Assignment 3

Chapter 04 (page 168): 10, 11, 13

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)

## Warning: package 'ISLR' was built under R version 4.0.5

data("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)

Looking at the pairs chart there seemes to be a linear relationship between Year and Volume.
When looking at all the other Lag variables it doesnt seem to be any relationship

(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?

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

## 
## 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

When reviewing the significance level of .05 only Lag2 variable seem to be statistically significance.

(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.

probs = predict(lr.1, type="response")
preds = rep("Down", 1089)
preds[probs > 0.5] = "Up"
table(preds, Weekly$Direction)

##       
## preds  Down  Up
##   Down   54  48
##   Up    430 557

Formula to determine the percentage of current predictions: (54 + 557) / (54+ 48 + 430 + 557) = .5611 or 56.11%
This states that the weekly market is correctly predicted 56.11% of the time
Up Trends correctly predicts: (557) / (48 + 557) = .9207 or 92.07%
Down trend correctly predicts: (54) / (430 +54) = .1115 or 11.15%

(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).

library(caret)

## Warning: package 'caret' was built under R version 4.0.5

## Loading required package: lattice

## Loading required package: ggplot2

train=Weekly$Year <= 2008
Weekly.test=Weekly[!train,]
logit.fit = glm(Direction ~ Lag2, family=binomial, data=Weekly, subset=train)
contrasts(Weekly$Direction)

##      Up
## Down  0
## Up    1

summary(logit.fit)

## 
## Call:
## glm(formula = Direction ~ Lag2, family = binomial, data = Weekly, 
##     subset = 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

glm.probs=predict(logit.fit,Weekly.test,type="response")
glm.pred=rep("Down",nrow(Weekly.test))
glm.pred[glm.probs > 0.50]="Up"
table(glm.pred,Weekly.test$Direction)

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

mean(glm.pred==Weekly.test$Direction)

## [1] 0.625

Accuracy : 0.625
Positive test
P-value: 0.2439 above .05 significance level - Not statistically significant

(e) Repeat (d) using LDA.

library(caret)
library(MASS)
lda.fit = lda(Direction ~ Lag2, data=Weekly, subset=train)
lda.fit

## Call:
## lda(Direction ~ Lag2, data = Weekly, subset = train)
## 
## Prior probabilities of groups:
##      Down        Up 
## 0.4477157 0.5522843 
## 
## Group means:
##             Lag2
## Down -0.03568254
## Up    0.26036581
## 
## Coefficients of linear discriminants:
##            LD1
## Lag2 0.4414162

lda.class=predict(lda.fit,Weekly.test)$class
table(lda.class,Weekly.test$Direction)

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

mean(lda.class==Weekly.test$Direction)

## [1] 0.625

Accuracy : 0.625
Positive test

(f) Repeat (d) using QDA.

require(MASS)
library(caret)

qda.fit = qda(Direction ~ Lag2, data=Weekly, subset=train)
qda.fit

## Call:
## qda(Direction ~ Lag2, data = Weekly, subset = train)
## 
## Prior probabilities of groups:
##      Down        Up 
## 0.4477157 0.5522843 
## 
## Group means:
##             Lag2
## Down -0.03568254
## Up    0.26036581

qda.class=predict(qda.fit,Weekly.test)$class
table(qda.class,Weekly.test$Direction)

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

mean(qda.class==Weekly.test$Direction)

## [1] 0.5865385

Accuracy : 0.5865

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

library(class)
train.X=Weekly[train,"Lag2",drop=F]
test.X=Weekly[!train,"Lag2",drop=F]
train.Direction=Weekly[train,"Direction",drop=T]
test.Direction=Weekly[!train,"Direction",drop=T]
set.seed(1)
knn.pred=knn(train.X,test.X,train.Direction,k=1)
table(knn.pred,test.Direction)

##         test.Direction
## knn.pred Down Up
##     Down   21 30
##     Up     22 31

mean(knn.pred==test.Direction)

## [1] 0.5

Accuracy : 0.5
Lower than the previous LDA & Logistic Regression

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

LDA & Logistic Regression both have the highest accuracy scores of 0.625

(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.

set.seed(1)
knn.pred=knn(train.X,test.X,train.Direction,k=4)
table(knn.pred,test.Direction)

##         test.Direction
## knn.pred Down Up
##     Down   20 17
##     Up     23 44

mean(knn.pred==test.Direction)

## [1] 0.6153846

When K is equal to 4 the accuracy level is high at .615, but still not as high as the previous LDA & Logistic Regression

qda.fit = qda(Direction ~ Lag2, data=Weekly, subset=train)
qda.fit

## Call:
## qda(Direction ~ Lag2, data = Weekly, subset = train)
## 
## Prior probabilities of groups:
##      Down        Up 
## 0.4477157 0.5522843 
## 
## Group means:
##             Lag2
## Down -0.03568254
## Up    0.26036581

qda.class=predict(qda.fit,Weekly.test)$class
table(qda.class,Weekly.test$Direction)

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

mean(qda.class==Weekly.test$Direction)

## [1] 0.5865385

Accuracy: 0.586
QDA doesn’t preform better then the previous LDA &Logistic Regression or when k is equal to 4

train=Weekly$Year <= 2008
Weekly.test=Weekly[!train,]
logit.fit = glm(Direction ~ Lag1+Lag2+Volume, family=binomial, data=Weekly, subset=train)
contrasts(Weekly$Direction)

##      Up
## Down  0
## Up    1

summary(logit.fit)

