Homework3
Obehi Ikpea
2023-02-22
Question 13
This question should be answered using the Weekly data set, which is part of the ISLR2 package. This data is similar in nature to the Smarket data from this chapter’s lab, except that it contains 1, 089 weekly returns for 21 years, from the beginning of 1990 to the end of 2010. (a) Produce some numerical and graphical summaries of the Weekly data. Do there appear to be any patterns?
#Solution
str(Weekly)## 'data.frame': 1089 obs. of 9 variables:
## $ Year : num 1990 1990 1990 1990 1990 1990 1990 1990 1990 1990 ...
## $ Lag1 : num 0.816 -0.27 -2.576 3.514 0.712 ...
## $ Lag2 : num 1.572 0.816 -0.27 -2.576 3.514 ...
## $ Lag3 : num -3.936 1.572 0.816 -0.27 -2.576 ...
## $ Lag4 : num -0.229 -3.936 1.572 0.816 -0.27 ...
## $ Lag5 : num -3.484 -0.229 -3.936 1.572 0.816 ...
## $ Volume : num 0.155 0.149 0.16 0.162 0.154 ...
## $ Today : num -0.27 -2.576 3.514 0.712 1.178 ...
## $ Direction: Factor w/ 2 levels "Down","Up": 1 1 2 2 2 1 2 2 2 1 ...summary(Weekly)## Year Lag1 Lag2 Lag3
## Min. :1990 Min. :-18.1950 Min. :-18.1950 Min. :-18.1950
## 1st Qu.:1995 1st Qu.: -1.1540 1st Qu.: -1.1540 1st Qu.: -1.1580
## Median :2000 Median : 0.2410 Median : 0.2410 Median : 0.2410
## Mean :2000 Mean : 0.1506 Mean : 0.1511 Mean : 0.1472
## 3rd Qu.:2005 3rd Qu.: 1.4050 3rd Qu.: 1.4090 3rd Qu.: 1.4090
## Max. :2010 Max. : 12.0260 Max. : 12.0260 Max. : 12.0260
## Lag4 Lag5 Volume Today
## Min. :-18.1950 Min. :-18.1950 Min. :0.08747 Min. :-18.1950
## 1st Qu.: -1.1580 1st Qu.: -1.1660 1st Qu.:0.33202 1st Qu.: -1.1540
## Median : 0.2380 Median : 0.2340 Median :1.00268 Median : 0.2410
## Mean : 0.1458 Mean : 0.1399 Mean :1.57462 Mean : 0.1499
## 3rd Qu.: 1.4090 3rd Qu.: 1.4050 3rd Qu.:2.05373 3rd Qu.: 1.4050
## Max. : 12.0260 Max. : 12.0260 Max. :9.32821 Max. : 12.0260
## Direction
## Down:484
## Up :605
##
##
##
## pairs(Weekly)
cor(Weekly[, -9])## Year Lag1 Lag2 Lag3 Lag4
## Year 1.00000000 -0.032289274 -0.03339001 -0.03000649 -0.031127923
## Lag1 -0.03228927 1.000000000 -0.07485305 0.05863568 -0.071273876
## Lag2 -0.03339001 -0.074853051 1.00000000 -0.07572091 0.058381535
## Lag3 -0.03000649 0.058635682 -0.07572091 1.00000000 -0.075395865
## Lag4 -0.03112792 -0.071273876 0.05838153 -0.07539587 1.000000000
## Lag5 -0.03051910 -0.008183096 -0.07249948 0.06065717 -0.075675027
## Volume 0.84194162 -0.064951313 -0.08551314 -0.06928771 -0.061074617
## Today -0.03245989 -0.075031842 0.05916672 -0.07124364 -0.007825873
## Lag5 Volume Today
## Year -0.030519101 0.84194162 -0.032459894
## Lag1 -0.008183096 -0.06495131 -0.075031842
## Lag2 -0.072499482 -0.08551314 0.059166717
## Lag3 0.060657175 -0.06928771 -0.071243639
## Lag4 -0.075675027 -0.06107462 -0.007825873
## Lag5 1.000000000 -0.05851741 0.011012698
## Volume -0.058517414 1.00000000 -0.033077783
## Today 0.011012698 -0.03307778 1.000000000The correlation matrix shows that the only significant correlation is the one between volumes and shares.
attach(Weekly)
plot(Volume)
From the plot above, we see that there is a positive association between
volume and year,as the volume variable is increasing with time. This
implies that the average number of daily shares traded in billions
increased from 1990 to 2010.
