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.

Calculation table

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

$=-22-\left(6\right)(-9)/3$

$=-22+54/3$

= -22+18

= -4

$S_{xx}=\sum x^{2}i-{\left(\sum x_i\right)}^2/n$

$=14-(6{)}^2/3$

$=14-36/3$

= 14-12

= 2

The total sum of squares SST is given by,

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

$=41-(-9{)}^2/3$

$=41-81/3$

= 41 - 27

= 14

The regression sum of squares SSR is given by,

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

$=\frac{(-4{)}^2}{2}=\frac{16}{2}=8$

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

= 14 - 8

= 6

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

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

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

= 2.449489

$\approx 2.45$

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

$b_1=\frac{s_{xp}}{s_{ux}}$

$=\frac{-4}{2}$

= -2

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

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

$=\frac{1}{3}(-9+2(6\left)\right)$

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

= 1

$\text{So, the regression equation is}{\hat{y}}_p=1-x_p\text{.}$

The formula for calculating the value of the point estimate is obtained by substituting the value of $x_p=2$in the regression equation.

${\hat{y}}_p=1-2x_p$

$=1-2\left(2\right)$

= - 3

$\text{The point estimate is}{\hat{y}}_p=-3$

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

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

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

$\text{We have}{x}_p=2\text{,}$

${\hat{y}}_p=-3,$

$s_e=2.45$and

$S_{xx}=2$

$\text{Then,}-3\pm 12.706\times (2.45)\sqrt{\frac{1}{3}+\frac{(2-6/3{)}^2}{2}}$

$-3\pm 31.1297\sqrt{0.3333}$

$\text{Thatis}-3\pm 17.97274067\text{,}$

$-20.97\text{to}14.97$

$\text{The}95\%\text{confidence interval for the conditional mean is}-20.97\text{to}14.97$

$\text{The regression equation is}{\hat{y}}_p=1-2x_p\text{.}$

$\text{The predicted value is obtained by substituting the value of}{x}_p=2\text{in the regression equation.}$

${\hat{y}}_p=1-2x_p$

$=1-2\left(2\right)$

= - 3

$\text{The predicted value is}{\hat{y}}_p=-3\text{.}$

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

df = n -2

= 3 - 2

= 1

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

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

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

$\text{We have}{x}_p=2$

${\hat{y}}_p=-3$

$s_e=2.45$and

$S_{xx}=2$