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.

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

library(ISLR)
attach(Weekly)

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)

plot(Volume, Year, main = "Volume vs Year")

Not much can be discerned from the correlation plots, however year does appear to have a relationship with volume as shown in the scatterplot.

(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?

fit.direction <- glm(Direction ~ . - Year - Today, family = binomial, data = Weekly)
summary(fit.direction)
## 
## Call:
## glm(formula = Direction ~ . - Year - Today, 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

Lag 2 is the only significant predictor. Its coefficient tells us that as lag 2 increases, direction increases by an average of 0.05844.

(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.

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

table(predict.direction, Direction)
##                  Direction
## predict.direction Down  Up
##              Down   54  48
##              Up    430 557

The confusion matrix tells us how well the classification model performs. With the matrix, we can calculate how accurate the model is (how often it is correct), the precision rate, the true positive, true negative, the false positive rate.

(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[Weekly$Year <= 2008,]
test <- Weekly[Weekly$Year > 2008,]
lag2.glm <- glm(Direction ~ Lag2, family = 'binomial', data = train)

summary(lag2.glm)
## 
## Call:
## glm(formula = Direction ~ Lag2, family = "binomial", data = train)
## 
## Deviance Residuals: 
##    Min      1Q  Median      3Q     Max  
## -1.536  -1.264   1.021   1.091   1.368  
## 
## Coefficients:
##             Estimate Std. Error z value Pr(>|z|)   
## (Intercept)  0.20326    0.06428   3.162  0.00157 **
## Lag2         0.05810    0.02870   2.024  0.04298 * 
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## (Dispersion parameter for binomial family taken to be 1)
## 
##     Null deviance: 1354.7  on 984  degrees of freedom
## Residual deviance: 1350.5  on 983  degrees of freedom
## AIC: 1354.5
## 
## Number of Fisher Scoring iterations: 4
probability2 <- predict(lag2.glm, newdata = test, type = 'response')
predict.glm2 <- ifelse(probability2 >= 0.5, 'Up', 'Down')

table(predict.glm2, test$Direction)
##             
## predict.glm2 Down Up
##         Down    9  5
##         Up     34 56
mean(predict.glm2 != test$Direction)
## [1] 0.375

(e) Repeat (d) using LDA.

library(MASS)

lda.model <- lda(Direction ~ Lag2, data = train)
probability3 <- predict(lda.model, newdata = test)
predict.lda <- probability3$class

table(predict.lda, test$Direction)
##            
## predict.lda Down Up
##        Down    9  5
##        Up     34 56
mean(predict.lda != test$Direction)
## [1] 0.375

(f) Repeat (d) using QDA.

qda.model <- qda(Direction ~ Lag2, data = train)
probability4 <- predict(qda.model, newdata = test)
predict.qda <- probability4$class

table(predict.qda, test$Direction)
##            
## predict.qda Down Up
##        Down    0  0
##        Up     43 61
mean(predict.qda != test$Direction)
## [1] 0.4134615

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

library(class)

knn.model <- knn(train = data.frame(train$Lag2),
test = data.frame(test$Lag2),
cl = train$Direction, k = 1)

table(knn.model, test$Direction)
##          
## knn.model Down Up
##      Down   21 30
##      Up     22 31
mean(knn.model != test$Direction)
## [1] 0.5

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

The Logistic Regression and LDA model gave the best results with the lowest error rates of 37.5%.

(i) 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.

