Question

6. The new mini-laptop computers can deliver as much computing power as machines ten times their size, but they weigh in at less than 3 lb. Experts predict they will soon replace their traditional 6- to 8-lb older laptop brothers as tech computer industry continues its relentless quest for smallness. If the standard deviation of the weight of all mini-laptops is 0.4 lb, how large a sample would be needed to estimate the population mean weight if the maximum error of estimate is to be 0.15 lb with 95% confidence? 3

Answer 1

Given that, standard deviation $(\sigma)$ = 0.4 lb

Margin of error ( E ) = 0.15 lb

Confidence level = 95%

A 95% confidence level has significance level $(\alpha)$ = 0.05 and has critical value is, $Z_{\alpha/2} = 1.96$

We want to find, the sample size ( n ),

$n = (Z_{\alpha/2} * \frac { \sigma}{E})^2 = (1.96 * \frac { 0.4}{0.15})^2= 27.3180$

$n \approx 27$

Required sample size is 27

Note: If you want the sample size is nearest to next whole number then it is equal to 28.