title: “Assignment_3” author: “Monique Villarreal” date: “2022-09-24” output: html_document =========================================================================
Question 13. Using the Weekly data set.
a) Produce some numerical and graphical summaries of the Weekly data. Are there any patterns?
In reviewing the output, the majority of the correlations between the lag variables and today’s returns are close to zero (i.e. little correlation between today’s return and the previous day’s). There is a larger correlation between year and volume. This relationship can be seen in the scatter plot matrix.
library(ISLR2)
dim(Weekly)## [1] 1089 9summary(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 1.000000000attach(Weekly)
plot(Weekly)
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?
In reviewing the output, Lag 2 appears to be the only predictor with a significant p-value.
glm.wfit=glm(Direction~Lag1+Lag2+Lag3+Lag4+Lag5+Volume, data=Weekly, family=binomial)
summary(glm.wfit)##
## 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: 4summary(glm.wfit)$coef[,4]## (Intercept) Lag1 Lag2 Lag3 Lag4 Lag5
## 0.001898848 0.118144368 0.029601361 0.546923890 0.293653342 0.583348244
## Volume
## 0.537674762c) Compute the confusion matrix and overall fraction of correct prediction. Explain what the confusion matrix is telling you about the types of mistakes made by logic regression.
The model will have correctly predicted movement in the market 56% of the time.
## Direction
## glm.pred Down Up
## Down 54 48
## Up 430 557mean(glm.pred==Direction)## [1] 0.5610652d) Now fit the logistic regression model using a training data period from 1990 to 2008, with Lag2 as the only predictor. Compute the confusion matrix and the overall fraction of correct prediction for the held out data (data from 2009 - 2010).
Has 62.5% accuracy rate
train=(Year<2009)
Weekly2.0.2009=Weekly[!train, ]
Direction2.0.2009=Direction[!train]
Weekly.fit=glm(Direction~Lag2,data=Weekly,family=binomial,subset=train)
Weekly.prob=predict(Weekly.fit,Weekly2.0.2009,type="response")
Weekly.pred=rep("Down",length(Weekly.prob))
Weekly.pred[Weekly.prob>0.5]="Up"
table(Weekly.pred,Direction2.0.2009)## Direction2.0.2009
## Weekly.pred Down Up
## Down 9 5
## Up 34 56mean(Weekly.pred==Direction2.0.2009)## [1] 0.625e) repeat using LDA
Has 62.5% accuracy rate
library(MASS)##
## Attaching package: 'MASS'## The following object is masked from 'package:ISLR2':
##
## Bostonlda.fit = lda(Direction~Lag2, data = Weekly, subset = train)
lda.fit## Call:
## lda(Direction ~ Lag2, data = Weekly, subset = train)
##
## Prior probabilities of groups:
## Down Up
## 0.4477157 0.5522843
##
## Group means:
## Lag2
## Down -0.03568254
## Up 0.26036581
##
## Coefficients of linear discriminants:
## LD1
## Lag2 0.4414162lda.pred = predict(lda.fit, Weekly2.0.2009)
names(lda.pred)## [1] "class" "posterior" "x"lda.class = lda.pred$class
table(lda.class, Direction2.0.2009)## Direction2.0.2009
## lda.class Down Up
## Down 9 5
## Up 34 56mean(lda.class==Direction2.0.2009)## [1] 0.625f) repeat using QDA
Has 58.7% accuracy rate
qda.fit = qda(Direction~Lag2, data = Weekly, subset = train)
qda.fit## Call:
## qda(Direction ~ Lag2, data = Weekly, subset = train)
##
## Prior probabilities of groups:
## Down Up
## 0.4477157 0.5522843
##
## Group means:
## Lag2
## Down -0.03568254
## Up 0.26036581qda.class = predict(qda.fit, Weekly2.0.2009)$class
table(qda.class, Direction2.0.2009)## Direction2.0.2009
## qda.class Down Up
## Down 0 0
## Up 43 61mean(qda.class==Direction2.0.2009)## [1] 0.5865385g) repeat using KNN with K=1
Has a 50% accuracy rate
library(class)
train.X = as.matrix(Lag2[train])
test.X = as.matrix(Lag2[!train])
train.direction = Direction[train]
set.seed(1)
KNN.pred = knn(train.X, test.X, train.direction, k = 1)
