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.

Given,

Computation table:

\(S_{xy}=\sum x_{i}y_{i}-(\sum x_{i})(\sum y_{i})/n\)

\(=-22-(6)(-9)/3\)

\(=-22+54/3\)

\(=-22+18\)

\(=-4\)

\(S_{xx}=\sum x^{2}_{i}-(\sum x_{i})^{2}/n\)

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

\(=14-36/3\)

\(=14-12\)

\(=2\)

The total sum of squares SST is given by,

\(S_{yy}=\sum y^{2}_{i}-(\sum y_{i})^{2}/n\)

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

\(=41-81/3\)

\(=41-27\)

\(=14\)

The regression sum of squares SSR is given by,

\(SSR=\frac{S_{xy}^{2}}{S_{xx}}\)

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

\(SSE=SST-SSR\)

\(=14-8\)

\(=6\)

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

\(s_{e}=\sqrt{\frac{SSE}{n-2}}\)

\(=\sqrt{\frac{6}{3-2}}\)

\(=2.449489743\)

\(\approx 2.45\)

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

\(b_{1}=\frac{S_{xy}}{S_{xx}}\)

\(=\frac{-4}{2}\)

\(=-2\)

The formula for calculating the value of y-intercept is

\(b_{0}=\frac{1}{n}(\sum y_{i}-b\sum x_{i})\)

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

\(=\frac{1}{3}(3)\)

\(=1\)

So, the regression equation is \(\hat{y_{p}}=1-x_{p}\)

The formula for calculating the value of the point estimate is obtained by substituting the value of \(x_{p}=2\) in the regression equation.

\(\hat{y_{p}}=1-2x_{p}\)

\(=1-2(2)\)

\(=-3\)

The point estimate is \(\hat{y_{p}}=-3\)

STEP 1: For a \(95%\) confidence interval, \(\alpha=0.05\). Because \(n=3\),

\(df=n-2\)

\(=3-2\)

\(=1\)

From technology, \(t_{\alpha/2}=t_{0.05/2}=t_{0.025}=12.706\)

STEP 2:

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

\(\hat{y_{p}}\pm t_{\alpha/2}\times s_{e}\sqrt{\frac{1}{n}+\frac{(x_{p}-\sum x_{i}/n)^{2}}{S_{xx}}}\)

We have, \(x_{p}=2\),

\(\hat{y_{p}} =-3\),

\(s_{e}=2.45\),

\(S_{xx}=2\).

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

\(-3\pm 31.1297 \sqrt{0.3333}\)

Or \(-3\pm 17.97274067\)

Or \(-20.97\) to \(14.97\)

Therefore, the \(95%\) confidence interval for the conditional mean is \(-20.97\) to \(14.97\).

The regression equation is \(\hat{y_{p}}=1-2x_{p}\)

The predicted value is obtained by substituting the value of \(x_{p}=2\) in the regression equation.

\(\hat{y_{p}}=1-2x_{p}\)

\(=1-2(2)\)

\(=-3\)

The predicted value is \(\hat{y_{p}}= -3\)

STEP 1: For a \(95%\) confidence interval, \(\alpha=0.05\). Because \(n=3\),

\(df=n-2\)

\(=3-2\)

\(=1\)

From technology, \(t_{\alpha/2}=t_{0.05/2}=t_{0.025}=12.706\)

STEP 2:

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

\(\hat{y_{p}}\pm t_{\alpha/2}\times s_{e}\sqrt{\frac{1}{n}+\frac{(x_{p}-\sum x_{i}/n)^{2}}{S_{xx}}}\)

We have, \(x_{p}=2\),

\(\hat{y_{p}} =-3\),

\(s_{e}=2.45\),

\(S_{xx}=2\).

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

\(-5\pm 31.1297 \sqrt{0.3333}\)

Or \(-3\pm 35.94548135\)

Or \(-38.94\) to \(32..94\)

Therefore, the \(95%\) prediction interval is \(-38.94\) to \(32.94\).