6. In this exercise, you will further analyze the “Wage” data set considered throughout this chapter.

  1. 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 fit to the data.
library(ISLR)
## Warning: package 'ISLR' was built under R version 4.0.5
library(boot)
library("gam")
## Warning: package 'gam' was built under R version 4.0.5
## Loading required package: splines
## Loading required package: foreach
## Warning: package 'foreach' was built under R version 4.0.5
## Loaded gam 1.20
set.seed(1)
deltas <- rep(NA, 10)
for (i in 1:10) {
    fit <- glm(wage ~ poly(age, i), data = Wage)
    deltas[i] <- cv.glm(Wage, fit, K = 10)$delta[1]
}
plot(1:10, deltas, xlab = "Degree", ylab = "Test MSE", type = "l")
d.min <- which.min(deltas)
points(which.min(deltas), deltas[which.min(deltas)], col = "red", cex = 2, pch = 20)

Above you can see we did a K-fold cross-validation using \(K=10\). This tells us that the optimal degree for the polynomial is \(d=4\).

fit1 <- lm(wage ~ age, data = Wage)
fit2 <- lm(wage ~ poly(age, 2), data = Wage)
fit3 <- lm(wage ~ poly(age, 3), data = Wage)
fit4 <- lm(wage ~ poly(age, 4), data = Wage)
fit5 <- lm(wage ~ poly(age, 5), data = Wage)
anova(fit1, fit2, fit3, fit4, fit5)
## Analysis of Variance Table
## 
## Model 1: wage ~ age
## Model 2: wage ~ poly(age, 2)
## Model 3: wage ~ poly(age, 3)
## Model 4: wage ~ poly(age, 4)
## Model 5: wage ~ poly(age, 5)
##   Res.Df     RSS Df Sum of Sq        F    Pr(>F)    
## 1   2998 5022216                                    
## 2   2997 4793430  1    228786 143.5931 < 2.2e-16 ***
## 3   2996 4777674  1     15756   9.8888  0.001679 ** 
## 4   2995 4771604  1      6070   3.8098  0.051046 .  
## 5   2994 4770322  1      1283   0.8050  0.369682    
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

The ANOVA has given a p-value that tells us a cubic or quadratic fit provides an acceptable fit to the data.

plot(wage ~ age, data = Wage, col = "darkgrey")
agelims <- range(Wage$age)
age.grid <- seq(from = agelims[1], to = agelims[2])
fit <- lm(wage ~ poly(age, 3), data = Wage)
preds <- predict(fit, newdata = list(age = age.grid))
lines(age.grid, preds, col = "red", lwd = 2)

  1. 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.
cvs <- rep(NA, 10)
for (i in 2:10) {
    Wage$age.cut <- cut(Wage$age, i)
    fit <- glm(wage ~ age.cut, data = Wage)
    cvs[i] <- cv.glm(Wage, fit, K = 10)$delta[1]
}
plot(2:10, cvs[-1], xlab = "Cuts", ylab = "Test MSE", type = "l")
d.min <- which.min(cvs)
points(which.min(cvs), cvs[which.min(cvs)], col = "red", cex = 2, pch = 20)

We can see that the above K-fold cross-validation with K=10, that the optimal number of cuts is 8. Now, we can make a plot of the fit.

plot(wage ~ age, data = Wage, col = "darkgrey")
agelims <- range(Wage$age)
age.grid <- seq(from = agelims[1], to = agelims[2])
fit <- glm(wage ~ cut(age, 8), data = Wage)
preds <- predict(fit, data.frame(age = age.grid))
lines(age.grid, preds, col = "red", lwd = 2)

