STA 6543 Assignment 3
Abigail Davis mjv135
2025-03-05
Chapter 04 (pg 189): 13, 14, 16
Chapter 04 Problem 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(MASS)##
## Attaching package: 'MASS'## The following object is masked from 'package:ISLR2':
##
## Bostonlibrary(e1071)
library(class)
attach(Weekly)(a) Produce some numerical and graphical summaries of the Weekly data. Do there appear to be any patterns?
head(Weekly)## Year Lag1 Lag2 Lag3 Lag4 Lag5 Volume Today Direction
## 1 1990 0.816 1.572 -3.936 -0.229 -3.484 0.1549760 -0.270 Down
## 2 1990 -0.270 0.816 1.572 -3.936 -0.229 0.1485740 -2.576 Down
## 3 1990 -2.576 -0.270 0.816 1.572 -3.936 0.1598375 3.514 Up
## 4 1990 3.514 -2.576 -0.270 0.816 1.572 0.1616300 0.712 Up
## 5 1990 0.712 3.514 -2.576 -0.270 0.816 0.1537280 1.178 Up
## 6 1990 1.178 0.712 3.514 -2.576 -0.270 0.1544440 -1.372 Downsummary(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 1.000000000Answer:
Because the “Up’ direction occurs more frequently than”Down”, this data could suggest that there is a upward trend in the Weekly data during the 1990-2010 period. All the Lags have similar statistics, indicating that the market may be relatively stable during this period. Volume and Year are strongly correlated, however the lag variables and today’s returns have very little correlation.
(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?
full_model <- glm(Direction ~ Lag1 + Lag2 + Lag3 + Lag4 + Lag5 + Volume, data = Weekly, family = binomial)
summary(full_model)##
## 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: 4Answer:
There is only one predictor that is significant at the 0.05 level, which is Lag2.
(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(full_model, type = "response")
probs[1:10]## 1 2 3 4 5 6 7 8
## 0.6086249 0.6010314 0.5875699 0.4816416 0.6169013 0.5684190 0.5786097 0.5151972
## 9 10
## 0.5715200 0.5554287contrasts(Direction)## Up
## Down 0
## Up 1pred = rep("Down", 1089)
pred[probs >.5] ="Up"table(pred, Direction)## Direction
## pred Down Up
## Down 54 48
## Up 430 557mean(pred == Direction)## [1] 0.5610652Answer:
Based on the confusion matrix, the model has an accuracy of ~56%. It does a good job predicting true positives, however it does struggle with false positives. It seems there is a bias toward the “Up” direction.
(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<2009)
Weekly.test = Weekly[!train,]
Direction.test = Direction[!train]
dim(Weekly.test)## [1] 104 9model2 <- glm(Direction ~ Lag2, data = Weekly, family = binomial, subset = train)
probs = predict(model2, Weekly.test, type = "response")pred = rep("Down", 104)
pred[probs>.5]="Up"
table(pred, Direction.test)## Direction.test
## pred Down Up
## Down 9 5
## Up 34 56mean(pred == Direction.test)## [1] 0.625Answer:
This means that this model performed better than the first model with a ~62.5 accuracy. This model still favors the “Up” direction and has more false positives. This suggests that even with delegated training period and only using the significant predictor, there is still a class imbalance causing bias.
(e) Repeat (d) using LDA.
ldamodel <- lda(Direction ~ Lag2, data = Weekly, family = binomial, subset = train)
ldamodel## Call:
## lda(Direction ~ Lag2, data = Weekly, family = binomial, subset = train)
##
## Prior probabilities of groups:
## Down Up
## 0.4477157 0.5522843
##
## Group means:
## Lag2
## Down -0.03568254
## Up 0.26036581
##
## Coefficients of linear discriminants:
## LD1
## Lag2 0.4414162lda.pred = predict(ldamodel, Weekly.test)
lda.class=lda.pred$class
table(lda.class, Direction.test)## Direction.test
## lda.class Down Up
## Down 9 5
## Up 34 56mean(lda.class == Direction.test)## [1] 0.625This means that this model performed better than the first model with a ~62.5% accuracy. It also performed exactly the same as the logistic regression with only Lag2. This model still favors the “Up” direction and has more false positives. This suggests that even with delegated training period and only using the significant predictor, there is still a class imbalance causing bias.
