91影视

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

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

The formula for forming a confidence interval is given by,

$\frac{(n-1)s^2}{{蠂}_{{\displaystyle \frac{伪}{2}}}^2}鈮�{蟽}^2鈮�\frac{(n-1)s^2}{{蠂}_{(1-{\displaystyle \frac{伪}{2}})}^2}$............(i)

We will find the values of ${蠂}_{\frac{伪}{2}}^2$and ${蠂}_{(1-\frac{伪}{2})}^2$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,

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

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" $\begin{array}{rcl}\frac{(15-1){(0.02)}^2}{23.6848}& 鈮�& {蟽}^2鈮�\frac{(15-1){(0.02)}^2}{6.57063}\\ \frac{0.0056}{23.6848}& 鈮�& {蟽}^2鈮�\frac{0.0056}{6.57063}\\ 0.00023644& 鈮�& {蟽}^2鈮�0.00085228\end{array}$

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" $\begin{array}{rcl}\frac{(22-1){\left(31\right)}^2}{32.6705}& 鈮�& {蟽}^2鈮�\frac{(22-1){\left(31\right)}^2}{11.5913}\\ \frac{20181}{32.6705}& 鈮�& {蟽}^2鈮�\frac{20181}{11.5913}\\ 617.713228& 鈮�& {蟽}^2鈮�1741.04716\end{array}$

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

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