Assignment 3 / Chapter 4
Tyler Crawford
2022-09-18
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)
attach(Weekly)a) 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
##
##
##
## 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 1.000000000pairs(Weekly)
- Viewing the correlation between all variables, except for Direction, shows us the coefficients between all the variable combinations. There does not seem to be any sort of correlation between any of the variables except for Year and Volume. This can be excepted though given the nature of the data set. The pairs functions show us each pair graphically and echoes my finding.
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)- Any predictor with a P-Value < 0.05 will be considered statistically significant. Lag2 has a P-Value of 0.0296 which means statistically significant. This predictor is the percentage returns from the 2 weeks previous.
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.
glm.prob=predict(glm.fit,type='response')
glm.pred=rep("Down",1089)
glm.pred[glm.prob>0.5]="Up"
table(glm.pred,Direction)## Direction
## glm.pred Down Up
## Down 54 48
## Up 430 557- Based on the confusion matrix, we have a prediction of ((54+557)/1089) 56.12% chance. This is just over half of the time which is OK, but not great, especially if investing in the stock market. We also have 430 false positives vs 54 true negatives which is not a good conversion.
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.2009=Weekly[!train,]
Direction.2009=Direction[!train]
glm.fit2=glm(Direction~Lag2,data=Weekly,family=binomial,subset=train)
glm.prob2=predict(glm.fit2,Weekly.2009,type='response')
glm.pred2=rep("Down",104)
glm.pred2[glm.prob2>0.5]="Up"
table(glm.pred2,Direction.2009)## Direction.2009
## glm.pred2 Down Up
## Down 9 5
## Up 34 56- Based on the confusion matrix formed through our held-out training data, we see a much better fraction of correct predictions than we did previously. An increase from 56.12% to 62.5% shows that we have a much better prediction than we did.
e) Repeat (d) using LDA.
library(MASS)##
## Attaching package: 'MASS'## The following object is masked from 'package:ISLR2':
##
## Bostonlda.fit=lda(Direction~Lag2,data=Weekly,subset=train)
lda.pred=predict(lda.fit,Weekly.2009)
lda.class=lda.pred$class
table(lda.class,Direction.2009)## Direction.2009
## lda.class Down Up
## Down 9 5
## Up 34 56- The confusion matrix for linear discriminant analysis gives us similar results to that of the logistic regression.
f) Repeat (d) using QDA.
qda.fit=qda(Direction~Lag2,data=Weekly,subset=train)
qda.class=predict(qda.fit,Weekly.2009)$class
table(qda.class,Direction.2009)## Direction.2009
## qda.class Down Up
## Down 0 0
## Up 43 61mean(qda.class==Direction.2009)## [1] 0.5865385- Quadratic Linear Analysis created a model with an accuracy of 58.65%, which is lower than the previous methods. Also, this model only predicted the correctness of “Up” trends and not “Down” trends.
g) Repeat (d) using KNN with K = 1.
library(class)
train.x=cbind(Lag1,Lag2)[train,]
test.x=cbind(Lag1,Lag2)[!train,]
train.Direction=Direction[train]
knn.pred=knn(train.x,test.x,train.Direction,k=1)
table(knn.pred,Direction.2009)## Direction.2009
## knn.pred Down Up
## Down 18 29
## Up 25 32mean(knn.pred==Direction.2009)## [1] 0.4807692- K-Nearest Neighbors created a model with an accuracy of 48.07%, which is worse than picking at random and the least accurate model thus far.
h) Repeat (d) using naive Bayes.
library(e1071)
nb.fit=naiveBayes(Direction~Lag2,data=Weekly,subset=train)
nb.pred=predict(nb.fit,Weekly.2009)
table(nb.pred,Direction.2009)## Direction.2009
## nb.pred Down Up
## Down 0 0
## Up 43 61mean(nb.pred==Direction.2009)## [1] 0.5865385- naïve Bayes created a model with an accuracy of 58.65%. This model performs much better than KNN (when k=1), equal to QDA, but not as well as LDA.
i) Which of these methods appears to provide the best results on this data?
