Assignment 3
Alexandra Holen
2026-06-24
Chapter 4
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 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(ISLR2)
library(class)
library(MASS)##
## Attaching package: 'MASS'## The following object is masked from 'package:ISLR2':
##
## Bostonlibrary(e1071)- 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)
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 1.000000000The only notable correlation is between Year and Volume with the lag variables and today’s return are all very close to 0.
- 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_fit <- glm(Direction ~ Lag1 + Lag2 + Lag3 + Lag4 + Lag5 + Volume,
data = Weekly,
family = binomial)
summary(log_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: 4The predictor that appears to be statistically significant is Lag2 with 0.0296 as the p-value
- 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.
log_probs <- predict(log_fit, type = "response")
log_pred <- rep("Down", length(log_probs))
log_pred[log_probs > 0.5] <- "Up"
table(Predicted = log_pred, Actual = Weekly$Direction)## Actual
## Predicted Down Up
## Down 54 48
## Up 430 557mean(log_pred == Weekly$Direction)## [1] 0.5610652Checking the matrix reveals the model predicts “Up” the vast majority of the time. The overall training accuracy is roughly 56.1%
- 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_idx <- (Weekly$Year <= 2008)
weekly_train <- Weekly[train_idx, ]
weekly_test <- Weekly[!train_idx, ]
log_fit_sub <- glm(Direction ~ Lag2, data = weekly_train, family = binomial)
log_probs_sub <- predict(log_fit_sub, weekly_test, type = "response")
log_pred_sub <- rep("Down", length(log_probs_sub))
log_pred_sub[log_probs_sub > 0.5] <- "Up"
table(Predicted = log_pred_sub, Actual = weekly_test$Direction)## Actual
## Predicted Down Up
## Down 9 5
## Up 34 56mean(log_pred_sub == weekly_test$Direction)## [1] 0.625The test accuracy using just Lag2 climbs to approximately 62.5%
- Repeat (d) using LDA.
lda_fit <- lda(Direction ~ Lag2, data = weekly_train)
lda_pred <- predict(lda_fit, weekly_test)
table(Predicted = lda_pred$class, Actual = weekly_test$Direction)## Actual
## Predicted Down Up
## Down 9 5
## Up 34 56mean(lda_pred$class == weekly_test$Direction)## [1] 0.625The test accuracy matches the logistic regression model exactly.
- Repeat (d) using QDA.
qda_fit <- qda(Direction ~ Lag2, data = weekly_train)
qda_pred <- predict(qda_fit, weekly_test)
table(Predicted = qda_pred$class, Actual = weekly_test$Direction)## Actual
## Predicted Down Up
## Down 0 0
## Up 43 61mean(qda_pred$class == weekly_test$Direction)## [1] 0.5865385The test accuracy is roughly 58.7%. The QDA predicts “Up” for every single week in the test set. Because the test set contains more “Up” weeks than “Down”, it achieves 58.7% by default choice, failing to capture any true downward variations.
- Repeat (d) using KNN with K =1.
train_X <- as.matrix(weekly_train$Lag2)
test_X <- as.matrix(weekly_test$Lag2)
train_Y <- weekly_train$Direction
knn_pred <- knn(train_X, test_X, train_Y, k = 1)
table(Predicted = knn_pred, Actual = weekly_test$Direction)## Actual
## Predicted Down Up
## Down 21 29
## Up 22 32mean(knn_pred == weekly_test$Direction)## [1] 0.5096154K=1 yields a test accuracy of 50% this indicates that a single nearest neighbor is too flexible and fits too much noise in the return data
- Repeat (d) using naive Bayes.
nb_fit <- naiveBayes(Direction ~ Lag2, data = weekly_train)
nb_pred <- predict(nb_fit, weekly_test)
table(Predicted = nb_pred, Actual = weekly_test$Direction)## Actual
## Predicted Down Up
## Down 0 0
## Up 43 61mean(nb_pred == weekly_test$Direction)## [1] 0.5865385Results mirror QDA with 58.7%
- Which of these methods appears to provide the best results on this data?
Comparing test accuracy, Logistic Regression and LDA tied for the highest test accuracy at 62.5%.
Question 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.
- 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.
median_mpg <- median(Auto$mpg)
mpg01 <- ifelse(Auto$mpg > median_mpg, 1, 0)
Auto_df <- data.frame(Auto, mpg01)- 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.
par(mfrow = c(2, 3))
boxplot(cylinders ~ mpg01, data = Auto_df, main = "Cylinders")
boxplot(displacement ~ mpg01, data = Auto_df, main = "Displacement")
boxplot(horsepower ~ mpg01, data = Auto_df, main = "Horsepower")
boxplot(weight ~ mpg01, data = Auto_df, main = "Weight")
boxplot(acceleration ~ mpg01, data = Auto_df, main = "Acceleration")
boxplot(year ~ mpg01, data = Auto_df, main = "Model Year")
Strong separation in the medians exists for weight, horsepower, and displacement, and cylinders.
