Assignment 3

Exercise 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.

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  
##            
##            
##            
##

Produce some numerical and graphical summaries of the Weekly data.

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  1.000000000

pairs(Weekly)

Positive correlation between year and volume.

Use the full data set to perform a logistic regression with Direction as the response and the five lag variables plus Volume as predictors.

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

Yes, Lag2 appears to be statistically significant.

Compute the confusion matrix and overall fraction of correct predictions.

glm.probs = predict(glm.fit, type = "response")
glm.preds = rep("Down", 1089)
glm.preds[glm.probs > .5]="Up"
table(glm.preds, Direction)

##          Direction
## glm.preds Down  Up
##      Down   54  48
##      Up    430 557

(54+557)/1089

## [1] 0.5610652

54/(54+430)

## [1] 0.1115702

557/(48+557)

## [1] 0.9206612

The overall fraction of correct predictions is 56.1%. The sensitivity is 92.1%, which means when Direction is up the model is very accurate. The specificity is 11.2%, which means when Direction is down the model is not very accurate.

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 = Year < 2009
test = Weekly[!train, ]
glm.fit = glm(Direction ~ Lag2, data=Weekly, family=binomial, subset=train)
glm.probs = predict(glm.fit, test, type = "response")
glm.preds = rep("Down", nrow(test))
glm.preds[glm.probs > .5]="Up"
table(glm.preds, test$Direction)

##          
## glm.preds Down Up
##      Down    9  5
##      Up     34 56

mean(glm.preds == test$Direction)

## [1] 0.625

The overall fraction of correct predictions for the held out data is 62.5%.

Repeat (d) using LDA

lda.fit = lda(Direction ~ Lag2, data=Weekly, subset=train)
lda.preds = predict(lda.fit, test)$class
table(lda.preds, test$Direction)

##          
## lda.preds Down Up
##      Down    9  5
##      Up     34 56

mean(lda.preds == test$Direction)

## [1] 0.625

Repeat (d) using QDA

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

##          
## qda.preds Down Up
##      Down    0  0
##      Up     43 61

mean(qda.preds == test$Direction)

## [1] 0.5865385

Repeat (d) using KNN with K = 1

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

set.seed(1)
knn.preds = knn(train.X, test.X, train.Direction, k=1)

table(knn.preds, test$Direction)

##          
## knn.preds Down Up
##      Down   21 30
##      Up     22 31

mean(knn.preds == test$Direction)

## [1] 0.5

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

The LDA method appears to provide the best results - it has the highest percentage of correct predictions at 62.5%.

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.

glm.fit = glm(Direction ~ Lag1+Lag2, data=Weekly, family=binomial, subset=train)
glm.probs = predict(glm.fit, test, type = "response")
glm.preds = rep("Down", nrow(test))
glm.preds[glm.probs > .5]="Up"
table(glm.preds, test$Direction)

##          
## glm.preds Down Up
##      Down    7  8
##      Up     36 53

mean(glm.preds == test$Direction)

## [1] 0.5769231

lda.fit = lda(Direction ~ Lag1*Lag2, data=Weekly, subset=train)
lda.preds = predict(lda.fit, test)$class
table(lda.preds, test$Direction)

##          
## lda.preds Down Up
##      Down    7  8
##      Up     36 53

mean(lda.preds == test$Direction)

## [1] 0.5769231

qda.fit = qda(Direction ~ I(Lag1^2) + I(Lag2^2), data=Weekly, subset=train)
qda.preds = predict(qda.fit, test)$class
table(qda.preds, test$Direction)

##          
## qda.preds Down Up
##      Down    7  7
##      Up     36 54

mean(qda.preds == test$Direction)

## [1] 0.5865385

knn.preds = knn(train.X, test.X, train.Direction, k=10)

table(knn.preds, test$Direction)

##          
## knn.preds Down Up
##      Down   17 18
##      Up     26 43

mean(knn.preds == test$Direction)

## [1] 0.5769231

Of these models, QDA offers the best performance which indicates that the inclusion of Lag1 introduced some non-linearity into the model. However, this model does not perform as well as the logistic regression and LDA that just included Lag 1 as the only predictor.

Exercise 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. 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.

data(Auto)
attach(Auto)

## The following object is masked from package:ggplot2:
## 
##     mpg

mpg01 = ifelse(median(mpg) < mpg, 1, 0)
Auto = data.frame(Auto, mpg01)

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.

pairs(Auto[1-9:10])

cor(Auto[1-9:10])

##                     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
## mpg01         0.8369392 -0.7591939   -0.7534766 -0.6670526 -0.7577566
##              acceleration       year      mpg01
## mpg             0.4233285  0.5805410  0.8369392
## cylinders      -0.5046834 -0.3456474 -0.7591939
## displacement   -0.5438005 -0.3698552 -0.7534766
## horsepower     -0.6891955 -0.4163615 -0.6670526
## weight         -0.4168392 -0.3091199 -0.7577566
## acceleration    1.0000000  0.2903161  0.3468215
## year            0.2903161  1.0000000  0.4299042
## mpg01           0.3468215  0.4299042  1.0000000

The pairs plot is not very telling when evaluating associations between mpg01 and other predictors. This may be due to the fact that mpg01 is a binary variable. Looking at the correlation matrix, the predictors that seem most associated with mpg01 is mpg (of course), cylinders, displacement, horsepower, and weight.

