Find a$95\%$prediction interval for the value of the response variable corresponding to the specified value of the predictor variable.

a. Obtain a point estimate for the mean tax efficiency of all mutual fund portfolios with$6\%$of their investments in energy securities.

b. Determine a$95\%$confidence interval for the mean tax efficiency of all mutual fund portfolios with$6\%$of their investments in energy securities.

c. Find the predicted tax efficiency of a mutual fund portfolio with $6\%$of its investments in energy securities.

d. Determine a $95\%$prediction interval for the tax efficiency of a mutual fund portfolio with $6\%$of its investments in energy securities.

$S_{xy}=\sum x_i{y}_i-\left(\sum x_i\right)\left(\sum y_i\right)/n$

$=47-\left(9\right)\left(15\right)/3$

$=47-135/3$

=47-45

=2

The total sum of squares SST is given by,

$S_{yy}=\sum y_i^2-{\left(\sum y_i\right)}^2/n$

$=83-(15{)}^2/3$

$=83-225/3$

=83-75

=8

The regression sum of squares SSR is given by,

$SSR=\frac{{S_{x_y}}^2}{S_{xx}}$

$=\frac{(2{)}^2}{2}=\frac{4}{2}=2$

$\mathrm{SSE}=\mathrm{SST}-\mathrm{SSR}$

=8-2

=6

The formula for calculating the standard error of the estimate is,

$s_e=\sqrt{\frac{SSE}{n-2}}$

$=\sqrt{\frac{6}{3-2}}$

=2.4498

$\approx 2.45$

The formula for calculating the slope of the regression line is.

$b_1=\frac{s_{xy}}{s_{xx}}$

$=\frac{2}{2}$

=1

The formula for calculating the value of $y$-intercept is

$b_0=\frac{1}{n}\left(\sum y_i-b\sum x_i\right)$

$=\frac{1}{3}(15-1(9\left)\right)$

$=\frac{1}{3}\left(6\right)$

=2

$\text{So, the regression equation is}{\hat{y}}_p=2+x_p\text{.}$

The formula for calculating the value of the point estimate is obtained by substituting the value of $x_p=3$in the regression equation.

${\hat{y}}_p=2+x_p$

=2+3

=5

$\text{The point estimate is}{\hat{y}}_p=5\text{.}$

$\text{For a 95\% confidence interval,}\alpha =0.05\text{.}\text{Since}n=3\text{,}$

df=n-2

=3-2

=1

$\text{Technologicaly,}{t}_{\alpha /2}=t_{0.05/2}=t_{0.025}=12.706$

The formula for calculating the end points of the confidence interval for the conditional mean of the response variable are

${\hat{y}}_p\pm t_{\alpha /2}\times s_e\sqrt{\frac{1}{n}+\frac{{\left(x_p-\sum x_i/n\right)}^2}{S_{xx}}}$

$\text{We have}{x}_p=3\text{,}$

${\hat{y}}_p=5$

$s_e=2.45$

$S_{xx}=2$

$\text{So,}5\pm 12.706\times (2.45)\sqrt{\frac{1}{3}+\frac{(3-9/3{)}^2}{2}}$

$5\pm 31.1297\sqrt{0.3333}$

$\text{Or}5\pm 17.97274067$

$\text{Or}-12.97\text{to}22.97$

$\text{Therefore, the}95\%\text{confidence interval for the conditional mean is}-12.97\text{to}22.97$

$\text{The regression equation is}{\hat{y}}_p=2+x_p\text{.}$

The predicted value is obtained by substituting the value of$x_p=3$ in the regression equation.

${\hat{y}}_p=2+x_p$

=2+3

=5

$\text{The predicted value is}{\hat{y}}_p=5\text{.}$

$\text{For a}95\%\text{confidence interval,}\alpha =0.05\text{.}\text{Since}n=3\text{,}$

df=n-2

=3-2

=1

$\text{Technologicaly,}{t}_{\alpha /2}=t_{0.05/2}=t_{0.025}=12.706$

The formula for calculating the end points of the prediction interval for the value of the response variable are

${\hat{y}}_p\pm t_{\alpha /2}\times s_e\sqrt{1+\frac{1}{n}+\frac{{\left(x_p-\sum x_i/n\right)}^2}{S_{xx}}}$

We have $x=3$

${\hat{y}}_p=5,$

$s_e=2.45$

$S_{xx}=2$

$\text{So}5\pm 12.706\times (2.45)\sqrt{1+\frac{1}{3}+\frac{(3-9/3{)}^2}{2}}$

$5\pm 31.1297\sqrt{1+0.3333}$

$\text{Or}5\pm 35.94548135$

$\text{Or}-30.94\text{to}40.94$

$\text{The}95\%\text{prediction interval is}-30.94\text{to}40.94$