Assignment 3
Valerie Ceciliano
2026-06-24
#Chapter 04: 13, 14, 16
Q13.
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?
library(ISLR2)
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)
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(a) Answer:
- Correlation model show’s correlations #between Lag1-Lag5 similiar and close to 0 or less than 1.
- For the boxplot, I didnt notice any obvious patterns from the historical lags to predict the market’s Direction.
- 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.fits <- glm(Direction ~ Lag1 + Lag2 + Lag3 + Lag4 + Lag5 + Volume, data = Weekly, family = binomial)
summary(glm.fits)##
## 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: 4(b) Answer
- Lag2 seems to be statistically significant, with a pvalue = 0.0296
- 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.fits, type = "response")
glm.pred <- rep("Down", length(glm.probs))
glm.pred[glm.probs > 0.5] <- "Up"
table(glm.pred, Weekly$Direction)##
## glm.pred Down Up
## Down 54 48
## Up 430 557mean(glm.pred == Weekly$Direction)## [1] 0.5610652(c) Answer
- The fraction of predictions or Training accuracy is 56.1%, training error rate is 43.9%
- Confusion matrix is tellling us: When the market went Down, the model predicted Down 54 times, but incorrectly predicted Up 430 times. When the market went Up, the model predicted Up 557 times but also incorrectly predicted Down 48 times. This model behaves optimistically most of the time with 56.1% accuracy and less of the time predicting poorly in last 21 years.
- 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$Year <= 2008)
Weekly.test <- Weekly[!train, ]
Direction.test <- Weekly$Direction[!train]
glm.fits2 <- glm(Direction ~ Lag2, data=Weekly, family = binomial, subset = train)
glm.probs2 <- predict(glm.fits2, Weekly.test, type = "response")
glm.pred2 <- rep("Down", length(glm.probs2))
glm.pred2[glm.probs2 > 0.5] <- "Up"
table(glm.pred2, Direction.test)## Direction.test
## glm.pred2 Down Up
## Down 9 5
## Up 34 56mean(glm.pred2 ==Direction.test)## [1] 0.625(d) Answer
- Test prediction accuracy is 62.5%, while test error rate is 37.5%. This is an improvement over the model trained on all variables, meaning that removing less important features gives you more accuracy or less model variance.
- Repeat (d) using LDA (Linear Discreet Analysis)
library(MASS)##
## Attaching package: 'MASS'## The following object is masked from 'package:ISLR2':
##
## Bostonlda.fit <- lda(Direction ~ Lag2,data = Weekly, subset = train)
lda.pred <- predict(lda.fit, Weekly.test)
table(lda.pred$class, Direction.test)## Direction.test
## Down Up
## Down 9 5
## Up 34 56mean(lda.pred$class ==Direction.test)## [1] 0.625(e) Answer
- LDA test accuracy is also 62.5%
- Repeat (d) using QDA.
qda.fit <- qda (Direction ~ Lag2, data = Weekly, subset = train)
qda.pred <- predict(qda.fit, Weekly.test)
table(qda.pred$class, Direction.test)## Direction.test
## Down Up
## Down 0 0
## Up 43 61mean(qda.pred$class == Direction.test)## [1] 0.5865385(f) Answer
- The QDA test accuracy drops to 58.7%, I dont see any Downward movement in the unseen test data subset.
- Repeat (d) using KNN with K = 1.
library(class)
train.X <- as.matrix(Weekly$Lag2[train])
test.X <- as.matrix(Weekly$Lag2[!train])
train.Direction <- Weekly$Direction[train]
set.seed(1)
knn.pred <- knn(train.X, test.X, train.Direction, k = 1)
table(knn.pred, Direction.test)## Direction.test
## knn.pred Down Up
## Down 21 30
## Up 22 31mean(knn.pred == Direction.test)## [1] 0.5(g) Answer
- KNN(k=1) test accuracy is 50%
- With k=1, the model is overly flexible, picking up randon sample data
- Repeat (d) using naive Bayes.
library(e1071)
nb.fit <- naiveBayes(Direction ~ Lag2, data = Weekly, subset = train)
nb.class <- predict(nb.fit, Weekly.test)
table(nb.class, Direction.test)## Direction.test
## nb.class Down Up
## Down 0 0
## Up 43 61mean(nb.class == Direction.test)## [1] 0.5865385(h) Answer
- Naive Bayes teset accuracy is 58.7%, this model is similiar to QDA, giving a high false positive rate, even on weeks that had declined.
