3. Consider the Gini index, classification error, and cross-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 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.

Hint: In a setting with two classes, pˆm1 = 1 − pˆm2. You could make this plot by hand, but it will be much easier to make in R.

p <- seq(0, 1, 0.001)
gini.index <- 2 * p * (1 - p)
class.error <- 1 - pmax(p, 1 - p)
cross.entropy <- - (p * log(p) + (1 - p) * log(1 - p))
matplot(p, cbind(gini.index, class.error, cross.entropy), pch=c(15,17,19),ylab = "Gini index, classification error, and cross-entropy",col = c("#FF6633" , "#6666ff", "#9900ff"), type = 'b')
legend('bottom', inset=.01, legend = c('Gini index', 'classification error', 'cross-entropy'), col = c("#FF6633" , "#6666ff", "#9900ff"),pch=c(15,17,19))

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)
attach(Carseats)
data = Carseats
# 70% of the sample size
smp_size <- floor(.7 * nrow(Carseats))

# set the seed to make partition reproducible
set.seed(1)
train_ind <- sample(seq_len(nrow(Carseats)), size = smp_size)

train <- Carseats[train_ind, ]
test <- Carseats[-train_ind, ]

(b) Fit a regression tree to the training set. Plot the tree, and interpret the results. What test error rate do you obtain?

Based on the results on the regression tree, the variables ShelveLoc , Price, Age, Income, CompPrice, and Advertising were used in construction the tree and the test error rate is 4.20. The plot of the regression tree suggests that the quality of the shelving location for the car seats, bad or medium, is a major factor to decrease in sales.

library(tree)
set.seed(1)
regtree.fit <- tree(Sales ~ ., data = train)
summary(regtree.fit)
## 
## Regression tree:
## tree(formula = Sales ~ ., data = train)
## Variables actually used in tree construction:
## [1] "ShelveLoc"   "Price"       "Age"         "Income"      "CompPrice"  
## [6] "Advertising"
## Number of terminal nodes:  18 
## Residual mean deviance:  2.409 = 631.1 / 262 
## Distribution of residuals:
##     Min.  1st Qu.   Median     Mean  3rd Qu.     Max. 
## -4.77800 -0.96100 -0.08865  0.00000  1.01800  4.14100
set.seed(1)
regtree.pred <- predict(regtree.fit, newdata = test)
mean((regtree.pred - test$Sales)^2)
## [1] 4.208383
plot(regtree.fit)
text(regtree.fit, pretty=0)

(c) Use cross-validation in order to determine the optimal level of tree complexity. Does pruning the tree improve the test error rate?

The results of the pruned tree with a cross validation of 11 nodes did not improve the test error rate which is 4.69

set.seed(1)
regtree.cv = cv.tree(regtree.fit)
plot(regtree.cv$size, regtree.cv$dev, type = "b")
regtree.min <- which.min(regtree.cv$dev)
points(regtree.min, regtree.cv$dev[regtree.min], col = "red", cex = 2, pch = 20)

set.seed(1)
prune.fit <- prune.tree(regtree.fit, best = 11)
set.seed(1)
prune.pred <- predict(prune.fit, newdata = test)
mean((prune.pred - test$Sales)^2)
## [1] 4.691032

(d) Use the bagging approach in order to analyze this data. What test error rate do you obtain? Use the importance() function to determine which variables are most important.

The test error rate is 2.57 and the top 3 most important variables are price, good shelf location, and competitor price.

library(randomForest)
set.seed(1)
bag.fit=randomForest(Sales~., data = train, mtry = 10, importance = TRUE)
set.seed(1)
bag.pred <- predict(bag.fit, newdata = test)
mean((bag.pred - test$Sales)^2)
## [1] 2.573252
library(caret)
varImp(bag.fit)

(e) Use random forests to analyze this data. What test error rate 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.

The effect of m either increased or decreased the error rate, it seems that as the number of m is increased then the error rate is decreased. The m=6 produced the least error rate in comparison to m=5 and m=4. The top 3 most important variables remained the same.

set.seed(1)
rf.fit.4=randomForest(Sales~., data = train, mtry = 4, importance = TRUE)
rf.fit.5=randomForest(Sales~., data = train, mtry = 5, importance = TRUE)
rf.fit.6=randomForest(Sales~., data = train, mtry = 6, importance = TRUE)
rf.pred.4 <- predict(rf.fit.4, newdata = test)
mean((rf.pred.4 - test$Sales)^2)
## [1] 2.662923
rf.pred.5 <- predict(rf.fit.5, newdata = test)
mean((rf.pred.5 - test$Sales)^2)
## [1] 2.614307
rf.pred.6 <- predict(rf.fit.6, newdata = test)
mean((rf.pred.6 - test$Sales)^2)
## [1] 2.496953
varImp(rf.fit.6)

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)
data = OJ
set.seed(1)
inTrain <- sample(nrow(OJ), 800)
train <- OJ[inTrain,]
test <- OJ[-inTrain,]

(b) Fit a tree to the training data, with Purchase as the response and the other variables except for Buy 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?

