In: Statistics and Probability
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 | |||||||
$ | 85,613 | $ | 1,003.10 | 1 | $ | 28,525 | ||||
113,255 | 929.31 | 36 | 40,575 | |||||||
110,315 | 749.28 | 7 | 45,250 | |||||||
91,935 | 726.17 | 8 | 16,600 | |||||||
97,600 | 831.85 | 21 | 40,700 | |||||||
104,400 | 983.27 | 18 | 63,100 | |||||||
113,800 | 1,075.54 | 20 | 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 | |||||||
Click here for the Excel Data File
Determine the regression equation. (Round your answers to 3 decimal places. Negative amounts should be indicated by a minus sign.)
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.)
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.)
Do a test of the null hypothesis to check if the interaction is significant. (Use the 0.05 significance level.)
we will solve it by using excel and the steps are
Enter the Data into excel
take extraa column as interaction between Loan Amound and Payment made.
Click on Data tab
Click on Data Analysis
Select Regression
Select input Y Range as Range of dependent variable.
Select Input X Range as Range of independent variable
click on labels if your selecting data with labels
click on ok.
So this is the output of Regression in Excel.
SUMMARY OUTPUT | ||||||||
Regression Statistics | ||||||||
Multiple R | 0.851 | |||||||
R Square | 0.724 | |||||||
Adjusted R Square | 0.650 | |||||||
Standard Error | 10094.198 | |||||||
Observations | 20.000 | |||||||
ANOVA | ||||||||
df | SS | MS | F | Significance F | ||||
Regression | 4.000 | 4001985190.187 | 1000496297.547 | 9.819 | 0.000 | |||
Residual | 15.000 | 1528392434.813 | 101892828.988 | |||||
Total | 19.000 | 5530377625.000 | ||||||
Coefficients | Standard Error | t Stat | P-value | Lower 95% | Upper 95% | Lower 95.0% | Upper 95.0% | |
Intercept | -142843.143 | 46373.230 | -3.080 | 0.008 | -241685.342 | -44000.943 | -241685.342 | -44000.943 |
Loan | 1.547 | 0.417 | 3.712 | 0.002 | 0.659 | 2.435 | 0.659 | 2.435 |
Monthly Payments | 36.446 | 21.153 | 1.723 | 0.105 | -8.639 | 81.532 | -8.639 | 81.532 |
Payments Made | 1353.611 | 1447.674 | 0.935 | 0.365 | -1732.033 | 4439.254 | -1732.033 | 4439.254 |
Loan*Payments Made | -0.013 | 0.014 | -0.932 | 0.366 | -0.042 | 0.017 | -0.042 | 0.017 |
Determine the regression equation
auction price = -142843.143+1.547*Loan +36.446*Monthly Payments+1353.611*Payments Made - 0.013*Loan*Payments Made
Complete the following table.
Coefficients | Standard Error | t Stat | P-value | Lower 95% | Upper 95% | |
Intercept | -142843.143 | 46373.230 | -3.080 | 0.008 | -241685.342 | -44000.943 |
Loan | 1.547 | 0.417 | 3.712 | 0.002 | 0.659 | 2.435 |
Monthly Payments | 36.446 | 21.153 | 1.723 | 0.105 | -8.639 | 81.532 |
Payments Made | 1353.611 | 1447.674 | 0.935 | 0.365 | -1732.033 | 4439.254 |
Loan*Payments Made | -0.013 | 0.014 | -0.932 | 0.366 | -0.042 | 0.017 |
Compute the t-value corresponding to the interaction term.
t-value= - 0.932
Do a test of the null hypothesis to check if the interaction is significant.
Since the p-value for interaction term is 0.366 > 0.05 so we reject the null hypothesis and conclude that interaction is not significant.