Assignment 3 - dmn335
2026-06-27
Question 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 market 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(ISLR2)## Warning: package 'ISLR2' was built under R version 4.6.1attach(Weekly)
library(MASS)## Warning: package 'MASS' was built under R version 4.6.1##
## Attaching package: 'MASS'## The following object is masked from 'package:ISLR2':
##
## Bostonlibrary(class)
library(e1071)## Warning: package 'e1071' was built under R version 4.6.1- 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
##
##
##
## pairs(Weekly)
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?
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)
##
## 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: 4Lag 2 appears to be statistically significant.
- 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.
glm.probs <- predict(glm.fit, type = "response")
glm.pred <- ifelse(glm.probs > 0.5, "Up", "Down")
table(Predicted = glm.pred, Actual = Weekly$Direction)## Actual
## Predicted Down Up
## Down 54 48
## Up 430 557mean(glm.pred == Weekly$Direction)## [1] 0.5610652We have a 56.1% chance to predict correctly. Model predicts up 92.1% of the time, which makes it bias towards predicting up.
- 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$Year <= 2008
test <- Weekly[!train, ]
glm.fit2 <- glm(Direction ~ Lag2,data = Weekly,subset = train,family = binomial)
glm.probs2 <- predict(glm.fit2, newdata = test, type = "response")
glm.pred2 <- ifelse(glm.probs2 > 0.5, "Up", "Down")
table(Predicted = glm.pred2, Actual = test$Direction)## Actual
## Predicted Down Up
## Down 9 5
## Up 34 56mean(glm.pred2 == test$Direction)## [1] 0.625- Repeat (d) using LDA.
lda.fit <- lda(Direction ~ Lag2, data = Weekly, subset = train)
lda.pred <- predict(lda.fit, newdata = test)
table(Predicted = lda.pred$class, Actual = test$Direction)## Actual
## Predicted Down Up
## Down 9 5
## Up 34 56mean(lda.pred$class == test$Direction)## [1] 0.625- Repeat (d) using QDA.
qda.fit <- qda(Direction ~ Lag2, data = Weekly, subset = train)
qda.pred <- predict(qda.fit, newdata = test)
table(Predicted = qda.pred$class, Actual = test$Direction)## Actual
## Predicted Down Up
## Down 0 0
## Up 43 61mean(qda.pred$class == test$Direction)## [1] 0.5865385- Repeat (d) using KNN with K = 1.
train.X <- matrix(Weekly$Lag2[ train], ncol = 1)
test.X <- matrix(Weekly$Lag2[!train], ncol = 1)
train.Y <- Weekly$Direction[train]
set.seed(1)
knn.pred1 <- knn(train.X, test.X, train.Y, k = 1)
table(Predicted = knn.pred1, Actual = test$Direction)## Actual
## Predicted Down Up
## Down 21 30
## Up 22 31mean(knn.pred1 == test$Direction)## [1] 0.5- Repeat (d) using naive Bayes.
nb.fit <- naiveBayes(Direction ~ Lag2, data = Weekly, subset = train)
nb.pred <- predict(nb.fit, newdata = test)
table(Predicted = nb.pred, Actual = test$Direction)## Actual
## Predicted Down Up
## Down 0 0
## Up 43 61mean(nb.pred == test$Direction)## [1] 0.5865385- Which of these methods appears to provide the best results on this data?
Logistic Regression and LDA appear to be the best at .625 successful prediction rate.
- 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.
Direction.2009 <- Weekly$Direction[!train]
qda.fit <- qda(Direction ~ Lag2, data = Weekly, subset = train)
qda.pred <- predict(qda.fit, test)
table(qda.pred$class, Direction.2009)## Direction.2009
## Down Up
## Down 0 0
## Up 43 61mean(qda.pred$class == Direction.2009)## [1] 0.5865385Question 14
In this problem, you will develop a model to predict whether a given car gets high or low gas mileage based on the Auto data set.
library(ISLR2)
library(MASS)
library(class)
library(e1071)
data(Auto)- Create a binary variable, mpg01, that contains a 1 if mpg contains a value above its median, and a 0 if mpg contains a value below its median. You can compute the median using the median() function. Note you may find it helpful to use the data.frame() function to create a single data set containing both mpg01 and the other Auto variables.
