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 of94members of the​ website's premium program who recently made a purchase on the website yielded a mean of$1700and a standard deviation of​$250

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

Answer 1

Solution :

Given that,

Point estimate = sample mean = = $1700

Population standard deviation =    =$250
Sample size = n =94

At 99% confidence level the z is ,

  = 1 - 99% = 1 - 0.99 = 0.01

/ 2 = 0.01 / 2 = 0.005

Z/2 = Z0.005 = 2.576   ( Using z table )

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

= 2.576* ( 250/ 94)

= 66.4235

At 99% confidence interval estimate of the population mean is,

- E < < + E

1700- 66.4235< < 1700+66.4235

1633.5765< < 1766.4235

(1633.5765, 1766.4235 )