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 95% confident that your sample mean is within 10 minutes of the population mean? Assume that the standard deviation of the population of monthly time spent on the internet is 211 min. What is a major obstable to getting a good estimate of the population mean? Use technology to find the estimated minimum required sample size.

The minimum sample size required is ___ computer users.

What is a major obstacle to getting a good estimate of the population mean?
a) There may not be 1,809 computer users to survey
b) It is difficult to precisely measure the amount of time spent on the internet,invidating some data values
c) The data does not provide information on what the computers users did while on the internet
d) There are no obstacles to getting a good estimate of the population mean

Answer 1

Explanation:

Given that confidence level is 95% and sample mean is within 10 minutes of the population mean. i.e. margin of error = 10

Std deviation of population= 211 minutes

Margin of error = Z critical * std error

Std error = sigma/sq rt n =
`(211)/(√(n) )`

Hence we have

\frac{211}{\sqrt{n} }
`(211)/(√(n) )*1.96<10\\√(n) >41.356\\n>1710`

The minimum sample size is 1710

a) There may not be 1,809 computer users to survey