- The LDA model when using solely the Lag2 predictor is the most accurate
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.
knn.pred=knn(train.x,test.x,train.Direction,k=3)
table(knn.pred,Direction.2009)## Direction.2009
## knn.pred Down Up
## Down 22 29
## Up 21 32mean(knn.pred==Direction.2009)## [1] 0.5192308knn.pred=knn(train.x,test.x,train.Direction,k=5)
table(knn.pred,Direction.2009)## Direction.2009
## knn.pred Down Up
## Down 22 32
## Up 21 29mean(knn.pred==Direction.2009)## [1] 0.4903846- Increasing our K value from 1 to 3 helped increase the accuracy of the KNN model from 48.07% to 51.92%. Increasing further from 3 to 5 decreased the accuracy of the model so 3 would be our ideal K value for KNN.
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.
detach(Weekly)
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=rep(0,length(mpg))
mpg01[mpg>median(mpg)]=1
Auto=data.frame(Auto,mpg01)
summary(Auto)## mpg cylinders displacement horsepower weight
## Min. : 9.00 Min. :3.000 Min. : 68.0 Min. : 46.0 Min. :1613
## 1st Qu.:17.00 1st Qu.:4.000 1st Qu.:105.0 1st Qu.: 75.0 1st Qu.:2225
## Median :22.75 Median :4.000 Median :151.0 Median : 93.5 Median :2804
## Mean :23.45 Mean :5.472 Mean :194.4 Mean :104.5 Mean :2978
## 3rd Qu.:29.00 3rd Qu.:8.000 3rd Qu.:275.8 3rd Qu.:126.0 3rd Qu.:3615
## Max. :46.60 Max. :8.000 Max. :455.0 Max. :230.0 Max. :5140
##
## acceleration year origin name
## Min. : 8.00 Min. :70.00 Min. :1.000 amc matador : 5
## 1st Qu.:13.78 1st Qu.:73.00 1st Qu.:1.000 ford pinto : 5
## Median :15.50 Median :76.00 Median :1.000 toyota corolla : 5
## Mean :15.54 Mean :75.98 Mean :1.577 amc gremlin : 4
## 3rd Qu.:17.02 3rd Qu.:79.00 3rd Qu.:2.000 amc hornet : 4
## Max. :24.80 Max. :82.00 Max. :3.000 chevrolet chevette: 4
## (Other) :365
## mpg01
## Min. :0.0
## 1st Qu.:0.0
## Median :0.5
## Mean :0.5
## 3rd Qu.:1.0
## Max. :1.0
## 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.
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,main="Cylinders vs mpg01")
boxplot(displacement~mpg01,main="displacement vs mpg01")
boxplot(horsepower~mpg01,main="horsepower vs mpg01")
boxplot(weight~mpg01,main= "Weight vs mpg01")
- One can safely assume that mpg is related to cylinders, displacement, horsepower, and weight of a vehicle. In plotting our newly created mpg binomial data frame, we can confirm that assumption to be true.
c) Split the data into a training set and a test set.
train=(year<76)
training=Auto[!train,]
test=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.fit=lda(mpg01~cylinders+displacement+horsepower+weight,data=Auto,subset=train)
lda.pred=predict(lda.fit,training)
lda.class=lda.pred$class
table(lda.class,mpg01.test)## mpg01.test
## lda.class 0 1
## 0 66 15
## 1 8 123mean(lda.class==mpg01.test)## [1] 0.8915094- The LDA model on the training data, vehicles from 1976 to 1982, created a test error of 89.15%. This is an excellent result for predicting MPG using cylinders, displacement, horsepower, and weight.
e) Perform QDA on the training data in order to predict mpg01 using the variables that seemed most associated with mpg01 in (b). What is the test error of the model obtained?
qda.fit=qda(mpg01~cylinders+displacement+horsepower+weight,data=Auto,subset=train)
qda.class=predict(qda.fit,training)$class
table(qda.class,mpg01.test)## mpg01.test
## qda.class 0 1
## 0 70 21
## 1 4 117mean(qda.class==mpg01.test)## [1] 0.8820755- The QDA model created a test error of 88.21%, less than that of the LDA model, but still very good odds of predicting MPG.