- Which of these methods appears to provide the best results on this data? # (i) Answer
- Logistic regression and LDA seem to provide the highest accuracy results
- 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.interact <- glm(Direction ~ Lag1 * Lag2, data = Weekly, family = binomial, subset = train)
glm.prob.interact <- predict(glm.fit.interact, Weekly.test, type = "response")
glm.pred.interact <- rep("Down", length(glm.prob.interact))
glm.pred.interact[glm.prob.interact > 0.5] <- "Up"
table(glm.pred.interact, Direction.test)## Direction.test
## glm.pred.interact Down Up
## Down 7 8
## Up 36 53mean(glm.pred.interact == Direction.test)## [1] 0.5769231(j) Answer: Logistic Regression with Direction of the market for last week and two weeks ago.
- Accuracy at 57.6 % is worst using both Lags instead of Lag2 alone.
lda.fit.multi <- lda(Direction ~ Lag1 + Lag2, data = Weekly, subset = train)
lda.pred.multi <- predict(lda.fit.multi, Weekly.test)
table(lda.pred.multi$class, Direction.test)## Direction.test
## Down Up
## Down 7 8
## Up 36 53mean(lda.pred.multi$class == Direction.test)## [1] 0.5769231(j) Answer: Logistic Regression combining Lag1 and Lag2
- Accuracy is also at 57.6 % similiar to above.
Q14.
In this problem, you will develop a model to predict whether a given car gets high or low gas mileage based on the Auto data set.
- Create a binary variable, mpg01, that contains a 1 if mpg contains a value above its median, and a 0 if mpg contains a value below its median. You can compute the median using the median() function. Note you may find it helpful to use the data.frame() function to create a single data set containing both mpg01 and the other Auto variables.
library(ISLR2)
mpg_median <- median(Auto$mpg)
Auto$mpg01 <- ifelse(Auto$mpg > mpg_median, 1, 0)pairs(Auto[, -9])
boxplot(cylinders ~ mpg01, data=Auto, main="Cylinders vs mpg01")
boxplot(displacement ~ mpg01, data=Auto, main="Displacement vs mpg01")
boxplot(horsepower ~ mpg01, data = Auto, main="Horsepower vs mpg01")
boxplot(weight ~ mpg01, data = Auto, main="Weight vs mpg01")
(a) Answer
- If mpg01=1, that means that there is a higher probability of gas mileage. 1 is on the positive side and 0 is on the lower negative side.
- 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])
boxplot(cylinders ~ mpg01, data = Auto, main="Cylinders vs mpg01")
boxplot(displacement ~ mpg01, data = Auto, main="Displacement vs mpg01")
boxplot(horsepower ~ mpg01, data = Auto, main="Horsepower vs mpg01")
boxplot(weight ~ mpg01, data=Auto, main="Weight vs mpg01")
(b) Answer
- Boxplots show that these lower categories are strongly associated with a higher probability of high gas mileage, mpg01=1.
- Split the data into a training set and a test set.
set.seed(42)
train_indices <- sample(1:nrow(Auto), 0.8 * nrow(Auto))
train_set <- Auto[train_indices, ]
test_set <- Auto[-train_indices, ]
mpg01_test <- test_set$mpg01- Perform LDA on the training data in order to predict mpg01 using the variables that seemed most associated with mpg01 in (b). What is the test error of the model obtained?
lda.fit <- lda(mpg01 ~ cylinders + displacement + horsepower + weight, data = train_set)
lda.pred <- predict(lda.fit, test_set)
table(lda.pred$class, mpg01_test)## mpg01_test
## 0 1
## 0 29 3
## 1 2 45mean(lda.pred$class !=mpg01_test)## [1] 0.06329114(d) Answer
- Lda is assuming normal distributions with shared variance across these groups.
- Perform QDA on the training data in order to predict mpg01 using the variables that seemed most associated with mpg01 in (b). What is the test error of the model obtained?
qda.fit <- qda(mpg01 ~ cylinders + displacement + horsepower + weight, data = train_set)
qda.pred <-predict(qda.fit, test_set)
table(qda.pred$class, mpg01_test)## mpg01_test
## 0 1
## 0 29 4
## 1 2 44mean(qda.pred$class !=mpg01_test)## [1] 0.07594937(e) Answer
- QDA allows for a quadratic decision boundary by relaxing the assumption of shared variance.
