plot for
Power Outage Time (min)
34
ASBE 6e Solutions for Instructors stem unit = leaf unit = Frequency 3 6 5 4 1 3 2 0 1
10 1 Ste m 0 1 2 3 4 5 6 7 8
1
9
Leaf 249 227778 14557 0025 4 013 26 4 9
26
b.
Object 73
c. The distribution is skewed right. Learning Objective: 03-1 Learning Objective: 03-3 Learning Objective: 03-4
3.26
a. Stem and Leaf plot for
35
Weight
ASBE 6e Solutions for Instructors stem unit = leaf unit =
s 10 1 Lea f 8 129 5
Frequency 1 3 1 0 5
Stem 2 3 4 5 6
2 5
7 8
23457 2 4 04666
3
9
012
20
b.
Object 75
c. The distribution appears to be bimodal with one central tendency approximately at 30 ounces and one central tendency approximately 80 ounces. Learning Objective: 03-1 Learning Objective: 03-3 Learning Objective: 03-4
3.27
a. 36
ASBE 6e Solutions for Instructors
b.
Object 77
c. The distribution is skewed to the right. Almost half the observations are between 1.2 and 1.6. There are two unusual values at 2.4 and 2.76. Learning Objective: 03-1 Learning Objective: 03-2 Learning Objective: 03-3 Learning Objective: 03-4
3.28
a. Stem and Leaf plot for
Win
37
ASBE 6e Solutions for Instructors stem unit = leaf unit = Frequency 21 29 18 7 7 8 4 4 0 2 100
Margin 10 1 Stem 0 0 1 1 2 2 3 3 4 4
Leaf 111111111223333333444 00055566777777777778888889999 000001112333444444 6677788 0001134 55678899 1233 5669 99
b.
38
ASBE 6e Solutions for Instructors
Object 79
c. Distribution is heavily skewed to the right, ranging from 1 to 49 points. Learning Objective: 03-1 Learning Objective: 03-2 Learning Objective: 03-3 Learning Objective: 03-4
3.29
a.
Object 81
b. The MegaStat default frequency distribution has 29 bins with a width of 1 minute. Many of the bins have a frequency of zero. (The table is too long to produce here.) 39
ASBE 6e Solutions for Instructors
Object 84
If choosing a bin width equal to 3 the shape is still the same but fewer bins with zero observations.
Object 87
c. The distribution is skewed to the right. Three-fourths of the calls are 5 minutes or less in duration. Learning Objective: 03-1 Learning Objective: 03-2 Learning Objective: 03-3 Learning Objective: 03-4
40
ASBE 6e Solutions for Instructors 3.30
a. The range of batting averages is from 0 to .42. Bin widths of .1 would result in 5 bins, bin widths of .05 would result in 9 bins. 9 bins seems a better choice.
b.
Object 89
The distribution appears to be skewed to the left with central tendency of .25. c.
Object 92
d. The histogram above has bin widths of .1. The distribution appears more symmetrical. Learning Objective: 03-2 Learning Objective: 03-3 Learning Objective: 03-4
41
ASBE 6e Solutions for Instructors 3.31
a.
b. The scatter plot shows a strong, positive linear relationship. Cockpit noise is related to airspeed: the faster the airspeed, the higher the sound level will be. Learning Objective: 03-8 3.32
a.
Object 94
b. There is a strong positive relationship between Dolana and ZalParm Stock Prices. c. The line chart below shows that the stock prices are moving together however there is much more weekly variation in the ZalParm price.
42
ASBE 6e Solutions for Instructors
Object 97
Learning Objective: 03-8
3.33
a.
Object 99
b. The relationship is negative and nonlinear. It is a moderate relationship. Learning Objective: 03-8 3.34
a. Column chart with 3D visual effect. b. Strengths: Column chart shows increasing trend. Weaknesses: distracting picture, Title doesn’t make sense; Non-zero origin; 3D effect does not add to presentation; tick marks do not stand out, measurement units on Y axis missing; vague source c. Correct weaknesses as noted. A line chart would be the preferred choice by most analysts. A zero origin on Y axis would show realistic differences in years. Learning Objective: 03-6 Learning Objective: 03-10
3.35
a. Bar chart with 3D visual effect. 43
ASBE 6e Solutions for Instructors b. Strengths: Good proportions, no distracting pictures. Weaknesses: No labels on X and Y axes, Title unclear, 3D effect does not add to presentation. c. Correct weaknesses as noted. 2D bar chart with zero origin on X axis would improve chart. Learning Objective: 03-6 Learning Objective: 03-10 3.36
a. Line chart b. Strengths: labels on X and Y axes, measurement included on Y axis, Good title; No distracting pictures; good use of gridlines. Weaknesses: large gap between time series due to difference in magnitude, no source. c. Correct weaknesses as noted, use of logarithmic scale would correct proportion issue and show growth rates more clearly. Learning Objective: 03-6 Learning Objective: 03-10
3.37
a. Exploded pie chart. b. Strengths include details on names and values, differentiating between types of countries, good use of color. c. Improvement might be percentage shares instead of actual values, and include just the total level of imports. Might consider a sorted column chart with OPEC and Non-OPEC countries color coded. Learning Objective: 03-7 Learning Objective: 03-10
3.38
a. Line chart. b. Strengths: labels on Y axes, use of gridlines, and source provided. Weaknesses: non-zero origin, distracting pictures, dramatic title. c. Correct weaknesses as noted, keep graph as line graph. Learning Objective: 03-6 Learning Objective: 03-10
3.39
a. Standard pie chart b. Strengths: source identified, answers the question posed. Weaknesses: “Other” category quite large. c. Might change title: Distribution of Advertising Dollars in the United States, 2001. Would add the total dollars spent on advertising as well. Learning Objective: 03-7 Learning Objective: 03-10
3.40
a. Column chart using pictures for columns b. Strengths: label on Y axis, zero origin, Short descriptive title, detailed, accurate source. Weaknesses: use of Area Trick to make 2000 appear much larger than 1980 values. c. Correct weaknesses as noted, regular column chart would be better choice. Learning Objective: 03-6 Learning Objective: 03-10
3.41
a. Line Chart with pictures 44
ASBE 6e Solutions for Instructors b. Strengths: label on X and Y axes, use of gridlines. Weaknesses: sensational title, distracting pictures, non-zero origin. c. Correct weaknesses as noted, keep graph as line graph. Learning Objective: 03-6 Learning Objective: 03-10 3.42
a. Column chart using pictures for columns b. Strengths: zero origin, short, descriptive title, detailed. Weaknesses: use of area trick to make 2010 appear much larger than 2005 values, no source. c. Correct weaknesses as noted, regular column chart would be better choice. Learning Objective: 03-6 Learning Objective: 03-10
3.43
a.
Object 102
b. A table, bar, or pie chart would also work. A line chart would not be appropriate because this is not a time series. Learning Objective: 03-6
3.44
a. Pareto Chart shown. 45
ASBE 6e Solutions for Instructors
Object 105
b. A bar or pie chart would work. Learning Objective: 03-6 3.45
a.
Object 107
b. A pie chart would also work. Learning Objective: 03-6 Learning Objective: 03-7
3.46
a. 46
ASBE 6e Solutions for Instructors
Object 109
3.47
b. We used a line chart because we are comparing two sets of data and want to see how both have changed over time. A column chart could be used as well with the "United States" set of bars a different color from the "Rest of the World" set of bars but side by side. Learning Objective: 03-5 Learning Objective: 03-6 a.
Object 111
b. A bar chart or column chart would also work. Learning Objective: 03-6 Learning Objective: 03-7
3.48
a. 47
ASBE 6e Solutions for Instructors
Object 113
b. A bar chart or pie chart would also work. Learning Objective: 03-6 Learning Objective: 03-7 3.49
a.
Object 115
b. Verizon, AT&T and T-Mobile account for 80 percent of all service providers for this sample group. Learning Objective: 03-5 Learning Objective: 03-6
48
ASBE 6e Solutions for Instructors
Chapter 4 Descriptive Statistics 4.1
a. Mean = 2.83, Median = 1.5, Mode = 0. Any one of these would be a useful measure of center. Data appears skewed right with many values equal to zero so one might choose median or mode over mean. b. Mean = 68.33, Median = 72, Mode = 40. This is continuous numerical data so either the mean or the median would be best. c. Mean = 304.38, Median = 303, No mode. This is continuous numerical data so either the mean or the median would be best. Learning Objective: 04-1 Learning Objective: 04-2
4.2
a. This is attribute data so mode is the only measure of central tendency possible. Mode = M, which occurs 9 out of 12 observations. b. This is discrete data with a small range so mode could be an appropriate measure of central tendency. Mode = 18, which occurs 6 out of 10 observations. c. There is no mode in this case because there is no value that occurs more than once. Therefore, either mean or median would be more appropriate. Mean = 29.5 and median = 29. Learning Objective: 04-1 Learning Objective: 04-2
4.3
a. No, the mode would not be a good measure of center because all data values are unique. b. Yes, the mode would work well because this is categorical data with repeated values. The median would also work as a measure of center. (Mode = C) c. No, the mode would not be the best choice. The data is continuous numerical and either the median or the mean would be a better choice. Learning Objective: 04-1 Learning Objective: 04-2
4.4
a. Either of the measures would work because the data set is symmetrical. Mean = Median =3 b. The median is a better choice because the data set is slightly skewed left. Mean = 6.53 and median = 6.95. c. The median is a better choice than the mean because the data is skewed right. Mean = 1.11 and median = 0. Learning Objective: 04-1 Learning Objective: 04-2 Learning Objective: 04-11
49
ASBE 6e Solutions for Instructors 4.5
a. Continuous data, skewed right, no mode. Median is the best choice. Median = 26.2 mpg and mean = 28.1 mpg. b. Mostly 1 rider therefore the median or the mode would be the best choice. Median = mode = 1. c. Symmetric distribution. The mean and median are the same and there are two modes. The mean or the median are the best choice. Mean = median = 3. (Modes = 2 and 4.) Learning Objective: 04-1 Learning Objective: 04-2 Learning Objective: 04-11
4.6
a. Mean = 56.22, Median = 47, Mode = 44. b. The distribution is most likely skewed right because the mean is larger than the median and mode. c. The mode is not a useful measure of central tendency because there are few values that repeat. The value 44 appears only three times out of 36 observations. Learning Objective: 04-1 Learning Objective: 04-2 Learning Objective: 04-11
4.7
a. Mean = 75.5, Median = 80.5, Mode = 93. b. The distribution is most likely skewed left because the mean is less than the median and mode. c. The mode is not a useful measure of central tendency because there are few values that repeat. The value 93 appears only three times out of 24 observations. Learning Objective: 04-1 Learning Objective: 04-2 Learning Objective: 04-11
4.8
a.
Count Mean Media n
Monday 10 72.00
Tuesday 10 72.00
Wednesday 10 76.00
Thursday 10 76.00
72.00 60.00
72.00 65.00
72.00 72.00
86.50 none
Note that both Tuesday and Wednesday have multiple modes. Tuesday has modes at 65 and 79 (4 each.) Wednesday has modes at 72 and 74 (2 each.) b. No, they don’t agree for all days. The mean and the median are the same for Monday and Tuesday, and the mode and the median are the same for Wednesday. c. The mode is an unreliable measure of central tendency for quantitative data. Where the mean and median disagree, one should look at the shape of the distribution to see which measure is more appropriate. d. Monday and Tuesday have a symmetric distribution. Even though the mode is not equal to the mean and median it is sufficient to say they are symmetrical given that the mean equals the median. Wednesday is skewed right because the mean is to the right of (or 50
ASBE 6e Solutions for Instructors greater than) the median. Thursday is skewed left because the mean is to the left of (or less than) the median. e. On average the facility processed more cars on Wednesday and Thursday. Learning Objective: 04-1 Learning Objective: 04-2 Learning Objective: 04-11 4.9
a. Descriptive Statistic count mean mode median
Data 32 27.34 26.00 26.00
b. The mode and median are the same value, but the mean is greater than the median. c.
Histogram 30 25 20 15 Percent 10 5 0
ATM customers
d. One might say data is slightly skewed to the right because the mean is greater than the median. It is difficult to conclude shape by looking at the graphs. Learning Objective: 04-1 Learning Objective: 04-2 Learning Objective: 04-11
51
ASBE 6e Solutions for Instructors 4.10
a. Descriptiv e Statistics count mean median mode
Data 28 107.25 106.00 95.00
b. The mean and median are close in value but the mode is more than $10 lower. c.
Histogram 35 30 25 20 15 Percent 10 5 0
Dinner Amount
d. When looking at the dot plot one might say data is slightly skewed to the right. This would also be logical because the mean is greater than the median. It is difficult to conclude shape by looking at the graphs. Learning Objective: 04-1 Learning Objective: 04-2 Learning Objective: 04-11 4.11
b. Descriptive Statistics count mean median mode
52
Data 65 4.27 2.00 1.00
ASBE 6e Solutions for Instructors c. No. The mean is more than twice the median and the median is twice the mode. d. The data are skewed to the right with a few high outliers at 20, 26, and 29 minutes. Learning Objective: 04-1 Learning Objective: 04-2 Learning Objective: 04-11 4.12
a. Median = 68, Midrange = 68, Geometric Mean = 67.37 b. Median = 303, Midrange = 295.5, Geometric Mean = 300.9 c. Median = 1.5, Midrange = 7.5, Geometric Mean is undefined. The midrange is not very robust because it is sensitive to extreme data such as the 94 for one exam score and the 15 for number of absences. The median and geometric mean appear to be close in value for the first two data sets. However, we might want to rely on the median because the data do not appear to have a steep increasing trend and the median is a more easily understood and recognized measure of central tendency. Learning Objective: 04-1 Learning Objective: 04-2 Learning Objective: 04-11
4.13
a. =TRIMMEAN(A1:A50,0.20) b. There are 50 observations. Ten percent of 50 equals five. So five values will be dropped from each tail. c. If five values are dropped from each tail this is a total of 10 values. Learning Objective: 04-2
4.14
Excel’s TRIMMEAN function multiplies the trimmed percentage by the number of observations and then truncates to the next lower integer. For the function TRIMMEAN (Data, .10) Excel will multiply by .05. a. For n = 41, .05×41 = 2.05. Excel would trim 2 values from each tail. b. For n = 66, .05×66 = 3.3. Excel would trim 3 values from each tail. c. For n = 83, .05×83 = 4.15. Excel would trim 4 values from each tail Learning Objective: 04-2
4.15
a. Mean = 100, Median = 0, Mode = 0, Midrange = 325, Geometric Mean is undefined because there are values equal to 0. b. Choose the mean. If planning for expected work-related medical expenses of $100 for eight officers the budget should be fairly close. Using the midrange would overestimate total expected expenses by quite a bit. Learning Objective: 04-1 Learning Objective: 04-2 Learning Objective: 04-11
53
ASBE 6e Solutions for Instructors
4.16
a. Mon 1 1 1 1 5 5 5 6 6 9
Tue 1 1 3 3 4 6 6 7 7 9
Wed 0 0 0 0 4 6 6 6 6 10
Thu 1 1 1 1 1 1 1 1 1 10
b. Descriptive Statistics mean median mode midrange geometric mean
Mon 4 5 1 5 2.89
Tue 4.7 5 1 5 3.76
Wed 3.8 5 0 5 NA
Thu 1.9 1 1 5.5 1.26
trimmed mean
3.75
4.62 5
3.5
1
Tuesday has multiple modes at 1, 3, 6 and 7 and Wednesday has multiple modes at 6 and 0. c. The geometric mean and mode are very different from the other measures. The mean, median, and midrange are close in value. d. The mean or median are better measures of central tendency for this type of data. The midrange is sensitive to extreme values and ignores most of the data. The geometric mean is less familiar to most people and the trimmed mean may exclude relevant data values. Learning Objective: 04-1 Learning Objective: 04-2 Learning Objective: 04-11 4.17
a. mean midrange geometric mean trim mean
27.34 25.50 26.08 27.46
b. The mean and the trimmed mean are similar in value and they are both greater than the midrange and the geometric range. Learning Objective: 04-1 54
ASBE 6e Solutions for Instructors Learning Objective: 04-2 4.18
a. mean midrange geometric mean trimmed mean
107.25 114 102.48 106
b. The mean and trimmed mean are similar and fall between the geometric mean and midrange. Learning Objective: 04-1 Learning Objective: 04-2 4.19
a. mean median mode midrange geometric mean trimmed mean
4.27 2.00 1.00 15.00 2.48 2.75
b. The data are skewed to the right. The mean is greater than the median. The midrange is much greater than the mean. Note: for the 10% trimmed mean use Excel’s function =TRIMMEAN(array, .20). Learning Objective: 04-1 Learning Objective: 04-2 Learning Objective: 04-11 4.20
8.78%. See calculation below.
GR = n-1
xn 139.1 - 1 = 8-1 - 1 = .0878 or 8.78%. x1 77.2
Learning Objective: 04-2 4.21
a.
Mean Sample Standard Deviation
Sample A: 7 1
55
Sample B: 62 1
Sample C: 1001 1
ASBE 6e Solutions for Instructors b. The midpoint of each sample is the mean. The other 2 data points are exactly 1 standard deviation from the mean. The idea is to illustrate that the standard deviation is not a function of the value of the mean. Learning Objective: 04-1 Learning Objective: 04-2 Learning Objective: 04-3 4.22
a. Mean b. Sample Standard Deviation c. Population Standard Deviation
Data Set A: 7.0000 1.0000 0.8165
Data Set B: 7.0000 2.1602 2.0000
Data Set C: 7.0000 3.8944 3.7417
d. The sample standard deviation is larger than the population standard deviation for the same data set. This makes sense because in the formula for sample standard deviation we divide by n-1 instead of n. This exercise shows us that samples can have similar means, but different standard deviations. We cannot get a sense of what the standard deviation is just by looking at the mean. Learning Objective: 04-1 Learning Objective: 04-2 Learning Objective: 04-3 4.23
The coefficient of variation for the hybrid vehicle is 2.2/43.2 = .051 or 5.1%. The coefficient of variation for the gasoline vehicle is 1.9/27.2 = .070 or 7%. The hybrid had more consistent gas mileage relative to the mean than the gasoline vehicle. Learning Objective: 04-3
4.24
The coefficient of variation for Bob is 9.1/182 = 0.05 or 5%. The coefficient of variation for Cedric is 7.6/152 = .05 or 5%. Bob and Cedric have the same relative consistency. Learning Objective: 04-3
4.25
=AVEDEV(1,2,3,4,5,6,7,8,9,10) = 2.5 Learning Objective: 04-3
4.26 n
å | xi - x |
MAD = i =1
n
.
|12 - 20 | + |18 - 20 | + | 21 - 20 | + | 22 - 20 | + | 27 - 20 | 5 8 + 2 + 1 + 2 + 7 20 = = = 4. 5 5 Using x = 20, MAD =
56
ASBE 6e Solutions for Instructors Learning Objective: 04-3
4.27
a. Stock Stock A Stock B Stock C
s/x 5.25/24.50 12.25/147.25 2.08/5.75
CV 21.43% 8.32% 36.17%
b. Stock C, the one with the smallest standard deviation and smallest mean, has the greatest relative variation. c. The stocks have different average values therefore directly comparing the standard deviations is not a good comparison of risk. The variation relative to the mean value is more appropriate. Learning Objective: 04-3 4.28
a. The ATR coefficient of variation for Retail: 23.18%, for Utilities: 18.82%, for Financial: 39.97% b. The financial sector has the highest relative variation. c. If one were to compare standard deviation instead of CV then it would appear that the retail sector has the highest variability. Learning Objective: 04-2 Learning Objective: 04-3
4.29
a.
4.30
a. Chebychev's theorem states that at least 75% of the data will fall within ± 2 standard deviations so: .75×200 = 150. b. The Empirical Rule states that 95.44% would fall in this range so: .9544×200 = 191. Learning Objective: 04-4 Learning Objective: 04-5
4.31
90-70 = 70-50 = 20. 20/10 = 2. Chebychev’s theorem states that at least 75% of the data will fall within ± 2 standard deviations so: .75×400 = 300. Learning Objective: 04-4
xA = 6.86, sA = 1.497, xB = 7.24, sB = 1.209 b. CVA = 1.497/6.86 = 0.218 or 21.8%, CVB = 1.209/7.24 = 0.167 or 16.7%. c. Consumers preferred sauce B. Sauce B also had more consistent ratings relative to the average. Learning Objective: 04-2 Learning Objective: 04-3
57
ASBE 6e Solutions for Instructors
4.32
a. z = (1325-875)/219 = 2.055. b. Because John’s standardized score is greater than 2 but less than 3 we would consider his rent unusual but not an outlier. c. John’s rent would be considered an outlier if |John’s z score| > 3. Set and rent - 875 z= 219 solve for rent. Rent = $1532. John’s rent would have to be $1532 or higher to be considered an outlier. Learning Objective: 04-5 Learning Objective: 04-6
4.33
a. z = (30-18.5)/4.8 = 2.396. b. Because the account’s standardized score is greater than 2 but less than 3 we would consider his rent unusual but not necessarily an outlier. c. An account would be considered an outlier if |z score| > 3. Set and days - 18.5 z =3= 4.8 solve for days. Days = 32.9. The account would have to be 33 days or longer to be considered an outlier. Learning Objective: 04-5 Learning Objective: 04-6
4.34
a. z = (92-46)/13 = 3.538. Yes this is an outlier because z > 3. b. z = (583-723)/69 = -2.02. Yes, this is an unusual observation because z < -2.0. c. z = (28-22)/7 = 0.857. No, this is not an outlier nor is it unusual because -2.0 < z < 2.0. Learning Objective: 04-5 Learning Objective: 04-6
4.35
a. z = (91-79)/5 = 2.4. Sam’s grocery bill of $91 was 2.4 standard deviations greater than his average of $79. b. z = (3.18-2.87)/.31 = 1. Mary’s GPA of 3.18 was 1 standard deviation above the mean of 2.87. c. z = (18-15)/5 = 0.6. Jamie’s weekly study hours of 18 hours was .6 standard deviations above the mean of 15 hours. Learning Objective: 04-5 Learning Objective: 04-6
4.36
The Empirical Rule states that 99.73% of the data will fall within ± 3 standard deviations. Therefore, we know that almost the entire range of data falls within 6 standard deviations. We use the formula:
s=
R (126.2 - 109.7) 16.5 = = = 2.75 6 6 6
58
ASBE 6e Solutions for Instructors
4.37
4.38
and conclude that our estimate of σ is 2.75. Learning Objective: 04-5 Learning Objective: 04-6 For each part below use the formula . Plug in the values for z, μ, and σand solve x-m z= s for x. a. Bob’s GPA: . x - 2.98 1.71 = , x = 1.71(.36) + 2.98 = 3.596 .36 b. Sarah’s weekly work hours: . x - 21.6 1.18 = , x = 1.18(7.1) + 21.6 = 29.978 7.1 c. Dave’s bowling score: . x - 150 -1.35 = , x = -1.35(40) + 150 = 96 40 Learning Objective: 04-5 Learning Objective: 04-6 a. The z scores are in the table below: Standardized Value for Number of customers during the noon hour (n = 30 days). -0.2985 1.2298 -0.5532 -0.1711 0.3383 1.6119 -0.2985 -0.1711 1.4845 0.5930 -0.8079 -0.1711 -1.0626 -0.0438 0.5930 -0.2985 -1.1900 -0.1711 0.8477 -1.1900 0.4657 0.9751 -0.8079 0.7204 0.7204 -2.3362 -1.4447 0.5930 0.9751 1.8666 -1.5721 -0.4259
b. From Megastat: empirical rule mean - 1s mean + 1s percent in interval (68.26%)
19.49 35.20 68.8%
mean - 2s mean + 2s percent in interval (95.44%)
11.64 43.05 96.9%
mean - 3s mean + 3s percent in interval (99.73%)
3.79 50.90 100.0%
There are no standardized values greater than 3 or less than 3 so no official outliers. There is one standardized value less than 2. Observation 9 has a z score = 2.3362 so we would consider this an unusual value. c. 68.8% of the data values fall within 1σ, more than 96.9% fall within 2σ, and 100% fall within 3σ. These percentages are slightly greater than the percentages from the Empirical rule but not too different. A normal distribution seems reasonable. Learning Objective: 04-5 Learning Objective: 04-6 59
ASBE 6e Solutions for Instructors
4.39
a. The z scores are in the table below: Standardized Value of Lengths of 60 calls initiated during the last week of July -0.5558 -0.3857 0.9756 0.1248 -0.2155 -0.2155 -0.3857 2.6772 -0.5558 0.2949 -0.2155 1.4860 -0.3857 -0.3857 -0.5558 3.6981 -0.2155 -0.5558 -0.5558 -0.3857 -0.5558 0.4651 -0.5558 -0.3857 -0.2155 -0.5558 -0.3857 -0.5558 -0.0454 -0.3857 -0.3857 4.2086 -0.5558 -0.5558 -0.5558 0.6353 -0.5558 -0.0454 -0.3857 -0.5558 -0.5558 -0.5558 -0.5558 0.2949 -0.5558 -0.2155 -0.2155 0.2949 -0.5558 -0.2155 -0.5558 -0.5558 0.1248 -0.5558
-0.5558 -0.2155 1.3159 0.1248 -0.3857 2.3368
b. From MegaStat: empirical rule mean - 1s mean + 1s percent in interval (68.26%)
-1.61 10.14 90.0%
mean - 2s mean + 2s percent in interval (95.44%)
-7.49 16.02 93.3%
mean - 3s mean + 3s percent in interval (99.73%)
-13.36 21.90 96.7%
There are two observations that would be considered outliers according to the Empirical Rule. The values 26 (z = 3.67) and 29 (z = 4.18) are both more than 3 standard deviations beyond the mean. The values 18 (z = 2.30) and 20 (z = 2.64) are both more than 2 standard deviations beyond the mean. These would be considered unusual values. c. There are more observations within 1σ of the mean than the empirical rule would indicate, 90% vs. 68.26%. There are fewer observations within 2σ the mean that the empirical rule would indicate, 93.3% vs. 95.44%. Data do not seem to be from a normal distribution. Learning Objective: 04-5 Learning Objective: 04-6 4.40
a. 60
ASBE 6e Solutions for Instructors
b. The long left whisker suggests left-skewness. Learning Objective: 04-7 Learning Objective: 04-8 Learning Objective: 04-11 4.41
a.
b. Strongly skewed right. Learning Objective: 04-7 Learning Objective: 04-8 Learning Objective: 04-11 4.42
Lower inner fence: 21 – 1.5(33 – 21) = 3. Upper inner fence: 33 + 1.5(33 – 21) = 51. The value 45 would not be considered an outlier because 3 < 45 < 51. Learning Objective: 04-8
4.43
Lower inner fence: 160 – 1.5(170 – 160) = 145. Upper inner fence: 170 + 1.5(170 – 160) = 185. The value 149 would not be considered an outlier because 145 < 149 < 185. Learning Objective: 04-8
4.44
a. Estimate Q1 = 32, Q2 = 38, and Q3 = 46 customers. b. Approximately 64 customers were served on the busiest day and 20 customers on the slowest day. c. The distribution appears fairly symmetric because the boxes are approximately equal in width and the whiskers are approximately equal in length. Learning Objective: 04-8 Learning Objective: 04-11
4.45
a. Estimate Q1 = 3300, Q2 = 3900, and Q3 = 4300 vehicles per day. b. xmin ≈ 2400 and xmax ≈ 4800. 61
ASBE 6e Solutions for Instructors c. The distribution appears left skewed because the left whisker is much longer than the right whisker. Learning Objective: 04-8 Learning Objective: 04-11
4.46
a. count 1st quartile median 3rd quartile interquartile range
Data 32 21.50 26.00 33.00 11.50
The first quartile tells us that 25% of the data is at or below 21.50 and the third quartile tells us that 75% of the data is at or below 33.00. The median is the same as the 2nd quartile which tells us that 50% of our data is at or below 26.00. b. The midhinge . This midhinge value indicates we have Q1 + Q3 21.5 + 33.0 = = = 27.25 2 2 right skewed data because it is greater than the median. c.
This boxplot shows us that our range of data is from 9 to 42 and that the median number of customers is 26. Days with 22 or fewer customers are in the bottom quartile. Days with 33 or more customers are in the upper quartile. The box plot indicates that there are no extreme values or outliers and that our data is right skewed. The longer whisker on the right indicates right-skewness just as the midhinge value showed. Note that the boxplot was created on MegaStat and the quartiles in part (a) were calculated using Excel 2010 functions. There may be slight discrepancies in quartile values. Learning Objective: 04-2 Learning Objective: 04-7 62
ASBE 6e Solutions for Instructors Learning Objective: 04-8 Learning Objective: 04-11 4.47
a. count 1st quartile median 3rd quartile interquartile range
Data 60 1 2 4.75 3.25
The first quartile tells us that 25% of the data is at or below 1 minute and the third quartile tells us that 75% of the data is at or below 4.75 minutes. Note: Be sure to use Excel’s =QUARTILE.EXC function. The older =QUARTILE function will give the wrong value for the 3rd quartile. b. The midhinge . The midhinge indicates the data could be Q1 + Q3 1 + 4.75 = = = 2.875 2 2 skewed to the right because it is greater than the median. BoxPlot
0
5
10
15
20
25
30
35
Data
c. The box plot confirms the quartile calculations and reveals that there are 7 calls of unusually large duration, 4 of them extreme in length. Learning Objective: 04-2 Learning Objective: 04-7 Learning Objective: 04-8 Learning Objective: 04-11 4.48
a.
63
ASBE 6e Solutions for Instructors
20 15 Y
10 5 0 20
30
40
50
60
70
80
90
100
X
To calculate the sample correlation coefficient we can use the formula or use the =CORREL(XData, Y Data) n
r=
å ( x - X )( y - Y ) i =1
n
i
i
n
å (x - X ) å ( y - Y ) i =1
function in
2
i
i =1
excel. This gives us
negative, linear
2
i
r = -.8841
. There appears to be a strong,
relationship.
b.
25 20 15 Y 10 5 0 10
20
30
40
50 X
64
60
70
80
90
100
ASBE 6e Solutions for Instructors Again we can use the equation or excel and the result is
r=
.90875. There appears
to be a strong, positive, linear relationship. c.
25 20 15 Y 10 5 0 10
20
30
40
50
60
70
80
X
Again, we can use the formula or excel and the result is
r=
.1704. There appears
to be a weak, positive, linear relationship. Learning Objective: 04-9 4.49
a. r=
sxy sx s y
=
-36.111 = -.8571 4.638 ´ 9.084
b. There is a strong, negative linear association between the two stock prices. Learning Objective: 04-9 4.50
a.
sxy
48.724 = .5401 sx s y 11.724 ´ 8.244 b. The relationship between X and Y is a moderate positive linear relationship. c. The correlation coefficient is unit-free and therefore is easier to interpret than the covariance. The covariance magnitude is dependent on the magnitude of the variables so you cannot compare covariance of different pairs of variables. Learning Objective: 04-9 r=
4.51
=
a.
65
ASBE 6e Solutions for Instructors
900 800 Selling Price ($1000)
700 600 500 2,000 2,500 3,000 3,500 4,000 Square Feet
4.52
b. Using Excel’s correlation function, =CORREL(Xdata, Ydata), r =.8338. c. There appears to be a strong, positive, linear relationship. Learning Objective: 04-9 a. Mean = 44.09 From 0 20 40 60
To 20 40 60 80 Total
frequency (f) 5 12 18 9 44
midpoint (m) 10 30 50 70
Average = 44.09 Standard Deviation = 18.38
f(m− ) x 2 f×m 50 5810.95 360 2382.64 900 628.51 630 6041.53 1940 14863.64 CV =
41.69%
Learning Objective: 04-10 4.53
a. Mean = 8.48 From 2 6 10 14 18
To 6 10 14 18 22 Total
frequency (f) 7 12 3 2 1 25
midpoint (m) 4 8 12 16 20
Average = 8.48 Standard Deviation = 4.13
Learning Objective: 04-10 4.54
a. Number of email accounts by each student: 66
f(m− f×m 28 96 36 32 20 212 CV =
x
)
2
140.49 2.76 37.17 113.10 132.71 426.24 48.69%
ASBE 6e Solutions for Instructors
Mean =
1+1+1+1+ 2 + 2 + 2 + 3 + 3 + 3 + 3 + 3 = 2.083 12 Median = 2 Mode = 3
Number of siblings: Mean =
0 + 1 + 2 + 2 + 10 =3 5 Median = 2 Mode = 2
Asset ratios: Mean =
1.85 + 1.87 + 2.02 + 2.05 + 2.11 + 2.18 + 2.29 + 3.01 = 2.1725 8 Median = 2.05 + 2.11 = 2.08 2 There is no mode.
b. Number of email accounts: The mode is typically not the best measure of center for numerical data. In this example the mean, median, and mode are similar in value but the mean or median is probably best. Number of siblings: The mean is probably the weakest because of the high outlier at 10. Asset ratios: The mode is the weakest because there are 8 unique values so no mode exists. Learning Objective: 04-1 Learning Objective: 04-2 4.55
Using Chebyshev’s Theorem, at least 75% fall within two standard deviations of the mean. So 2.02 ± 2(0.22) gives the range [1.58, 2.46]. Learning Objective: 04-4
4.56
a. 100m dash times: Mean =
9.87 + 9.98 + 10.02 + 10.15 + 10.36 + 10.36 = 10.12 6
67
ASBE 6e Solutions for Instructors Median =
10.02 + 10.15 = 10.085 2 Mode = 10.36 Number of children:
Mean =
0 +1+1+ 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 6 =2 13 Median = 2 Mode = 2
Number of cars in driveway: Mean =
0 + 0 +1+1+ 2 + 2 + 3 + 5 = 1.75 8 Median = 1+ 2 = 1.5 2 Mode = 0, 1, and 2
b. 100m dash times: The mode is the weakest because all of the values fall at or below 10.36. We also know that the mode should be used for a small range of discrete data or attribute data. This is a small range of continuous data. Number of children: All measures of central tendency in this case have the same value of 2 which is a strong indicator of a "typical" data value Number of cars in driveway: The mode is the weakest because there are three different values for mode and only five unique values in the entire data set. Learning Objective: 04-1 Learning Objective: 04-2 4.57
Using Chebychev’s theorem: 1− (1/32) = .8889. There is at least 88.89% within 3 standard deviations which is the range from 71 to 119. Learning Objective: 04-4
4.58
The mean of 396 is 10 gm away from the upper and lower range values. Because the standard deviation is 5 gm, we are looking for the percent falling within 2 standard deviations of the mean (10/5 = 2). Chebychev’s theorem: 1 – (1/22) = .75, .75 × 200 = 150. At least 150 bags will weigh between 386 and 406 gm. Learning Objective: 04-4
68
ASBE 6e Solutions for Instructors 4.59
a. z = (0.2761 – 0.2731)/ 0.000959 = 3.128. b. This shipment would be considered an outlier because the standardized score is greater than 3. Learning Objective: 04-5 Learning Objective: 04-6
4.60
a. Bob's standardized z score is
1430 - 1340 =1 90 b. No, his score is not unusual because it is within 2 standard deviations of the mean. The Empirical Rule states that a z score would have to be outside of 2 standard deviations from the mean to be unusual. Learning Objective: 04-5 Learning Objective: 04-6 For each part below use the formula . Plug in the values for z, μ, and σ and solve x-m z= s for x. z=
4.61
a. b.
x = 74 + (2.30) ´ 7 = 90.1 x = 53 + ( - 1.45) ´ 12 = $35.60
c.
x = 4 + ( - 0.79) ´ 1.15 = 3.09 hours Learning Objective: 04-6 4.62
a. Nolan's standardized
z
score is
48 - 30 = 2.57 7 b. Yes, his time is unusual because it falls above 2 standard deviations when standardized. Learning Objective: 04-5 Learning Objective: 04-6 z=
4.63
a. To estimate sigma use (xmax – xmin)/6 = (30-18)/6 = 2. b. Assume a normal distribution. Learning Objective: 04-5 Learning Objective: 04-6
4.64
a. Mean = 724.67, Median = 720, Mode = 730 b. Yes, the measures do tend to agree. The mean falls between the median and mode and there is only a $5 difference between mean and median and between mean and mode. c. Using the =STDEV.S function in excel, we get a standard deviation of 114.28. d. Standardized scores for monthly rents
69
ASBE 6e Solutions for Instructors 0.04666 8
-1.44089
0.04666 8 2.67176 8 0.13417 2
-1.09087 0.04666 8
-0.91587 0.65919 2
-0.30334
0.04666 8 0.13417 2 -0.65336 0.30917 8 0.83419 8
1.79673 5 -0.91587 -0.56585 -0.30334 1.00920 5
-0.21584
-1.35338
-0.04083 1.09670 8
-0.47835 1.79673 5
-0.12834
-1.96591
-0.04083
-0.21584
e. There is one unusual value. The observation 1030 has a z score = 2.67. f. It is possible that the data are normally distributed based on the empirical rule because close to 68% of the data fall within ± 1 standard deviation, close to 95.44% fall within ± 2 standard deviations, and close to 99.73% of the data fall within ± 3 standard deviations.
empirical rule mean - 1s mean + 1s percent in interval (68.26%)
610.39 838.95 70.0%
mean - 2s mean + 2s percent in interval (95.44%)
496.10 953.23 96.7%
mean - 3s mean + 3s percent in interval (99.73%)
381.82 1,067.51 100.0%
Learning Objective: 04-1 Learning Objective: 04-2 Learning Objective: 04-3 Learning Objective: 04-4 Learning Objective: 04-5 Learning Objective: 04-6 4.65
a. Mean = 26.71. Median = 14.5. Mode = 11. Midrange = 124.5. b. Q1 = 7.25, Q3 = 20.75, Midhinge = 14. c. The geometric mean is only valid for data greater than zero. Learning Objective: 04-1 Learning Objective: 04-2 Learning Objective: 04-3 Learning Objective: 04-7
4.66
a. From Excel: mean = 33.31, median = 26, and mode = 17. 70
ASBE 6e Solutions for Instructors b. The mean is greater than the median by 7 minutes. c. No the mode is not a good measure of central tendency because this is continuous numerical data. There are only 3 power outages out of 26 that lasted 17 minutes. d. This distribution is skewed right because the mean is greater than the median. Learning Objective: 04-1 Learning Objective: 04-2 Learning Objective: 04-11 4.67
a. From Excel: mean = 66.85, median = 69.5, and mode = 86. b. The mean and median are fairly close in value. c. No the mode is not a good measure of central tendency because this is continuous numerical data. There are only 2 packages out of 20 that weighed 86 ounces. d. It is difficult to describe the shape of this distribution based solely on the values of the mean, median, and mode, however, because the mean and median are close in value one might hypothesize that the shape is somewhat symmetric. Learning Objective: 04-1 Learning Objective: 04-2 Learning Objective: 04-11
4.68
With a skewness coefficient greater than zero (0.773 > 0), we would describe the distribution as right skewed. With a kurtosis coefficient greater than one (1.277 > 1) we would describe the distribution as sharply peaked or leptokurtic. Learning Objective: 04-11
4.69
a. Stock funds:
4.70
The coefficient of variation for this year is 3026/12104 = .25 or 25%. The coefficient of variation for last year is 1701/6804 = .25 or 25%. They have the same relative variation. Learning Objective: 04-3
4.71
Bob’s z score is (580-620)/12 = -3.33. This order of chips and salsa is an outlier because the z score < -3. Learning Objective: 04-6
4.72
The coefficient of variation for plumbing supplier’s vinyl washers is:
, median = 1.22. Bond funds: , median = 0.85. x = 1.329 x = 0.875 b. Stock funds: s = 0.5933, CV = (0.5933/1.329)×100 = 44.65%. Bond funds: s = 0.4489, CV = (0.4489/0.875)×100 = 51.32%. c. The bond funds have more variability relative to the mean because the CV is greater than for stock funds. Learning Objective: 04-1 Learning Objective: 04-2 Learning Objective: 04-3
71
æ 6,053 ö 100 ç ÷ = 25% è 24, 212 ø
ASBE 6e Solutions for Instructors The coefficient of variation for steam boilers is
æ 1.7 ö 100 ç ÷ = 25% è 6.8 ø
Learning Objective: 04-1 Learning Objective: 04-3 4.73
a. Histogram 40 35 30 25 20 15 Percent 10 5 0
Rice Krispies Sales
b. The data is skewed to the right. c. From Excel: Mean = 20.12, Standard Deviation = 7.64. d. There are two unusual values. Sales of 36 have a z score of 2.08: z = (36 – 20.12)/7.64 = 2.08. Sales of 37 have a z score of 2.21: z = (37 – 20.12)/7.64 = 2.21. There is one outlier value. Sales of 49 have a z score of 3.78: z = (49 – 20.12)/7.64 = 3.78. Learning Objective: 04-1 Learning Objective: 04-2 Learning Objective: 04-3 Learning Objective: 04-5 Learning Objective: 04-6 4.74
a. See table below for CV values.
CV
values are calculated using
æ s ö CV = 100 ç ÷ èXø
Comparative Returns on Four Types of Investments Mean Standard Coefficient of Return Deviation Variation Venture funds (adjusted) 19.2 14.0 72.92% All common stocks 15.6 14.0 89.74% Real estate 11.5 16.8 146.09% Federal short term 6.7 1.9 28.36% paper Investment
b. The standard deviations are “absolute” not relative measures of dispersion. It is best to use the CV when comparing across variables that have different means. 72
ASBE 6e Solutions for Instructors c. The risk and returns are captured by the CV. Federal short term paper has the lowest CV and hence lowest risk, real estate the greatest risk. Venture funds have lower risk and greater return than common stocks based on the CV. In other words, there is more risk when there is more variation in the returns. Learning Objective: 04-1 Learning Objective: 04-2 Learning Objective: 04-3 4.75
a. CV Tuition Plans = 100%×(2.7/6.3) = 42.86%. CV SP500 = 100%×(15.8/12.9) = 122.48%. The tuition plans have lower returns than the SP 500, but less risk as measured by the CV. This is not surprising because the goal of a tuition plan is to ensure that a minimum amount of money is available at the time the plan matures, thus parents and students are willing to take a lower return in exchange for lower risk. b. We use the CV to compare the risk relative to the average return. The standard deviation alone cannot be used to compare distributions because the means are different. Learning Objective: 04-1 Learning Objective: 04-3
4.76
a. Midrange = (180+60)/2 = 120. b. x -x 180 - 60 s = max min ¶ = = 20 6 6 c. Assuming normality is important because a normal distribution is symmetric and this allows us to estimate the mean with the midrange and to estimate the standard deviation using the assumption that the range will be approximately 6 standard deviations. d. Caffeine levels in brewed coffee are dependent on many factors including brand of coffee, grind of coffee beans, and brew time. Causes of variation in caffeine level are unpredictable so a well-behaved bell-shaped curve may not be a reasonable assumption. Learning Objective: 04-1 Learning Objective: 04-2 Learning Objective: 04-3 Learning Objective: 04-5 Learning Objective: 04-6
4.77
a. Midrange = (.92+.79)/2 = .855 b. x -x 0.92 - 0.79 s = max min = = 0.0217 6 6 b. A normal distribution is plausible here because there are likely to be controls on the level of chlorine added to the water. There will be some variation around the mean but it will be predictable. Learning Objective: 04-1 Learning Objective: 04-2 Learning Objective: 04-3 73
ASBE 6e Solutions for Instructors Learning Objective: 04-5 4.78
Because the skewness coefficient (-.35 > -.40) and kurtosis coefficient (+.75 < +.88) fall within the 90% range for a sample size equal to 100 we can conclude that the distribution is approximately normal. Learning Objective: 04-11
4.79
Because the skewness coefficient (-.35 < -.28) and kurtosis coefficient (+.75 > +.62) fall outside the 90% range for a sample size equal to 200 we can conclude that the distribution is left-skewed and leptokurtic (sharply peaked.) Learning Objective: 04-11
4.80
a. The distribution should be skewed to the right because the mean is greater than the median and the mode. b. Most ATM transaction times will tend to be low in value but a few will be of longer duration. Learning Objective: 04-1 Learning Objective: 04-11
4.81
a. The distribution should be skewed to the right because the mean is greater than the median and the mode. b. Most patrons keep books out for a week or so. There will be a few patrons that keep a book out much longer. Learning Objective: 04-1 Learning Objective: 04-11
4.82
a. The distribution should be skewed to the left because the mean is less than the median and the mode. b. It appears that most students scored a C or higher but there were a few students that may not have studied for the exam which pulled the mean down. Learning Objective: 04-1 Learning Objective: 04-11
4.83
a. One would expect the mean to be close in value to the median, or slightly higher. b. In general, the life span would have a normal distribution. If skewed, the distribution is more likely skewed right than left. Life span is bounded below by zero but is unbounded in the positive direction. Learning Objective: 04-1 Learning Objective: 04-3 Learning Objective: 04-11
4.84
The mean would be greater than the median. There are likely to be a few waiting times that are extremely long. Learning Objective: 04-1 Learning Objective: 04-3 74
ASBE 6e Solutions for Instructors Learning Objective: 04-11
4.85
a. It is the midrange, not the median. b. The midrange is influenced by outliers. Because salaries tend to be skewed to the right, the midrange will be greater than the median. The community should use the median to base charges. Learning Objective: 04-1 Learning Objective: 04-3 Learning Objective: 04-11
4.86
a. The distribution would be skewed right. b. Switching from the mean to the median would trigger a penalty sooner because the median is less than the mean. c. The union would oppose this change because they would probably have to pay more penalties. Learning Objective: 04-1 Learning Objective: 04-3 Learning Objective: 04-11
4.87
a. and c. Week 1Week 2Week 3Week 4 mean 50.00 50.00 50.00 50.00 sample standard deviation 10.61 10.61 10.61 10.61 median 50.00 52.00 56.00 47.00
b. Based on the mean and standard deviation it appears that the distributions are the same. One might conclude that the weeks occupancies are the same. c. See table above for median values. d.
e. Based on the medians and dot plots, the distributions are quite different. 75
ASBE 6e Solutions for Instructors Learning Objective: 04-1 Learning Objective: 04-2 Learning Objective: 04-3
4.88
Answers will vary. Learning Objective: 04-1 Learning Objective: 04-5
4.89
a. and b. From 0 2 4 8 16 32
To (not incl) 2 4 8 16 32 64 Total
Average = 9.458
frequency (f) 7 42 33 21 11 6 120
midpoint (m) 1 3 6 12 24 48
Standard Deviation = 10.855
f(m−
x
)
2
f×m 7 500.764 126 1751.642 198 394.606 252 135.697 264 2326.167 288 8912.915 1135 14021.791 CV = 114.77%
c. No, the distribution appears to be skewed right. The frequencies of the lower ranges are much greater than the higher ranges. One would expect that the distribution within each interval to also be skewed right, i.e., more values close to the lower end of the range. This would make a difference in the estimate for the mean because the method used with grouped data assumes the average value within each range is the midpoint. We give equal weight to the midpoints of each interval and our estimate of the mean could be too low. d. Unequal bin sizes allowed frequencies greater than 0. Because the times are skewed right, there would be many bins with frequencies equal to zero in the higher end of the range. Learning Objective: 04-10 4.90
a. From 119 120 121 122 123 124 125
To 120 121 122 123 124 125 126
frequency (f) 1 5 20 27 14 10 5
76
midpoint (m) 119.5 120.5 121.5 122.5 123.5 124.5 125.5
f(m− f×m 119.5 602.5 2430 3307.5 1729 1245 627.5
x
)
11.81 29.69 41.29 5.15 4.44 24.44 32.85
2
ASBE 6e Solutions for Instructors 126 127
127 128 Total
3 2 87
126.5 127.5
Average = 122.94 Standard Deviation = 1.624
379.5 38.09 255 41.65 10695.5 226.70 CV =
1.32%
b. The raw data would show us the years when the winning times were much longer than the average. c. Because the overall distribution on time is slightly skewed right it is possible that the times within an interval are also skewed right. We give equal weight to the midpoints of each interval and our estimate of the mean could be too low. Learning Objective: 04-10 4.91
a. From 40 50 60 80
To 50 60 80 100
frequency (f) 12 116 74 2
Total
204
midpoint (m) 45 55 70 90
Average = 60.196 Standard Deviation = 8.536
f(m− ) x 2 f×m 540 2771.050 6380 3131.911 5180 7112.649 180 1776.547 12,28 0 14792.157 CV = 14.2%
b. No the unequal class sizes don’t hamper the calculations. Class sizes are unequal to ensure that no class size has a zero value. Learning Objective: 04-10 4.92 a. midpoint To frequency (f) (m) 150 1 145 160 25 155 170 24 165 180 4 175 190 2 185 Total 56 161.6 Average = 1 Standard Deviation = 7.93 From 140 150 160 170 180
f(m− ) x 2 f×m 145 275.892 3875 1092.303 3960 275.810 700 717.168 370 1094.184 9050 3455.358 CV = 4.9%
b. Grouped estimates came pretty close. c. The distribution of times within the class 150-160 may be skewed left with more observations closer to 160 (which is close to the mean) and the distribution of times within the class 160-170 may be skewed right with more observations closer to 160 as well. This observation is based on the apparent shape of the distribution of the entire 77
ASBE 6e Solutions for Instructors data set: somewhat bell shaped with a peak at approximately 160 with a possible slight right skew. Learning Objective: 04-10 4.93
Answers will vary. Learning Objective: 04-1 Learning Objective: 04-2 Learning Objective: 04-3 Learning Objective: 04-4 Learning Objective: 04-5 Learning Objective: 04-6 Learning Objective: 04-7 Learning Objective: 04-8
4.94
Answers will vary. Learning Objective: 04-1 Learning Objective: 04-2 Learning Objective: 04-3 Learning Objective: 04-4 Learning Objective: 04-5 Learning Objective: 04-6 Learning Objective: 04-7 Learning Objective: 04-8
4.95
a. Assault Rate per 100,000 in 1990 vs. 2004
2004
1990
b. r =.8332. c. Graph says that these two years are strongly correlated. If a state had a high assault rate in 1990, they also had a high rate in 2004. d. 1990: Mean = 331.92. Median = 307. Standard Deviation = 172.914 2004: Mean = 256.6. Median = 232. Standard Deviation = 131.259 The comparison shows that the summary measures for 1990 were greater than the summary measures for 2004. 78
ASBE 6e Solutions for Instructors Learning Objective: 04-1 Learning Objective: 04-2 Learning Objective: 04-3 Learning Objective: 04-9
4.96
a. Airspeed (nautical miles per hour) vs. Cockpit Noise (decibels)
Noise Level (db)
Airspeed (knots)
b. r = .9459 c. The graph and the correlation coefficient both show that there is a strong, positive, linear relationship between airspeed and cockpit noise. This relationship most likely exists because the plane becomes louder as it goes faster. Optional: Airspeed (nautical miles per hour) vs. Cockpit Noise (decibels) f(x) = 0.08x + 64.23 R² = 0.89 Noise Level (db)
Airspeed (knots)
Noise = 64.23 + 0.0765(Airspeed). As airspeed increases, noise level increases at a rate of 0.0765 per knot. Learning Objective: 04-1 79
ASBE 6e Solutions for Instructors Learning Objective: 04-2 Learning Objective: 04-3 Learning Objective: 04-9
4.97
a. % Change in US GDP and US Consumption 2000-2015 6.0 5.0 4.0 3.0 ChgCons
2.0 1.0 0.0 -4.0 -3.0 -2.0 -1.0 -1.00.0 1.0 2.0 3.0 4.0 5.0 -2.0 ChgGDP
b. The scatter plot shows a strong positive linear relationship. c. As the US GDP increases, consumers have more spending power which will result in an increase in personal consumption expenditures. d. r = .9314 Learning Objective: 04-9 4.98
a.
80
ASBE 6e Solutions for Instructors
45.00 43.00 41.00 39.00 37.00 35.00 Zalparm 33.00 31.00 29.00 27.00 25.00 10.00 12.00 14.00 16.00 18.00 20.00 22.00 Dolona
b. The relationship appears to be strong, positive linear. c. r= .861 d.
Weekly closing price of two stocks 50.00 45.00 40.00 35.00 30.00 25.00 20.00 15.00 10.00 5.00 0.00
Dolona
Zalparm
The line shows that the stocks move together although Zalparm has more variation week to week than Dolona. Learning Objective: 04-9
81
ASBE 6e Solutions for Instructors
Chapter 5 Probability 5.1
a. S = {(V,B), (V,E), (V,O), (M,B), (M,E), (M,O), (A,B), (A,E), (A,O)} b. Events are not equally likely. Barnes and Noble probably carries more books than other merchandise. Learning Objective: 05-1
5.2
a. S = {(S,L), (S,T), (S,B), (P,L), (P,T), (P,B), (C,L), (C,T), (C,B)} b. There are different likelihoods of risk levels among the 3 types of business forms; therefore the different elementary events will have different probabilities. Learning Objective: 05-1
5.3
a. S = {(L,B), (L,B’), (R,B), (R,B’)} b. Events are not equally likely. There are more right handed people than left handed people. Learning Objective: 05-1
5.4
a. S ={(1,H), (2,H), (3,H), (4,H), (5,H), (6,H), (1,T), (2,T), (3,T), (4,T), (5,T), (6,T)} b. Yes, assuming that we have a fair die and fair coin. Learning Objective: 05-1
5.5
Subjective (opinion of experienced stock brokers) or Empirical (based on previous IPOs) Learning Objective: 05-2
5.6
Subjective (opinion of industry experts) Learning Objective: 05-2
5.7
Empirical (based on previous launches) Learning Objective: 05-2
5.8
Classical (equally likely dice roll outcomes) Learning Objective: 05-2
5.9
Classical (each day of the year equally likely) Learning Objective: 05-2
5.10
Empirical (results from a study) Learning Objective: 05-2
5.11
Empirical (frequency based on observations) Learning Objective: 05-2
81
ASBE 6e Solutions for Instructors 5.12
Empirical (based on success of other new restaurants) or Subjective (opinion of business experts in Nashville) Learning Objective: 05-2
5.13
a. Not mutually exclusive, you can both work 20 hours or more and be an accounting major. b. Mutually exclusive, you cannot be born in Canada and in the United States. You have to be one or the other. c. Not mutually exclusive, you can own two different cars. Learning Objective: 05-3
5.14
a. Collectively exhaustive, you have to be either a college grad, have been in college some, or have never attended college. b. Not collectively exhaustive, you can be born somewhere besides the U.S., Canada, or Mexico. c. Collectively exhaustive, you have to either be a full-time student, a part-time student, or not a student. Learning Objective: 05-3
5.15
a. P(A B) = P(A) + P(B) – P(AB) = .4 + .5 − .05 = .85. b. P(A | B) = P(AB) / P(B) = .05/.50 = .10. c. P(B | A) = P(AB) / P(A) .05/.4 = .125. Learning Objective: 05-3
5.16
a. P(AB) = P(A) + P(B) − P(AB) = .7 + .3 – 0 = 1.0. b. P(AB) = P(AB) / P(B) = .00 / .30 = .00. c. P(B | A) = P(AB) / P(A) .00/.7 = .00. Learning Objective: 05-3
5.17
a. P(S) = .217. b. P(S’) = 1 − .217 = .783. c. Odds in favor of = P(S) / P(S) = .217/.783 = .277. d. Odds against S = P(S) / P(S) = .783/.217 = 3.61 Learning Objective: 05-3 a. Odds in favor of being audited: P (audit ) .017 = = .0173 to 1 1 - P(audit ) 1 - .017 b. Odds in favor of not being audited: P (no audit ) .983 = = 57.82 to 1 1 - P(no audit ) .017 Learning Objective: 05-4
5.18
5.19
a. X = 1 if the drug is approved, 0 otherwise. b. X = 1 if batter gets a hit, 0 otherwise. c. X = 1 if breast cancer detected, 0 otherwise. Learning Objective: 05-3 82
ASBE 6e Solutions for Instructors
5.20
a. (admitted unconditionally, admitted conditionally, not admitted) b. (completed pass, incomplete pass, intercepted pass) c. (deposit, withdrawal, bill payment, funds transfer) Learning Objective: 05-1
5.21
a. P(S) = 1−.246. There is a 75.4% chance that a female aged 18-24 is a nonsmoker. b. P(S C) = .246+ .830 − .232 = .844. There is an 84.4% chance that a female aged 1824 is a smoker or is Caucasian. c. P(S | C) = .232/.830 = .2795. Given that the female aged 18-24 is a Caucasian, there is a 27.95% chance that they are a smoker. d. P(S C) = P(S) – P(S C) = .246 − .232 = .014. P(S | C) = .014/.17 = .0824. Given that the female aged 18-24 is not Caucasian, there is an 8.24% chance that she smokes. Learning Objective: 05-3
5.22
a. P(C) = 1 – P(C) = 1 – .4 = .6. b. P(C M) = P(C) + P(M) – P(C M) = .4 + .5 – .24 = .66. c. P(M | C) = P(C M)/ P(C) = .24/.4 = .6. d. P(C | M) = P(C M)/ P(M) = .24/.5 = .48. Learning Objective: 05-3
5.23
P(JK) = P(J) × P(K) = .26×.48 = .1248 so P(JK) = P(J) + P(K) – P(JK) = .26 + .48 – .1248 = .6152 Learning Objective: 05-5
5.24
P(AB) = P(A) × P(B) = .40×.50 = .20 Learning Objective: 05-5
5.25
a. P(AB) = P(A B) / P(B) = .05/.50 = .10 b. No, A and B are not independent because P(AB) ≠ P(A). Learning Objective: 05-5
5.26
a. P(A) ×P(B) = .40×.60 =.24 and P(A B) = .24, therefore A and B are independent. b. P(A) × P(B) = .90×.20 =.18 and P(A B) = .18, therefore A and B are independent. c. P(A) × P(B) = .50×.70 = .35 and P(A B) = .25, therefore A and B are not independent. Learning Objective: 05-5
5.27
a. P(JK) = P(J K) / P(K) = .15/.4 = .375 b. No, J and K are not independent because P(JK) ≠ P(J). Learning Objective: 05-5
5.28
a. P(J) ×P(K) = .5 ×.4 =.2 and P(J K) = .3, therefore J and K are not independent. b. P(J) × P(K) = .6 ×.2 =.12 and P(J K) = .12, therefore J and K are independent. c. P(J) × P(K) = .15 ×.5 = .075 and P(J K) = .1, therefore J and K are not independent. 83
ASBE 6e Solutions for Instructors Learning Objective: 05-5 5.29
a. P(V M) = P(V) + P(M) – P(VM) = .73 + .18 – .03 = .88. b. P(V M) ≠ P(V) × P(M) therefore V and M are not independent. Learning Objective: 05-5
5.30
a. There is 25% chance that a clock will not wake Bob (a failure, F). Both clocks would have to fail in order for him to oversleep. Assuming independence: P(F1F2) = P(F1) × P(F2) = .25×.25 = .0625. There is a 6.25% chance that Bob will oversleep. b. The probability that at least one of the clocks wakes Bob is 1 – P(F1)×P(F2)×P(F3) = 1– (.25×.25×.25) = .9844, which is less than 99%. Learning Objective: 05-3 Learning Objective: 05-5
5.31
“Five nines” reliability means P(not failing) = .99999. P(power system failure) = 1 − (.05)3 = .999875. The system does not meet the test. Learning Objective: 05-3 Learning Objective: 05-5 a. P(Success | At least 60) = = . There is a 12.5%
5.32
P( Success Ç At least 60) P( At least 60)
.005 = .125 .04
chance of success given that he or she is at least 60 years old. b. which does not equal P( Success ) ´ P ( At least 60) = (.31)(.04) = .0124 therefore the two events are not independent.
P( Success Ç At least 60) = .005
Learning Objective: 05-3 Learning Objective: 05-5 5.33
Ordering a soft drink is independent of ordering a square pizza. P(ordering a soft drink)×P(ordering a square pizza) = .5(.8) = .4. This is equal to P(ordering both a soft drink and a square pizza). Learning Objective: 05-3 Learning Objective: 05-5
5.34
a. P(a viewer is aged 18-34) = 69/100=.69 b. P(a viewer prefers watching TV videos) = 48/100=.48 c. P (a viewer is aged 18-34 and prefers watching videos on mobile/laptop device) = 39/100=.39 or 39% d. P(a viewer prefers videos on mobile/laptop device | they are 18-34) = P (User created Ç Age 18 - 34) .39 = = .5652 or 56.52%. P ( Age 18 - 34) .69 e. P (aged 35 - 54 È mobileorlaptop device) = P (aged 35 - 54) + P ( mobileorlaptop device ) - P(aged 35 - 54 Ç mobileorlaptop device) = 84
ASBE 6e Solutions for Instructors (.20+.52)–.10=.62 or 62% Learning Objective: 05-6 5.35
a. P(Recycles) = 340/1000 = .34 b. P(Don’t Recycle | Lives in Deposit Law State) = 66/220= .30 c. P(Recycle and Live in Deposit Law State) = 154/1000 = .154 d. P(Recycle | Lives in Deposit Law State) = 154/220 = .70 Learning Objective: 05-6
5.36
a. P(V) = 116/158 =.7342 b. P(A) = 47/158 = .2975 c. P(A V) = 32/158 = .2025 d. P(A V) = P(A) + P(V) – P(A V)= 47/158 + 116/158 – 32/158 = .8291 e. P(A | V) = P(V A) / P(V) = 32/116 = .2759 f. P(V | A) = P(V A) / P(A) = 32/47 = .6809 Learning Objective: 05-6
5.37
a. P(D) = 79/156 =.5064 b. P(R) = 22/156 = .1410 c. P(D R) = 8/156 = .0513 d. P(D R) = P(D) + P(R) − P(D R)= 79/156 + 22/156 – 8/158 = .5962 e. P(R | D) = P(R D) / P(D) = 8/79= .1013 f. P(R | P) = P(R P) / P(P) = 7/43 = .1628 Learning Objective: 05-6
5.38
a. P(A) = 100/200 = .50. There is a 50% chance that a student is an accounting major. b. P(M) =102/200 = .51. There is a 51% chance that a student is male. c. P(A M) = 56/200 = .28. There is a 28% chance that a student is a male accounting major. d. P(F S) = 24/200 = .12. There is a 12% chance that a student is a female statistics major. e. P(A | M) = P(M A) / P(A)= 56 /102 = .549. There is 54.9% chance that a male student is an accounting major. f. P(A | F) = P(F A) / P(F)= (44/200)/(98/200) = .4490. There is a 44.9% chance that a female student is an accounting major. g. P(F | S) = P(F S) / P(S) (24/200) /(40/200) = .60. There is a 60% chance that a statistics student is female. h. P(E F) = P(E) + P(F) – P(F E) 60/200 + 98/100 – 30/100 = 128/200 = 64%. There is 64% chance that a student is an economics major or a female. Learning Objective: 05-6
5.39
Gender and Major are not independent. For example, P(A F) = .22. P(A)×P(F) = .245. Because the values are not equal, the events are not independent. Learning Objective: 05-5 85
ASBE 6e Solutions for Instructors Learning Objective: 05-6 5.40
a. P(D3) = 15/38 = .3947. b. P(Y3) = 15/38 = .3947. c. P(Y3 | D1) = P(Y3 D1) / (D1) = (2/38)/(11/38) =.1818. Another way is 2/11=.1818 because the two 38s cancel each other out. d. P(D1 | Y3) = P(Y3 D1) / P(Y3) = (2/38)/(15/38) =.1333. Another way is 2/15=.1333 because the two 38s cancel each other out. Learning Objective: 05-6
5.41
Learning Objective: 05-7
86
ASBE 6e Solutions for Instructors 5.42
a.
Learning Objective: 05-7 5.43
Let A = using the drug. P(A) = .04. P(A) = .96. Let T be a positive result. False positive: P(T | A) = .05. False negative: P(T | A) = .10. P(T | A) = 1 − .10 = .90. P(T) = P(T ∩ A) + P(T ∩ A) = P(T | A) P(A) + P(T | A) P(A) = (.90)(.04) + (.96)(.05) = .084. P(A | T) = P(T ∩ A) / P(T ) = [P(T | A) P(A)] / P(T) = (.9)(.04)/.084 = .4286. Learning Objective: 05-8
5.44
Given: P(A) = .5, P(B) = .5, P(D) = .04, P(D ) = .96, P(D|A) = .06. Find P(A | D). . P( D | A ) ´ P( A ) .06 ´ .5 P (A| D )= = = .75 P( D ) .04 Learning Objective: 05-8
5.45
Let W = suitcase contains a weapon. P(W) = .001. P(W ) = .999. Let A be the alarm trigger. False positive: P(A | W ) = .02. False negative: P(A | W) = .02. P(A | W) = 1 − .02 = . 98. P(A) = P(A ∩W) + P(A ∩ W ) = P(A |W) P(W) + P(A | W ) P(W ) = (.98)(.001) + (.999)(.02) = .02096. P(W | A) = P(A ∩ W) / P(W) = [P(A | W) P(W)] / P(A) = (.98)(.001)/.02096 = .04676. Learning Objective: 05-8
87
ASBE 6e Solutions for Instructors
5.46
a. 8×7=56×6=336×5=1680×4=6720×3=20160×2=40320×1=40,320. b. 32!=2.6313×1035 d. Google displays it more clearly but a calculator may be more accessible. Learning Objective: 05-9
5.47
a. 20C5 =
5.48
=15,504. 20! 5!15! Learning Objective: 05-9 n! nCr = r !(n - r )!
Given n = 31 and r = 5: 31C5 = 169,911. P(picking the winner) = 1/169911
= .00000588. Or using the hint given: There are 31×30×29×28×27 ways to pick the winning combination. There are 5! permutations of those five numbers. so P(picking the winner) = 1/169911 = .00000589. 31´ 30 ´ 29 ´ 28 ´ 27 = 169,911 5! Learning Objective: 05-3 Learning Objective: 05-9 5.49
a. 266 =308,915,776 b. 366 =2,176,782,336 c. 326 =1,073,741,824 Learning Objective: 05-9
5.50
a. 26×26×10×10×10×10 = 6,760,000 unique IDs. b. No, that only yields 26,000 unique IDs. c. As growth occurs over time, you would not have to worry about a duplicate ID nor would you have to generate new ones. Learning Objective: 05-9
5.51
a. 106 = 1,000,000. b. 105 = 100,000. c. 106 = 1,000,000 Learning Objective: 05-9
5.52
a.
n! n Pr (n r )!
, Given n = 4 and r = 4: 4P4 = 24.
b. ABCD, ABDC, ACBD, ACDB, ADBC, ADCB, BACD, BADC, BCAD, BCDA, BDAC, BDCA, CABD, CADB, CBAD, CBDA, CDBA, CDAB, DABC, DACB, DBAC, DBCA, DCAB, DCBA Learning Objective: 05-9 88
ASBE 6e Solutions for Instructors
5.53
a. 7! = 5,040 ways. b. No, too many! Learning Objective: 05-9
5.54
a. 8P3 = b. 8P5 =
8! 5! 8! 3!
= 336. = 6720.
c. 8P 1 =
= 8. 8! 7! d. 8P8 = = 40320. 8! 0! Learning Objective: 05-9 5.55
a. 8C3 = b. 8C5 = c. 8C1 =
8! 3!5! 8! 5!3!
8! 1!7!
= 56. = 56. = 8.
d. 8C8 =
= 1. 8! 8!0! Learning Objective: 05-9 5.56
a. b.
n! n Pr (n r )!
. Given n = 10 and r = 4: 10P4 = 5040.
n! nCr r !(n r )!
. Given n = 10 and r = 4: 4C4 = 210.
c. The number of combinations is smaller than the number of permutations because order doesn't matter and each combination is only counted once. If order mattered, we would count the same combination in every order possible. For example, when order doesn't matter the combination {Sales agent A, B, C, D} is the same as {Sales agent B, C, D, A}. Learning Objective: 05-9 89
ASBE 6e Solutions for Instructors 5.57
Answers may vary Learning Objective: 05-1 Learning Objective: 05-2 Learning Objective: 05-3 Learning Objective: 05-5
5.58
Answers may vary Learning Objective: 05-1 Learning Objective: 05-2 Learning Objective: 05-3 Learning Objective: 05-5
5.59
Answers may vary Learning Objective: 05-1 Learning Objective: 05-2 Learning Objective: 05-3 Learning Objective: 05-5
5.60
a. S = {Brown/Brown, Brown/Yellow, Brown/Red, Brown/Blue, Brown/Orange, Brown/Green, Yellow/Brown, Yellow/Yellow, Yellow/Red, Yellow/Blue, Yellow/Orange, Yellow/Green, Red/Brown, Red/Yellow, Red/Red, Red/Blue, Red/Orange, Red/Green, Blue/Brown, Blue/Yellow, Blue/Red, Blue/Blue, Blue/Orange, Blue/Green, Orange/Brown, Orange/Yellow, Orange/Red, Orange/Blue, Orange/Orange, Orange/Green, Green/Brown, Green/Yellow, Green/Red, Green/Blue, Green/Orange, Green/Green}. b. P(BR)×P(BR) = .13×.13 = .0169. c. P(BL)×P(BL) = .24×.24 = .0576. d. P(G)×P(G) = .16×.16 = .0256. e. P(BR)×P(G) = .13×.16 = .0208. Learning Objective: 05-1 Learning Objective: 05-3 Learning Objective: 05-5
5.61
Empirical. These are most likely based on historical data. Learning Objective: 05-2
5.62
The judge most likely has past information and could have easily calculated this probability. Therefore, it is empirical. Learning Objective: 05-2
5.63
Response frequencies from the survey would be used to calculate an empirical probability. Learning Objective: 05-2
5.64
Subjective. The marketing director is most likely making this estimate based on the reported results of other companies that have offered free shipping. 90
ASBE 6e Solutions for Instructors Learning Objective: 05-2 5.65
Empirical or subjective. Most likely estimated by interviewing ER doctors. Learning Objective: 05-2
5.66
Empirical. From a sample of births in the U.S. Learning Objective: 05-2
5.67
Subjective. Simulated experiment using a computer model. Learning Objective: 05-2
5.68
104/118 = .8814 Learning Objective: 05-2
5.69
Empirical or subjective. Determined either by observation or survey. Learning Objective: 05-2
5.70
Not independent. P(A) = .80, P(B) = .60. P(A) ×P(B)=.80×.60 = .48 ≠ .40. Because .48 does not equal .40, these are not independent. Learning Objective: 05-5
5.71
The system won’t fail unless both fail and the chance that both servers fail is less than the chance that one server fails. So the chance of failure has actually decreased. Learning Objective: 05-5
5.72
The odds that a lie will be detected are: P(lie detected)/[1−P(lie detected)]=.65/.35=1.8571 so the odds in favor of detecting a lie are 1.8571 to 1. Learning Objective: 05-4
5.73
Odds against a 2011 Audi being stolen:
P (notstolen) P ( stolen)
= .9952/.0048 ≈ 207 to 1.
Learning Objective: 05-4 5.74
a. Odds against being struck by lightning:
P ( notbeingstruck ) P (beingstruck )
= .99984/.00016 = 6249 to 1.
Learning Objective: 05-4 5.75
P(Butler making it to the NCAA finals) = 1/(125+1) = .008. Learning Objective: 05-4
5.76
a. 29 = 512 separate codes. b. (1/512) × 1000 = 1.95 times (approximately 2) or (1/1024)× 1000 = 0.977 (approximately 1). We assume that each combination is selected independently. 91
ASBE 6e Solutions for Instructors Learning Objective: 05-5 Learning Objective: 05-9 5.77
a. 263103 = 17,576,000. b. 366 = = 2,176,782,336. c. 0 and 1 might be disallowed because they are similar in appearance to letters like O and I. d. Yes, 2.1 billion unique plates should be enough. e. 346 = 1,544,804,416. Learning Objective: 05-9
5.78
Suppose the correct order for the meals is ABC. There are 6 possible permutations and the possibilities for incorrect orders include: ACB (2 incorrect meals), BAC (2 incorrect meals), BCA (3 incorrect meals), CAB (3 incorrect meals), and CBA (2 incorrect meals). a. P(No diner gets the correct meal) = 2/6 or 1/3 b. P(Exactly one diner gets the correct meal) = 3/6 or 1/2 c. P(Exactly two diners get the correct meal) = 0 d. P(All three diners get the correct meal) = 1/6 Learning Objective: 05-1 Learning Objective: 05-3 Learning Objective: 05-9
5.79
If the order in which students are selected for the teams matters then we would use a permutation formula: 7P3 = 210. Most likely order does not matter so it would be a combination of 7 things taking 3 at a time: 7C3 = 35. Learning Objective: 05-9
5.80
a. Given P(C) = .78, P(H) = .37, P(C | H) = .95. P(C and H) = P(C | H)× P(H) = (.95)(.37) = .3515. b. P(H | C) = P(C and H) / P(C) = .3515/.78 = .4506 c. No, C and H are not independent because neither of the marginal probabilities is equal to the conditional probabilities. Learning Objective: 05-3 Learning Objective: 05-5
5.81
a. Given P(D) = .25, P(P) = .57, P(D and P) = .12. P(D | P) = P(D and P) / P(P) = .12/.57 = .2105 c. No, D and P are not independent because the marginal probability of event D is not equal to the conditional probability of D given P. Learning Objective: 05-3 Learning Objective: 05-5
5.82
Let F denote a failure and S denote a non-failure. Use the multiplicative rule of probability for independent events: 92
ASBE 6e Solutions for Instructors a. P(F1 F2) = P(F1)×P(F2) = .02×.02 = .0004. b. P(S1 S2) = P(S1) ×P(S2) = .98×.98 = .9604. c. . Using the General P( F1 È F2 ) = P ( F1 ) + P ( F2 ) - P ( F1 Ç F2 ) = (.02 + .02) - .0004 = .0396 Law of Addition we can find the union of the first alternator failing or the second alternator failing. That probability is 3.96%. Learning Objective: 05-3 Learning Objective: 05-5 5.83
No, P(A)×P(B) = .4 × .5 ≠ .05. Learning Objective: 05-5
5.84
Let B1 = first child is a boy and B2 = second child is a boy. Find P(B1B2 | B1). P(B1B2 | B1) = . Because B1 and B2 are independent of each other: P( B1 Ç B2 ) P( B1 ) P( B1 Ç B2 ) P( B1 )P( B2 ) = = P( B2 ) = .5 P( B1 ) P( B1 ) Learning Objective: 05-3 Learning Objective: 05-5
5.85
P(at least one gyro will operate) = 1 – P(both gyros will fail) = 1 – (.0008)2 = .99999936. This is greater than .99999 so yes they’ve achieved “five-nines” reliability. Learning Objective: 05-5
5.86
a. Independent – gender of one baby doesn’t affect gender of another baby. b. These are typically considered dependent. Insurance rates are higher for most men because they are involved in more accidents. c. Dependent - most calls would be either first thing in the day or during lunch hours. Many folks will call as soon as the office opens because they are anxious to get their questions answered. Or – some folks will call during the lunch hour because that is when they are free to make calls. Learning Objective: 05-5
5.87
Assuming independence, P(3 cases won out of next 3) = .73 = .343. Learning Objective: 05-5
5.88
a. In order to guarantee 99.999% reliability, the probability of a system failure should be no more than 0.00001. Because the servers are considered independent we can set up the equation (.01)k = .00001. ln[(.01)k ] = ln(.00001). k = ln(.00001)/ln(.01) = 2.5. Round up to the next higher integer. 3 are required. P(system works) = 1 – P(system fails) = 1 – (.01)3 = .999999 or 99.9999% reliability. b. For P(a server failure) = .10, k = 5, so 5 servers are required. 93
ASBE 6e Solutions for Instructors Learning Objective: 05-5 5.89
Assuming independence, P(4 adults say yes) = .564 = 0.0983. Learning Objective: 05-5
5.90
a. P(fatal accident over a lifetime) = 1 − P(no fatal accident over a lifetime) = 1 − (3,999,999/4,000,000)50,000 = .012422. Over the average U.S. driver's lifetime, there is a 1.24% chance of having a fatal accident. b. Independence might not hold because young drivers are more prone to accidents and very old drivers are more prone to accidents. c. The probability of an accident each time you get behind the wheel is so small that an individual might take the risk. Learning Objective: 05-5
5.91
See the Excel Spreadsheet in Learning Stats: 05-13 Birthday Problem.xls. For 2 riders: P(no match) = .9973. For 10 riders: P(no match) = 0.8831. For 20 riders: P(no match) = 0.5886. For 50 riders: P(no match) = 0.0296. Learning Objective: 05-5
5.92
See the Excel Spreadsheet in Learning Stats: 05-13 Birthday Problem.xls. If there are 23 riders, P(match) = .50730. If there are 32 riders, P(match) = .75. Learning Objective: 05-5
5.93
a. i. P(C) = 193/400 = .4825. The probability of seeing a car in a shopping mall parking lot is .4825. ii. P(G) = 100/400 = .25. The probability of seeing a vehicle in the Great Lakes shopping mall is .25. iii. P(V | S) = 19/100 = .19. The probability of seeing a parked SUV at the Somerset mall is .19. iv. P(C | J) = 64/100 = .64. The probability of seeing a parked car at the Jamestown mall is .64. v. P(C and G) = 36/400 = .09. The probability that a parked vehicle is a car and is at the Great Lakes mall is .09. vi. P(T and O) = 6/400 = .015. The probability a parked vehicle is a truck and is at the Oakland mall is .015. b. Yes, the vehicle type and mall location are dependent. For example, P(T)×P(O) = (.115) (.25) = .02875. P(T and O) = .015. Because .02875 ≠ .015, the events are dependent. Learning Objective: 05-5 Learning Objective: 05-6
5.94
a. i. P(S) = 320/1000 = .32. The likelihood of a male 18-24 smoking is .32. 94
ASBE 6e Solutions for Instructors
5.95
5.96
ii. P(W) =850/1000 = .85. The likelihood of a male 18-24 being white is .85. iii. P(S | W) = P(S and W)/ P(W) = .29/.85 = .3412. The likelihood of a white male 1824 being a smoker is .3412. iv. P(S | B) = P(S and B)/ P(B) = (30/1000)/(150/1000) = .200. The likelihood of a black male 18-24 being a smoker is .20. v. P(S and W) = 290/1000 = .290. The likelihood of a male 18-24 being a smoker and being white is .290. vi. P(N and B) = 120/1000 = .12 The likelihood of a male 18-24 not smoking and being black is .12. b. P(S and W) = .29 and the P(S)×P(W) = .32×.85 = .272. P(S and B) = .030 and P(B)×P(S) = .32×.15 = .048. Yes, the smoking rates suggest that race and smoking are dependent. d. If smoking is dependent on race, then health officials might target or design special programs based on race. Learning Objective: 05-5 Learning Objective: 05-6 a. i. P(F) = 19/34 = .5588. The probability that the forecasters predicted a decline in interest rates is .5588. ii. P(A+) = 18/34 = .5294. The probability there was a rise in interest rates is .5294. iii. P(A | F) = 7/19 = .3684. Given that the forecasters predicted a decline in interest rates, the probability that there was an actual decline is .3684. iv. P(A+ | F+) = 6/15 = .4. Given that the forecasters predicted an increase in interest rates, the probability that there was an actual increase is .4. v. P(A+ and F+) = 6/34 = .1765. The probability that in a given year there was both a forecasted increase and actual increase in interest rates is .1765. vi. P(A and F) = 7/34 = .2059. The probability that in a given year there was both a forecasted decline and actual decline in interest rates is .2059. b. No, P(A−) = .4705 and P(A− | F−) = .3684. Interest rates moved down 47% of the time and yet the forecasters’ predictions of a decline showed a 37% accuracy rate. Learning Objective: 05-6 a.
i. P(B) = 25/61 = .4098. The likelihood of a climb is .4098. ii. P(L) = 14/61= .2295. The likelihood of a low noise level is .2295. iii. P(H | C) = 3/8 = .3750. The likelihood of a high noise given you are in the cruise phase is .3750. iv. P(H | D) = 14/28 = .5. The likelihood of a high noise given you are in the descent phase is .5. v. P(L and B) = 6/61=.0984. The likelihood of both a low noise and climbing is .0984. vi. P(L and C)= 2/61=.0328. The likelihood of both a low noise and cruising is .0328. b. Flight noise is dependent on flight phase. For example: P(H) = .2951 and P(H | C) = . 375. If the events were independent the two probabilities would be the same. Learning Objective: 05-5 Learning Objective: 05-6
5.97 Cancer
95
No Cancer
Totals
ASBE 6e Solutions for Instructors Positive Test Negative Test Totals
4 0 4
500 9496 9996
504 9496 10000
P(Cancer | Positive Test) = 4/504 = 0.00794. Learning Objective: 05-8
5.98
P(Authorized) = .95, P(Unauthorized) = .05, P(Denied Access | Authorized) = .001, P(Allowed Access|Unauthorized) = .000001, P(Denied Access|Unauthorized) = . 999999. Using Bayes’ Theorem:
P( Authorized Ç DeniedAccess ) = P( DeniedAccess ) P( DeniedAccess | Authorized ) P ( Authorized ) = P( DeniedAccess | Authorized ) P( Authorized ) + P( DeniedAccess | Unauthorized ) P(Unauthorized ) (.001)(.95) = .01865. (.001)(.95) + (.999999)(.05) Learning Objective: 05-8
P( Authorized | DeniedAccess ) =
5.99 Positive Test Negative Test Totals
Used Drugs 18 2 20
Has Not Used Drugs 72 408 480
P(Used Drugs | Positive Test) = 18/90 = .20. Learning Objective: 05-8
96
Totals 90 410 500
ASBE 6e Solutions for Instructors
Chapter 6 Discrete Distributions 6.1
Example A is a probability distribution because the sum of P(x) is 1 (.8 + .2 = 1) and all probabilities are nonnegative. Examples B and C are not probability distributions because the sum of P(x) is .95 for B and 1.30 for C. Learning Objective: 06-1
6.2
a. P(X = 75) = .3 b. P(X ≤ 75) = .2 + .4 + .3 = .9 c. P(X > 50) = .3 + .1 = .4 d. P(X < 100) = .2 + .4 + .3 = .9 e. (b) is a CDF because a CDF is defined by the inequality “≤”. Learning Objective: 06-1
6.3
a. P(X ≥ 3) = .2 + .15 + .05 = .4 b. P(X ≤ 2) = .05 + .3 + .25 = .6 c. P(X > 4) = .05 + .3 + .25 + .2 = .8 d. P(X =1) = .3 e. (b) is a CDF because a CDF is defined by the inequality “≤”. Learning Objective: 06-1
6.4
a. Based on the table below we see that: E(X) = 57.5, V(X) = 506.25, = 22.5. b. The distribution is skewed to the right. x
P(x)
xP(x)
xE(X)
P(x)[xE(X)]2
25 50 75 100 Total
0.20 0.40 0.30 0.10 1.00
5.00 20.00 22.50 10.00 57.50
-32.50 -7.50 17.50 42.50
211.25 22.5 91.875 180.625 506.25
Learning Objective: 06-2
97
ASBE 6e Solutions for Instructors 6.5
a. Based on the table below: E(X) = 2.25, V(X) = 1.6875, = 1.299. b. The distribution is skewed to the right. x
P(x)
xP(x)
xE(X)
P(x)[xE(X)]2
0 1 2 3 4 5 Total
0.05 0.30 0.25 0.20 0.15 0.05 1.00
0.00 0.30 0.50 0.60 0.60 0.25 2.25
-2.25 -1.25 -0.25 0.75 1.75 2.75
0.25 0.47 0.02 0.11 0.46 0.38 1.6875
Learning Objective: 06-2 6.6
X is the amount of the “value” of a bottle. X can be either $215,000, the value of the Lamborghini, or $0. To find the expected value multiply each value of X by the associated probability. E(X) = ($215,000)(.00000884) + ($0)(.99999116) = 1.9006. Learning Objective: 06-2
6.7
X is the amount of the claim. Assume a student who files a claim will claim the maximum amount of $5000. The values of X are either $5000 or $0. Expected payout = E(X) = $5000(.01) + ($0)(.999) = $50, so company adds $25 and charges $75. Learning Objective: 06-2
6.8
X is the lottery ticket winnings. The values of X are either $28 million or $0. The expected winning is E(X) = $28,000,000(.000000023) + $0(.999999977) = $0.644. Because the lottery ticket costs more than the expected winnings ($1.00 > $0.644) the ticket is too high to buy. Learning Objective: 06-2
6.9
Let X equal the loss during a hurricane. The values of X are either $250 million, $950 million, or $0. Expected Loss = E(X ) = $250(.3) + $950(.3) + $0(.4) = $360 million. Learning Objective: 06-2
10
6.
Using a = 2 and b = 8, μ =(2 + 8)/2 = 5.
s = [(8 - 2 + 1) 2 - 1] / 12 = 2 Learning Objective: 06-3
6.11
a. With a = 20 and b = 60,
m = (20 + 60) / 2 = 40
b. P(X 40) = 1 P(X ≤ 39) = 1 P(X 30) = 1 P(X ≤ 29) = 1
39 - 20 + 1 60 - 20 + 1 29 - 20 + 1 60 - 20 + 1 98
and
s = [(60 - 20 + 1) 2 - 1] /12 = 11.83 = .5122.
= .7561.
ASBE 6e Solutions for Instructors Learning Objective: 06-3 6.12
a. Answers will vary but the excel function is =RANDBETWEEN(1,2) and the mean and standard deviation of the numbers generated should be close to = 1.5 and = .50 because and a + b 1+ 2 [(b - a) + 1]2 - 1 [(2 - 1) + 1]2 - 1 = = 1.5 = = .5 2 2 12 12 b. Answers will vary but the excel function is =RANDBETWEEN() and the mean and standard deviation of the numbers generated should be close to = 3.0 and = 1.41. c. Answers will vary but the excel function is =RANDBETWEEN() and the mean and standard deviation of the numbers generated should be close to = 49.5 and = 28.87. d. See parts a thru c. e. See parts a thru c. Learning Objective: 06-3
6.13
a. X = 0, 1, or 2 b. X = 4, 5, 6, or 7 c. X = 4, 5, 6, 7, 8, 9, or 10 Learning Objective: 06-4
6.14
a.
6.15
a. = (8)(.1) = 0.8,
P( X ³ 7) b. < 4) P( X c. P( X £ 2) Learning Objective: 06-4
b. = (10)(.4) = 4, c.
s = (8)(.1)(1 - .1) = 0.8485 s = (10)(.4)(1 - .4) = 1.5492
= (12)(.5) = 6,
s = (12)(.5)(1 - .5) = 1.7321 Learning Objective: 06-4 6.16
a. = (30)(.9) = 27, b. = (80)(.7) = 56,
s = (30)(.9)(1 - .9) = 1.6432
s = (80)(.7)(1 - .7) = 4.0988
c. = (20)(.8) = 16,
s = (20)(.8)(1 - .8) = 1.7889 Learning Objective: 06-4 99
ASBE 6e Solutions for Instructors
6.17
a. P( X = 5) =
9! .95 (1 - .9)9-5 = .0074 5!(9 - 5)!
P ( X = 0) =
6! .20 (1 - .2)6-0 = .2621 0!(6 - 0)!
b.
c.
9! .89 (1 - .8)9-9 = .1342 9!(9 - 9)! Learning Objective: 06-4 P( X = 9) =
6.18
a. P( X = 2) =
b. P ( X = 1) =
8! .102 (1 - .10)8-2 = .1488 2!(8 - 2)!
10! .401 (1 - .40)10-1 = .0403 1!(10 - 1)!
c.
12! .703 (1 - .70)12-3 = .0015 3!(12 - 3)! Learning Objective: 06-4 P( X = 3) =
6.19
Use the Excel function =BINOM.DIST(x,n,π,1) where “1” stands for cumulative. a. P(X ≤ 3) =BINOM.DIST(3,8,.2,1) = .9437 b. P(X > 7) = 1 – P(X ≤ 7) = 1-BINOM.DIST(7,10,.5,1) = .0547 c. P(X < 3) = P(X ≤ 2) =BINOM.DIST(2,6,.7.1) = .0705 Learning Objective: 06-4
6.20
Use the Excel function =BINOM.DIST(x,n,π,1) where “1” stands for cumulative. a. P(X ≤ 10) =BINOM.DIST(10,14,.95,1) = .00417 b. P(X > 2) = 1 – P(X ≤ 2) = 1-BINOM.DIST(2,5,.45,1) = .4069 c. P(X ≤ 1) =BINOM.DIST(1,10,.15,1) = .5443 Learning Objective: 06-4
6.21
Use the Excel function =BINOM.DIST(x,n,π,1) where “1” stands for cumulative. a. P(X > 10) = 1 – P(X ≤ 10) =1-BINOM.DIST(10,16,.8,1) = .9183 b. P(X ≥ 4) = 1 – P(X ≤ 3) =1-BINOM.DIST(3,8,.4,1) = .4059 c. P(X ≤ 2) =BINOM.DIST(2,6,.2,1) = .9011 Learning Objective: 06-4
6.22
Use the Excel function =BINOM.DIST(x,n,π,1) where “1” stands for cumulative. a. P(X < 4 ) = P(X ≤ 3) =BINOM.DIST(3,12,.10,1) = .9744. b. P(X ≥ 3) = 1 – P(X ≤ 2) =1-BINOM.DIST(2,7,.40,1) = .5801. c. P(X ≤ 9) =BINOM.DIST(9,14,.60,1) = .7207.
100
ASBE 6e Solutions for Instructors Learning Objective: 06-4 6.23
The number of customers out of the next 10 who pay with American Express is a binomial random variable with n = 10 and π = .2. a. P(X = 0) =BINOM.DIST(0,10,.2,0) = .1074. b. P(X ≥ 2) = 1 – P(X ≤ 1) =1-BINOM.DIST(1,10,.2,1) = .6242. c. P(X < 3) = P(X ≤ 2) =BINOM.DIST(2,10,.2,1) = .6778 d. μ = nπ =(10)(.2) = 2 e. σ = = 1.2649 (10)(.2)(1 - .2) f. See below. g. Skewed to the right.
Learning Objective: 06-4 6.24
The number of customers out of 12 who have an incorrect address in the database is a binomial random variable with n = 12 and π = .05. a. or use Excel: =BINOM.DIST(0,12,.05,0). 12! P ( X = 0) = .050 (1 - .05)12 -0 = .5404 0!(12 - 0)! b. or use Excel: =BINOM.DIST(1,12,.05,0). 12! P( X = 1) = .051 (1 - .05)12-1 = .3413 1!(12 - 1)! c. or use Excel: =BINOM.DIST(2,12,.05,0). 12! 2 12-2 P ( X = 2) = .05 (1 - .05) = .0988 2!(12 - 2)! d. = .9804. Either add parts a-c P( X < 3) = P( X £ 2) = P( X = 0) + P ( X = 1) + P( X = 2) above or use the Excel function =BINOM.DIST(2,12,.05,1) = .9804. e. See below, skewed to the right.
101
ASBE 6e Solutions for Instructors
Learning Objective: 06-5
6.25
The number of customers out of the next 5 who order a nonalcoholic beverage is a binomial random variable with n = 5 and π = .38. a. P(X =0) =BINOM.DIST(0,5,.38,0) = .0916 b. P(X ≥ 2) = 1–P(X ≤ 1) =1- BINOM.DIST(1,5,.38,1) = .6276 c. P(X < 4) = P(X ≤ 3) =BINOM.DIST(3,5,.38,1) = .9274 d. P(X =5) =BINOM.DIST(5,5,.38,0) = .0079 Learning Objective: 06-4
6.26
The number of car buyers out of a sample of 8 who use the Internet for research is a binomial random variable with n = 8 and π = .6. a. or use Excel: =BINOM.DIST(8,8,.6,0). 8! P ( X = 8) = .608 (1 - .60)8-8 = .0168 8!(8 - 8)! b. =1-BINOM.DIST(4,8,.60,) = .5941. P( X ³ 5) = 1 - P( X £ 4) c. > =1-BINOM.DIST(4,8,.60,1) = .5941. Note that we could also P( X 4) = 1 - P( X £ 4) use the answer from part b because . P( X ³ 5) = P( X > 4) d. and . m = np = (8)(.60) = 4.8 s = 8(.6)(.4) = 1.3856 e. It is almost symmetric (slightly left-skewed).
102
ASBE 6e Solutions for Instructors
Learning Objective: 06-4 6.27
The number of passengers out of a sample of 16 who check their bags is a binomial random variable with n = 16 and π = .7. a. P(X = 16) = = .0033 or use Excel: =BINOM.DIST(16,16,.7,0). 16! 16 0 (.7) (1-.7) 0!16! b. P(X < 10) = P(X ≤ 9) =BINOM.DIST(9,16,.7,1) = .1753. c. P(X ≥ 10) = 1 P(X ≤ 9) =1-BINOM.DIST(9,16,.7,1) = .8247. Learning Objective: 06-4
6.28
The number of drivers stopped for speeding out of a sample of 12 who have an invalid drivers license is a binomial random variable with n = 12 and π = .15. a. or use Excel: =BINOM.DIST(0,12,.15,0). 12! P ( X = 0) = .150 (1 - .15)12-0 = .1422 0!(12 - 0)! b. or use Excel: =BINOM.DIST(1,12,.15,0). 12! P( X = 1) = .151 (1 - .15)12-1 = .3012 1!(12 - 1)! c. =1-BINOM.DIST(1,12,.15,1) = .5565. P( X ³ 2) = 1 - P( X £ 1) Learning Objective: 06-4
6.29
a. = 1, = = 1.0 and b. = 2, = = 2.0 and c. = 4, = = 4.0 and
l l l
= 1. = 1.414. = 2.0
Learning Objective: 06-5 6.30
a. = 9, = = 9.0 and
l
= 3.
