Chapter 8, Problem 5SE

In a simulation of 30 mobile computer networks, the average speed, pause time, and number of neighbor were measured. A “neighbor” is a computer within the transmission range of another. The data are presented in the following table.

1. a. Fit the model with Neighbors as the dependent variable, and independent variables Speed, Pause, Speed,·Pause, Speed², and Pause².
2. b. Construct a reduced model by dropping any variables whose P-values are large, and test the plausibility of the model with an F test.
3. c. Plot the residuals versus the fitted values for the reduced model. Are there any indications that the model is inappropriate? If so, what are they?
4. d. Someone suggests that a model containing Pause and Pause² as the only dependent variables is adequate. Do you agree? Why or why not?
5. e. Using a best subsets software package, find the two models with the highest R² value for each model size from one to five variables. Compute C_p and adjusted R² for each model.
6. f. Which model is selected by minimum C_p? By adjusted R²? Are they the same?

To determine

Construct a multiple linear regression model with neighbor as the dependent variable, speed, pause, $\text{speed}\times \text{pause}$, ${\text{speed}}^2$ and ${\text{pause}}^2$ as the independent variables for the given data.

Answer to Problem 5SE

A multiple linear regression model for the given data is:

$\underset{\_}{\widehat{y}\text{=}10.840-0.0739x_1-0.1274x_2+0.001110x_1^2-0.000243x_1{x}_2+0.001674x_2^2}$.

Explanation of Solution

Calculation:

The data represents the values of the variables number of neighbors, average speed and pause time for a simulation of 30 mobile network computers.

Multiple linear regression model:

A multiple linear regression model is given as $y_i={\beta }_0+{\beta }_1{x}_{1i}+\mathrm{...}+{\beta }_k{x}_{ki}+{\epsilon }_i$ where $y_i$ is the response variable, and $x_{1i},x_{2i},\mathrm{...},x_{ki}$ are the k predictor variables. The quantities ${\beta }_0,{\beta }_1,\mathrm{...},{\beta }_k$ are the slopes corresponding to $x_{1i},x_{2i},\mathrm{...},x_{ki}$ respectively.${\widehat{\beta }}_0$ is the estimated intercept of the line, from the sample data.

Let $x_1,x_2$ be speed and pause. The response variable is $y=\text{neighbors}$.

Regression:

Software procedure:

Step by step procedure to obtain regression using MINITAB software is given as,

• Choose Stat > Regression > General Regression.
• In Response, enter the numeric column containing the response data Y.
• In Model, enter the numeric column containing the predictor variables X1, X2, X1*X2, X1*X1 and X2*X2.
• Click OK.

Output obtained from MINITAB is given below:

The ‘Coefficient’ column of the regression analysis MINITAB output gives the slopes corresponding to the respective variables stored in the column ‘Term’.

A careful inspection of the output shows that the fitted model is:

$\widehat{y}\text{=}10.840-0.0739x_1-0.1274x_2+0.001110x_1^2-0.000243x_1{x}_2+0.001674x_2^2$.

Hence, the multiple linear regression model for the given data is:

$\underset{\_}{\widehat{y}\text{=}10.840-0.0739x_1-0.1274x_2+0.001110x_1^2-0.000243x_1{x}_2+0.001674x_2^2}$.

To determine

Construct a reduced model by dropping the variables with large P- values.

Check whether the reduced model is plausible or not.

Answer to Problem 5SE

A multiple linear regression model for the given data is:

$\underset{\_}{\widehat{y}=10.967-0.0799x_1-0.1325x_2+0.001110x_1^2+0.001674x_2^2}$.

Yes, there is enough evidence to conclude that the reduced model is plausible.

Explanation of Solution

Calculation:

From part (a), it can be seen that the ‘P’ column of the regression analysis MINITAB output gives the slopes corresponding to the respective variables stored in the column ‘Term’.

By observing the P- values of the MINITAB output, it is clear that the largest P-value is 0.390 corresponding to the predictor variable $x_1{x}_2$. Remaining all P- values are reasonable.

Now, the new regression has to be fitted after dropping the predictor variable $x_1{x}_2$.

