Question

You want to estimate the mean amount of time Internet users spend on Facebook each month. How many Internet users must be surveyed in order to be 95% confident that your sample mean is within 15 minutes of the population mean? Based on a previous study, it was found that the population standard deviation spent on Facebook is 210 minutes.

Answer 1

Solution

standard deviation =s =   = 210

Margin of error = E = 15

At 95% confidence level the z is ,

= 1 - 95% = 1 - 0.95 = 0.05

/ 2 = 0.05 / 2 = 0.025

Z/2 = Z0.025 = 1.96

sample size = n = [Z/2* / E] 2

n = ( 1.96*210 / 15 )2

n =752.95

Sample size = n =753 ROUNDED