- Use the full data set to perform a logistic regression with Direction as the response and the five lag variables plus Volume as predictors. Use the summary function to print the results. Do any of the predictors appear to be statistically significant? If so, which ones?
#Solution
logistic.model <- glm(Direction~Lag1+Lag2+Lag3+Lag4+Lag5+Volume, data=Weekly, family="binomial")
summary(logistic.model)##
## Call:
## glm(formula = Direction ~ Lag1 + Lag2 + Lag3 + Lag4 + Lag5 +
## Volume, family = "binomial", data = Weekly)
##
## Deviance Residuals:
## Min 1Q Median 3Q Max
## -1.6949 -1.2565 0.9913 1.0849 1.4579
##
## Coefficients:
## Estimate Std. Error z value Pr(>|z|)
## (Intercept) 0.26686 0.08593 3.106 0.0019 **
## Lag1 -0.04127 0.02641 -1.563 0.1181
## Lag2 0.05844 0.02686 2.175 0.0296 *
## Lag3 -0.01606 0.02666 -0.602 0.5469
## Lag4 -0.02779 0.02646 -1.050 0.2937
## Lag5 -0.01447 0.02638 -0.549 0.5833
## Volume -0.02274 0.03690 -0.616 0.5377
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## (Dispersion parameter for binomial family taken to be 1)
##
## Null deviance: 1496.2 on 1088 degrees of freedom
## Residual deviance: 1486.4 on 1082 degrees of freedom
## AIC: 1500.4
##
## Number of Fisher Scoring iterations: 4The lag2 variable appears to be statistically significant as it has a small p-value.
- Compute the confusion matrix and overall fraction of correct predictions. Explain what the confusion matrix is telling you about the types of mistakes made by logistic regression.
#Solution
direction.prob <- predict(logistic.model, type="response")
direction.pred <- rep("Down", 1089)
direction.pred[direction.prob>0.5]="Up"
table(direction.pred, Direction)## Direction
## direction.pred Down Up
## Down 54 48
## Up 430 557mean(direction.pred==Direction)## [1] 0.5610652From the confusion matrix, we see that our model correctly predicted that the market had a positive return on 557 weeks and and would have a negative return on 54 weeks. From the output, we see that the model correctly predicted the movement of the market 56.1% of the time. The mistake logistic regression made was train and test the model using the same set of observations. We achieved a 43.9% training error rate.
- Now fit the logistic regression model using a training data period from 1990 to 2008, with Lag2 as the only predictor. Compute the confusion matrix and the overall fraction of correct predictions for the held out data (that is, the data from 2009 and 2010).
#Solution
train <- (Year<2009)
Weekly.2008 <- Weekly[!train,]
Direction.2008 <- Direction[!train]train.model <- glm(Direction~Lag2,data=Weekly, family="binomial",subset=train )
train.probs <- predict(train.model, Weekly.2008, type="response")
train.pred <- rep("Down",104 )
train.pred[train.probs>0.5] <- "Up"
table(train.pred, Direction.2008)## Direction.2008
## train.pred Down Up
## Down 9 5
## Up 34 56mean(train.pred==Direction.2008)## [1] 0.625- Repeat (d) using LDA.
library(MASS)##
## Attaching package: 'MASS'## The following object is masked from 'package:ISLR2':
##
## Boston## The following object is masked from 'package:dplyr':
##
## selectdirection.lda <- lda(Direction~Lag2, data=Weekly, subset = train)
lda.pred <- predict(direction.lda, Weekly.2008)
lda.class <- lda.pred$class
table(lda.class,Direction.2008)## Direction.2008
## lda.class Down Up
## Down 9 5
## Up 34 56mean(lda.class==Direction.2008)## [1] 0.625- Repeat (d) using QDA.
#Solution
direction.qda <- qda(Direction~Lag2, data=Weekly, subset = train)
qda.class <- predict(direction.qda, Weekly.2008)$class
table(qda.class,Direction.2008)## Direction.2008
## qda.class Down Up
## Down 0 0
## Up 43 61mean(qda.class==Direction.2008)## [1] 0.5865385- Repeat (d) using KNN with K = 1.
#Solution
library(class)
train.X <- as.matrix(Lag2[train])
test.X <- as.matrix((Lag2[!train]))
train.Direction <- Direction[train]
set.seed(1)
knn.pred <- knn(train.X, test.X, train.Direction,k=1)