(f) Repeat (d) using QDA.
qdamodel <- qda(Direction ~ Lag2, data = Weekly, family = binomial, subset = train)
qdamodel## Call:
## qda(Direction ~ Lag2, data = Weekly, family = binomial, subset = train)
##
## Prior probabilities of groups:
## Down Up
## 0.4477157 0.5522843
##
## Group means:
## Lag2
## Down -0.03568254
## Up 0.26036581qda.class = predict(qdamodel, Weekly.test)$class
table(qda.class, Direction.test)## Direction.test
## qda.class Down Up
## Down 0 0
## Up 43 61mean(qda.class == Direction.test)## [1] 0.5865385Answer:
This means that this model performed better than the first model with a ~58.7% accuracy. However, it did perform worse than the logistic regression with only Log2 and LDA. This model still favors the “Up” direction and has a lot of false positives. This suggests that even with delegated training period and only using the significant predictor, there is still a class imbalance causing bias.
(g) Repeat (d) using KNN with K = 1
train.X = as.matrix(Lag2[train])
test.X = as.matrix(Lag2[!train])
train.Direction = Direction[train]
test.Direction = Direction[!train]set.seed(1)
knn.pred = knn(train.X, test.X, train.Direction, k=1)
table(knn.pred, test.Direction)## test.Direction
## knn.pred Down Up
## Down 21 30
## Up 22 31mean(knn.pred == test.Direction)## [1] 0.5Answer:
This means that this model performed worse than any other model with a ~50% accuracy. This model still favors the “Up” direction and has the most of false negatives. This suggests that even with delegated training period and only using the significant predictor, there is still a class imbalance causing bias.
(h) Repeat (d) using naive Bayes.
nb <- naiveBayes(Direction~Lag2, data = Weekly, subset=train)
nb##
## Naive Bayes Classifier for Discrete Predictors
##
## Call:
## naiveBayes.default(x = X, y = Y, laplace = laplace)
##
## A-priori probabilities:
## Y
## Down Up
## 0.4477157 0.5522843
##
## Conditional probabilities:
## Lag2
## Y [,1] [,2]
## Down -0.03568254 2.199504
## Up 0.26036581 2.317485pred.nb <- predict(nb, Weekly.test)
table(pred.nb, Direction.test)## Direction.test
## pred.nb Down Up
## Down 0 0
## Up 43 61mean(pred.nb == Direction.test)## [1] 0.5865385This means that this model performed better than the first model and the KNN model with a ~58.7 accuracy. It does the same as the QDA. It performs worse than the logistic regression with only Lag2 and the LDA. This model still favors the “Up” direction and has more false positives. This suggests that even with delegated training period and only using the significant predictor, there is still a class imbalance causing bias.
(i) Which of these methods appears to provide the best results on this data?
The logistic regression with only Lag2 and the LDA have the best results on this data and the highest accuracy. This is followed by the QDA and Naive Bayes, and lastly the full logistic model and KNN.
(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.
Here is an interaction term between Lag 2 and Lag 3:
model_interact1 <- glm(Direction ~ Lag2 * Lag3, data = Weekly, family = binomial, subset = train)
probs_interact1 = predict(model_interact1, Weekly.test, type = "response")
pred_interact1 = rep("Down", 104)
pred_interact1[probs_interact1 > .5] = "Up"
table(pred_interact1, Direction.test)## Direction.test
## pred_interact1 Down Up
## Down 8 4
## Up 35 57mean(pred_interact1 == Direction.test)## [1] 0.625This means that this model performed worse than the logistic model with only Lag2 with a ~60.58 accuracy. This model still favors the “Up” direction and has more false positives. This suggests that even with delegated training period and adding an interaction term between Lag2 and Lag3, there is still a class imbalance causing bias.
