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 97

members of the​ website's premium program who recently made a purchase on the website yielded a mean of $2000 and a standard deviation of $150.

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).

Answer 1

Solution :

Given that,

Point estimate = sample mean = = $2000

Population standard deviation =    = $150

Sample size = n =97

At 95% confidence level the z is ,

= 1 - 95% = 1 - 0.95 = 0.05

/ 2 = 0.05 / 2 = 0.025

Z_/2 = Z_0.025 = 1.96

Margin of error = E = Z/2 * ( /n)

= 1.96 * ( 150 /  97 )

=29.85
At 95% confidence interval estimate of the population mean
is,

- E < < + E

2000 - 29.85   <   < 2000+ 29.85

1970.15 <   < 2029.85

( 1970.15 , 2029.85)

interpretation = At 95% confidence interval estimate of the population mean is fall between lower bound and upper bound( 1970.15 , 2029.85)