Assignment 3
Lauren Hall
2025-06-20
Question 13
library(ISLR2)## Warning: package 'ISLR2' was built under R version 4.3.3library(MASS)## Warning: package 'MASS' was built under R version 4.3.2##
## Attaching package: 'MASS'## The following object is masked from 'package:ISLR2':
##
## Bostonlibrary(class)## Warning: package 'class' was built under R version 4.3.3library(e1071)## Warning: package 'e1071' was built under R version 4.3.3data("Weekly")
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
##
##
##
## Q13 (a) Produce some numerical and graphical summaries of the Weekly data. Do there appear to be any patterns?
pairs(Weekly[, -9]) # Exclude Direction (factor)
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 1.000000000plot(Weekly$Volume, type='l', main="Volume Over Time")
table(Weekly$Direction)##
## Down Up
## 484 605Patterns are that there are more down than ups in the data. There is a gradual increase in weekly volume over time. There are weak relationships between Lag1 to Lag2 and Today as well as in Lag1 to Lag2 and Year. Strong correlation between year and volume.
Q13 (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: 4The statistically significant predictor is Lag2 with a p-value less than 0.05.
Q13 (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.
probs <- predict(glm.fit, type = "response")
pred <- ifelse(probs > 0.5, "Up", "Down")
table(Predicted = pred, Actual = Weekly$Direction)## Actual
## Predicted Down Up
## Down 54 48
## Up 430 557accuracy <- (54 + 557) / 1089 # TP + TN / total
accuracy ## [1] 0.5610652This is telling us that this model often predicts “Up” more than “down.” The false negative is very high. There is a moderate accuracy of 56.11% with a bias towards predicting “Up” direction.
Q13 (d) Now fit the logistic regression model using a training data period from 1990 to 2008, with Lag2 as the only predictor. Compute the confusion matrix and the overall fraction of correct predictions for the held out data (that is, the data from 2009 and 2010).
train <- Weekly$Year <= 2008
test <- Weekly$Year > 2008
glm.fit2 <- glm(Direction ~ Lag2, data = Weekly, family = binomial, subset = train)
probs2 <- predict(glm.fit2, newdata = Weekly[test, ], type = "response")
pred2 <- ifelse(probs2 > 0.5, "Up", "Down")
table(Predicted = pred2, Actual = Weekly$Direction[test])## Actual
## Predicted Down Up
## Down 9 5
## Up 34 56mean(pred2 == Weekly$Direction[test])## [1] 0.625Q13 (e) Repeat (d) using LDA.
lda.fit <- lda(Direction ~ Lag2, data = Weekly, subset = train)
lda.pred <- predict(lda.fit, Weekly[test, ])
table(Predicted = lda.pred$class, Actual = Weekly$Direction[test])## Actual
## Predicted Down Up
## Down 9 5
## Up 34 56mean(lda.pred$class == Weekly$Direction[test])## [1] 0.625Q13 (f) Repeat (d) using QDA.
qda.fit <- qda(Direction ~ Lag2, data = Weekly, subset = train)
qda.pred <- predict(qda.fit, Weekly[test, ])
table(Predicted = qda.pred$class, Actual = Weekly$Direction[test])## Actual
## Predicted Down Up
## Down 0 0
## Up 43 61mean(qda.pred$class == Weekly$Direction[test])## [1] 0.5865385Q13 (g) Repeat (d) using KNN with K =1.
train.X <- as.matrix(Weekly[train, "Lag2"])
test.X <- as.matrix(Weekly[test, "Lag2"])
train.Direction <- Weekly$Direction[train]
set.seed(1)
knn.pred <- knn(train.X, test.X, train.Direction, k = 1)
table(Predicted = knn.pred, Actual = Weekly$Direction[test])## Actual
## Predicted Down Up
## Down 21 30
## Up 22 31mean(knn.pred == Weekly$Direction[test])## [1] 0.5Q13 (h) Repeat (d) using naive Bayes.
nb.fit <- naiveBayes(Direction ~ Lag2, data = Weekly, subset = train)
nb.pred <- predict(nb.fit, Weekly[test, ])
table(Predicted = nb.pred, Actual = Weekly$Direction[test])## Actual
## Predicted Down Up
## Down 0 0
## Up 43 61mean(nb.pred == Weekly$Direction[test])## [1] 0.5865385Q13 (i) Which of these methods appears to provide the best results on this data?
