Assignment 3
Linda Davila
2/13/2023
library(class)
library(MASS)library(ISLR2)##
## Attaching package: 'ISLR2'## The following object is masked from 'package:MASS':
##
## BostonProblem 13
This question should be answered using the Weekly data set, which is part of the ISLR2 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? The minimums and maximums are the same for lag1-4 , the median values start to differ by a little bit in lag 4 and 5. The scatter plot also shows that the percentage returns stay constant throughout the year. Most of the returns stay between -5 to 5 percent throughout the years.
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
##
##
##
## par(mfrow = c(4,4))
plot(Weekly)
attach(Weekly)
plot(Lag1, Year)
plot(Lag5,Year)
(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? Lag2 is statistically significant
LogR <- glm(Direction ~ Lag1 + Lag2 + Lag3 + Lag4 + Lag5 + Volume, data=Weekly, family = binomial)
summary(LogR)##
## 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(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. The confusion matrix tells us how many correct or incorrect predictions our model produced. The model correctly predicted that the market would go up 557 times and correctly predicted that the market would go down 54 times.
LogR.prob <- predict(LogR, type = "response")
LogR.predict =rep("Down", length(LogR.prob))
LogR.predict[LogR.prob > 0.5] = "Up"
table(LogR.predict, Direction)## Direction
## LogR.predict Down Up
## Down 54 48
## Up 430 557(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 <- (Year < 2008)
weekly.2008 <- Weekly[!train,]
dim(weekly.2008)## [1] 156 9direction.2008 <- Direction[!train]LogR.fit <- glm(Direction ~ Lag2, data=Weekly, family = binomial, subset = train)
LogR.fit.prob <- predict(LogR.fit, weekly.2008, type = "response")
LogR.fit.predict =rep("Down", length(LogR.fit.prob))
LogR.fit.predict[LogR.fit.prob > 0.5] = "Up"
table(LogR.fit.predict, direction.2008)## direction.2008
## LogR.fit.predict Down Up
## Down 7 5
## Up 65 79(e) Repeat (d) using LDA.
library(MASS)LogR.lda <- lda(Direction ~ Lag2, data=Weekly, family = binomial, subset = train)
LogR.lda.prob <- predict(LogR.lda, weekly.2008)$class
table(LogR.lda.prob, direction.2008)## direction.2008
## LogR.lda.prob Down Up
## Down 6 5
## Up 66 79mean(LogR.lda.prob == direction.2008)## [1] 0.5448718(f) Repeat (d) using QDA.
LogR.qda <- qda(Direction ~ Lag2, data=Weekly, family = binomial, subset = train)
LogR.qda.prob <- predict(LogR.qda, weekly.2008)$class
table(LogR.qda.prob, direction.2008)## direction.2008
## LogR.qda.prob Down Up
## Down 0 0
## Up 72 84mean(LogR.qda.prob==direction.2008)## [1] 0.5384615(g) Repeat (d) using KNN with K = 1.
library(class)knn.train=cbind(Lag2[train])
knn.test=cbind(Lag2[!train])
train.Direction =Direction[train]
set.seed(1)
knn.pred=knn(knn.train,knn.test,train.Direction,k=1)
table(knn.pred,direction.2008)## direction.2008
## knn.pred Down Up
## Down 32 38
## Up 40 46mean(knn.pred==direction.2008)## [1] 0.5(h) Repeat (d) using naive Bayes.
library(e1071)
nb.fit <- naiveBayes(Direction ~ Lag2 , data = Weekly, subset = train)
nb.class <- predict(nb.fit, weekly.2008)
table(nb.class, direction.2008)mean(nb.class==direction.2008)## [1] 0.5384615(i) Which of these methods appears to provide the best results on this data? Linear discriminant analysis has the best results with a 54% prediction accuracy
(j) 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.
train <- (Year < 2003)
weekly.2003 <- Weekly[!train,]
dim(weekly.2003)## [1] 417 9direction.2003 <- Direction[!train]Using LDA
lda2 <- lda(Direction ~ Lag5, data=Weekly, family = binomial, subset = train)
lda.prob2 <- predict(lda2, weekly.2003)$class
table(lda.prob2, direction.2003)## direction.2003
## lda.prob2 Down Up
## Down 9 12
## Up 174 222mean(lda.prob2==direction.2003)## [1] 0.5539568Using KNN
knn.train2=cbind(Lag2[train])
knn.test2=cbind(Lag2[!train])
train.Direction2 =Direction[train]
set.seed(1)
knn.pred2=knn(knn.train2,knn.test2,train.Direction2,k=4)
table(knn.pred2,direction.2003)## direction.2003
## knn.pred2 Down Up
## Down 83 85
## Up 100 149mean(knn.pred2== direction.2003)## [1] 0.5563549Problem 14
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.
