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 greatest confidence level, given that the margin of error is to be kept at half year. Recall that$\sigma =12.1$ years.

Short answer

The confidence level is$78.5\%$.

Step-by-step solution

Given Information

It is given that

$n=900$

$\sigma =12.1$

$E=0.5$

Determination of Confidence Level

We know that confidence level is given by

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

$0.5=z_{\frac{a}{2}}\left(\frac{12.1}{\sqrt{900}}\right)$

$z_{\frac{a}{2}}=0.5\left(\frac{\sqrt{900}}{12.1}\right)$

role="math" localid="1652726923958" $z_{\frac{a}{2}}=1.24$

From tables, confidence level for $z_{\frac{a}{2}}=1.24$is $0.1075$.

Also,

Confidence level, $\frac{\alpha }{2}=0.1075$

$\alpha =0.215$

Also,

$100(1-\alpha )\%=100(1-0.215)\%$

$=78.5\%$

Therefore, required confidence level is$78.5\%$