10. This question relates to the “College” data set.

  1. 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.
library(leaps)
## Warning: package 'leaps' was built under R version 4.0.5
set.seed(1)
attach(College)
train <- sample(length(Outstate), length(Outstate) / 2)
test <- -train
College.train <- College[train, ]
College.test <- College[test, ]
fit <- regsubsets(Outstate ~ ., data = College.train, nvmax = 17, method = "forward")
fit.summary <- summary(fit)
par(mfrow = c(1, 3))
plot(fit.summary$cp, xlab = "Number of variables", ylab = "Cp", type = "l")
min.cp <- min(fit.summary$cp)
std.cp <- sd(fit.summary$cp)
abline(h = min.cp + 0.2 * std.cp, col = "red", lty = 2)
abline(h = min.cp - 0.2 * std.cp, col = "red", lty = 2)
plot(fit.summary$bic, xlab = "Number of variables", ylab = "BIC", type='l')
min.bic <- min(fit.summary$bic)
std.bic <- sd(fit.summary$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(fit.summary$adjr2, xlab = "Number of variables", ylab = "Adjusted R2", type = "l", ylim = c(0.4, 0.84))
max.adjr2 <- max(fit.summary$adjr2)
std.adjr2 <- sd(fit.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)

Cp, BIC and Adjusted R2 show that the subset levels out around 6 variables.

fit <- regsubsets(Outstate ~ ., data = College, method = "forward")
coeffs <- coef(fit, id = 6)
names(coeffs)
## [1] "(Intercept)" "PrivateYes"  "Room.Board"  "PhD"         "perc.alumni"
## [6] "Expend"      "Grad.Rate"
  1. 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.
fit <- gam(Outstate ~ Private + s(Room.Board, df = 2) + s(PhD, df = 2) + s(perc.alumni, df = 2) + s(Expend, df = 5) + s(Grad.Rate, df = 2), data=College.train)
par(mfrow = c(2, 3))
plot(fit, se = T, col = "blue")

  1. Evaluate the model obtained on the test set, and explain the results obtained.
preds <- predict(fit, College.test)
err <- mean((College.test$Outstate - preds)^2)
err
## [1] 3349290
tss <- mean((College.test$Outstate - mean(College.test$Outstate))^2)
rss <- 1 - err / tss
rss
## [1] 0.7660016

R squared test = 0.77 from using GAM with 6 predictors.

  1. For which variables, if any, is there evidence of a non-linear relationship with the response ?
summary(fit)
## 
## Call: gam(formula = Outstate ~ Private + s(Room.Board, df = 2) + s(PhD, 
##     df = 2) + s(perc.alumni, df = 2) + s(Expend, df = 5) + s(Grad.Rate, 
##     df = 2), data = College.train)
## Deviance Residuals:
##      Min       1Q   Median       3Q      Max 
## -7402.89 -1114.45   -12.67  1282.69  7470.60 
## 
## (Dispersion Parameter for gaussian family taken to be 3711182)
## 
##     Null Deviance: 6989966760 on 387 degrees of freedom
## Residual Deviance: 1384271126 on 373 degrees of freedom
## AIC: 6987.021 
## 
## Number of Local Scoring Iterations: NA 
## 
## Anova for Parametric Effects
##                         Df     Sum Sq    Mean Sq F value    Pr(>F)    
## Private                  1 1778718277 1778718277 479.286 < 2.2e-16 ***
## s(Room.Board, df = 2)    1 1577115244 1577115244 424.963 < 2.2e-16 ***
## s(PhD, df = 2)           1  322431195  322431195  86.881 < 2.2e-16 ***
## s(perc.alumni, df = 2)   1  336869281  336869281  90.771 < 2.2e-16 ***
## s(Expend, df = 5)        1  530538753  530538753 142.957 < 2.2e-16 ***
## s(Grad.Rate, df = 2)     1   86504998   86504998  23.309 2.016e-06 ***
## Residuals              373 1384271126    3711182                      
## ---
## 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  1.9157    0.1672    
## s(PhD, df = 2)               1  0.9699    0.3253    
## s(perc.alumni, df = 2)       1  0.1859    0.6666    
## s(Expend, df = 5)            4 20.5075 2.665e-15 ***
## s(Grad.Rate, df = 2)         1  0.5702    0.4506    
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

The non-parametric ANOVA test shows us strong evidence of a non-linear relationship between “PhD” and “Expend”, and a mildly strong non-linear relationship (p-value of 0.05) between”Room.Board" and “Grad.Rate”.