103
ASBE 6e Solutions for Instructors b. = 12, = = 12.0 and c. = 7, = = 7.0 and
6.31
6.32
= 3.464. l = 2.646.
l Learning Objective: 06-5 a. = 4, . Can also use Excel: =POISSON.DIST(6,4,0). (4.0) 6 (e) -4.0 P( X = 6) = = .1042 6! b. = 12, . Can also use Excel: 10 -12.0 (12.0) (e) P ( X = 10) = = .1048 10! =POISSON.DIST(10,12,0). c. = 7, . Can also use Excel: =POISSON.DIST(4,7,0). (7.0) 4 (e) -7.0 P ( X = 4) = = .0912 4! Learning Objective: 06-5 a. = 0.1, P ( X = 2) = b. = 2.2, P( X = 1) = c. = 1.6,
2
(.1) (e) 2! 1
-.1
(2.2) (e) 1!
. Can also use Excel: =POISSON.DIST(2,.1,0). = .0045
-2.2
Can also use Excel: =POISSON.DIST(1,1.1,0). = .2438
(1.6) 3 ( e) -1.6 P ( X = 3) = = .1378 3! Learning Objective: 06-5
Can also use Excel: =POISSON.DIST(3,1.6,0).
6.33
Use the Excel function =POISSON.DIST(x,λ,1) where “1” stands for cumulative. a. = 4.3, P(X 3) =POISSON.DIST(3,4.3,1) = .3772. b. = 5.2, P(X > 7) = 1 – P(X ≤ 7) =1-POISSON.DIST(7,5.2,1) =.1551. c. = 2.7, P(X < 3) = P(X ≤ 2) =POISSON.DIST(2,2.7,1) = .4936. Learning Objective: 06-5
6.34
Use the Excel function =POISSON.DIST(x,λ,1) where “1” stands for cumulative. a. = 11.0, P(X 10) =POISSON.DIST(10,11,1) = .4599. b. = 5.2, P(X > 3) = 1 − P(X ≤ 3) =1-POISSON.DIST(3,5.2,1) = .7619. c. = 3.7, P(X < 2) = P(X ≤ 1) =POISSON.DIST(1,3.7,1) = .1162. Learning Objective: 06-5
6.35
Use the Excel function =POISSON.DIST(x,λ,1) where “1” stands for cumulative. a. = 8.0, P(X > 10) = 1 − P(X 10) =1-POISSON.DIST(10,8,1) = .1841. b. = 4.0, P(X ≤ 5) =POISSON.DIST(5,4,1) = .7851. c. = 5.0, P(X ≥ 2) = 1 − P(X ≤ 1) =1-POISSON.DIST(1,5,1) = .9596. 104
ASBE 6e Solutions for Instructors Learning Objective: 06-5 6.36
Use the Excel function =POISSON.DIST(x,λ,1) where “1” stands for cumulative. a. = 5.8, P(X < 4) = P(X ≤ 3) =POISSON.DIST(3,5.8,1) =.1700. b. = 4.8, P(X ≥ 3) = 1 – P(X ≤ 2) =1-POISSON.DIST (2,4.8,1) = .8574. c. = 7.0, P(X ≤ 9) =POISSON.DIST (9,7,1) =.8305. Learning Objective: 06-5
6.37
The number of problems per vehicle is a Poisson random variable with λ = 1.7. a. P(X ≥ 1) = 1 – P(X = 0) =1-POISSON.DIST(0,1.7,0) = .8173. b. P(X = 0) =POISSON.DIST(0,1.7,0) = .1827. c. P(X > 3) = 1 P(X ≤ 3) =1-POISSON.DIST(3,1.7,1) = .0932 d. Skewed right.
Learning Objective: 06-5 6.38
a. Cancellations are independent, come one at a time. We also know that cancellations are discrete and that we are given a mean rate per unit of time. b. Can also use Excel: =POISSON.DIST(0,1.5,0). (1.5)0 ( e) -1.5 P ( X = 0) = = .2231 0! c. Can also use Excel: =POISSON.DIST(1,1.5,0). (1.5)1 ( e) -1.5 P ( X = 1) = = .3347 1! d. P(X > 2) =1 P(X ≤ 2) =1-POISSON.DIST(2,1.5,1) = .1912. e. P(X 5) =1 P(X ≤ 4) =1-POISSON.DIST(4,1.5,1) = .0186. Learning Objective: 06-5
6.39
a. The number of add-ons is discrete and we’ll assume they “arrive” one at a time and are independent. b. P(X ≥ 2) = 1 – P(X ≤ 1) =1-POISSON.DIST(1,1.4,1) = .4082. c. P(X = 0) = POISSON.DIST(0,1.0,0) = .2466. d. Skewed right.
105
ASBE 6e Solutions for Instructors
Learning Objective: 06-5 6.40
a. Not independent events, the warm room leads to yawns from all. b. Answers will vary. Possible example: The occurrence of a contagious illness in a particular geographic area. The fact the illness is contagious suggests the “arrivals” are not independent. Learning Objective: 06-5
6.41
Let = n = (500)(.003) = 1.5 a. P(X ≥ 2) = 1 – P(X ≤ 1) =1- POISSON.DIST(1,1.5,1) = .4422. b. P(X < 4) = P(X ≤ 3) = .9344. c. Yes, based on our rule of thumb n 20 and .05. Learning Objective: 06-6
6.42
Let = n = (100000)(.000002) = 2 a. P(X ≥ 1) = 1 P(X=0) =1- POISSON.DIST(0,2,0) = .8647. b. P(X ≥ 2) = 1 – P(X ≤ 1) =1-POISSON.DIST(1,2,1) = .5940. c. Yes, based on our rule of thumb n 20 and .05. Learning Objective: 06-6
6.43
a. μ = n = (200)(.03) = 6 letters. b. = 2.412. s = (200)(.03)(1 - .03) c. Using the Poisson approximation to the binomial: P(X ≥ 10) = 1 – P(X ≤ 9) =1-POISSON.DIST(9,6,1) = .0839. If using Excel to calculate the exact binomial: P(X ≥ 10) =1-BINOM.DIST(9,200,.03,1) = .0808. d. Using the Poisson approximation to the binomial: P(X ≤ 4) =POISSON.DIST(4,6,1) = .2851. If using Excel to calculate the exact binomial: P(X ≤ 4) =BINOM.DIST(4,200,.03,1) = .2810. e. Yes, based on our rule of thumb n 20 and .05 Learning Objective: 06-6
6.44
Let = n = (100)(.01) = 1. P(X ≥ 2) = 1 – P(X ≤ 1) =1-POISSON.DIST(1,1,1) = .2642. Learning Objective: 06-6
106
ASBE 6e Solutions for Instructors
6.45
a. Expected Value = μ = n b. Poisson approximation: P(X = 0) = POISSON.DIST(0,2.3,0) = .1003 P(X > 2) =1 P(X ≤ 2) =1-POISSON.DIST(2,2.3,1) = .4040. c. The approximation is fairly accurate because n 20 and .05. Learning Objective: 06-6
6.46
a. i. X = 0, 1, 2, or 3 ii. X = 0, 1, 2, or 3 iii. X = 0, 1, 2, 3, or 4 iv. X = 0, 1, 2, 3, 4, 5, 6, or 7 b. i. P(X = 3) = .0333 C C C C s x N - s n- x = 4 3 10-4 3-3 = C N n 10 C3 ii. P(X = 2) = .1316 C C C C s x N -s n - x = 3 2 20-3 5-2 = N Cn 20 C5 iii. P(X = 1) = .4469 s C x N - s Cn - x 9 C1 36-9 C4-1 = = N Cn 36 C4 iv. P(X = 3) = .1098 C C C C s x N -s n - x = 10 3 50-10 7-3 = C N n 50 C7 c. i. =HYPGEOM.DIST(3,3,4,10,0) = .0333 ii. =HYPGEOM.DIST(2,5,3,20,0) = .1316 iii. =HYPGEOM.DIST(1,4,9,36,0) = .4469 iv. =HYPGEOM.DIST(3,7,10,50,0) = .1098 Learning Objective: 06-7
6.47
a.
b. The distribution is symmetric with a small range (2 to 4). Learning Objective: 06-7
107
ASBE 6e Solutions for Instructors 6.48
a. Let X = number of essay questions the student receives. Using Excel with N = 60, n = 10, and s = 6: P(X = 0) =HYPGEOM.DIST(0,10,6,60,0) = .3174 P(X = 1) =HYPGEOM.DIST(1,10,6,60,0) = .4232 P(X = 2) =HYPGEOM.DIST(2,10,6,60,0) = .2070 P(X = 3) =HYPGEOM.DIST(3,10,6,60,0) = .0470 P(X = 4) =HYPGEOM.DIST(4,10,6,60,0) = .0051 P(X = 5) =HYPGEOM.DIST(5,10,6,60,0) = .0000 P(X = 6) =HYPGEOM.DIST(6,10,6,60,0) = .0000 and all of these added together = 1. b. P(X = 0) = =HYPGEOM.DIST(0,10,6,60,0) = .3174 c. P(X 1) = 1 P(X = 0) =1-HYPGEOM.DIST(0,10,6,60,0) = .6826. d. P(X 2) = 1 P(X 1) =1-HYPGEOM.DIST(1,10,6,60,1) = .2594. e. Skewed right.
Learning Objective: 06-7 6.49
a. Let X = number of incorrect vouchers in sample. Use Excel with N = 50, n = 5, and s = 20. b. P(X = 0) =HYPGEOM.DIST(0,5,20,50,0) = .0673 c. P(X = 1) =HYPGEOM.DIST(1,5,20,50,0) = .2587 d. P(X 3) = 1 P(X 2) =1-HYPGEOM.DIST(2,5,20,50,1) = .3100 e. Fairly symmetric or slightly skewed right.
Learning Objective: 06-7
108
ASBE 6e Solutions for Instructors 6.50
a. Let X = number of HIV specimens in sample. Using Excel with N = 40, n = 5 and s = 8: P(X = 0) =HYPGEOM.DIST(0,5,8,40,0) = .3060 P(X = 1) =HYPGEOM.DIST(1, 5,8,40,0) = .4372 P(X = 2) =HYPGEOM.DIST(2, 5,8,40,0) = .2111 P(X = 3) =HYPGEOM.DIST(3, 5,8,40,0) = .0422 P(X = 4) =HYPGEOM.DIST(4, 5,8,40,0) = .0034 P(X = 5) =HYPGEOM.DIST(5, 5,8,40,0) = .0000 b. P(X = 0) =HYPGEOM.DIST(0,5,8,40,0) = .3060 c. P(X < 3) = P(X 2) =HYPGEOM.DIST(2,5,8,40,1) =.9543. d. P(X 2) = 1 P(X 1) =1-HYPGEOM.DIST(1, 5,8,40,1) = .2568. e. Skewed right.
Learning Objective: 06-7 6.51
a. n/N = 3/100 = .03 < .05 therefore it is safe to use binomial approximation with π = s/N = 40/100 = .4. Binomial approximation: P(X = 3) =BINOM.DIST(3,3,.4,0) = .0640. Exact probability: P(X = 3) =HYPGEOM.DIST(3,3,40,100,0) = .0611. b. n/N = 10/200 = .05 therefore it is probably safe to use binomial approximation with π = s/N = 60/200 = .3. Binomial approximation: P(X = 2) =BINOM.DIST(2,10,.3,0) = .2335. Exact probability: =HYPGEOM.DIST(2,10,60,200,0) = .2354. c. n/N = 12/160 = .075 > .05 therefore it isn’t safe to use binomial approximation. If one were to use the binomial approximate π = s/N = 16/160 = .10. Binomial approximation: P(X = 1) =BINOM.DIST(1,12,.1,0) = .3766. Exact probability: =HYPGEOM.DIST(1,12,16,160,0) = .3885. d. n/N = 7/500 = .014 < .05, therefore it is safe to use binomial approximation with π = s/N = 350/100 = .7. Binomial approximation: P(X = 5) =BINOM.DIST(5,7,.7,0) = .3177. Exact probability: =HYPGEOM.DIST(5,7,350,500,0) = .3198. Learning Objective: 06-8
6.52
a. X = the number of vouchers with errors. X has a hypergeometric distribution with N = 200, s = 20, and n = 5. Because n/N < .05 it is safe to use the binomial approximation with π = s/N = 20/200 = .1. b. P(X = 0) =BINOMDIST(0,5,.1,0) = .5905 109
ASBE 6e Solutions for Instructors c. P(X 2) = 1–P(X 1) =1-BINOMDIST(1,5,.1,1) = .0815. Learning Objective: 06-8 6.53
a. X = the number of firearms backgrounds checks that show a felony conviction. X has a hypergeometric distribution with N = 500, s = 50, and n = 10. Because n/N = 10/500 = . 02 < .05 it is safe to use the binomial with = s/N= 50/500 = .1. b. P(X=0) =BINOM.DIST(0,10,.1,0) = .3487. c. P(X 2) = 1 – P(X ≤ 1) =1-BINOM.DIST(1,10,.1,1) = .2639. d. P(X < 4) = P(X ≤ 3) =BINOM.DIST(3,10,.1,1) = .9872. Learning Objective: 06-8
6.54
a. X = the number of vehicles that comply with California emissions law. X has a hypergeometric distribution with N = 400, s = 320, and n = 6. Because n/N = 6/400 = . 015 < 0.05 it is safe to use the binomial approximation with π = s/N = 320/400 = .8. b. P(X = 6) =BINOM.DIST(6,6,.8,0) = .2621. c. P(X ≥ 4) = 1 P(X ≤ 3) =1-BINOMDIST(3,6,.8,1) = .9011. Learning Objective: 06-8
6.55
a. P(X = 5) = (.5)(1−.5)5-1 = .0313. b. P(X = 3) = (.25)(1−.25)3-1 = .1406. c. P(X = 4) = (.6)(1−.6)4-1 = .0384. Learning Objective: 06-9
6.56
a. Geometric mean is 1/ = 1/(.20) = 5 b. Using the geometric CDF, P(X ≤ 10) = 1 − (1 − )x = 1 − (1 − .20)10 = .8926. Learning Objective: 06-9
6.57
a. Geometric mean is 1/ = 1/(.50) = 2 b. Using the geometric CDF, P(X > 10) = 1 − [1 − (1−.50)10 ]= .5010 = .00098 Learning Objective: 06-9
6.58
a. = 79.62.54 = 202.184 cm b. = 3.242.54 = 8.2296 cm c. Rule 1 for the mean and Rule 2 for the standard deviation. Learning Objective: 06-10
6.59
a. Applying Rule 3, we add the means for each month to get = 9500+7400 + 8600 = $25,500. Applying Rule 4, we add the variances for each month and then take the square root of this sum to find the standard deviation for the quarter: 2 = 1250+1425+1610 = 4285. σ = (4285).5 = 65.4599. b. Rule 4 assumes that the sales for each month are independent of each other. This may not be valid, given that a prior month’s sales usually influence the next month’s sales. Learning Objective: 06-10
6.60
= (5/9)(37.1)–17.78 = 2.83° C, = (5/9)(10.3) = 5.72° C
110
ASBE 6e Solutions for Instructors Learning Objective: 06-10 6.61
a. Let Y = Bob’s point total. .
Y (5)(80) 400, Y (25)(5) 11.18
b. No, given that the standard deviation is 11.18, 450 is more than three standard deviations from the mean. Learning Objective: 06-10 6.62
The probability of a payout is 1 .99863 = .00137. The expected payout is (.00137) (1,000,000) = $1,370 dollars. To break even, the company would charge $1,370. Learning Objective: 06-2
6.63
The expected loss is E(X) = ($250)(.02) + ($0)(.98) = $5. This exceeds the $4 cost of insurance (assuming you would lose the entire cost of the PDA). Statistically, it is worth it to insure to obtain “worry-free” shipping, despite the small likelihood of a loss. Learning Objective: 06-2
6.64
E(# rentals) = .1(30)+.4(60)+.4(120)+.1(180) = 93. $300/93 = $3.23. Jane should price her rentals at $4 per tube in order to meet her expected revenue goal. Learning Objective: 06-3
6.65
a. b. c. d. e.
6.66
This is a discrete binomial probability problem with "success" as "order bread" and "failure" as "do not order bread". The probability of success, π, is .14 and the number of trials, n, is 10. a. P(X > 5) = 1–P(X ≤ 5). Using Excel: =1-BINOM.DIST(5,10,.14,1) = 1– .99905 = .00095. b. P(X ≤ 2). Using Excel: =BINOM.DIST(2,10,.14,1) = .8455. c. P(X = 0). Using Excel: =BINOM.DIST(0,10,.14,0) = .2213. d. From MegaStat we obtain the probability distribution and probability histogram. The distribution is skewed to the right. Or we could recognize that because π < .5 the distribution is skewed right.
= .80 (answers will vary). = .300 (answers will vary). = .50 (answers will vary). = .80 (answers will vary). Outcomes of one trial might influence the next. For example, if I fail to make a free throw because I shot the ball “long”, I will adjust my next shot to be a little “shorter,” hence, violating the independence rule. Learning Objective: 06-4
111
ASBE 6e Solutions for Instructors
Binomial distribution 10 0.14
X 0 1 2 3 4 5 6 7 8 9 10
P(X) 0.22130 0.36026 0.26391 0.11457 0.03264 0.00638 0.00086 0.00008 0.00000 0.00000 0.00000
n p cumulative probability 0.22130 0.58156 0.84547 0.96004 0.99267 0.99905 0.99991 0.99999 1.00000 1.00000 1.00000
Learning Objective: 06-4 6.67
Define X to be the number of vehicles that fail the emissions test. X is binomial with n =10 and π = .1. a. P(X = 0) = .5905. Using Excel: =BINOM.DIST(0,5,.1,0) . b. P(X = 1) = .3281. Using Excel: =BINOM.DIST(1,5,.1,0). c. Strongly skewed to the right.
Learning Objective: 06-4 6.68
This is a discrete binomial probability problem with "success" as "can transact business in a foreign language" and "failure" as "cannot transact business in a foreign language". Probability of success, π, is .20 and the number of trials, n, is 10. a. P(X = 0) = BINOM.DIST(0,10,.20,0) = .1074. b. P(X ≥ 2) = 1 – P(X ≤ 1). Using Excel: =1-BINOM.DIST(1,10,.20,1) = .6242. c. P(X = 10) =BINOM.DIST(10,10,.20,0) = .00000 Learning Objective: 06-4
112
ASBE 6e Solutions for Instructors 6.69
This is a discrete binomial probability problem with "success" as "crispy" and "failure" as "original". Probability of success, π, is .50 and the number of trials, n = 4. a. P(X = 0) =BINOM.DIST(0,4,.5,0) = .0625 b. P(X ≥ 2) = 1 – P(X ≤ 1) =1-BINOM.DIST(1,4,.5,1) = .6875 c. P(X ≤ 2) =BINOM.DIST(2,4,.5,1) = .6875 d. Symmetric.
Learning Objective: 06-4 6.70
This is a discrete binomial probability problem with "success" as "order light beer" and "failure" as "do not order light beer". The probability of success, π, is .40 and the number of trials, n = 8. a. P(X = 0). Using Excel: =BINOM.DIST(0,8,.40,0) = .0168. b. P(X = 1). Using Excel: =BINOM.DIST(1,8,.40,0) = .0896. c. P(X = 2). Using Excel: =BINOM.DIST(2,8,.40,0) = .2090. d. P(X ≤ 2). Using Excel: =BINOM.DIST(2,8,.40,1) = .3154. e. Slightly skewed right.
Learning Objective: 06-4 6.71
a. P(X = 3) =BINOM.DIST(3,20,0.3,0) = .0716.
113
ASBE 6e Solutions for Instructors b. P(X = 7) =BINOM.DIST(7,50,0.1,0) = .1076. c. P(X ≤ 6) =BINOM.DIST(6,80,0.05,1) = .8947. d. P(X ≥ 30) = 1 – P(X ≤ 29) =1-BINOM.DIST(29,120,0.2,1) = .1067. Learning Objective: 06-4 6.72
a. This is a discrete binomial probability problem where "success" can be "got a sentence correct" and failure can be "got a sentence wrong". There are 20 sentences so we can say there are 20 trials and the probability of getting one correct if randomly guessing is .50. P(X ≥ 14) = 1 – P(X ≤ 13). Using Excel: =1-BINOM.DIST(13,20,.50,1) = .0577. The probability of passing is about 6%; not very likely to pass when you guess. b. Using the same approach as in part a, we can plug in various values of a "passing score" and we can see that P(X ≥ 15) = .0207 therefore a score of 15 would be needed in order to get a 5% or less probability that someone can pass by guessing. Learning Objective: 06-4
6.73
Define X to be the number of defaulted student loans out of a sample of 10. X is binomial with n =10 and π = .07. a. P(X = 0) =BINOM.DIST(0,10,.07,0) = .4840. b. P(X ≥ 3) = 1 – P(X ≤ 2) =1-BINOM.DIST(2,10,.07,1) = .0283. c. For this binomial, = n = (10)(.07) = 0.7 defaults. Learning Objective: 06-4
6.74
This is a discrete binomial probability problem where "success" is "something in the pocket" and failure is "nothing in the pocket". The probability of success, π, is .08 and the number of trials, n, is 14. Using Excel: =BINOM.DIST(0,14,0.08,0) = 0.3112. P( X = 0) Learning Objective: 06-4
6.75
Define X to be the number of movie commercials advertising ‘R” rated movies. X is binomial with n = 16 and = .8 a. P(X ≥ 10) = 1 – P(X ≤ 9) =1-BINOM.DIST(9,16,.8,1) = .9733. b. P(X < 8) = P(X ≤ 7) =BINOM.DIST(7,16,.8,1) = .0015. Learning Objective: 06-5
6.76
a. P(X = 7) =POISSON.DIST(7,10,0) = .0901. b. P(X = 3) =POISSON.DIST (3,10,0) = .0076. c. P(X < 5) = P(X ≤ 4) =POISSON.DIST (4,10,1) = .0293. d. P(X ≥ 11) = 1 – P(X ≤ 10) =1-POISSON.DIST (10,10,1) = .4170. Learning Objective: 06-5
6.77
Let X = the number of no shows. a. If n = 10 and = .10, then P(X = 0) =BINOM.DIST(0,10,.1,0) = .3487. b. If n = 11 and = .10, then P(X ≥ 1) = 1 – P(X=0) =1-BINOM.DIST(0,11,.1,0) = 1 – . 3138 = .6862. c. If they sell 11 seats, there is no way that more than 1 will be bumped. P(X > 1) = 0.
114
ASBE 6e Solutions for Instructors d. Let X = the number who do show up and set = .90. We want P(X 10) .95 so we use Excel’s function = 1BINOM.DIST(9,n,.9,TRUE) for various values of n. It turns out that n = 13 will suffice because P(X 13) = .9658 > .95. P(X 9) 0.30264 0.11087 0.03416
n 11 12 13
P(X 10) 0.69736 0.88913 0.96584
Learning Objective: 06-4 6.78
a. Let X be the number that are not working. As long as no more than 2 are not working, he will have enough. P(X ≤ 2) =BINOM.DIST(2,10,0.2,1) = .6778. b. Let X be the number that are working and set = .8. We want P(X 8) .95 so we use Excel’s function =1BINOM.DIST(7,n,0.8,TRUE) for various values of n. It turns out that n = 13 will suffice. n 10 11 12 13
P(X ≤ 7) 0.32220 0.16114 0.07256 0.03004
P(X 8) 0.67780 0.83886 0.92744 0.96996
Learning Objective: 06-4 6.79
a. Because calls to a fire station within a minute are most likely all about the same fire, the calls are not independent. b. Answers will vary. Learning Objective: 06-5
6.80
a. This is a binomial probability with "success" as "catch fraudulent e-mail" and "failure" as "do not catch fraudulent e-mail". The probability of success, π, is .20 and the number of trials, n, is 16. The expected number that would be caught by such a filter is nπ = .20×16 = 3.2. b. = .02815. Using Excel: =BINOM.DIST(0,16,.2,0). 16! 0 16 - 0 P( X = 0) = .20 (1 - .20) 0!(16 - 0)! Learning Objective: 06-5
6.81
a. P(X = 5) =POISSON.DIST(5, 2.8, 0) = .0872. b. P(X ≤ 5) =POISSON.DIST(5, 2.8, 1) = .9349. c. arrivals/5 minute interval d. Independent time intervals Learning Objective: 06-5
115
ASBE 6e Solutions for Instructors 6.82
a. This is a Poisson probability because broken bats are rare and independent events. Given that λ = 1.0: = .3679. Using Excel: =POISSON.DIST(0,1,0). 0 -1 (1 )(e ) P ( x = 0) = 0! b. P(X ≥ 2) = 1 P(X ≤ 1) = 1 [P(X = 0) + P(X = 1)] = 1–.7358 = .2642.
P ( x = 0) =
0
-1
= .3679
(1 )(e ) 0!
(11 )(e -1 ) P ( x = 1) = 1!
= .3679
Using Excel: =1-POISSON.DIST(1,1,1) = .2642. There is about a 26% chance of having at least 2 broken bats. Learning Objective: 06-5 6.83
X is a Poisson random variable with λ = .3. a. P(X = 0) =POISSON.DIST(0,.3,0) = .7408 b. P(X ≥ 2) =1-POISSON.DIST(1,.3,1) = .0369 Learning Objective: 06-5
6.84
a. b. c. d.
Near collisions are random and independent events that occur one at a time. P(X ≥ 1) = 1 – P(X = 0). Using Excel: =1-POISSON.DIST(0,1.2,0) = .6988. P(X > 3) = 1 P(X ≤ 3). Using Excel: =1-POISSON.DIST(3,1.2,1) = .0338. See below.
Learning Objective: 06-5
116
ASBE 6e Solutions for Instructors
6.85
a. Assume that cancellations are independent of each other, occur randomly and one at a time. b. P(X = 0) =POISSON.DIST(0,1.5,0) = .2231. c. P(X = 1) =POISSON.DIST(1,1.5,0) = .3347. d. P(X > 2) = 1 – P(X ≤ 2) =1-POISSON.DIST(2,1.5,1) = .1912. e. P(X ≥ 5) = 1 – P(X ≤ 4) =1-POISSON.DIST(4,1.5,1) = .0186. Learning Objective: 06-5
6.86
X is a Poisson random variable with λ = 3.8/hour. a. P(X = 0) =POISSON.DIST(0,3.8,0) = .0224. b. P(X < 4) = P(X ≤ 3) =POISSON.DIST(3,3.8,1) = .4735. c. P(X > 5) = 1 – P(X ≤ 5) =1-POISSON.DIST(5,3.8,1) = .1844. Learning Objective: 06-5
6.87
a. We assume that paint defects are independent events, distributed randomly over the surface and occur one at a time. For this problem, we would use a mean of = 2.4 defects per 3 square meter area. b. P(X = 0) =POISSON.DIST(0,2.4,0) = .0907. c. P(X = 1) =POISSON.DIST(1,2.4,0) = .2177. d. P(X ≤ 1) =POISSON.DIST(1,2.4,1) = .3084. Learning Objective: 06-5
6.88
X = number of claimed dependents that are ineligible for insurance. X is a binomial random variable with n = 7 and π = .02. a. P(X =0) =BINOM.DIST(0,7,.02,0) = .8681. b. P(X ≥ 1) = 1 P(X = 0) =1-BINOM.DIST(0,7,.02,0) = .1319. Learning Objective: 06-4
6.89
Assume the arrival of a rogue wave follows a Poisson distribution. The ship is out for five days so the average number of rogue waves per voyage will be: = = 4.524 per 5-day voyage. .0377 waves 24hours 5days ´ ´ hour day voyage P(X ≥ 1) = 1 – P(X = 0) =1-POISSON.DIST(0,4.524,0) = .9892. Learning Objective: 06-5
6.90
a. Earthquakes are random and independent events. No one can predict when they will occur and we assume they occur one at a time. We are also given a mean rate of discrete events per unit of time. b. P(X < 3) = P(X ≤ 2). Using Excel: =POISSON.DIST(2,1.2,1) = .8795. c. P(X > 5) = 1 P(X ≤ 5). Using Excel: =1-POISSON.DIST(5,1.2,1) = .0015.
117
ASBE 6e Solutions for Instructors
Learning Objective: 06-5 6.91
a. Crashes are unrelated events, can’t predict them, so they do happen randomly. We assume crashes occur independently and one at a time. b. P(X ≥ 1) = 1 – P(X = 0) =1-POISSON.DIST(0,2,0) = .8647. c. P(X < 5) = P(X ≤ 4) =POISSON.DIST(4,2,1) = .9474. d. Skewed to the right.
Learning Objective: 06-5 *6.92
Binomial n = 2500, = .001 or Poisson with = .001(2500) = 2.5 leaks per 2500 meters. Using the Poisson approximation to the binomial: a. P(X = 0). Using Excel: =POISSON.DIST(0,2.5,0) = .0821 b. P(X ≥ 3) = 1 – P(X ≤ 2). Using Excel: =1–POISSON.DIST(2,2.5,1) = .4562. c. µ = 2.5 leaks per 2500 meters. Learning Objective: 06-6
118
ASBE 6e Solutions for Instructors Binomial n = 200, = .02. Define X to be the number of twin births in 200 deliveries. E(X) = (200)(.02) = 4. b. Using the Poisson approximation: P(X = 0) =POISSON.DIST(0,4,0) = .0183. Using the binomial function: P(X = 0) =BINOM.DIST(0,200,.02,0) = .0176. c. Using the Poisson approximation: P(X = 1) =POISSON.DIST(1,4,0) = .0733. Using the binomial function: P(X = 1) =BINOM.DIST(1,200,.02,0) = .0718. d. Yes, the approximation is justified. Our rule of thumb is n 20 and .05 which is met here and the probabilities from the Poisson are similar to the binomial. Learning Objective: 06-6
*6.93
a.
*6.94
a. P(X = 0). Using Excel: =BINOM.DIST(0,200,.03,0) = .0023 or =POISSON.DIST(0,6,0) = .0025. b. P(X = 1). Using Excel: =BINOM.DIST(1,200,.03,0) = .0140 or =POISSON.DIST(1,6,0) = .0149. c. P(X = 2). Using Excel: =BINOM.DIST(2,200,.03,0) = .0430 or =POISSON.DIST(2,6,0) = .0446. d. Set = n = (200)(.03) = 6.0 (see parts a through c) e. Yes, n 20 and .05 and probabilities are similar. Learning Objective: 06-6
*6.95
a. For the binomial, = n = (5708)(.00128) = 7.31 is the expected number of “bumped” passengers per hour. b. Using the Poisson approximation: P(X < 10) = P(X ≤ 9) =POISSON.DIST(9,7.31,1) = .7977. P(X > 5) =1 P(X ≤ 5) =1-POISSON.DIST(5,7.31,1) = .7371. c. Yes, the approximation is justified. Our rule of thumb is n 20 and .05 which is met. Learning Objective: 06-6
*6.96
a. For the binomial, µ = n = (500)(.02) = 10 is the expected number of "passes". b. Using the Poisson approximation with = 10: P(X 5) =POISSON.DIST(5,10,1). = .0671. Learning Objective: 06-6
*6.97
a. Using the Poisson approximation with = (150)(.025) = 3.75: P(X ≥ 4) = 1 P(X ≤ 3) =1-POISSON.DIST(3,3.75,1) = .5162 b. Assume calls are independent. Learning Objective: 06-6
*6.98
a. For the binomial, µ = n = (100,000)(.00014) = 14 is the expected number of "retained foreign bodies". b. Using the Poisson approximation with = n = (100,000)(.00014) = 14 we get .0055. The excel function is =POISSON(5,14,1). Less than 1% P( X £ 5) = chance (.55%) of getting 5 or fewer retained foreign bodies. Learning Objective: 06-6
119
ASBE 6e Solutions for Instructors *6.99
a. Geometric mean: 1/ = 1/.25 = 4. b. Using geometric CDF, P(X ≤ 6) = 1 – (1 − π)6 = 1 – (1 − .25)6 = .8220. Approximately 82.2% change of landing first job offer by the sixth interview. Learning Objective: 06-9
*6.100
a. Geometric mean is 1/ = 1/(.04) = 25 customers. b. Using geometric CDF, 1–.4420 = .5580. x 20 P ( X £ 20) = 1 - (1 - p ) = 1 - (1 - .04) = Approximately 55.8% likelihood of the first birthday cake order coming within the first 20 customers. Learning Objective: 06-9
*6.101
a. Geometric mean is 1/ = 1/(.08) = 12.5 cars. b. Using geometric CDF, P(X 5) = 1(1)x = 1 (1.08)5 = .3409. Approximately 34% change of finding the first car with a burned out headlight within the first five cars inspected. Learning Objective: 06-9
*6.102
a. Geometric mean is 1/ = 1/(.07) = 14.29 operations b. Using geometric CDF, P(X 20) = 1 P(X 19) = 1 [1(1)x] = 1–[1–(1)19]= 1– (1–.2519) = .2519. Approximately 25% likelihood of conducting 20 or more operations before the first fatality. Learning Objective: 06-9
*6.103
a. Geometric mean is 1/ = 1/(.05) = 20. b. Using geometric CDF, P(X ≥ 30) = 1 − P(X 29) = 1 – (1 (1)x) = 1 – (1 (1.05)29) = 1 – (1 .2259) = .2259. Learning Objective: 06-9
*6.104
Geometric mean is 1/ = 1/(.02) = 50 Learning Objective: 06-9
*6.105
a.
m = 233.1 lb ´
.4536 kg = 105.73kg lb
b.
.4536 kg = 15.85kg lb c. Rule 1 for the mean and Rule 2 for the standard deviation. Learning Objective: 06-10
s = 34.95 lb ´
*6.106
a. By Rule 1, expected total cost is vQ+F = vQ+F = (8)(25000) + 15000 = $350,000 By Rule 2, std dev. of total cost is vQ+F = vQ = (8)(2000) = $16,000 b. To break even, we want TR TC = 0 where TR = expected total revenue and TC = expected total cost. Since TR = (Price)(Quantity) = PQ we set PQ (vQ+F) = 0 and solve
120
ASBE 6e Solutions for Instructors for P to get P(25000) 350000 = 0 or P = $14. For a profit of $20,000 we have P(25000) 370000 = 0 or P = $14.80. Learning Objective: 06-10 6.107
, 2 2 2 2 m X1 + X 2 = m1 + m 2 = 3420 + 390 = 3810ml s X 1 + X 2 = s 1 + s 2 = 10 + 2 = 10.2ml Learning Objective: 06-10
*6.108
a. Using Rule 3: total time = 20+10+14+6+48 = 98 hours b. Using Rule 4: total time = . The 2-sigma interval 2 2 2 2 2 4 + 2 + 3 + 2 + 6 = 8.31 around the mean is ± 2 or 98 ± (2)(8.31). The range is 81.4 to 114.6 hours. Learning Objective: 06-10
*6.109
a. By Rule 1, mean of total cost: vQ+F = vQ+F = (2225)(7) + 500 = $16,075 By Rule 2, std dev. of total cost: vQ+F = vQ = (2225)(2) = $4,450 By Rule 1, expected revenue is E(PQ) = PQ = (2850)(7) = $19,950 Expected profit is TR – TC = 19,950 – 16,075 = $3,875 Learning Objective: 06-10
*6.110
a.
*
total time = 15+30+25+45+20 = 135 minutes, total time =
.
4 + 6 + 5 + 10 + 5 = 14.2 b. Independence might not hold if the outcome of one step affects the duration of the next procedure. Learning Objective: 06-10 2
*6.111
2
2
a. X+Y = $70 + $200 = $270 b. σ X+Y = 102 + 302 + 2 ´ 400 = $42.43 c. The variance of the total is greater than either of the individual variances. Learning Objective: 06-10
121
2
2
ASBE 5e Solutions for Instructors
Chapter 7 Continuous Distributions Calculations are done on Excel except where noted that Appendix Tables are used. Probabilities are rounded to four decimals. Z scores are typically rounded to two decimals when using Appendix tables and to no more than four decimals when using Excel. 7.1
a. Discrete. We can count the number of passengers with children. b. Continuous. A proportion can be anything between 0 and 1 inclusive. c. Continuous. Weight can take on fractional values. Learning Objective: 07-1
7.2
a. Continuous. Temperature can take on fractional values. b. Discrete. We can count the number who order only coffee. c. Continuous. Time can take on fractional values. Learning Objective: 07-1
7.3
In order to be a valid PDF, total area under f(x) must equal 1. a. Area = .25(1) = .25 therefore this is not a PDF. b. This is a valid PDF. Area = 4(.25) = 1. c. Area = ½(2)(2) = 2 therefore this is not a PDF. Learning Objective: 07-1
7.4
For a continuous PDF, we use the area under the curve to measure the probability. The area above a single point is defined to be zero so if we summed up all the point probabilities we would have a sum equal to zero. In addition, by definition there are an infinite number of points in the interval over which a continuous random variable is defined. This is why we take an integral, rather than summing. Learning Objective: 07-1
7.5
a. = (0+10)/2 =5
= 2.8868 (10 - 0) 12
b. = (200+100)/2 = 150
2
= 28.8675 (200 - 100) 12
c. = (1+99)/2= 50
2
= 28.2902 (99 - 1)2 12
Learning Objective: 07-2
121
ASBE 5e Solutions for Instructors
7.6
a. Use the CDF: P(X ≤ x) =
b. Use P(X > x) =
b- x b-a
c. Use P(c ≤ X ≤ d) =
x-a b-a
. P(X < 10) for U(0,50) = (10-0)/(50-0) = 0.2
. P(X > 500) for U(0,1000) = (1000-500)/(1000-0) = 0.5
d -c b-a
. (25 < X < 45) for U(15,65) = (45-25)/(65-15) = .4
Learning Objective: 07-2 7.7
P(X=25) = 0 and P(X=45) = 0 for a continuous uniform distribution. Therefore using a < or ≤ yields the same result. Learning Objective: 07-2
7.8
a. The equation is:
a+b 2
. = (2500+4500)/2 = 3500 .
b. The equation is:
(b - a ) 12
2
= 577.3503 (4500 - 2500) 12
122
2
ASBE 5e Solutions for Instructors c. P(X < 3000) = (3000-2500)/(4500-2500) =0.25. d. P(X > 4000) = (4500-4000)/(4500-2500) = 0.25. e. P(3000< X < 4000) = (4000-3000)/(4500-2500) =0.50. Learning Objective: 07-2 7.9
a. Define X to be the time at which the bus arrives within the five minute window starting at 10:18:00. Let a = 0 and b = 5. μ = (0+5)/2 = 2.5 minutes from the lower limit. μ = 10:20:30. 2 2 (5−0) b. σ = =1.4434 12 c. P(bus is early) = P(X < 2 min) = (2-0)/(5-0) = .4 d. P(bus arrives between 10:19 and 10:21) = P(1 < X < 3) = 2/5 = .4 Learning Objective: 07-2
√
7.10
a. Define X to be Jill’s resting heart rate. Let a = 74 and b = 77. μ = (74+77)/2 = 75.5 bpm. μ = 10:20:30. 2 2 (77−74) b. σ = =0.866 12 c. P(X < 75.5 bpm) = (75.5-74)/(77-74) = .5 d. P(X > 76.3) = (77-76.3)/(77-74) = .2333 Learning Objective: 07-2
√
7.11
The Empirical means and standard deviations differ (X axis scales are different) and so do f(x) heights. Learning Objective: 07-3
7.12
a. The maximum height of a normal distribution is the mean. In this case, the maximum height is at x = 75. b. No, f(x) does not touch the X axis at any point. The distribution is asymptotic to the X axis. Learning Objective: 07-3
7.13
The Empirical Rule states that for data from a normal distribution we expect about 68.26% will lie within ± 1 about 95.44% will lie within ± 2 about 99.73% will lie within ± 3 Learning Objective: 07-3
7.14
a. Yes. Most sizes tend toward the same value with equal percentages above and below this value. b. No, distribution could be skewed. Most likely the distribution is skewed right. Most 30 year old women have a high school degree with some women having several years of
123
ASBE 5e Solutions for Instructors higher education and a few women having advanced degrees which might be six or more years of higher education. c. No, distribution could be skewed right. Most bill payments will be delivered within a week but there may be a few that take much longer. d. No, distribution could be skewed right. Most insurance claims can be settled in about the same amount of time but some will take much longer. Learning Objective: 07-3 7.15
These values were found using Appendix C-1. a. P(0 < Z < .50) = .1915. b. Due to symmetry : P(−.50 < Z < 0) = P(0 < Z < .50) = .1915. c. Because 0 is the mean of the standard normal distribution P(Z > 0) = .5000. d. Because the standard normal is a continuous distribution P(Z = 0) = 0. Learning Objective: 07-4
7.16
These values were found using Appendix C-2. a. P(1.22 < Z < 2.15) = P(Z < 2.15) – P(Z < 1.22) = .9842 − .8888 = .0954 b. P(2.00 < Z < 3.00) = P(Z < 3.00) – P(Z < 2.00) = .99865 − .9772 = .0215 c. P(−2.00 < Z < 2.00) = (Z < 2.00) – P(Z < −2.00) = .9772 − .0228 = .9544 d. P(Z > .50) = 1 – P(Z < .50) = 1 − .6915 = .3085 Learning Objective: 07-4
7.17
These values were found using Appendix C-2. a. P(−1.22 < Z < 2.15) = P(Z < 2.15) – P(Z < −1.22) = .9842 − .1112 = .8730 b. P(−3.00 < Z < 2.00) = P(Z < 2.00) – P(Z < −3.00) = .9772 − .00135 = .9759 c. P(Z < 2.00) = .9772 d. Because the standard normal is a continuous distribution P(Z = 0) = 0. Learning Objective: 07-4
7.18
a. Using the table in Appendix C-2,
P( Z < -1.96) = .0250
b. Using the table in Appendix C-2:
= 1–.9750 = .0250. We P( Z > 1.96) = 1 - P( Z < 1.96) also know that the area under the curve for the lower tail at −1.96 will be the same as the area under the curve for the upper tail at +1.96 so we could have just used part a to get the answer.
124
ASBE 5e Solutions for Instructors c. Using the table in Appendix C-2:
P( Z < 1.65) = .9505 d. Using the table in Appendix C-2: P(Z > 1.65) = 1–P(Z < 1.65) = 1–.0495 = .9505. Learning Objective: 07-4 7.19
Using the table in Appendix C-2: a. P(Z < −1.28) = .1003 b. P(Z > 1.28) = 1 − P(Z < 1.28) = 1 − .8997 = .1003 c. P(−1.96 < Z < 1.96) = P(Z < 1.96) P(Z < 1.96) = .975 .025 = .95 d. P(−1.65 < Z < 1.65) = P(Z < 1.65) P(Z < 1.65) = .9505 .0485 = .902 Learning Objective: 07-4
7.20
The proportion of students scoring higher than Bob is the area under the standard normal to the right of 2.17 or P(Z > 2.17). Using the table in Appendix C-2 : P(Z > 2.17) = 1 − P(Z < 2.17) = 1 − .9850 = .0150. 1.5% of the 200 students, or .015×200 = 3 students, scored higher than Bob. Learning Objective: 07-4
7.21
If Joan ran 1.75 standard deviations faster than the other women in her age group then she was 1.75 standard deviations below the average time. The proportion of women who ran faster than Joan is the area under the standard normal curve to the left of 1.75 or P(Z < −1.75). Using the table in Appendix C-2: P(Z < −1.75) = .0401 or 4.01%. . 0401×405 = 16.24. About 16 women ran faster than Joan. Learning Objective: 07-4
7.22
a. If we use Appendix C-2 to find the z-score associated with a given area we want to look inside the body of the table first instead of the far left column. To find the z-score associated with the highest 10%: the table gives areas less than z so we use .90 as our probability. The value closest to .90 in table C-2 is .8997. Reading to the left and up we find z = 1.28. The exact value from Excel can be found using =NORM.S.INV(.90) = 1.2816. b. Search for the area .50 in the body of the table in Appendix C-2. The z-score associated with this area is 0. This makes sense because we know the standardized normal distribution has mean = 0 and it is symmetric so half is above zero and half is below zero. c. The highest 7% is the same as the lowest 93%. Using Appendix C-2, search for the value closest to .93. The value closest is .9306 and its z-score is 1.48. The exact value from Excel using =NORM.S.INV(.93) is 1.4758. Learning Objective: 07-5
7.23
Using Appendix C-2: a. Searching for the area closest to .06 we find both the area .06 is halfway between .0606 and .0594. We would choose z = 1.555. From Excel using = NORM.S.INV(.06), z = 1.5488. b. Searching for the area closest to .60 we find .5987. The z-score = 0.25. From Excel using =NORM.S.INV(.60), z = .2533. 125
ASBE 5e Solutions for Instructors c. Searching for the area closest to .07 we find .0694. The z-score = 1.48. From Excel using =NORM.S.INV(.07), z = 1.4758. Learning Objective: 07-5 7.24
a. We are looking for z-scores associated with the middle 50%. Because the curve is symmetric this translates into an area of .25 below and .25 above. We want the z-scores associated with the lower .25 tail and the upper .25 tail. From Appendix C-2, the approximate z- score associated with a lower tail of .25 is .67. The corresponding zscore for the upper .25 tail is +.67. From Excel using =NORM.S.INV(.25), z = .6745. To find the upper z-score use =NORM.S.INV(.75) = .6745. b. The z-score associated with a lower tail area of .05 is 1.645. Using Excel: =NORM.S.INV(.05) = −1.6449. c. The middle 90% means the lower tail area is .05 and the upper tail area is .05. From part c and using the symmetry of the normal curve, the z-scores are 1.645 and +1.645. Using Excel: =NORM.S.INV(.05) = −1.6449 and =NORM.S.INV(.95) = 1.6449. Learning Objective: 07-5
7.25
Using Appendix C-2: a. The middle 60% means the lower tail area is .20 and the upper tail area is .20. The area in Appendix C-2 closest to .20 is .2005 resulting in a z score = −0.84. For the upper tail the z-score = +0.84 Using Excel: =NORM.S.INV(.20) = −0.8416 and =NORM.S.INV(.80) = 0.8416. b. he highest 2% means the area to the left is .98. Using Appendix C-2, the area closest to . 98 is .9798 which results in a z-score = 2.05. Using Excel =NORM.S.INV(.98) gives z = 2.0537. c. The middle 95% means the lower tail area is .025 and the upper tail area is .025. From Appendix C-2, the area .025 is associated with a z-score 1.96. The z-score for the upper tail is +1.96. Using Excel: =NORM.S.INV(.025) = −1.95996 and =NORM.S.INV(.975) = 1.95996. Learning Objective: 07-5
7.26
The z score associated with the highest 20 percent is found using =NORM.S.INV(.80) = 0.8416. The students who score in the top 20 percent score approximately 0.84 standard deviations above the mean score. Learning Objective: 07-5
7.27
The z score associated with the fastest 10 percent is found using =NORM.S.INV(.10) = −1.28. The runners must finish 1.28 standard deviations below the mean to win a gift certificate. Learning Objective: 07-5
7.28
a. =NORM.DIST(52.3,56,4,1) b. =1−NORM.DIST(58,56,4,1) c. = NORM.DIST(63.7,56,4,1) – NORM.DIST(50,56,4,1) Learning Objective: 07-4 126
ASBE 5e Solutions for Instructors 7.29
a. =1−NORM.DIST(592,600,5,1) b. =NORM.DIST(603,600,5,1) c. = NORM.DIST(603,600,5,1) – NORM.DIST(592,600,5,1) Learning Objective: 07-4
7.30
a.
7.31
a. P(X < 300) = NORM.DIST(300,290,14,1) = 0.7625 b. P(X > 250) = 1 − P(X > 250) = 1 – NORM.DIST(250,290,14,1) = 0.9979 c. P(275 < X < 310) =NORM.DIST(310,290,14,1)-NORMDIST(275,290,14,1) = 0.7814 Learning Objective: 07-4
= 1– = 1–NORM.DIST(232000,232000,7000,1) = P( X > 232,000) P( X < 232,000) 1–.50=.50. We can also use the standardized z score: . 232000 - 232000 z= =0 7000 so = 1–.5 = .5. P( X < 232,000) = P( Z < 0) = .5 P( X > 232,000) b. = NORM.DIST(239000,232000,7000,1) – P( X < 239,000) - P( X < 232,000) NORM.DIST(232000,232000,7000,1) = .8413–.5 = .3413. We can also use the standardized z-score: and . 239000 - 232000 232000 - 232000 z= =1 z= =0 7000 7000 P(232,000 < X < 239,000) = P(0 < Z < 1) which is .3413 (from Appendix C-1). c. =NORM.DIST(239000,232000,7000,1) = .8413. We can also use the P( X < 239,000) standardized z-score: . = .8413 (from Appendix CP( Z < 1) 239000 - 232000 z= =1 7000 2). d. = NORM.DIST(245000,232000,7000,1) = .9684. We can also use the P( X < 245,000) standardized z-score: . = .9686 (from P( Z < 1.86) 245000 - 232000 z= = 1.86 7000 Appendix C-2). e. P(X > 225,000) = 1– = 1 – NORM.DIST(225000,232000,7000,1) = P( X < 225,000) 1–.1587 = .8413. We can also use the standardized z-score: 225000 - 232000 z= = -1 7000 .1 = .8413. P( Z < -1) Learning Objective: 07-4
127
ASBE 5e Solutions for Instructors
7.32
a. Using the Empirical Rule we know that approximately 95% of all observations will be within 2 standard deviations of the mean. The approximate range is 3.3 ± 2×0.13 or (3.04, 3.56). b. . 1–NORMDIST(3.5,3.3,.13,TRUE) = 1–.9380 = 0.062. P( X > 3.5) = 1 - P( X < 3.5) We can also use the standardized z-score: . = .9382. P( Z < 1.54) 3.5 - 3.3 z= = 1.54 .13 P(Z > 1.54) = 1–.9382 = .0618. Learning Objective: 07-4
7.33
If solving using a hand held calculator and Appendix C-2: P(X > 24) = P(Z >
24 - 19.2 2.5
)=
P(Z > 1.92) = 1 − P(Z < 1.92) = 1 .9726 = .0274. When using Excel: P(X > 24) =1-NORM.DIST(24,19.2,2.5,1) = .0274. Learning Objective: 07-4 7.34
I want to know the probability of getting a business traveler that is taller than 5' 9". To make it easier we can convert everything to inches and use a mean of 70 inches and use 69 inches instead of 5'9". = 1–NORM.DIST(69,70,2.7, 1) P( X > 69) = 1 - P( X < 69) = .6444. We can also use the standardized z-score: . =. P( Z < -.37) 69 - 70 z= = -.37 2.7 3557. P(Z > .37) = 1–.3557 = .6443. Learning Objective: 07-4
7.35
Find P(X >.5) = P(Z >
.5 - .48 .008
) = P(Z > 2.5) =1 − P(Z < 2.5) =1-NORM.S.DIST(2.5,1) = .
0062. 6.2% of BigBash bats will exceed the new standard. Alternatively one could use the Excel function =1-NORM.DIST(.5,.48,.008,1) = .0062. Learning Objective: 07-4 7.36
Find P(X < $4200) = P(Z <
4200 - 3456 478
) = P(Z < 1.56) =NORM.S.DIST(1.56,1) = .9406.
94% of merit scholarship students will not receive enough to cover their full tuition. Alternatively one could use the Excel function =1-NORM.DIST(4200,3456,478,1) = . 9402. (The difference is due to the rounding of the z score to 1.56.) Learning Objective: 07-4 7.37
a. =NORM.INV(.8,56,4) b. Xupper =NORM.INV(.8,56,4), Xlower =NORM.INV(.2,56,4) c. =NORM.INV(.3,56,4) 128
ASBE 5e Solutions for Instructors Learning Objective: 07-5 7.38
a. =NORM.INV(.6,600,5) b. Xupper =NORM.INV(.7,600,5), Xlower =NORM.INV(.3,600,5) c. =NORM.INV(.2,600,5) Learning Objective: 07-5
7.39
Use Excel’s NORM.INV(probability, µ,σ) function to find the X value associated with the cumulative probability. a. P(X > x) = .1. Highest 10% is the lowest 90%:, x =NORM.INV(.9,10,3) =13.84 min. b. P(xL < X < xU) = .50. Middle 50% means 25% below the shortest time and 25% above the longest time. Shortest time: =NORM.INV(.25,10,3) = 7.98 min. Longest time: =NORM.INV(.75,10,3) = 12.02 min. c. P(X > x) = .80. Highest 80% is the lowest 20%: x =NORM.INV(.2,10,3) = 7.48 min. d. P(X < x) = .10. Lowest 10%: x =NORM.INV(.1,10,3) = 6.16 min. Learning Objective: 07-5
7.40
Use Excel’s NORM.INV(probability, µ,σ) function to find the X value associated with the cumulative probability. a. P(X > x) = .05. Highest 5% is the lowest 95%., x =NORM.INV(.95,12,2) = 15.29 min. b. P(X > x) = .50. Lowest 50%: x =NORM.INV(.5,12,2) = 12 min., c. P(xL < X < xU) = .95. Middle 95% means 2.5% below the shortest time and 2.5% above the longest time. Shortest time: =NORM.INV(.025,12,2) = 8.08 min. Longest time: =NORM.INV(.975,12,2) = 15.92 min. d. P(X < x) = .80. Lowest 80%: =NORM.INV(.80,12,2) = 13.68 min. Learning Objective: 07-5
7.41
Use Excel’s NORM.INV(probability, 114, 7) or NORM.S.INV(probability) function to find the X or Z value associated with the cumulative probability. a. P(X > x) = .05, z =NORM.S.INV(.95) = 1.645, , x = 124.52 oz. 1.645 =
b. P(X < x) = .50, z =NORM.S.INV(.5) = 0,
x - 114 0= 7
, x = 114 oz.
c. P(xL < X < xU) = .95, z = NORM.S.INV (.025) = 1.96,
x - 114 1.96 = U 7
x - 114 7
x - 114 -1.96 = L 7
, xL = 100.28 oz.,
, xU = 127.72 oz. 95% of weights will fall between 100.28 and 127.72 oz.
d. P(X < x) = .80, z =NORM.S.INV(.80) = .842, Learning Objective: 07-5
129
x - 114 0.842 = 7
, x = 119.89 oz.
ASBE 5e Solutions for Instructors 7.42
Use Excel’s NORM.INV(probability, µ,σ) function to get the X value associated with the cumulative probability. We can also obtain X values using the standardized z-score formula and Appendix C, as illustrated in parts a, c, and d. a. P(X > x) = .10. Highest 10% is also the lowest 90%: x =NORM.INV(.9,360,9) = 371.53g. We can also find the z-score associated with an area of .90 which is 1.282 (Appendix C). Use the z-score formula to solve for x: , x = 371.54 g. x - 360 1.282 = 9 b. P(xL < X < xU) = .50. Middle 50% means 25% below the lightest weight and 25% above the heaviest weight. Lightest weight: =NORM.INV(.25,360,9) = 353.93 g. Heaviest weight: =NORM.INV(.75,360,9) = 366.07 g. c. P(X > x) = .80. Highest 80% is the lowest 20%: x =NORM.INV(.20,360,9) = 352.43 g. We can also find the z-score associated with an area of .2 which is -.845. Use the zscore formula to solve for x: , x = 352.40 g. x - 360 -.845 = 9 d. P(X < x) = .10. Lowest 10%: x =NORM.INV(.1,360,9) = 348.47 g. The z-score associated with an area of .1 is 1.282 (Table 7.7). Use the z-score formula to solve for x: , x = 348.46 g. x - 360 -1.282 = 9 Learning Objective: 07-5
7.43
a. P(X > 8) =1NORM.DIST(8,6.9,1.2,1) = 0.17966. This probability indicates that the event is not common but not unlikely. One could also say that an 8 pound baby is at approximately the 80th percentile which also indicates a pretty high weight. b. 90th percentile or P(X < x) = .90. NORM.INV(.9,6.9,1.2) = 8.44 pounds c. 95% of birth weights would be between 4.55 and 9.25 pounds. NORM.INV(0.025,6.9,1.2) = 4.55 pounds, NORM.INV(.975,6.9,1.2) = 9.25 Learning Objective: 07-5
7.44
a. We are looking for the top 5% which is the same as the bottom 95%. Because our Excel functions give us cumulative values and because our table is the area less than z, we want to use the lower 95%. NORM.INV(.95,600,100) = 764.49. We can also use the standardized z-score formula. Using Appendix C, we know that an area of .95 is associated with a z-score of 1.645. Use the z-score formula to solve for x: , x = 764.5 . x - 600 1.645 = 100 b. We are looking for the lower 25%. NORM.INV(.25,600,100) = 532.5510. We know that an area of .25 is associated with a z-score of .675 (Appendix C). Use the z-score formula to solve for x: , x = 532.5. x - 600 -.675 = 100 c. The middle 80% means 20% is outside our range and that is split between the upper and lower tails. Therefore, we want the values associated with a lower tail of .10 and an 130
ASBE 5e Solutions for Instructors upper tail of .10. NORM.INV(.1,600,100) = 471.84, NORM.INV(.9,600,100) = 728.16. We know that an area of .1 is associated with a z-score of -1.282 (Appendix C or Table 7.7). Use the z-score formula to solve for x: , x = 471.80. x - 600 -1.282 = 100 The area .9 is associated with a z-score of 1.282 and using the z-score formula we can solve for x: , x = 728.20. x - 600 1.282 = 100 Learning Objective: 07-5 7.45
P(X < xL) = .25 and P(X > xU) = .25. Solve for xL using =NORM.INV(.25,19.2,2.5) = 18 and and xU = NORM.INV(.75,19.2,2.5) = 21. The middle 50% of occupied beds falls between 18 and 21. Learning Objective: 07-5
7.46
Use the standardized z-score formula to solve for µ.
7.47
Given that P(X < $171) = .70, use z =NORM.S.INV(.7) = 0.5244 . Solve the following for :: 0.5244 = ($171 $157)/= $26.70. When using Appendix C: z ≈ .52 and $ Learning Objective: 07-5
7.48
a. P(X < 110) =NORM.DIST(110,100,15,1) = 0.7475 b. P(X < 2) =NORM.DIST(2,0,1,1) = 0.9772 c. P(X < 5000) =NORM.DIST(5000,6000,1000,1) = 0.1587 d. P(X < 450) =NORM.DIST(450,600,100,1) = 0.0668 Learning Objective: 07-4
7.49
a. P(80 < X < 110) = P(X < 110) – P(X < 80) =NORM.DIST(110,100,15,1)NORM.DIST(80,100,15,1) = .6563 b. P(1.50 < X < 2.00) = P(X < 2.00) – P(X < 1.50) =NORM.DIST(2,0,1,1)NORM.DIST(1.5,0,1,1) = .0441 c. P(4500 < X < 7000) = P(X < 7000) – P(X < 4500) =NORM.DIST(7000,6000,1000,1)NORM.DIST(4500,6000,1000,1) = .7745 d. P(225 < X < 450) = P(X < 450) – P(X < 225) =NORM.DIST(450,600,100,1)NORM.DIST(225,600,100,1) = .0667 Learning Objective: 07-4
, and σ = $3. The P( X > $13.16) = .20 z-score associated with an upper tail of .20 is the same as looking for a z-score for the lower area of .80. Using Excel: =NORM.S.INV(.8) = 0.84162. Solve for µ: 0.84162 = , µ = $10.64. 13.16 - m 3 Learning Objective: 07-5
131
ASBE 5e Solutions for Instructors 7.50
a. 10th percentile: P(X < x) = .10 =NORM.INV(.1,360,9) = 348.47 g b. 32nd percentile: P(X < x) = .32 =NORM.INV(.32,360,9) = 355.79 g c. 75th percentile: P(X < x) = .75 =NORM.INV(.75,360,9) = 366.07 g d. 90th percentile: P(X < x) = .90 =NORM.INV(.9,360,9) = 371.53 g e. 99.9th percentile: P(X < x) = .999 =NORM.INV(.999,360,9) = 387.81 g f. 99.99th percentile: P(X < x) = .9999 =NORMINV(.9999,360,9) = 393.47 g Learning Objective: 07-5
7.51
a. P(wait more than an hour) = P(X > 60 min) = 1 – P(X < 60 min) =1NORM.DIST(60,40,28,1) = 0.2375 b. P(X < 20 min) =NORM.DIST(20,40,28,1) = 0.2375 c. P(X ≥ 10 min) = 1 – P(X < 10 min) =1-NORMDIST(10,40,28,1) = 0.8580 Learning Objective: 07-4
7.52
P(X ≤ 6000) =NORM.DIST(6000,7000,420,10 = .00863. It seems to be an adequate margin of safety because there is only a .87% (i.e., less than 1%) chance of failing to meet the high strength criterion. Learning Objective: 07-4
7.53
Using Appendix C-2: n 10 and n(1-) 10 so we can use the normal approximation to the binomial. = n = 8.07 np (1 - p )
a. P(X < 50) ≈ P(X ≤ 49.5) = P(Z <
49.5 - 70 8.07
b. P(X > 100) ≈ P(X ≥ 100.5) = P(Z >
) = P(Z < −2.54) = .0055
100.5 - 70 8.07
) = P(Z > 3.78) = 1 − P(Z < 3.78) = 1 − .
99992 = .00008 Learning Objective: 07-6 7.54
Using Appendix C-2: n 10 and n(1-) 10 (800×.03 = 24, 800×.97 = 776) so we can use the normal approximation. a. = n24 , = = = 4.8249 np (1 - p ) (24)(.97) b. P(X ≥ 20) ≈ P(X ≥ 19.5) = P(Z > = .8238. c. P(X > 30) ≈ P(X ≥ 30.5) = P(Z >
19.5 - 24 4.8249
30.5 - 24 4.8249
) = P(Z > .93) = 1 − P(Z < .93) = 1.1762
) = P(Z > 1.35) = 1 − P(Z < 1.35) = 1.9115
= .0885 Learning Objective: 07-6 132
ASBE 5e Solutions for Instructors
7.55
Using Appendix C-2: n 10 and n(1-) 10 (200×.90 = 180, 200×.1 = 20) so we can use the normal approximation. = n180, = = = 4.2426 np (1 - p ) (200)(.9)(.1) a. P(X ≥ 175) ≈ P(X ≥ 174.5) = P(Z ≥ 1.0968 = .9032 b. P(X < 190) ≈ P(X ≤ 189.5) = P(Z ≤
174.5 - 180 4.2426
189.5 - 180 4.2426
) = P(Z ≥ 1.30) = 1 − P(Z < 1.30) =
) = P(Z ≤ 2.24) = .9875
Learning Objective: 07-6 7.56
Using Appendix C-2: n 10 and n(1-) 10 (8465×.048 = 406.32, 8465×.952 = 776) so we can use the normal approximation. a. n = (8,465)(.048) = 406.32 b. P(X 400) ≈ P(X 399.5) = P(Z ≥ ) = P(Z ≥ 0.35) = 1 − P(Z < 0.35) = 399.5 - 406.32 19.6677
1 .3632 = .6368 c. P(X < 450) ≈ P(X ≤ 449.5) = P(Z ≤
449.5 - 406.32 19.6677
) = P(Z ≤ 2.20) = .9861
Learning Objective: 07-6 7.57
Using Appendix C-2: Let = = 7.071. s= l a. P(X ≥ 60) ≈ P(X 59.5) = P(Z ≥ = .0901. b. P(X < 35) ≈ P(X ≤ 34.5) = P(Z ≤
59.5 - 50 7.071
34.5 - 50 7.071
) = P(Z ≥ 1.34) = 1 − P(Z < 1.34) = 1 .9099
) = P(Z ≤ −2.19) = .0143.
c. 20 therefore the normal approximation is appropriate. d. P(X ≥ 60) = 1 − P(X < 59) =1-POISSON.DIST(59,50,1) = .09227 and P(X < 35) = P(X ≤ 34) =POISSON.DIST(34,50,1) = .01078. The approximations are fairly close. Learning Objective: 07-6
133
ASBE 5e Solutions for Instructors 7.58
Using Appendix C-2: For 100 vehicles, let = a. P(X ≥ 150) ≈ P(X 149.5) = P(Z ≥ 8944 = .1056. b. P(X < 100) ≈ P(X ≤ 99.5) = P(Z ≤
= 11.62.
s= l ) = P(Z ≥ 1.25) = 1 − P(Z < 1.25) = 1 .
149.5 - 135 11.62
99.5 - 135 11.62
) = P(Z ≤ −3.06) = .00111.
c. P(X ≥ 150) = 1 − P(X ≤ 149) =1-POISSON.DIST(149,135,1) = .10736 and P(X < 100) = P(X ≤ 99) =POISSON.DIST(99,135,1) = .00071. The approximations are fairly close. Learning Objective: 07-6 7.59
Using Appendix C-2: Let = = 5.29. s= l a. P(X > 35) ≈ P(X 35.5) = P(Z ≥ = .0778 b. P(X < 25) ≈ P(X ≤ 24.5) = P(Z ≤
35.5 - 28 5.29
24.5 - 28 5.29
) = P(Z ≥ 1.42) = 1 − P(Z < 1.42) = 1 .9222
) = P(Z ≤ −0.66) = .2546
c. 20 therefore the normal approximation is appropriate. d. P(X > 35) = 1 − P(X ≤ 35) =1-POISSON.DIST(35,28,1) = .0822 and P(X < 25) = P(X ≤ 24) =POISSON.DIST(24,28,1) = .2599. The approximations are fairly close. Learning Objective: 07-6 7.60
Let =
= 12.25 s= l a. P(X 175) ≈ P(X 174.5) = P(Z ≥ 9772 = .0228 b. P(X < 125) ≈ P(X ≤ 124.5) = P(Z ≤
174.5 - 150 12.25
124.5 - 150 12.25
) = P(Z ≥ 2.00) = 1 − P(Z < 2.00) = 1 .
) = P(Z ≤ −2.08) = .0188.
c. 10 therefore the normal approximation is appropriate. d. From Excel: P(X 175) = 1 − P(X ≤ 174) = 1 – POISSON.DIST(174, 150,1) = .0248 and P(X < 125) = P(X ≤ 124) =POISSON.DIST(124, 150,1) = .0165. The approximations are fairly close. Learning Objective: 07-6 7.61
a. P(X > 7) =
e- l x = e- (0.3)(7) = .1225 134
ASBE 5e Solutions for Instructors b. P(X < 2) =
1 - e- l x = 1 - e- (0.3)(2) = 1 - .5488 = .4512 Learning Objective: 07-7 7.62
a. P(X > 30 minutes) = b. P(X < 15 minutes) = c. P(15 < X < 30) =
e- l x = e- (4.2)(.5) = .1225 1 - e- (4.2)(.25) = 1 - .3499 =
.6501
P ( X < 30) - P ( X < 15) = (1 - e - (4.2)(.5) ) - (1 - e - (4.2)(.25) ) =
.8775 − .6501
= .2274 Learning Objective: 07-7 7.63
= 2.1 alarms/minute or = .035 alarms/second a. P(X < 60 seconds) = 1ex = 1 e(0.035)(60) =1- .1225 = .8775 b. P(X > 30 seconds) = ex =e(0.035)(30) = .3499 c. P(X > 45 seconds) = ex =e(0.035)(45) = .2070 Learning Objective: 07-7
7.64
Use = 1/30 a. P((X 40 days) = ex =e(1/30)(40) = e-1.333 = .2636 b. P(X ≤ 20 days) = 1 – ex = 1– e(1/30)(20) = 1 – e-.667 = 1 – .5134 = .4866 Learning Objective: 07-7
7.65
Use = (2.8/min)(1min/60sec) = .0467/sec a. (X 30 seconds) = ex = e(.0467)(30) = e−1.4 = .2466 b. P(X ≤ 15 seconds) = 1 − ex == 1- e(.0467)(15) = 1 − e-.70 = 1 − .4966 = .5034 c. P(X > 60 seconds) = ex = e(.0467)(60) = e−2.8 = .0608 Learning Objective: 07-7
7.66
a. .9 = 1− et. Solve for t using 1/5: .9 = 1− e(1/5)t , .1 = e(1/5)t, ln(.1) = ln[e(1/5)t ], −2.3026 = (1/5)t, t = −2.3026/-0.2 = 11.51 minutes b. .9 = 1− et. Solve for using t = 10 minutes: .9 = 1− e(λ)10 , .1 = e(λ)10, ln(.1) = ln[e(λ)10], −2.3026 = (λ)10, λ = −2.3026/-10 = .23 responses per minute or 4.35 minutes on average Learning Objective: 07-8
7.67
a. .5 = et. Solve for t using .5 = e(4.2)t , ln(.5) = ln[e(4.2)t ], −0.6931 = (4.2)t, t = −0.6931/−4.2 = 0.1650 hours b. .25 = et . Solve for t using 5 = e(4.2)t , ln(.25) = ln[e(4.2)t ], −1.3863 = (4.2)t, t = −1.3863/−4.2 = 0.3301 hours c. .10 = et . Solve for t using = e(4.2)t , ln(.10) = ln[e(4.2)t ], −2.3026 = (4.2)t, t = −2.3026/−4.2 = 0.5482 hours Learning Objective: 07-8 135
ASBE 5e Solutions for Instructors
7.68
a. P(X < t) = .50 so P(X > t) = .50 = et. Solve for t using λ = 0.5: .5 = e(0.5)t , ln(.5) = ln[e(0.5)t ], −0.6931 = (0.5)t, t = −0.6931/−0.5 =1.3862 minutes b. P(X < t) = .25 so P(X > t) = .75 = et. Solve for t using λ = 0.5: .75 = e(0.5)t , ln(.75) = ln[e(0.5)t ], −0.2877 = (0.5)t, t = −0.2877/−0.5 = 0.5754 minutes c. P(X < t) = .30 so P(X > t) = .70 = et. Solve for t using λ = 0.5: .7 = e(0.5)t , ln(.7) = ln[e(0.5)t ], −0.3567 = (0.5)t, t = −0.3567/−0.5 =0.7134 minutes Learning Objective: 07-8
7.69
a. P(X > t) = .5. Use = 1/20 = .05. To solve for t: .5 = e(0.5)t , ln(.5) = ln[e(0.5)t ], −0.6931 = (0.05)t, t = −0.6931/−0.05 =13.862 minutes b. The distribution on time is skewed to the right therefore the median < mean. c. P(X > t) = .25. Use = 1/20 = .05. To solve for t: .25 = e(0.5)t , ln(.25) = ln[e(0.5)t ], −1.3863 = (0.05)t, t = −1.3863/−0.05 = 27.726 minutes Learning Objective: 07-8
7.70
a. P(X < t) = .1 so P(X > t) = .90 = et. Solve for t using λ = 1/8 = .125: .9 = e(0.125)t , ln(.9) = ln[e(0.125)t ], −0.1054 = (0.125)t, t = −0.1054/−0.125 = 0.8432 years b. P(X < t) = .2 so P(X > t) = .80 = et. Solve for t using λ = 1/8 = .125: .8 = e(0.125)t , ln(.8) = ln[e(0.125)t ], −0.2231 = (0.125)t, t = −0.2231/−0.125 = 1.7848 years Learning Objective: 07-8
7.71
a. = (0+25+75)/3 =33.33 b. = = 15.59 2 2 2 0 + 25 + 75 - 0 ´ 25 - 0 ´ 75 - 25 ´ 75 18 c. $25 falls between $0 and $25 so use the equation P(X < $25) = = .3333 ( x - a)2 (25 - 0) 2 = (b - a )(c - a ) (25 - 0)(75 - 0) d. Shaded area represents the probability.
0
25
75
Learning Objective: 07-9 7.72
a. = (50+105+65)/3 = 73.33
136
ASBE 5e Solutions for Instructors b. =
= 11.61 502 + 652 + 1052 - (50)(65) - (50)(105) - (65)(105) 18 c. $75 falls between b and c so use the equation: P(X > 75) = = .4091 (c - x ) 2 (105 - 75) 2 = (c - a )(c - b) (105 - 50)(105 - 65) d. Shaded area represents the probability.
Learning Objective: 07-9 7.73
a. Discrete, can be counted. b. Continuous, can take on fractional values. c. Continuous, can take fractional values. Learning Objective: 07-1
7.74
a. Area = .5(2) = 1 therefore this is a valid PDF. b. Area = ½(2)(2) = 2 therefore this is not a valid PDF. c. Area = ½(.5)(2)(2) = 1 therefore this is a valid PDF. Learning Objective: 07-1
7.75
a. = (25+65)/2 = 45. b. = = 11.547 (65 - 25) 2 12 c. P(X > 45) = (65-45)/(65-25) = 0.5 d. P(X > 55) = (65-55)/(65-25) = 0.25 e. P(30< X <60) = (60-30)/(65-25) = 0.75 Learning Objective: 07-2
7.76
a. = b. =
5 + 50 55 = = 27.5 2 2
inches per second.
inches per second
(50 - 5) = 12.99 12 2
137
ASBE 5e Solutions for Instructors c. Use the CDF to solve for the quartiles. P(X ≤ Q1 ) = .25.
, Q1 = 16.25 inches Q1 - 5 = .25 50 - 5 , Q3 = 38.75 inches per second.
per second. P(X ≤ Q3 ) = .75.
Q3 - 5 = .75 50 - 5 d. P(re-swiping) = 1 P(10 < X < 40) = 1 − .6667 = .3333. 33.33% must re-swipe. Learning Objective: 07-2
7.77
Answers will vary. a. Would expect distribution to be skewed to the right. b. Would expect distribution to be skewed to the right. c. Normal d. Normal Learning Objective: 07-3
7.78
All can be right-skewed by outliers. Learning Objective: 07-3
7.79
a. z = (82-75)/7 = 1, =NORM.S.DIST(1,1) = .8413, 84th Percentile b. z = (93-75)/7 = 2.57, =NORM.S.DIST(2.57,1) = .9949, 99th Percentile c. z = (63-75)/7 = 1.714, =NORM.S.DIST(-1.714,1) = .0433, 4th Percentile Learning Objective: 07-4
7.80
a. = b. =
.74 + .98 = .86 2
ppm.
ppm.
(.98 - .74) = .0693 12 c. We want P(X > .80) so use either 1 – P(X < .80) = 1 – (.80–.74)/(.98–.74) = 1 – .25 = .75 OR (.98–.80)/(.98–.74) = .75 P(.80 < x < .98) = d. P(X < 85) =(.85–.74)/(.98–.74) = .4583 e. P(.8 < X <.9) = (.9–.8)/(.98–.74) = .4167 Learning Objective: 07-2 2
7.81
a. P(X > x) = .50 = P(X < x) = .50. x =NORM.INV(.5,450,80) = 450 bags b. P(X < x) = .25 =NORM.INV(.25,450,80) = 396.04 bags c. P(xL < X < xU) = .80, xL =NORM.INV(.1, 450,80) = 347.48 bags and xU =NORM.INV(.9, 450,80) = 552.52 bags d. 5th percentile: P(X < x) = .05. x =NORM.INV(.05,450,80) = 318.42 bags Learning Objective: 07-4
7.82
a. P(X < x) = .50. x =NORM.INV(.5,235,10) = 235 pizzas 138
ASBE 5e Solutions for Instructors b. P(X > x) = .25 = P(X < x) = .75. x =NORM.INV(.75,235,10) = 241.74 pizzas c. 90th percentile: P(X < x) = .90. x =NORM.INV(.90,235,10) = 247.82 pizzas d. P(xL < X < xU) = .80, xL =NORM.INV(.1, 235,10) = 222.18 pizzas and xU =NORM.INV(.9, 235,10) = 247.82 pizzas Learning Objective: 07-5 7.83
a. The first quartile is associated with the lowest 25%. P(X < Q1) = .25, z =NORM.S.INV(.25) = − .67, , Q1 = $6.77 Q1 - 7.00 -0.67 = 0.35 b. The second quartile is associated with the lowest 50%. P(X < Q2) = .5, z =NORM.S.INV(.5) = 0, , Q2 = $7.00 Q2 - 7.00 0= 0.35 th c. The 90 percentile is z =NORM.S.INV(.90) = 1.2816, , x = $7.45 1.2816 =
x - 7.00 0.35
Learning Objective: 07-5 7.84
a. The likelihood of a value greater than the mean is .50 because the whole area under the curve is 1 and the mean is the midway point. b. This corresponds to P(Z > 1) or 1 – .1587 (Use Appendix C-2 to P( Z < 1) = 1 - .8413 = get ) P( Z < 1) c. This corresponds to P(Z > 2) or .9772 0.0228 (Use Appendix C-2 to = 1 - P( Z < 2) = 1get ) P( Z < 2) d. This corresponds to P(–2 < Z <2) = (Use P( Z < 2) - P(Z < -2) = .9772 - .0228 = .9544 Appendix C-2) Learning Objective: 07-4
7.85
a. P(X>130) =1-NORM.DIST(130,115,20,1) = .2266 b. P(X<100) =NORM.DIST(100,115,20,1) = .2266 c. P(X<91) =NORM.DIST(91,115,20,1) = .1151 Learning Objective: 07-4
7.86
a. P(X < 579)
579 - 579 = P( Z < ) = P ( Z < 0) = .5 14
139
.
ASBE 5e Solutions for Instructors b. P(X > 590) = 1–
P( X < 590)
590 - 579 = 1 - P( Z < ) = 1 - P( Z < .79) 14
=
1-
.7852 = .
2148. (From App C-2) From Excel: P(X > 590) =1-NORM.DIST(590,579,14,1) = .21602 c. P(X < 600) = = .9332 600 - 579 P( Z < ) = P( Z < 1.5) 14 Learning Objective: 07-4 7.87
a. The TPMS will trigger a warning when the tire pressure is below 22.5 psi. .75(30) = 22.5 b. P(X<22.5) =NORM.DIST(22.5,30,2,1) = 8.84E-05 c. P(28<X<32) =NORM.DIST(32,30,2,1)-NORM.DIST(28,30,2,1) = .8413 − .1587 = . 6826 Learning Objective: 07-4
7.88
a. P(X > 120 seconds) = 1–P(X<120) = .7881 b. P(X > 180 seconds) = 1– P(X<180)
120 - 140 = 1 - P( Z < ) = 1 - P ( Z < -.80) 25 180 - 140 = 1 - P( Z < ) = 1 = P( Z < 1.6) 25
=
1-
.2119
= 1 – .9452 =
.0548 c. Using Appendix C-2, the z-score associated with a probability of .95 is 1.645. Use the zscore to solve for x. , x - 140 x = 181.125 1.645 = 25 Using Excel: NORM.INV(.95, 140,25) =181.121 d. Using Appendix C-2, the z-score associated with a probability of .99 is 2.325. Use the zscore to solve for x. , 198.125. x - 140 x = 2.325 = 25 Using Excel: NORM.INV(.99,140,25) = 198.159. Learning Objective: 07-4 Learning Objective: 07-5 7.89
P(a tennis ball fails to meet the specifications) = 1P(1.975 < X < 2.095) =1(NORM.DIST(2.095,2.035,.03,1)-NORM.DIST(1.975,2.035,.03,1)) =1 – (.9773 − . 0228) = .0455. Learning Objective: 07-4
7.90
a.
99 + 107 m= = 103 2
degrees
140
ASBE 5e Solutions for Instructors b.
degrees
(107 - 99) = 2.3094 12 c. Use the CDF to find the third quartile. P(X < Q3) =
s=
2
Q3 - 99 = .75. Q3 = 105 107 - 99
degrees.
Learning Objective: 07-2 7.91
Using Appendix C-2: P(X ≥ 230) = P(Z ≥
230 - 207 14
) = P(Z ≥ 1.64) = 1− P(Z ≥ 1.64) .
0505. Learning Objective: 07-4 7.92
a. P(X > 30) = 1–
7.93
P(X ≤ 90) =NORM.DIST(90,84,10,1) = .7258 Learning Objective: 07-4
7.94
7.95
1 – .9962 = .0038. 30 - 22 = 1 - P( Z < ) = 1 = P( Z < 2.67) = 3 b. Assuming finishing times are independent: P(all three men finish in time) = (1–.0038)3 = .9886. Learning Objective: 07-4
P( X < 30)
P( X < 453)
453 - 470 = P( Z < ) = P( Z < -3.4) 5 contain less than 453 grams. Learning Objective: 07-4 Using App C-2: P(X>5200) = P(Z >
5200 - 4905 355
= .00034. It is very unlikely that a box will
) = P(Z > .83) = 1 − P(Z < .83) = .2033.
Using Excel: 1NORM.DIST(5200,4905,355,1)=.2030 Learning Objective: 07-4 7.96
a. P(X< 135)
= .3085. 135 - 145 = P( Z < ) = P ( Z < -.5) 20 b. P(X > 175) = = 1 – .9332 = .0668. 1 - P( X < 175) 175 - 145 = 1 - P( Z < ) = 1 = P( Z < 1.5) 20 c. P(125 < X < 165) = = P( X < 165) - P( X < 125) 165 - 145 125 - 145 P( Z < ) - P( Z < )= 20 20 P(Z < 1) – P(Z < 1) = .8413 − .1587 = .6826. 141
ASBE 5e Solutions for Instructors d. With variability, physicians run the risk of not treating a patient with dangerous blood pressure or treating a patient with healthy blood pressure. Understanding variability allows physicians to minimize the chances of making these two types of errors. Learning Objective: 07-4 7.97
a. =NORM.S.DIST(-1.62,1) = .0526. John scored better than only 5.26% of the others. b. =NORM.S.DIST(.50,1) = .6915. Mary scored above average, better than approximately 69.15% of others. c. =NORM.S.DIST(1.79,1) = .9633. Zak scored better than 96.33% of others. d. =NORM.S.DIST(2.48,1) = .9934. Frieda scored better than 99.34% of others. Learning Objective: 07-3 Learning Objective: 07-4
7.98
a. False, the normal distribution is asymptotic. Thus, a value outside the given interval is possible. b. False, the standardized values do allow for meaningful comparison. Z scores are unit free. c. False, the normal distribution is a “family” of distributions, each having the same shape, but different means and standard deviations. Learning Objective: 07-3
*7.99
a. For route A: P(X<54) =NORM.DIST(54,54,6,1) = .5. For route B: P(X<54) =NORM.DIST(54,60,3,1) = .0228. He should take route A. b. For route A: P(X<60) =NORM.DIST(60,54,6,1) = .8413. For route B: P(X<60) =NORM.DIST(60,60,33,1) = .5. He should take route A. c. For route A: P(X<66) =NORM.DIST(66,54,6,1) = .9722. For route B: P(X<66) =NORM.DIST(66,60,3,1) = .9772. He could take either route. Because the standard deviation is smaller for route B, the chance of getting to the airport in less than 66 minutes is the same for each route. Learning Objective: 07-4
7.100
a. Under filling the bottle means putting less than 500 ml in the bottle. Find the value of for which P(X > 500) = .95. This corresponds to a z = −1.645. Use the z-score formula to solve for − µ 508.225 ml. 500 - m 1.645 = 5 b. To ensure that 99% contain at least 500 ml, solve − for µ, µ = 511.63. 500 - m 2.326 = 5 c. To ensure that 99.9% contain at least 500 ml, solve − for µ, µ = 515.45. 500 - m 3.09 = 5 Learning Objective: 07-5
142
ASBE 5e Solutions for Instructors 7.101
Find the value for X such that P(X > x) = .80. This corresponds to a z = −.842. Use the zscore formula to solve for x. x = 11.49 inches. x - 12.5 -.842 = 1.2 Learning Objective: 07-5
7.102
a. For method A: P(X<28) =
= .5. For method B: P(X<28) = 28 - 28 P( Z < ) = P ( Z < 0) 4 = .0228. Method A is preferred because there is a greater
28 - 32 ) = P ( Z < -2) 2 chance of completion within 28 minutes. b. For method A: P(X<38) = .9938. For method B: 38 - 28 = P( Z < ) = P( Z < 2.5) 4 P(X<38)= = .9987. Method B is preferred because there is 38 - 32 P( Z < ) = P ( Z < 3) 2 a greater chance of completion within 38 minutes. c. For method A: P(X<36) = = .9772. For method B: P(X<36) = 36 - 28 P( Z < ) = P ( Z < 2) 4 = .9772. Either method is acceptable because they both give 36 - 32 P( Z < ) = P ( Z < 2) 4 the same probability. Learning Objective: 07-4 P( Z <
7.103
a. P(X > μ) = .5 (property of the normal distribution). Assuming independence, the probability that both exceed the mean is: .5×.5 =.25. b. P(X < μ) = .5 (property of the normal distribution). Assuming independence, the probability that both are less than the mean is: .5×.5 =.25. c. P(X<μ) = .5 (property of the normal distribution). Assuming independence, the probability that one is greater than and one is less than the mean is: .5×.5 =.25. There are two combinations that yield this, so the likelihood is: .25+.25 = .50 that one exceeds the mean and one is less than the mean. d. P(X = μ) = 0, this is a property of a continuous random variable. The probability that both equal the mean is zero. Learning Objective: 07-4
7.104
Use the normal approximation to the binomial distribution because we clearly meet the requirement that n 10 and n(1–) 10. , . (2000)(.02) = 40 (2000)(1 - .02) = 1,960 143
ASBE 5e Solutions for Instructors The normal parameters are:
m = (2000)(.02) = 40, s = (40)(.98) = 6.26
. Using
appendix C-2: a. P(X ≥ 50) ≈ P(X ≥ 49.5) and the continuity corrected Z-value is
. 49.5 - 40 z= = 1.52 6.26 P(Z > 1.52) = 1 .9357 = .0643. There is a 6.43% chance of at least 50 twin births. Using Excel: P(X ≥ 49.5) =1-NORM.DIST(49.5,40,6.26,1) = .0646 b. P(X < 35) ≈ P(X ≤ 34.5) and the continuity corrected Z-value is . 34.5 - 40 z= = -.88 6.26 P(Z < .88) = .1894. There is an 18.94% chance of fewer than 35 twin births. Using Excel: P(X ≤ 34.5) =NORM.DIST(34.5,40,6.26,1) = .1898 Learning Objective: 07-6 7.105
Use µ = 100(.25) = 25 and σ =
=4.33. P(X < 20) ≈ P(X ≤ 19.5)
100(.25)(.75) =NORM.DIST(19.5,25,4.33,1) = .1020 Learning Objective: 07-6 7.106
Use the normal approximation to the binomial distribution because we clearly meet the requirement that n 10 and n(1-) 10. The parameters are: . Use the NORM.INV function in Excel m = (1, 000)(.06) = 60, s = (60)(.94) = 7.51 with probability of .25 for the first quartile, .75 for the third quartile, = 60 and = 7.51. Q1 =NORM.INV(.25,60,7.51) = 54.93 cars, Q3 =NORM.INV(.75,60,7.51) = 65.07 cars. Alternatively, use the z-score equation where z = .675 for Q1 and z = .675 for Q3 , . , Q3 = 65.07. Q3 - 60 Q1 - 60 Q1 = 54.93 .675 = -.675 = 7.51 7.51 Learning Objective: 07-6
7.107
Use the normal approximation to the binomial distribution because we clearly meet the requirement that n 10 and n(1-) 10. nn(1-) = 100(.75) = 75. a. Find the value of x such that P(X ≥ x) = .05 or P(X < x) = .95. =NORM.INV(.95,25,4.33) = 32.12. Set passing at 33 correct answers. Less than 5% of the time a guesser will correctly guess the answer 33 out of 100 questions. b. Find the value of X such that P(X ≥ x) = .01 or P(X < x) = .99. =NORM.INV(.99,25,4.33) = 35.07. Set passing at 36 correct answers. Less than 1% of the time a guesser will correctly guess the answer 36 out of 100 questions.
144
ASBE 5e Solutions for Instructors c. Use the NORM.S.INV function in Excel to find the z-scores for the quartiles. For the first quartile use =NORM.S.INV(.25) = −0.675, for the third quartile use =NORM.S.INV(.75) = 0.675. , Q1 = 22.08. , Q3 = 27.92. Q3 - 25 Q1 - 25 -.675 = .675 = 4.33 4.33 Learning Objective: 07-6 7.108
Use the normal approximation of the binomial distribution because we clearly meet the requirement that n 10 and n(1-) 10. The parameters are: m= n . s = (160)(.2) = 5.6569 a. P(X < 150) ≈ P(X ≤ 149.5). The continuity-corrected z-score is 149.5 - 160 z= = -1.86 5.6569 . Using Appendix C-2: .0314 P( Z < -1.86) = Using Excel P(X ≤ 149.5) =NORM.DIST(149.5,160,5.6569,1) = .0317 b. P(X ≥ 150) ≈ P(X ≥ 149.5) = 1 − .0314 = .9686. Or 1 − .0317 = .9683 Learning Objective: 07-6
7.109
Use the normal approximation of the binomial distribution because we clearly meet the requirement that n 10 and n(1-) 10. The parameters are: m= n . s = (30)(.98) = 5.422 a. P(X ≥ 25) ≈ P(X ≥ 24.5). The continuity-corrected z-score is . 24.5 - 30 z= = -1.01 5.422 Using Appendix C-2: 1−.1562 = .8438 P(Z > -1.01) = Using Excel P(X ≥ 24.5) =1-NORM.DIST(24.5,30,5.422,1) = .8448 b. P(X > 40) ≈ P(X ≥ 40.5). The continuity-corrected z-score is . 40.5 - 30 z= = 1.94 5.422 Using Appendix C-2: 1−.9738 = .0262 P( Z > 1.94) = Using Excel: P(X ≥ 40.5) =1-NORM.DIST(40.5,30,5.422,1) = .0264 Learning Objective: 07-6
7.110
Converting the rate from days to years, 73. Let = 73 and
= s= l 8.544. It is appropriate to use the normal approximation given that 20. P(X < 60) ≈
145
ASBE 5e Solutions for Instructors . From Appendix C-2: P(Z < 1.58) = .0571. Using 59.5 - 73 z= = -1.58 8.544 Excel: P(X ≤ 59.5) =NORM.DIST(59.5,73,8.544,1) = .0571 Learning Objective: 07-6 P(X ≤ 59.5).
7.111
a. P(X > 100,000) = ex =e(1/70000)(100000) = .2397 b. P(X ≤ 50,000) = 1ex =1e(1/70000)(50000) = 1 - .4894 = .5105 c. P(50,000 ≤ X ≤ 80,000) =(1e(1/70000)(80000)) – (1e(1/70000)(50000)) = (1 – .3189) – (1 – .4895) = .1706 Learning Objective: 07-7
7.112
a. P(X>6) = e(.1)(6) = .5488. 54.88% chance they will wait at least 6 months until the next claim. b. P(X>12) = e(.1)(12) = .3012. 30.12% chance they will wait at least a year until the next claim. c. P(X>24) = e(.1)(24) = .0907. 9.07% chance they will wait at least 2 years until the next claim. d. P(6 < X < 12) = P(X < 12) − P(X < 6) = (1 − .3012) − (1 − .5488) = .2476. 24.76% chance they will wait between 6 months to a year. Learning Objective: 07-7
7.113
a. P(X ≤ 3 minutes) =1 - ex =1 - e(1/3)(3) = 1 - .3679 = .6321 b. The distribution is skewed to the right so the mean is greater than the median. Learning Objective: 07-7 Convert 30 years to hours: 30 years × 365 days × 24 hours = 262,800. MTBF =
7.114
400,000 hours . a. P(time to failure > 30 years) = P(time to failure > 262,800) = b.
P( X < 3 years ) = P( X < 26, 280hours)
=
e
-lx
=
æ 26,280 ö -ç ÷ è 400,000 ø
e- l x =
e
æ 262,800 ö -ç ÷ è 400,000 ø
1 l
= .5184.
1- e = 1 - .9364 = .0636 probability of failure within the first three years is only .0636. Learning Objective: 07-7 7.115
a. = (300 + 350 + 490)/3 = 380. b. = 3002 + 3502 + 4902 - 300(350) - 300(490) - 350(490) = 40.21 18
c. P(X > 400) =
(490 - 400)2 = .3045 (490 - 300)(490 - 350)
.
Learning Objective: 07-9 146
=
. Yes, the
ASBE 5e Solutions for Instructors
7.116
a. = (50 + 95 + 60)/3 = 68.33 b. = 502 + 602 + 952 - 50(60) - 50(95) - 60(95) = 9.65 18
c. P(X <75) = 1-
(95 - 75)2 = .7460 (96 - 50)(95 - 60)
d.
Learning Objective: 07-9 7.117
a. = (500 + 700 + 2100)/3 = 1100. b. = 5002 + 7002 + 21002 - 500*700 - 500* 2100 - 700 * 2100 = 355.90 18
c. P(X > 750) =
.
2
(2100 - 750) = .8136 (2100 - 500)(2100 - 700)
d. Shaded area is probability in part c.
500
750
Learning Objective: 07-9 7.118
a. The Excel formula could be: NORM.S.INV(RAND()). b. Answers will vary. c. Answers will vary. Learning Objective: 07-3 Learning Objective: 07-4
7.119
a. The Excel formula could be: NORM.INV(RAND(),4000,200). b. Answers will vary. c. Answers will vary. Learning Objective: 07-3 147
2100
ASBE 5e Solutions for Instructors Learning Objective: 07-4 7.120
a. The z scores were, respectively, 5.75, 4.55, 5.55, and 5.45. 195 - 80 z= = 5.75 20 171 - 80 z= = 4.55 20 191 - 80 z= = 5.55 20 189 - 80 z= = 5.45 20 b. If the exams scores had an historical mean and standard deviation of 80 and 20 with a normal distribution then the exam scores reported by the four officers were highly unlikely. All scores were outliers because they were higher than 80 + 3 = 80 + (3)
s
(20) = 140. Learning Objective: 07-3 Learning Objective: 07-4
148
ASBE 6e Solutions for Instructors
Chapter 8 Sampling Distributions and Estimation 8.1
For a 90% interval use z.05 =NORM.S.INV(.95) = 1.645 a. For μ = 100, σ = 12, n = 36: or [96.71, 103.29] 12 100 ± 1.645 36 b. For μ = 2,000, σ = 150, n = 9: or [1917.75, 2082.25] 150 2000 ± 1.645 9 c. For μ = 500, σ = 10, n = 25: or [496.71, 503.29] 10 500 ± 1.645 25 Learning Objective: 08-3
8.2
For a 95% interval use z.025 =NORM.S.INV(.975) = 1.96 a. = or (196.08, 203.92). s 12 m ± 1.96 200 ± 1.960 n 36 b.
s m ± 1.96 n
= 1000 ± 1.960
15
or (990.2, 1009.80).