table(knn.pred, Direction.2008)## Direction.2008
## knn.pred Down Up
## Down 21 30
## Up 22 31mean(knn.pred==Direction.2008)## [1] 0.5- Repeat (d) using naive Bayes.
#Solution
library(e1071)## Warning: package 'e1071' was built under R version 4.1.3nb.fit <- naiveBayes(Direction~Lag2, data=Weekly, subset=train)
nb.class <- predict(nb.fit, Direction.2008)## Warning in predict.naiveBayes(nb.fit, Direction.2008): Type mismatch between
## training and new data for variable 'Lag2'. Did you use factors with numeric
## labels for training, and numeric values for new data?table(nb.class, Direction.2008)## Direction.2008
## nb.class Down Up
## Down 0 0
## Up 43 61mean(nb.class==Direction.2008)## [1] 0.5865385- Which of these methods appears to provide the best results on this data?
#solution
mean(train.pred!=Direction.2008)## [1] 0.375mean(lda.class!=Direction.2008)## [1] 0.375mean(qda.class!=Direction.2008)## [1] 0.4134615mean(knn.pred!=Direction.2008)## [1] 0.5Based on the results of the test error rate, we see that the logistic regression model and the linear discriminant analysis classifier provided the best results on the data.
- Experiment with different combinations of predictors, including possible transformations and interactions, for each of the methods. Report the variables, method, and associated confusion matrix that appears to provide the best results on the heldout data. Note that you should also experiment with values for K in the KNN classifier
#Solution
#1: The combination of lag1 and lag2 except naives bayes
#LG model
train.model1 <- glm(Direction~Lag1*Lag2,data=Weekly, family="binomial",subset=train )
train.probs1 <- predict(train.model1, Weekly.2008, type="response")
train.pred1 <- rep("Down",104 )
train.pred1[train.probs1>0.5] <- "Up"
table(train.pred1, Direction.2008)## Direction.2008
## train.pred1 Down Up
## Down 7 8
## Up 36 53mean(train.pred1==Direction.2008)## [1] 0.5769231#LDA MODEL
library(MASS)
direction.lda1 <- lda(Direction~Lag1+Lag2+Lag1:Lag2, data=Weekly, subset = train)
lda.pred1 <- predict(direction.lda1, Weekly.2008)
lda.class1 <- lda.pred1$class
table(lda.class1,Direction.2008)## Direction.2008
## lda.class1 Down Up
## Down 7 8
## Up 36 53mean(lda.class1==Direction.2008)## [1] 0.5769231#QDA MODEL
direction.qda1 <- qda(Direction~Lag1+Lag2+Lag1:Lag2, data=Weekly, subset = train)
qda.class1 <- predict(direction.qda1, Weekly.2008)$class
table(qda.class1,Direction.2008)## Direction.2008
## qda.class1 Down Up
## Down 23 36
## Up 20 25mean(qda.class1==Direction.2008)## [1] 0.4615385#KNN model
library(class)
train.X1 <-cbind(Lag1, Lag2)[train,]
test.X1 <- cbind(Lag1, Lag2)[!train,]
train.Direction1 <- Direction[train]
set.seed(3)
knn.pred1 <- knn(train.X1, test.X1, train.Direction1,k=1)
table(knn.pred1, Direction.2008)## Direction.2008
## knn.pred1 Down Up
## Down 18 29
## Up 25 32mean(knn.pred1==Direction.2008)## [1] 0.4807692# Naives Bayes Model
library(e1071)
nb.fit1 <- naiveBayes(Direction~Lag1+Lag2, data=Weekly, subset=train)
nb.class1 <- predict(nb.fit1, Direction.2008)## Warning in predict.naiveBayes(nb.fit1, Direction.2008): Type mismatch between
## training and new data for variable 'Lag1'. Did you use factors with numeric
## labels for training, and numeric values for new data?## Warning in predict.naiveBayes(nb.fit1, Direction.2008): Type mismatch between
## training and new data for variable 'Lag2'. Did you use factors with numeric
## labels for training, and numeric values for new data?table(nb.class1, Direction.2008)## Direction.2008
## nb.class1 Down Up
## Down 0 0
## Up 43 61mean(nb.class1==Direction.2008)## [1] 0.5865385From the output above, we see that the logistic regression model and LDA classifier provides the best output for the data.