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~cylinders+displacement+horsepower+weight,data=Auto,family=binomial)
glm.pred=predict(glm.fit,training,type='response')
glm.prob=rep("Down",212)
glm.prob[glm.pred>0.5]="Up"
table(glm.prob,mpg01.test)## mpg01.test
## glm.prob 0 1
## Down 67 15
## Up 7 123(67+123)/212## [1] 0.8962264- The GLM model created a test error of 89.62% which is much better than either the LDA or QDA models previously.
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(mpg01~cylinders+displacement+horsepower+weight,data=Auto,subset=train)
nb.pred=predict(nb.fit,training)
table(nb.pred,mpg01.test)## mpg01.test
## nb.pred 0 1
## 0 68 15
## 1 6 123mean(nb.pred==mpg01.test)## [1] 0.9009434- The naiveBayes model has created a test error at 90.09%, predicting mpg on the training data 90% of the time!
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.mpg=mpg01[train]
set.seed(1)
knn.pred=knn(train.x,test.x,train.mpg,k=1)
table(knn.pred,mpg01.test)## mpg01.test
## knn.pred 0 1
## 0 72 29
## 1 2 109mean(knn.pred==mpg01.test)## [1] 0.8537736- The KNN model where K=1 created a test error of 85.38%.
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.
detach(Auto)
library(MASS)
attach(Boston)#Create a binary variable for “crim”
crim01=rep(0,length(crim))
crim01[crim>median(crim)]=1
Boston=data.frame(Boston,crim01)#Split the first half of data set to train
train=1:(length(crim)/2)
test=(length(crim) / 2 + 1):length(crim)
Boston.train=Boston[train, ]
Boston.test=Boston[test, ]
crim01.test=crim01[test]#Logistic Regression and Summary, based on crime data
glm.fit=glm(crim01~.-crim01-crim,data=Boston,family=binomial,subset = train)## Warning: glm.fit: fitted probabilities numerically 0 or 1 occurredsummary(glm.fit)##
## Call:
## glm(formula = crim01 ~ . - crim01 - crim, family = binomial,
## data = Boston, subset = train)
##
## Deviance Residuals:
## Min 1Q Median 3Q Max
## -2.83229 -0.06593 0.00000 0.06181 2.61513
##
## Coefficients:
## Estimate Std. Error z value Pr(>|z|)
## (Intercept) -91.319906 19.490273 -4.685 2.79e-06 ***
## zn -0.815573 0.193373 -4.218 2.47e-05 ***
## indus 0.354172 0.173862 2.037 0.04164 *
## chas 0.167396 0.991922 0.169 0.86599
## nox 93.706326 21.202008 4.420 9.88e-06 ***
## rm -4.719108 1.788765 -2.638 0.00833 **
## age 0.048634 0.024199 2.010 0.04446 *
## dis 4.301493 0.979996 4.389 1.14e-05 ***
## rad 3.039983 0.719592 4.225 2.39e-05 ***
## tax -0.006546 0.007855 -0.833 0.40461
## ptratio 1.430877 0.359572 3.979 6.91e-05 ***
## black -0.017552 0.006734 -2.606 0.00915 **
## lstat 0.190439 0.086722 2.196 0.02809 *
## medv 0.598533 0.185514 3.226 0.00125 **
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## (Dispersion parameter for binomial family taken to be 1)
##
## Null deviance: 329.367 on 252 degrees of freedom
## Residual deviance: 69.568 on 239 degrees of freedom
## AIC: 97.568
##
## Number of Fisher Scoring iterations: 10glm.probs=predict(glm.fit,Boston.test,type = "response")
glm.pred=rep(0,length(glm.probs))
glm.pred[glm.probs>0.5]=1
table(glm.pred,crim01.test)## crim01.test
## glm.pred 0 1
## 0 68 24
## 1 22 139mean(glm.pred==crim01.test)## [1] 0.8181818- We may conclude that, for this logistic regression, we have a test error rate of 81.81%
#Logistic Regression and Summary, based on crime, chas, nox, and tax