Here is KNN with K= 3, 10, 15:
set.seed(1)
knn_pred_3 = knn(train.X, test.X, train.Direction, k = 3)
mean(knn_pred_3 == test.Direction)## [1] 0.5480769knn_pred_10 = knn(train.X, test.X, train.Direction, k = 10)
mean(knn_pred_10 == test.Direction)## [1] 0.5865385knn_pred_15 = knn(train.X, test.X, train.Direction, k = 15)
mean(knn_pred_15 == test.Direction)## [1] 0.5865385This means that all 3 of these KNNs performed better than the k=1 model with the best being k=10 and k=15 a ~58.65% accuracy.
Here is an LDA with Lag1, Lag2, and Volume:
lda_model2 <- lda(Direction ~ Lag1 + Lag2 + Volume, data = Weekly, subset = train)
lda_pred2 = predict(lda_model2, Weekly.test)$class
table(lda_pred2, Direction.test)## Direction.test
## lda_pred2 Down Up
## Down 27 33
## Up 16 28mean(lda_pred2 == Direction.test)## [1] 0.5288462This means that this model performed worse than the first LDA model with a ~58.65% accuracy.. This model still favors the “Up” direction and has more false positives. This suggests that even with delegated training period and only using the significant predictor, there is still a class imbalance causing bias.
This proves that our original logistic regression with only Lag2 and the LDA have the best results on this data and the highest accuracy.
Chapter 04 Problem 14
In this problem, you will develop a model to predict whether a given car gets high or low gas mileage based on the Auto data set.
attach(Auto)(a) Create a binary variable, mpg01, that contains a 1 if mpg contains a value above its median, and a 0 if mpg contains a value below its median. You can compute the median using the median() function. Note you may find it helpful to use the data.frame() function to create a single data set containing both mpg01 and the other Auto variables.
mpg01<- median(Auto$mpg)
mpg01 <- ifelse(Auto$mpg > mpg01, 1, 0)
Auto<-data.frame(Auto,mpg01)
head(Auto)## mpg cylinders displacement horsepower weight acceleration year origin
## 1 18 8 307 130 3504 12.0 70 1
## 2 15 8 350 165 3693 11.5 70 1
## 3 18 8 318 150 3436 11.0 70 1
## 4 16 8 304 150 3433 12.0 70 1
## 5 17 8 302 140 3449 10.5 70 1
## 6 15 8 429 198 4341 10.0 70 1
## name mpg01
## 1 chevrolet chevelle malibu 0
## 2 buick skylark 320 0
## 3 plymouth satellite 0
## 4 amc rebel sst 0
## 5 ford torino 0
## 6 ford galaxie 500 0(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.
pairs(Auto)
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 1.0000000par(mfrow = c(2,2))
boxplot(cylinders ~ mpg01, data = Auto, main = "MPG01 by Cylinders")
boxplot(displacement ~ mpg01, data = Auto, main = "MPG01 by Displacement")
boxplot(horsepower ~ mpg01, data = Auto, main = "MPG01 by Horsepower")
boxplot(weight ~ mpg01, data = Auto, main = "MPG01 by Weight")
The correlation matrix for this dataset provides us insight that MPG01
is negatively correlated with variables: cylinders, displacement,
horsepower, and weight. After producing boxplots of these variables, we
can also tell that when the mpg is above the median (1), then these
variables will be lower than when mpg is below the median (0).
(c) Split the data into a training set and a test set.
set.seed(1)
train <- sample(nrow(Auto), size = 0.7*nrow(Auto))
train_set <- Auto[train,]
test_set <- Auto[-train,]
mpg01.test<-mpg01[-train](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?
lda_mpg01 <- lda(mpg01 ~ cylinders + displacement + horsepower + weight, data = train_set)
lda_pred_mpg01<- predict(lda_mpg01, test_set)
lda_class_mpg01<-lda_pred_mpg01$class
table(lda_class_mpg01, mpg01.test)## mpg01.test
## lda_class_mpg01 0 1
## 0 50 3
## 1 11 541-mean(lda_class_mpg01 == mpg01.test)## [1] 0.1186441Answer:
There is an approximate 11.86% test error of the model.
