In the book Business Research Methods, Donald R. Cooper and C. William Emory (1995) discuss a manager who wishes to compare the effectiveness of two methods for training new salespeople. The authors describe the situation as follows: The company selects 22 sales trainees who are randomly divided into two equal experimental groupsone receives type A and the other type Btraining. The salespeople are then assigned and managed without regard to the training they have received. At the year's end, the manager reviews the performances of salespeople in these groups and nds the following results: A Group B Group Average Weekly Sales 331 = $1,350 352 = $1,086 Standard Deviation 51 = 233 52 = 263 (a) Set up the null and alternative hypotheses needed to attempt to establish that type Atraining results in higher mean weekly sales than does type Btraining. H0:uA uBS ' 'Ha: uAuB> r ' (b) Because different sales trainees are assigned to the two experimental groups, it is reasonable to believe that the two samples are independent. Assuming that the normality assumption holds, and using the equal variances procedure, test the hypotheses you set up in part a at level of significance .10, .05, .01 and .001. How much evidence is there that type A training produces results that are superior to those of type B? (Round your answer to 3 decimal places.) H0 with u equal to 0.10. H0 with u equal to 0.05 H0 with u equal to 0.01 H0 with u equal to 0.001 evidence that [M - p B > 0 (c) Use the equal variances procedure to calculate a 95 percent confidence interval for the difference between the mean weekly sales obtained when type A training is used and the mean weekly sales obtained when type B training is used. Interpret this interval. (Round your answer to 2 decimal places.) Confidence interval [A marketing research rm wishes to compare the prices charged by two supermarket chainsMiller's and Albert's. The research firm, using a standardized oneweek shopping plan (grocery list), makes identical purchases at 10 of each chain's stores. The stores for each chain are randomly selected, and all purchases are made during a single week. It is found that the mean and the standard deviation of the shopping expenses at the 10 Miller's stores are $121.92 and $1.40, respectively. It is also found that the mean and the standard deviation of the shopping expenses at the 10 Albert's stores are $114.81 and $1.84, respectively. Assuming normality, test to see if the corresponding population variances differ by setting (1 equal to .05. Is it reasonable to use the equal variances procedure to compare population means? (a) Calculate the value of the test statistic. (Do not round intermediate calculations. Round your answer to 2 decimal places.) Test statistic I ' (b) Calculate the critical value. (Round your answer to 2 decimal places.) Critical value I ' (c) At the .05 significance level, what it the conclusion? 0 Reject HO O Fail to reject H0 Standard deviation of returns is often used as a measure of a mutual funds volatility (risk). A larger standard deviation of returns is an indication of higher risk. According to Morningstarcom (June 17, 2010), the American Century Equity Income Institutional Fund, a large cap fund, has a standard deviation of returns equal to 19.68 percent. Morningstancom also reported that the Fidelity Small Cap Discovery Fund has a standard deviation of returns equal to 21.21 percent. Each standard deviation was computed using a sample of size 31. (a) Calculate the value of the test statistic. (Do not round intermediate calculations. Round your answer to 2 decimal places.) Test statistic I I (b) Calculate the critical value by using excel. (Round your answer to 2 decimal places.) Critical value I I (c) At the .10 significance level, what it the conclusion? 0 Fail to reject O Reject In an article in the Journal ofAdvertising, Weinberger and Spotts compare the use of humor in television ads in the United States and in the United Kingdom. Suppose that independent random samples of television ads are taken in the two countries. A random sample of 400 television ads in the United Kingdom reveals that 141 use humor, while a random sample of 500 television ads in the United States reveals that 122 use humor. (a) Set up the null and alternative hypotheses needed to determine whether the proportion of ads using humor in the United Kingdom differs from the proportion of ads using humor in the United States. H0: p1 - p2 0 versus Ha: p1 - p2 0. (b) Test the hypotheses you set up in part a by using critical values and by setting (1 equal to .10, .05, .01, and .001. How much evidence is there that the proportions of UK. and US. ads using humor are different? (Round the proportion values to 3 decimal places. Round your answer to 2 decimal places.) 2 I' r H0 at each value of a; r I' evidence. (c) Set up the hypotheses needed to attempt to establish that the difference between the proportions of U.K. and US. ads using humor is more than .05 (ve percentage points). Test these hypotheses by using a pvalue and by setting 0. equal to .10, .05, .01, and .001. How much evidence is there that the difference between the proportions exceeds .05? (Round the proportion values to 3 decimal places. Round your 2 value to 2 decimal places and p-value to 4 decimal places.) pval ue r H0 at each value of u = .10 and u = .05; r 'evidenoe. (d) Calculate a 95 percent confidence interval for the difference between the proportion of U.K. ads using humor and the proportion of us. ads using humor. Interpret this interval. Can we be 95 percent condent that the proportion of U.K. ads using humor is greater than the proportion of us. ads using humor? (Round the proportion values to 3 decimal places. Round your answers to 4 decimal places.) 95% of Condence Interval [E __;lu A marketing research rm wishes to compare the prices charged by two supermarket chains Miller's and Albert's. The research rm, using a standardized one-week shopping plan (grocery list), makes identical purchases at 10 of each chain's stores. The stores for each chain are randomly selected, and all purchases are made during a single week. The shopping expenses obtained at the two chains, along with box plots of the expenses, are as follows: Miller's $119.25 $121.32 $122.34 $120.14 $122.19 $123.71 $121.72 $122.42 $123.63 $122.44 Albert's $111.99 $114.88 $115.11 $117.02 $116.89 $116.62 $115.38 $114.40 $113.91 $111.87 13 Click here for the Excel Data File 124 119 Expense 114 Albert Miller Market Because the stores in each sample are different stores in different chains, it is reasonable to assume that the samples are independent, and we assume that weekly expenses at each chain are normally distributed. Twosample T for Millers vs Alberts N Mean StDeV SE Mean Millers 10 121.92 1.40 0.44 Alberts 10 114.81 1.84- 0.58 (b) Using the tstatistic given on the output and critical values, test H0 versus Ha by setting 0 equal to .10, .05, .01, and .001. How much evidence is there that the mean weekly expenses at Miller's and Albert's differ? (Round your answer to 2 decimal places.) (c) Above gure gives the p-value for testing Ho : ,uM 11A = 0 versus Ha : um ,uA at 0. Use the pvalue to test Ho versus Ha by setting a equal to .10, .05, .01, and .001. How much evidence is there that the mean weekly expenses at Miller's and Albert's differ? (d-1) Above figure gives a 95 percent confidence interval for UM - MA. Use this confidence interval to describe the size of the difference between the mean weekly expenses at Miller's and Albert's. (Round your answers to 2 decimal places.) The mean weekly expenses differ between and (d-2) Do you think that these means differ in a practically important way? O Not probably O Probably (e) Set up the null and alternative hypotheses needed to attempt to establish that the mean weekly expense for the shopping plan at Miller's exceeds the mean weekly expense at Albert's by more than $5. Test the hypotheses at the .10, .05, .01, and .001 levels of significance. How much evidence is there that the mean weekly expense at Miller's exceeds that at Albert's by more than $5? (Round your answer so- to 4 decimal places and t-value to 3 decimal places.) HO: UM - HA versus Ha: UM - HA s2p = 2.6728 t= 2.886 Reject HO at a = but not at a = evidence