Assignment 6
Jimmy De La Fuente
4/23/2021
Question 6
In this exercise, you will further analyze the Wage data set considered throughout this chapter.
(a) Perform polynomial regression to predict wage using age. Use cross-validation to select the optimal degree d for the polynomial. What degree was chosen, and how does this compare to the results of hypothesis testing using ANOVA? Make a plot of the resulting polynomial to fit the data.
library(ISLR)
library(boot)
set.seed(1)
all.deltas = rep(NA, 10)
for (i in 1:10) {
glm.fit = glm(wage~poly(age, i), data = Wage)
all.deltas[i] = cv.glm(Wage, glm.fit, K=10)$delta[2]
}
plot(1:10, all.deltas, xlab="Degree", ylab = "CV error", type="l", pch=20, lwd = 2, ylim=c(1590,1700))
min.point = min(all.deltas)
sd.points = sd(all.deltas)
abline(h=min.point + 0.2 * sd.points, col="red", lty="dashed")
abline(h=min.point - 0.2 * sd.points, col='red', lty="dashed")
legend("topright", "0.2-standard deviation lines", lty="dashed", col="red")
The plot with standard deviation lines appears to show that d = 3 is the smallest degree which gives a small cross-validation error.
fit.1 = lm(wage~poly(age, 1), data=Wage)
fit.2 = lm(wage~poly(age, 2), data=Wage)
fit.3 = lm(wage~poly(age, 3), data=Wage)
fit.4 = lm(wage~poly(age, 4), data=Wage)
fit.5 = lm(wage~poly(age, 5), data=Wage)
fit.6 = lm(wage~poly(age, 6), data=Wage)
fit.7 = lm(wage~poly(age, 7), data=Wage)
fit.8 = lm(wage~poly(age, 8), data=Wage)
fit.9 = lm(wage~poly(age, 9), data=Wage)
fit.10 = lm(wage~poly(age, 10), data=Wage)
anova(fit.1, fit.2, fit.3, fit.4, fit.5, fit.6, fit.7, fit.8, fit.9, fit.10)## Analysis of Variance Table
##
## Model 1: wage ~ poly(age, 1)
## Model 2: wage ~ poly(age, 2)
## Model 3: wage ~ poly(age, 3)
## Model 4: wage ~ poly(age, 4)
## Model 5: wage ~ poly(age, 5)
## Model 6: wage ~ poly(age, 6)
## Model 7: wage ~ poly(age, 7)
## Model 8: wage ~ poly(age, 8)
## Model 9: wage ~ poly(age, 9)
## Model 10: wage ~ poly(age, 10)
## Res.Df RSS Df Sum of Sq F Pr(>F)
## 1 2998 5022216
## 2 2997 4793430 1 228786 143.7638 < 2.2e-16 ***
## 3 2996 4777674 1 15756 9.9005 0.001669 **
## 4 2995 4771604 1 6070 3.8143 0.050909 .
## 5 2994 4770322 1 1283 0.8059 0.369398
## 6 2993 4766389 1 3932 2.4709 0.116074
## 7 2992 4763834 1 2555 1.6057 0.205199
## 8 2991 4763707 1 127 0.0796 0.777865
## 9 2990 4756703 1 7004 4.4014 0.035994 *
## 10 2989 4756701 1 3 0.0017 0.967529
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1The ANOVA model shows that all above d = 3 are significant.
plot(wage~age, data=Wage, col="darkgrey")
agelim = range(Wage$age)
age.grid = seq(from=agelim[1], to=agelim[2])
lm.fit = lm(wage~poly(age,3), data=Wage)
lm.pred = predict(lm.fit, data.frame(age=age.grid))
lines(age.grid, lm.pred, col="blue", lwd=2)
(b) Fit a step function to predict wage using age, and perform cross-validation to choose the optimal number of cuts. Make a plot of the fit obtained.
all.cvs = rep(NA, 10)
for (i in 2:10){
Wage$age.cut = cut(Wage$age, i)
lm.fit = glm(wage~age.cut, data=Wage)
all.cvs[i] = cv.glm(Wage, lm.fit, K=10)$delta[2]
}
plot(2:10, all.cvs[-1], xlab="Number of Cuts", ylab="CV Error", type = "l", pch=20, lwd=2)
The plot shows the number of cuts to be K = 8
lm.fit = glm(wage~cut(age, 8), data=Wage)
agelim = range(Wage$age)
age.grid = seq(from=agelim[1], to=agelim[2])
lm.pred = predict(lm.fit, data.frame(age=age.grid))
plot(wage~age, data=Wage, col='darkgrey')
lines(age.grid, lm.pred, col="red", lwd=2)
Question 10
This question relates to the College data set.
set.seed(1)
library(ISLR)
library(gam)## Loading required package: splines## Loading required package: foreach## Loaded gam 1.20library(leaps)
attach(College)(a) Split the data into a training set and a test set. Using out-of-state tuition as the response and the other variables as the predictors, perform forward stepwise selection on the training set in order to identify a satisfactory model that uses just a subset of the predictors.
