Assignment 6
Abel Garcia Chavez
2023-05-03
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 fit to the data.
library(ISLR)
library(boot)
set.seed(69)
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)$delta[1]
}
plot(1:10, deltas, xlab = "Degree", ylab = "Test MSE", type = "l")
d.min <- which.min(deltas)
points(which.min(deltas), deltas[which.min(deltas)], col = "red", cex = 2, pch = 20)
fit1 <- lm(wage ~ age, data = Wage)
fit2 <- lm(wage ~ poly(age, 2), data = Wage)
fit3 <- lm(wage ~ poly(age, 3), data = Wage)
fit4 <- lm(wage ~ poly(age, 4), data = Wage)
fit5 <- lm(wage ~ poly(age, 5), data = Wage)
anova(fit1, fit2, fit3, fit4, fit5)## Analysis of Variance Table
##
## Model 1: wage ~ age
## Model 2: wage ~ poly(age, 2)
## Model 3: wage ~ poly(age, 3)
## Model 4: wage ~ poly(age, 4)
## Model 5: wage ~ poly(age, 5)
## Res.Df RSS Df Sum of Sq F Pr(>F)
## 1 2998 5022216
## 2 2997 4793430 1 228786 143.5931 < 2.2e-16 ***
## 3 2996 4777674 1 15756 9.8888 0.001679 **
## 4 2995 4771604 1 6070 3.8098 0.051046 .
## 5 2994 4770322 1 1283 0.8050 0.369682
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1plot(wage ~ age, data = Wage, col = "black")
agelims <- range(Wage$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)
The optimal degree chosen here was 6 also we can analyze the p values the cubic and quartic polynomial are good while the others are not.
(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.
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)$delta[1]
}
plot(2:10, cvs[-1], xlab = "Cuts", ylab = "Test MSE", type = "l")
d.min <- which.min(cvs)
points(which.min(cvs), cvs[which.min(cvs)], col = "red", cex = 2, pch = 20)
plot(wage ~ age, data = Wage, col = "black")
agelims <- range(Wage$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)
The optimal number of cuts is 8.
Question 10
This question relates to the College data set.
(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 step wise selection on the training set in order to identify a satisfactory model that uses just a subset of the predictors.
library(leaps)
set.seed(69)
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$adjr2, xlab = "Number of variables", ylab = "Adjusted R2", type = "l", ylim = c(0.4, 0.84))
fit <- regsubsets(Outstate ~ ., data = College, method = "forward")
coeffs <- coef(fit, id = 6)
names(coeffs)## [1] "(Intercept)" "PrivateYes" "Room.Board" "PhD" "perc.alumni"
## [6] "Expend" "Grad.Rate"(b) Fit a GAM on the training data, using out-of-state tuition as the response and the features selected in the previous step as the predictors. Plot the results, and explain your findings.
library(gam)## Loading required package: splines## Loading required package: foreach## Loaded gam 1.22-2fit <- gam(Outstate ~ Private + s(Room.Board, df = 2) + s(PhD, df = 2) + s(perc.alumni, df = 2) + s(Expend, df = 5) + s(Grad.Rate, df = 2), data=College.train)
par(mfrow = c(2, 3))
plot(fit, se = T, col = "black")
There is a non-linear relationship with Expend.
(c) Evaluate the model obtained on the test set, and explain the results obtained.
preds <- predict(fit, College.test)
err <- mean((College.test$Outstate - preds)^2)
err## [1] 3432452tss <- mean((College.test$Outstate - mean(College.test$Outstate))^2)
rss <- 1 - err / tss
rss## [1] 0.8059409The \(R^2\) is 0.81 with 6 predictors.
(d) 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
## -7501.69 -1193.61 -16.52 1323.38 7766.54
##
## (Dispersion Parameter for gaussian family taken to be 3633640)
##
## Null Deviance: 5678347309 on 387 degrees of freedom
## Residual Deviance: 1355349106 on 373.0003 degrees of freedom
## AIC: 6978.828
##
## Number of Local Scoring Iterations: NA
##
## Anova for Parametric Effects
## Df Sum Sq Mean Sq F value Pr(>F)
## Private 1 1464993950 1464993950 403.175 < 2.2e-16 ***
## s(Room.Board, df = 2) 1 887489099 887489099 244.242 < 2.2e-16 ***
## s(PhD, df = 2) 1 387644469 387644469 106.682 < 2.2e-16 ***
## s(perc.alumni, df = 2) 1 151517695 151517695 41.699 3.322e-10 ***
## s(Expend, df = 5) 1 529888991 529888991 145.829 < 2.2e-16 ***
## s(Grad.Rate, df = 2) 1 81236134 81236134 22.357 3.216e-06 ***
## Residuals 373 1355349106 3633640
## ---
## 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 2.7114 0.10048
## s(PhD, df = 2) 1 2.7451 0.09839 .
## s(perc.alumni, df = 2) 1 0.6834 0.40892
## s(Expend, df = 5) 4 16.9093 9.43e-13 ***
## s(Grad.Rate, df = 2) 1 2.4200 0.12064
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1Expend, Room.board, Grad.Rate, perc.alumni show evidence and PhD shows small evidence.