28 Quiz 3
1. Consider the mtcars data set. Fit a model with mpg as the outcome that includes number of cylinders as a factor variable and weight as confounder. Give the adjusted estimate for the expected change in mpg comparing 8 cylinders to 4.
mcyl <- relevel(factor(mtcars$cyl), "4")
model <- lm(mpg ~ mcyl+wt, data = mtcars)
summary(model)$coef[3]## [1] -6.070862. Consider the mtcars data set. Fit a model with mpg as the outcome that includes number of cylinders as a factor variable and weight as a possible confounding variable. Compare the effect of 8 versus 4 cylinders on mpg for the adjusted and unadjusted by weight models. Here, adjusted means including the weight variable as a term in the regression model and unadjusted means the model without weight included. What can be said about the effect comparing 8 and 4 cylinders after looking at models with and without weight included?.
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) 33.990794 1.8877934 18.005569 6.257246e-17
## mcyl6 -4.255582 1.3860728 -3.070244 4.717834e-03
## mcyl8 -6.070860 1.6522878 -3.674214 9.991893e-04
## wt -3.205613 0.7538957 -4.252065 2.130435e-04## Estimate Std. Error t value Pr(>|t|)
## (Intercept) 26.663636 0.9718008 27.437347 2.688358e-22
## mcyl6 -6.920779 1.5583482 -4.441099 1.194696e-04
## mcyl8 -11.563636 1.2986235 -8.904534 8.568209e-103. Consider the mtcars data set. Fit a model with mpg as the outcome that considers number of cylinders as a factor variable and weight as confounder. Now fit a second model with mpg as the outcome model that considers the interaction between number of cylinders (as a factor variable) and weight. Give the P-value for the likelihood ratio test comparing the two models and suggest a model using 0.05 as a type I error rate significance benchmark.
model1 <- lm(mpg ~ mcyl+wt, data = mtcars)
model2 <- lm(mpg ~ mcyl+wt + mcyl*wt, data = mtcars)
lrtest(model1, model2)## Likelihood ratio test
##
## Model 1: mpg ~ mcyl + wt
## Model 2: mpg ~ mcyl + wt + mcyl * wt
## #Df LogLik Df Chisq Pr(>Chisq)
## 1 5 -73.311
## 2 7 -70.741 2 5.1412 0.07649 .
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1As the P value is above 0.05, we would then fail to reject the hypothesis, suggesting that the interaction terms may not be necessary.
4. Consider the mtcars data set. Fit a model with mpg as the outcome that includes number of cylinders as a factor variable and weight included in the model as lm(mpg ~ I(wt * 0.5) + factor(cyl), data = mtcars) How is the weight coefficient interpreted?
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) 33.990794 1.887793 18.005569 6.257246e-17
## I(wt * 0.5) -6.411227 1.507791 -4.252065 2.130435e-04
## factor(cyl)6 -4.255582 1.386073 -3.070244 4.717834e-03
## factor(cyl)8 -6.070860 1.652288 -3.674214 9.991893e-04The expected change in mpg per one ton increase in weight for a specific number of cylinders (4,6,8).
5. Consider the following data set; give the hat diagonal for the most influential point
x <- c(0.586, 0.166, -0.042, -0.614, 11.72)
y <- c(0.549, -0.026, -0.127, -0.751, 1.344)
model <- lm(y~x)
max(influence(model)$hat)## [1] 0.9945734## 1 2 3 4 5
## 0.2286650 0.2438146 0.2525027 0.2804443 0.9945734## showing how it's actually calculated
xm <- cbind(1, x)
diag(xm %*% solve(t(xm) %*% xm) %*% t(xm))## [1] 0.2286650 0.2438146 0.2525027 0.2804443 0.99457346. Consider the following data set; Give the slope dfbeta for the point with the highest had value
x <- c(0.586, 0.166, -0.042, -0.614, 11.72)
y <- c(0.549, -0.026, -0.127, -0.751, 1.344)
model <- lm(y~x)
influence.measures(model)## Influence measures of
## lm(formula = y ~ x) :
##
## dfb.1_ dfb.x dffit cov.r cook.d hat inf
## 1 1.0621 -3.78e-01 1.0679 0.341 2.93e-01 0.229 *
## 2 0.0675 -2.86e-02 0.0675 2.934 3.39e-03 0.244
## 3 -0.0174 7.92e-03 -0.0174 3.007 2.26e-04 0.253 *
## 4 -1.2496 6.73e-01 -1.2557 0.342 3.91e-01 0.280 *
## 5 0.2043 -1.34e+02 -149.7204 0.107 2.70e+02 0.995 *## [1] -133.8226