A recent study found that 64 children who watched a commercialfor potato chips featuring a celebrity endorser ate a mean of 36grams of potato chips as compared to a mean of 26 grams for 54children who watched a commercial for an alternative food snack.Suppose that the sample standard deviation for the children whowatched the​ celebrity-endorsed commercial was 21.1 grams and thesample standard deviation for the children who watched thealternative food snack commercial was 12.9 grams. Complete parts​(a) through​ (c) below.
a. Assuming that the population variances are equal andalphaequals0.05​, is there evidence that the mean amount of potatochips eaten was significantly higher for the children who watchedthe​ celebrity-endorsed commercial? Let population 1 be the weightsof potato chips eaten by children who watched the​celebrity-endorsed commercial and let population 2 be the weightsof potato chips eaten by children who watched the alternative foodsnack commercial. What are the correct null and alternative​hypotheses? a)What is the test​ statistic? (Round to two decimalplaces)
what is the p value? (round to three decimal places)
b) Assuming that the population variances are​ equal, constructa 95​% confidence interval estimate of the difference mu 1 minus mu2 between the mean amount of potato chips eaten by the children whowatched the​ celebrity-endorsed commercial and children who watchedthe alternative food snack commercial.determine the 95​% confidenceinterval using​ technology, rounding to two decimal places.
c) Compare and discuss the results of​ (a) and​ (b). Check tomake sure the results of​ (a) and​ (b) can be compared. Recall thatconfidence intervals are comparable only to the results of a​two-tail hypothesis test. Since the test done in part​ (a) is a​one-tail hypothesis​ test, these two results cannot be meaningfullycompared.