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

set.seed(1)
train = sample(1:nrow(Auto), 0.75*nrow(Auto), replace = FALSE)
Auto.train = Auto[train,]
Auto.test = Auto[-train,]
mpg01.test = Auto.test$mpg01

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?

lda.fit = lda(mpg01 ~ cylinders + displacement + horsepower + weight, data = Auto, subset = train)
lda.pred = predict(lda.fit, Auto.test)
table(lda.pred$class, mpg01.test)

##    mpg01.test
##      0  1
##   0 42  2
##   1 11 43

mean(lda.pred$class != mpg01.test)

## [1] 0.1326531

The test error rate of the LDA model is 13.26%.

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?

qda.fit = qda(mpg01 ~ cylinders + displacement + horsepower + weight, data = Auto, subset = train)
qda.pred = predict(qda.fit, Auto.test)
table(qda.pred$class, mpg01.test)

##    mpg01.test
##      0  1
##   0 45  4
##   1  8 41

mean(qda.pred$class != mpg01.test)

## [1] 0.122449

The test error rate of the QDA model is 12.24%.

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, family = binomial, subset = train)
glm.probs = predict(glm.fit, Auto.test, type = "response")
glm.pred = rep(0, length(glm.probs))
glm.pred[glm.probs > 0.5] = 1
table(glm.pred, mpg01.test)

##         mpg01.test
## glm.pred  0  1
##        0 45  2
##        1  8 43

mean(glm.pred != mpg01.test)

## [1] 0.1020408

The test error rate of the logistic regression model is 10.20%.

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(Auto$cylinders, Auto$displacement, Auto$horsepower, Auto$weight)[train,]
test.X = cbind(Auto$cylinders, Auto$displacement, Auto$horsepower, Auto$weight)[-train,]
train.mpg01 = mpg01[train]
knn.pred = knn(train.X, test.X, train.mpg01, k = 1)
table(knn.pred, mpg01.test)

##         mpg01.test
## knn.pred  0  1
##        0 43  4
##        1 10 41

mean(knn.pred != mpg01.test)

## [1] 0.1428571

The test error rate of KNN when k = 1 is 14.29%.

knn.pred2 = knn(train.X, test.X, train.mpg01, k = 10)
table(knn.pred2, mpg01.test)

##          mpg01.test
## knn.pred2  0  1
##         0 42  4
##         1 11 41

mean(knn.pred2 != mpg01.test)

## [1] 0.1530612

The test error rate of KNN when k = 10 is 15.31%.

knn.pred3 = knn(train.X, test.X, train.mpg01, k = 100)
table(knn.pred3, mpg01.test)

##          mpg01.test
## knn.pred3  0  1
##         0 42  3
##         1 11 42

mean(knn.pred3 != mpg01.test)

## [1] 0.1428571

The test error rate of KNN when k = 100 is 14.29%. For our KNN models, the model that performed the best on this dataset is when k = 1.

Exercise 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.

View(Boston)
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)
plot(Boston)

After exploring data, we will create the new variable and data set.

crim1 <- rep(0, length(crim)) 
crim1[crim > median(crim)] <- 1
Boston <- data.frame(Boston, crim1)

train = 1:(length(crim) / 2) 
test = (length(crim) / 2 + 1):length(crim)
Boston.train = Boston[train, ]
Boston.test = Boston[-train, ]
crim1.test = crim1[-train]

newBoston <- glm(crim1 ~ crim+zn+indus+chas+nox+rm+age+dis+rad+tax+black+lstat+medv, data = Boston, family = binomial, subset = train)

## Warning: glm.fit: algorithm did not converge

## Warning: glm.fit: fitted probabilities numerically 0 or 1 occurred

newBoston.probs <- predict(newBoston, Boston.test, type = "response")
newBoston.pred <- rep(0, length(newBoston.probs))
newBoston.pred[newBoston.probs > 0.5] <- 1
table(newBoston.pred, crim1.test)

##               crim1.test
## newBoston.pred   0   1
##              0  83   8
##              1   7 155

mean(newBoston.pred == crim1.test)

## [1] 0.9407115

Bostonlda <- lda(crim1 ~ crim+zn+indus+chas+nox+rm+age+dis+rad+tax+black+lstat+medv, data = Boston)
Bostonlda.pred <- predict(Bostonlda, Boston.test)
table(Bostonlda.pred$class, Boston.test$crim1)

##    
##       0   1
##   0  88  19
##   1   2 144

mean(Bostonlda.pred$class == Boston.test$crim1)

## [1] 0.916996

set.seed(1)
Bostonknn=as.matrix(crim1[train])
Bostonknn.test=as.matrix(crim1[!train])
Bostonknn1 =crim1[train]
Bostonknn.pred=knn(Bostonknn,Bostonknn.test,Bostonknn1, Bostonknn,k=1)
summary(Bostonknn)

##        V1        
##  Min.   :0.0000  
##  1st Qu.:0.0000  
##  Median :0.0000  
##  Mean   :0.3557  
##  3rd Qu.:1.0000  
##  Max.   :1.0000

mean(Bostonknn.pred != crim1.test)

## [1] NaN

My findings showed that the best model to use for this data is likely the linear regression model. This has a model accuracy of 94.07%. Then the LDA showed an accuracy of 91.67% which is also very good, but not quite as good as the linear regression model. I found that the linear regression model would be the best.

Assignment 3

Arsalan Bhojani

7/8/2021