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 15 hours, with a standard deviation of 3.5 hours. It is desired to estimate the mean viewing time within one-quarter hour. The 98% 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,

standard deviation = = 3.5

margin of error = E = 0.25

At 98% confidence level the z is ,

= 1 - 98% = 1 - 0.98 = 0.02

/ 2 = 0.02 / 2 = 0.01

Z_/2 = Z_0.01 = 2.326

Sample size = n = ((Z_/2 * ) / E)²

= ((2.326 * 3.5) / 0.25)²

= 1061

Sample size = 1061