9
c.
= or (49.608, 50.392). s 1 m ± 1.96 50 ± 1.960 n 25 Learning Objective: 08-3 8.3
a. b.
m ± 1.96s
=
4.035 ± 1.96 ´ 0.005
s m ± 1.96 n
= 4.035 ± 1.96
or (4.0252, 4.0448).
0.005
or (4.03304, 4.03696).
25
c. In either case, we would conclude that our sample came from a population that did not have a population mean equal to 4.035. Learning Objective: 08-3
146
ASBE 6e Solutions for Instructors 8.4
a. .9480
or (.9460, .9500) 0.006 = .9480 ± .002 35 b. Our sample size is greater than 30 so as we take random samples of 35, our distribution will approach normal. It is bell-shaped with a mean of .9480 and standard deviation of .001014. c. Central Limit Theorem Learning Objective: 08-3
8.5
a. (a)
(b)
±
s n s n
1.96
32 4
32 16
s 32 n 64 b. The standard error is reduced by half. Learning Objective: 08-4 (c)
8.6
a. (a)
(b)
s n s n
24 9
24 26
24 s n 144 b. The standard error is reduced by half. Learning Objective: 08-4 (c)
8.7
a.
σ X = σ / n = 0.25 / 10 = 0.0791 b. or (3.345, 3.655) 0.25 3.50 ± 1.96 = 3.50 ± 0.1550 10 Learning Objective: 08-3 Learning Objective: 08-4
147
ASBE 6e Solutions for Instructors 8.8
a.
8.9
a.
8.10
a.
1.25 = 0.3125 16 b. 27 or (26.486, 27.514) ± 1.25 1.645 = 27 ± 0.514 16 Learning Objective: 08-3 Learning Objective: 08-4 ´ σ 4 x ± 1.645 =14 ± 1.645 or (11.06, 16.94) √n √5 ´ σ 5 =37 ± 2.576 b. x ± 2.576 or (33.68, 40.33) √n √ 15 ´ σ 15 =121 ± 1.96 c. x ± 1.96 or (115.12, 126.88) √n √ 25 Learning Objective: 08-5
´ σ 4 3 24 ± 1.645 = 24 ± 1.56 x ± 1.645 √ n =14 ± 1.645 √ 5 10
b. ´ σ 5 x ± 2.576 =37 ± 2.576 √n √ 15 c.
´ σ 15 x ± 1.96 =121 ± 1.96 √n √ 25
125 ± 2.576
or (22.44, 25.56)
or (120.88, 129.12) 8 = 125 ± 4.12 25
or (12.167, 12.833) 1.2 = 12.5 ± 0.333 50
12.5 ± 1.96
Learning Objective: 08-5 8.11
a.
´ σ 4 0.15 2.4 ± 1.645 = 2.4 ± 0.0823 x ± 1.645 √ n =14 ± 1.645 √ 5 9
b. ´ σ 5 x ± 2.576 =37 ± 2.576 √n √ 15 c.
´ σ 15 x ± 1.96 =121 ± 1.96 √n √ 25
2.4 ± 1.96
or (2.3177, 2.4823)
or (2.302, 2.498) 0.15 = 2.4 ± 0.098 9
2.4 ± 2.576
or (2.2712, 2.5288) 0.15 = 2.4 ± 0.1288 9
d. As the confidence increases, the interval gets wider. In order to be more confident one has to give up some precision.
148
ASBE 6e Solutions for Instructors Learning Objective: 08-5 8.12
a.
or (34.8763, 39.1237) 5 37 ± 1.645 = 37 ± 2.1237 15 b. or (34.4696, 39.5304) 5 37 ± 1.96 = 37 ± 2.5304 15 c. or (33.6744, 40.3256) 5 37 ± 2.576 = 37 ± 3.3256 15 d. As the confidence increases, the interval gets wider. In order to be more confident one has to give up some precision. Learning Objective: 08-5
8.13
a.
b.
40 270 ± 1.96 = 270 ± 15.68 25 40 270 ± 1.96 = 270 ± 11.09 50
or (254.32, 285.68)
(258.91, 281.09)
c.
(262.16, 277.84) 40 270 ± 1.96 = 270 ± 7.84 100 d. Width decreases as n increases. Learning Objective: 08-4 Learning Objective: 08-5
8.14
a.
b.
50 850 ± 1.96 = 850 ± 9.8 100
or (840.2, 859.8)
100 850 ± 1.96 = 850 ± 19.6 100
or (830.4, 869.6)
c.
or (810.8, 889.2) 200 850 ± 1.96 = 850 ± 39.2 100 d. The confidence interval gets wider as increases. s Learning Objective: 08-4 Learning Objective: 08-5
149
ASBE 6e Solutions for Instructors 8.15
8.16
8.17
8.18
1.25 28.0 ± 1.96 = 28.0 ± 0.775 10 Learning Objective: 08-5
or (27.225, 28.775)
a. The shape of the distribution will be close to normal because our sample size is greater than 30 (n = 40). b. or (30.4941, 42.3059) 14.5 36.4 ± 2.576 = 36.4 ± 5.9059 40 Learning Objective: 08-3 Learning Objective: 08-5 0.000959 0.2731 ± 2.576 = 0.2731 ± 0.0003245 58 Learning Objective: 08-5 a.
b.
c.
s 3 x ±t = 24 ± 1.9432 = 24 ± 2.2034 n 7
or (0.2727755, 0.2734245)
or (21.7966, 26.2034).
or (37.9013, 46.0987). ´ s 6 x±t =¿ x ± t s = 42 ± 2.8982 = 42 ± 4.0987 √n n 18 x ±t
or (113.5714, 124.4286). ´ s 14 s =¿ 119 ± 2.0518 = 119 ± 5.4286 = x±t 28 √n n
Learning Objective: 08-6 Note: t values are found using the Excel formula =T.INV.2T(1cc, d.f.) where cc is the confidence level. For a. this would be =tinv(1-.90, 6). The t values can also be found using Appendix D. 8.19
a. Appendix D = 2.262, Excel =T.INV.2T(.05, 9) = 2.2622 b. Appendix D = 2.602, Excel =T.INV.2T (.02, 15) = 2.6025 c. Appendix D = 1.678, Excel =T.INV.2T (.10, 47) =1.6779 Learning Objective: 08-6 Learning Objective: 08-6
8.20
a. Appendix D = 2.021, Excel =T.INV.2T (.05, 40) = 2.0211. z-score is 1.96 b. Appendix D = 1.990, Excel =T.INV.2T (.05, 80) = 1.9901. z-score is 1.96 c. Appendix D = 1.984, Excel =T.INV.2T (.05, 100) = 1.984. z-score is 1.96
150
ASBE 6e Solutions for Instructors All are fairly close to 1.96. Larger samples mean larger degrees of freedom which means the t-distribution is closer to normal. Learning Objective: 08-5 Learning Objective: 08-6 8.21
a. d.f. = 9, t = 2.262,
or (255.694, 284.306) 270 ± 2.262
20 10
= 270 ± 14.306
b. d.f. = 19, t = 2.093,
or (260.64, 279.36) 270 ± 2.093
20 20
= 270 ± 9.36
c. d.f. = 39, t = 2.023,
or (263.603, 276.397) 270 ± 2.023
20 40
= 270 ± 6.397
d. The confidence interval width decreases as n increases. Learning Objective: 08-4 Learning Objective: 08-6 8.22
t.025 =T.INV.2T(.05,24) = 2.064 a. or (843.808, 856.192) 15 850 ± 2.064 = 850 ± 6.192 25 b. or (837.616, 862.384) 30 850 ± 2.064 = 850 ± 12.384 25 c. or (825.232, 874.768) 60 850 ± 2.064 = 850 ± 24.768 25 d. As s increases our interval gets wider. If the variation or dispersion of our sample goes up and our confidence level stays the same, we give up some precision and our interval needs to be wider. Learning Objective: 08-4 Learning Objective: 08-6
8.23
a. t.025 =T.INV.2T(.05,20) = 2.086,
´ s s 27.79 x±t =¿ x ± t = 45.66 ± 2.0860 n √ n 21
or
(33.01, 58.31). b. The confidence interval could be made narrower by increasing the size of the sample or decreasing the confidence level. Learning Objective: 08-4 Learning Objective: 08-6
151
ASBE 6e Solutions for Instructors
8.24 a.
or (18.2757, 21.4743). ´ s s 3.6492 x ± t =¿ x ±t = 19.875 ± 1.753 = 19.875 ± 1.5993 √n n 16 Note: t values are found using the Excel formula =tinv(1-cc, n-1) where cc is the confidence coefficient. For a. this, this would be = tinv(1-.90, 15). They can also be found using Appendix D. Learning Objective: 08-6
8.25
s 78.407 ´ s = 812.5 ± 3.1058 x±t =¿ x ± t n √n 12
or (742.20, 882.80).
Learning Objective: 08-6 8.26
a.
8.27
a. 1.
´ s 17541.8 s x ±t = x ± t √ n =¿ 24520 ± 2.2622 = 24520 ± 12548.9 n 10 37068.9). b. Increase the sample size or decrease the confidence level. Learning Objective: 08-4 Learning Objective: 08-6
2.
3.
or (81.873, 88.127).
´ s s 4.3716 x ±t = x ± t √ n =¿ 85 ± 2.2622 n 10 ´ s s 8.127 =¿ 88.6 ± 2.2622 x ±t = x±t √n n 10 ´ s s 3.712 =¿ 76 ± 2.2622 x ±t = x±t √n n 10
or (11971.1,
or (82.787, 94.414).
or (73.345, 78.655).
b. Confidence intervals 1 and 2 overlap. The scores on exam 3 are very different than the first two. There was a decrease in the average exam score on the third exam. c. Here the standard deviation is not known, so use the t-distribution. Learning Objective: 08-6 8.28
a.
b.
p (1 - p) n p (1 - p) n
=
=
.50(1 - .50) 30 .20(1 - .20) 50
= .0913
= .0566
152
ASBE 6e Solutions for Instructors c.
p (1 - p) n
=
.10(1 - .10) 100
= .0300
d.
= = .0032 p (1 - p) .005(1 - .005) n 500 Learning Objective: 08-7
8.29
a.
b.
c.
d.
p (1 - p) n p (1 - p) n p (1 - p) n
=
=
=
=
.30(1 - .30) 40 .10(1 - .10) 200 .40(1 - .40) 30
p (1 - p) .03(1 - .03) n 400 Learning Objective: 08-7
= .0725
= .0212
= .0894
= .0085
8.30
a. No, nπ = 200×.02 = 4 < 10. b. No, nπ = 100×.05 = 5 < 10. c. Yes, nπ = 50×.50 = 25 and n(1−π) = 50×.50 = 25. Both are greater than 10. Learning Objective: 08-7
8.31
a. Yes, nπ = 25×.50 = 12.5 and n(1−π) = 25×.50 = 12.5. Both are greater than 10. b. Yes, nπ = 60×.20 = 12 and n(1−π) = 60×.20 = 48. Both are greater than 10. c. No, nπ =.08×100 = 8 < 10. Learning Objective: 08-7
8.32
The confidence level for most polls is assumed to be 95%. Use z.025 = 1.96 a. 1.96 =1.96 p (1 - p) .50(1 - .50) = .1386 n 50 b. 1.96 =1.96 p (1 - p) .50(1 - .50) = .0694 n 200
153
ASBE 6e Solutions for Instructors c. 1.96
p (1 - p) n
=1.96
.50(1 - .50) = .0439 500
d. 1.96
= 1.96 p (1 - p) .50(1 - .50) = .022 n 2000 Learning Objective: 08-7 8.33
8.34
For a confidence level of 95% use z.025 = 1.96. a. = = .0620 p(1 - p) .5(1 - .5) 1.96 1.96 n 250 b. = = .0877 p(1 - p) .5(1 - .5) 1.96 1.96 n 125 c. = = .1216 p(1 - p) .5(1 - .5) 1.96 1.96 n 65 Learning Objective: 08-7 p = 24 / 500 = .048
.
= p (1 - p ) p± z n
or (.0293, .0667) .048(1 - .048) .048 ± 1.96 = .048 ± .0187 500
b. We can assume normality because .048×500 = 24
³ 10
and 500(1–.048) = 476
³
10.
Learning Objective: 08-7 8.35
a.
p = 19 / 52 = .3654
.
=
or (.2556, 4752)
.3654(1 - .3654) p (1 - p ) .3654 ± 1.96 n 52 b. np = .3654×52 = 19 and n(1−p) = 33. Both values are greater than 10 so normality assumption is met. Learning Objective: 08-7
8.36
a.
8.37
a.
p±z
= or (.2948, .5424) p (1 - p ) .419(1 - .419) p±z .419 ± 1.645 n 43 b. We can assume normality because 43×.419 = 18 10 and 43(1.419) = 24.983 10. ³ ³ Learning Objective: 08-7 p = 18 / 43 = .419
p = .048
.
p±z
.
p (1 - p ) n
=
.048 ± 2.576
154
.048(1 - .048) 250
or (.013, .083)
ASBE 6e Solutions for Instructors b. Given np = 250×.048 = 12 and n(1p) = 250×.952 = 238 we can assume normality. Learning Objective: 08-7 8.38
a. FPCF =
8.39
. Use the FPCF. n 90 = = .09 > .05 N 1000 a. For 90% confidence use z.05 = 1.645.
. Yes, the population can
N -n 450 - 10 n 10 = = .9899, = = .0222 < .05 N -1 450 - 1 N 450 be considered effectively infinite. b. FPCF = . No, the population cannot N -n 300 - 25 n 25 = = .959, = = .0833 > .05 N -1 300 - 1 N 300 be considered effectively infinite. c. FPCF = . No, the population N -n 1800 - 280 n 280 = = .9192, = = .1556 > .05 N -1 1800 - 1 N 1800 cannot be considered effectively infinite. Learning Objective: 08-8
s n
N -n 15 1000 - 90 ,50 ± 1.645 N -1 90 1000 - 1
s x±z n
N -n 15 1000 - 90 ,50 ± 1.96 N -1 90 1000 - 1
x±z 50 ± 2.482 or [47.518, 52.482] b. For 95% confidence use z.025 = 1.96.
50 ± 2.958 or [47.042, 52.958] c. For 99% confidence use z.005 = 2.576. x±z
s n
. Use the FPCF. n 1200 = = .21 > .05 N 5800 a. For 90% confidence use t.05 with d.f. = 1199. t = 1.646. , s N -n 0.2 5800 - 1200 x ±t ,3.7 ± 1.646 5800 - 1 n N -1 1200 3.7 ± 0.0085 or [3.6915, 3.7085]
155
,
N -n 15 1000 - 90 ,50 ± 2.576 N -1 90 1000 - 1
50 ± 3.887 or [46.113, 53.887] Learning Objective: 08-8 8.40
,
,
ASBE 6e Solutions for Instructors b. For 95% confidence use t.025 with d.f. = 1199. t = 1.962. , s N -n 0.2 5800 - 1200 x ±t ,3.7 ± 1.962 5800 - 1 n N -1 1200 3.7 ± 0.01 or [3.69, 3.71] c. For 99% confidence use t.005 with d.f. = 1199. t = 2.5799. , s N -n 0.2 5800 - 1200 x ±t ,3.7 ± 2.5799 5800 - 1 n N -1 1200 3.7 ± 0.0133 or [3.6867, 3.7133] Learning Objective: 08-8 8.41
. Use the FPCF. For a 90% confidence level use z.05 = 1.645. n 500 = = .22 > .05 N 2300 p (1 - p) N - n 245 p±z ,p= = .49, n N -1 500 .49(1 - .49) 2300 - 500 500 2300 - 1 .49 ± .033 or [.457, .523] Learning Objective: 08-8 .49 ± 1.645
8.42
a. æ zs ö n=ç ÷ è E ø
2
b. æ zs ö n=ç ÷ è E ø
2
=
and the sample size is rounded up to 55.
2
æ (1.96)(7,500) ö ç ÷ = 54.02 2, 000 è ø =
and the sample size is rounded up to 217.
2
æ (1.96)(7,500) ö n=ç ÷ = 216.09 1,000 è ø
c.
= and the sample size is rounded up to 865. 2 2 z s (1.96)(7,500) æ ö æ ö n=ç ÷ n=ç ÷ = 864.36 500 è E ø è ø Using Megastat: >CONFIDENCE INTERVAL/SAMPLE SIZE>SAMPLE SIZE MEAN and just plug in the values for E, standard deviation, and confidence level. Learning Objective: 08-9 8.43
2
=
2
and the sample size is rounded up to 25.
æ zs ö æ (1.96)(10) ö n=ç ÷ ç ÷ = 24.01 4 è E ø è ø Using MegaStat: 25 Learning Objective: 08-9
8.44
Assume a normal distribution.
156
ASBE 6e Solutions for Instructors a. Solve for sigma using
1.3333. (b - a ) (28 - 20) s= = = 6 6 b. = and our sample size is rounded up to 5. Using 2 2 (1.645)(1.3333) ö æ zs ö æ n=ç ÷ = 4.8105 ÷ ç 1 ø è E ø è Megastat: >CONFIDENCE INTERVAL/SAMPLE SIZE>SAMPLE SIZE MEAN and just plug in the values for E, standard deviation, and confidence level. Learning Objective: 08-9 8.45
Assume a normal distribution. a. Solve for sigma using 16.67. ( b - a ) (200 - 100) s= = = 6 6 b. = and the sample size is rounded up to 43. 2 2 z s (1.96)(16.67) æ ö æ ö n=ç ÷ ç ÷ = 42.7 5 è E ø è ø From Megastat: n = 43. Learning Objective: 08-9
8.46
Assume a Poisson distribution.
s= m
= 2.1213.
2
æ (2.3264)(2.1213) ö n=ç ÷ = 97.4166 .5 è ø
and our sample size is rounded up to 98. Learning Objective: 08-9 8.47
2
=
= 385 2 æ (1.96)(2.5) ö
æ zs ö n=ç ÷ ç ÷ è E ø è .25 ø Learning Objective: 08-9
8.48
a. Assume a normal distribution. We can estimate b.
2
(b - a ) (51.96 - 43.89) s= = = 1.345 6 6 and our sample size is rounded up to 28.
æ (1.96)(1.345) ö n=ç ÷ = 27.7982 .5 è ø Learning Objective: 08-9 8.49
a. (22-17)/6 = 0.833 b. and our sample size is rounded up to 31. 2 æ (1.645)(0.833) ö n=ç ÷ = 30.04 .25 è ø
157
.
ASBE 6e Solutions for Instructors Learning Objective: 08-9 8.50
We can use π =.5.
2
. Round the sample size
2
æzö æ -2.326 ö n = ç ÷ p (1 - p ) = ç ÷ .5(.5) = 2,164.11 èEø è .025 ø up to 2,165. Using Megastat: >CONFIDENCE INTERVAL/SAMPLE SIZE>SAMPLE SIZE p and just plug in the values for E, p, and confidence level. Learning Objective: 08-10
8.51
2
. Round up to n = 1692. We use =.5.
2
æzö æ 1.645 ö n = ç ÷ p (1 - p ) = ç ÷ .5(.5) = 1691.27 èEø è .02 ø Learning Objective: 08-10 8.52
We can use π =.5.
2
. Round the sample size up
2
æzö æ 1.96 ö n = ç ÷ p (1 - p ) = ç ÷ .5(.5) = 600.25 èEø è .04 ø to 601. Using Megastat: >CONFIDENCE INTERVAL/SAMPLE SIZE>SAMPLE SIZE p and just plug in the values for E, p, and confidence level. Learning Objective: 08-10
8.53
2
2
. No need to round up. Using Megastat: 2401.
æzö æ 1.96 ö n = ç ÷ p (1 - p ) = ç ÷ .5(.5) = 2401 èEø è .02 ø Learning Objective: 08-10
8.54
We can use π =.5.
and our sample size is 2,401. 2 2 z 1.96 æ ö æ ö n = ç ÷ p (1 - p ) = ç ÷ .5(.5) = 2, 401 èEø è .02 ø Using Megastat: >CONFIDENCE INTERVAL/SAMPLE SIZE>SAMPLE SIZE MEAN and just plug in the values for E, p, and confidence level. Learning Objective: 08-10
8.55
Use an initial estimate for π = 15/19 = .79.
2
2
æzö æ 1.645 ö n = ç ÷ p (1 - p ) = ç ÷ .79(.21) = 124.7 èEø è .06 ø
Round the sample size to 125. Learning Objective: 08-10
158
.
ASBE 6e Solutions for Instructors 8.56
a.
( n - 1) s 2 ( n - 1) s 2 (15 - 1)102 (15 - 1)102 2 2 <s < , <s < ,53.5988 < s 2 < 248.712 2 2 cU cL 26.12 5.629
b.
( n - 1) s 2 ( n - 1) s 2 (18 - 1)12 2 (18 - 1)12 2 2 2 < s < , < s < ,81.0865 < s 2 < 323.6383 2 2 cU cL 30.19 7.564 Learning Objective: 08-11 8.57
8.58
8.59
(n - 1) s 2 ( n - 1) s 2 (20 - 1)1.33252 (20 - 1)1.33252 2 2 < s < , < s < ,1.0269 < s 2 < 3.7877 cU2 c L2 32.8523 8.9065 1.0134 < σ < 1.9462 Learning Objective: 08-11 , 2 (n - 1) s 2 ( n 1) s (25 - 1)1.09532 (25 - 1)1.09532 2 2 < s < < s < ,0.7315 < s 2 < 2.3220 2 2 cU cL 39.36 12.4 Learning Objective: 08-11 ( n - 1) s 2 (n - 1) s 2 (16 - 1)7.5932 (16 - 1)7.5932 2 2 < s < , < s < ,34.598 < s 2 < 119.104 cU2 c L2 24.9958 7.2609 5.882 < σ < 10.913 Learning Objective: 08-11
8.60
a.
or (8.3774, 12.3426). Using Megastat: s 5.31 x ±t = 10.36 ± 2.045 = 10.36 ± 1.9826 n 30 <CONFIDENCE INTERVALS/SAMPLE SIZE<CONFIDENCE INTERVAL - MEAN and then plug in the mean, standard deviation, n, and confidence level. Also select "t" instead of "z" because we do not have the population variance. Learning Objective: 08-6
8.61
a. Because the diameter is continuous there will always be slight variation in values from nickel to nickel. b. or (0.833172, 0.835514) s 0.001886 x ±t = 0.834343 ± 2.8453 = 0.834343 ± 0.001171 n 21 From Megastat: (.8332, .8355) c. The t distribution assumes a normal population, but in practice, this assumption can be relaxed, as long as the population is not badly skewed. We assume that here. d. Use to estimate the sample size. z = 2.576 so n = 95 æ zs ö n=ç ÷ è E ø
2
159
ASBE 6e Solutions for Instructors Learning Objective: 08-6 Learning Objective: 08-9 8.62
8.63
8.64
a.
or (3.2283, s .1320 x ±t = 3.3048 ± 1.833 = 3.3048 ± .0765 n 10 3.3813). Using Megastat: <CONFIDENCE INTERVALS/SAMPLE SIZE<CONFIDENCE INTERVAL MEAN and then plug in the mean, standard deviation, n, and confidence level. Also select "t" instead of "z" because we do not have the population variance. b. and our sample size is rounded up to 53. 2 2 æ zs ö (1.645)(.1320) æ ö n=ç ÷ =ç ÷ = 52.3886 è E ø .03 è ø Learning Objective: 08-6 Learning Objective: 08-9
x=
3.3048,
s=
.1320.
s 10.3087 x ±t = 34.538 ± 1.7823 = 34.538 ± 5.096 n 13 From Megastat: (29.4427, 39.6342) Learning Objective: 08-6 a. FPCF =
b.
or (29.442, 39.634)
.
N -n 1591 - 20 = = .994 N -1 1591 - 1 . Yes, the population can be considered effectively infinite.
n 20 = = .0126 < .05 N 1591 Learning Objective: 08-8
8.65
a. FPCF =
b.
.
N -n 187 - 25 = = .9333 N -1 187 - 1 . No, the population cannot be considered effectively infinite.
n 25 = = .134 > .05 N 187 Learning Objective: 08-8
8.66
a.
or (266.7614, 426.2386). Using s 170.3784 x ±t = 346.5 ± 2.093 = 346.5 ± 79.7386 n 20 Megastat: <CONFIDENCE INTERVALS/SAMPLE SIZE<CONFIDENCE INTERVAL - MEAN and then plug in the mean, standard deviation, n, and confidence level. Also select "t" instead of "z" because we do not have the population variance.
160
ASBE 6e Solutions for Instructors b. æ zs ö n=ç ÷ è E ø
so our sample size is rounded up to 279.
2
2
æ (1.96)(170.3784) ö =ç ÷ = 278.7926 20 è ø Learning Objective: 08-6 Learning Objective: 08-9 8.67
a.
or (19.249, 20.689). s 1.351 x ±t = 19.969 ± 2.1315 = 19.969 ± 0.720 n 16 From Megastat: (19.249, 20.689) b. Fuel economy can also vary due to tire pressure and weather. There may be more than sampling variability contributing to differences in sample means. Learning Objective: 08-6
8.68
a.
or (7.3821, 9.8899).
s 3.9265 187 - 25 = 8.636 ± 1.711 = 8.636 ± 1.2539 187 - 1 n 25 b. and our sample size is rounded up to 19. 2 2 æ zs ö æ (1.645)(3.9265) ö n=ç ÷ =ç ÷ = 18.5422 1.5 è E ø è ø Learning Objective: 08-6 Learning Objective: 08-9 x ±t
8.69
a.
or (133.013, 158.315). Because s 27.793 x ±t = 145.664 ± 2.086 = 145.664 ± 12.651 n 21 the sample size is greater than 5% of the population size, if we use the FPCF = .9165 the confidence interval is (134.0690, 157.2590). b. Use with z = 1.96 to get n = 119. æ zs ö n=ç ÷ è E ø
c.
2
(21 - 1)27.793 (21 - 1)27.793 <s2 < , 452.122 < s 2 < 1610.783 34.170 9.591 Learning Objective: 08-6 Learning Objective: 08-9 Learning Objective: 08-10 8.70
a.
2
2
or 21.26 < σ < 40.14
or (3230.4544, s 97.0535 x ±t = 3278.7222 ± 2.110 = 3278.7222 ± 48.2678 n 18 3326.9900). Using Megastat: <CONFIDENCE INTERVALS/SAMPLE SIZE<CONFIDENCE INTERVAL - MEAN and then plug in the mean, standard
161
ASBE 6e Solutions for Instructors deviation, n, and confidence level. Also select "t" instead of "z" because we do not have the population variance. b. and our sample size is rounded up to 91. 2 2 æ zs ö æ (1.96)(97.0535) ö n=ç ÷ =ç ÷ = 90.4637 E è ø 20 è ø c. The line chart shows a decrease in the number of steps over time.
Learning Objective: 08-6 Learning Objective: 08-9 8.71
a.
s 1.0953 x ±t = 29.53 ± 2.0639 = 29.53 ± 0.452 n 25 From Megastat: (29.078, 29.982) b. Use with z = 1.96 to get n = 116. æ zs ö n=ç ÷ è E ø
or (29.078, 29.982).
2
Learning Objective: 08-6 Learning Objective: 08-9 8.72
Using Appendix E with d.f. = 9,
c L2 = 2.7
and
cU2 = 19.02
.
( n - 1) s ( n - 1) s (10 - 1)7.2 (10 - 1)7.2 <s2 < , <s2 < , 24.5 < s 2 < 172.8 2 2 cU cL 19.02 2.7 4.38 < σ < 13.15 Learning Objective: 08-11 2
8.73
2
2
2
or
2L =CHISQ.INV(.05,29) = 17.7084, 2U =CHISQ.INV(.95,29) =42.557 (n - 1) s (n - 1) s (30 - 1)5.31 (30 - 1)5.31 <s2 < , <s 2 < ,19.214 < s 2 < 46.175 2 2 cU cL 42.557 17.7084 4.95 < σ < 6.80 Learning Objective: 08-11 2
2
2
162
2
or
ASBE 6e Solutions for Instructors 8.74
8.75
s 12.3322 x ±t = 80.4167 ± 1.796 = 80.4167 ± 6.3938 12 n Learning Objective: 08-6
or (74.02, 86.81).
a.
or (3.442, 3.642). s 0.311 x ±t = 3.542 ± 2.0227 = 3.542 ± 0.0995 n 40 b. The distribution is more likely to be skewed to the right with a few songs having very long playing times. c. Use with z = 1.96 to get n = 38. æ zs ö n=ç ÷ è E ø
2
Learning Objective: 08-6 Learning Objective: 08-9 8.76
a. Uniform Distribution:
Chromium Barium Fluoride
Normal Distribution: b-a 6
s=
(b - a )2 12
s=
(.69 - .47)2 .69 - .47 = .0635 s = = .0367 12 6
s=
(.019 - .004) 2 .019 - .004 = .0043 s= = .0025 12 6
s=
(1.17 - 1.07)2 1.17 - 1.07 = .0289 s= = .0167 12 6
s=
Learning Objective: 08-9 8.77
a.
p = 19 /100 = .19
.
p (1 - p ) .19(1 - .19) = .19 ± 1.645 = .19 ± .065 n 100 b. Normality can be assumed. np = 19 > 10 and n(1−p) = 81 > 10.. c. Use with z = 1.645 to get n = 463.
or (.125, .255)
p±z
2
æ zö n = ç ÷ p (1 - p ) èEø
d. A quality control manager needs to understand that the sample proportion will usually be different from the population proportion but that the way the sample proportion varies is predictable. Learning Objective: 08-7 Learning Objective: 08-10
163
ASBE 6e Solutions for Instructors 8.78
p = 14 /180 = .0778
.
p (1 - p ) .0778(1 - .0778) = .0778 ± 1.96 = .0778 ± .0391 n 180 (.0387, .1169). Using Megastat: <CONFIDENCE INTERVALS/SAMPLE SIZE<CONFIDENCE INTERVAL p and then plug in the values for p, n, and confidence level. Learning Objective: 08-7
8.79
2
=
or
p±z
2
= 136
æ zs ö æ (2.33)(.5) ö n=ç ÷ ç ÷ .1 è E ø è ø Learning Objective: 08-10
8.80
2
2
and our sample size is rounded up to 97 . Using
æzö æ 1.96 ö n = ç ÷ p (1 - p ) = ç ÷ .5(.5) = 96.04 èEø è .10 ø Megastat: <CONFIDENCE INTERVALS/SAMPLE SIZE<SAMPLE SIZE p and then plug in the values for E, p, and confidence level. Learning Objective: 08-10
8.81
8.82
8.83
a.
= .29 ± 1.96
= (.258, .322) . 29 (. 71 ) p (1 - p ) p±z n 787 b. Yes, np = 228.23 and n(1–p) = 558.77 which are both greater than 10. Learning Objective: 08-7 s 3.1 x ±t = 798.3 ± 2.045 = 798.3 ± 1.158 n 30 Learning Objective: 08-6 p = 83 /125 = .664
.
p±z
or (797.142, 799.458).
p (1 - p ) .664(.336) = .664 ± 1.645 = .664 ± .069 n 125
or (.595, .733) 8.84
Learning Objective: 08-7 = .58 ± 1.645 = (.568, .592). .58(.42) p (1 - p ) p±z = .58 ± .0120 n 4581 Learning Objective: 08-7
164
ASBE 6e Solutions for Instructors 8.85
a. Margin of error = ± z
8.86
a.
= ± 1.96
= ± .035. Must assume π = .5 and 95%
p (1 p ) .5(.5) n 787 confidence. b. Margin of error would be greater if subgroup consisted of only males because the sample size is now smaller. Learning Objective: 08-7 p (1 - p ) .1113(1 - .1113) = .1113 ± 1.645 = .1113 ± .0186 n 773 b. We can assume normal because and np = (773)(.1113) = 86 ³ 10 . n(1 - p ) = 773(1 - .1113) = 686.9651 ³ 10 Learning Objective: 08-7
or (.0927, .1299)
p±z
8.87
a. p = 98/213 = .46,
= .46 ± 1.96
.46(.54) p (1 - p ) n 213 b. Normality assumption holds. np = 98 and n(1p) = 115 Learning Objective: 08-7
= (.393, .527)
p±z
8.88
p (1 - p ) .72(1 - .72) = .72 ± 1.645 = .72 ± .1045 n 50 Learning Objective: 08-7
or (.6155, .8245).
p±z
8.89
a. Normality is not assumed, np = 4 which is less than 10. b. Using MINITAB (.0044, .0405) Learning Objective: 08-7
8.90
a. Margin of error =
1.96
.5(1 - .5) = .0205 2277
b.
or (.4229, .4571)
p (1 - p ) .44(1 - .44) = .44 ± 1.645 = .44 ± .0171 n 2277 c. Because the interval falls below .5 we would conclude that it is unlikely 50% of the voters approve of the President’s performance. Learning Objective: 08-7 p±z
165
ASBE 6e Solutions for Instructors 8.91
a. Margin of error =
1.96
.5(1 - .5) = .04 600
Assume that p = .5 and a 95% confidence level.
Learning Objective: 08-7 8.92
( n - 1) s 2 (n - 1) s 2 (40 - 1)0.0967 (40 - 1)0.0967 2 < s < , <s2 < 2 2 cU cL 58.12 23.65 . 2 2 0.0649 < s < 0.1595, 0.2548 < s < 0.3994
, 0.0649 <
s2
< 0.1595.
Learning Objective: 08-11 8.93
a.
2
2
2
2
Round the sample size up to 385.
æzö æ 1.96 ö n = ç ÷ p (1 - p ) = ç ÷ .5(.5) = 384.16 èEø è .05 ø
b.
, n would increase to 1537
æzö æ 1.96 ö n = ç ÷ p (1 - p ) = ç ÷ .5(.5) = 1536.64 èEø è .025 ø Learning Objective: 08-10
166
ASBE 6e Solutions for Instructors
Chapter 9 One-Sample Hypothesis Tests 9.1
a. (.05)(1000) = 50 times b. (.01)(1000) = 10 times c. (.001)(1000) = 1 time Learning Objective: 09-2
9.2
a. Null hypothesis: The man is not having a heart attack. Type I error: I admit him when he does not have a heart attack. Type II error: I fail to admit him, he does have a heart attack. Better to make a Type I than Type II error. b. Type I error: I reject the null and let them land even though they could have stayed up for 15 minutes (or more). Type II: Don’t let the plane land and the plane runs out of fuel. It is more costly to make a Type II error. c. Type I error: I reject the null and rush out to Staples, get caught in the snow and fail to finish the report (when if I had stayed I would have finished it). Type II error: I run out of ink and can’t finish the report. Better to stay and try to finish the report, in fact better to print out some of it than none of it. Better to make a Type I error. Learning Objective: 09-2
9.3
a. Null hypothesis: Employee is not using illegal drugs. Alternative hypothesis: Employee is using illegal drugs. b. Type I error: Test is positive for drug use when no drugs are being used. Type II error: Test is negative for drug use when the person is using drugs. c. I might dismiss or discipline someone who is a non-drug user (Type I error). They could sue for wrongful damages. I might keep on someone who should be dismissed and they cause serious injury via a work related accident to themselves or others (Type II error). Learning Objective: 09-2
9.4
a. Null hypothesis: There is no fire. Alternative hypothesis: There is a fire. b. Type I error: A smoke detector sounds an alarm when there is no fire. Type II error: A smoke detector does not sound an alarm when there is a fire. c. Consequence of making a type I error is that some guests will be inconvenienced by a false alarm and there is the cost of having the fire department summoned. Consequence of making a Type II error is that the hotel will burn down and perhaps kill or injure many. Learning Objective: 09-2
164
ASBE 6e Solutions for Instructors 9.5
A false negative is a Type II error. If the null hypothesis is that the brakes are good and we fail to reject this hypothesis when in fact the brakes are not good then the consequence would be a possible car accident when the brakes fail. Learning Objective: 09-2
9.6
A false positive is a Type I error. If the null hypothesis is that the nuclear plant’s cooling system is good and we reject this hypothesis when in fact the system is good then the consequence is that we might evacuate people who live near the plant unnecessarily while we perform a more thorough inspection and conduct repairs. Learning Objective: 09-2
9.7
H0: 2.5 mg vs. H1: ≠2.5 mg Learning Objective: 09-3
9.8
H0:
9.9
H0: ≤4 min vs. H1: > 4 min Learning Objective: 09-3
9.10
H0:
9.11
Graphs should show a normal distribution with a mean of 80. a. Rejection region in the lower tail.
min vs. H1: < 1.2 min. ³ 1.2 Learning Objective: 09-3
yrs vs. H1: < 37 yrs. ³ 37 Learning Objective: 09-3
b. Rejection region in both tails.
c. Rejection region in the upper tail.
165
ASBE 6e Solutions for Instructors
Learning Objective: 09-4 9.12
H0: mm vs. H1: ≠ mm Learning Objective: 09-3
9.13
a. zcalc =
b. zcalc =
x - m0 242 - 230 = s 18 n 20
= 2.98
x - m0 3.44 - 3.50 = s 0.24 n 40
= 1.58
c. zcalc =
x - m0 21.02 - 20.00 = s 2.52 n 30 Learning Objective: 09-6
= 2.22
9.14
a. The positive z-score is found using =NORM.S.INV(.95). z.05 = ±1.645 (two-tailed) b. z.01 =NORM.S.INV(.99) = 2.326 (right-tailed) c. z.05 =NORM.S.INV(.05) = –1.645 (left-tailed) Learning Objective: 09-4
9.15
a. The positive z-score is found using =NORM.S.INV(.975). z.025 = ±1.96 (two-tailed) b. z.10 =NORM.S.INV(.90) = 1.28 (right-tailed) c. z.01 =NORM.S.INV(.01) = −2.326 (left-tailed) Learning Objective: 09-4
9.16
a. a. zcalc =
x - m0 423 - 420 = s 6 n 9
= 1.50
166
ASBE 6e Solutions for Instructors b. zcalc =
x - m0 8330 - 8344 = s 48 n 36
= 1.75
c. zcalc =
x - m0 3.102 - 3.110 = s 0.250 n 25 Learning Objective: 09-4
= −0.16
9.17
a. H0: ≤3.5 mg vs. H1: > 3.5 mg b. zcalc = = 2.50 x - m0 3.59 - 3.50 = s 0.18 n 25 c. Yes, zcrit =NORM.S.INV(.99) = 2.33 and 2.50 > 2.33. d. p-value =1-NORM.S.DIST(2.5,1) = .0062 Learning Objective: 09-4 Learning Objective: 09-5 Learning Objective: 09-6
9.18
a. H0:
= 140
mg vs. H1:
¹
140 mg.
b.
x - m0 139.4 - 140 = = -1.3416 s 2 n 20 c. Decision Rule: Reject H0 if or if . –1.3416 does not fall in zcalc > +1.645 zcalc < -1.645 the rejection region so we cannot reject the null. The sample does not contradict the manufacturer's claim. d. =NORM.S.DIST(-1.3416,1) = .0899 P ( zcalc < -1.3416) p-value = . The p-value is greater than a 2 ´ P( zcalc < -1.3416) = 2 ´ .0899 = .1798 > .10 . Do not reject the null. Learning Objective: 09-4 Learning Objective: 09-5 Learning Objective: 09-6 zcalc =
167
ASBE 6e Solutions for Instructors 9.19
a.
1.5, p-value =
2 ´ .0668 =
.1336 (from Appendix C-2).
x - m0 63 - 60 =z= = s 8 n 16 Using Excel: =2×NORM.S.DIST( 1.5,1) or =2×(1–NORM.S.DIST(1.5,1)) = .1336. b. −2.0, p-value = .0228. 58 - 60 zcalc = = 5 25 Using Excel: =NORM.S.DIST( 2.0,1) = .02275. c. 3.75, p-value = .0001. 65 - 60 zcalc = = 8 36 Using Excel =1–NORM.S.DIST(3.75,1) = .0000884. zcalc =
Learning Objective: 09-6
9.20
From Appendix C-2 a. p-value = P(Z > 1.34) = .0901 b. p-value = P(Z < −2.07) = .0192 c. p-value = 2×P(Z < −1.69) = 2×.0455 = .091 Learning Objective: 09-6
9.21
H0: 2.035 oz vs. H1: > 2.035 oz.,
= 2.50 and p-value 2.036 - 2.035 zcalc = = 0.002 25 =NORM.S.DIST(2.5,1) = .0062 < .025. Reject H0. The mean weight is heavier than it is supposed to be. Learning Objective: 09-4 Learning Objective: 09-5 Learning Objective: 09-6
9.22
a. H0: 195 flights/hour vs. H1: > 195 flights/hour. Reject H0 if z > 1.96. b. > 1.96 so reject the null hypothesis and conclude that x - m0 200 - 195 zcalc = = = 2.11 s 13 n 30 the average number of arrivals has increased. If we had used an = .01, we would 168
ASBE 6e Solutions for Instructors have failed to reject the null hypothesis because the decision rule would have been: Reject H0 if zcalc > 2.326. c. We have assumed a normal population or at least one that is not badly skewed. Learning Objective: 09-4 Learning Objective: 09-5 Learning Objective: 09-6
9.23
a. H0: =10 oz vs. H1: ≠ 10 oz. Reject H0 if zcalc > 1.96 or zcalc < −1.96. b. < 1.96 and zcalc > −1.96 so we fail to reject the x - m0 10.01583 - 10.0 zcalc = = = 0.7834 s 0.07 n 12 null hypothesis (p-value =2×(1-NORM.S.DIST(0.7834,1)) = .4333). c. We assume the population is normally distributed. Learning Objective: 09-4 Learning Objective: 09-5 Learning Objective: 09-6
9.24
H0: ≤4 min vs. H1: > 4 min. Reject H0 if zcalc > 2.326. < 2.326 so there is not enough evidence to reject the null hypothesis, 4.5 - 4 zcalc = = 2.12 1 18 p-value =NORM.S.DIST(2.12,1) = .017 > .01. There is not enough evidence to show the goal is not being met. Learning Objective: 09-4 Learning Objective: 09-5 Learning Objective: 09-6
9.25
H0: 3000 hours vs. H1: > 3000 hours. Reject H0 if zcalc > 2.326. =3.26 > 2.326 so there is enough evidence to reject the null hypothesis. 3515 - 3000 zcalc = 500 10 We can conclude the new lamp’s mean life exceeds the current average. p-value =1-NORM.S.DIST(3.26,1) = .0006. 169
ASBE 6e Solutions for Instructors Learning Objective: 09-4 Learning Objective: 09-5 Learning Objective: 09-6 9.26
a. Although statistically significant, the improvement may lack practical importance because customers may not notice and may be unwilling to pay the extra cost. b. Although statistically significant, the improvement may lack practical importance because customers may not notice and may be unwilling to pay the extra cost. c. Although statistically significant, the improvement may lack practical importance because customers may not notice and may be unwilling to pay the extra cost. Learning Objective: 09-2 Learning Objective: 09-4
9.27
H0: = 1.967 newtons vs. H1: ≠ 1.967 newtons. Reject H0 if zcalc > 1.96 or zcalc < −1.96. = −1.80. Because zcalc falls between −1.96 and 1.96 there is not enough 1.88 - 1.967 zcalc = 0.145 9 evidence to reject the null hypothesis. We conclude there is not a significant deviation from the target force needed to activate the entry clicker. p-value =2×NORM.S.DIST(-1.8,1) = .0719. Learning Objective: 09-4 Learning Objective: 09-5 Learning Objective: 09-6
9.28
a. tcalc = b. tcalc =
x - m0 14.7 - 13.0 = = 3.27 s 1.8 n 12 241 - 250 = -2.12 12 8
c.
2,102 - 2, 000 = 1.74 242 17 Learning Objective: 09-7 tcalc =
170
ASBE 6e Solutions for Instructors 9.29
a. Using d.f. = 20 the positive value of t =T.INV(.95,20) = 1.725, t.05 = ±1.725 (two-tailed) b. Using d.f. = 8 t.01 =T.INV(.99, 8) = 2.896 (right-tailed) c. Using d.f. = 27, t.05 =T.INV(.05,27) = 1.703 (left-tailed) Learning Objective: 09-7
9.30
a. Using d.f. = 17 the positive value of t =T.INV(.975,17) = 2.110, t.025 = ±2.110 (twotailed) b. Using d.f. = 14 t.10 =T.INV(.90, 14) = 1.345 (right-tailed) c. Using d.f. = 30, t.01 =T.INV(.01,30) = 2.457 (left-tailed) Learning Objective: 09-7
9.31
a. tcalc = b. tcalc =
x - m0 347 - 349 = = -3.33 s 1.8 n 9 x - m0 45 - 50 = = -1.67 s 12 n 16
c.
x - m0 4.103 - 4.004 = = 2.02 s 0.245 n 25 Learning Objective: 09-7 tcalc =
9.32
a. From Appendix D using d.f. = 14, t.05 = 1.761 and t.10 = 1.345. t = 1.457 falls between these two values so .05 < p-value < .10 (right-tailed) b. From Appendix D using d.f. = 8, t.02 = 2.896 and t.05 = 2.306. t = 2.601 falls between these two values so.02 < p-value < .05 (two-tailed) c. Because the t distribution is symmetrical about the mean we can treat the positive values in Appendix D like negative values. From Appendix D using d.f. = 22, t.025 = 2.074 and t.05 = 1.717. t = 1.847 falls between these two values so.025 < p-value < .05 (left-tailed) Learning Objective: 09-7
9.33
a. Using Excel: =T.DIST.RT(1.457,14) = .0836 b. =T.DIST.2T(2.601,8) = .0316 c. =T.DIST(-1.847,22,1) = .0391 Learning Objective: 09-7
9.34
a. Using Excel: =T.DIST.RT(1.677, 12) = .0597 b. =T.DIST(-2.107,4,1) = .0514 c. =T.DIST.2T(1.865,33) = .0711 171
ASBE 6e Solutions for Instructors Learning Objective: 09-7 9.35
a.
, p-value =T.DIST.2T(1.5,15) = .1544 > .025. Fail to reject the null 203 - 200 tcalc = = 1.5 8 16 hypothesis. b. , p-value =T.DIST(-2.0,24,1) = .0285 < .05. Reject the null 198 - 200 tcalc = = -2.0 5 25 c. , p-value =T.DIST.RT(3.75,35) = .0003 < .05. Reject the null 205 - 200 tcalc = = 3.75 8 36 hypothesis. Learning Objective: 09-7
9.36
H0: 530 bags/hour vs. H1: < 530 bags/hour. tcrit =T.INV(.05,15) = −1.753. Reject H0 if tcalc < −1.753. > −1.753 so fail to reject the null hypothesis. 510 - 530 tcalc = = -1.6 50 16 No significant evidence that the claim is overstated. Learning Objective: 09-7
9.37
a. H0: μ ≥ 400 sf/gal vs. H1: μ < 400 sf/gal. tcrit = T.INV(.1,5) = −1.476. Reject H0 if tcalc < −1.476. < −1.476 therefore reject H0. 383.33 - 400 tcalc = = -1.977 20.656 6 b. Yes, if were less than .05 our decision would be different. t.05 =T.INV(.05,5) = −2.015. c. p-value = .0525. Because .0525 < .10, we would reject the null hypothesis. d. A significant result in a hypothesis does not always translate to a practically important difference. In this case, if the painter plans his or her paint purchase based on coverage of 400 square feet per gallon, but in reality the paint covers 5% less, the painter may run short on large jobs. A difference of 5% may not matter on a small job. Learning Objective: 09-7 172
ASBE 6e Solutions for Instructors
9.38
a. H0: μ ≥ 18 oz vs. H1: μ < 18 oz. tcrit =T.INV(.05,17) = −1.740. Reject H0 if tcalc < −1.740. < −1.740 therefore reject H0. The true mean is 17.78 - 18 tcalc = = -2.277 .41 18 smaller than the specification. b. Yes, if =.01, we would have failed to reject the null hypothesis. T.01= T.INV(.01,17) = −2.567. c. Using Excel: =T.DIST(-2.277,17,1) gives p-value = .0180. Because the p-value < .05 we reject the null hypothesis. Learning Objective: 09-7
9.39
a. H0: ≤$318 vs. H1: > $318. tcrit =T.INV(.95,19) = 1.729. Reject H0 if tcalc > 1.729. > 1.729 so reject H0. 341 - 318 tcalc = = 4.472 23 20 b. p-value =1-T.DIST(4.472,19,1) = .00013. Or one could use the function T.DIST.RT(4.472,19). Learning Objective: 09-7 Learning Objective: 09-8
9.40
H0: μ = 37 yrs vs. H1: μ or tcalc > +1.677.
¹
37 yrs. tcrit =T.INV.2T(.1,49) = 1.677. Reject H0 if tcalc < –1.677
therefore fail to reject the null hypothesis. 38.5 - 37 tcalc = = 0.663 16 50 There is no significant evidence that the average age has changed in 2010. Learning Objective: 09-7 9.41
H0: ≤1.6 vs. H1: > 1.6 tcrit =T.INV(.95,39) = 1.685.
1.80 - 1.60 tcalc = = 1.14 1.11 40 and the p-value =T.DIST.RT(1.14,39) = .1306 > .05. Fail to reject H0.
Learning Objective: 09-7
173
< 1.685
ASBE 6e Solutions for Instructors 9.42
H0: μ ≤ 30,000 miles vs. H1: μ > 30,000 miles. tcrit =T.INV(.90,20) = 1.325. Reject H0 if tcalc > 1.325. > 1.325 therefore reject H0. This 33,950 - 30, 000 tcalc = = 1.53 11,866 21 dealer shows a significantly greater mean number of miles than the national average. Learning Objective: 09-7
9.43
a.
H0: μ = 3.25 vs. H1: μ ≠ 3.25. tcrit =T.INV.2T(.05,17) = 2.11. Reject H0 if tcalc > 2.11 or tcalc < 2.11. < 2.11 therefore fail to reject H0. 3.35 - 3.25 tcalc = = 1.70 0.25 18 b. The 95% confidence interval is: or (3.23, 3.47). Because this interval 0.25 3.35 ± 2.11 18 does contain 3.25 we would fail to reject the null hypothesis. c. We constructed the 95% confidence interval using the t statistic associated with .025 in the tail areas. This is the same area we used to determine the critical value for the t statistic. A two-tailed test and a confidence should always result in the same conclusion as long as is the same for both. Learning Objective: 09-7
9.44
a. The p-value tells us that there is a 38.7% chance that we would make our sample observation (or something more extreme) given the null is true. This is not a small probability therefore we fail to reject the null hypothesis. b. The p-value tells us that there is a 4.3% chance that we would make our sample observation (or something more extreme) given the null is true. This is a small probability therefore we would most likely reject the null hypothesis. c. The p-value tells us that there is a .12% chance that we would make our sample observation (or something more extreme) given the null is true. This is an extremely small probability therefore we would reject the null hypothesis. Learning Objective: 09-6
9.45
a.
. From Appendix C-2: p-value = 2×.0228 = .0456. Using Excel: p.28 - .2 zcalc = = 2.0 .2 ´ .8 100 value =2*(1-NORM.S.DIST(2.0,1)) = .0455.
174
ASBE 6e Solutions for Instructors b.
.60 - .50 zcalc = = 1.897 .5 ´ .5 90
c.
, p-value =1-NORM.S.DIST(1.897,1) = .0289.
.82 - .75 zcalc = = 1.143 .75(.25) 50 Learning Objective: 09-8
, p-value =1-NORM.S.DIST(1.143,1) = .1265.
9.46
a.
. From Appendix C-2: p-value = 1–.9664=.0336. p -p0 .70 - .60 z= = = 1.83 p 0 (1 - p 0 ) .60(.40) n 80 Using Excel: =1-NORM.S.DIST(1.83,1) = .0336 b. , From Appendix C-2: p-value = 2(1–.9808) = .0384 .45 - .30 z= = 2.07 .30(.70) 40 Using Excel: =2*(1-NORM.S.DIST(2.07,1)) = .0385 c. , From Appendix C-2: p-value = .0099 .03 - .10 z= = -2.33 .10(.90) 100 Using Excel: =NORM.S.DIST(-2.33,1) = .00099 Learning Objective: 09-8
9.47
a. .30×20 = 6 < 10 and .7×20 = 14 > 10 (cannot assume normality) b. .05×50 = 2.5 < 10 and .95×50 = 47.5 >10 (cannot assume normality) c. .10×400 = 40 > 10 and .9×400 = 360 > 10 (normality can be assumed) Learning Objective: 09-9
9.48
H0: ≤ .10 vs. H1: 10.
, p-value =1-NORM.S.DIST(2.67,1) = .18 - .10 z= = 2.67 .10(.90) 100 1–.9962=.0038 and .0038 < .05 so reject H0 and conclude that the proportion who believe Pepsi is concerned with consumers’ health has increased. Learning Objective: 09-8
175
ASBE 6e Solutions for Instructors 9.49
a. H0: ≥ .997 vs. H1: 997. zcrit =NORM.S.INV(.05) = −1.645.Reject the null hypothesis if z < −1.645. > −1.645 so fail to reject H0. .996 - .997 zcalc = = -1.082 .997(.003) 3500 One could also define π to be the proportion of syringes that are defective. H0: ≤ .003 vs. H1: 003. zcrit =NORM.S.INV(.05) = 1.645.Reject the null hypothesis if z > 1.645. < 1.645 so fail to reject H0. .004 - .003 zcalc = = 1.082 .003(.997) 3500 b. Yes, .997×3500 = 3489.5 and .003×3500 = 10.5, both are greater than 10. c. A Type I error would be sending back an acceptable shipment. This could be a problem if the hospital runs low on insulin syringes. A Type II error would be accepting a bad shipment. The will be a problem if a defective syringe is used for an insulin injection. d. p-value =NORM.S.DIST(-1.082,1) = .1396. Learning Objective: 09-2 Learning Objective: 09-8 Learning Objective: 09-9
9.50
a. H0: ≥ .40 vs. H1: 40. zcrit =NORM.S.INV(.05) = −1.645.Reject the null hypothesis if z < −1.645. > −1.645 so fail to reject H0. The .30 - .40 zcalc = = -1.58 .40(.60) 60 proportion of children in Colorado who were given a throat swab for diagnosing strep throat is not significantly below 40 percent. b. nπ = 60(.4) = 24 and n(1−π) = 60(.6) = 36. Both are greater than 10 so we can assume a normal distribution for p. Learning Objective: 09-8 Learning Objective: 09-9
9.51
a. H0: ≥ .50 vs. H1: .50. zcrit =NORM.S.INV(.05) = −1.645. Reject the null hypothesis if z < 1.645. p = 24/64 = .375, <−1.645 so reject H0. .375 - .50 zcalc = = -2.0 50(.50) 64 b. p-value =NORM.S.DIST(-2.0,1) = .0228 c. The proportion of calls lasting less than 2 minutes is less than .5. The call center is not being as efficient as the management would like. d. Yes, a difference of 12.5% is important. 176
ASBE 6e Solutions for Instructors
Learning Objective: 09-8
9.52
a. H0: .052 vs. H1: <052 . zcrit =NORM.S.INV(.01) = −2.326. Reject the null hypothesis if z < −2.326. > −2.326 therefore fail to reject .03 - .052 zcalc = = -1.72 .052(.948) 300 the null hypothesis. b. p-value =NORM.S.DIST(-1.72,1) = .0427 c. n15.6 > 10 and so normality is justified. n(1 - p 0 ) = 284.4 > 10 Learning Objective: 09-8 Learning Objective: 09-9
9.53
H0: .50 vs. H1: >50 zcrit =NORM.S.INV(.95) = 1.645. Reject the null hypothesis if z > 1.645. > 1.645 so reject H0. .63 - .50 zcalc = = 6.89 .50(.50) 702 Learning Objective: 09-8
9.54
a. Normality is not justified. n = (12)(.5) = 6 < 10. b. Using the normal assumption, H0: .50 vs. H1: >50
.8333 - .50 = 2.309 .5(.5) 12 and the p-value =1-NORM.S.DIST(2.309,1) = .0105. The p-value is < .05 so we reject the null and conclude this small sample does provide evidence that more than half of the homes sold were sold for less than the asking price. c. =1-BINOM.DIST(9,12,.5,TRUE) = .0193. Learning Objective: 09-4 Learning Objectives: 09-9 Learning Objective: 09-10 zcalc =
9.55
a. Using the binomial distribution, P(X ≥ 4 | n = 2000, = .001) =1BINOM.DIST(3,2000,.001,1) = .1428 > .10. The standard is being met. b. Because there are only 4 defective items observed we cannot assume normality. Learning Objective: 09-8 Learning Objective: 09-9 177
ASBE 6e Solutions for Instructors 9.56
a. P(X 2 | n = 20, = .0005) =1-BINOM.DIST(1,20,.0005) = .00005. The students at Julliard do differ from the national norm. b. Normality cannot be assumed because there were less than 10 students in the sample with perfect pitch. Learning Objective: 09-8 Learning Objective: 09-9
9.57
a. Power = P(pcalc > pcritical | π = .04) Step 1:
pcritical = .03 + 1.28 Step 2: z=
(.03)(1 - .03) = .0409 400
.0409 - .04 = .0919 (.04)(.96) 400
Step 3:
Power = P( Z > .0919) = 1-NORM.S.DIST(.0919,1) = .4634 Using LearningStats: Plug in the values, .03, 400, .10, right, .04 which returns a similar value of .4622. Differences are due to rounding. b. Step 1: (.03)(1 - .03) pcritical = .03 + 1.28 = .0409 400 Step 2: .0409 - .05 z= = -0.8351 (.05)(.95) 400 Step 3: Power = P( Z > -.8351) = 1-NORM.S.DIST(-.8351,1) = .7982 Using LearningStats: Plug in the values, .03, 400, .05, right, .05 which returns a similar value of .7974. c. Step 1: (.03)(1 - .03) pcritical = .03 + 1.28 = .0409 400 Step 2: .0409 - .06 z= = -1.609 (.06)(.94) 400 Step 3: Power = P( Z > -1.609) = 1-NORM.S.DIST(-1.609,1) = .9462 Using LearningStats: Plug in the values .03, 400, .05, right, .06 which returns a similar value of .9459. Learning Objective: 09-10
178
ASBE 6e Solutions for Instructors
9.58
a. Power = P(pcalc > pcritical | π = .04) Step 1:
pcritical = .03 + 1.645 Step 2: z=
(.03)(1 - .03) = .0440 400
.0440 - .04 = .4082 (.04)(.96) 400
Step 3:
Power = P( Z > .4082) = .3416 Using LearningStats: Plug in the values, .03, 400, .05, right, .04 which returns a similar value of .3404. Differences are due to rounding. b. Step 1: (.03)(1 - .03) pcritical = .03 + 1.645 = .0440 400 Step 2: .0440 - .05 z= = -.5455 (.05)(.95) 400 Step 3: Power = P( Z > -.5455) = .7073 Using LearningStats: Plug in the values, .03, 400, .05, right, .05 which returns a similar value of .7081. c. Step 1: (.03)(1 - .03) pcritical = .03 + 1.645 = .0440 400 Step 2: .0440 - .06 z= = -1.3445 (.06)(.94) 400 Step 3: Power = P( Z > -1.3445) = .9106 Using LearningStats: Plug in the values .03, 400, .05, right, .06 which returns a similar value of .9107. When you increase the true value of , you also increase the power. We can see this
p
with the power curve produced from LearningStats:
179
ASBE 6e Solutions for Instructors
Learning Objective: 09-10 9.59
a. Power = P( Step 1:
x < xcritical | m = 3.48)
æ .10 ö xcritical = 3.5 - 2.326 ç ÷ = 3.4535 è 25 ø
Step 2: z=
3.4535 - 3.48 = -1.325 .10 25
Step 3:
Power = P( Z < -1.325) = NORM.S.DIST(-1.325,1) = .0926 Using LearningStats: Plug in the values 3.5, .1, 25, .01, left, 3.48 which returns a similar value .0924. b. Power = P( x < xcritical | m = 3.46) Step 1: æ .10 ö xcritical = 3.5 - 2.326 ç ÷ = 3.4535 è 25 ø Step 2: 3.4535 - 3.46 z= = -.325 .10 25 Step 3: Power = P( Z < -.325) = NORM.S.DIST(-.325,1) = .3726 Using LearningStats: Plug in the values 3.5, .1, 25, .01, left, 3.46 which returns a similar value .3721. c. Power = P( x < xcritical | m = 3.44)
180
ASBE 6e Solutions for Instructors Step 1:
æ .10 ö xcritical = 3.5 - 2.326 ç ÷ = 3.4535 è 25 ø
Step 2: z=
3.4535 - 3.44 = 0.675 .10 25
Step 3:
Power = P ( Z < 0.675) = NORM.S.DIST(0.675,1) = .7502 Using LearningStats: Plug in the values 3.5, .1, 25, .01, left, 3.44 which returns a similar value .7497.
Results from Learning Stats using Learning Stats Excel file 09-08 PowerCurvesDIY.xls:
X Value 3.44000 3.46000 3.48000
zLeft BetaLeft 0.674 -0.326 -1.326
Learning Objective: 09-10 9.60
a. Power = P( Step 1:
x < xcritical | m = 3.48)
æ .10 ö xcritical = 3.5 - 1.645 ç ÷ = 3.4671 è 25 ø
Step 2:
z=
3.4671 - 3.48 = -.645 .10 25 181
ASBE 6e Solutions for Instructors Step 3:
Power = P( Z < -.645) = .2595 Using LearningStats: Plug in the values 3.5, .1, 25, .05, left, 3.48 which returns the same value .2595. b. a. Power = P( x < xcritical | m = 3.46) Step 1: æ .10 ö xcritical = 3.5 - 1.645 ç ÷ = 3.4671 è 25 ø Step 2: 3.4671 - 3.46 z= = .355 .10 25 Step 3: Power = P( Z < 0.355) = NORM.S.DIST(.355,1) =.6387 Using LearningStats: Plug in the values 3.5, .1, 25, .05, left, 3.46 which returns a power of .6388. c. a. Power = P( x < xcritical | m = 3.44) Step 1: æ .10 ö xcritical = 3.5 - 1.645 ç ÷ = 3.4671 è 25 ø Step 2: 3.4671 - 3.44 z= = 1.355 .10 25 Step 3: Power = P( Z < 1.355) = NORM.S.DIST(1.355,1) = .9123 Using LearningStats: Plug in the values 3.5, .1, 25, .05, left, 3.44 which returns a power of .9123. As the true value of decreases, the power increases. Results from Learning Stats: m
182
ASBE 6e Solutions for Instructors
X Value 3.44000 3.46000 3.48000
zLeft BetaLeft 1.355 0.355 -0.645
As the true value of
decreases, the power increases. We can see this with the power m curve produced from LearningStats:
Learning Objective: 09-10 9.61
H0: 2 50 vs. H1: 2 < 50. Reject the null hypothesis if 2 < 4.660 (=CHISQ.INV(.01,14)). therefore we fail to 2 ( n 1) s (14)35 c 2calc = = = 9.80 s 02 50 reject H0. Learning Objective: 09-11
9.62
H0: 2 = 24 vs. H1: 2
¹
24. Reject the null hypothesis if
(=CHISQ.INV(..95,9)) or
c2
(n - 1) s 2 (9)16 c 2calc = = =6 s 02 24 reject H0. Learning Objective: 09-11
> 16.92 c2 < 3.325 (=CHISQ.INV(.05,9)).
which falls between the critical values therefore we fail to
183
ASBE 6e Solutions for Instructors 9.63
H0: 2 1.21 vs. H1: 2 > 1.21. Reject the null hypothesis if 2 > 28.87 (=CHISQ.INV(.95,18)). therefore we reject 2 (n - 1) s (18)1.96 c 2calc = = = 29.16 2 s0 1.21 H0. Learning Objective: 09-11
9.64
H0: 2 = 0.01 vs. H1: 2 ≠ 0.01. Reject the null hypothesis if 2 > 27.49 (=CHISQ.INV(.95,15)) or 2 < 6.262 (=CHISQ.INV(.05,15)). which falls between the critical values therefore 2 ( n 1) s (15).009767 c 2calc = = = 14.65 s 02 .01 we fail to reject H0. Learning Objective: 09-11
9.65
H0: 2 = 625 vs. H1: 2 ≠ 625. Reject the null hypothesis if 2 > 21.92 (=CHISQ.INV(.975,11)) or 2 < 3.816 (=CHISQ.INV(.025,11)). which falls between the critical values therefore 2 2 (n - 1) s (11)21.66 c 2calc = = = 8.26 2 s0 252 we fail to reject H0. Learning Objective: 09-11
9.66
a. P(Type II error) = P(fail to reject
| is false) = 0 because you would never "fail to H0 H0
reject". b. This is bad policy because reducing the risk of Type II error increases the risk of Type I error. Learning Objective: 09-2 9.67
a. P(Type I error) =
reject | is true) = 0. P( H0 H0 b. This is bad policy because reducing the risk of Type I error increases the risk of Type II error. Learning Objective: 09-2
9.68
a. H0: ≤90 vs. H1: > 90 b. Type I error occurs when the physician concludes a patient has high blood pressure when they do not. The consequence is unnecessary treatment and worry. Type II error occurs when the physician concludes that a patient’s blood pressure is okay when it is
184
ASBE 6e Solutions for Instructors too high. The consequence is untreated high blood pressure which can lead to serious health complications. c. A Type II error would have a more serious consequence. The patient could have severe health problems if high blood pressure is undiagnosed. Learning Objective: 09-1 Learning Objective: 09-2 9.69
a. H0: User is authorized vs. H1: User is unauthorized. b. Type I error occurs when the scanner fails to admit an authorized user. Type II error occurs when the scanner admits an unauthorized user. c. A Type II error has a more serious consequence. Allowing entry to an unauthorized user could result in damage to the plant or possibly even a terrorist attack. Learning Objective: 09-1 Learning Objective: 09-2
9.70
a.
9.71
H0: ≤.02 vs. H1: > .02 Learning Objective: 09-1
9.72
P(Type II error) = 0. We’ve rejected the null hypothesis therefore it is impossible to make a Type II error. Learning Objective: 09-2
9.73
P(Type I error) = 0. There can be no Type I error if we fail to reject the null hypothesis. Learning Objective: 09-2
9.74
a. H0: A patient does not have cancerous cells vs. H1: A patient has cancerous cells. A false negative is a Type II error and means that the test shows no cancerous cells are present when in fact there are. A false positive is a Type I error and means that the test shows cancerous cells are present when they are not. b. In this case “null” stands for absence of cancerous cells.
vs. H 0 : m ³ 30 H1 : m < 30 b. Type I error would mean you reject the hypothesis that the training raises SAT scores by at least 30 points when in fact it does. Learning Objective: 09-1 Learning Objective: 09-2
185
ASBE 6e Solutions for Instructors c. The patient bears the cost of a false negative. If their health problems are not diagnosed early they will not seek treatment. The insurance company bears the costs of a false positive. Typically more tests will need to be done to check the results. Learning Objective: 09-1 Learning Objective: 09-2 9.75
a. Type I error: You should have been accepted, but the scanner rejected you. Type II error: You should have been rejected, but the scanner accepted you. b. The consequence of falsely rejecting someone is not as severe as falsely accepting someone. Or it could be that the scanner is dirty and cannot read the fingerprint accurately. Learning Objective: 09-2
9.76
This is the probability of making a Type I error. This means that half of the women who test positive for cancer don't actually have it. Learning Objective: 09-2
9.77
a. H0: ≤20 min vs. H1: > 20 min b. tcalc = 2.545 24.77 - 20 7.26 15 c. Yes. Using d.f. = 14, tcrit =T.INV(.95,14) = 1.761. Reject H0 if tcalc > 1.761. 2.545 > 1.761 therefore reject H0. d. p-value =T.DIST.RT(2.545,14) = .0117 < .05. Yes. Learning Objective: 09-7
9.78
a.
H 0 : m = 220
vs.
H1 : m ¹ 220
b.
x - m0 228.2 - 220 = = 1.4248 s 18.2 n 10 c. Decision Rule: tcrit =T.INV(.975,9) = 2.262. Reject H0 if tcalc < –2.262 or tcalc > 2.262. Therefore, do not reject H0. d. Using Excel: =T.DIST.2T(1.4248,9) = .1880 > .05 so do not reject. We come to the same conclusion as we did in part c. Learning Objective: 09-7 tcalc =
9.79
a. A two-tailed test would be used. You would not want to overfill or under-fill the can. b. Overfilling costs you money and under-filling cheats the customer. 186
ASBE 6e Solutions for Instructors c. Because the weight is normally distributed and the population standard deviation is known the sample mean will have a normal distribution. d. zcrit =NORM.S.INV(.995) = 2.576. Reject the null hypothesis if z > 2.576 or z < −2.576. Learning Objective: 09-4 9.80
a. Because the population distribution is normal and you know the population standard deviation, you should use the normal distribution for the sampling distribution on the sample mean. b. H0: = 520 vs. H1: ≠ 520. zcrit =NORM.S.INV(.975) = 1.96. Reject H0 if zcalc > 1.96 or z < −1.96. c. < −1.96 therefore reject the null hypothesis. The sample x - m0 515 - 520 zcalc = = = -5 s 4 n 16 result is highly significant showing there is a difference in the mean fill. Learning Objective: 09-4
9.81
a. H0: ≥ 90 vs. H1: < 90. b. tcrit =T.INV(.01,7) = −2.998. Reject H0 if tcalc < −2.998. x - m0 tcalc = s n c. Because −0.92 > −2.998 we fail to reject the null hypothesis. 88.375 - 90 tcalc = = -0.92 4.984 8 The sample does not give enough evidence to reject Bob’s claim that he is a 90+ student. d. We assume that the population distribution is normal. e. The p-value =T.DIST(-0.92,7,1) = .1941. Because .1941 > .01 we fail to reject the null hypothesis. Learning Objective: 09-7
9.82
a. H0: ≤ 10 pages vs. H1: > 10 pages. tcrit =T.INV(.99,34) = 2.441. Reject H0 if tcalc > 2.441. so reject the null hypothesis and conclude x - m0 14.44 - 10 tcalc = = = 5.9028 s 4.45 n 35 that the true mean is greater than 10 pages. b. The p-value =T.DIST.RT(5.9028,34) ≈ 0 so we would reject the null hypothesis. 187
ASBE 6e Solutions for Instructors
Learning Objective: 09-7 9.83
a. H0: ≥ 2.268 grams vs. H1: < 2.268 grams. tcrit =T.INV(.05,34) = −1.761. Reject H0 if tcalc < −1.761. < −1.761 so reject the null x - m0 2.256 - 2.268 tcalc = = = -1.79 s .026 n 15 hypothesis and conclude that the true mean is less than 2.268 grams. b. With use, the metal could erode slightly so that the average weight is less than the newly minted dimes. Learning Objective: 09-7
9.84
a. H0: ≤ .50 vs. H1: > .50. zcrit =NORM.S.INV(.1) = 1.282. Reject H0 if zcalc > 1.282. so reject the null hypothesis and conclude that p -p0 .6333 - .5 zcalc = = = 2.07 p 0 (1 - p 0 ) .5(.5) 60 n the true proportion is greater than .5. b. Using Appendix C: p-value = 1–.9808=.0192. Using Excel: =1– NORM.S.DIST(2.07,1) = .0192 < .10 so we would reject the null hypothesis. The coin does seem to be biased toward heads. Learning Objective: 09-8
9.85
a. H0: ≤ .10 vs. H1: > .10. zcrit =NORM.S.INV(.95) = 1.645. Reject H0 if zcalc > 1.645. > 1.645 so reject the null hypothesis and conclude p -p0 .16 - .10 zcalc = = = 2.00 p 0 (1 - p 0 ) .1(.9) 100 n that the true proportion is greater than .1. b. Yes, if were less than .0228, our decision would be different. c. p-value =1-NORM.S.DIST(2.0,1) = .0228. Conclude that more than 10% of all onedollar bills have something extra written on them. Learning Objective: 09-8
188
ASBE 6e Solutions for Instructors 9.86
a. H0: ≤ .25 vs. H1: > .25. zcrit =NORM.S.INV(.95) = 1.645. Reject H0 if zcalc > 1.645. < 1.645 so fail to reject the null hypothesis. p -p0 .31 - .25 zcalc = = = 1.3856 p 0 (1 - p 0 ) .25(.75) 100 n There is no significant evidence that it has risen. b. This is not a close decision. c. We assume a normal distribution on the sample statistic, p. This makes sense because both n > 10 and n(1−) > 10. Learning Objective: 09-8 Learning Objective: 09-9
9.87
a. H0: ≤ .05 vs. H1: > .05. zcrit =NORM.S.INV(.975) = 1.96. Reject H0 if zcalc > 1.96. < 1.96 so we fail to reject the null hypothesis at p -p0 .08 - .05 zcalc = = = 1.95 p 0 (1 - p 0 ) .05(.95) 200 n the .025 level of significance. The standard is not being violated. b. p-value =1-NORM.S.DIST(1.95,1) = .0258. .0258 > .025 therefore fail to reject the null hypothesis. This decision is very close. Learning Objective: 09-8
9.88
a. H0: ≤ 30 years vs. H1: > 30 years. tcrit =T.INV(.95,11) = 1.796. Reject H0 if tcalc > 1.796. < 1.796 so do not reject the null hypothesis and we 30.1667 - 30 tcalc = = .0983 5.8750 12 cannot conclude that the true mean age is greater than 30 years. b. There is very little difference. Even if we had found statistical significance, it is questionable whether the difference would be "practically important" c. Using Excel: and p-value =T.DIST.RT(.0983,11) = .4617 which is much greater than . 05 so the result is not statistically significant. Learning Objective: 09-7
9.89
a. H0: ≤ .10 vs. H1: > .10. zcrit =NORM.S.INV(.95) = 1.645. Reject H0 if zcalc > 1.645. < 1.645 so fail to reject the null hypothesis. We p -p0 .125 - .10 zcalc = = = 0.882 p 0 (1 - p 0 ) .1(.9) 112 n 189
ASBE 6e Solutions for Instructors do not have strong evidence to conclude that more than 10% of all flights have contaminated drinking water. b. p-value =1-NORM.S.DIST(0.882,1) = .1889. Learning Objective: 09-8 9.90
H0: ≤ .95 vs. H1: > .95. zcrit =NORM.S.INV(.975) = 1.96. Reject H0 if zcalc > 1.96. > 1.96 so reject the null hypothesis and p -p0 .97 - .95 zcalc = = = 2.0519 p 0 (1 - p 0 ) .95(.05) 500 n conclude that the true proportion is greater than .95. The company is exceeding its goal. Learning Objective: 09-8
9.91
a. H0: ≥ .50 vs. H1: < .50. zcrit =NORM.S.INV(.05) = −1.645. Reject H0 if zcalc < −1.645. < −1.645so reject the null hypothesis p -p0 .4572 - .50 zcalc = = = -2.07 p 0 (1 - p 0 ) .5(.5) 584 n and conclude that the true proportion is less than .5. b. p-value =NORM.S.DIST(-2.07,1) = .0192. .0192 < .05 therefore we would reject the null hypothesis. c. The sample proportion was .46. This is a difference of 4%. This is an important difference. There are thousands of college athletes in the US. Increasing the graduation rate for college athletes is a goal that many universities are striving for today. Learning Objective: 09-8
9.92
a. H0: ≤ $250 vs. H1: > $250. tcrit =T.INV(.95,24) = 1.711. Reject H0 if tcalc > 1.711. < 1.711 so we fail to reject the null hypothesis. x - m0 275.66 - 250 tcalc = = = 1.6426 s 78.11 n 25 It does not appear that the average out of pocket expense is greater than $250. b. The decision is fairly close. Learning Objective: 09-7
190
ASBE 6e Solutions for Instructors 9.93
H0: ≥ 3.5 minutes vs. H1: < 3.5 minutes. tcrit =T.INV(.05,19) = −1.729. Reject H0 if tcalc < −1.729. < −1.729 so we reject the null hypothesis, x - m0 3.2 - 3.5 tcalc = = = -3.35 s 0.4 n 20 the measures were effective. p-value =T.DIST(-3.35,19,1) = .0017. Learning Objective: 09-7
9.94
H0: ≤ 5 days vs. H1: > 5 days. tcrit =T.INV(.95,11) = 1.796. Reject H0 if tcalc > 1.796. < 1.796 so we fail to reject the null hypothesis. It x - m0 5.0833 - 5 tcalc = = = .0962 s 2.9987 n 12 does not appear that the average repair time is longer than 5 days so the goal is being met. p-value =T.DIST.RT(.0962,19) = .4622. Learning Objective: 09-7
9.95
a. H0: ≤.67 vs. H1: > .67. zcrit =NORM.S.INV(.95) = 1.645. Reject H0 if zcalc > 1.645. > 1.645 therefore reject the null hypothesis. The p -p0 .8 - .67 zcalc = = = 1.95 p 0 (1 - p 0 ) .67(.33) 50 n proportion has increased. p-value =1-NORM.S.DIST(1.95,1) = .0253. b. The normality condition has been met. Learning Objective: 09-8 Learning Objective: 09-9
9.96
H0: ≤ 6 vs. H1: > 6. tcrit =T.INV(.95,11) = 1.796. Reject H0 if tcalc > 1.796. > 1.796 so reject the null hypothesis. The mean x - m0 6.7283 - 6 tcalc = = = 2.3408 s 1.0778 n 12 furfuryl ether content exceeds the taste threshold. p-value =T.DIST.RT(2.3408,11) = . 0196. Learning Objective: 09-7
191
ASBE 6e Solutions for Instructors 9.97
a. The p-value is .042. A sample proportion as extreme would occur by chance about 42 times in 1,000 samples if in fact the null hypothesis were true. This is fairly convincing evidence that the drug is effective. b. A p-value of .087 is approximately twice .042. This sample is less convincing of the effectiveness of the drug. Learning Objective: 09-6
9.98
The p-value tells us the chance of making this particular sample observation if in fact the null hypothesis is true. A small p-value says that there is a very small chance of making this sample observation assuming the null hypothesis is true therefore our assumption about the null hypothesis must be false. Learning Objective: 09-6
9.99
a. b. c.
vs. H 0 : m ³ 880 H1 : m < 880 tcrit =T.INV(.05,8) = -1.86. Reject H0 if tcalc < -1.86.
x - m0 871 - 880 tcalc = = = -1.731 s 15.6 n 9
Because -1.731 > -1.86, do not reject H0.
Learning Objective: 09-7
9.100
H 0 : m £ 100
vs.
H1 : m > 100
tcrit =T.INV(.95,7) = 1.895. Reject H0 if tcalc > 1.895.
x - m0 106 - 100 tcalc = = = 2.357 s 7.2 n 8 true.
Because 2.357 > 1.895, reject H0. Thetis’s claim is not
Learning Objective: 09-7 9.101
a. P(X ≥ 3 | n = 100, = .01) =1-BINOM.DIST(2,100,.01,1) = .0794. Because .0794 > . 025 we fail to reject the null hypothesis. b. The p-value is .0794. This sample does not contradict the automaker’s claim. Learning Objective: 09-8
192
ASBE 6e Solutions for Instructors 9.102
a. P(X ≥ 2 | n = 36, = .02) =1-BINOM.DIST(1,36,.02,1) = .1618 Because .1618 > .10 we fail to reject the null hypothesis. b. p-value = .1618. This sample does not show that the standard is exceeded. Learning Objective: 09-8
9.103
H0: ≤ .50 vs. H1: > .50. Let n = 16 and x = 10. Find P(X ≥ 10 | n = 16, = .5) =1BINOM.DIST(9,16,.5,1) = .2272. Because .2272 > .1, we cannot conclude that more than 50% feel better with the experimental medication. Learning Objective: 09-8
9.104
H0: ≥ .10 vs. H1: < .10. P(X = 0 | n = 31, = .1) = .0382 (Using Excel: =BINOMDIST(0,31,0.1,TRUE)). Because .0382 < .10, we can reject the null hypothesis. It appears that the on-time percentage has fallen. Learning Objective: 09-9
9.105
a. From MINITAB: 95% confidence interval is (0, .0125). b. A binomial distribution should be used because there were no patients in the sample who actually received a stent. c. Yes, this sample shows that the proportion of patients who experience restenosis is less than 5%. Note that if we used the proportion of .05 to evaluate the assumption of normality of p that we would meet the criteria. nπ = 11.9 and n(1−π) = 226.1. Learning Objective: 09-8
9.106
a. H0: µ ≥ 400 vs. H1: µ < 400 Power = P( x < xcritical | m = 380) For µ = 380: Step 1:
, H0: µ ≥ 400 vs. H1: µ < 400β = 1 Power
æ 20 ö xcritical = 400 - 1.645 ç ÷ = 391.775 è 16 ø
Step 2:
z= Step 3:
391.775 - 380 = 2.355 20 16
Power = P( Z < 2.355) = .99074
and β = 1.99074 = .00926
b. To construct the Power Curve use LearningStats worksheet 09-08 PowerCurvesDIY.xls:
193
ASBE 6e Solutions for Instructors
X Value 380 385 390 395 400
zLeft BetaLeft 2.355 1.355 0.355 -0.645 -1.645
Learning Objective: 09-10 9.107
H0: µ ≥ 500 vs. H1: µ < 500 Power = P( x < xcritical | m = 495) For µ = 495: Step 1:
, β = 1 Power
æ 12 ö xcritical = 500 - 1.645 ç ÷ = 490.13 è 4ø
Step 2: z=
490.13 - 495 = -0.8117 12 4
Step 3:
Power = P( Z < -0.8117) = NORM.S.DIST(-0.8117,1) =.2085 =.7915 For µ = 490: Step 1: æ 12 ö xcritical = 500 - 1.645 ç ÷ = 490.13 è 4ø
194
and β = 1.2085
ASBE 6e Solutions for Instructors Step 2: z= Step 3:
490.13 - 490 = 0.0217 12 4
Power = P(Z < 0.0217) = NORM.S.DIST(.0217,1) =.5086
.4914. For µ = 485: Step 1:
Step 2: z= Step 3:
Step 2: z= Step 3:
æ 12 ö xcritical = 500 - 1.645 ç ÷ = 490.13 è 4ø
490.13 - 485 = 0.855 12 4
Power = P( Z < .855) = NORM.S.DIST(.855,1) =.8037
1963. For µ = 480: Step 1:
and β = 1.5086 =
and β = 1.8037 =
.
æ 12 ö xcritical = 500 - 1.645 ç ÷ = 490.13 è 4ø 490.13 - 480 = 1.688 12 4
Power = P( Z < 1.688) = NORM.S.DIST(1.688,1) =.9543
and β = 1.9543 =
0457. To construct the Power Curve use LearningStats worksheet 09-08 PowerCurvesDIY.xls. The second curve was constructed using n =16. The Power and β values are also shown from Learning Stats.
195
.
ASBE 6e Solutions for Instructors
X Value 480 485 490 495
X Value 480 485 490 495
zLeft BetaLeft 1.688 0.855 0.022 -0.812
BetaLeft
Learning Objective: 09-10 9.108
H0: 2 = 64 vs. H1: 2 ≠ 64. Reject the null hypothesis if 2 > 39.36 (=CHISQ.INV(.975,24)) or 2 < 12.40 (=CHISQ.INV(.025,24)). which falls between the critical values 2 ( n 1) s (25 1)65.8267 c 2calc = = = 24.685 s 02 64 therefore we fail to reject H0 and our sample is consistent with the hypothesis that the true variance is 64. Learning Objective: 09-11
9.109
a. H0: ≤ 106 vs. H1: > 106. tcrit =T.INV(.9975,23) = Reject the null hypothesis if tcalc > 2.807. > 2.807 so reject the null hypothesis. x - m0 106.997 - 106 tcalc = = = 130.95 s 0.0373 n 24 The mean brightness is significantly greater than 106. 196
ASBE 6e Solutions for Instructors b. H0: 2 ≥ .0025 vs. H1: 2 < .0025. Reject the null hypothesis if 2 < 9.26 (=CHISQ.INV(.005,23)) . therefore we 2 2 (n - 1) s (24 - 1).0373 c 2calc = = = 12.8 2 s0 .0025 would fail to reject the null hypothesis. This sample does not provide evidence that the variance is less than .0025. Learning Objective: 09-11
197
ASBE 6e Solutions for Instructors
Chapter 10 Two-Sample Hypothesis Tests 10.1
Use the following formulas for each test in (a) – (c), substituting the appropriate values. Check work with LearningStats file 10-03 Calculator - Two Means. You can also use MegaStat to solve. and d.f. = n1 + n2 – 2. 2 2 x1 - x2 ( n 1) s + ( n 1) s 1 2 2 tcalc = where s p 2 = 1 2 2 n1 + n2 - 2 sp s + p n1 n2 a. H0: µ1µ2 ≥ 0 vs. H0: µ1µ2 < 0, d.f. = 15+15−2 = 28, tcrit =T.INV(.025,28) = 2.0484. Reject H0 if tcalc < 2.0484. 3.05 - 3.25
(15 - 1)(0.20 2 ) + (15 - 1)(0.30 2 ) tcalc = = - 2.1483, where s p 2 = = 0.065 15 + 15 - 2 0.065 0.065 + 15 15 −2.1483 < 2.0484 therefore reject the null hypothesis and conclude there are differences in GPA for juniors and seniors. The p-value = .0202 using Excel formula: =T.DIST(-2.1483,28,1).
.
b. H0: µ1µ2 = 0 vs. H0: µ1µ2 ≠ 0, d.f. = 22+19−2 = 39, tcrit =T.INV.2T(.05,39) = 2.0227. Reject H0 if tcalc < 2.0227 or tcalc > +2.0227. 15 - 18 (22 - 1)(52 ) + (19 - 1)(7 2 ) 2 tcalc = = - 1.5948, where s p = = 36.077 22 + 19 - 2 36.077 36.077 + 22 19 1.5948 > 2.0227 and 1.5948 < 2.0227 therefore fail to reject the null hypothesis and conclude there is no difference in the commute miles for the two community colleges. The p-value = .1188 using Excel formula: =T.DIST.2T(1.5948,39). c. H0: µ1µ2 ≤ 0 vs. H0: µ1µ2 > 0, d.f. = 12+17−2 = 27, tcrit =T.INV(.95,27) = 1.7033. Reject H0 if tcalc > 1.7033. 139 - 137 (12 - 1)(2.82 ) + (17 - 1)(2.7 2 ) tcalc = =1.9351, where s p 2 = = 7.514 12 + 17 - 2 7.514 7.514 + 12 17 1.9351 > 1.7033 therefore reject the null hypothesis. The p-value = .0318 using Excel formula: =T.DIST.RT(1.9351,27). Learning Objective: 10-1 Learning Objective: 10-2
195
ASBE 6e Solutions for Instructors
10.2
Use the following formulas for each test in (a) – (c), substituting the appropriate values. Check work with LearningStats file 10-03 Calculator - Two Means. You can also use MegaStat to solve.
tcalc =
x1 - x2 s12 s22 + n1 n2
with Welch's d.f.
[ s12 / n1 + s22 / n2 ]2 = 2 ( s1 / n1 ) 2 ( s22 / n2 ) 2 + n1 - 1 n2 - 1
a. H0: µ1µ2 ≥ 0 vs. H0: µ1µ2 < 0, d.f. =
[.04 / 15 + .09 /15] = (.04 / 15) 2 (.09 /15)2 + 15 - 1 15 - 1 tcrit =T.INV(.025,24) = 2.0639. Reject H0 if tcalc < 2.0639. 2
=
24,
< 2.0639 therefore reject the null and conclude there are 3.05 - 3.25 tcalc = = -2.1483 .04 .09 + 15 15 differences in GPA for juniors and seniors. Using Excel the p-value =T.DIST(-2.1436,24,1) = = .0212 < .025. b. H0: µ1µ2 = 0 vs. H0: µ1µ2 ≠ 0, d.f. =
,
[25 / 22 + 49 /19] = 32 (25 / 22) 2 (49 / 19) 2 + 22 - 1 19 - 1 tcrit =T.INV.2T(.05,32) = 2.0369. Reject H0 if tcalc < 2.0369 or tcalc > +2.0369. which is between the critical values therefore do not reject 15 - 18 tcalc = = -1.5564 25 49 + 22 19 the null and conclude there is no difference in the commute miles for the two community colleges. Using Excel for p-value: =T.DIST.2T(1.5564,32,2) and p-value = .1294 > .05. =
c. H0: µ1µ2 ≤ 0 vs. H0: µ1µ2 > 0, d.f. =
2
[7.84 /12 + 7.29 /17]2 = = 23 (7.84 /12) 2 (7.29 /17) 2 + 12 - 1 17 - 1 tcrit =T.INV(.95,23) = 1.7139. Reject H0 if tcalc > 1.7139.
196
,
ASBE 6e Solutions for Instructors > 1.7139 therefore reject the null and conclude there is a 139 - 137 tcalc = = 1.9226 7.84 7.29 + 12 17 difference in credits.. Using Excel for p-value: =T.DIST.RT(1.9226,23) and p-value = .0335 < .05. Learning Objective: 10-1 Learning Objective: 10-2 10.3
Use the following formulas assuming unequal variances, substituting the appropriate values. Check work with LearningStats file 10-03 Calculator - Two Means. You can also use MegaStat to solve.
tcalc =
x1 - x2 s12 s22 + n1 n2
with Welch's d.f. =
[ s12 / n1 + s22 / n2 ]2 ( s12 / n1 )2 ( s22 / n2 ) 2 + n1 - 1 n2 - 1
a. H0: 12 = 0 vs. H1: 12 ≠ 0. Using the quick rule d.f. = min(n1 −1, n2 − 1) = min(190, 832) = 190, tcrit =T.INV.2T(.01,190) = 2.602. Reject H0 if tcalc <2.602 or if tcalc > +2.602. (If using Welch’s formula d.f. = 423 and tcrit = 2.588) < 2.602 therefore reject H0 and conclude that the 4.9 - 7.9 tcalc = = -6.184 5.42 8.32 + 191 833 average age is different for the two groups of employees. b. Using d.f. = 190, p-value = 3.713E09 < .01. If using Welch’s d.f. = 423, p-value = 1.48E09 < .01. Learning Objective: 10-1 Learning Objective: 10-2 10.4
a. H0:
vs. H1: , d.f.= 10+10−2 = 18. tcrit =T.INV(.99,18) = mtoy - m ford 0 mtoy - m ford > 0 2.552. Reject the null hypothesis if tcalc > 2.552. and so we reject 45.5 - 42 (10 - 1)3.24 + (10 - 1)5.29 2 tcalc = = 3.7895 sp = = 4.265 10 + 10 - 2 4.265 4.265 + 10 10 the null hypothesis. The average MPG is significantly lower for the Ford Fusion. Using Megastat: >Hypothesis Testing>Compare Two Independent Groups. b. Using Excel: =T.DIST.RT(3.7895,18) and the p-value = .0007. Learning Objective: 10-1 Learning Objective: 10-2
197
ASBE 6e Solutions for Instructors 10.5
a. H0: µ1µ2 ≤ 0 vs. H0: µ1µ2 > 0. d.f.= 17+14−2 = 29. tcrit =T.INV(.99,29) = 2.462. Reject the null hypothesis if tcalc > 2.462. and 2 2 30.47 - 21.62 (17 - 1)15.1 + (14 - 1)9.5 tcalc = = 1.902 s 2p = = 166.256 166.256 166.256 17 + 14 - 2 + 17 14 1.902 < 2.462 therefore we fail to reject the null hypothesis. The average amount of purchases when the music is slow is not less than when the music is fast. b. The p-value =T.DIST.RT(1.902,29) = .0336. Learning Objective: 10-1 Learning Objective: 10-2
10.6
Assume
is the average of the 1960 shoe sizes and is the average of the 1980 shoe m1 m2 sizes. H0: 1 µ2 ≥0 vs. H1: 1 µ, d.f. = 12 + 12 2 = 22, tcrit =T.INV(.025,22) = -
2.074 . Reject the null hypothesis if tcalc < 2.074. and < 2.074 7.7083 8.2083 (12 1).3845 + (12 1).2027 tcalc = = -2.2604 s 2p = = .2936 12 + 12 - 2 .2936 .2936 + 12 12 therefore reject the null hypothesis and conclude that the average shoe size appears to have increased. Using Megastat: >Hypothesis Testing>Compare Two Independent Groups. Learning Objective: 10-1 Learning Objective: 10-2 10.7
H0: 12 = 0 vs. H1: 12 ≠ 0, Using the quick rule d.f. = min(n1 −1, n2 − 1) = min(11, 11) = 11, t.005 =T.INV(.005,11) = 3.106. Reject the null hypothesis if tcalc< −3.106 or tcalc > 3.106. . (Because the sample sizes are the same Welch’s gives the same degrees of freedom. ) < −3.106 therefore we reject the null hypothesis, there is a 9.4 - 12.7 tcalc = = -3.55 3.22 .352 + 12 12 significant difference in caffeine content between these two beverages. Learning Objective: 10-1 Learning Objective: 10-2
10.8
Assume
is the average temperature at Panera and is the average temperature at m1 m2 Bruegger’s. H0: 1 µ2 ≤0 vs. H1: 1 µ, d.f. = 12 + 12 2 = 22, tcrit -
198
ASBE 6e Solutions for Instructors =T.INV(.99,22) = 2.508. Reject the null hypothesis if tcalc > 2.508. and 2 2 170.33 - 161.08 (12 1)5.03 + (12 1)5.57 tcalc = = 4.270 s 2p = = 28.163 28.163 28.163 12 + 12 - 2 + 12 12 4.270 > 2.508 therefore reject the null hypothesis and conclude that the average temperature is higher at Panera. Using Megastat: >Hypothesis Testing>Compare Two Independent Groups. Learning Objective: 10-1 Learning Objective: 10-2 10.9
Assume
is the average monthly number of fast food restaurant visits for seniors and is m1 m2 the average number for freshmen. H0: 1 µ2 ≥0 vs. H1: 1 µ, d.f. = 11 + 11 2 -
= 20, tcrit =T.INV(.05,20) = −1.725. Reject the null hypothesis if tcalc < −1.725. and 2 2 10.64 - 14.09 (11 - 1)4.92 + (11 - 1)3.96 tcalc = = -1.812 s 2p = = 19.944 19.944 19.944 11 + 11 - 2 + 11 11 −1.812 < −1.725 therefore reject the null hypothesis and conclude that seniors eat at fast food restaurants, on average, fewer times than freshmen each month. Using Megastat: >Hypothesis Testing>Compare Two Independent Groups. Learning Objective: 10-1 Learning Objective: 10-2 10.10
a. Assuming equal variances: d.f. =
( n1 - 1) + (n2 - 1) = (12 - 1) + (9 - 1) = 19
and t.05
=T.INV(.05,19) = −1.729 . . The interval is (12 - 1)484, 416 + (9 - 1)702, 244 1 1 (1,101 - 1, 766) ± 1.729 + 12 + 9 - 2 12 9 . The interval does not include zero so we conclude there are ( -1243.745, -86.255) significant differences in the means. Using Megastat: <Hypothesis Tests<Compare Two Independent Groups and check "pooled variance". b. Assuming unequal variances: Using the quick rule d.f. = min(12 1, 9 1) = 8 and t.05 =T.INV(.05,8) = −1.860 . . The interval is ( 484, 416 702, 244 (1,101 - 1,766) ± 1.860 + 12 9 . This interval does not include 0 so we conclude there is a -$1305.00, -$25.00) significant difference in means. If we were to use Welch’s formula d.f. = 15 and t.05 =T.INV(.05,15) = −1.7531. This would result in a narrower interval (−$1268.20, − $61.80) and the same conclusion. Using Megastat: <Hypothesis Tests<Compare Two Independent Groups and check "unequal variances". 199
ASBE 6e Solutions for Instructors c. The assumption about variances did not change the conclusion. d. Test Group: or ($740.15, $1,461.85) Control Group: 696 1101 ± 1.796 = 1101 ± 360.85 12 or ($1,246.64, $2,285.56) Yes, the confidence 838 1766 ± 1.860 = 1766 ± 519.56 9 intervals overlap. Learning Objective: 10-3 10.11
a. Assuming equal variances: d.f. =
(n1 - 1) + ( n2 - 1) = (22 - 1) + (17 - 1) = 37
and t.05
=T.INV(.05,37) = −1.687.
