Assignment 7

Exercise 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 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.

p <- seq(0, 1, 0.01)
gini.index <- 2 * p * (1 - p)
class.error <- 1 - pmax(p, 1 - p)
entropy <- - (p * log(p) + (1 - p) * log(1 - p))
matplot(p, cbind(class.error, gini.index, entropy), col = c("blue", "green", "red" ))

Exercise 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.

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

data(Carseats)
attach(Carseats)

set.seed(1)
train = sample(1:nrow(Carseats), nrow(Carseats)/2)
Carseats.train = Carseats[train, ]
Carseats.test = Carseats[-train, ]

Fit a regression tree to the training set. Plot the tree, and inter-pret 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)

regpred.carseats = predict(tree.carseats, Carseats.test)
mean((Carseats.test$Sales - regpred.carseats)^2)

## [1] 4.922039

Test MSE 4.922039

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

set.seed(2)
cv.carseats = cv.tree(tree.carseats)
plot(cv.carseats$size, cv.carseats$dev, xlab = "Tree size", ylab = "Classification Error Rate")

Optimal level of tree complexity = 15. Since the number is still very large we will use the pruning method to determine the best number.

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

The test MSE we obtain from pruning tree is 4.918134

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.

set.seed(3)
bag.carseats = randomForest(Sales~ ., data = Carseats, subset = train, mtry = 10, importance = TRUE)
bag.carseats

## 
## Call:
##  randomForest(formula = Sales ~ ., data = Carseats, mtry = 10,      importance = TRUE, subset = train) 
##                Type of random forest: regression
##                      Number of trees: 500
## No. of variables tried at each split: 10
## 
##           Mean of squared residuals: 2.90505
##                     % Var explained: 63.06

bagpred.carseats = predict(bag.carseats, newdata = Carseats.test) 
mean((Carseats.test$Sales - bagpred.carseats)^2)

## [1] 2.630702

importance(bag.carseats)

##                %IncMSE IncNodePurity
## CompPrice   25.6368749    169.596380
## Income       5.5092406     93.989153
## Advertising 13.0320872     99.531672
## Population  -1.3427420     56.478915
## Price       52.4411414    498.870909
## ShelveLoc   47.4964524    384.768230
## Age         16.7659587    153.617412
## Education   -0.4141856     44.838301
## Urban        0.2914200      9.449681
## US           5.1460185     15.033204

The test MSE we get from the bagging approach is 2.630702. The most important predictors of Sales is Price, ShelveLoc, and CompPrice.

Use random forests to analyze this data. What test MSE do you obtain? Use the importance() function to determine which vari-ables are most important. Describe the eﬀect of m,the numberof variables considered at each split, on the error rate obtained.

set.seed(4)
rf.carseats = randomForest(Sales~ ., data = Carseats.train, mtry = 8, impotance = T)
rfpred.carseats = predict(rf.carseats, Carseats.test)
mean((Carseats.test$Sales - rfpred.carseats)^2)

## [1] 2.612723

importance(rf.carseats)

##             IncNodePurity
## CompPrice      166.998303
## Income          97.792200
## Advertising    102.178454
## Population      62.289204
## Price          499.049584
## ShelveLoc      361.148227
## Age            164.647876
## Education       48.259514
## Urban            8.941162
## US              19.884424

The test MSE we obtain from the rf is 2.612723. The most important predictors of Sales is Price, ShelveLoc, Age, and CompPrice. As we change the value of m in our random forest model, the test MSE decreases from 4.5 to 2.62. When m = 8, we have a test MSE of 2.612723, which is the lowest between all m values, where 0 <= m <= 10.

Exercise 9 - This problem involves the OJ data set which is part of the ISLR package.

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

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?

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"     "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 9 terminal nodes and a training error rate of 0.1588

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.

OJ.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 < 0.196197 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 ) *

I select node #8. The split criterion is LoyalCH < 0.0356415, the number of observations in that branch is 59 with a deviance of 10.14 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.

Create a plot of the tree, and interpret the results.

plot(OJ.tree)
text(OJ.tree,pretty=0, cex = .50)

The most important indicator appears to be LoyalCH. The top three nodes contain this indicator.

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(OJ.tree, newdata = OJ.test, type = "class")
table(tree.pred,OJ.test$Purchase)

##          
## tree.pred  CH  MM
##        CH 160  38
##        MM   8  64

test_error = 1 - ((160 + 64) / 270)
phrase = "The test error rate is: "
paste(phrase, test_error)

## [1] "The test error rate is:  0.17037037037037"

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.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"

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")

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

7-node size corresponds to the lowest cross-validated classi-ﬁcation error rate.

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(OJ.tree, best=7)
plot(prune.OJ)
text(prune.OJ,pretty=0, cex = 0.50)

Assignment 7

Arsalan Bhojani

8/7/2021