Number 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.

p <- seq(0, 1, 0.01)
gini <- 2 * p * (1 - p)
classerror <- 1 - pmax(p, 1 - p)
entropy <- - (p * log(p) + (1 - p) * log(1 - p))
matplot(p, cbind(gini, classerror, entropy), col = c("blue", "lightblue", "black"))

### Number 8 ## (a) Split the data set into a training set and a test set.

library(ISLR2)
carseats <- Carseats
set.seed(1)
train <- sample(1:nrow(carseats), 200)
test <- Carseats[-train, ]

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

library(tree)
## Warning: package 'tree' was built under R version 4.2.2
tree <- tree(Sales ~., data = carseats, subset = train)
summary(tree)
## 
## Regression tree:
## tree(formula = Sales ~ ., data = carseats, subset = 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)
text(tree, pretty = 0)

treepred <- predict(tree, test)
plot(treepred , test$Sales)
abline (0, 1)

mean((test$Sales - treepred)^2)
## [1] 4.922039
# The test MSE is 4.92

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

set.seed(3)
cv <- cv.tree(tree)
min <- which.min(cv$dev)
min
## [1] 1
plot(cv$size, cv$dev, type = "b")
points(min, cv$dev[min], col = "red", cex = 2, pch = 20)

# optimal level is 9

prune <- prune.tree(tree, best = 9)
plot(prune)
text(prune, pretty = 0)

predprune = predict(prune, test)
plot(predprune , test$Sales)
abline (0, 1)

mean((test$Sales - predprune)^2)
## [1] 4.918134
# Pruning does not decrease the MSE, it increases by about .01. We can also see by the plot that pruning did not improve the test.

(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.

library(randomForest)
## randomForest 4.7-1.1
## Type rfNews() to see new features/changes/bug fixes.
bag <- randomForest(Sales ~., data = carseats, subset = train, mtry =10, importance = TRUE)
predbag <- predict(bag, newdata = test)
plot(predbag , test$Sales)
abline (0, 1)

mean((predbag - test$Sales)^2)
## [1] 2.617413
importance(bag)
##                %IncMSE IncNodePurity
## CompPrice   25.8395521    173.028616
## Income       3.7400180     88.246346
## Advertising 11.5586859    102.750315
## Population  -4.3487076     60.585131
## Price       53.0886784    510.532384
## ShelveLoc   46.4358875    372.977805
## Age         16.4441978    156.566479
## Education    2.3618726     45.308113
## Urban       -0.5679812      9.234006
## US           4.9344961     16.531234
# The MSE improved and is now 2.61. The variables that are most important are shown with the importance() function. The plot shows a relatively even scatter around the line. However, some variables are poorly predicted.

(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.

set.seed (1)
rf <- randomForest(Sales ~., data = carseats, subset = train, mtry = 5, importance = TRUE)
predrf <- predict(rf, newdata = test)
plot(predrf, test$Sales)
abline (0, 1)

mean((predrf - test$Sales)^2)
## [1] 2.714168
importance(rf)
##                %IncMSE IncNodePurity
## CompPrice   17.4126238     157.53631
## Income       2.9969399     110.40731
## Advertising 11.0485672     105.75049
## Population  -1.5321044      80.73318
## Price       43.3572135     452.02367
## ShelveLoc   44.4474163     331.64508
## Age         14.5322339     176.64252
## Education    0.8237454      55.91141
## Urban       -2.7805788      11.07321
## US           3.7773881      23.75322
# The test MSE is 2.71, and this is shown by the predicted values in the plot. There is a more even spread (still outliers), but it has good looking residuals. The variable ShelveLoc shows the highest % when using the importance function. Behind that variable is Price, CompPrice, and Advertising, respectively.