#2 Transforming the predictors with quadratic transformation
#LG model
train.model2 <- glm(Direction~Lag1+I(Lag1^2),data=Weekly, family="binomial",subset=train )
train.probs2 <- predict(train.model2, Weekly.2008, type="response")
train.pred2 <- rep("Down",104 )
train.pred2[train.probs2>0.5] <- "Up"
table(train.pred2, Direction.2008)## Direction.2008
## train.pred2 Down Up
## Down 4 5
## Up 39 56mean(train.pred2==Direction.2008)## [1] 0.5769231#LDA MODEL
library(MASS)
direction.lda2 <- lda(Direction~Lag1+I(Lag1^2), data=Weekly, subset = train)
lda.pred2 <- predict(direction.lda2, Weekly.2008)
lda.class2 <- lda.pred2$class
table(lda.class2,Direction.2008)## Direction.2008
## lda.class2 Down Up
## Down 4 6
## Up 39 55mean(lda.class2==Direction.2008)## [1] 0.5673077#QDA MODEL
direction.qda2 <- qda(Direction~Lag1+I(Lag1^2), data=Weekly, subset = train)
qda.class2 <- predict(direction.qda2, Weekly.2008)$class
table(qda.class2,Direction.2008)## Direction.2008
## qda.class2 Down Up
## Down 32 54
## Up 11 7mean(qda.class2==Direction.2008)## [1] 0.375#KNN model
library(class)
train.X2 <-cbind(Lag1, I(Lag1^2))[train,]
test.X2 <- cbind(Lag1, I(Lag1^2))[!train,]
train.Direction2 <- Direction[train]
set.seed(4)
knn.pred2 <- knn(train.X2, test.X2, train.Direction2,k=1)
table(knn.pred2, Direction.2008)## Direction.2008
## knn.pred2 Down Up
## Down 18 30
## Up 25 31mean(knn.pred2==Direction.2008)## [1] 0.4711538# Naives Bayes Model
library(e1071)
nb.fit2 <- naiveBayes(Direction~Lag1+I(Lag1^2), data=Weekly, subset=train)
nb.class2 <- predict(nb.fit2, Direction.2008)## Warning in predict.naiveBayes(nb.fit2, Direction.2008): Type mismatch between
## training and new data for variable 'Lag1'. Did you use factors with numeric
## labels for training, and numeric values for new data?## Warning in predict.naiveBayes(nb.fit2, Direction.2008): Type mismatch between
## training and new data for variable 'I(Lag1^2)'. Did you use factors with numeric
## labels for training, and numeric values for new data?table(nb.class2, Direction.2008)## Direction.2008
## nb.class2 Down Up
## Down 0 0
## Up 43 61mean(nb.class2==Direction.2008)## [1] 0.5865385Here, we see that the naives bayes classifier produced the best result for the data.
K=2
library(class)
train.X3 <- as.matrix(Lag2[train])
test.X3 <- as.matrix((Lag2[!train]))
train.Direction3 <- Direction[train]
set.seed(1)
knn.pred3 <- knn(train.X3, test.X3, train.Direction3,k=2)
table(knn.pred3, Direction.2008)## Direction.2008
## knn.pred3 Down Up
## Down 19 27
## Up 24 34mean(knn.pred3==Direction.2008)## [1] 0.5096154K=5
library(class)
train.X4 <- as.matrix(Lag2[train])
test.X4 <- as.matrix((Lag2[!train]))
train.Direction4 <- Direction[train]
set.seed(1)
knn.pred4 <- knn(train.X4, test.X4, train.Direction4,k=2)
table(knn.pred4, Direction.2008)## Direction.2008
## knn.pred4 Down Up
## Down 19 27
## Up 24 34mean(knn.pred4==Direction.2008)## [1] 0.5096154#K=10
library(class)
train.X5 <- as.matrix(Lag2[train])
test.X5 <- as.matrix((Lag2[!train]))
train.Direction5 <- Direction[train]
set.seed(1)
knn.pred5 <- knn(train.X5, test.X5, train.Direction5,k=2)
table(knn.pred5, Direction.2008)## Direction.2008
## knn.pred5 Down Up
## Down 19 27
## Up 24 34mean(knn.pred5==Direction.2008)## [1] 0.5096154From the results above, we see that increasing K did not improve the prediction accuracy of the KNN classifer.
QUESTION 14
- In this problem, you will develop a model to predict whether a given car gets high or low gas mileage based on the Auto data set.
- Create a binary variable, mpg01, that contains a 1 if mpg contains a value above its median, and a 0 if mpg contains a value below its median. You can compute the median using the median() function. Note you may find it helpful to use the data.frame() function to create a single data set containing both mpg01 and the other Auto variables.
