Question

How large a sample is needed to estimate the population mean monthly gasoline expenditure within $10 with 95% confidence? The population standard deviation is $63.4. (Remember that your answer must be a whole number.)

Answer 1

For your exercise, we have:

E=10

`\sigma=63.4`

n=?

E= margin of error

`E=z.\frac{\sigma}{\sqrt[]{n}}`

For 95% confidence, we have z=1.96 .

Now, let's replace the known data.

`\begin{gathered} 10=1.96.\frac{63.4}{\sqrt[]{n}} \\ 10.\sqrt[]{n}=1.96\cdot63.4 \\ \sqrt[]{n}=(124.264)/(10) \\ \sqrt[]{n}=12.4264 \\ (\sqrt[]{n})^2=(12.4264)^2 \\ n=154.4154 \\ n=155 \end{gathered}`