library(ISLR2)data("Auto")
summary(Auto)## mpg cylinders displacement horsepower weight
## Min. : 9.00 Min. :3.000 Min. : 68.0 Min. : 46.0 Min. :1613
## 1st Qu.:17.00 1st Qu.:4.000 1st Qu.:105.0 1st Qu.: 75.0 1st Qu.:2225
## Median :22.75 Median :4.000 Median :151.0 Median : 93.5 Median :2804
## Mean :23.45 Mean :5.472 Mean :194.4 Mean :104.5 Mean :2978
## 3rd Qu.:29.00 3rd Qu.:8.000 3rd Qu.:275.8 3rd Qu.:126.0 3rd Qu.:3615
## Max. :46.60 Max. :8.000 Max. :455.0 Max. :230.0 Max. :5140
##
## acceleration year origin name
## Min. : 8.00 Min. :70.00 Min. :1.000 amc matador : 5
## 1st Qu.:13.78 1st Qu.:73.00 1st Qu.:1.000 ford pinto : 5
## Median :15.50 Median :76.00 Median :1.000 toyota corolla : 5
## Mean :15.54 Mean :75.98 Mean :1.577 amc gremlin : 4
## 3rd Qu.:17.02 3rd Qu.:79.00 3rd Qu.:2.000 amc hornet : 4
## Max. :24.80 Max. :82.00 Max. :3.000 chevrolet chevette: 4
## (Other) :365library(ISLR2)
attach(Auto)
mpg01 <- ifelse(mpg > 22.75, 1 , 0)
Auto2 <- 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. There is an association between cylinders and mpg, the cars that have an mpg less than 22.75 have cylinders between 6-8. If they have an mpg of more than 22.75 they are 4 cylinder vehicles. There is an association between horsepower and mpg, the cars that have an mpg less than 22.75 have horsepower between 100-150. If they have an mpg of more than 22.75 they have horsepower between 80-100. There is also an association between weight and mpg as well as acceleration and year.
par(mfrow = c(4,4))
plot(Auto2)
par(mfrow = c(3,3))
boxplot(cylinders ~ mpg01, data = Auto2)
boxplot(displacement ~ mpg01, data = Auto2)
boxplot(horsepower ~ mpg01, data = Auto2)
boxplot(weight ~ mpg01, data = Auto2)
boxplot(acceleration ~ mpg01, data = Auto2)
boxplot(year ~ mpg01, data = Auto2)
(c) Split the data into a training set and a test set.
set.seed(1)
auto.train <- sample(1:dim(Auto2)[1], dim(Auto2)[1]*.75, rep=FALSE)
auto.test <- -auto.train
training_data<- Auto2[auto.train, ]
testing_data <- Auto2[auto.test, ]
mpg01.test <- mpg01[auto.test](d) Perform LDA on the training data in order to predict mpg01 using the variables that seemed most associated with mpg01 in
auto.lda <- lda(mpg01 ~ cylinders, data = training_data)
auto.predict<- predict(auto.lda, testing_data)$class
table(auto.predict, mpg01.test)## mpg01.test
## auto.predict 0 1
## 0 45 3
## 1 8 42(b). What is the test error of the model obtained?
mean(auto.predict != mpg01.test)## [1] 0.1122449(e) Perform QDA on the training data in order to predict mpg01 using the variables that seemed most associated with mpg01 in
auto.qda <- qda(mpg01 ~ cylinders, data = training_data)
auto.qda.prob <- predict(auto.qda, testing_data)$class
table(auto.qda.prob, mpg01.test)## mpg01.test
## auto.qda.prob 0 1
## 0 45 3
## 1 8 42(b). What is the test error of the model obtained?
mean(auto.qda.prob != mpg01.test)## [1] 0.1122449(f) Perform logistic regression on the training data in order to predict mpg01 using the variables that seemed most associated with mpg01 in
LogR.auto<- glm(mpg01 ~ cylinders, data=training_data, family = binomial)
LogR.auto.prob <- predict(LogR.auto, training_data, type = "response")
LogR.auto.predict <- predict(LogR.auto, testing_data)
