Question

You are a new hire at Laurel Woods Real Estate, which specializes in selling foreclosed homes via public auction. Your boss has asked you to use the following data (mortgage balance, monthly payments, payments made before default, and final auction price) on a random sample of recent sales in order to estimate what the actual auction price will be. Add a new variable that describes the potential interaction between the loan amount and the number of payments made.

Loan			Monthly Payments			Payments Made		Auction Price
$	85,602		$	1,010.12		1		$	71,125
	114,144			953.51		33			74,550
	110,851			738.28		6			47,125
	84,989			939.74		6			16,600
	97,600			853.79		21			40,700
	104,400			983.27		13			63,100
	113,800			1,075.54		21			72,600
	116,400			1,087.16		35			72,300
	100,000			900.01		33			58,100
	92,800			683.11		36			37,100
	105,200			915.24		34			52,600
	105,900			905.67		38			51,900
	94,700			810.70		25			43,200
	105,600			891.33		20			52,600
	104,100			864.38		7			42,700
	85,700			1,074.73		30			22,200
	113,600			871.61		24			77,000
	119,400			1,021.23		58			69,000
	90,600			836.46		3			35,600
	104,500			1,056.37		22			63,000

  Determine the regression equation. (Round your answers to 3 decimal places. Negative amounts should be indicated by a minus sign.)

Auction Price = + Loan + Monthly Payment + Payments Made + x1x3

Complete the following table. (Round your answers to 3 decimal places. Leave no cells blank - be certain to enter "0" wherever required. Negative amounts should be indicated by a minus sign.)

	Predictor Coefficient SE Coefficient t p-value Constant Loan Monthly Payment Payments Made (Loan)(Payments Made)

Compute the t-value corresponding to the interaction term. (Round your answer to 2 decimal places. Negative amount should be indicated by a minus sign.)

	t-value

Do a test of the null hypothesis to check if the interaction is significant. (Use the 0.05 significance level.)

	This is , so we conclude that there is

Answer 1

Excel output:

SUMMARY OUTPUT
Regression Statistics
Multiple R	0.780620406
R Square	0.609368218
Adjusted R Square	0.505199743
Standard Error	12319.36614
Observations	20
ANOVA
	df	SS	MS	F	Significance F
Regression	4	3551241518	887810379.5	5.849833323	0.004832061
Residual	15	2276501732	151766782.1
Total	19	5827743250
	Coefficients	Standard Error	t Stat	P-value	Lower 95%	Upper 95%
Intercept	-73059.23822	57177.94067	-1.277752178	0.220758282	-194931.1333	48812.65686
Loan	0.943686905	0.470276767	2.006662825	0.063160439	-0.058684291	1.946058101
Monthly Payments	34.95690627	27.65739773	1.263926079	0.225546015	-23.99344129	93.90725383
Payment made	-1320.788645	1745.558697	-0.756656678	0.460975682	-5041.358919	2399.781628
X1*X3	0.011387324	0.016667207	0.683217299	0.504886433	-0.024137987	0.046912635

a) Regression equation:

b)

	Coefficients	Standard Error	t Stat	P-value
Intercept	-73059.23822	57177.94067	-1.277752178	0.220758282
Loan	0.943686905	0.470276767	2.006662825	0.063160439
Monthly Payments	34.95690627	27.65739773	1.263926079	0.225546015
Payment made	-1320.788645	1745.558697	-0.756656678	0.460975682
X1*X3	0.011387324	0.016667207	0.683217299	0.504886433

c) Computed t-value:

d) P-value:0.504886433

The test statistic is not significant and fail to reject H0.

This is not significant and so we conlude that there is no relationship between interaction variable and output variable.