Question 10

This question should be answered using the Weekly data set, which is part of the ISLR 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.

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

    Year and Volume appear to have a strong relationship with a positive correlation of 0.84. Based on the scatterplot below, the number of shares traded has significantly grown over time. The rest of the variables have very weak correlations with coefficients close to zero, indicating there are no other relationships among the variables in the data set.

library(ISLR)
attach(Weekly)
library(GGally)
ggpairs(Weekly, ggplot2::aes(color = Direction),
        upper = list(combo = wrap("box_no_facet", alpha=0.5), continuous = wrap("points", alpha=0.5)),
        diag = list(discrete= wrap("barDiag", alpha=0.7), continuous = wrap("densityDiag", alpha=0.5)),
        lower = list(combo = wrap("box_no_facet", alpha=0.5),continuous = wrap("cor", size=4)))

  1. 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?

    Lag 2 has a significant positive association with Direction (p-value = 0.03).

glm.fit <- glm(Direction ~ Lag1 + Lag2 + Lag3 + Lag4 + Lag5 + Volume, 
               data = Weekly, family = binomial)
summary(glm.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
  1. 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.

    Our logistic regression model has an accuracy of 56.1%. The sensitivity is quite high at 92.1%, which is the percentage of correct predictions that the stock market is up. However, the specificity of the model is much lower, which is the percentage of correct predictions that the stock market is down, at 11.2%.

glm.prob <- predict(glm.fit, type = "response")
glm.pred <- ifelse(glm.prob > 0.5, 1, 0)
Weekly$Direction01 <- ifelse(Weekly$Direction == "Up", 1, 0)
caret::confusionMatrix(as_factor(glm.pred), as_factor(Weekly$Direction01), positive = '1')
## Confusion Matrix and Statistics
## 
##           Reference
## Prediction   0   1
##          0  54  48
##          1 430 557
##                                          
##                Accuracy : 0.5611         
##                  95% CI : (0.531, 0.5908)
##     No Information Rate : 0.5556         
##     P-Value [Acc > NIR] : 0.369          
##                                          
##                   Kappa : 0.035          
##                                          
##  Mcnemar's Test P-Value : <2e-16         
##                                          
##             Sensitivity : 0.9207         
##             Specificity : 0.1116         
##          Pos Pred Value : 0.5643         
##          Neg Pred Value : 0.5294         
##              Prevalence : 0.5556         
##          Detection Rate : 0.5115         
##    Detection Prevalence : 0.9063         
##       Balanced Accuracy : 0.5161         
##                                          
##        'Positive' Class : 1              
## 
  1. Now fit the logistic regression model using a training data period from 1990 to 2008, with Lag2 as the only predictor. Compute the confusion matrix and the overall fraction of correct predictions for the held out data (that is, the data from 2009 and 2010).
library(tidyverse)
Weekly.train <- Weekly %>% 
  filter(Year < 2009)
dim(Weekly.train)
## [1] 985  10
Weekly.test <- Weekly %>% 
  filter(Year >= 2009)
dim(Weekly.test)
## [1] 104  10
glm.fit <- glm(Direction ~ Lag2, data = Weekly.train, family = binomial)
glm.probs <- predict(glm.fit, Weekly.test, type = "response")
glm.pred <- ifelse(glm.probs > 0.5, "Up", "Down")
direction.test <- Weekly.test$Direction
table(glm.pred, direction.test)
##         direction.test
## glm.pred Down Up
##     Down    9  5
##     Up     34 56
mean(glm.pred == direction.test)
## [1] 0.625
  1. Repeat (d) using LDA.
lda.fit <- lda(Direction ~ Lag2, data = Weekly.train)
lda.pred <- predict(lda.fit, Weekly.test)
table(lda.pred$class, direction.test)
##       direction.test
##        Down Up
##   Down    9  5
##   Up     34 56
mean(lda.pred$class == direction.test)
## [1] 0.625
  1. Repeat (d) using QDA.
qda.fit <- qda(Direction ~ Lag2, data = Weekly.train)
qda.class <- predict(qda.fit, Weekly.test)$class
table(qda.class, direction.test)
##          direction.test
## qda.class Down Up
##      Down    0  0
##      Up     43 61
mean(qda.class == direction.test)
## [1] 0.5865385
  1. Repeat (d) using KNN with K = 1.
train <- Weekly.train[c("Lag2")]
test <- Weekly.test[c("Lag2")]
cl <- Weekly.train$Direction
knn.pred <- knn(train, test, cl, k=1)
table(knn.pred, direction.test)
##         direction.test
## knn.pred Down Up
##     Down   21 30
##     Up     22 31
mean(knn.pred == direction.test)
## [1] 0.5
  1. Which of these methods appears to provide the best results on this data?

