Question

A large on-demand, video streaming company is designing a large-scale survey to determine the mean amount of time corporate executives watch on-demand television. A small pilot survey of 10 executives indicated that the mean time per week is 19 hours, with a standard deviation of 5.5 hours. The estimate of the mean viewing time should be within one-quarter hour. The 98% level of confidence is to be used. (Use z Distribution Table.)

Answer 1

Solution :

Given that,

Population standard deviation = = 5.5

Margin of error = E = 0.25

At 98% confidence level the z is,

= 1 - 98%

= 1 - 0.98 = 0.02

/2 = 0.01

Z_/2 = 2.326

sample size = n = [Z_/2* / E] ²

n = [2.326 * 5.5 / 0.25]²

n = 2618.57

Sample size = n = 2619