Question

It is known that the population variance equals 484. With a .95 probability, the sample size that needs to be taken to estimate the population mean if the desired margin of error is 5 or less is

Answer 1

Answer:

The minimum sample size is n = 75 so that the desired margin of error is 5 or less.                          

Step-by-step explanation:

We are given the following in the question:

Population variance = 484

Standard deviation =

$\sigma^2 = 484\\\sigma =\sqrt{484} = 22$

Confidence level = 0.95

Significance level = 0.05

Margin of error = 5

Formula:

Margin of error =

$E = z\times \dfrac{\sigma}{\sqrt{n}}\\\\n = (\dfrac{z\times \sigma}{E})^2$

$z_{critical}\text{ at}~\alpha_{0.05} = 1.96$

Putting values, we get

$E\leq 5\\\\1.96\times \dfrac{22}{\sqrt{n}} \leq 5\\\\\sqrt{n} \geq 1.96\times \dfrac{22}{5}\\\\n \geq (1.96\times \dfrac{22}{5})^2\\\\n \geq 74.373$

Thus, the minimum sample size is n = 75 so that the desired margin of error is 5 or less.