You want to estimate the mean amount of time Internet users

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 results from a prior Nielson survey, assume that the standard deviation of the population of monthly times spent on Facebook is 210 min.    Show work, answer question, and give units

Solution

Compute Sample Size
n = (Z a/2 * S.D / ME ) ^2
Z/2 at 0.05% LOS is = 1.96 ( From Standard Normal Table )
Standard Deviation ( S.D) = 210
ME =15
n = ( 1.96*210/15) ^2
= (411.6/15 ) ^2
= 752.954 ~ 753      

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

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