In: Statistics and Probability
Jewel, the branch manager of an outlet store of a nationwide chain of pet supply stores, wants to study characteristics of her customers. In particular, she decides to focus on the amount of money spent by her customers. A sample of 50 customers spent an average of $21.34 with a standard deviation of $9.25 on pet supplies.
a. What is the sampling distribution of the point estimate?
b. Calculate the margin of error for a 99% confidence interval estimating the actual average amount spent in the pet supply store.
c. Construct and interpret the confidence interval.
Solution
Given that,
a ) = 21.34
= 9.25
n =50
b ) 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
Margin of error = E = Z/2* (/n)
= 2.576 * (9.25 / 50 )
= 3.37
c ) At 99% confidence interval estimate of the population mean is,
- E < < + E
21.34 - 3.37 < < 21.34 + 3.37
17.97 < < 24.71
(17.97, 24.71 )