Chapter 13, Problem 50CE

(a)

To determine

State the conclusion about the overall fit of the model referring F statistic.

State the conclusion about the overall fit of the model referring p-value.

Answer to Problem 50CE

The conclusion about the overall fit of the model referring F statistic is at least one of the coefficients is not zero and the overall regression is significant.

The conclusion about the overall fit of the model referring p-value is at least one of the coefficients is not zero and the overall regression is significant.

Explanation of Solution

Calculation:

The given information is that, a regression model for a sample of 43 vehicles is fitted for predicting the City MPG (miles per gallon in city driving) based on the predictors Length (length of car in inches), Width (width of car in inches), and Weight (weight of car in pounds).

From the reported regression results, the value of F statistic is 27.90, the degrees of freedom for regression are 3, degrees of freedom for residual are 39 and the p-value is 8.35E-10 which is approximately 0.0000. The considered level of significance is $\alpha =0.05$.

The formula for F statistic is,

$F_{\text{calc}}=\frac{MSR}{MSE}$

In the formula, MSR denotes the mean square regression and MSE denotes the mean square error.

Rejection rules based on F statistic:

• • If the test statistic value is greater than the critical value, then reject the null hypothesis. The regression is significant.
• • If the test statistic value is smaller than the critical value, then retain the null hypothesis. The regression is not significant.

Rejection rules based on p-value:

• • If p-value is less than the level of significance then the null hypothesis is rejected. The predictor is significant.
• • If p-value is greater than the level of significance then the null hypothesis is not rejected. The predictor is not significant.

Let ${\beta }_1$ denote the coefficient for predictor X₁ = Length (length of car in inches), let ${\beta }_2$ denote the coefficient for predictor X₂ = Width (width of car in inches), let ${\beta }_3$ denote the coefficient for predictor X₃ =Weight (weight of car in pounds).

Null hypothesis:

$H_0:$ All the coefficients are equal to zero $\left({\beta }_1={\beta }_2={\beta }_3=0\right)$.

Alternative hypothesis:

$H_1:$ At least one of the coefficients is not zero.

Critical value:

The considered significance level is $\alpha =0.05$.

The degrees of freedom for numerator are 3, the degrees of freedom for denominator are 39 from completed F table.

From the Appendix F: Critical values of $F_{.05}$:

• • Locate the value 3 in numerator degrees of freedom $\left(d{f}_1\right)$ row.
• • Locate the value 39 in denominator degrees of freedom $\left(d{f}_2\right)$ column. This value is not in the column. Consider the smaller value than 39. The value is 30.
• • The intersecting value that corresponds to the (3, 30) with level of significance 0.05 is 2.92.

Hence, the critical value for $df=\left(3,39\right)$ with 0.05, level of significance is 2.92.

Conclusion referring F statistic:

The value of test statistic is 27.90.

The critical value is 2.92.

The test statistic value is greater than the critical value.

The null hypothesis is rejected.

The decision is that at least one of the coefficients is not zero and the overall regression is significant.

Conclusion referring p-value:

The p-value for overall regression is 0.000.

The level of significance is 0.05.

The p-value is less than the level of significance.

That is, $p\text{-value}\left(=0.000\right)<\alpha \left(=0.05\right)$.

The null hypothesis is rejected.

The decision is that at least one of the coefficients is not zero and the overall regression is significant.

(b)

To determine

Identify whether there is evidence that some predictors were unhelpful.

Answer to Problem 50CE

There is evidence that the predictors length and width were unhelpful.

Explanation of Solution

Calculation:

From the reported results, the p-value for predictor length is 0.9725, the p-value for width is 0.7379 and p-value for predictor weight is 0.000. The level of significance considered is $\alpha =0.05$.

For length:

Let ${\beta }_1$ is the parameter for the predictor length.

Null hypothesis:

$H_0:{\beta }_1=0$

The predictor variable length is not related to City MPG.

Alternative hypothesis:

$H_1:{\beta }_1\ne 0$

The predictor variable length is related to City MPG.

Conclusion:

The p-value for predictor length is 0.9725.

The level of significance is 0.05.

The p-value is greater than the level of significance.

That is, $p\text{-value}\left(=0.9725\right)>\alpha \left(=0.05\right)$.

The null hypothesis is not rejected.

The predictor variable length is not related to City MPG.

The predictor length is not significant and is not helpful.

For width:

Let ${\beta }_2$ is the parameter for the predictor width.

Null hypothesis:

$H_0:{\beta }_2=0$

The predictor variable width is not related to City MPG.

Alternative hypothesis:

$H_1:{\beta }_2\ne 0$

The predictor variable width is related to City MPG.

Conclusion:

The p-value for predictor width is 0.7379.

The level of significance is 0.05.

The p-value is greater than the level of significance.

That is, $p\text{-value}\left(=0.7379\right)>\alpha \left(=0.05\right)$.

The null hypothesis is not rejected.

The predictor variable width is not related to City MPG.

The predictor width is not significant and is not helpful.

For weight:

Let ${\beta }_3$ is the parameter for the predictor weight.

Null hypothesis:

$H_0:{\beta }_3=0$

The predictor variable weight is not related to City MPG.

Alternative hypothesis:

$H_1:{\beta }_3\ne 0$

The predictor variable weight is related to City MPG.

Conclusion:

The p-value for predictor weight is 0.000.

The level of significance is 0.05.

The p-value is greater than the level of significance.

That is, $p\text{-value}\left(=0.000\right)<\alpha \left(=0.05\right)$.

The null hypothesis is rejected.

The predictor variable weight is related to City MPG.

The predictor weight is significant and is helpful.

(c)

To determine

Identify whether there is any suspect that the multicollinearity is a problem or not.

Answer to Problem 50CE

There is suspect that mild multicollinearity might be a problem in the regression.

Explanation of Solution

Calculation:

The given information is that, the value of VIF for length is 2.669, for width is 2.552 and for weight is 2.836.

Variance inflation factor (VIF):

The general test of multicollinearity is done using the variance inflation factor. The formula for VIF for the predictor $X_j$ is,

$VI{F}_j=\frac{1}{1-R_j^2}$

In the formula, $R_j^2$ is the coefficient of determination when predictor $X_j$ is regressed against all the remaining predictors without including Y.

Detecting multicollinearity using variance inflation factor:

• • If value of VIF is 1, there is no multicollinearity in the regression.
• • If value of VIF is 2, there is mild multicollinearity in the regression.
• • If value of VIF is 10, there is strong multicollinearity in the regression.
• • If value of VIF is 100, there is severe multicollinearity in the regression.

VIF for length:

The value of VIF for length is 2.669; this value is greater than 2 indicating that there is suspect of mild multicollinearity in the regression.

VIF for width:

The value of VIF for length is 2.552; this value is greater than 2 indicating that there is suspect of mild multicollinearity in the regression.

VIF for weight:

The value of VIF for length is 2.836; this value is greater than 2 indicating that there is suspect of mild multicollinearity in the regression.

Hence, there is suspect that mild multicollinearity might be a problem.