#Solution
attach(Auto)## The following object is masked from package:ggplot2:
##
## mpgAuto <- Auto|>mutate(mpg01=case_when(mpg>median(mpg)~1, mpg<median(mpg)~0))- Explore the data graphically in order to investigate the association between mpg01 and the other features. Which of the other features seem most likely to be useful in predicting mpg01? Scatterplots and boxplots may be useful tools to answer this question. Describe your findings.
cor(Auto[, -9])## mpg cylinders displacement horsepower weight
## mpg 1.0000000 -0.7776175 -0.8051269 -0.7784268 -0.8322442
## cylinders -0.7776175 1.0000000 0.9508233 0.8429834 0.8975273
## displacement -0.8051269 0.9508233 1.0000000 0.8972570 0.9329944
## horsepower -0.7784268 0.8429834 0.8972570 1.0000000 0.8645377
## weight -0.8322442 0.8975273 0.9329944 0.8645377 1.0000000
## acceleration 0.4233285 -0.5046834 -0.5438005 -0.6891955 -0.4168392
## year 0.5805410 -0.3456474 -0.3698552 -0.4163615 -0.3091199
## origin 0.5652088 -0.5689316 -0.6145351 -0.4551715 -0.5850054
## mpg01 0.8369392 -0.7591939 -0.7534766 -0.6670526 -0.7577566
## acceleration year origin mpg01
## mpg 0.4233285 0.5805410 0.5652088 0.8369392
## cylinders -0.5046834 -0.3456474 -0.5689316 -0.7591939
## displacement -0.5438005 -0.3698552 -0.6145351 -0.7534766
## horsepower -0.6891955 -0.4163615 -0.4551715 -0.6670526
## weight -0.4168392 -0.3091199 -0.5850054 -0.7577566
## acceleration 1.0000000 0.2903161 0.2127458 0.3468215
## year 0.2903161 1.0000000 0.1815277 0.4299042
## origin 0.2127458 0.1815277 1.0000000 0.5136984
## mpg01 0.3468215 0.4299042 0.5136984 1.0000000par(mfrow=c(2,2))
boxplot(cylinders~mpg01, data=Auto)
boxplot(displacement~mpg01, data=Auto)
boxplot(horsepower~mpg01, data=Auto)
boxplot(weight~mpg01, data=Auto)
boxplot(acceleration~mpg01, data=Auto)
boxplot(year~mpg01, data=Auto)
boxplot(origin~mpg01, data=Auto)
From the plot and pairwisw correlation matrix, There seem to be an
association between mpg01 and cylinders, displacement,
horsepower,weight,and origin.
- Split the data into a training set and a test set.
library(caTools)## Warning: package 'caTools' was built under R version 4.1.3set.seed(20)
split <- sample.split(Auto, SplitRatio = 0.8)
Autotrain <- subset(Auto, split==TRUE)
Autotest<- subset(Auto, split==FALSE)- Perform LDA on the training data in order to predict mpg01 using the variables that seemed most associated with mpg01 in (b). What is the test error of the model obtained?
library(MASS)
mpg01.lda <- lda(mpg01 ~ cylinders + weight + displacement + horsepower+origin, data = Autotrain)
mpg01.pred <- predict(mpg01.lda, Autotest)
table(mpg01.pred$class, Autotest$mpg01)##
## 0 1
## 0 36 4
## 1 4 34mean(mpg01.pred$class != Autotest$mpg01)## [1] 0.1025641The test error is 10.26%
- Perform QDA on the training data in order to predict mpg01 using the variables that seemed most associated with mpg01 in (b). What is the test error of the model obtained?
fit.qda <- qda(mpg01 ~ cylinders + weight + displacement + horsepower +origin, data = Autotrain)
pred.qda <- predict(fit.qda, Autotest)
table(pred.qda$class, Autotest$mpg01)##
## 0 1
## 0 37 7
## 1 3 31mean(pred.qda$class !=Autotest$mpg01 )## [1] 0.1282051The test error is 12.82%
- Perform logistic regression on the training data in order to predict mpg01 using the variables that seemed most associated with mpg01 in (b). What is the test error of the model obtained?
log.glm <- glm(mpg01 ~ cylinders + weight + displacement + horsepower+origin, data = Autotrain, family = binomial)
probs.log <- predict(log.glm, Autotest, type = "response")
pred.log <- rep(0, length(probs.log))
pred.log[probs.log > 0.5] <- 1
table(pred.log, Autotest$mpg01)##
## pred.log 0 1
## 0 38 4
## 1 2 34mean(pred.log != Autotest$mpg01)## [1] 0.07692308The test error is 7.69%
- Perform naive Bayes on the training data in order to predict mpg01 using the variables that seemed most associated with mpg01 in (b). What is the test error of the model obtained?
library(e1071)
nb.train <- naiveBayes(mpg01 ~ cylinders + weight + displacement + horsepower+origin, data = Autotrain)
class.train <- predict(nb.train, Autotest)
table(class.train, Autotest$mpg01)##
## class.train 0 1
## 0 36 3
## 1 4 35mean(class.train!=Autotest$mpg01)## [1] 0.08974359The test error is 8.97%
- Perform KNN on the training data, with several values of K, in order to predict mpg01. Use only the variables that seemed most associated with mpg01 in (b). What test errors do you obtain? Which value of K seems to perform the best on this data set?