    Logistic regression and LDA give us the best results with 62.5% accuracy.

  2. 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.

# Logistic regression with interaction
glm.fit <- glm(Direction ~ Lag2:Volume, data = Weekly.train, family = binomial)
glm.probs <- predict(glm.fit, Weekly.test, type = "response")
glm.pred <- ifelse(glm.probs > 0.5, "Up", "Down")
table(glm.pred, direction.test)
##         direction.test
## glm.pred Down Up
##     Down    9  6
##     Up     34 55
mean(glm.pred == direction.test)
## [1] 0.6153846
# LDA with interactions
lda.fit <- lda(Direction ~ Lag2:Volume + Lag2:Lag3, data = Weekly.train)
lda.class <- predict(lda.fit, Weekly.test)$class
table(lda.class, direction.test)
##          direction.test
## lda.class Down Up
##      Down   10  6
##      Up     33 55
mean(lda.class == direction.test)
## [1] 0.625
# QDA with transformations
qda.fit <- qda(Direction ~ sqrt(abs(Lag1)) + log(abs(Lag2)), data = Weekly.train)
qda.class <- predict(qda.fit, Weekly.test)$class
table(qda.class, direction.test)
##          direction.test
## qda.class Down Up
##      Down    1  0
##      Up     42 61
mean(qda.class == direction.test)
## [1] 0.5961538
# KNN k=3
knn.pred <- knn(train, test, cl, k=3)
table(knn.pred, direction.test)
##         direction.test
## knn.pred Down Up
##     Down   16 20
##     Up     27 41
mean(knn.pred == direction.test)
## [1] 0.5480769
# KNN k=5
knn.pred <- knn(train, test, cl, k=5)
table(knn.pred, direction.test)
##         direction.test
## knn.pred Down Up
##     Down   16 20
##     Up     27 41
mean(knn.pred == direction.test)
## [1] 0.5480769
# KNN k=10
knn.pred <- knn(train, test, cl, k=10)
table(knn.pred, direction.test)
##         direction.test
## knn.pred Down Up
##     Down   17 20
##     Up     26 41
mean(knn.pred == direction.test)
## [1] 0.5576923
# KNN k=25
knn.pred <- knn(train, test, cl, k=25)
table(knn.pred, direction.test)
##         direction.test
## knn.pred Down Up
##     Down   19 23
##     Up     24 38
mean(knn.pred == direction.test)
## [1] 0.5480769

After trying different combinations of predictors, the LDA model with Lag2 interactions between Volume and Lag3 achieved the highest accuracy of 62.5%, which is the same as the original LDA and logistic regression models.

detach(Weekly)

Question 11

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.

  1. 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)
Auto <- Auto %>% 
  mutate(mpg01 = as_factor(ifelse(mpg > median(mpg), 1, 0)))
head(Auto[,c(1,ncol(Auto))])
##   mpg mpg01
## 1  18     0
## 2  15     0
## 3  18     0
## 4  16     0
## 5  17     0
## 6  15     0
  1. 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.

    The variables most associated with mpg01 (correlations above 0.6) are cylinders, displacement, horsepower, weight, and mpg itself.

Auto %>% 
  select(-name) %>% 
  ggpairs(ggplot2::aes(color = mpg01), 
        diag = list(continuous = wrap("densityDiag", alpha=0.5)),
        lower = list(continuous = wrap("points", alpha=0.5)))

