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 15 hours, with a standard deviation of 2.5 hours. The estimate of the mean viewing time should be within 30 minutes. The 99% level of confidence is to be used. (Use z Distribution Table.)

Answer 1

Let, number of executives surveyed = n

ME = Z*SD/(n)^.5

Value of Z at 99% (two tail) = 2.575

So,

.5 = 2.575*2.5/(n)^.5

n = ((2.575*2.5)/.5)^2

n = 165.76 or 166

So, 166 executives should be surveyed.