Question

Weights of women in one age group are normally distributed with a standard deviation σ of 23 lb. A researcher wishes to estimate the mean weight of all women in this age group. Find how large a sample must be drawn in order to be 90 percent confident that the sample mean will not differ from the population mean by more than 2.4 lb.

Answer 1

The confidence interval is given by

`\begin{gathered} \bar{x}\pm z*\frac{\sigma}{\sqrt[]{n}} \\ \text{ Where} \\ \bar{x}=\text{ the sample mean} \\ z=\text{ the z-score} \\ \sigma=\text{ the standard deviation} \\ n=\text{ the sample size} \end{gathered}`

We will now find the alpha level, α.

`\alpha=(1-0.9)/(2)=(0.1)/(2)=0.05`

The z-score for a 90% confidence interval is 1.645.

The question gives the margin of error as 2.4 lb.

Therefore,

`2.4=z*\frac{\sigma}{\sqrt[]{n}}`

Substituting the values, we get

`2.4=1.645*\frac{23}{\sqrt[]{n}}`

Making n the subject of formula:

`\begin{gathered} \sqrt[]{n}=(1.645*23)/(2.4) \\ \sqrt[]{n}=(37.835)/(2.4)=15.685 \\ \therefore \\ n=15.685^2 \\ n=246 \end{gathered}`

The sample size is 246.