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

Step-by-step explanation:

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

$n=(\dfrac{z_{\alpha/2}\cdot \sigma}{E})^2$

, 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 :_

$n=(\dfrac{1.96\cdot 5}{2.5})^2$

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

Hence, the required sample size = 16