Assignment #3
Ebony Gaines
February 18, 2019
Q10.
This question should be answered using the Weekly data set, 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?
library(ISLR)
library(corrplot)## corrplot 0.84 loadedsummary(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
## Min. :-18.1950 Min. :-18.1950 Min. :0.08747
## 1st Qu.: -1.1580 1st Qu.: -1.1660 1st Qu.:0.33202
## Median : 0.2380 Median : 0.2340 Median :1.00268
## Mean : 0.1458 Mean : 0.1399 Mean :1.57462
## 3rd Qu.: 1.4090 3rd Qu.: 1.4050 3rd Qu.:2.05373
## Max. : 12.0260 Max. : 12.0260 Max. :9.32821
## Today Direction
## Min. :-18.1950 Down:484
## 1st Qu.: -1.1540 Up :605
## Median : 0.2410
## Mean : 0.1499
## 3rd Qu.: 1.4050
## Max. : 12.0260corrplot(cor(Weekly[,-9]), method="square")
* The only variables that appear to have any significant linear relation are Year and Volume. The correlational plot doesn’t illustrate that any other variables are linearly related.
(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?
attach(Weekly)
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: 4- The only variable that was statistically significant at the level of significance \(\alpha\) =0.05 is Lag2. Otherwise the other variables fail to reject the null hypothesis; \(\beta\) = 0.
(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.prob= predict(Weekly.fit, type='response')
logWeekly.pred =rep("Down", length(logWeekly.prob))
logWeekly.pred[logWeekly.prob > 0.5] = "Up"
table(logWeekly.pred, Direction)## Direction
## logWeekly.pred Down Up
## Down 54 48
## Up 430 557- To determine the percentage of current predictions, the followin formula is used: \[\frac{(54+557)}{(54+48+430+557)} = 0.5611\] This illustrates that the model predicted the weekly market trend correctly 56.11% of the time. Separating in how the model correctly predicts the Up and Down trends. While the model correctly predicted the Up weekly trends \(\frac{557}{48+557} =0.9207\); 92.07% correct. In contrast Down weekly trends were predicted at a lower rate, \(\frac{54}{430+54} = 0.1115\); or only 11.15% correctly predicted.
(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.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 = Direction[!train]
table(logWeekly.pred, Direction.0910)## Direction.0910
## logWeekly.pred Down Up
## Down 9 5
## Up 34 56mean(logWeekly.pred == Direction.0910)## [1] 0.625- When spliting up the whole Weekly dataset into a training and test dataset, the model correctly predicted weekly trends at rate of 62.5%, which is a moderate improvement from the model that utilized the whole dataset. Also this model such as the previous one did better at predicting upward trends(91.80%) compared to downward trends(20.93%); although this model was able to improve significantly on correctly predicting downward trends.
(e) Repeat (d) using LDA.
library(MASS)
Weeklylda.fit<-lda(Direction~Lag2, data=Weekly,family=binomial, subset=train)
Weeklylda.pred<-predict(Weeklylda.fit, Weekly.0910)
table(Weeklylda.pred$class, Direction.0910)## Direction.0910
## Down Up
## Down 9 5
## Up 34 56mean(Weeklylda.pred$class==Direction.0910)## [1] 0.625- Using Linear Discriminant Analysis to develop a classifying model yielded similar results as the logistic regression model created in part D.
(f) Repeat (d) using QDA.
Weeklyqda.fit = qda(Direction ~ Lag2, data = Weekly, subset = train)
Weeklyqda.pred = predict(Weeklyqda.fit, Weekly.0910)$class
table(Weeklyqda.pred, Direction.0910)## Direction.0910
## Weeklyqda.pred Down Up
## Down 0 0
## Up 43 61mean(Weeklyqda.pred==Direction.0910)## [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 considered predicting the correctness of weekly upward trends disregrading the downward weekly trends.
(g) Repeat (d) using KNN with K = 1.
library(class)
Week.train=as.matrix(Lag2[train])
Week.test=as.matrix(Lag2[!train])
train.Direction =Direction[train]
set.seed(1)
Weekknn.pred=knn(Week.train,Week.test,train.Direction,k=1)
table(Weekknn.pred,Direction.0910)## Direction.0910
## Weekknn.pred Down Up
## Down 21 30
## Up 22 31mean(Weekknn.pred == Direction.0910)## [1] 0.5- The K-Nearest neighbors resulted in a classifying model with an accuracy rate of 50% which is equal to random chance.
(h) Which of these methods appears to provide the best results on this data?
