library(MASS)
library(tidyverse)
library(ISLR2)
library(corrplot)
library(caret)
library(e1071)
library(class)Classification
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.
data(Weekly)13A
Produce some numerical and graphical summaries of the Weekly data. Do there appear to be any patterns?
plot(Weekly)
There appears to be a logarithmic/exponential relationship between Year and Volume.
Weekly |>
select(-Direction) |>
cor() |>
corrplot(method = "number")
No variable are highly correlated with one another.
13B
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?
mod.log1 <- glm(Direction ~ Lag1 + Lag2 + Lag3 + Lag4 +Lag5 + Volume, data = Weekly, family = binomial)
summary(mod.log1)
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: 4Lag2 is the only significant variable used.
13C
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(mod.log1, type = "response")
preds <- rep("Down",nrow(Weekly))
preds[probs > .5] <- "Up"
preds <- as.factor(preds)
confusionMatrix(preds, Weekly$Direction)Confusion Matrix and Statistics
Reference
Prediction Down Up
Down 54 48
Up 430 557
Accuracy : 0.5611
95% CI : (0.531, 0.5908)
No Information Rate : 0.5556
P-Value [Acc > NIR] : 0.369
Kappa : 0.035
Mcnemar's Test P-Value : <2e-16
Sensitivity : 0.11157
Specificity : 0.92066
Pos Pred Value : 0.52941
Neg Pred Value : 0.56434
Prevalence : 0.44444
Detection Rate : 0.04959
Detection Prevalence : 0.09366
Balanced Accuracy : 0.51612
'Positive' Class : Down
mean(preds == Weekly$Direction)[1] 0.561065256% Accuracy. The models is predicting “Up” far more often than “Down”, resulting in poor accuracy and sensitivity.
13D
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[Weekly$Year < 2009,]
test <- Weekly[Weekly$Year > 2008,]
mod.log2 <- glm(Direction ~ Lag2, data = train, family = binomial)
probs <- predict(mod.log1, newdata = test, type = "response")
preds <- ifelse(probs < .5, "Down","Up")
confusionMatrix(as.factor(preds), test$Direction)Confusion Matrix and Statistics
Reference
Prediction Down Up
Down 17 13
Up 26 48
Accuracy : 0.625
95% CI : (0.5247, 0.718)
No Information Rate : 0.5865
P-Value [Acc > NIR] : 0.24395
Kappa : 0.1907
Mcnemar's Test P-Value : 0.05466
Sensitivity : 0.3953
Specificity : 0.7869
Pos Pred Value : 0.5667
Neg Pred Value : 0.6486
Prevalence : 0.4135
Detection Rate : 0.1635
Detection Prevalence : 0.2885
Balanced Accuracy : 0.5911
'Positive' Class : Down
Accuracy improves to 62.5% and sensitivity improves to 40%.
13E
Repeat (d) using LDA.
mod.lda <- lda(Direction ~ Lag2, data = train)
preds <- predict(mod.lda, test)
confusionMatrix(preds$class, test$Direction)Confusion Matrix and Statistics
Reference
Prediction Down Up
Down 9 5
Up 34 56
Accuracy : 0.625
95% CI : (0.5247, 0.718)
No Information Rate : 0.5865
P-Value [Acc > NIR] : 0.2439
Kappa : 0.1414
Mcnemar's Test P-Value : 7.34e-06
Sensitivity : 0.20930
Specificity : 0.91803
Pos Pred Value : 0.64286
Neg Pred Value : 0.62222
Prevalence : 0.41346
Detection Rate : 0.08654
Detection Prevalence : 0.13462
Balanced Accuracy : 0.56367
'Positive' Class : Down
LDA results in 62% accuracy
13F
Repeat (d) using QDA.
mod.qda <- qda(Direction ~ Lag2, data = train)
preds <- predict(mod.qda, test)
confusionMatrix(preds$class, test$Direction)Confusion Matrix and Statistics
Reference
Prediction Down Up
Down 0 0
Up 43 61
Accuracy : 0.5865
95% CI : (0.4858, 0.6823)
No Information Rate : 0.5865
P-Value [Acc > NIR] : 0.5419
Kappa : 0
Mcnemar's Test P-Value : 1.504e-10
Sensitivity : 0.0000
Specificity : 1.0000
Pos Pred Value : NaN
Neg Pred Value : 0.5865
Prevalence : 0.4135
Detection Rate : 0.0000
Detection Prevalence : 0.0000
Balanced Accuracy : 0.5000
'Positive' Class : Down
QDA results in an accuracy of 58%
13H
Repeat (d) using naive Bayes.
