Question

Customers who participate in a store's free loyalty card program save money on their purchases, but allow the store to keep track of the customer's shopping habits and potentially sell these data to third parties. A Survey revealed that 225 of a random sample of 250 U.S. adults would agree to participate in a store loyalty card program, despite the potential for information sharing. Let p represent the true proportion of all customers who would participate in a store loyalty card program. A store owner claims that more than 80% of all customers would participate in a loyalty card program. If the probability of a type I error is set at .01, test the store owners claim that the true proportion of all customers who would participate in a store loyalty card program exceeds .8 using the survey data. What decision and conclusion should be made?

A. Reject H₀ as the test statistic of 5.27 is greater than the criitcal value of .326. We conclude the true proportion of all customers who would participate in a store loyalty card program doe snot significantly exceed 80%. There is not sufficient evidence to support the store owners claim.

B. Fail to reject H₀ as the sample proportion of .9 is greater than the Type I error of .01. We conclude the true proportion of all customers who would participate in a store loyalty card program exceeds 80%. There is sufficient evidence to support the store owners claim.

C. Reject H₀ as the test statistic of 3.95 is greater than the critical value of 2.326. We conclude the true proportion of all customers who would participate in a store loyalty card program significantly exceeds 80%. There is sufficient evidence to support the store owners claim.

D. Fail to reject H₀ as the test statistic of 3.95 is greater than the critical value of 2.326. We conclude the true proportion of all customers who would participate in a store loyalty card program does not significantly exceed 80%. There is not sufficient evidence to support the store owners claim.

Answer 1

TS = (p^ - p)/sqrt(pq/n)
= (225/250 - 0.8)/sqrt(0.8*0.2/250)
= 3.95284
critical value for right-tailed test is 2.326
TS > critical value
we reject the null
hence
option C) is correct

C. Reject H0 as the test statistic of 3.95 is greater than the critical value of 2.326. We conclude the true proportion of all customers who would participate in a store loyalty card program significantly exceeds 80%. There is sufficient evidence to support the store owners claim.