The best method and the highest accuracy was LDA with a 62.50% accuracy.
Q13 (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 <- Weekly$Year <= 2008
test <- Weekly$Year > 2008
glm.fit <- glm(Direction ~ Lag2 * Lag1 + Volume,
data = Weekly, family = binomial, subset = train)
probs <- predict(glm.fit, Weekly[test, ], type = "response")
pred <- ifelse(probs > 0.5, "Up", "Down")
table(Predicted = pred, Actual = Weekly$Direction[test])## Actual
## Predicted Down Up
## Down 27 32
## Up 16 29mean(pred == Weekly$Direction[test])## [1] 0.5384615lda.fit <- lda(Direction ~ Lag2 + Lag1 + I(Lag1*Lag2), data = Weekly, subset = train)
lda.pred <- predict(lda.fit, Weekly[test, ])
table(Predicted = lda.pred$class, Actual = Weekly$Direction[test])## Actual
## Predicted Down Up
## Down 7 8
## Up 36 53mean(lda.pred$class == Weekly$Direction[test])## [1] 0.5769231qda.fit <- qda(Direction ~ Lag2 + I(Lag2^2) + I(Lag1*Lag2), data = Weekly, subset = train)
qda.pred <- predict(qda.fit, Weekly[test, ])
table(Predicted = qda.pred$class, Actual = Weekly$Direction[test])## Actual
## Predicted Down Up
## Down 18 29
## Up 25 32mean(qda.pred$class == Weekly$Direction[test])## [1] 0.4807692train.X <- scale(as.matrix(Weekly[train, c("Lag1", "Lag2")]))
test.X <- scale(as.matrix(Weekly[test, c("Lag1", "Lag2")]))
train.Direction <- Weekly$Direction[train]
set.seed(1)
knn.pred <- knn(train.X, test.X, train.Direction, k = 3)
table(Predicted = knn.pred, Actual = Weekly$Direction[test])## Actual
## Predicted Down Up
## Down 23 32
## Up 20 29mean(knn.pred == Weekly$Direction[test])## [1] 0.5nb.fit <- naiveBayes(Direction ~ Lag2 + I(Lag2^2), data = Weekly, subset = train)
nb.pred <- predict(nb.fit, Weekly[test, ])## Warning in predict.naiveBayes(nb.fit, Weekly[test, ]): Type mismatch between
## training and new data for variable 'I(Lag2^2)'. Did you use factors with
## numeric labels for training, and numeric values for new data?table(Predicted = nb.pred, Actual = Weekly$Direction[test])## Actual
## Predicted Down Up
## Down 0 0
## Up 43 61mean(nb.pred == Weekly$Direction[test])## [1] 0.5865385After these transformations performed on this subset of the data, the naive bayes method proved to have the highest accuracy of 58.65% therefore it is considered the best method in this case.
Question 14: In this problem, you will develop a model to predict whether a given car gets high or low gas mileage based on the Auto data set
Q14: (a) Create a binary variable, mpg01, that contains a 1 if mpg contains a value above its median, and a 0 if mpg contains a value below its median. You can compute the median using the median() method of the data frame. Note you may find it helpful to add a column mpg01 to the data frame by assignment. Assuming you have stored the data frame as Auto, this can be done as follows: Auto[‘mpg01’] = mpg01
Auto <- read.csv("C:/Users/lclha/Documents/MSDA Program 2024-2026/Summer 2025/Predictive Modeling/Datasets/Auto.csv", header = TRUE, na.strings = "?")
