Question

A U.S. labor economist interviews 581 workers and finds that the mean number of hours they worked in the last week was 36.9 with a standard deviation of 3.6 hours. Make a 95% confidence interval for the mean of all workers in the U.S.

Answer 1

Solution :

Given that,

Point estimate = sample mean = = 36.9

Population standard deviation = = 3.6

Sample size = n = 581

At 95% confidence level the z is ,

= 1 - 95% = 1 - 0.95 = 0.05

/ 2 = 0.05 / 2 = 0.025

Z_/2 = Z_0.025 = 1.96

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

= 1.96 * (3.6 / 581)

= 0.3

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

- E < < + E

36.9 - 0.3 < < 36.9 + 0.3

36.6 < < 37.2

The 95% confidence interval for the mean 36.6 , 37.2