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

Answer 1

We have given sample mean x bar = 19 , n = 10 , standard deviation = 5.5 , C = 0.98

Also we have given margin of error E = one-quarter hour.= 0.15

We have asked to find the sample size (n ).

We know that Sample size n = [ (Z_0.98 * ) / E ]²

  = [ (2.33 * 5.5) / 0.15 ]²

= 7275.99

7276

Hence the required sample size n is 7276 (Rounded to the next whole integer)

That is 7276 executives should be surveyed

Hope this will help you. Thank you :)