lag2.glm <- glm(Direction ~ Lag2:Lag1, family = 'binomial', data = train)
probability2 <- predict(lag2.glm, newdata = test, type = 'response')
predict.glm2 <- ifelse(probability2 >= 0.5, 'Up', 'Down')
table(predict.glm2, test$Direction)
##             
## predict.glm2 Down Up
##         Down    1  1
##         Up     42 60
mean(predict.glm2 != test$Direction)
## [1] 0.4134615
lda.model <- lda(Direction ~ Lag2:Lag1, data = train)
probability3 <- predict(lda.model, newdata = test)
predict.lda <- probability3$class
table(predict.lda, test$Direction)
##            
## predict.lda Down Up
##        Down    0  1
##        Up     43 60
mean(predict.lda != test$Direction)
## [1] 0.4230769
qda.model <- qda(Direction ~ Lag2:Lag1, data = train)
probability4 <- predict(qda.model, newdata = test)
predict.qda <- probability4$class
table(predict.qda, test$Direction)
##            
## predict.qda Down Up
##        Down   16 32
##        Up     27 29
mean(predict.qda != test$Direction)
## [1] 0.5673077
knn.model <- knn(train = data.frame(train$Lag2),
test = data.frame(test$Lag2),
cl = train$Direction, k = 10)
table(knn.model, test$Direction)
##          
## knn.model Down Up
##      Down   18 19
##      Up     25 42
mean(knn.model != test$Direction)
## [1] 0.4230769

The original LDA model and logistic regression model for Lag2 performed better than with the interaction between Lag2 and Lag1. The error for the logistic regression of Lag2 alone was 37.5% while the error for the logistic regression of Lag2*Lag1 was 41.35%.Increasing the k for the knn model decreased its error from 49% to 44%.

detach(Weekly)

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.

(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)
mpg01 <- rep(0, length(mpg))
mpg01[mpg > median(mpg)] <- 1
Auto <- data.frame(Auto, mpg01)

(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.

head(Auto)
##   mpg cylinders displacement horsepower weight acceleration year origin
## 1  18         8          307        130   3504         12.0   70      1
## 2  15         8          350        165   3693         11.5   70      1
## 3  18         8          318        150   3436         11.0   70      1
## 4  16         8          304        150   3433         12.0   70      1
## 5  17         8          302        140   3449         10.5   70      1
## 6  15         8          429        198   4341         10.0   70      1
##                        name mpg01
## 1 chevrolet chevelle malibu     0
## 2         buick skylark 320     0
## 3        plymouth satellite     0
## 4             amc rebel sst     0
## 5               ford torino     0
## 6          ford galaxie 500     0
pairs(Auto)

par(mfrow=c(2,3))
boxplot(cylinders ~ mpg01, data = Auto, main = "Cylinders vs mpg01")
boxplot(displacement ~ mpg01, data = Auto, main = "Displacement vs mpg01")
boxplot(horsepower ~ mpg01, data = Auto, main = "Horsepower vs mpg01")
boxplot(weight ~ mpg01, data = Auto, main = "Weight vs mpg01")
boxplot(acceleration ~ mpg01, data = Auto, main = "Acceleration vs mpg01")
boxplot(year ~ mpg01, data = Auto, main = "Year vs mpg01")

Based on the scatterplots, there is a lot of linearity between the variables. Mpg appears to be linearly correlated to displancement, horsepower, weight, and year. The boxplots shows how the predictors change depending on whether a car has low or high gas mileage. The weight boxplot indicates that heavier cars tend to have lower mileage. The year boxplot shows that older cars tend to have worse mileage than newer ones. Both weight and year seem to be good predictors of mpg01.

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

train <- (year %% 2 == 0) #split data into training set
Auto.train <- Auto[train, ]
Auto.test <- Auto[!train, ]
mpg01.test <- 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?

fit.lda <- lda(mpg01 ~ cylinders + displacement + horsepower + weight, data = Auto, subset = train)
fit.lda
## Call:
## lda(mpg01 ~ cylinders + displacement + horsepower + weight, data = Auto, 
##     subset = train)
## 
## Prior probabilities of groups:
##         0         1 
## 0.4571429 0.5428571 
## 
## Group means:
##   cylinders displacement horsepower   weight
## 0  6.812500     271.7396  133.14583 3604.823
## 1  4.070175     111.6623   77.92105 2314.763
## 
## Coefficients of linear discriminants:
##                        LD1
## cylinders    -0.6741402638
## displacement  0.0004481325
## horsepower    0.0059035377
## weight       -0.0011465750
predict.lda <- predict(fit.lda, Auto.test)
table(predict.lda$class, mpg01.test)
##    mpg01.test
##      0  1
##   0 86  9
##   1 14 73
mean(predict.lda$class != mpg01.test)
## [1] 0.1263736

The LDA model gives an error rate of 12.63%.

