A survey is being planned to estimate with a 98 confidence i

A survey is being planned to estimate, with a 98% confidence interval, the mean amount of time per week that seniors spend watching television. The population standard deviation is known to be 5.2 hours. What sample size will be required to obtain an estimate that is within 0.5 hours of the population mean?

Solution

Given a=0.02, |Z(0.01)|=2.33 (From standard normal table)

So n=(Z*s/E)^2

=(2.33* 5.2/0.5 )^2

= 587.1898

Take n=588