Question

A recent study found that 63 children who watched a commercial for potato chips featuring a celebrity endorser ate a mean of 38 grams of potato chips as compared to a mean of 25 grams for 53 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 21.3 grams and the sample standard deviation for the children who watched the alternative food snack commercial was 12.7 grams.

A.   Assuming that the population variances are equal and α=0.05​, is there evidence that the mean amount of potato chips eaten was significantly higher for the children who watched the​ celebrity-endorsed commercial? Let population 1 be the weights of potato chips eaten by children who watched the​ celebrity-endorsed commercial and let population 2 be the weights of potato chips eaten by children who watched the alternative food snack commercial.

null and alternative​ hypotheses is

H0​:μ1−μ2 ≠0

H1​:μ1−μ2=0

What is the test​ statistic?

tSTAT=

p-value=

b. Assuming that the population variances are​ equal, construct a 90%, 95%, and 99% confidence interval estimate of the difference μ1−μ2 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. The formula for the confidence interval estimate for the difference in the means of two independent samples is given​ below, where tα/2 is the critical value of the t​ distribution, with n1+n2−2 degrees of​ freedom, for an area of α/2 in the upper​ tail, andX1​,X2​, S2p​, n1​, and n2 are all defined the same as they are for the test statistic.

Answer 1

Here in this scenario our claim is that the mean amount of potato chips eaten was significantly higher for the children who watched the​ celebrity-endorsed commercial.

To test this claim we have to use two sample t test because here the population standard deviations is unknown. Now based on given sample information.

Further the test is performed at 0.05 % level of significance as below,

The t critical value is calculated using t table.

The t cal test Statistic to = 3.897.

The p value is 0.0001.

B) Assuming that the population Variance are equal we constructed the 90% , 95% and 99% Confidence Interval for difference in means as using t distribution as below,

The formula for calculating the Confidence Interval for difference in means as below,

The t critical value is calculated using t table.

Hope it helps.

Thank you.