Chapter 18, Problem 15E

a.

To determine

Provide the plot of residuals in the order in which the data are presented.

Answer to Problem 15E

The plot for the ordered residuals is as follows:

Explanation of Solution

Residual:

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

From Exercise 24163-14-9E, the regression equation for predicting the job performance is $\widehat{y}=29.28+5.22\text{Aptitude}+22.13\text{Union}$.

Performance(y)	$\widehat{y}$	$\text{Residual}=y-\widehat{y}$
58	55.38	2.62
53	50.16	2.84
33	81.48	−48.48
97	81.48	15.52
36	39.72	−3.72
83	65.82	17.18
67	60.6	6.4
84	76.26	7.74
98	98.39	−0.39
45	61.85	−16.85
97	93.17	3.83
90	82.73	7.27
96	87.95	8.05
66	67.07	−1.07
82	82.73	−0.73

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 0.05 significance level.

Answer to Problem 15E

There is no autocorrelation among the residuals at the 0.05 significance level.

Explanation of Solution

The null and alternative hypotheses are given as follows:

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, $e_{t-1}$	${\left(e_t-e_{t-1}\right)}^2$	$e_t^2$
58	55.38	2.62	-	-	6.8644
53	50.16	2.84	2.62	0.0484	8.0656
33	81.48	$-48.48$	2.84	2633.74	2350.31
97	81.48	15.52	$-48.48$	4096	240.87
36	39.72	$-3.72$	15.52	370.178	13.8384
83	65.82	17.18	$-3.72$	436.81	295.152
67	60.6	6.4	17.18	116.208	40.96
84	76.26	7.74	6.4	1.7956	59.9076
98	98.39	$-0.39$	7.74	66.0969	0.1521
45	61.85	$-16.85$	$-0.39$	270.932	283.923
97	93.17	3.83	$-16.85$	427.662	14.6689
90	82.73	7.27	3.83	11.8336	52.8529
96	87.95	8.05	7.27	0.6084	64.8025
66	67.07	$-1.07$	8.05	83.1744	1.1449
82	82.73	$-0.73$	$-1.07$	0.1156	0.5329
				${\displaystyle \sum {\left(e_t-e_{t-1}\right)}^2}=8,515.21$	${\displaystyle \sum e_t^2}=3,434.05$

The test statistic is as follows:

$\begin{array}{c}d=\frac{{\displaystyle \sum _{t=2}^n{\left(e_t-e_{t-1}\right)}^2}}{{\displaystyle \sum _{t=1}^n{\left(e_t\right)}^2}}\\ =\frac{8,515.21}{3,434.05}\\ =2.48\end{array}$

Thus, the Durbin–Watson statistic is 2.48.

Critical value:

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

The level of significance is 0.05 and the sample size is 15.

From Table Appendix B.9A: Critical values for the Durbin–Watson d Statistic (α=.05) for $k=2$ and $n=15$, The value of $d_L$ is 0.95 and the value of $d_U$ is 1.54.

Rejection Rule:

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

Conclusion:

The value of d is 2.48 that is greater than 1.54.

That is, $d\left(=2.48\right)>d_U\left(=1.54\right)$.

From the rejection rule, the null hypothesis is not rejected.

It can be concluded that there is no autocorrelation among the residuals.