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 96 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) Construct a 95​% 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

Solution :

Given that,

Point estimate = sample mean = = 1400

sample standard deviation = s = 200

sample size = n = 96

Degrees of freedom = df = n - 1 = 96 - 1 = 95

At 95% confidence level

= 1 - 95%

=1 - 0.95 =0.05

/2 = 0.025

t/2,df = t_0.025,95 = 1.985

Margin of error = E = t_/2,df * (s /n)

= 1.985 * (200 / 96)

Margin of error = E = 40.52

The 95% confidence interval estimate of the population mean is,

- E + E

1400 - 40.52   1400 + 40.52

1359.48    1440.52