Question

Let’s consinder a mortgage application using HMDA (The Home Mortgage Disclosure Act). Here is a sample from 30 mortgage applications.

ID	loanamt	income	hprice
1	109	63	155
2	185	137	264
3	121	53	128
4	125	78	125
5	119	37	149
6	153	65	171
7	380	188	484
8	100	58	125
9	110	78	158
10	41	31	116.5
11	115	54	128
12	248	117	280
13	126	60	157.5
14	260	192	325
15	90	40	145
16	50	36	230
17	125	45	125
18	125	55	145
19	158	62	175
20	130	29	209
21	204	77	260
22	30	28	150
23	114	60	143
24	188	91	253
25	187	85	285
26	84	44	105
27	450	265	650
28	108	49	120
29	100	53	125
30	53	24	66

loanamt = Amount of Mortgage Loan Application (in $1000)

income = Applicant’s Annual Income (in $1000)

hprice = House Price to buy (in $1000)

Regression Analyis

Let’s consider the following regression model. Estimate the model using Minitab and answer the questions using the output.

Loanamt_i = b₀ + b₁ * income_i + e_t

6) t test for the coefficient of income (Ho: B1 = 0 )

7) F statistics and perform the test for the model

8) Variance of e_t

9) According to the model what are the predicted loan amount if applicants have annual income of $50,000, $100,000, and $200,000 and their confidence intervals?

10) List and explain the assumptions you made for a simple regression model

Answer 1

As asked, using minitab'17 the analysis would be as followed:

Using the above model as; loanamt= b0+ b1 income

6)

In minitab, computing of the linear regression model also gives the t-value for both the constant as well as coefficient.

Here, hypothesis would be as followed:

H₀: b₀=0

H₁: b₀≠0

The analysis from minitab is :

t-value corresponding to income is 13.74 and p-value corresponding to it is <0.0001 , which means we would reject the null hypothesis.

Session Coefficients VIF Term Constant income Coef SE Coef T-Value 29.4 10.5 2.81 1.556 0.113 13.74 P-value 0.009 0.000 1.00 Regression Equation loanamt = 29.4 + 1.556 income Fits and Diagnostics for Unusual Observations Obs loanamt Fit Resid Std Resid 08 09 010 0 11 012 013 014 06 Variancel 1077.96 07 N1 30 109 8347. Worksheet 1 *** + 01 02 loanamt income 63 185 137 121 53 125 119 153 380 100 110 78 03 hprice 155.0 264.0 128.0 125.0 149.0 171.0 484.0 125.0 158.0 116.5 04 05 VARIANCES1 RESI1 8347.5 -18.3894 3003.2 -57.5202 14431.0 9.1689 -25.7267 32.0620 22.4990 58.1329 - 19.6102 -40.7267 -36.6031 41

7) By using ANOVA table, the computed F-statistic would be as followed:

Hypothesis would be as followed:

Method Null hypothesis All means are equal Alternative hypothesis At least one mean is different significance level a = 0.05 Equal variances were assumed for the analysis.

il Minitab - Untitled file Edit Data Calc Stat Graph Editor Tools Window Help Assistant #H% 11 03 SO GOODSOD GIF | -- A3 KG + VE || XQOTOO 11M Session Regression Analysis: loanamt versus income Analysis of Variance P-value 0.000 0.000 Source DF Adj ss Adj MS F-Value Regression 1 210817 210817 188.83 income 1 210817 210817 188.83 Error 28 31261 1116 Lack-of-Fit 25 30856 1234 Pure Error 3 405 Total 29 242078 0.046 Model Summary

As the p-value is too low, so we may reject the null hypothesis.

8) Variance of et refers to the variance of the residuals; residuals calculated using the minitab would be as followed:

Worksheet 1 *** C8 C3 hprice C6 Variance1 1077.96 C7 N1 30 137 78 37 C1 C2 loanamt income 109 63 185 121 125 119 153 65 380 188 100 58 110 41 115 248 117 126 60 260 192 90 264.0 128.0 125.0 149.0 171.0 484.0 125.0 158.0 116.5 128.0 280.0 157.5 325.0 145.0 230.0 125.0 145.0 175.0 209.0 260.0 150.0 143.0 253.0 285.0 10rn C4 C5 VARIANCES1 RESII 8347.5 -18.3894 3003.2 -57.5202 14431.0 9.1689 -25.7267 32.0620 22.4990 58.1329 -19.6102 -40.7267 -36.6031 1.6130 36.5963 3.2781 -68.0904 -1.6055 -35.3822 25.6154 10.0572 32.1665 55.5086 54.8291 -42.9356 -8.7219 17.0476 25.3826 190007 50 125 125 158 130 204 20 29 21 22 30 60 114 188 187

It's variance is 1077.96

9)

predicted loan amount would be calculated by using the following equation:

Model Summary S R -sq 33.4135 87.09% R-sq(adj) 86.63% R-sq (pred) 84.308 Coefficients VIF Term Constant income Coef SE Coef 29.4 10.5 1.556 0.113 T-Value 2.81 13.74 P-Value 0.009 0.000 1.00 Regression Equation Loanamt = 29.4 + 1.556 income Fits and Diagnostics for Unusual Observations Obs loanamt 14 260.0 27 450.0 Fit Resid Std Resid 328.1 -68.1 -2.26 441.7 8.3 0.34 R X Large residual Unusual x x

Put the values in the place of income you'll get your predicted amount from the equation as followed:

loanamt= 29.4+1.556*50000= 77829.4

loanamt= 29.4+1.556*100000= 155629.4

loanamt= 29.4+1.556*200000= 311229.4