## 
## Call:
## glm(formula = Direction ~ Lag1 + Lag2 + Volume, family = binomial, 
##     data = Weekly, subset = train)
## 
## Deviance Residuals: 
##     Min       1Q   Median       3Q      Max  
## -1.4681  -1.2581   0.9929   1.0840   1.5339  
## 
## Coefficients:
##             Estimate Std. Error z value Pr(>|z|)   
## (Intercept)  0.29792    0.09136   3.261  0.00111 **
## Lag1        -0.05975    0.02917  -2.048  0.04054 * 
## Lag2         0.04774    0.02941   1.624  0.10446   
## Volume      -0.07093    0.05263  -1.348  0.17777   
## ---
## 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: 1345.1  on 981  degrees of freedom
## AIC: 1353.1
## 
## Number of Fisher Scoring iterations: 4

glm.probs=predict(logit.fit,Weekly.test,type="response")
glm.pred=rep("Down",nrow(Weekly.test))
glm.pred[glm.probs > 0.50]="Up"
table(glm.pred,Weekly.test$Direction)

##         
## glm.pred Down Up
##     Down   27 33
##     Up     16 28

mean(glm.pred==Weekly.test$Direction)

## [1] 0.5288462

Accuracy: 0.5288
Logistic Regressions doesn’t preform better then the previous LDA & Logistic Regression or when k is equal to 4

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.

data(Auto)

(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.

library(MASS)
library(ISLR)

Auto$mpg01 <- ifelse(Auto$mpg > median(Auto$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.

pairs(Auto[,-9])

Comparing the variables to mpg - horsepower and weight look to have the most correlation. However, displacement seems to have some correlation as well, but not as highly correlated as the other two variables.

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

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

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

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

(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?

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

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

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

## [1] 0.08439898

Accuracy: 0.0843

(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?

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

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

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

## [1] 0.09974425

Accuracy: .0997
QDA performs better than LDA

(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?

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

Accuracy: .0843
Logistic Regression preforms the same as LDA

(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?

#K=1
library(ISLR)
set.seed(1)
train.Auto = train.auto[,c("horsepower","weight","acceleration")]
test.Auto =  test.auto[,c("horsepower","weight","acceleration")]
knn.pred=knn(train.Auto,test.Auto,train.auto$mpg01,k=1)
table(knn.pred,test.auto$mpg01)

##         
## knn.pred   0   1
##        0 179  13
##        1  16 183

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

## [1] 0.9258312

Accuracy .925

#k=2
knn.pred=knn(train.Auto,test.Auto,train.auto$mpg01,k=2)
table(knn.pred,test.auto$mpg01)

##         
## knn.pred   0   1
##        0 168  19
##        1  27 177

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

## [1] 0.8823529

Accuracy .882

#k=3
knn.pred=knn(train.Auto,test.Auto,train.auto$mpg01,k=3)
table(knn.pred,test.auto$mpg01)

##         
## knn.pred   0   1
##        0 168  15
##        1  27 181

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

## [1] 0.8925831

Accuracy .892

#k=5
knn.pred=knn(train.Auto,test.Auto,train.auto$mpg01,k=5)
table(knn.pred,test.auto$mpg01)

##         
## knn.pred   0   1
##        0 166  15
##        1  29 181

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

## [1] 0.887468

Accuracy .887

#k=10
knn.pred=knn(train.Auto,test.Auto,train.auto$mpg01,k=10)
table(knn.pred,test.auto$mpg01)

##         
## knn.pred   0   1
##        0 162  11
##        1  33 185

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

## [1] 0.887468

Accuracy .887
When k=1 it looks like it has the highest accuracy score with 92.5%

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.

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

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

Logistic Regression:

set.seed(1)
Boston.fit <-glm(crime_rate~ indus+nox+age+dis+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, crime_rate.test)

##            crime_rate.test
## Boston.pred   0   1
##           0  75   8
##           1  15 155

mean(Boston.pred != crime_rate.test)

## [1] 0.09090909

Accuracy: .0909
LDA:

Boston.ldafit <-lda(crime_rate~ indus+nox+age+dis+rad+tax, data=Boston.train,family=binomial)
Bostonlda.pred = predict(Boston.ldafit, Boston.test)
table(Bostonlda.pred$class, crime_rate.test)

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

mean(Bostonlda.pred$class != crime_rate.test)

## [1] 0.1067194

Accuracy: .1067
KNN:

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

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

mean(Bosknn.pred !=crime_rate.test)

## [1] 0.8458498

Accuracy: .8458

#K=3
train.K=cbind(indus,nox,age,dis,rad,tax)[train,]
test.K=cbind(indus,nox,age,dis,rad,tax)[test,]
Bosknn.pred=knn(train.K, test.K, crime_rate.test, k=3)
table(Bosknn.pred,crime_rate.test)

##            crime_rate.test
## Bosknn.pred   0   1
##           0  31  17
##           1  59 146

mean(Bosknn.pred !=crime_rate.test)

## [1] 0.3003953

Accuracy: .3003

#K=10
train.K=cbind(indus,nox,age,dis,rad,tax)[train,]
test.K=cbind(indus,nox,age,dis,rad,tax)[test,]
Bosknn.pred=knn(train.K, test.K, crime_rate.test, k=10)
table(Bosknn.pred,crime_rate.test)

##            crime_rate.test
## Bosknn.pred   0   1
##           0  41  10
##           1  49 153

mean(Bosknn.pred !=crime_rate.test)

## [1] 0.2332016

Accuracy: .2332
Reviewing all of the accuracy scores the K nearest neighbors (KNN) with k=1 has the highest rate with 84.58%.
Logistic regression had the lowest accuracy score with 9.09%

Assignment 3_svw192

Brenda Parnin

6/29/2021

Chapter 04 (page 168): 10, 11, 13

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?

(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?

(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.

(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).

(e) Repeat (d) using LDA.

(f) Repeat (d) using QDA.

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

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

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.

(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.

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

(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?

(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?

(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?

(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?

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.