(f) Now analyze the data using BART, and report your results.

test <- sample(1:nrow(test), 200)
library(BART)
## Loading required package: nlme
## Loading required package: nnet
## Loading required package: survival
x <- carseats[, c(1:11)]
y <- carseats[, "Sales"]
xtrain <- x[train, ]
ytrain <- y[train]
xtest <- x[-test, ]
ytest <- y[-test]
set.seed (1)
bartfit <- gbart(xtrain, ytrain, x.test = xtest)
## *****Calling gbart: type=1
## *****Data:
## data:n,p,np: 200, 15, 200
## y1,yn: 2.781850, 1.091850
## x1,x[n*p]: 10.360000, 1.000000
## xp1,xp[np*p]: 5.560000, 1.000000
## *****Number of Trees: 200
## *****Number of Cut Points: 100 ... 1
## *****burn,nd,thin: 100,1000,1
## *****Prior:beta,alpha,tau,nu,lambda,offset: 2,0.95,0.273474,3,4.01382e-30,7.57815
## *****sigma: 0.000000
## *****w (weights): 1.000000 ... 1.000000
## *****Dirichlet:sparse,theta,omega,a,b,rho,augment: 0,0,1,0.5,1,15,0
## *****printevery: 100
## 
## MCMC
## done 0 (out of 1100)
## done 100 (out of 1100)
## done 200 (out of 1100)
## done 300 (out of 1100)
## done 400 (out of 1100)
## done 500 (out of 1100)
## done 600 (out of 1100)
## done 700 (out of 1100)
## done 800 (out of 1100)
## done 900 (out of 1100)
## done 1000 (out of 1100)
## time: 3s
## trcnt,tecnt: 1000,1000
yhat.bart <- bartfit$yhat.test.mean
mean((ytest - yhat.bart)^2)
## [1] 0.08306884
plot(yhat.bart, ytest)
abline (0, 1)

ord <- order(bartfit$varcount.mean , decreasing = T)
bartfit$varcount.mean[ord]
##       Sales  ShelveLoc3  ShelveLoc2         US2  ShelveLoc1      Urban2 
##      32.954      22.459      21.159      18.784      18.752      18.390 
##      Urban1         US1 Advertising       Price   Education   CompPrice 
##      15.512      14.402      13.118      10.007       8.873       7.820 
##         Age  Population      Income 
##       7.394       6.673       6.099
# The BART gave us incredibly good results. The MSE is only 0.083. The residual plot looks very good. Towards the higher yhat.bart values, there is a little bit of dispersing, but the overall model looks great.

Number 9

Part A

set.seed(1)
library(ISLR2)
oj <- OJ
train <- sample(1:nrow(oj), 800)
test <- 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?

library(tree)
tree <- tree(Purchase~., data = oj, subset = train)
summary(tree)
## 
## Classification tree:
## tree(formula = Purchase ~ ., data = oj, subset = 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
# There are 8 terminal nodes and the test error rate is 16.75.

(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
## 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 ) *
# The terminal node 4 is the one I choose to describe. It has 153 observations and a deviance of 106.70. The branch takes on 11% of yes values and 89% of no values.

(d) Create a plot of the tree, and interpret the results.

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

# the visual of the tree shows each branch, which signifies what would happen when choosing 'yes' or 'no' or greater than or less than the value shown.

(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?

purchasetesting <- oj$Purchase[-train]
treepred <- predict(tree, test, type = "class")
table(treepred, purchasetesting)
##         purchasetesting
## treepred  CH  MM
##       CH 160  38
##       MM   8  64
(27+19)/270
## [1] 0.1703704
# the decision tree correctly predicted 145 CH indicators and 80 MM. The test error rate is 17.04.

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

cv <- cv.tree(tree, FUN = prune.misclass)
cv
## $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"
#The optimal tree size is 5.

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

plot(cv$size, cv$dev, type = "b", xlab = "size", ylab = "dev")

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

# The tree size corresponding to the lowest classification error rate is 6.

(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 <- prune.tree(tree,best=6)

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

summary(tree)
## 
## Classification tree:
## tree(formula = Purchase ~ ., data = oj, subset = 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)
## 
## Classification tree:
## snip.tree(tree = tree, nodes = c(10L, 4L, 12L))
## Variables actually used in tree construction:
## [1] "LoyalCH"       "PriceDiff"     "ListPriceDiff"
## Number of terminal nodes:  6 
## Residual mean deviance:  0.7919 = 628.8 / 794 
## Misclassification error rate: 0.1788 = 143 / 800
#the error rate of the pruned tree and unpruned tree are the same.

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

mean(treepred!= test$Purchase)
## [1] 0.1703704
mean(predict(prune, test, type = "class")!= test$Purchase)
## [1] 0.1851852
# The error rate of the unpruned tree is 17.04 and the error rate of the pruned tree is 16.3. The error rate of the unpruned tree is higher.