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 99 members of the​ website's premium program who recently made a purchase on the website yielded a mean of ​$1900 and a standard deviation of ​$200. Complete parts​ (a) and​ (b) below.

a. Construct a 95​% 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 95​% 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. 95​% 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.

C.The sample mean spending in 95​% of all samples of 99 members of the​ website's premium program will be between the lower and upper limits of the confidence interval.

D.There is a 95​% 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.

Answer 1

Solution:

a) Confidence interval, CI = mean -/+ z-value*(standard deviation/n^(1/2))

Z-value at 95% confidence level = 1.96, so

CI = 1900 -/+ 1.96*(200/99^(1/2))

CI = 1900 -/+ 39.397

CI = (1900 - 39.397, 1900 + 39.397)

CI = (1860.603, 1939.397)

b) Confidence interval is interpreted as follows: when repetitive samples (of 99) are taken, 95% of the times the sample mean value will lie within the interval range.

Thus, the correct option is (C).