How much evidence is there that the mean bad debt ratio for Ohio banks exceeds 3.5 percent?

How much evidence is there that the mean bad debt ratio for Ohio banks exceeds 3.5 percent?

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The p-value for the hypothesis test of Exercise 9.53 can be computer calculated to be .0200. How much evidence is there that μ is less than eight seconds?

9.55 The bad debt ratio for a financial institution is defined to be the dollar value of loans defaulted divided by the total dollar value of all loans made. Suppose that a random sample of seven Ohio banks is selected and that the bad debt ratios (written as percentages) for these banks are 7%, 4%, 6%, 7%, 5%, 4%, and 9%. BadDebt

a Banking officials claim that the mean bad debt ratio for all Midwestern banks is 3.5 percent and that the mean bad debt ratio for Ohio banks is higher. Set up the null and alternative hypotheses needed to attempt to provide evidence supporting the claim that the mean bad debt ratio for Ohio banks exceeds 3.5 percent.

b Assuming that bad debt ratios for Ohio banks are approximately normally distributed, use critical values and the given sample information to test the hypotheses you set up in part a by setting α equal to .10, .05, .01, and .001. How much evidence is there that the mean bad debt ratio for Ohio banks exceeds 3.5 percent? What does this say about the banking official’s claim?

c Are you qualified to decide whether we have a practically important result? Who would be? How might practical importance be defined in this situation?

d The p-value for the hypothesis test of part (b) can be computer calculated to be .006. What does this p-value say about whether the mean bad debt ratio for Ohio banks exceeds 3.5 percent?

9.56 In the book Business Research Methods, Donald R. Cooper and C. William Emory (1995) discuss using hypothesis testing to study receivables outstanding. To quote Cooper and Emory:

…the controller of a large retail chain may be concerned about a possible slowdown in payments by the company’s customers. She measures the rate of payment in terms of the average number of days receivables outstanding. Generally, the company has maintained an average of about 50 days with a standard deviation of 10 days. Since it would be too expensive to analyze all of a company’s receivables frequently, we normally resort to sampling.

a Set up the null and alternative hypotheses needed to attempt to show that there has been a slowdown in payments by the company’s customers (there has been a slowdown if the average days outstanding exceeds 50).

b Assume approximate normality and suppose that a random sample of 25 accounts gives an average days outstanding of with a standard deviation of s = 8. Use critical values to test the hypotheses you set up in part a at levels of significance α = .10, α = .05, α = .01, and α = .001. How much evidence is there of a slowdown in payments?

c Are you qualified to decide whether this result has practical importance? Who would be?

9.57 Consider a chemical company that wishes to determine whether a new catalyst, catalyst XA-100, changes the mean hourly yield of its chemical process from the historical process mean of 750 pounds per hour. When five trial runs are made using the new catalyst, the following yields (in pounds per hour) are recorded: 801, 814, 784, 836, and 820. ChemYield

a Let μ be the mean of all possible yields using the new catalyst. Assuming that chemical yields are approximately normally distributed, the MegaStat output of the test statistic and p-value, and the Excel output of the p-value, for testing H0: μ = 750 versus Ha: μ ≠ 750 are as follows:

(Here we had Excel calculate twice the area under the t distribution curve having 4 degrees of freedom to the right of 6.942585.) Use the sample data to verify that the values of , s, and t given on the output are correct.

b Use the test statistic and critical values to test H0 versus Ha by setting α equal to .10, .05, .01, and .001.

9.58 Consider Exercise 9.57. Use the p-value to test H0: μ = 750 versus Ha: μ ≠ 750 by setting α equal to .10, .05, .01, and .001. How much evidence is there that the new catalyst changes the mean hourly yield?

9.59 Whole Foods is an all-natural grocery chain that has 50,000 square foot stores, up from the industry average of 34,000 square feet. Sales per square foot of supermarkets average just under $400 per square foot, as reported by USA Today in an article on “A whole new ballgame in grocery shopping.” Suppose that sales per square foot in the most recent fiscal year are recorded for a random sample of 10 Whole Foods supermarkets. The data (sales dollars per square foot) are as follows: 854, 858, 801, 892, 849, 807, 894, 863, 829, 815. Let μ denote the mean sales dollars per square foot for all Whole Foods supermarkets during the most recent fiscal year, and note that the historical mean sales dollars per square foot for Whole Foods supermarkets in previous years has been $800. Below we present the MINITAB output obtained by using the sample data to test H0: μ = 800 versus Ha: μ > 800. WholeFoods

a Use the p-value to test H0 versus Ha by setting α equal to .10, .05, and .01.

b How much evidence is there that μ exceeds $800?

9.60 Consider Exercise 9.59. Do you think that the difference between the sample mean of $846.20 and the historical average of $800 has practical importance?

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