glm.fit=glm(crim01~.-crim01-crim-chas-nox-tax,data=Boston, family=binomial,subset=train)## Warning: glm.fit: fitted probabilities numerically 0 or 1 occurredsummary(glm.fit)##
## Call:
## glm(formula = crim01 ~ . - crim01 - crim - chas - nox - tax,
## family = binomial, data = Boston, subset = train)
##
## Deviance Residuals:
## Min 1Q Median 3Q Max
## -3.04443 -0.24461 -0.00114 0.38919 2.72999
##
## Coefficients:
## Estimate Std. Error z value Pr(>|z|)
## (Intercept) -17.291707 6.019497 -2.873 0.004071 **
## zn -0.478891 0.104276 -4.593 4.38e-06 ***
## indus 0.362719 0.082969 4.372 1.23e-05 ***
## rm -2.364642 0.967625 -2.444 0.014535 *
## age 0.063371 0.015457 4.100 4.14e-05 ***
## dis 1.494535 0.397249 3.762 0.000168 ***
## rad 1.756498 0.357330 4.916 8.85e-07 ***
## ptratio 0.575045 0.161917 3.551 0.000383 ***
## black -0.018916 0.006754 -2.801 0.005102 **
## lstat 0.057632 0.053051 1.086 0.277326
## medv 0.237282 0.081326 2.918 0.003527 **
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## (Dispersion parameter for binomial family taken to be 1)
##
## Null deviance: 329.37 on 252 degrees of freedom
## Residual deviance: 139.59 on 242 degrees of freedom
## AIC: 161.59
##
## Number of Fisher Scoring iterations: 9glm.probs=predict(glm.fit,Boston.test,type="response")
glm.pred=rep(0,length(glm.probs))
glm.pred[glm.probs>0.5]=1
table(glm.pred, crim01.test)## crim01.test
## glm.pred 0 1
## 0 78 28
## 1 12 135mean(glm.pred==crim01.test)## [1] 0.8418972- We may conclude that, for this logistic regression, we have a test error rate of 84.19%.
#LDA
lda.fit=lda(crim01~.-crim01-crim,data=Boston,subset=train)
lda.pred=predict(lda.fit,Boston.test)
table(lda.pred$class,crim01.test)## crim01.test
## 0 1
## 0 80 24
## 1 10 139mean(lda.pred$class==crim01.test)## [1] 0.8656126- We may conclude that, for this LDA, we have a test error rate of 86.56%.
#LDA on chas, non, and tax
lda.fit=lda(crim01~.-crim01-crim-chas-nox- tax,data=Boston,subset=train)
lda.pred=predict(lda.fit,Boston.test)
table(lda.pred$class,crim01.test)## crim01.test
## 0 1
## 0 83 28
## 1 7 135mean(lda.pred$class==crim01.test)## [1] 0.8616601- We may conclude that, for this LDA, we have a test error rate of 86.17%.
#Next we will do a few KNN models across different K values
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)[test,]
train.crim01=crim01[train]
set.seed(1)
knn.pred=knn(train.X,test.X,train.crim01,k=1)
table(knn.pred,crim01.test)## crim01.test
## knn.pred 0 1
## 0 85 111
## 1 5 52mean(knn.pred==crim01.test)## [1] 0.541502- Where K=1, we have a test error rate of 54.15%,
knn.pred=knn(train.X,test.X,train.crim01,k=10)
table(knn.pred,crim01.test)## crim01.test
## knn.pred 0 1
## 0 83 23
## 1 7 140mean(knn.pred==crim01.test)## [1] 0.8814229- Where K=10, we have a test error rate of 88.14%!
knn.pred=knn(train.X,test.X,train.crim01,k=10)
table(knn.pred,crim01.test)## crim01.test
## knn.pred 0 1
## 0 83 22
## 1 7 141mean(knn.pred==crim01.test)## [1] 0.8853755- Where K=10, our test error rate begins to flatten and we have 88.53%.
knn.pred=knn(train.X,test.X,train.crim01,k=100)
table(knn.pred,crim01.test)## crim01.test
## knn.pred 0 1
## 0 86 119
## 1 4 44mean(knn.pred==crim01.test)## [1] 0.513834- Where K=100, we become much less effective and have a test error rate of 51%.
After comparing our crime rates against multiple logistic regression and LDA models, naive Bayes, and KNN models where K = 1, 10, and 100, we can conclude that our most effective model is KNN where K=10.