91Ó°ÊÓ

Sample size \(n=36\)

Sample Mean \(\bar{x}=58.53\)

Sample standard deviation \(\sigma=13.36\)

We have to find the sample size for a confidence interval for \(\mu\) is to have a margin of error of \(2\) years and a confidence level of \(95%\).

The population mean age is \(\mu\)

Sample size \(n=36\)

Sample mean \(\bar{x}=58.53\)

Margin of error \(E=2\)years

Sample Standard deviation \(s=\frac{1}{n-1}\sum_{i-1}^{n}(x_{i}-\bar{x})^{2}\)

\(=13.36\)

For \(100(1-\alpha)%\) confidence interval \(\mu\)

\(n=\left ( \frac{Z_{\alpha/2}\times \sigma}{E} \right )^{2}\)

But here \(\sigma\) is unknown. So, we estimate it by sample standard deviation \(s\).

Here the confidence level \(=95%\)

\(=100\times 0.95%\)

\(1-\alpha=0.95\)

\(\Rightarrow =0.05\)

\(\Rightarrow \frac{\alpha}{2}=0.025\)

\(Z_{\alpha/2}=Z_{0.025}=1.96\) [From standard normal table values]

Sample Size \(n=\left ( \frac{Z_{\alpha/2}\times \sigma}{E} \right )^{2}\)

\(=\left ( \frac{1.96\times 13.36}{2} \right )\)

\(=171.42\)

\(=172\)

Hence, the required sample size is \(172\).

The population standard deviation, \(\sigma\)is unknown, we used sample standard deviation \(s\) in place of \(\sigma\) because sample standard deviation \(s(division n-1)\) is an unbiased estimator of population standard deviation \(\sigma\) i.e. mean of all possible sample standard deviations(with divisor \(n-1\))is population mean.

Hence, sample standard deviation \(s(division n-1)\) is an unbiased estimator of population standard deviation \(\sigma\).