A 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 12 hours, with a standard deviation of 4.0 hours. It is desired to estimate the mean viewing time within two-quarter hour. The 95 percent level of confidence is to be used.

Solution

Compute Sample Size
n = (Z a/2 * S.D / ME ) ^2
Z/2 at 0.05% LOS is = 1.96 ( From Standard Normal Table )
Standard Deviation ( S.D) = 4
ME =0.45
n = ( 1.96*4/0.45) ^2
= (7.84/0.45 ) ^2
= 303.534 ~ 304