. (22 - 1)1.882 + (17 - 1)1.7 2 1 1 (8.64 - 8.82) ± 1.687 + 22 + 17 - 2 22 17 The interval is (−1.163, 0.830). The interval does include zero so we conclude there is no significant difference in the means. Using Megastat: <Hypothesis Tests<Compare Two Independent Groups and check "pooled variance". b. Assuming unequal variances: Using the quick rule d.f. = min(22 1, 17 1) = 16 and t.05 =T.INV(.05,16) = −1.746 . . The interval is 1.882 1.702 (8.64 - 8.82) ± 1.746 + 22 17 (−1.184, 0.824). The interval does include zero so we conclude there is no significant difference in the means. Using Megastat: <Hypothesis Tests<Compare Two Independent Groups and check "unequal variances". If we were to use Welch’s formula d.f. = 36 and t.05 =T.INV(.05,36) = −1.688. This would result in a narrower interval and the same conclusion: (−1.151, 0.791). c. The intervals are similar. Assumptions about variances did not make a big difference. d. Test Group: or (7.95, 9.33) Control Group: 1.88 8.64 ± 1.721 = 8.64 ± 0.690 22 or (8.1, 9.54) Yes, the confidence intervals overlap. 1.70 8.82 ± 1.746 = 8.82 ± 0.720 17 Learning Objective: 10-3 10.12
Assuming unequal variances: d.f. = min(10 1, 12 1) = 9 using the quick rule, t.025 =T.INV(.025,9) = 2.262 . The interval is 9093.5296 8136.04 (795 - 894.17) ± 2.262 + 10 12 (-191.0433, -7.2901). This interval does not include zero so we conclude there is a significant difference in means for undergraduate and graduate rent. It appears that undergraduates pay less on average than graduate students. Using Megastat: the
200
ASBE 6e Solutions for Instructors interval with Welch’s d.f. = 18 is
<Hypothesis Tests<Compare Two (-182.872, -15.461) Independent Groups and check "unequal variances" and display confidence interval at 95%. Assuming equal variances: d.f. = min(10 + 12 − 2) = 20, t.025 =T.INV(.025,20) = . The interval is ( (10 - 1)9093.5296 + (12 - 1)8136.04 1 1 (795 - 894.17) ± 2.086 + 10 + 12 - 2 10 12 . This interval does not include zero so we conclude there is a -181.845, -16.495) significant difference in means for undergraduate and graduate rent. It appears that undergraduates pay less on average than graduate students. Using Megastat: <Hypothesis Tests<Compare Two Independent Groups and check "equal variances" and display confidence interval at 95%. Learning Objective: 10-3 10.13
a. Define the difference as daughter’s height–mother’s height. H0: d ≤ 0 vs. H0: d > 0. tcrit =T.INV(.95,6) = 1.943 Reject the null hypothesis if tcalc > 1.943. , < 1.943 so we fail to reject the d - md 5.857 - 0 d = 5.857 sd = 8.03 tcalc = = = 1.93 sd 8.03 7 n null hypothesis. There is not a significant difference in height between mothers and daughters. Using Megastat: >Hypothesis tests>Paired Observations b. The decision is close. The p-value is .0509 (=T.DIST.RT(1.93,6) )which is slightly greater than .05. c. A daughter’s height is affected by her father’s height as well as her grandparents. Nutrition also plays a role in a person’s development. Learning Objective: 10-4
10.14
a. Define the difference as Old – New. H0: d ≤ 0 versus H0: d > 0. tcrit =T.INV(.95,4) = 2.132. Reject the null hypothesis if tcalc > 2.132 (d.f. = 5 – 1 = 4). , so . We reject the null and s = 5.4129 d = 6.4 d d - md 6.4 - 0 tcalc = = = 2.6439 sd 5.4129 n 5 conclude the new method shows a faster average. Using Megastat: >Hypothesis tests>Paired Observations b. The decision is not close. The p-value is .0287 (=T.DIST.RT(2.6439,4) ) which is less than .05 and the tcalc was well within the rejection region. Learning Objective: 10-4
10.15
a. Define the difference as rentals with new price – rentals with old price. H0: d ≤ 0 versus H0: d > 0. tcrit =T.INV(.90,9) = 1.383. Reject the null hypothesis if tcalc > 1.383.
201
ASBE 6e Solutions for Instructors ,
d = 2.0 sd = 2.211
so
so we reject the null hypothesis.
d - md 2-0 = = 2.86 sd 2.211 10 n The number of rentals under the new price is greater.. b. The decision is not close. The p-value is .0094 ( =T.INV(2.86,9)) which is much less than .10. c. Yes, this is an important difference. The sample showed an average increase of 2 rentals per month which is a 20% increase. This means more revenue for the store. Learning Objective: 10-4 tcalc =
10.16
a. Define the difference as New car mpg – Old car mpg. H0: d ≤ 5 versus H0: d > 5. tcrit =T.INV(.95,13) = 1.771. Reject the null hypothesis if tcalc > 1.771. , so > 1.771 therefore we d - md 6.5071 - 5 d = 6.5071 sd = 3.0254 tcalc = = = 1.864 sd 3.0254 14 n reject the null hypothesis. The increase in efficiency exceeds 5 mpg. Learning Objective: 10-4
10.17
a. Define the difference as Seed time – Prelim time. H0: d ≤ 0.5 sec versus H0: d > 0.5 sec. tcrit =T.INV(.95,25) = 1.708. Reject the null hypothesis if tcalc > 1.708. , so > 1.708 therefore we reject s = 2.1735 d - md 1.276 - 0.5 d = 1.276 d tcalc = = = 1.82 sd 2.1735 26 n the null hypothesis. The difference between seed times and prelim times is greater than ½ second. Learning Objective: 10-4
10.18
a. Define the difference as daughter – mother. H0: d ≤ 0 versus H0: d > 0, tcrit =T.INV(.99,11) = 2.718. Reject the null hypothesis if tcalc > 2.718 . , d = .7917 so . We reject the null and conclude shoe sd = .8649 d - md .7917 - 0 tcalc = = = 3.1706 sd .8649 n 12 sizes have increased because the daughter sizes are significantly larger than the mother sizes. Using Megastat: >Hypothesis tests>Paired Observations b. The decision is not close. The p-value is .0045 (=T.DIST.RT(3.1706,11)) which is less than .01 and the tcalc was well within the rejection region. c. The statistics are convincing because we see evidence of an increase and it was not a close decision. However, we should be aware that the sample showed a difference of less than a whole shoe size. d. In general, adults are showing a trend of increasing size. 202
ASBE 6e Solutions for Instructors Learning Objective: 10-4 10.19
Define the difference as Entry − Exit. H0: d = 0 versus H0: d ≠ 0, t.005 =T.INV(.005,7) = −3.499. Reject the null hypothesis if tcalc > 3.499 or tcalc < −3.499. , so > −3.499 therefore we fail s = 4.957 d - md . - 3.0 - 0 d = -3.0 d tcalc = = = -1.712 sd 4.957 8 n to reject the null hypothesis. There is no difference between the number of entry failures and exit failures. The decision is not close. The p-value is .1306 (=T.DIST.2T(1.712,7))which is much greater than .01. Learning Objective: 10-4
10.20
The following formula and Excel function are used to calculate zcalc and find the p-value: x +x p1 - p2 pc = 1 2 , zcalc = n1 + n2 æ1 1 ö p (1 - p ) ç + ÷ è n1 n2 ø p-value =NORM.S.DIST(zcalc,1) a. Right-tailed test: 228 + 703 .95 - .925 pc = , zcalc = = 1.332 240 + 760 1 ö æ 1 .931(1 - .931) ç + ÷ è 240 760 ø p-value =1-NORM.S.DIST(1.332,1) = .0914 < .10. Reject H0. b. Left-tailed test: 36 + 66 .45 - .55 pc = , zcalc = = -1.386 80 + 120 1 ö æ 1 .51(1 - .51) ç + ÷ è 80 120 ø p-value =NORM.S.DIST(-1.386,1) = .0829 > .05. Fail to reject H0. c. Two-tailed test: 52 + 56 .65 - .80 pc = , zcalc = = -2.041 80 + 70 1 ö æ 1 .72(1 - .72) ç + ÷ è 80 70 ø p-value =2*NORM.S.DIST(-2.041,1) = .0413 < .05. Reject H0. Learning Objective: 10-5
10.21
The following formula and Excel function are used to calculate zcalc and find the p-value:
203
ASBE 6e Solutions for Instructors
pc =
x1 + x2 , zcalc = n1 + n2
p1 - p2
æ1 1 ö p (1 - p ) ç + ÷ è n1 n2 ø p-value =NORM.S.DIST(zcalc,1) a. Left-tailed test: 28 + 14 pc = , zcalc = 336 + 112
.0833 - .125
= -1.311 1 ö æ 1 .0938(1 - .0938) ç + ÷ è 336 112 ø p-value =NORM.S.DIST(-1.311,1) = .0949 < .10. Reject H0. b. Right-tailed test: 276 + 440 .92 - .88 pc = , zcalc = = 1.787 300 + 500 1 ö æ 1 .895(1 - .895) ç + ÷ è 300 500 ø p-value =1-NORM.S.DIST(1.787,1) = .0370 < .05. Reject H0. c. Two-tailed test: 35 + 42 .70 - .56 pc = , zcalc = = 1.577 50 + 75 1 ö æ 1 .616(1 - .616) ç + ÷ è 50 75 ø p-value =2*(1-NORM.S.DIST(1.577,1)) = .1148 > .10. Fail to reject H0. Learning Objective: 10-5 10.22
For each problem, the following formulas are used: x +x p1 - p2 pc = 1 2 , zcalc = n1 + n2 æ1 1ö p (1 - p ) ç + ÷ è n1 n2 ø a.
vs. . Reject if or H 0 : p1 - p 2 = 0 H1 : p 1 - p 2 ¹ 0 zcalc < -1.96 zcalc > 1.96 .4 - .3 40 + 30 = 1.4825 pc = = .35 zcalc = 100 + 100 1 ù é 1 .35(1 - .35) ê + ë100 100 úû Fail to reject the null and conclude there is no difference in the proportion of dissatisfied workers. The decision is not close and the p-value is .1382. Using Megastat: >Hypothesis Testing>Compare Two Independent Proportions. Using Excel to get the p-value: =2*(1-NORM.S.DIST(1.4825,1)). Using Appendix C-2: (2)(.0694) = .1388. The p-value is not less than .05 so we do not reject the null. b.
H0: π1 π2 ≥ 0 vs. H0: π1 π2 < 0. Reject if zcalc < 2.3263. .12 - .24 24 + 12 = -2.1617 pc = = .144 zcalc = 200 + 50 1ù é 1 .144(1 - .144) ê + ë 200 50 úû 204
ASBE 6e Solutions for Instructors Fail to reject the null and conclude the proportion of rooms rented at the second hotel is not less than the proportion rented at the first hotel. The decision is somewhat close because the p-value is .0153. Using Megastat: >Hypothesis Testing>Compare Two Independent Proportions. Using Excel to get the p-value: =NORM.S.DIST(-2.1617,1). Using Appendix C-2: .0154. The p-value is not less than .01 so do not reject the null. c. vs. . Reject if H 0 : p1 - p 2 £ 0 H1 : p 1 - p 2 > 0 zcalc > 1.645 .075 - .05 36 + 26 = 1.6379 pc = = .062 zcalc = 480 + 520 1 ù é 1 .062(1 - .062) ê + ë 480 520 úû Fail to reject the null and conclude home equity default rates are not greater for the second bank. Using Megastat: >Hypothesis Testing>Compare Two Independent Proportions. The decision is very close because the p-value is .0507. Using Excel to get the p-value: =1-NORM.S.DIST(1.6379,1). Using Appendix C-2: 1-.9495 = .0505. The p-value is not less than .05 so do not reject the null but it is very close. Learning Objective: 10-5 10.23
For each problem, the following formulas were used: x +x p1 - p2 pc = 1 2 , zcalc = n1 + n2 æ1 1ö p(1 - p) ç + ÷ è n1 n2 ø a.
vs. . Reject if H 0 : p1 - p 2 ³ 0 H1 : p 1 - p 2 < 0 zcalc < -2.33 .30 - .54 .3(50) + .54(50) = -2.43 pc = = .42 zcalc = 50 + 50 1ù é1 .42(1 - .42) ê + ú ë 50 50 û −2.43 < −2.33 therefore reject the null and conclude there is a greater proportion of repeat buyers at the second dealership. Using Megastat: >Hypothesis Testing>Compare Two Independent Proportions. Using Excel to get the p-value: =NORM.S.DIST(2.43,1) = .0076. b. vs. . Reject H0 if of if H 0 : p1 - p 2 = 0 H1 : p 1 - p 2 ¹ 0 zcalc < -1.645 zcalc > 1.645 .45 - .25 .45(80) + .25(48) = 2.26 pc = = .375 zcalc = 80 + 48 1ù é1 .375(1 - .375) ê + ú ë 80 48 û 2.26 > 1.645 therefore reject the null and conclude there is a difference in the proportion of honor students between the two sororities. Using Megastat: >Hypothesis Testing>Compare Two Independent Proportions. Using Excel to get the p-value: =2*(1-NORM.S.DIST(2.26,1)) = .0238.
205
ASBE 6e Solutions for Instructors c.
vs. . Reject if H 0 : p1 - p 2 ³ 0 H1 : p 1 - p 2 < 0 zcalc < -1.645 .20 - .32 .2(80) + .32(75) = -1.706 pc = = .258 zcalc = 80 + 75 1ù é1 .258(1 - .258) ê + ú ë 80 75 û −1.706 < −1.645 therefore reject the null and conclude there is a greater proportion of first time Hawaii visitors at the second hotel. Using Megastat: >Hypothesis Testing>Compare Two Independent Proportions. Using Excel to get the p-value: =NORM.S.DIST(-1.706,1) = .0440. Learning Objective: 10-5
10.24
a. Define 1 = proportion of hurricanes in U.S. from 1990-1998. Define 2 = proportion of hurricanes in U.S. from 1999-2006 H0: vs. H1: . This is a left-tailed test. Reject the null hypothesis if p1 - p 2 ³ 0 p1 - p 2 < 0 zcalc < −2.33. .413 - .6429 19 + 45 zcalc = = -2.4354 pc = = .5517 46 + 70 1ù é1 .5517(1 - .5517) ê + ú ë 46 70 û −2.4354 < −2.33. Reject the null and conclude the proportion of hurricanes has increased. Using Megastat: >Hypothesis Testing>Compare Two Independent Proportions. b. Normality is assumed because n1p1 > 10, n2p2 > 10, n1 (1-p1) > 10, and n2 (1-p2) > 10. Learning Objective: 10-5 Learning Objective: 10-6
10.25
a. Define 1 = proportion of shoppers that paid by debit card in 1999. Define 2 = proportion of shoppers that paid by debit card in 2004. H0: vs. H1: . This is a left-tailed test. Reject the null hypothesis if p1 - p 2 ³ 0 p1 - p 2 < 0 z < −2.33. b. > −2.33 so we fail to .21 - .31 42 + 62 = -2.28 pc = = .26 zcalc = 200 + 200 1 ù é 1 .26(1 - .26) ê + ë 200 200 úû reject the null hypothesis (although the decision is close.) The sample does not provide strong enough evidence to conclude that there is a difference in the two proportions. c. p-value =NORM.S.INV(-2.28,1) = .0113 > .01. d. Normality is assumed because n1p1 > 10, n2p2 > 10, n1 (1-p1) > 10, and n2 (1-p2) > 10. Learning Objective: 10-5 Learning Objective: 10-6
10.26
Define 1 = proportion of loyal mayonnaise purchasers and 2 = proportion of loyal soap purchasers. 206
ASBE 6e Solutions for Instructors H0:
vs. H1: . This is a two-tailed test. Reject the null hypothesis if p1 - p 2 = 0 p1 - p 2 ¹ 0 zcalc < −1.96 or zcalc > 1.96. .65 - .53 65 + 53 = 1.7252 pc = = .59 zcalc = 100 + 100 1 ù é 1 .59(1 - .59) ê + ë100 100 úû 1.7252 is between the critical values. Fail to reject the null hypothesis. The sample evidence does not show a significant difference in the proportion of loyal purchasers. Using Megastat: >Hypothesis Testing>Compare Two Independent Proportions. Learning Objective: 10-5 10.27
Define 1 = proportion of respondents in first group (the group given the gift certificate.) Define 2 = proportion of respondents in the second group. H0: vs. H1: . This is a two-tailed test. Reject the null hypothesis if p1 - p 2 = 0 p1 - p 2 ¹ 0 zcalc < −1.96 or zcalc > 1.96. > 1.96 therefore we reject .13 - .09 65 + 45 = 2.02 pc = = .11 zcalc = 500 + 500 1 ù é 1 .11(1 - .11) ê + ë 500 500 úû the null hypothesis. The sample shows a significant difference in response rates. Learning Objective: 10-5
10.28
a. Define 1 = proportion of flights with contaminated water in August and September 2004. Define 2 = proportion of flights with contaminated water in November and December 2004. H0: vs. H1: . Reject the null hypothesis if p1 - p 2 ³ 0 p1 - p 2 < 0 zcalc < −1.645. .1266 - .1716 20 + 29 = -1.1395 pc = = .1498 zcalc = 158 + 169 1 ù é 1 .1498(1 - .1498) ê + ë158 169 úû −1.1395 > −1.645. Fail to reject the null hypothesis. The level of contamination was not lower in the first sample. Using Megastat: >Hypothesis Testing>Compare Two Independent Proportions. b. Using Excel: =NORM.S.DIST(-1.1395,1) and the p-value = .1273. Using Appendix C: p-value = .1271. Not less than .05 so fail to reject the null. c. From the public health perception, importance outweighs significance. Our sample information did not allow us to conclude that the contamination proportion has gone down after sanitation improvements. d. Normality is assumed because n1p1 > 10, n2p2 > 10, n1 (1-p1) > 10, and n2 (1-p2) > 10. Learning Objective: 10-5 Learning Objective: 10-6
207
ASBE 6e Solutions for Instructors 10.29
a. H0: B C = 0 vs. H1: B C ≠ 0. Reject the null hypothesis if zcalc < −1.96 or zcalc > 1.96. which is between the .0778 - .1045 14 + 7 = -0.669 pc = = .085 zcalc = 180 + 67 1ù é 1 .085(1 - .085) ê + ë180 67 úû critical values so we fail to reject the null hypothesis. This sample does not give enough evidence to conclude that the proportions are different. b. Normality should not be assumed because n2p2 < 10. Learning Objective: 10-5 Learning Objective: 10-6
10.30
a. Assume
10.31
a. H0: 1 − 2 ≤ .10 versus H1: 1 − 2 > .10. Reject the null hypothesis if zcalc > 1.645. (.2813 - .1458) - (.10) zcalc = = 0.66 .2813(1 - .2813) .1458(1 - .1458) + 128 96 0.66 < 1.645 therefore we fail to reject the null hypothesis. b. zcalc = 0.66 and the p-value =1-NORM.S.DIST(0.66,1) = .2545 so we fail to reject the null hypothesis. The proportion of calls lasting at least five minutes has not decreased by 10%. Learning Objective: 10-5
10.32
a. H0: 1 − 2 ≤ .20 versus H1: 1 − 2 > .20. Reject the null hypothesis if zcalc > 1.96.
is the proportion in 2012 and is the proportion in 2009. p1 p2 H0: 1 − 2 ≤ .05 versus H1: 1 − 2 > .05. Reject the null hypothesis if zcalc > 1.28. . (.21 - .31) - (.05) zcalc = = 1.1468 .21(1 - .21) .31(1 - .31) + 200 200 1.1468 < 1.28 therefore we fail to reject the null hypothesis. The percentage of shoppers paying by debit card did not increase by 5%. Using Megastat: >Hypothesis Testing>Compare Two Independent Proportions and use ".05" as the "hypothesized difference". b. zcalc = 1.1468 and the p-value =1-NORM.S.DIST(1.1468,1) = .1257 so we fail to reject the null hypothesis. The percentage of shoppers paying by debit card did not increase by 5%. Learning Objective: 10-5
208
ASBE 6e Solutions for Instructors . Using Megastat: >Hypothesis .37 - .09 - .20 zcalc = = 3.1878 .37(1 - .37) .09(1 - .09) + 500 500 Testing>Compare Two Independent Proportions and use ".2" as the "hypothesized difference". b. zcalc = 3.1878 and the p-value =1-NORM.S.DSIT(3.1878,1) = .0007 so we reject the null hypothesis. The response rate did increase by at least 20%. Learning Objective: 10-5 10.33
p1 = 60/120 = .5, p2 = 26/50 = .52. For a 90% confidence level use z.05 = 1.645. or −.02 ± .1384 or (−.1584, .1184). Because .5(1 - .5) .52(1 - .52) (.5 - .52) ± 1.645 + 120 50 zero falls in the interval, we cannot conclude there is a difference in proportions. Learning Objective: 10-7
10.34
p1 = 36/72 = .5, p2 = 12/72 = .1667. For a 95% confidence level use z.025 = 1.96. or .3333± .1441 or (.1892, .4774). Since .5(1 - .5) .1667(1 - .1667) (.5 - .1667) ± 1.96 + 72 72 the interval does not include zero, we can conclude that . Using Megastat: p1 > p 2 >Hypothesis Tests>Compare Two Independent Proportions. Learning Objective: 10-7
10.35
p1 = 71/100 = .71, p2 = 122/200 = .61. For a 90% confidence level use z.05 = 1.645. or .10 ± .0938 or (.0062, .1938). Because .71(1 - .71) .61(1 - .61) (.71 - .61) ± 1.645 + 100 200 zero does not fall in the interval, we can conclude there is a difference in proportion of cigarette smokers who are loyal to their brand versus toothpaste users. Learning Objective: 10-7
Note: in each of the exercises 10.36-10.40, the Excel function =F.INV(α,df1,df2) could be used for finding FL instead of taking the reciprocal of the right tailed F value. 10.36
a.
. df1 = 21, df2 = 15. = ; FR F21,15 » F20,15 = 2.33 H1 : s 12 s 22 ¹ 1 . Reject if Fcalc > 2.33 or F < 0.459. 1 1 H0 FL = = = 0.459 F15,21 2.18
H 0 : s 12 s 22 = 1
vs.
209
ASBE 6e Solutions for Instructors so we reject the null hypothesis and conclude there are 104.04 Fcalc = = 2.54 > 2.33 40.96 unequal variances. b. vs. . df1 = 24, df2 = 17. = . Reject 2 2 2 2 F » F = 2.23 F H0 : s1 s 2 £ 1 H1 : s 1 s 2 > 1 24,17 20,17 R if Fcalc > 2.23. so we do not reject the null hypothesis .7921 H0 Fcalc = = 1.7645 < 2.23 .4489 and conclude variances are not different. c. vs. . df1 = 11, df2 = 9. = . Reject 1 1 FL H 0 : s 12 s 22 ³ 1 H1 : s 12 s 22 < 1 = = .3448 F9,11 2.90 if Fcalc < .3448. so we reject the null hypothesis and 15,376 H0 Fcalc = = .2275 < .3448 67, 600 conclude the first variance is less than the second variance. Learning Objective: 10-8 10.37
a.
vs. . df1 = 10, df2 = 7. Using Excel: FR H 0 : s 12 s 22 = 1 H1 : s 12 s 22 ¹ 1 =F.INV.RT(.025,10,7) = 4.76; . Reject H0 if Fcalc > 4.76 or 1 1 FL = = = 0.253 F7,10 3.95 Fcalc < 0.253. Fcalc =
2
Because 2.54 is between the critical values so we fail
5.1 = 2.54 3. 2 2 to reject the null hypothesis and conclude the variances are not different. b. vs. . df1 = 7, df2 = 7. Using Excel: H 0 : s 12 s 22 ³ 1 H1 : s 12 s 22 < 1 . Reject H0 if Fcalc < .264. (df1= 7, df2 = 7.) Fcalc = 1 1 FL = = = 0.264 F7,7 3.787 < .264 so we fail to reject the null hypothesis. 2212 = 0.247 4452 c.
vs. . df1= 9, df2 = 12. Using Excel: FR H 0 : s 12 s 22 £ 1 H1 : s 12 s 22 > 1 =F.INV.RT(.05,9,12) = 2.80. Reject H0 if Fcalc > 2.80 Fcalc = > 2.8 so we 67 2 = 19.95 152 reject the null hypothesis and conclude that the variances are different. Learning Objective: 10-8 210
ASBE 6e Solutions for Instructors
10.38
a. H0: 1
≥ 0 vs. H1: 1 - 2 < 0. Reject the null hypothesis if tcalc < −1.86 (d.f. = 5 + - 2 5 2 = 8). 40.8 - 48.6 (5 - 1)6.7 + (5 - 1)9.8 = -4.2938 s 2p = = 8.25 tcalc = 5+5-2 8.25 8.25 + 5 5 so we reject the null hypothesis. The sample provides evidence that the mean sound level has been reduced with the new flooring. b. vs. . df1 = 4, df2 = 4. = ; FR F4,4 = 9.60 H 0 : s 12 s 22 = 1 H1 : s 12 s 22 ¹ 1 . Reject if Fcalc > 9.6 or Fcalc < .1042. 1 1 6.7 H0 FL = = = .1042 Fcalc = = 0.6837 F4,4 9.60 9.8 so we do not reject the null hypothesis and conclude the variance has not changed. Using Megastat for both (a) and (b): >Hypothesis Tests>Compare Two Independent Groups and check the box "Test for equality of variances". Learning Objective: 10-1 Learning Objective: 10-8
10.39
H0: 12 ≥ 22 versus 12 < 22. df1 = 11, df2 = 11.
Fcalc < .3549. Fcalc =
0.001807 2 = 0.103 0.0056422 The new drill has a reduced variance. Learning Objective: 10-8 10.40
. Reject H0 if 1 FL = = = 0.355 F11,11 2.818 . 0.103 < 0.355 so we reject the null hypothesis.
a. H0: 1
1
- 2 ≤ 0 vs. H1: 1 - 2 > 0. Reject the null hypothesis if tcalc > 1.833 (d.f. = min(10 1, 15 1) = 9). > 1.833 so we reject the 48.6 - 36.13 tcalc = = 3.1637 141.3778 20.981 + 10 15 null hypothesis. The sample provides evidence that the mean weight of an international bag is greater than a domestic bag. b. vs. . df1 = 9, df2 = 14. = . Reject if Fcalc FR F9,14 = 2.65 H0 H 0 : s 12 s 22 £ 1 H1 : s 12 s 22 > 1 > 2.65. > 2.65 so we reject the null hypothesis. The variance 141.3778 Fcalc = = 6.7387 20.9801 of international bag weight is greater than domestic bag weight. Learning Objective: 10-1 Learning Objective: 10-8
211
ASBE 6e Solutions for Instructors 10.41
Assume
is the proportion of men who ranked fresh fruit at the top of their snack list and p1 is the proportion of women.
p2 a. H0: M W = 0 vs H1: M W ≠ 0. Z.05 =NORM.S.INV(.05) = −1.645 Reject the null hypothesis if zcalc < −1.645 or zcalc > 1.645. b. pM = 15/25 =.60 and pW = 22/32 = .6875 c. , p-value .6 - .6875 15 + 22 = -.687 pc = = .649 zcalc = 25 + 32 1ù é1 .649(1 - .649) ê + ú ë 25 32 û =2*NORM.S.DIST(-.687,1) = .492. The sample does not show a significant difference in proportions. d. Normality can be assumed because n1p1 > 10, n1(1−p1) ≥ 10, n2p2 > 10, and n2(1−p2) ≥ 10. Learning Objective: 10-5 Learning Objective: 10-6 10.42
a. Assume
is the early proportion of free throws and is the later proportion. p1 p2 H0: 1 − 2 0 vs. H1: 1 − 2 < 0. Reject the null hypothesis if zcalc < –1.28. ³ .7021 - .764 66 + 68 zcalc = = -.9452 pc = = .7322 94 + 89 1ù é1 .7322(1 - .7322) ê + ú ë 94 89 û
Fail to reject the null hypothesis. They do not improve their free throw percentage as the season progresses. Using Megastat: >Hypothesis Testing>Compare Two Independent Proportions. b. Using Excel: =NORM.S.DIST(-.9452,1) and the p-value = .1723. This is not less than .10 so fail to reject the null. Learning Objective: 10-5 10.43
Assume
is the proportion of college students who eat cereal and is the proportion p1 p2 young children. a. H0: 1 2 ≤ 0 versus H1: 1 2 > 0. b. zcrit =NORM.S.INV(.95) = 1.645. Reject the null hypothesis if zcalc > 1.645. c. p1 = 833/850 = .980, p2 = 692/740 = .9351, .980 - .9351 833 + 692 = 4.509 pc = = .9591 zcalc = 850 + 740 1 ù é 1 .9591(1 - .9591) ê + ë 850 740 úû
212
ASBE 6e Solutions for Instructors d. Because 4.509 > 1.645 we reject the null hypothesis. The sample evidence shows a significant difference in the two proportions. e. The p-value ≈ .0000. (=1-NORM.S.DIST(4.509,1)) This result is not due to chance. f. Normality assumption is valid because both n1p1 > 10, n1(1−p1) ≥ 10, n2p2 > 10, and n2(1−p2) ≥ 10. Learning Objective: 10-5 Learning Objective: 10-6 10.44
a. H0:
vs. H1: . Reject the null hypothesis if zcalc > 1.645. p1 - p 2 £ 0 p1 - p 2 > 0 b. p1 = .1690, p2 = .1360. c. 202 + 779 .1690 - .1360 pc = = .1417 zcalc = = 2.9751 1,195 + 5, 727 1 ù é 1 .1417(1 - .1417) ê + ú ë1,195 5, 727 û From Appendix C-2 p-value = .0014 or Using Excel: =1-NORM.S.DIST(2.98,1). Reject the null hypothesis and conclude the Fortune 100 has a greater proportion of women board members than the Fortune 500. Using Megastat: >Hypothesis Testing>Compare Two Independent Proportions. d. The increase is most likely due to an increase in women executives and an increased awareness of the benefit of having more diverse boards. Learning Objective: 10-5
10.45
a. p1 = 36/202 = .1782, p2 = 142/993 = .143. b. , p-value =2*(136 + 142 .1782 - .143 pc = = .1490 zcalc = = 1.281 202 + 993 1 ù é 1 .149(1 - .149) ê + ë 202 993 úû NORM.S.DIST(1.281,1)) = .2002. Because the p-value is greater than .05, we fail to reject the null hypothesis. There is not enough evidence in this sample to conclude that there is a difference in the proportion of minority men (out of all males) and minority women (out of all females) on Fortune 100 boards. Learning Objective: 10-5
10.46 a.
H0:
p1 - p 2 £ 0 3333.
vs. H1:
p1 - p 2 > 0
. Reject the null hypothesis if zcalc > 1.28. p1 = .40, p2 = .
10 + 8 = .3673 zcalc = 25 + 24
.4 - .3333
= .4842 1ù é1 .3673(1 - .3673) ê + ú ë 25 24 û From Appendix C-2 p-value = .3156 or Using Excel: =1-NORM.S.DIST(.4842,1) = . 3141. We fail to reject the null hypothesis and conclude there is no difference between
pc =
213
ASBE 6e Solutions for Instructors early finishers and late finishers. Using Megastat: >Hypothesis Testing>Compare Two Independent Proportions. b. No, the normality assumption is not valid because n2p2 = 8 < 10. c. Early finishers might know the material better and finish faster. On the other hand, if a student has not studied they might quickly write an answer down and turn in their exam just to get it over with. Learning Objective: 10-5 Learning Objective: 10-6 10.47
a. H0:
p1 - p 2 = 0 −2.576.
vs. H1:
p1 - p 2 ¹ 0
97 + 167 pc = = .4444 zcalc = 252 + 342
. Reject the null hypothesis if zcalc > 2.576 or zcalc <
.3849 - .4883
which falls
= -2.507 1 ù é 1 .4444(1 - .4444) ê + ë 252 342 úû between the critical values so we fail to reject the null hypothesis. The decision is very close. b. The p-value =2*NORM.S.DIST(-2.507,1) = .0122. c. Normality assumption is valid because both n1p1 > 10, n1(1−p1) ≥ 10, n2p2 > 10, and n2(1−p2) ≥ 10. d. Gender differences may imply different marketing strategies. Learning Objective: 10-5 Learning Objective: 10-6
10.48
Define π1 to be the response rate of the group with the word “discount” in the email. H0: vs. H1: . Reject the null hypothesis if zcalc > 2.326. p1 - p 2 £ 0 p1 - p 2 > 0 . .086 - .06 129 + 90 = 2.737 pc = = .073 zcalc = 1500 + 1500 1 ù é 1 .073(1 - .073) ê + ë1500 1500 úû Because 2.737 > 2.326 we reject the null hypothesis. The response rate for the group that received emails with the word discount in the subject line was higher than for the group with the word free. Learning Objective: 10-5
10.49
a. H0: 1 2 ≥ 0 versus H1: 1 2 < 0. Reject the null hypothesis if zcalc < 2.33. < 2.33so we .1491 - .5714 61 + 40 = -8.003 pc = = .2109 zcalc = 409 + 70 1ù é 1 .2109(1 - .2109) ê + ë 409 70 úû reject the null hypothesis. 214
ASBE 6e Solutions for Instructors b. p-value =NORM.S.DIST(-8.003,1) = .0000. This is less than .01 so the difference is quite significant. c. Normality can be assumed because n1p1 > 10, n1(1−p1) ≥ 10, n2p2 > 10, and n2(1−p2) ≥ 10. Learning Objective: 10-5 Learning Objective: 10-6 10.50
a. b.
pE = 56/304 = .1842, pW = 145/562 = .2580. H0: E W = 0 vs. H1: E W ≠ 0. Reject the null hypothesis if zcalc > 1.96 or zcalc < −1.96. .1842 - .2580 56 + 145 = -2.4554 pc = = .2321 zcalc = 304 + 562 1 ù é 1 .2321(1 - .2321) ê + ë 304 562 úû so we reject the null hypothesis and conclude that there is a greater proportion of large gloves sold on the west side of Vail. Using Megastat: >Hypothesis Testing>Compare Two Independent Proportions. c. There could be a different type of skier on the east side of Vail, perhaps more children ski on the east side as opposed to the west side. Learning Objective: 10-5
10.51
a. H0:
10.52
a. H0: F
vs. H1: . Using the quick rule d.f. = min(25-1, 25-1) = 24. tcrit m1 - m 2 ³ 0 m1 - m 2 < 0 =T.INV(.05,24) = −1.711. Reject the null hypothesis if tcalc < −1.711. If using Welch’s formula d.f. =43 and tcrit = −1.681. > −1.711 or −1.681 therefore we fail to reject the null 7.12 - 8.29 tcalc = = -1.581 2.142 3.022 + 25 25 hypothesis. We cannot conclude that there was an increase in the average concession stand purchases with the coupon. Learning Objective: 10-1
- P = 0 vs. H1: F - P ≠ 0. Reject the null hypothesis if zcalc > 1.645 or zcalc < −1.645. .521 - .4896 434 + 189 = 1.020 pc = = .5111 zcalc = 833 + 386 1 ù é 1 .5111(1 - .5111) ê + ë 833 386 úû 1.02 is between the critical values. Do not reject the null and conclude there is no difference in proportion of rehires. Using Megastat: >Hypothesis Testing>Compare Two Independent Proportions. b. Using Excel: =2*(1-NORM.S.DIST(1.02,1)) and p-value = .3077. This is not less than .10 so we do not reject the null. Learning Objective: 10-5 215
ASBE 6e Solutions for Instructors
10.53
Define 1 = proportion of magazine subscription renewals without a reminder. Define 2 = proportion of magazine subscription renewals with a reminder. H0: 1 2 ≥ 0 versus H1: 1 2 < 0. This is a left-tailed test. Reject the null hypothesis if z < −1.645. a. .925 - .95 703 + 228 = -1.332 pc = = .931 zcalc = 760 + 240 1 ù é 1 .931(1 - .931) ê + ë 760 240 úû −1.332 > −1.645 therefore fail to reject the null hypothesis. The renewal rate is not higher with the reminder. b. Normality can be assumed because n1p1 > 10, n1(1−p1) ≥ 10, n2p2 > 10, and n2(1−p2) ≥ 10. Learning Objective: 10-5 Learning Objective: 10-6
10.54
a. Assume
is the proportion of those that received the information and is those that p1 p2 did not. H0: 1 − 2 ≥ 0 vs. H1: 1 − 2 < 0.
b.
zcalc =
.315 - .385
= -2.075 pc =
126 + 154 = .35 400 + 400
1 ù é 1 .35(1 - .35) ê + ë 400 400 úû and the p-value =NORM.S.DIST(-2.075,1) = .019. c. The p-value is less than.05 so we reject the null hypothesis for α = .05. Conclude that the admission rate is lower for those who received information. However, for α = .01 we would fail to reject H0 and conclude the readmission rates were not significantly different. Using Megastat: >Hypothesis Tests> Compare Two Independent Proportions. Learning Objective: 10-5 10.55
H0: 1 2 ≤ 0 versus H1: 1 2 > 0. Assume unequal variances. Using the quick rule d.f. = min( 44-1, 42-1) = 41. Reject the null hypothesis if tcalc > 2.42. (If using Welch’s d.f. = 83.) > 2.42 so we reject the null hypothesis and conclude that 2.73 - 1.90 tcalc = = 4.09 0.97 2 0.912 + 44 42 the virtual team mean is higher. Learning Objective: 10-1
216
ASBE 6e Solutions for Instructors is the proportion for San Francisco and for Seattle. H0: 1 − 2 ≤ 0 vs. H1: p1 p2 1 − 2 > 0. b. Reject if zcalc > 1.96. H0 .41 - .38 1312 + 684 = 2.079 pc = = .3992 zcalc = 3200 + 1800 1 ù é 1 .3992(1 - .3992) ê + ë 3200 1800 úû 2.079 > 1.96 and the p-value =.0188 so we do reject the null hypothesis and conclude the proportion of encryption is higher in San Francisco. Using Megastat: >Hypothesis Tests>Compare Two Independent Proportions. Learning Objective: 10-5
10.56
a. Assume
10.57
a. Assume
is the proportion that died with the implant and the proportion that died p1 p2 without the implant. H0: 1 2 ≥ 0 versus H1: 1 2 < 0.
b.
pc =
104 + 98 = .164 zcalc = 742 + 490
.1402 - .20
= -2.77 1 ù é 1 .164(1 - .164) ê + ë 742 490 úû p-value =NORM.S.DIST(-2.77,1) = .0028. Reject the null hypothesis. A greater proportion died without the defibrillator implanted. c. Normality can be assumed because n1p1 > 10, n1(1−p1) ≥ 10, n2p2 > 10, and n2(1−p2) ≥ 10. d. Exercise habits, nutrition, weight, smoking etc., might influence the decision. Also, many people cannot afford them and lack insurance to pay the costs. Learning Objective: 10-5 Learning Objective: 10-6 10.58
a.
H 0 : md = 0
vs.
H 0 : md ¹ 0
b.
-0.385714 = -2.3458 1.3757 70 c. tcrit =T.INV.2T(.05,69) = 1.995. Reject the null hypothesis is tcalc < −1.995 or tcalc > 1.995. d. p-value =T.DIST.2T(2.348,69) = .0218 e. Because −2.3458 < −1.995 reject the null hypothesis. There is a significant difference in the taste of the two sauces. Note that the p-value < .05 also. Learning Objective: 10-4 tcalc =
217
ASBE 6e Solutions for Instructors 10.59
a. Assume µ1 is the average cost to perform a background check in 2010 and µ2 is the average cost to perform a background check in 2012. H0: 1 2 = 0 versus H1: 1 2 ≠ 0. Assuming unequal variances and using the quick rule d.f. = 9. Reject H0 if tcalc < −2.262 or tcalc > 2.262. If using Welch’s d.f. = 16. which falls between the critical values therefore we fail to 105 - 75 tcalc = = 1.718 322 452 + 10 10 reject the null hypothesis. There has not been a significant change in the average cost to perform a background check. b. A paired sample test may have made more sense. By comparing the costs from one year to the next for the same 10 companies we would have eliminated a source of variation due to different businesses. Learning Objective: 10-1 Learning Objective: 10-4
10.60
a. Assume µ1 is the average call length in July and µ2 is the average call length in August. H0: 1 – 2 vs. H1: 1 – 2 > 0. £0 b. 2.640. Using the quick rule d.f. = 47. The p-value 4.48 - 2.396 tcalc = = 34.457 4.0723 + 64 48 =T.DIST.RT(2.64,47) = .0056. Because the p-value < .01 and because tcalc falls in the rejection region, we would reject the null hypothesis and conclude the length of calls had been reduced. c. The distribution could be skewed to the right by one or two extremely long calls. A heavily skewed distribution could make the t distribution an unwise choice. Learning Objective: 10-1
10.61
a. New Bumper:
Control Group:
218
ASBE 6e Solutions for Instructors b. Assume µ1 is the average repair incident down time for the new bumper and µ2 is the average for the control. H0: 1 2 ≥ 0 versus H1: 1 2 < 0. c. Assuming equal variances, tcrit =T.INV(.05,19) = −1.729. Reject H0 if tcalc < −1.729. d. and . 2 2 5.92 8.89 (12 1)3.42 + (9 1)4.96 tcalc = = -1.627 s 2p = = 17.13 17.13 17.13 12 + 9 - 2 + 12 9 e. Because −1.627 > −1.729 we fail to reject the null hypothesis. f. The p-value =T.DIST(-1.627,19) = .0601 > .05. This decision was close. g. A sample difference of approximately 3 days downtime would be considered important but the variation in the downtimes is large enough that we cannot conclude the true means are different. Learning Objective: 10-1
10.62
a. H0: N – S ≥ vs. H1: N – S < 0. Assuming unequal variances and using the quick rule d.f. = min(14 . Reject the null hypothesis if tcalc < −2.650. -1,16 - 1) = 13 b. −5.29 < −2.650 3123 - 8456 tcalc = = 2390116 13527684 + 14 16 c. This sample provides strong evidence that the average spending in the northern region is much higher than average spending in the southern region. d. Folks in the south may use services differently or may be older. Also, the cost of health care may be much higher in the north than the south. Learning Objective: 10-1
10.63
a. Assume µ1 is the average particulate count in Athens and µ2 is the average in London. : 1 2 ≤ 0 vs. H1 : 1 > 2. Assuming equal variances d.f. = 28 and 2 2 39.5 - 31.5 (15 1)3.75 + (15 1)2.25 tcalc = = 7.085 s 2p = = 9.5625 9.5625 9.5625 15 + 15 - 2 + 15 15 p-value ≈ 0. Reject H0. b. H0: 12/22 = 1 versus 12/22 ≠ 1. FL =F.INV(.025,14,14) = 0.3357, FR =F.INV.RT(.025,14,14) = 2.9786 . Reject H0 if Fcalc < .3357 or Fcalc > 2.9786. (df1 = 14, df2 = 14.) Fcalc = 3.752/2.252 = 2.78 which is between the two critical values so we fail to reject the null hypothesis of equal variances. Learning Objective: 10-1 Learning Objective: 10-8
219
ASBE 6e Solutions for Instructors 10.64
a. H0: 1
= 0 vs. H1: 1 - 2 ≠ 0. - 2 b. Reject the null hypothesis if tcalc < −1.686 or tcalc> 1.686 (d.f. = 20 + 20
- 2 = 38). Assume equal variances because the standard deviations are similar. However, case 2 (assuming equal) and case 3 (assuming unequal) will give us the same because the tcalc sample sizes are the same.
c.
9.1 - 10.3 = -1.5485 5.76 6.25 + 20 20 d. We fail to reject the null hypothesis because −1.5485 falls between the two critical values and conclude there was no difference in mean scores. e. The p-value =T.DIST.2T(1.5485, 38) = .1298. Because the p-value > .10, we fail to reject the null hypothesis. Learning Objective: 10-1 tcalc =
10.65
a. H0: M W ≤ 0 vs H1: M W > 0. b. Reject the null hypothesis if tcalc > 2.438 with d.f. = 35. c. Assuming equal variances, and 2 2 (20 1)10114.71 + (17 1)14540.66 s 2p = = 12,336.632 20 + 17 - 2 117853 - 98554 tcalc = = 4.74 12336.632 12336.632 + 20 17 d. 4.74 > 2.438 therefore reject the null hypothesis. Men are paid more on average. e. p-value =T.DIST.RT(4.74,35) ≈ .0000. This shows that the sample result would be unlikely if H0 were true. Learning Objective: 10-1
10.66
a. H0: 1 – 2 = 0 vs. H1: 1 – 2 ≠ 0. b. Reject the null hypothesis if tcalc > 2.045 or tcalc < −2.045 (df = 13 + 18 – 2 = 29). c. 22.32 - 25.56 (13 - 1)18.9225 + (18 - 1)37.9456 = -1.6232 s 2p = = 30.074 tcalc = 13 + 18 - 2 30.074 30.074 + 13 18 d. −1.6232 falls between the critical values so fail to reject the null hypothesis. There appears to be no difference in the average order size between Friday and Saturday night. Using Megastat: >Hypothesis Tests>Compare Two Independent Groups and select "pooled variance". e. p-value =T.DIST.2T(1.6232,29) = .1154 which is > .05 so do not reject H0. Learning Objective: 10-1 220
ASBE 6e Solutions for Instructors
10.67 a.
The distributions appear skewed to the right.
b. H0: F M =0 versus H1: F M ≠ 0. c. Assume equal variances. d.f. = 30+30−2 = 58. Reject the null hypothesis tcalc > 2.663 or if tcalc < −2.663. d. 50.33 - 50.00 (30 - 1)81.682 + (30 - 1)71.632 2 = 0.017 sp = = 5901.24 tcalc = 5901.24 5901.24 30 + 30 - 2 + 30 30 e. Because 0.017 falls between the critical values we fail to reject the null hypothesis. It does not appear that the means are different. f. The p-value =T.DIST.2T(.017,58) = .9865. This indicates that the sample result shows no significant difference. Learning Objective: 10-1 10.68
a.
vs.
.
10.69
a. H 0: 1 2 = 0 versus H1: 1 2 ≠ 0. b. Assume equal variances. d.f. = 28 + 29 − 2 =55. t.025 =T.INV(.025,55) = −2.004. Reject the null hypothesis if tcalc < −2.004 or tcalc > 2.004. c. 328 - 435 (28 - 1)1042 + (29 - 1)147 2 = -3.162 s 2p = = 16311 tcalc = 16311 16311 28 + 29 - 2 + 28 29 Because tcalc = −3.162 < −2.004 we reject the null hypothesis. Mean sales are lower on the east side. Learning Objective: 10-1
H0 : s s = 1 H1 : s s ¹ 1 b. df1 = 17, df2 = 16. =F.INV.RT(.05,17,16) = 2.317, FL =F.INV(.05,17,16) = 0.437. FR Reject if Fcalc > 2.317 or Fcalc < 0.437. H0 c. Fcalc = 147.22/237.92 = 0.383. d. Because 0.383 < 0.437 we reject the null hypothesis and conclude that the variances are not equal. Using Megastat: >Hypothesis Tests>Compare Two Independent Groups and select "Test for equality of variances". Learning Objective: 10-8 2 1
2 2
2 1
2 2
221
ASBE 6e Solutions for Instructors 10.70
a. H 0: d = 0 versus H1: d ≠ 0. This is a paired sample test. Treat the five Saturdays as the sample and the store sales as the two treatments. b. Reject the null hypothesis if tcalc < −2.132 or tcalc > 2.132 (Use Appendix D, df = 5 – 1 = 4). c. −1.31. Fail to reject the null hypothesis. The average sales appear to -66.4 - 0 tcalc = = 113.1473 5 be the same. Learning Objective: 10-4
10.71
H0: 12/22 = 1 versus 12/22 ≠ 1. df1 = 30, df2 = 29. FR =F.INV.RT(.025,30,29) = 2.09, FL =F.INV(.025,30,29) = 0.48. Reject the null hypothesis if Fcalc > 2.09 or Fcalc < 0.48. Fcalc =13.482052/15.42712 = .76 which falls between the critical values. We fail to reject the null hypothesis. The variances are not different. Learning Objective: 10-8
10.72
a. H0: d ≥ 0 versus H1: d < 0 where d = after – before. b. -1.444 - 0 tcalc = = -1.98 2.186 9 c. Use d.f. = 8. tcrit =T.INV(.05,8) = −1.8595. Reject the null hypothesis if tcalc < −1.8595. d. p-value =T.DIST(-1.98,8,1) = .0415. e. Because −1.98 < −1.8595 we reject the null hypothesis and conclude that the number of graffiti incidents has decreased. Learning Objective: 10-4
10.73
This is a paired observation test. H0: d = 0 versus H1: d ≠ 0 where d = company assessed value – employee assessed value. Using d.f. = 7, t.005 =T.INV(.005,7) = − 3.50. Reject the null hypothesis if tcalc > 3.50 or if tcalc < −3.50. . Because −0.87 falls between the critical values we fail to reject -4.0 - 0 tcalc = = -0.87 13.071 8 the null hypothesis. There is not a difference between the company’s assessed home values and the employees’ assessed home values. Learning Objective: 10-4
10.74
H 0: 1 – 2 = 0 vs. H1: 1 – 2 ≠ 0. Assuming unequal variances and using the quick rule d.f. = 8. t.05 =T.INV(.05,8) = − 1.86. Reject the null hypothesis if the p-value is less than
222
ASBE 6e Solutions for Instructors .10 or if
−1.336 and the p-value 2430 - 2607 tcalc = = 60450 97025 + 9 9 =T.DIST.2T(-1.336,8) = .2183 so we fail to reject the null hypothesis. The average size of the homes in the two neighborhoods is the same. Using Megastat: >Hypothesis Tests>Compare Two Independent Groups. Assume unequal variances. If we had assumed equal variances our conclusion would have been the same. tcalc = −1.3356 and using 16 d.f. the p-value = .2004. Fail to reject H0. Learning Objective: 10-1 tcalc < -1.86
or
tcalc >
1.86.
10.75
a. H0 : 1 2 = 0 vs. H1 : 1 2 ≠ 0, Assuming equal variances d.f. =12 + 12 – 2 = 22. t.005 =T.INV(.005,22) = −2.819. Reject the null hypothesis if tcalc > 2.819 or tcalc < −2.819. 406.75 - 430.33 (12 - 1)63.162 + (12 - 1)39.262 2 = -1.098 sp = = 2765.27 tcalc = 2765.27 2765.27 12 + 12 - 2 + 12 12 Because −1.098 falls between the critical values we fail to reject H0. There is no difference in the average number of defects. b. H0: 12/22 = 1 versus 12/22 ≠ 1. FR =F.INV.RT(.005,11,11) = 5.3197, FL =F.INV(.005,11,11) = 0.188. Reject H0 if Fcalc < .1880 or Fcalc > 5.3197. (df1 = 11, df2 = 11.) Fcalc = 63.1922/39.262 = 2.59 falls between the critical values so we fail to reject the null hypothesis and conclude there is no difference between the variances. Learning Objective: 10-1 Learning Objective: 10-8
10.76
H 0: d = 0 vs. H1: d ≠ 0. Reject the null hypothesis if the p-value < .10 or if or
1.796 (df = 12
- 1 = 11).
tcalc < -1.796
-2.5 - 0 = -1.7644 4.9082 12 and the p-value = .1054. We fail to reject the null hypothesis but the decision is quite close. We conclude there is no difference in estimates from the two expert teams. Learning Objective: 10-4
10.77
tcalc >
tcalc =
a. H0: A2/B2 ≤ 1 versus A2/B2 > 1. df1 = 11, df2 = 11. FR =F.INV.RT(.025,11,11) = 3.53. Reject the null hypothesis if Fcalc > 3.53. Fcalc = 2.938622/0.93592 = 9.86 so we reject the null hypothesis. Portfolio A has greater variance than portfolio B. b. H 0: 1 − 2 = 0 versus H1: 1 − 2 ≠ 0. Assume unequal variances (from part a.) Using the quick rule d.f. = 11. tcrit =T.INV.2T(.025,11) = 2.593. Reject the null hypothesis if tcalc > 2.593 or tcalc < −2.593.
223
ASBE 6e Solutions for Instructors which falls between the critical values. We fail to
8.5358 - 8.10
tcalc =
= 0.49 2.93862 0.93592 + 12 12 reject the null hypothesis. The portfolio means are equal. Learning Objective: 10-1 Learning Objective: 10-8 10.78
a.
H 0 : s 12 s 22 = 1
FL =
1 FR ,77,61
=
vs.
. df1 = 61, df2 = 77. = ; FR F61,77 » F60,60 = 1.53 H1 : s 12 s 22 ¹ 1 . Reject if Fcalc > 1.53 or Fcalc < .6536. 1 H0
1.53
= .6536
so we do not reject the null hypothesis and conclude no 265.69 Fcalc = = 1.3951 190.44 difference in variances. b. Yes, the pooled statistic assumes equal variances and that is what we concluded in part t a. Learning Objective: 10-8 10.79
p1 = 65/100 = .65, p2 = 53/100 = .53. For a 95% confidence level use z.025 = 1.96. or (.0153, .2553). Yes, the .65(1 - .65) .53(1 - .53) (.65 - .53) ± 1.96 + = .12 ± .1353 100 100 interval includes zero. There is no difference in the two proportions. Learning Objective: 10-7
10.80
p1 = 65/500 = .13, p2 = 45/500 = .09. For a 95% confidence level use z.025 = 1.96. and the interval is (.0013, .0787). It does .13(1 - .13) .09(1 - .09) (.13 - .09) ± 1.96 + 500 500 not include zero so it appears there is a difference in the two groups. Learning Objective: 10-7
10.81
a. Assuming equal variances because 2.4 and 2.5 are relatively close to each other: d.f. = and t.05 =T.INV(.05,38) = −1.686. (n1 - 1) + ( n2 - 1) = (20 - 1) + (20 - 1) = 38 = −1.2 ± 1.326 or (2.51, (9.1 - 10.3) ± 1.686
(20 - 1)2.4 + (20 - 1)2.5 20 + 20 - 2 2
2
1 1 + 20 20
0.11) Note: a formal test for equal variances would confirm our assumption.
224
ASBE 6e Solutions for Instructors b. No because the estimated difference in mean scores could be zero, according to the confidence interval in part a. Learning Objective: 10-3 10.82
a.
and the interval is (9.25, 28.75).
(15 - 1)196 + (15 - 1)144 1 1 + 15 + 15 - 2 15 15 (Use df = 15+15 −2 = 28). The interval does not include zero so we conclude there is a difference between DVD users and non-DVD users. b. Without doing a formal test for equal variances we assume equal variances because 12 and 14 are relatively close. A formal test would confirm this assumption. Learning Objective: 10-3 (146 - 127) ± 2.048
10.83
a. p1 = 3/32 = .0938, p2 = 11/32 = .3438. For a 95% confidence level use z.025 = 1.96. or (.4431, .0938(1 - .0938) .3438(1 - .3438) (.0938 - .3438) ± 1.96 + = -.25 ± .1931 32 32 .0569) b. Normality is not met therefore the interval could be wider than what we calculated. Learning Objective: 10-7
10.84
a. p1 = 14/50 = .28, p2 = 23/56 = .41. For a 95% confidence level use z.025 = 1.96. and the interval is ( , .0491). The .3091 .28(1 - .28) .41(1 - .41) (.28 - .41) ± 1.96 + 50 56 interval includes zero so we conclude there is no difference between those with positive versus negative emotions. b. We can assume normality because and for each sample. np ³ 10 n(1 - p ) ³ 10 Learning Objective: 10-7
10.85
a. H0 : M F ≤ 0 vs. H1 : M F > 0, tcalc =
43.20 - 36.60
= 2.616 8.302 3.102 + 13 9 b. Assuming unequal variances and using the quick rule d.f. = min(13−1, 9−1) = 8. tcrit =T.INV(.99,8) = 2.896. Reject the null hypothesis if tcalc > 2.896. c. Because 2.616 < 2.896 we fail to reject H0. Males did not spend more on average than females for a tank of gas. d. H0: 12/22 = 1 versus 12/22 ≠ 1. FR =F.INV.RT(.005,12,8) = 7.015, FL =F.INV(.005,12,8) = 0.187. Reject H0 if Fcalc < .187 or Fcalc > 7.015. (df1 = 12, df2 = 8.) Fcalc = 8.302/3.102 = 7.169 > 7.015 so we reject the null hypothesis and conclude there is
225
ASBE 6e Solutions for Instructors a difference between the variances. The assumption of unequal variances was reasonable. Learning Objective: 10-1 Learning Objective: 10-8 10.86
a. H0 : S G ≤ 0 vs. H1 : S G > 0, (11 - 1)2.692 + (16 - 1)2.662 2 sp = = 7.1398 tcalc = 11 + 16 - 2
14.51 - 11.88 = 2.513 7.1398 7.1398 + 11 16
b. Assuming equal variances d.f. = 11 + 16 − 2 = 25. tcrit =T.INV(.99,25) = 2.485. Reject the null hypothesis if tcalc > 2.485. c. Because 2.513 < 2.485 we reject H0. On average, Sonando High students ate at fast food restaurants more than Gedacht High students during the past month. d. H0: 12/22 = 1 versus 12/22 ≠ 1. FR =F.INV.RT(.005,10,15) = 4.424, FL =F.INV(.005,10,15) = 0.183. Reject H0 if Fcalc < .183 or Fcalc > 4.424. (df1 = 10, df2 = 15.) Fcalc = 2.692/2.662 = 1.023 which falls between the critical values. We fail to reject the null hypothesis and conclude there is no difference between the variances. The assumption of equal variances was reasonable. Learning Objective: 10-1 Learning Objective: 10-8 10.87
H0: A
- B ≤ 0 vs. H1: A - B > 0. Reject the null hypothesis if zcalc > 2.326. .075 - .05 57 + 62 = 2.294 pc = = .0595 zcalc = 760 + 1240 1 ù é 1 .0595(1 - .0595) ê + ë 760 1240 úû 2.294 < 2.326 so do no reject the null and conclude there is no difference in proportion of returned items. Using Megastat: >Hypothesis Testing>Compare Two Independent Proportions. Learning Objective: 10-5
10.88
H0: B
- A ≤ 0 vs. H1: B - A > 0. Reject the null hypothesis if zcalc > 1.28. .4 - .3 32 + 36 = 1.463 pc = = .34 zcalc = 80 + 120 1 ù é1 .34(1 - .34) ê + ë 80 120 úû 1.463 > 1.28 so reject the null and conclude that the proportion of employees hearing inappropriate comments after the training went down. Using Megastat: >Hypothesis Testing>Compare Two Independent Proportions. Learning Objective: 10-5
226
ASBE 6e Solutions for Instructors 10.89
H 0: d ≤ 0 vs. H1: d > 0. Reject the null hypothesis if the p-value < .05 or if (df = 5
tcalc >
2.132
< 2.132. 55 - 0 tcalc = = 1.851 66.453 5 The p-value = .0689 > .05. We fail to reject the null hypothesis but the decision is quite close. We conclude there is no difference in attendance between Sunday and Saturday matinees. Learning Objective: 10-4 10.90
- 1 = 4).
a. H0: C2/U2 = 1 versus C2/U2 ≠ 1. df1 = 7, df2 = 7. Reject the null hypothesis if p-value < 05. Fcalc = 3560.0822/8108.1022 = 0.1928, p-value =2*F.DIST(.1928,7,7) = .0453 < . 05. We reject the null hypothesis. The variances in tire life between checked and unchecked tires are different. b. H 0: C − U = 0 versus H1: C − U ≠ 0. Assume unequal variances (from part a.) tcrit =T.INV.2T(.05,9) = 2.262. Reject the null hypothesis if tcalc > 2.262. < 2.262. We fail to reject the null hypothesis. 41381.25 - 35490 tcalc = = 1.882 3560.0822 8108.1022 + 8 8 The average tire life between checked and unchecked tires is not significantly different. c. Testing equality of the variances first allows one to use the appropriate formula for the t statistic when testing equality of the means. Learning Objective: 10-1 Learning Objective: 10-8
227
ASBE 6e Solutions for Instructors
Chapter 11 Analysis of Variance 11.1
a. b. c.
n = total df +1 = 39 + 1 = 40 The number of groups is equal to the between groups d.f. plus 1 or 4 + 1 =5. H0: A = B = C = D means are the same H1: Not all the means are equal at least one pair of means differ d. Fcrit = F4,35 =F.INV.RT(.05,4,35) = 2.64 c. Fcalc = MSE/MSR = 29.9954/16.6634 = 1.80 e. 1.80 < 2.64 therefore we fail to reject H0. The population means do not differ. Learning Objective: 11-03 Learning Objective: 11-04 Learning Objective: 11-05 Learning Objective: 11-06
11.2
a. b. c.
11.3
a.
11.4
a.
n = total df +1 = 23 + 1 = 24 The number of groups is equal to the between groups d.f. plus 1 or 3 + 1 =4. H0: A = B = C means are the same H1: Not all the means are equal at least one pair of means differ d. Fcrit = F3,20 =F.INV.RT(.10,3,20) = 2.38 c. Fcalc = MSE/MSR = 40.29018/16.25101 = 2.48 e. 2.48 > 2.38 therefore we reject H0. At least one pair of means differ. Learning Objective: 11-03 Learning Objective: 11-04 Learning Objective: 11-05 Learning Objective: 11-06 H0: 1 = 2 = 3 = 4 = 5 means are the same H1: Not all the means are equal at least one pair of means differ b. The between groups d.f. is equal to c −1 or 5 −1 = 4. This is the numerator d.f. The within groups d.f. is equal to n –c or 30–5 = 25. This is the denominator d.f. c. Fcrit = F4,25 =F.INV.RT(.10,4,25) = 2.18 d. 2.447 > 2.18 therefore we reject H0. At least one pair of means differ. e. p-value =F.DIST.RT(2.447,4,25) = .0726 Learning Objective: 11-03 Learning Objective: 11-04 Learning Objective: 11-05 Learning Objective: 11-06 H0: 1 = 2 = 3 = 4 means are the same H1: Not all the means are equal at least one pair of means differ b. The between groups d.f. is equal to c −1 or 4 −1 = 3. This is the numerator d.f. The within groups d.f. is equal to n –c or 25–4 = 21. This is the denominator d.f. c. Fcrit = F3,21 =F.INV.RT(.05,3,21) = 3.072 d. 3.251 > 3.072 therefore we reject H0. At least one pair of means differ. e. p-value =F.DIST.RT(3.251,3,21) = .0422 Learning Objective: 11-03 Learning Objective: 11-04 Learning Objective: 11-05 Learning Objective: 11-06 225
ASBE 6e Solutions for Instructors 11.5
a.
b. c. d. e. f.
The hypotheses to be tested are: H0: A = B = C mean scrap rates are the same H1: Not all the means are equal at least one pair of means differ Treatment has 2 degrees of freedom and error has 12 degrees of freedom. One factor, F = 5.31 and critical value for = .05 is F2,12 = 3.89. We reject the null hypothesis because the test statistic exceeds the critical value. The decision is not close. The p-value of .0223 is less than .05. The difference between the mean of the scrap rates most likely did not occur by chance. From the dot plot, we see Plant B above the overall mean and Plant C below the overall mean. Mean 12.30 13.96 9.58 11.95
Source Treatment Error Total
n 5 5 5 15
Std. Dev 1.573 2.077 2.651 2.728
Treatment Plant A Plant B Plant C Total
One-Factor ANOVA SS df MS 48.897 2 24.4487 55.260 12 4.6050 104.157 14
F 5.31
p-value .0223
Learning Objective: 11-03 Learning Objective: 11-04 Learning Objective: 11-05 Learning Objective: 11-06 11.6
a.
b. c. d. e. f.
The hypotheses to be tested are: H0: 1 = 2 = 3 = 4 physician means are the same H1: Not all the means are equal at least one pair of means is differ Treatment has 3 degrees of freedom and error has 24 degrees of freedom. One factor, F = 3.50 and critical value for = .05 is F3,24 = 3.01. We reject the null hypothesis because the test statistic exceeds the critical value. This is a fairly close decision. The p-value of .0310 is less than .05. The difference between the physician means most likely did not occur by chance. From the dot plot, we see Physician 1 and Physician 3 below the overall mean and Physician 2 above the overall mean. 226
ASBE 6e Solutions for Instructors
Mean 28.3 34.2 27.3 32.0 30.2
n 7 6 8 7 28
Std. Dev 4.89 4.12 4.62 4.24 5.08
Treatment Physician 1 Physician 2 Physician 3 Physician 4 Total
One-Factor ANOVA SS df MS 212.35 3 70.782 485.76 24 20.240 698.11 27
Source Treatment Error Total
F 3.50
p-value .0310
Learning Objective: 11-03 Learning Objective: 11-04 Learning Objective: 11-05 Learning Objective: 11-06 11.7
a.
b. c. d. e. f.
The hypotheses to be tested are: H0: 1 = 2 = 3 = 4 mean GPAs are the same H1: Not all the means are equal at least one pair of means differ Treatment has 3 degrees of freedom and error has 24 degrees of freedom. One factor, F = 3.52 and critical value for = .05 is F3,24 = 3.01. We reject the null hypothesis since the test statistic exceeds the critical value. This is a fairly close decision. The p-value of .0304 is less than .05. The difference between the GPA means most likely did not occur by chance. From the dot plot, we see the GPA for Accounting below the overall mean and Human Resources and Marketing above the overall mean. Mean 2.834 3.024 3.241 3.371 3.118
n 7 7 7 7 28
Std. Dev 0.5053 0.1776 0.3077 0.2575 0.3785
227
Treatment Accounting Finance Human Resources Marketing Total
ASBE 6e Solutions for Instructors
One-Factor ANOVA SS df MS 1.1812 3 0.39372 2.6867 24 0.11195 3.8679 27
Source Treatment Error Total
F 3.52
p-value .0304
Learning Objective: 11-03 Learning Objective: 11-04 Learning Objective: 11-05 Learning Objective: 11-06 11.8
a.
b. c. d. e. f.
The hypotheses to be tested are: H0: 1 = 2 = 3 = 4 mean sales are the same H1: Not all the means are equal at least one pair of means differ Treatment has 3 degrees of freedom and error has 16 degrees of freedom. One factor, F = 4.71 and critical value for = .05 is F3,16 = 3.24. We reject the null hypothesis since the test statistic exceeds the critical value. This is not a close decision. The p-value of .0153 is less than .05. The difference between the mean sales most likely did not occur by chance. From the dot plot, we see the weekly sales for Stores 2 and 3 below the overall mean and Store 1 above the overall mean. Mean 108.0 87.4 91.0 101.0 96.9
Source Treatment Error Total
n 5 5 5 5 20
Std. Dev 5.34 10.83 11.11 10.30 12.20
Treatment Store 1 Store 2 Store 3 Store 4 Total
One-Factor ANOVA SS df MS 1,325.35 3 441.783 1,501.20 16 93.825 2,826.55 19
228
F 4.71
p-value .0153
ASBE 6e Solutions for Instructors
Learning Objective: 11-03 Learning Objective: 11-04 Learning Objective: 11-05 Learning Objective: 11-06 11.9
a. There are c(c−1)/2 or 4(4−1)/2 = 6 different pairs of means to compare. b. The d.f. are c for the numerator and n−c for the denominator: 4 and 34−4 =30. c. T4,30 = 2.72 (see Table 11.4 Learning Objective: 11-07
11.10
a. There are c(c−1)/2 or 5(5−1)/2 = 10 different pairs of means to compare. b. The d.f. are c for the numerator and n−c for the denominator: 5 and 25−5 =20. c. T5,20 = 2.99 (see Table 11.4 Learning Objective: 11-07
11.11
a. 3. b. c = 3, n – c = 12. c. T3,12 = 2.68. This value was found by interpolating between T3,10 and T3,15. d. Only Plant B and Plant C differ at α = .05 using MegaStat Tukey test. Using the pairwise t-tests, Plant B and Plant C differ. Post hoc analysis Tukey simultaneous comparison t-values (d.f. = 12) Plant C Plant A Plant B 9.58 12.30 13.96 Plant C 9.58 Plant A 12.30 2.00 Plant B 13.96 3.23 1.22 critical values for experimentwise error rate: 0.05 2.67 0.01 3.56 p-values for pairwise t-tests Plant C 9.58 Plant C 9.58 Plant A 12.30 .0682 Plant B 13.96 .0073
229
Plant A 12.30
.2448
Plant B 13.96
ASBE 6e Solutions for Instructors Learning Objective: 11-07 11.12
a. 6. b. c = 4, n – c = 24. c. T4,20 = 2.80. (Table 11.4 used.) d. Using Tukey’s simultaneous comparison t-values, Physicians 2 and 3 differ. Using the pairwise t-tests, Physicians 2 and 3 differ at = .01 and Physicians 1 and 2 differ at = .05. Post hoc analysis Tukey simultaneous comparison t-values (d.f. = 24) Physician Physician Physician 3 1 4 27.3 28.3 32.0 Physician 3 27.3 Physician 1 28.3 0.44 Physician 4 32.0 2.04 1.54 Physician 2 34.2 2.85 2.35 0.87
Physician 2 34.2
critical values for experiment wise error rate: 0.05 2.76 0.01 3.47 p-values for pairwise t-tests
Physician 3 Physician 1 Physician 4 Physician 2 Learning Objective: 11-07 11.13
27.3 28.3 32.0 34.2
Physician 3 27.3
Physician 1 28.3
Physician 4 32.0
.6604 .0525 .0089
.1355 .0274
.3953
Physician 2 34.2
a. 6. b. c = 4, n – c = 24. c. T4,20 = 2.80. (Table 11.4 used.) d. Using Tukey’s simultaneous comparison t-values, Marketing and Accounting differ. Using the pairwise ttests, Marketing and Accounting differ at = .01 and Human Resources and Accounting differ at = .05. Post hoc analysis Tukey simultaneous comparison t-values (d.f. = 24) Accounting Finance 2.834 3.024 Accounting 2.834 Finance 3.024 1.06 Human Resources 3.241 2.28 1.21 Marketing 3.371 3.00 1.94 critical values for experimentwise error rate: 0.05 0.01
230
2.76 3.47
Human Resources 3.241
0.73
Marketing 3.371
ASBE 6e Solutions for Instructors
Accounting Finance Human Resources Marketing
2.834 3.024 3.241 3.371
Accounting 2.834
Finance 3.024
Human Resources 3.241
.2986 .0320 .0062
.2365 .0641
.4743
Marketing 3.371
Learning Objective: 11-07 11.14
a. 6. b. c = 4, n – c = 16. c. T4,15 = 2.88. (Table 11.4 used) d. Using Tukey’s simultaneous comparison t-values, Store 1 and Store 2 differ. Using the pairwise t-tests, Store 1 and Store 2 differ at = .01, Store 4 and Store 2 differ at = .05, and Store 1 and Store 3 differ at = .05. Post hoc analysis Tukey simultaneous comparison t-values (d.f. = 16) Store 2 Store 3 Store 4 87.4 91.0 101.0 Store 2 87.4 Store 3 91.0 0.59 Store 4 101.0 2.22 1.63 Store 1 108.0 3.36 2.77 1.14
Store 1 108.0
critical values for experiment wise error rate: 0.05 2.86 0.01 3.67
p-values for pairwise t-tests Store 2 Store 3 Store 4 Store 1 87.4 91.0 101.0 108.0 Store 2 Store 3 Store 4 Store 1
87.4 91.0 101.0 108.0
.5650 .0412 .0040
.1221 .0135
Learning Objective: 11-07 11.15 a.
H0: 12 = 22 = 32 H1: Not all the j2 are equal b. df1 = c = 3, df2 = n/c – 1 = 15/3 – 1 = 4 c. From Table 11.5, H3,4 = 15.5 d. Hcalc = < 15.5 s 2max 929 = = 7.68 s 2min 121 e. Fail to reject the null hypothesis. The variances are not different. Learning Objective: 11-8 231
.2700
ASBE 6e Solutions for Instructors 11.16 a.
H0: 12 = 22 = 32 = 42 = 52 H1: Not all the j2 are equal b. df1 = c = 5, df2 = n/c – 1 = 30/5 – 1 = 5 c. From Table 11.5, H5,5 = 16.3 d. Hcalc = < 16.3 s 2max 462 = = 14.69 s 2min 122 e. Fail to reject the null hypothesis. The variances are not different. Learning Objective: 11-8
For Exercises 11.17 through 11.20, the hypotheses to be tested are: H0: 12 = 22 = ... = c2 H1: Not all the j2 are equal where c = the number of groups. The test statistic is: . smax 2 H calc = smin 2 Hcritical can be found in Table 11.5 using degrees of freedom given by: Numerator df1 = c Denominator df2 = (n/c)–1 (round down to next lower integer if necessary). a. df1 = c = 3, df2 = n/c − 1 = 4. b. From Table 11. 5 we find Hcritical = 15.5 (df1 = c = 3, df2 = (n/c)1 = 4). c. Hcalc = 7.027/2.475 = 2.839. d. We fail to reject the null hypothesis of equal variances because the test statistic Hcalc is less than the critical value. e. This result agrees with Levene’s test (p-value = .843) and the confidence intervals overlap. Mean 12.30
n 5
Std. Dev 1.573
13.96 9.58 11.95
5 5 15
2.077 2.651 2.728
Varianc e 2.475
Treatmen t Plant A
4.313 7.027
Plant B Plant C Total
Test for Equal Variances for Scrap Rate Bartlett's Test Test Statistic P-Value
Plant A
0.95 0.622
Levene's Test Test Statistic P-Value
Plant
11.17
Plant B
Plant C
0
2 4 6 8 10 95% Bonferroni Confidence Intervals for StDevs
Learning Objective: 11-8 232
0.17 0.843
ASBE 6e Solutions for Instructors
11.18
a. df1 = c = 4, df2 = (n/c)1 = 6. b. From Table 11. 5 we find Hcritical = 10.4 (df1 = c = 4, df2 = (n/c)1 = 6). c. Hcalc = 23.90/16.97 = 1.41. d. We fail to reject the null hypothesis of equal variances because the test statistic Hcalc is less than the critical value. e. This result agrees with Levene’s test (p-value = .885) and the confidence intervals overlap.
Mean 28.3 34.2 27.3 32.0 30.2
n 7 6 8 7 28
Std. Dev 4.89 4.12 4.62 4.24 5.08
Variance 23.90 16.97 21.36 18.00
Treatment Physician 1 Physician 2 Physician 3 Physician 4 Total
Test for Equal Variances for Wait Time Bartlett's Test Test Statistic P-Value
Physician 1
0.20 0.978
Physician
Levene's Test Test Statistic P-Value
Physician 2
0.21 0.885
Physician 3
Physician 4
2 4 6 8 10 12 14 95% Bonferroni Confidence Intervals for StDevs
Learning Objective: 11-8 11.19
a. df1 = c = 4, df2 = n/c1 = 6. b. From Table 11. 5 we find Hcritical = 10.4 (df1 = c = 4, df2 = n/c1 = 6). c. Hcalc = (0.2553)/(0.0315) = 8.10. d. We fail to reject the null hypothesis of equal variances because the test statistic Hcalc is less than the critical value. e. This result agrees with Levene’s test (p-value = .145). However, both tests are closer than in Exercises 11.9 and 11.10 (the high variance in accounting is striking, even though the confidence intervals do overlap.). Mea n 2.834 3.024 3.241 3.371 3.118
n 7 7 7 7 28
Std. Dev 0.5053 0.1776 0.3077 0.2575 0.3785
Variance 0.2553 0.0315 0.0947 0.0663
233
Treatment Accounting Finance Human Resources Marketing Total
ASBE 6e Solutions for Instructors Test for Equal Variances for GPA Bartlett's Test Test Statistic P-Value
Accounting
6.36 0.095
Levene's Test Test Statistic P-Value
1.98 0.145
Major
Finance
Human Resources
Marketing
0.0 0.2 0.4 0.6 0.8 1.0 1.2 1.4 1.6 95% Bonferroni Confidence Intervals for StDevs
Learning Objective: 11-8 11.20
a. df1 = c = 4, df2 = (n/c)1 = 4. b. From Table 11. 5 we find Hcritical = 20.6 (df1 = c = 4, df2 = (n/c)1 = 4). c. Hcalc = 123.5/28.5 = 4.33. d. We fail to reject the null hypothesis of equal variances because the test statistic Hcalc is less than the critical value. e. This result agrees with Levene’s test (p-value = .810) and the confidence intervals overlap. Mean 108.0 87.4 91.0 101.0 96.9
n 5 5 5 5 20
Std. Dev 5.34 10.83 11.11 10.30 12.20
Variance 28.50 117.30 123.50 106.00
Treatment Store 1 Store 2 Store 3 Store 4 Total
Test for Equal Variances for Sales Bartlett's Test Test Statistic P-Value
Store 1
2.07 0.558
Levene's Test Test Statistic P-Value
0.32 0.810
Store
Store 2
Store 3
Store 4
0
10 20 30 40 50 95% Bonferroni Confidence Intervals for StDevs
Learning Objective: 11-8 11.21
a.
b.
Date is the blocking factor and Plant is the treatment or research interest. Rows (Date): H0: A1 = A2 = A3 H1: Not all the Aj are equal to zero Columns: (Plant) H0: B1 = B2 = B3 = B4 = 0 H1: Not all the Bk are equal to zero Plant means differ at = .05, F = 41.19, p-value = .0002. Blocking factor (date) also significant, F = 234
ASBE 6e Solutions for Instructors
c. d.
8.62, p-value = .0172. A test statistic of this magnitude would arise about 2 times in 10,000 samples if the null were true. Plot suggests that Plants 1 and 2 are below overall mean, Plants 3 and 4 above. ANOVA table: Two factor without replication Source SS df MS F Treatments (Plant) 216.25 3 72.083 41.19 Blocks (Date) 30.17 2 15.083 8.62 Error 10.50 6 1.750 Total 256.92 11
Mean 20.333 18 29 25
n 3 3 3 3
Std. Dev 1.528 2 2.646 2.646
Factor Level Plant 1 Plant 2 Plant 3 Plant 4
21.5 25.25 22.5
4 4 4
4.041 5.377 5.508
Mar 4 Mar 11 Mar 18
Post hoc analysis Tukey simultaneous comparison t-values (d.f. = 6) Plant 2 Plant 1 18.000 20.333 Plant 2 18.000 Plant 1 20.333 2.16 Plant 4 25.000 6.48 4.32 Plant 3 29.000 10.18 8.02 critical values for experimentwise error rate: 0.05 3.46 0.01 4.97
235
Plant 4 25.000
3.70
p-value .0002 .0172
Plant 3 29.000
ASBE 6e Solutions for Instructors p-values for pairwise t-tests
Plant 2 Plant 1 Plant 4 Plant 3
18.000 20.333 25.000 29.000
Plant 2 18.000
Plant 1 20.333
Plant 4 25.000
.0741 .0006 .0001
.0050 .0002
.0100
Plant 3 29.000
Learning Objective: 11-9 Learning Objective: 11-10 11.22
a.
Vehicle Size is the blocking factor and Fuel Type is the treatment or research interest. Rows (Vehicle Size): H0: A1 = A2 = A3 = A4 H1: Not all the Aj are equal to zero Columns (Fuel Type) H0: B1 = B2 = B3 = B4 = 0 H1: Not all the Bk are equal to zero b. Fuel type means differ at = .05, F = 6.94, p-value = .0039. Blocking factor (Vehicle Size) also significant, F = 34.52, p-value = .0000. c. A test statistic of this magnitude would arise about 39 times in 10,000 samples if the null were true. d. Plot suggests that 89 Octane and 91 Octane are somewhat above the overall mean. The Tukey tests show a significant difference in fuel economy between Ethanol 10% and 89 Octane, Ethanol 10% and 91 Octane, and 87 Octane and 91 Octane. The pairwise t-tests confirm this plus a couple of weaker differences. ANOVA table: Two factor without replication Source SS df MS F p-value Treatments (Fuel Type) 54.065 4 13.5163 6.94 .0039 Blocks (Vehicle Size) 201.612 3 67.2040 34.52 0.0000 Error 23.363 12 1.9469 Total 279.040 19 Mean 22.5750 25.5500 25.8000 22.7500 21.8250 28.0200 25.4200 21.1600 20.2000 23.7000
n 4 4 4 4 4 5 5 5 5 20
Std. Dev 3.5575 3.2254 4.2716 4.5625 3.5874 2.0130 2.3392 1.4293 2.7911 3.8323
Group 87 Octane 89 Octane 91 Octane Ethanol 5% Ethanol 10% Compact Mid-Size Full-Size SUV Total
236
ASBE 6e Solutions for Instructors Post hoc analysis Tukey simultaneous comparison t-values (d.f. = 12) Ethanol 10% 87 Octane 21.8250 22.5750 Ethanol 10% 21.8250 87 Octane 22.5750 0.76 Ethanol 5% 22.7500 0.94 0.18 89 Octane 25.5500 3.78 3.02 91 Octane 25.8000 4.03 3.27
Ethanol 5% 22.7500
89 Octane 25.5500
2.84 3.09
0.25
87 Octane 22.5750
Ethanol 5% 22.7500
89 Octane 25.5500
.8622 .0108 .0067
.0150 .0093
.8043
91 Octane 25.8000
critical values for experiment wise error rate: 0.05 3.19 0.01 4.13
p-values for pairwise t-tests Ethanol 10% 21.8250 Ethanol 10% 21.8250 87 Octane 22.5750 .4618 Ethanol 5% 22.7500 .3670 89 Octane 25.5500 .0026 91 Octane 25.8000 .0017
Learning Objective: 11-9 Learning Objective: 11-10
237
91 Octane 25.8000
ASBE 6e Solutions for Instructors 11.23
a.
b. c. d.
Hotel is the blocking factor and Driver is the treatment or research interest. Rows (Hotel): H0: A1 = A2 = A3 = A4 H1: Not all the Aj are equal to zero Columns (Driver) H0: B1 = B2 = B3 = B4 = B5= 0 H1: Not all the Bk are equal to zero Driver means are on the borderline at = .05, F = 3.26, p-value = .0500. Blocking factor (Hotel) is not significant, F = 1.11, p-value = .3824. A test statistic of this magnitude would arise about 5 times in 100 samples if the null were true. Driver averages appear to be significantly different. In particular drivers C and E have the largest time difference. ANOVA table: Two factor without replication Source SS df MS Treatments (Driver) 134.403 4 33.6008 Blocks (Hotel) 34.404 3 11.4680 Error 123.721 12 10.3101 Total 292.528 19
4 4 4 4 4
Std. Dev 4.8321 2.3977 3.1859 3.1591 1.8655
Group Driver A Driver B Driver C Driver D Driver E
5 5 5 5 20
6.1064 3.0971 2.5530 3.3366 3.9238
Hotel 1 Hotel 2 Hotel 3 Hotel 4 Total
Mean 26.3250 25.2250 30.2500 26.7000 22.2000
n
28.0400 25.1200 26.6600 24.7400 26.1400
F 3.26 1.11
p-value .0500 .3824
Learning Objective: 11-9 Learning Objective: 11-10 11.24
a.
b. c. d.
Qtr is the blocking factor and Store is the treatment or research interest. Rows (Qtr): H0: A1 = A2 = A3 = A4 H1: Not all the Aj are equal to zero Columns (Store) H0: B1 = B2 = B3 = 0 H1: Not all the Bk are equal to zero Store means do not differ at = .05, F = 1.60, p-value = .2770. Blocking factor (Qtr) is significant, F = 15.58, p-value = .0031. A test statistic of this magnitude would arise about 28 times in 100 samples if the null were true. Plot shows no consistent differences in means for stores. The Tukey tests and the pairwise tests are not calculated since the treatments do not significantly affect the sales.
238
ASBE 6e Solutions for Instructors
Source Treatments (Store) Blocks (Qtr) Error Total
ANOVA table: Two factor without replication SS df MS 41,138.67 2 20,569.333 601,990.92 3 200,663.639 77,277.33 6 12,879.556 720,406.92 11
Mean 1,456.250 1,375.250 1,518.250
n 4 4 4
Std. Dev 231.073 261.153 323.770
Store 1 Store 2 Store 3
1,509.000 1,423.333 1,120.667 1,746.667 1,449.917
3 3 3 3 12
205.263 59.878 63.760 97.079 255.913
Qtr 1 Qtr 2 Qtr 3 Qtr 4 Total
F 1.60 15.58
p-value .2779 .0031
Learning Objective: 11-9 Learning Objective: 11-10
11.25 Factor A: Row Effect (Year) H0: A1 = A2 = A3 = 0 year means are the same H1: Not all the Aj are equal to zero year means differ Factor B: Column Effect (Portfolio Type) H0: B1 = B2 = B3 = B4 = 0 stock portfolio type means are the same H1: Not all the Bk are equal to zero stock portfolio type means differ Interaction Effect (YearPortfolio) H0: All the ABjk are equal to zero there is no interaction effect H1: Not all ABjk are equal to zero there is an interaction effect b. See tables. c. Years differ at = .05, F = 66.82, p-value < .0001. Portfolios differ at = .05, F = 5.48, p-value = .0026. 239
ASBE 6e Solutions for Instructors
d. e.
Interaction is significant at = .05, F = 4.96, p-value = .0005. The small p-values indicate that the sample would be unlikely if the null were true. The interaction plot lines do cross and support the interaction found and reported above. The visual indications of interaction are strong for the portfolio returns data.
Factor 1 (Year) 2004 2005 2006
Source Factor 1 (Year) Factor 2 (Portfolio) Interaction Error Total
Table of Means Factor 2 (Portfolio Type) Health Energy Retail 15.74 22.20 18.36 22.84 27.98 23.92 13.24 12.62 19.90 17.27 20.93 20.73
Leisure 18.52 25.46 10.98 18.32
Two-Factor ANOVA with Replication SS df MS 1,191.584 2 595.7922 146.553 3 48.8511 265.192 6 44.1986 427.980 48 8.9162 2,031.309 59
18.71 25.05 14.19 19.31
F 66.82 5.48 4.96
p-value 1.34E-14 .0026 .0005
Post hoc analysis for Factor 1 Tukey simultaneous comparison t-values (d.f. = 48) Row 3 Row 1 Row 2 14.19 18.71 25.05 Row 3 14.19 Row 1 18.71 4.79 Row 2 25.05 11.51 6.72 critical values for experiment wise error rate: 0.05 2.42 0.01 3.07
Post hoc analysis for Factor 2 Tukey simultaneous comparison t-values (d.f. = 48) Health Leisure Retail 17.27 18.32 20.73 Health 17.27 Leisure 18.32 0.96 Retail 20.73 3.17 2.21 Energy 20.93 3.36 2.40 0.19 critical values for experiment wise error rate: 0.05 2.66 0.01 3.29 240
Energy 20.93
ASBE 6e Solutions for Instructors
Learning Objective: 11-11 11.26 Factor A: Row Effect (Year) H0: A1 = A2 = A3 = 0 year means are the same H1: Not all the Aj are equal to zero year means differ Factor B: Column Effect (Department) H0: B1 = B2 = B3 = 0 department means are the same H1: Not all the Bk are equal to zero department type means differ Interaction Effect (YearDepartment) H0: All the ABjk are equal to zero there is no interaction effect H1: Not all ABjk are equal to zero there is an interaction effect b. See tables. c. Years do not differ at = .05, F = 0.64, p-value = .5365. Departments differ at = .05, F = 12.66, pvalue = .0004. Interaction is not significant at = .05, F = 2.38, p-value = .0899. d. The p-values range from highly significant (Department) to insignificant (Year). The interaction effect, if any, is weak since about 9 samples in 100 would show an F statistic this large in the absence of interaction. e. The interaction plot lines do cross for Department, but are approximately parallel for Year and support the lack of interaction found and reported above. The visual indications of interaction are, therefore, non-existent for the team ratings.
Factor 1 (Year) 2004 2005 2006
Source Factor 1 (Year) Factor 2 (Department) Interaction Error Total
Table of Means Factor 2 (Department) Marketing Engineering Finance 84.7 73.0 89.3 79.0 77.0 89.7 88.7 79.7 84.7 84.1 76.6 87.9 Two factor ANOVA with Replication SS df MS 30.52 2 15.259 599.41 2 299.704 225.48 4 56.370 426.00 18 23.667 1,281.41 26
241
F 0.64 12.66 2.38
82.3 81.9 84.3 82.9
p-value .5365 .0004 .0899
ASBE 6e Solutions for Instructors Post hoc analysis for Factor 2 Tukey simultaneous comparison t-values (d.f. = 18) Engineering Marketing 76.6 84.1 Engineering 76.6 Marketing 84.1 3.29 Finance 87.9 4.94 1.65
Finance 87.9
critical values for experimentwise error rate: 0.05 2.55 0.01 3.32
Learning Objective: 11-11
11.27 Factor A: Row Effect (Age Group) H0: A1 = A2 = A3 = A4 = 0 age group means are the same H1: Not all the Aj are equal to zero age group means differ Factor B: Column Effect (Region) H0: B1 = B2 = B3 = B4= 0 region means are the same H1: Not all the Bk are equal to zero region means differ Interaction Effect (Age GroupRegion) H0: All the ABjk are equal to zero there is no interaction effect H1: Not all ABjk are equal to zero there is an interaction effect b. See tables. c. Age groups differ at = .05, F = 36.96, p-value < .0001. Regions do not differ at = .05, F = 0.55, pvalue = .6493. Interaction is significant at = .05, F = 3.66, p-value = .0010. d. The p-values range from highly significant (Age Group) to insignificant (Region). The interaction effect is significant since only about 1 sample in 1000 would show an F statistic this large in the absence of interaction. e. The interaction plot lines do cross (e.g., MidWest crosses the others by age group) but visually there is not a strong indication of interaction. This is perhaps because the data range is not large (data appear to be rounded to nearest .1 so there is only 2-digit accuracy).