Regression:

Software procedure:

Step by step procedure to obtain regression using MINITAB software is given as,

• Choose Stat > Regression > General Regression.
• In Response, enter the numeric column containing the response data Y.
• In Model, enter the numeric column containing the predictor variables X1, X2, X1*X1 and X2*X2.
• Click OK.

Output obtained from MINITAB is given below:

The ‘Coefficient’ column of the regression analysis MINITAB output gives the slopes corresponding to the respective variables stored in the column ‘Term’.

A careful inspection of the output shows that the fitted model is:

$\widehat{y}=10.967-0.0799x_1-0.1325x_2+0.001110x_1^2+0.001674x_2^2$.

Hence, the multiple linear regression model for the given data is:

$\underset{\_}{\widehat{y}=10.967-0.0799x_1-0.1325x_2+0.001110x_1^2+0.001674x_2^2}$.

The full model is,

$\widehat{y}\text{=}10.840-0.0739x_1-0.1274x_2+0.001110x_1^2-0.000243x_1{x}_2+0.001674x_2^2$

The reduced model is,

$\widehat{y}=10.967-0.0799x_1-0.1325x_2+0.001110x_1^2+0.001674x_2^2$

The test hypotheses are given below:

Null hypothesis:

$H_0:{\beta }_4=0$

That is, the dropped predictor of the full model is not significant to predict y.

Alternative hypothesis:

$H_1:{\beta }_4\ne 0$

That is, the dropped predictor of the full model is significant to predict y.

Test statistic:

$F=\frac{\frac{\left(SS{E}_{\text{Reduced}}-SS{E}_{\text{Full}}\right)}{p-k}}{\frac{SS{E}_{\text{Full}}}{\left[n-\left(p+1\right)\right]}}$

Where,

$SS{E}_{\text{Full}}$ represents the sum of squares due to error obtained from the full model.

$SS{E}_{\text{Reduced}}$ represents the sum of squares due to error obtained from the reduced model.

n represents the total number of observations.

p represents the number of predictors on the full model.

k represents the number of predictors on the reduced model.

From the obtained MINITAB outputs, the value of error sum of squares for full model is ${\text{SSE}}_{\text{Full}}=2.2642$ and the value of error sum of squares for the reduced model is ${\text{SSE}}_{\text{Reduced}}=2.7307$.

The total number of observations is $n=30$.

Number of predictors on the full model is $p=5$ and the number of predictors on the reduced model is $k=4$.

Degrees of freedom of F-statistic for reduced model:

In a reduced multiple linear regression analysis, the F-statistic is $f=\frac{\frac{\left(SS{E}_{\text{Reduced}}-SS{E}_{\text{Full}}\right)}{p-k}}{\frac{SS{E}_{\text{Full}}}{\left[n-\left(p+1\right)\right]}}$.

In the ratio, the numerator is obtained by dividing the quantity $SS{E}_{\text{Reduced}}-SS{E}_{\text{Full}}$ by its degrees of freedom, $p-k$. The denominator is obtained by dividing the error sum of squares of full model by the error degrees of freedom, $n-\left(p+1\right)$.

Thus, the degrees of freedom for the F-statistic in a reduced multiple regression analysis are $p-k$ and $n-\left(p+1\right)$.

Hence, the numerator degrees of freedom is $p-k=5-4=1$ and the denominator degrees of freedom is $n-\left(p+1\right)=30-6=24$.

Test statistic under null hypothesis:

Under the null hypothesis, the test statistic is obtained as follows:

$\begin{array}{c}f=\frac{\frac{\left(SS{E}_{\text{Reduced}}-SS{E}_{\text{Full}}\right)}{p-k}}{\frac{SS{E}_{\text{Full}}}{\left[n-\left(p+1\right)\right]}}\\ =\frac{\frac{\left(2.7307-2.6462\right)}{5-4}}{\frac{2.6462}{\left[30-\left(5+1\right)\right]}}\\ =\frac{0.0845}{0.11026}\\ =0.76638\end{array}$

Thus, the test statistic is $\underset{\_}{F=0.76638}$

Since, the level of significance is not specified. The prior level of significance $\alpha =0.05$ can be used.