Auto <- na.omit(Auto)Auto$mpg01 <- ifelse(Auto$mpg > median(Auto$mpg), 1, 0)Q14 (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.
par(mfrow = c(2, 2))
boxplot(horsepower ~ mpg01, data = Auto, main = "Horsepower vs mpg01")
boxplot(weight ~ mpg01, data = Auto, main = "Weight vs mpg01")
boxplot(displacement ~ mpg01, data = Auto, main = "Displacement vs mpg01")
boxplot(cylinders ~ mpg01, data = Auto, main = "Cylinders vs mpg01")
cor(Auto[, sapply(Auto, is.numeric)])## 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 1.0000000Useful predictors would be horsepower, weight,displacement,cylinder have far apart medians between the binary responses for mpg01 and they all have strong negative correlations against mpg01 variable.
Q14 (c) Split the data into a training set and a test set.
set.seed(1)
train_index <- sample(1:nrow(Auto), nrow(Auto) / 2)
train <- Auto[train_index, ]
test <- Auto[-train_index, ]Q14 (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?
vars <- c("horsepower", "weight", "displacement", "cylinders")library(MASS)
lda.fit <- lda(mpg01 ~ horsepower + weight + displacement + cylinders, data = train)
lda.pred <- predict(lda.fit, test)
lda_error <- mean(lda.pred$class != test$mpg01)
lda_error## [1] 0.127551Q14(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?
qda.fit <- qda(mpg01 ~ horsepower + weight + displacement + cylinders, data = train)
qda.pred <- predict(qda.fit, test)
qda_error <- mean(qda.pred$class != test$mpg01)
qda_error## [1] 0.1173469Q14 (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?
glm.fit <- glm(mpg01 ~ horsepower + weight + displacement + cylinders, data = train, family = binomial)
probs <- predict(glm.fit, test, type = "response")
glm.pred <- ifelse(probs > 0.5, 1, 0)
logit_error <- mean(glm.pred != test$mpg01)
logit_error## [1] 0.122449Q14 (g) Perform naive Bayes on the training data in order to predict mpg01 using the variables that seemed most associated with mpg01 in (b). What is the test error of the model obtained?
library(e1071)
nb.fit <- naiveBayes(as.factor(mpg01) ~ horsepower + weight + displacement + cylinders, data = train)
nb.pred <- predict(nb.fit, test)
nb_error <- mean(nb.pred != test$mpg01)
nb_error## [1] 0.1173469Q14 (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 (b). What test errors do you obtain? Which value of K seems to perform the best on this data set?
library(class)
train.X <- scale(train[, vars])
test.X <- scale(test[, vars], center = attr(train.X, "scaled:center"), scale = attr(train.X, "scaled:scale"))
train.Y <- train$mpg01
# Try different K values
set.seed(1)
knn_errors <- sapply(1:10, function(k) {
knn.pred <- knn(train.X, test.X, train.Y, k = k)
mean(knn.pred != test$mpg01)
})
knn_errors## [1] 0.1530612 0.1275510 0.1122449 0.1173469 0.1173469 0.1275510 0.1173469
## [8] 0.1224490 0.1173469 0.1173469best_k <- which.min(knn_errors)
best_knn_error <- knn_errors[best_k]
best_k## [1] 3best_knn_error## [1] 0.1122449cat("LDA error:", lda_error, "\n")## LDA error: 0.127551cat("QDA error:", qda_error, "\n")## QDA error: 0.1173469cat("Logistic Regression error:", logit_error, "\n")## Logistic Regression error: 0.122449cat("Naive Bayes error:", nb_error, "\n")## Naive Bayes error: 0.1173469cat("Best KNN error:", best_knn_error, "at K =", best_k, "\n")## Best KNN error: 0.1122449 at K = 3Question 16: 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, naive Bayes, and KNN models using various subsets of the predictors. Describe your findings.
