Question 3

Consider the Gini index, classification error, and entropy in asimple classification setting with two classes. Create a single plot that displays each of these quantities as a function of ˆpm1. The xaxis should display ˆpm1, ranging from 0 to 1, and the y-axis should display the value of the Gini index, classification error, and entropy.

p = seq(0,1,0.01)
gini = p*(1-p)*2
entropy = -(p*log(p) + (1-p)*log(1-p))
error = 1-pmax(p,1-p)
plot(p,entropy,cex=.5,col='red')
lines(p,gini,lwd=2,col='darkblue',lty=3)
lines(p,error,lwd=2,col='green',lty=3)

Question 8

In the lab, a classification tree was applied to the Carseats data set after converting Sales into a qualitative response variable. Now we will seek to predict Sales using regression trees and related approaches, treating the response as a quantitative variable.

(a) Split the data set into a training set and a test set.

library(ISLR)
## Warning: package 'ISLR' was built under R version 3.6.3
attach(Carseats)
library(caret)
## Warning: package 'caret' was built under R version 3.6.3
## Loading required package: lattice
## Loading required package: ggplot2
library(rpart)
library(rpart.plot)
## Warning: package 'rpart.plot' was built under R version 3.6.3
library(tree)
## Warning: package 'tree' was built under R version 3.6.3
set.seed(1)
inTrain=createDataPartition(Carseats$Sales,p=0.5,list=FALSE)
train=Carseats[inTrain,]
test=Carseats[-inTrain,]

(b) Fit a regression tree to the training set. Plot the tree, and interpret the results. What test MSE do you obtain? The test MSE is 4.54 and only 7 of the 11 variables were used.

tree.carseats = tree(Sales~.,data=train)
plot(tree.carseats)
text(tree.carseats, pretty = 0)

summary(tree.carseats)
## 
## Regression tree:
## tree(formula = Sales ~ ., data = train)
## Variables actually used in tree construction:
## [1] "ShelveLoc"   "Price"       "Income"      "Age"         "CompPrice"  
## [6] "Advertising"
## Number of terminal nodes:  17 
## Residual mean deviance:  2.272 = 418 / 184 
## Distribution of residuals:
##     Min.  1st Qu.   Median     Mean  3rd Qu.     Max. 
## -4.76200 -0.95950  0.08568  0.00000  1.00600  3.55000
pred.carseats = predict(tree.carseats, test)
mean((test$Sales - pred.carseats)^2)
## [1] 4.540765

(c) Use cross-validation in order to determine the optimal level of tree complexity. Does pruning the tree improve the test MSE? The optimal level of complexity is 10 as found by the plot comparing size and dev. In this case the test MSE increased to 4.70.

cv.carseats=cv.tree(tree.carseats)
plot(cv.carseats$size,cv.carseats$dev,type='b')

prune.carseats=prune.tree(tree.carseats,best=10)
plot(prune.carseats)
text(prune.carseats,pretty=0)

pred.prune=predict(prune.carseats, test)
mean((test$Sales - pred.prune)^2)
## [1] 4.700748

(d) Use the bagging approach in order to analyze this data. What test MSE do you obtain? Use the importance() function to determine which variables are most important. The test MSE obtained from bagging is 2.55 and the variables of importance are listed below.

library(randomForest)
## Warning: package 'randomForest' was built under R version 3.6.3
## randomForest 4.6-14
## Type rfNews() to see new features/changes/bug fixes.
## 
## Attaching package: 'randomForest'
## The following object is masked from 'package:ggplot2':
## 
##     margin
set.seed(1)
bag.carseats=randomForest(Sales~.,data=train,  mtry=10, ntree=500, importance = TRUE)
pred.bag = predict(bag.carseats,test)
mean((test$Sales - pred.bag)^2)
## [1] 2.558038
varImpPlot(bag.carseats)

(e) Use random forests to analyze this data. What test MSE do you obtain? Use the importance() function to determine which variables are most important. Describe the effect of m, the number of variables considered at each split, on the error rate obtained. In using a smaller m value for the random forest model increases the test MSE to 2.73.

rf.carseats=randomForest(Sales~.,data=train,mtry=6,importance=TRUE)
pred.rf=predict(rf.carseats,test)
mean((test$Sales - pred.rf)^2)
## [1] 2.776207
importance(rf.carseats)
##                %IncMSE IncNodePurity
## CompPrice   18.6627888    155.602761
## Income       8.8872053    116.586678
## Advertising 15.4243033    128.996157
## Population  -1.0388128     64.207620
## Price       45.1933827    385.836459
## ShelveLoc   42.0388186    331.772115
## Age         15.1750039    145.114710
## Education   -1.1244668     56.813716
## Urban       -0.8126023      9.026471
## US           3.1571079     15.329066

Question 9

This problem involves the OJ data set which is part of the ISLR package.