- The methods that have the highest accuracy rates are the Logistic Regression and Linear Discriminant Analysis; both having rates 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:Lag4+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 = Direction[!train]
table(logWeekly.pred, Direction.0910)## Direction.0910
## logWeekly.pred Down Up
## Down 3 4
## Up 40 57mean(logWeekly.pred == Direction.0910)## [1] 0.5769231#LDA with Interaction Lag2:Lag4
Weeklylda.fit<-lda(Direction~Lag2:Lag4+Lag2, data=Weekly,family=binomial, subset=train)
Weeklylda.pred<-predict(Weeklylda.fit, Weekly.0910)
table(Weeklylda.pred$class, Direction.0910)## Direction.0910
## Down Up
## Down 3 3
## Up 40 58mean(Weeklylda.pred$class==Direction.0910)## [1] 0.5865385Weeklyqda.fit = qda(Direction ~ poly(Lag2,2), data = Weekly, subset = train)
Weeklyqda.pred = predict(Weeklyqda.fit, Weekly.0910)$class
table(Weeklyqda.pred, Direction.0910)## Direction.0910
## Weeklyqda.pred Down Up
## Down 7 3
## Up 36 58mean(Weeklyqda.pred==Direction.0910)## [1] 0.625#K=10
Week.train=as.matrix(Lag2[train])
Week.test=as.matrix(Lag2[!train])
train.Direction =Direction[train]
set.seed(1)
Weekknn.pred=knn(Week.train,Week.test,train.Direction,k=10)
table(Weekknn.pred,Direction.0910)## Direction.0910
## Weekknn.pred Down Up
## Down 17 21
## Up 26 40mean(Weekknn.pred == Direction.0910)## [1] 0.5480769#K=100
Week.train=as.matrix(Lag2[train])
Week.test=as.matrix(Lag2[!train])
train.Direction =Direction[train]
set.seed(1)
Weekknn.pred=knn(Week.train,Week.test,train.Direction,k=100)
table(Weekknn.pred,Direction.0910)## Direction.0910
## Weekknn.pred Down Up
## Down 10 11
## Up 33 50mean(Weekknn.pred == Direction.0910)## [1] 0.5769231detach(Weekly)- After trying different transformations for the different methods, the original Logistic regression and LDA models had the best accuraccy rates. Although squaring Lag2 and then using the QDA method resulted in the same accuracy model as the original Logistic regression and LDA models. Bt this makes sense because this is similar as taking the linear discriminant analysis of the original model.
Q11.
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(ISLR)
attach(Auto)
summary(Auto)## mpg cylinders displacement horsepower
## Min. : 9.00 Min. :3.000 Min. : 68.0 Min. : 46.0
## 1st Qu.:17.00 1st Qu.:4.000 1st Qu.:105.0 1st Qu.: 75.0
## Median :22.75 Median :4.000 Median :151.0 Median : 93.5
## Mean :23.45 Mean :5.472 Mean :194.4 Mean :104.5
## 3rd Qu.:29.00 3rd Qu.:8.000 3rd Qu.:275.8 3rd Qu.:126.0
## Max. :46.60 Max. :8.000 Max. :455.0 Max. :230.0
##
## weight acceleration year origin
## Min. :1613 Min. : 8.00 Min. :70.00 Min. :1.000
## 1st Qu.:2225 1st Qu.:13.78 1st Qu.:73.00 1st Qu.:1.000
## Median :2804 Median :15.50 Median :76.00 Median :1.000
## Mean :2978 Mean :15.54 Mean :75.98 Mean :1.577
## 3rd Qu.:3615 3rd Qu.:17.02 3rd Qu.:79.00 3rd Qu.:2.000
## Max. :5140 Max. :24.80 Max. :82.00 Max. :3.000
##
## name
## amc matador : 5
## ford pinto : 5
## toyota corolla : 5
## amc gremlin : 4
## amc hornet : 4
## 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(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? Scatterplots and boxplots may be useful tools to answer this question.Describe your findings.
corrplot(cor(Auto[,-9]), method="square")
- The variables that appear to correlate strongly with mpg01 are Cylinders,Displacement, and Weight; these variables appear to correlate negatively with this variable. Also Horsepower and Origin appear to correlate moderately with mpg01.
(c) Split the data into a training set and a test set.
train <- (year %% 2 == 0)
train.auto <- Auto[train,]
test.auto <- Auto[-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?
autolda.fit <- lda(mpg01~displacement+horsepower+weight+year+cylinders+origin, data=train.auto)
autolda.pred <- predict(autolda.fit, test.auto)
table(autolda.pred$class, test.auto$mpg01)##
## 0 1
## 0 169 7
## 1 26 189mean(autolda.pred$class != test.auto$mpg01)## [1] 0.08439898- Using LDA method to create a classifying model resulted in a test error rate of 8.44%.
(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?
autoqda.fit <- qda(mpg01~displacement+horsepower+weight+year+cylinders+origin, data=train.auto)
autoqda.pred <- predict(autoqda.fit, test.auto)
table(autoqda.pred$class, test.auto$mpg01)##
## 0 1
## 0 176 20
## 1 19 176mean(autoqda.pred$class != test.auto$mpg01)## [1] 0.09974425- Using QDA method for classification resulted in a model with a test error rate of 9.97%
(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?
auto.fit<-glm(mpg01~displacement+horsepower+weight+year+cylinders+origin, data=train.auto,family=binomial)
auto.probs = predict(auto.fit, test.auto, type = "response")
auto.pred = rep(0, length(auto.probs))
auto.pred[auto.probs > 0.5] = 1
table(auto.pred, test.auto$mpg01)##
## auto.pred 0 1
## 0 174 12
## 1 21 184mean(auto.pred != test.auto$mpg01)## [1] 0.08439898*The logistic regression method resulted in a model with a test error rate of 8.44%.
