Assignment #3

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.

library(ISLR2)
library(MASS)

## 
## Attaching package: 'MASS'

## The following object is masked from 'package:ISLR2':
## 
##     Boston

library(corrplot)

## corrplot 0.92 loaded

library(class)
library(e1071)

(a) Produce some numerical and graphical summaries of the Weekly data. Do there appear to be any patterns?

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  
##            
##            
##            
## 

plot(Weekly)

(b) 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?

“Yes, the intercept and lag 2 are statistically significant, due to the fact that they have a p-value < 0.05, therefore less significant.”

attach(Weekly)

Weekly.fit<-glm(Direction~Lag1 + Lag2 + Lag3 + Lag4 + Lag5 + Volume, data = Weekly, family = binomial)
summary(Weekly.fit)

## 
## 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: 4

(c) 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.

“The model here tends to predict market Up trends better, so on average, the responses are shown to be predicted correctly 56% of the time."

Weekly.probs = predict(Weekly.fit, type = "response")
Weekly.pred = rep("Down", length(Weekly.probs)) 
Weekly.pred[Weekly.probs > 0.5] <- "Up"

table(Weekly.pred, Weekly$Direction)

##            
## Weekly.pred Down  Up
##        Down   54  48
##        Up    430 557

mean(Weekly.pred == Weekly$Direction)

## [1] 0.5610652

(d) 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).

train = (Weekly$Year < 2009)
Weekly.test <- Weekly[!train,]
Weekly.fit <- glm(Direction~Lag2, data = Weekly,family = binomial, subset = train)
logWeekly.prob= predict(Weekly.fit, Weekly.test, type = "response")
logWeekly.pred = rep("Down", length(logWeekly.prob))
logWeekly.pred[logWeekly.prob > 0.5] = "Up"
Direction.test = Direction[!train]
table(logWeekly.pred, Direction.test)

##               Direction.test
## logWeekly.pred Down Up
##           Down    9  5
##           Up     34 56

mean(logWeekly.pred == Direction.test)

## [1] 0.625

(e) Repeat (d) using LDA.

Weekly.LDA.fit <- lda(Direction~Lag2, data = Weekly,family = binomial, subset = train)
Weekly.LDA.pred <- predict(Weekly.LDA.fit, Weekly.test)
table(Weekly.LDA.pred$class, Direction.test)

##       Direction.test
##        Down Up
##   Down    9  5
##   Up     34 56

mean(Weekly.LDA.pred$class == Direction.test)

## [1] 0.625

(f) Repeat (d) using QDA.

Weekly.QDA.fit = qda(Direction ~ Lag2, data = Weekly, subset = train)
Weekly.QDA.pred = predict(Weekly.QDA.fit, Weekly.test)$class
table(Weekly.QDA.pred, Direction.test)

##                Direction.test
## Weekly.QDA.pred Down Up
##            Down    0  0
##            Up     43 61

mean(Weekly.QDA.pred==Direction.test)

## [1] 0.5865385

(g) Repeat (d) using KNN with K = 1.

Week.train=as.matrix(Lag2[train])
Week.test=as.matrix(Lag2[!train])
train.Direction =Direction[train]

set.seed(1)

Week.KNN.pred=knn(Week.train,Week.test,train.Direction,k=1)

table(Week.KNN.pred,Direction.test)

##              Direction.test
## Week.KNN.pred Down Up
##          Down   21 30
##          Up     22 31

(h) Repeat (d) using naive Bayes.

Weekly.NaiveBayes.fit = naiveBayes(Direction~Lag2, data = Weekly, subset = train)
Weekly.NaiveBayes.pred = predict(Weekly.NaiveBayes.fit,Weekly.test)
table(Weekly.NaiveBayes.pred, Direction.test)

##                       Direction.test
## Weekly.NaiveBayes.pred Down Up
##                   Down    0  0
##                   Up     43 61

mean(Weekly.NaiveBayes.pred==Direction.test)

## [1] 0.5865385

(i) Which of these methods appears to provide the best results on this data?

"In this specific case, Logistic Regression and Linear Discriminant Analysis gave the best result due to the 62.5% of the observations being correctly classified. On the other hand, the KKN method gave the worst results due to the 50% of the observations being correctly classified."

(j) 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 held out data. Note that you should also experiment with values for K in the KNN classifier.

