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

Write the equations for the following statistics, find or calculate them from the Minitab or Excel output, and explain the meanings of the statistics

1.Equations for R² and r (correlation coefficient), and perform the t test for the correlation coefficient between Loanamt and Income is zero.

2.Standard Error of b1 and variance of b1
3. Standard deviation and variance of e_t
4. Plot the residuals, and explain if you find any possible violations of assumptions on the regression model.

Answer 1

Loanamti = 29.3726 +1.5558 * Incomei + et

The hypothesis being tested is:

H0: ρ = 0

Ha: ρ ≠ 0

The p-value is 0.0000.

Since the p-value (0.0000) is less than the significance level (0.05), we can reject the null hypothesis.

Therefore, we can conclude that the correlation coefficient between Loanamt and Income is not zero.

The residual plot is:

All the assumptions are met.

The calculations are:

	r²	0.871
	r	0.933
	Std. Error	33.414
	n	30
	k	1
	Dep. Var.	Loanamt
ANOVA table
Source	SS	df	MS	F	p-value
Regression	2,10,816.8918	1	2,10,816.8918	188.83	5.72E-14
Residual	31,260.9749	28	1,116.4634
Total	2,42,077.8667	29
Regression output				confidence interval
variables	coefficients	std. error	t (df=28)	p-value	95% lower	95% upper
Intercept	29.3726
Income	1.5558	0.1132	13.741	5.72E-14	1.3239	1.7877
Observation	Loanamt	Predicted	Residual
1	109.0	127.4	-18.4
2	185.0	242.5	-57.5
3	121.0	111.8	9.2
4	125.0	150.7	-25.7
5	119.0	86.9	32.1
6	153.0	130.5	22.5
7	380.0	321.9	58.1
8	100.0	119.6	-19.6
9	110.0	150.7	-40.7
10	41.0	77.6	-36.6
11	115.0	113.4	1.6
12	248.0	211.4	36.6
13	126.0	122.7	3.3
14	260.0	328.1	-68.1
15	90.0	91.6	-1.6
16	50.0	85.4	-35.4
17	125.0	99.4	25.6
18	125.0	114.9	10.1
19	158.0	125.8	32.2
20	130.0	74.5	55.5
21	204.0	149.2	54.8
22	30.0	72.9	-42.9
23	114.0	122.7	-8.7
24	188.0	171.0	17.0
25	187.0	161.6	25.4
26	84.0	97.8	-13.8
27	450.0	441.7	8.3
28	108.0	105.6	2.4
29	100.0	111.8	-11.8
30	53.0	66.7	-13.7