(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_mpg01 <- qda(mpg01 ~ cylinders + displacement + horsepower + weight, data = train_set)
qda_pred_mpg01 <- predict(qda_mpg01, test_set)
qda_class_mpg01 <- qda_pred_mpg01$class
table(qda_class_mpg01, mpg01.test)## mpg01.test
## qda_class_mpg01 0 1
## 0 52 5
## 1 9 521-mean(qda_class_mpg01 == mpg01.test)## [1] 0.1186441Answer:
Same as the LDA, the QDA approximated an 11.86% test error of the model.
(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?
log_mpg01 <- glm(mpg01 ~ cylinders + displacement + horsepower + weight, data = train_set, family = binomial)
log_pred_mpg01 <- predict(log_mpg01, test_set, type = "response")
log_class_mpg01 <- ifelse(log_pred_mpg01 > 0.5, 1, 0)
table(log_class_mpg01,mpg01.test)## mpg01.test
## log_class_mpg01 0 1
## 0 53 3
## 1 8 541-mean(log_class_mpg01 == mpg01.test)## [1] 0.09322034Answer:
The logistic regression model approximated an 9.32% test error of the model.
(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?
nb_mpg01 <- naiveBayes(mpg01 ~ cylinders + displacement + horsepower + weight, data = train_set)
nb_pred_mpg01 <- predict(nb_mpg01, test_set)
table(nb_pred_mpg01, mpg01.test)## mpg01.test
## nb_pred_mpg01 0 1
## 0 52 4
## 1 9 531-mean(nb_pred_mpg01 == mpg01.test)## [1] 0.1101695Answer:
The naive Bayes model approximated an 11.02% test error of the model.
(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
train.X <- cbind(cylinders, displacement, horsepower, weight)[train,]
test.X <- cbind(cylinders, displacement, horsepower, weight)[-train,]
train.mpg01 <- mpg01[train]
set.seed(1)
pred.knn <-knn(train.X, test.X, train.mpg01, k=1)
table(pred.knn, mpg01.test)## mpg01.test
## pred.knn 0 1
## 0 51 6
## 1 10 511 - mean(pred.knn == mpg01.test)## [1] 0.1355932The KNN model for K = 1 has a test error rate of approximately 13.56%.
pred.knn <-knn(train.X, test.X, train.mpg01, k=100)
table(pred.knn, mpg01.test)## mpg01.test
## pred.knn 0 1
## 0 50 7
## 1 11 501 - mean(pred.knn == mpg01.test)## [1] 0.1525424The KNN model for K = 100 has a test error rate of approximately 15.25%.
pred.knn <-knn(train.X, test.X, train.mpg01, k=50)
table(pred.knn, mpg01.test)## mpg01.test
## pred.knn 0 1
## 0 49 7
## 1 12 501 - mean(pred.knn == mpg01.test)## [1] 0.1610169The KNN model for K = 50 has a test error rate of approximately 15.25%.
pred.knn <-knn(train.X, test.X, train.mpg01, k=15)
table(pred.knn, mpg01.test)## mpg01.test
## pred.knn 0 1
## 0 49 5
## 1 12 521 - mean(pred.knn == mpg01.test)## [1] 0.1440678The KNN model for K = 15 has a test error rate of approximately 14.41%.
pred.knn <-knn(train.X, test.X, train.mpg01, k=5)
table(pred.knn, mpg01.test)## mpg01.test
## pred.knn 0 1
## 0 51 5
## 1 10 521 - mean(pred.knn == mpg01.test)## [1] 0.1271186The KNN model for K = 5 has a test error rate of approximately 12.71%.
pred.knn <-knn(train.X, test.X, train.mpg01, k=3)
table(pred.knn, mpg01.test)## mpg01.test
## pred.knn 0 1
## 0 52 4
## 1 9 531 - mean(pred.knn == mpg01.test)## [1] 0.1101695The KNN model for K = 3 has a test error rate of approximately 11.02%.
pred.knn <-knn(train.X, test.X, train.mpg01, k=2)
table(pred.knn, mpg01.test)## mpg01.test
## pred.knn 0 1
## 0 52 6
## 1 9 511 - mean(pred.knn == mpg01.test)## [1] 0.1271186The KNN model for K = 3 has a test error rate of approximately 14.41%.