Weekly.fit <- glm(Direction~Lag2:Lag5+Lag2, data = Weekly,family = binomial, subset = train)
logWeekly.prob= predict(Weekly.fit, Weekly.test, type = "response")
logWeekly.pred = rep("Down", length(logWeekly.prob))
logWeekly.pred[logWeekly.prob > 0.5] = "Up"
Direction.test = Direction[!train]
table(logWeekly.pred, Direction.test)

##               Direction.test
## logWeekly.pred Down Up
##           Down    9  4
##           Up     34 57

mean(logWeekly.pred == Direction.test)

## [1] 0.6346154

LDA [63%]

Weekly.LDA.fit <- lda(Direction~Lag2:Lag5+Lag2, data = Weekly,family = binomial, subset = train)
Weekly.LDA.pred <- predict(Weekly.LDA.fit, Weekly.test)
table(Weekly.LDA.pred$class, Direction.test)

##       Direction.test
##        Down Up
##   Down    9  4
##   Up     34 57

mean(Weekly.LDA.pred$class == Direction.test)

## [1] 0.6346154

QDA [50%]

Weekly.QDA.fit = qda(Direction ~ Lag2:Lag5+Lag2, data = Weekly, subset = train)
Weekly.QDA.pred = predict(Weekly.QDA.fit, Weekly.test)$class
table(Weekly.QDA.pred, Direction.test)

##                Direction.test
## Weekly.QDA.pred Down Up
##            Down    5 13
##            Up     38 48

mean(Weekly.QDA.pred==Direction.test)

## [1] 0.5096154

KNN

Week.train=as.matrix(Lag2[train])
Week.test=as.matrix(Lag2[!train])
train.Direction =Direction[train]

set.seed(1)

Week.KNN.pred=knn(Week.train,Week.test,train.Direction,k=15)
table(Week.KNN.pred,Direction.test)

##              Direction.test
## Week.KNN.pred Down Up
##          Down   20 20
##          Up     23 41

Naive Bayes [48%]

Weekly.NaiveBayes.fit = naiveBayes(Direction~Lag5, data = Weekly, subset = train)
Weekly.NaiveBayes.pred = predict(Weekly.NaiveBayes.fit,Weekly.test)
table(Weekly.NaiveBayes.pred, Direction.test)

##                       Direction.test
## Weekly.NaiveBayes.pred Down Up
##                   Down    2 13
##                   Up     41 48

mean(Weekly.NaiveBayes.pred==Direction.test)

## [1] 0.4807692

detach(Weekly)

"In all of these instances, LDA had the best prediction rate of 63% after testing all different methods."

---

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.

library(ISLR)

## Warning: package 'ISLR' was built under R version 4.2.3

## 
## Attaching package: 'ISLR'

## The following objects are masked from 'package:ISLR2':
## 
##     Auto, Credit

library(tibble)
library(dplyr)

## 
## Attaching package: 'dplyr'

## The following object is masked from 'package:MASS':
## 
##     select

## The following objects are masked from 'package:stats':
## 
##     filter, lag

## The following objects are masked from 'package:base':
## 
##     intersect, setdiff, setequal, union

(a) 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.

attach(Auto)
summary(Auto)

##       mpg          cylinders      displacement     horsepower        weight    
##  Min.   : 9.00   Min.   :3.000   Min.   : 68.0   Min.   : 46.0   Min.   :1613  
##  1st Qu.:17.00   1st Qu.:4.000   1st Qu.:105.0   1st Qu.: 75.0   1st Qu.:2225  
##  Median :22.75   Median :4.000   Median :151.0   Median : 93.5   Median :2804  
##  Mean   :23.45   Mean   :5.472   Mean   :194.4   Mean   :104.5   Mean   :2978  
##  3rd Qu.:29.00   3rd Qu.:8.000   3rd Qu.:275.8   3rd Qu.:126.0   3rd Qu.:3615  
##  Max.   :46.60   Max.   :8.000   Max.   :455.0   Max.   :230.0   Max.   :5140  
##                                                                                
##   acceleration        year           origin                      name    
##  Min.   : 8.00   Min.   :70.00   Min.   :1.000   amc matador       :  5  
##  1st Qu.:13.78   1st Qu.:73.00   1st Qu.:1.000   ford pinto        :  5  
##  Median :15.50   Median :76.00   Median :1.000   toyota corolla    :  5  
##  Mean   :15.54   Mean   :75.98   Mean   :1.577   amc gremlin       :  4  
##  3rd Qu.:17.02   3rd Qu.:79.00   3rd Qu.:2.000   amc hornet        :  4  
##  Max.   :24.80   Max.   :82.00   Max.   :3.000   chevrolet chevette:  4  
##                                                  (Other)           :365

