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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

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

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

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

To obtain \(90\% \)confidence interval for the ratio of the variances \((\alpha = 0.1)\)the sample standard deviations of the \(m = n = 4\) sized samples are needed.

The Sample Variance \({s^2}\)is

\({s^2} = \frac{1}{{n - 1}} \cdot {S_{xx}}\)

where

\({S_{xx}} = \sum {{{\left( {{x_i} - \bar x} \right)}^2}} = \sum {x_i^2} - \frac{1}{n} \cdot {\left( {\sum {{x_i}} } \right)^2}\)

The Sample Standard Deviation \(s\) is

\(s = \sqrt {{s^2}} = \sqrt {\frac{1}{{n - 1}} \cdot {S_{xx}}} \)

Thus, the sample standard deviations are, for the Epoxy

\(\begin{array}{c}{s_1} = \sqrt {\frac{1}{{4 - 1}} \cdot \left( {\left( {{{1.75}^2} + {{2.12}^2} + {{2.05}^2} + {{1.97}^2}} \right) - \frac{1}{4} \cdot {{(1.75 + 2.12 + 2.05 + 1.97)}^2}} \right)} \\ = 0.16\end{array}\)

and, for the MMA prepolymer sample

\(\begin{array}{c}{s_1} = \sqrt {\frac{1}{{4 - 1}} \cdot \left( {\left( {{{1.77}^2} + {{1.59}^2} + {{1.7}^2} + {{1.69}^2}} \right) - \frac{1}{4} \cdot {{(1.77 + 1.59 + 1.7 + 1.69)}^2}} \right)} \\ = 0.074.\end{array}\)

Also, for\(\alpha = 0.1(\alpha /2 = 0.05)\), the \(F\) values are

\(\begin{array}{c}{F_{0.05,3,3}} = 9.28\\{F_{0.95,3,3}} = \frac{1}{{{F_{0.05,3,3}}}}\\ = 0.108\end{array}\)

Finally, the \(90\% \)confidence interval for the proportion is

\(\left( {\frac{{s_2^2}}{{s_1^2}} \cdot {F_{1 - \alpha /2,m - 1,n - 1}},\frac{{s_2^2}}{{s_1^2}} \cdot {F_{\alpha /2,m - 1,n - 1}}} \right)\)

\(\begin{array}{l} = \left( {\frac{{{{0.074}^2}}}{{{{0.16}^2}}} \cdot 1.108,\frac{{{{0.074}^2}}}{{{{0.16}^2}}} \cdot 9.28} \right)\\ = (0.023,1.99).\end{array}\)