Question

A corporation monitors time spent by office workers browsing the web on their computers instead of working. In a sample of computer records of 39 workers, the average amount of time spent browsing in an eight-hour work day was 35 minutes with standard deviation 2.9 minutes. Construct a 99% confidence interval for the mean number of minutes the office workers spend browsing the web.

Report the answer accurate to four decimal places

_____ < μμ < _______

Answer 1

Solution :

Given that,

Point estimate = sample mean = = 35

Population standard deviation =    = 2.9

Sample size = n =39

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 * (2.9 / 39)

= 1.1962

At 99% confidence interval is,

- E < < + E

35-1.1962 < < 35+1.1962

33.8038< < 36.1962