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 92 members of the​ website's premium program who recently made a purchase on the website yielded a mean of ​$1400 and a standard deviation of ​$200. Complete parts​ (a) and​ (b) below.

A. Construct a 99​% confidence interval estimate for the mean spending for all shoppers who are members of the​ website's premium program.

____≤μ≤_____

​(Round to two decimal places as​ needed.)

b. Interpret the interval constructed in​ (a).

Choose the correct answer below.

A.With 99​% ​confidence, the mean spending in dollars for all shoppers who are members of the​ website's premium program is between the lower and upper limits of the confidence interval.

B.There is a 99​% probability that mean spending in dollars for all shoppers who are members of the​ website's premium program is between the lower and upper limits of the confidence interval.

C. 99​% of all shoppers who are members of the​ website's premium program have spent an amount in dollars that is between the lower and upper limits of the confidence interval.

D.The sample mean spending in 99​% of all samples of 92 members of the​ website's premium program will be between the lower and upper limits of the confidence interval.

Answer 1

2.576

Since population standard deviation is unknown use t distribution df=n-1=92-1=91

For 99%percent of confidence t*=2.629

The 99 percent confidence interval is xbar+-(t*)*s/(n^1/2)

(1400+-2.629)*200/(92)1/2

1400+-54.81

1454.81,1335.19