91Ó°ÊÓ

The sample number of observations is\(n = 12\).

The level of confidence is 95%.

The degrees of freedom are computed asfollows:

\(\begin{array}{c}df = n - 1\\ = 12 - 1\\ = 11\end{array}\)

The level of confidence is 95%,whichimplies the level of significance is 0.05.

Using the Chi-square table, the critical values at 0.05 level of significance and 11 degrees of freedom are \(\chi _L^2 = 3.8157\) and \(\chi _R^2 = 21.92\).

The confidence interval for the standard deviation is given asfollows:

\(\sqrt {\frac{{\left( {n - 1} \right){s^2}}}{{\chi _R^2}}} < \sigma < \sqrt {\frac{{\left( {n - 1} \right){s^2}}}{{\chi _L^2}}} \)

Let x represents the sample observations.

The mean value is computed asfollows:

\(\begin{array}{c}\bar x = \frac{{\sum x }}{n}\\ = \frac{{62 + 61 + 61 + 57 + ... + 60 + 67}}{{12}}\\ = 60.667\end{array}\)

The standard deviation is computed asfollows:

\(\begin{array}{c}s = \sqrt {\frac{{\sum {{{\left( {x - \bar x} \right)}^2}} }}{{n - 1}}} \\ = \sqrt {\frac{{{{\left( {62 - 60.667} \right)}^2} + {{\left( {61 - 60.667} \right)}^2} + {{\left( {61 - 60.667} \right)}^2} + ... + {{\left( {67 - 60.667} \right)}^2}}}{{12 - 1}}} \\ = 4.075\end{array}\)

The 95% confidence interval estimate of the standard deviation of the population from which the sample was obtained is computed asfollows:

\(\begin{array}{c}\sqrt {\frac{{\left( {n - 1} \right){s^2}}}{{\chi _R^2}}} < \sigma < \sqrt {\frac{{\left( {n - 1} \right){s^2}}}{{\chi _L^2}}} \\\sqrt {\frac{{\left( {12 - 1} \right){{4.075}^2}}}{{21.92}}} < \sigma < \sqrt {\frac{{\left( {12 - 1} \right){{4.075}^2}}}{{3.8157}}} \\2.887 < \sigma < 6.919\\2.9 < \sigma < 6.9\end{array}\)

Therefore, the 95% confidence interval estimate of the standard deviation of the population from which the sample was obtained is\({\bf{2}}{\bf{.9}}\;{\bf{mi/h < \sigma < 6}}{\bf{.9}}\;{\bf{mi/h}}\).