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

pm1 = seq(0, 1, 0.01)
gini = pm1 * (1 - pm1) * 2
entropy = -(pm1 * log(pm1) + (1 - pm1) * log(1 - pm1))
class.err = 1 - pmax(pm1, 1 - pm1)
matplot(pm1, cbind(gini, entropy, class.err), col = c("magenta", "red", "turquoise"))

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

8A) Split the data set into training and test

library(ISLR)
## Warning: package 'ISLR' was built under R version 4.0.5
library(randomForest)
## Warning: package 'randomForest' was built under R version 4.0.5
## randomForest 4.6-14
## Type rfNews() to see new features/changes/bug fixes.
library(tree)
## Warning: package 'tree' was built under R version 4.0.5
attach(Carseats)
set.seed(1)

train = sample(dim(Carseats)[1], dim(Carseats)[1]/2)
Carseats.train = Carseats[train, ]
Carseats.test = Carseats[-train, ]

8B)Fit a regression tree to the training set. Plot the tree, and interpret the results. What test MSE do you obtain?

tree.carseats = tree(Sales ~ ., data = Carseats.train)
summary(tree.carseats)
## 
## Regression tree:
## tree(formula = Sales ~ ., data = Carseats.train)
## Variables actually used in tree construction:
## [1] "ShelveLoc"   "Price"       "Age"         "Advertising" "CompPrice"  
## [6] "US"         
## Number of terminal nodes:  18 
## Residual mean deviance:  2.167 = 394.3 / 182 
## Distribution of residuals:
##     Min.  1st Qu.   Median     Mean  3rd Qu.     Max. 
## -3.88200 -0.88200 -0.08712  0.00000  0.89590  4.09900
plot(tree.carseats)
text(tree.carseats, pretty = 0)

pred.carseats = predict(tree.carseats, Carseats.test)
mean((Carseats.test$Sales - pred.carseats)^2)
## [1] 4.922039

The test MSE is 4.9220391 or almost 5

8C)Use cross-validation in order to determine the optimal level of tree complexity. Does pruning the tree improve the test MSE?

cv.carseats = cv.tree(tree.carseats, FUN = prune.tree)
par(mfrow = c(1, 2))
plot(cv.carseats$size, cv.carseats$dev, type = "b")
plot(cv.carseats$k, cv.carseats$dev, type = "b")

# Best size = 9
pruned.carseats = prune.tree(tree.carseats, best = 9)
par(mfrow = c(1, 1))
plot(pruned.carseats)
text(pruned.carseats, pretty = 0)

pred.pruned = predict(pruned.carseats, Carseats.test)
mean((Carseats.test$Sales - pred.pruned)^2)
## [1] 4.918134

8D)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.

bag.carseats = randomForest(Sales ~ ., data = Carseats.train, mtry = 10, ntree = 500, 
    importance = T)
bag.pred = predict(bag.carseats, Carseats.test)
mean((Carseats.test$Sales - bag.pred)^2)
## [1] 2.657296
importance(bag.carseats)
##                 %IncMSE IncNodePurity
## CompPrice   23.07909904    171.185734
## Income       2.82081527     94.079825
## Advertising 11.43295625     99.098941
## Population  -3.92119532     59.818905
## Price       54.24314632    505.887016
## ShelveLoc   46.26912996    361.962753
## Age         14.24992212    159.740422
## Education   -0.07662320     46.738585
## Urban        0.08530119      8.453749
## US           4.34349223     15.157608

Baggin improves the test MSE to 23.079099, 2.8208153, 11.4329562, -3.9211953, 54.2431463, 46.26913, 14.2499221, -0.0766232, 0.0853012, 4.3434922, 171.1857341, 94.0798245, 99.0989412, 59.8189051, 505.8870156, 361.9627526, 159.7404223, 46.7385854, 8.4537487, 15.1576076. We see Price, ShelveLoc, and CompPrice are the top three predictors of Sale

8E) 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.

rf.carseats = randomForest(Sales ~ ., data = Carseats.train, mtry = 5, ntree = 500, 
    importance = T)
rf.pred = predict(rf.carseats, Carseats.test)
mean((Carseats.test$Sales - rf.pred)^2)
## [1] 2.701665
importance(rf.carseats)
##                %IncMSE IncNodePurity
## CompPrice   19.8160444     162.73603
## Income       2.8940268     106.96093
## Advertising 11.6799573     106.30923
## Population  -1.6998805      79.04937
## Price       46.3454015     448.33554
## ShelveLoc   40.4412189     334.33610
## Age         12.5440659     169.06125
## Education    1.0762096      55.87510
## Urban        0.5703583      13.21963
## US           5.8799999      25.59797

Random forest increases MSE on test to 2.7016647. The top three predictors here are Price, ShelveLoc, and Age. M changes MSE by around 2 to 3.