SOLUTION
#STANDARDIZE THE DATA
library(caTools)
set.seed(2055)
standardized.X <- scale(Auto[,c(-1,-6,-7,-9,-10)])
standard.split<- sample(1:nrow(standardized.X), size=nrow(standardized.X)*0.8,replace=F)
standard.train <- standardized.X[standard.split,]
standard.test<- standardized.X[-standard.split,]
mpg09.train <- Auto[standard.split, 10]
mpg09.test <- Auto[-standard.split, 10]K Model
library(class)
set.seed(689)
knn.1 <- knn(standard.train,standard.test,mpg09.train,k=1)
knn.3 <- knn(standard.train, standard.test,mpg09.train,k=3)
knn.5 <- knn(standard.train, standard.test,mpg09.train,k=5)
knn.7 <- knn(standard.train, standard.test,mpg09.train,k=7)
knn.9 <- knn(standard.train, standard.test,mpg09.train,k=9)
knn.11 <- knn(standard.train, standard.test,mpg09.train,k=11)
knn.13 <- knn(standard.train, standard.test,mpg09.train,k=13)
knn.15 <- knn(standard.train, standard.test,mpg09.train,k=15)
knn.17 <- knn(standard.train, standard.test,mpg09.train,k=17)table (knn.1, mpg09.test)## mpg09.test
## knn.1 0 1
## 0 37 3
## 1 5 34table (knn.3, mpg09.test)## mpg09.test
## knn.3 0 1
## 0 36 3
## 1 6 34table (knn.5, mpg09.test)## mpg09.test
## knn.5 0 1
## 0 37 3
## 1 5 34table (knn.7, mpg09.test)## mpg09.test
## knn.7 0 1
## 0 36 2
## 1 6 35table (knn.9, mpg09.test)## mpg09.test
## knn.9 0 1
## 0 36 1
## 1 6 36table (knn.11, mpg09.test)## mpg09.test
## knn.11 0 1
## 0 36 1
## 1 6 36table (knn.13, mpg09.test)## mpg09.test
## knn.13 0 1
## 0 36 2
## 1 6 35table (knn.15, mpg09.test)## mpg09.test
## knn.15 0 1
## 0 36 2
## 1 6 35table (knn.17, mpg09.test)## mpg09.test
## knn.17 0 1
## 0 36 2
## 1 6 35TEST ERROR RATE
mean(knn.1!=mpg09.test)## [1] 0.1012658mean(knn.3!=mpg09.test)## [1] 0.1139241mean(knn.5!=mpg09.test)## [1] 0.1012658mean(knn.7!=mpg09.test)## [1] 0.1012658mean(knn.9!=mpg09.test)## [1] 0.08860759mean(knn.11!=mpg09.test)## [1] 0.08860759mean(knn.13!=mpg09.test)## [1] 0.1012658mean(knn.15!=mpg09.test)## [1] 0.1012658mean(knn.17!=mpg09.test)## [1] 0.1012658We see that the KNN classifier with k=9 and k=11 had the lowest test error rate of 8.86%
Exercise 16
Using the Boston data set, fit classification models in order to predict whether a given census tract has a crime rate above or below the median. Explore logistic regression, LDA, naive Bayes, and KNN models using various subsets of the predictors. Describe your findings. Hint: You will have to create the response variable yourself, using the variables that are contained in the Boston data set.
