Chapter 14, Problem 23CE

Great Plains Distributors, Inc. sells roofing and siding products to home improvement retailers, such as Lowe’s and Home Depot, and commercial contractors. The owner is interested in studying the effects of several variables on the sales volume of fiber-cement siding products.

The company has 26 marketing districts across the United States. In each district, it collected information on the following variables: sales volume (in thousands of dollars), advertising dollars (in thousands), number of active accounts, number of competing brands, and a rating of market potential.

Conduct a multiple regression analysis to find the best predictors of sales.

1. a. Draw a scatter diagram comparing sales volume with each of the independent variables. Comment on the results.
2. b. Develop a correlation matrix. Do you see any problems? Does it appear there are any redundant independent variables?
3. c. Develop a regression equation. Conduct the global test. Can we conclude that some of the independent variables are useful in explaining the variation in the dependent variable?
4. d. Conduct a test of each of the independent variables. Are there any that should be dropped?
5. e. Refine the regression equation so the remaining variables are all significant.
6. f. Develop a histogram of the residuals and a normal probability plot. Are there any problems?
7. g. Determine the variance inflation factor for each of the independent variables. Are there any problems?

To determine

Make a scatter diagram that compares sales with each of the independent variables.

Explain the results.

Answer to Problem 23CE

The scatter diagram is obtained below,

Explanation of Solution

Step by step procedure to obtain the correlation matrix using MINITAB software is given below:

• Choose Graph > Matrix plot.
• Under Each Y versus each X, select Simple.
• Under Y variable select the column of Sales.
• Under X variables select the columns of Ad Dollars, Number of accounts, Number of Competitors, and Potential.
• Click OK.

The MINITAB output is obtained.

According to the output, as the number of competitors increases the sales decreases. In other hand, as the number of accounts and the rating of market potential increases, the sales also increase. However, there is no linear relationship between the Ad Dollars with the sales.

To determine

Make the correlation matrix.

Explain whether there is any problem due to redundant independent variables.

Answer to Problem 23CE

The correlation matrix is obtained as,

Explanation of Solution

Multiple linear regression model:

A multiple linear regression model is given as $\widehat{y}=a+b_1{x}_1+b_2{x}_2+b_3{x}_3+\mathrm{...}+b_k{x}_k$ where y is the response or dependent variable, and $x_1,x_2,\mathrm{...},x_k$ are the k quantitative independent variables where k is a positive integer.

Here, a is the intercept term of the regression model, that is, the value of predicted value of y when X’s are 0 and $b_i$’s are the slopes, that is, the amount of change of the predicted value of y for one unit increase in $x_i$ when all other independent variables are constant.

In the given problem the predicted dependent variable y is the sales. The years of advertising dollars, the number of accounts, the number of competitors and the rating of market potential, are defined as $x_1,x_2,x_3\text{and}{x}_4$, respectively.

Step by step procedure to obtain the correlation matrix using MINITAB software is given below:

• Choose Stat > Basic Statistics > Correlation.
• Select the columns of Sales, Ad Dollars, Number of accounts, Number of Competitors, and Potential under Variables tab.
• Click OK.

The MINITAB output is obtained.

Multicollinearity:

In a multiple regression model, when there is high correlation between two or more independent variables, then multicollinearity occurs.

There is moderate correlation between the independent variables “Potential” and “Number of Accounts”.

Thus, there is no chance of occurrence of multicollinearity in the regression model.

To determine

Find the regression equation and perform a global test.

Answer to Problem 23CE

The regression equation is $\underset{\_}{\widehat{y}=178.3+1.81x_1+3.318x_2-21.185x_3+0.325x_4}$.

Some of the independent variables are useful in explaining the variation in the dependent variable at 0.05 significance level.

Explanation of Solution

Calculation:

Step by step procedure to obtain the regression equation using MINITAB software:

• Choose Stat > Regression > Regression > Fit Regression Model.
• Under Responses, enter the column of Sales.
• Under Continuous predictors, enter the columns of Ad Dollars, Number of accounts, Number of Competitors, and Potential.
• Click OK.

Output using MINITAB software is given below:

Thus, the regression equation is $\underset{\_}{\widehat{y}=178.3+1.81x_1+3.318x_2-21.185x_3+0.325x_4}$.

Consider that y is dependent variable and $x_i\text{'}\text{s}$ are the independent variables where ${\beta }_i\text{'}\text{s}$ are the corresponding population regression coefficient for all $i=1,2,3,4$.

State the hypotheses:

Null hypothesis:

$H_0:{\beta }_1={\beta }_2={\beta }_3={\beta }_4=0$.