mpg01 <- ifelse(Auto$mpg > median(Auto$mpg), 1, 0)
Auto <- data.frame(Auto, mpg01)mpg_median <- median(Auto$mpg)- 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.
cor(Auto[, -9])## 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
## origin 0.5652088 -0.5689316 -0.6145351 -0.4551715 -0.5850054
## mpg01 0.8369392 -0.7591939 -0.7534766 -0.6670526 -0.7577566
## acceleration year origin mpg01
## mpg 0.4233285 0.5805410 0.5652088 0.8369392
## cylinders -0.5046834 -0.3456474 -0.5689316 -0.7591939
## displacement -0.5438005 -0.3698552 -0.6145351 -0.7534766
## horsepower -0.6891955 -0.4163615 -0.4551715 -0.6670526
## weight -0.4168392 -0.3091199 -0.5850054 -0.7577566
## acceleration 1.0000000 0.2903161 0.2127458 0.3468215
## year 0.2903161 1.0000000 0.1815277 0.4299042
## origin 0.2127458 0.1815277 1.0000000 0.5136984
## mpg01 0.3468215 0.4299042 0.5136984 1.0000000pairs(Auto[, -9])
boxplot(cylinders ~ mpg01, data = Auto, main = "Cylinders")
boxplot(displacement ~ mpg01, data = Auto, main = "Displacement")
boxplot(horsepower~ mpg01, data = Auto, main = "Horsepower")
boxplot(weight~ mpg01, data = Auto, main = "Weight")
cylinders, displacement, horsepower, and weight show a strong association between the variables. Each metric correlates closer to 1 with mpg01 hovering around .066-0.77
- Split the data into a training set and a test set.
set.seed(1)
n <- nrow(Auto)
train <- sample(n, n * 0.8)
Auto.train <- Auto[ train,]
Auto.test <- Auto[-train,]
mpg01.test <- Auto$mpg01[-train]- Perform LDA on the training data in order to predict mpg01 using the variables that seemed most associated with mpg01 in (b). What is the test error of the model obtained?
lda.fit <- lda(mpg01 ~ cylinders + displacement + horsepower + weight, data = Auto, subset = train)
lda.pred <- predict(lda.fit, Auto.test)
table(lda.pred$class, mpg01.test)## mpg01.test
## 0 1
## 0 35 0
## 1 7 37mean(lda.pred$class != mpg01.test)## [1] 0.08860759The test error is .0887, which is good.
- Perform QDA on the training data in order to predict mpg01 using the variables that seemed most associated with mpg01 in (b). What is the test error of the model obtained?
qda.fit <- qda(mpg01 ~ cylinders + displacement + horsepower + weight, data = Auto, subset = train)
qda.pred <- predict(qda.fit, Auto.test)
table(qda.pred$class, mpg01.test)## mpg01.test
## 0 1
## 0 37 2
## 1 5 35mean(qda.pred$class != mpg01.test)## [1] 0.08860759The test error is .0887, similar to LDA
- Perform logistic regression on the training data in order to predict mpg01 using the variables that seemed most associated with mpg01 in (b). What is the test error of the model obtained?
glm.fit <- glm(mpg01 ~ cylinders + displacement + horsepower + weight, data = Auto, subset = train, family = binomial)
glm.prob <- predict(glm.fit, Auto.test, type = "response")
glm.pred <- ifelse(glm.prob > 0.5, 1, 0)
table(glm.pred, mpg01.test)## mpg01.test
## glm.pred 0 1
## 0 38 1
## 1 4 36mean(glm.pred != mpg01.test)## [1] 0.06329114test error of the logistic model is .0632, which is better than lda and qda so far because it’s error rate is lower.