##Solution
attach(Boston)
Boston <- Boston|>mutate(crime=case_when(crim>median(crim)~1, crim<median(crim)~0))
cor(Boston[, ])## crim zn indus chas nox
## crim 1.00000000 -0.20046922 0.40658341 -0.055891582 0.42097171
## zn -0.20046922 1.00000000 -0.53382819 -0.042696719 -0.51660371
## indus 0.40658341 -0.53382819 1.00000000 0.062938027 0.76365145
## chas -0.05589158 -0.04269672 0.06293803 1.000000000 0.09120281
## nox 0.42097171 -0.51660371 0.76365145 0.091202807 1.00000000
## rm -0.21924670 0.31199059 -0.39167585 0.091251225 -0.30218819
## age 0.35273425 -0.56953734 0.64477851 0.086517774 0.73147010
## dis -0.37967009 0.66440822 -0.70802699 -0.099175780 -0.76923011
## rad 0.62550515 -0.31194783 0.59512927 -0.007368241 0.61144056
## tax 0.58276431 -0.31456332 0.72076018 -0.035586518 0.66802320
## ptratio 0.28994558 -0.39167855 0.38324756 -0.121515174 0.18893268
## black -0.38506394 0.17552032 -0.35697654 0.048788485 -0.38005064
## lstat 0.45562148 -0.41299457 0.60379972 -0.053929298 0.59087892
## medv -0.38830461 0.36044534 -0.48372516 0.175260177 -0.42732077
## crime 0.40939545 -0.43615103 0.60326017 0.070096774 0.72323480
## rm age dis rad tax ptratio
## crim -0.21924670 0.35273425 -0.37967009 0.625505145 0.58276431 0.2899456
## zn 0.31199059 -0.56953734 0.66440822 -0.311947826 -0.31456332 -0.3916785
## indus -0.39167585 0.64477851 -0.70802699 0.595129275 0.72076018 0.3832476
## chas 0.09125123 0.08651777 -0.09917578 -0.007368241 -0.03558652 -0.1215152
## nox -0.30218819 0.73147010 -0.76923011 0.611440563 0.66802320 0.1889327
## rm 1.00000000 -0.24026493 0.20524621 -0.209846668 -0.29204783 -0.3555015
## age -0.24026493 1.00000000 -0.74788054 0.456022452 0.50645559 0.2615150
## dis 0.20524621 -0.74788054 1.00000000 -0.494587930 -0.53443158 -0.2324705
## rad -0.20984667 0.45602245 -0.49458793 1.000000000 0.91022819 0.4647412
## tax -0.29204783 0.50645559 -0.53443158 0.910228189 1.00000000 0.4608530
## ptratio -0.35550149 0.26151501 -0.23247054 0.464741179 0.46085304 1.0000000
## black 0.12806864 -0.27353398 0.29151167 -0.444412816 -0.44180801 -0.1773833
## lstat -0.61380827 0.60233853 -0.49699583 0.488676335 0.54399341 0.3740443
## medv 0.69535995 -0.37695457 0.24992873 -0.381626231 -0.46853593 -0.5077867
## crime -0.15637178 0.61393992 -0.61634164 0.619786249 0.60874128 0.2535684
## black lstat medv crime
## crim -0.38506394 0.4556215 -0.3883046 0.40939545
## zn 0.17552032 -0.4129946 0.3604453 -0.43615103
## indus -0.35697654 0.6037997 -0.4837252 0.60326017
## chas 0.04878848 -0.0539293 0.1752602 0.07009677
## nox -0.38005064 0.5908789 -0.4273208 0.72323480
## rm 0.12806864 -0.6138083 0.6953599 -0.15637178
## age -0.27353398 0.6023385 -0.3769546 0.61393992
## dis 0.29151167 -0.4969958 0.2499287 -0.61634164
## rad -0.44441282 0.4886763 -0.3816262 0.61978625
## tax -0.44180801 0.5439934 -0.4685359 0.60874128
## ptratio -0.17738330 0.3740443 -0.5077867 0.25356836
## black 1.00000000 -0.3660869 0.3334608 -0.35121093
## lstat -0.36608690 1.0000000 -0.7376627 0.45326273
## medv 0.33346082 -0.7376627 1.0000000 -0.26301673
## crime -0.35121093 0.4532627 -0.2630167 1.00000000From the pairwise correlation matrix, we see there is an association between crime and the indus, nox, age, dis, rad, tax predictors. So our classifiers will be built based on those predictors.