  1. Split the data into a training set and a test set.
set.seed(1)
trainInd <- sample(1:nrow(Auto), 0.75*nrow(Auto)) 
Auto.train <- Auto[trainInd, ]
Auto.test <- Auto[-trainInd, ]
mpg01.test <- Auto.test$mpg01
  1. Perform LDA on the training data in order to predict mpg01 using the variables that seemed most associated with mpg01 in (b). What is the test error of the model obtained?
library(MASS)
lda.fit <- lda(mpg01 ~ cylinders + displacement + horsepower + weight, data = Auto.train)
lda.pred <- predict(lda.fit, Auto.test)
mean(lda.pred$class != mpg01.test)
## [1] 0.1326531
  1. 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?
qda.fit <- qda(mpg01 ~ cylinders + displacement + horsepower + weight, data = Auto.train)
qda.pred <- predict(qda.fit, Auto.test)
mean(qda.pred$class != mpg01.test)
## [1] 0.122449
  1. 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?
glm.fit <- glm(mpg01 ~ cylinders + displacement + horsepower + weight, 
               data = Auto.train, family = binomial)
glm.probs <- predict(glm.fit, Auto.test, type = "response")
glm.pred <- ifelse(glm.probs > 0.5, 1, 0)
mean(glm.pred != mpg01.test)
## [1] 0.1020408
  1. 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 KNN model with 4 nearest neighbors has the lowest test error of 10.2%, which is the same error rate as the logistic model.

train <- Auto.train[c(2:5)]
test <- Auto.test[c(2:5)]
cl <- Auto.train$mpg01
knn.pred <- knn(train, test, cl, k=3)
mean(knn.pred != mpg01.test)
## [1] 0.122449
knn.pred <- knn(train, test, cl, k=4)
mean(knn.pred != mpg01.test)
## [1] 0.1020408
knn.pred <- knn(train, test, cl, k=5)
mean(knn.pred != mpg01.test)
## [1] 0.1326531
knn.pred <- knn(train, test, cl, k=10)
mean(knn.pred != mpg01.test)
## [1] 0.1530612
knn.pred <- knn(train, test, cl, k=25)
mean(knn.pred != mpg01.test)
## [1] 0.1428571
detach(Auto)

Question 13

Using the Boston data set, fit classification models in order to predict whether a given suburb has a crime rate above or below the median. Explore logistic regression, LDA, and KNN models using various subsets of the predictors. Describe your findings.

library(MASS)
attach(Boston)
Boston <- Boston %>% 
  mutate(crime01 = ifelse(crim > median(crim), 1, 0))
head(Boston[,c(1,ncol(Boston))])
##      crim crime01
## 1 0.00632       0
## 2 0.02731       0
## 3 0.02729       0
## 4 0.03237       0
## 5 0.06905       0
## 6 0.02985       0
library(ggcorrplot)
ggcorrplot(cor(Boston), type = "upper", lab=TRUE, hc.order = TRUE, 
           colors = rev(colorspace::diverge_hsv(n=3)))

Boston$crime01 <- as_factor(Boston$crime01)
pairs(Boston)

set.seed(2)
trainInd <- sample(1:nrow(Boston), 0.8*nrow(Boston))
Boston.train <- Boston[trainInd, ]
Boston.test <- Boston[-trainInd, ]
crime01.test <- Boston.test$crime01

Logistic Regression

glm.fit <- glm(crime01 ~ . -crim, data = Boston.train, family = binomial)
summary(glm.fit)
## 
## Call:
## glm(formula = crime01 ~ . - crim, family = binomial, data = Boston.train)
## 
## Deviance Residuals: 
##     Min       1Q   Median       3Q      Max  
## -1.9239  -0.1979  -0.0061   0.0020   3.2468  
## 
## Coefficients:
##               Estimate Std. Error z value Pr(>|z|)    
## (Intercept) -31.951406   6.736023  -4.743 2.10e-06 ***
## zn           -0.060678   0.033100  -1.833   0.0668 .  
## indus        -0.040889   0.048136  -0.849   0.3956    
## chas          0.560825   0.819833   0.684   0.4939    
## nox          42.436265   7.707026   5.506 3.67e-08 ***
## rm           -0.237023   0.794074  -0.298   0.7653    
## age           0.015697   0.013222   1.187   0.2351    
## dis           0.551900   0.227557   2.425   0.0153 *  
## rad           0.657003   0.167022   3.934 8.37e-05 ***
## tax          -0.005635   0.002821  -1.998   0.0457 *  
## ptratio       0.318066   0.133227   2.387   0.0170 *  
## black        -0.009705   0.006065  -1.600   0.1095    
## lstat         0.068844   0.053456   1.288   0.1978    
## medv          0.143262   0.072217   1.984   0.0473 *  
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## (Dispersion parameter for binomial family taken to be 1)
## 
##     Null deviance: 558.64  on 403  degrees of freedom
## Residual deviance: 169.55  on 390  degrees of freedom
## AIC: 197.55
## 
## Number of Fisher Scoring iterations: 9
glm.probs <- predict(glm.fit, Boston.test, type = "response")
glm.pred <- ifelse(glm.probs > 0.5, 1, 0)
mean(glm.pred == crime01.test)
## [1] 0.9117647
# Logistic regression using variables with correlations above 0.6
glm.fit <- glm(crime01 ~ rad + tax + age + indus + nox + dis, data = Boston.train, family = binomial)
glm.probs <- predict(glm.fit, Boston.test, type = "response")
glm.pred <- ifelse(glm.probs > 0.5, 1, 0)
mean(glm.pred == crime01.test)
## [1] 0.8137255