P-value:

Software procedure:

• Choose Graph > Probability Distribution Plot choose View Probability > OK.
• From Distribution, choose F, enter 1 in numerator df and 24 in denominator df.
• Click the Shaded Area tab.
• Choose X-Value and Right Tail for the region of the curve to shade.
• Enter the X-value as 0.76638.
• Click OK.

Output obtained from MINITAB is given below:

From the output, the P- value is 0.39.

Thus, the P- value is 0.39.

Decision criteria based on P-value approach:

If $P\text{-value}\le \alpha$, then reject the null hypothesis $H_0$.

If $P\text{-value}>\alpha$, then fail to reject the null hypothesis $H_0$.

Conclusion:

The P-value is 0.39 and $\alpha$ value is 0.05.

Here, P-value is greater than the $\alpha$ value.

That is $0.39\left(=P\right)>0.05\left(=\alpha \right)$.

By the rejection rule, fail to reject the null hypothesis.

Hence, there is sufficient evidence to conclude that the dropped predictor variable is not significant to predict the response variable y.

Thus, the reduced model is useful than the full model to predict the response variable y.

To determine

Plot the residuals versus fitted line plot for the reduced model.

Check whether the model is appropriate.

Answer to Problem 5SE

Residual plot:

Yes, the model seems to be appropriate.

Explanation of Solution

Calculation:

Residual plot:

Software procedure:

Step by step procedure to obtain regression using MINITAB software is given as,

• Choose Stat > Regression > General Regression.
• In Response, enter the numeric column containing the response data Y.
• In Model, enter the numeric column containing the predictor variables X1, X2, X1*X1 and X2*X2.
• In Graphs, Under Residuals for plots, select Regular.
• Under Residual plots select box Residuals versus fits.
• Click OK.

Conditions for the appropriateness of regression model using the residual plot:

• The plot of the residuals vs. fitted values should fall roughly in a horizontal band contended and symmetric about x-axis. That is, the residuals of the data should not represent any bend.
• The plot of residuals should not contain any outliers.
• The residuals have to be scattered randomly around “0” with constant variability among for all the residuals. That is, the spread should be consistent.

Interpretation:

In residual plot there is high bend or pattern, which can violate the straight line condition and there is change in the spread of the residuals from one part to another part of the plot.

However, it is difficult to determine about the violation of the assumptions without the data.

Thus, the model seems to be appropriate.

To determine

Check whether the model with only two dependent variables $x_2,x_2^2$ is adequate.

Answer to Problem 5SE

No, the model with only two dependent variables $x_2,x_2^2$ is not adequate.

Explanation of Solution

Calculation:

Regression:

Software procedure:

Step by step procedure to obtain regression using MINITAB software is given as,

• Choose Stat > Regression > General Regression.
• In Response, enter the numeric column containing the response data Y.
• In Model, enter the numeric column containing the predictor variables X2 and X2*X2.
• Click OK.

Output obtained from MINITAB is given below:

The ‘Coefficient’ column of the regression analysis MINITAB output gives the slopes corresponding to the respective variables stored in the column ‘Term’.

A careful inspection of the output shows that the fitted model is:

$\widehat{y}=9.960-0.1325x_2+0.001674x_2^2$.

Hence, the multiple linear regression model for the given data is:

$\underset{\_}{\widehat{y}=9.960-0.1325x_2+0.001674x_2^2}$.

The full model is,

$\widehat{y}\text{=}10.840-0.0739x_1-0.1274x_2+0.001110x_1^2-0.000243x_1{x}_2+0.001674x_2^2$

The reduced model is,

$\underset{\_}{\widehat{y}=9.960-0.1325x_2+0.001674x_2^2}$

The test hypotheses are given below:

Null hypothesis:

$H_0:{\beta }_1={\beta }_3={\beta }_4$

That is, the dropped predictors of the full model are not significant to predict y.

Alternative hypothesis:

$H_1:$ At least one ${\beta }^{\prime }s\ne 0$

That is, at least one of the dropped predictors of the full model are significant to predict y.