library(MASS)
library(class)
library(e1071)
# Load Boston data
data("Boston")
# Create binary target variable: 1 if crime rate > median, else 0
Boston$crim01 <- ifelse(Boston$crim > median(Boston$crim), 1, 0)
?Boston## starting httpd help server ... donecorrelations <- cor(Boston[, sapply(Boston, is.numeric)])
correlations## crim zn indus chas nox
## crim 1.00000000 -0.20046922 0.40658341 -0.055891582 0.42097171
## zn -0.20046922 1.00000000 -0.53382819 -0.042696719 -0.51660371
## indus 0.40658341 -0.53382819 1.00000000 0.062938027 0.76365145
## chas -0.05589158 -0.04269672 0.06293803 1.000000000 0.09120281
## nox 0.42097171 -0.51660371 0.76365145 0.091202807 1.00000000
## rm -0.21924670 0.31199059 -0.39167585 0.091251225 -0.30218819
## age 0.35273425 -0.56953734 0.64477851 0.086517774 0.73147010
## dis -0.37967009 0.66440822 -0.70802699 -0.099175780 -0.76923011
## rad 0.62550515 -0.31194783 0.59512927 -0.007368241 0.61144056
## tax 0.58276431 -0.31456332 0.72076018 -0.035586518 0.66802320
## ptratio 0.28994558 -0.39167855 0.38324756 -0.121515174 0.18893268
## black -0.38506394 0.17552032 -0.35697654 0.048788485 -0.38005064
## lstat 0.45562148 -0.41299457 0.60379972 -0.053929298 0.59087892
## medv -0.38830461 0.36044534 -0.48372516 0.175260177 -0.42732077
## crim01 0.40939545 -0.43615103 0.60326017 0.070096774 0.72323480
## rm age dis rad tax ptratio
## crim -0.21924670 0.35273425 -0.37967009 0.625505145 0.58276431 0.2899456
## zn 0.31199059 -0.56953734 0.66440822 -0.311947826 -0.31456332 -0.3916785
## indus -0.39167585 0.64477851 -0.70802699 0.595129275 0.72076018 0.3832476
## chas 0.09125123 0.08651777 -0.09917578 -0.007368241 -0.03558652 -0.1215152
## nox -0.30218819 0.73147010 -0.76923011 0.611440563 0.66802320 0.1889327
## rm 1.00000000 -0.24026493 0.20524621 -0.209846668 -0.29204783 -0.3555015
## age -0.24026493 1.00000000 -0.74788054 0.456022452 0.50645559 0.2615150
## dis 0.20524621 -0.74788054 1.00000000 -0.494587930 -0.53443158 -0.2324705
## rad -0.20984667 0.45602245 -0.49458793 1.000000000 0.91022819 0.4647412
## tax -0.29204783 0.50645559 -0.53443158 0.910228189 1.00000000 0.4608530
## ptratio -0.35550149 0.26151501 -0.23247054 0.464741179 0.46085304 1.0000000
## black 0.12806864 -0.27353398 0.29151167 -0.444412816 -0.44180801 -0.1773833
## lstat -0.61380827 0.60233853 -0.49699583 0.488676335 0.54399341 0.3740443
## medv 0.69535995 -0.37695457 0.24992873 -0.381626231 -0.46853593 -0.5077867
## crim01 -0.15637178 0.61393992 -0.61634164 0.619786249 0.60874128 0.2535684
## black lstat medv crim01
## crim -0.38506394 0.4556215 -0.3883046 0.40939545
## zn 0.17552032 -0.4129946 0.3604453 -0.43615103
## indus -0.35697654 0.6037997 -0.4837252 0.60326017
## chas 0.04878848 -0.0539293 0.1752602 0.07009677
## nox -0.38005064 0.5908789 -0.4273208 0.72323480
## rm 0.12806864 -0.6138083 0.6953599 -0.15637178
## age -0.27353398 0.6023385 -0.3769546 0.61393992
## dis 