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 300.

Complete parts​ (a) and​ (b) below.

a. Construct a 90% confidence interval estimate for the mean spending for all shoppers who are members of the​ website's premium program.

b. Interpret the interval of (a)

Answer 1

Solution :

Given that,

Point estimate = sample mean = = $1900

Population standard deviation =    = $300

Sample size = n =99

At 90% confidence level

= 1 - 90%  

= 1 - 0.90 =0.10

/2 = 0.05

Z/2 = Z_{0.05 = 1.645}

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

= 1.645* ( 300/  99 )

= 49.5986
At 90% confidence interval estimate of the population mean
is,

- E < < + E

1900 - 49.5986 <   < 1900+ 49.5986

1850.4014<   < 1949.5986

( 1850.4014 , 1949.5986 )