Linear Discriminant Analysis

# LDA full model
lda.fit <- lda(crime01 ~ . -crim, data = Boston.train)
lda.pred <- predict(lda.fit, Boston.test)
mean(lda.pred$class == crime01.test)
## [1] 0.7941176
# LDA with correlated variables
lda.fit <- lda(crime01 ~ rad + tax + age + indus + nox + dis, data = Boston.train)
lda.pred <- predict(lda.fit, Boston.test)
mean(lda.pred$class == crime01.test)
## [1] 0.7843137
# LDA using significant variables from logistic model
lda.fit <- lda(crime01 ~ nox + dis + rad + tax + ptratio + medv, data = Boston.train)
lda.pred <- predict(lda.fit, Boston.test)
mean(lda.pred$class == crime01.test)
## [1] 0.7647059

Quadratic Discriminant Analysis

# QDA full model
qda.fit <- qda(crime01 ~ . -crim, data = Boston.train)
qda.pred <- predict(qda.fit, Boston.test)
mean(qda.pred$class == crime01.test)
## [1] 0.8627451
# QDA with correlated variables
qda.fit <- qda(crime01 ~ rad + tax + age + indus + nox + dis, data = Boston.train)
qda.pred <- predict(qda.fit, Boston.test)
mean(qda.pred$class == crime01.test)
## [1] 0.8333333
# QDA using significant variables
qda.fit <- qda(crime01 ~ nox + dis + rad + tax + ptratio + medv, data = Boston.train)
qda.pred <- predict(qda.fit, Boston.test)
mean(qda.pred$class == crime01.test)
## [1] 0.8235294

K-Nearest Neighbors

# KNN using significant variables
train <- Boston.train[c("nox", "dis", "rad", "tax", "ptratio", "medv")]
test <- Boston.test[c("nox", "dis", "rad", "tax", "ptratio", "medv")]
cl <- Boston.train$crime01
knn.pred <- knn(train, test, cl, k=2)
mean(knn.pred == crime01.test)
## [1] 0.9509804
knn.pred <- knn(train, test, cl, k=3)
mean(knn.pred == crime01.test)
## [1] 0.9705882
confusionMatrix(as_factor(knn.pred), as_factor(crime01.test), positive = '1')
## Confusion Matrix and Statistics
## 
##           Reference
## Prediction  0  1
##          0 38  2
##          1  1 61
##                                           
##                Accuracy : 0.9706          
##                  95% CI : (0.9164, 0.9939)
##     No Information Rate : 0.6176          
##     P-Value [Acc > NIR] : <2e-16          
##                                           
##                   Kappa : 0.938           
##                                           
##  Mcnemar's Test P-Value : 1               
##                                           
##             Sensitivity : 0.9683          
##             Specificity : 0.9744          
##          Pos Pred Value : 0.9839          
##          Neg Pred Value : 0.9500          
##              Prevalence : 0.6176          
##          Detection Rate : 0.5980          
##    Detection Prevalence : 0.6078          
##       Balanced Accuracy : 0.9713          
##                                           
##        'Positive' Class : 1               
## 
knn.pred <- knn(train, test, cl, k=10)
mean(knn.pred == crime01.test)
## [1] 0.9313725
knn.pred <- knn(train, test, cl, k=15)
mean(knn.pred == crime01.test)
## [1] 0.9411765
knn.pred <- knn(train, test, cl, k=25)
mean(knn.pred == crime01.test)
## [1] 0.8921569

The KNN classifier had the best performance on the Boston data set out of all the models. The KNN with 3 nearest neighbors achieved the highest accuracy in predicting whether a suburb has a crime rate above the median at 97.1%, with 96.8% sensitivity and 97.4% specificity.

detach(Boston)