242
ASBE 6e Solutions for Instructors
Factor 1 (Age Group) Youth (under 18) College (18-25) Adult (25-64) Senior (65 +)
Source Factor 1 (Age Group) Factor 2 (Region) Interaction Error Total
Northeast 4.00 3.86 3.50 3.42 3.70
Table of Means Factor 2 (Region) Southeast Midwest 4.12 3.68 3.70 3.88 3.42 3.76 3.52 3.18 3.69 3.63
Two factor ANOVA with Replication SS df MS 4.193 3 1.3975 0.062 3 0.0208 1.245 9 0.1383 2.420 64 0.0378 7.920 79
Post hoc analysis for Factor 1 Tukey simultaneous comparison t-values (d.f. = 64) Senior (65 +) 3.37 Senior (65 +) 3.37 Adult (25-64) 3.55 2.93 College (18-25) 3.81 7.07 Youth (under 18) 3.97 9.68 critical values for experimentwise error rate: 0.05 0.01
Learning Objective: 11-11
243
West 4.06 3.78 3.52 3.36 3.68
3.97 3.81 3.55 3.37 3.67
F 36.96 0.55 3.66
p-value 5.56E-14 .6493 .0010
Adult (25-64) 3.55
College (18-25) 3.81
Youth (Under 18) 3.97
4.15 6.75
2.60
2.64 3.25
ASBE 6e Solutions for Instructors 11.28 Factor A: Row Effect (Quarter) H0: A1 = A2 = A3 = A4 = 0 quarter means are the same H1: Not all the Aj are equal to zero quarter means differ Factor B: Column Effect (Supplier) H0: B1 = B2 = B3 = 0 supplier means are the same H1: Not all the Bk are equal to zero supplier means differ Interaction Effect (QuarterSupplier) H0: All the ABjk are equal to zero there is no interaction effect H1: Not all ABjk are equal to zero there is an interaction effect b. See tables. c. Quarters differ at = .05, F = 6.01, p-value < .0020. Suppliers differ at = .05, F = 4.30, p-value = . 0211. Interaction is not significant at = .05, F = 0.44, p-value = .8446. d. The p-values indicate that both main effects are significant. The interaction effect is not significant, since about 84 samples in 100 would show an F statistic this large in the absence of interaction. e. The interaction plot lines do not cross to a noticeable degree, so we see no evidence of interaction.
Factor 1 (Quarter) Qtr 1 Qtr 2 Qtr 3 Qtr 4
Table of Means Factor 2 (Supplier) Supplier 1 Supplier 2 Supplier 3 12.3 10.8 14.3 12.3 11.0 12.3 10.3 8.5 10.0 10.5 9.5 10.8 11.3 9.9 11.8
12.4 11.8 9.6 10.3 11.0
Two factor ANOVA with Replication Source Factor 1 (Quarter) Factor 2 (Supplier) Interaction Error Total
SS 63.23 30.17 9.33 126.25 228.98
df 3 2 6 36 47
MS 21.076 15.083 1.556 3.507
Post hoc analysis for Factor 1 Tukey simultaneous comparison t-values (d.f. = 36) Qtr 3 Qtr 4 9.6 10.3 Qtr 3 9.6 Qtr 4 10.3 0.87 Qtr 2 11.8 2.94 2.07 Qtr 1 12.4 3.71 2.83 critical values for experimentwise error rate: 0.05 2.70 0.01 3.35
244
F 6.01 4.30 0.44
Qtr 2 11.8
0.76
pvalue .0020 .0211 .8446
Qtr 1 12.4
ASBE 6e Solutions for Instructors Post hoc analysis for Factor 2 Tukey simultaneous comparison t-values (d.f. = 36) Supplier 2 Supplier 1 Supplier 3 9.9 11.3 11.8 Supplier 2 9.9 Supplier 1 11.3 2.08 Supplier 3 11.8 2.83 0.76 critical values for experimentwise error rate: 0.05 2.45 0.01 3.11
Learning Objective: 11-11
11.29
We fail to reject the null hypothesis of equal means. The p-value (.1000) exceeds .05. There is no significant difference among the GPAs. We ignore importance, since the results are not significant. The dot plot comparison confirms that differences are not strong. For tests of equal variances, from table 11.5 Hcritical = 13.7 (df1 = c = 4, df2 = (n/c)1 = 25/4 1 = 5). We fail to reject the null hypothesis of homogeneous variances since the test statistic Hcalc = (0.3926)/(.0799) = 4.91 is less than the critical value. This result agrees with Levene’s test (p-value = .606) and the confidence intervals overlap. Mean 2.484 2.916 3.227 3.130 2.968
n 5 7 7 6 25
Std. Dev 0.6240 0.6265 0.2826 0.4447 0.5477
Variance 0.3894 0.3926 0.0799 0.1978
Group Freshman Sophomore Junior Senior Total
One factor ANOVA Source Treatment Error Total
SS 1.8180 5.3812 7.1992
df 3 21 24
245
MS 0.60599 0.25625
F 2.36
pvalue .1000
ASBE 6e Solutions for Instructors
Test for Equal Variances for GPA Bartlett's Test Test Statistic P-Value
Freshman
3.75 0.290
Levene's Test Test Statistic P-Value
0.63 0.606
Class
Junior
Senior
Sophomore
0.0 0.5 1.0 1.5 2.0 2.5 95% Bonferroni Confidence Intervals for StDevs
Learning Objective: 11-1 Learning Objective: 11-2 Learning Objective: 11-3 Learning Objective: 11-4 Learning Objective: 11-5 Learning Objective: 11-6 Learning Objective: 11-7 Learning Objective: 11-8 Learning Objective: 11-9 Learning Objective: 11-10 11.30
We fail to reject the null hypothesis of equal means. The p-value (.0523) exceeds .05, although it is a very close decision. We ignore importance, since the results are not significant. The dot plot does suggest that differences exist. A larger sample might be in order. For tests of equal variances, from table 11.5 Hcritical = 8.38 (df1 = c = 3, df2 = (n/c)1 = 23/3 1 = 6). We fail to reject the null hypothesis of homogeneous variances since the test statistic Hcalc = (451.11)/(89.41) = 5.05 is less than the critical value. This result agrees with Levene’s test (p-value = .140) and the confidence intervals overlap. However, the variances would have been judged unequal had we used = .10. Mean 261.2 238.0 244.4 245.3
n 5 10 8 23
Std. Dev 11.95 21.24 9.46 17.91
246
Variances 142.70 451.11 89.41
Group Budgets Payables Pricing Total
ASBE 6e Solutions for Instructors One factor ANOVA Source Treatment Error Total
SS 1,803.76 5,256.68 7,060.43
df 2 20 22
MS 901.880 262.834
pvalue .0523
F 3.43
Test for Equal Variances for Days Bartlett's Test Test Statistic P-Value
Budgets
4.77 0.092
Levene's Test
Dept
Test Statistic P-Value
2.17 0.140
Payables
Pricing
0
10 20 30 40 50 95% Bonferroni Confidence Intervals for StDevs
Learning Objective: 11-1 Learning Objective: 11-2 Learning Objective: 11-3 Learning Objective: 11-4 Learning Objective: 11-5 Learning Objective: 11-6 Learning Objective: 11-7 Learning Objective: 11-8 Learning Objective: 11-9 Learning Objective: 11-10 11.31
We reject the null hypothesis of equal means. The p-value (.0022) is less than .05. Even a small difference in output could be important in a large array of solar cells. The dot plot does suggest that differences exist. Cell Type C is above the overall mean, while Cell Type B is below the overall mean. For tests of equal variances, from table 11.5 Hcritical = 10.8 (df1 = c = 3, df2 = (n/c)1 = 18/3 1 = 5). We fail to reject the null hypothesis of homogeneous variances since the test statistic Hcalc = (4.57)/(4.00) = 1.14 is less than the critical value. This result agrees with Levene’s test (p-value = .975) and the confidence intervals overlap. Tukey tests show that C differs from A and B.
Mean
n
Std. Dev 247
Variances
Group
ASBE 6e Solutions for Instructors 123.8 123.0 127.8 124.9
6 6 6 18
2.04 2.00 2.14 2.91
4.17 4.00 4.57
Cell Type A Cell Type B Cell Type C Total
One factor ANOVA Source Treatment Error Total
SS 80.11 63.67 143.78
df 2 15 17
MS 40.056 4.244
F 9.44
Post hoc analysis Tukey simultaneous comparison t-values (d.f. = 15) Cell Type Cell Type Cell Type B A C 123.0 123.8 127.8 Cell Type B 123.0 Cell Type A 123.8 0.70 Cell Type C 127.8 4.06 3.36 critical values for experimentwise error rate: 0.05 2.60 0.01 3.42 p-values for pairwise t-tests Cell Type B 123.0 Cell Type B 123.0 Cell Type A 123.8 .4943 Cell Type C 127.8 .0010
248
Cell Type A 123.8
.0043
Cell Type C 127.8
pvalue .0022
ASBE 6e Solutions for Instructors Test for Equal Variances for Watts Bartlett's Test Test Statistic P-Value
Cell Type A
0.02 0.989
Levene's Test
Cell Type
Test Statistic P-Value
0.03 0.975
Cell Type B
Cell Type C
1 2 3 4 5 6 7 95% Bonferroni Confidence Intervals for StDevs
Learning Objective: 11-1 Learning Objective: 11-2 Learning Objective: 11-3 Learning Objective: 11-4 Learning Objective: 11-5 Learning Objective: 11-6 Learning Objective: 11-7 Learning Objective: 11-8 Learning Objective: 11-9 Learning Objective: 11-10 11.32
We cannot reject the null hypothesis of equal means. The p-value (.4188) exceeds .05. Since the means do not differ significantly, the issue of importance is moot. The dot plot does not suggest any differences. For tests of equal variances, from table 11.5 Hcritical = 15.5 (df1 = c = 3, df2 = (n/c)1 = 15/3 1 = 4). We fail to reject the null hypothesis of homogeneous variances since the test statistic Hcalc = (246,324)/ (103,581) = 2.38 is less than the critical value. This result agrees with Levene’s test (p-value = .569) and the confidence intervals overlap. Mean 1,282.0 1,376.0 1,638.0 1,432.0
Source Treatment Error Total
n 5 5 5 15
Std. Dev 496.31 321.84 441.78 424.32
Variance 246,324 103,581 195,170
One factor ANOVA d SS f MS 340,360.00 2 170,180.000 2,180,280.00 12 181,690.000 2,520,640.00 14
249
Group Goliath Varmint Weasel Total
F 0.94
p-value .4188
ASBE 6e Solutions for Instructors
Test for Equal Variances for Damage Bartlett's Test Test Statistic P-Value
Goliath
0.67 0.715
Levene's Test
Vehicle
Test Statistic P-Value
0.59 0.569
Varmint
Weasel
0
500 1000 1500 2000 95% Bonferroni Confidence Intervals for StDevs
Learning Objective: 11-1 Learning Objective: 11-2 Learning Objective: 11-3 Learning Objective: 11-4 Learning Objective: 11-5 Learning Objective: 11-6 Learning Objective: 11-7 Learning Objective: 11-8 Learning Objective: 11-9 Learning Objective: 11-10 11.33
We cannot reject the null hypothesis of equal means. The p-value (.1857) exceeds .05. Since the means do not differ significantly, the issue of importance is moot. The dot plot does not suggest any differences. For tests of equal variances, from table 11.5 Hcritical = 20.6 (df1 = c = 4, df2 = (n/c)1 = 22/4 1 = 4). We fail to reject the null hypothesis of homogeneous variances since the test statistic Hcalc = (141.610)/ (45.428) = 3.12 is less than the critical value. This result agrees with Levene’s test (p-value = .645) and the confidence intervals overlap. Mean 14.2 21.5 16.9 9.3 15.0
n 5 4 7 6 22
Std. Dev 8.29 11.90 7.95 6.74 8.98
250
Variances 68.724 141.610 63.203 45.428
Group Hospital A Hospital B Hospital C Hospital D Total
ASBE 6e Solutions for Instructors One factor ANOVA SS df MS 388.96 3 129.655 1,305.99 18 72.555 1,694.95 21
Source Treatment Error Total
F 1.79
p-value .1857
Test for Equal Variances for Wait Bartlett's Test Test Statistic P-Value
Hospital A
1.26 0.739
Levene's Test Test Statistic P-Value
0.56 0.645
Hospital
Hospital B
Hospital C
Hospital D
0 10 20 30 40 50 60 70 80 95% Bonferroni Confidence Intervals for StDevs
Learning Objective: 11-1 Learning Objective: 11-2 Learning Objective: 11-3 Learning Objective: 11-4 Learning Objective: 11-5 Learning Objective: 11-6 Learning Objective: 11-7 Learning Objective: 11-8 Learning Objective: 11-9 Learning Objective: 11-10 11.34
We reject the null hypothesis of equal means. The p-value (.0029) is less than .05. Productivity differences could be important in a competitive market, and might signal a need for additional worker training. The dot plot suggests that Plant B has lower productivity. For tests of equal variances, from table 11.5 Hcritical = 6.94 (df1 = c = 3, df2 = (n/c)1 = 25/3 1 = 7). We fail to reject the null hypothesis of homogeneous variances since the test statistic Hcalc = (2.9791)/(0.68558) = 4.35 is less than the critical value. This result agrees with Levene’s test (p-value = .122) and the confidence intervals overlap.
251
ASBE 6e Solutions for Instructors Mean 3.97 3.02 5.57 4.38
n 9 6 10 25
Variance 0.685584 1.196836 2.979076
Std. Dev 0.828 1.094 1.726 1.647
Group Plant A Plant B Plant C Total
One factor ANOVA Source Treatment Error Total
SS 26.851 38.269 65.120
df 2 22 24
MS 13.4253 1.7395
F 7.72
Post hoc analysis Tukey simultaneous comparison t-values (d.f. = 22) Plant B Plant A Plant C 3.02 3.97 5.57 Plant B 3.02 Plant A 3.97 1.37 Plant C 5.57 3.75 2.65 critical values for experimentwise error rate: 0.05 2.52 0.01 3.25 p-values for pairwise t-tests Plant B 3.02 Plant B 3.02 Plant A 3.97 .1855 Plant C 5.57 .0011
252
Plant A 3.97
.0148
Plant C 5.57
pvalue .0029
ASBE 6e Solutions for Instructors Test for Equal Variances for Output Bartlett's Test Test Statistic P-Value
Plant A
4.21 0.122
Levene's Test
Plant
Test Statistic P-Value
1.69 0.208
Plant B
Plant C
0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0 95% Bonferroni Confidence Intervals for StDevs
Learning Objective: 11-1 Learning Objective: 11-2 Learning Objective: 11-3 Learning Objective: 11-4 Learning Objective: 11-5 Learning Objective: 11-6 Learning Objective: 11-7 Learning Objective: 11-8 Learning Objective: 11-9 Learning Objective: 11-10 11.35
It appears that the researcher is not treating this as a randomized block, since both factors appear to be of research interest. Hence, this will be referred to as a two-factor ANOVA without replication. Factor A (Method): H0: A1 = A2 = A3 H1: Not all the Aj are equal to zero Factor B: (Road Condition) H0: B1 = B2 = B3 H1: Not all the Bk are equal to zero Mean stopping distance is significantly affected by surface (p = 0.0002) but not by road condition (p = 0.5387). Tukey tests show significant differences between Ice and the other two surfaces. To test for homogeneous variances use Hcritical = 87.5 (df1 = 3 and df2 = 2). Since Hcalc = 1.37 (for Method) and Hcalc = 14.5 (for Surface) we cannot reject the hypothesis of equal variances.
Mean 452.000 184.667 154.000 271.000 249.667 270.000 263.556
Table of Means n Std. Dev Group 3 9.849 Ice 3 37.528 Split Traction 3 11.358 Packed Snow 3 3 3 9
253
151.803 177.827 164.739 143.388
Pumping Locked ABS Total
ASBE 6e Solutions for Instructors
Source Column (Surface) Row (Method) Error Total
Two-Factor ANOVA Without Replication SS df MS 161,211.56 2 80,605.778 869.56 2 434.778 2,399.11 4 599.778 164,480.22 8
F 134.39 0.72
Post hoc analysis Tukey simultaneous comparison t-values (d.f. = 4) Packed Snow Split Traction 154.000 184.667 Packed Snow 154.000 Split Traction 184.667 1.53 Ice 452.000 14.90 13.37
p-value .0002 .5387
Ice 452.000
critical values for experimentwise error rate: 0.05 0.01
3.56 5.74
p-values for pairwise t-tests Packed Snow Split Traction 154.000 184.667 Packed Snow Split Traction Ice
154.000 184.667 452.000
.1999 .0001
Ice 452.000
.0002
Learning Objective: 11-1 Learning Objective: 11-2 Learning Objective: 11-3 Learning Objective: 11-4 Learning Objective: 11-5 Learning Objective: 11-6 Learning Objective: 11-7 Learning Objective: 11-8 Learning Objective: 11-9 Learning Objective: 11-10 11.36
We cannot reject the null hypothesis of equal means. The p-value (.9666) exceeds .05. The dot plot does not show large differences between the quarters. For tests of homogeneity of variances, from table 11.5 Hcritical = 6.31 (df1 = c = 4, df2 =( n/c)1 = 40/4 1 = 9). We fail to reject the null hypothesis of homogeneous variances since the test statistic Hcalc = (19,359.94)/(5692.70) = 3.401 is less than the critical value. This result agrees with Levene’s test (p-value = .412). 254
ASBE 6e Solutions for Instructors Table of Means Mean 522.0 536.0 546.0 541.0 536.3
Source Treatment Error Total
n 10 10 10 10 40
Std. Dev 84.30 139.14 75.45 130.08 106.81
Variances 7,106.49 19,359.94 5,692.70 16,920.81
One factor ANOVA SS df MS 3,207.50 3 1,069.167 441,730.00 36 12,270.278 444,937.50 39
Group Fall Winter Spring Summer Total
F 0.0871
p-value .9666
Comparison of Groups 800.0 700.0 600.0 500.0 400.0 300.0 200.0
Fall
Winter
Spring
Summer
Learning Objective: 11-1 Learning Objective: 11-2 Learning Objective: 11-3 Learning Objective: 11-4 Learning Objective: 11-5 Learning Objective: 11-6 Learning Objective: 11-7 Learning Objective: 11-8 Learning Objective: 11-9 Learning Objective: 11-10 11.37
We reject the null hypothesis of equal means. The p-value (.0019) is less than .05. The average number of days each city’s offices were open do differ significantly. The dot plot suggests that Chalmers is open more than the other cities. For tests of homogeneity of variances, from table 11.5 Hcritical = 25.2 (df1 = c = 5, df2 = (n/c)1 = 26/5 1 = 4). We fail to reject the null hypothesis of homogeneous variances since the test statistic Hcalc = (124.77)/(46.92) = 2.659 is less than the critical value. This result agrees with Levene’s test (p-value = .910).
255
ASBE 6e Solutions for Instructors Table of Means Mean 218.8 198.3 210.0 192.0 199.4 204.7
Source Treatment Error Total
SS 2,480.33 2,104.78 4,585.12
n 6 4 6 5 5 26
Std. Dev Group 10.83 Chalmers 6.85 Greenburg 11.17 Villa Nueve 9.97 Ulysses 9.42 Hazeltown 13.54 Total
One Factor ANOVA df MS 4 620.083 21 100.228 25
Post hoc analysis p-values for pairwise t-tests Ulysses 192.0 Ulysses 192.0 Greenburg 198.3 .3626 Hazeltown 199.4 .2556 Villa Nueve 210.0 .0073 Chalmers 218.8 .0002
Greenburg 198.3
.8657 .0833 .0045
Tukey simultaneous comparison t-values (d.f. = 21) Ulysses Greenburg 192.0 198.3 Ulysses 192.0 Greenburg 198.3 0.93 Hazeltown 199.4 1.17 0.17 Villa Nueve 210.0 2.97 1.82 Chalmers 218.8 4.43 3.19 critical values for experimentwise error rate: 0.05 0.01
2.98 3.72
Learning Objective: 11-1 Learning Objective: 11-2 Learning Objective: 11-3 Learning Objective: 11-4 Learning Objective: 11-5 Learning Objective: 11-6 Learning Objective: 11-7 Learning Objective: 11-8 Learning Objective: 11-9 Learning Objective: 11-10
256
F 6.19
p-value .0019
Hazeltown Villa Nueve 199.4 210.0
.0950 .0042
.1414
Hazeltown Villa Nueve 199.4 210.0
1.75 3.21
Chalmers 218.8
1.53
Chalmers 218.8
ASBE 6e Solutions for Instructors 11.38
Factor A: Row Effect (Well Depth) H0: A1 = A2 = A3 = 0 well depth means are the same H1: Not all the Aj are equal to zero well depth means differ Factor B: Column Effect (Age of Well) H0: B1 = B2 = B3 = 0 age of well means are the same H1: Not all the Bk are equal to zero age of well means differ Interaction Effect (DepthAge) H0: All the ABjk are equal to zero there is no interaction effect H1: Not all ABjk are equal to zero there is an interaction effect Depth of wells differ at = .05, F = 25.78, p-value < .0001. Age of wells differ at = .05, F = 3.84, pvalue = .0409. Interaction is not significant at = .05, F = .017 p-value = .9500. The interaction plot lines do not cross and so do not support that an interaction effect exists. Table of Means
Factor 1: Depth of well
Shallow Medium Deep
ANOVA table Source Factor 1 Factor 2 Interaction Error Total
Factor 2: Age of Well 10 to 19 20 and Over 5.33 5.97 4.40 4.87 3.30 3.60 4.34 4.81
Under 10 5.27 3.80 2.67 3.91
SS 24.500 3.647 0.327 8.553 37.027
df 2 2 4 18 26
MS 12.2500 1.8233 0.0817 0.4752
Post hoc analysis p-values for pairwise t-tests for Factor 1 Deep 3.19 Deep 3.19 Medium 4.36 .0021 Shallow 5.52 1.10E-06 Tukey simultaneous comparison t-values (d.f. = 18) Deep 3.19 Deep 3.19 Medium 4.36 3.59 Shallow 5.52 7.18 critical values for experimentwise error rate: 0.05 0.01
257
F 25.78 3.84 0.17
Medium 4.36
5.52 4.36 3.19 4.36
p-value 5.20E-06 .0409 .9500
Shallow 5.52
.0021
Medium 4.36
3.59
2.55 3.32
Shallow 5.52
ASBE 6e Solutions for Instructors Post hoc analysis p-values for pairwise t-tests for Factor 2 Under 10 3.91 Under 10 3.91 10 to 19 4.34 .1990 20 and Over 4.81 .0126
Tukey simultaneous comparison t-values (d.f. = 18) Under 10 3.91 Under 10 3.91 10 to 19 4.34 1.33 20 and Over 4.81 2.77 critical values for experimentwise error rate: 0.05 0.01
Learning Objective: 11-1 Learning Objective: 11-2 258
10 to 19 4.34
20 and Over 4.81
.1681
10 to 19 4.34
1.44
2.55 3.32
20 and Over 4.81
ASBE 6e Solutions for Instructors Learning Objective: 11-3 Learning Objective: 11-4 Learning Objective: 11-5 Learning Objective: 11-6 Learning Objective: 11-7 Learning Objective: 11-8 Learning Objective: 11-9 Learning Objective: 11-10 11.39
We cannot reject the null hypothesis of equal means. The p-value (.8166) exceeds .05. The dot plot does not show large differences among groups, although the fourth quintile seems to have smaller variance. For tests of homogeneity of variances, from table 11.5 Hcritical = 7.11 (df1 = c = 5, df2 = (n/c)1 = 50/5 1 = 9). We reject the null hypothesis of homogeneous variances since the test statistic Hcalc = 112.04/14.13 = 7.93 exceeds the critical value. This result agrees with Levene’s test (p-value = .018) even though the confidence intervals do overlap.
Mean 30.00 29.12 31.17 28.71 26.66 29.13
Source Treatment Error Total
n 10 10 10 10 10 50
Table of Means Std. Dev 9.548 10.213 10.585 3.759 6.305 8.286
Variance
One Factor ANOVA SS df MS 111.959 4 27.9897 3,252.570 45 72.2793 3,364.529 49
259
Group Quintile 1 Quintile 2 Quintile 3 Quintile 4 Quintile 5 Total
F 0.39
p-value .8166
ASBE 6e Solutions for Instructors Test for Equal Variances for Dropout Bartlett's Test
Quintile 1
Test Statistic P-Value
10.28 0.036
Levene's Test Test Statistic P-Value
Quintile
Quintile 2
3.33 0.018
Quintile 3
Quintile 4
Quintile 5 0
5 10 15 20 25 95% Bonferroni Confidence Intervals for StDevs
Learning Objective: 11-1 Learning Objective: 11-2 Learning Objective: 11-3 Learning Objective: 11-4 Learning Objective: 11-5 Learning Objective: 11-6 Learning Objective: 11-7 Learning Objective: 11-8 Learning Objective: 11-9 Learning Objective: 11-10 11.40
This is a replicated experiment with two factors and interaction. Based on the p-values, we conclude that means differ for Angle (p = .0088) and for Vehicle(p = .0007). However, there is no significant interaction for AngleVehicle (p = .6661). The interaction plots support this conclusion, as the lines do not cross. The Tukey tests say that pairwise means differ for Rear End and Slant, and that Goliath differs from Varmint and Weasel.
Factor 1 (Angle) Head-On Slant Rear end
Table of Means Factor 2 (Vehicle) Goliath Varmint 983.3 1,660.0 1,470.0 1,733.3 973.3 1,220.0 1,142.2 1,537.8
Weasel 1,896.7 1,996.7 1,513.3 1,802.2
1,513.3 1,733.3 1,235.6 1,494.1
Two Factor ANOVA with Replication Source Factor 1 (Angle) Factor 2 (Vehicle) Interaction Error Total
SS 1,120,029.63 1,985,985.19 216,637.04 1,619,400.00 4,942,051.85
df 2 2 4 18 26
Post hoc analysis for Factor 1 260
MS 560,014.815 992,992.593 54,159.259 89,966.667
F 6.22 11.04 0.60
pvalue .0088 .0007 .6661
ASBE 6e Solutions for Instructors Tukey simultaneous comparison t-values (d.f. = 18) Rear end Head-On 1,235.6 1,513.3 Rear end 1,235.6 Head-On 1,513.3 1.96 Slant 1,733.3 3.52 1.56 critical values for experimentwise error rate: 0.05 0.01
2.55 3.32
Post hoc analysis for Factor 2 Tukey simultaneous comparison t-values (d.f. = 18) Goliath Varmint 1,142.2 1,537.8 Goliath 1,142.2 Varmint 1,537.8 2.80 Weasel 1,802.2 4.67 1.87 critical values for experimentwise error rate: 0.05 0.01
Learning Objective: 11-1 Learning Objective: 11-2 Learning Objective: 11-3 Learning Objective: 11-4 Learning Objective: 11-5 Learning Objective: 11-6 Learning Objective: 11-7 Learning Objective: 11-8 Learning Objective: 11-9 Learning Objective: 11-10
261
Slant 1,733.3
2.55 3.32
Weasel 1,802.2
ASBE 6e Solutions for Instructors
11.41
This is a replicated experiment with two factors and interaction. Based on the p-values, the means differ for Brand (p = .0284) and for Store (p = .0021). However, while the interaction plots show some interaction it is not highly significant for BrandStore (p = .0712). The pairwise Tukey tests indicate that there is a difference between Brand A and Brand D, and between Store 1 and Store 3.
Factor 1 (Brand of Chips) Brand A Brand B Brand C
Table of Means Factor 2 (Store) Store 1 Store 2 10.0 21.5 17.5 11.0 14.5 19.5 15.3 35.0
Store 3 15.5 35.8 25.8 36.3
15.7 21.4 19.9 28.8
Two Factor ANOVA with Repetition Source Factor 1 Factor 2 Interaction Error Total
SS 1,083.75 1,570.04 1,376.13 3,840.00 7,869.92
df 3 2 6 36 47
MS 361.250 785.021 229.354 106.667
Post hoc analysis p-values for pairwise t-tests for Factor 1 Brand A 15.7 Brand A 15.7 Brand C 19.9 .3202 Brand B 21.4 .1811 Brand D 28.8 .0035
Brand C 19.9
.7241 .0414
Tukey simultaneous comparison t-values (d.f. = 36) Brand A 15.7 Brand A 15.7 Brand C 19.9 1.01 Brand B 21.4 1.36 Brand D 28.8 3.12 critical values for experimentwise error rate: 0.05 0.01
Brand C 19.9
0.36 2.11
2.70 3.35
262
F 3.39 7.36 2.15
p-value .0284 .0021 .0712
Brand B Brand D 21.4 28.8
.0871
Brand B Brand D 21.4 28.8
1.76
ASBE 6e Solutions for Instructors Post hoc analysis p-values for pairwise t-tests for Factor 2 Store 1 14.3 Store 1 14.3 Store 2 21.8 .0491 Store 3 28.3 .0005
Store 2 21.8
Store 3 28.3
.0807
Tukey simultaneous comparison t-values (d.f. = 36) Store 1 14.3 Store 1 14.3 Store 2 21.8 2.04 Store 3 28.3 3.83 critical values for experimentwise error rate: 0.05 0.01
Store 2 21.8
Store 3 28.3
1.80
2.45 3.11
Learning Objective: 11-1 Learning Objective: 11-2 Learning Objective: 11-3 Learning Objective: 11-4 Learning Objective: 11-5 Learning Objective: 11-6 Learning Objective: 11-7 Learning Objective: 11-8 Learning Objective: 11-9 Learning Objective: 11-10 11.42
This is a replicated experiment with two factors and interaction. Based on the p-values, the means differ for Temperature (p = .0000) and for PVC Type(p = .0013). However, there is no interaction for TemperaturePVC Type (p = .9100). We conclude that the burst strength is affected by temperature, by PVC type, but not by the interaction between temperature and PVC type. The dot plots suggest that PVC2 is the best brand. The pairwise Tukey tests indicate that there is a difference between PVC2 and PVC3, but no difference between PVC1 and PVC2 or PVC1 and PVC3.
Factor 1 (Temperature) Hot (70 Degrees C) Warm (40 Degrees C) Cool (10 Degrees C))
Table of Means Factor 2 (PVC Typr) PVC1 PVC2 268.0 287.0 314.0 334.3 354.0 361.3 312.0 327.6
Two Factor ANOVA with Replication 263
PVC3 258.0 306.0 335.3 299.8
271.0 318.1 350.2 313.1
ASBE 6e Solutions for Instructors Source Factor 1 (Temperature) Factor 2 (PVC Type) Interaction Error Total
SS 28,580.22 3,488.89 171.56 3,174.00 35,414.67
df 2 2 4 18 26
Post hoc analysis for Factor 1 Tukey simultaneous comparison t-values (d.f. = 18) Hot (70o C) 271.0 Hot (70o C) 271.0 Warm (40o C) 318.1 7.53 Cool (10o C)) 350.2 12.66 critical values for experimentwise error rate: 0.05 0.01 Post hoc analysis for Factor 2 Tukey simultaneous comparison t-values (d.f. = 18) PVC3 299.8 PVC3 299.8 PVC1 312.0 1.95 PVC2 327.6 4.44 critical values for experimentwise error rate: 0.05 0.01
Learning Objective: 11-1 Learning Objective: 11-2 Learning Objective: 11-3 Learning Objective: 11-4 Learning Objective: 11-5 Learning Objective: 11-6 Learning Objective: 11-7 Learning Objective: 11-8 264
MS 14,290.111 1,744.444 42.889 176.333
Warm (40o C) 318.1
F 81.04 9.89 0.24
Cool (10o C) 350.2
5.13
2.55 3.32
PVC1 312.0
2.48
2.55 3.32
PVC2 327.6
p-value 0.0000 .0013 .9100
ASBE 6e Solutions for Instructors Learning Objective: 11-9 Learning Objective: 11-10 11.43
This is a two-factor ANOVA (randomized block model) without replication. The response (trucks produced per shift) is weakly related to Plant (row factor, Fcalc = 2.72, p = .0912) and strongly related to Day (column factor, Fcalc = 9.18, p = .0012). Productivity is important to car companies. Sample sizes could be increased because daily production is routinely recorded for each shift. Randomized blocks ANOVA
ANOVA table Source Treatments Blocks Error Total
Mean 155.500 198.750 245.500 237.500 194.750
n 4 4 4 4 4
Std. Dev 33.091 Mon 30.587 Tue 33.601 Wed 21.393 Thu 15.777 Fri
187.600 226.400 196.200 215.400 206.400
5 5 5 5 20
41.920 Plant A 25.745 Plant B 33.789 Plant C 58.701 Plant D 41.500 Total
SS 21,124.30 4,692.40 6,906.10 32,722.80
df MS 4 5,281.075 3 1,564.133 12 575.508 19
Post hoc analysis p-values for pairwise t-tests Mon 155.500 Mon 155.500 Fri 194.750 .0392 Tue 198.750 .0255 Thu 237.500 .0004 Wed 245.500 .0002
F 9.18 2.72
p-value .0012 .0912
Fri 194.750
Tue 198.750
Thu 237.500
.8176 .0269 .0112
.0414 .0174
.6457
Tukey simultaneous comparison t-values (d.f. = 12) 265
Wed 245.500
ASBE 6e Solutions for Instructors
Mon Fri Tue Thu Wed
155.500 194.750 198.750 237.500 245.500
Mon 155.500
Fri 194.750
Tue 198.750
Thu 237.500
2.31 2.55 4.83 5.31
0.24 2.52 2.99
2.28 2.76
0.47
Wed 245.500
critical values for experimentwise error rate: 0.05 3.19 0.01 4.13 Learning Objective: 11-1 Learning Objective: 11-2 Learning Objective: 11-3 Learning Objective: 11-4 Learning Objective: 11-5 Learning Objective: 11-6 Learning Objective: 11-7 Learning Objective: 11-8 Learning Objective: 11-9 Learning Objective: 11-10 11.44
This is a two-factor ANOVA with replication and interaction. Based on the p-values, we conclude that the means differ by Weight (p = .0009) and by Medication (p = .0119). There is no significant interaction effect WeightMedication (p = .9798).
Means: Factor 1 (Weight) 1.1 or Less 1.1 to 1.3 1.3 to 1.5
Source Factor 1 (Weight) Factor 2 (Medication) Interaction Error Total
Table of Means Factor 2 (Medication) Med 1 Med 2 Med 3 133.0 141.0 136.0 140.5 141.5 140.5 148.5 153.0 148.5 140.7 145.2 141.7 SS 717.58 459.00 27.75 325.00 1,529.33
df 2 3 6 12 23
MS 358.792 153.000 4.625 27.083
Post hoc analysis for Factor 1 Tukey simultaneous comparison t-values (d.f. = 12) 1.1 or Less 1.1 to 1.3 266
Med 4 127.5 132.0 140.0 133.2 F 13.25 5.65 0.17
1.3 to 1.5
134.4 138.6 147.5 140.2 p-value .0009 .0119 .9798
ASBE 6e Solutions for Instructors
1.1 or Less 1.1 to 1.3 1.3 to 1.5
134.4 138.6 147.5
134.4
138.6
1.63 5.04
3.41
147.5
critical values for experimentwise error rate: 0.05 2.67 0.01 3.56 Post hoc analysis for Factor 2 Tukey simultaneous comparison t-values (d.f. = 12) Med 4 Med 1 133.2 140.7 Med 4 133.2 Med 1 140.7 2.50 Med 3 141.7 2.83 0.33 Med 2 145.2 3.99 1.50
Med 3 141.7
Med 2 145.2
1.16
critical values for experimentwise error rate: 0.05 2.97 0.01 3.89
Learning Objective: 11-1 Learning Objective: 11-2 Learning Objective: 11-3 Learning Objective: 11-4 Learning Objective: 11-5 Learning Objective: 11-6 Learning Objective: 11-7 Learning Objective: 11-8 Learning Objective: 11-9 Learning Objective: 11-10
11.45
This is a two-factor ANOVA with replication and interaction. We conclude that means do not differ by Instructor Gender (p = .43) or by Student Gender (p = .24) but there is an interaction effect between the 267
ASBE 6e Solutions for Instructors two factors Instructor GenderStudent Gender (p = .03). The sample size is very large, so it is unlikely that any effect was overlooked (the test should have excellent power). Learning Objective: 11-1 Learning Objective: 11-2 Learning Objective: 11-3 Learning Objective: 11-9 Learning Objective: 11-10 11.46
This is an unreplicated two-factor ANOVA. Although MegaStat calls it a randomized block ANOVA, the wording of the problem suggests that both factors are of research interest. We conclude that texture is not significantly affected by age group (p = 0.2999) or by surface type (p = 0.2907). The dot plots support these conclusions, since there are no strong or consistent differences in the groups. No interaction is estimated since there is no replication.
Mean 5.1500 5.3750 6.1250 4.9500
Table of Means n Std. Dev 4 1.2261 Shiny 4 0.8846 Satin 4 0.4924 Pebbled 4 0.8851 Pattern
5.7750 5.7250
4 4
1.1236 0.4031
5.4500 4.6500 5.4000
4 4 16
0.9292 0.9983 0.9345
Youth (Under 21) Adult (21 to 39) Middle-Age (40 to 61) Senior (62 and over) Total
Two Factor ANOVA without Replication Source Columns (Surface) Rows (Age Group) Error Total
SS 3.165 3.245 6.690 13.100
df 3 3 9 15
Learning Objective: 11-1 Learning Objective: 11-2 Learning Objective: 11-3 268
MS 1.0550 1.0817 0.7433
F 1.42 1.46
pvalue .2999 .2907
ASBE 6e Solutions for Instructors Learning Objective: 11-4 Learning Objective: 11-5 Learning Objective: 11-6 Learning Objective: 11-7 Learning Objective: 11-8 Learning Objective: 11-9 Learning Objective: 11-10 11.47
a. Fcalc = MSE/MSR = (1069.17/12270.28) = 0.0871 b. p-value =F.DIST.RT(.0871,3,36) = .9666 c. F.05 w/df 3.36 =F.INV.RT(.05,3,36) = 2.87 d. Based on the p-value, there is no significant difference in means. Learning Objective: 11-1 Learning Objective: 11-3 Learning Objective: 11-4 Learning Objective: 11-5
11.48
a. b.
This is a two-factor ANOVA with replication. Factor A: Fcalc = MSE/MSR = (12199.52/2987.81) = 4.083, p-value =F.DIST.RT(4.083,3,36) = .0136, Factor B: Fcalc = MSE/MSR = (11355.15/2987.81) = 3.800, p-value =F.DIST.RT(3.8,2,36) = .0318, Interaction: Fcalc = MSE/MSR = (29502.56/2987.81) = 9.874, p-value =F.DIST.RT(9.874,6,36) = .0000. c. There are significant effects due to Factor A, Factor B, and significant interaction between the two factors. Learning Objective: 11-1 Learning Objective: 11-2 Learning Objective: 11-3 Learning Objective: 11-4 Learning Objective: 11-5 Learning Objective: 11-9 Learning Objective: 11-10
11.49
a. This is a two-factor ANOVA without replication. b. The original data table showed four rows (plant location df = 3), four columns (Noise level df = 3) and 1 observation per cell (error df = 9) and there were 16 total observations. c. At = .05, plant location is not significant (p = .1200) while noise level is quite significant (p = .0093). Learning Objective: 11-2 Learning Objective: 11-9 Learning Objective: 11-10
11.50
a. This is a two Factor ANOVA with replication. b. There are 4 friends since df = 3 (df = r1) and 3 months since df =2 (df = c1). The total number of observations is 36 since df = 35 (df = n1). Thus, because the data matrix is 43 (12 cells) there must have been 36/12 = 3 observations per cell (i.e., 3 bowling scores per friend per month). c. Based on the p-values, we see Month (p = .0002) is significant at = .01, Friend (p < .0001) is significant at = .01, and there is only a weak interaction because MonthFriend (p = .0786) is only significant at = .10. We conclude that mean bowling scores are influenced by the month, friend and possibly by an interaction between the month (time of year) and the bowler. Learning Objective: 11-2 Learning Objective: 11-9 Learning Objective: 11-10
11.51
a. This is a randomized block (unreplicated two-factor) ANOVA. 269
ASBE 6e Solutions for Instructors b. Based on the p-values, air pollution is significantly affected by car type (p < .0001) and time of day (p < . 0001). c. Variances may appear to be unequal. Equal variances are important because analysis of variance assumes that observations on the response variable are from normally distributed populations that have the same variance. However, we cannot rely on our eyes alone to judge variances, and we should do a test for homogeneity. d. In Hartley’s test, for freeway, we get Hcalc = (14333.7)/(2926.7) = 4.90 which is less than the critical value from Table 11.5. Hcritical = 20.6, and so we fail to reject the hypothesis of equal variances. Similarly, we fail to reject the null of equal variances for time of day, since Hcalc = (14333.6)/872.9 = 16.4 is less than the critical value from Table 11.5. Hcritical = 50.7. Learning Objective: 11-2 Learning Objective: 11-8 Learning Objective: 11-9 Learning Objective: 11-10 11.52
a. This is a two factor ANOVA with replication. b. There are 5 suppliers since df = 4 (df = r1) and 4 quarters since df =3 (df = c1). The total number of observations is 100 since df = 99 (df = n1). Therefore, we have a 54 data matrix (20 cells) which implies 100/20 = 5 observations per cell (i.e., 5 observations per supplier per quarter). c. Based on the p-values Quarter (p = .0009) and Supplier (p < .0001), we conclude that both main effects are significant at = .01. However, there is also a very strong interaction effect QuarterSupplier (p = . 0073). We conclude that shipment times are influenced by the quarter, supplier and the interaction between the quarter (time of year) and the supplier. However, in view of the interaction effect, the main effects may be problematic. Learning Objective: 11-2 Learning Objective: 11-9 Learning Objective: 11-10
11.53
a. This is a one Factor ANOVA. b. The number of bowlers is 5 because d.f. = 4 (df = c1). The sample size is 67 since df = 66 (df = n1). c. Based on the p-value from the ANOVA table (p < .0001) we reject the null hypothesis of no difference between the mean scores and conclude there is a difference. d. Hcritical = 5.30 e. Hcalc = (200.797)/(77.067) = 2.61 f. The sample variances range from 77.067 to 200.797. Hcalc < Hcritical so we fail to reject the hypothesis of equal variances. Learning Objective: 11-2 Learning Objective: 11-8 Learning Objective: 11-9 Learning Objective: 11-10
11.54
a. This is a one Factor ANOVA b. Based on the p-value from the ANOVA table (essentially zero) we strongly reject the null hypothesis of no difference in mean profit/asset ratios. c. The plots indicate that company size (as measured by employees) does affect profitability per dollar of assets. There are possible outliers in several of the groups. d. Variances may be unequal, based on the dot plots and possible outliers. e. To test the hypothesis of homogeneity, we compare Hartley’s critical value (using df1 = 4 and df2 ≈ 60) Hcritical = 1.96 with the sample statistic Hcalc = (34.351)/ (8.108) = 4.24 and reject the hypothesis of equal variances. There isn’t anything we can do about it, though. f. Specifically, the Tukey tests show that small companies differ significantly from medium large, and huge companies (although the latter three categories are not different at = .05). Learning Objective: 11-2 Learning Objective: 11-3 Learning Objective: 11-7 270
ASBE 6e Solutions for Instructors Learning Objective: 11-8
271
ASBE 6e Solutions for Instructors
Chapter 12 Bivariate Regression 12.1
For each sample: H0: ρ = 0 versus H1: ρ ≠ 0. The following formula is used to calculate the test statistic: and tcrit =T.INV(α/2,df). Because these are all two-tailed n-2 tcalc = r 1- r2 tests the decision rule is Reject H0 if tcalc > +tcrit or tcalc < −tcrit Summary Table r .45
tcalc 2.138
Decision 2.138 > 2.101, Reject H0
−.35
−1.977
−1.977 < −1.701, Reject H0
.60
1.677
−.30
−2.416
1.677 is between the critical values, Fail to Reject H0 −2.416 < −2.390, Reject H0
Learning Objective: 12-1 12.2
a. The scatter plot shows a positive correlation between hours worked and weekly pay.
b. Hours Worked (X) Weekly Pay (Y) 10 93 15 171 20 204 20 156 35 261 20 177 x
r=
y
2130 = .9199 350 15318 271
( xi - x )( yi - y )
840 30 0 0 1260 2130 SSxy
ASBE 6e Solutions for Instructors c. tcrit = t.025 =T.INV(.025, 3) = ±3.182, using d.f. = n−2 = 3 d. . We reject the null hypothesis of zero correlation 5-2 tcalc = .9199 = 4.063 1 - (.9199) 2 because 4.063 > 3.182. e. p-value =T.DIST.2T(4.063,3) = .0269. Learning Objective: 12-1 12.3
a. The scatter plot shows a negative correlation between operators and wait time.
b. Operators (X) 4 5 6 7 8 6
Wait (Y) 385 335 383 344 288 347
x
y
( xi - x )( yi - y )
−76 12 0 −3 −118 −185 SSxy
-185 = -.7328 10 6374 c. tcrit = t.025 =T.INV(.025,3) = ±3.182, using d.f. = 3 d. . We fail to reject the null hypothesis of zero 5- 2 tcalc = -.7328 = -1.865 1 - (-.7328) 2 correlation because 1.865 > 3.182. e. p-value = T.DIST.2T(1.865,3) = .159. Learning Objective: 12-1 r=
272
ASBE 6e Solutions for Instructors
12.4
a. The scatter plot shows little correlation between age and amount spent.
Object 3
b. rcalc = −.292 c. tcrit = t.025 =T.INV(.025,8) = ±2.306, using d.f. = 8 d. 10 - 2 tcalc = -.292 = -.864 1 - (-.292) 2 e. Because tcalc (−.864) > −2.306, we fail to reject the null hypothesis of zero correlation. Learning Objective: 12-1 12.5
a. The scatter plot shows a positive correlation between returns from last year and returns from this year.
Object 6
b. rcalc = .5313 c. tcrit = t.025 =T.INV(.025,15) = ±2.131, using d.f. = 15 d. 17 - 2 tcalc = .5313 = 2.429 1 - (.5313) 2 273
ASBE 6e Solutions for Instructors e. Because tcalc (2.429) > 2.131, we reject the null hypothesis of zero correlation. Learning Objective: 12-1 12.6
a. The scatter plot shows a positive correlation between orders and ship cost.
b. rcalc = .820 c. tcrit = t.025 =T.INV(.025,10) = ±2.228, using d.f. = 10 d. 12 - 2 tcalc = .820 = 4.530 1 - (.820)2 e. Because tcalc (4.53) > 2.228, we reject the null hypothesis of zero correlation. Learning Objective: 12-1 12.7
a. Increasing the size of a home by 1 square foot increases the price by $150. b. HomePrice = $125000 + $150×(2000) = $425,000 c. The intercept might be interpreted as the value of the lot without a home. But the range of values for X does not include zero so it would be dangerous to extrapolate for x = 0. Learning Objective: 12-2
12.8
a. An increase in the price of the item of $1 reduces its expected sales by 37.5 units. b. Sales = 842 – ($20)×37.5 = 92 c. From a practical point of view no. A zero price is unrealistic. Learning Objective: 12-2
12.9
a. An increase in the median age of one year means the number of car thefts decreases by 35.3. b. CarTheft = 1,667 35.3×40 = 255 c. The intercept would not be meaningful because you would not have a median age of zero for any state. Learning Objective: 12-2
12.10
a. An increase in the microprocessor speed of one MHz means the computer power dissipation increases by 0.032 watts. b. Computer power dissipation = 15.73 + 0.032×3000 = 111.73 watts c. The intercept would not be meaningful because you would not have a computer with zero speed. Learning Objective: 12-2 274
ASBE 6e Solutions for Instructors 12.11
a. An increase in a country’s Power distance index of one unit means the number of international franchises increases by 1.75. b. International franchises = 47.5 + 1.75×85 = 101.25 c. The intercept would not be meaningful because you cannot have a negative number of franchises. While the range for the index does include zero, it is unlikely that a country’s index value will be close to zero. Learning Objective: 12-2
12.12
a. Increasing the average revenue by $1 million raises the net income by $30,700. b. If revenue is zero, then net income is $2,277 million which suggests that the firm has net income when revenue is zero. This does not seem logical. c. Revenue = $2,277 + 0.0307×($20,000) = $2,891 million Learning Objective: 12-2 Learning Objective: 12-3
12.13
a. Increasing the median income by $1,000 raises the median home price by $2610. b. If median income is zero, then the model suggests that median home price is $51,300. While it does not seem logical that the median family income for any city is zero, it is unclear what the lower bound would be. c. HomePrice = $51.3 + 2.61×($50) = $181.8 or $181,800 Homeprice = $51.3 + 2.61× ($100) = $312.3 or $312,300 Learning Objective: 12-2 Learning Objective: 12-3
12.14
a. Increasing the number of hours worked per week by 1 hour reduces the expected number of credits by .07. b. Yes, the intercept makes sense in this situation. It is possible that a student does not have a job outside of school. c. Credits = 15.4 .07×0 = 15.4 credits Credits = 15.4 .07×40 = 12.6 credits The more hours a student works, the less credits (courses) he will take on average. Learning Objective: 12-2 Learning Objective: 12-3
12.15
a. Chevy Blazer: a one year increase in vehicle age reduces the price by $1050. Chevy Silverado: a one year increase in vehicle age reduces the price by $1339. b. Chevy Blazer: If age = 0 then price = $16,189. This could be the price of a new Blazer. Chevy Silverado: If age = 0 then price = $22,591. This could be the price of a new Silverado. c. $16,189 – $1,050×5 = $10,939 $22,591 $1,339×5 = $15,896 Learning Objective: 12-2 Learning Objective: 12-3
275
ASBE 6e Solutions for Instructors 12.16
a.
yˆ i = $2, 277 + 0.0307($41,078) = $3538.0946 ei = yi - yˆ i = 8301 - 3538.0946 = 4762.9054 the net income.
, . The regression equation underestimated
b.
, yˆ i = $2,277 + 0.0307($61,768) = $4173.2776 . The regression equation overestimated the ei = yi - yˆi = 893 - 4173.2776 = -3280.2776 net income. Learning Objective: 12-2 Learning Objective: 12-3 12.17
a.
, . The regression equation yˆ i = 15.4 - 0.07(14) = 14.42 ei = yi - yˆ i = 18 - 14.42 = 3.58 underestimated the number of credits. b. , . The regression equation yˆ i = 15.4 - 0.07(30) = 13.3 ei = yi - yˆ i = 6 - 13.3 = -7.3 overestimated the number of credits. Learning Objective: 12-2 Learning Objective: 12-3
12.18
a.
b.
Hours Worked (X) 10 15 20 20 35 20
Weekly Pay (Y) 93 171 204 156 261 177
x
y
( xi - x )( yi - y )
840 30 0 0 1260 2130 SSxy
, , = 55.286 + 6.086x 2130 b0 = 177 - 6.086(20) = 55.286 ŷ b1 = = 6.086 350
c.
Hours Weekly Worked (xi) Pay (yi) 10 93 15 171 20 204 20 156 35 261
Estimated Pay ( ) yˆ i yi - yˆi 116.14623.146 146.57624.424 177.00626.994 177.00621.006 268.2967.296 276
( yi - yˆi )2
( yˆi - y )(2yi - y )2
535.7373 596.5318 728.676 441.252 53.23162
3703.2097056 925.6198 36 3.6E-05 729 3.6E-05 441 8334.967056
ASBE 6e Solutions for Instructors 20
177
x
y
20
177
d. R2 =
177.0060.006
3.6E-05 3.6E-05 0 SSE SSR SST 2355.429 12963.79
12,963 = .8462 15,318
e.
Learning Objective: 12-4 12.19
a.
b.
Operators (X) 4 5 6 7 8 6
Wait (Y) 385 335 383 344 288 347
x
y
( xi - x )( yi - y )
−76 12 0 −3 −118 −185 SSxy
, , = 458 − 18.5x -185 b0 = 347 + 18.5(6) = 458 ŷ b1 = = -18.5 10
c. Operators 4 5 6 7 8
Wait Time (yi) 385 335 383 344 288
Estimated Time ( ) yˆ i yi - yˆi
( yi - yˆi ) 2
( yˆi - y ) 2
( yi - y ) 2
384 1 365.5 30.5 347 36 328.5 15.5 310 22
1 930.25 1296 240.25 484
1369 342.25 0 342.25 1369
1444 144 1296 9 3481
277
ASBE 6e Solutions for Instructors 6
347
x
y
d. R2 =
2951.5
3422.5
6374
3, 422.5 = .5369 6,374.0
e.
Learning Objective: 12-4 12.20
a. and b.
Object 8
c. An increase in age of 10 years leads to an average decrease in spending of $0.53. d. The intercept is not meaningful in this case because age wouldn’t be zero. e. R2 = .0851 8.51% of the variation in spending is due to the variation in age. Age of the consumer has little impact on the amount spent. Learning Objective: 12-4
278
ASBE 6e Solutions for Instructors
12.21
a. and b.
c. An increase of 1% in last year’s return leads to an increase, on average, of .458% for this year’s return. d. If last year’s return is zero, this year’s return is 11.155%. Yes, this is meaningful, returns can be zero. e. R2 = .2823. Only 28.23% of the variation in this year’s return is explained by last year’s return. Learning Objective: 12-4 12.22
a. and b.
c. An increase of 100 orders leads to an average increase in shipping cost of $493.22. d. The intercept is not meaningful in this case. 279
ASBE 6e Solutions for Instructors e. R2 = .6717. 67.17% of the variation in shipping costs is explained by number of orders. Learning Objective: 12-4
12.23
a. MegaStat regression output: Regression Analysis r² 0.846 r 0.920 Std. Error 28.020 ANOVA table Source SS Regression 12,962.5714 Residual 2,355.4286 Total 15,318.0000 Regression output variables Intercept Hours Worked (X)
n 5 k 1 Dep. Var. Weekly Pay (Y)
df 1 3 4
MS 12,962.5714 785.1429
F 16.51
p-value .0269
confidence interval t (df=3) p-value 95% lower 95% upper 1.703 .1872 -48.0500 158.6214 4.063 .0269 1.3192 10.8522
coefficients std. error 55.2857 32.4705 6.0857 1.4978
b. H0: 1 = 0 vs. H1: 1 ≠ 0. c. p-value = .0269, (1.3192, 10,8522) d. The slope is significantly different from zero because the p-value is less than .05. Learning Objective: 12-5 12.24
a. MegaStat regression output: Regression Analysis r² 0.537 r -0.733 Std. Error 31.366 ANOVA table Source Regression Residual Total
SS 3,422.5000 2,951.5000 6,374.0000
n 5 k 1 Dep. Var. Wait Time (Y) df 1 3 4
Regression output variables coefficients std. error Intercept 458.0000 61.1438 Operators (X) -18.5000 9.9188
MS 3,422.5000 983.8333
F 3.48
p-value .1590
confidence interval t (df=3) p-value 95% lower 95% upper 7.491 .0049 263.4131 652.5869 -1.865 .1590 -50.0662 13.0662
b. H0: 1 = 0 vs. H1: 1 ≠ 0. c. p-value = .1590, (50.0662, 13.0662) 280
ASBE 6e Solutions for Instructors d. The slope is not significantly different from zero because the p-value is greater than .05. Learning Objective: 12-5 12.25
a.
= 557.4511 + 3.0047x ŷ b. For a 95% confidence level use t.025 =T.INV(.025,30) = −2.042. The 95% confidence interval is 3.0047 ± 2.042(0.8820) or (1.203, 4.806). c. H0: β1 ≤ 0 versus H1: β1 > 0. tcrit =T.INV(.95,30) = 1.697. Reject the null hypothesis if tcalc > 1.697. tcalc = 3.0047/0.8820 = 3.407 > 1.697 so we reject the null hypothesis. d. p-value =T.DIST.RT(3.407,30) = .0009 < .05 so we reject the null hypothesis. The slope is positive. Increased debt is correlated with increased NFL team value. Learning Objective: 12-5
12.26
a.
12.27
a.
12.28
a.
= 7.6425 + 0.9467x ŷ b. For a 95% confidence level use t.025 =T.INV(.025,14) = −2.145. The 95% confidence interval is 0.9467 ± 2.145(0.0936) or (0.7460, 1.1473). c. H0: β1 ≤ 0 versus H1: β1 > 0. tcrit =T.INV(.95,14) = 1.761. Reject the null hypothesis if tcalc > 1.761. tcalc = 0.9467/0.0936 = 10.114 > 1.761 so we reject the null hypothesis. d. p-value =T.DIST.RT(10.114,14) = .0000 < .05 so we reject the null hypothesis. The slope is positive. Increased revenue is correlated with increased expenses. Learning Objective: 12-5 = 1.8064 + .0039x ŷ b. Intercept: tcalc = 1.8064/0.6116 = 2.954, Slope: tcalc = 0.0039/0.0014 = 2.786 (Excel value is slightly different due to internal rounding.) c. d.f. = 10, t.025 =T.INV(.025, 10) = −2.228 so tcrit = ±2.228. Or one could use =T.INV.2T(.05,10) d. Intercept: p-value =T.DIST.2T(2.954,10) = .0144. Slope: p-value =T.DSIT.2T(2.869,10) = .0167. e. Fcalc = (2.869)2 = 8.23 f. This model fits the data fairly well. The F statistic is highly significant (p-value = . 0167). Also, R2 = .452 indicating almost half of the variation in annual taxes is explained by home price. Learning Objective: 12-6
b. c. d. e.
= 614.930 − 1.09.11x ŷ Intercept: tcalc = 614.930/51.2343 = 12.002. Slope: tcalc = −109.112/51.3623 = −2.124. d.f. = 18, t.025 = T.INV(.025, 18) = − 2.101 so tcrit = ±2.101. Intercept: p-value =T.DIST.2T(12.002,18) = .0000, Slope: p-value =T.DIST.2T(2.124,18) = .0478 Fcalc = (−2.124)2 = 4.51
281
ASBE 6e Solutions for Instructors f. This model has a poor fit. The F statistic is barely significant at a level of .05 (p-value = .0478) and R2 = .2. Only 20% of the variation in units sold can be explained by average price. Learning Objective: 12-6 12.29
a. MegaStat regression output: Regression Analysis r² r
0.085 -0.292
Std. Error
2.128
ANOVA table Source Regression Residual Total
n k Dep. Var.
10 1 Spent (Y)
SS 3.3727 36.2396 39.6123
df 1 8 9
MS 3.3727 4.5299
F 0.745
p-value .4133
Regression output coefficient variables s Intercept 10.9609 Age (X) -0.0530
std. error 2.0885 0.0614
t (df=8) 5.248 -0.863
pvalue .0008 .4133
confidence interval 95% 95% lower upper 6.1449 15.7770 -0.1946 0.0886
b. (−0.1946, 0.0886) This interval does contain zero therefore we cannot conclude that the slope is greater than zero. c. The t statistic is −0.863 and the p-value is .4133. Because the p-value is greater than 0.05, we cannot conclude that the slope is different from zero. d. Fcalc = 0.745 with a p-value = .4133. This indicates that the model does not fit the data. e. The p-values match. Fcalc = (−0.863)2 = 0.745. f. This model does not fit the data. The F statistic is not significant. Learning Objective: 12-6
12.30 a. MegaStat regression output: Regression Analysis r² 0.282 r 0.531 Std. Error 4.335 ANOVA table Source Regression Residual Total
SS 110.8585 281.8321 392.6906
n 17 k 1 Dep. Var. This Year (Y) df 1 15 16
Regression output variables coefficients std. error Intercept 11.1549 2.1907
MS 110.8585 18.7888
F 5.90
p-value .0282
confidence interval t (df=15) p-value 95% lower 95% upper 5.092 .0001 6.4854 15.8243
282
ASBE 6e Solutions for Instructors Last Year (X)
0.4580
0.1885
2.429
.0282
0.0561
0.8598
b. (0.0561, 0.8598) This interval does not contain zero and falls on the positive side therefore we can conclude that the slope is greater than zero. c. The t statistic is 2.429 and the p-value is .0282. Because the p-value is less than 0.05, we can conclude that the slope is positive. d. Fcalc = 5.90 with a p-value = .0282. This indicates that the model does provide some fit to the data. e. The p-values match. Fcalc = (2.429)2 = 5.90. f. This model provides modest fit to the data. Although the F statistic is significant, R2 shows that only 28% of the variation in this year’s return is explained by last year’s return. Learning Objective: 12-6 12.31
a. MegaStat regression output: Regression Analysis r² 0.672 r 0.820 Std. Error 599.029 ANOVA table Source SS Regression 7,340,819.5514 Residual 3,588,357.1152 Total 10,929,176.6667 Regression output variables Intercept Orders (X)
n 12 k 1 Dep. Var. Ship Cost (Y) df 1 10 11
MS 7,340,819.5514 358,835.7115
F 20.46
p-value .0011
confidence interval t (df=10) p-value 95% lower 95% upper -0.029 .9771 -2,392.7222 2,330.3432 4.523 .0011 2.5024 7.3619
coefficients std. error -31.1895 1,059.8678 4.9322 1.0905
b. (2.5024, 7.3619) This interval does not contain zero therefore we can conclude that the slope is greater than zero. c. The t statistic is 4.523 and the p-value is 0.0011. Because the p-value is less than 0.05, we can conclude that the slope is positive. d. Fcalc = 20.46 with a p-value = .0011. This indicates that the model does provide some fit to the data. e. The p-values match. Fcalc = (4.523)2 = 20.46. f. This model provides a good fit to the data. The F statistic is highly significant and R2 shows that 67% of the variation in shipping cost is explained by number of orders. Learning Objective: 12-6 12.32
a. MegaStat Predictions: Predicted values for: Weekly Pay (Y) 95% Confidence Intervals
Hours Worked (X) 12 17 21
Predicted lower 128.314 73.138 158.743 116.377 183.086 142.922
upper 183.491 201.109 223.249
283
95% Prediction Intervals
lower 23.451 60.017 85.285
upper 233.178 257.469 280.887
ASBE 6e Solutions for Instructors 25 30
207.429 160.970 237.857 175.709
253.887 106.879 300.005 129.164
307.978 346.551
b. x = 17: 95% confidence interval (116.377, 201.109), 95% prediction interval (60.017, 257.469) c. The 95% confidence interval for µY: 177 ± 2.776 or (100.174, 253.826).
61.883 5
d. The margin of error for the confidence interval in part (c) is 76.826 whereas the margin of error for the confidence interval in part b is less than 76.826. Knowing the number of hours a student works helps us better estimate the average credit hours a student takes. Learning Objective: 12-7 12.33
a. MegaStat Predictions: Predicted values for: Profit (Y) 95% Confidence Intervals
Revenue (X) 1.8 15.0 30.0
95% Prediction Intervals
Predicted lower upper lower -0.070148 -0.554318 0.414023 -1.326790 0.757399 0.367076 1.147723 -0.466153 1.697794 1.106753 2.288834 0.396234
upper 1.186495 1.980952 2.999353
b. x = 15: 95% confidence interval (0.3671, 1.1477), 95% prediction interval (0.4662,1.9810) c. The 95% confidence interval for µY: 0.628 ± 2.306 or (0.2045, 1.4605).
1.083 9
d. The margin of error for the confidence interval in part (c) is 0.8325 whereas the margin of error for the confidence interval in part b is less than 0.8325. Knowing the revenue an entertainment company brings in helps us better estimate their average profit. Learning Objective: 12-7 12.34
No, these plots do not show that regression error assumptions of normality or constant variance have been violated. The plot on the left is a normplot and shows a fairly straight line on the diagonal. This indicates that the assumption of a normal distribution for residuals is reasonable. The plot on the right shows residual values plotted against the corresponding x value. The plot suggests that the residuals are homoscedastic because there is no increase or decrease in residual magnitude. Learning Objective: 12-8
12.35
The plot on the left is a normplot and shows a fairly straight line on the diagonal. This indicates that the assumption of a normal distribution for residuals is reasonable. The plot on the right shows residual values plotted against the corresponding x value. The
284
ASBE 6e Solutions for Instructors plot suggests that the residuals are heteroscedastic because there is an increase in residual magnitude as the x values increase. Learning Objective: 12-8 12.36
a.
b.
c.
e 25 ei* = i » = 3.125 sei 8
> 3 so this is an outlier.
e -15 ei* = i » = -2.143 sei 7 e 112 ei* = i » = 1.455 sei 77
< −2 so this is unusual.
< 2 so this is not unusual nor an outlier.
Learning Objective: 12-9 12.37
a. b. c.
ei = sei ´ ei* = 14 ´ 2.00 = 28.00
.
ei = sei ´ ei* = 14 ´ -1.50 = -21.00
ei = sei ´ ei* = 14 ´ 0.50 = 7.00
.
.
Learning Objective: 12-9 12.38
a. Predicted Defects = 3.2 + 0.045(100) = 7.7 defects per million parts b. ei = yi − = 4.4 – 7.7 = −3.3
ŷi
c. ei* =
ei -3.3 » = -3.084 sei 1.07
d. Yes, this residual is considered an outlier because ei* < −3. Learning Objective: 12-9 12.39
a. Predicted MPG = 49.22 – 0.081(200) = 33.02 MPG b. ei = yi − = 38.15 – 33.02 = 5.13
ŷi
c. ei* =
ei 5.13 » = 2.527 sei 2.03
d. No, this residual is not considered an outlier because ei* < 3 but it is considered unusual because ei* > 2. Learning Objective: 12-9 12.40
a. 4/16 = 0.25. hi = 0.13 < .25 therefore leverage is not unusual. b. 4/50 = 0.08. hi = 0.13 > .08 therefore leverage is unusual. 285
ASBE 6e Solutions for Instructors b. 4/100 = 0.04. hi = 0.36 > .04 therefore leverage is unusual. Learning Objective: 12-10 12.41
a. For n = 25 the cutoff for high leverage is 4/25 = 0.16. b. For n = 40 the cutoff for high leverage is 4/40 = 0.10. c. For n = 200 the cutoff for high leverage is 4/200 = 0.02. Learning Objective: 12-10
12.42
a.
12.43
a.
12.44
a.
. 4/n = 4/29 = 0.1379. Because 1 ( xi - x ) 2 1 (2382 - 2004)2 hi = + = + = 0.1774 n SS XX 29 999, 603 0.1774 > 0.1379 this would be considered a high leverage observation. b. . 4/n = 4/29 = 0.1379. Because 2 2 ( x x ) 1 1 (2125 - 2004) hi = + i = + = 0.0491 n SS XX 29 999, 603 0.0491 < 0.1379 this would not be considered a high leverage observation. c. . 4/n = 4/29 = 0.1379. Because 0.1820 2 2 1 (x - x ) 1 (1620 - 2004) hi = + i = + = 0.1820 n SS XX 29 999,603 > 0.1379 this would be considered a high leverage observation. Learning Objective: 12-10 . 4/n = 4/74 = 0.0541. Because 0.185 1 ( xi - x ) 2 1 (0.072 - 2.027) 2 hi = + = + = 0.185 n SS XX 74 22.285 > 0.0541 this would be considered a high leverage observation. b. . 4/n = 4/74 = 0.0541. Because 2 2 1 (x - x ) 1 (1.413 - 2.027) hi = + i = + = 0.0304 n SS XX 74 22.285 0.0304 < 0.0541 this would not be considered a high leverage observation. c. . 4/n = 4/74 = 0.0541. Because 1 ( xi - x ) 2 1 (3.376 - 2.027) 2 hi = + = + = 0.0952 n SS XX 74 22.285 0.0952 > 0.0541 this would be considered a high leverage observation. Learning Objective: 12-10
b.
exp( -6.13 + 0.000881(3000)) P(401k ) = = 0.029684 1 + exp(-6.13 + 0.000881(3000)) exp( -6.13 + 0.000881(5000)) P(401k ) = = 0.151228 1 + exp(-6.13 + 0.000881(5000)) 286
.
.
ASBE 6e Solutions for Instructors c.
exp( -6.13 + 0.000881(8000)) P(401k ) = = 0.714634 1 + exp(-6.13 + 0.000881(8000))
.
Learning Objective: 12-12 12.45
a.
b.
c.
exp(-5.88 + 1.055(0)) P(Uninsured) = = 0.002787 1 + exp(-5.88 + 1.055(0)) exp(-5.88 + 1.055(3)) P(Uninsured) = = 0.062094 1 + exp(-5.88 + 1.055(3)) exp(-5.88 + 1.055(6)) P(Uninsured) = = 0.610639 1 + exp(-5.88 + 1.055(6))
.
.
.
Learning Objective: 12-12 Questions 12.46 through 12.61 refer to 10 different data sets labeled A-J. The answers to each question are listed for each data set in turn. Note that one can find the tcrit values using either =T.INV(α/2, df) or =T.INV.2T(α, df). DATA SET A 12.46
Cross-sectional. Learning Objective: 02-3
12.47
Answers will vary. Most likely from a survey similar to the Current Population Survey conducted by The Bureau of Labor Statistics. Learning Objective: 02-7
12.48
Answers will vary. Sample size is sufficient for educational purposes but most government studies will have much larger sample sizes. Learning Objective: 02-6
12.49
A positive slope would be logical. It makes sense that higher income is associated with higher home values. Cause and effect cannot be assumed. An increase in income does not automatically compel a family to purchase a more expensive home. Learning Objective: 12-2
287
ASBE 6e Solutions for Instructors
12.50
Object 10
There is a moderate positive relationship between Median Income and Median Home Price. Learning Objective: 03-8 12.51
See graph above. Yes, a linear relationship is plausible. Learning Objective: 03-8
12.52
An increase in median income of $1000, increases median home price by $4,528. No, the intercept does not have meaning because it seems unlikely that the median income for a family will be equal to zero. Learning Objective: 12-2
12.53
MegaStat output is provided. Regression Analysis 288
ASBE 6e Solutions for Instructors
ANOVA table Source Regression Residual Total
r² r Std. Error
0.188 0.433 71355.998
n k Dep. Var.
51 1 MedPrice
SS 57,576,975,879.30 249,492,242,100.62 307,069,217,979.92
df 1 49 50
MS 57,576,975,879.30 5,091,678,410.22
F 11.31
std. error 68,091.420 2 1.3466
t (df=49)
p-value
-0.623 3.363
.5365 .0015
Residual 98,782.4 -90,594.0 -49,409.6 15,324.7 126,617.2 -12,804.9 10,376.7 196,744.4 1,505.0 -12,653.6 -27,939.6 272,616.7 -52,630.9 -46,885.9 -14,154.7 -66,153.5 -47,469.8 -13,397.9 16,840.0 131,706.7 8,080.0 -58,362.2 -70,764.1 -30,430.4 -6,656.8 -38,599.6 33,649.1 -40,883.0 -55,986.2 -138,928.7 -22,122.7
Leverag e 0.049 0.041 0.023 0.066 0.026 0.057 0.111 0.029 0.037 0.032 0.032 0.052 0.023 0.020 0.025 0.020 0.025 0.048 0.061 0.021 0.091 0.062 0.025 0.021 0.070 0.026 0.047 0.022 0.020 0.118 0.079
Regression output variables
coefficients
Intercept Income
-42,387.3735 4.5283
Observatio n 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31
MedPrice 241,750.0 128,969.0 120,560.0 147,669.0 330,037.0 217,558.0 266,845.0 404,380.0 216,902.0 144,501.0 129,446.0 499,837.0 118,036.0 140,436.0 152,387.0 113,416.0 118,687.0 130,344.0 152,413.0 306,363.0 256,411.0 175,175.0 96,398.0 164,105.0 123,754.0 126,484.0 178,188.0 154,481.0 133,473.0 120,415.0 220,625.0
Predicted 142,967.6 219,563.0 169,969.6 132,344.3 203,419.8 230,362.9 256,468.3 207,635.6 215,397.0 157,154.6 157,385.6 227,220.3 170,666.9 187,321.9 166,541.7 179,569.5 166,156.8 143,741.9 135,573.0 174,656.3 248,331.0 233,537.2 167,162.1 194,535.4 130,410.8 165,083.6 144,538.9 195,364.0 189,459.2 259,343.7 242,747.7
289
p-value .0015
confidence interval 95% lower 95% upper -179,222.21 1.82
Studentized
94,447.46 7.23 Studentize d Deleted
Residual 1.420 -1.297 -0.701 0.222 1.798 -0.185 0.154 2.798 0.021 -0.180 -0.398 3.924 -0.746 -0.664 -0.201 -0.936 -0.674 -0.192 0.243 1.866 0.119 -0.845 -1.004 -0.431 -0.097 -0.548 0.483 -0.579 -0.793 -2.073 -0.323
Residual 1.435 -1.306 -0.697 0.220 1.841 -0.183 0.153 3.022 0.021 -0.178 -0.395 4.689 -0.743 -0.660 -0.199 -0.935 -0.670 -0.191 0.241 1.916 0.118 -0.842 -1.004 -0.428 -0.096 -0.544 0.479 -0.575 -0.790 -2.148 -0.320
ASBE 6e Solutions for Instructors 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51
282,633.0 161,176.0 153,364.0 245,858.0 113,730.0 118,771.0 206,977.0 153,439.0 230,354.0 135,855.0 156,243.0 123,613.0 134,854.0 194,147.0 229,420.0 203,970.0 248,555.0 153,935.0 129,369.0 183,202.0
161,990.8 183,033.6 156,086.0 188,580.7 165,396.1 152,793.9 186,751.3 176,390.7 191,374.6 146,431.7 162,978.0 132,362.4 171,645.0 214,369.1 210,868.8 230,969.7 211,932.9 151,317.7 185,614.7 193,992.0
120,642.2 -21,857.6 -2,722.0 57,277.3 -51,666.1 -34,022.9 20,225.7 -22,951.7 38,979.4 -10,576.7 -6,735.0 -8,749.4 -36,791.0 -20,222.1 18,551.2 -26,999.7 36,622.1 2,617.3 -56,245.7 -10,790.0
0.028 0.020 0.033 0.020 0.026 0.037 0.020 0.021 0.021 0.044 0.027 0.066 0.022 0.036 0.032 0.058 0.033 0.038 0.020 0.021
1.715 -0.309 -0.039 0.811 -0.734 -0.486 0.286 -0.325 0.552 -0.152 -0.096 -0.127 -0.521 -0.289 0.264 -0.390 0.522 0.037 -0.796 -0.153
Learning Objective: 12-4 12.54
a. 95% confidence interval for β1: (1.82, 7.23). This interval does not contain zero which means that we are confident the slope is not zero. b. H0: β1 = 0 vs. H1: β1 ≠ 0, d.f. = 49, t.025 =T.INV(.025, 49) = ±2.0096, tcalc = 3.363 > 2.0096. Therefore reject H0 and conclude that the slope is significantly different from zero. c. The p-value = .0015. This means that 15 times out of 10,000 we will see a sample this extreme if the slope is actually equal to zero. d. Yes, this sample supports our a priori hypothesis about the slope. Learning Objective: 12-5
12.55
a. R2 = .188. 18.8% of the variation in median home price can be explained by median household income. Closer to 100% is always desirable but this is still decent considering we are using only one predictor variable. b. Fcalc = 11.31 and its p-value = .0015. The F statistic is significant which means the linear model provides significant fit. c. This model provides a good enough fit for practical use. Learning Objective: 12-6
12.56
Observations 8, 12, and 30 have unusual residual values. Observation 8 and 12 are Washington DC and Hawaii. The model underestimated the median home price. Observation 30 is Nebraska. The model overestimated the median home price. More research would be needed to better explain the unusual observations. Learning Objective: 12-9
12.57
a. The normplot of residuals shows a dip in the line on the diagonal. The histogram of residuals shows a right skewed distribution. 290
1.751 -0.306 -0.038 0.808 -0.730 -0.482 0.284 -0.322 0.548 -0.150 -0.095 -0.126 -0.518 -0.286 0.262 -0.386 0.518 0.037 -0.793 -0.151
ASBE 6e Solutions for Instructors
Object 13
Object 15
b. There is a departure from normality. Learning Objective: 12-8
291
ASBE 6e Solutions for Instructors 12.58
Object 18
The residual plot does show signs of heteroscedasticity. Learning Objective: 12-8 12.59
An autocorrelation test is not appropriate because this is not a time series data set. Learning Objective: 12-8
12.60
Answers will vary. Confidence and prediction intervals for x = $50,000 and $75,000 are shown in the table below. Predicted values for: MedPrice 95% Confidence Intervals
Income 50,000 75,000
Predicted 184,025.288 297,231.618
lower 163,945.876 226,710.038
upper 204,104.699 367,753.198
95% Prediction Intervals
lower 39,231.022 137,433.340
Learning Objective: 12-7 12.61
Observations 7, 21, 30, and 31all have high leverage. These states are Connecticut, Maryland, Nebraska, and Maryland. All states have higher than average median income. Learning Objective: 12-9
DATA SET B 12.46
Cross-sectional. Learning Objective: 02-3 292
upper
328, 457,
ASBE 6e Solutions for Instructors 12.47
Data were most likely from a semester’s class of 58 students. This can be treated as a sample from the larger population of students who take statistics over several years. Learning Objective: 02-7
12.48
A sample size of 58 should be sufficient to draw significant conclusions about the relationship between midterm and final exam scores. Learning Objective: 02-6
12.49
A positive slope would be logical. It makes sense that higher midterm scores are associated with higher final exam scores and vice versa. While cause and effect cannot be assumed it would be reasonable to hypothesize that if a student performs well on their midterm exam due to a solid understanding of the material they will then be able to comprehend the material in the last half of the semester and will be well-prepared for their final exam. Learning Objective: 12-2
12.50
Object 20
There is a moderate positive relationship between Midterm Exam Score and Final Exam Score. Learning Objective: 03-8 12.51
See graph above. Yes, a linear relationship is plausible. Learning Objective: 03-8
12.52
A one point increase in the midterm score is associated with a 1.014 point increase in the final exam score. While theoretically a student could score zero on the midterm, the yintercept is negative which is not feasible. Learning Objective: 12-2
293
ASBE 6e Solutions for Instructors 12.53
MegaStat output is provided. Regression Analysis r² r Std. Error
0.430 0.656 7.517
n k Dep. Var.
58 1 Final Exam Score
MS 2,385.544 6 56.5011
F
p-value
42.22
2.33E-08
ANOVA table Source
SS
df
Regression Residual Total
2,385.5446 3,164.0588 5,549.6034
1 56 57
coefficients -1.5568
std. error 12.0004
t (df=56) -0.130
1.0138
0.1560
6.498
Regression output variables Intercept Midterm Exam Score
Observation 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26
confidence interval
Final Exam Score 78.0 85.0 81.0 54.0 70.0 73.0 89.0 84.0 86.0 79.0 75.0 63.0 72.0 69.0 86.0 78.0 75.0 68.0 77.0 78.0 77.0 73.0 79.0 74.0 79.0 73.0
Predicte d 79.5 86.6 71.4 68.4 85.6 82.6 77.5 74.5 73.5 74.5 83.6 72.4 73.5 72.4 79.5 74.5 71.4 76.5 75.5 65.4 73.5 70.4 84.6 73.5 75.5 75.5
Residual -1.5 -1.6 9.6 -14.4 -15.6 -9.6 11.5 9.5 12.5 4.5 -8.6 -9.4 -1.5 -3.4 6.5 3.5 3.6 -8.5 1.5 12.6 3.5 2.6 -5.6 0.5 3.5 -2.5
294
p-value .8972 2.33E08
Leverag e 0.022 0.063 0.027 0.042 0.055 0.035 0.018 0.018 0.020 0.018 0.040 0.023 0.020 0.023 0.022 0.018 0.027 0.017 0.017 0.066 0.020 0.031 0.047 0.020 0.017 0.017
95% lower -25.5965
95% upper 22.4829
0.7012
1.3263 Studentize d
Studentize d
Deleted
Residual -0.208 -0.226 1.290 -1.957 -2.139 -1.298 1.541 1.279 1.685 0.607 -1.168 -1.272 -0.197 -0.464 0.868 0.473 0.481 -1.141 0.203 1.741 0.475 0.348 -0.765 0.072 0.471 -0.334
Residual -0.206 -0.224 1.297 -2.009 -2.212 -1.306 1.561 1.286 1.714 0.604 -1.172 -1.279 -0.195 -0.461 0.866 0.470 0.477 -1.145 0.201 1.774 0.472 0.346 -0.762 0.071 0.468 -0.332
ASBE 6e Solutions for Instructors 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58
72.0 81.0 86.0 76.0 83.0 83.0 86.0 71.0 83.0 79.0 68.0 90.0 89.0 83.0 81.0 79.0 58.0 77.0 85.0 67.0 70.0 79.0 59.0 74.0 86.0 85.0 79.0 81.0 31.0 82.0 70.0 69.0
83.6 76.5 77.5 85.6 80.6 77.5 84.6 72.4 82.6 82.6 71.4 82.6 83.6 72.4 85.6 75.5 57.2 72.4 80.6 71.4 76.5 81.6 63.3 78.5 78.5 81.6 74.5 77.5 57.2 79.5 67.4 72.4
-11.6 4.5 8.5 -9.6 2.4 5.5 1.4 -1.4 0.4 -3.6 -3.4 7.4 5.4 10.6 -4.6 3.5 0.8 4.6 4.4 -4.4 -6.5 -2.6 -4.3 -4.5 7.5 3.4 4.5 3.5 -26.2 2.5 2.6 -3.4
0.040 0.017 0.018 0.055 0.025 0.018 0.047 0.023 0.035 0.035 0.027 0.035 0.040 0.023 0.055 0.017 0.167 0.023 0.025 0.027 0.017 0.030 0.086 0.020 0.020 0.030 0.018 0.018 0.167 0.022 0.050 0.023
-1.576 0.603 1.139 -1.318 0.329 0.736 0.189 -0.195 0.056 -0.486 -0.463 1.004 0.733 1.420 -0.633 0.471 0.110 0.612 0.598 -0.598 -0.873 -0.348 -0.602 -0.609 1.003 0.463 0.607 0.467 -3.826 0.330 0.357 -0.464
-1.597 0.600 1.142 -1.327 0.326 0.733 0.187 -0.193 0.055 -0.482 -0.460 1.004 0.730 1.433 -0.630 0.468 0.109 0.609 0.595 -0.595 -0.871 -0.345 -0.599 -0.606 1.003 0.459 0.604 0.464 -4.411 0.327 0.355 -0.461
Learning Objective: 12-4 12.54
a. 95% confidence interval for β1: (0.7012, 1.3263). This interval does not contain zero which means that we are confident the slope is not zero. b. H0: β1 = 0 vs. H1: β1 ≠ 0, d.f. = 56, t.025 =T.INV(.025,56) = ±2.003, tcalc = 4.059 > 2.003. Therefore reject H0 and conclude that the slope is significantly different from zero. c. The p-value = .0000. This means that there is very little chance of observing this sample if there is no correlation between midterm scores and final scores. d. Yes, this sample supports our a priori hypothesis about the slope. Learning Objective: 12-4
12.55
a. R2 = .43. 43% of the variation in final exam scores can be explained by midterm exam scores. Closer to 100% is always desirable but this is still decent considering we are using only one predictor variable. b. Fcalc = 42.22 and its p-value = .0000. The F statistic is significant which means the linear model provides significant fit. 295
ASBE 6e Solutions for Instructors c. This model provides a good enough fit for practical use. Learning Objective: 12-6 12.56
Students 4 and 5 have unusual residual values with residuals that are negative. The model overestimated the final exam score for these two students. Student 55 had a residual beyond −4 which means the model significantly overestimated the final exam score. This observation would be considered an outlier. Low exam scores are typically due to students who do not study or attend class. On occasion the material is very difficult which can also result in a low exam score. Learning Objective: 12-9
12.57
a. The normplot of residuals show a fairly straight line on the diagonal. The histogram of residuals shows a skewed left distribution which is fairly common for exam scores. Most students do well on exams with a few students in the low range.
Object 22
296
ASBE 6e Solutions for Instructors
Object 24
b. The residuals do not show strong departure from normality. The assumption of residual normality is valid. Learning Objective: 12-8 12.58
Object 26
The residual plot does not show signs of heteroscedasticity although the low final exam score can be seen in the lower part of the graph. Learning Objective: 12-8 297
ASBE 6e Solutions for Instructors 12.59
An autocorrelation test is not appropriate because this is not a time series data set. Learning Objective: 12-8
12.60
Answers will vary. Confidence and prediction intervals for x = 75 and 95 are shown in the table below. Predicted values for: Final Exam Score 95% Confidence Intervals
Midterm Exam Score
Predict ed
75
74.477
95
94.753
lower 72.43 3 88.68 8
95% Prediction Intervals
upper
lower
upper
76.521
59.281
89.673
100.818
78.520
110.986
Learning Objective: 12-7 12.61
Students 43, 49, and 55 have high leverage. These three students had very low midterm scores. Learning Objective: 12-9
298
ASBE 6e Solutions for Instructors
DATA SET C 12.46
Cross-sectional. Learning Objective: 02-3
12.47
Answers will vary. Hospital data bases will contain this type of information. Learning Objective: 02-7
12.48
Answers will vary. Sample size is sufficient for educational purposes but most studies conducted by a hospital will use larger sample sizes because the data is available. Learning Objective: 02-6
12.49
A positive slope would be logical. The estimated length of stay of a patient should be based on their condition upon admission and should be related to their actual length of stay. Although estimating a patient’s time in the hospital should not cause their time to be longer or shorter. Learning Objective: 12-2
12.50
There is a strong positive relationship between ELOS and ALOS. Learning Objective: 03-8 12.51
See graph above. Yes, a linear relationship is plausible. Learning Objective: 03-8
12.52
An increase in ELOS of one month increases ALOS by 1.03 months. No, the intercept does not have meaning because by definition admission to a hospital implies staying in the hospital. Learning Objective: 12-2
299
ASBE 6e Solutions for Instructors 12.53
MegaStat output is provided. Regression Analysis r² 0.625 r 0.791 Std. Error 2.282 ANOVA table Source Regression Residual Total
SS 121.5515 72.8860 194.4375
n 16 k 1 Dep. Var. ALOS df 1 14 15
Regression output variables coefficients std. error Intercept 0.5147 1.5903 ELOS 1.0293 0.2130
Observation 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16
ALOS Predicted 10.00 11.32 2.00 5.15 4.00 8.23 11.00 12.87 11.00 8.23 11.00 9.78 6.50 6.69 5.00 5.66 8.00 6.69 16.00 12.87 6.50 7.72 6.00 5.15 3.50 4.12 10.00 6.69 7.00 8.23 5.50 3.60
MS 121.5515 5.2061
F 23.35
t (df=14) 0.324 4.832
p-value .7510 .0003
Residual Leverage -1.32 0.171 -3.15 0.116 -4.23 0.065 -1.87 0.283 2.77 0.065 1.22 0.098 -0.19 0.071 -0.66 0.096 1.31 0.071 3.13 0.283 -1.22 0.063 0.85 0.116 -0.62 0.167 3.31 0.071 -1.23 0.065 1.90 0.200
p-value .0003
confidence interval 95% lower 95% upper -2.8961 3.9255 0.5724 1.4862 Studentized Studentized Deleted Residual Residual -0.636 -0.622 -1.466 -1.536 -1.919 -2.154 -0.966 -0.963 1.254 1.282 0.564 0.550 -0.087 -0.083 -0.305 -0.295 0.595 0.581 1.622 1.735 -0.552 -0.538 0.398 0.385 -0.296 -0.287 1.505 1.584 -0.559 -0.545 0.930 0.925
Learning Objective: 12-4 12.54
a. 95% confidence interval for β1: (0.5724, 1.4862). This interval does not contain zero which means that we are confident the slope is not zero. b. H0: β1 = 0 vs. H1: β1 ≠ 0, d.f. = 14, t.025 =T.INV(.025,14) = ±2.145, tcalc = 4.832 > 2.145. Therefore reject H0 and conclude that the slope is significantly different from zero. c. The p-value = .0003. This means that 3 times out of 10,000 we will see a sample this extreme if the slope is actually equal to zero. d. Yes, this sample supports our a priori hypothesis about the slope. Learning Objective: 12-5
12.55
a. R2 = .625. 62.5% of the variation in actual length of stay of a patient can be explained by their expected length of stay. This shows a moderately strong relationship.
300
ASBE 6e Solutions for Instructors b. Fcalc = 23.35 and its p-value = .0003. The F statistic is significant which means the linear model provides significant fit. c. This model provides a good enough fit for practical use. Learning Objective: 12-6 12.56
There are no unusual residuals. Learning Objective: 12-9
12.57
a. The histogram of residuals does not show a clear normal distribution; however, the normal plot of residuals has a modest straight line.
Object 29
Object 32
301
ASBE 6e Solutions for Instructors
b. While the residuals show some departure from normality, this is a very small sample so it is difficult to see clear normality. There are no strong outliers in the data set so at this point the slight departure from normality is not too troublesome. Learning Objective: 12-8 12.58
Object 34
12.59
The residual plot does not show signs of heteroscedasticity. Learning Objective: 12-8 An autocorrelation test is not appropriate because this is not a time series data set. Learning Objective: 12-8
12.60
Answers will vary. Prediction and confidence intervals for x = 5 and 9 days is shown. Predicted values for: ALOS 95% Confidence Intervals
ELOS 4 9
Predicted 4.6318 9.7782
lower 2.8052 8.2426
upper 6.4584 11.3138
95% Prediction Intervals
lower -0.5917 4.6492
upper 9.8554 14.9072
Learning Objective: 12-7 12.61
Observations 4 and 10 have high leverage. These patients had unusually long estimated lengths of stay. Learning Objective: 12-9
DATA SET D 302
ASBE 6e Solutions for Instructors
12.46
Cross-sectional. Learning Objective: 02-3
12.47
Answers will vary. This type of information could probably be collected from the airplane manufacturer. Learning Objective: 02-7
12.48
Answers will vary. Sample size of 52 should be sufficient to observe important relationships. Learning Objective: 02-6
12.49
A positive slope would be logical. Cruise speed should be related to engine size. And yes, a cause and effect relationship would make sense Learning Objective: 12-2
12.50
There is a strong positive relationship between TotalHP and CruiseSpeed. 303
ASBE 6e Solutions for Instructors Learning Objective: 03-8 12.51
See graph above. Yes, a linear relationship is plausible. Learning Objective: 03-8
12.52
An increase in one unit of horsepower, increases cruise speed by .1931 mph. No, the intercept does not have meaning because an engine cannot have zero horsepower. Learning Objective: 12-2
12.53
MegaStat output is provided. Regression Analysis r² 0.684 r 0.827 Std. Error 20.596
n 52 k 1 Dep. Var. Cruise
ANOVA table Source SS Regression 45,896.1706 Residual 21,209.2717 Total 67,105.4423
df 1 50 51
Regression output variables coefficients Intercept 103.0870 TotalHP 0.1931
std. error 6.2426 0.0186
t (df=50) p-value 16.513 6.92E-22 10.402 4.20E-14
Predicted 125.5 219.0 228.6 213.2 161.0 172.6 172.6 161.0 124.3 131.1 134.0 137.9 147.5 213.2 184.2 222.8 247.9 137.9 158.1 158.1 199.7
Residual Leverage -25.5 0.046 -19.0 0.093 12.4 0.119 -14.2 0.079 13.0 0.019 -8.6 0.022 -31.6 0.022 -0.0 0.019 -17.3 0.048 -27.1 0.038 -12.0 0.035 -8.9 0.031 -3.5 0.023 -19.2 0.079 -14.2 0.031 0.2 0.103 -13.9 0.185 -13.9 0.031 27.9 0.019 31.9 0.019 -9.7 0.052
Observation 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21
Cruise 100.0 200.0 241.0 199.0 174.0 164.0 141.0 161.0 107.0 104.0 122.0 129.0 144.0 194.0 170.0 223.0 234.0 124.0 186.0 190.0 190.0
MS 45,896.1706 424.1854
304
F 108.20
p-value 4.20E-14
confidence interval 95% lower 95% upper 90.5483 115.6256 0.1558 0.2304 Studentized Studentized Deleted Residual Residual -1.267 -1.275 -0.967 -0.966 0.640 0.636 -0.717 -0.714 0.636 0.632 -0.423 -0.420 -1.552 -1.575 -0.001 -0.001 -0.863 -0.860 -1.341 -1.353 -0.593 -0.589 -0.437 -0.433 -0.172 -0.171 -0.970 -0.969 -0.701 -0.697 0.009 0.009 -0.749 -0.746 -0.683 -0.679 1.366 1.379 1.563 1.586 -0.481 -0.478
ASBE 6e Solutions for Instructors 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52
159.0 160.0 148.0 143.0 160.0 140.0 235.0 191.0 132.0 115.0 170.0 175.0 156.0 188.0 128.0 107.0 148.0 129.0 191.0 147.0 213.0 186.0 148.0 180.0 186.0 100.0 176.0 151.0 98.0 163.0 143.0
148.5 148.5 163.0 161.0 141.7 127.2 170.7 163.0 127.2 137.9 143.6 150.2 141.7 157.2 134.0 127.2 161.0 137.9 199.7 148.5 170.7 161.0 161.0 188.1 188.1 132.1 161.0 153.3 118.7 151.4 137.9
10.5 11.5 -15.0 -18.0 18.3 12.8 64.3 28.0 4.8 -22.9 26.4 24.8 14.3 30.8 -6.0 -20.2 -13.0 -8.9 -8.7 -1.5 42.3 25.0 -13.0 -8.1 -2.1 -32.1 15.0 -2.3 -20.7 11.6 5.1
0.023 0.023 0.019 0.019 0.027 0.044 0.021 0.019 0.044 0.031 0.026 0.022 0.027 0.020 0.035 0.044 0.019 0.031 0.052 0.023 0.021 0.019 0.019 0.035 0.035 0.037 0.019 0.020 0.058 0.021 0.031
0.517 0.566 -0.733 -0.884 0.900 0.634 3.157 1.375 0.237 -1.127 1.296 1.217 0.703 1.512 -0.296 -1.004 -0.639 -0.437 -0.432 -0.072 2.077 1.224 -0.639 -0.399 -0.102 -1.586 0.734 -0.113 -1.037 0.571 0.254
0.513 0.562 -0.730 -0.882 0.898 0.630 3.492 1.388 0.235 -1.130 1.305 1.223 0.700 1.532 -0.293 -1.005 -0.635 -0.433 -0.428 -0.072 2.151 1.231 -0.635 -0.395 -0.101 -1.611 0.731 -0.112 -1.038 0.567 0.252
Learning Objective: 12-4 12.54
a. 95% confidence interval for β1: (0.1558, 0.2304). This interval does not contain zero which means that we are confident the slope is not zero. b. H0: β1 = 0 vs. H1: β1 ≠ 0, d.f. = 50, t.025 =T.INV(.025,50) = ±2.009, tcalc = 10.402 > 2.009. Therefore reject H0 and conclude that the slope is significantly different from zero. c. The p-value = 4.20×10-14. This means that it is highly unlikely to obtain a slope estimate of this value if the true slope is equal to zero. d. Yes, this sample supports our a priori hypothesis about the slope. Learning Objective: 12-5
12.55
a. R2 = .684. 68.4% of the variation in cruise speed can be explained by engine horsepower. This shows a moderately strong relationship. b. Fcalc = 108.20 and its p-value = 4.20×10-14. The F statistic is significant which means the linear model provides significant fit. 305
ASBE 6e Solutions for Instructors c. This model provides a good enough fit for practical use. Learning Objective: 12-6 12.56
Observation 28 is an outlier residual. This corresponds to the ExtraExtra400 model. Observation 42 is an unusual residual. This is the Piper Malibu Mirage model. Learning Objective: 12-8
12.57
a. The residual normplot is somewhat linear. The histogram of residuals shows a very slight right skewed distribution.
