Chapter 13.4, Problem 13.4E

i.

To determine

To identify:

The independent variable $x_1,x_2,x_3$ used for prediction of $y$

Answer to Problem 13.4E

Independent variable $x_1,x_2,x_3$ contribute the information to predict $y$ ,

Explanation of Solution

Given information:

  $\begin{array}{l}{b}_0=1.04\\ b_1=1.29\\ SE\left(b_1\right)=0.42\\ b_2=2.72\\ SE\left(b_2\right)=0.65\\ b_3=0.41\\ SE\left(b_3\right)=0.17\end{array}$

Formula used:

The student’s t- statistics is given by the following expression:

  $t=\frac{b_i-{\beta }_t}{SE\left(b_i\right)}$

The values:

  $\begin{array}{l}\left(t\le 3.0714\right)=0.9947\\ \left(t\le 4.1846\right)=0.9992\\ \left(t\le 2.4117\right)=0.9827\end{array}$

Calculation:

To identify which independent variable $x_1,x_2,x_3$ contribute the information to predict $y$ , need to use the significance of partial regression co-efficient.

The $F$ test with the hypothesis:

To test the null hypothesis:

  $H_0:{\beta }_t=0$

Vs

  $H_0:{\beta }_t\ne 0$

And, $df=\left(n-k-1\right)=15-3-1=11$

1. At $b_1=1.29,SE\left(b_1\right)=0.42$:

The student’s t- statistics is calculated as:

  $\begin{array}{l}t=\frac{b_1-{\beta }_1}{SE\left(b_1\right)}\\ t=\frac{0.129-0}{0.42}\\ t=3.071428\end{array}$

The corresponding $P$ -value is calculated as:

  $\begin{array}{l}2P\left(t\ge 3.0714\right)=2P\left(1-\left(t\le 300714\right)\right)\\ 2P\left(t\ge 3.0714\right)=2P\left(1-0.9947\right)\\ 2P\left(t\ge 3.0714\right)=2\left(0.0053\right)\\ 2P\left(t\ge 3.0714\right)=0.0106\end{array}$

Reject the null hypothesis $H_0$ as the $P-value=0.0106$ is less than $\alpha =0.05$

Hence, independent variable $x_1$ contribute the information to predict $y$ .

2. At $b_2=2.72,SE\left(b_2\right)=0.65$:

The student’s t- statistics is calculated as:

  $\begin{array}{l}t=\frac{b_2-{\beta }_2}{SE\left(b_2\right)}\\ t=\frac{2.72-0}{0.65}\\ t=4.1846\end{array}$

The corresponding $P$ -value is calculated as:

  $\begin{array}{l}2P\left(t\ge 4.1846\right)=2P\left(1-\left(t\le 4.1846\right)\right)\\ 2P\left(t\ge 4.1846\right)=2P\left(1-0.9992\right)\\ 2P\left(t\ge 4.1846\right)=2\left(0.0008\right)\\ 2P\left(t\ge 4.1846\right)=0.0016\end{array}$

Reject the null hypothesis $H_0$ as the $P-value=0.0016$ is less than $\alpha =0.05$

Hence, independent variable $x_2$ contribute the information to predict $y$ .

3. At $b_3=0.41,SE\left(b_3\right)=0.17$:

The student’s t- statistics is calculated as:

  $\begin{array}{l}t=\frac{b_3-{\beta }_3}{SE\left(b_3\right)}\\ t=\frac{0.41-0}{0.17}\\ t=2.4117\end{array}$

The corresponding $P$ -value is calculated as:

  $\begin{array}{l}2P\left(t\ge 2.4117\right)=2P\left(1-\left(t\le 2.4117\right)\right)\\ 2P\left(t\ge 2.4117\right)=2P\left(1-0.9827\right)\\ 2P\left(t\ge 2.4117\right)=2\left(0.0173\right)\\ 2P\left(t\ge 2.4117\right)=0.0346\end{array}$

Reject the null hypothesis $H_0$ as the $P-value=0.0346$ is less than $\alpha =0.05$

Hence, independent variable $x_3$ contribute the information to predict $y$ .