Boston$crime <- as.factor(Boston$crime)Splitting the Data
library(caTools)
set.seed(27)
split1<- sample.split(Boston, SplitRatio = 0.8)
Bostontrain <- subset(Boston, split1==TRUE)
Bostontest<- subset(Boston, split1==FALSE)Using indus, nox, age, dis, rad, tax predictors
Logistic regression
Boston.glm <- glm(crime ~ indus+ nox + age +dis +rad +tax, data = Bostontrain, family = binomial)
probs.Boston<- predict(Boston.glm, Bostontest, type = "response")
pred.Boston<- rep(0, length(probs.Boston))
pred.Boston[probs.Boston> 0.5] <- 1
table(pred.Boston, Bostontest$crime)##
## pred.Boston 0 1
## 0 48 9
## 1 2 42mean(pred.Boston == Bostontest$crime)## [1] 0.8910891mean(pred.Boston != Bostontest$crime)## [1] 0.1089109LDA
library(MASS)
Boston.lda <- lda(crime ~ indus+ nox + age +dis +rad +tax, data = Bostontrain)
Boston.predict<- predict(Boston.lda, Bostontest)
table(Boston.predict$class, Bostontest$crime)##
## 0 1
## 0 48 14
## 1 2 37mean(Boston.predict$class==Bostontest$crime)## [1] 0.8415842mean(Boston.predict$class!=Bostontest$crime)## [1] 0.1584158Naive Bayes
library(e1071)
nb.Boston <- naiveBayes(crime ~ indus+ nox + age +dis +rad +tax, data = Bostontrain)
predict.Boston <- predict(nb.Boston, Bostontest)
table(predict.Boston, Bostontest$crime)##
## predict.Boston 0 1
## 0 48 13
## 1 2 38mean(predict.Boston==Bostontest$crime)## [1] 0.8514851mean(predict.Boston!=Bostontest$crime)## [1] 0.1485149KNN
SOLUTION
Boston1 <- Boston
Boston1$crime <- as.numeric(Boston1$crime)#STANDARDIZE THE DATA
library(caTools)
set.seed(2955)
standardized.K <- scale(Boston1[,c(3,5,7,8,9,10)])
standard.splits<- sample(1:nrow(standardized.K), size=nrow(standardized.K)*0.8,replace=F)
Boston.train <- standardized.K[standard.splits,]
Boston.test<- standardized.K[-standard.splits,]
crime.train <- Boston1[standard.splits, 15]
crime.test <- Boston1[-standard.splits, 15]K Model
library(class)
set.seed(689)
knna.1 <- knn(Boston.train,Boston.test,crime.train,k=1)
knnb.3 <- knn(Boston.train, Boston.test,crime.train,k=3)
knnc.5 <- knn(Boston.train, Boston.test,crime.train,k=5)
knnd.7 <- knn(Boston.train, Boston.test,crime.train,k=7)
knne.9 <- knn(Boston.train, Boston.test,crime.train,k=9)
knnf.11 <- knn(Boston.train, Boston.test,crime.train,k=11)
knng.13 <- knn(Boston.train, Boston.test,crime.train,k=13)
knnh.15 <- knn(Boston.train,Boston.test,crime.train,k=15)
knni.17 <- knn(Boston.train, Boston.test,crime.train,k=17)table (knna.1, crime.test)## crime.test
## knna.1 1 2
## 1 51 3
## 2 5 43table (knnb.3,crime.test)## crime.test
## knnb.3 1 2
## 1 52 3
## 2 4 43table (knnc.5, crime.test)## crime.test
## knnc.5 1 2
## 1 52 3
## 2 4 43table (knnd.7, crime.test)## crime.test
## knnd.7 1 2
## 1 52 4
## 2 4 42table (knne.9, crime.test)## crime.test
## knne.9 1 2
## 1 50 3
## 2 6 43table (knnf.11, crime.test)## crime.test
## knnf.11 1 2
## 1 50 3
## 2 6 43table (knng.13, crime.test)## crime.test
## knng.13 1 2
## 1 50 3
## 2 6 43table (knnh.15, crime.test)## crime.test
## knnh.15 1 2
## 1 49 3
## 2 7 43table (knni.17, crime.test)## crime.test
## knni.17 1 2
## 1 49 3
## 2 7 43TEST ERROR RATE
mean(knna.1!=crime.test)## [1] 0.07843137mean(knnb.3!=crime.test)## [1] 0.06862745mean(knnc.5!=crime.test)## [1] 0.06862745mean(knnd.7!=crime.test)## [1] 0.07843137mean(knne.9!=crime.test)## [1] 0.08823529mean(knnf.11!=crime.test)## [1] 0.08823529mean(knng.13!=crime.test)## [1] 0.08823529mean(knnh.15!=crime.test)## [1] 0.09803922mean(knni.17!=crime.test)## [1] 0.09803922FINDINGS
After building the classifiers from the training dataset, their performance was assessed using the confusion matrix and the corresponding test error rate. From the output, we see that the KNN classifiers performed the best as they have the smallest error rate, and the naives bayes classifier performed the worst.
We also see that the KNN classifier with K=3 and K=5 performed the best.