Homework SourseA survey is being planned to determine the mean amount of ti
A survey is being planned to determine the mean amount of time corporation executives watch television. A pilot survey indicated that the mean time per week is 13 hours, with a standard deviation of 2.0 hours. It is desired to estimate the mean viewing time within one-quarter hour. The 90 percent level of confidence is to be used.
| A survey is being planned to determine the mean amount of time corporation executives watch television. A pilot survey indicated that the mean time per week is 13 hours, with a standard deviation of 2.0 hours. It is desired to estimate the mean viewing time within one-quarter hour. The 90 percent level of confidence is to be used. |
Solution
Given a=1-0.9=0.1, Z(0.05) = 1.645 (from standard normal table)
So n=(Z*s/E)^2
=(1.645*2/0.25)^2
=173.1856
Take n=174
