Question

Data has been gathered to explain executive salaries from a variety of factors.

Perform a Multiple Regression to predict SALARY from EXPerience, EDUCation, GENDER, NUMberSupported and ASSETS.

d. Test for violation of the Model Assumptions. How did you do this?

e. Test for multicollinearity. How did you do this?

SALARY	EXP	EDUC	GENDER	NUMSUP	ASSETS
93300	12	15	1	240	170
130000	25	14	1	510	160
88200	20	14	0	370	170
74400	3	19	1	170	170
115300	19	12	1	520	150
70400	14	13	0	420	160
114200	18	18	1	290	170
72600	2	17	1	200	180
108600	14	13	1	560	180
68600	4	16	1	230	160
102000	8	18	1	540	150
101400	19	15	1	90	180
149400	23	16	1	560	180
57100	5	15	0	470	150
87400	3	16	1	340	190
131000	22	17	1	70	200
90300	24	14	0	160	180
115600	22	16	1	160	190
102800	13	18	1	110	180
141900	21	16	1	410	180
90900	10	13	1	370	190
73400	11	12	1	180	170
101000	12	19	1	60	200
85400	10	19	1	60	180
138300	26	17	1	110	200
82300	7	15	1	280	190
85500	7	19	1	110	180
75300	10	19	0	300	170
87500	23	14	0	220	170
127100	12	15	1	570	200
80100	6	16	1	240	180
90900	15	16	0	300	150
109600	15	18	1	260	170
70700	8	13	1	150	160
104400	18	19	0	350	160
71200	2	13	1	370	190
85400	13	14	1	150	160
89300	12	17	0	480	190
124800	21	15	1	310	180
42800	3	12	0	340	150
125000	20	16	1	520	160
122200	20	19	1	200	170
107100	20	17	0	490	160
61000	1	15	0	570	180
59800	2	17	1	70	160
95700	9	17	1	300	160
85600	11	17	0	190	160
88900	21	13	0	500	160
143000	20	20	1	390	170
109200	17	16	0	520	180
156700	24	12	1	530	200
65100	2	17	0	590	190
105900	9	13	1	560	170
74300	2	18	0	600	190
79300	13	12	0	390	170
106600	14	18	1	110	170
106400	18	13	1	190	190
77400	10	14	1	110	160
129400	21	13	1	430	190
82600	11	14	0	440	150
126100	26	15	1	210	190
121900	22	18	1	320	160
96200	3	16	1	560	180
128900	17	18	1	450	190
72200	2	16	1	410	180
58800	4	18	0	70	150
79300	8	17	1	90	190
96100	13	15	1	290	160
94900	3	18	1	530	180
89000	13	16	0	420	170
108800	25	19	0	150	200
95300	11	15	1	500	190
71200	2	17	0	430	190
173400	26	17	1	570	190
107000	20	20	1	90	150
100000	19	12	1	340	160
100700	12	13	1	440	170
152800	22	18	1	500	160
95300	13	13	0	570	180
77300	2	15	1	560	190
84600	15	14	1	160	170
92600	12	13	1	390	190
85900	13	19	0	370	200
79400	5	17	1	330	160
80100	8	17	0	560	170
114100	21	20	0	590	180
78500	5	16	1	290	200
87300	9	18	0	440	180
102900	19	15	0	480	190
116300	23	19	1	130	150
51500	3	12	0	440	190
106500	13	19	1	310	150
109000	22	17	0	370	200
66600	9	12	0	180	160
111100	7	19	1	520	200
83100	10	18	0	90	180
159500	25	18	1	590	160
122500	10	19	1	480	200
67300	3	19	1	80	160
97900	16	17	0	380	160

Answer 1

d. We need to check the following assumptions:

1. Homoscedasticity of residuals or equal variance

R code :

par(mfrow=c(2,2))  
mod_1 <- lm(SALARY ~ ., data=data)  
plot(mod_1)

Output :

From the Top left graph and bottom right graph , there is no certain pattern. Hence we can say that Homoscedasticity of residuals can be accepted.

2. No autocorrelation of residuals

This can be tested by using dwtest( Durbin Watson test) in R

R code:

library(lmtest)
dwtest(data[,1] ~ data[,2])
dwtest(data[,1] ~ data[,3])
dwtest(data[,1] ~ data[,4])
dwtest(data[,1] ~ data[,5])

Output :

dwtest(data[,1] ~ data[,2])

Durbin-Watson test

data: data[, 1] ~ data[, 2]

DW = 2.2839, p-value = 0.9244

alternative hypothesis: true autocorrelation is greater than 0

> dwtest(data[,1] ~ data[,3])

Durbin-Watson test

data: data[, 1] ~ data[, 3]

DW = 2.3343, p-value = 0.9555

alternative hypothesis: true autocorrelation is greater than 0

> dwtest(data[,1] ~ data[,4])

Durbin-Watson test

data: data[, 1] ~ data[, 4]

DW = 2.1778, p-value = 0.819

alternative hypothesis: true autocorrelation is greater than 0

> dwtest(data[,1] ~ data[,5])

Durbin-Watson test

data: data[, 1] ~ data[, 5]

DW = 2.3297, p-value = 0.9515

alternative hypothesis: true autocorrelation is greater than 0

Since, P-value ( in each case ) > 0.05 , we can say that Autocorrelation is not there in the data.

3. Normality of residuals

This can be visually checked using the qqnorm() plot.

R code:

mod <- lm(SALARY ~ ., data=data)
plot(mod)

Output:

The qqnorm() plot evaluates this assumption. If points lie exactly on the line, it is perfectly normal distribution. In out case the points lie on the line approximately , we can say that Residuals are Normally distributed.

e. Test for multicollinearity

Using VIF function in R we can check for multicollinearity.

VIF is a metric computed for every X variable that goes into a linear model. If the VIF of a variable is high, it means the information in that variable is already explained by other X variables present in the given model, which means, more redundant is that variable.So, lower the VIF (<2) the better.

R Code:

library(car)
mod2 <- lm(SALARY ~ ., data=data)
vif(mod2)

Output:

EXP EDUC GENDER NUMSUP ASSETS
1.002071 1.037777 1.063135 1.101590 1.029408

We can see that VIF value for each variable is less than 2 , we can say that No multicollinearity is present in the data.