Assignment 3
R.Gudiry
8/3/2021
Chapter 4
Exercise 10
This question should be answered using the Weekly dataset, which is part of the ISLR 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.
(A) Produce some numerical and graphical summaries of the Weekly data. Do there appear to be any patterns?
data= Weekly
names(Weekly)## [1] "Year" "Lag1" "Lag2" "Lag3" "Lag4" "Lag5"
## [7] "Volume" "Today" "Direction"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 1.000000000The numerical and graphical data indicate that the data is normall distributed, there is no clear patterns among the data. From the summary we see that there is very little difference between the median and means between lags and the scatter plots indicate that there is no significant relationships between the variables with the exception of Year and Volume. Volume increase over time.
(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?
weekly.fit=glm(Direction~Lag1+Lag2+Lag3+Lag4+Lag5+Volume, data=Weekly,family=binomial)
summary (weekly.fit)##
## 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: 4The smallest p-value here is associated with Lag2. The positive coefficient for this predictor suggests that if the market had a positive return yesterday, then it is likely to go up today. The p-value is 0.0296, demonstrates evidence of a real association between Lag2 and Direction.
(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.
logWeekly.probs=predict(weekly.fit, type="response")
logWeekly.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(Weekly$Direction)## Up
## Down 0
## Up 1logWeekly.pred=rep("Down" ,1089)
logWeekly.pred[logWeekly.probs >.5]="Up"
table(logWeekly.pred , Weekly$Direction )##
## logWeekly.pred Down Up
## Down 54 48
## Up 430 557mean(logWeekly.pred==Weekly$Direction)## [1] 0.5610652The diagonal elements of the confusion matrix indicate the correct predictions while the off-diagonals represent incorrect predictions. The model correctly predicted that the market would go down on 54 weeks and that it would go up on 557 weeks, for a total of 611 correct predictions. In this case, the logistic regression correctly predicted the movement of the market 56.1% of the time.
(D) Now fit the logistic regression model using a training data period from 1990 to 2008, with Lag2 as the only predictor. Compute the confusion matrix and the overall fraction of correct predictions for the held out data (that is, the data from 2009 and 2010).
train = (Weekly$Year<2009)
Weekly.0910<-Weekly[!train,]
Weekly.fit<-glm(Direction~Lag2, data=Weekly,family=binomial, subset=train)
logWeekly.prob= predict(Weekly.fit, Weekly.0910, type = "response")
logWeekly.pred = rep("Down", length(logWeekly.prob))
logWeekly.pred[logWeekly.prob > 0.5] = "Up"
Direction.0910 = Weekly$Direction[!train]
table(logWeekly.pred, Direction.0910)## Direction.0910
## logWeekly.pred Down Up
## Down 9 5
## Up 34 56mean(logWeekly.pred == Direction.0910)## [1] 0.625When splitting up the whole Weekly datasheet into a training and test data set, the model correctly predicted weekly movement of the market 62.5% of the time. This is a moderate improvement from the model that utilized the whole data set.
(E) Repeat (d) using LDA.
ldaWeekly.fit<-lda(Direction~Lag2, data=Weekly,family=binomial, subset=train)
ldaWeekly.pred<-predict(ldaWeekly.fit, Weekly.0910)
table(ldaWeekly.pred$class, Direction.0910)## Direction.0910
## Down Up
## Down 9 5
## Up 34 56mean(ldaWeekly.pred$class==Direction.0910)## [1] 0.625The Linear Discriminant Analysis yielded similar results to the logistic regression analysis in section D. The model correctly predicted weekly movement of the market 62.5% of the time.
(F) Repeat (D) using QDA
qdaWeekly.fit = qda(Direction ~ Lag2, data = Weekly, subset = train)
qdaWeekly.pred = predict(qdaWeekly.fit, Weekly.0910)$class
table(qdaWeekly.pred, Direction.0910)## Direction.0910
## qdaWeekly.pred Down Up
## Down 0 0
## Up 43 61mean(qdaWeekly.pred==Direction.0910)## [1] 0.5865385The Quadratic Linear Analysis results were lower than the Logistic Regression Analysis in section D. The model correctly predicted weekly movement of the market 58.6% of the time.