Auto <- as_tibble(ISLR::Auto)
mpg01 <- Auto %>% 
  mutate(mpg01 = ifelse(mpg > median(mpg), 1, 0))

(b) 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

corrplot(cor(Auto[,-9]), method="circle")

(c) Split the data into a training set and a test set.

train <- (mpg01$year %% 2 == 0)
train.auto <- mpg01[train,]
test.auto <- mpg01[-train,]

(d) 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?

auto.LDA.fit <- lda(mpg01~displacement+horsepower+weight+year+cylinders+origin, data=train.auto)
auto.LDA.pred <- predict(auto.LDA.fit, test.auto)
table(auto.LDA.pred$class, test.auto$mpg01)

##    
##       0   1
##   0 169   7
##   1  26 189

mean(auto.LDA.pred$class != test.auto$mpg01)

## [1] 0.08439898

(e) 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?

"The test error is 9.97%"

auto.QDA.fit <- qda(mpg01~displacement+horsepower+weight+year+cylinders+origin, data=train.auto)
auto.QDA.pred <- predict(auto.QDA.fit, test.auto)
table(auto.QDA.pred$class, test.auto$mpg01)

##    
##       0   1
##   0 176  20
##   1  19 176

mean(auto.QDA.pred$class != test.auto$mpg01)

## [1] 0.09974425

(f) 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?

"The test error is 8.4%"

auto.fit<-glm(mpg01~displacement+horsepower+weight+year+cylinders+origin, data=train.auto,family=binomial)
auto.probs = predict(auto.fit, test.auto, type = "response")
auto.pred = rep(0, length(auto.probs))
auto.pred[auto.probs > 0.5] = 1
table(auto.pred, test.auto$mpg01)

##          
## auto.pred   0   1
##         0 174  12
##         1  21 184

mean(auto.pred != test.auto$mpg01)

## [1] 0.08439898

(g) 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?

"The test error is 9.97%"

auto.NaiveBayes.fit <- naiveBayes(mpg01~displacement+horsepower+weight+year+cylinders+origin, data=train.auto)
auto.NaiveBayes.pred <- predict(auto.NaiveBayes.fit, test.auto)
table(auto.NaiveBayes.pred, test.auto$mpg01)

##                     
## auto.NaiveBayes.pred   0   1
##                    0 171  15
##                    1  24 181

mean(auto.NaiveBayes.pred != test.auto$mpg01)

## [1] 0.09974425

(h) 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?

The test errors are:
K1 = 7%
K4 = 13%
K8 = 12%

attach(Auto)

## The following objects are masked from Auto (pos = 3):
## 
##     acceleration, cylinders, displacement, horsepower, mpg, name,
##     origin, weight, year

train.K= cbind(displacement,horsepower,weight,cylinders)[train,]
test.K=cbind(displacement,horsepower,weight,cylinders)[-train,]

set.seed(1)

autok.pred=knn(train.K,test.K,train.auto$mpg01,k=1)
mean(autok.pred != test.auto$mpg01)

## [1] 0.07161125

autok.pred=knn(train.K,test.K,train.auto$mpg01,k=4)
mean(autok.pred != test.auto$mpg01)

## [1] 0.1278772

autok.pred=knn(train.K,test.K,train.auto$mpg01,k=8)
mean(autok.pred != test.auto$mpg01)

## [1] 0.1227621

detach(Auto)

---

Question 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.

summary(Boston)