Test statistic:

$F=\frac{\frac{\left(SS{E}_{\text{Reduced}}-SS{E}_{\text{Full}}\right)}{p-k}}{\frac{SS{E}_{\text{Full}}}{\left[n-\left(p+1\right)\right]}}$

Where,

$SS{E}_{\text{Full}}$ represents the sum of squares due to error obtained from the full model.

$SS{E}_{\text{Reduced}}$ represents the sum of squares due to error obtained from the reduced model.

n represents the total number of observations.

p represents the number of predictors on the full model.

k represents the number of predictors on the reduced model.

From the obtained MINITAB outputs, the value of error sum of squares for full model is ${\text{SSE}}_{\text{Full}}=2.2642$ and the value of error sum of squares for the reduced model is ${\text{SSE}}_{\text{Reduced}}=2.7307$.

The total number of observations is $n=30$.

Number of predictors on the full model is $p=5$ and the number of predictors on the reduced model is $k=2$.

Degrees of freedom of F-statistic for reduced model:

In a reduced multiple linear regression analysis, the F-statistic is $F=\frac{M\left(SS{E}_{\text{Reduced}}-SS{E}_{\text{Full}}\right)}{MSS{E}_{\text{Full}}}$.

In the ratio, the numerator is obtained by dividing the quantity $SS{E}_{\text{Reduced}}-SS{E}_{\text{Full}}$ by its degrees of freedom, $p-k$. The denominator is obtained by dividing the error sum of squares of full model by the error degrees of freedom, $n-\left(p+1\right)$.

Thus, the degrees of freedom for the F-statistic in a reduced multiple regression analysis are $p-k$ and $n-\left(p+1\right)$.

Hence, the numerator degrees of freedom is $p-k=5-2=3$ and the denominator degrees of freedom is $n-\left(p+1\right)=30-6=24$.

Test statistic under null hypothesis:

Under the null hypothesis, the test statistic is obtained as follows:

$\begin{array}{c}F=\frac{\frac{\left(SS{E}_{\text{Reduced}}-SS{E}_{\text{Full}}\right)}{p-k}}{\frac{SS{E}_{\text{Full}}}{\left[n-\left(p+1\right)\right]}}\\ =\frac{\frac{\left(7.840-2.6462\right)}{5-2}}{\frac{2.6462}{\left[30-\left(5+1\right)\right]}}\\ =\frac{1.73127}{0.11026}\\ =15.702\end{array}$

Thus, the test statistic is $\underset{\_}{F=15.702}$.

Since, the level of significance is not specified. The prior level of significance $\alpha =0.05$ can be used.

P-value:

Software procedure:

• Choose Graph > Probability Distribution Plot choose View Probability > OK.
• From Distribution, choose F, enter 3 in numerator df and 24 in denominator df.
• Click the Shaded Area tab.
• Choose X-Value and Right Tail for the region of the curve to shade.
• Enter the X-value as 15.702.
• Click OK.

Output obtained from MINITAB is given below:

From the output, the P- value is $7.3\times {10}^{-6}$.

Thus, the P- value is $\underset{\_}{7.3\times {10}^{-6}}$.

Decision criteria based on P-value approach:

If $P\text{-value}\le \alpha$, then reject the null hypothesis $H_0$.

If $P\text{-value}>\alpha$, then fail to reject the null hypothesis $H_0$.

Conclusion:

The P-value is $\underset{\_}{7.3\times {10}^{-6}}$ and $\alpha$ value is 0.05.

Here, P-value is less than the $\alpha$ value.

That is $\underset{\_}{7.3\times {10}^{-6}}\left(=P\right)<0.05\left(=\alpha \right)$.

By the rejection rule, reject the null hypothesis.

Hence, there is sufficient evidence to conclude that at least one of the dropped predictors of the full model are significant to predict y.

Thus, the model with only two dependent variables $x_2,x_2^2$ is not adequate.

To determine

Find the two models with the highest $R^2$ value.

Obtain the values of mallows $C_p$ and adjusted $R^2$ for each model.

