Assignment 6
Julius Cervantes
2026-07-18
##Question 6
In this exercise, you will further analyze the Wage data set considered throughout this chapter.
- 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. - ten fold CV selects polynomial of 9. The ANOVA results showed that adding higher order polynomial beyond 9 didn’t significantly improve model.
library(ISLR2)## Warning: package 'ISLR2' was built under R version 4.6.1library(boot)
data(Wage)
set.seed(1)
cv.errors <- rep(NA, 10)
for(i in 1:10){
glm.fit <- glm(wage ~ poly(age, i), data = Wage)
cv.errors[i] <- cv.glm(Wage, glm.fit, K = 10)$delta[1]
}
best.degree <- which.min(cv.errors)
best.degree## [1] 9plot(1:10, cv.errors, type = "b", xlab = "Polynomial Degree", ylab = "CV Error")
fit.poly <- lm(wage ~ poly(age, best.degree), data = Wage)
age.grid <- seq(min(Wage$age), max(Wage$age))
preds <- predict(fit.poly, newdata = data.frame(age = age.grid))
plot(wage ~ age, data = Wage, col = "gray")
lines(age.grid, preds, col = "blue", lwd = 2)
fit1 <- lm(wage ~ age, data = Wage)
fit2 <- lm(wage ~ poly(age,2), data = Wage)
fit3 <- lm(wage ~ poly(age,3), data = Wage)
fit4 <- lm(wage ~ poly(age,4), data = Wage)
fit5 <- lm(wage ~ poly(age,5), data = Wage)
anova(fit1, fit2, fit3, fit4, fit5)## Analysis of Variance Table
##
## Model 1: wage ~ age
## Model 2: wage ~ poly(age, 2)
## Model 3: wage ~ poly(age, 3)
## Model 4: wage ~ poly(age, 4)
## Model 5: wage ~ poly(age, 5)
## Res.Df RSS Df Sum of Sq F Pr(>F)
## 1 2998 5022216
## 2 2997 4793430 1 228786 143.5931 < 2.2e-16 ***
## 3 2996 4777674 1 15756 9.8888 0.001679 **
## 4 2995 4771604 1 6070 3.8098 0.051046 .
## 5 2994 4770322 1 1283 0.8050 0.369682
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1- Fit a step function to predict wage using age, and perform cross validation to choose the optimal number of cuts. Make a plot of the fit obtained.
set.seed(1)
cv.errors <- rep(NA, 10)
for(k in 2:10){
Wage$age.cut <- cut(Wage$age, k)
glm.fit <- glm(wage ~ age.cut, data = Wage)
cv.errors[k] <- cv.glm(Wage, glm.fit, K = 10)$delta[1]
}
best.cuts <- which.min(cv.errors)
best.cuts## [1] 8plot(2:10, cv.errors[2:10], type = "b", xlab = "Number of Cuts", ylab = "CV Error")
Wage$age.cut <- cut(Wage$age, best.cuts)
fit.step <- glm(wage ~ age.cut, data = Wage)
age.grid <- seq(min(Wage$age), max(Wage$age))
preds <- predict(fit.step, newdata = data.frame(age.cut = cut(age.grid, best.cuts)))
plot(wage ~ age, data = Wage, col = "gray")
lines(age.grid, preds, col = "blue", lwd = 2)
Question 10
This question relates to the College data set.
- 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 4.6.1set.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$adjr2)
abline(h = max.adjr2 + 0.2 * std.adjr2, col = "red", lty = 2)
abline(h = max.adjr2 - 0.2 * std.adjr2, col = "red", lty = 2)
#Cp, BIC and adjr2 show that size 6 is the minimum size for the subset for which the scores are within 0.2 standard devitations of optimum.
lm1 <- regsubsets(Outstate ~ ., data = College, method = "forward")
coeffs <- coef(fit, id = 6)
names(coeffs)## [1] "(Intercept)" "PrivateYes" "Room.Board" "Terminal" "perc.alumni"
## [6] "Expend" "Grad.Rate"- Fit a GAM on the training data, using out-of-state tuition as the response and the features selected in the previous step as the predictors. Plot the results, and explain your findings. - GAM plots show expend has the strongest nonlinear relationship with out of state tuition. Room.board, perc.alumni and grad.rate show mostly increasing trends.
library(gam)## Warning: package 'gam' was built under R version 4.6.1## Loading required package: splines## Loading required package: foreach## Warning: package 'foreach' was built under R version 4.6.1## Loaded gam 1.22-7gam1 <- gam(Outstate ~ Private + s(Room.Board, df = 2) + s(Terminal, df = 2) + s(perc.alumni, df = 2) + s(Expend, df = 5) + s(Grad.Rate, df = 2), data=College.train)
par(mfrow = c(2, 3))
plot(gam1, se = T, col = "blue")
- Evaluate the model obtained on the test set, and explain the results obtained. -We obtain a test R^2 of 0.763972 using GAM with 6 predictors.
preds <- predict(gam1, newdata = College.test)
err <- mean((College.test$Outstate - preds)^2)
err## [1] 3378340tss <- mean((College.test$Outstate - mean(College.test$Outstate))^2)
r_squared <- 1 - err / tss
r_squared## [1] 0.763972- For which variables, if any, is there evidence of a non-linear relationship with the response? -ANOVA shows a strong evidence of non-linear relationship between “Outstate” and “Expend”“, and a moderately strong non-linear relationship (using p-value of 0.05) between”Outstate” and “Grad.Rate”” or “PhD”.
summary(gam1)##
## Call: gam(formula = Outstate ~ Private + s(Room.Board, df = 2) + s(Terminal,
## 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
## -7318.32 -1123.66 -28.31 1259.62 7338.91
##
## (Dispersion Parameter for gaussian family taken to be 3701607)
##
## Null Deviance: 6989966760 on 387 degrees of freedom
## Residual Deviance: 1380699954 on 373.0002 degrees of freedom
## AIC: 6986.018
##
## Number of Local Scoring Iterations: NA
##
## Anova for Parametric Effects
## Df Sum Sq Mean Sq F value Pr(>F)
## Private 1 1782594667 1782594667 481.573 < 2.2e-16 ***
## s(Room.Board, df = 2) 1 1604874368 1604874368 433.562 < 2.2e-16 ***
## s(Terminal, df = 2) 1 289618280 289618280 78.241 < 2.2e-16 ***
## s(perc.alumni, df = 2) 1 349447569 349447569 94.404 < 2.2e-16 ***
## s(Expend, df = 5) 1 578389738 578389738 156.254 < 2.2e-16 ***
## s(Grad.Rate, df = 2) 1 90976435 90976435 24.578 1.086e-06 ***
## Residuals 373 1380699954 3701607
## ---
## 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 1.7567 0.1858
## s(Terminal, df = 2) 1 1.2035 0.2733
## s(perc.alumni, df = 2) 1 0.1715 0.6790
## s(Expend, df = 5) 4 21.6541 4.441e-16 ***
## s(Grad.Rate, df = 2) 1 0.4668 0.4948
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1