8 Nielsen MediaResearch wants to estimate the mean amount of

8.    Nielsen MediaResearch wants to estimate the mean amount of time (in min) thatfull-time college students spend watching television each weekday.Find the sample size necessary to estimate that mean with a 15-minmargin of error. Assume that a 96% confidence level is desired.Also assume that a pilot study showed that the standard deviationis estimated to be 112.2 min.

Solution

let sample size = n standard deviation, = 112.2 standard error, e = /n at 96% confidence, value of z from z table = 2.06 margin of error = 15 margin of error= z * e hence, 15 = 2.06 *(112.2/n) hence, n = 237.43 ˜ 238