(G) Repeat (D) using KNN with K = 1
knnWeek.train=as.matrix(Weekly$Lag2[train])
knnWeek.test=as.matrix(Weekly$Lag2[!train])
train.Direction=Weekly$Direction[train]
set.seed(1)
knnWeek.pred=knn(knnWeek.train,knnWeek.test,train.Direction,k=1)
table(knnWeek.pred,Direction.0910)## Direction.0910
## knnWeek.pred Down Up
## Down 21 30
## Up 22 31mean(knnWeek.pred == Direction.0910)## [1] 0.5This KNN analysis results were lower than the QDA, LDA, and the Logistic analysis. This model’s accuracy rate is 50%.
(H) Which of these methods appears to provide the best results on this data?
The method(s) that appear to provide the best results on this data are the Logistic Regression and Linear Discriminant Analyses. The methods both had an accuracy rate of 62.5%.
(I) Experiment with different combinations of predictors, including possible transformations and interactions, for each of the methods. Report the variables, method, and associated confusion matrix that appears to provide the best results on the held out data. Note that you should also experiment with values for K in the KNN classifier.
#Logistic Regression with interaction Lag2:Lag4
weekly.fit=glm(Direction~Lag2:Lag2+Lag4, data=Weekly,family=binomial)
logWeekly.probs=predict(weekly.fit, type="response")
logWeekly.pred=rep("Down" ,1089)
logWeekly.pred[logWeekly.probs >.5]="Up"
table(logWeekly.pred , Weekly$Direction )##
## logWeekly.pred Down Up
## Down 33 25
## Up 451 580mean(logWeekly.pred==Weekly$Direction)## [1] 0.5629017The accuracy of this model is 56.2%
#LDA with Interaction Lag2:Lag4
ldaWeekly.fit<-lda(Direction~Lag2:Lag2+Lag4, data=Weekly,family=binomial, subset=train)
ldaWeekly.pred<-predict(ldaWeekly.fit, Weekly.0910)
table(ldaWeekly.pred$class, Direction.0910)## Direction.0910
## Down Up
## Down 8 4
## Up 35 57mean(ldaWeekly.pred$class==Direction.0910)## [1] 0.625The accuracy of this model is 62.5%.
#QDA with interactions lag2:lag4
qdaWeekly.fit = qda(Direction ~ Lag2:Lag2+Lag4, data = Weekly, subset = train)
qdaWeekly.pred = predict(qdaWeekly.fit, Weekly.0910)$class
table(qdaWeekly.pred, Direction.0910)## Direction.0910
## qdaWeekly.pred Down Up
## Down 9 14
## Up 34 47mean(qdaWeekly.pred==Direction.0910)## [1] 0.5384615The accuracy of this model is 53.8%.
#KNN with K=4
knnWeek.train=as.matrix(Weekly$Lag2[train])
knnWeek.test=as.matrix(Weekly$Lag2[!train])
train.Direction=Weekly$Direction[train]
set.seed(1)
knnWeek.pred=knn(knnWeek.train,knnWeek.test,train.Direction,k=4)
table(knnWeek.pred,Direction.0910)## Direction.0910
## knnWeek.pred Down Up
## Down 20 17
## Up 23 44(mean(knnWeek.pred == Direction.0910))## [1] 0.6153846The accuracy of this model is 61.5%.
Using the Logistic Regression to test several transformations it was determined that using the interaction between Lag2 and Lag4 provided the best results. This interaction was also used in the LDA and QDA models. The KNN model produced the best result when K=4. Comparing the accuracy of these new models with the original accuracy results identified that none of the transformation results were better then the original results. In fact, the LDA model with Lag2 and Lag4 interactions provided the same accuracy as the LDA model without the interaction: 62.5%.
