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.

library(ISLR)

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

library(MASS)
library(class)

data(Weekly)

10 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  
##            
##            
##            
##

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)

By observing the above output, we can see that Volume and Year are clearly positively correlated. From 1990 to 2010, there appears to be an exponential increase in the volume of trades over time.

10 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?

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

summary(log.reg)

## 
## 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

The above output shows that Lag2 is statistically significant as it returned a p-value of 0.0296, which is below our significance level of 0.05.

10 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

weekly.prob <- predict(log.reg, Weekly, type = 'response')

weekly.preds <- rep("Down", length(weekly.prob) )

weekly.preds[weekly.prob > 0.5] = "Up"

table(weekly.preds, Weekly$Direction)

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

The above output shows that the percentage of correct predictions is 56.11% \[((54+557)/1089) = 0.56106\].
The model was highly accurate at predicting Up trends with a score of 92.07% \[(557/(48+557)) = 0.92066\].
However, the model was very inaccurate at predicting Down with a score of 11.16% \[(54/(430+54)) = 0.11157\].

10 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).

weekly.train <- (Weekly$Year < 2009)
weekly.latest <- Weekly[!weekly.train, ]
weekly.glm <- glm(Direction ~ Lag2, data = Weekly, family = 'binomial', subset = weekly.train)


weekly.prob <- predict(weekly.glm, weekly.latest, type = "response")
weekly.preds <- rep("Down", length(weekly.prob))
weekly.preds[weekly.prob > 0.5] = "Up"

weekly.direction <- Weekly$Direction[!weekly.train]
table(weekly.preds, weekly.direction)

##             weekly.direction
## weekly.preds Down Up
##         Down    9  5
##         Up     34 56

The above output shows that the percentage of correct predictions is 62.50% \[((9+56)/104) = 0.625\].
The model was highly accurate at predicting Up trends with a score of 91.80% \[(56/(56+5)) = 0.9180\].
However, the model was relatively inaccurate at predicting Down with a score of 20.93% \[(9/(9+34)) = 0.2093\].

10 E) Repeat (d) using LDA.

weekly.lda <- lda(Direction ~ Lag2, data = Weekly, family = 'binomial', subset = weekly.train)
weekly.lda.preds <- predict(weekly.lda, weekly.latest)

table(weekly.lda.preds$class, weekly.direction)

##       weekly.direction
##        Down Up
##   Down    9  5
##   Up     34 56

The above output shows that using LDA resulted in identical calculations.

10 F) Repeat (d) using QDA.

weekly.qda <- qda(Direction ~ Lag2, data = Weekly, subset = weekly.train)
weekly.qda.preds <- predict(weekly.qda, weekly.latest)$class

table(weekly.qda.preds, weekly.direction)

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

The above output shows that the percentage of correct predictions is 58.65% \[((0+61)/104) = 0.5865\].
The model was perfectly accurate at predicting Up trends with a score of 100% \[(61/(61+0)) = 1\].
However, the model was unable to predict Down with a score of 0% \[(0/(0+43) = 0\].

10 G) Repeat (d) using KNN with K = 1.

weekly.train.X <- as.matrix(Weekly$Lag2[weekly.train])
weekly.test.X <- as.matrix(Weekly$Lag2[!weekly.train])
weekly.train.direction = Weekly$Direction[weekly.train]

set.seed(1)
weekly.knn.preds <- knn(weekly.train.X, weekly.test.X, weekly.train.direction, k=1)

table(weekly.knn.preds, weekly.direction)

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

The above output shows that KNN with k=1 resulted in a model with an accuracy rate of exactly 50% \[((21+31)/104) = 0.5\]

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

The models with the highest accuracy rates are the Logistic Regression (62.50%) and LDA (62.50%).

10 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.

weekly.glm.2 <- glm(Direction ~ Lag1:Lag2, data = Weekly, family = "binomial", subset = weekly.train)
weekly.glm.2.prob <- predict(weekly.glm.2, weekly.latest, type = "response")
weekly.glm.2.preds <- rep("Down", length(weekly.glm.2.prob))
weekly.glm.2.preds[weekly.glm.2.prob > 0.5] = "Up"

weekly.direction = Weekly$Direction[!weekly.train]
table(weekly.glm.2.preds, weekly.direction)

##                   weekly.direction
## weekly.glm.2.preds Down Up
##               Down    1  1
##               Up     42 60

The above Logistic Regression model shows an accuracy rate of 58.65%.

weekly.lda.2 <- lda(Direction ~ Lag1:Lag2, data = Weekly, family = 'binomial', subset = weekly.train)
weekly.lda.2.preds <- predict(weekly.lda.2, weekly.latest)

table(weekly.lda.2.preds$class, weekly.direction)

