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:

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

\(=30-(10)(8)/4\)

\(=30-80/4\)

\(=30-20\)

\(=10\)

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

\(=30-(10)^{2}/4\)

\(=30-100/4\)

\(=30-25\)

\(=5\)

The total sum of squares SST is given by,

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

\(=42-(8)^{2}/4\)

\(=42-64/4\)

\(=42-16\)

\(=26\)

The regression sum of squares SSR is given by,

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

\(=\frac{(10)^{2}}{5}=\frac{100}{5}=20\)

\(SSE=SST-SSR\)

\(=26-20\)

\(=6\)

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

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

\(=\sqrt{\frac{6}{4-2}}\)

\(=1.73205\)

\(\approx 1.73\)

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

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

\(=\frac{10}{5}\)

\(=2\)

The formula for calculating the value of y-intercept is

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

\(=\frac{1}{4}(8-2(10))\)

\(=\frac{1}{4}(-12)\)

\(=-3\)

So, the regression equation is \(\hat{y_{p}}=-3+2x_{p}\)

The formula for calculating the value of the point estimate is obtained by substituting the value of \(x_{p}=4\) in the regression equation.

\(\hat{y_{p}}=-3+2x_{p}\)

\(=-3+2(4)\)

\(=5\)

The point estimate is \(\hat{y_{p}}=-5\)

STEP 1: For a \(95%\) confidence interval, \(\alpha=0.05\). Because \(n=4\),

\(df=n-2\)

\(=4-2\)

\(=2\)

From technology, \(t_{\alpha/2}=t_{0.05/2}=t_{0.088}=4.303\)

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, \(x_{p}=4\),

\(\hat{y_{p}} =5\),

\(s_{e}=1.732\),

\(S_{xx}=5\).

So, \(5\pm 4.303\times (1.732) \sqrt{\frac{1}{4}+\frac{(4-10/4)^{2}}{5}}\)

\(5\pm 7.452796 \sqrt{0.25+0.45}\)

Or \(5\pm 6.235456301\)

Or \(-1.235\) to \(11.235\)

Therefore, the \(95%\) confidence interval for the conditional mean is \(-1.235\) to \(11.235 \).

The regression equation is \(\hat{y_{p}}=-3+2x_{p}\)

The predicted value is obtained by substituting the value of \(x_{p}=4\) in the regression equation.

\(\hat{y_{p}}=-3+2x_{p}\)

\(=-3+2(4)\)

\(=5\)

The predicted value is \(\hat{y_{p}}= 5\)

STEP 1: For a \(95%\) confidence interval, \(\alpha=0.05\). Because \(n=4\),

\(df=n-2\)

\(=4-2\)

\(=2\)

From technology, \(t_{\alpha/2}=t_{0.05/2}=t_{0.088}=4.303\)

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, \(x_{p}=4\),

\(\hat{y_{p}} =5\),

\(s_{e}=1.732\),

\(S_{xx}=5\).

So, \(5\pm 4.303\times (1.732) \sqrt{\frac{1}{4}+\frac{(4-10/4)^{2}}{5}}\)

\(5\pm 7.452796 \sqrt{1+0.25+0.45}\)

Or \(5\pm 9.717257122\)

Or \(-4.717\) to \(14.717\)

Therefore, the \(95%\) prediction interval is \(-4.717\) to \(14.717\).