Answer to Problem 5SE

The two models with the highest $R^2$ are:

First model with $X_1,X_2,X_1^2,X_2^2$ predictors and the second model with $X_1,X_2,X_2^2$ predictors.

The values of M Mallows’ $C_p$ and adjusted $R^2$ for the various subsets are as follows:

Predictor variables	Mallows’ $C_p$	Adjusted $R^2$
$X_2$	92.5	60.1
$X_1{X}_2$	97	58.6
$X_2,X_2^2$	47.1	75.2
$X_1,X_2$	53.3	73
$X_1,X_2,X_2^2$	7.9	89.2
$X_2,X_1{X}_2,X_2^2$	15.5	86.4
$X_1,X_2,X_1^2,X_2^2$	4.8	90.7
$X_1,X_2,X_1{X}_2,X_2^2$	9.2	89
$X_1,X_2,X_1{X}_2,X_1^2,X_2^2$	6	90.6

Explanation of Solution

Calculation:

Coefficient of multiple determination $R^2$:

The coefficient of multiple determination, $R^2$, is given by:

$R^2=1-\frac{SSE}{SST}$, where SST and SSE are the total sum of squares and error sum of squares respectively.

The subset with larger $R^2$ is considered to be best subset for prediction.

Regression:

Software procedure:

Step by step procedure to obtain regression using MINITAB software is given as,

• Choose Stat > Regression > Regression> Best subsets.
• In Response, enter the numeric column containing the response data Y.
• In Model, enter the numeric column containing the predictor variables X1, X2, X1*X2, X1*X1 and X2*X2.
• Click OK.

Output obtained from MINITAB is given below:

For the one predictor case, the highest value of $R^2$ is 61.5, corresponding to $X_2$.

For the two predictor case, the highest value of $R^2$ is 76.9, corresponding to $X_2,X_2^2$.

For the three predictor case, the highest value of $R^2$ is 90.3, corresponding to $X_1,X_2,X_2^2$.

For the four predictor case, the highest value of $R^2$ is 92.0, corresponding to $X_1,X_2,X_1^2,X_2^2$.

For the five predictor case, the value of $R^2$ is 90.6.

The value of $R^2$ is the highest for predictors $X_1,X_2,X_1^2,X_2^2$. However, the subset with highest value of $R^2$ is considered to be best subset for prediction.

Thus, depending upon the factors affecting the analysis it would be most preferable to use the regression equation corresponding to the predictors $X_1,X_2,X_1^2,X_2^2$.

The second highest value of $R^2$ is 90.6 for the five predictor case and there is not much difference in the value of $R^2$ for the full model and the model with $X_1,X_2,X_2^2$ predictors.

That is, 90.6 and 90.3 are not much distinct.

Therefore, the model with $X_1,X_2,X_2^2$ predictors is the second best model.

Thus, the two best models are:

First model with $X_1,X_2,X_1^2,X_2^2$ predictors and the second model with $X_1,X_2,X_2^2$ predictors.

From the accompanying MINITAB output, the values of Mallows’ $C_p$ and adjusted $R^2$ for the various subsets are as follows:

Predictor variables	Mallows’ $C_p$	Adjusted $R^2$
$X_2$	92.5	60.1
$X_1{X}_2$	97	58.6
$X_2,X_2^2$	47.1	75.2
$X_1,X_2$	53.3	73
$X_1,X_2,X_2^2$	7.9	89.2
$X_2,X_1{X}_2,X_2^2$	15.5	86.4
$X_1,X_2,X_1^2,X_2^2$	4.8	90.7
$X_1,X_2,X_1{X}_2,X_2^2$	9.2	89
$X_1,X_2,X_1{X}_2,X_1^2,X_2^2$	6	90.6

To determine

Select the variables for the model, using the Mallows’ $C_p$ criterion and adjusted-$R^2$ criterion.

Check whether both the models are same.

Answer to Problem 5SE

The variables for the model using the Mallows’ $C_p$ criterion are $X_1,X_3\text{and}{X}_4$.

The variables for the model using the adjusted-$R^2$ criterion is $X_1,X_2,X_3,X_4$.