Answer: From the KNN models I tried, the K = 3 model had the lowest error rate of 11.02% making it the best model.
Chapter 04 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.: 0.08205 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 black
## Min. : 1.000 Min. :187.0 Min. :12.60 Min. : 0.32
## 1st Qu.: 4.000 1st Qu.:279.0 1st Qu.:17.40 1st Qu.:375.38
## Median : 5.000 Median :330.0 Median :19.05 Median :391.44
## Mean : 9.549 Mean :408.2 Mean :18.46 Mean :356.67
## 3rd Qu.:24.000 3rd Qu.:666.0 3rd Qu.:20.20 3rd Qu.:396.23
## Max. :24.000 Max. :711.0 Max. :22.00 Max. :396.90
## lstat medv
## Min. : 1.73 Min. : 5.00
## 1st Qu.: 6.95 1st Qu.:17.02
## Median :11.36 Median :21.20
## Mean :12.65 Mean :22.53
## 3rd Qu.:16.95 3rd Qu.:25.00
## Max. :37.97 Max. :50.00high_crim<- median(Boston$crim)
high_crim <- ifelse(Boston$crim > high_crim, 1, 0)
Boston<-data.frame(Boston,high_crim)
head(Boston)## crim zn indus chas nox rm age dis rad tax ptratio black lstat
## 1 0.00632 18 2.31 0 0.538 6.575 65.2 4.0900 1 296 15.3 396.90 4.98
## 2 0.02731 0 7.07 0 0.469 6.421 78.9 4.9671 2 242 17.8 396.90 9.14
## 3 0.02729 0 7.07 0 0.469 7.185 61.1 4.9671 2 242 17.8 392.83 4.03
## 4 0.03237 0 2.18 0 0.458 6.998 45.8 6.0622 3 222 18.7 394.63 2.94
## 5 0.06905 0 2.18 0 0.458 7.147 54.2 6.0622 3 222 18.7 396.90 5.33
## 6 0.02985 0 2.18 0 0.458 6.430 58.7 6.0622 3 222 18.7 394.12 5.21
## medv high_crim
## 1 24.0 0
## 2 21.6 0
## 3 34.7 0
## 4 33.4 0
## 5 36.2 0
## 6 28.7 0plotting
cor(Boston)## 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
## high_crim 0.40939545 -0.43615103 0.60326017 0.070096774 0.72323480
## rm age dis rad tax
## crim -0.21924670 0.35273425 -0.37967009 0.625505145 0.58276431
## zn 0.31199059 -0.56953734 0.66440822 -0.311947826 -0.31456332
## indus -0.39167585 0.64477851 -0.70802699 0.595129275 0.72076018
## chas 0.09125123 0.08651777 -0.09917578 -0.007368241 -0.03558652
## nox -0.30218819 0.73147010 -0.76923011 0.611440563 0.66802320
## rm 1.00000000 -0.24026493 0.20524621 -0.209846668 -0.29204783
## age -0.24026493 1.00000000 -0.74788054 0.456022452 0.50645559
## dis 0.20524621 -0.74788054 1.00000000 -0.494587930 -0.53443158
## rad -0.20984667 0.45602245 -0.49458793 1.000000000 0.91022819
## tax -0.29204783 0.50645559 -0.53443158 0.910228189 1.00000000
## ptratio -0.35550149 0.26151501 -0.23247054 0.464741179 0.46085304
## black 0.12806864 -0.27353398 0.29151167 -0.444412816 -0.44180801
## lstat -0.61380827 0.60233853 -0.49699583 0.488676335 0.54399341
## medv 0.69535995 -0.37695457 0.24992873 -0.381626231 -0.46853593
## high_crim -0.15637178 0.61393992 -0.61634164 0.619786249 0.60874128
## ptratio black lstat medv high_crim
## crim 0.2899456 -0.38506394 0.4556215 -0.3883046 0.40939545
## zn -0.3916785 0.17552032 -0.4129946 0.3604453 -0.43615103
## indus 0.3832476 -0.35697654 0.6037997 -0.4837252 0.60326017
## chas -0.1215152 0.04878848 -0.0539293 0.1752602 0.07009677
## nox 0.1889327 -0.38005064 