Q3 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 pˆm1. The x- axis should display pˆm1, 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, 0.01)
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), col = c("red", "green", "blue"))

Q8. In the lab, a classification tree was applied to the Carseats data set af- ter 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.

  1. Split the data set into a training set and a test set.
library(ISLR)
set.seed(1)
train <- sample(1:nrow(Carseats), nrow(Carseats) / 2)
Carseats.train <- Carseats[train, ]
Carseats.test <- Carseats[-train, ]
  1. Fit a regression tree to the training set. Plot the tree, and inter- pret the results. What test MSE do you obtain?
library(tree)
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)

yhat <- predict(tree.carseats, newdata = Carseats.test)
mean((yhat - Carseats.test$Sales)^2)
## [1] 4.922039
  We may conclude that the Test MSE is about 4.15.
  1. Use cross-validation in order to determine the optimal level of tree complexity. Does pruning the tree improve the test MSE?
cv.carseats <- cv.tree(tree.carseats)
plot(cv.carseats$size, cv.carseats$dev, type = "b")
tree.min <- which.min(cv.carseats$dev)
points(tree.min, cv.carseats$dev[tree.min], col = "red", cex = 2, pch = 20)

prune.carseats <- prune.tree(tree.carseats, best = 8)
plot(prune.carseats)
text(prune.carseats, pretty = 0)

yhat <- predict(prune.carseats, newdata = Carseats.test)
mean((yhat - Carseats.test$Sales)^2)
## [1] 5.113254
We may see that pruning the tree increases the Test MSE to 5.1.
  1. Use the bagging approach in order to analyze this data. What test MSE do you obtain? Use the importance() function to de- termine which variables are most important.
library(randomForest)
## randomForest 4.6-14
## Type rfNews() to see new features/changes/bug fixes.
set.seed(1)
bag.car = randomForest(Sales~.,data=Carseats.train,mtry = 10, importance = TRUE)
yhat.bag = predict(bag.car,newdata=Carseats.test)
mean((yhat.bag-Carseats.test$Sales)^2)
## [1] 2.605253
We may see that bagging decreases the Test MSE to 2.6.
importance(bag.car)
##                %IncMSE IncNodePurity
## CompPrice   24.8888481    170.182937
## Income       4.7121131     91.264880
## Advertising 12.7692401     97.164338
## Population  -1.8074075     58.244596
## Price       56.3326252    502.903407
## ShelveLoc   48.8886689    380.032715
## Age         17.7275460    157.846774
## Education    0.5962186     44.598731
## Urban        0.1728373      9.822082
## US           4.2172102     18.073863
varImpPlot(bag.car)

The most important variables are the price that company charges for car seats at each site and the quality of the shelving location for the car seats at each site. The test MSE associated with the bagegd regression tree is 2.55, almost half that obtained using an optimally-pruned single tree. we can conclude that price and shelveLoc are the most important tools
  1. 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.
library(randomForest)
set.seed(1)
rf.car = randomForest(Sales~.,data=Carseats.train,mtry = 3, importance = TRUE)
yhat.rf = predict(rf.car,newdata=Carseats.test)
mean((yhat.rf-Carseats.test$Sales)^2)
## [1] 2.960559
In this case, with m=p‾√, we have a Test MSE of 2.9

Q9. This problem involves the OJ data set which is part of the ISLR package.

  1. Create a training set containing a random sample of 800 obser- vations, and a test set containing the remaining observations.
set.seed(1)
train <- sample(1:nrow(OJ), 800)
OJ.train <- OJ[train, ]
OJ.test <- OJ[-train, ]
  1. 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?
tree.oj <- tree(Purchase ~ ., data = OJ.train)
summary(tree.oj)
## 
## Classification tree:
## tree(formula = Purchase ~ ., data = OJ.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

The fitted tree has 8 terminal nodes and a training error rate of 0.165.

(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.

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 < 0.196196 55   73.14 CH ( 0.61818 0.38182 ) *
##         25) PctDiscMM > 0.196196 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 ) *
We pick the node labelled 8, which is a terminal node because of the asterisk. The split criterion is LoyalCH < 0.035, the number of observations in that branch is 57 with a deviance of 10.07 and an overall prediction for the branch of MM. Less than 2% of the observations in that branch take the value of CH, and the remaining 98% take the value of MM.
  1. Create a plot of the tree, and interpret the results.
plot(tree.oj)
text(tree.oj, pretty = 0)

We may see that the most important indicator of “Purchase” appears to be “LoyalCH”, since the first branch differentiates the intensity of customer brand loyalty to CH. In fact, the top three nodes contain “LoyalCH”.

(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 ?

tree.pred <- predict(tree.oj, OJ.test, type = "class")
table(tree.pred, OJ.test$Purchase)
##          
## tree.pred  CH  MM
##        CH 160  38
##        MM   8  64
1 - (160 + 64) / 270
## [1] 0.1703704
We may conclude that the test error rate is about 17%.
  1. Apply the cv.tree() function to the training set in order to determine the optimal tree size.
cv.oj <- cv.tree(tree.oj, FUN = prune.misclass)
cv.oj
## $size
## [1] 9 8 7 4 2 1
## 
## $dev
## [1] 150 150 149 158 172 315
## 
## $k
## [1]       -Inf   0.000000   3.000000   4.333333  10.500000 151.000000
## 
## $method
## [1] "misclass"
## 
## attr(,"class")
## [1] "prune"         "tree.sequence"
  1. Produce a plot with tree size on the x-axis and cross-validated classification error rate on the y-axis.
plot(cv.oj$size, cv.oj$dev, type = "b", xlab = "Tree size", ylab = "Deviance")

  1. Which tree size corresponds to the lowest cross-validated classi- fication error rate?

We may see that the 2-node tree is the smallest tree with the lowest classification error rate.

  1. 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.misclass(tree.oj, best = 2)
plot(prune.oj)
text(prune.oj, pretty = 0)

  1. Compare the training error rates between the pruned and un- pruned trees. Which is higher?
summary(tree.oj)
## 
## Classification tree:
## tree(formula = Purchase ~ ., data = OJ.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
summary(prune.oj)
## 
## Classification tree:
## snip.tree(tree = tree.oj, nodes = 3:2)
## Variables actually used in tree construction:
## [1] "LoyalCH"
## Number of terminal nodes:  2 
## Residual mean deviance:  0.9768 = 779.5 / 798 
## Misclassification error rate: 0.205 = 164 / 800
The misclassification error rate is slightly higher for the pruned tree (0.1825 vs 0.165).
  1. Compare the test error rates between the pruned and unpruned trees. Which is higher?
prune.pred <- predict(prune.oj, OJ.test, type = "class")
table(prune.pred, OJ.test$Purchase)
##           
## prune.pred  CH  MM
##         CH 142  24
##         MM  26  78
1 - (142 + 78) / 270
## [1] 0.1851852
 In this case, the pruning process increased the test error rate to about 18%, but it produced a way more interpretable tree.