Chapter 18, Problem 16E

a.

To determine

The given plot of residuals in the order in which the data are presented.

Answer to Problem 16E

The plot for the ordered residuals is as follows:

Explanation of Solution

Step-by-step procedure to obtain the regression using EXCEL:

• Enter the data for Commissions, Calls, and Driven in EXCEL sheet.
• Go to Data Menu.
• Click on Data Analysis.
• Select Regression and click on OK.
• Select the column of Commissions under Input Y Range.
• Select the column of Calls and Driven under Input X Range.
• Click on OK.

Output for the Regression obtained using Excel is as follows:

From the Excel output, the regression equation is $\widehat{y}=-101.32+0.63\text{Calls}+0.02\text{Driven}$. Note that the answer will be different based on round off the decimal places.

Residual:

Formula for residual is $\text{Residual}=\text{Actual value}-\text{Predicted value}$.

Commissions	$\widehat{y}$	$\text{Residual}=y-\widehat{y}$
22	33.67	−11.67
13	26.36	−13.36
33	44.02	−11.02
38	55.16	−17.16
23	33.92	−10.92
47	57.12	−10.12
29	47.9	−18.9
38	53.09	−15.09
41	46.24	−5.24
32	35.6	−3.6
20	26.15	−6.15
13	29.37	−16.37
47	59.92	−12.92
38	56.46	−18.46
44	51.46	−7.46
29	33.73	−4.73
38	45.22	−7.22
37	56.73	−19.73
14	26.95	−12.95
34	46.36	−12.36
25	35.64	−10.64
27	40.5	−13.5
25	32.11	−7.11
43	55.46	−12.46
34	47.87	−13.87

Step-by-step procedure to obtain the plot for Residuals using EXCEL:

• Enter the data for Residuals in Excel sheet.
• Select the column of Residuals.
• Go to Insert Menu.
• Select line chart.

Thus, the plot for the ordered residuals is obtained.

b.

To determine

Test the autocorrelation at the 0.01 significance level.

Answer to Problem 16E

There is a positive autocorrelation among the residuals at the 0.01 significance level.

Explanation of Solution

The null and alternative hypotheses are given below:

H₀: There is no autocorrelation among the residuals.

H₁: There is a positive residual autocorrelation.

Test Statistic:

The Durbin–Watson statistic for testing the hypothesis is as follows:

$d=\frac{{\displaystyle \sum _{t=2}^n{\left(e_t-e_{t-1}\right)}^2}}{{\displaystyle \sum _{t=1}^n{\left(e_t\right)}^2}}$

y	$\widehat{y}$	$e_t=y-\widehat{y}$	Lagged Residual, $e_{t-1}$	${\left(e_t-e_{t-1}\right)}^2$	$e_t^2$
22	33.67	–11.67			136.189
13	26.36	–13.36	–11.67	2.8561	178.49
33	44.02	–11.02	–13.36	5.4756	121.44
38	55.16	–17.16	–11.02	37.6996	294.466
23	33.92	–10.92	–17.16	38.9376	119.246
47	57.12	–10.12	–10.92	0.64	102.414
29	47.9	–18.9	–10.12	77.0884	357.21
38	53.09	–15.09	–18.9	14.5161	227.708
41	46.24	–5.24	–15.09	97.0225	27.4576
32	35.6	–3.6	–5.24	2.6896	12.96
20	26.15	–6.15	–3.6	6.5025	37.8225
13	29.37	–16.37	–6.15	104.448	267.977
47	59.92	–12.92	–16.37	11.9025	166.926
38	56.46	–18.46	–12.92	30.6916	340.772
44	51.46	–7.46	–18.46	121	55.6516
29	33.73	–4.73	–7.46	7.4529	22.3729
38	45.22	–7.22	–4.73	6.2001	52.1284
37	56.73	–19.73	–7.22	156.5	389.273
14	26.95	–12.95	–19.73	45.9684	167.703
34	46.36	–12.36	–12.95	0.3481	152.77
25	35.64	–10.64	–12.36	2.9584	113.21
27	40.5	–13.5	–10.64	8.1796	182.25
25	32.11	–7.11	–13.5	40.8321	50.5521
43	55.46	–12.46	–7.11	28.6225	155.252
34	47.87	–13.87	–12.46	1.9881	192.377
				${\displaystyle \sum {\left(e_t-e_{t-1}\right)}^2}=850.521$	${\displaystyle \sum e_t^2}=3,924.62$

The test statistic is as follows:

$\begin{array}{c}d=\frac{{\displaystyle \sum _{t=2}^n{\left(e_t-e_{t-1}\right)}^2}}{{\displaystyle \sum _{t=1}^n{\left(e_t\right)}^2}}\\ =\frac{850.521}{3,924.62}\\ =0.22\end{array}$

Thus, the Durbin–Watson statistic is 0.22.

Critical value:

From the given information table, there are two independent variables. That is, $k=2$.

The level of significance is 0.01 and the sample size is 25.

From Table Appendix B.9C: Critical values for the Durbin–Watson d Statistic (α=.01), for $k=2$ and $n=25$, The value of $d_L$ is 0.98 and the value of $d_U$ is 1.30.

Rejection Rule:

• If $d<d_L$, then reject the null hypothesis.
• If $d>d_U$, then do not reject the null hypothesis.
• If $d_L<d<d_U$, provide inconclusive results.

Conclusion:

The value of d is 0.22 that is less than 0.98.

That is, $d\left(=0.22\right)<d_L\left(=0.98\right)$.

From the rejection rule, reject the null hypothesis.

It can be concluded that there is a positive autocorrelation among the residuals.