That is, the model is not significant.

Alternative hypothesis:

$H_1:\text{At least one }{\beta }_i\text{ is not equal to }0$.

That is, the model is significant.

In case of global test the F test statistic is defined as,

$F=\frac{\frac{\text{SSR}}{k}}{\frac{\text{SSE}}{n-k-1}}$, where SSR, SSE, n and k are the regression sum of square, error sum of square, sample size and the number of independent variables.

According to the output the value of F statistic is 479.1 with numerator degrees of freedom 4 and denominator degrees of freedom 21.

Consider that, the level of significance is $\alpha =0.05$.

Decision rule:

• If $p\text{-value}\le \alpha$, then reject the null hypothesis.
• Otherwise failed to reject the null hypothesis.

Conclusion:

Here, p-value corresponding to the global test is 0.

Hence, $p\text{-value}\left(=0\right)<\alpha \left(=0.05\right)$.

That is, the p-value is less than the level of significance.

Therefore, reject the null hypothesis.

Hence, it can be concluded that some of the independent variables are useful in explaining the variation in the dependent variable at 0.05 significance level.

To determine

Perform individual tests of each independent variables.

Explain whether any variable should be dropped.

Answer to Problem 23CE

There is significant relationship between y and $x_2\text{and}{x}_3$, whereas there is no significant relationship between y and $x_1\text{and}{x}_4$.

It is better to drop these two variables $x_1\text{and}{x}_4$ and perform the regression analysis using $x_2\text{and}{x}_3$ as independent variables.

Explanation of Solution

Calculation:

For independent variable $x_1$:

Consider that ${\beta }_1$ is the population regression coefficient of independent variable $x_1$.

State the hypotheses:

Null hypothesis:

$H_0:{\beta }_1=0$.

That is, there is no significant relationship between y and $x_1$.

Alternative hypothesis:

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

That is, there is significant relationship between y and $x_1$.

In case of individual regression coefficient test the t test statistic is defined as,

$t=\frac{b_i}{s_{b_i}}$, where $b_i$ and $s_{b_i}$ are the i^th regression coefficient and the standard deviation of the i^th regression coefficient.

According to the output in Part (c) the t statistic value corresponding to $x_1$ is 1.67 with 21 degrees of freedom.

Conclusion:

Here, p-value corresponding to the “Ad Dollars”$\left(x_1\right)$ is 0.109.

Hence, $p\text{-value}\left(=0.109\right)>\alpha \left(=0.05\right)$.

That is, the p-value is greater than the level of significance.

Therefore, fail to reject the null hypothesis.

Hence, it can be concluded that there is no significant relationship between y and $x_1$.

For independent variable $x_2$:

Consider that ${\beta }_2$ is the population regression coefficient of independent variable $x_2$.

State the hypotheses:

Null hypothesis:

$H_0:{\beta }_2=0$.

That is, there is no significant relationship between y and $x_2$.

Alternative hypothesis:

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

That is, there is significant relationship between y and $x_2$.

According to the output in Part (c) the value of t test statistic corresponding to $x_2$ is 20.37 with 21 degrees of freedom.

Conclusion:

Here, p-value corresponding to the “Number of accounts”$\left(x_2\right)$ is 0.

Hence, $p\text{-value}\left(=0\right)<\alpha \left(=0.05\right)$.

That is, the p-value is less than the level of significance.

Therefore, reject the null hypothesis.

Hence, it can be concluded that there is significant relationship between y and $x_2$.

For independent variable $x_3$:

Consider that ${\beta }_3$ is the population regression coefficient of independent variable $x_3$.

State the hypotheses:

Null hypothesis:

$H_0:{\beta }_3=0$.

That is, there is no significant relationship between y and $x_3$.

Alternative hypothesis:

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

That is, there is significant relationship between y and $x_3$.

According to the output in Part (c) the value of t test statistic corresponding to $x_3$ is –26.89 with 21 degrees of freedom.

Conclusion:

Here, p-value corresponding to the “Number of competitors”$\left(x_3\right)$ is 0.

Hence, $p\text{-value}\left(=0\right)<\alpha \left(=0.05\right)$.

That is, the p-value is less than the level of significance.

Therefore, reject the null hypothesis.

Hence, it can be concluded that there is significant relationship between y and $x_3$.

For independent variable $x_4$:

Consider that ${\beta }_4$ is the population regression coefficient of independent variable $x_4$.

State the hypotheses:

Null hypothesis:

$H_0:{\beta }_4=0$.

That is, there is no significant relationship between y and $x_4$.