- Perform naive Bayes on the training data in order to predict mpg01 using the variables that seemed most associated with mpg01 in (b). What is the test error of the model obtained?
nb.fit <- naiveBayes(mpg01 ~ cylinders + displacement + horsepower + weight, data = Auto, subset = train)
nb.pred <- predict(nb.fit, Auto.test)
table(nb.pred, mpg01.test)## mpg01.test
## nb.pred 0 1
## 0 37 1
## 1 5 36mean(nb.pred != mpg01.test)## [1] 0.07594937Test error rate is .07594
- Perform KNN on the training data, with several values of K, in order to predict mpg01. Use only the variables that seemed most associated with mpg01 in (b). What test errors do you obtain? Which value of K seems to perform the best on this data set?
predictors <- c("cylinders", "displacement", "horsepower", "weight")train.X <- as.matrix(Auto.train[, predictors])
test.X <- as.matrix(Auto.test[, predictors])
train.Y <- Auto.train$mpg01
test.Y <- Auto.test$mpg01
set.seed(1)
knn.pred <- knn(train.X, test.X, train.Y, k = 3)
table(knn.pred, test.Y)## test.Y
## knn.pred 0 1
## 0 38 3
## 1 4 34mean(knn.pred != test.Y)## [1] 0.08860759k-fold = 3 returned the lowest error rate of .0886 after testing 1-22.
Question 16
Using the Boston data set, fit classification models in order to predict whether a given census tract has a crime rate above or below the median. Explore logistic regression, LDA, naive Bayes, and KNN models using various subsets of the predictors. Describe your findings. Hint: You will have to create the response variable yourself, using the variables that are contained in the Boston data set.
data(Boston)?Boston## starting httpd help server ... donecrim01 <- ifelse(Boston$crim > median(Boston$crim), 1, 0)
Boston <- data.frame(Boston, crim01)cor(Boston[, -1])[, "crim01"]## zn indus chas nox rm age
## -0.43615103 0.60326017 0.07009677 0.72323480 -0.15637178 0.61393992
## dis rad tax ptratio black lstat
## -0.61634164 0.61978625 0.60874128 0.25356836 -0.35121093 0.45326273
## medv crim01
## -0.26301673 1.00000000boxplot(nox ~ crim01, data = Boston, main = "NOX")
boxplot(rad ~ crim01, data = Boston, main = "RAD")
boxplot(tax ~ crim01, data = Boston, main = "TAX")
boxplot(age ~ crim01, data = Boston, main = "AGE")
boxplot(dis ~ crim01, data = Boston, main = "DIS")
boxplot(indus ~ crim01, data = Boston, main = "INDUS")
set.seed(1)
n <- nrow(Boston)
train <- sample(n, n * 0.7)
Boston.train <- Boston[ train,]
Boston.test <- Boston[-train,]
crim01.test <- Boston$crim01[-train]glm.fit <- glm(crim01 ~ nox + rad + tax + age + dis + indus, data = Boston, subset = train, family = binomial)
glm.prob <- predict(glm.fit, Boston.test, type = "response")
glm.pred <- ifelse(glm.prob > 0.5, 1, 0)
table(glm.pred, crim01.test)## crim01.test
## glm.pred 0 1
## 0 58 6
## 1 15 73mean(glm.pred != crim01.test)## [1] 0.1381579lda.fit <- lda(crim01 ~ nox + rad + tax + age + dis + indus,data = Boston, subset = train)
lda.pred <- predict(lda.fit, Boston.test)
table(lda.pred$class, crim01.test)## crim01.test
## 0 1
## 0 71 20
## 1 2 59mean(lda.pred$class != crim01.test)## [1] 0.1447368nb.fit <- naiveBayes(crim01 ~ nox + rad + tax + age + dis + indus,data = Boston, subset = train)
nb.pred <- predict(nb.fit, Boston.test)
table(nb.pred, crim01.test)## crim01.test
## nb.pred 0 1
## 0 65 19
## 1 8 60mean(nb.pred != crim01.test)## [1] 0.1776316predictors <- c("nox", "rad", "tax", "age", "dis", "indus")
train.X <- as.matrix(Boston.train[, predictors])
test.X <- as.matrix(Boston.test[, predictors])
train.Y <- Boston.train$crim01
set.seed(1)
knn.pred3 <- knn(train.X, test.X, train.Y, k = 3)
mean(knn.pred3 != crim01.test)## [1] 0.08552632