table(LogR.auto.predict, mpg01.test)## mpg01.test
## LogR.auto.predict 0 1
## -4.34469714129446 25 0
## -1.20201612021664 20 3
## 0.369324390322266 1 0
## 1.94066490086117 7 41
## 3.51200541140008 0 1(b). What is the test error of the model obtained?
mean(LogR.auto.predict != mpg01.test)## [1] 1(g) Perform naive Bayes on the training data in order to predict mpg01 using the variables that seemed most associated with mpg01 in
library(e1071)
nb.auto <- naiveBayes(mpg01 ~ cylinders , data = training_data)
nb.class.auto <- predict(nb.auto, testing_data)
table(nb.class.auto, mpg01.test)## mpg01.test
## nb.class.auto 0 1
## 0 45 3
## 1 8 42(b). What is the test error of the model obtained?
mean(nb.class.auto != mpg01.test)## [1] 0.1122449(h) 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
library(class)
attach(Auto2)## The following object is masked _by_ .GlobalEnv:
##
## mpg01## The following objects are masked from Auto:
##
## acceleration, cylinders, displacement, horsepower, mpg, name,
## origin, weight, yearknn.train.X <-cbind(cylinders,horsepower)[auto.train,]
knn.test.X <-cbind(cylinders, horsepower)[auto.test,]
knn.train.mpg01 <- mpg01[auto.train]
set.seed(1)
knn.pred.auto <- knn(knn.train.X, knn.test.X, knn.train.mpg01, k = 5)
mean(knn.pred.auto != mpg01.test)## [1] 0.1428571knn.train.X <-cbind(cylinders,horsepower)[auto.train,]
knn.test.X <-cbind(cylinders, horsepower)[auto.test,]
knn.train.mpg01 <- mpg01[auto.train]
set.seed(1)
knn.pred.auto <- knn(knn.train.X, knn.test.X, knn.train.mpg01, k = 8)
mean(knn.pred.auto != mpg01.test)## [1] 0.1530612knn.train.X <-cbind(cylinders,horsepower)[auto.train,]
knn.test.X <-cbind(cylinders, horsepower)[auto.test,]
knn.train.mpg01 <- mpg01[auto.train]
set.seed(1)
knn.pred.auto <- knn(knn.train.X, knn.test.X, knn.train.mpg01, k = 1)
mean(knn.pred.auto != mpg01.test)## [1] 0.1326531(b). What test errors do you obtain? Which value of K seems to perform the best on this data set? With K= 5 there is test error of .14, K=8 has a test error rate of .15, and K=1 has a test error rate of .13. The K=1 value performs the best with the data.
Problem 16
Using the Boston data set, fit classification models in order to predict whether a given census tract has a crime rate above or below the median. Explore logistic regression, LDA, naive Bayes, and KNN models using various subsets of the predictors. Describe your findings. Hint: You will have to create the response variable yourself, using the variables that are contained in the Boston data set.
attach(Boston)summary(Boston)## crim zn indus chas
## Min. : 0.00632 Min. : 0.00 Min. : 0.46 Min. :0.00000
## 1st Qu.: 0.08204 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 lstat
## Min. : 1.000 Min. :187.0 Min. :12.60 Min. : 1.73
## 1st Qu.: 4.000 1st Qu.:279.0 1st Qu.:17.40 1st Qu.: 6.95
## Median : 5.000 Median :330.0 Median :19.05 Median :11.36
## Mean : 9.549 Mean :408.2 Mean :18.46 Mean :12.65
## 3rd Qu.:24.000 3rd Qu.:666.0 3rd Qu.:20.20 3rd Qu.:16.95
## Max. :24.000 Max. :711.0 Max. :22.00 Max. :37.97
## medv
## Min. : 5.00
## 1st Qu.:17.02
## Median :21.20
## Mean :22.53
## 3rd Qu.:25.00
## Max. :50.00crim2 <- ifelse(crim > .25651, 1 , 0)
boston2 <- data.frame(Boston, crim2)set.seed(3)
boston2.train <- sample(1:dim(boston2)[1], dim(boston2)[1]*.65, rep=FALSE)
boston2.test <- -boston2.train
training_boston2<- boston2[boston2.train, ]
testing_boston2 <- boston2[boston2.test, ]
crim2.test <- crim2[boston2.test]using LDA
boston2.lda <- lda(crim2 ~ medv, data = boston2)
boston2.predict <- predict(boston2.lda, boston2)$class
table(boston2.predict, crim2)## crim2
## boston2.predict 0 1
## 0 139 70
## 1 114 183mean(boston2.predict != crim2)## [1] 0.3636364Using Bayes
nb.boston2 <- naiveBayes(crim2 ~ age , data = boston2)
nb.class.boston2 <- predict(nb.boston2, boston2)
table(nb.class.boston2, crim2)## crim2
## nb.class.boston2 0 1
## 0 177 31
## 1 76 222mean(nb.class.boston2 != crim2)## [1] 0.2114625