Question

Civilian Labor Force. Consider again the problem of estimating the mean age, $\mu$, of all people in the civilian labor force. In Example $8.7$on page 328 , we found that a sample size of 2250 is required to have a margin of error of $0.5$year and a $95\%$ confidence level. Suppose that, due to financial constraints, the largest sample size possible is 900 . Determine the smallest margin of error, given that the confidence level is to be kept at $95\%$. Recall that$\sigma =12.1$ years.

Short answer

The margin of error is$0.791$year.

Step-by-step solution

Given Information

It is given that $\sigma =12.1$

$n=900$

Confidence level is $95\%$

$z_{\frac{a}{2}}=1.96$

Determination of Margin of Error

The margin of error is given by

$E=z_{\frac{a}{2}}\left(\frac{\sigma }{\sqrt{n}}\right)$

$=1.96\left(\frac{12.1}{\sqrt{900}}\right)$

$=0.791$

Hence, margin of error is$0.791$