(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?

fit.qda <- qda(mpg01 ~ cylinders + weight + displacement + horsepower, data = Auto, subset = train)
fit.qda
## Call:
## qda(mpg01 ~ cylinders + weight + displacement + horsepower, data = Auto, 
##     subset = train)
## 
## Prior probabilities of groups:
##         0         1 
## 0.4571429 0.5428571 
## 
## Group means:
##   cylinders   weight displacement horsepower
## 0  6.812500 3604.823     271.7396  133.14583
## 1  4.070175 2314.763     111.6623   77.92105
predict.qda <- predict(fit.qda, Auto.test)
table(predict.qda$class, mpg01.test)
##    mpg01.test
##      0  1
##   0 89 13
##   1 11 69
mean(predict.qda$class != mpg01.test)
## [1] 0.1318681

The QDA gives an test error of 13.19%.

(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?

fit.glm <- glm(mpg01 ~ cylinders + weight + displacement + horsepower, data = Auto, family = binomial, subset = train)
summary(fit.glm)
## 
## Call:
## glm(formula = mpg01 ~ cylinders + weight + displacement + horsepower, 
##     family = binomial, data = Auto, subset = train)
## 
## Deviance Residuals: 
##      Min        1Q    Median        3Q       Max  
## -2.48027  -0.03413   0.10583   0.29634   2.57584  
## 
## Coefficients:
##               Estimate Std. Error z value Pr(>|z|)    
## (Intercept)  17.658730   3.409012   5.180 2.22e-07 ***
## cylinders    -1.028032   0.653607  -1.573   0.1158    
## weight       -0.002922   0.001137  -2.569   0.0102 *  
## displacement  0.002462   0.015030   0.164   0.8699    
## horsepower   -0.050611   0.025209  -2.008   0.0447 *  
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## (Dispersion parameter for binomial family taken to be 1)
## 
##     Null deviance: 289.58  on 209  degrees of freedom
## Residual deviance:  83.24  on 205  degrees of freedom
## AIC: 93.24
## 
## Number of Fisher Scoring iterations: 7
probability <- predict(fit.glm, Auto.test, type = "response")
predict.glm <- rep(0, length(probability))
predict.glm[probability > 0.5] <- 1
table(predict.glm, mpg01.test)
##            mpg01.test
## predict.glm  0  1
##           0 89 11
##           1 11 71
mean(predict.glm != mpg01.test)
## [1] 0.1208791

Linear regression gives an test error rate of 12.09%.

(g) 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?

train.X <- cbind(cylinders, weight, displacement, horsepower)[train, ]
test.X <- cbind(cylinders, weight, displacement, horsepower)[!train, ]
train.mpg01 <- mpg01[train]
set.seed(1)

predict.knn <- knn(train.X, test.X, train.mpg01, k = 1)
table(predict.knn, mpg01.test)
##            mpg01.test
## predict.knn  0  1
##           0 83 11
##           1 17 71
mean(predict.knn != mpg01.test)
## [1] 0.1538462
predict.knn <- knn(train.X, test.X, train.mpg01, k = 5)
table(predict.knn, mpg01.test)
##            mpg01.test
## predict.knn  0  1
##           0 82  9
##           1 18 73
mean(predict.knn != mpg01.test)
## [1] 0.1483516
predict.knn <- knn(train.X, test.X, train.mpg01, k = 10)
table(predict.knn, mpg01.test)
##            mpg01.test
## predict.knn  0  1
##           0 79  7
##           1 21 75
mean(predict.knn != mpg01.test)
## [1] 0.1538462

K=5 gives the lowest error rate compared to k=1 and k=10. K=1 had an error rate of 15.4%, K=5 had an error rate of 14.8%, and k=10 had an error rate of 15.4%.

detach(Auto)

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.

