Chapter 11.14, Problem 88E

a.

To determine

State whether the given statement is true or false.

Answer to Problem 88E

The given statement is false.

Explanation of Solution

The statement is, if Model I is fit, the estimate for ${\sigma }^2$ is based on 16 degrees of freedom.

Model I is as follows:

$Y={\beta }_0+{\beta }_1{x}_1+{\beta }_2{x}_2+{\beta }_3{x}_3+{\beta }_4{x}_4+\epsilon$

An unbiased estimator of ${\sigma }^2$ is as follows:

$s^2=\frac{SSE}{n-\left(k+1\right)}$

According to this, the estimate for ${\sigma }^2$ is based on $n-\left(k+1\right)$ degrees of freedom.

It is given that $n=20$

From Model I, the value of $k=3$.

The degrees of freedom is as follows:

$\begin{array}{c}n-\left(k+1\right)=20-\left(4+1\right)\\ =15\end{array}$

This indicates that the estimate for ${\sigma }^2$ is based on 15 degrees of freedom.

Since the estimate for ${\sigma }^2$ is based on 15 degrees of freedom, the given statement is false.

b.

To determine

Explain whether the given statement is true or false.

Answer to Problem 88E

The given statement is true.

Explanation of Solution

The fitted Model II is as follows:

$\widehat{y}={\widehat{\beta }}_0+{\widehat{\beta }}_1{x}_1+{\widehat{\beta }}_2{x}_2$

To determine whether $x_2$ contributes to a better fit of the model to the data, one needs to test the null hypothesis $H_0:{\beta }_2=0$ against $H_a:{\beta }_2\ne 0$.

The significance level is 0.01.

The test statistic is as follows:

$t=\frac{{\widehat{\beta }}_3}{\left(s\sqrt{c_{33}}\right)}$

Where,

$s=\sqrt{\frac{SSE}{n-\left(k+1\right)}}=\sqrt{\frac{Y^{\prime }Y-{\widehat{\beta }}^{\prime }{X}^{\prime }Y}{n-\left(k+1\right)}}$, $\widehat{\beta }={\left(X^{\prime }X\right)}^{-1}{X}^{\prime }Y$

Here, $k+1$ is the number of unknown ${\beta }_i$ values and n is the number of data values.

Also, it is noticed that $c_{ii}$ is the element in row $\left(i+1\right)$ and column $\left(i+1\right)$ of ${\left(X^{\prime }X\right)}^{-1}$.

Where, $0\le i\le k$

The rejection region for this test is as follows:

$RR:\left\{\left|t\right|>t_{\alpha /2}\right\}=\left\{t<-t_{\alpha /2}\text{ or }t>t_{\alpha /2}\right\}$

Where, $P\left(t<-t_{\alpha /2}\right)=\alpha /2$

The value of $t_{\alpha /2}$ can be found using Student’s t-distribution with $n-2$ degrees of freedom.

According to this, if $\left|t\right|<t_{0.005}$, fail to reject the null hypothesis $H_0:{\beta }_2=0$. This indicates that $x_2$ does not contribute to a better fit of the model to the data.

If $\left|t\right|>t_{0.005}$, reject the null hypothesis $H_0:{\beta }_2=0$. It can be concluded that $x_2$ contributes to a better fit of the model to the data.

Thus, the given statement is true.

c.

To determine

Check whether $SS{E}_I\le SS{E}_{II}$ when models I and II are both fit.

Answer to Problem 88E

The given statement is true.

Explanation of Solution

If there are k independent variables, then the fitted model is as follows:

$\widehat{y}={\widehat{\beta }}_0+{\widehat{\beta }}_1{x}_1+{\widehat{\beta }}_2{x}_2+\cdots {\widehat{\beta }}_k{x}_k$

It is known that $SSE={\displaystyle \sum _{i=1}^n{\left(y_i-{\widehat{y}}_i\right)}^2}$

Where ${\widehat{y}}_i$ is the predicted value of $y_i$ and $1\le i\le n$

Here, Model I contains four independent variables $x_1$, $x_2$, $x_3$ and $x_4$.

Also Model II contains two independent variables $x_1$ and $x_2$.

This indicates that the predicted value for $y_i$ cannot be worse for Model I than Model II.

Where, $1\le i\le 20$

This means that ${\left(y_i-{\widehat{y}}_i\right)}^2$ cannot be greater when Model I is used than Model II. Therefore, in this case, SSE cannot be greater for Model I than Model II. That is, $SS{E}_I\le SS{E}_{II}$

According to this, the given statement is true.

d.

To determine

Check whether ${\widehat{\sigma }}^2{}_I\le {\widehat{\sigma }}^2{}_{II}$ when Models I and II are both fit.

Answer to Problem 88E

The given statement is false.

Explanation of Solution

It is known that an unbiased estimator for ${\sigma }^2$ is as follows:

$s^2=\frac{SSE}{n-\left(k+1\right)}or{\widehat{\sigma }}^2=\frac{SSE}{n-\left(k+1\right)}$

Where, $SSE={\displaystyle \sum _{i=1}^n{\left(y_i-{\widehat{y}}_i\right)}^2}$, ${\widehat{y}}_i$ is the predicted value of $y_i$ and $1\le i\le n\left(=20\right)$.

Also $k+1$ is the number of unknown ${\beta }_i$ values and n is the number of data values.

For Model I, it is seen that $k_1=4$ and for Model II, one can have $k_2=2$.

Now, compute the values of ${\widehat{\sigma }}^2{}_I$ and ${\widehat{\sigma }}^2{}_{II}$ as follows:

${\widehat{\sigma }}^2{}_I=\frac{SS{E}_I}{n-\left(k_1+1\right)}=\frac{SS{E}_I}{15}$

${\widehat{\sigma }}^2{}_{II}=\frac{SS{E}_{II}}{n-\left(k_2+1\right)}=\frac{SS{E}_{II}}{17}$

From Part (c), it is seen that $SS{E}_I\le SS{E}_{II}$.  Hence it cannot be concluded that ${\widehat{\sigma }}^2{}_I\le {\widehat{\sigma }}^2{}_{II}$.

Therefore, the given statement is false.

e.

To determine

Check whether Model II is a reduction of Model I.

Answer to Problem 88E

The statement istrue.

Explanation of Solution

Models I and II are as follows:

$\begin{array}{l}Y={\beta }_0+{\beta }_1{x}_1+{\beta }_2{x}_2+{\beta }_3{x}_3+{\beta }_4{x}_4+\epsilon \\ Y={\beta }_0+{\beta }_1{x}_1+{\beta }_2{x}_2+\epsilon \end{array}$

It is observed that Model II is a subset of Model I.

That is, $\text{Model I}=\text{Model II}+{\beta }_3{x}_3+{\beta }_4{x}_4+\epsilon$

Therefore, the given statement is true.

f.

To determine

Check whether Model III is a reduction of Model I.

Answer to Problem 88E

The given statement isfalse.

Explanation of Solution

Models I and III are as follows:

$\begin{array}{l}Y={\beta }_0+{\beta }_1{x}_1+{\beta }_2{x}_2+{\beta }_3{x}_3+{\beta }_4{x}_4+\epsilon \\ Y={\beta }_0+{\beta }_1{x}_1+{\beta }_2{x}_2+{\beta }_3{x}_1{x}_2+\epsilon \end{array}$

Here, Model III contains $x_1{x}_2$ and Model I do not contain $x_1{x}_2$. This indicates that Model III is not a reduction of Model I. Therefore, the given statement is false.