table(KNN.pred, Direction2.0.2009)## Direction2.0.2009
## KNN.pred Down Up
## Down 21 30
## Up 22 31(21+31)/104## [1] 0.5h) repeat using naive Bayes.
Has a 58.6% accuracy rate
library(e1071)
NB.fit = naiveBayes(Direction~Lag2, data = Weekly, subset = train)
NB.fit##
## 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 2.317485NB.class = predict(NB.fit, Weekly2.0.2009)
table(NB.class, Direction2.0.2009)## Direction2.0.2009
## NB.class Down Up
## Down 0 0
## Up 43 61mean(NB.class==Direction2.0.2009)## [1] 0.5865385i) Which provides the best results on the data?
The LDA and logistic regression both have higher accuracy rates at 62.5%.
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 have the best results on the held out data. Note that you should also experiment with the values for K in the KNN classifier.
The LDA and logistic regression continued to have the same accuracy rate when introducing an interaction term. Increasing K improved the accuracy rate for KNN from 50% to about 54%.
Weekly.fit = glm(Direction~Lag2:Lag3 + Lag2, family = binomial, data = Weekly, subset = train)
Weekly.prob = predict(Weekly.fit, Weekly2.0.2009, type = "response")
Weekly.pred = rep("Down", length(Weekly.prob))
Weekly.pred[Weekly.prob > 0.5] = "Up"
Direction2.0.2009 = Direction[!train]
table(Weekly.pred, Direction2.0.2009)## Direction2.0.2009
## Weekly.pred Down Up
## Down 9 5
## Up 34 56mean(Weekly.pred==Direction2.0.2009)## [1] 0.625lda.fit = lda(Direction~Lag2:Lag3 + Lag2, data = Weekly, subset = train)
lda.pred = predict(lda.fit, Weekly2.0.2009)
lda.class = lda.pred$class
table(lda.class, Direction2.0.2009)## Direction2.0.2009
## lda.class Down Up
## Down 9 5
## Up 34 56mean(lda.class == Direction2.0.2009)## [1] 0.625train.X = as.matrix(Lag2[train])
test.X = as.matrix(Lag2[!train])
train.direction = Direction[train]
set.seed(1)
KNN.pred = knn(train.X, test.X, train.direction, k = 7)
table(KNN.pred, Direction2.0.2009)## Direction2.0.2009
## KNN.pred Down Up
## Down 15 20
## Up 28 41mean(KNN.pred == Direction2.0.2009)## [1] 0.5384615Question 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.
library(ISLR2)
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.
mpg01 = rep(0, length(mpg))
mpg01[mpg>median(mpg)] = 1
Auto = data.frame(Auto, mpg01)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? Scatter plots and box plots may be useful.
An association can be seen with displacement, horsepower, weight, and cylinders.
plot(Auto[,-9])
c) Split the data into a training set and a test set
Auto = data.frame(mpg01, apply(cbind(cylinders, weight, displacement, horsepower, acceleration), 2, scale), year)
train.A = (year %% 2 ==0)
test.A = !train.A
Auto.train = Auto[train.A,]
Auto.test = Auto[test.A,]
mpg01.test = mpg01[test.A]d) Perform LDA on the training data in order to predict mpg01 using the variables that seemed most associated with mpg01 in part (b). What is the test error of the model obtained?
Test error is 12.6%
lda.Afit = lda(mpg01~cylinders + displacement + weight + horsepower, data = Auto, subset = train.A )
autolda.pred = predict(lda.Afit, Auto.test)
table(autolda.pred$class, mpg01.test)## mpg01.test
## 0 1
## 0 86 9
## 1 14 73mean(autolda.pred$class != mpg01.test)## [1] 0.1263736e) Perform QDA on the training data in order to predict mpg01 using the variables that seemed most associated with mpg01 in part (b). What is the test error of the model obtained?