Exercise 11
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
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(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(Auto$mpg))
mpg01[Auto$mpg > median(Auto$mpg)] <- 1
Auto2 = 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 tools to answer this question. Describe your findings.
corrplot(cor(Auto2[,-9]), method="circle")
There are three variables that have a strong negative correlation with mpg01: cylinders, displacement, and weight. Two variables have a moderate correlation with mpg01: horsepower(negative correlation) and origin (positive correlation).
(C) Split the data into a training set and a test set.
# Split data into 80% training, 20% test and make sure my test data has same proportion
set.seed(1)
split = initial_split(Auto2, strata = mpg01, prop = .8)
train.Auto2 = training(split)
test.Auto2 = testing (split)(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?
ldaAuto2.fit <- lda(mpg01~displacement+horsepower+weight+year+cylinders+origin, data=train.Auto2)
ldaAuto2.pred <- predict(ldaAuto2.fit, test.Auto2)
lda_cm = table(ldaAuto2.pred$class, test.Auto2$mpg01)
(error= 1-(sum (diag(lda_cm))/sum(lda_cm)))## [1] 0.1125The test error rate is 11.25% for this 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?
qdaAuto2.fit <- qda(mpg01~displacement+horsepower+weight+year+cylinders+origin, data=train.Auto2)
qdaAuto2.pred <- predict(qdaAuto2.fit, test.Auto2)
(qda_cm=table(qdaAuto2.pred$class, test.Auto2$mpg01))##
## 0 1
## 0 32 5
## 1 8 35(error= 1-(sum (diag(qda_cm))/sum(qda_cm)))## [1] 0.1625The test error rate is 16.25% for this 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?
logAuto2.fit<-glm(mpg01~displacement+horsepower+weight+year+cylinders+origin, data=train.Auto2,family=binomial)
logAuto2.probs = predict(logAuto2.fit, test.Auto2, type = "response")
logAuto2.pred = rep(0, length(logAuto2.probs))
logAuto2.pred[logAuto2.probs > 0.5] = 1
(log_cm=table(logAuto2.pred, test.Auto2$mpg01))##
## logAuto2.pred 0 1
## 0 33 6
## 1 7 34(error= 1-(sum (diag(log_cm))/sum(log_cm)))## [1] 0.1625The test error rate is 16.25% for this model.
(G) Perform KNN on the training data, with several values of K, in order to predict mpg01. Use only the variables that seemed most associated with mpg01 in (b). What test errors do you obtain? Which value of K seems to perform the best on this data set?
k_values = c(3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15)
my_accuracy = double()
for (i in k_values){
kknn_Auto2=kknn(mpg01~cylinders + displacement + weight, train=train.Auto2, test=test.Auto2, k=i)
CM_tempauto = table(kknn_Auto2$fitted, test.Auto2$mpg01)
accuracy_tempauto = sum(diag(CM_tempauto))/sum(CM_tempauto)
my_accuracy = c(my_accuracy,accuracy_tempauto)
}
(myresults = data.frame( k = k_values, accuracy = my_accuracy))## k accuracy
## 1 3 0.3750
## 2 4 0.3625
## 3 5 0.3625
## 4 6 0.3625
## 5 7 0.3500
## 6 8 0.3500
## 7 9 0.3375
## 8 10 0.3250
## 9 11 0.3250
## 10 12 0.3250
## 11 13 0.2750
## 12 14 0.2625
## 13 15 0.2625The K Nearest Neighbors results identified k=3 has the highest accuracy at 37.5%. As K increases the accuracy decreases in this model.