(a) Create a training set containing a random sample of 800 observations, and a test set containing the remaining observations.

attach(OJ)
set.seed(1)
train = sample(1:nrow(OJ),800)
test = !train
train.oj=OJ[train,]
test.oj = OJ[-train,]

(b) Fit a tree to the training data, with Purchase as the response and the other variables as predictors. Use the summary() function to produce summary statistics about the tree, and describe the results obtained. What is the training error rate? How many terminal nodes does the tree have? There are 5 variables and 9 terminal nodes with a training error rate of .156.

tree.oj = tree(Purchase~.,data=train.oj)
summary(tree.oj)
## 
## Classification tree:
## tree(formula = Purchase ~ ., data = train.oj)
## Variables actually used in tree construction:
## [1] "LoyalCH"       "PriceDiff"     "SpecialCH"     "ListPriceDiff"
## [5] "PctDiscMM"    
## Number of terminal nodes:  9 
## Residual mean deviance:  0.7432 = 587.8 / 791 
## Misclassification error rate: 0.1588 = 127 / 800

(c) Type in the name of the tree object in order to get a detailed text output. Pick one of the terminal nodes, and interpret the information displayed. Looking at terminal node 8, the split variable is LoyaCH with values less than 0.036. There are 59 observations in this node with a deviance of 10.14. The prediction of Purchase in this node is Minute Maid with a probability of 98%.

tree.oj
## node), split, n, deviance, yval, (yprob)
##       * denotes terminal node
## 
##  1) root 800 1073.00 CH ( 0.60625 0.39375 )  
##    2) LoyalCH < 0.5036 365  441.60 MM ( 0.29315 0.70685 )  
##      4) LoyalCH < 0.280875 177  140.50 MM ( 0.13559 0.86441 )  
##        8) LoyalCH < 0.0356415 59   10.14 MM ( 0.01695 0.98305 ) *
##        9) LoyalCH > 0.0356415 118  116.40 MM ( 0.19492 0.80508 ) *
##      5) LoyalCH > 0.280875 188  258.00 MM ( 0.44149 0.55851 )  
##       10) PriceDiff < 0.05 79   84.79 MM ( 0.22785 0.77215 )  
##         20) SpecialCH < 0.5 64   51.98 MM ( 0.14062 0.85938 ) *
##         21) SpecialCH > 0.5 15   20.19 CH ( 0.60000 0.40000 ) *
##       11) PriceDiff > 0.05 109  147.00 CH ( 0.59633 0.40367 ) *
##    3) LoyalCH > 0.5036 435  337.90 CH ( 0.86897 0.13103 )  
##      6) LoyalCH < 0.764572 174  201.00 CH ( 0.73563 0.26437 )  
##       12) ListPriceDiff < 0.235 72   99.81 MM ( 0.50000 0.50000 )  
##         24) PctDiscMM < 0.196197 55   73.14 CH ( 0.61818 0.38182 ) *
##         25) PctDiscMM > 0.196197 17   12.32 MM ( 0.11765 0.88235 ) *
##       13) ListPriceDiff > 0.235 102   65.43 CH ( 0.90196 0.09804 ) *
##      7) LoyalCH > 0.764572 261   91.20 CH ( 0.95785 0.04215 ) *

(d) Create a plot of the tree, and interpret the results. *LoyalCH is is found quite a bit in this tree, being split in the first two levels.

plot(tree.oj)
text(tree.oj,pretty = 1)

(e) Predict the response on the test data, and produce a confusion matrix comparing the test labels to the predicted test labels. What is the test error rate? Test error rate is about 17%

pred.oj=predict(tree.oj,test.oj,type="class")
table(test.oj$Purchase,pred.oj)
##     pred.oj
##       CH  MM
##   CH 160   8
##   MM  38  64

(f) Apply the cv.tree() function to the training set in order to determine the optimal tree size. (g) Produce a plot with tree size on the x-axis and cross-validated classification error rate on the y-axis.

cv.oj=cv.tree(tree.oj)
plot(cv.oj$size, cv.oj$dev, type = "b")

(h) Which tree size corresponds to the lowest cross-validated classification error rate? The optimal tree size appears to be 9. (i) Produce a pruned tree corresponding to the optimal tree size obtained using cross-validation. If cross-validation does not lead to selection of a pruned tree, then create a pruned tree with five terminal nodes.

prune.oj = prune.tree(tree.oj, best = 5)

(j) Compare the training error rates between the pruned and unprunedvtrees. Which is higher? The pruned tree has a higher error rate of .205 compared to .159 of the unpruned tree.

summary(prune.oj)
## 
## Classification tree:
## snip.tree(tree = tree.oj, nodes = c(4L, 12L, 5L))
## Variables actually used in tree construction:
## [1] "LoyalCH"       "ListPriceDiff"
## Number of terminal nodes:  5 
## Residual mean deviance:  0.8239 = 655 / 795 
## Misclassification error rate: 0.205 = 164 / 800

(k) Compare the test error rates between the pruned and unpruned trees. Which is higher? The pruned tree has a higher test error rate of 18.8 compared to 17 for the unpruned tree.

pred.prune=predict(prune.oj,test.oj,type="class")
table(test.oj$Purchase,pred.prune)
##     pred.prune
##       CH  MM
##   CH 133  35
##   MM  17  85
table(test.oj$Purchase,pred.oj)
##     pred.oj
##       CH  MM
##   CH 160   8
##   MM  38  64