Conclusion:

Therefore, all independent variable $x_1,x_2,x_3$ contribute the information to predict $y$ .

ii.

To determine

The least-square prediction equation

Answer to Problem 13.4E

The least-square prediction equation is $\sout{Y}=1.04+1.29x_1+2.72x_2+0.41x_3$

Explanation of Solution

Given information:

  $\begin{array}{l}{b}_0=1.04\\ b_1=1.29\\ SE\left(b_1\right)=0.42\\ b_2=2.72\\ SE\left(b_2\right)=0.65\\ b_3=0.41\\ SE\left(b_3\right)=0.17\end{array}$

Calculation:

The line which makes the vertical distance from the data points to the regression line is known as least-square regression. This distance is as small as possible.

The least-square prediction equation is given by the following eq:

  $\widehat{y}=b_0+b_1{x}_1+b_2{x}_2+b_3{x}_3$

Put the given values:

  $\begin{array}{l}\widehat{y}=b_0+b_1{x}_1+b_2{x}_2+b_3{x}_3\\ \widehat{y}=1.04+1.29x_1+2.72x_2+0.41x_3\end{array}$

Conclusion:

Hence, least-square prediction equation is derived as $\widehat{y}=1.04+1.29x_1+2.72x_2+0.41x_3$

iii.

To determine

To explain:

The relationship between lines shown by the graph the relationship between $x_1$ and $y$ when $\left(x_2=1,x_3=0\right)$ and $\left(x_2=1,x_3=0.5\right)$

Answer to Problem 13.4E

Lines shown by the graph is appearing to be parallel with each other

Explanation of Solution

Given information:

Predictor variables: $x_1,x_2,x_3$

Calculation:

The given equation is $\widehat{y}=b_0+b_1{x}_1+b_2{x}_2+b_3{x}_3$ related to three predictor variables $x_1,x_2,x_3$ .

When $\left(x_2=1,x_3=0\right)$ the equation becomes:

  $\begin{array}{l}\widehat{y}=b_0+b_1{x}_1+b_2{x}_2+b_3{x}_3\\ \widehat{y}=1.04+1.29x_1+2.72x_2+0.41x_3\\ \widehat{y}=1.04+1.29x_1+2.72\left(1\right)+0.41\left(0\right)\\ \widehat{y}=3.76+1.29x_1\end{array}$

When $\left(x_2=1,x_3=0.5\right)$ the equation becomes:

  $\begin{array}{l}\widehat{y}=b_0+b_1{x}_1+b_2{x}_2+b_3{x}_3\\ \widehat{y}=1.04+1.29x_1+2.72x_2+0.41x_3\\ \widehat{y}=1.04+1.29x_1+2.72\left(1\right)+0.41\left(0.5\right)\\ \widehat{y}=3.965+1.29x_1\end{array}$

The below graph depicting the relationship between $x_1$ and $y$ when $\left(x_2=1,x_3=0\right)$ and $\left(x_2=1,x_3=0.5\right)$

  

Two lines shown by graph are seems to be parallel with each other.

Conclusion:

The above graph depicting the relationship between the two lines and they appear to be parallel to each other.

iv.

To determine

To explain:

The practical interpretation of ${\beta }_1$

Answer to Problem 13.4E

The measurement of changes in $x_1$when all other independent variables are held to be constant is done by${\beta }_1$.

Explanation of Solution

Three-dimensional extension line of means is depicted by the given eq.

  $E\left(y\right)={\beta }_0+{\beta }_1{x}_1+{\beta }_2{x}_2$

When $x_1$ and $x_2$ both are zero, the intercept-the average value of $y$ is given by ${\beta }_0$

Partial slopes of the model is denoted by ${\beta }_1$ and ${\beta }_0$ . They are also known as partial regression coefficients.

The measurement of changes occurs in $y$ for one unit change in $x_1$ when all other independent variables are held to be constant is done by ${\beta }_1$ .

The slope estimated by a fit line with $x_1$ alone is not same as partial regression coefficients ${\beta }_1$ with $x_1$ and $x_2$ .

Conclusion:

Hence, the unknown constant values are estimated by using the sample data.