9A)This problem involves the OJ data set which is part of the ISLR package starting with creating a training and test set.

library(ISLR)
attach(OJ)
set.seed(1013)

train = sample(dim(OJ)[1], 800)
OJ.train = OJ[train, ]
OJ.test = OJ[-train, ]

9B)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?

library(tree)
oj.tree = tree(Purchase ~ ., data = OJ.train)
summary(oj.tree)
## 
## Classification tree:
## tree(formula = Purchase ~ ., data = OJ.train)
## Variables actually used in tree construction:
## [1] "LoyalCH"       "PriceDiff"     "ListPriceDiff" "SalePriceMM"  
## Number of terminal nodes:  7 
## Residual mean deviance:  0.7564 = 599.8 / 793 
## Misclassification error rate: 0.1612 = 129 / 800

The tree uses four vairbales LoyalCH, PriceDiff, ListPriceDiff, and SalePriceMM. It has 7 nodes and misclassification error rate is 0.1612

9C)ype 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.

oj.tree
## node), split, n, deviance, yval, (yprob)
##       * denotes terminal node
## 
##  1) root 800 1069.00 CH ( 0.61125 0.38875 )  
##    2) LoyalCH < 0.5036 344  407.30 MM ( 0.27907 0.72093 )  
##      4) LoyalCH < 0.276142 163  121.40 MM ( 0.12270 0.87730 ) *
##      5) LoyalCH > 0.276142 181  246.30 MM ( 0.41989 0.58011 )  
##       10) PriceDiff < 0.065 75   75.06 MM ( 0.20000 0.80000 ) *
##       11) PriceDiff > 0.065 106  144.50 CH ( 0.57547 0.42453 ) *
##    3) LoyalCH > 0.5036 456  366.30 CH ( 0.86184 0.13816 )  
##      6) LoyalCH < 0.753545 189  224.30 CH ( 0.71958 0.28042 )  
##       12) ListPriceDiff < 0.235 79  109.40 MM ( 0.48101 0.51899 )  
##         24) SalePriceMM < 1.64 22   20.86 MM ( 0.18182 0.81818 ) *
##         25) SalePriceMM > 1.64 57   76.88 CH ( 0.59649 0.40351 ) *
##       13) ListPriceDiff > 0.235 110   75.81 CH ( 0.89091 0.10909 ) *
##      7) LoyalCH > 0.753545 267   85.31 CH ( 0.96255 0.03745 ) *

Picking node 4 the splitting value is LoyalCH with a value of .28. There are 163 points below this node, and the deviance below this region is 121. This node is in fact a terminal node marked by the asterisks. The prediction of this node is Sales = MM. 12% of the value in this node is CH as a value of Sales. About 87% have MM as a value of Sales.

9D)Create a plot tree

plot(oj.tree)
text(oj.tree, pretty = 0)

LoyalCH is the important value of the tree. If LoyalCH < .27, the tree predicts MM, if it is less than .75 the tree predicts CH. The decision also depends on PriceDif, ListPrice, and SalePrice to determine if it is MM or CH.

9E)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?

oj.pred = predict(oj.tree, OJ.test, type = "class")
table(OJ.test$Purchase, oj.pred)
##     oj.pred
##       CH  MM
##   CH 149  15
##   MM  30  76

9F)Apply the cv.tree() function to the training set in order to determine the optimal tree size.

cv.oj = cv.tree(oj.tree, FUN = prune.tree)

9G)Produce a plot with tree size on the x-axis and cross-validated classification error rate on the y-axis.

plot(cv.oj$size, cv.oj$dev, type = "b", xlab = "Tree Size", ylab = "Deviance")

9H) Which tree size corresponds to the lowest cross-validated classification error rate?

6 gives the lowest cross-validated classification

9i)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.

oj.pruned = prune.tree(oj.tree, best = 6)

9J)Compare the training error rates between the pruned and unpruned trees. Which is higher

summary(oj.pruned)
## 
## Classification tree:
## snip.tree(tree = oj.tree, nodes = 12L)
## Variables actually used in tree construction:
## [1] "LoyalCH"       "PriceDiff"     "ListPriceDiff"
## Number of terminal nodes:  6 
## Residual mean deviance:  0.7701 = 611.5 / 794 
## Misclassification error rate: 0.175 = 140 / 800

The misclassification error rate is .175 which is about .01 higher than the original

9k)Compare the test error rates between the pruned and unpruned trees. Which is higher?

pred.unpruned = predict(oj.tree, OJ.test, type = "class")
misclass.unpruned = sum(OJ.test$Purchase != pred.unpruned)
misclass.unpruned/length(pred.unpruned)
## [1] 0.1666667
pred.pruned = predict(oj.pruned, OJ.test, type = "class")
misclass.pruned = sum(OJ.test$Purchase != pred.pruned)
misclass.pruned/length(pred.pruned)
## [1] 0.2

The higher test error rate is the pruned tree at .2 over the unpruned tree .16