0.29151167 -0.4969958 0.2499287 -0.61634164
## rad -0.44441282 0.4886763 -0.3816262 0.61978625
## tax -0.44180801 0.5439934 -0.4685359 0.60874128
## ptratio -0.17738330 0.3740443 -0.5077867 0.25356836
## black 1.00000000 -0.3660869 0.3334608 -0.35121093
## lstat -0.36608690 1.0000000 -0.7376627 0.45326273
## medv 0.33346082 -0.7376627 1.0000000 -0.26301673
## crim01 -0.35121093 0.4532627 -0.2630167 1.00000000boxplot(Boston$tax ~ Boston$crim01, main = "Tax vs Crime01")
boxplot(Boston$nox ~ Boston$crim01, main = "NOx vs Crime01")
boxplot(Boston$dis ~ Boston$crim01, main = "Distance vs Crime01")
boxplot(Boston$indus ~ Boston$crim01, main = "Industry vs Crime01")
boxplot(Boston$age ~ Boston$crim01, main = "Age of Units Prior 1940 vs Crime01")
boxplot(Boston$rad ~ Boston$crim01, main = "Radical Highways vs Crime01")
Useful predictors would be tax, nox, dis,indus,age, rad variables have
far apart medians between the binary responses for crim01 and they all
have strong correlations against crim01 variable.
set.seed(1)
train_index <- sample(1:nrow(Boston), nrow(Boston) / 2)
train <- Boston[train_index, ]
test <- Boston[-train_index, ]glm.fit <- glm(crim01 ~ nox + rad + dis + tax + indus + age, data = train, family = binomial)
glm.probs <- predict(glm.fit, newdata = test, type = "response")
glm.pred <- ifelse(glm.probs > 0.5, 1, 0)
logit_error2 <- mean(glm.pred != test$crim01)
logit_error2## [1] 0.1146245lda.fit <- lda(crim01 ~ nox + rad + dis + tax + indus + age, data = train)
lda.pred <- predict(lda.fit, test)
lda_error2 <- mean(lda.pred$class != test$crim01)
lda_error2## [1] 0.1581028nb.fit <- naiveBayes(as.factor(crim01) ~nox + rad + dis + tax + indus + age, data = train)
nb.pred <- predict(nb.fit, test)
nb_error2 <- mean(nb.pred != test$crim01)
nb_error2## [1] 0.1778656vars <- c("nox", "rad", "dis", "tax", "age","indus")
train.X <- scale(train[, vars])
test.X <- scale(test[, vars], center = attr(train.X, "scaled:center"), scale = attr(train.X, "scaled:scale"))
train.Y <- train$crim01
# Try K = 1 to 10
knn.errors <- sapply(1:10, function(k) {
pred <- knn(train.X, test.X, cl = train.Y, k = k)
mean(pred != test$crim01)
})
# Best K
best_k2 <- which.min(knn.errors)
best_knn_error2 <- knn.errors[which.min(knn.errors)]
best_k2## [1] 3best_knn_error2## [1] 0.07509881cat("LDA error:", lda_error2, "\n")## LDA error: 0.1581028cat("Logistic Regression error:", logit_error2, "\n")## Logistic Regression error: 0.1146245cat("Naive Bayes error:", nb_error2, "\n")## Naive Bayes error: 0.1778656cat("Best KNN error:", best_knn_error2, "at K =", best_k2, "\n")## Best KNN error: 0.07509881 at K = 3Based on test error rates, KNN (with K = 3) performed best on this dataset. Variables like nox, rad, dis, tax, age,indus were the most predictive of whether a suburb’s crime rate is above the median. LDA and Naive Bayes also performed reasonably well, while logistic regression was slightly better.