train = sample(length(Outstate), length(Outstate)/2)
test = -train
College.Train = College[train, ]
College.Test = College[test, ]
reg.fit = regsubsets(Outstate ~ ., data = College.Train, nvmax = 17, method = "forward")
reg.summary = summary(reg.fit)
par(mfrow = c(1, 3))
plot(reg.summary$cp, xlab = "Number of Variables", ylab = "Cp", type = "l")
min.cp = min(reg.summary$cp)
std.cp = sd(reg.summary$cp)
abline(h = min.cp + 0.2 * std.cp, col = "red", lty = 2)
abline(h = min.cp - 0.2 * std.cp, col = "red", lty = 2)
plot(reg.summary$bic, xlab = "Number of Variables", ylab = "BIC", type = "l")
min.bic = min(reg.summary$bic)
std.bic = sd(reg.summary$bic)
abline(h = min.bic + 0.2 * std.bic, col = "red", lty = 2)
abline(h = min.bic - 0.2 * std.bic, col = "red", lty = 2)
plot(reg.summary$adjr2, xlab = "Number of Variables", ylab = "Adjusted R2",
type = "l", ylim = c(0.4, 0.84))
max.adjr2 = max(reg.summary$adjr2)
std.adjr2 = sd(reg.summary$adjr2)
abline(h = max.adjr2 + 0.2 * std.adjr2, col = "red", lty = 2)
abline(h = max.adjr2 - 0.2 * std.adjr2, col = "red", lty = 2)
Each plot shows that the number of variables for the minimum size for the subset is 6.
reg.fit = regsubsets(Outstate ~ ., data = College, method="forward")
coef = coef(reg.fit, id=6)
names(coef)## [1] "(Intercept)" "PrivateYes" "Room.Board" "PhD" "perc.alumni"
## [6] "Expend" "Grad.Rate"(b) Fit a GAM on the training data, using out-of-state tuition as the response and the features selected in the previous step as the predictors. Plot the results, and explain your findings.
gam_model <- gam(Outstate ~ Private + s(Room.Board, df=2) + s(perc.alumni, df=2) + s(PhD, df=2) + s(Expend, df=2) + s(Grad.Rate, df=2), data=College.Train)
par(mfrow=c(2, 3))
plot(gam_model, se=TRUE, col="green")
(c) Evaluate the model obtained on the test set, and explain the results obtained.
gam.pred = predict(gam_model, College.Test)
gam.err = mean((College.Test$Outstate - gam.pred)^2)
gam.err## [1] 3438192gam.tss = mean((College.Test$Outstate - mean(College.Test$Outstate))^2)
test.rss = 1 - gam.err/gam.tss
test.rss## [1] 0.7597905This gives ups the test R-Squared, and we can see that it is about .76 using GAM with 6 predictors.
(d) For which variables, if any, is there evidence of a non-linear relationship with the response?
summary(gam_model)##
## Call: gam(formula = Outstate ~ Private + s(Room.Board, df = 2) + s(perc.alumni,
## df = 2) + s(PhD, df = 2) + s(Expend, df = 2) + s(Grad.Rate,
## df = 2), data = College.Train)
## Deviance Residuals:
## Min 1Q Median 3Q Max
## -6500.0 -1344.2 -104.2 1350.9 8792.9
##
## (Dispersion Parameter for gaussian family taken to be 3946537)
##
## Null Deviance: 6989966760 on 387 degrees of freedom
## Residual Deviance: 1483898477 on 376.0001 degrees of freedom
## AIC: 7007.986
##
## Number of Local Scoring Iterations: NA
##
## Anova for Parametric Effects
## Df Sum Sq Mean Sq F value Pr(>F)
## Private 1 1843025455 1843025455 466.998 < 2.2e-16 ***
## s(Room.Board, df = 2) 1 1692725509 1692725509 428.914 < 2.2e-16 ***
## s(perc.alumni, df = 2) 1 541335260 541335260 137.167 < 2.2e-16 ***
## s(PhD, df = 2) 1 206873519 206873519 52.419 2.549e-12 ***
## s(Expend, df = 2) 1 419096845 419096845 106.194 < 2.2e-16 ***
## s(Grad.Rate, df = 2) 1 81869673 81869673 20.745 7.101e-06 ***
## Residuals 376 1483898477 3946537
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Anova for Nonparametric Effects
## Npar Df Npar F Pr(F)
## (Intercept)
## Private
## s(Room.Board, df = 2) 1 2.005 0.15763
## s(perc.alumni, df = 2) 1 0.314 0.57572
## s(PhD, df = 2) 1 3.781 0.05258 .
## s(Expend, df = 2) 1 47.156 2.725e-11 ***
## s(Grad.Rate, df = 2) 1 1.057 0.30446
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1There is evidence of a non-linear relationship between the response and Expend