##       weekly.direction
##        Down Up
##   Down    0  1
##   Up     43 60

The above LDA model shows an accuracy rate of 57.69%.

weekly.qda.2 = qda(Direction ~ Lag1:Lag2, data = Weekly, subset = weekly.train)
weekly.qda.2.preds = predict(weekly.qda.2, weekly.latest)$class

table(weekly.qda.2.preds, weekly.direction)

##                   weekly.direction
## weekly.qda.2.preds Down Up
##               Down   16 32
##               Up     27 29

The above QDA model shows an accuracy rate of 43.27%.

set.seed(1)

weekly.knn.preds.2 <- knn(weekly.train.X, weekly.test.X, weekly.train.direction, k=10)

table(weekly.knn.preds.2, weekly.direction)

##                   weekly.direction
## weekly.knn.preds.2 Down Up
##               Down   17 21
##               Up     26 40

The above KNN model with k=10 shows an accuracy rate of 54.81%.

set.seed(1)

weekly.knn.preds.3 <- knn(weekly.train.X, weekly.test.X, weekly.train.direction, k=15)

table(weekly.knn.preds.3, weekly.direction)

##                   weekly.direction
## weekly.knn.preds.3 Down Up
##               Down   20 20
##               Up     23 41

The above KNN model with k=15 shows an accuracy rate of 58.65%.

After running multiple different models, our highest achieved accuracy rate was tied between Logistic Regression with Lag1 and Lag2 (58.65%) and KNN with k=15 (58.65%).

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)

11 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.

mpg01 <- rep(0, length(Auto$mpg))
mpg01[Auto$mpg > median(Auto$mpg)] = 1

Auto <- data.frame(Auto, mpg01)

11 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)

par(mfrow=c(1,2))

plot(Auto$year, Auto$acceleration, xlab = "Year", ylab = "Acceleration")
plot(Auto$year, Auto$horsepower, xlab = "Year", ylab = "Horsepower")

Our above scatter plots seem to show a positive correlation between acceleration and year. Conversely, we also observe a slight negative correlation between horsepower and year.

par(mfrow=c(1,2))

boxplot(cylinders ~ mpg01, data = Auto, main = "Cylinders")
boxplot(displacement ~ mpg01, data = Auto, main = "Displacement")

The above box plots show a negative correlation between cylinders and mpg01 as a negative correlation between displacement and mpg01.

11 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, ]
test.mpg01 <- Auto$mpg01[!train]

11 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 <- lda(mpg01 ~ cylinders + displacement + horsepower + weight + year, data = train.auto)

auto.lda.preds <- predict(auto.lda, test.auto)

table(auto.lda.preds$class, test.mpg01)

##    test.mpg01
##      0  1
##   0 87  6
##   1 13 76

The above output of our LDA model returns an error rate of 10.44% \[1 - ((87+76)/182) = 0.1044\]

11 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?

auto.qda <- qda(mpg01 ~ cylinders + displacement + horsepower + weight + year, data = train.auto)

auto.qda.preds <- predict(auto.qda, test.auto)

table(auto.qda.preds$class, test.auto$mpg01)

##    
##      0  1
##   0 89 12
##   1 11 70

The above output of our QDA model returns an error rate of 12.64% \[1 - ((89+70)/182) = 0.1264\]

11 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.lm <- glm(mpg01 ~ cylinders + displacement + horsepower + weight + year, data = train.auto, family = 'binomial')
auto.lm.prob <- predict(auto.lm, test.auto, type = "response")
auto.lm.preds = rep(0, length(auto.lm.prob))
auto.lm.preds[auto.lm.prob > 0.5] = 1

table(auto.lm.preds, test.auto$mpg01)

##              
## auto.lm.preds  0  1
##             0 87  8
##             1 13 74

The above output of our Logistic Regression model returns an error rate of 11.54% \[1 - ((87+74)/182) = 0.1154\]

11 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?

attach(Auto)

## The following object is masked _by_ .GlobalEnv:
## 
##     mpg01

set.seed(1)

#K = 1

auto.train.X = cbind(cylinders, displacement, horsepower, weight, year)[train, ]
auto.test.X = cbind(cylinders, displacement, horsepower, weight, year)[!train, ]
train.mpg01 = mpg01[train]

auto.knn.preds.1 = knn(auto.train.X, auto.test.X, train.mpg01, k = 1)
table(auto.knn.preds.1, test.mpg01)

##                 test.mpg01
## auto.knn.preds.1  0  1
##                0 83 11
##                1 17 71

#K = 10

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

##                  test.mpg01
## auto.knn.preds.10  0  1
##                 0 77  7
##                 1 23 75

#K = 100

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

##                   test.mpg01
## auto.knn.preds.100  0  1
##                  0 81  7
##                  1 19 75

Observed test error rates:

K=1: 15.38% \[1 - (83+71)/182) = 0.1538\]
K=10: 16.48% \[1 - (77+75)/182) = 0.1648\]
K=100: 14.29% \[1 - (81+75)/182) = 0.1429\]

Our best KNN model was where K=100, which resulted in a test error rate of 14.29%.