##       crim                zn             indus            chas        
##  Min.   : 0.00632   Min.   :  0.00   Min.   : 0.46   Min.   :0.00000  
##  1st Qu.: 0.08205   1st Qu.:  0.00   1st Qu.: 5.19   1st Qu.:0.00000  
##  Median : 0.25651   Median :  0.00   Median : 9.69   Median :0.00000  
##  Mean   : 3.61352   Mean   : 11.36   Mean   :11.14   Mean   :0.06917  
##  3rd Qu.: 3.67708   3rd Qu.: 12.50   3rd Qu.:18.10   3rd Qu.:0.00000  
##  Max.   :88.97620   Max.   :100.00   Max.   :27.74   Max.   :1.00000  
##       nox               rm             age              dis        
##  Min.   :0.3850   Min.   :3.561   Min.   :  2.90   Min.   : 1.130  
##  1st Qu.:0.4490   1st Qu.:5.886   1st Qu.: 45.02   1st Qu.: 2.100  
##  Median :0.5380   Median :6.208   Median : 77.50   Median : 3.207  
##  Mean   :0.5547   Mean   :6.285   Mean   : 68.57   Mean   : 3.795  
##  3rd Qu.:0.6240   3rd Qu.:6.623   3rd Qu.: 94.08   3rd Qu.: 5.188  
##  Max.   :0.8710   Max.   :8.780   Max.   :100.00   Max.   :12.127  
##       rad              tax           ptratio          black       
##  Min.   : 1.000   Min.   :187.0   Min.   :12.60   Min.   :  0.32  
##  1st Qu.: 4.000   1st Qu.:279.0   1st Qu.:17.40   1st Qu.:375.38  
##  Median : 5.000   Median :330.0   Median :19.05   Median :391.44  
##  Mean   : 9.549   Mean   :408.2   Mean   :18.46   Mean   :356.67  
##  3rd Qu.:24.000   3rd Qu.:666.0   3rd Qu.:20.20   3rd Qu.:396.23  
##  Max.   :24.000   Max.   :711.0   Max.   :22.00   Max.   :396.90  
##      lstat            medv      
##  Min.   : 1.73   Min.   : 5.00  
##  1st Qu.: 6.95   1st Qu.:17.02  
##  Median :11.36   Median :21.20  
##  Mean   :12.65   Mean   :22.53  
##  3rd Qu.:16.95   3rd Qu.:25.00  
##  Max.   :37.97   Max.   :50.00

attach(Boston)

Binary Crim Variable

Crime1 <- rep(0, length(crim))
Crime1[crim > median(crim)] <- 1
Boston= data.frame(Boston,Crime1)

train = 1:(dim(Boston)[1]/2)
test = (dim(Boston)[1]/2 + 1):dim(Boston)[1]
Boston.train = Boston[train, ]
Boston.test = Boston[test, ]
Crime1.test = Crime1[test]

corrplot(cor(Boston), method="circle")

Logistic Regression “The test error is 9.5%”

set.seed(1)

Boston.fit <-glm(Crime1~ indus+nox+age+rad+tax, data=Boston.train,family=binomial)
Boston.probs = predict(Boston.fit, Boston.test, type = "response")
Boston.pred = rep(0, length(Boston.probs))
Boston.pred[Boston.probs > 0.5] = 1

table(Boston.pred, Crime1.test)

##            Crime1.test
## Boston.pred   0   1
##           0  72   6
##           1  18 157

mean(Boston.pred != Crime1.test)

## [1] 0.09486166

LDA “The test error is 10.7%”

Boston.LDA.fit <-lda(Crime1~ indus+nox+age+rad+tax, data=Boston.train,family=binomial)
Boston.LDA.pred = predict(Boston.LDA.fit, Boston.test)
table(Boston.LDA.pred$class, Crime1.test)

##    Crime1.test
##       0   1
##   0  81  18
##   1   9 145

mean(Boston.LDA.pred$class != Crime1.test)

## [1] 0.1067194

Naive Bayes “The test error is 10.7%”

Boston.NaiveBayes.fit <- naiveBayes(Crime1~ indus+nox+age+rad+tax, data=Boston.train)
Boston.NaiveBayes.pred <- predict(Boston.NaiveBayes.fit, Boston.test)
table(Boston.NaiveBayes.pred, Boston.test$Crime1)

##                       
## Boston.NaiveBayes.pred   0   1
##                      0  81  18
##                      1   9 145

mean(Boston.NaiveBayes.pred != Boston.test$Crime1)

## [1] 0.1067194

KNN “The test error is 84.6%”

train.K=cbind(indus,nox,age,rad,tax)[train,]
test.K=cbind(indus,nox,age,rad,tax)[test,]
Bosknn.pred=knn(train.K, test.K, Crime1.test, k=1)

table(Bosknn.pred,Crime1.test)

##            Crime1.test
## Bosknn.pred   0   1
##           0  31 155
##           1  59   8

mean(Bosknn.pred != Crime1.test)

## [1] 0.8458498

KNN 2 “The test error is 25.7%”

train.K=cbind(indus,nox,age,rad,tax)[train,]
test.K=cbind(indus,nox,age,rad,tax)[test,]
Bosknn.pred=knn(train.K, test.K, Crime1.test, k=50)

table(Bosknn.pred,Crime1.test)

##            Crime1.test
## Bosknn.pred   0   1
##           0  38  13
##           1  52 150

mean(Bosknn.pred != Crime1.test)

## [1] 0.256917

…