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 13 hours, with a standard deviation of 2.5 hours. The estimate of the mean viewing time should be within 30 minutes. The 98% level of confidence is to be used. (Use z Distribution Table.) How many executives should be surveyed? (Round the z-score to 2 decimal places and final answer to the next whole number.)

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 13 hours, with a standard deviation of 2.5 hours. The estimate of the mean viewing time should be within 30 minutes. The 98% level of confidence is to be used. (Use z Distribution Table.) How many executives should be surveyed? (Round the z-score to 2 decimal places and final answer to the next whole number.) Number of executives

Answer 1

Here n = 10

xbar = 13 hours

s = 2.5 hours

E = margin of error = 0.5 hours

c = confidence level = 0.98

we have to find n sample size

n = ( Z_c * $\sigma$ ^ )²/ E²

for c = 0.98 , Z_c = Z_0.98 = 2.33

n = ( 2.33 * 2.5 )² / 0.5²

n = 136

Number of executives = 136