##Exercise 13
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, and KNN models using various subsets of the predictors. Describe your findings.
data=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. :50.00Create the variable Crime.
crime01 <- rep(0, length(Boston$crim))
crime01[Boston$crim > median(Boston$crim)] <- 1
Boston= data.frame(Boston,crime01)Split the data set into 80% training and 20% test.
set.seed(1)
split_Boston = initial_split(Boston, strata = crime01, prop = .8)
train_Boston = training(split_Boston)
test_Boston = testing (split_Boston)Review and identify correlations to the variable crime01
corrplot(cor(Boston), method="circle")
The variables with the strongest correlation to crime01 are nox, rad, tax, indus, age and dis
###Logistic Regression
Boston_fit <-glm(crime01~indus+nox+age+dis+rad+tax, data=train_Boston, family=binomial)
Boston_probs = predict(Boston_fit, test_Boston, type = "response")
Boston_pred = rep(0, length(Boston_probs))
Boston_pred[Boston_probs > 0.5] = 1
(Boston_CM=table(Boston_pred, test_Boston$crime01))##
## Boston_pred 0 1
## 0 47 10
## 1 4 41(sum (diag(Boston_CM))/sum(Boston_CM))## [1] 0.8627451summary(Boston_fit)##
## Call:
## glm(formula = crime01 ~ indus + nox + age + dis + rad + tax,
## family = binomial, data = train_Boston)
##
## Deviance Residuals:
## Min 1Q Median 3Q Max
## -2.03855 -0.29638 -0.01332 0.00746 2.72335
##
## Coefficients:
## Estimate Std. Error z value Pr(>|z|)
## (Intercept) -25.185374 4.178905 -6.027 1.67e-09 ***
## indus -0.060518 0.046628 -1.298 0.19433
## nox 44.361188 7.730513 5.738 9.55e-09 ***
## age 0.004101 0.009408 0.436 0.66293
## dis 0.191673 0.159352 1.203 0.22904
## rad 0.602396 0.131556 4.579 4.67e-06 ***
## tax -0.006678 0.002483 -2.689 0.00716 **
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## (Dispersion parameter for binomial family taken to be 1)
##
## Null deviance: 560.06 on 403 degrees of freedom
## Residual deviance: 200.21 on 397 degrees of freedom
## AIC: 214.21
##
## Number of Fisher Scoring iterations: 8###Linear Discriminat Analysis
Boston_lda_fit <-lda(crime01~ indus+nox+age+dis+rad+tax, data=train_Boston,family=binomial)
Boston_lda_pred = predict(Boston_lda_fit, test_Boston)
(Boston_lda_CM=table(Boston_lda_pred$class, test_Boston$crime01))##
## 0 1
## 0 49 14
## 1 2 37(sum (diag(Boston_lda_CM))/sum(Boston_lda_CM))## [1] 0.8431373###K Nearest Neighbors
k_values = c(3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15)
my_accuracy = double()
for (i in k_values){
kknn_Boston=kknn(crime01~indus+nox+age+dis+rad+tax, train=train_Boston, test=test_Boston, k=i)
Boston_temp_CM = table(kknn_Boston$fitted, test_Boston$crime01)
Boston_temp_accuracy = sum(diag(Boston_temp_CM))/sum(Boston_temp_CM)
my_accuracy = c(my_accuracy,Boston_temp_accuracy)
}
(myresults = data.frame( k = k_values, accuracy = my_accuracy))## k accuracy
## 1 3 0.4607843
## 2 4 0.4411765
## 3 5 0.3921569
## 4 6 0.3823529
## 5 7 0.3431373
## 6 8 0.3333333
## 7 9 0.3137255
## 8 10 0.3039216
## 9 11 0.2647059
## 10 12 0.2450980
## 11 13 0.2352941
## 12 14 0.2352941
## 13 15 0.2352941All of the modeling methods used the same variables. The variables were obtained using corrplot which provided a visual representation of all variables and their correlations to one another. In this case the focus was on significant correlations to the variable the crime01. Out of the Logistic Regression, LDA, and KNN models the one with the greatest accuracy is the Logistic Regression model. It achieved an accuracy of 86.27%. Looking closer at this model we see that variables that are statistically significant are tax, rad, and nox. For classification modeling, KNN identified that K=3 had the highest accuracy rate of 46.07%.