0.5908789 -0.4273208 0.72323480
## rm -0.3555015 0.12806864 -0.6138083 0.6953599 -0.15637178
## age 0.2615150 -0.27353398 0.6023385 -0.3769546 0.61393992
## dis -0.2324705 0.29151167 -0.4969958 0.2499287 -0.61634164
## rad 0.4647412 -0.44441282 0.4886763 -0.3816262 0.61978625
## tax 0.4608530 -0.44180801 0.5439934 -0.4685359 0.60874128
## ptratio 1.0000000 -0.17738330 0.3740443 -0.5077867 0.25356836
## black -0.1773833 1.00000000 -0.3660869 0.3334608 -0.35121093
## lstat 0.3740443 -0.36608690 1.0000000 -0.7376627 0.45326273
## medv -0.5077867 0.33346082 -0.7376627 1.0000000 -0.26301673
## high_crim 0.2535684 -0.35121093 0.4532627 -0.2630167 1.00000000crim, indus, nox, age, dis, rad, and tax seem somewhat correlated with high_crim.
splitting
set.seed(1)
train <- sample(nrow(Boston), size = 0.7*nrow(Boston))
train_set <- Boston[train,]
test_set <- Boston[-train,]
high_crim.test<-high_crim[-train]Logistic regression:
log_high_crim <- glm(high_crim ~ . -high_crim - crim, data = train_set, family = binomial)
log_pred_high_crim <- predict(log_high_crim, test_set, type = "response")
log_class_high_crim <- ifelse(log_pred_high_crim > 0.5, 1, 0)
table(log_class_high_crim,high_crim.test)## high_crim.test
## log_class_high_crim 0 1
## 0 62 5
## 1 11 741- mean(log_class_high_crim == high_crim.test)## [1] 0.1052632Answer:
The full logistic regression model approximated an 10.53% test error of the model.
log_high_crim2 <- glm(high_crim ~ . -high_crim - crim -rm -lstat -medv, data = train_set, family = binomial)
log_pred_high_crim2 <- predict(log_high_crim2, test_set, type = "response")
log_class_high_crim2 <- ifelse(log_pred_high_crim2 > 0.5, 1, 0)
table(log_class_high_crim2,high_crim.test)## high_crim.test
## log_class_high_crim2 0 1
## 0 61 5
## 1 12 741- mean(log_class_high_crim2 == high_crim.test)## [1] 0.1118421Answer:
The logistic model without the variables approximated an 11.18% test error of the model, indicating that the full model had a slightly higher accuracy.
LDA
lda_high_crim <- lda(high_crim ~ . -high_crim - crim, data = train_set)
lda_pred_high_crim<- predict(lda_high_crim, test_set)
lda_class_high_crim<-lda_pred_high_crim$class
table(lda_class_high_crim, high_crim.test)## high_crim.test
## lda_class_high_crim 0 1
## 0 70 20
## 1 3 591- mean(lda_class_high_crim == high_crim.test)## [1] 0.1513158The full LDA model approximated an 15.13% test error of the model.
lda_high_crim2 <- lda(high_crim ~ . -high_crim - crim -rm -lstat -medv, data = train_set)
lda_pred_high_crim2<- predict(lda_high_crim2, test_set)
lda_class_high_crim2<-lda_pred_high_crim2$class
table(lda_class_high_crim2, high_crim.test)## high_crim.test
## lda_class_high_crim2 0 1
## 0 71 20
## 1 2 591- mean(lda_class_high_crim2 == high_crim.test)## [1] 0.1447368The LDA model without the weaker correlated variables approximated a 14.47% test error of the model. This indicates that this model is slightly better than the full model unlike the logistic model.
