Question

2. Develop a simple linear regression model to predict a person’s income (INCOME) based on their age (AGE) using a 95% level of confidence.

a. Write the regression equation.

2. Discuss the statistical significance of the model as whole using the appropriate regression statistic at a 95% level of confidence.
3. Discuss the statistical significance of the coefficient for the independent variable using the appropriate regression statistic at a 95% level of confidence.
4. Interpret the coefficient for the independent variable.

What percentage of the observed variation in a person’s income is explained by the model?

5. Predict the value of a person’s income who is 45 years old, using this regression model.

INCOME	AGE	EARNRS	EDUC	CHILDS	HRS1
500	27	3	12	0	56
500	23	3	12	1	10
500	78	0	16	2	0
500	64	0	17	0	0
500	54	1	14	3	0
500	22	2	13	1	0
500	21	1	12	2	24
500	53	2	14	4	0
500	57	1	16	0	4
500	22	2	12	0	0
500	21	2	14	0	0
500	23	2	12	0	57
500	45	2	14	0	9
500	21	1	14	0	0
500	18	2	12	0	35
500	38	3	15	3	20
500	43	2	12	1	50
500	71	0	12	2	12

Answer 1

a) Regression Line -

Income = 500 - 0.00000 * Age

Output from R -

lm(formula = Income ~ Age)

Coefficients:
(Intercept) Age
5.000e+02 -8.709e-16

b) Following is the model significance

F-value = 17.34

p-value = 0.0007 < 0.05

Hence, model is not significant at 95% confidence level.

Call:
lm(formula = Income ~ Age)

Residuals:
Min 1Q Median 3Q Max
-4.504e-14 -4.155e-14 -2.544e-14 -1.172e-14 4.451e-13

Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept) 5.000e+02 6.182e-14 8.088e+15 <2e-16 ***
Age -8.709e-16 1.424e-15 -6.110e-01 0.549
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

Residual standard error: 1.158e-13 on 16 degrees of freedom
Multiple R-squared: 0.5201, Adjusted R-squared: 0.4901
F-statistic: 17.34 on 1 and 16 DF, p-value: 0.0007319

Warning message:
In summary.lm(Lin_Mod) : essentially perfect fit: summary may be unreliable

c) siginificance of the independent variable

t-value for variable Age = -0.611

p-value = 0.549 > 0.05

Hence, independent variable Age is significant at 95% confidence level.

d) Interpretation of independent variables

average unit change in Age changes income by unit.

e) At Age = 45, income = 500 + 0.00000 * 45 = 500

At age 45, income = 500