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 15 minutes of the population mean? Assume that the standard deviation of the population of monthly time spent on the internet is 224 min. What is a major obstacle 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 (Round up to the nearest whole number.)

Answer 1

Given that,

standard deviation = $\sigma$ = 224

margin of error = E = 15

At 90% confidence level the z is ,

$\alpha$ = 1 - 90% = 1 - 0.90 = 0.1

$\alpha$ / 2 = 0.1/ 2 = 0.05

Z$\alpha$/2 = Z0.05 =1.645

Sample size = n = [(Z$\alpha$/2 * $\sigma$ ) / E]2

= ((1.645*224) /15)2

= 603.45

Sample size = 604