In: Statistics and Probability
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.
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 )