5. In recent years, several companies have been formed to compete with AT&T in long distance...

 

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5. In recent years, several companies have been formed to compete with AT&T in long distance calls. All advertise that their rates are lower than AT&T and as a result their bills will be lower. AT&T has responded by arguing that there will be no difference in billing for the average consumer. Suppose that a statistics practitioner working for AT&T determines that the mean and standard deviation of monthly long-distance bills for all its residential customers are $17.09 and $3.87, respectively. He then takes a random sample of 100 customers and recalculates their last month's bill using the rates quoted by a leading competitor. Assuming that the standard deviation of this population is the same as for AT&T, and using, E xi=1,754.99. a. Under normality of the population of long distance calls, what is the proportion of AT&T residential customers who pay more than $25. b. We want to know is the average leading competitor's bill different from the average AT&T bill. Set up the null and alternative hypotheses. Using the rejection region method, test the hypothesis with 5% significance level c. Calculate the p-value of the test and interpret its meaning. What is the conclusion of the test using the p-value? d. Explain the difference between a rejection region and the p-value in hypothesis testing. e. Explain what happens to the p-value in (b) if the sample size is decreased to 50 customers. f. The power of a test is the probability to correctly reject the null hypothesis when a specific alternative hypothesis is true. Compute the power of the test in (a) when the true population mean value is $18. What do you conclude about the test? 5. In recent years, several companies have been formed to compete with AT&T in long distance calls. All advertise that their rates are lower than AT&T and as a result their bills will be lower. AT&T has responded by arguing that there will be no difference in billing for the average consumer. Suppose that a statistics practitioner working for AT&T determines that the mean and standard deviation of monthly long-distance bills for all its residential customers are $17.09 and $3.87, respectively. He then takes a random sample of 100 customers and recalculates their last month's bill using the rates quoted by a leading competitor. Assuming that the standard deviation of this population is the same as for AT&T, and using, E xi=1,754.99. a. Under normality of the population of long distance calls, what is the proportion of AT&T residential customers who pay more than $25. b. We want to know is the average leading competitor's bill different from the average AT&T bill. Set up the null and alternative hypotheses. Using the rejection region method, test the hypothesis with 5% significance level c. Calculate the p-value of the test and interpret its meaning. What is the conclusion of the test using the p-value? d. Explain the difference between a rejection region and the p-value in hypothesis testing. e. Explain what happens to the p-value in (b) if the sample size is decreased to 50 customers. f. The power of a test is the probability to correctly reject the null hypothesis when a specific alternative hypothesis is true. Compute the power of the test in (a) when the true population mean value is $18. What do you conclude about the test?

Answer rating: 100% (QA)

a To find the proportion of ATT residential customers who pay more than 25 we need to calculate the zscore and then find the corresponding area under the normal distribution curve First calculate the ... View the full answer