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?