91Ó°ÊÓ

Given,

\(x=2\)

Computation table:

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

\(=34-(12)(10)/4\)

\(=34-120/4\)

\(=34-30\)

\(=4\)

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

\(=52-(12)^{2}/4\)

\(=52-144/4\)

\(=52-36\)

\(=16\)

The total sum of squares SST is given by,

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

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

\(=30-100/4\)

\(=30-25\)

\(=5\)

The regression sum of squares SSR is given by,

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

\(=\frac{(4)^{2}}{16}=\frac{16}{16}=1\)

\(SSE=SST-SSR\)

\(=5-1\)

\(=4\)

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

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

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

\(=1.414213562\)

\(\approx 1.414\)

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

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

\(=\frac{4}{16}\)

\(=0.25\)

The formula for calculating the value of y-intercept is

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

\(=\frac{1}{4}(10-0.25(12))\)

\(=\frac{1}{4}(7)\)

\(=1.75\)

So, the regression equation is \(\hat{y_{p}}=1.75+0.25x_{p}\)

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

\(\hat{y_{p}}=1.75+0.25x_{p}\)

\(=1.75+0.25(2)\)

\(=2.25\)

The point estimate is \(\hat{y_{p}}=2.25\)

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

\(\hat{y_{p}} =2.25\),

\(s_{e}=1.414\),

\(S_{xx}=16\).

So, \(2.25\pm 4.303\times (1.414) \sqrt{\frac{1}{4}+\frac{(2-12/4)^{2}}{16}}\)

\(2.25\pm 6.0853026 \sqrt{0.25+0.0625}\)

Or \(2.25\pm 3.401787569\)

Or \(-1.152\) to \(5.652\)

Therefore, the \(95%\) confidence interval for the conditional mean is \(-1.152\) to \(5.652 \).

The regression equation is \(\hat{y_{p}}=1.75+0.25x_{p}\)

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

\(\hat{y_{p}}=1.75+0.25x_{p}\)

\(=1.75+0.25(2)\)

\(=2.25\)

The predicted value is \(\hat{y_{p}}= 2.25\)

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

\(\hat{y_{p}} =2.25\),

\(s_{e}=1.414\),

\(S_{xx}=16\).

So, \(2.25\pm 4.303\times (1.414) \sqrt{\frac{1}{4}+\frac{(2-12/4)^{2}}{16}}\)

\(2.25\pm 6.0853026 \sqrt{1+0.25+0.0625}\)

Or \(2.25\pm 6.971589948\)

Or \(-4.721\) to \(9.221\)

Therefore, the \(95%\) confidence interval for the conditional mean is \(--4.721\) to \(9.221\).