mod.bayes <- naiveBayes(Direction ~ Lag2, data = train)
preds <- predict(mod.bayes, test)
confusionMatrix(preds, test$Direction)Confusion Matrix and Statistics
Reference
Prediction Down Up
Down 0 0
Up 43 61
Accuracy : 0.5865
95% CI : (0.4858, 0.6823)
No Information Rate : 0.5865
P-Value [Acc > NIR] : 0.5419
Kappa : 0
Mcnemar's Test P-Value : 1.504e-10
Sensitivity : 0.0000
Specificity : 1.0000
Pos Pred Value : NaN
Neg Pred Value : 0.5865
Prevalence : 0.4135
Detection Rate : 0.0000
Detection Prevalence : 0.0000
Balanced Accuracy : 0.5000
'Positive' Class : Down
Naive Bayes has an accuracy of 58%
13I
Which of these methods appears to provide the best results on this data?
LDA and Logistic Regression achieve the same results of about 62% accuracy.
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.
data(Auto)14A
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.
df <- Auto |>
mutate(mpg01 = if_else(mpg <= median(mpg), 0, 1)) |>
mutate(mpg01 = as.factor(mpg01))
table(df$mpg01)
0 1
196 196 14B
Explore the data graphically in order to investigate the association between mpg01 and the other features. Which of the other features seem most likely to be useful in predicting mpg01? Scatterplots and boxplots may be useful tools to answer this question. Describe your findings.
df |> dplyr::select(-c(mpg, name)) |>
plot()
Displacement, horsepower, acceleration, weight, and cylinders all appear to separate the data fairly well.
df |>
ggplot(aes(x = horsepower, y = acceleration, color =mpg01)) +
geom_point()
14C
Split the data into a training set and a test set.
set.seed(333)
idx <- sample(nrow(df),nrow(df)*.75)
train <- df[idx,]
test <- df[-idx,]14D
Perform LDA on the training data in order to predict mpg01 using the variables that seemed most associated with mpg01 in (b). What is the test error of the model obtained?
lda.mod <- lda(mpg01 ~ weight + acceleration + horsepower + year, data = train)
preds <- predict(lda.mod, test)
confusionMatrix(preds$class, test$mpg01)Confusion Matrix and Statistics
Reference
Prediction 0 1
0 45 4
1 9 40
Accuracy : 0.8673
95% CI : (0.7838, 0.9274)
No Information Rate : 0.551
P-Value [Acc > NIR] : 1.976e-11
Kappa : 0.7347
Mcnemar's Test P-Value : 0.2673
Sensitivity : 0.8333
Specificity : 0.9091
Pos Pred Value : 0.9184
Neg Pred Value : 0.8163
Prevalence : 0.5510
Detection Rate : 0.4592
Detection Prevalence : 0.5000
Balanced Accuracy : 0.8712
'Positive' Class : 0
87% accuracy
14E
Perform QDA on the training data in order to predict mpg01 using the variables that seemed most associated with mpg01 in (b). What is the test error of the model obtained?
qda.mod <- qda(mpg01 ~ weight + acceleration + horsepower + year, data = train)
preds <- predict(qda.mod, test)
confusionMatrix(preds$class, test$mpg01)Confusion Matrix and Statistics
Reference
Prediction 0 1
0 47 5
1 7 39
Accuracy : 0.8776
95% CI : (0.7959, 0.9351)
No Information Rate : 0.551
P-Value [Acc > NIR] : 3.596e-12
Kappa : 0.7536
Mcnemar's Test P-Value : 0.7728
Sensitivity : 0.8704
Specificity : 0.8864
Pos Pred Value : 0.9038
Neg Pred Value : 0.8478
Prevalence : 0.5510
Detection Rate : 0.4796
Detection Prevalence : 0.5306
Balanced Accuracy : 0.8784
'Positive' Class : 0
88% accuracy.
14F
Perform logistic regression on the training data in order to predict mpg01 using the variables that seemed most associated with mpg01 in (b). What is the test error of the model obtained?
log.mod <- glm(mpg01 ~ weight + acceleration + horsepower + year, data = train, family = "binomial")
preds <- predict(log.mod, test, type = "response")
preds <- if_else(preds >= .5, 1, 0) |>
as.factor()
confusionMatrix(preds, test$mpg01)Confusion Matrix and Statistics
Reference
Prediction 0 1
0 47 4
1 7 40
Accuracy : 0.8878
95% CI : (0.808, 0.9426)
No Information Rate : 0.551
P-Value [Acc > NIR] : 5.976e-13
Kappa : 0.7746
Mcnemar's Test P-Value : 0.5465
Sensitivity : 0.8704
Specificity : 0.9091
Pos Pred Value : 0.9216
Neg Pred Value : 0.8511
Prevalence : 0.5510
Detection Rate : 0.4796
Detection Prevalence : 0.5204
Balanced Accuracy : 0.8897
'Positive' Class : 0
89%
14G
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?
