Question

The manager of High Point Sofa and Chair, a large furniture manufacturer located in North Carolina, is studying the job performance ratings of a sample of 15 electrical repairmen employed by the company. An aptitude test is required by the human resources department to become an electrical repairman. The manager was able to get the score for each repairman in the sample. In addition, he determined which of the repairmen were union members (code = 1) and which were not (code = 0). The sample information is reported below.

Worker	Job Performance Score		Aptitude Test Score		Union Membership
Abbott	58		5		0
Anderson	53		4		0
Bender	33		10		0
Bush	97		10		0
Center	36		2		0
Coombs	83		7		0
Eckstine	67		6		0
Gloss	84		9		0
Herd	98		9		1
Householder	45		2		1
Iori	97		8		1
Lindstrom	90		6		1
Mason	96		7		1
Pierse	66		3		1
Rohde	82		6		1

1. Use a statistical software package to develop a multiple regression equation using the job performance score as the dependent variable and aptitude test score and union membership as independent variables. (Round your answers to 3 decimal places.)

Performance = ____ + _____ aptitude + _____ union

2. Complete the table. (Round your answers to 3 decimal places.)

	Predictor Coefficients Standard Error t p-value Constant Aptitude Union

3. Find the Standard error, R² and R²(adj).

	S R2 % R-Sq(adj) %

4. Complete the ANOVA table

	Analysis of Variance Source DF SS MS F p-value Constant 3 Aptitude 11 Union 14

5. How effective are these two variables in explaining the variation in job performance? (Round your answer to 1 decimal place.)

Adjusting for the number of independent variables, these variables explain _____ percent of the variation in performance

6. Compute the test statistic for union membership. (Round your answer to 3 decimal places.)

Value of test statistic _____

7. What is the effect of union membership on job performance?

On average, union membership increases the job performance score by _____

8. If you include an interaction, what is the regression equation? (Negative amounts should be indicated by a minus sign. Round your answers to 2 decimal places.)

Performance = _____ + ______ aptitude + ____ union + ____ x1 x2

9. Complete the following table. (Negative amounts should be indicated by a minus sign. Round your answers to 3 decimal places.)

	Predictor Coefficient SE Coefficient t p-value Constant    Aptitude Union X1X2

10. Compute the value of the test statistic for the interaction term.

Value of the test statistic ______

Answer 1

1)

SUMMARY OUTPUT
Regression Statistics
Multiple R	0.730065169
R Square	0.532995151
Adjusted R Square	0.45516101
Standard Error	16.91656565
Observations	15
ANOVA
	df	SS	MS	F	Significance F
Regression	2	3919.291	1959.645506	6.847832338	0.010373572
Residual	12	3434.0423	286.1701935
Total	14	7353.3333
	Coefficients	Standard Error	t Stat	P-value	Lower 95%
Intercept	29.27618014	12.766599	2.293185446	0.040696356	1.460149387
aprtitude score	5.222463375	1.7024835	3.067555902	0.009761027	1.513070435
union membership	22.13510581	8.8522127	2.50051671	0.02788895	2.847791175

y^ = 29.276 + 5.222 *aptitude +22.135*union

2)

	Coefficients	Standard Error	t Stat	P-value
Intercept	29.27618014	12.766599	2.293185446	0.040696356
aprtitude score	5.222463375	1.7024835	3.067555902	0.009761027
union membership	22.13510581	8.8522127	2.50051671	0.02788895

3)

S	16.916566
R2	53.3	%
R-Sq(adj)	45.52	%

4)

ANOVA
	df	SS	MS	F	Significance F
Regression	2	3919.291	1959.645506	6.847832338	0.010373572
Residual	12	3434.0423	286.1701935
Total	14	7353.3333

5)

45.52%

6)

6)
TS = 2.5005

7)
22.135

8)

SUMMARY OUTPUT
Regression Statistics
Multiple R	0.758129265
R Square	0.574759982
Adjusted R Square	0.458785432
Standard Error	16.86020489
Observations	15
ANOVA
	df	SS	MS	F	Significance F
Regression	3	4226.401736	1408.800579	4.95591473	0.020455502
Residual	11	3126.931598	284.2665089
Total	14	7353.333333
	Coefficients	Standard Error	t Stat	P-value	Lower 95%
Intercept	38.68893528	15.61766138	2.477255355	0.030723104	4.314694357
aprtitude score	3.801670146	2.178913977	1.744754582	0.108863499	-0.994087182
union membership	-0.100699988	23.14073331	-0.004351633	0.996605827	-51.03311059
x1*x2	3.61009456	3.473233124	1.039404621	0.320924371	-4.034440003

y^ = 38.6889 + 3.8017 * aptitude -0.1007*union + 3.6101 x1 x2

9)

	Coefficients	Standard Error	t Stat	P-value
Intercept	38.68893528	15.61766138	2.477255355	0.030723104
aprtitude score	3.801670146	2.178913977	1.744754582	0.108863499
union membership	-0.100699988	23.14073331	-0.004351633	0.996605827
x1*x2	3.61009456	3.473233124	1.039404621	0.320924371

10)

TS = 1.0394

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