Question

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 3 hours. It is desired to estimate the mean viewing time within one quarter hour. The 0.95 degree of confidence is to be used. How many executives should be surveyed?

Answer 1

Answer:  554

Step-by-step explanation:

Formula we use to find the sample size :

$n=(\dfrac{z_{\alpha/2}\cdot\sigma}{E})^2$

Given : Standard deviation : $\sigma=3\ hour$

Margin of error : E=one quarter hour = 0.25 hour

Two-tailed , Critical value use for 0.95 degree of confidence:$z_{\alpha/2}=1.96$

i.e. $n=(\dfrac{(1.96)\cdot(3)}{0.25})^2=553.1904\approx554$

Hence, the required sample size = 554

i.e. 554 executives should be surveyed.