Question

A report in a research journal states that the average weight loss of people on a certain drug is 27 lbs with a standard deviation of 5 lbs. If this study is repeated, how large should the sample size be so that the margin of error would be no more than 2.5 lbs with 95% confidence?

Answer 1

Answer: 16

Explanation:

We know that the formula to find the sample size is given by :-

`n=((z_(\alpha/2)\cdot \sigma)/(E))^2`

, where
`\sigma` = population standard deviation.

E = Margin of error

`z_(\alpha/2)`= Two-tailed z-value for significance level of
`\alpha.`

As per given , we have

`\alpha=1-0.95=0.05`

Using z-value , table

`z_(\alpha/2)=1.96`

`\sigma= 5\ lbs`

Margin of error = 2.5 lbs

Then, the required sample size would be :_

`n=((1.96\cdot 5)/(2.5))^2`

`n=(3.92)^2=15.3664\approx16`

Hence, the required sample size = 16