Object 37
Object 39
b. The residuals do not show significant departure from normality. 306
ASBE 6e Solutions for Instructors Learning Objective: 12-8
12.58
Object 41
The residual plot does not show obvious signs of heteroscedasticity. We can see a possible high outlier on the graph. Learning Objective: 12-8 12.59
An autocorrelation test is not appropriate because this is not a time series data set. Learning Objective: 12-8
12.60
Answers will vary. Confidence and prediction intervals for x = 150 and 250 are shown below. Predicted values for: Cruise 95% Confidence Intervals
TotalHP 150 250
Predicted 132.057 151.371
lower 124.072 145.350
Learning Objective: 12-7
307
upper 140.043 157.391
95% Prediction Intervals
lower 89.926 109.567
upper 174.189 193.174
ASBE 6e Solutions for Instructors 12.61
Observations 2-4 and 14, 16, and 17 have high leverage. Learning Objective: 12-9
DATA SET E 12.46
Cross-sectional. Learning Objective: 02-3
12.47
Answers will vary. This type of information would probably be collected from a researcher who has obtained different types of processors. Learning Objective: 02-7
12.48
Answers will vary. Sample size of 14 is fairly small and can open one up to Type II error. Learning Objective: 02-6
12.49
A positive slope would be logical. Microprocessor speed should be related to Power dissipation. Learning Objective: 12-2
12.50
Object 43
There is a strong positive relationship between microprocessor speed and power dissipation. Learning Objective: 03-8 308
ASBE 6e Solutions for Instructors
12.51
See graph above. Yes, a linear relationship is plausible. Learning Objective: 03-8
12.52
An increase in one unit of microprocessor speed increases power dissipation by 0.032 watts. No, the intercept does not have meaning a speed of zero is not logical. Learning Objective: 12-2
12.53
MegaStat output is provided. Regression Analysis r² r Std. Error
0.925 0.962 13.109
ANOVA table Source
SS 25,561.915 Regression 7 Residual 2,062.0843 27,624.000 Total 0 Regression output coefficient variables s Intercept 15.7299 Speed
Observatio n 1 2 3 4 5 6 7 8 9 10 11 12 13 14
0.0319
Power 3.0 10.0 35.0 20.0 42.0 50.0 51.0 73.0 115.0 130.0 95.0 136.0 95.0 125.0
n k Dep. Var.
df 1 12
MS 25,561.915 7 171.8404
14 1 Power
F
p-value
148.75
4.03E-08
13 confidence interval std. error 5.6634
t (df=12) 2.777
0.0026
12.196
Predicted 16.4 18.9 23.2 25.3 34.8 34.8 57.1 82.6 136.8 117.7 89.0 117.7 108.1 117.7
Residual -13.4 -8.9 11.8 -5.3 7.2 15.2 -6.1 -9.6 -21.8 12.3 6.0 18.3 -13.1 7.3
Learning Objective: 12-4
309
p-value .0167 4.03E08
Leverag e 0.184 0.174 0.157 0.150 0.120 0.120 0.078 0.078 0.246 0.160 0.086 0.160 0.128 0.160
95% lower 3.3905
95% upper 28.0693
0.0262
0.0375 Studentize d
Studentize d
Deleted
Residual -1.129 -0.748 0.985 -0.437 0.582 1.233 -0.488 -0.764 -1.912 1.027 0.479 1.527 -1.071 0.611
Residual -1.143 -0.734 0.983 -0.422 0.565 1.263 -0.472 -0.750 -2.196 1.030 0.463 1.629 -1.078 0.594
ASBE 6e Solutions for Instructors 12.54
a. 95% confidence interval for β1: (0.0262, 0.0375). This interval does not contain zero which means that we are confident the slope is not zero. b. H0: β1 = 0 vs. H1: β1 ≠ 0, d.f. = 50, t.025 =T.INV(.025,12) = ±2.179, tcalc = 12/196 > 2.179. Therefore reject H0 and conclude that the slope is significantly different from zero. c. The p-value = .0000. This means that it is highly unlikely to obtain a slope estimate of this value if the true slope is equal to zero. d. Yes, this sample supports our a priori hypothesis about the slope. Learning Objective: 12-5
12.55
a. R2 = .925. 92.5% of the variation in power dissipation can be explained by microprocessor speed. This shows a strong relationship. b. Fcalc = 148.75 and its p-value = .0000. The F statistic is significant which means the linear model provides significant fit. c. This model provides a good enough fit for practical use. Learning Objective: 12-6
12.56
There are no unusual or outlier standardized residuals. Learning Objective: 12-9
12.57
a. The residual normplot is somewhat linear. The histogram of residuals shows a left skewed distribution.
Object 45
310
ASBE 6e Solutions for Instructors
Object 48
b. The sample size is small so it is difficult to determine normality but there is no obvious evidence to assume non-normality. Learning Objective: 12-8 12.58
Object 50
The residual plot does not show obvious signs of heteroscedasticity. Learning Objective: 12-8 12.59
An autocorrelation test is not appropriate because this is not a time series data set. 311
ASBE 6e Solutions for Instructors Learning Objective: 12-8 12.60
Answers will vary. Confidence and prediction intervals for x = 1000 and 2500 are shown below. Predicted values for: Power Speed 1,000 2,500
Predicted 47.583 95.362
95% Confidence Intervals
95% Prediction Intervals
lower 38.962 86.485
lower 17.748 65.452
Learning Objective: 12-7 12.61
There are no high leverage observations. Learning Objective: 12-9
DATA SET F 312
upper 56.203 104.238
upper 77.417 125.271
ASBE 6e Solutions for Instructors
12.46
Cross-sectional. Learning Objective: 02-3
12.47
Answers will vary. This could probably be collected through surveys. Learning Objective: 02-7
12.48
Answers will vary. Sample size of 10 is fairly small and can open one up to Type II error. But this information could be difficult to obtain. Learning Objective: 02-6
12.49
A positive slope would be logical. Increased website hits should be associated with increased revenue. Learning Objective: 12-2
12.50
Object 52
There is a weak positive relationship between website hits and restaurant revenue. Learning Objective: 03-8 12.51
See graph above. Yes, a linear relationship is plausible. Learning Objective: 03-8
12.52
An increase of one website visit increases weekly revenue by $1.67. The intercept could be interpreted as the weekly revenue with no website hits. Although for this particular sample there were no restaurants with an x near zero so it would be dangerous to extrapolate. Learning Objective: 12-2
313
ASBE 6e Solutions for Instructors 12.53
MegaStat output is provided. Regression Analysis r² r Std. Error
0.128 0.357 1078.961
n k Dep. Var.
ANOVA table Source
SS
df
F
p-value
Regression
1,361,311.7375
1
1.17
.3111
Residual 9,313,262.2625 Total 10,674,574.0000 Regression output variables coefficients
8 9
MS 1,361,311.737 5 1,164,157.782 8
t (df=8)
p-value
10,382.3728
std. error 2,353.059 2
4.412
.0022
1.6695
1.5439
1.081
.3111
Intercept Website Hits
Observatio n 1 2 3 4 5 6 7 8 9 10
Restaurant Revenue 12,113.0 11,409.0 14,579.0 11,605.0 12,308.0 12,320.0 13,225.0 13,652.0 13,893.0 13,896.0
Predicted 12,407.5 12,869.9 12,661.3 12,811.5 12,501.0 13,107.0 12,591.1 13,496.0 13,036.9 13,517.7
Residual -294.5 -1,460.9 1,917.7 -1,206.5 -193.0 -787.0 633.9 156.0 856.1 378.3
10 1 Restaurant Revenue
Leverage 0.278 0.101 0.142 0.106 0.217 0.131 0.170 0.361 0.114 0.380
confidence interval 95% lower 95% upper 15,808.537 4,956.2085 1 -1.8907
5.2297 Studentize d
Studentize d
Deleted
Residual -0.321 -1.428 1.919 -1.182 -0.202 -0.783 0.645 0.181 0.843 0.445
Residual -0.302 -1.547 2.443 -1.218 -0.190 -0.762 0.620 0.170 0.826 0.422
Learning Objective: 12-4 12.54
a. 95% confidence interval for β1: (−1.8907, 5.2297). This interval does contain zero which means that we are not confident the slope is not zero. b. H0: β1 = 0 vs. H1: β1 ≠ 0, d.f. = 50, t.025 =T.INV(.025,8) = ±2.306, tcalc = 1.081 falls between the critical values. Therefore fail to reject H0 and conclude that the slope isnot significantly different from zero. c. The p-value = .3111. This means that it is likely to obtain a slope estimate of this value even if the true slope is equal to zero. d. No, this sample does not support our a priori hypothesis about the slope. Learning Objective: 12-5
314
ASBE 6e Solutions for Instructors 12.55
a. R2 = .128. Because the slope is not significant R2 does not have meaning. b. Fcalc = 1.17 and its p-value = .3111. The F statistic is not significant which means the linear model does not provide significant fit. c. This model is not fit for practical use. Learning Objective: 12-6
12.56
Restaurant 3 shows an unusual residual. It appears the model is underestimating the revenue. Learning Objective: 12-9
12.57
a. The residual normplot is somewhat linear. The histogram of residuals shows a fairly uniform distribution.
Object 54
315
ASBE 6e Solutions for Instructors
Object 57
b. The sample size is small so it is difficult to determine normality from the histogram. Based on the normplot there is no obvious evidence to assume non-normality. Learning Objective: 12-8 12.58
Object 59
The residual plot does not show obvious signs of heteroscedasticity. Learning Objective: 12-8 12.59
An autocorrelation test is not appropriate because this is not a time series data set. 316
ASBE 6e Solutions for Instructors Learning Objective: 12-8 12.60
Answers will vary. Confidence and prediction intervals for x = 1200 and 1500 are shown below. Keep in mind that these intervals are unreliable because there is not significant relationship between website hits and revenue. Predicted values for: Restaurant Revenue 95% Confidence Intervals
95% Prediction Intervals
Website Hits
Predicted
lower
upper
lower
1,200
12,385.790
11,036.168
13,735.411
1,500
12,886.644
12,099.326
13,673.962
9,555.230 10,276.95 8
upper 15,216.34 9 15,496.33 0
Learning Objective: 12-7 12.61
There are no high leverage observations. Learning Objective: 12-9
DATA SET G 12.46
Cross-sectional. Learning Objective: 02-3
12.47
Answers will vary. This information can be gathered from manufacturers’ specification information which is listed on their websites. They manufacturers use sophisticated sampling techniques for estimating these values. Learning Objective: 02-7
12.48
Answers will vary. Sample size of 73 is reasonable. Learning Objective: 02-6
12.49
A negative slope would be logical. It would make sense that the heavier a car is the lower the MPG. Learning Objective: 12-2
317
ASBE 6e Solutions for Instructors 12.50
Object 61
There is a fairly strong negative relationship between the Weight and CityMPG. Learning Objective: 03-8 12.51
See graph above. Yes, a linear relationship is plausible. Learning Objective: 03-8
12.52
An increase in the weight of a car by one pound reduces its city mpg by 0.005 mpg. No, the intercept does not make sense. Learning Objective: 12-2
12.53
MegaStat output is provided. Regression Analysis r² r Std. Error
0.800 -0.894 2.238
n k Dep. Var.
73 1 City MPG
ANOVA table Source Regression Residual Total
SS 1,421.035 355.650 1,776.685
df 1 71 72
MS 1,421.035 5.009
F 283.69
Regression output variables Intercept
coefficients 40.0832
std. error 1.1526
t (df=71) 34.777
p-value 2.49E-46
318
p-value 1.67E-26
confidence interval 95% lower 95% upper 37.7850 42.3814
ASBE 6e Solutions for Instructors Weight
Observation 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45
-0.0052
City MPG 20.0 22.0 22.0 23.0 18.0 17.0 21.0 22.0 20.0 14.0 16.0 26.0 19.0 25.0 31.0 15.0 20.0 19.0 19.0 23.0 31.0 29.0 27.0 15.0 19.0 24.0 23.0 19.0 15.0 18.0 19.0 17.0 20.0 21.0 17.0 25.0 24.0 21.0 13.0 21.0 16.0 18.0 17.0 22.0 14.0
0.00030678
Predicted 19.6 21.6 22.2 21.9 19.5 15.7 21.0 23.0 21.4 11.5 20.9 24.1 20.4 21.8 28.4 10.8 22.5 19.3 19.4 23.3 27.5 26.8 24.8 21.4 19.1 23.6 23.0 21.5 18.2 19.8 21.3 20.5 22.1 23.2 15.9 25.7 26.6 23.6 13.5 22.2 18.2 18.5 15.8 20.9 10.1
319
-16.843
Residual 0.4 0.4 -0.2 1.1 -1.5 1.3 -0.0 -1.0 -1.4 2.5 -4.9 1.9 -1.4 3.2 2.6 4.2 -2.5 -0.3 -0.4 -0.3 3.5 2.2 2.2 -6.4 -0.1 0.4 -0.0 -2.5 -3.2 -1.8 -2.3 -3.5 -2.1 -2.2 1.1 -0.7 -2.6 -2.6 -0.5 -1.2 -2.2 -0.5 1.2 1.1 3.9
1.67E-26
-0.0058
Leverage 0.015 0.014 0.014 0.014 0.016 0.035 0.014 0.016 0.014 0.079 0.014 0.020 0.014 0.014 0.050 0.090 0.015 0.016 0.016 0.017 0.042 0.036 0.023 0.014 0.017 0.018 0.016 0.014 0.020 0.015 0.014 0.014 0.014 0.017 0.033 0.028 0.034 0.018 0.055 0.014 0.020 0.019 0.034 0.014 0.099
Studentize d Residual 0.189 0.194 -0.069 0.513 -0.661 0.603 -0.003 -0.430 -0.629 1.153 -2.190 0.866 -0.652 1.425 1.211 1.983 -1.128 -0.119 -0.196 -0.149 1.595 1.011 1.000 -2.874 -0.061 0.185 -0.003 -1.105 -1.433 -0.793 -1.040 -1.575 -0.953 -1.013 0.511 -0.335 -1.169 -1.182 -0.248 -0.540 -0.998 -0.223 0.551 0.496 1.815
-0.0046 Studentize d Deleted Residual 0.188 0.192 -0.069 0.511 -0.658 0.600 -0.003 -0.427 -0.627 1.155 -2.252 0.864 -0.649 1.436 1.215 2.026 -1.130 -0.119 -0.195 -0.148 1.612 1.011 1.000 -3.036 -0.061 0.184 -0.003 -1.107 -1.444 -0.791 -1.040 -1.592 -0.952 -1.014 0.508 -0.333 -1.172 -1.185 -0.247 -0.537 -0.998 -0.222 0.548 0.493 1.845
ASBE 6e Solutions for Instructors 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73
28.0 26.0 22.0 12.0 17.0 29.0 25.0 13.0 25.0 19.0 18.0 27.0 15.0 25.0 36.0 27.0 24.0 24.0 25.0 13.0 24.0 30.0 22.0 25.0 17.0 24.0 21.0 16.0
28.2 23.6 22.4 14.8 15.5 26.6 23.1 12.9 25.7 21.6 19.4 27.9 17.4 25.6 29.1 26.3 22.4 22.4 23.1 10.3 22.2 28.2 24.4 24.2 15.7 23.4 21.9 16.0
-0.2 2.4 -0.4 -2.8 1.5 2.4 1.9 0.1 -0.7 -2.6 -1.4 -0.9 -2.4 -0.6 6.9 0.7 1.6 1.6 1.9 2.7 1.8 1.8 -2.4 0.8 1.3 0.6 -0.9 0.0
0.048 0.018 0.015 0.042 0.036 0.035 0.016 0.062 0.028 0.014 0.016 0.045 0.024 0.027 0.058 0.032 0.015 0.015 0.016 0.097 0.014 0.049 0.021 0.020 0.034 0.017 0.014 0.033
-0.077 1.091 -0.167 -1.276 0.667 1.081 0.877 0.061 -0.305 -1.186 -0.612 -0.399 -1.067 -0.265 3.179 0.312 0.713 0.731 0.840 1.272 0.819 0.813 -1.077 0.352 0.572 0.273 -0.384 0.014
-0.076 1.093 -0.166 -1.282 0.664 1.082 0.876 0.060 -0.303 -1.190 -0.609 -0.397 -1.068 -0.263 3.409 0.310 0.710 0.729 0.838 1.277 0.817 0.811 -1.079 0.350 0.570 0.272 -0.382 0.014
Learning Objective: 12-4 12.54
a. 95% confidence interval for β1: (0.0058, 0.0046). This interval does not contain zero which means that we are confident the slope is not zero. b. H0: β1 = 0 vs. H1: β1 ≠ 0, d.f. = 41, t.025 =T.INV(.025,71) = ±1.994, tcalc = 16.843 < 2.020. Therefore reject H0 and conclude that the slope is significantly different from zero. c. The p-value = 1.67×10-26. This means that it is highly unlikely to obtain a slope estimate of this value if the true slope is equal to zero. d. Yes, this sample supports our a priori hypothesis about the slope. Learning Objective: 12-5
12.55
a. R2 = .80. 80% of the variation in the City MPG of a vehicle can be explained by its weight. This shows a fairly strong relationship. b. Fcalc = 283.69 and its p-value = 1.67×10-26. The F statistic is significant which means the linear model provides significant fit. c. This model provides a good enough fit for practical use. Learning Objective: 12-6
12.56
Observations 11, 16, 24, and 60 are unusual residuals. 320
ASBE 6e Solutions for Instructors Learning Objective: 12-9 12.57
a. The normplot of residuals is not perfect. The high outlier can be seen and the line sags in the middle. The histogram of residuals shows a somewhat bell-shaped distribution with one high outlier.
Object 63
Object 65
b. The residuals do not show significant departure from normality in spite of a few unusual values. Learning Objective: 12-8 12.58 321
ASBE 6e Solutions for Instructors
Object 68
The residual plot does not show obvious signs of heteroscedasticity. Learning Objective: 12-8 12.59
An autocorrelation test is not appropriate because this is cross sectional data. Learning Objective: 12-8
12.60
Answers will vary. Confidence and prediction intervals for x = 3000 and 4000 are shown below.
Learning Objective: 12-7 12.61
Observations 9, 16, 45, 53, 60, and 65 have high leverage. Learning Objective: 12-9
322
ASBE 6e Solutions for Instructors DATA SET H 12.46
Cross-sectional. Learning Objective: 02-3
12.47
Answers will vary. This information can be gathered from food labels or manufacturers’ websites. Learning Objective: 02-7
12.48
Answers will vary. Sample size is reasonable although a large sample would be better. Learning Objective: 02-6
12.49
A positive slope would be logical. It would make sense that the more fat calories the sauce has the more calories are in the sauce in general. Learning Objective: 12-2
12.50
There is a fairly strong positive relationship between the FatCalories/Gram and Calories/Gram. Learning Objective: 03-8 12.51
See graph above. Yes, a linear relationship is plausible. Learning Objective: 03-8
12.52
An increase in the fat calories per gram by 1, increases total calories per gram by 2.2179. The intercept might make sense for sauce labeled “fat free.” Learning Objective: 12-2
323
ASBE 6e Solutions for Instructors 12.53
MegaStat output is provided. Regression Analysis r² 0.843 r 0.918 Std. Error 0.102
n 20 k 1 Dep. Var. Calories Per Gram
ANOVA table Source Regression Residual Total Regression output variables Intercept Fat Calories Per Gram
SS 1.0076 0.1880 1.1955
df 1 18 19
coefficients 0.3054
std. error 0.0460
2.2179
0.2258
Observation Calories Per Gram Predicted 1 0.640 0.749 2 0.560 0.572 3 0.400 0.483 4 0.480 0.394 5 0.640 0.572 6 0.400 0.305 7 0.560 0.483 8 0.550 0.483 9 0.480 0.572 10 0.480 0.572 11 1.250 1.148 12 1.000 1.037 13 1.000 0.949 14 1.170 1.037 15 0.920 0.860 16 0.670 0.860 17 0.860 0.749 18 0.700 0.727 19 0.560 0.749 20 0.640 0.660
MS 1.0076 0.0104
F 96.49
t (df=18) p-value 6.635 3.16E-06 9.823
1.17E-08
p-value 1.17E-08
confidence interval 95% lower 95% upper 0.2087 0.4021 1.7436
2.6923 Studentized Studentized Deleted Residual Leverage Residual Residual -0.109 0.053 -1.096 -1.103 -0.012 0.066 -0.117 -0.114 -0.083 0.096 -0.853 -0.846 0.086 0.142 0.907 0.902 0.068 0.066 0.693 0.682 0.095 0.203 1.037 1.039 0.077 0.096 0.794 0.785 0.067 0.096 0.691 0.681 -0.092 0.066 -0.927 -0.923 -0.092 0.066 -0.927 -0.923 0.102 0.251 1.151 1.162 -0.037 0.164 -0.400 -0.390 0.051 0.112 0.534 0.523 0.133 0.164 1.420 1.465 0.060 0.076 0.612 0.601 -0.190 0.076 -1.933 -2.111 0.111 0.053 1.116 1.124 -0.027 0.051 -0.270 -0.262 -0.189 0.053 -1.900 -2.066 -0.020 0.051 -0.204 -0.198
Learning Objective: 12-4
12.54
a. 95% confidence interval for β1: (1.7436, 2.6923). This interval does not contain zero which means that we are confident the slope is not zero. b. H0: β1 = 0 vs. H1: β1 ≠ 0, d.f. = 18, t.025 =T.INV(.025,18) = ±2.101, tcalc = 9.823 > 2.101. Therefore reject H0 and conclude that the slope is significantly different from zero. c. The p-value = 1.17×10-8. This means that it is highly unlikely to obtain a slope estimate of this value if the true slope is equal to zero. 324
ASBE 6e Solutions for Instructors d. Yes, this sample supports our a priori hypothesis about the slope. Learning Objective: 12-5 12.55
a. R2 = .843. 84.3% of the variation in the total calories/gram of a pasta sauce can be explained by the number of fat calories/gram. This shows a fairly strong relationship. b. Fcalc = 96.49 and its p-value = 1.17×10-8. The F statistic is significant which means the linear model provides significant fit. c. This model provides a good enough fit for practical use. Learning Objective: 12-6
12.56
Observation 16 is an unusual residual. This is the Ragu Old World Style with meat sauce. Learning Objective: 12-9
12.57
a. The normal probability plot of residuals does not show a clear normal pattern.
Object 70
325
ASBE 6e Solutions for Instructors
Object 72
b. The residuals show some departure from normality but this is a small sample size and a larger sample might help. Learning Objective: 12-8 12.58
Object 74
The residual plot does not show obvious signs of heteroscedasticity. Learning Objective: 12-8 12.59
An autocorrelation test is not appropriate because this is cross-sectional data. Learning Objective: 12-8 326
ASBE 6e Solutions for Instructors 12.60
Answers will vary. Confidence and prediction intervals for x = 0.10 and 0.20 are shown below. Predicted values for: Calories Per Gram 95% Confidence Intervals
Fat Calories Per Gram
Predict ed
0.10
0.52722
0.20
0.74901
lower 0.4669 0 0.6997 8
upper 0.58754 0.79824
95% Prediction Intervals
lower 0.3042 2 0.5287 6
upper 0.75021 0.96927
Learning Objective: 12-7 12.61
Observations 6 and 11 show high leverage. These correspond to Healthy Choice Traditional and Prego Hearty Meat Peperoni. Learning Objective: 12-9
DATA SET I 12.46
Time-series. Learning Objective: 02-3
12.47
Answers will vary. This information can be gathered by taking a random household and observing their energy usage. Learning Objective: 02-7
12.48
Answers will vary. Two years of data is good but when looking for seasonal influences more years would be better. Learning Objective: 02-6
12.49
A negative slope would be logical. It would make sense that the lower the temperature the more energy a household would use. Learning Objective: 12-2
327
ASBE 6e Solutions for Instructors 12.50
Object 77
There is a fairly strong negative relationship between the Average Daily Temperature and Energy Consumption. Learning Objective: 03-8 12.51
See graph above. Yes, a linear relationship is plausible. Learning Objective: 03-8
12.52
An increase in 1° in average temperature decreases the monthly energy use by 9.661 kwh. Yes, the intercept does make sense. There can be a month with an average temperature of 0°. Learning Objective: 12-2
12.53
MegaStat output is provided. Regression Analysis r² 0.766 r -0.875 Std. Error 84.951
n 24 k 1 Dep. Var. Electric Consumption (KWH)
ANOVA table Source Regression Residual Total Regression output variables Intercept Avg Daily Temp (deg F)
SS 520,420.3570 158,766.1430 679,186.5000
df MS 1 520,420.3570 22 7,216.6429 23
coefficients std. error 1,165.7901 62.9671 -9.6609 1.1376
Observation Energy Use (KWH) Predicted 1 436.0 566.8
328
F 72.11
p-value 2.16E-08
confidence interval t (df=22) p-value 95% lower 95% upper 18.514 6.64E-15 1,035.2043 1,296.3760 -8.492 2.16E-08 -12.0202 -7.3016 Studentized Studentized Deleted Residual Leverage Residual Residual -130.8 0.056 -1.585 -1.645
ASBE 6e Solutions for Instructors 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24
464.0 446.0 391.0 444.0 608.0 885.0 821.0 830.0 750.0 617.0 598.0 597.0 528.0 477.0 562.0 658.0 690.0 862.0 1,008.0 840.0 867.0 606.0 657.0
479.9 431.6 499.2 557.2 663.4 808.3 779.4 789.0 827.7 731.0 644.1 499.2 470.2 402.6 489.5 586.1 702.1 798.7 837.3 924.3 798.7 702.1 653.8
-15.9 14.4 -108.2 -113.2 -55.4 76.7 41.6 41.0 -77.7 -114.0 -46.1 97.8 57.8 74.4 72.5 71.9 -12.1 63.3 170.7 -84.3 68.3 -96.1 3.2
0.098 0.135 0.086 0.059 0.042 0.089 0.073 0.078 0.101 0.054 0.042 0.086 0.105 0.161 0.092 0.050 0.047 0.083 0.108 0.184 0.083 0.047 0.042
-0.197 0.183 -1.332 -1.373 -0.667 0.945 0.509 0.502 -0.964 -1.380 -0.554 1.205 0.719 0.956 0.895 0.868 -0.145 0.778 2.127 -1.098 0.840 -1.158 0.039
-0.192 0.179 -1.358 -1.403 -0.658 0.943 0.500 0.494 -0.963 -1.411 -0.545 1.218 0.711 0.954 0.891 0.863 -0.142 0.771 2.332 -1.104 0.834 -1.168 0.038
Learning Objective: 12-4 12.54
a. 95% confidence interval for β1: (12.0202, 7.3016). This interval does not contain zero which means that we are confident the slope is not zero. b. H0: β1 = 0 vs. H1: β1 ≠ 0, d.f. = 22, t.025 =T.INV(.025,22) = ±2.074, tcalc = 8.492 < 2.074. Therefore reject H0 and conclude that the slope is significantly different from zero. c. The p-value = 2.16×10-8. This means that it is highly unlikely to obtain a slope estimate of this value if the true slope is equal to zero. d. Yes, this sample supports our a priori hypothesis about the slope. Learning Objective: 12-5
12.55
a. R2 = .766. 76.6% of the variation in the energy usage can be explained by average daily temperature. This shows a fairly strong relationship. b. Fcalc = 72.11 and its p-value = 2.16×10-8. The F statistic is significant which means the linear model provides significant fit. c. This model provides a good enough fit for practical use. Learning Objective: 12-6
12.57
Observation 20 is an unusual residual. The model underestimates the energy usage for that month. Learning Objective: 12-9
329
ASBE 6e Solutions for Instructors 12.58
a. The normplot of residuals does not show a clear normal pattern. The histogram also shows a non-normality.
Object 79
Object 81
b. The residuals show some departure from normality but this is a small sample size and a larger sample might help. Learning Objective: 12-8 12.58
330
ASBE 6e Solutions for Instructors
Object 83
The residual plot does not show obvious signs of heteroscedasticity. Learning Objective: 12-8 12.59
An autocorrelation test is appropriate because this is time series data. The residual plot below does not show obvious increasing or decreasing trends nor does it show signs of negative autocorrelation. The DW statistic = 1.53 which indicates slight positive autocorrelation. This is not unexpected because temperature for one month is related to temperature for the previous month. The level of autocorrelation does not invalidate the regression results.
Object 85
Learning Objective: 12-8 331
ASBE 6e Solutions for Instructors
12.60
Answers will vary. Confidence and prediction intervals for x = 50 and 70 are shown below. Predicted values for: Electric Consumption (KWH) 95% Confidence Intervals
Avg Daily Temp (deg F) 50 70
Predicted 682.745 489.527
lower 645.995 436.022
Learning Objective: 12-7 12.61
Observation 21 shows high leverage. Learning Objective: 12-9
332
upper 719.495 543.033
95% Prediction Intervals
lower 502.776 305.405
upper 862.715 673.650
ASBE 6e Solutions for Instructors DATA SET J 12.46
Time-series. Learning Objective: 02-3
12.47
Answers will vary. This information is collected by The Bureau of Labor Statistics. Learning Objective: 02-7
12.48
Answers will vary. 47 years of data is a good sample. Methods of measurement can vary over time. It is important to consider how the measures were calculated and compare years in which the calculations were similar. Learning Objective: 02-6
12.49
A positive slope would be logical. It would make sense that the change in Commodities CPI would move in the same direction as change in Services CPI. Learning Objective: 12-2
12.50
Object 88
There is a fairly strong positive relationship between the Commodities% and Services%. Learning Objective: 03-8 12.51
See graph above. Yes, a linear relationship is plausible. Learning Objective: 03-8
12.52
An increase in the change in Commodities CPI of 1% increases the change in Service CPI by .830%. Yes, it is possible that there is no change in the CPI between two years. Learning Objective: 12-2
333
ASBE 6e Solutions for Instructors 12.53
MegaStat output is provided. Regression Analysis r² 0.727 r 0.853 Std. Error 1.574 ANOVA table Source Regression Residual Total
SS 297.2715 111.4532 408.7247
Regression output variables coefficients Intercept 2.2068 Commodities% 0.8302
Observation 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32
Services% 3.40 1.70 2.00 2.00 2.00 2.30 3.80 4.30 5.20 6.90 8.00 5.70 3.80 4.40 9.20 9.60 8.30 7.70 8.60 11.00 15.40 13.10 9.00 3.50 5.20 5.10 5.00 4.20 4.60 4.90 5.50 5.10
n 47 k 1 Dep. Var. Services% df 1 45 46
MS 297.2715 2.4767
F 120.03
std. error 0.3506 0.0758
t (df=45) 6.293 10.956
Predicted 2.95 2.70 2.95 2.95 3.20 3.12 4.37 3.78 5.11 6.11 5.94 5.20 4.70 8.35 12.09 9.51 5.78 7.02 8.18 11.59 12.42 9.18 5.61 4.61 5.03 3.95 1.46 4.86 5.11 6.11 6.52 4.78
Residual Leverage 0.45 0.037 -1.00 0.041 -0.95 0.037 -0.95 0.037 -1.20 0.034 -0.82 0.035 -0.57 0.023 0.52 0.027 0.09 0.021 0.79 0.025 2.06 0.024 0.50 0.021 -0.90 0.022 -3.95 0.057 -2.89 0.185 0.09 0.086 2.52 0.023 0.68 0.034 0.42 0.053 -0.59 0.162 2.98 0.201 3.92 0.077 3.39 0.022 -1.11 0.022 0.17 0.021 1.15 0.026 3.54 0.066 -0.66 0.021 -0.51 0.021 -1.21 0.025 -1.02 0.028 0.32 0.022
334
p-value 1.14E-07 2.76E-14
p-value 2.76E-14
confidence interval 95% lower 95% upper 1.5005 2.9130 0.6776 0.9828 Studentized Studentized Deleted Residual Residual 0.289 0.286 -0.652 -0.648 -0.618 -0.613 -0.618 -0.613 -0.778 -0.774 -0.530 -0.526 -0.363 -0.360 0.332 0.329 0.056 0.056 0.509 0.505 1.323 1.334 0.324 0.321 -0.577 -0.572 -2.584 -2.769 -2.031 -2.107 0.058 0.058 1.622 1.653 0.438 0.434 0.272 0.269 -0.408 -0.404 2.120 2.209 2.592 2.779 2.178 2.277 -0.716 -0.712 0.110 0.108 0.740 0.736 2.328 2.454 -0.426 -0.422 -0.329 -0.326 -0.778 -0.774 -0.660 -0.656 0.205 0.203
ASBE 6e Solutions for Instructors 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47
3.90 3.90 3.30 3.40 3.20 3.00 2.70 2.50 3.40 4.10 3.10 3.20 2.90 3.30 3.80
3.87 3.78 3.62 3.78 4.37 3.37 2.29 3.70 4.95 3.04 1.63 3.04 4.12 5.20 4.20
0.03 0.12 -0.32 -0.38 -1.17 -0.37 0.41 -1.20 -1.55 1.06 1.47 0.16 -1.22 -1.90 -0.40
0.026 0.027 0.029 0.027 0.023 0.031 0.048 0.028 0.021 0.036 0.062 0.036 0.025 0.021 0.024
0.021 0.075 -0.205 -0.247 -0.749 -0.238 0.267 -0.774 -0.993 0.688 0.967 0.106 -0.782 -1.217 -0.257
0.021 0.074 -0.203 -0.245 -0.745 -0.236 0.264 -0.771 -0.993 0.684 0.967 0.104 -0.779 -1.224 -0.254
Learning Objective: 12-4 12.54
a. 95% confidence interval for β1: (0.6776, 0.9828). This interval does not contain zero which means that we are confident the slope is not zero. b. H0: β1 = 0 vs. H1: β1 ≠ 0, d.f. = 45, t.025 =T.INV(.025,45) = ±2.014, tcalc = 10.956 > 2.014. Therefore reject H0 and conclude that the slope is significantly different from zero. c. The p-value = 2.76×10-14. This means that it is highly unlikely to obtain a slope estimate of this value if the true slope is equal to zero. d. Yes, this sample supports our a priori hypothesis about the slope. Learning Objective: 12-5
12.55
a. R2 = .727. 72.7% of the variation in the Services CPI change can be explained by Commodities CPI change. This shows a fairly strong relationship. b. Fcalc = 120.03 and its p-value = 2.76×10-14. The F statistic is significant which means the linear model provides significant fit. c. This model provides a good enough fit for practical use. Learning Objective: 12-6
12.56
Observations 14, 15, 21-23 and 27 are unusual residuals. Learning Objective: 12-9
12.57
a. The histogram shows a slight bell-shaped curve although there is more concentration in the middle of the graph than you would see in a true normal distribution. The normplot shows a similar pattern.
335
ASBE 6e Solutions for Instructors
Object 91
Object 93
b. The histogram of residuals does not show obvious departure from normality. Learning Objective: 12-8
12.58
336
ASBE 6e Solutions for Instructors
Object 95
The residual plot does not show significant signs of heteroscedasticity although there is a slight fan out pattern as X increases. Learning Objective: 12-8 12.59
An autocorrelation test is appropriate because this is time series data. The residual plot below does not show obvious increasing or decreasing trends nor does it show signs of negative autocorrelation. The DW statistic = 1.08 which indicates slight positive autocorrelation. This is not unexpected because CPI indexes are economic data which one would expect to be correlated month by month. The level of autocorrelation does not invalidate the regression results.
Object 97
337
ASBE 6e Solutions for Instructors Learning Objective: 12-8 12.60
Answers will vary. Confidence and prediction intervals for x = 1.5 and 2.5 are shown below. Predicted values for: Services% 95% Confidence Intervals
Commodities%
Predicte d
1.5
3.4520
2.5
4.2822
lower 2.898 2 3.795 4
upper
lower
upper
4.0059
0.2343
6.6698
4.7690
1.0753
7.4891
Learning Objective: 12-7 12.61
Observations 15, 16, 20 and 21 show high leverage. Learning Objective: 12-9
338
95% Prediction Intervals
ASBE 6e Solutions for Instructors 12.62
No, r measures the strength and direction of the linear relationship, but not the amount of variation explained by the explanatory variable. Learning Objective: 12-1
12.63
H0: ρ = 0 versus H1: ρ ≠ 0. α = .025 so tcrit = t.0125 =T.INV(.0125,53) = ±2.3069. tcalc = 2.3256 > 2.3069 so we reject the null hypothesis. The correlation is 55 - 2 .3043 = 1 - .30432 not equal to zero. Learning Objective: 12-1
12.64
The correlation coefficient, r, is only .13, indicating that there exists a very weak positive correlation between prices on successive days. The fact that it is a highly significant result stems from a large sample size which increases power of the test. This means that very small correlations will show statistical significance even though the correlation is not truly important. Learning Objective: 12-1
12.65
a.
ŷ
= 55.2 +.73(2000) = 1515.2 total free throws expected.
b. No, the intercept is not meaningful. You can’t make free throws without attempting them. c. Quick rule: or (1406.03, 1624.37) 1515.2 ± 2.052(53.2) = 1515.2 ± 109.17
Learning Objective: 12-2 Learning Objective: 12-7 12.66
a.
ŷ
= 30.7963+ .0343x (R2 = .202, se = 6.816)
b. d.f. = 33, α = .05 so tcrit = t.025 =T.INV(.025,33) = ±2.035 c. tcalc = 2.889 > 2.035 so we will reject the null hypothesis that the slope is zero. d. We are 95% confident that the slope is contained in the interval (.0101, .0584). This CI does not contain zero, hence, there is a relationship between the weekly pay and the income tax withheld. e. Fcalc = (2.889)2 = 8.3463 f. The value of R2 assigns only 20% of the variation in income withholding to the weekly pay. While the F statistic is significant, the fit is only a modest fit. Learning Objective: 12-4 Learning Objective: 12-5 Learning Objective: 12-6 12.67
a.
ŷ
= 1743.57 − 1.2163x (R2 = .370, s = 286.793)
b. d.f. = 13, α = .05 so tcrit = t.025 =T.INV(.025,13) = ±2.160 c. tcalc = −2.764 < −2.160 so we will reject the null hypothesis that the slope is zero. 339
ASBE 6e Solutions for Instructors d. We are 95% confident that the slope is contained in the interval (−2.1617, −0.2656). This CI does not contain zero, hence, there is a relationship between the weekly pay and monthly machine downtime. e. Fcalc = (−2.764)2 = 7.639696 f. The value of R2 assigns only 37% of the variation in monthly machine downtime to the monthly maintenance spending (dollars). Thus, throwing more “money” at the problem of downtime will not completely resolve the issue. Indicates that there are most likely other reasons why machines have the amount of downtime incurred. Learning Objective: 12-4 Learning Objective: 12-5 Learning Objective: 12-6 12.68
a.
ŷ
= 6.5763 +0.0452x (R2 = .519, s = 6.977)
b. d.f.= 62, α = .05 so tcrit = t.025 =T.INV(.025,62) = ±1.999 c. tcalc = 8.183 > 1.999 so we will reject the null hypothesis that the slope is zero. d. We are 95% confident that the slope is contained in the interval (0.0342, 0.0563). This CI does not contain zero, hence, there is a relationship between the total assets (billions) and total revenue (billions). e. Fcalc = (8.183)2 = 66.96 f. The value of R2 assigns 51.9% of the variation in total revenue (billions) to the total assets (billions). Thus, increasing assets will lead to an increase in income. However, the results also indicate that there are most likely other reasons why companies earn the revenue they do. Learning Objective: 12-4 Learning Objective: 12-5 Learning Objective: 12-6 12.69
a. r = −.387 (from Excel) b. H0: ρ = 0 versus H1: ρ ≠ 0. Using d.f. = 41, α = .01 so tcrit = t.005 =T.INV(.005,41) = ±2.701. . Because −2.687 is between the critical 43 - 2 tcalc = -.387 = -2.687 1 - (-.387) 2 values we do not reject the hypothesis of zero correlation. The sample does not provide evidence at the .01 significance level that the stock prices move together. c. The scatterplot shows a slight negative correlation between IBM and HPQ stock prices. (Note that if α = .05, t.025 = ±2.0195, and we would reject the hypothesis of zero correlation and conclude that there is correlation between the stock prices.)
340
ASBE 6e Solutions for Instructors
Object 99
Learning Objective: 12-1 12.70
a.
b. r = −.297. This shows a weak negative linear relationship between loyalty card use and sales growth. c. H0: ρ = 0 versus H1: ρ ≠ 0. Using d.f. = 72 and = .05, α = .05 so tcrit = t.025 =T.INV(.025,72) = ±1.9935. . Because 2.639 < 74 - 2 tcalc = -.297 = -2.639 1 - .297 2 1.9935, we reject the hypothesis of no correlation and the sample evidence supports the notion of negative correlation. d. It appears that a higher loyalty card usage is associated with lower sales growth. Learning Objective: 12-1
341
ASBE 6e Solutions for Instructors 12.71
a. The scatter plot indicates that there is a positive correlation between the fertility rates in 1990 and 2000.
Object 102
b. r = .749. There is a strong positive linear relationship between the fertility rates in 1990 and 2000. c. H0: ρ = 0 versus H1: ρ ≠ 0. Using d.f. = 13 and = .05, t.025 =T.INV(.025,13) = 2.160. . Because 4.076 > 2.160, we reject the hypothesis of no 15 - 2 tcalc = .749 = 4.076 1 - .7492 correlation and the sample evidence supports the notion of positive correlation. There is a positive correlation. Learning Objective: 12-1 12.72
a. The regression output shows the p-value on the Price coefficient is equal to .6019 which is greater than any alpha one might choose. Therefore the slope is NOT significant. b. R2 = .01104. If there had been a significant relationship between Price and Sound Quality then R2 would tell us that only 1.1% of the variation in Sound Quality could be explained by Price. But because Price is not a significant predictor of Sound Quality, R2 does not have any meaning. c. It appears there is very little relationship between price and sound quality of speakers and it is unlikely that the typical consumer will be able to tell the difference in sound quality between various priced speakers. Learning Objective: 12-4 Learning Objective: 12-5
12.73
For each of these, the scatter plot will contain the answers to (a), (b), and (d) with respect to the fitted equation. c. Salary: The fit is good. Assessed: The fit is excellent. HomePrice2: The fit is good. d. Salary: An increase in the age by 1 year increases salary by $1447.4. Assessed: An increase in 1 sq. ft. of floor space increases assessed value by $313.30. 342
ASBE 6e Solutions for Instructors HomePrice2: An increase in 1 sq. ft. of home size increases the selling price by $209.20. e. The intercept is not meaningful for any of these data sets as a zero value for any of X’s respectively cannot realistically result in a positive Y value.
Learning Objective: 12-2 Learning Objective: 12-4 Learning Objective: 12-6 12.74
a.
estimated slope t= standard error
See table below for calculations.
b. Answers shown in right column in table below. Dependent Variable Highest grade achieved Reading grade equivalent Class standing Absence from school Grammatical reasoning Vocabulary Hand-eye coordination
Estimated Slope
p-value
Differ from
0.027
0.008
Yes
0.07
0
Yes
0.006 4.8
0.048 0.006
No Yes
0.159
0.012
No
0.124 0.041
0 0.02
Yes No
0?
343
ASBE 6e Solutions for Instructors Reaction time Minor antisocial behavior
11.8 0.639
0.08 0.082
No No
c. It would be inappropriate to assume cause and effect without a better understanding of how the study was conducted. Learning Objective: 12-5 12.75
a.
c. The fit of this regression is weak as given by R2 = 0.2474. 24% of the variation in % Operating Margin is explained by % Equity Financing. Learning Objective: 03-8 Learning Objective: 12-6 12.76
a.
c. The fit of this regression is very good as given by R2 = 0.8216. The regression line does show a strong positive linear relationship between molecular w.r.t. and retention time, indicating that the greater the molecular w.r.t. the greater is the retention time. Learning Objective: 03-8 Learning Objective: 12-6
344
ASBE 6e Solutions for Instructors
12.77
a. Because the p-value for the state SAT participation rate coefficient is approximately zero we can conclude that there is a significant relationship between participation and state average SAT score. Because the coefficient is negative we conclude that as participation in the SAT goes up, the average score goes down. b. This would be considered a strong relationship because the R2 = .8419. If preferred one could interpret the correlation coefficient, R = sqrt(.8419) = .9178 as indicating a strong relationship between the two variables. c. Some states require their high school students take the SAT which means there is a wider range of ability levels taking the test. This increases the participation rate and also brings the average down. Learning Objective: 12-4 Learning Objective: 12-5
12.78
a. The scatter plot shows a positive relationship.
c. The fit of this regression is very good as given by R2 = .8206. The regression line shows a strong positive linear relationship between revenue and profit, indicating that the greater revenue is associated with higher profit. Learning Objective: 03-8 Learning Objective: 12-4 Learning Objective: 12-5 12.79
a. The slope of each model indicates the impact an additional year in vehicle age has on the price. This relationship for each model is negative indicating that an additional year of age reduces the asking price. This impact ranges from a low for the Taurus (an additional year reduces the asking price by $906) to a high for the Ford Explorer (an additional year reduces the asking price by $2,452). b. The intercepts could indicate the price of a new vehicle.
345
ASBE 6e Solutions for Instructors c. Based on the R2 values: The fit is very good for the Explorer, the F-150 Pickup and the Taurus. The fit is weak for the Mustang. One reason for the seemingly poor fit for the Mustang is the fact that this is a collector item (if in good condition) so that the age is less important of a factor in determining the asking price. d. Answers will vary, but a bivariate model for 3 of the vehicles explains approximately 2/3 of the variation in asking price at a minimum. Other factors: condition of the car, collector status, proposed usage, price of a new vehicle. Learning Objective: 12-2 Learning Objective: 12-4 Learning Objective: 12-5 12.80
a. The regression results are not significant, based on the p-value, for the 1-Year holding period. The results for the 2-Year period are significant at the 5% level, while for 5-, 8-, and 10-Years the results are significant at the 1% level. For each regression there is an inverse relationship between P/E and the stock return. For the 8-Year and 10-Year period the relationship is approximately 1. The R2 increases as the holding period increases. This indicates that P/E ratio explains a greater portion of the variation in stock return, the longer the stock is held. b. Yes, given the data are time series, the potential for autocorrelation is present. Also, it is commonly recognized that stock returns do exhibit a high degree of autocorrelation, as do most financial series. Learning Objective: 12-4 Learning Objective: 12-8
12.81
a. P(SatRad) =
exp(-11.96 + 0.02421(350)) = 0.0297 1 + exp(-11.96 + 0.02421(350))
P(SatRad) =
exp(-11.96 + 0.02421(450)) = 0.25626 1 + exp(-11.96 + 0.02421(450))
b.
c.
exp(-11.96 + 0.02421(550)) = 0.79503 1 + exp(-11.96 + 0.02421(550)) Learning Objective: 12-10 P(SatRad) =
346
ASBE 6e Solutions for Instructors
Chapter 13 Multiple Regression 13.1
a.
ŷ
= 4.306 − 0.082ShipCost + 2.265PrintAds + 2.498WebAds + 16.697Rebate%
b.
The coefficient of ShipCost says that each additional $1 of shipping cost reduces about $ 0.082 from net revenue. The coefficient of PrintAds says that each additional $1000 of printed ads adds about $2,265 to net revenue. The coefficient of WebAds says that each additional $1000 of printed ads adds about $2,498 to net revenue. The coefficient of Rebate% says that each additional percentage in the rebate rate adds about $16,700 to net revenue. c. The intercept is meaningless. You have to supply some product, so shipping cost can’t be zero. You don’t have to have a rebate or ads, they can be zero. d. NetRevenue = 4.306 − 0.082(10) + 2.265(50) + 2.498(40) + 16.697(15) = 467 thousands or $467,111. Learning Objective: 13-1 13.2
a.
ŷ
= 1225.44 + 11.52FloorSpace − 6.935CompetingAds − 0.1496Price
b.
The coefficient of FloorSpace says that each additional square foot of floor space adds about $11,520 to average sales. The coefficient of CompetingAds says that each additional $1000 of CompetingAds reduces sales by $6,935. The coefficient of Price says that each additional $1 of Advertised Price reduces sales by $149.60. c. No. If all of these variables are zero, you wouldn’t sell a bike (no one will advertise a bike for zero). d. Sales = 1225.44 +11.52(80) − 6.935(100) − 0.1496(1200) = 1274.02 thousands or $1,274,020. Learning Objective: 13-1 13.3
a.
ŷ
= 2.8931 + 0.1542LiftWait + 0.2495AmountGroomed + 0.0539SkiPatrolVisibility
0.1196FriendlinessHosts Overall satisfaction increases with an increase in satisfaction for each individual predictor except for friendliness of hosts. This counterintuitive result could be due to an interaction effect. Interaction effects will be explored later in the chapter. c. No. Satisfaction scores of zero for the individual predictors is out of the range of the variable values. It is unwise to extrapolate. d. Overall satisfaction score = 2.8931 + 0.1542(5) + 0.2495(5) + 0.0539(5) 0.1196(5) = 4.5831 Learning Objective: 13-1 b.
348
ASBE 6e Solutions for Instructors 13.4
a.
ŷ
= 4198.5808 27.3540AgeMed + 17.4893Bankrupt 0.0124FedSpend
29.0314HSGrad% b. The 2005 state by state crime rate per 100,000 decreases by about 27 as the state median age increases, increases by about 17 for every 1000 new bankruptcies filed, decreases by .0124 for each dollar increase in federal funding per person, and decreases by about 29 for each 1% increase in high school graduations. c. No, a state would not have 0 median age or 0 values for any of the other predictor variables. d. Burglary Rate = 4198.5808 27.3540(35) + 17.4893(7) 0.0124(6000) 29.0314(80) = 966.7039 Learning Objective: 13-1 13.5
a. b.
df1 = 4 and df2 = 45 From Appendix F: F.05 = 2.61, using df1 = 4 and df2 = 40. From Excel with df1 = 4 and df2 = 45: F.05 =F.INV.RT(.05,4,45) = 2.5787. c. Fcalc = 64,853/4990 = 12.997. Yes, the overall regression is significant. H0: All the coefficients are zero (1 = 2 = 3 = 0) H1: At least one coefficient is not zero d. R2 = 259,412/483,951 = .536 R2adj = 1 –[224539/45)/(483951/49)] = .4948 Learning Objective: 13-2
13.6
a. b. c.
13.7
a. b.
df1 = 3 and df2 = 26 From Appendix F: F.05 = 2.98. From Excel: =F.INV.RT(.05,3,26) = 2.975. Fcalc = 398802/14590 = 27.334. Yes, the overall regression is significant. H0: All the coefficients are zero (1 = 2 = 3 = 0) H1: At least one coefficient is not zero d. R2 = 1196410/1575741 = .7593 R2adj = 1 – [(379332/26)/(1575742/29)]= .7315 Learning Objective: 13-2 df1 = 4 and df2 = 497 From Appendix F: F.05 = 2.42 using df1 = 4 and df2 = 200. From Excel with df1 = 4 and df2 = 497: F.05 =F.INV.RT(.05,4,497) = 2.39. c. Fcalc = 8.2682/0.6398 = 12.9231. Yes, the overall regression is significant. H0: All the coefficients are zero (1 = 2 = 3 = 0) H1: At least one coefficient is not zero d. R2 = 33.0730/351.0598 = .0942 R2adj = 1 – [(317.0598/497)/(351.0598/501)]= .0896 Learning Objective: 13-2
349
ASBE 6e Solutions for Instructors
13.8
a. b.
df1 = 4 and df2 = 45 From Appendix F: F.05 = 2.61 using df1 = 4 and df2 = 40. From Excel with df1 = 4 and df2 = 45: F.05 =F.INV.RT(.05,4,45) = 2.5787. c. Fcalc = 295683.3212/35221.1519 = 8.3951. Yes, the overall regression is significant. H0: All the coefficients are zero (1 = 2 = 3 = 0) H1: At least one coefficient is not zero d. R2 = 1182733.285/2767685.12 = .4273 R2adj = 1 – [(1584951.835/45)/(2767685.12/49)]= .3764 Learning Objective: 13-2
13.9
a. tcalc
bj - 0
, p-value =T.DIST.2T(|tcalc|, 45)
sj
Predictor Intercept ShipCost PrintAds
Coef, bj sj 4.31 -0.082 2.265 2.498 16.697
Rebate%
tcalc 0.0608585 -0.0175289 2.1571429 2.9537661 4.6770308
p-value 0.9517414 0.9860922 0.0363725 0.0049772 .0003
b.
t.005 =T.INV(.005,45) = ±2.69. Web Ads and Rebate% differ significantly from zero (pvalue < .01 and tcalc > 2.69.) c. See table in part a. Learning Objective: 13-3 13.10
a. tcalc
bj - 0
, p-value =T.DIST.2T(|tcalc|, 26)
sj
Predictor Intercept FloorSpace CmpetingAds
Coef, bj sj
tcalc
1225.4 11.522 -6.935 -0.14955
3.0843192 8.6631579 -1.7759283 -1.6752548
b.
p-value 0.0048 0.0000 0.0874 0.1059
t.005 =T.INV(.005,26) = ±2.779. Only Floor Space differs significantly from zero (pvalue < .01 and tcalc > 2.779). c. See table in part a. Learning Objective: 13-3
350
ASBE 6e Solutions for Instructors
13.11
a. tcalc
bj - 0
, p-value =T.DIST.2T(|tcalc|, 497)
sj Predictor Coef, bj sj tcalc Intercept 2.8931 0.3680 7.8617 LiftWait 0.1542 0.0440 3.5045 AmountGroom 0.2495 0.0529 4.7164 ed SkiPatrolVisibil 0.0539 0.0443 1.2167 ity FriendlinessHo 0.0623 0.1196 1.9197 st
p-value 2.37E-14 .0005 3.07E-06 .2245 .0557
b.
t.005 =T.INV(.005,497) = ±2.69. Coefficients on LiftWait and AmountGroomed differ significantly from zero. c. See table in part a. Learning Objective: 13-3 13.12
a. tcalc
bj - 0
, p-value =T.DIST.2T(|tcalc|, 45)
sj Predictor
Coef, bj sj
Intercept
4,198.5808
AgeMed Bankrupt FedSpend HSGrad%
27.3540 17.4893 0.0124 29.0314
tcalc 799.339 5 12.5687 12.4033 0.0176 7.1268
p-value
5.2526 2.1764 1.4101 0.7045 4.0736
3.95E-06 0.0348 0.1654 0.4848 0.0002
b.
t.005 = T.INV(.005,45) = ± 2.586. Only the coefficient on HSGrad% differs significantly from zero at a significance level of .01. c. See table in part a. Learning Objective: 13-3 13.13
Use
ŷi
± t/2se with 34 df, t.025 =T.INV(.025,34) = 2.032. (Use the positive value.)
Half width of 95% prediction interval = 2.032(3620) = 7355.84 Using the quick rule the half width = 2se = 2(3620) = 7240 Yes, the quick rule gives similar results. Learning Objective: 13-4
351
ASBE 6e Solutions for Instructors 13.14
Use
ŷi
± t/2se with 20 df, t.025 = T.INV(.025,34) = 2.086. (Use the positive value.)
Half width of 95% prediction interval = 2.086 (1.17) = 2.4406 Using the quick rule the half width = 2se = 2(1.17) = 2.34 Yes, the quick rule gives similar results. Learning Objective: 13-4 13.15
a.
Number of nights (NumNight) needed and number of bedrooms (NumBedrooms) are both discrete. b. Two: SwimPool = 1 if there is a swimming pool and ParkGarage = 1 if there is a parking garage c. CondoPrice = 0 + 1NumNights + 2NumBedrooms + 3SwimPool + 4ParkGarage Learning Objective: 13-5
13.16
a. b.
13.17
a.
13.18
a.
Weight of stone (continuous) Nine: There are 6 different values for Color rating so the model would need 5 (6-1) indicator variables for color rating. There are 5 different Clarity rating values so the model would need 4 (5-1) indicator variables for clarity rating. c. Price = 0 + 1Weight + 2ColorD + 3ColorE + 4ColorF + 5ColorG + 6ColorH + 7ClarityIF + 8ClarityVVS1+ 9ClarityVVS2+ 10ClarityVS1 Learning Objective: 13-5 ln(Price) = 5.4841 0.0733SalePrice + 1.1196Sub-Zero + 0.0696Capacity + 0.04662DoorFzBot 0.34322DoorFzTop 0.70961DoorFz 0.88201DoorNoFz b. Use p-value =T.DIST.2T(|tcalc|, 319) SalePrice: p-value = .0019, Sub-Zero: p-value = 2.24E-13, Capacity: p-value = 2.69E-31, 2DoorFzBot: p-value = .5623, 2DoorFzTop = p-value = 3.57E-19, 1DoorFz: p-value = 1.18E-07, 1DoorNoFz: p-value = 8.32E09 The only variable that is not a significant predictor is 2DoorFzBot. This is an indicator variable which means that there is not a significant difference in price between two door refrigerators that have the freezer compartment on the side or on the bottom. c. The coefficient on two door, top freezer is −0.3432 so the natural log of the price decreases by 0.3432. d. The side freezer model demands a higher price because there is a negative coefficient on the 1DoorFz model indicator variable. Learning Objective: 13-5
b. c. d.
SentenceLength = 3.2563 + 0.5219Age + 7.7412Convictions 6.0852Married 14.3402Employed Use p-value =T.DIST.2T(|tcalc|, 45). Age: p-value = 9.54E-06, Convictions: p-value = 2.03E-09, Married: p-value = .0228, Employed: p-value = 1.00E-06 A married male convicted of assault will receive a sentence that is about 6 years shorter than an unmarried male assault convict. About 14 years. 352
ASBE 6e Solutions for Instructors SentenceLength = 3.2563+ 0.5219(25) + 7.7412(1) 6.0852(0) 14.3402(0) = 24.045. Learning Objective: 13-5 e.
13.19
a.
The scatter plot shows an obvious increasing trend but it is nonlinear rather than linear. The increase in salary is much steeper in the earlier years than in the later years. A nonlinear model would be appropriate.
Object 3
b.
MegaStat Output is below: R2 = .915, Fcalc =194.99, p-value = .0000. Yes, the model is significant. Regression Analysis R² Adjusted R² R Std. Error ANOVA table Source Regression Residual Total
SS 29,901.972 8 2,760.3861 32,662.359 0
0.915 0.911 0.957 8.757
n k Dep. Var.
39 2 Salary ($1000)
df
MS
F
p-value
2 36
14,950.9864 76.6774
194.99
4.84E-20
38
Regression output variables Intercept Years YearsSq
coefficients 45.3322 5.6218 -0.0945
std. error 3.0644 0.4742 0.0139
t (df=36) 14.793 11.856 -6.789
353
p-value 7.13E-17 5.45E-14 6.21E-08
confidence interval 95% 95% lower upper 39.1172 51.5472 4.6601 6.5835 -0.1227 -0.0663
ASBE 6e Solutions for Instructors c.
Years: p-value = .0000, YearsSq: p-value = .0000. Both of these predictors are significant. Learning Objective: 13-6 13.20
MegaStat Output is below: Male: p-value = .5009, YearsxMale: p-value = .0505. The binary variable Male is not significant. The interaction variable YearsxMale is significant because the p-value is less than .10. The coefficient on the interaction term is positive which means that as men gain more years of experience their salaries tend to be higher than females. Regression Analysis R² Adjusted R² R Std. Error ANOVA table Source Regression Residual Total
SS 30,865.754 2 1,796.6048 32,662.359 0
0.945 0.939 0.972 7.269
n k Dep. Var.
39 4 Salary ($1000)
df
MS
F
p-value
4 34
7,716.4386 52.8413
146.03
6.59E-21
38
Regression output variables Intercept Years YearsSq Male YearsxMale
coefficients 44.6252 4.7166 -0.1033 3.2301 1.0938
std. error 3.7277 0.5098 0.0129 4.7476 0.5395
t (df=34) 11.971 9.251 -8.019 0.680 2.027
Learning Objective: 13-6
354
p-value 9.62E-14 8.23E-11 2.40E-09 .5009 .0505
confidence interval 95% 95% lower upper 37.0495 52.2008 3.6805 5.7527 -0.1295 -0.0771 -6.4182 12.8783 -0.0026 2.1902
ASBE 6e Solutions for Instructors 13.21
a.
All but one pair of variables is significantly correlated at α = .01. See the matrix below. LiftOps and Scanners (r = .635), Crowds and LiftWait (r = .577), AmountGr and TrailGr (r = .531), SkiSafe and SpSeen (r = .488) Correlation Matrix scanner liftwai snosur crowd amountg trailg skisaf spsee s liftops t trailv f s r r e n hosts scanners 1.000 liftops .635 1.000 liftwait .146 .180 1.000 1.00 trailv .115 .206 .128 0 snosurf .190 .242 .227 .373 1.000 crowds .245 .299 .577 .235 .348 1.000 amountg r .245 .271 .251 .221 .299 .372 1.000 1.00 trailgr .266 .337 .205 .360 .358 .362 .531 0 skisafe .200 .306 .196 .172 .200 .332 .274 .323 1.000 spseen .145 .190 .207 .172 .184 .230 .149 .172 .488 1.000 1.00 hosts .245 .278 .046 .140 .119 .133 .128 .156 .212 .350 0 502 sample size ± .088 critical value .05 (two-tail) ± .115 critical value .01 (two-tail)
b.
All VIFs are less than 2. No cause for concern. variables Intercept scanners liftops liftwait trailv snosurf crowds amountgr trailgr skisafe spseen hosts
Learning Objective: 13-7
355
VIF 1.718 1.887 1.531 1.270 1.337 1.838 1.504 1.676 1.518 1.477 1.224
ASBE 6e Solutions for Instructors 13.22
a. Al and Si (r = .456), Cr and Zn (r = .529) are significantly correlated at an = .01. Al and Ti (r = .389), Si and Zn (r = .365), and Ti and Pb (r = .345) are significantly correlated at an = .05. Correlation Matrix Al Si Cr Ti Zn 1.000 .456 1.000 .133 -.073 1.000 .389 .278 .011 1.000 .286 .365 .529 -.083 1.000 -.202 -.053 -.114 -.345 .180 33 sample size ± .344 critical value .05 (two-tail) ± .442 critical value .01 (two-tail)
Al Si Cr Ti Zn Pb
b.
Pb
1.000
All VIFs are low in value. No cause for concern. variables Intercept Al Si Cr Ti Zn Pb
VIF 1.500 1.674 1.761 1.390 2.197 1.281
Learning Objective: 13-7 13.23
If hi > a. b. c.
2(k + 1) n
then the observation is considered to be a high leverage observation.
2(5 + 1) = .1667 72 2(4 + 1) = .10 100
. hi = .15 < .1667 therefore this is not a high leverage observation.
. hi = .18 > .10 therefore this is a high leverage observation.
2(7 + 1) = .0667 240
. hi = .08 > .0667 therefore this is a high leverage observation.
Learning Objective:13-10 13.24
Assumption #1: Residuals are normally distributed. It appears this assumption has been violated because the histogram shows a skewed left distribution. Because the data set is fairly small the normplot is not as useful for detecting non-normality.
356
ASBE 6e Solutions for Instructors Assumption #2: Residuals have constant variance. It appears this assumption has been violated. The residuals plotted against the predicted Y values show a fan out pattern which indicates heteroscedasticity or non-constant variance. Assumption #3: Residuals are independent. It appears this assumption has been violated. Applying the runs test (see section 12.8) to the Runs plot we see there are 10 crossing points. If autocorrelation did not exist we would expect approximately 15/2 or 7 to 8 crossings. We have more than 8 crossings so negative autocorrelation is a concern. Learning Objective:13-8 Questions 13.25 through 13.41 refer to 10 different data sets labeled A-J. The answers to each question are listed for each data set in turn. DATA SET A Response Variable: Vehicle City Mileage 13.25
Cross-sectional. Unit of observation: vehicle model. Learning Objective: 02-3
13.26
The variable magnitudes are not too different. Weight is approximately 100 times the magnitude of the other variables but this should not cause problems in the analysis. Learning Objective: 13-11
13.27
The intercept would not have meaning. It would not be logical to have a car with zero values for any of the predictors. A priori reasoning for the relationship between each predictor and the response variable are listed in the table below. Predictor
Relationship with Response
Length Width Weight
Negative Negative Negative
ManTran
Positive
Reason? Bigger size, lower mileage Bigger size, lower mileage Bigger size, lower mileage Cars with manual transmission are thought to get higher MPG than automatic transmission cars.
Learning Objective: 12-2 13.28
73/4 = 10.75 > 18.25. The data set meets both Evans’ and Doane’s Rules. Learning Objective: 13-2
13.29
The estimated regression equation is
ŷ
= 63.9418 − 0.0360length − 0.325width −
0.0034weight − 1.0056ManTran. The signs on the coefficients match our a priori reasoning except for vehicles with manual transmission. In reality, today’s automatic transmission vehicles have better technology and are now more fuel efficient than manual transmission vehicles. MegaStat Output: Regression Analysis 357
ASBE 6e Solutions for Instructors R² Adjusted R² R Std. Error
0.834 0.825 0.913 2.080
n k Dep. Var.
73 4 CityMPG
df
MS
F
p-value
4 68
370.6114 4.3270
85.65
8.26E-26
Total
SS 1,482.445 8 294.2392 1,776.684 9
variables
coefficient s
std. error
t (df=68)
p-value
Intercept Length Width Weight ManTran
63.9418 -0.0360 -0.3250 -0.0034 -1.0056
7.5533 0.0157 0.1307 0.0006 0.6979
8.465 -2.293 -2.486 -5.243 -1.441
3.13E-12 .0249 .0154 1.69E-06 .1542
ANOVA table Source Regression Residual
72
Regression output
confidence interval 95% 95% lower upper 79.014 48.8694 2 -0.0673 -0.0047 -0.5859 -0.0641 -0.0047 -0.0021 -2.3982 0.3870
VIF
1.855 4.985 5.098 1.274 3.303 mean VIF
Learning Objective: 13-2 13.30
Refer to the output in the question 13.29. The coefficient confidence intervals for length, width, and weight are all negative. This means that these variables are all significant predictors of MPT. ManTran has a coefficient confidence interval that contains zero and is therefore NOT a significant predictor in the model (at a significance level of . 05.) Learning Objective: 13-3
13.31
For each coefficient test the following hypotheses: H0: βj = 0 vs. H1: βj ≠ 0. Using df = 68, t.025 =T.INV(.025, 68) = ±1.995. Only ManTran is NOT significant with tcalc = 1.441 > −2.024. Learning Objective: 13-3
13.32
a.
13.33
Fcalc = 85.65 with a p-value = 8.26E-26. R2 = .834 and R2adj = .825. The model provides significant fit with a fairly strong prediction of city mileage. Learning Objective: 12-6
Length: p-value = .0249 < .05, Width: p-value = .0154 < .05, Weight: p-value = 1.69E-06 < .05. b. This is consistent with the answer in 13.31. c. The tests conclude the same thing. Most analysts prefer the p-value approach because it tells the strength of the predictor significance. Learning Objective: 13-3
358
ASBE 6e Solutions for Instructors Learning Objective: 13-2 13.34
Prediction interval:
yˆi
± t.025se =
yˆi
± 1.996(2.08) =
yˆi
± 4.15. Yes, this model does have
practical value. Learning Objective: 13-4 13.35
a. Correlation Matrix CityMP G CityMP G Length Width Weight
1.000 -.670 -.860 -.894 73 ± .230 ± .300
b.
Lengt h
Width
Weigh t
1.000 .635 1.000 .644 .890 1.000 sample size critical value .05 (twotail) critical value .01 (twotail)
Both length and width are significantly correlated with Weight. Collinearity could be a problem but according to Klein’s Rule we shouldn’t be overly concerned. All values of r are less than .)
.834 = .913
Learning Objective: 13-7 13.36
a. variables Intercept Length Width Weight ManTran
VIF 1.855 4.985 5.098 1.274
b.
The VIF values are all under 10 which suggest that multicollinearity has not caused instability. Learning Objective: 13-7 13.37
Observations 16, 24, 32, and 60, are the observations that had an unusual residual. Learning Objective: 13-9
13.38
Observation 1, 2, 15, 45, 49, 50, 60 and 65 had high leverage values. To determine if any of the observations were influential, remove them from the data set and rerun the regression. If the regression statistics change significantly then the observation could be considered influential. 359
ASBE 6e Solutions for Instructors Learning Objective: 13-10 13.39
Assuming normally distributed residuals appears reasonable with a high outlier. Although running a normality test of the residuals (A-D and moment tests) results in a failure; more than likely because of the outlier.
Object 5
Object 8
Learning Objective: 13-8
13.40
Assuming homoscedastic residuals appears reasonable.
360
ASBE 6e Solutions for Instructors
Object 10
Learning Objective: 13-8 13.41
This is cross-sectional data. A test for autocorrelation is not warranted. Learning Objective: 13-8
361
ASBE 6e Solutions for Instructors DATA SET B: Response Variable: Noodles & Company Sales/SqFt 13.25
Cross-sectional. Unit of observation: restaurant. Learning Objective: 02-3
13.26
The variable magnitudes are all similar. Learning Objective: 13-11
13.27
The intercept would not have meaning. It would not be logical to have a restaurant with zero values for any of the predictors. A priori reasoning for the relationship between each predictor and the response variable are listed in the table below. Predictor
Relationship with Response
SeatsInside Seats-Patio MedIncome MedAge
Reason? The larger the size of the restaurant, the greater the sales.
Positive
The larger the size of the restaurant, the greater the sales. The higher the income of the potential customers, the higher the sales The older the potential customers, the higher the sales More education would be positively correlated with higher income and therefore higher the sales
Positive Positive Positive
BachDeg% Positive
Learning Objective: 12-2 13.28
74/5 = 14.8 > 10. The data set meets both Evans’ and Doane’s Rules. Learning Objective: 13-2
13.29
The estimated regression equation is
ŷ
= 429.5114 − 1.8149Seats-Inside + 1.2719Seats-
Patio − 2.1021MedIncome − 0.0158MedAge + 8.6604BachDeg% The signs on the coefficients do not match our a priori reasoning for Seats-Inside, MedIncome, and MedAge. MegaStat Output: Regression Analysis R² 0.233 Adjusted R² 0.177 R 0.483 Std. Error 124.529 ANOVA table Source SS Regression 320,276.8169 Residual 1,054,515.7777 Total 1,374,792.5946 Regression output variables Intercept
coefficients 429.5114
n 74 k 5 Dep. Var. Sales/SqFt
df 5 68 73
MS 64,055.3634 15,507.5850
F 4.13
std. error 182.1907
t (df=68) 2.357
p-value .0213
362
p-value .0025
confidence interval 95% lower 95% upper 65.9556 793.0672
ASBE 6e Solutions for Instructors Seats-Inside Seats-Patio MedIncome MedAge BachDeg%
-1.8149 1.2719 -2.1021 -0.0158 8.6604
0.9975 1.0614 1.0941 4.4891 2.6187
-1.819 1.198 -1.921 -0.004 3.307
.0733 .2350 .0589 .9972 .0015
-3.8054 -0.8462 -4.2853 -8.9737 3.4348
0.1757 3.3900 0.0811 8.9420 13.8860
Learning Objective: 13-2 13.30
Refer to the output in question 13.29. The 95% coefficient confidence intervals for the variables Seats-Inside, Seats-Patio, MedIncome, and MedAge all contain zero. Only BachDeg% is a significant predictor in the model at a significance level of .05. Learning Objective: 13-3
13.31
For each coefficient test the following hypotheses: H0: βj = 0 vs. H1: βj ≠ 0. Using df = 68, t.025 =T.INV(.025,68) = ±1.995. Only BachDeg% has a significant result at = .05 with tcalc = 3.307 > 1.995. Learning Objective: 13-3
13.32
a.
13.33
Fcalc = 4.13 with a p-value = .0025. R2 = .233 and R2adj = .177. The model provides significant fit but does not provide strong prediction of restaurant sales/sqft. Learning Objective: 12-6 Learning Objective: 13-2
13.34
Prediction interval:
BachDeg%: p-value = .0015 < .05. Note that both MedIncome and Seats-Inside are both significant at = .10. b. This is consistent with the answer in 13.31. c. The tests conclude the same thing. Most analysts prefer the p-value approach because it tells the strength of the predictor significance. Learning Objective: 13-3
yˆi
± t.025se =
yˆi
± 1.995(124.529) =
yˆi
± 248.4354. No, this model does
not have practical value because the width of the interval is so wide. Learning Objective: 13-4
13.35
a. Seats-Inside
Seats-Patio
MedIncome MedAge
363
BachDeg%
ASBE 6e Solutions for Instructors Seats-Inside Seats-Patio MedIncome MedAge BachDeg%
b.
1.000 .007 1.000 -.047 -.009 1.000 -.102 -.065 .416 -.158 .151 .552 74sample size ± .229 critical value .05 (two-tail) ± .298 critical value .01 (two-tail)
1.000 .097
1.000
Both MedAge and BachDeg% are significantly correlated with MedIncome. Collinearity could be a problem according to Klein’s Rule because .552 > .
.233 = .483
Learning Objective: 13-7 13.36
a. variables Intercept SeatsInside SeatsPatio MedIncom e MedAge BachDeg%
VIF
1.045 1.039 1.807 1.267 1.584
b.
The VIF values are all under 2 which suggests that multicollinearity has not caused instability. Learning Objective: 13-7 13.37
Observations 6, 19, 22, 46, and 69 would be considered unusual or outlier residuals. Learning Objective: 13-9
13.38
Observation 14, 19, 23, and 69 had high leverage values. To determine if any of the observations were influential, remove them from the data set and rerun the regression. If the regression statistics change significantly then the observation could be considered influential. Learning Objective: 13-10
13.39
Assuming normally distributed residuals appears reasonable.
364
ASBE 6e Solutions for Instructors
Object 12
Object 15
Learning Objective: 13-8
13.40
Assuming homoscedastic residuals appears reasonable. There is no obvious fan out or funnel pattern in the plot below. 365
ASBE 6e Solutions for Instructors
Object 17
Learning Objective: 13-8 13.41
This is cross-sectional data. A test for autocorrelation is not warranted. Learning Objective: 13-8
366
ASBE 6e Solutions for Instructors DATA SET C Response Variable: Medical Office Building Assessed Value 13.25
Cross-sectional. Unit of observation: office building. Learning Objective: 02-3
13.26
The variable magnitudes are not too different. Floor space is approximately 1000 times the magnitude of the other predictor variables but is similar to the response variable. Learning Objective: 13-11
13.27
The intercept would not have meaning. It would not be logical to have a building with zero values for any of the predictors. A priori reasoning for the relationship between each predictor and the response variable are listed in the table below. Predictor
Relationship with Response
Floor Offices Entrances
Positive Positive Positive
Age Freeway
Negative Positive
Reason? Bigger size, higher value More offices, higher value More entrances, higher value Increase in age means more maintenance, lower value Closer access to freeway, higher value
Learning Objective: 12-2 13.28
32/5 = 6.5. The data set meets Doane’s Rule but not Evans’. Learning Objective: 13-2
13.29
The estimated regression equation is
ŷ
= −59.3894 + 0.2509Floor + 97.7927Offices +
72.8405Entrances − 0.4570Age + 116.1786Freeway. The signs on the coefficients match our a priori reasoning. MegaStat Output: Regression Analysis R² 0.967 Adjusted R² 0.961 R 0.983 Std. Error 90.189 ANOVA table Source SS Regression 6,225,261.2561 Residual 211,486.6189 Total 6,436,747.8750 Regression output variables Intercept Floor Offices Entrances Age
coefficients -59.3894 0.2509 97.7927 72.8405 -0.4570
n 32 k 5 Dep. Var. Assessed df 5 26 31
MS 1,245,052.2512 8,134.1007
std. error 71.9826 0.0218 30.8056 38.7501 1.2011
t (df=26) -0.825 11.494 3.175 1.880 -0.380
367
F 153.07
p-value .4168 1.08E-11 .0038 .0714 .7067
p-value 2.01E-18
confidence interval 95% lower 95% upper -207.3519 88.5730 0.2060 0.2957 34.4709 161.1145 -6.8114 152.4924 -2.9258 2.0118
ASBE 6e Solutions for Instructors Freeway
116.1786
34.7721
3.341
.0025
44.7036
187.6535
Learning Objective: 13-2 13.30
Refer to the output in question 13.29. The coefficient confidence intervals contain zero for the variables Entrances and Age. This means that Floor, Offices, and Freeway are significant predictors in the model. Learning Objective: 13-3
13.31
For each coefficient test the following hypotheses: H0: βj = 0 vs. H1: βj ≠ 0. Using df = 26, t.025 =T.INV(.025, 26) = ±2.056. Floor, Offices, and Freeway both have significant results with tcalc = 11.494, 3.175, and 3.341, respectively (both are greater than 2.056.) Learning Objective: 13-3
13.32
a.
13.33
Fcalc = 153.07 with a p-value = 2.01E-18. R2 = .967 and R2adj = .961. The model provides significant fit with a very strong prediction of building assessed value. Learning Objective: 12-6 Learning Objective: 13-2
13.34
Prediction interval:
Floor: p-value = 1.08E-11, Offices: p-value = .0038, and Freeway Weight: p-value = . 0025. All p-values are less than .05 b. This is consistent with the answer in 13.31. c. The tests conclude the same thing. Most analysts prefer the p-value approach because it tells the strength of the predictor significance. Learning Objective: 13-3
yˆi
± t.025se =
yˆi
± 2.056(90.189) =
yˆi
± 185.429. Yes, this model does
have practical value. The prediction interval width should provide valuable information. Learning Objective: 13-4 13.35
a. Floor Offices Entrances Age Freeway
b.
Floor Offices Entrances 1.000 .823 1.000 .567 .444 1.000 -.189 -.241 .136 -.331 -.368 -.082 32sample size ± .349 critical value .05 (two-tail) ± .449 critical value .01 (two-tail)
Age
Freeway
1.000 .175
1.000
Both Offices and Entrances are significantly correlated with Floor. Collinearity is most likely not a problem according to Klein’s Rule. The correlation coefficient values are less than
.967. 368
ASBE 6e Solutions for Instructors Learning Objective: 13-7 13.36
a. variables Intercept Floor Offices Entrances Age Freeway
VIF 3.757 3.267 1.638 1.169 1.185
b.
The VIF values are all under 4 which suggests that multicollinearity has not caused instability. Learning Objective: 13-7 13.37
Building 5 was the only observation that had an unusual residual. Learning Objective: 13-9
13.38
There were no high leverage values. Learning Objective: 13-10
13.39
The histogram shows a slight left skewed distribution but is unimodal with no obvious outliers. The normplot is a fairly straight line on the diagonal. Assuming normally distributed residuals appears reasonable.
Object 19
369
ASBE 6e Solutions for Instructors
Object 21
Learning Objective: 13-8
13.40
Assuming homoscedastic residuals appears reasonable. The residual plot does not show a fan out or funnel pattern.
370
ASBE 6e Solutions for Instructors
Object 23
Learning Objective: 13-8 13.41
This is cross-sectional data. A test for autocorrelation is not warranted. Learning Objective: 13-8
371
ASBE 6e Solutions for Instructors DATA SET D Response Variable: Percent Change in Consumer Price Index 13.25
Time-series. Unit of observation: one year. Learning Objective: 02-3
13.26
The variable magnitudes are similar. Learning Objective: 13-11
13.27
The intercept would not have meaning. While it might be possible to have 0% change in currency demands and deposits, it would not be logical to have a zero unemployment rate or zero utilization of manufacturing capacity. A priori reasoning for the relationship between each predictor and the response variable are listed in the table below. Predictor
Relationship with Response
CapUtil ChgM1 ChgM2 Unem
Positive Negative Positive Positive
Reason? Greater utilization, increase in CPI Increase in deposits, CPI stays stable Increase in small deposits, CPI increases Unemployment increases, CPI increases
Learning Objective: 12-2 13.28
41/4 = 10.25. The data set meets both Evans’ and Doane’s Rules. Learning Objective: 13-2
13.29
The estimated regression equation is
ŷ
= −25.2195 + 0.2806CapUtil − 0.0847ChgM1 +
0.2205ChgM2 + 1.0511Unem. The signs on the coefficients match our a priori reasoning. MegaStat Output: Regression Analysis R² 0.225 Adjusted R² 0.139 R 0.474 Std. Error 2.623 ANOVA table Source Regression Residual Total
n 41 k 4 Dep. Var. ChgCPI
SS 71.8691 247.5957 319.4649
df 4 36 40
MS 17.9673 6.8777
F 2.61
Regression output variables coefficients Intercept -25.2195
std. error 11.7919
t (df=36) -2.139
p-value .0393
372
p-value .0514
confidence interval 95% lower 95% upper -49.1346 -1.3044
ASBE 6e Solutions for Instructors CapUtil ChgM1 ChgM2 Unem
0.2806 -0.0847 0.2205 1.0511
0.1258 0.1117 0.1383 0.4086
2.231 -0.758 1.594 2.572
.0320 .4531 .1197 .0144
0.0255 -0.3112 -0.0601 0.2224
0.5357 0.1418 0.5011 1.8798
Learning Objective: 13-2 13.30
Refer to the output in question 13.29. The 95% coefficient confidence intervals for ChgM1 and ChgM2 both contain zero. This means that neither of these predictors is significant at = .05. The confidence intervals for CapUtil and Unem do not contain zero so both predictors are significant at a .05 level. Learning Objective: 13-3
13.31
For each coefficient test the following hypotheses: H0: βj = 0 vs. H1: βj ≠ 0. Using df = 36, t.025 =T.INV(.025, 36) = ±2.028. CapUtil has a tcalc = 2.231 and Unemp has a tcalc = 2.572, which are both greater than 2.028 so this indicates significance. Learning Objective: 13-3
13.32
a. b. c.
13.33
Fcalc = 2.61 with a p-value = .0514. R2 = .225 and R2adj = .139. The model does not provide significant fit at = .05. Learning Objective: 12-6 Learning Objective: 13-2
13.34
Prediction interval:
CapUtil: p-value = .0320 and Unemp p-value = .0144. Both are significant at = .05. This is consistent with the answer in 13.31. The tests conclude the same thing. Most analysts prefer the p-value approach because it tells the strength of the predictor significance. Learning Objective: 13-3
yˆi
± t.025se =
yˆi
± 2.028(2.623) =
yˆi
± 5.319. Based on the response to
question 13.33 and the wide prediction interval, no, this model does not have practical value. Learning Objective: 13-4 13.35
a.
CapUtil ChgM1 ChgM2 Unem
CapUtil 1.000 -.257 -.284 -.649
ChgM1
ChgM2
Unem
1.000 .316 .504
1.000 .303
1.000
41sample size ± .308 critical value .05 (two-tail)
373
ASBE 6e Solutions for Instructors ± .398 critical value .01 (two-tail)
b.
Unemployment is significantly correlated with manufacturing capacity utilization and the variable CHGM1. Collinearity could be a problem according to Klein’s Rule. Both .649 and .504 are greater than .225 = .474. Learning Objective: 13-7 13.36
a. variables Intercept CapUtil ChgM1 ChgM2 Unem
VIF 1.785 1.420 1.171 2.192
b.
The VIF values are all under 3 which suggests that multicollinearity has not caused instability. Learning Objective: 13-7 13.37
1974, 1979, and 1980 were years that had unusual or outlier residuals. Learning Objective: 13-9
13.38
1992 and 2001 were high leverage years. To determine if any of the observations were influential, remove them from the data set and rerun the regression. If the regression statistics change significantly then the observation could be considered influential. Learning Objective: 13-10
13.39
The normplot is not as straight of a line as one would like to see. The histogram of residuals is right skewed with several possible outliers. Assuming normally distributed residuals is questionable. 374
ASBE 6e Solutions for Instructors
Object 25
Object 27
Learning Objective: 13-8
13.40
Assuming homoscedastic residuals appears reasonable although one might question the slight increase in residual magnitude for the predictions of greater positive change.
375
ASBE 6e Solutions for Instructors
Object 30
Learning Objective: 13-8 13.41
A test for autocorrelation is warranted. DW = 0.75 which suggests positive autocorrelation.
Object 33
Learning Objective: 13-8
376
ASBE 6e Solutions for Instructors DATA SET E Response Variable: College Graduation Rate by State 13.25
Cross-sectional. Unit of observation: state. Learning Objective: 02-3
13.26
The variable magnitudes are not too different. Education spending by state is approximately 100 times the magnitude of the other variables but this should not cause a problem. Learning Objective: 13-11
13.27
The intercept would not have meaning. None of the quantitative predictor variables would logically be zero. A priori reasoning for the relationship between each predictor and the response variable are listed in the table below. Predictor
Relationship with Response
Dropout
Negative
EdSpend
Positive
Metro% Age
Positive Negative
LPRFem
Positive
Neast
Positive
Seast
Positive
West
Positive
Reason? Higher dropout rate, lower college graduation rate Higher spending, higher college graduation rate Greater urban population, higher college graduation rate Older population, fewer attending college More women in workforce, more college graduates With Midwest as the base, more college graduates in the northeast With Midwest as the base, more college graduates in the southeast With Midwest as the base, more college graduates in the west
Learning Objective: 12-2 13.28
50/8 = 6.25 > 5. The data set meets Doane’s Rule but does not meet Evans’. Learning Objective: 13-2
13.29
The estimated regression equation is
= −21.7629 − 0.2579Dropout + 0.0025EdSpend + ŷ 0.2036Metro% − 0.1458Age + 0.5477LPRFem + 5.26Neast + 2.0008Seast + 2.6378West. The signs on the coefficients match our a priori reasoning. MegaStat Output: Regression Analysis R² Adjusted R² R Std. Error
0.692 0.632 0.832 3.099
n k Dep. Var.
377
50 8 ColGrad%
ASBE 6e Solutions for Instructors ANOVA table Source Regression Residual
SS 885.4526 393.7026 1,279.155 Total 2 Regression output coefficient variables s Intercept -21.7629 Dropout -0.2579 EdSpend 0.0025 Metro% 0.2036 AgeMed -0.1458 LPRFem 0.5477 Neast 5.2600 Seast 2.0008 West 2.6378
df 8 41
MS 110.6816 9.6025
F 11.53
p-value 2.16E-08
49
std. error 21.9424 0.1846 0.0018 0.0562 0.2730 0.1880 1.5626 1.7252 1.3417
t (df=41) -0.992 -1.398 1.396 3.621 -0.534 2.913 3.366 1.160 1.966
p-value .3271 .1697 .1703 .0008 .5963 .0058 .0017 .2529 .0561
confidence interval 95% 95% lower upper -66.0766 22.5507 -0.6307 0.1148 -0.0011 0.0062 0.0900 0.3171 -0.6971 0.4056 0.1680 0.9274 2.1042 8.4157 -1.4834 5.4849 -0.0718 5.3474
Learning Objective: 13-2 13.30
Refer to the output in question 13.29. The coefficient confidence intervals contain zero except for the variables Metro%, LPRFem, and Neast. These three variables are the only predictors significant at = .05. Learning Objective: 13-3
13.31
For each coefficient test the following hypotheses: H0: βj = 0 vs. H1: βj ≠ 0. Using df = 41, t.025 =T.INV(.025,41) = ±2.020. The t statistics for Metro%, LPRFem, and Neast are 3.621, 2.913, and 3.366, respectively. Each value is greater than 2.02. Learning Objective: 13-3
13.32
a. b. c.
13.33
Fcalc = 11.53 with a p-value = 2.16E-09. R2 = .692 and R2adj = .632. The model provides significant fit with a fairly strong prediction of state college graduation rates. Learning Objective: 12-6 Learning Objective: 13-2
13.34
Prediction interval:
Metro%: p-value =.0008, LPRFem: p-value = .0058, and Neast: p-value = .0017. This is consistent with the answer in 13.31. The tests conclude the same thing. Most analysts prefer the p-value approach because it tells the strength of the predictor significance. Learning Objective: 13-3
± t.025se = ± 2.02(3.099) = ± 6.26. Yes, this model does have yˆi yˆi yˆi practical value. Learning Objective: 13-4 378
ASBE 6e Solutions for Instructors 13.35
a.
Dropout EdSpen d Metro% AgeMed LPRFem
Dropou t 1.000
EdSpen d
Metro %
AgeMe d
LPRFe m
-.323 .156 -.101 -.741
1.000 .183 .119 .228
1.000 -.274 -.291
1.000 -.002
1.000
50 ± .279 ± .361
sample size critical value .05 (two-tail) critical value .01 (two-tail)
b.
The only two variables that have an r value that might raise a red flag are LPRFem and Dropout. However this should not be a problem according to Klein’s Rule because .741 < .692 = .832. Learning Objective: 13-7 13.36
a. variables Intercept Dropout EdSpend Metro% AgeMed LPRFem Neast Seast West
VIF 2.637 1.447 1.684 1.886 3.174 2.182 2.827 1.890
b.
The VIF values are all 3 or less (except for LPRFEM which is slightly greater than 3) which suggests that multicollinearity has not caused instability. Learning Objective: 13-7 13.37
Delaware and Wyoming are the only two states that showed unusual or outlier residuals. Learning Objective: 13-9
13.38
Utah and West Virginia had high leverage values. To determine if any of the observations were influential, remove them from the data set and rerun the regression. If the regression statistics change significantly then the observation could be considered influential. Learning Objective: 13-10
379
ASBE 6e Solutions for Instructors 13.39
Although the normplot is not as straight of a line as one would like and the histogram is slightly skewed to the left, the distribution is unimodel with tapering tails and there are no obvious outliers. Assuming normally distributed residuals appears reasonable.
Object 35
Object 37
Learning Objective: 13-8
380
ASBE 6e Solutions for Instructors 13.40
Assuming homoscedastic residuals is not reasonable. There is a clear fan out pattern. It appears that the residual variation increases as the percentage of a state’s population living in metropolitan areas increases. Taking a log transform of the dependent variable would not be a good solution here because the magnitudes of the variables are similar. There may be lurking variables we could add to the model to correct this problem.
Object 40
Object 43
Learning Objective: 13-8 13.41
This is cross-sectional data. A test for autocorrelation is not warranted. Learning Objective: 13-8 381
ASBE 6e Solutions for Instructors DATA SET F Response Variable: Cruise Speed of Piston Aircraft 13.25
Cross-sectional. Unit of observation: aircraft model. Learning Objective: 02-3
13.26
The variable magnitudes are all similar. Learning Objective: 13-11
13.27
The intercept would not have meaning. It would not be logical to have an aircraft with zero values for any of the predictors. A priori reasoning for the relationship between each predictor and the response variable are listed in the table below. Note that the variable Age is calculated by subtracting the year of manufacture from 2010. Predictor
Relationship with Response
Age TotalHP NumBlades Turbo
Negative Positive Positive Positive
Reason? Older engine, lower speed Bigger engine, higher speed More blades, higher speed Stronger engine, higher speed
Learning Objective: 12-2 13.28
55/4 = 13.75. The data set meets both Evans’ and Doane’s Rules. Learning Objective: 13-2
13.29
The estimated regression equation is
= 92.4431 + 0.1787TotalHP + 8.8269NumBlades + ŷ 15.9752Turbo −0.3927Age. The signs on the coefficients match our a priori reasoning. MegaStat Output: Regression Analysis R² Adjusted R² R Std. Error
0.768 0.750 0.876 18.097
SS 54,232.905 Regression 0 16,375.204 Residual 1 70,608.109 Total 1 Regression output
df
ANOVA table Source
n k Dep. Var.
