Question

A marketing researcher wants to estimate the mean amount spent​ ($) on a certain retail website by members of the​ website's premium program. A random sample of 98 members of the​ website's premium program who recently made a purchase on the website yielded a mean of ​$1200 and a standard deviation of ​$300. Construct a 99​% confidence interval estimate for the mean spending for all shoppers who are members of the​ website's premium program. ​(Round to two decimal places as​ needed.)

Answer 1

Let the sample size be n=98, mean be 7 = 1200 and a standard deviation sd - 300 Compute 99% confidence interval estimate for the mean spending for all shoppers who are members of the website's premium program. That is, Here, te is the critical t-value obtained from the t-table at a=0.01 and degrees of freedom (n-1)=(98-1)=97 - (1200(2.62754 Yok ). 1200+(2.6275x ) =(1120.37,1279.63) Therefore, the 99% confidence interval estimate for the mean spending for all shoppers who are members of the website's premium program is (1120.37,1279.63).