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.

data(Boston)

attach(Boston)

# Create variable

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

# Split data

train = 1:(length(crime) / 2)
test = (length(crime) / 2 + 1):length(crime)

Boston.train = Boston[train, ]
Boston.test = Boston[test, ]
crime.test = crime[test]

# Logistic Regression 1

crime.lm.1 = glm(crime ~ . - crime - crim, data = Boston, family = "binomial", subset = train)

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

crime.prob.1 = predict(crime.lm.1, Boston.test, type = "response")
crime.preds.1 = rep(0, length(crime.prob.1))
crime.preds.1[crime.prob.1 > 0.5] = 1
table(crime.preds.1, crime.test)

##              crime.test
## crime.preds.1   0   1
##             0  68  24
##             1  22 139

Our first Logistic Regression model returned an error rate of 18.19% \[1 - ((68+139)/253) = .1819\]

#Logistic Regression 2

crime.lm.2 = glm(crime ~ . - crime - crim - chas - nox, data = Boston, family = 'binomial', subset = train)

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

crime.prob.2 = predict(crime.lm.2, Boston.test, type = "response")
crime.preds.2 = rep(0, length(crime.prob.2))
crime.preds.2[crime.prob.2 > 0.5] = 1
table(crime.preds.2, crime.test)

##              crime.test
## crime.preds.2   0   1
##             0  78  28
##             1  12 135

Our second Logistic Regression model returned an error rate of 15.81% \[1 - ((78+135)/253) = .1581\]

# LDA 1

crime.lda.1 = lda(crime ~ . - crime - crim, data = Boston, subset = train)
crime.lda.preds.1 = predict(crime.lda.1, Boston.test)
table(crime.lda.preds.1$class, crime.test)

##    crime.test
##       0   1
##   0  80  24
##   1  10 139

Our first LDA model returned an error rate of 13.44% \[1 - ((80+139)/253) = .1344\]

# LDA 2

crime.lda.2 = lda(crime ~ . - crime - crim - chas - nox - tax, data = Boston, subset = train)
crime.lda.preds.2 = predict(crime.lda.2, Boston.test)
table(crime.lda.preds.2$class, crime.test)

##    crime.test
##       0   1
##   0  83  28
##   1   7 135

Our second LDA model returned an error rate of 13.83% \[1 - ((83+135)/253) = .1383\]

#KNN, k = 10

set.seed(1)

Boston.train.X = cbind(zn, indus, chas, nox, rm, age, dis, rad, tax, ptratio, black, lstat, medv)[train, ]
Boston.test.X = cbind(zn, indus, chas, nox, rm, age, dis, rad, tax, ptratio, black, lstat, medv)[test, ]
crime.train = crime[train]

crime.knn.preds.10 = knn(Boston.train.X, Boston.test.X, crime.train, k = 10)
table(crime.knn.preds.10, crime.test)

##                   crime.test
## crime.knn.preds.10   0   1
##                  0  83  21
##                  1   7 142

Our KNN model with k=10 returned an error rate of 11.07% \[1 - ((83+142)/253) = .1107\]

#KNN, k = 100

crime.knn.preds.100 = knn(Boston.train.X, Boston.test.X, crime.train, k = 100)
table(crime.knn.preds.100, crime.test)

##                    crime.test
## crime.knn.preds.100   0   1
##                   0  86 121
##                   1   4  42

Our KNN model with k=100 returned an error rate of 49.01% \[1 - ((86+43))/253 = .4901\]

Our best performing model was KNN with k=10 which resulted in an error rate of approximately 11%

Assignment 3

Michael King

6/29/2021

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.

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

10 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?

10 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

10 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).

10 E) Repeat (d) using LDA.

10 F) Repeat (d) using QDA.

10 G) Repeat (d) using KNN with K = 1.

10 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.

11 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.

11 C) Split the data into a training set and a test set.

11 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?

11 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?

11 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?

11 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.

Assignment 3

Michael King

6/29/2021

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.

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

10 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?

10 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

10 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).

10 E) Repeat (d) using LDA.

10 F) Repeat (d) using QDA.

10 G) Repeat (d) using KNN with K = 1.

10 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.

11 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.

11 C) Split the data into a training set and a test set.

11 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?

11 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?

11 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?

11 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.

10 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?

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.

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.