mod.bayes <- naiveBayes(mpg01 ~ weight + acceleration + horsepower + year, data = train)
preds <- predict(mod.bayes, test)
confusionMatrix(preds, test$mpg01)Confusion Matrix and Statistics
Reference
Prediction 0 1
0 45 5
1 9 39
Accuracy : 0.8571
95% CI : (0.7719, 0.9196)
No Information Rate : 0.551
P-Value [Acc > NIR] : 9.971e-11
Kappa : 0.7137
Mcnemar's Test P-Value : 0.4227
Sensitivity : 0.8333
Specificity : 0.8864
Pos Pred Value : 0.9000
Neg Pred Value : 0.8125
Prevalence : 0.5510
Detection Rate : 0.4592
Detection Prevalence : 0.5102
Balanced Accuracy : 0.8598
'Positive' Class : 0
86%
14H
Perform KNN on the training data, with several values of K, in order to predict mpg01. Use only the variables that seemed most associated with mpg01 in (b). What test errors do you obtain? Which value of K seems to perform the best on this data set?
train.X <- cbind(train$weight,train$acceleration,train$horsepower,train$year)
test.X <- cbind(test$weight,test$acceleration,test$horsepower,test$year)
train.Direction <- df$mpg01[idx]
knn.pred <- knn(train = train.X, test = test.X, cl = train.Direction, k = 5)
confusionMatrix(as.factor(knn.pred), test$mpg01)Confusion Matrix and Statistics
Reference
Prediction 0 1
0 46 8
1 8 36
Accuracy : 0.8367
95% CI : (0.7484, 0.9037)
No Information Rate : 0.551
P-Value [Acc > NIR] : 2.007e-09
Kappa : 0.67
Mcnemar's Test P-Value : 1
Sensitivity : 0.8519
Specificity : 0.8182
Pos Pred Value : 0.8519
Neg Pred Value : 0.8182
Prevalence : 0.5510
Detection Rate : 0.4694
Detection Prevalence : 0.5510
Balanced Accuracy : 0.8350
'Positive' Class : 0
86% with k=5
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)
str(Boston)'data.frame': 506 obs. of 13 variables:
$ crim : num 0.00632 0.02731 0.02729 0.03237 0.06905 ...
$ zn : num 18 0 0 0 0 0 12.5 12.5 12.5 12.5 ...
$ indus : num 2.31 7.07 7.07 2.18 2.18 2.18 7.87 7.87 7.87 7.87 ...
$ chas : int 0 0 0 0 0 0 0 0 0 0 ...
$ nox : num 0.538 0.469 0.469 0.458 0.458 0.458 0.524 0.524 0.524 0.524 ...
$ rm : num 6.58 6.42 7.18 7 7.15 ...
$ age : num 65.2 78.9 61.1 45.8 54.2 58.7 66.6 96.1 100 85.9 ...
$ dis : num 4.09 4.97 4.97 6.06 6.06 ...
$ rad : int 1 2 2 3 3 3 5 5 5 5 ...
$ tax : num 296 242 242 222 222 222 311 311 311 311 ...
$ ptratio: num 15.3 17.8 17.8 18.7 18.7 18.7 15.2 15.2 15.2 15.2 ...
$ lstat : num 4.98 9.14 4.03 2.94 5.33 ...
$ medv : num 24 21.6 34.7 33.4 36.2 28.7 22.9 27.1 16.5 18.9 ...df <- Boston |>
mutate(response = if_else(crim >= median(crim), 1, 0)) |>
dplyr::select(-crim) |>
mutate(response = as.factor(response))idx <- sample(nrow(df),nrow(df)*.8)
train <- df[idx,]
test <- df[-idx,]
log.mod <- glm(response ~ ., data = train, family = "binomial")
preds <- predict(log.mod, test, type = "response")
preds <- if_else(preds >= .5, 1, 0) |>
as.factor()
confusionMatrix(preds, test$response)Confusion Matrix and Statistics
Reference
Prediction 0 1
0 47 4
1 5 46
Accuracy : 0.9118
95% CI : (0.8391, 0.9589)
No Information Rate : 0.5098
P-Value [Acc > NIR] : <2e-16
Kappa : 0.8235
Mcnemar's Test P-Value : 1
Sensitivity : 0.9038
Specificity : 0.9200
Pos Pred Value : 0.9216
Neg Pred Value : 0.9020
Prevalence : 0.5098
Detection Rate : 0.4608
Detection Prevalence : 0.5000
Balanced Accuracy : 0.9119
'Positive' Class : 0
91% accuracy is achieved with a logit model