Alternative hypothesis:

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

That is, there is significant relationship between y and $x_4$.

According to the output in Part (c) the value of t test statistic corresponding to $x_4$ is 0.69 with 21 degrees of freedom.

Conclusion:

Here, p-value corresponding to the “Potential”$\left(x_4\right)$ is 0.495.

Hence, $p\text{-value}\left(=0.495\right)>\alpha \left(=0.05\right)$.

That is, the p-value is greater than the level of significance.

Therefore, fail to reject the null hypothesis.

Hence, it can be concluded that there is no significant relationship between y and $x_4$.

Hence, it can be said that as there is no significant relationship between “Sales” and “Ad Dollars” and between “Sales” and “Potential”, thus it is better to omit these two variables and perform the regression analysis using “Number of accounts” and “Number of competitors” as independent variables.

To determine

Refine the regression equation so the remaining variables are all significant.

Answer to Problem 23CE

The refined regression equation is $\widehat{y}=186.7+3.408x_1-21.193x_2$.

Explanation of Solution

Calculation:

In this part the dependent variable is sales (y) and the independent variables are the number of accounts $\left(x_1\right)$ and the number of competitors $\left(x_2\right)$.

Step by step procedure to obtain the regression equation using MINITAB software:

• Choose Stat > Regression > Regression > Fit Regression Model.
• Under Responses, enter the column of Sales.
• Under Continuous predictors, enter the columns of Number of accounts, and Number of Competitors.
• Choose Graphs.
• Under Residual plot select Histogram of residuals, Normal probability plot of residuals.
• Click OK.
• Click OK.

Output using MINITAB software is given below:

Hence, the regression equation is $\widehat{y}=186.7+3.408x_1-21.193x_2$.

For independent variable $x_1$:

Consider that ${\beta }_1$ is the population regression coefficient of independent variable $x_1$.

State the hypotheses:

Null hypothesis:

$H_0:{\beta }_1=0$.

That is, there is no significant relationship between y and $x_1$.

Alternative hypothesis:

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

That is, there is significant relationship between y and $x_1$.

According to the output the t statistic value corresponding to $x_1$ is 23.37 with 23 degrees of freedom.

Conclusion:

Here, p-value corresponding to the “Number of Accounts”$\left(x_1\right)$ is 0.

Hence, $p\text{-value}\left(=0\right)<\alpha \left(=0.05\right)$.

That is, the p-value is less than the level of significance.

Therefore, reject the null hypothesis.

Hence, it can be concluded that there is significant relationship between y and $x_1$.

For independent variable $x_2$:

Consider that ${\beta }_2$ is the population regression coefficient of independent variable $x_2$.

State the hypotheses:

Null hypothesis:

$H_0:{\beta }_2=0$.

That is, there is no significant relationship between y and $x_2$.

Alternative hypothesis:

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

That is, there is significant relationship between y and $x_2$.

According to the output in the value of t test statistic corresponding to $x_2$ is –26.4 with 23 degrees of freedom.

Conclusion:

Here, p-value corresponding to the “Number of competitors”$\left(x_2\right)$ is 0.

Hence, $p\text{-value}\left(=0\right)<\alpha \left(=0.05\right)$.

That is, the p-value is less than the level of significance.

Therefore, reject the null hypothesis.

Hence, it can be concluded that there is significant relationship between y and $x_2$.

Hence, it can be said that as there is significant relationship between “Sales” and “Number of accounts” and “Number of competitors”.

To determine

Provide a histogram and normal probability plot for residuals.

Also explain whether there are any problems.

Explanation of Solution

Calculation:

Histogram:

From Part (e), the histogram and normal probability plot is obtained as,

Assumption of normality from histogram:

• The majority of the observation in the middle and centered on the mean of 0.
• There are lower frequencies on the tails of the distributions.

According to the given histogram, the most of the observations are centered on the mean of 0 and there are less frequencies on the tails of the distributions. In addition, it can be considered as perfect symmetric.

According to the normal probability plot, all of the residuals are linear.

Hence, there are no problems in normality assumptions.

To determine

Find the variation inflection factor for each of the independent variables.

Explain whether there is any problem.

Explanation of Solution

Variation inflation factor(VIF):

The variation inflation factor is defined as,

$VIF=\frac{1}{1-R_i^2}$ , where $R_i^2$ is the coefficient of determination of i^th independent variable.

It is used to detect multicollinearity in the regression. If $VIF<10$, then three is no presence of multicollinearity.

From Part (e), the variation inflection factor corresponding to both variables are 1.12, which is less than 10.

Hence, there is no presence of multicollinearity.