Question

A research worker wishes to estimate the mean of a population by using sufficiently large sample. The probability is 0.95 that the sample mean will not differ from the true mean by more that 25% of the standard deviation. How large sample should be taken?

Answer 1

Solution:

We have $P(|\bar{X}-\mu|<\sigma / 4)=0.95$

Divide by $\sigma / \sqrt{n}$

${P}\left(\frac{\bar{X}-\mu}{\sigma / \sqrt{n}}<\frac{\sqrt n}{4}\right)=0.95$

But $\frac{\bar{X}-\mu}{\sigma / \sqrt{n}}$ has a standard normal distribution.

And, ${P}(-1.96< \text{Standard Normal} <1.96)=0.95$

So, $1.96=\sqrt{n} / 4$

$\Rightarrow 1.96\times4=\sqrt n \\ \Rightarrow n=(7.84)^2=61.4656 \\\Rightarrow n=62$

So, a sample of 62 should be taken.