Statistical Techniques in Business and Economics
Statistical Techniques in Business and Economics
16th Edition
ISBN: 9780077639723
Author: Lind
Publisher: Mcgraw-Hill Course Content Delivery
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Chapter 18, Problem 15E

a.

To determine

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

a.

Expert Solution
Check Mark

Answer to Problem 15E

The plot for the ordered residuals is as follows:

Statistical Techniques in Business and Economics, Chapter 18, Problem 15E

Explanation of Solution

Residual:

Formula for residual is Residual=Actual valuePredicted value.

From Exercise 24163-14-9E, the regression equation for predicting the job performance is y^=29.28+5.22Aptitude+22.13Union.

Performance(y)y^Residual=yy^
5855.382.62
5350.162.84
3381.48−48.48
9781.4815.52
3639.72−3.72
8365.8217.18
6760.66.4
8476.267.74
9898.39−0.39
4561.85−16.85
9793.173.83
9082.737.27
9687.958.05
6667.07−1.07
8282.73−0.73

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.

b.

Expert Solution
Check Mark

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=t=2n(etet1)2t=1n(et)2

yy^et=yy^Lagged Residual, et1(etet1)2et2
5855.382.62--6.8644
5350.162.842.620.04848.0656
3381.4848.482.842633.742350.31
9781.4815.5248.484096240.87
3639.723.7215.52370.17813.8384
8365.8217.183.72436.81295.152
6760.66.417.18116.20840.96
8476.267.746.41.795659.9076
9898.390.397.7466.09690.1521
4561.8516.850.39270.932283.923
9793.173.8316.85427.66214.6689
9082.737.273.8311.833652.8529
9687.958.057.270.608464.8025
6667.071.078.0583.17441.1449
8282.730.731.070.11560.5329
    (etet1)2=8,515.21et2=3,434.05

The test statistic is as follows:

d=t=2n(etet1)2t=1n(et)2=8,515.213,434.05=2.48

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 dL is 0.95 and the value of dU is 1.54.

Rejection Rule:

  • If d<dL, then reject the null hypothesis.
  • If d>dU, then do not reject the null hypothesis.
  • If dL<d<dU, it leads to inconclusive results.

Conclusion:

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

That is, d(=2.48)>dU(=1.54).

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

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

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Statistical Techniques in Business and Economics

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