The variables that were used to construct the tree is LoyalCH, PriceDiff, SpecialCH, ListPriceDiff, and PctDiscMM. The training error rate is 0.1588 and the tree has 9 nodes.

library(tree)
set.seed(1)
regtree.fit <- tree(Purchase ~ ., data = train)
summary(regtree.fit)
## 
## Classification tree:
## tree(formula = Purchase ~ ., data = train)
## 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.

The node label 3 displays that the split criterion is LoyalCH > 0.5036, the number of observations of the branch is 435, the deviance is 337.90, the prediction is CH, and the fraction of the observations that is CH is 86.9%.

regtree.fit
## 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.

Based on the plot, the most important variable in splitting the tree is LoyalCH.

set.seed(1)
plot(regtree.fit)
text(regtree.fit, pretty=0)

?Carseats

(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?

The test error rate is 17.04% ((8+38)/270)

set.seed(1)
regtree.pred <- predict(regtree.fit, test, type = 'class')
caret::confusionMatrix(regtree.pred, test$Purchase)
## Confusion Matrix and Statistics
## 
##           Reference
## Prediction  CH  MM
##         CH 160  38
##         MM   8  64
##                                           
##                Accuracy : 0.8296          
##                  95% CI : (0.7794, 0.8725)
##     No Information Rate : 0.6222          
##     P-Value [Acc > NIR] : 8.077e-14       
##                                           
##                   Kappa : 0.6154          
##                                           
##  Mcnemar's Test P-Value : 1.904e-05       
##                                           
##             Sensitivity : 0.9524          
##             Specificity : 0.6275          
##          Pos Pred Value : 0.8081          
##          Neg Pred Value : 0.8889          
##              Prevalence : 0.6222          
##          Detection Rate : 0.5926          
##    Detection Prevalence : 0.7333          
##       Balanced Accuracy : 0.7899          
##                                           
##        'Positive' Class : CH              
##

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

set.seed(1)
regtree.cv = cv.tree(regtree.fit, FUN = prune.misclass)
regtree.cv
## $size
## [1] 9 8 7 4 2 1
## 
## $dev
## [1] 145 145 146 146 167 315
## 
## $k
## [1]       -Inf   0.000000   3.000000   4.333333  10.500000 151.000000
## 
## $method
## [1] "misclass"
## 
## attr(,"class")
## [1] "prune"         "tree.sequence"

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

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

(h) Which tree size corresponds to the lowest cross-validated classification error rate?

The tree with 9 or 8 terminal nodes results in lowest cross-validation error rate, with 145 cross-validation errors. However, nodes 7 and 4 produced a 146 cross-validation errors.

(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.

set.seed(1)
prune.fit = prune.misclass(regtree.fit, best=4)
summary(prune.fit)
## 
## Classification tree:
## snip.tree(tree = regtree.fit, nodes = c(4L, 10L, 3L))
## Variables actually used in tree construction:
## [1] "LoyalCH"   "PriceDiff"
## Number of terminal nodes:  4 
## Residual mean deviance:  0.8922 = 710.2 / 796 
## Misclassification error rate: 0.1788 = 143 / 800

(j) Compare the training error rates between the pruned and unpruned trees. Which is higher?

The training error rate of the pruned tree is 17.88% and unpruned tree is 15.88%, pruned tree produced a higher training error rate.

(k) Compare the test error rates between the pruned and unprunedtrees. Which is higher?

The test error rate of the pruned tree 17.78% ((41+7)/270) is higher in comparison to the unpruned tree test error rate os 17.04%.

set.seed(1)
prune.pred <- predict(prune.fit, test, type = 'class')
caret::confusionMatrix(prune.pred, test$Purchase)
## Confusion Matrix and Statistics
## 
##           Reference
## Prediction  CH  MM
##         CH 161  41
##         MM   7  61
##                                           
##                Accuracy : 0.8222          
##                  95% CI : (0.7713, 0.8659)
##     No Information Rate : 0.6222          
##     P-Value [Acc > NIR] : 6.769e-13       
##                                           
##                   Kappa : 0.5954          
##                                           
##  Mcnemar's Test P-Value : 1.906e-06       
##                                           
##             Sensitivity : 0.9583          
##             Specificity : 0.5980          
##          Pos Pred Value : 0.7970          
##          Neg Pred Value : 0.8971          
##              Prevalence : 0.6222          
##          Detection Rate : 0.5963          
##    Detection Prevalence : 0.7481          
##       Balanced Accuracy : 0.7782          
##                                           
##        'Positive' Class : CH              
##