Run full regression diagnostics on the second regression model. These include diagnostics for functional form, heteroscedasticity, normality, multicollinearity, and influential data points. Attempt to make corrections for any violations.
Using the 2016 ANES, replicate the analysis with feelings towards Trump using feelings towards Hillary
Clinton. The variable for Clinton is V161086 and use the same 8 predictor variables.
Functional Form
x11()
plot(y=model.2$residuals,x=model.2$fitted.values, xlab="Fitted Values", ylab="Residuals")
abline(h=0, col="red")
It does not look like the local means are 0, thus we may have violated functional form.
library(lmtest)
Loading required package: zoo
Attaching package: 'zoo'
The following objects are masked from 'package:base':
as.Date, as.Date.numeric
resettest(model.2, power=2:3, type="fitted")
RESET test
data: model.2
RESET = 9.5918, df1 = 2, df2 = 1202, p-value = 7.366e-05
We find that p .05, thus rejecting the null and concluding we have violated the functional form assumptions.
x11()
crPlots(model.2)
It appears that education and partisan identification are the least linearly related to our outcome variable.
dwtest(model.2, order.by=nes2$gender)
Durbin-Watson test
data: model.2
DW = 1.9785, p-value = 0.3426
alternative hypothesis: true autocorrelation is greater than 0
dwtest(model.2, order.by=nes2$education)
Durbin-Watson test
data: model.2
DW = 1.9288, p-value = 0.1014
alternative hypothesis: true autocorrelation is greater than 0
dwtest(model.2, order.by=nes2$pid.num)
Durbin-Watson test
data: model.2
DW = 1.9279, p-value = 0.09921
alternative hypothesis: true autocorrelation is greater than 0
wtest(model.2, order.by=nes2$ideology.num)
Durbin-Watson test
data: model.2
DW = 2.0368, p-value = 0.7294
alternative hypothesis: true autocorrelation is greater than 0
dwtest(model.2, order.by=nes2$hc.law.num)
Durbin-Watson test
data: model.2
DW = 1.9537, p-value = 0.2012
alternative hypothesis: true autocorrelation is greater than 0
dwtest(model.2, order.by=nes2$economy.num)
Durbin-Watson test
data: model.2
DW = 2.0153, p-value = 0.5931
alternative hypothesis: true autocorrelation is greater than 0
dwtest(model.2, order.by=nes2$wall.num)
Durbin-Watson test
data: model.2
DW = 1.9788, p-value = 0.3455
alternative hypothesis: true autocorrelation is greater than 0
dwtest(model.2, order.by=nes2$isis.num)
Durbin-Watson test
data: model.2
DW = 2.0122, p-value = 0.5721
alternative hypothesis: true autocorrelation is greater than 0
According to the DW tests, none of our predictors appear problematic. But, let’s use a Box-Tidwell test to check education and partisan identification.
We need to create a new Clinton variable that does not have 0. 1
nes2$clinton2 <- nes2$clinton +1
boxTidwell(clinton2 ~ education + pid.num, data=nes2, na.action=na.exclude)
Warning in boxTidwell.default(y, X1, X2, max.iter = max.iter, tol = tol, :
1Notice the warning message that the maximum iterations were exceeded and thus the test did not converge on a final set of estimates. We can set the maximum number of iterations by including max.iter=. In this specific case, there was still a warning message after increasing the number of iterations mutiple times, but in all scenarios the results wind-up roughly the same.
maximum iterations exceeded
MLE of lambda Score Statistic (z) Pr(>|z|)
education 10.97818 1.8484 0.064547 .
pid.num 0.74844 2.6679 0.007633 **
---
Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
iterations = 26
We see that we do not need to transform education, but we do need to transform pid.num. We need to
raise pid.num by .748.
summary(model.2a <- lm(clinton ~ gender + education + pid.num + I(pid.num^.748) +
ideology.num + hc.law.num + economy.num + wall.num + isis.num,
data=nes2))
Call:
lm(formula = clinton ~ gender + education + pid.num + I(pid.num^0.748) +
ideology.num + hc.law.num + economy.num + wall.num + isis.num,
data = nes2)
Residuals:
Min 1Q Median 3Q Max
-76.782 -12.464 -0.868 12.149 98.315
Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept) 74.1997 7.2859 10.184 < 2e-16 ***
gender1. Female 2.4292 1.1804 2.058 0.039809 *
education -0.3420 0.2657 -1.287 0.198315
pid.num 8.6415 4.4493 1.942 0.052345 .
I(pid.num^0.748) -28.4419 8.0732 -3.523 0.000443 ***
ideology.num -0.6363 0.5473 -1.163 0.245243
hc.law.num 3.4649 0.3160 10.964 < 2e-16 ***
economy.num 3.8339 0.6712 5.712 1.40e-08 ***
wall.num -1.7655 0.3110 -5.676 1.73e-08 ***
isis.num -0.3671 0.3007 -1.221 0.222395
---
Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
Residual standard error: 20.12 on 1203 degrees of freedom
Multiple R-squared: 0.6439, Adjusted R-squared: 0.6413
F-statistic: 241.7 on 9 and 1203 DF, p-value: < 2.2e-16
The transformed partisan identification is significant, but the coefficient is very strange. This indicates that
we should be suspicious of using this model.
x11()
crPlots(model.2a)
9.2 Heteroscedasticity
x11()
plot(y=model.2$residuals,x=model.2$fitted.values, xlab="Fitted Values", ylab="Residuals")
abline(h=0, col="red")
As you may have concluded when we first created the plot, there appears to be heteroscedasticity present; the downwards-slanting residuals clearly make for a pattern.
bptest(model.2, studentize=FALSE)
Breusch-Pagan test
data: model.2
BP = 45.018, df = 8, p-value = 3.651e-07
Since p .05, we reject the null of constant error variance and conclude that we have heteroscedasticity. To
deal with heteroscedasticity, we will re-run our model with robust standard errors using the coeftest()
function from the sandwich library.
library(sandwich)
coeftest(model.2, vcov = vcovHC)
t test of coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept) 56.32244 5.70546 9.8717 < 2.2e-16 ***
gender1. Female 2.90697 1.17780 2.4681 0.01372 *
education -0.30113 0.27013 -1.1148 0.26517
pid.num -6.96075 0.51496 -13.5170 < 2.2e-16 ***
ideology.num -0.45376 0.62835 -0.7221 0.47035
hc.law.num 3.53857 0.37419 9.4566 < 2.2e-16 ***
economy.num 3.86514 0.70384 5.4915 4.857e-08 ***
wall.num -1.65448 0.36514 -4.5311 6.451e-06 ***
isis.num -0.29145 0.32757 -0.8897 0.37380
---
Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
We find the same significance results as we did before.
9.3 Normality
x11()
hist(model.2$residuals,xlab="Residuals",main="")
The residuals are somewhat normally distributed, but there are long tails. This suggests we may have
violated the normality assumption.
x11()
qqnorm(model.2$residuals)
qqline(model.2$residuals,col="red")
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