A random sample of 40 companies with assets over $10 million was surveyed and asked to indicate their industry and annualcomputer technology expense. The ANOVA comparing the average computer technology expense among three industriesrejected the null hypothesis. The mean square error (MSE) was 195. The following table summarized the results:EducationTax ServicesFood ServicesNumber Sampled101416Mean Expense (1,000,000s)15.520Based on the comparison between the mean annual computer technology expense for companies in the food services and taxservices industries, the 95% confidence interval shows an interval of –5.85 to 14.85 for the difference. This result indicates that Multiple Choicethe interval contains a difference of 20.7companies in the food service industry spend significantly less than companies in the tax service industrythere is not enough evidence of a significant difference between the two industry technology expensescompanies in the food service industry spend significantly more than companies in the tax service industry

Solution Based on the comparison between the mean annual computer technology expense for companies…

Answered: A random sample of 40 companies with…

MATLAB: An Introduction with Applications

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

Section: Chapter Questions

Problem 1P

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A veterinarian wants to perform a study at a local animal shelter where 200 dogs need training on behavior modification. The purpose of the study would be to determine if using a firm voice during training for 10 minutes improves a dog's learning behavior compared to not speaking during training for 10 minutes. Part A: Describe an appropriate design for the study. Part B: The hypotheses for this study are as follows: Ho: There is no difference in the mean improvement of dogs during behavior modification for either treatment. Ha: The mean improvement of dogs during behavior modification is greater for the firm voice treatment. The shelters will allow trainers to use a firm voice during training if the null hypothesis is rejected. What are the possible Type I and II errors? Describe the consequences of each in the context of this study and discuss which type you think is more serious.n
Suppose the mean price for used cars is $10,198. A manager of a Kansas City used car dealership reviewed a sample of 50 recent used car sales at the dealership in an attempt to determine whether the population mean price for used cars at this particular dealership differed from the national mean. The prices for the sample of 50 cars are shown in the file named UsedCars. 12,400 10,400 12,100 10,000 11,000 8,895 7,675 9,975 6,350 10,470 9,895 11,250 8,795 12,500 9,340 10,150 9,200 9,395 11,000 10,640 10,000 7,500 8,000 10,440 10,200 10,300 9,740 9,280 10,930 8,000 9,000 7,680 9,400 10,730 7,350 12,240 11,970 8,240 9,910 10,080 9,440 8,970 9,500 10,050 10,130 11,400 8,500 7,500 9,090 10,500 (a) Formulate the hypotheses that can be used to determine whether a difference exists in the mean price for used cars at the dealership. H0: ? ≤ 10,198 Ha: ? > 10,198 H0: ? < 10,198 Ha: ? ≥ 10,198 H0: ? = 10,198 Ha: ? ≠ 10,198 H0: ? > 10,198 Ha: ? ≤ 10,198 H0: ? ≥ 10,198…
(3) The table contains hypothetical data from a study that a bank conducted on the credit card balances of male and female credit card holders (credit cards owned by both men and women were left out of the study). Male Female Mean $3,550 $2,670 Median $2,175 $2,500 Std Dev $1,240 $825 a. What conclusions can we come to by comparing the medians of the males and females? b. What conclusions can we come to by comparing the standard deviations of the males and females? c. Notice that the mean and median of the males are fairly far apart while the mean and median of the females are not. What does this tell us about the credit card balances of the males vs the credit card balances of the females?
An adventure company runs two obstacle courses, Fundash and Coolsprint. The designer of the courses suspects that the mean completion time of Fundash is not equal to the mean completion time of Coolsprint. To test this, she selects 240 Fundash runners and 200 Coolsprint runners. (Consider these as random samples of the Fundash and Coolspring runners.) The 240 Fundash runners complete the course with a mean time of 76.7 minutes and a standard deviation of 5.2 minutes. The 200 Coolsprint runners complete the course with a mean time of 77.5 minutes and a standard deviation of 5.4 minutes. Assume that the population standard deviations of the completion times can be estimated to be the sample standard deviations, since the samples that are used to compute them are quite large. At the 0.05 level of significance, is there enough evidence to support the claim that the mean completion time, μ1 , of Fundash is not equal to the mean completion time, μ2 , of Coolsprint?…
A psychology student conducted a study on using a chief executive officer's facial structure to predict a firm's financial performance. The facial width-to-height ratio (WHR) for each in a sample of 55 CEOs at publicly traded firms was determined. The sample resulted in x=1.74 and s=2.02. The student wants to predict the financial performance of a firm based on the value of the true mean facial WHR of CEOs. The student wants to use the value of μ=2.3. Do you recommend he use this value? Conduct a test of hypothesis for μ to help you answer the question. Specify all the elements of the test, including H0, Ha, test statistic, p-value, and your conclusion. Test at α=0.10. Find the p-value. p-value=enter your response here (

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