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 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