In order to estimate the mean amoutn of tiem computer users spend on the interne

Question

In order to estimate the mean amoutn of tiem 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 12 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 obstacle to getting a good estimate of the poulation mean? Use technology to find the estimated minimum required sample size.

The minimum sampel size required is ___ computer users.

Explanation / Answer

E = z * s / square root(n) where z is the z-score we use to calculate the confidence interval, s is the standard deviation, n is sample size and ME is the desired margin of error

we want a 90% confidence interval(CI)

alpha(a) = 1 - (90/100) = 0.10

critical probability(p*) = 1 - (a/2) = 0.95

z-score associated with p* is 1.644853
12 = (1.644853) * 211) / square root(n)

square root(n) = 347.063983 / 12


n = 836.4820021 approximately 836

A major obstacle to getting a good estimate of the poulation mean can be that there may not be 836 computer users to survey.

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