Yes, both the models are same.

Explanation of Solution

Mallows’ $C_p$:

An important utility of the Mallows’ $C_p$ criterion is to compare between regression equations of subsets having different sizes, all taken from the same all-subsets regression.

Mallows’ $C_p$ criterion is given as:

$C_p=\frac{SS{E}_{subset}}{MS{E}_{all}}-\left(n-2p\right)$, where $SS{E}_{subset}$ denotes the error sum of squares of the current model and $MS{E}_{all}$ denotes the error mean square for the set of all potential predictors, n is the sample size and $p=k+1$, with k being the number of predictors.

The predictor with the lowest value of $C_p$ or the value of $C_p$ closest to p is chosen to predict the response variable.

From part (e), the values of Mallows’ $C_p$ and adjusted $R^2$ for the various subsets are as follows:

Predictor variables	Mallows’ $C_p$	Adjusted $R^2$
$X_2$	92.5	60.1
$X_1{X}_2$	97	58.6
$X_2,X_2^2$	47.1	75.2
$X_1,X_2$	53.3	73
$X_1,X_2,X_2^2$	7.9	89.2
$X_2,X_1{X}_2,X_2^2$	15.5	86.4
$X_1,X_2,X_1^2,X_2^2$	4.8	90.7
$X_1,X_2,X_1{X}_2,X_2^2$	9.2	89
$X_1,X_2,X_1{X}_2,X_1^2,X_2^2$	6	90.6

For the one predictor case, the lowest value of $C_p$ is 92.5, corresponding to $X_2$.

For the two predictor case, the lowest value of $C_p$ is 47.1, corresponding to $X_2,X_2^2$.

For the three predictor case, the lowest value of $C_p$ is 7.9, corresponding to $X_1,X_2,X_2^2$.

For the four predictor case, the lowest value of $C_p$ is 4.8, corresponding to $X_1,X_2,X_1^2,X_2^2$.

For the five predictor case, the value of $C_p$ is 6.0.

The value of $C_p$ is the lowest for predictors $X_1,X_2,X_1^2,X_2^2$. However, the subset with lowest value of $C_p$ is considered to be best subset for prediction.

Thus, depending upon the factors affecting the analysis it would be most preferable to use the regression equation corresponding to the predictors $X_1,X_2,X_1^2,X_2^2$.

Hence, the variables for the model using the Mallows’ $C_p$ criterion are $X_1,X_2,X_1^2,X_2^2$.

Adjusted $R^2$ or $R_a^2$:

An important utility of the adjusted coefficient of multiple determination or $R_a^2$ is to find the best subset of the predictors, that can predict the response variable. The best subset may be a smaller subset of all the predictors and need not necessarily be a larger subset, as long as it predicts the response variable accurately. The subset with larger $R_a^2$ is considered to be best subset for prediction.

The adjusted coefficient of multiple determination, $R_a^2$, is given by:

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

For the one predictor case, the highest value of $R_a^2$ is 60.1, corresponding to $X_2$.

For the two predictor case, the highest value of $R_a^2$ is 75.2, corresponding to $X_2,X_2^2$.

For the three predictor case, the highest value of $R_a^2$ is 89.2, corresponding to $X_1,X_2,X_2^2$.

For the four predictor case, the highest value of $R_a^2$ is 90.7, corresponding to $X_1,X_2,X_1^2,X_2^2$.

For the five predictor case, the value of $R^2$ is 90.6.

The value of adjusted $R^2$ is the highest for predictors $X_1,X_2,X_1^2,X_2^2$. However, the subset with highest value of adjusted $R^2$ is considered to be best subset for prediction.

Thus, provided other factors do not affect the analysis it could be most preferable to use the regression equation corresponding to the predictors, $X_1,X_2,X_1^2,X_2^2$.

Hence, the variables for the model using the adjusted-$R^2$ criterion is $X_1,X_2,X_1^2,X_2^2$.

Both Mallows’ $C_p$ and adjusted-$R^2$ suggests that the best model contains the predictor variables $X_1,X_2,X_1^2,X_2^2$.