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 13 hours, with a standard deviation of 2.0 hours. It is desired to estimate the mean viewing time within one-quarter hour. The 95% level of confidence is to be used. (Use z Distribution Table.)

How many executives should be surveyed? (Round your z-score to 2 decimal places and round up your final answer to the next whole number.)

Answer 1

Solution :

Given that,

Population standard deviation = = 2

Margin of error = E = 0.15

At 95% confidence level the t is ,

= 1 - 95% = 1 - 0.95 = 0.05

/ 2 = 0.05 / 2 = 0.025

Z_/2 = Z_0.025 = 1.96

sample size = n = (Z_/2* / E) ²

n = (1.96 *2 /0.15 )²

n = 682.95

n = 683

Sample size = 683

683 executives should be surveyed.