a) Given: Sample size, \(\displaystyle{n}_{{{1}}}={30}\) Sample size, \(\displaystyle{n}_{{{2}}}={39}\) Sample standard deviation 1, \(\displaystyle{s}_{{{1}}}={16.37}\) Sample standard deviation 2, \(\displaystyle{s}_{{{2}}}={9.88}\) Let's calculate \(\displaystyle{95}\%\) confidence interval for the ratio of two population variances. \(\displaystyle{C}{I}=\ {\left({\frac{{{{s}_{{{1}}}^{{{2}}}}}}{{{{s}_{{{2}}}^{{{2}}}}}}}\ {F}_{{{1}\ -\ \frac{\alpha}{{2}},\ {n}_{{{2}}}\ -\ {1},\ {n}_{{{1}}}\ -\ {1},}}\ {\frac{{{{s}_{{{1}}}^{{{2}}}}}}{{{{s}_{{{2}}}^{{{2}}}}}}}{F}_{{\frac{\alpha}{{2}},\ {n}_{{{2}}}\ -\ {1},\ {n}_{{{1}}}\ -\ {1}}}\right)}\)

\(\displaystyle{C}{I}=\ {\left({\frac{{{16.37}^{{{2}}}}}{{{9.88}^{{{2}}}}}}\ {F}_{{{1}\ -\ \frac{{0.05}}{{2.39}}\ -\ {1.30}\ -\ {1},}}\ {\frac{{{16.37}^{{{2}}}}}{{{9.88}^{{{2}}}}}}{F}_{{\frac{{0.5}}{{2.39}}\ -\ {1.30}\ -\ {1}}}\right)}\)

\(\displaystyle{C}{I}=\ {\left({\frac{{{16.37}^{{{2}}}}}{{{9.88}^{{{2}}}}}}\ {F}_{{\frac{{0.95}}{{2}},\ {38},\ {29},}}\ {\frac{{{16.37}^{{{2}}}}}{{{9.88}^{{{2}}}}}}{F}_{{\frac{{0.5}}{{2}},\ {38},\ {29}}}\right)}\) Using the critical value table, \(\displaystyle{C}{I}=\ {\left({\frac{{{16.37}^{{{2}}}}}{{{9.88}^{{{2}}}}}}\ \times{0.507},\ \ {\frac{{{16.37}^{{{2}}}}}{{{9.88}^{{{2}}}}}}{2.038}\right)}\)

\(\displaystyle{C}{I}=\ {\left({1.3919},\ {5.5949}\right)}\) Hence, it is \(\displaystyle{95}\%\) confidence that the true ratio of population variances lies in interval \(\displaystyle{\left({1.3919},\ {5.5949}\right)}\)

b) Given: Sample size, \(\displaystyle{n}_{{{1}}}={25}\) Sample size, \(\displaystyle{n}_{{{2}}}={20}\) Sample standard deviation 1, \(\displaystyle{s}_{{{1}}}={5.2}\) Sample standard deviation 2, \(\displaystyle{s}_{{{2}}}={6.8}\) Let's calculate \(\displaystyle{99}\%\) confidence interval for the ratio of two population variances. \(\displaystyle{C}{I}=\ {\left({\frac{{{{s}_{{{1}}}^{{{2}}}}}}{{{{s}_{{{2}}}^{{{2}}}}}}}\ {F}_{{{1}\ -\ \frac{\alpha}{{2}},\ {n}_{{{2}}}\ -\ {1},\ {n}_{{{1}}}\ -\ {1},}}\ {\frac{{{{s}_{{{1}}}^{{{2}}}}}}{{{{s}_{{{2}}}^{{{2}}}}}}}{F}_{{\frac{\alpha}{{2}},\ {n}_{{{2}}}\ -\ {1},\ {n}_{{{1}}}\ -\ {1}}}\right)}\)

\(\displaystyle{C}{I}=\ {\left({\frac{{{5.2}^{{{2}}}}}{{{6.8}^{{{2}}}}}}\ {F}_{{{1}\ -\ \frac{{0.01}}{{2}},\ {39}\ -\ {1},\ {30}\ -\ {1},}}\ {\frac{{{5.2}^{{{2}}}}}{{{6.8}^{{{2}}}}}}{F}_{{\frac{{0.01}}{{2}},\ {39}\ -\ {1},\ {30}\ -\ {1}}}\right)}\)

\(\displaystyle{C}{I}=\ {\left({\frac{{{5.2}^{{{2}}}}}{{{6.8}^{{{2}}}}}}\ {F}_{{\frac{{0.99}}{{2}},\ {38},\ {29},}}\ {\frac{{{5.2}^{{{2}}}}}{{{6.8}^{{{2}}}}}}{F}_{{\frac{{0.01}}{{2}},\ {38},\ {29}}}\right)}\) Using the critical value table, \(\displaystyle{C}{I}=\ {\left({\frac{{{5.2}^{{{2}}}}}{{{6.8}^{{{2}}}}}}\ \times{0.3025},\ \ {\frac{{{5.2}^{{{2}}}}}{{{6.8}^{{{2}}}}}}{3.092}\right)}\)

\(\displaystyle{C}{I}={\left({0.2128},\ {2.175}\right)}\) Hence, it is \(\displaystyle{99}\%\) confidience that the ratio of population variances lies in the interval \(\displaystyle{\left({0.2128},\ {2.175}\right)}\)