naive Bayes
nb_high_crim <- naiveBayes(high_crim ~ . -high_crim - crim, data = train_set)
nb_pred_high_crim <- predict(nb_high_crim, test_set)
table(nb_pred_high_crim, high_crim.test)## high_crim.test
## nb_pred_high_crim 0 1
## 0 68 20
## 1 5 591- mean(nb_pred_high_crim == high_crim.test)## [1] 0.1644737The full naive Bayes model approximated an 16.45% test error of the model.
nb_high_crim2 <- naiveBayes(high_crim ~ . -high_crim - crim -rm -lstat -medv, data = train_set)
nb_pred_high_crim2 <- predict(nb_high_crim2, test_set)
table(nb_pred_high_crim2, high_crim.test)## high_crim.test
## nb_pred_high_crim2 0 1
## 0 65 22
## 1 8 571- mean(nb_pred_high_crim2 == high_crim.test)## [1] 0.1973684The naive Bayes model without the weaker correlated variables approximated a 19.73% test error of the model. This indicates that this model is slightly better than the full model unlike the logistic model.
KNN models
train.X <- cbind(zn, indus, chas, nox, rm, age, dis, rad, tax,ptratio,black,lstat,medv)[train,]
test.X <- cbind(zn, indus, chas, nox, rm, age, dis, rad, tax,ptratio,black,lstat,medv)[-train,]
train.high_crim <- high_crim[train]
set.seed(1)
pred.knn_high_crim <-knn(train.X, test.X, train.high_crim, k=1)
table(pred.knn_high_crim, high_crim.test)## high_crim.test
## pred.knn_high_crim 0 1
## 0 65 6
## 1 8 731 - mean(pred.knn_high_crim == high_crim.test)## [1] 0.09210526The full KNN model with k=1 gave us a test error rate of approximately 11.18%.
pred.knn_high_crim <-knn(train.X, test.X, train.high_crim, k=10)
table(pred.knn_high_crim, high_crim.test)## high_crim.test
## pred.knn_high_crim 0 1
## 0 59 9
## 1 14 701 - mean(pred.knn_high_crim == high_crim.test)## [1] 0.1513158The full KNN model with k=10 gave us a test error rate of approximately 14.47%
pred.knn_high_crim <-knn(train.X, test.X, train.high_crim, k=6)
table(pred.knn_high_crim, high_crim.test)## high_crim.test
## pred.knn_high_crim 0 1
## 0 61 6
## 1 12 731 - mean(pred.knn_high_crim == high_crim.test)## [1] 0.1184211The full KNN model with k=6 gave us a test error rate of approximately 11.84%.
train.X <- cbind(zn, indus, chas, nox, age, dis, rad, tax,ptratio,black)[train,]
test.X <- cbind(zn, indus, chas, nox, age, dis, rad, tax,ptratio,black)[-train,]
train.high_crim <- high_crim[train]
set.seed(1)
pred.knn_high_crim <-knn(train.X, test.X, train.high_crim, k=1)
table(pred.knn_high_crim, high_crim.test)## high_crim.test
## pred.knn_high_crim 0 1
## 0 64 4
## 1 9 751 - mean(pred.knn_high_crim == high_crim.test)## [1] 0.08552632The KNN model with high correlated variables removed, produced a test error rate of approximately 8.55% at k=1.
pred.knn_high_crim <-knn(train.X, test.X, train.high_crim, k=5)
table(pred.knn_high_crim, high_crim.test)## high_crim.test
## pred.knn_high_crim 0 1
## 0 62 5
## 1 11 741 - mean(pred.knn_high_crim == high_crim.test)## [1] 0.1052632The KNN model with high correlated variables removed, produced a test error rate of approximately 10.53% at k=5.
pred.knn_high_crim <-knn(train.X, test.X, train.high_crim, k=3)
table(pred.knn_high_crim, high_crim.test)## high_crim.test
## pred.knn_high_crim 0 1
## 0 64 4
## 1 9 751 - mean(pred.knn_high_crim == high_crim.test)## [1] 0.08552632The KNN model with high correlated variables removed, produced a test error rate of approximately 8.55% at k=3 which is the same as k=1.