Question

In order to estimate the mean amount of time computer users spend on the internet each​ month, how many computer users must be surveyed in order to be 90​% confident that your sample mean is within 13 minutes of the population​ mean? Assume that the standard deviation of the population of monthly time spent on the internet is 228 min

Answer 1

832

Explanation:

standard deviation =228 minute

error =13 minute given

confidence level =905% =0.90

α=1-0.90=0.1

`z_(\alpha )/(2)=z_(0.1)/(2)=1.645`

we know that sample size should be greater than

`n\geq \left ( z_(\alpha )/(2)* (\sigma )/(E) \right )^2`

`n\geq \left ( 1.645* (228)/(13) \right )^(2)`

`n\geq 28.850^2`

`n\geq 832.3668`

n=832