Given the following values of $\overline{\mathbf{x}}$, $\mathit{s}$, and $\mathit{n}$, form a 90% confidence interval for${\mathit{\sigma }}^{\mathbf{2}}$

a. $\overline{\mathbf{x}}\mathbf{=}\mathbf{21}\mathbf{,}\mathit{s}\mathbf{=}\mathbf{2}\mathbf{.}\mathbf{5}\mathbf{,}\mathit{n}\mathbf{=}\mathbf{50}$

b. $\overline{\mathbf{x}}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{3}\mathbf{,}\mathit{s}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{02}\mathbf{,}\mathit{n}\mathbf{=}\mathbf{15}$

c. $\overline{\mathbf{x}}\mathbf{=}\mathbf{167}\mathbf{,}\mathit{s}\mathbf{=}\mathbf{31}\mathbf{,}\mathit{n}\mathbf{=}\mathbf{22}$

d.$\overline{\mathbf{x}}\mathbf{=}\mathbf{9}\mathbf{.}\mathbf{4}\mathbf{,}\mathit{s}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{5}\mathbf{,}\mathit{n}\mathbf{=}\mathbf{5}$

We know that, for a 90% confidence interval that is for a 0.90 confidence coefficient,

$\alpha =0.10and\frac{\alpha }{2}=0.05.So,(1-\frac{\alpha }{2})=0.95.$

The formula for forming a confidence interval is given by,

$\frac{(n-1)s^2}{{\chi }_{{\displaystyle \frac{\alpha }{2}}}^2}\le {\sigma }^2\le \frac{(n-1)s^2}{{\chi }_{(1-{\displaystyle \frac{\alpha }{2}})}^2}$............(i)

We will find the values of ${\chi }_{\frac{\alpha }{2}}^2$and ${\chi }_{(1-\frac{\alpha }{2})}^2$from table IV, Appendix D of the book as per the degrees of freedom which is calculated as$(n-1)$.

a. Substituting the given values $\overline{x}=21,s=2.5,n=50$in equation (i), we get the required interval as,

$\begin{array}{rcl}\frac{(50-1){(2.5)}^2}{67.5048}& \le & {\sigma }^2\le \frac{(50-1){(2.5)}^2}{34.7642}\\ \frac{306.25}{67.5048}& \le & {\sigma }^2\le \frac{306.25}{34.7642}\\ 4.536& \le & {\sigma }^2\le 8.809\end{array}$

b. Substituting the given values $\overline{\mathrm{x}}=1.3,s=0.02,n=15$ in equation (i), we get the required interval as,

role="math" localid="1661407112611" $\begin{array}{rcl}\frac{(15-1){(0.02)}^2}{23.6848}& \le & {\sigma }^2\le \frac{(15-1){(0.02)}^2}{6.57063}\\ \frac{0.0056}{23.6848}& \le & {\sigma }^2\le \frac{0.0056}{6.57063}\\ 0.00023644& \le & {\sigma }^2\le 0.00085228\end{array}$

c. Substituting the given values $\overline{\mathrm{x}}=167,s=31,n=22$in equation (i), we get the required interval as,

role="math" localid="1661407085643" $\begin{array}{rcl}\frac{(22-1){\left(31\right)}^2}{32.6705}& \le & {\sigma }^2\le \frac{(22-1){\left(31\right)}^2}{11.5913}\\ \frac{20181}{32.6705}& \le & {\sigma }^2\le \frac{20181}{11.5913}\\ 617.713228& \le & {\sigma }^2\le 1741.04716\end{array}$

d. Substituting the given values $\overline{\mathrm{x}}=9.4,s=1.5,n=5$in equation (i), we get the required interval as,

$\begin{array}{rcl}\frac{(5-1){(1.5)}^2}{9.48773}& \le & {\sigma }^2\le \frac{(5-1){(1.5)}^2}{0.710721}\\ \frac{9}{9.48773}& \le & {\sigma }^2\le \frac{9}{0.710721}\\ 0.9485936& \le & {\sigma }^2\le 12.663197\end{array}$