91Ó°ÊÓ

The confidence interval for the ratio of the two variances with approximate \(100(1 - \alpha )\) confidence level can be obtained from

\(P\left( {{F_{1 - \alpha /2,m - 1,n - 1}} \le \frac{{S_1^2/\sigma _1^2}}{{S_2^2/\sigma _2^2}} \le {F_{\alpha /2,m - 1,n - 1}}} \right) = 1 - \alpha \)Look at

\({F_{1 - \alpha /2,m - 1,n - 1}} \le \frac{{S_1^2/\sigma _1^2}}{{S_2^2/\sigma _2^2}} \le {F_{\alpha /2,m - 1,n - 1}}\)

This is obviously equivalent to

\(\frac{{S_2^2}}{{S_1^2}} \cdot {F_{1 - \alpha /2,m - 1,n - 1}} \le \frac{{\sigma _2^2}}{{\sigma _1^2}} \le \frac{{S_2^2}}{{S_1^2}} \cdot {F_{\alpha /2,m - 1,n - 1}}\)

By substituting corresponding values with sample standard deviations \({s_1}\) and \({s_2}\)the confidence interval for the proportion of the two variances is obtained

\(\frac{{s_2^2}}{{s_1^2}} \cdot {F_{1 - \alpha /2,m - 1,n - 1}} \le \frac{{\sigma _2^2}}{{\sigma _1^2}} \le \frac{{s_2^2}}{{s_1^2}} \cdot {F_{\alpha /2,m - 1,n - 1}}\)

Thus, the confidence interval for the proportion of the two standard deviations is square root of the confidence interval for the proportion of the two variances

\(\frac{{{s_2}}}{{{s_1}}} \cdot \sqrt {{F_{1 - \alpha /2,m - 1,n - 1}}} \le \frac{{{\sigma _2}}}{{{\sigma _1}}} \le \frac{{{s_2}}}{{{s_1}}} \cdot \sqrt {{F_{\alpha /2,m - 1,n - 1}}} \)

To obtain the upper bound, the value \({F_{\alpha /2,m - 1,n - 1}}\) needs to be replaces with\({F_{\alpha ,m - 1,n - 1}}\); thus the upper bound is

\(\frac{{{s_2}}}{{{s_1}}} \cdot \sqrt {{F_{\alpha ,m - 1,n - 1}}} \cdot \)

Given values in the example \(9.6\) are \({s_1} = 0.79\)and \({s_2} = 3.59,{\rm{ }}m = n = 10.\)The \(95\% \)upper bound \((\alpha = 0.05)\) for the proportion \({\sigma _2}{\sigma _1}\)is

\(\frac{{{s_2}}}{{{s_1}}} \cdot \sqrt {{F_{\alpha ,m - 1,n - 1}}} = \frac{{3.59}}{{0.79}} \cdot \sqrt {3.18} = 8.1\)

where \({F_{\alpha ,m - 1,n - 1}} = {F_{0.05,10 - 1,10 - 1}} = {F_{0.05,9,9}} = 3.18\)which was obtained using computer software (you can use the table in the appendix.