Set-Up

library(MASS)
attach(Boston)

crim01 <- rep(0, length(crim)) #make crim binary 0 or 1
crim01[crim > median(crim)] <- 1 #split into 0 or 1 by median 
Boston <- data.frame(Boston, crim01) #incorportate crim01 (binary) into data set

train <- 1:(length(crim) / 2) #first half of data into training set
test <- (length(crim) / 2 + 1):length(crim)
Boston.train <- Boston[train, ]
Boston.test <- Boston[test, ]
crim01.test <- crim01[test]

GLM

glm.fit <- glm(crim01 ~ . - crim01 - crim, family = binomial, data = Boston, subset = train)
## Warning: glm.fit: fitted probabilities numerically 0 or 1 occurred
glm.prob <- predict(glm.fit, Boston.test, type = "response")
glm.predict <- rep(0, length(glm.prob))
glm.predict[glm.prob > 0.5] <- 1
table(glm.predict, crim01.test)
##            crim01.test
## glm.predict   0   1
##           0  68  24
##           1  22 139
mean(glm.predict != crim01.test) #error rate for glm 
## [1] 0.1818182
glm.fit <- glm(crim01 ~ . - crim01 - crim - zn - nox - dis, family = binomial, data = Boston, subset = train)

glm.prob <- predict(glm.fit, Boston.test, type = "response")
glm.predict <- rep(0, length(glm.prob))
glm.predict[glm.prob > 0.5] <- 1
table(glm.predict, crim01.test)
##            crim01.test
## glm.predict   0   1
##           0  85  30
##           1   5 133
mean(glm.predict != crim01.test) #error rate for glm -zn -nox -dis
## [1] 0.1383399

LDA

lda.fit <- lda(crim01 ~ . - crim01 - crim, data = Boston, subset = train)
lda.predict <- predict(lda.fit, Boston.test)
table(lda.predict$class, crim01.test)
##    crim01.test
##       0   1
##   0  80  24
##   1  10 139
mean(lda.predict$class != crim01.test) #error rate for lda
## [1] 0.1343874
lda.fit <- lda(crim01 ~ . - crim01 - crim - zn - nox - dis, data = Boston, subset = train)
lda.predict <- predict(lda.fit, Boston.test)
table(lda.predict$class, crim01.test)
##    crim01.test
##       0   1
##   0  85  30
##   1   5 133
mean(lda.predict$class != crim01.test) #error rate for lda -zn -nox -dis
## [1] 0.1383399

KNN

train.X <- cbind(zn, indus, chas, nox, rm, age, dis, rad, tax, ptratio, black, lstat, medv)[train, ]
test.X <- cbind(zn, indus, chas, nox, rm, age, dis, rad, tax, ptratio, black, lstat, medv)[test, ]
train.crim01 <- crim01[train]
set.seed(1)

knn.predict <- knn(train.X, test.X, train.crim01, k = 1)
table(knn.predict, crim01.test)
##            crim01.test
## knn.predict   0   1
##           0  85 111
##           1   5  52
mean(knn.predict != crim01.test) 
## [1] 0.458498
knn.predict <- knn(train.X, test.X, train.crim01, k = 5)
table(knn.predict, crim01.test)
##            crim01.test
## knn.predict   0   1
##           0  84  37
##           1   6 126
mean(knn.predict != crim01.test)
## [1] 0.1699605
knn.predict <- knn(train.X, test.X, train.crim01, k = 10)
table(knn.predict, crim01.test)
##            crim01.test
## knn.predict   0   1
##           0  83  22
##           1   7 141
mean(knn.predict != crim01.test)
## [1] 0.1146245

Our results show that the KNN with k=10 gives the lowest error rate of 11.86%. We were also able to lower the error rates of the GLM by removing the variables zn (proportion of residential land zoned for lots over 25,000 sq.ft), nox (nitrogen oxides concentration) and ds (weighted mean of distances to five Boston employment centres) with an error rate of 13.83% rather than 18.18%. Removing these variables did not have much of a difference in the error rates of the LDA models.