Test error is 13%
qda.Afit = qda(mpg01~cylinders + displacement + weight + horsepower, data = Auto, subset = train.A)
qda.Aclass = predict(qda.Afit, Auto.test)$class
table(qda.Aclass, mpg01.test)## mpg01.test
## qda.Aclass 0 1
## 0 89 13
## 1 11 69mean(qda.Aclass!=mpg01.test)## [1] 0.1318681f) 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?
Test error is 12.6%
glm.Afit = glm(mpg01~cylinders + displacement + weight + horsepower, data = Auto, subset = train.A)
glm.autoprob = predict(glm.Afit, Auto.test, type = "response")
glm.autopred = rep(0, length(glm.autoprob))
glm.autopred[glm.autoprob>0.5] = 1
table(glm.autopred, mpg01.test)## mpg01.test
## glm.autopred 0 1
## 0 86 9
## 1 14 73mean(glm.autopred!=mpg01.test)## [1] 0.1263736## 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?
```r
nb.Afit = naiveBayes(mpg01~cylinders + displacement + weight + horsepower, data = Auto, subset = train)
nb.Aclass = predict(nb.Afit, Auto.test)
mean(nb.Aclass!= mpg01.test)## [1] 0.1263736h) 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.auto = cbind(cylinders, displacement, weight, horsepower)[train.A,]
test.auto = cbind(cylinders, displacement, weight, horsepower)[!train.A,]
train.mpg01 = mpg01[train.A]
set.seed(1)
knn.autopred = knn(train.auto, test.auto, train.mpg01, k = 1)
mean(knn.autopred != mpg01.test)## [1] 0.1538462knn.autopred = knn(train.auto, test.auto, train.mpg01, k = 25)
mean(knn.autopred != mpg01.test)## [1] 0.1428571knn.autopred = knn(train.auto, test.auto, train.mpg01, k = 50)
mean(knn.autopred != mpg01.test)## [1] 0.1428571knn.autopred = knn(train.auto, test.auto, train.mpg01, k = 100)
mean(knn.autopred != mpg01.test)## [1] 0.1428571Question 16. Using 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.
attach(Boston)crime1 = rep(0, length(crim))
crime1[crim>median(crim)] = 1
Boston = data.frame(Boston, crime1)train = 1:(dim(Boston)[1]/2)
test = (dim(Boston)[1]/2+1):dim(Boston)[1]
Boston.train = Boston[train,]
Boston.test = Boston[test,]
crime1.test = crime1[test]plot(Boston)
### Can see a correlation between nox, tax, dis, medv, and lstat.
set.seed(1)
Boston.fit = glm(crime1~nox + tax + dis + medv + lstat, data = Boston, family = binomial)
Boston.probs = predict(Boston.fit, Boston.test, type ="response")
Boston.pred = rep(0, length(Boston.probs))
Boston.pred[Boston.probs > 0.5] = 1
table(Boston.pred, crime1.test)## crime1.test
## Boston.pred 0 1
## 0 75 15
## 1 15 148mean(Boston.pred == crime1.test)## [1] 0.8814229Logistic regression shows an 88% accuracy rate based on these associated predictors.
ldaBoston.fit = lda(crime1~nox+tax+dis+medv+lstat, data = Boston.train)
ldaBoston.pred = predict(ldaBoston.fit, Boston.test)
table(ldaBoston.pred$class, crime1.test)## crime1.test
## 0 1
## 0 80 16
## 1 10 147mean(ldaBoston.pred$class == crime1.test)## [1] 0.8972332The LDA shows an 89.7% accuracy rate.
nbBoston.fit = naiveBayes(crime1~nox+tax+dis+lstat, data = Boston, subset = train)
nbBoston.class = predict(nbBoston.fit, Boston.test)
table(nbBoston.class, crime1.test)## crime1.test
## nbBoston.class 0 1
## 0 75 18
## 1 15 145mean(nbBoston.class == crime1.test)## [1] 0.8695652The naive Bayes results in an 87% accuracy rate.
train.B = cbind(nox,tax,dis,lstat)[train,]
test.B = cbind(nox,tax,dis,lstat)[test,]
train.crime = crime1.test
set.seed(1)
Bostonknn.pred = knn(train.B, test.B, train.crime, k=1)
table(Bostonknn.pred, crime1.test)## crime1.test
## Bostonknn.pred 0 1
## 0 33 143
## 1 57 20mean(Bostonknn.pred == crime1.test)## [1] 0.2094862Bostonknn.pred = knn(train.B, test.B, train.crime, k=10)
table(Bostonknn.pred, crime1.test)## crime1.test
## Bostonknn.pred 0 1
## 0 43 20
## 1 47 143mean(Bostonknn.pred==crime1.test)## [1] 0.7351779Bostonknn.pred = knn(train.B, test.B, train.crime, k=25)
table(Bostonknn.pred, crime1.test)## crime1.test
## Bostonknn.pred 0 1
## 0 38 14
## 1 52 149mean(Bostonknn.pred==crime1.test)## [1] 0.7391304