Question

What is the minimum sample size needed to be 90% confident that the sample mean is within two units of the population mean? A previous study show the population standard deviation is 8.4.

Answer 1

Answer: 48

Explanation:

Formula to find the minimum sample size :-

`n=((z^*\cdot \sigma)/(E))^2`

, where
`\sigma` = population standard deviation

z* = Critical value.

E = Margin of error .

As per given , we have

`\sigma=8.4`

E=2 units

Confidence level= 90%

By z-vale table , the critical value for 90% confidence =
`z^*=1.645`

Then, the minimum sample size would be :-

`n=(((1.645)\cdot (8.4))/(2))^2\\\\=(6.909)^2\\\\=47.734281\approx48`

Hence, the minimum sample size needed = 48