Question: National data show that, on average, college freshman spend 7.5 hours a week going to parties. President DeRosa doesn’t believe that these figures apply at his school; he thinks freshman at his school party less. He takes a simple random sample of 50 freshmen and interviews them. He asks you to analyze his sample data. You determine that a 95% confidence interval for the mean number of hours freshmen spend each week going to parties is (5.7, 7.2).

(a) Explain to the President what is meant by the “95% confidence interval of (5.7, 7.2).”

(b) The President wants to test the hypothesis that the mean for his school is different from the national mean at a 5% significance level. Specify the null and alternative hypotheses for this test.

(c) Which test procedure should he use to conduct this test? Why?

(d) Eager to gain favor with the president, you tell him that you can save him lots of time because, based on the confidence interval results already presented, you know what he will conclude and he doesn’t have to perform any other calculations. Will he reject or fail to reject the null hypothesis at the 5% significance level? Explain.