Question

A population is normal with a variance of 13. suppose you wish to estimate the population mean. find the sample size needed to assure with 68.26 percent confidence that the sample mean will not differ from the population mean by more than 2 units

Answer 1

The margin of error is given by the formula

$ME=z_{\alpha /2}\frac{\sigma}{\sqrt{n}}$

Solving for n,

$n=z^2_{\alpha /2}\frac{\sigma^2}{ME^2}$

Here $1-\alpha =0.6826\\ \alpha =0.3174$

Using standard normal table $z_{\alpha /2}=-1$

$\sigma^2=13,ME=2$

Substituting numerical values,

$n=(-1)^2\frac{13}{2^2} \approx 4$