4
MS 13,558.226 2
50
327.5041
55 4 Cruise F
p-value
41.40
2.75E-15
54
variables
coefficients
std. error
t (df=50)
p-value
Intercept TotalHP NumBlades Turbo
92.4431 0.1787 8.8269 15.9752
13.2196 0.0195 5.7530 6.2959
6.993 9.167 1.534 2.537
6.16E-09 2.76E-12 .1313 .0143
382
confidence interval 95% 95% lower upper 65.890 7 118.9955 0.1396 0.2179 -2.7284 20.3823 3.3296 28.6208
ASBE 6e Solutions for Instructors Age
-0.3927
0.1991
-1.972
.0541
-0.7927
0.0073
Learning Objective: 13-2 13.30
Refer to the output in question 13.29. The coefficient confidence intervals contain zero except for the variables TotalHP and Turbo. Only TotalHP and Turbo are significant predictors at = .05. Learning Objective: 13-3
13.31
For each coefficient test the following hypotheses: H0: βj = 0 vs. H1: βj ≠ 0. Using df = 50, t.025 =T.INV(.025, 50) = ±2.009. The predictor TotalHP has a tcalc = 9.167 and Turbo has a tcalc = 2.537 which are both greater than 2.009 therefore we reject the null hypotheses and conclude their coefficients are not equal to zero. Learning Objective: 13-3
13.32
a.
13.33
Fcalc = 41.40 with a p-value = 2.75E-15. R2 = .768 and R2adj = .750. The model provides significant fit with a fairly strong prediction of aircraft cruising speed. Learning Objective: 12-6 Learning Objective: 13-2
13.34
Prediction interval:
13.35
a.
TotalHP: p-value = 2.76E-12 and Turbo: p-value =.0143. Both p-values are less than . 05. Note that the variable Age has a p-value = .0541 which is significant at = .10 b. This is consistent with the answer in 13.31. c. The tests conclude the same thing. Most analysts prefer the p-value approach because it tells the strength of the predictor significance. Learning Objective: 13-3
± t.025se = ± 2.009(18.097) = ± 36.357. Yes, this model does yˆi yˆi yˆi have practical value. Learning Objective: 13-4
TotalHP NumBlade s Turbo Age
TotalHP 1.000 .491 .096 .154 55 ± .266 ± .345
NumBlade s
Turbo
1.000 .388 1.000 -.180 -.030 sample size critical value .05 (two-tail) critical value .01 (two-tail)
383
Age
1.000
ASBE 6e Solutions for Instructors b.
Collinearity should not be a problem according to Klein’s Rule. The correlation coefficient values are all less than . 2 R Learning Objective: 13-7 13.36
a. variables Intercept Age TotalHP NumBlade s Turbo
VIF 1.131 1.459 1.716 1.201
b.
The VIF values are all under 2 which suggest that multicollinearity has not caused instability. Learning Objective: 13-7 13.37
Aircraft 23, 39 and 46 have unusual residual values. Learning Objective: 13-9
13.38
Observations 3, 8, 43, and 46 have high leverage values. To determine if any of the observations were influential, remove them from the data set and rerun the regression. If the regression statistics change significantly then the observation could be considered influential. Learning Objective: 13-10
13.39
Assuming normally distributed residuals appears reasonable.
Object 45
384
ASBE 6e Solutions for Instructors
Object 47
Learning Objective: 13-8 13.40
Assuming homoscedastic residuals appears reasonable.
Object 50
Learning Objective: 13-8 13.41
This is cross-sectional data. A test for autocorrelation is not warranted. Learning Objective: 13-8
385
ASBE 6e Solutions for Instructors DATA SET G Response Variable: Chromatographic Retention Time 13.25
Cross-sectional. Unit of observation: chemical compound. Learning Objective: 02-3
13.26
The variable magnitudes are all similar. Learning Objective: 13-11
13.27
The intercept would not have meaning. It would not be logical to for a particular compound to have zero values for any of the predictors. A priori hypotheses for the relationship between each predictor and the response variable are listed in the table below. Predictor
Relationship with Response
MW BP RI H1 H2 H3 H4
Negative Positive Negative Negative Negative Negative Negative
Learning Objective: 12-2 13.28
35/7 = 5. The data set meets Doane’s Rule but not Evans’. Learning Objective: 13-2
13.29
The estimated regression equation is
= 51.3827 − 0.1772MW + 1.4901BP − 13.1620RI − ŷ 13.8067H1 − 6.4334H2 − 12.2297H3 − 0.5823H4. The signs on the coefficients match our a priori reasoning. MegaStat Output: Regression Analysis R² 0.987 Adjusted R² 0.983 R 0.993 Std. Error 8.571
n 35 k 7 Dep. Var. Ret
ANOVA table Source SS Regression 146,878.2005 Residual 1,983.3648 Total 148,861.5653
df 7 27 34
Regression output variables coefficients Intercept 51.3827 MW -0.1772 BP 1.4901 RI -13.1620 H1 -13.8067
std. error 162.7418 0.3083 0.1831 107.2293 9.7452
MS 20,982.6001 73.4580
F 285.64
t (df=27) p-value 0.316 .7546 -0.575 .5701 8.139 9.64E-09 -0.123 .9032 -1.417 .1680
386
p-value 1.27E-23
confidence interval 95% lower 95% upper -282.5359 385.3012 -0.8097 0.4553 1.1144 1.8657 -233.1784 206.8544 -33.8022 6.1888
ASBE 6e Solutions for Instructors H2 H3 H4
-6.4334 -12.2297 -0.5823
8.6848 8.1138 4.8499
-0.741 -1.507 -0.120
.4652 .1434 .9053
-24.2531 -28.8779 -10.5335
11.3864 4.4184 9.3689
Learning Objective: 13-2 13.30
Refer to the output in question 13.29. The coefficient confidence intervals contain zero except for the variable BP (Boiling Point). Only BP is a significant predictor at = . 05. Learning Objective: 13-3
13.31
For each coefficient test the following hypotheses: H0: βj = 0 vs. H1: βj ≠ 0. Using df = 27, t.025 =T.INV(.025,27) = ±2.052. The predictor BP has a tcalc = 8.139. Learning Objective: 13-3
13.32
a. b. c.
13.33
Fcalc = 285.64 with a p-value = 1.27E-23. R2 = .987 and R2adj = .983. The model provides significant fit with a fairly strong prediction of compound retention time. However, residual assumptions should be verified. Learning Objective: 12-6 Learning Objective: 13-2
13.34
Prediction interval:
13.35
a.
BP: p-value = 9.64E-09. This is consistent with the answer in 13.31. The tests conclude the same thing. Most analysts prefer the p-value approach because it tells the strength of the predictor significance. Learning Objective: 13-3
± t.025se = ± 2.052(8.571) = ± 17.588 Yes, this model does have yˆi yˆi yˆi practical value, provided the residual assumptions are verified. Learning Objective: 13-4
MW BP RI H1 H2 H3 H4
MW BP RI H1 H2 1.000 .906 1.000 .240 .580 1.000 .065 -.218 -.747 1.000 -.233 -.336 -.202 -.194 1.000 -.214 -.185 -.145 -.194 -.094 .117 .167 .202 -.316 -.153 35sample size ± .334 critical value .05 (two-tail) ± .430 critical value .01 (two-tail)
387
H3
H4
1.000 -.153
1.000
ASBE 6e Solutions for Instructors b.
Collinearity could be a problem. The correlation coefficient for BP and MW is close to .987 = .9935. Learning Objective: 13-7 13.36
a. variables Intercept MW BP RI H1 H2 H3 H4
VIF 21.409 31.113 13.115 9.235 2.816 2.458 1.793
b.
The VIF values MW, BP, RI, and H1 are all high. It is possible that they are causing variance inflation which could cause some instability. If the regression analysis is rerun using only BP as the predictor variable, very little changes in the fit statistics and the scatterplot shows a very strong linear relationship. Learning Objective: 13-7 13.37
Observations 15, 17, and 25 have unusual residual values. Learning Objective: 13-9
13.38
Observation 24 has a high leverage value. To determine if any of the observations were influential, remove them from the data set and rerun the regression. If the regression statistics change significantly then the observation could be considered influential. Learning Objective: 13-10
388
ASBE 6e Solutions for Instructors 13.39
While the normplot and histogram are not a perfect representation of a normal distribution, there are no strong departures from normality. Assuming normally distributed residuals appears reasonable.
Object 53
Object 55
Learning Objective: 13-8
389
ASBE 6e Solutions for Instructors 13.40
The model appears to underestimate retention at the low end and the high end of the range of values. It might be prudent to explore a nonlinear relationship with retention and BP.
Object 57
Learning Objective: 13-8 13.41
This is cross-sectional data. A test for autocorrelation is not warranted. Learning Objective: 13-8
390
ASBE 6e Solutions for Instructors DATA SET H Response Variable: 2007 State Foreclosure Rate 13.25
Cross-sectional. Unit of observation: state. Learning Objective: 02-3
13.26
The variable magnitudes are all similar. Learning Objective: 13-11
13.27
The intercept would not have meaning. It would not be logical to for a particular state to have zero values for any of the predictors. A priori hypotheses for the relationship between each predictor and the response variable are listed in the table below. Predictor
Relationship with Response
MassLayoff SubprimeShar e PriceIncomeR atio Homeownersh ip 5YrApp UnempChang e %HousMoved
Positive Positive Positive Negative Positive Positive Positive
Learning Objective: 12-2 13.28
50/7 = 7.14. The data set meets Doane’s Rule but not Evans’. Learning Objective: 13-2
13.29
The estimated regression equation is
= 51.2829 + 0.0751MassLayoff + ŷ 18.5385SubPrimeShare − .6965PriceIncomeRatio − .0587Homeownership − . 04415YrApp – 16.4618UnEmpChange – 99.2433%HouseMoved. The signs on the coefficients match our a priori logic for only Mass Layoffs and Share of mortgages that were subprime MegaStat Output: Regression Analysis R² 0.739 Adjusted R²0.696 R 0.860 Std. Error 3.732 ANOVA table Source SS Regression 1,657.8345 Residual 584.9417 Total 2,242.7762
n 50 k 7 Dep. Var. Foreclosure df 7 42 49
MS 236.8335 13.9272
391
F 17.01
p-value 2.01E-10
ASBE 6e Solutions for Instructors Regression output variables coefficients Intercept 51.2829 MassLayoff 0.0751 SubprimeShare 18.5385 PriceIncomeRatio -0.6965 Homeownership -0.0587 5YrApp 0.0441 UnempChange 16.4618 %HousMoved -99.2433
std. error 13.8099 0.2056 17.7785 0.7438 0.1309 0.0343 7.7223 14.3588
t (df=42) p-value 3.713 .0006 0.365 .7167 1.043 .3030 -0.936 .3544 -0.448 .6561 1.284 .2062 2.132 .0389 -6.912 1.94E-08
confidence interval 95% lower 95% upper 23.4134 79.1524 -0.3398 0.4900 -17.3400 54.4169 -2.1976 0.8045 -0.3228 0.2054 -0.0252 0.1134 0.8775 32.0461 -128.2206 -70.2661
Learning Objective: 13-2 13.30
Refer to the output in question 13.29. The coefficient confidence intervals contain zero except for the variables UnempChange and %HousMoved. These variables were the only two significant at = .05. Learning Objective: 13-3
13.31
For each coefficient test the following hypotheses: H0: βj = 0 vs. H1: βj ≠ 0. Using df = 42, t.025 =T.INV(.025,42) = ±2.018. The predictors UnempChange and %HousMoved have t statistics equal to 2.132 > 2.018 and 6.912 < 2.018, respectively. Learning Objective: 13-3
13.32
a. b. c.
13.33
Fcalc = 17.01 with a p-value = 2.01E-10. R2 = .739 and R2adj = .696. The model provides significant fit with a fairly strong prediction of foreclosure rates. However, residual assumptions should be verified. Learning Objective: 12-6 Learning Objective: 13-2
13.34
Prediction interval:
UnempChange: p-value = .0389 and %HousMoved: p-value = 1.94E-08. This is consistent with the answer in 13.31. The tests conclude the same thing. Most analysts prefer the p-value approach because it tells the strength of the predictor significance. Learning Objective: 13-3
± t.025se = ± 2.018(3.732) = ± 7.531. Yes, this model does have yˆi yˆi yˆi practical value, provided the residual assumptions are verified. Learning Objective: 13-4
392
ASBE 6e Solutions for Instructors
13.35
a. Subprime PriceIncom MassLayoff Share Ratio MassLayoff 1.000 SubprimeShare -.022 1.000 PriceIncomeRatio .073 -.119 1.000 Homeownership -.045 .130 -.501 5YrApp .045 -.105 .786 UnempChange .280 .124 .200 %HousMoved -.144 -.507 -.314 50sample size ± .279 critical value .05 (two-tail) ± .361 critical value .01 (two-tail)
Home ownership
5YrApp
Unemp Change
%Hous Moved
1.000 -.434 -.066 .229
1.000 .211 -.119
1.000 -.143
1.000
b.
Collinearity might be a problem. There are several pairs of variables that have significant correlation. In particular 5YrApp and PriceIncomRatio have an r close in value to .739 = .860. Learning Objective: 13-7
13.36
a. variables Intercept MassLayoff SubprimeSh are PriceIncome Ratio Homeowners hip 5YrApp UnempChan ge %HousMove d
VIF 1.123 1.653 3.447 1.405 2.868 1.170 1.901
b. The VIF values are all 4 or less. Concern about multicollinearity is not high. Learning Objective: 13-7 13.37
Colorado, Nevada, and Vermont have outlier/unusual residual values. Learning Objective: 13-9
13.38
California, Mississippi, Nevada, and Vermont have high leverage values. To determine if any of the observations were influential, remove them from the data set and rerun the regression. If the regression statistics change significantly then the observation could be considered influential. 393
ASBE 6e Solutions for Instructors Learning Objective: 13-10 13.39
Residuals do not appear to be normally distributed. A histogram shows a right skewed distribution.
Object 59
Object 61
Learning Objective: 13-8
394
ASBE 6e Solutions for Instructors
13.40
The residual plot shows some indication that the model is overestimating foreclosure rates in the middle range of values. A plot of residuals against unemployment shows heteroscedasticity.
Object 63
Object 65
Learning Objective: 13-8 13.41
This is cross-sectional data. A test for autocorrelation is not warranted. 395
ASBE 6e Solutions for Instructors Learning Objective: 13-8
396
ASBE 6e Solutions for Instructors DATA SET I Response Variable: Body Fat % 13.25
Cross-sectional. Unit of observation: an individual male. Learning Objective: 02-3
13.26
The variable magnitudes are similar. Learning Objective: 13-11
13.27
The intercept would not have meaning. A priori logic for the relationship between each predictor and the response variable are listed in the table below. Predictor
Relationship with Response
Age Weight Height Neck Chest Abdomen Hip Thigh
Positive Positive Neutral Positive Positive Positive Positive Positive
Learning Objective: 12-2 13.28
50/8 = 6.25. The data set meets Doane’s Rule but not Evans’. Learning Objective: 13-2
13.29
The estimated regression equation is y = −35.4309 +0.0905Age − 0.1928Weight − 0.0642Height − 0.3348Neck + 0.0239Chest + 0.9132Abdomen −0.3107Hip +0.7787Thigh. The signs on the coefficients match our a priori logic for Age, Chest, and Abdomen only. MegaStat Output: Regression Analysis R² 0.841 Adjusted R² 0.810 R 0.917 Std. Error 3.957
n 50 k 8 Dep. Var. Fat%
ANOVA table Source SS Regression 3,399.1446 Residual 641.8882 Total 4,041.0328
df 8 41 49
MS 424.8931 15.6558
F 27.14
Regression output variables coefficients Intercept -35.4309 Age 0.0905 Weight -0.1928 Height -0.0642 Neck -0.3348
std. error 24.9040 0.0880 0.0783 0.1160 0.4023
t (df=41) -1.423 1.028 -2.462 -0.554 -0.832
p-value .1624 .3099 .0181 .5827 .4100
397
p-value 4.82E-14
confidence interval 95% lower 95% upper -85.7256 14.8638 -0.0872 0.2682 -0.3510 -0.0346 -0.2984 0.1700 -1.1472 0.4776
ASBE 6e Solutions for Instructors Chest Abdomen Hip Thigh
0.0239 0.9132 -0.3107 0.7787
0.1788 0.1640 0.2749 0.2907
0.133 5.570 -1.130 2.678
.8945 1.77E-06 .2649 .0106
-0.3373 0.5821 -0.8658 0.1915
0.3850 1.2444 0.2445 1.3658
Learning Objective: 13-2 13.30
Refer to the output in question 13.29. The 95% coefficient confidence intervals contain zero except for the variables Weight, Abdomen, and Thigh. This means that these three variables are the only significant predictors in the model. Learning Objective: 13-3
13.31
For each coefficient test the following hypotheses: H0: βj = 0 vs. H1: βj ≠ 0. Using df = 41, t.025 =T.INV(.025,41) = ±2.020. Weight, Abdomen, and Thigh have t statistics equal to 2.462 < 2.02, 5.570 > 2.02, and 2.678 > 2.02, and are all significant predictors. Learning Objective: 13-3
13.32
a.
13.33
Fcalc = 27.14 with a p-value = 4.82E-14. R2 = .841 and R2adj = .810. The model provides significant fit with a fairly strong prediction of percent body fat. Learning Objective: 12-6 Learning Objective: 13-2
13.34
Prediction interval:
13.35
a.
Weight: p-value = .0181, Abdomen: p-value = 1.77E-06, and Thigh Weight: p-value = .0106. b. This is consistent with the answer in 13.31. c. The tests conclude the same thing. Most analysts prefer the p-value approach because it tells the strength of the predictor significance. Learning Objective: 13-3
± t.025se = ± 2.020(3.957) = ± 7.993. Yes, this model does have yˆi yˆi yˆi practical value. Learning Objective: 13-4
Age Weight Height Neck Chest Abdomen Hip
Age Weight Height 1.000 .265 1.000 -.276 .109 1.000 .176 .882 .201 .376 .912 .014 .442 .915 -.052 .314 .959 -.045 50sample size ± .279 critical value .05 (two-tail) ± .361 critical value .01 (two-tail)
Neck
Chest
Abdomen
Hip
1.000 .820 .781 .804
1.000 .942 .911
1.000 .942
1.000
398
ASBE 6e Solutions for Instructors b.
According to Klein’s Rule there are many pairs of data that are causing concern for collinearity. The correlation coefficient values are greater than . .841 = .917. Learning Objective: 13-7 13.36
a. variables Intercept Age Weight Height Neck Chest Abdomen Hip Thigh
VIF 1.712 31.111 1.689 5.472 11.275 17.714 25.899 11.931
b.
The VIF values are high except for Age, Height, and Neck. Multicollinearity is a concern. Learning Objective: 13-7 13.37
There are no unusual standardized residuals. Learning Objective: 13-9
13.38
Observation 5, 15, 36, 39, 42 had high leverage values. To determine if any of the observations were influential, remove them from the data set and rerun the regression. If the regression statistics change significantly then the observation could be considered influential. Learning Objective: 13-10
13.39
Assuming normally distributed residuals appears reasonable although the histogram appears slightly right skewed. 399
ASBE 6e Solutions for Instructors
Object 67
Object 69
Learning Objective: 13-8
13.40
Assuming homoscedastic residuals appears reasonable. 400
ASBE 6e Solutions for Instructors
Object 71
Learning Objective: 13-8 13.41
This is cross-sectional data. A test for autocorrelation is not warranted. Learning Objective: 13-8
401
ASBE 6e Solutions for Instructors DATA SET J Response Variable: Used Vehicle Price 13.25
Cross-sectional. Unit of observation: vehicle model. Learning Objective: 02-3
13.26
The variable magnitudes are different. The response variable magnitude is in tens of thousands and the predictor variables are integers between 0 and 50. Learning Objective: 13-11
13.27
The intercept would not have meaning. It would not be logical to have a car with zero values for any of the predictors. A priori reasoning for the relationship between each predictor and the response variable are listed in the table below. Predictor
Relationship with Response
Age
Negative
Car
Negative
Truck
Positive
SUV
Positive
Reason? Older car, lower price Cars are less expensive than Vans which is the base indicator variable Trucks are more expensive than Vans which is the base indicator variable SUVs are more expensive than Vans which is the base indicator variable
Learning Objective: 12-2 13.28
637/4 = 159.25. The data set meets both Evans’ and Doane’s Rules. Learning Objective: 13-2
13.29
The estimated regression equation is
= 15,340.7233 − 693.9768Age − 533.5731Car + ŷ 5,748.1799Truck + 3,897.5375SUV. The signs on the coefficients match our a priori reasoning. MegaStat Output: Regression Analysis R² 0.139 Adjusted R² 0.134 R 0.373 Std. Error 8573.178 ANOVA table Source Regression Residual Total
SS 7,512,691,865.5390 46,451,606,047.1737 53,964,297,912.7127
Regression output variables Intercept Age Car Truck SUV
df 4 632 636
n 637 k 4 Dep. Var. Price MS 1,878,172,966.3848 73,499,376.6569
F 25.55
p-value 1.20E-19
confidence interval t (df=632) p-value 95% lower 95% upper 12.381 1.12E-31 12,907.5585 17,773.8881 -5.897 6.02E-09 -925.0680 -462.8855 -0.435 .6635 -2,940.8241 1,873.6780 4.359 1.52E-05 3,158.7909 8,337.5689 2.963 .0032 1,314.2852 6,480.7899
coefficients std. error 15,340.7233 1,239.0560 -693.9768 117.6801 -533.5731 1,225.8598 5,748.1799 1,318.6111 3,897.5375 1,315.4861
402
ASBE 6e Solutions for Instructors Learning Objective: 13-2 13.30
Refer to the output in question 13.29. The 95% coefficient confidence interval for the indicator variable Car contains zero. The other three variables are significant predictors in the model. Learning Objective: 13-3
13.31
For each coefficient test the following hypotheses: H0: βj = 0 vs. H1: βj ≠ 0. Using df = 632, t.025 =T.INV(.025,632) = ±1.964. Age, Truck, and SUV have t statistic values equal to 5.897, 4.359, and 2.963, respectively. Learning Objective: 13-3
13.32
a.
13.33
Fcalc = 25.55 with a p-value = 1.20E-19. R2 = .139 and R2adj = .134. The model provides significant fit but the model is not a strong predictive equation. Learning Objective: 12-6 Learning Objective: 13-2
13.34
Prediction interval:
Age: p-value = 6.02E-09, Truck: p-value = 1.52E-05, and SUV: p-value = .0032. All three p-values are less than .05 b. This is consistent with the answer in 13.31. c. The tests conclude the same thing. Most analysts prefer the p-value approach because it tells the strength of the predictor significance. Learning Objective: 13-3
± t.025se = ± 1.964(8573.178) = ± 16837.722. No, this model yˆi yˆi yˆi does not have practical value. Learning Objective: 13-4
13.35
a. Correlation Matrix Age Car Truck SUV
Age Car Truck SUV 1.000 .003 1.000 .017 -.478 1.000 -.092 -.495 -.308 1.000 637sample size ± .078 critical value .05 (two-tail) ± .102 critical value .01 (two-tail)
b.
There is no concern for collinearity. While there is significant correlation between the indicator variables, this is to be expected because of the way they are defined. We would expect to see correlation between indicator variables defined on the same characteristic. Learning Objective: 13-7
403
ASBE 6e Solutions for Instructors 13.36
a. variables Intercept Age Car Truck SUV
VIF 1.017 3.201 2.662 2.749
b.
The VIF values are all 4 or less which suggests that multicollinearity has not caused instability. Learning Objective: 13-7 13.37
Vehicles 212, 342, and 502 were extremely high outlier residuals. Vehicles 246, 397, 631, and 632 were unusual/outlier residuals. The next step would be to investigate the three high outlier residuals for possible exclusion from the data set. It is possible their prices were mistyped or the vehicles do not fit the profile of a vehicle for which the model is being developed. Learning Objective: 13-9
13.38
There were many observations with high leverage, too many to list here. To determine if any of the observations were influential, remove them from the data set and rerun the regression. If the regression statistics change significantly then the observation could be considered influential. Learning Objective: 13-10
13.39
Residuals do not appear to be normally distributed. Outliers should be investigated for possible removal from data set.
404
ASBE 6e Solutions for Instructors
Object 73
Object 75
Learning Objective: 13-8
405
ASBE 6e Solutions for Instructors 13.40
Assuming homoscedastic residuals is not reasonable. There are obviously outliers in the data set. Removing the outliers and rerunning the analysis will most likely show heteroscedasticity in the residual plot.
Object 77
Learning Objective: 13-8 13.41
This is cross-sectional data. A test for autocorrelation is not warranted. Learning Objective: 13-8
406
ASBE 6e Solutions for Instructors DATA SET K Response Variable: 500 yard freestyle time 13.25
Cross-sectional. Unit of observation: individual swimmer. Learning Objective: 02-3
13.26
The variable magnitudes are similar. Learning Objective: 13-11
13.27
The intercept would not have meaning. It would not be logical to have a freestyle time when the age of the swimmer is zero. A priori reasoning for the relationship between each predictor and the response variable are listed in the table below. Predictor
Relationship with Response
Seed
Positive
Gender
Positive
Age
Positive
Reason? The lowest seeded swimmers should have the lowest times and vice versa If Gender = 1 indicates female it is possible the women have slower times than men The older a swimmer is the slower their times will be
Learning Objective: 12-2 13.28
198/3 = 66. The data set meets both Evans’ and Doane’s Rules. Learning Objective: 13-2
13.29
The estimated regression equation is
= −35.3401 + 0.9286Seed +13.4401Gender + ŷ 0.8105Age . The signs on the coefficients match our a priori reasoning. MegaStat Output: Regression Analysis R² Adjusted R² R Std. Error
0.942 0.941 0.970 39.015
n k Dep. Var.
SS 4,755,790.720 Regression 5 Residual 295,300.5055 5,051,091.226 Total 0 Regression output
df
MS
F
p-value
3 194
1,585,263.5735 1,522.1676
1041.45
2.64E-119
p-value .1005 8.29E-90 .0440 .0463
confidence interval 95% 95% lower upper -77.5708 6.8907 0.8791 0.9781 0.3675 26.5126 0.0134 1.6076
ANOVA table Source
variables Intercept Seed Gender Age
coefficients -35.3401 0.9286 13.4401 0.8105
198 3 Time
197
std. error 21.4123 0.0251 6.6282 0.4042
t (df=194) -1.650 36.985 2.028 2.005
407
ASBE 6e Solutions for Instructors Learning Objective: 13-2 13.30
Refer to the output in question 13.29. None of the 95% coefficient confidence interval for all three variables contain zero. All three variables are significant predictors in the model. Learning Objective: 13-3
13.31
For each coefficient test the following hypotheses: H0: βj = 0 vs. H1: βj ≠ 0. Using df = 194, t.025 =T.INV(.025,194) = ±1.972. All variables have t statistic values greater than 1.972 therefore all predictors are significant. Learning Objective: 13-3
13.32
a.
13.33
Fcalc = 1041.45 with a p-value = 2.64E-119. R2 = .942 and R2adj = .941. The model provides significant fit and is a strong predictor of finishing times. Learning Objective: 12-6 Learning Objective: 13-2
13.34
Prediction interval:
13.35
Seed: p-value = 8.29E-90,Gender: p-value = .0440, and Age: p-value = .0463. All three p-values are less than .05 b. This is consistent with the answer in 13.31. c. The tests conclude the same thing. Most analysts prefer the p-value approach because it tells the strength of the predictor significance. Learning Objective: 13-3
± t.025se = ± 1.972(39.015) = ± 76.938. Yes, this model does yˆi yˆi yˆi have have practical value. Learning Objective: 13-4 a. Correlation Matrix Seed Gender Age
Seed 1.000 .347 .580 198 ± .139 ± .183
Gender
Age
1.000 -.136 1.000 sample size critical value .05 (twotail) critical value .01 (twotail)
b.
There is no concern for collinearity. While there is significant correlation between the indicator variables, the correlation is not high enough to cause concern. Learning Objective: 13-7 13.36
a. variables
408
VIF
ASBE 6e Solutions for Instructors Seed Gender Age
2.086 1.411 1.870
b.
The VIF values are all 3 or less which suggests that multicollinearity has not caused instability. Learning Objective: 13-7 13.37
Seven swimmers had residuals that were either unusual (greater than 2 but less than 3) or outliers (greater than 3.) This means the model underestimated their finishing times. Six swimmers had residuals that were either unusual (less than −2 but greater than −3) or outliers (less than −3.) This means the model overestimated their finishing times. Learning Objective: 13-9
13.38
There were nine observations with high leverage. To determine if any of the observations were influential, remove them from the data set and rerun the regression. If the regression statistics change significantly then the observation could be considered influential. Learning Objective: 13-10
13.39
Residuals do not appear to be normally distributed. Outliers should be investigated for possible removal from data set. The histogram shows outliers on both the low and high end.
Object 79
409
ASBE 6e Solutions for Instructors
Object 82
Learning Objective: 13-8 13.40
Assuming homoscedastic residuals is not reasonable. There is a clear fan out pattern. It appears that as the seed time increases, the variation in residuals increases.
Object 84
410
ASBE 6e Solutions for Instructors
Object 86
Learning Objective: 13-8 13.41
This is cross-sectional data. A test for autocorrelation is not warranted. Learning Objective: 13-8
411
ASBE 6e Solutions for Instructors 13.42
a.
Each coefficient measures the additional revenue earned by selling one more unit (one more car, truck, or SUV, respectively.) b. The intercept is not meaningful. Ford has to sell at least one car, truck or SUV to earn revenue. No sales means no revenue. Learning Objective: 12-2 Learning Objective: 13-1
13.43
There are no quantitative predictor variables. A better approach would be to use an ANOVA procedure that compares means within groups. In addition, the sample size is too small relative to the number of predictors. There would have to be 6 binary variables to cover the suppliers and substrate categories. With only 11 observations this would violate both Evans’ Rule and Doane’s Rule. Learning Objective: 13-5
13.44
a. b.
13.45
a. b. c.
13.46
a.
One binary must be omitted to prevent perfect multicollinearity. Same reasoning as in (a). The effect of the missing category is captured in the model intercept. c. Monday is the busiest day. The coefficient on Monday is positive meaning the occupancy rates go up that day relative to the base day of Sunday. All other days had negative coefficients. d. Shift 3 is captured in the intercept. Both Shift 1 and Shift 2 have lower AvgOccupancy given that they have negative coefficients. e. The intercept represents the AvgOccupancy on Sundays during Shift 3. f. The fit is poor because R2 = .094. Learning Objective: 13-2 Learning Objective: 13-5 Cost per Load = 26.0 – 6.3 Top-Load – 0.2714 Powder The regression as a whole is not significant based on the Fcalc p-value = .3710. Examination of the individual regression coefficients indicates that the two binary variables are not significantly different from zero, p-values >.10. Conclusion: cost per average load does not differ based on whether or not it is a topload washer or whether or not powder was used. No apparent cost savings based on washer type or detergent type. Learning Objective: 13-2 Learning Objective: 13-5 The best model in terms of fit as measured by s, R2, and R2adj is the model with three variables (InfMort, GDPCap, and Literate). Note that the three variable model is only marginally better in terms of fit statistics than the one or two variable models. b. No gain in fit is achieved by adding LifeExp and Density. c. Examination of the individual regression coefficients indicates that the InfMort and Literate have p-values < .01 and GDPCap has a p-value < 0.05. Conclusion: Infant mortality and literacy rate have the greatest impact on birth rates. Learning Objective: 13-2 Learning Objective: 13-3 412
ASBE 6e Solutions for Instructors 13.47
a.
Yes, the coefficients make sense, except for TrnOvr. One would think that turnovers would actually reduce the number of wins, not increase them. b. No. It is negative and the number of games won is limited to zero or greater. c. One needs either 5 observations or 10 observations per predictor. There are 6 predictor variables in the model, so we need a minimum of 30 observations to meet Doane’s rule. The fact there were only 23 teams and therefore only 23 observations means the sample size was probably too small to make the model reliable. d. Rebounds and points are highly correlated. We don’t need both of them in the model. This could be inflating the variance of the predictor estimates, causing the predictors to appear insignificant. Learning Objective: 13-2 Learning Objective: 13-3 Learning Objective: 13-7
13.48
a.
13.49
a.
Both men and women who had prior marathon experience had lower times on average than those who were running for the first time. b. No the intercept does not have any meaning. If all predictor/binary variables were 0 then you wouldn’t have an individual racer. c. It is suspected non-linearity is present among age, weight, and height. In this model we see increases in age decreases times, but at an increasing rate, increases in weight decreases time, but at an increasing rate and increasing height increases time, but at a decreasing rate. d. The model predicted that I would run the marathon in about 12 and ½ hours. And that could be right. I can walk 4 mph so it would take at least 6 to 7 hours minimum! Learning Objective: 13-2 Learning Objective: 13-3 Learning Objective: 13-5
13.50
a. b.
Examination of the individual regression coefficients indicates that all of the variables are significantly different from zero, p-values <0.01. The regression as a whole indicates a very strong fit. R2 = .811. The predictor variables as a group explain 81.1% of the variation in Salary. b. The ethnicity of a professor does matter. A professor who is African-American earns on average $2,093 less than one who is not. c. Rank matters. Assistant professors earn on average $6,438 less than higher ranking professors. New hires earn less than those who have been there for some time. To stay competitive universities often have to offer high salaries to the top candidates so this seems counter-intuitive. Learning Objective: 13-2 Learning Objective: 13-3 Learning Objective: 13-5
The regression as a whole is significant based on the Fcalc p-value = 8.35E-10. The variables Length and Width are not significant predictors of gas mileage. The variable Weight is a significant predictor (p-value = .0000) and according to the R2 value of .682, explains approximately 68% of the variation in CityMPG. 413
ASBE 6e Solutions for Instructors c.
The three predictors of CityMPG are most likely strongly correlated with each other. The VIF values do not show any concern (all less than 10.) Learning Objective: 13-2 Learning Objective: 13-3 Learning Objective: 13-7 13.51
While the four predictor model gives the highest R2 (.474) and lowest standard error (143.362), the predictor Divorce is not significant so one might consider the three predictor model to be the best choice. b. The decrease in R2adj from .427 to .418 after removing Divorce is quite small and justifies the three predictor model. c. The two most significant predictors are Income (p-value = .0013) and Pupil/Tea (pvalue = .0001.) Learning Objective: 13-2 Learning Objective: 13-3 Learning Objective: 13-5 a.
414
ASBE 6e Solutions for Instructors
Chapter 14 Time Series Analysis Note: For questions that deal with “best model” the student should consider four criteria: 14.1
Occam’s Razor Overall fit Believability Fit to recent data
Would a simpler model suffice? How does the trend fit the past data? Does the extrapolated trend “look right?” Does the fitted trend match the last few data points
a. See graphs. b. Increased bird strikes might be explained by an increase in the number of flights and more aircraft in the skies. c. See graphs. d. Linear model is simple and has a good fit. Both quadratic and exponential model appear to increase rapidly in future year. e. As expected both the exponential and quadratic models forecast higher than the linear model. The linear model seems to fit the past data but lags what appears to be an increase in bird strikes in recent years. t 17 18 19
Linear 12,511 13,011 13,511
Quadratic 14,170 15,256 16,406
Object 3
Learning Objective: 14-2 414
Exponentia l 13,140 13,942 14,794
ASBE 6e Solutions for Instructors Learning Objective: 14-3 14.2
a. Fairly linear increase b. A growing interest in organic produce and willingness to pay for the produce plus the desire by consumers for certification c. See graphs below showing the linear, quadratic, and exponential models d. It appears that the quadratic model is capturing the growth in organic farming the best; although the model should be adjusted after three years. e. Exponentia Quadrati t Linear l c 9 12654.8 12994.4 14692.5 10 13450.8 14140.1 16847.0 11 14246.8 15386.9 19273.2 Notice that the quadratic model has much higher forecasts than either the linear or the exponential models. It would be prudent to research what is actually driving this growth to determine if the quadratic model reflects a continuing trend.
Object 5
Object 7
415
ASBE 6e Solutions for Instructors
Object 9
Learning Objective: 14-2 Learning Objective: 14-3
14.3
a. b. c. d.
Seems cyclical with a 3-5 year cycle. Peaks in 2004, 2007 and 2012. Erosion of unskilled jobs, globalization, tougher bargaining. See graph below showing the linear, quadratic, and exponential models The quadratic model is clearly not suitable, predicting almost no strikers within three years. Both linear and exponential predict a drop in strikes. Using a trend model without considering cyclical effects is not the best forecasting approach
e. t 16 17 18
Linear 57.73 53.86 49.98
Quadratic 29.94 15.65 0.13
416
Exponentia l 47.07 44.59 42.25
ASBE 6e Solutions for Instructors
Object 11
Learning Objective: 14-2 Learning Objective: 14-3 14.4
a. The line plot shows a noticeable decreasing trend from 2003 to 2009. See graphs below. b. There has been a decrease in car dealerships in the U.S. since 2003 due to various factors such as lower car sales due to increasing gas prices. There may have been consolidation of smaller dealerships into large dealerships which could also account for the decrease in the number of dealerships. c. See graphs below showing the linear, quadratic, and exponential models d. The quadratic fit has the best R2 and seems to provide a sensible fit. t 8 9 10
Linear 18829.7 18331.6 17833.6
Quadratic 17074.7 15260.4 13153.6
Object 14
417
Exponentia l 18776.8 18313.2 17861.0
ASBE 6e Solutions for Instructors
Object 16
Object 18
Learning Objective: 14-2 Learning Objective: 14-3 14.5
14.6
a. y15 = 926e−.026(15) = 626.95 b. y15 = 2,217 – 8(15) = 2,097 c. y15 = 447 – 29(15) + 7(15)2 = 1,587 Learning Objective: 14-3 a. See graphs. A cyclical pattern is observed (positive autocorrelation). b. As m increases (i.e., more smoothing) the trendline does not describe the data as well. m = 2 gives the best “fit” while m = 6 gives the most smoothing. A larger m is actually helpful if we are trying to reduce the impact of day-to-day fluctuations (if you want to track the data exactly, why smooth at all). c. Can help anticipate the next day’s rate. Trend models aren’t much help.
418
ASBE 6e Solutions for Instructors
419
ASBE 6e Solutions for Instructors
Learning Objective: 14-5 14.7
a. Yields seem to exhibit a cyclical pattern that is not like any standard trend model (looks like positive autocorrelation) so exponential smoothing seems like a good choice for making a one-period-ahead forecast.
b-f.MegaStat output is lengthy. Here is a summary table: = .10 = .20 Mean Squared Error 0.039 0.028 Mean Absolute Percent Error 3.8% 3.1% Percent Positive Errors 42.3% 46.2% Forecast for Period 53 4.3 4.37 Learning Objective: 14-4 Learning Objective: 14-6 420
= .30 0.021 2.7% 51.9% 4.43
ASBE 6e Solutions for Instructors 14.8
a. Seasonality is present as well as a positive trend. Calculation of Seasonal Indexes 1 2 2004 2005 2006 2007 2008 2009 mean: adjusted :
0.836 0.797 0.787 0.773 0.738 0.786
0.983 0.998 1.017 0.986 1.068 1.011
0.788
1.013
Object 20
b. Time and seasonal binaries are significant. 421
3 0.996 1.017 1.018 1.041 1.012
4 1.201 1.161 1.201 1.184 1.145
1.017
1.178
3.992
1.019
1.181
4.000
ASBE 6e Solutions for Instructors Regression Analysis
ANOVA table Source Regressio n Residual Total
R² Adjusted R² R Std. Error
0.860 0.831 0.928 1171.700
SS 160,850,280.033 3 26,084,750.5917 186,935,030.625 0
df 4 19
n k Dep. Var.
24 4 Revenue
MS 40,212,570.008 3 1,372,881.6101
F
p-value
29.29
6.87E-08
23
Regression output variables coefficients
std. error
t (df=19)
p-value
Intercept Time Index QTR1 QTR2 QTR3
684.8853 35.0112 684.5869 680.0958 677.3869
13.102 7.877 -6.072 -2.807 -2.512
5.80E-11 2.11E-07 7.72E-06 .0112 .0212
8,973.3083 275.7875 -4,156.4708 -1,909.2583 -1,701.3792
confidence interval 95% lower 95% upper 10,406.789 7,539.8269 8 202.5081 349.0669 -5,589.3278 -2,723.6139 -3,332.7153 -485.8014 -3,119.1662 -283.5921
c. Forecasts for 2011: Period 2011 Q1 2011 Q2 2011 Q3 2011 Q4
Forecasts $11,711.525 million $14,234.525 million $14,718.192 million $16,695.358 million
Learning Objective: 14-7 Learning Objective: 14-8
14.9
a. No trend present, but there is seasonality. It is important to emphasize that 1.00 is the key reference point for a multiplicative seasonal index. The creative student might plot the seasonal indexes. A chart can show that March, April, May appear to be
422
ASBE 6e Solutions for Instructors significantly above average (seasonal index exceeding 1.00). MegaStat’s trend fitted to deseasonalized data (shown below) shows virtually no trend (R2 = .0006). Calculation of Seasonal Indexes 2004 2005 2006 2007 mean: adjusted :
Jan
Feb
Mar
Apr
May
Jun
0.874 0.875 0.766 0.838
0.888 1.013 0.962 0.954
1.230 1.186 1.096 1.171
1.234 1.134 1.128 1.165
1.169 1.079 1.164 1.137
0.898 0.963 1.086 0.982
0.836
0.951
1.167
1.162
1.134
0.979
Jul 1.198 0.689 0.922
Aug 0.716 0.800 0.996
Sep 1.280 0.847 1.029
Oct 0.898 1.052 0.940
Nov 0.734 1.105 0.948
Dec 1.022 1.133 1.052
0.937
0.837
1.052
0.963
0.929
1.069
0.934
0.835
1.049
0.961
0.926
1.066
Object 22
b. Fit of the multiple regression model is not very good (R2 = .416, R2adj = .216) and no predictors are significant at = .05. 423
ASBE 6e Solutions for Instructors Regression Analysis R² Adjusted R² R Std. Error
0.416 0.216 0.645 389.888
n k Dep. Var.
SS 3,791,194.066 Regression 7 5,320,448.600 Residual 0 9,111,642.666 Total 7 Regression output variables coefficients Intercept 3,056.2500 Time Index -0.5167 Jan -501.1833 Feb -398.6667 March 219.8500 April 227.3667 May 234.3833 June -91.8500 July -409.5833 Aug -508.8167 Sept -49.8000 Oct -333.5333 Nov -419.2667
df
MS
F
p-value
12
315,932.8389
2.08
.0459
35
152,012.8171
ANOVA table Source
48 12 Corvette Sales
47 std. error 232.0299 4.1945 279.5269 278.8652 278.2652 277.7273 277.2517 276.8389 276.4892 276.2027 275.9796 275.8202 275.7245
t (df=35) 13.172 -0.123 -1.793 -1.430 0.790 0.819 0.845 -0.332 -1.481 -1.842 -0.180 -1.209 -1.521
p-value 3.99E-15 .9027 .0816 .1617 .4348 .4185 .4036 .7420 .1475 .0739 .8578 .2347 .1373
confidence interval 95% lower 95% upper 2,585.2043 3,527.2957 -9.0320 7.9987 -1,068.6531 66.2865 -964.7932 167.4598 -345.0584 784.7584 -336.4496 791.1830 -328.4676 797.2343 -653.8629 470.1629 -970.8862 151.7195 -1,069.5379 51.9046 -610.0684 510.4684 -893.4781 226.4114 -979.0172 140.4838
c. Forecasts for 2008 are shown. These forecasts involve plugging the correct binary value (0, 1) for each month into the fitted regression equation. It is simpler than it seems at first glance, because all binaries are zero except the month being forecast. Period January Februar y March April May June
14.10
Forecast 2,529.8 2,631.8 3,249.8 3,256.8 3,263.3 2,936.5
Period July August September October November December
Learning Objective: 14-7 Learning Objective: 14-8 a. See graphs in part c below.
424
Forecast 2,618.3 2,518.5 2,977.0 2,692.8 2,606.5 3,025.3
ASBE 6e Solutions for Instructors b. We see a positive upward trend. JetBlue was a new airline in 2002. As air travel becomes more and more common, air travelers are looking for more options. JetBlue has been successful using a customer-focused strategy. c.
Object 25
d. The linear trend appears to be the better model for forecasting future revenue. The exponential model would appear to forecast greater revenue than the historical trend would indicate. e. t
Linear
Exponentia l
9 10 11
$6,807.86 $7,275.90 $7,743.94
$7,218.56 $7,988.92 $8,841.49
Learning Objective: 14-2 Learning Objective: 14-3
14.11
a. Dual scale graph is helpful, due to differing magnitudes. (After creating a line chart it is possible to create a secondary scale by clicking on one of the data series and choosing to format the data series. You can then choose a secondary axis.)
425
ASBE 6e Solutions for Instructors
Object 27
b. Electronic sales are much higher than mechanical but the trends for the two types of watches are the same. Sales have been fairly steady over the past six years with a dip in 2009. 2009 most likely shows a drop due to the recession. c. Electronic Watch Sales: = 21,668e−.013t, R2 = .105. Mechanical Watch Sales: = ŷ ŷ .055t 2 3320.6e , R = .581. d. Watches seem to be a stable product with an explainable dip in sales in 2009. The exponential trend would not be appropriate for level product sales. Either a moving average or exponential smoothing would actually make more sense. e. Mechanic t al Electronic 7 4,844.1 19,938.2 8 5,069.6 19,696.8 9 5,295.1 19,455.4 f. The recession in 2008 could have resulted in fewer sales of Swiss watches in 2009. Learning Objective: 14-2 Learning Objective: 14-3
426
ASBE 6e Solutions for Instructors
14.12
a.
Object 29
b. Trend is positive. There seems to be an increase in the rate of growth over the past few years. c. See graph above. Either the quadratic or the exponential trend will capture the increased growth but the linear trend lags behind the growth. d. t 11 12 13
Linear 158,608 159,192 159,776
Quadratic 161,301 163,354 165,653
Exponential 158,541 159,129 159,719
Learning Objective: 14-2 Learning Objective: 14-3 14.13
a. See graph in part c below. b. The trend appeared to be decreasing up until 2000. After 2000 there has been an increase in participation rate although it has leveled off over the past two or three elections. c.
427
ASBE 6e Solutions for Instructors
Object 31
d. The quadratic model will overestimate the voter participation while the exponential and linear will underestimate. Perhaps the best forecast would be to average the past two or three years. e. Quadrati Exponenti t Linear c al 16 51.4 58.8 51.7 Learning Objective: 14-2 Learning Objective: 14-3 14.14
a. b. c.
ŷ ŷ
=2286e0.076(7) = 3891.5 = 1149 + 12.78(7) = 1238.5 = 501 +18.2(7) – 7.1(72) = 280.5
ŷ Learning Objective: 14-3
428
ASBE 6e Solutions for Instructors
14.15
a.
Object 33
b. All three are increasing steadily. The slight dip in growth in 2008-2010 is most likely due to the 2008 recession. As the economy has recovered we see the growth resume. c. See graph in part a. d. Forecasts for 2017, 2018 and 2019 in $ billions. Linea Quadrat Exponenti t r ic al 18 3,625 3,790 3,770 19 3,738 3,957 3,937 20 3,850 4,130 4,111 Learning Objective: 14-2 Learning Objective: 14-3
429
ASBE 6e Solutions for Instructors
14.16
a.
Object 35
b. The graph shows low shipments between 1986-1996 with a strong increase from 19962000. More variability can be see between 2000-2015. Airplanes are capital goods, affected by business cycles (stronger demand expected during periods of growth), interest rates (credit availability for financing), and foreign demand (exchange rates, economic conditions abroad, foreign competition). c. There is no clear trend over the past 30 years so a fitted trend model would fail to capture the whole pattern of aircraft sales. d The 2 year moving average is shown by the dotted line in the graph in part a. It appears to fit the data well. Given the lack of real trend, a two year or three year moving average would be an acceptable forecasting method for this time series. A better method would be a causal model that incorporates business variables that an industry expert identifies. e. The two year moving average forecast for 2016 is 1,612. Learning Objective: 14-2 Learning Objective: 14-5 14.17 a. y12 = 372e−.041(12) = 227.4 b. y12 = 719 + 10(12) = 839 c. y12 = 1299 – 51(12) + 7(12)2 = 1695 Learning Ojective: 14-3
430
ASBE 6e Solutions for Instructors
14.18
a. See graphs below.
Object 37
Object 40
b. All four time series show an increasing trend. Federal taxes and spending are political variables that are affected by societal needs, the balance of political power in Congress, and leadership from the executive branch. Economic growth, inflation, business cycles, and economic policies underlie changes in the GDP and federal debt. c. See graphs in part a.
431
ASBE 6e Solutions for Instructors d. Outlays were growing faster than receipts leading to a rising deficit. Federal debt was growing faster than the GDP over this entire period.
e. The ratio has increased continuously since 2001. The largest increase in shown from 2007 to 2009.
Object 43
Learning Objective: 14-2 Learning Objective: 14-3
14.19
a. and b. Both have slight downward trends. The forecast for the women’s finish time in 2046 is 142.72 minutes. The forecast for men’s finish time in 2036 is 128.77 minutes. No the finish times for men and women are not likely to converge based on the historical information.
432
ASBE 6e Solutions for Instructors
Object 45
c. and d. Moving average gives a good fit. Moving averages work well when there is no obvious trend. The dotted line on each series shows a 3 period moving average.
Object 47
Learning Objective: 14-2 Learning Objective: 14-3 Learning Objective: 14-5 14.20
a.
433
ASBE 6e Solutions for Instructors
Object 49
b. A positive linear trend is seen. As the population ages there is higher demand for leisure activities which requires more workers in these fields. c. See graph in part a. d. t 12 13 14
Linear 15,515 15,767 16,018
Quadrati c 16,578 17,361 18,226
Learning Objective: 14-2 Learning Objective: 14-3 14.21
a.
434
Exponenti al 15,547 15,825 16,107
ASBE 6e Solutions for Instructors
Object 51
b. There is no apparent trend between 2006-2016, although 2016 seems to be a significant increase over 2015. c. A fitted trend is not helpful if no trend exists. d. When no trend exists a moving average forecast is appropriate. The forecast for 2017 using a 3 period moving average is 106 deaths. Learning Objective: 14-2 Learning Objective: 14-3 Learning Objective: 14-5
14.22
a. See graph in part c below. b. Decreasing trend in deaths due to lightning. c.
435
ASBE 6e Solutions for Instructors
Object 54
d. Based on the exponential model the forecast for 2020 would be approximately 25 deaths. Learning Objective: 14-2 Learning Objective: 14-3 14.23
a.
Object 56
b. There was an increasing trend from the 1999/2000 season until 2010/2011 season. The drop in 2011/2012 halted this increase in visits and the industry has not fully recovered. c. With no apparent trend in recent years the best forecasting method is a 3 period moving average. Forecast for the 2016/2017 season is 54.7 million visits. Learning Objective: 14-2 Learning Objective: 14-5 14.24
a. and b.
436
ASBE 6e Solutions for Instructors
Object 58
c. Women’s winning time forecast for 2040: -0.0629(27) + 11.85 = 10.15 seconds. Men’s winning time forecast for 2040: -0.0408(27) + 10.533 = 9.43 seconds. Yes, theoretically it seems the times might converge but it will be beyond 2040. See calculations below. d. Set 0.0629t + 11.85 = 0.0408t + 10.533 and solve for t. t = 59.6 (round up to 60). Time period 60 is the year 2172. The rate at which both men and women are improving their race times is slowing down. It will take many years for the times to converge if they continue this way. Learning Objective: 14-2 Learning Objective: 14-3
14.25
a.
437
ASBE 6e Solutions for Instructors
Object 60
b. We see an overall increasing trend in imports until 2005. c. All trend lines forecast an increase in imports in the short term. If the last ten years can be used to predict our future then there will be a continued decrease in imports. The quadratic model might be the best of the three. d. Quadrat Exponenti t Linear ic al 13 3,943 3,468 5,953 . Learning Objective: 14-2 Learning Objective: 14-3
438
ASBE 6e Solutions for Instructors 14.26
a.
2 period moving average Object 62
3 period moving average Object 64
4 period moving average Object 66
439
ASBE 6e Solutions for Instructors
5 period moving average Object 68
m-period Next period forecast 2 1.8350 3 1.8280 4 1.8244 5 1.8177 b. The 5-period moving average creates a smoother curve. Using this allows one to forecast the weekly or monthly trends without overreacting to daily fluctuations. c. Yes a moving average is appropriate when trying to forecast short time periods, one period ahead. Forecasts should account for trends and cycles without overreacting to noise (random variation.) Learning Objective: 14-5 14.27
a. See graphs. b. See graphs. The degree of smoothing varies dramatically as is increased. c. For this data = .20 seems to track the recent data well, yet provides enough smoothing to iron out the “blips.” It gives enough weight to recent data to bring its forecasts above 1.80 (lagging but reflecting the recent rise in rates). In contrast, = .05 or = .10 are not responsive to recent data (too much smoothing) so they give a forecast below 1.80. While = .50 gives a good “fit” it does not smooth the data very much. Forecasters generally use smaller values (the default of = .20 is a common choice).
440
ASBE 6e Solutions for Instructors d. Yes, based on graphs and fitted trendline. = .05
= .10
= .20
= .50
Learning Objective: 14-6 14.28
a.
Object 70
441
ASBE 6e Solutions for Instructors b. Yes. Seasonality is quite apparent. There are highs and lows, depending on the month. There is also an upward trend. The seasonal swings seem to increase in amplitude. c. Month Index Month Index Jan 1.820 Jul 0.679 Feb 1.882 Aug 0.252 Mar 1.416 Sep 0.333 Apr 1.057 Oct 0.649 May 0.574 Nov 1.324 Jun 0.287 Dec 1.725 It is important to note that 1.00 is the key reference point for a multiplicative seasonal index. The creative student might plot the seasonal indexes to show that Nov through Mar (the winter months) are above 1.00, Apr is near 1.00, and the summer and fall months are below 1.00. d. December is the highest, August is the lowest. This is logical, based on the seasons’ impact on heating degree-days (presumably the residence is in a northern climate). e. There is some upward trend. Logical, since generally, fossil fuels are getting more expensive. See graph below.
Object 73
f. We used December as our norm. January and February binaries are not significant meaning that these two months are not significantly different from December. All other binaries are significant and negative meaning that December, January, and February are the highest months.
Regression Analysis 442
ASBE 6e Solutions for Instructors R² Adjusted R² R Std. Error ANOVA table Source SS Regression 78,693.0817 Residual 13,244.5909 Total 91,937.6726 Regression output variables coefficients Intercept 110.6567 Time Index 0.7482 Jan -11.2099 Feb -8.9406 Mar -37.4263 Apr -56.6644 May -90.3026 Jun -108.6883 Jul -93.1165 Aug -111.1022 Sep -105.7554 Oct -75.9086 Nov -38.0693
0.856 0.807 0.925 19.453 df 12 35 47 std. error 11.5768 0.2093 13.9466 13.9136 13.8837 13.8568 13.8331 13.8125 13.7950 13.7808 13.7696 13.7617 13.7569
n k Dep. Var. MS 6,557.7568 378.4169
t (df=35) 9.558 3.575 -0.804 -0.643 -2.696 -4.089 -6.528 -7.869 -6.750 -8.062 -7.680 -5.516 -2.767
48 12 Bills F 17.33
p-value 2.73E-11 .0010 .4270 .5247 .0107 .0002 1.57E-07 3.00E-09 8.05E-08 1.72E-09 5.17E-09 3.36E-06 .0090
p-value 2.70E-11
confidence interval 95% lower 95% upper 87.1545 134.1589 0.3233 1.1731 -39.5230 17.1033 -37.1867 19.3056 -65.6116 -9.2409 -84.7953 -28.5336 -118.3853 -62.2200 -136.7292 -80.6475 -121.1220 -65.1111 -139.0786 -83.1258 -133.7092 -77.8016 -103.8463 -47.9709 -65.9973 -10.1413
Learning Objective: 14-7 Learning Objective: 14-8 14.29
a.
Object 75
b. Yes, it appears that delays increase from January to July and then decrease towards the end of each year. There is an overall increasing trend as well. c. From MegaStat: 443
ASBE 6e Solutions for Instructors Monthly Seasonal Index Jan 0.798 Feb 0.794 March 0.984 April 0.806 May 1.076 June 1.247 July 1.363 Aug 1.242 Sept 0.935 Oct 0.925 Nov 0.930 Dec 0.901
d.
It appears that the summer months have the most delays and the winter months have the fewest. Many more people travel over the summer than during the winter so this does make sense. e. Yes, there is a trend in the deseasonalized data. See graph above. Learning Objective: 14-7
14.30
a. 444
ASBE 6e Solutions for Instructors
Object 77
b. It is possible there is a seasonal pattern with quarters 2 and 4 having higher shipments than quarters 1 and 3.
c. See output and graph. Qtr1 Index 0.800
Qtr2 Index 0.979
Qtr3 Index 0.901
Qtr4 Index 1.320
Object 79
d. The seasonal indexes indicate below average shipments in the first and third quarters, close to average in the second quarters and above average shipments in the fourth 445
ASBE 6e Solutions for Instructors quarters. Reasons for higher shipments in fourth quarter could relate to tax incentives for getting product shipped by end of year. The aviation industry is most likely linked to trends around corporate purchases, recent business models that promote fractional shares of jets, and new airlines that are entering the market. e. Referring to the graph in part c above there is a slight increasing trend in the deseasonalized data. Learning Objective: 14-7 14.31
a.
Object 81
b. See output and graph. There is an upward trend in the deseasonalized data until 2008 followed by a decrease. This is logical, given consumers’ dependence on credit until the recession hit. Calculation of Seasonal Indexes 1 2 3 4 2005 2006 2007 2008 2009 2010 mean: adjusted :
5
6
1.014 1.010 1.017 1.022 1.010 1.015
0.999 0.994 1.002 1.000 0.991 0.997
0.985 0.985 0.991 0.985 0.979 0.985
0.989 0.985 0.992 0.986 0.978 0.986
0.994 0.990 0.992 0.986 0.985 0.989
0.990 0.991 0.996 0.989 0.990 0.991
1.014
0.997
0.985
0.986
0.989
0.991
446
7 0.990 0.989 0.991 1.001 0.997
8 0.997 0.997 0.999 1.010 1.004
9 0.997 0.996 0.997 1.010 1.003
10 0.995 0.995 1.001 1.009 1.005
11 1.008 1.010 1.013 1.018 1.011
12 1.033 1.033 1.035 1.037 1.031
0.994
1.001
1.001
1.001
1.012
1.034
0.993
1.001
1.000
1.001
1.011
1.033
ASBE 6e Solutions for Instructors
Object 84
c. Highest: August through January are the highest borrowing months. Yes this is logical. Credit increases due to the start of the new school year followed by the Christmas buying season. d. The difficulty of finding a forecasting model to describe this time period is that a recession hit our economy in 2008 resulting in a decrease in borrowing. The trend in borrowing will change when national events such a recession occur. Learning Objective: 14-7 14.32
a. From MegaStat, the seasonal indexes are shown below as well as the deseasonalized data. The seasonal influence is not strong but is consistent. The graph shows slightly higher than average sales in the second and third quarters and lower than average sales in the first and fourth quarters. Calculation of Seasonal Indexes 1 2 1005 1006 1007 1008 1009 1010 mean: adjusted :
0.890 0.910 0.928 0.932 0.942 0.920
1.084 1.099 1.129 1.074 1.032 1.083
0.923
1.087
447
3 1.045 1.053 1.043 1.054 1.032
4 0.957 0.927 0.952 0.909 0.952
1.045
0.940
1.048
0.942
ASBE 6e Solutions for Instructors
Object 86
b. Regression Analysis using quarterly binaries. Quarter 2 would be considered significantly different from quarter 4 if α = .05 (p-value = .0609 < .10). Regression Analysis R² Adjusted R² R Std. Error
0.805 0.764 0.897 628.372
SS 31,045,224.061 Regression 9 Residual 7,502,180.4381 38,547,404.500 Total 0 Regression output
df
ANOVA table Source
variables
coefficients
Intercept Time Index
5,227.3000 149.7643
QTR1
-440.0405
QTR2
726.5286
4 19
n k Dep. Var.
24 4 Revenue
MS 7,761,306.015 5 394,851.6020
F
p-value
19.66
1.53E-06
23
std. error 367.297 7 18.7762 367.137 7 364.729 2
confidence interval 95% 95% lower upper 5,996.06 4,458.5370 30 110.4652 189.0633
t (df=19)
p-value
14.232 7.976
1.38E-11 1.75E-07
-1.199
.2454
-1,208.4686
1.992
.0609
-36.8584
448
328.3876 1,489.91 55
ASBE 6e Solutions for Instructors
QTR3
333.0976
363.276 4
0.917
.3707
-427.2486
1,093.44 39
c. The predictions for 2011 are: Predicted values for: Revenue 95% Confidence Intervals
Time Index
QTR1
QTR2
QTR3
25
1
0
0
26
0
1
0
27
0
0
1
28
0
0
0
Predicte d 8,531.36 7 9,847.70 0 9,604.03 3 9,420.70 0
95% Prediction Intervals
lower
upper
7,762.604
9,300.130 10,616.46 3 10,372.79 6 10,189.46 3
9,078.937 8,835.270 8,651.937
lower 7,007.96 8 8,324.30 2 8,080.63 5 7,897.30 2
upper 10,054.76 5 11,371.09 8 11,127.43 2 10,944.09 8
Learning Objective: 14-7 Learning Objective: 14-8 14.33
a. Using December as the base month the output below shows that all months are significantly different from December at α = .01. November is significantly different from December at α = .05. The time index is not significant which indicates there is no decreasing or increasing trend to the data. This is also confirmed when looking at the line chart below the MegaStat output. From MegaStat: Regression Analysis R² Adjusted R² R Std. Error
0.754 0.704 0.869 525.514
n k Dep. Var.
SS 50,021,989.138 Regression 1 16,293,750.736 Residual 9 66,315,739.875 Total 0 Regression output variables coefficients Intercept 3,229.4250 Time Index 2.1685 Jan 1,523.5196 Feb 1,016.5179 Mar 1,536.6827 Apr 1,380.5143 May 1,816.1792
df
MS
F
p-value
12
4,168,499.0948
15.09
8.06E-14
59
276,165.2667
ANOVA table Source
72 12 Student Pilot Certificates
71 std. error 249.2733 3.0220 305.2214 304.9071 304.6224 304.3675 304.1424
t (df=59) 12.955 0.718 4.992 3.334 5.045 4.536 5.971
449
p-value 6.67E-19 .4759 5.62E-06 .0015 4.63E-06 2.87E-05 1.44E-07
confidence interval 95% lower 95% upper 2,730.6303 3,728.2197 -3.8785 8.2154 912.7730 2,134.2663 406.4002 1,626.6355 927.1347 2,146.2308 771.4764 1,989.5522 1,207.5917 2,424.7666
ASBE 6e Solutions for Instructors Jun Jul Aug Sep Oct Nov
2,464.5107 2,587.0089 3,194.6738 2,027.8387 1,498.8369 652.3351
303.9471 303.7818 303.6465 303.5412 303.4660 303.4209
8.108 8.516 10.521 6.681 4.939 2.150
3.59E-11 7.38E-12 3.71E-15 9.37E-09 6.80E-06 .0357
1,856.3139 1,979.1429 2,587.0785 1,420.4541 891.6028 45.1914
3,072.7075 3,194.8750 3,802.2691 2,635.2233 2,106.0710 1,259.4789
Object 88
b. Monthly forecasts for 2010: Mont h Jan Feb Mar Apr May Jun Jul Aug Sep Oct Nov Dec
Predict ed 4,911.2 4,406.4 4,928.7 4,774.7 5,212.6 5,863.1 5,987.7 6,597.6 5,432.9 4,906.1 4,061.7 3,411.6
Learning Objective: 14-8 14.34
yt = 213 + 11 t − 9 Qtr1 + 12 Qtr2 − 15 Qtr3 a. For t = 21 this corresponds to quarter 1. y = 213 + 11(21) – 9(1) + 12(0) – 15(0) = 435. b. For t = 8 this corresponds to quarter 4. y = 213 + 11(8) – 9(0) + 12(0) – 15(0) = 301. a. For t = 15 this corresponds to quarter 3. y = 213 + 11(15) – 9(0) + 12(0) – 15(1) = 363. Learning Objective: 14-8 450
ASBE 6e Solutions for Instructors 14.35
yt = 491 + 19 t + 29 Qtr1 − 18 Qtr2 + 12 Qtr3 a. For t = 14 this corresponds to quarter 2. y = 491 + 19(14) + 29(0) – 18(1) + 12(0) = 739. b. For t = 17 this corresponds to quarter 1. y = 491 + 19(17) + 29(1) – 18(0) + 12(0) = 843. a. For t = 20 this corresponds to quarter 4. y = 491 + 19(20) + 29(0) – 18(0) + 12(0) = 871. Learning Objective: 14-8
451
ASBE 6e Solutions for Instructors
Chapter 15 Chi-Square Tests 15.1
a. b. c. d. e. f. g.
H0: Choice of vehicle and Buyer’s age are independent. Degrees of Freedom = (r1)(c1) = (41)(31) = 6 CHISQ.INV.RT(.1,6) = 10.64 and test statistic = 10.667. Test statistic is 10.667 (p-value = .0992) so reject the null at = .10. Gasoline and Under 30 contributes the most. All expected frequencies exceed 5. p-value is close to level of significance but less so we would reject the null hypothesis.
Diesel
Gasolin e
Hybrid
Electric
Total
Observed Expected (O - E)² / E Observed Expected (O - E)² / E Observed Expected (O - E)² / E Observed Expected (O - E)² / E Observed Expected (O - E)² / E
Under 30 5 7.50 0.83
30 < 50 10 10.00 0.00
50 and Over 15 12.50 0.50
Total 30 30.00 1.33
15 22.50 2.50 25 18.75 2.08 15 11.25 1.25 60 60.00 6.67 10.67 6 .0992
30 30.00 0.00 25 25.00 0.00 15 15.00 0.00 80 80.00 0.00 chi-square df p-value
45 37.50 1.50 25 31.25 1.25 15 18.75 0.75 100 100.00 4.00
90 90.00 4.00 75 75.00 3.33 45 45.00 2.00 240 240.00 10.67
Learning Objective: 15-1 Learning Objective: 15-2 Learning Objective: 15-3 15.2
a. b. c. d. e. f. g.
H0: Age Group and Ownership are independent. Degrees of Freedom = (r1)(c1) = (21)(41) = 3 CHISQ.INV.RT(.01,3) = 11.34 and test statistic = 19.31. Because the p-value (.0002) is less than .01, we reject the null and find dependence. Adults and Europe and Adults and Latin America contribute the most. All expected frequencies exceed 5. The p-value from MegaStat shows that observed difference would arise by chance only 2 times in 10,000 samples if the two variables really were independent. 451
ASBE 6e Solutions for Instructors
Observed Expected (O - E)² / E Observed Expected (O - E)² / E Observed Expected (O - E)² / E
U.S. 80 75.75 0.24 20 24.25 0.74 100 100.00 0.98 19.31 3 .0002
Europe 89 75.75 2.32 11 24.25 7.24 100 100.00 9.56
Asia 69 75.75 0.60 31 24.25 1.88 100 100.00 2.48
Latin. America 65 75.75 1.53 35 24.25 4.77 100 100.00 6.29
Total 303 303.00 4.68 97 97.00 14.63 400 400.00 19.31
chi-square df p-value
Learning Objective: 15-1 Learning Objective: 15-2 Learning Objective: 15-3 15.3
a. b. c. d. e. f. g.
H0: Verbal and Quantitative are independent Degrees of Freedom = (r1) (c1) = (31)(31) = 4 CHISQ.INV.RT(.005,4) = 14.86 and test statistic = 55.88. Test statistic is 55.88 (p-value < .0001) so we reject the null at = .005. The upper left cell (Under 25 and Under 25) contributes the most. Expected frequency is less than 5 in two cells. p-value is nearly zero (observed difference not due to chance). Under Observed 25 Expected (O - E)² / E 25 toObserved 35 Expected (O - E)² / E 35 orObserved More Expected (O - E)² / E TotalObserved Expected (O - E)² / E
Under 25 25 10.50 20.02 4 15.00 8.07 1 4.50 2.72 30 30.00 30.81
25 to 35 9 14.00 1.79 28 20.00 3.20 3 6.00 1.50 40 40.00 6.49
55.88 4 .0000
chi-square df p-value
Learning Objective: 15-1 Learning Objective: 15-2 Learning Objective: 15-3 452
35 or More 1 10.50 8.60 18 15.00 0.60 11 4.50 9.39 30 30.00 18.58
Total 35 35.00 30.40 50 50.00 11.87 15 15.00 13.61 100 100.00 55.88
ASBE 6e Solutions for Instructors 15.4
a. b. c. d. e. f. g.
HDTVs Owned None
One
Two or More
Total
H0: HDTVs Owned and In Store Purchases are independent. Degrees of Freedom = (r1)(c1) = (31)(31) = 4 CHISQ.INV.RT(.1,4) = 7.779 and test statistic = 8.57. Because the p-value (.0729) is less than .10, we reject the null and find dependence. None and None contributes the most. All expected frequencies exceed 5. The p-value from MegaStat shows that observed difference would arise by chance only 72.9 times in 1000 samples if the two variables really were independent. In Store Purchases Observed Expected (O - E)² / E Observed Expected (O - E)² / E Observed Expected (O - E)² / E Observed Expected (O - E)² / E
None 13 8.00 3.13 18 16.00 0.25
One 14 14.40 0.01 31 28.80 0.17
19 26.00 1.88 50 50.00 5.26 8.57 4 .0729
45 46.80 0.07 90 90.00 0.25 chi-square df p-value
More Than One 13 17.60 1.20 31 35.20 0.50
Total 40 40.00 4.34 80 80.00 0.92
66 57.20 1.35 110 110.00 3.06
130 130.00 3.31 250 250.00 8.57
Learning Objective: 15-1 Learning Objective: 15-2 Learning Objective: 15-3 15.5
a. b. c. d. e. f. g.
H0: Completion Rate and Notification are independent. Degrees of Freedom = (r1)(c1) = (21)(21) = 1 CHISQ.INV.RT(.025,1) = 5.024 and test statistic = 5.42 (close decision). Since the p-value (.0199) is less than .025, we reject the null and find dependence. Completed and No, Completed and Yes contribute the most. All expected frequencies exceed 5. The p-value from MegaStat shows that observed difference would arise by chance only 20 times in 1000 samples if the two variables really were independent.
Observed
Completed 39
453
Not Completed 155
Total 194
ASBE 6e Solutions for Instructors Expected (O - E)² / E Observed Expected (O - E)² / E Observed Expected (O - E)² / E
30.66 2.27 22 30.34 2.29 61 61.00 4.56 5.42 1 .0199
163.34 0.43 170 161.66 0.43 325 325.00 0.86
194.00 2.70 192 192.00 2.72 386 386.00 5.42
chi-square df p-value
Hypothesis test for two independent proportions p1 0.6393 39/61 39 61
p2 0.4769 155/325 155 325 0.1624 0. 0.0698 2.33 .0199 5.419795 0.0302 0.2946 0.1322
pc 0.5026 194/386 194 386
p (as decimal) p (as fraction) X n
difference hypothesized difference std. error z p-value (two-tailed) z-squared confidence interval 95.% lower confidence interval 95.% upper half-width
h. In a two-tailed two-sample z test for H0: 1 2 = 0 we verify that z2 is the same as the chi-square statistic, as presented in the table above. The p-value is the same (.0199) and z2 = 5.42 (after rounding) is equal to the chi-square value from the previous table. Learning Objective: 15-1 Learning Objective: 15-2 Learning Objective: 15-3
15.6
H0: College students’ ownership of smartphones follows the national trend. H1: College students’ ownership of smartphones does not follow the national trend. 454
ASBE 6e Solutions for Instructors Choose α = .05. Smartphone Screen Size 4 inches or more 3.5 to 3.9 inches less than 3.5 inches
Observed National Trend Frequency Expected Observed Chi-Square Proportion (n = 148) Frequency Expected Contribution 0.24 30 35.52 -5.520 0.858 0.4 75 59.2 15.800 4.217 0.36
43
53.28
-10.280
1.983
χ2calc = 0.858 + 4.217 + 1.983 = 7.058, p-value =CHISQ.DIST.RT(7.058,2) = .0293. Because .0293 < .05 we reject H0 and conclude that college students’ ownership of smartphones does not follow the national trend. There are a greater percentage of college students who own smartphones with screens from 3.5-3.9 inches than the national percentage. Learning Objective: 15-4 15.7
H0: The current distribution of restaurant visits by type of restaurant is the same as it was during the last recession. H1: The current distribution of restaurant visits by type of restaurant is not the same as it was during the last recession. Choose α = .05.
Restaurant Type Fast Food Quick Casual Casual Dining Family Style Fine Dining
Proportio n of Visits during Recession 0.36 0.29 0.15 0.15 0.05
Observed Frequenc Expected Chi-Square y (n = Frequenc Observed Contributio 250) y Expected n 95 90 5.000 0.278 68 72.5 -4.500 0.279 38 37.5 0.500 0.007 30 37.5 -7.500 1.500 19 12.5 6.500 3.380
χ2calc = 0.278 + 0.279 + 0.007 + 1.500 + 3.380 = 5.444, p-value =CHISQ.DIST.RT(5.444,4) = .2447. Because .2447 > .05 we fail to reject H0 and conclude that restaurant visits by type of restaurant have not changed since the last survey. Learning Objective: 15-4
15.8
Most purchasers are 18-24 years of age. Fewest bought by those 65 and over. At = .01, this sample does contradict the assumption that readership is uniformly distributed among these six age groups, since the p-value is less than .01.
455
ASBE 6e Solutions for Instructors
Goodness of Fit Test observed 38 28 19 16 10 9 120 31.30 5 8.17E-06
expected 20.000 20.000 20.000 20.000 20.000 20.000 120.000
O-E 18.000 8.000 -1.000 -4.000 -10.000 -11.000 0.000
(O - E)² / E 16.200 3.200 0.050 0.800 5.000 6.050 31.300
% of chisq 51.76 10.22 0.16 2.56 15.97 19.33 100.00
chi-square df p-value
Learning Objective: 15-5
15.9
Vanilla and Mocha are the leading flavors. Coffee least favorite as measured by sales. However, at = .05, this sample does not contradict the assumption that sales are the same for each beverage, since the p-value (.8358) is greater than .05.
456
ASBE 6e Solutions for Instructors
Goodness of Fit Test observed 18 23 23 20 84 .86 3 .8358
expected 21.000 21.000 21.000 21.000 84.000
O-E -3.000 2.000 2.000 -1.000 0.000
(O - E)² / E 0.429 0.190 0.190 0.048 0.857
% of chisq 50.00 22.22 22.22 5.56 100.00
chi-square df p-value
Learning Objective: 15-5
15.10
Graph reveals that 0 and 8 occur the most frequently, while 1 has the smallest occurrence rate. At = .05, we cannot reject the hypothesis that the digits are from a uniform population, because the p-value (.5643) is greater than .05.
Goodness of Fit Test observed 33 17
expected 27.000 27.000
O-E 6.000 -10.000
457
(O - E)² / E 1.333 3.704
% of chisq 17.31 48.08
ASBE 6e Solutions for Instructors 25 30 31 28 24 25 32 25 270 7.70 9 .5643
27.000 27.000 27.000 27.000 27.000 27.000 27.000 27.000 270.000
-2.000 3.000 4.000 1.000 -3.000 -2.000 5.000 -2.000 0.000
0.148 0.333 0.593 0.037 0.333 0.148 0.926 0.148 7.704
1.92 4.33 7.69 0.48 4.33 1.92 12.02 1.92 100.00
chi-square df p-value
Learning Objective: 15-5 15.11
At = .05, we cannot reject the hypothesis that the movie goers are from a uniform population, because the p-value (.1247) is greater than .05 Goodness of Fit Test Age observed 10 < 20 5 20 < 30 6 30 < 40 10 40 < 50 3 50 < 60 14 60 < 70 9 70 < 80 9 56 10.00 6 .1247
expected 8.000 8.000 8.000 8.000 8.000 8.000 8.000 56.000
O-E -3.000 -2.000 2.000 -5.000 6.000 1.000 1.000 0.000
(O - E)² / E 1.125 0.500 0.500 3.125 4.500 0.125 0.125 10.000
% of chisq 11.25 5.00 5.00 31.25 45.00 1.25 1.25 100.00
chi-square df p-value
Learning Objective: 15-5 15.12
a. The sample mean and standard deviation are close to those used to generate the values. Mean Std Dev
Sample Generated 5.20 2.4075
5.00 2.2361
b. and c. See table below. Because the p-value (.7853) is greater than 0.05, we don’t reject the null hypothesis that the observations were coming from a Poisson distribution. Note that the end categories were collapsed so that their expected frequencies would be at least 5. A common error that students make is to fail to check that their probabilities sum to 1 and that the expected frequencies sum to n. If these sum to less than expected, it is an indication that they forgot the Poisson probabilities beyond the highest observed value (X = 12, 13, 14, ... etc.).
458
ASBE 6e Solutions for Instructors x 2 or less 3 4 5 6 7 8 or more
P(X=x) 0.12465 0.14037 0.17547 0.17547 0.14622 0.10444 0.13337 1.00000
observed 6 7 7 8 10 3 9 50
expected 6.2326 7.0187 8.7734 8.7734 7.3111 5.2222 6.6686 50.0000
O-E -0.2326 -0.0187 -1.7734 -0.7734 2.6889 -2.2222 2.3314 0.0000
(O - E)² / E 0.0087 0.0000 0.3585 0.0682 0.9889 0.9456 0.8151 3.1850
3.185chi-square 6df .7853p-value
d. If we use = 5.2 instead of = 5, the test statistic changes and we lose one degree of freedom (because is being estimated from the sample). However, in this case, the pvalue (.7610) is about the same, so we still fail to reject the hypothesis of a Poisson distribution. x 2 or less 3 4 5 6 7 8 or more
P(X=x) 0.10879 0.12928 0.16806 0.17479 0.15148 0.11253 0.15508 1.00000
observed 6 7 7 8 10 3 9 50
expected 5.4393 6.4639 8.4031 8.7393 7.5740 5.6264 7.7539 50.0000
O-E 0.5607 0.5361 -1.4031 -0.7393 2.4260 -2.6264 1.2461 0.0000
(O - E)² / E 0.0578 0.0445 0.2343 0.0625 0.7771 1.2260 0.2002 2.602
2.602chi-square 5df .7610p-value
Learning Objective: 15-6 15.13
Using sample mean = 4.948717949 the test statistic is 3.483 (p-value = .4805) with d.f. = 611 = 4. The critical value for = .05 is 9.488 so we cannot reject the hypothesis of a Poisson distribution. Note that the end categories were collapsed so that their expected frequencies would be at least 5. A common error that students make is to fail to check that their probabilities sum to 1 and that the expected frequencies sum to n. If these sum to less than expected, it is an indication that they forgot the Poisson probabilities beyond the highest observed value (X = 11, 12, 13, ... etc.). x 2 or less 3 4 5 6
P(X=x) 0.12904 0.14326 0.17724 0.17542 0.14468
Obs 3 5 9 10 5
Exp 5.032 5.587 6.912 6.841 5.643
459
OE -2.032 -0.587 2.088 3.159 -0.643
(OE)² / E 0.821 0.062 0.631 1.458 0.073
ASBE 6e Solutions for Instructors 7 or more
0.23036 1.00000
7 39
8.984 39.000
-1.984 0.000
0.438 3.483
Learning Objective: 15-6 15.14
At = .05 you cannot reject the hypothesis that truck arrivals per day follow a Poisson process, because the p-value (.2064) is greater than .05. For this test, we use the estimated sample mean = 2.6. Note that the top categories were collapsed so that their expected frequencies would be at least 5. A common error that students make is to fail to check that their probabilities sum to 1 and that the expected frequencies sum to n. If these sum to less than expected, it is an indication that they forgot the Poisson probabilities beyond the highest observed value (X = 8, 9, 10, ... etc.). x 0 1 2 3 4 5 or more Total
Days 4 23 28 22 8 15 100
P(X=x) Exp 0.07427 7.4274 0.19311 19.3111 0.25104 25.1045 0.21757 21.7572 0.14142 14.1422 0.12258 12.2577 1.00000 100.0000
Obs-Exp -3.4274 3.6889 2.8955 0.2428 -6.1422 2.7423 0.0000
Chi-Square 1.5816 0.7047 0.3340 0.0027 2.6677 0.6135 5.9041
df p-value
4 0.2064
Learning Objective: 15-6 15.15
From sample,
x
= 75.375 and s = 8.943376. Set ej = 40/8 = 5. Students might form
categories somewhat differently, so results may vary slightly depending on rounding. Using 8 classes with class limits to ensure equal expected frequencies (the “optimal expected frequencies” option) the test statistic is 6.000 (p-value = .3062) using d.f. = 821 = 5. The critical value for = .05 is 11.07 so we cannot reject the hypothesis of a normal distribution.
Score Under 65.09 65.09 < 69.34 69.34 < 72.53 72.53 < 75.38
Obs 5 3 5 3
Exp 5.000 5.000 5.000 5.000
460
ObsExp 0.000 -2.000 0.000 -2.000
Chi-Square 0.000 0.800 0.000 0.800
ASBE 6e Solutions for Instructors 75.38 < 78.22 78.22 < 81.41 81.41 < 85.66 85.66 or more Total
9 7 4 4 40
5.000 5.000 5.000 5.000 40.000
4.000 2.000 -1.000 -1.000 0.000
3.200 0.800 0.200 0.200 6.000
Learning Objective: 15-7 15.16
For this test, we use the estimated sample mean 31.1512 and standard deviation 9.890436. Set ej = 42/8 = 5.25. Students might form categories somewhat differently, so results may vary slightly depending on rounding. At = .025, you cannot reject the hypothesis that carry-out orders follow a normal population, since the p-value (.7074) is greater than .025. Cost of Order Under 19.77 19.77 < 24.48 24.48 < 28.00 28.00 < 31.15 31.15 < 34.30 34.30 < 37.82 37.82 < 42.53 42.53 or more Total
Obs 6 6 6 3 5 6 3 7
Exp 5.2500 5.2500 5.2500 5.2500 5.2500 5.2500 5.2500 5.2500
Obs-Exp 0.7500 0.7500 0.7500 -2.2500 -0.2500 0.7500 -2.2500 1.7500
Chi-Square 0.1070 0.1070 0.1070 0.9640 0.0120 0.1070 0.9640 0.5830
42
42.0000 d.f. p-value
0.0000 5 .7074
2.9520
Learning Objective: 15-7 15.17
The probability plot looks rather linear, yet the p-value (.033) for the Anderson-Darling test is less than = .05. This tends to contradict the chi-square test used in Exercise 15.13. However, the Kolmogorov-Smirnov test (DMax = .158) = does not reject normality (p > . 20). The data are a borderline case, having some characteristics of a normal distribution. If we have to choose one test, the A-D is the most powerful.
461
ASBE 6e Solutions for Instructors
Probability Plot of Exam Score Normal 99
Mean StDev N AD P-Value
95 90
75.38 8.943 40 0.811 0.033
Percent
80 70 60 50 40 30 20 10 5
1
50
60
70 80 Exam Score
90
100
Learning Objective: 15-8 The probability plot looks linear and the p-value (.404) for the Anderson-Darling test exceeds = .05. The Kolmogorov-Smirnov test (DMax = .085) = does not reject normality (p > .20). Therefore, we cannot reject the hypothesis of normality. Probability Plot of Cost of Order Normal 99
Mean StDev N AD P-Value
95 90 80
Percent
15.18
70 60 50 40 30 20 10 5
1
10
20
30 40 Cost of Order
462
50
60
31.15 9.890 42 0.373 0.404
ASBE 6e Solutions for Instructors
Learning Objective: 15-8 15.19
a. b. c. d.
H0: Pay Category and Job Satisfaction are independent. Degrees of Freedom = (r1)(c1) = (21)(31) = 2 CHISQ.INV.RT(.05,2) = 5.991 and test statistic = 9.69. Since the p-value (.0079) is less than .05, we reject the null and find that Pay Category and Job Satisfaction are not independent. e. Highlighted cells contribute the most — Satisfied and Salaried. f. No small expected frequencies g. The p-value from MegaStat shows that observed difference would arise by chance only 7.9 times in 1000 samples if the two variables really were independent.
Salarie d
Hourly
Total
Observed Expected (O - E)² / E Observed Expected (O - E)² / E Observed Expected (O - E)² / E
Satisfied
Neutral
Dissatisfied
Total
24 15.50 4.66 131 139.50 0.52 155 155.00 5.18 9.69 2 .0079
10 14.50 1.40 135 130.50 0.16 145 145.00 1.55 chi-square df p-value
2 6.00 2.67 58 54.00 0.30 60 60.00 2.96
36 36.00 8.72 324 324.00 0.97 360 360.00 9.69
Learning Objective: 15-1 Learning Objective: 15-2 Learning Objective: 15-3 463
ASBE 6e Solutions for Instructors
15.20
a. b. c. d.
H0: Credits Earned and Certainty of Major are independent. Degrees of Freedom = (r1)(c1) = (31)(31) = 4 CHISQ.INV.RT(.01,4) = 13.28 Since the p-value (.0443) is greater than .01, we fail to reject the null and conclude independence. e. Highlighted cell contributes the most (see table). f. No expected frequencies less than 5. g. The p-value from MegaStat shows that observed difference would arise by chance only 44.3 times in 1000 samples if the two variables really were independent.
0-9
10 59
60 or more
Total
Observed Expected (O - E)² / E
Very Uncertain 12 9.00 1.00
Somewhat Certain 8 6.00 0.67
Very Certain 4 9.00 2.78
Total 24 24.00 4.44
Observed Expected (O - E)² / E
8 7.50 0.03
2 5.00 1.80
10 7.50 0.83
20 20.00 2.67
Observed Expected (O - E)² / E Observed Expected (O - E)² / E
4 7.50 1.63 24 24.00 2.67
6 5.00 0.20 16 16.00 2.67 chi-square df p-value
10 7.50 0.83 24 24.00 4.44
20 20.00 2.67 64 64.00 9.78
9.78 4 .0443
Learning Objective: 15-1 Learning Objective: 15-2 Learning Objective: 15-3 15.21
a. b. c. d. e. f. g. h.
H0: Order Handed In and Grade are independent. Degrees of Freedom = (r1)(c1) = (21)(21) = 1 CHISQ.INV.RT(.1,1) = 2.706 and test statistic = 0.76. Because the p-value is greater than .10, we cannot reject the null and find independence. Highlighted cells contribute the most—see table and (O - E)2 / E. No small expected frequencies The p-value from MegaStat shows that observed difference would arise by chance 382 times in 1000 samples if the two variables really were independent, so the sample result is not convincing. See table below. The z2 does equal the chi-squared value and gives the same two-tailed p-value. 464
ASBE 6e Solutions for Instructors
“B” or better
“C” or worse
Total
Earlier Hand-In
Later Hand-In
Total
Observed Expected (O - E)² / E
11 9.50 0.24
8 9.50 0.24
19 19.00 0.47
Observed Expected (O - E)² / E Observed Expected (O - E)² / E
14 15.50 0.15 25 25.00 0.38 .76 1 .3821
17 15.50 0.15 25 25.00 0.38 chi-square df p-value
31 31.00 0.29 50 50.00 0.76
p1 0.44 11/ 25 11. 25
p2 0.32 8/25 8. 25 0.12 0. 0.137 3 0.87 .3821
pc 0.38 p (as decimal) 19/ 50 p (as fraction) 19. X 50 n difference hypothesized difference std. error z p-value (two-tailed)
Learning Objective: 15-1 Learning Objective: 15-2 Learning Objective: 15-3 15.22
a. b. c. d.
H0: Per person spending and Potsticker % of sales are independent. Degrees of Freedom = (r1)(c1) = (31)(31) = 4 CHISQ.INV.RT(.05,4) = 9.488 and test statistic = 25.24. Since the p-value (.0000) is less than .05, we reject the null and conclude the variables are dependent. e. Highlighted cells contribute the most — see table and (O - E)2 / E. f. There are no cells with expected frequency less than 5. g. The p-value from MegaStat shows that observed difference would arise by chance only 4.5 times in 100,000 samples if the two variables really were independent. From MegaStat: Chi-square Contingency Table Test for Independence
465
ASBE 6e Solutions for Instructors
Low
Medium
High
Total
Low 14 7.78 4.96 7 9.08 0.48 3 7.14 2.40 24 24.00 7.84
Observed Expected (O - E)² / E Observed Expected (O - E)² / E Observed Expected (O - E)² / E Observed Expected (O - E)² / E
Medium 7 8.43 0.24 15 9.84 2.71 4 7.73 1.80 26 26.00 4.75
High 3 7.78 2.94 6 9.08 1.05 15 7.14 8.67 24 24.00 12.65
Total 24 24.00 8.15 28 28.00 4.23 22 22.00 12.87 74 74.00 25.24
25.24 chi-square 4 df 4.49E-05 p-value
Conclusion: Based on the p-value of the Chi-Square statistic, it appears that per person spending is NOT independent of potsticker % of sales. Learning Objective: 15-1 Learning Objective: 15-2 Learning Objective: 15-3 15.23
a. b. c. d.
H0: Non-English Proficiency and Daily Newspaper Reading are independent. Degrees of Freedom = (r1)(c1) = (41)(31) = 6 CHISQ.INV.RT(.10,6) = 10.645 and test statistic = 4.14. Since the p-value (.6577) is greater than .10, we cannot reject the null and find independence. e. Yellow highlighted values show cells that contribute the most — see table and (O - E)2 / E. f. There are four cells that are highlighted in blue that have small expected frequency but they are all greater than 2. g. The p-value from MegaStat shows that observed difference would arise by chance only 66 times in 100 samples if the two variables really were independent.
From MegaStat:
Never
Occasionally
466
Regularly
Total
ASBE 6e Solutions for Instructors None
Observed Expected (O - E)² / E Slight Observed Expected (O - E)² / E Moderate Observed Expected (O - E)² / E Fluent Observed Expected (O - E)² / E Total Observed Expected (O - E)² / E
4 3.62 0.04 11 10.70 0.01 6 7.57 0.33 5 4.11 0.19 26 26.00 0.56
13 15.32 0.35 45 45.25 0.00 33 32.03 0.03 19 17.41 0.15 110 110.00 0.53
5 3.06 1.22 9 9.05 0.00 7 6.41 0.06 1 3.48 1.77 22 22.00 3.05
22 22.00 1.61 65 65.00 0.01 46 46.00 0.41 25 25.00 2.11 158 158.00 4.14
4.14 chi-square 6 df .6577 p-value
Conclusion: Based on the p-value, we cannot reject the null hypothesis that non-English proficiency and daily newspaper reading are independent of each other. Learning Objective: 15-1 Learning Objective: 15-2 Learning Objective: 15-3 15.24
a. b. c. d. e. f. g.
H0: Vehicle Type and Mall Location are independent. Degrees of Freedom = (r1)(c1) = (51)(41) = 12 CHISQ.INV.RT(.05,12) = 21.03 and test statistic = 24.53. Since the p-value (.0172) is less than .05, we can reject the null, we find dependence. Highlighted cells contribute the most—see table and (O - E)2 / E. Small expected frequencies in the full size van row. The p-value (.0172) from MegaStat shows that observed difference would arise by chance about 17 times in 1,000 samples if the two variables really were independent.
Car
Minivan
Observed Expected O-E (O - E)² / E Observed
Somerset 44 48.25 -4.25 0.37 21
Oakland 49 48.25 0.75 0.01 15
467
Great Lakes 36 48.25 -12.25 3.11 18
Jamestown 64 48.25 15.75 5.14 13
Total 193 193.00 0.00 8.64 67
ASBE 6e Solutions for Instructors
Full-size Van
SUV
Truck
Total
Expected O-E (O - E)² / E Observed Expected O-E (O - E)² / E Observed Expected O-E (O - E)² / E Observed Expected O-E (O - E)² / E Observed Expected O-E (O - E)² / E
16.75 4.25 1.08 2 2.50 -0.50 0.10 19 21.00 -2.00 0.19 14 11.50 2.50 0.54 100 100.00 0.00 2.29
16.75 -1.75 0.18 3 2.50 0.50 0.10 27 21.00 6.00 1.71 6 11.50 -5.50 2.63 100 100.00 0.00 4.64
16.75 1.25 0.09 3 2.50 0.50 0.10 26 21.00 5.00 1.19 17 11.50 5.50 2.63 100 100.00 0.00 7.12
16.75 -3.75 0.84 2 2.50 -0.50 0.10 12 21.00 -9.00 3.86 9 11.50 -2.50 0.54 100 100.00 0.00 10.48
67.00 0.00 2.19 10 10.00 0.00 0.40 84 84.00 0.00 6.95 46 46.00 0.00 6.35 400 400.00 0.00 24.53
24.53chi-square 12df .0172p-value
Learning Objective: 15-1 Learning Objective: 15-2 Learning Objective: 15-3 15.25
a. H0: Smoking and Race are independent. b. Degrees of Freedom = (r1)(c1) = (21)(21) = 1 c. CHISQ.INV.RT(.005,1) = 7.879 and test statistic = 5.84 (for males) and 14.79 (for females). d. For males, the p-value (.0157) is not less than .005, so we cannot reject the hypothesis of independence. However, for females, the p-value (.0001) is less than .005 so we conclude dependence. e. Highlighted cells contribute the most—see table and (O - E)2 / E. f. No small expected frequencies. g. The p-value for males is just within the chance level, while the female p-value indicates significance.
