In: Finance
MRE2. The file Credit Approval Decisions Coded provides information on credit history for a sample of banking customers. (18 pts)
a. Use regression analysis to identify the best model to predict credit score based on the other variables.
b. Report and interpret R2 and Standard Error (SE) for the model developed in a.
c. Interpret the regression coefficient(s) for the model developed in a.
DATA:
Coded Credit Approval Decisions | |||||
Homeowner | Credit Score | Years of Credit History | Revolving Balance | Revolving Utilization | Decision |
1 | 725 | 20 | $ 11,320 | 25% | 1 |
1 | 573 | 9 | $ 7,200 | 70% | 0 |
1 | 677 | 11 | $ 20,000 | 55% | 1 |
0 | 625 | 15 | $ 12,800 | 65% | 0 |
0 | 527 | 12 | $ 5,700 | 75% | 0 |
1 | 795 | 22 | $ 9,000 | 12% | 1 |
0 | 733 | 7 | $ 35,200 | 20% | 1 |
0 | 620 | 5 | $ 22,800 | 62% | 0 |
1 | 591 | 17 | $ 16,500 | 50% | 0 |
1 | 660 | 24 | $ 9,200 | 35% | 1 |
1 | 700 | 19 | $ 22,000 | 18% | 1 |
1 | 500 | 16 | $ 12,500 | 83% | 0 |
1 | 565 | 6 | $ 7,700 | 70% | 0 |
0 | 620 | 3 | $ 37,400 | 87% | 0 |
1 | 774 | 13 | $ 6,100 | 7% | 1 |
1 | 802 | 10 | $ 10,500 | 5% | 1 |
0 | 640 | 7 | $ 17,300 | 59% | 0 |
0 | 523 | 14 | $ 27,000 | 79% | 0 |
1 | 811 | 20 | $ 13,400 | 3% | 1 |
0 | 763 | 2 | $ 11,200 | 70% | 0 |
0 | 555 | 4 | $ 2,500 | 100% | 0 |
0 | 617 | 9 | $ 8,400 | 34% | 0 |
1 | 642 | 13 | $ 16,000 | 25% | 1 |
0 | 688 | 3 | $ 3,300 | 11% | 1 |
1 | 649 | 12 | $ 7,500 | 5% | 1 |
1 | 695 | 15 | $ 20,300 | 22% | 1 |
1 | 701 | 9 | $ 11,700 | 15% | 1 |
0 | 635 | 7 | $ 29,100 | 85% | 0 |
0 | 507 | 2 | $ 2,000 | 100% | 0 |
1 | 677 | 12 | $ 7,600 | 9% | 1 |
0 | 485 | 5 | $ 1,000 | 80% | 0 |
0 | 582 | 3 | $ 8,500 | 65% | 0 |
1 | 699 | 17 | $ 12,800 | 27% | 1 |
1 | 703 | 22 | $ 10,000 | 20% | 1 |
0 | 585 | 18 | $ 31,000 | 78% | 0 |
1 | 620 | 8 | $ 16,200 | 55% | 0 |
1 | 695 | 16 | $ 9,700 | 11% | 1 |
1 | 774 | 13 | $ 6,100 | 7% | 1 |
1 | 802 | 10 | $ 10,500 | 5% | 1 |
0 | 640 | 7 | $ 17,300 | 59% | 0 |
0 | 536 | 14 | $ 27,000 | 79% | 0 |
1 | 801 | 20 | $ 13,400 | 3% | 1 |
0 | 760 | 2 | $ 11,200 | 70% | 0 |
0 | 567 | 4 | $ 2,200 | 95% | 0 |
0 | 600 | 10 | $ 12,050 | 81% | 0 |
1 | 702 | 11 | $ 11,700 | 15% | 1 |
1 | 636 | 8 | $ 29,100 | 85% | 0 |
0 | 509 | 3 | $ 2,000 | 100% | 0 |
0 | 595 | 18 | $ 29,000 | 78% | 0 |
1 | 733 | 15 | $ 13,000 | 24% | 1 |
A-
Regression Analysis - Best Model (Y - Credit score X - Revolving Utilization )
SUMMARY OUTPUT | |||||
Regression Statistics | |||||
Multiple R | 0.796420254 | ||||
R Square | 0.63428522 | ||||
Adjusted R Square | 0.626666162 | ||||
Standard Error | 55.00605103 | ||||
Observations | 50 | ||||
ANOVA | |||||
df | SS | MS | F | Significance F | |
Regression | 1 | 251886.1288 | 251886.1288 | 83.2498 | 0.0000 |
Residual | 48 | 145231.9512 | 3025.6656 | ||
Total | 49 | 397118.0800 | |||
Coefficients | Standard Error | t Stat | P-value | ||
Intercept | 757.9223 | 13.9489 | 54.3357 | 0.0000 | |
Revolving Utilization | -220.7319 | 24.1921 | -9.1241 | 0.0000 |
So Model Will be, Credit Score = 757.9223 - 220.7319 * Revolving Utilization
B - R Square - 0.63428522 or 63.428522%
Standard Error - 55.00605103
Interpretation
R square - 63.428533% change in Credit Score is explained by revolving utilization
Standard Error - The Standard error of 55.00605103 indicates the variability of predicted values from actual ones.
C.
Anova / F stat
Null Hypothesis - Credit Score is not dependent on the Independent Variables (i.e. Homeowner, Revolving Utilization, Revolving Balance, Years of credit history and decision)
Alternate Hypothesis - Credit Score is dependent on all or any
of the Independent Variables (i.e. Homeowner, Revolving
Utilization, Revolving Balance, Years of credit history and
decision) |
If P value of F stat > 0.05 - We Cannot Reject the Null Hypothesis
If P value of F stat < 0.05 - We Can Reject the Null Hypothesis
Since the P value of F stat is 0.0000, which is less than 0.05, so can reject the null hypothesis, Meaning that we should accept the alternate hypothesis. So we can say that Credit Score is dependent on all or any of the Independent Variables (i.e. Homeowner, Revolving Utilization, Revolving Balance, Years of credit history and decision).
Model is Significant
T stat
Null Hypothesis - Credit Score is not dependent on revolving utilization
Alternate Hypothesis - Credit Score is dependent on revolving utilization. |
If P value of T stat > 0.05 - We Cannot Reject the Null Hypothesis
If P value of T stat < 0.05 - We Can Reject the Null Hypothesis
Since the P value of T stat is 0.0000, which is less than 0.05, so can reject the null hypothesis, Meaning that we should accept the alternate hypothesis. So we can say that Credit Score is dependent on revolving Utilization.
There is a 0% chance that the coefficient of revolving utilization is 0.
Revolving utilization is a significant variable for credit score