- Split the data into a training set and a test set.
sample_size <- floor(0.70 * nrow(Auto_df))
set.seed(1)
train_idx <- sample(seq_len(nrow(Auto_df)), size = sample_size)
auto_train <- Auto_df[train_idx, ]
auto_test <- Auto_df[-train_idx, ]- 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_auto <- lda(mpg01 ~ weight + horsepower + displacement + cylinders, data = auto_train)
lda_auto_pred <- predict(lda_auto, auto_test)
table(Predicted = lda_auto_pred$class, Actual = auto_test$mpg01)## Actual
## Predicted 0 1
## 0 50 3
## 1 11 54mean(lda_auto_pred$class != auto_test$mpg01)## [1] 0.1186441- 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_auto <- qda(mpg01 ~ weight + horsepower + displacement + cylinders, data = auto_train)
qda_auto_pred <- predict(qda_auto, auto_test)
table(Predicted = qda_auto_pred$class, Actual = auto_test$mpg01)## Actual
## Predicted 0 1
## 0 52 5
## 1 9 52mean(qda_auto_pred$class != auto_test$mpg01)## [1] 0.1186441- 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?
log_auto <- glm(mpg01 ~ weight + horsepower + displacement + cylinders,
data = auto_train,
family = binomial)
log_auto_probs <- predict(log_auto, auto_test, type = "response")
log_auto_pred <- ifelse(log_auto_probs > 0.5, 1, 0)
table(Predicted = log_auto_pred, Actual = auto_test$mpg01)## Actual
## Predicted 0 1
## 0 53 3
## 1 8 54mean(log_auto_pred != auto_test$mpg01)## [1] 0.09322034- 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_auto <- naiveBayes(mpg01 ~ weight + horsepower + displacement + cylinders, data = auto_train)
nb_auto_pred <- predict(nb_auto, auto_test)
table(Predicted = nb_auto_pred, Actual = auto_test$mpg01)## Actual
## Predicted 0 1
## 0 52 4
## 1 9 53mean(nb_auto_pred != auto_test$mpg01)## [1] 0.1101695- 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_auto <- scale(auto_train[, c("weight", "horsepower", "displacement", "cylinders")])
test_X_auto <- scale(auto_test[, c("weight", "horsepower", "displacement", "cylinders")])
train_Y_auto <- auto_train$mpg01
for (kv in c(1, 5, 10, 20, 50)) {
set.seed(1)
knn_auto_pred <- knn(train_X_auto, test_X_auto, train_Y_auto, k = kv)
error_rate <- mean(knn_auto_pred != auto_test$mpg01)
cat(stringr::str_interp("K = ${kv} | Test Error Rate: ${round(error_rate, 4)}\n"))
}## K = 1 | Test Error Rate: 0.1441
## K = 5 | Test Error Rate: 0.1356
## K = 10 | Test Error Rate: 0.1356
## K = 20 | Test Error Rate: 0.1102
## K = 50 | Test Error Rate: 0.1102For this data split, KNN with K=20 and K=50 performed optimally among the neighborhood models, yielding a test error rate of 11.02%.
Question 16
Using the Boston data set, fit classification models in order to predict whether a given census tract has a crime rate above or below the median. Explore logistic regression, LDA, naive Bayes, and KNN models using various subsets of the predictors. Describe your findings. Hint: You will have to create the response variable yourself, using the variables that are contained in the Boston data set
median_crim <- median(Boston$crim)
crim01 <- ifelse(Boston$crim > median_crim, 1, 0)
Boston_df <- data.frame(Boston, crim01)cor(Boston)[, "crim"]## crim zn indus chas nox rm
## 1.00000000 -0.20046922 0.40658341 -0.05589158 0.42097171 -0.21924670
## age dis rad tax ptratio black
## 0.35273425 -0.37967009 0.62550515 0.58276431 0.28994558 -0.38506394
## lstat medv
## 0.45562148 -0.38830461set.seed(123)
train_idx <- sample(1:nrow(Boston_df), 0.7 * nrow(Boston_df))
train_set <- Boston_df[train_idx, ]
test_set <- Boston_df[-train_idx, ]
actual_labels <- test_set$crim01Logistic Regression
log_fit <- glm(crim01 ~ nox + rad + tax + dis, data = train_set, family = binomial)
log_probs <- predict(log_fit, test_set, type = "response")
log_pred <- ifelse(log_probs > 0.5, 1, 0)
table(Predicted = log_pred, Actual = actual_labels)## Actual
## Predicted 0 1
## 0 69 10
## 1 6 67mean(log_pred != actual_labels)## [1] 0.1052632LDA
lda_fit <- lda(crim01 ~ nox + rad + tax + dis, data = train_set)
lda_pred <- predict(lda_fit, test_set)$class
table(Predicted = lda_pred, Actual = actual_labels)## Actual
## Predicted 0 1
## 0 70 15
## 1 5 62mean(lda_pred != actual_labels)## [1] 0.1315789QDA
qda_fit <- qda(crim01 ~ nox + rad + tax + dis, data = train_set)
qda_pred <- predict(qda_fit, test_set)$class
table(Predicted = qda_pred, Actual = actual_labels)## Actual
## Predicted 0 1
## 0 71 21
## 1 4 56mean(qda_pred != actual_labels)## [1] 0.1644737KNN
predictors <- c("nox", "rad", "tax", "dis")
scaled_X <- scale(Boston_df[, predictors])
train_X <- scaled_X[train_idx, ]
test_X <- scaled_X[-train_idx, ]
train_Y <- train_set$crim01
for (kv in c(1, 3, 5, 10, 20)) {
set.seed(123)
knn_pred <- knn(train_X, test_X, train_Y, k = kv)
cat(sprintf("K = %2d | Test Error Rate: %.2f%%\n", kv, mean(knn_pred != actual_labels) * 100))
}## K = 1 | Test Error Rate: 4.61%
## K = 3 | Test Error Rate: 5.26%
## K = 5 | Test Error Rate: 5.92%
## K = 10 | Test Error Rate: 4.61%
## K = 20 | Test Error Rate: 4.61%KNN with K=1, K=10 and K=20 performed optimally on the test set, achieving the lowest overall test error rate of 4.61%. The non-parametric model comfortably outperformed the parametric alternatives as Logistic Regression at 10.5%, suggesting that the true decision boundary separating high-crime census tracts based on industrial features (nox, rad, tax, dis) is highly non-linear