- Perform logistic regression on the training data in order to predict mpg01 using the variables that seemed most associated with mpg01 in (b). What is the test error of the model obtained?
glm.fit <- glm(mpg01 ~ cylinders + displacement + horsepower + weight, data = train_set, family=binomial)
glm.probs <- predict(glm.fit, test_set, type="response")
glm.pred <- ifelse(glm.probs > 0.5, 1,0)
table(glm.pred, mpg01_test)## mpg01_test
## glm.pred 0 1
## 0 29 3
## 1 2 45mean(glm.pred !=mpg01_test)## [1] 0.06329114(f) Answer
- Logistic regression models the probability directly using a link function.
- Perform naive Bayes on the training data in order to predict mpg01 using the variables that seemed most associated with mpg01 in (b). What is the test error of the model obtained?
library(e1071)
nb.fit <- naiveBayes(mpg01 ~ cylinders + displacement + horsepower + weight, data = train_set)
nb.pred <- predict(nb.fit, test_set)
table(nb.pred, mpg01_test)## mpg01_test
## nb.pred 0 1
## 0 29 3
## 1 2 45mean(nb.pred != mpg01_test)## [1] 0.06329114(g) Answer
- Naive Bayes assumes independence among the features conditional on the response class.
- Perform KNN on the training data, with several values of K, in order to predict mpg01. Use only the variables that seemed most associated with mpg01 in (b). What test errors do you obtain? Which value of K seems to perform the best on this data set?
train.X <- cbind(train_set$cylinders, train_set$displacement, train_set$horsepower, train_set$weight)
test.X <- cbind(test_set$cylinders, test_set$displacement, test_set$horsepower, test_set$weight)
train.Y <- train_set$mpg01
for (k in c(1, 5, 10, 50)) {
set.seed(42)
knn.pred <- knn(train.X, test.X, train.Y, k=k)
print(paste("K =", k, "Test Error:", mean(knn.pred != mpg01_test)))
}## [1] "K = 1 Test Error: 0.113924050632911"
## [1] "K = 5 Test Error: 0.113924050632911"
## [1] "K = 10 Test Error: 0.126582278481013"
## [1] "K = 50 Test Error: 0.113924050632911"(h) Answer
- For K = 1, the model overfits due to data noice, giving a high variance. It appears that higher levels generally stablize the boundary
Q16.
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.
crime_median <- median(Boston$crim)
Boston$crim01 <- ifelse(Boston$crim > crime_median, 1, 0)
set.seed(123)
train_idx <- sample(1: nrow(Boston), 0.8 * nrow(Boston))
boston_train <- Boston[train_idx, ]
boston_test <- Boston[-train_idx, ]
crim01_test <- boston_test$crim01
# Logistic Regression
fit.glm <- glm(crim01 ~ indus + nox + age + rad + tax, data=boston_train, family=binomial)
predict.glm <- ifelse(predict(fit.glm, boston_test, type="response") > 0.5, 1,0)
err.glm <- mean(predict.glm !=crim01_test)
#LDA
fit.lda <- lda(crim01 ~ indus + nox + age + rad + tax, data = boston_train)
pred.lda <- predict(fit.lda, boston_test)$class
err.lda <- mean(pred.lda != crim01_test)
#Naive Bayes
fit.nb <- naiveBayes(crim01 ~ indus + nox + age + rad + tax, data = boston_train)
pred.nb <- predict(fit.nb, boston_test)
err.nb <- mean(pred.nb != crim01_test)
#KNN
train.X.b <- as.matrix(boston_train[, c("indus", "nox", "age", "rad", "tax")])
test.X.b <- as.matrix(boston_test[, c("indus", "nox", "age", "rad", "tax")])
pred.knn <- knn(train.X.b, test.X.b, boston_train$crim01, k=10)
err.knn <- mean(pred.knn != crim01_test)
cat("Logistic Error:", err.glm, "\nLA Error:", err.lda, "\nNaive Bayes Error:", err.nb, "\nKNN (K=10) Error:", err.knn)## Logistic Error: 0.1078431
## LA Error: 0.1764706
## Naive Bayes Error: 0.1960784
## KNN (K=10) Error: 0.1078431Answer
- There are strong correlations to “indus”, “nox”, “age”, “rad”, “tax” with high crime areas in Boston.
- KNN and Logistic regression consistently are listed as high performers.