Observed Expected (O - E)² / E Observed
Males Smoker 145 136.00 0.60 15
468
Nonsmoker 280 289.00 0.28 60
Total 425 425.00 0.88 75
ASBE 6e Solutions for Instructors Expected (O - E)² / E Observed Expected (O - E)² / E
24.00 3.38 160 160.00 3.97
51.00 1.59 340 340.00 1.87
5.84 1 .0157
White
Black
Total
Observed Expected (O - E)² / E Observed Expected (O - E)² / E Observed Expected (O - E)² / E
75.00 4.96 500 500.00 5.84
chi-square df p-value
Females Smoker 116 102.09 1.90 7 20.91 9.25 123 123.00 11.15
Nonsmoker 299 312.91 0.62 78 64.09 3.02 377 377.00 3.64
Total 415 415.00 2.51 85 85.00 12.27 500 500.00 14.79
14.79chi-square 1df .0001p-value
Learning Objective: 15-1 Learning Objective: 15-2 Learning Objective: 15-3 15.26
a. b. c. d.
H0: Cockpit Noise Level and Flight Phase are independent. Degrees of Freedom = (r1)(c1) = (31)(31) = 4 CHISQ.INV.RT(.05,4) = 9.488 and test statistic = 15.16. Since the p-value (.0044) is less than .05, we can reject the null, i.e., we find dependence. e. Highlighted cells contribute the most—see table and (O - E)2 / E. f. Small expected frequencies in the Cruise column. g. The p-value (.0044) from MegaStat shows that observed difference would arise by chance about 44 times in 1,000 samples if the two variables really were independent.
Medium
Observed Expected (O - E)² / E Observed Expected (O - E)² / E
Climb 6 5.74 0.01 18 11.89 3.15
Cruise 2 1.84 0.01 3 3.80 0.17
469
Descent 6 6.43 0.03 8 13.31 2.12
Total 14 14.00 0.05 29 29.00 5.43
ASBE 6e Solutions for Instructors Observed Expected (O - E)² / E Observed Expected (O - E)² / E
1 7.38 5.51 25 25.00 8.67 15.16 4 .0044
3 2.36 0.17 8 8.00 0.36
14 8.26 3.98 28 28.00 6.13
18 18.00 9.67 61 61.00 15.16
chi-square df p-value
Learning Objective: 15-1 Learning Objective: 15-2 Learning Objective: 15-3 15.27
a. b. c. d.
H0: Actual Change and Forecasted Change are independent. Degrees of Freedom = (r1)(c1) = (21)(21) = 1 CHISQ.INV.RT(.10,1) = 2.706 and test statistic = 1.80. Since the p-value (.1792) exceeds .10, we cannot reject the null, i.e., we find independence. e. Highlighted cells contribute the most—see table and (O - E)2 / E. f. No small expected frequencies. g. The p-value (.1792) from MegaStat shows that observed difference would arise by chance about 18 times in 100 samples if the two variables really were independent. Observed Expected (O - E)² / E Observed Expected (O - E)² / E Observed Expected (O - E)² / E
Decline 7 8.94 0.42 9 7.06 0.53 16 16.00 0.96 1.80 1 .1792
15.28
Rise 12 10.06 0.37 6 7.94 0.47 18 18.00 0.85
Total 19 19.00 0.80 15 15.00 1.01 34 34.00 1.80
chi-square df p-value
Learning Objective: 15-1 Learning Objective: 15-2 Learning Objective: 15-3 a. H0: Smoking and Education Level are independent. b. Degrees of Freedom = (r1)(c1) = (41)(41) = 6 c. CHISQ.INV.RT(.005,6) = 18.55 and test statistic = 227.78. d. Since the p-value (less than .0001) is smaller than .005, we reject the null, i.e., we find dependence. e. First and fourth rows contribute the most—see table and (O - E)2 / E. f. No small expected frequencies. 470
ASBE 6e Solutions for Instructors g. The tiny p-value from MegaStat is highly significant. Point out to students that this is partly an artifact due to the huge sample size (i.e., in large samples, just about any deviation from independence would be significant). No Smoking 641 764.76 20.03 1370 1428.83 2.42 635 559.69 10.13 550 442.72 26.00 3196 3196.00 58.58
< HighObserved School Expected (O - E)² / E High School Observed Expected (O - E)² / E Some College Observed Expected (O - E)² / E College Observed Expected (O - E)² / E Total Observed Expected (O - E)² / E
227.78 6 2.28E-46
< 1/2 Pack 196 139.74 22.65 290 261.09 3.20 68 102.27 11.48 30 80.90 32.02 584 584.00 69.36
>= 1/2 Pack 196 128.50 35.46 270 240.08 3.73 53 94.04 17.91 18 74.39 42.74 537 537.00 99.84
Total 1033 1033.00 78.14 1930 1930.00 9.35 756 756.00 39.53 598 598.00 100.76 4317 4317.00 227.78
chi-square df p-value
Learning Objective: 15-1 Learning Objective: 15-2 Learning Objective: 15-3 15.29
a. H0: ROI and Sales Growth are independent. b. For 22 table: Degrees of Freedom = (r1)(c1) = (21)(21) = 1 For 33 table: Degrees of Freedom = (r1)(c1) = (31)(31) = 4 c. For 22 table: CHISQ.INV.RT(.05,1) = 3.841 and test statistic = 7.15. For 33 table: CHISQ.INV.RT(.05,4) = 9.488 and test statistic = 12.30. d. For 22 table: Conclude dependence since p-value = .0075 is smaller than .05. For 33 table: Conclude dependence since p-value = .0153 is smaller than .05. e. First column contributes the most—see table and (O - E)2 / E. f. No small expected frequencies. g. The tables agree. Both p-values are significant at = .05. 22 Cross-Tabulation of Companies Low High Observed 24 16 Expected 17.88 22.12 (O - E)² / E 2.09 1.69 Observed 14 31
471
Total 40 40.00 3.78 45
ASBE 6e Solutions for Instructors Expected (O - E)² / E Observed Expected (O - E)² / E
20.12 1.86 38 38.00 3.95 7.15 1 .0075
Medium
24.88 1.50 47 47.00 3.20 chi-square df p-value
33 Cross-Tabulation of Companies Low Medium Observed 9 12 Expected 5.27 12.52 (O - E)² / E 2.64 0.02 Observed 6 14 Expected 5.08 12.07 (O - E)² / E 0.17 0.31 Observed 1 12 Expected 5.65 13.41 (O - E)² / E 3.82 0.15 Observed 16 38 Expected 16.00 38.00 (O - E)² / E 6.63 0.48 12.30 4 .0153
45.00 3.36 85 85.00 7.15
High 7 10.21 1.01 7 9.85 0.82 17 10.94 3.36 31 31.00 5.19
Total 28 28.00 3.67 27 27.00 1.30 30 30.00 7.33 85 85.00 12.30
chi-square df p-value
Learning Objective: 15-1 Learning Objective: 15-2 Learning Objective: 15-3 15.30
a. b. c. d.
H0: Type of Cola Drinker and Correct Response are independent. Degrees of Freedom = (r1)(c1) = (21)(21) = 1 CHISQ.INV.RT(.05,1) = 3.841 and test statistic = 1.50. Because the p-value (.2207) exceeds .05, we cannot reject the null, i.e., we find independence. e. Highlighted cells contribute the most—see table and (O - E)2 / E. f. No small expected frequencies. g. The p-value shows that observed difference would arise by chance about 22 times in 100 samples if the two variables really were independent. We get the same p-value result using a two-tailed test of two proportions, and z2 = (-1.22)2 = 1.50—the same as the chi-square test statistic (except for rounding). Regular
472
Diet Cola
Total
ASBE 6e Solutions for Instructors
Yes
No
Total
Cola 6 8.00 0.50 18 16.00 0.25 24 24.00 0.75 1.50 1 .2207
Observed Expected (O - E)² / E Observed Expected (O - E)² / E Observed Expected (O - E)² / E
10 8.00 0.50 14 16.00 0.25 24 24.00 0.75 chi-square df p-value
16 16.00 1.00 32 32.00 0.50 48 48.00 1.50
Hypothesis test for two independent proportions p1
p2
0.25 6/24 6. 24
0.4167 10/24 10. 24 -0.1667 0. 0.1361 -1.22 .2207
pc 0.333 3 p (as decimal) 16/48 p (as fraction) 16. X 48 n difference hypothesized difference std. error z p-value (two-tailed)
Learning Objective: 15-1 Learning Objective: 15-2 Learning Objective: 15-3 15.31
a. b. c. d.
H0: Student Category and Reason for Choosing are independent. Degrees of Freedom = (r1)(c1) = (31)(31) = 4 CHISQ.INV.RT(.01,4) = 13.28 and test statistic = 66.40. Because the p-value (less than .0001) is less than .01, we reject the null, i.e., we find dependence. e. Highlighted cell shows highest contribution—see table and (O - E)2 / E. f. No small expected frequencies. g. Extremely small p-value indicates that the variables are not independent.
Freshme n
Transfers
Observed Expected (O - E)² / E Observed
Tuition
Location
Reputatio n
Total
51 30.63 13.56 16
32 35.00 0.26 31
36 53.38 5.66 21
119 119.00 19.47 68
473
ASBE 6e Solutions for Instructors
MBAs
Total
Expected (O - E)² / E Observed Expected (O - E)² / E Observed Expected (O - E)² / E
17.50 0.13 3 21.88 16.29 70 70.00 29.97 66.40 4 1.31E-13
20.00 6.05 17 25.00 2.56 80 80.00 8.87 chisquare df p-value
30.50 2.96 65 38.13 18.94 122 122.00 27.56
68.00 9.14 85 85.00 37.79 272 272.00 66.40
Learning Objective: 15-1 Learning Objective: 15-2 Learning Objective: 15-3 15.32
H0: Political Orientation and Age of Car Usually Driven are independent. From MegaStat: Chi-square Contingency Table Test for Independence
Liberal
Middle-of-Road
conservative
Total
Observed Expected (O - E)² / E Observed Expected (O - E)² / E Observed Expected (O - E)² / E Observed Expected (O - E)² / E
Under 3 19 15.83 0.63 33 33.10 0.00 16 19.07 0.49 68 68.00 1.13
6-Mar 7 or More 12 13 15.60 12.57 0.83 0.01 31 28 32.61 26.29 0.08 0.11 24 13 18.79 15.14 1.45 0.30 67 54 67.00 54.00 2.36 0.43
Total 44 44.00 1.48 92 92.00 0.19 53 53.00 2.24 189 189.00 3.91
3.91 chi-square 4 df .4178 p-value
Conclusion: Based on the p-value we cannot conclude that political orientation and age of car usually driven are not independent. Learning Objective: 15-1 Learning Objective: 15-2 Learning Objective: 15-3 15.33
H0: The 2016 distribution of car colors is the same as it was during 2006. H1: The 2016 distribution of car colors is not the same as it was during 2006. Choose α = .05.
474
ASBE 6e Solutions for Instructors
Car Color Silver/Gray Blue white Tan/Beige Black Green Red Other
2006 Percentage 0.22 0.12 0.16 0.11 0.15 0.05 0.13 0.06
2016 Sample Frequencies Expected Observed (n = 200) Frequency Expected 70 44 16 24 32 32 12 22 42 30 6 10 20 26 2 12
Chi-Square Contribution 26 -8 0 -10 12 -4 -6 -10
39.71 6.89 0.00 11.75 12.41 4.14 3.58 21.54
χ2calc = 38.69, p-value =CHISQ.DIST.RT(38.69,7) = .000. Because .000 < .05 we reject H0 and conclude that the distribution of car colors has changed since 2006. It appears that more people prefer silver/grey or black cars more than six years ago. Learning Objective: 15-2 15.34
At = .05, this sample does not contradict the assumption that the 50 answers are uniformly distributed since the p-value (.6268) is greater than .05. No parameters are estimated, so d.f. = c1m = 510 = 4. observed 8 8 9 11 14 50 2.60 4 .6268
expected 10.000 10.000 10.000 10.000 10.000 50.000
O-E -2.000 -2.000 -1.000 1.000 4.000 0.000
(O - E)² / E 0.400 0.400 0.100 0.100 1.600 2.600
% of chisq 15.38 15.38 3.85 3.85 61.54 100.00
chi-square df p-value
Learning Objective: 15-4 15.35
To obtain expected values, multiply the U.S. proportions by 50. At = .05, Oxnard employees do not differ significantly from the national distribution, since the p-value (.1095) exceeds .05. No parameters are estimated, so d.f. = c1m = 410 = 3. A common error that students may make is to treat percentages as if they were frequencies (i.e., to convert the Oxnard frequencies to percentages). Doing so is a serious error because it doubles the sample size. observed 4 20 15
expected 8.250 22.900 12.200
O-E -4.250 -2.900 2.800
475
(O - E)² / E 2.189 0.367 0.643
% of chisq 36.22 6.08 10.63
ASBE 6e Solutions for Instructors 11 50
6.650 50.000
4.350 0.000
2.845 6.045
47.07 100.00
6.045 chi-square 3 df .1095 p-value
Learning Objective: 15-4 15.36
At = .01, you cannot reject the hypothesis that the digits are from a uniform population since the p-value (.6570) is greater than .01. There are 356 occurrences since 894 = 356. No parameters are estimated, so d.f. = c1m = 1010 = 9. observed 39 27 35 39 35 35 27 42 36 41 356 6.81 9 .6570
expected 35.600 35.600 35.600 35.600 35.600 35.600 35.600 35.600 35.600 35.600 356.000
O-E 3.400 -8.600 -0.600 3.400 -0.600 -0.600 -8.600 6.400 0.400 5.400 0.000
(O - E)² / E 0.325 2.078 0.010 0.325 0.010 0.010 2.078 1.151 0.004 0.819 6.809
% of chisq 4.77 30.51 0.15 4.77 0.15 0.15 30.51 16.90 0.07 12.03 100.00
chi-square df p-value
Learning Objective: 15-5 15.37
At = .10, you cannot reject the hypothesis that the die is fair, since the p-value (.4934) is greater than .10. No parameters are estimated, so d.f. = c1m = 610 = 5. observed 7 14 9 13 7 10 60
expected 10.000 10.000 10.000 10.000 10.000 10.000 60.000
O-E -3.000 4.000 -1.000 3.000 -3.000 0.000 0.000
4.40chi-square 5df .4934p-value
Learning Objective: 15-4 476
(O - E)² / E 0.900 1.600 0.100 0.900 0.900 0.000 4.400
% of chisq 20.45 36.36 2.27 20.45 20.45 0.00 100.00
ASBE 6e Solutions for Instructors
15.38
At = .025, you cannot reject the hypothesis that goals per game follow a Poisson process, since the p-value (.9293) is greater than .025. One parameter is estimated, so d.f. = c1m = 711 = 5. A common error that students may make is to fail to define the top category as open ended (X = 6, 7, 8, ...) so that the last entry in the P(X=x) column actually is P(X 6) = 1P(X 5). If this error is made, the probabilities will sum to less than 1 and the expected frequencies will sum to less than 232. Another common mistake is not combining end categories to enlarge expected frequencies (e.g., Cochran’s rule requires ej 5). Goals 0 1 2 3 4 5 6 or more Total games Total goals Mean goals/game
fj 19 49 60 47 32 18 7 232 575 2.47844828
P(X=x) 0.08387 0.20788 0.25760 0.21282 0.13187 0.06536 0.04060 1.00000
ej 19.4586 48.2271 59.7642 49.3742 30.5928 15.1646 9.4185 232
(fj-ej)2 0.21031 0.59733 0.05559 5.63674 1.98010 8.03976 5.84900
fj-ej -0.4586 0.7729 0.2358 -2.3742 1.4072 2.8354 -2.4185 0.0000
df p-value
(fj-ej)2/ej 0.01081 0.01239 0.00093 0.11416 0.06472 0.53017 0.62101 1.35419
5 0.92926
Learning Objective: 15-6 15.39
Estimated mean is = 1.06666667. For d.f. = c1m = 411 = 2 the critical value is CHISQ.INV.RT(.025,2) = 7.378, test statistic is 4.947 (p= .0843) so we can’t reject the hypothesis of a Poisson distribution A common error that students may make is to fail to define the top category as open ended (X = 3, 4, 5, ...) so that the last entry in the P(X=x) column actually is P(X 3) = 1P(X 2). If this error is made, the probabilities will sum to less than 1 and the expected frequencies will sum to less than 60. Another common mistake is not combining the top categories to enlarge expected frequencies (e.g., Cochran’s rule requires ej 5). x 0 1 2 3 or more Total
fj 25 18 8 9 60
P(X=x)
ej
fj-ej
0.344154 0.367097 0.195785 0.092964 1.000000
20.64923 22.02584 11.74712 5.57781 60.00000
4.35077 -4.02584 -3.74712 3.42219 0.00000
(fj-ej)2/ej 0.917 0.736 1.195 2.100 4.947
Learning Objective: 15-5 15.40
Results may vary, depending on which software package was used, how the categories were defined, and which options were selected (e.g., equal expected frequencies versus 477
ASBE 6e Solutions for Instructors equal class widths). For the chi-square test, we use d.f. = c3 because two parameters are estimated,. i.e. c1m = c12 = c3. Note that the chi-square test’s p-value may not agree with the A-D test’s p-value. Point out to students that the chi-square test is based on grouped frequencies, whereas the A-D test is based on individual data values, and hence they may disagree. The A-D test is more powerful, but its methods are less intuitive for most students. Data Set A Kentucky Derby Winning Times, 1950-2016
The histogram shows a right-skewed distribution with a single peak. Kentucky Derby Winning Times, 1950-2016 45 40 35 30 25 20 Percent 15 10 5 0
Time
MegaStat Normal GOF test results: χ2 = 4.358, p-value = CHISQ.DIST.RT(4.358, 4) = .3597. Because .3597 > .01 we would conclude that the data is normally distributed.
From Minitab:
478
ASBE 6e Solutions for Instructors
Note that the AD statistic is 0.456 and the p-value = .259. Because the p-value > .01 this sample does not provide evidence against the normal distribution. Learning Objective: 15-6 Data Set B National League Runs Scored Leader, 1900-2016
The histogram shows single peaked, symmetrical distribution. This looks like a normal distribution. National League Runs Scored by Leader (1900-2016) 35 30 25 20 15 Percent 10 5 0
Runs
MegaStat Normal GOF test results: χ2 = 1.359, p-value =CHISQ.DIST.RT(1.359,5) = .9287. Because .9287 > .01 this sample does not provide evidence against the normal distribution. From Minitab: 479
ASBE 6e Solutions for Instructors
Note that the AD statistic is 0.323 and the p-value = .523. Because .523 > .01 this sample does not provide evidence against the normal distribution. Learning Objective: 15-6 Data Set C Weights (in grams) of Pieces of Halloween Candy (n = 78) Weight (gm) Under 1.120 1.120 < 1.269 1.269 < 1.385 1.385 < 1.492 1.492 < 1.607 1.607 < 1.757 1.757 or more Total Parameters from
Obs 9 9 14 13 8 17 8 78
Exp 11.14 11.14 11.14 11.14 11.14 11.14 11.14 78
sample
Obs-Exp -2.14 -2.14 2.86 1.86 -3.14 5.86 -3.14 0
d.f. = 4
Chi-Square 0.412 0.412 0.733 0.31 0.886 3.079 0.886 6.718 p < 0.152
χ2 = 6.718, p-value =CHISQ.DIST.RT(6.718,4) = .152. Because .152 > .01 this sample does not provide evidence against the normal distribution.
From Minitab: 480
ASBE 6e Solutions for Instructors
Probability Plot of Candy Wt (gm)
Histogram of Candy Wt (gm) Normal
Normal Mean StDev N
14
99.9
1.438 0.2985 78
12
95 90
Percent
10
Frequency
Mean StDev N AD P-Value
99
8 6
1.438 0.2985 78 0.555 0.148
80 70 60 50 40 30 20 10
4
5
2
1
0
0.1
0.6
0.9
1.2 1.5 Candy Wt (gm)
1.8
2.1
0.5
1.0
1.5 Candy Wt (gm)
2.0
2.5
Note that the AD statistic is 0.555 and the p-value = .148. Because .555 > .01 this sample does not provide evidence against the normal distribution. Learning Objective: 15-6 Data Set D Price-Earnings Ratios for Specialty Retailers (n = 58) PE Ratio Under 10.35 10.35 < 15.19 15.19 < 18.92 18.92 < 22.39 22.39 < 26.12 26.12 < 30.96 30.96 or more Total Parameters from
Obs 4 11 13 13 6 6 5 58
Exp 8.29 8.29 8.29 8.29 8.29 8.29 8.29 58
sample
Obs-Exp -4.29 2.71 4.71 4.71 -2.29 -2.29 -3.29 0
d.f. = 4
2
Chi-Square 2.217 0.889 2.682 2.682 0.631 0.631 1.303 11.034
p < 0.026
χ = 11.034, p-value =CHISQ.DIST.RT(11.034,4) = .026. Because .026 < .05 this sample does provide evidence against the normal distribution at the .05 level of significance. If we used level of significance = .01 then we would fail to reject H0 and conclude that we have a normal distribution
. From Minitab:
481
ASBE 6e Solutions for Instructors Probability Plot of PE Ratio
Histogram of PE Ratio Normal
Normal
30
Mean StDev N
25
99
20.66 9.651 58
Mean StDev N AD P-Value
95 90 80
Percent
20
Frequency
20.66 9.651 58 2.307 <0.005
15 10
70 60 50 40 30 20 10
5 0
5
0
16
32 PE Ratio
48
1
64
0
10
20
30
40 PE Ratio
50
60
70
80
Note that the AD statistic is 0.2.307 and the p-value = .005. Because .005 < .01 this sample does provide evidence against the normal distribution. Learning Objective: 15-6 Data Set E U.S. Presidents’ Ages at Inauguration (n = 45)
The histogram shows a slight right-skewed distribution. US President Ages at Inauguration 35 30 25 20 15 10 5 Percent 0
Age
MegaStat Normal GOF test results: χ2 = 2.067, p-value =CHISQ.DIST.RT(2.067,3) = .5587. Because .5587 > .01 this sample does not provide evidence against the normal distribution. From Minitab:
482
ASBE 6e Solutions for Instructors
Note that the AD statistic is 0.379 and the p-value = .392. Because .392 > .01 this sample does not provide evidence against the normal distribution. Learning Objective: 15-6 Data Set F Weights of 31 Randomly-Chosen Circulated Nickels (n = 31) Weight (gm) Under 4.908 4.908 < 4.943 4.943 < 4.972 4.972 < 5.000 5.000 < 5.036 5.036 or more Total Parameters from
Obs 4 4 6 4 8 5 31
Exp 5.17 5.17 5.17 5.17 5.17 5.17 31
sample
Obs-Exp -1.17 -1.17 0.83 -1.17 2.83 -0.17 0
d.f. = 3
2
Chi-Square 0.263 0.263 0.134 0.263 1.554 0.005 2.484
p < 0.478
χ = 2.484, p-value =CHISQ.DIST.RT(2.484,3) = .478. Because .478 > .01 this sample does not provide evidence against the normal distribution.
483
ASBE 6e Solutions for Instructors From Minitab: Probability Plot of Nickel Wt (gm)
Histogram of Nickel Wt (gm) Normal
Normal Mean StDev N
9 8
99
4.972 0.06623 31
90
4.972 0.06623 31 0.881 0.021
80
6 Percent
Frequency
7
5 4
70 60 50 40 30
3
20
2
10 5
1 0
Mean StDev N AD P-Value
95
4.80
4.88
4.96 Nickel Wt (gm)
5.04
1
5.12
4.80
4.85
4.90
4.95 5.00 Nickel Wt (gm)
5.05
5.10
5.15
Note that the AD statistic is 0.881 and the p-value = .021. Because .021 > .01 this sample does not provide evidence against the normal distribution. However, the histogram shows a highly skewed left distribution and the p-value would indicate a non-normal distribution at α = .05. The small sample size could be a problem. Learning Objective: 15-6 15.41
Results will vary, but should be close to the intended distribution in a chi-square test or ztest for the intended mean. However, students often are surprised at how much a “normal” sample can differ from a perfect bell shape, and even the mean and standard deviation may not be “on target.” That is the nature of random sampling, as we learned in Chapters 8 and 9. The following sample was created using Excel’s function =NORM.INV(RAND(),0,1). The histogram is reasonably bell shaped. The test statistic for zero mean is z = (10)(0.031808) = 0.3181 (p-value = .7504) so the mean is not significantly different from 0. The standard deviation is a little larger than 1.
Descriptive statistics count mean sample std dev minimum
100 0.031808 1.117752 -2.35182753
484
ASBE 6e Solutions for Instructors maximum 1st quartile median 3rd quartile Classes Under -1.641 -1.641 < -0.930 -0.930 < -0.218 -0.218 < 0.493 0.493 < 1.204 1.204 < 1.915 1.915 or more Total
3.337159939 -0.672417 -0.005821 0.824195 Obs 8 10 24 23 21 10 4 100
Exp 6.73 12.76 21.66 24.85 19.29 10.12 4.6 100
Obs-Exp Chi-Square 1.27 0.24 -2.76 0.596 2.34 0.254 -1.85 0.138 1.71 0.152 -0.12 0.001 -0.6 0.079 0 1.461 d.f. = 4 p-value < 0.834
Learning Objective: 15-6 15.42
Results will vary, but should be close to the intended distribution in a chi-square test or test for the desired mean. However, students often are surprised at how much a “uniform” sample can differ from a perfect uniform shape, and even the mean and standard deviation may not be “on target.” That is the nature of random sampling, as we learned in Chapters 8 and 9. The following sample was created using Excel’s function =RAND(). The histogram does not look quite uniform. The test statistic for = 0.50000 is z = (0.5334370.500000)/.0288675 = 1.1583 (p-value = .2467) so the mean is not significantly different from 0.5000. The standard deviation (.29142) is a little larger than .28868.
Descriptive statistics count 100 mean 0.533437 sample variance 0.084926 sample standard deviation 0.291421 minimum 0.017795
485
ASBE 6e Solutions for Instructors maximum 1st quartile median 3rd quartile Classes Under 0.1404 0.1404 < 0.2630 0.2630 < 0.3856 0.3856 < 0.5081 0.5081 < 0.6307 0.6307 < 0.7533 0.7533 < 0.8759 0.8759 or more Total
Parameters from
0.998479 0.275343 0.620489 0.772453
Obs 12 10 14 9 6 23 12 14 100
Exp 12.5 12.5 12.5 12.5 12.5 12.5 12.5 12.5 100
user d.f. = 7
p-value < 0.050
Obs-Exp -0.5 -2.5 1.5 -3.5 -6.5 10.5 -0.5 1.5 0
ChiSquare 0.02 0.5 0.18 0.98 3.38 8.82 0.02 0.18 14.08
Learning Objective: 15-4 15.43
Results will vary, but should be close to the intended distribution in a chi-square test or ztest for the intended mean. However, students often are surprised at how much a “Poisson” sample can differ from the expected shape, and even the mean and standard deviation may not be “on target.” That is the nature of random sampling, as we learned in Chapters 8 and 9. The following sample was created using Excel’s Tools > Data Analysis > Random Number Generation with a mean of = 4. The histogram looks fine. The test statistic for = 4 is z =(5)(4.17)20 = 0.85 (p-value = .3953) so the mean doe not differ significantly from 4. The standard deviation (2.09) is very close to 2 (the square root of = 4
Descriptive Statistics count mean sample standard deviation minimum
486
100 4.17 2.09 1
ASBE 6e Solutions for Instructors maximum 1st quartile median 3rd quartile interquartile range mode X Values 1 or less 2 3 4 5 6 7 8 9 or more Total
Obs 9 11 20 24 14 8 7 3 4 100
11 3.00 4.00 5.00 2.00 4.00 Exp 9.16 14.65 19.54 19.54 15.63 10.42 5.95 2.98 2.13 100
Obs-Exp Chi-Square -0.16 0.003 -3.65 0.91 0.46 0.011 4.46 1.02 -1.63 0.17 -2.42 0.562 1.05 0.184 0.02 0 1.87 1.631 0 4.49 d.f. = 7 p-value < 0.722
Learning Objective: 15-5 15.44
Assuming a Poisson distribution the sample mean is 58/117 or 0.50. The table showing observed frequencies and expected frequencies along with the Chi-square statistic is below. Because the p-value (.8218) is greater than 0.10, we fail to reject the null hypothesis that the number of justices appointed in one year follows a Poisson distribution. Note that the end categories were collapsed so that their expected frequencies would be at least 5. A common error that students make is to fail to check that their probabilities sum to 1 and that the expected frequencies sum to n. If these sum to less than expected, it is an indication that they forgot the Poisson probabilities beyond the highest observed value (X = 3, 4, 5 ... etc.). x
P(X=x)
observed
expected
O-E
(O - E)² / E
0
0.60653066
72
70.964087
1.036
0.0151
1 2 or more
0.30326533
35
35.482044
-0.482
0.0065
0.09020401
10
10.553869
-0.554
0.02907
1
117 0.05074 1 0.8218
117 chi-square df p-value
0
0.05074
Learning Objective: 15-6
487
ASBE 6e Solutions for Instructors
Chapter 16 Nonparametric Tests 16.1
H0: Events follow a random pattern H1: Events do not follow a random pattern Because the p-value (.8942) is greater than .05 we fail to reject the null hypothesis of randomness. Runs Test for Random Sequence n 12 15 27 14.333 2.515 -0.133 .8942
runs 7 7 14
A B total
expected value standard deviation z test statistic p-value (two-tailed)
Learning Objective: 16-2 16.2
H0: Events follow a random pattern H1: Events do not follow a random pattern Because the p-value (.7741) is greater than .10 we fail to reject the null hypothesis of randomness. Runs Test for Random Sequence n 10 14 24 12.667 2.326 -0.287 .7741 Learning Objective: 16-2
16.3
runs 6 6 12
X O total
expected value standard deviation z test statistic p-value (two-tailed)
H0: Events follow a random pattern H1: Events do not follow a random pattern Because the p-value (.4858) is greater than .05 we fail to reject the null hypothesis of randomness.
488
ASBE 6e Solutions for Instructors
Runs Test for Random Sequence n runs 14 8 T 11 7 F 25 15 total 13.320 2.411 0.697 .4858
expected value standard deviation z test statistic p-value (two-tailed)
Learning Objective: 16-2 16.4
H0: Events follow a random pattern H1: Events do not follow a random pattern Because the p-value (.1533) is greater than .01 we fail to reject the null hypothesis of randomness. Runs Test for Random Sequence n runs 21 10 N 12 10 H 33 20 total 16.273 2.610 1.428 .1533
expected value standard deviation z test statistic p-value (two-tailed)
Learning Objective: 16-2 16.5
a. At = .10, the population median does not differ from 50 (p-value = .4732). The worksheet and test statistic calculation are shown. Student 1 2 3 4 5 6 7 8 9 10 11 12 13
xi 74 5 87 26 60 99 37 45 7 78 70 84 97
xi-50 24 -45 37 -24 10 49 -13 -5 -43 28 20 34 47
| xi-50 | Rank 24 14.5 45 24 37 20 24 14.5 10 5.5 49 27 13 9 5 3 43 22 28 17 20 13 34 19 47 25
489
R+ 14.5
R24
20 14.5 5.5 27 9 3 22 17 13 19 25
ASBE 6e Solutions for Instructors 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28
93 54 24 62 32 60 66 2 43 62 7 100 64 17 48
43 4 -26 12 -18 10 16 -48 -7 12 -43 50 14 -33 -2
43 4 26 12 18 10 16 48 7 12 43 50 14 33 2
22 2 16 7.5 12 5.5 11 26 4 7.5 22 28 10 18 1 406
22 2 16 7.5 12 5.5 11 26 4 7.5 22 28 10
234.5
18 1 171.5
Test Statistic: Wz=
n ( n + 1)
234.5 -
4
n ( n + 1)(2 n + 1)
=
28(28 + 1) 4
28(28 + 1)(56 + 1)
24
=
234.5 - 203 43.9147
= 0.7173
24
b. The histogram appears platykurtic, but the A-D test statistic (p = .468) indicates that the hypothesis of normality should not be rejected. Summary for Score A nderson-D arling N ormality Test
0
20
40
60
80
100
A -S quared P -V alue
0.34 0.468
M ean S tD ev V ariance S kew ness Kurtosis N
53.679 30.283 917.041 -0.199844 -0.988643 28
M inimum 1st Q uartile M edian 3rd Q uartile M aximum
2.000 27.500 60.000 77.000 100.000
95% C onfidence Interv al for M ean 41.936
65.421
95% C onfidence Interv al for M edian 39.690
68.207
95% C onfidence Interv al for S tD ev 9 5 % C onfidence Inter vals
23.942
Mean Median 40
45
50
55
60
Learning Objective: 16-3
490
65
70
41.219
ASBE 6e Solutions for Instructors 16.6
a. In the Wilcoxon signed-rank test at = .05, there is a difference in the population median scores on the two exams (p-value = .00235). The worksheet is shown. Student 9 12 8 20 10 14 3 16 4 18 15 17 7 1 19 5 11 2 13 6
Exam 1 52 95 71 54 79 81 65 54 60 92 59 75 72 70 70 63 84 74 83 58
Exam 2 53 96 69 58 84 76 59 47 68 100 68 84 82 81 81 75 96 89 99 77
d -1 -1 2 -4 -5 5 6 7 -8 -8 -9 -9 -10 -11 -11 -12 -12 -15 -16 -19
|d| 1 1 2 4 5 5 6 7 8 8 9 9 10 11 11 12 12 15 16 19
Rank 1.5 1.5 3 4 5.5 5.5 7 8 9.5 9.5 11.5 11.5 13 14.5 14.5 16.5 16.5 18 19 20 210
R+
R1.5 1.5
3 4 5.5 5.5 7 8 9.5 9.5 11.5 11.5 13 14.5 14.5 16.5 16.5 18 19 20 186.5
23.5
Test Statistic: Wz=
n ( n + 1) 4
n ( n + 1)(2 n + 1)
23.5 =
20(20 + 1) 4
20(20 + 1)(40 + 1)
24
=
23.5 - 105 26.7862
= -3.04
24
b. In a t-test at = .05, there is a significant difference in the mean scores on the two exams since the p-value is less than .05. The MegaStat results are shown. Hypothesis Test: Paired Observations 0.000 hypothesized value 70.550 mean Exam 1 77.100 mean Exam 2 -6.550 mean difference (Exam 1 - Exam 2) 7.473 std. dev. 1.671 std. error 20 n 19 df -3.92 t .0009 p-value (two-tailed) Learning Objective: 16-3 491
ASBE 6e Solutions for Instructors 16.7
a. At = .05, there is no difference in the medians, because the p-value is greater than . 05. Wilcoxon - Mann/Whitney Test n sum of ranks 10 135 Bob's Portfolio 12 118 Tom’s Portfolio 22 253 total 115.00 expected value 15.16 standard deviation 1.319 z, uncorrected .1872 p-value (two-tailed) b. MegaStat’s results are shown, assuming equal variances (t = 1.618, p = .1213). At = .05, there is no difference in the means, since the p-value is greater than .05. If you assume unequal variances, the result is similar (t = 1.661, p = .113). Both tests lead to the same decision. Samples are too small for a meaningful test for normality. HypHothesis Test: Independent Groups (t-test, pooled variance) Bob's Portfolio Tom’s Portfolio 6.040 4.100 mean 2.352 3.119 std. dev. 10 12 n 20 1.9400 7.8392 2.7999 1.1988 0 1.618 .1213
df difference (Bob's Tom’s Portfolio) pooled variance pooled std. dev. standard error of diff hypothesized difference t p-value (two-tailed)
Learning Objective: 16-4
16.8
a. We fail to reject the null hypothesis that there is a difference in the medians since the p-value is greater than .05. MegaStat’s results are shown. 492
ASBE 6e Solutions for Instructors
Wilcoxon - Mann/Whitney Test n sum of ranks 9 125 Old Bumper 12 106 New Bumper 21 231 total 99.00 expected value 14.07 standard deviation 1.848 z p-value (two.0646 tailed) b. We fail to reject the null hypothesis that there is a difference in the means since the pvalue is greater than .05. MegaStat’s results are shown. We have the same decision as in (a). Samples are too small for a meaningful test for normality. Hypothesis Test: Independent Groups (t-test, pooled variance) Old Bumper New Bumper 1,766.11 1,101.42 mean 837.62 696.20 std. dev. 9 12 n 19 664.694 576,031.463 758.967 334.673 0 1.99 .0616 .0308
df difference (Old - New) pooled variance pooled std. dev. standard error of difference hypothesized difference t p-value (two-tailed) p-value (one-tailed)
Learning Objective: 16-4
16.9
MegaStat results are shown. At = .05, there is no difference in median volatility in these four portfolios (p-value = .1258). The ANOVA test gives the same conclusion
493
ASBE 6e Solutions for Instructors (p-value = .0923). Had we used = .10, the difference would have been significant in the ANOVA test. Media n 15.90 22.70 20.70 20.65 19.65
n 15 14 13 12 54
One factor ANOVA Mean 19.92037037 17.27 19.92037037 22.59 19.92037037 19.65 19.92037037 20.33 19.90
Avg. Rank 19.97 Health 33.68 Energy 28.23 Retail 28.92 Leisure Total 5.724 H 3 d.f. .1258 p-value
n Std. Dev 15 4.644 Health 14 6.395 Energy 13 5.601 Retail 12 5.400 Leisure 54 5.728 Total
ANOVA table Source Treatment Error Total
SS 208.016 1,530.733 1,738.750
df MS 3 69.3388 50 30.6147 53
F 2.26
pvalue .0923
Based on the four individual histograms, we would doubt normality. However, each sample is rather small for a normality test. Pooling the samples, we get a p-value of . 490 for MINITAB’s Anderson-Darling test statistic, so normality can’t be rejected.
494
ASBE 6e Solutions for Instructors
Summary for Volatility A nderson-D arling N ormality Test
6
12
18
24
A -S quared P -V alue
0.34 0.490
M ean S tD ev V ariance S kew ness Kurtosis N
19.920 5.716 32.674 0.014614 -0.137501 54
M inimum 1st Q uartile M edian 3rd Q uartile M aximum
30
4.900 15.250 19.650 24.275 32.500
95% C onfidence Interv al for M ean 18.360
21.481
95% C onfidence Interv al for M edian 17.636
22.064
95% C onfidence Interv al for S tD ev 9 5 % C onfidence Inter vals
4.805
7.057
Mean Median 17
18
19
20
21
22
Learning Objective: 16-5 16.10
a. At = .05, there is a difference in median productivity because the p-value is less than .05. Kruskal-Wallis Test Median n 4.10 9 2.90 6 5.40 10 4.50 25
Avg. Rank 11.61 Station A 6.67 Station B 18.05 Station C Total 9.472 H 2 d.f. .0088 p-value multiple comparison values for avg. ranks 8.63 (.05) 10.58(.01)
b. At = .05, there is a difference in median productivity because the p-value is less than .05. 495
ASBE 6e Solutions for Instructors One factor ANOVA Mean 4.38 3.97 4.38 3.02 4.38 5.57 4.38
n 9 6 10 25
Std. Dev 0.828 1.094 1.726 1.647
df 2 22 24
MS 13.4253 1.7395
Station A Station B Station C Total
ANOVA tabl e Source Treatment Error Total
SS 26.851 38.269 65.120
F 7.72
p-value .0029
The samples are rather small for a normality test. Pooling the samples them, we get a p-value of .392 for MINITAB’s Anderson-Darling test statistic, so normality can’t be rejected. Summary for Units Per Hour A nderson-D arling N ormality Test
2
3
4
5
6
7
8
A -S quared P -V alue
0.37 0.392
M ean S tD ev V ariance S kew ness Kurtosis N
4.3800 1.6472 2.7133 0.660729 0.072543 25
M inimum 1st Q uartile M edian 3rd Q uartile M aximum
1.9000 3.0000 4.5000 5.2500 8.4000
95% C onfidence Interv al for M ean 3.7001
5.0599
95% C onfidence Interv al for M edian 3.2396
4.9802
95% C onfidence Interv al for S tD ev 9 5 % C onfidence Inter vals
1.2862
2.2915
Mean Median 3.0
3.5
4.0
4.5
5.0
Learning Objective: 16-5
16.11
The median ratings of surfaces do not differ at = .05 because the p-value is greater than .05. Sum of Ranks Avg. Rank 496
ASBE 6e Solutions for Instructors 9.00 10.00 17.50 10.00 13.50 60.00
2.25 2.50 4.38 2.50 3.38 3.00
Shiny Satin Pebbled Pattern Embossed Total 4n 4.950 chi-square 4 d.f. .2925 p-value multiple comparison values for avg. ranks 3.14(.05) 3.68(.01)
Learning Objective: 16-6 The median sales of coffee sizes do not differ at = .05, because the p-value is greater than .05
16.12
Friedman Test Sum of Ranks 10.00 10.00 10.00 30.00
Avg. Rank 2.00 2.00 2.00 2.00 5 0.000 2 1.0000
Small Medium Large Total n chi-square (corrected for ties) d.f. p-value
multiple comparison values for avg. ranks 1.51 (.05) 1.86 Learning Objective: 16-6
16.13
a. Worksheet is shown for rank correlation. Obs 1 2
Company Campbell Soup ConAgra Foods
Profit in year: 2004 2005 595 647 775 880
497
2004 6 5
Rank in year: 2005 7 5
ASBE 6e Solutions for Instructors 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20
Dean Foods Del Monte Foods Dole Food Flowers Foods General Mills H. J. Heinz Hershey Foods Hormel Foods Interstate Bakeries J. M. Schmucker Kellogg Land O'Lakes McCormick Pepsico Ralcorp Holdings Sara Lee Smithfield Foods Wm. Wrigley, Jr.
356 134 105 15 917 566 458 186 27 96 787 107 211 3568 7 1221 26 446
285 165 134 51 1055 804 591 232 -26 111 891 21 215 4212 65 1272 227 493
10 13 15 19 3 7 8 12 17 16 4 14 11 1 20 2 18 9
10 14 15 18 3 6 8 11 20 16 4 19 13 1 17 2 12 9
Rank sum:210
210
b. Spearman rank correlation found by using the Excel function CORREL on the rank columns is 0.9338. c. . The p-value =1-NORM.S.DIST(4.07,1) = zcalc = rs n - 1 = .9338 20 - 1 = 4.07 2.4E-05. Clearly, we can reject the hypothesis of no correlation at = .01. d. MegaStat’s calculations are shown. Spearman Coefficient of Rank Correlation 2004 2005 2004 1.000 2005 .934 1.000 20 sample size ± .444 critical value .05 (two-tail) ± .561 critical value .01 (two-tail) e. Calculated using the CORREL function on the actual data (not the ranks) we get r = 0.9960: f. In this example, there is no strong argument for the Spearman test since the data are ratio. However, the assumption of normality may be dubious (samples are too small for a reliable normality test). Learning Objective: 16-7
16.14
a. Table of Ranks: 498
ASBE 6e Solutions for Instructors
Fund 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24
12-Mo 11.2 -2.4 8.6 3.4 3.9 10.3 16.1 6.7 6.5 11.1 8.0 11.2 14.0 11.6 13.2 -1.0 6.2 21.1 -1.2 8.7 9.7 0.4 0.9 12.7
5-Yr 10.5 5.0 8.6 3.7 -2.9 9.6 14.1 6.2 7.4 14.0 7.3 14.2 9.7 14.7 11.8 2.3 10.5 9.0 3.0 7.1 10.2 9.3 6.0 10.0 Total:
Rank 12-Mo 7.5 24 13 19 18 10 2 15 16 9 14 7.5 3 6 4 22 17 1 23 12 11 21 20 5 300
Rank 5-Yr 6.5 20 14 21 24 11 3 18 15 4 16 2 10 1 5 23 6.5 13 22 17 8 12 19 9 300
b. The Spearman’s rank correlation coefficient is .742. c. Yes, assumption of zero correlation can be rejected at both the .01 and .05 level of significance. See MegaStat output below. d. MegaStat output.
12-Mo 5-Yr
12-Mo 5-Yr 1.000 .742 1.000 24 sample size ± .404 ± .515
e. Pearson correlation found by using the Excel function CORREL is 0.6560. f. In this example, there is no strong argument for the Spearman test since the data are ratio. Despite the low outlier in 5-year returns, the sample passes the test for normality (p-value = .541).
499
ASBE 6e Solutions for Instructors Summary for 12-Mo
Summary for 5-Yr A nderson-Darling N ormality Test
0
5
10
15
20
A nderson-Darling N ormality Test
A -Squared P -V alue
0.31 0.541
A -Squared P -V alue
0.34 0.460
M ean S tDev V ariance S kew ness Kurtosis N
7.9542 5.9032 34.8478 -0.022436 -0.236113 24
M ean S tDev V ariance S kew ness Kurtosis N
8.3875 4.2216 17.8220 -0.696921 0.850848 24
M inimum 1st Q uartile M edian 3rd Q uartile M aximum
-2.4000 3.5250 8.6500 11.5000 21.1000
M inimum 1st Q uartile M edian 3rd Q uartile M aximum
-2.9000 6.0500 9.1500 10.5000 14.7000
0
5
10
15
95% C onfidence Interv al for M ean 5.4615
95% C onfidence Interv al for M ean
10.4469
6.6049
95% C onfidence Interv al for M edian 5.8016
11.2000
6.9441
95% C onfidence Interv al for StD ev 9 5 % Confidence Inter vals
4.5880
10.1701
95% C onfidence Interv al for M edian 10.2520
95% C onfidence Interv al for StD ev 9 5 % Confidence Inter vals
8.2808
Mean
3.2811
5.9219
Mean
Median
Median 5
6
7
8
9
10
11
7
8
9
10
Learning Objective: 16-7 16.15
Because the p-value (.4385) is greater than .05, we fail to reject the null hypothesis of randomness. H0: Events follow a random pattern H1: Events do not follow a random pattern n runs 21 14 B 29 14 A 50 28 total 25.360 expected value 3.408 standard deviation 0.775 z p-value (two.4385 tailed) Learning Objective: 16-2
16.16
Because the p-value (.7746) is greater than .01 we fail to reject the null hypothesis of randomness. H0: Events follow a random pattern H1: Events do not follow a random pattern n runs 21 9 H 14 8 M 35 17 total 17.800 expected value 2.794 standard deviation -0.286 z .7746 p-value (two-tailed) Learning Objective: 16-2
16.17
Because the p-value (.1560) is greater than .05 we fail to reject the null hypothesis of randomness. 500
ASBE 6e Solutions for Instructors H0: Events follow a random pattern H1: Events do not follow a random pattern n runs 18 11 C 16 11 X 34 22 total 17.941 expected value 2.861 standard deviation 1.419 z p-value (two.1560 tailed) Learning Objective: 16-2 16.18
Because the p-value (.1821) is greater than .05 we fail to reject the null hypothesis of randomness. H0: Events follow a random pattern H1: Events do not follow a random pattern n runs 34 13 Up 27 13 Dn 61 26 total 31.098 expected value 3.821 standard deviation -1.334 z p-value (two.1821 tailed) Learning Objective: 16-2
16.19
16.20
Since the p-value (.0996) is greater than .05 we fail to reject the null hypothesis of randomness. H0: Events follow a random pattern vs. H1: Events do not follow a random pattern n runs 13 5 Lo 11 4 Hi 24 9 total 12.917 expected value 2.378 standard deviation -1.647 z .0996 p-value (two-tailed) Learning Objective: 16-2 MegaStat results are shown. At = .10, the median ELOS does not differ for the two groups, since the p-value (.5529) is greater than .10. The hypotheses are: 501
ASBE 6e Solutions for Instructors
H0: M1 − M2 = 0 H1: M1 − M2 0
(no difference in ELOS) (ELOS differs for the two groups)
Wilcoxon - Mann/Whitney Test n sum of ranks 10 124 Clinic A 12 129 Clinic B 22 253 total 115.00 expected value 15.17 standard deviation 0.593 z .5529 p-value (two-tailed) Although the histogram is somewhat platykurtic in appearance, normality may be assumed at = .10 based on the Anderson-Darling p-value (.147). To perform this test, the two samples were pooled. Even so, the sample is rather small for a normality test. Summary for Weeks A nderson-D arling N ormality Test
20
30
40
A -S quared P -V alue
0.54 0.147
M ean S tD ev V ariance S kew ness Kurtosis N
30.364 10.913 119.100 0.460007 -0.632645 22
M inimum 1st Q uartile M edian 3rd Q uartile M aximum
50
16.000 20.000 30.000 40.000 52.000
95% C onfidence Interv al for M ean 25.525
35.202
95% C onfidence Interv al for M edian 23.891
36.109
95% C onfidence Interv al for S tD ev 9 5 % C onfidence Inter vals
8.396
15.596
Mean Median 25.0
27.5
30.0
32.5
35.0
37.5
Learning Objective: 16-4
16.21
MegaStat results are shown. At = .05, the median defect counts do not differ for the two groups, since the p-value (.2987) is greater than .05. The hypotheses are: H0: M1 − M2 = 0 H1: M1 − M2 0
(no difference in number of bad pixels) (number of bad pixels differs between the two labs) 502
ASBE 6e Solutions for Instructors
Wilcoxon - Mann/Whitney Test n sum of ranks 12 132 Lab A 12 168 Lab B 24 300 total 150.00 expected value 17.32 standard deviation -1.039 z .2987 p-value (two-tailed) Learning Objective: 16-4 LO4
16.22 MegaStat results are shown. At = .01, the median salaries do not differ for the two groups, since the p-value (.2871) is greater than .01. The hypotheses are: H0: M1 − M2 = 0 H1: M1 − M2 0
(no difference in salaries between the two industries) (Median salary differs for the two industries)
Wilcoxon - Mann/Whitney Test n sum of ranks 30 843 30 987 60 1830 915.00 67.64 -1.06 .2905
Industry A Industry B total expected value standard deviation z p-value (two-tailed)
Learning Objective: 16-4 16.23
MegaStat results are shown. At = .05, the median difference in heart rates does not differ from zero, since the p-value (.1386) is greater than .05. Note: The 5th observation is omitted because its difference is zero (heart rate of 82 before and after) which leaves only n = 29. The hypotheses are: H0: Md = 0 the median difference in pulse rate is zero H1: Md 0 the median difference in pulse rate is not zero
Wilcoxon Signed Rank Test variables:Before - After sum of positive 149 ranks 503
ASBE 6e Solutions for Instructors sum of negative 286 ranks 29 n 217.50 expected value 46.25 standard deviation -1.481 z .1386 p-value (two-tailed) The worksheet is shown: Student 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30
Before 60 70 77 80 82 82 41 65 58 50 82 56 71 67 66 70 69 64 70 59 62 66 81 56 64 78 75 66 59 98
After 62 76 78 83 82 83 66 63 60 54 93 55 67 68 75 64 66 69 73 58 65 68 77 57 62 79 74 67 63 82
d |d| -2 2 -6 6 -1 1 -3 3 0 0 -1 1 -25 25 2 2 -2 2 -4 4 -11 11 1 1 4 4 -1 1 -9 9 6 6 3 3 -5 5 -3 3 1 1 -3 3 -2 2 4 4 -1 1 2 2 -1 1 1 1 -1 1 -4 4 16 16 Rank sum:
Rank 12 24.5 5 16.5 5 29 12 12 20.5 27 5 20.5 5 26 24.5 16.5 23 16.5 5 16.5 12 20.5 5 12 5 5 5 20.5 28 435
R+
R12 24.5 5 16.5 5 29
12 12 20.5 27 5 20.5 5 26 24.5 16.5 23 16.5 5 16.5 12 20.5 5 12 5 5 5 20.5 28 149
286
The histograms are bell-shaped, and normality may be assumed at any common based on the Anderson-Darling p-values (.543 and .388).
504
ASBE 6e Solutions for Instructors Summary for Before
Summary for After A nderson-Darling N ormality Test
48
64
80
96
A nderson-Darling N ormality Test
A -S quared P-V alue
0.31 0.543
A -Squared P -V alue
0.38 0.388
M ean StDev V ariance Skew ness Kurtosis N
68.133 11.461 131.361 0.182087 0.832718 30
M ean S tDev V ariance S kew ness Kurtosis N
69.633 9.597 92.102 0.431633 -0.347835 30
M inimum 1st Q uartile M edian 3rd Q uartile M aximum
41.000 59.750 66.500 77.250 98.000
M inimum 1st Q uartile M edian 3rd Q uartile M aximum
60
70
80
90
95% C onfidence Interv al for M ean 63.854
95% C onfidence Interv al for M ean
72.413
66.050
95% C onfidence Interv al for M edian 64.000
70.771
9.128
73.217
95% C onfidence Interv al for M edian 64.229
95% C onfidence Interv al for S tDev 9 5 % Confidence Inter vals
54.000 62.750 67.500 77.250 93.000
74.771
95% C onfidence Interv al for S tDev 9 5 % C onfidence Intervals
15.408
Mean
7.643
12.901
Mean
Median
Median 64
66
68
70
72
65.0
67.5
70.0
72.5
75.0
Learning Objective: 16-3 16.24
MegaStat results are shown. At = .05, the median downtimes (days) do not differ for the two groups, since the p-value (.1658) is greater than .05. The hypotheses are: H0: M1 − M2 = 0 H1: M1 − M2 0
(no difference in number of repair incidents) (Number of repair incidents differs for the two groups)
Wilcoxon - Mann/Whitney Test n sum of ranks 12 112.5 New Bumper 9 118.5 Old Bumper 21 231 total 132.00 expected value 14.07 standard deviation -1.386 z .1658 p-value (two-tailed) The histogram is right-skewed, but the Anderson-Darling p-value (.088) would not lead to rejection of the hypothesis of normality at = .05. To perform this test, the two samples were pooled. Even so, the sample is rather small for a normality test. Summary for Days A nderson-D arling N ormality Test
4
8
12
16
A -S quared P -V alue
0.63 0.088
M ean S tD ev V ariance S kew ness Kurtosis N
7.1905 4.3084 18.5619 0.938235 0.530538 21
M inimum 1st Q uartile M edian 3rd Q uartile M aximum
1.0000 4.0000 7.0000 10.0000 18.0000
95% C onfidence Interv al for M ean 5.2293
9.1516
95% C onfidence Interv al for M edian 4.6735
8.3265
95% C onfidence Interv al for S tD ev 9 5 % C onfidence Inter vals
3.2961
Mean Median 5
6
7
8
505
9
6.2216
ASBE 6e Solutions for Instructors Learning Objective: 16-4 16.25
MegaStat results are shown. At = .01, the median square footage does differ for the two groups, since the p-value (.0020) is less than .01. The hypotheses are: H0: M1 = M2 H1: M1 M2
(no difference in square footage) (square footage differs for the two groups)
Wilcoxon - Mann/Whitney Test n sum of ranks 11 79.5 Grosse Hills 11 173.5 Haut Nez Estates 22 253 total 126.50 expected value 15.23 standard deviation -3.086 z p-value (two.0020 tailed) The histogram is bell-shaped, and normality may be assumed at any common based on the Anderson-Darling p-value (.359). To perform this test, the two samples were pooled. Summary for Sq Ft A nderson-D arling N ormality Test
3000
3500
4000
4500
A -S quared P -V alue
0.39 0.359
M ean S tD ev V ariance S kew ness Kurtosis N
3559.1 487.0 237208.7 0.890787 0.934743 22
M inimum 1st Q uartile M edian 3rd Q uartile M aximum
2800.0 3215.0 3450.0 3850.0 4850.0
95% C onfidence Interv al for M ean 3343.1
3775.0
95% C onfidence Interv al for M edian 3220.0
3752.7
95% C onfidence Interv al for S tD ev 9 5 % C onfidence Inter vals
374.7
696.0
Mean Median 3200
3300
3400
3500
3600
3700
3800
Learning Objective: 16-4 16.26
MegaStat results are shown. At = .01, the median GPA does not differ among the classes, since the p-value (.1794) is greater than .01. H0: All c population medians are the same H1: Not all the population medians are the same Kruskal-Wallis Test 506
ASBE 6e Solutions for Instructors Median 2.19 2.96 3.26 3.10 3.01
n 5 7 7 6 25
Avg. Rank 7.40 Frosh 11.93 Soph 16.21 Junior 15.17 Senior Total 4.898 H 3 d.f. .1794 p-value multiple comparison values for avg. ranks 10.98(.05) 13.09(.01)
The histogram is rather bimodal, but normality cannot be rejected at any common based on the Anderson-Darling p-value (.277). To perform this test, the four samples were pooled. Summary for GPA A nderson-D arling N ormality Test
2.0
2.4
2.8
3.2
A -S quared P -V alue
0.43 0.277
M ean S tD ev V ariance S kew ness Kurtosis N
2.9680 0.5477 0.3000 -0.388769 -0.543016 25
M inimum 1st Q uartile M edian 3rd Q uartile M aximum
3.6
1.9100 2.5800 3.0100 3.3250 3.8900
95% C onfidence Interv al for M ean 2.7419
3.1941
95% C onfidence Interv al for M edian 2.8339
3.3081
95% C onfidence Interv al for S tD ev 9 5 % C onfidence Inter vals
0.4277
0.7619
Mean Median 2.7
2.8
2.9
3.0
3.1
3.2
3.3
Learning Objective: 16-5 16.27
MegaStat results are shown. At = .01, the median crash damage does not differ among the cars, since the p-value (.4819) is greater than .01. Sample sizes are too small for a reasonable test for normality, even if the samples are combined. H0: All c population medians are the same H1: Not all the population medians are the same Kruskal-Wallis Test Mediann 1,220.005 1,390.005 1,830.005
Avg. Rank 6.40 Goliath 7.80 Varmint 9.80 Weasel 507
ASBE 6e Solutions for Instructors 1,390.0015
Total 1.460 H 2 d.f. .4819 p-value multiple comparison values for avg. ranks 6.77 (.05) 8.30
Learning Objective: 16-5
16.28
MegaStat results are shown. At = .05, the median waiting time does not differ among the hospitals, since the p-value (.1788) is greater than .05. H0: All c population medians are the same H1: Not all the population medians are the same Kruskal-Wallis Test Median n Avg. Rank 11.00 5 11.20 Hospital A 18.00 7 15.43 Hospital B 11.50 6 10.33 Hospital C 7.00 4 6.75 Hospital D 13.50 22 Total 4.907 H 3 d.f. .1788 p-value multiple comparison values for avg. ranks 10.33(.05) 12.31(.01) The histogram looks rather right-skewed, but normality may be assumed at any common based on the Anderson-Darling p-value (.561). To perform this test, the four samples were pooled.
508
ASBE 6e Solutions for Instructors Summary for Minutes A nderson-D arling N ormality Test
0
10
20
A -S quared P -V alue
0.30 0.561
M ean S tD ev V ariance S kew ness Kurtosis N
15.045 8.984 80.712 0.557667 -0.076613 22
M inimum 1st Q uartile M edian 3rd Q uartile M aximum
30
0.000 8.750 13.500 21.000 36.000
95% C onfidence Interv al for M ean 11.062
19.029
95% C onfidence Interv al for M edian 9.973
19.027
95% C onfidence Interv al for S tD ev 9 5 % C onfidence Inter vals
6.912
12.839
Mean Median 10
12
14
16
18
20
Learning Objective: 16-5
16.29
MegaStat results are shown. At = .05, the median output (watts) does differ among the three types of cells, since the p-value (.0110) is smaller than .05. H0: All c population medians are the same H1: Not all the population medians are the same Kruskal-Wallis Test Median n 123.50 6 122.00 6 128.00 6 125.00 18
Avg. Rank 7.75 A 6.00 B 14.75 C Total 9.026 H 2 d.f. .0110 p-value multiple comparison values for avg. ranks 7.38(.05) 9.05(.01)
The histogram appears right-skewed, but normality may be assumed at any common based on the Anderson-Darling p-value (.524). To perform this test, the three samples were pooled.
509
ASBE 6e Solutions for Instructors Summary for Watts A nderson-D arling N ormality Test
122
124
126
128
130
A -S quared P -V alue
0.31 0.524
M ean S tD ev V ariance S kew ness Kurtosis N
124.89 2.91 8.46 0.446497 -0.585434 18
M inimum 1st Q uartile M edian 3rd Q uartile M aximum
132
121.00 122.00 125.00 127.25 131.00
95% C onfidence Interv al for M ean 123.44
126.34
95% C onfidence Interv al for M edian 122.52
126.48
95% C onfidence Interv al for S tD ev 9 5 % C onfidence Inter vals
2.18
4.36
Mean Median 122
123
124
125
126
127
Learning Objective: 16-5
16.30
MegaStat results are shown. At = .01, the median stopping distance does not differ among the types of braking method, since the p-value (.2636) is greater than .01. Normality test is not practical with only nine data points. H0: All c populations have the same median (braking method not related to stopping distance). H1: Not all the populations have the same median (braking method related to stopping distance). Friedman Test Sum of Ranks 6.00 4.00 8.00 18.00
Avg. Rank 2.00 1.33 2.67 2.00 3 2.667 2 .2636
Pumping Locked ABS Total n chi-square d.f. p-value
multiple comparison values for avg. ranks 1.95 (.05) 2.40 Learning Objective: 16-6 16.31
MegaStat results are shown. At = .01, the median waiting time does not differ among the time of day, since the p-value (.6038) is greater than .01. 510
ASBE 6e Solutions for Instructors
H0: All c populations have the same median (waiting time not related to time of day). H1: Not all the populations have the same median (waiting time related to time of day). Friedman Test Sum of Ranks Avg. Rank 76.00 2.92 Mon 87.50 3.37 Tue 78.00 3.00 Wed 79.50 3.06 Thu 69.00 2.65 Fri 390.00 3.00 Total 26 n 2.731 chi-square 4 d.f. .6038 p-value multiple comparison values for avg. ranks 1.23 (.05) 1.44 All five histograms are skewed. Only one histogram (Thursday) might be considered normal, according to the Anderson-Darling test (individual tests not shown). Combining the five days (shown below) we reject normality (p < .005). There are high outliers. A non-parametric test is desirable. Summary for Waiting Time A nderson-D arling N ormality Test
0
30
60
90
120
150
A -S quared P -V alue <
5.13 0.005
M ean S tD ev V ariance S kew ness Kurtosis N
51.485 27.745 769.771 1.20873 1.35480 130
M inimum 1st Q uartile M edian 3rd Q uartile M aximum
0.000 34.000 42.000 63.750 152.000
95% C onfidence Interv al for M ean 46.670
56.299
95% C onfidence Interv al for M edian 37.365
47.000
95% C onfidence Interv al for S tD ev 9 5 % C onfidence Inter vals
24.733
31.598
Mean Median 35
40
45
50
55
Learning Objective: 16-6 16.32
The Spearman rank correlation is preferred with skewed data. At = .01, there is a significant rank correlation between revenue and profit since the Spearman coefficient of rank correlation is outside the critical region as given in the MegaStat output (.761 > .561). H0: Rank correlation is zero (s 0) 511
(no relationship exists)
ASBE 6e Solutions for Instructors H1: Rank correlation is positive (s > 0) (there is a relationship) Spearman Coefficient of Rank Correlation Revenue Profit Revenue 1.000 Profit .761 1.000 20 sample size ± .444 critical value .05 (two-tail) ± .561 critical value .01 (two-tail) Learning Objective: 16-7
16.33
At = .05, there is a significant rank correlation between fertility in 2000 and fertility in 2009 because the Spearman coefficient of rank correlation is outside the critical region as given in the MegaStat output (.670 > .374). H0: Rank correlation is zero (s = 0) (no relationship exists) H1: Rank correlation is positive (s ≠ 0) (there is a relationship) Spearman Coefficient of Rank Correlation #1 #2
#1 1.000 .670 28 ± .374 ± .479
16.34
#2 1.000 sample size critical value .05 (twotail) critical value .01 (twotail)
At = .01, there is not a significant rank correlation between calories and sodium since the Spearman coefficient of rank correlation is inside the critical region as given in the MegaStat output (.229 < .623). Samples are too small for reliable test for normality. However, there is one severe outlier in the calories (possibly a data recording error). The sodium histograms are somewhat right-skewed. All in all, the nonparametric test seems like a good idea. H0: Rank correlation is zero (s = 0) H1: Rank correlation is not zero (s ≠ 0)
Spearman Coefficient of Rank Correlation Fat (g) Calories 512
Sodium
ASBE 6e Solutions for Instructors (mg) Fat (g) Calories Sodium (mg)
1.000 .680 .559
1.000 .229
1.000
16 sample size ± .497 critical value .05 (two-tail) ± .623 critical value .01 (two-tail) Learning Objective: 16-7 16.35
At = .05, there is a significant agreement between the two reviewers’ movie ratings because the Spearman coefficient of rank correlation is outside the critical region as given in the MegaStat output (.696 > .468). If one can reject in a two-tailed test one can reject in a one-tailed test. H0: Rank correlation is zero (s ≤ 0) (no relationship or do not agree) H1: Rank correlation is not zero (s > 0) (the reviewers tend to agree) Spearman Coefficient of Rank Correlation A B A 1.000 B .696 1.000 18 sample size critical value .05 ± .468 (two-tail) critical value .01 ± .590 (two-tail) Learning Objective: 16-7
LO7
16.36 At = .05, there is a significant rank correlation between gas prices and carbon emission since the Spearman coefficient of rank correlation is outside the critical region as given in the MegaStat output (|.588| > .355). H0: Rank correlation is zero (s = 0) (no relationship) H1: Rank correlation is positive (s ≠ 0) (there is a relationship)
Spearman Coefficient of Rank Correlation Gas Price ($/L) CO2/GDP (kg/$) Gas Price ($/L) 1.000 CO2/GDP (kg/$) -.588 1.000 513
ASBE 6e Solutions for Instructors 31 sample size ± .355 critical value .05 (two-tail) ± .456 critical value .01 (two-tail) The gas price histogram appears left-skewed, but its Anderson-Darling p-value (.169).suggests normality at = .05. However, the CO2 histogram is strongly rightskewed and non-normal (p < .005). Summary for Gas Price ($/L)
Summary for CO2/GDP (kg/$) A nderson-Darling N ormality Test
0.4
0.6
0.8
1.0
1.2
A nderson-Darling N ormality Test
A -Squared P -V alue
0.52 0.169
A -Squared P -V alue <
2.55 0.005
M ean S tDev V ariance S kew ness Kurtosis N
0.83648 0.23623 0.05581 -0.414781 -0.915642 31
M ean S tDev V ariance S kew ness Kurtosis N
0.68581 0.58049 0.33697 1.39163 0.78585 31
M inimum 1st Q uartile M edian 3rd Q uartile M aximum
0.38100 0.61300 0.87900 1.03300 1.19800
M inimum 1st Q uartile M edian 3rd Q uartile M aximum
0.13000 0.31000 0.41000 0.97000 2.08000
0.5
1.0
1.5
2.0
95% C onfidence Interv al for M ean 0.74983
95% C onfidence Interv al for M ean
0.92314
0.47288
95% C onfidence Interv al for M edian 0.76341
0.99017
0.35000
95% C onfidence Interv al for StD ev 9 5 % Confidence Inter vals
0.18878
0.67949
95% C onfidence Interv al for StD ev 9 5 % Confidence Inter vals
0.31577
Mean
0.89873
95% C onfidence Interv al for M edian
0.46388
0.77593
Mean
Median
Median 0.75
0.80
0.85
0.90
0.95
1.00
0.3
0.4
0.5
0.6
0.7
0.8
0.9
Learning Objecitve: 16-7 16.37
At = .01, there is a significant rank correlation between this week’s points and last week’s points since the Spearman coefficient of rank correlation is outside the critical region as given in the MegaStat output (. 812 > .561). We are not surprised that team rankings usually do not change much from week to week. There is no reason to expect normality since ratings do not tend toward a common mean. H0: Rank correlation is zero (s = 0) (no relationship) H1: Rank correlation is positive (s ≠ 0) (there is a relationship) Spearman Coefficient of Rank Correlation This Week Last Week This Week 1.000 Last Week .812 1.000 20 sample size ± .444 critical value .05 (two-tail) ± .561 critical value .01 (two-tail) Learning Objecitve: 16-7
514
ASBE 6e Solutions for Instructors
Chapter 17 Quality Management 17.1
a. Productivity is a ratio of output to input and measures efficiency. b. Quality control refers to methods used to ensure product or service quality. c. Process control refers to methods used to ensure process consistency and conformance to process specifications. Learning Objective: 17-1
17.2
From the modern perspective productivity and quality move in the same direction. When you improve quality you also improve productivity. In the past, the relationship was thought to be the inverse. People believed that to improve quality they must slow down processes which would decrease productivity. This is not the case if one considers that rework and scrap due to poor quality actually decrease productivity. Learning Objective: 17-1
17.3
Common cause variation is normal and expected. It’s part of the process. Special cause variation is caused by something outside the normal process. Learning Objective: 17-1
17.4
Zero variation is a lofty goal but not achievable. Variation will always exist. We can attempt to minimize variation but cannot eliminate all sources of variation. Learning Objective: 17-1
17.5
Statisticians help organization define metrics used to measure quality and then help design data collection and process monitoring systems. Learning Objective: 17-1
17.6
Students may name Deming, Shewhart, Ishikawa, Taguchi, and others they’ve heard of. Learning Objective: 17-2
17.7
Deming’s 14 Points (abbreviated) 1. Maintain constancy of purpose. 2. Adopt a new philosophy. 3. Don’t rely on inspection—design quality in. 4. Don’t award contracts just on the basis of price. 5. Continuous improvement. 6. Institute training on the job. 7. Supervision should help people do a better job. 8. Drive out fear and create trust. 9. Break down barriers between departments. 10. Eliminate slogans, exhortations, and targets. 516
ASBE 6e Solutions for Instructors 11. Eliminate numerical goals. 12. Remove barriers to pride in work. 13. Continuing education for all. 14. Act to accomplish the transformation. See www.deming.org for a more detailed complete list. Learning Objective: 17-2 17.8
a. A car dealership typically surveys their customers and might track the number of customers who respond that they were satisfied with their service. b. A bank might track the number of transactions that are error free. They might track the time required to complete a transaction. c. The movie theater might track the time required to purchase a ticket or the number of customer complaints. Learning Objective: 17-3
17.9
SQC refers to a specific set of tools used to monitor, improve, and control product quality. Most of the tools rely on statistical techniques. SPC refers to the statistical tools used to monitor, improve, and control process characteristics. Learning Objective: 17-3
17.10
Total Quality Management or TQM Business Process Redesign or BPR Continuous Quality Improvement or CQI Six Sigma or 6-sigma uses the DMAIC cycle (define, measure, analyze, improve, control) PDCA cycle (plan, do, check, act) Learning Objective: 17-3
17.11
The improvement process is never-ending because once you eliminate one source of variation you move to the next source of variation. Two cycles: PDCA (plan, do, check, act) and DMAIC (define, measure, analyze, improve, control) Learning Objective: 17-3
17.12
SERVQUAL Reliability, Responsiveness, Assurance, Empathy, and Tangibles Learning Objective: 17-3
17.13
Service blueprints and Service transaction analysis Learning Objective: 17-3
17.14
Attribute control charts are for nominal data (e.g., proportion conforming) while variable control charts are for ratio or interval data (e.g., means). Learning Objective: 17-4
17.15
a. Sampling frequency depends on cost and physical possibility of sampling. 517
ASBE 6e Solutions for Instructors b. For normal data, small samples may suffice for a mean (Central Limit Theorem). c. Large samples may be needed for a proportion to get sufficient precision. Learning Objective: 17-4 17.16
a. We can estimate using the sample standard deviation (s), or using
where
R d2 R the average range and d2 is a control chart factor from Table 17.4, or using the average of the sample standard deviations of many samples ( ).
is
s is often used because we often know the range of values for our process. For
b. The
R historical reasons, this method is still used frequently. Before the advent of computers it was easier to calculate than s. c. MINITAB defaults to s because when a sample data set is provided it is the most direct way to estimate σ. Learning Objective: 17-5 17.17
This is the Empirical Rule (see chapters 4 and 7): a. Within ± 1 standard deviations 68.26 percent of the time b. Within ± 2 standard deviations 95.44 percent of the time c. Within ± 3 standard deviations 99.73 percent of the time Learning Objective: 17-5
17.18
Students may need to be reminded that “sigma” refers to the standard error of the mean . s
n
Rule 1. Single point outside 3 sigma Rule 2. Two of three successive points outside 2 sigma on same side of centerline Rule 3. Four of five successive points outside 1 sigma on same side of centerline Rule 4. Nine successive points on same side of centerline Learning Objective: 17-5 17.19 UCL = x + 3 LCL = x - 3
R d2 n R d2 n
0.42
= 12.5 + 3 = 12.5 - 3
2.326 5 0.42 2.326 5
= 12.742 = 12.258
Learning Objective: 17-5 17.20 UCL = m + 3 LCL = m - 3
s n s n
= 400 + 3 = 400 - 3
2 4 2 4
= 403 = 397
Learning Objective: 17-5 518
ASBE 6e Solutions for Instructors
17.21
Estimated is x = R=
R / d2
x1 + x2 + ... + x9 9 R1 + R2 + ... + R9 9
=
= 30/2.059 = 14.572, UCL = 98.37, LCL = 54.63
72.25 + 74.25 + ... + 82.25
=
9 43 + 31 + ... + 41 9
= 76.5
= 30
Control Chart for the Mean 120
Sample Mean
100
98.37
80
76.50
60
54.63
40 20 0 1
2
3
4
5
6
7
8
9
Sample Number
Control Chart for the Range 80 70
68.46
60 50 Sample Range
40 30
30.00
20 10 0
0.00 1
2
3
4
5
Sample Number
Learning Objective: 17-5 519
6
7
8
9
ASBE 6e Solutions for Instructors
17.22 x = R=
x1 + x2 + ... + x8 8 R1 + R2 + ... + R8
=
8
5.52 + 5.51 + ... + 5.51
= 5.50 8 0.13 + 0.11 + ... + 0.13 = = 0.110 8
Estimate of and estimate of = (0.110)/(2.326) = 0.0473 Control Chart for the Mean 5.60 5.55 5.50 Sample Mean
5.45
5.5614
5.4973
5.4331
5.40 5.35 1
2
3
4
5
Sample Number
520
6
7
8
ASBE 6e Solutions for Instructors
Control Chart for the Range 0.25
0.2352
0.20 0.15 Sample Range
0.10
0.1113
0.05 0.00
0.0000 1 2
3
4
5
6
7
8
Sample Number
Learning Objective: 17-5
17.23
= 0.82 (centerline) R UCL = D4 = (2.004)(0.82) = 1.64328 R LCL = D3 = (0)(0.82) = 0 R Learning Objective: 17-6
17.24
= 0.82 R UCL = D4 LCL = D3
centerline for R chart R
= (2.574)(12) = 30.888 upper control limit
= (0)(12) = 0 R Learning Objective: 17-6
lower control limit
17.25
Services are often assessed using percent conforming or acceptable quality, so we use p charts. Learning Objective: 17-4
17.26
Yes, it is OK to assume normality because n = (500)(.02) = 10 and n(1−) = (500)(.98) = 490 > 10.
521
ASBE 6e Solutions for Instructors
UCL = p + 3
p(1 - p)
(.02)(.98)
= .02 + 3
n
= .0388
500
=
LCL = p - 3
p(1 - p )
.02 - 3
n
(.02)(.98)
= .0012
500
Learning Objective: 17-7 17.27
Yes, safe to assume normality because n = (20)(.50) = 10 and n(1−) = (20)(.50) = 10. UCL = p + 3
p(1 - p)
(.5)(.5)
= .50 + 3
n
20
= .8354
=
LCL = p - 3
p(1 - p )
.50 - 3
n
(.50)(.50) 20
= .1646
Learning Objective: 17-7 17.28
Because n(1π) = 40(.10) =4 is not greater than 10, we can’t assume normality. (use 1.0 since UCL cannot exceed UCL = p + 3
p(1 - p)
= .90 + 3
n
(.90)(.10) 40
= 1.042302
1). =
LCL = p - 3
p(1 - p )
.90 - 3
n
(.90)(.10) 40
= .757698
Learning Objective: 17-7
17.29
By either criterion, process is within acceptable standard (Cp = 1.67, Cpk = 1.67). Cp index: CP =
USL - LSL 6s
=
725 - 715 6(1)
= 1.667
Cpk index: z USL =
zmin =
USL - m s
=
725 - 720 1
min( zUSL , zLSL )
and = 5.00
z LSL =
m - LSL s
=
720 - 715 1
= 5.00
= min(5.00, 5.00) = 5.00 and so C pk =
Learning Objective: 17-9 522
Z min 3
=
5.00 3
= 1.667
ASBE 6e Solutions for Instructors
17.30
If the minimum capability index is 1.33, this process meets the Cp but fails on the Cpk criterion. Cp index: USL - LSL
Cp =
6s
=
0.432 - 0.423 6(0.001)
= 1.50
Cpk index: z USL =
USL - m s
zmin =
=
0.432 - 0.426 0.001
min( zUSL , zLSL )
and = 6.00
z LSL =
m - LSL s
=
0.426 - 0.423 0.001
= 3.00
= min{6.00, 3.00} = 3.00 and so C pk =
Z min 3
=
3.00 3
= 1.00
Learning Objective: 17-9 17.31
If the minimum capability index is 1.33, the process fails on both criteria, especially Cpk due to bad centering (Cp = 1.17, Cpk = 0.67). Cp index: Cp =
USL - LSL 6s
=
55.9 - 55.2 6(0.1)
= 1.1667
Cpk index: and z
USL
zmin =
=
USL - m s
=
55.9 - 55.4
min( zUSL , zLSL )
= 5.00
.1
z
LSL
=
m - LSL s
=
55.4 - 55.2
= 2.00
.1
= min{5.00, 2.00} = 2.00 and so C pk =
Z min 3
=
2.00 3
= 0.667
Learning Objective: 17-9 17.32
The charts and their purposes are: a. chart monitors a process mean for samples of n items. Requires estimates of and x
(or
and control factor d2).
R b. R chart monitors variation around the mean for samples of n items. Requires estimate of or and control chart factor D4. R
523
ASBE 6e Solutions for Instructors c. p chart monitors the proportion of conforming items in samples of n items. Requires estimate of . d. I chart monitors individual items when inspection is continuous (n = 1). Requires estimates of and . Learning Objective: 17-4 17.33
Answers will vary. For example: a. GPA, number of classes re-taken, faculty recommendation letters (Likert). b. Knowledge of material, enthusiasm, organization, fairness (Likert scales for all). c. Number of bounced checks, size of monthly bank balance errors, unpaid VISA balance. d. Number of print errors, clarity of graphs, useful case studies (Likert scales for last two.) Learning Objective: 17-1
17.34
Answers will vary. For example: a. Percent of time “out of range,” frequency of poor reception, perceived ease of use of menus. b. Percent of time server is unavailable, frequency of spam or “pop-up” ads. c. Customer wait time in queue to pick up or drop off, rating of garment cleanliness, rating of staff courtesy. d. Waiting time in office, staff courtesy, percent of cost covered by insurance. e. Waiting time for service, perceived quality of haircut, rating of friendliness of haircutter. f. Waiting time for service, perceived quality of food, rating of staff courtesy. Learning Objective: 17-1
17.35
Answers will vary. For example: a. MPG, repair cost. b. Frequency of jams, ink cost. c. Frequency of re-flushes, water consumption. d. Battery life, ease of use (Likert scale). e. Cost, useful life, image sharpness (Likert scale). f. Cost, useful life, watts per lumen. Learning Objective: 17-1
17.36
a. Sampling (not cost effective to test every engine). b. 100% inspection (airlines record fuel usage and passenger load on every flight). c. 100% inspection (McDonald’s computers would have this information for each day). d. Sampling (you can’t test the life of every battery). e. Sampling (cost might prohibit hospitals from recording this in normal bookkeeping). Learning Objective: 17-1
524
ASBE 6e Solutions for Instructors 17.37
x
is normally distributed from the Central Limit Theorem for sufficiently large values of
n (i.e., symmetric distribution). However, the range and standard deviation do not follow a normal distribution (e.g., standard deviation has a chi distribution). Learning Objective: 17-2 17.38
Answers will vary. It is because
is normally distributed from the Central Limit x Theorem for sufficiently large values of n. However, some processes may not be normal, and subgroups typically are too small for the CLT to apply unless the data are at least symmetric (see Chapter 8). For small n, normality would exist if the underlying process generates normally-distributed data—a reasonable assumption for many, but not all, processes (especially in manufacturing). If non-normal, special techniques are required (beyond the scope of an introductory class in statistics). Learning Objective: 17-2
17.39
a. Variation and chance defects are inevitable in all human endeavors. b. Some processes have very few defects (maybe zero in the short run, but not in the long run). c. Quarterbacks cannot complete all their passes, etc. Learning Objective: 17-2 17.40 Answers will vary, depending on how diligent a web search is conducted. Learning Objective: 17-2
17.41
Answers will vary (e.g., forgot to set clock, clock set incorrectly, couldn’t find backpack, stopped to charge cell phone, had to shovel snow in driveway, alarm didn’t go off, traffic, car won’t start, can’t find parking). Learning Objective: 17-3
17.42
Answers will vary (addition or subtraction error, forgot to record a deposit or withdrawal, recorded data incorrectly e.g., $54.65 instead of $56.54, missing check number, lost debit card receipt). Learning Objective: 17-1
17.43
Answers will vary (e.g., weather, union slowdown, pilot arrived late, crew change required, de-icing planes in winter, traffic congestion at takeoff , no arrival gate available). Learning Objective: 17-1
17.44
a. We are given that LSL = 540 and USL = 550. If = 545 and = 1.25, Cp index: Cp =
USL - LSL 6s
=
550 - 540 6(1.25)
= 1.3333
Cpk index: 525
ASBE 6e Solutions for Instructors
z USL =
USL - m s
=
550 - 545 1.25
zmin =
and =4
z LSL =
m - LSL s
=
545 - 540
=4
1.25
= min{ 4.0, 4.0} = 4.0 and so min( z USL , z LSL )
C pk =
Z min
=
3
4 3
= 1.3333
b. If = 543 mils and = 1.25 mils, Cp index: Cp =
USL - LSL 6s
=
550 - 540 6(1.25)
= 1.33
Cpk index: z USL =
USL - m s
=
550 - 543 1.25
zmin =
and = 5.60
z LSL =
m - LSL s
=
543 - 540 1.25
= 2.40
= min{5.60, 2.40} = 4.00 and so min( z USL , z LSL )
C pk =
Z min 3
=
2.40 3
= 0.80
Learning Objective: 17-9 17.45
a. If = 1.00 mils and = 0.05 mils, and if the minimum capability index is 1.33, this process meets capability standards (Cp = Cpk = 1.33). Cp index: Cp =
USL - LSL 6s
=
1.20 - .80
= 1.33
6(0.05)
Cpk index: z USL =
USL - m s
=
1.20 - 1.00 0.05
zmin =
and = 4.00
z LSL =
m - LSL s
=
1.00 - 0.80 0.05
= 4.00
= min{4.00, 4.00} = 4.00 and so min( z USL , z LSL )
C pk =
Z min 3
=
4.00 3
= 1.333
b. If = 0.90 mils and = 0.05 mils, and if the minimum capability index is 1.33, this process meets capability standards (Cp = 1.33, Cpk = 0.67). Cp = 1.33, Cpk = 0.67 Cp index:
526
ASBE 6e Solutions for Instructors USL - LSL
Cp =
6s
=
1.20 - 0.80
= 1.33
6(0.05)
Cpk index: z USL =
USL - m
=
s
1.20 - 0.90 0.05
zmin =
and = 6.00
z LSL =
m - LSL s
=
0.90 - 0.80 0.05
= 2.00
= min{6.00, 2.00} = 2.00 and so min( z USL , z LSL )
C pk =
Z min 3
2.00
=
3
= 0.667
Learning Objective: 17-9 17.46
a. If = 140 mg and = 5 mg, and if the minimum capability index is 1.33, this process meets capability standards (Cp = Cpk = 1.33). Cp index: Cp =
USL - LSL
=
6s
160 - 120
= 1.33
6(5)
Cpk index: z USL =
USL - m s
=
160 - 140 5
zmin =
and = 4.00
z LSL =
m - LSL s
=
140 - 120
= 4.00
5
= min{4.00, 4.00} = 4.00 and so min( z USL , z LSL )
C pk =
Z min 3
=
4.00 3
= 1.333
b. If = 140 mg and = 3 mg, and if the minimum capability index is 1.33, this process exceeds capability standards (Cp = Cpk = 2.22). Cp index: Cp =
USL - LSL 6s
=
160 - 120
= 2.22
6(3)
Cpk index: z USL =
USL - m s
=
160 - 140
zmin =
3
and = 6.67
z LSL =
m - LSL s
=
140 - 120 3
= 6.67
= min{6.67, 6.67} = 6.67 and min( z USL , z LSL )
C pk =
Learning Objective: 17-9 527
Z min 3
=
6.67 3
= 2.22
ASBE 6e Solutions for Instructors 17.47
a. UCL = 6050 + 3
100 3
and = 6223.205
LCL = 6050 - 3
100 3
= 5876.795
b. Chart violates no rules. c. Process is in control.
Learning Objective: 17-8 17.48
a. Histogram is bell-shaped and probability plot is linear with one possible low outlier (the Anderson-Darling statistic has p-value = .296). Probability Plot of Pounds Normal 99
Mean StDev N AD P-Value
95 90
6073 85.93 24 0.422 0.296
Percent
80 70 60 50 40 30 20 10 5
1
5800
5900
6000
6100
6200
6300
Pounds
b. Yes, it approximates the normal distribution. c. Sample mean is 6072.625 and the sample standard deviation is 85.92505. They are both close to the process values. Learning Objective: 16-4 17.49
a. UCL = 1.00 + 3
.07 5
and = 1.0939
LCL = 1.00 - 3
b. Chart violates no rules. c. Process is in control. Learning Objective: 17-5 Learning Objective: 17-8 528
.07 5
= .9061
ASBE 6e Solutions for Instructors
17.50
a. Histogram is bell-shaped and probability plot is linear with one possible high outlier (the Anderson-Darling statistic has p-value = .656). Probability Plot of Mils Normal 99
Mean StDev N AD P-Value
95 90
1.006 0.06547 35 0.270 0.656
Percent
80 70 60 50 40 30 20 10 5
1
0.85
0.90
0.95
1.00
1.05 Mils
1.10
1.15
1.20
b. The distribution is approximately normal. c. Sample mean is 1.006 and the sample standard deviation is 0.0655, both close to the process values. Learning Objective: 16-4 17.51
a. Cp = 1.00 and Cpk = 0.83. Cp index: Cp =
USL - LSL 6s
=
30 - 18
= 1.00
6(2)
Cpk index: z USL =
USL - m s
=
30 - 23 2
zmin =
and = 3.50
z LSL =
m - LSL s
=
23 - 18 2
= 2.50
= min{3.50, 2.50} = 2.50 and so min( z USL , z LSL )
C pk =
Z min 3
=
2.50 3
= 0.833
b. If the minimum capability index is 1.33, this process is well below capability standards. c. The frequency of the door being opened. Door not being closed tightly. 529
ASBE 6e Solutions for Instructors Learning Objective: 17-9 17.52
a. UCL = 23.00 + 3
2 4
and = 26.00
LCL = 23.000 - 3
2 4
= 20.00
b. Control chart suggests a downward trend but does not violate Rule 4.
c. The sixth mean hits the UCL, so possibly not in control. Learning Objective: 17-5 Learning Objective: 17-8 17.53
a. The sample mean of 23.025 and the standard deviation of 2.006 are very close to the process values ( = 23, = 2). b. The histogram is symmetric, though perhaps platykurtic. Probability plot is linear but Anderson-Darling test statistic has a p-value below .005 so fails normality test. Probability Plot of Temperature Normal 99.9
Mean StDev N AD P-Value
99
Percent
95 90
23.03 2.006 80 1.348 <0.005
80 70 60 50 40 30 20 10 5 1 0.1
15.0
17.5
20.0
22.5 25.0 Temperature
27.5
30.0
Learning Objective: 16-4 17.54
a. Cp = 2.00 and Cpk = 2.00. If the minimum capability index is 1.33, this process is capable. Cp index: 530
ASBE 6e Solutions for Instructors
Cp =
USL - LSL 6s
=
14.6 - 13.4
= 2.00
6(.10)
Cpk index:
z USL =
USL - m s
=
14.60 - 14.00
zmin =
.10
and
= 6.00
z LSL =
m - LSL s
=
14.00 - 13.40 .10
= 6.00
= min{6.00, 6.00} = 6.00 and so min( z USL , z LSL )
C pk =
Z min 3
=
6.00 3
= 2.00
b. Because the process is capable, there is no reason to change unless the customers can see the variation. Learning Objective: 17-9 17.55
a. Histogram is arguably normal, though somewhat bimodal. However, the probability plot is linear and the Anderson-Darling test’s p-value of .795 indicates a good fit to a normal distribution. Probability Plot of Hours Normal 99
Mean StDev N AD P-Value
95 90
8785 216.1 20 0.224 0.795
Percent
80 70 60 50 40 30 20 10 5
1
8200
b. Center line = 8760, UCL = 8760 + 3
200 5
= 9028.33
UCL = 8784.75 + 3 216.1398 5
8600
8800 Hours
9000
9200
, and
c. Center line = 8784.75,
LCL = 8784.75 - 3
8400
216.1398 5
LCL = 8760 - 3
9400
200 5
= 8491.67
, and = 9074.73
= 8494.77
d. The UCL and LCL from the sample differ substantially from those based on the assumed process parameters. This small sample is perhaps not a reliable basis for setting the UCL and LCL. Learning Objective: 16-4 Learning Objective: 17-5 17.56
a. Cp = 2.00 and Cpk = 1.83. 531
ASBE 6e Solutions for Instructors
Cp index: Cp =
USL - LSL
=
6s
477 - 453
= 2.00
6(2)
Cpk index: z USL =
USL - m s
=
477 - 466 2
zmin =
and = 5.50
m - LSL
z LSL =
s
=
466 - 453 2
= 6.50
= min{5.50, 6.50} = 5.50 and so min( z USL , z LSL )
C pk =
Z min 3
=
5.50 3
= 1.83
b. If the minimum capability index is 1.33, this process is capable. c. Need to pack the box without crushing the Chex. Given the smallness and fragility of each Chex it would be difficult to attain. Learning Objective: 17-9 17.57
a. UCL = 465 + 3
3 3
= 470.20
LCL = 465 - 3
3 3
= 459.80
b. No rules are violated. The process is in control.
Learning Objective: 17-5 Learning Objective: 17-8 17.58
a. The histogram and probability plot do not appear grossly non-normal, but the p-value (.042) for the Anderson-Darling test suggests that the box fill may not be normal (our conclusion depends on ).
532
ASBE 6e Solutions for Instructors Probability Plot of Grams Normal 99
Mean StDev N AD P-Value
95 90
464.2 2.797 30 0.761 0.042
Percent
80 70 60 50 40 30 20 10 5
1
456
458
460
462
464 Grams
466
468
470
472
b. Sample mean is 464.2 which is very close to 465. Learning Objective: 16-4 a. From MegaStat: UCL = 12.22095, LCL = 11.75569, centerline = 11.98832 b. Process appears to be in control.
c. The histogram and probability plot (Anderson-Darling p-value = .871) suggest a normal distribution. Probability Plot of Weight in Grams Normal 99.9
Mean StDev N AD P-Value
99 95 90
Percent
17.59
80 70 60 50 40 30 20 10 5 1 0.1
Learning Objective: 16-4 Learning Objective: 17-5 Learning Objective: 17-8 533
11.2
11.4
11.6
11.8 12.0 12.2 Weight in Grams
12.4
12.6
12.8
11.99 0.2208 84 0.204 0.871
ASBE 6e Solutions for Instructors
17.60
a. UCL = p + 3
LCL = p - 3
p(1 - p)
= .06 + 3
n p(1 - p) n
= .06 - 3
(.06)(.94) 200 (.06)(.94) 200
= .11038
= .0096
b. Yes, 200(.06) = 12 and 200(.94) = 188 both are greater than 10. Learning Objective: 17-7 17.61
a. UCL = p + 3
p(1 - p) n
= .05 + 3
(.05)(.95) 100
= .1154
(set to .0000 since LCL can’t be LCL = p - 3
p(1 - p) n
= .05 - 3
(.05)(.95) 100
= -.0154
negative). b. Sample 7 hits the LCL, so the process may not be in control.
c. Samples are too small to assume normality (nπ = 5). Better to use MINITAB’s binomial option to set the control limits. Learning Objective: 17-7 Learning Objective: 17-8 17.62
Chart A: Trend Chart B: Oscillation Chart C: Level Shift Chart D: Instability 534
ASBE 6e Solutions for Instructors Chart E: None Chart F: Cyclical Learning Objective: 17-8 17.63
Chart A: Rule 4. Chart B: No rules violated. Chart C: Rule 4. Chart D: Rules 1, 4. Chart E: No rules violated. Chart F: Rules 1, 2. Learning Objective: 17-8
17.64
Each pattern is clearly evident, except possibly instability in the third series. See charts below.
Yes, upward trend present.
Yes, downward trend present
Yes, instability (Rule 1).
Yes, cyclical (only 6 centerline crossings).
535
ASBE 6e Solutions for Instructors
Maybe oscillation (14 centerline crossings). Learning Objective: 17-8
17.65
Each pattern is clearly evident, except possibly instability in the third series.
Yes, in control (no rules violated).
Yes, upward trend, but no rule violations.
Yes, downward trend, but no rule violations. Yes, unstable (Rule 1).
536
ASBE 6e Solutions for Instructors
Yes, cycle (only 7 centerline crossings). Learning Objective: 17-8
537