Question

A recent study found that 65 children who watched a commercial for potato chips featuring a celebrity endorser ate a mean of 39 grams of potato chips as compared to a mean of 25 grams for 55 children who watched a commercial for an alternative food snack. Suppose that the sample standard deviation for the children who watched the​ celebrity-endorsed commercial was 21.2 grams and the sample standard deviation for the children who watched the alternative food snack commercial was 12.7 grams. Complete parts​ (a) through​ (c) below.

a)What is the test​ statistic? What is the p-value?

b) Assuming that the population variances are​ equal, construct a 95​% confidence interval estimate of the difference mu 1 minus mu 2 between the mean amount of potato chips eaten by the children who watched the​ celebrity-endorsed commercial and children who watched the alternative food snack commercial.determine the 95​% confidence interval using​ technology, rounding to two decimal places

c) Compare and discuss the results of​ (a) and​ (b). Check to make sure the results of​ (a) and​ (b) can be compared. Recall that confidence 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 meaningfully compared.

Answer 1

potato chips for) pop-1 . (alternative food) I snack Pop"- 2 n = 65 n = 55 My- 25 gan x = 39 gan we will use 2-sample samples t-test for & Calculating pooled Ad. dern/estimate of Рp" хам и с Sp = (1, 1) si + (Mat) sa = (64)(212)+(54) (12.7)? n, + ₂-2 65+ 55-2 317.5747 Sp = 17.82063 Test Calculating Turn = sidence for diff of mean sp. 1 1 2 + 1 = 17.82063.41 = 3.264943

& Test Statistic (5.3.) is given by : 39-25) -o - 3.2649.43 + = 14 3-264943 i, t = 4.287977 Ho: l M.-M. -o 4 ,- >O a = 0-05 C one -tailed test (based on statement fc part) s t p-value = Pn T.S. 34.29) i = 0.000 Retete Z VACAA t-dixis f = 118 Pevalve 971 t = 4.29 •

118 6 95% ca in given by I from t-table, coresponding to df = n, they & crea of 0.025 in each fail critical t-values are given as to = t 1-980 = (x-r>) + (te) Cervo Uchaart. = - Li@ast. = (ay-two)- (to learn toe) = = 14 - (1.980 (3.26 4943) 7.535413 An . . In onder to compare 95% for lower calenlated a & b, bound ů to for difference be 2 (Co-To) - (to) (Teo-Fe) te n @ 954. 14 - (1652) 3.264943) t = -1.658 » = 8.586725 . - 95% CI For a purt - lower bound does not contains zero & also for b part does not contain zero so me reject Ho.