(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=1
train.K= cbind(displacement,horsepower,weight,cylinders,year, origin)[train,]
test.K=cbind(displacement,horsepower,weight,cylinders, year, origin)[-train,]
set.seed(1)
autok.pred=knn(train.K,test.K,train.auto$mpg01,k=1)
mean(autok.pred != test.auto$mpg01)## [1] 0.07161125#K=5
autok.pred=knn(train.K,test.K,train.auto$mpg01,k=5)
mean(autok.pred != test.auto$mpg01)## [1] 0.112532#K=10
autok.pred=knn(train.K,test.K,train.auto$mpg01,k=10)
mean(autok.pred != test.auto$mpg01)## [1] 0.1253197detach(Auto)- After reviewing the results of the K nearest neighbors,it appears that K=1 has the lowest error rate of 7.16%. As the value of K increases so does the error rate for this particular model.
Q13.
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.
summary(Boston)## crim zn indus chas
## Min. : 0.00632 Min. : 0.00 Min. : 0.46 Min. :0.00000
## 1st Qu.: 0.08204 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. :50.00attach(Boston)Creating binary crim variable.
crime01 <- rep(0, length(crim))
crime01[crim > median(crim)] <- 1
Boston= data.frame(Boston,crime01)Splitting the dataset
train = 1:(dim(Boston)[1]/2)
test = (dim(Boston)[1]/2 + 1):dim(Boston)[1]
Boston.train = Boston[train, ]
Boston.test = Boston[test, ]
crime01.test = crime01[test]Determination of any associations to crime01
corrplot(cor(Boston), method="square")
- It appears that the variables indus, nox,age,dis, rad and tax have the strongest association with the desired variable.
Logistic Regression
set.seed(1)
Boston.fit <-glm(crime01~ indus+nox+age+dis+rad+tax, data=Boston.train,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, crime01.test)## crime01.test
## Boston.pred 0 1
## 0 75 8
## 1 15 155mean(Boston.pred != crime01.test)## [1] 0.09090909summary(Boston.fit)##
## Call:
## glm(formula = crime01 ~ indus + nox + age + dis + rad + tax,
## family = binomial, data = Boston.train)
##
## Deviance Residuals:
## Min 1Q Median 3Q Max
## -1.97810 -0.21406 -0.03454 0.47107 3.04502
##
## Coefficients:
## Estimate Std. Error z value Pr(>|z|)
## (Intercept) -42.214032 7.617440 -5.542 2.99e-08 ***
## indus -0.213126 0.073236 -2.910 0.00361 **
## nox 80.868029 16.066473 5.033 4.82e-07 ***
## age 0.003397 0.012032 0.282 0.77772
## dis 0.307145 0.190502 1.612 0.10690
## rad 0.847236 0.183767 4.610 4.02e-06 ***
## tax -0.013760 0.004956 -2.777 0.00549 **
## ---
## 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: 144.44 on 246 degrees of freedom
## AIC: 158.44
##
## Number of Fisher Scoring iterations: 8Linear Discriminat Analysis
Boston.ldafit <-lda(crime01~ indus+nox+age+dis+rad+tax, data=Boston.train,family=binomial)
Bostonlda.pred = predict(Boston.ldafit, Boston.test)
table(Bostonlda.pred$class, crime01.test)## crime01.test
## 0 1
## 0 81 18
## 1 9 145mean(Bostonlda.pred$class != crime01.test)## [1] 0.1067194K Nearest Neighbors
#K=1
train.K=cbind(indus,nox,age,dis,rad,tax)[train,]
test.K=cbind(indus,nox,age,dis,rad,tax)[test,]
Bosknn.pred=knn(train.K, test.K, crime01.test, k=1)
table(Bosknn.pred,crime01.test)## crime01.test
## Bosknn.pred 0 1
## 0 31 155
## 1 59 8mean(Bosknn.pred !=crime01.test)## [1] 0.8458498#K=100
train.K=cbind(indus,nox,age,dis,rad,tax)[train,]
test.K=cbind(indus,nox,age,dis,rad,tax)[test,]
Bosknn.pred=knn(train.K, test.K, crime01.test, k=100)
table(Bosknn.pred,crime01.test)## crime01.test
## Bosknn.pred 0 1
## 0 21 6
## 1 69 157mean(Bosknn.pred !=crime01.test)## [1] 0.2964427- After reviewing the results of each classification method, logistic regression had the lowest test error rate of 9.09%.Closely reveiwing the logistic regression model, the only variables that were statistically significant were: indus, nox, rad and tax. All of the modeling methods contained the same variables so comparison would be the easiest; these variables were determined graphically to has the best association with crime01. K nearest neighbors with k=1 had the highest error rate (84.58%), thus making that model ineffective in classification. To also note, as the value of K increased the error rate did improve but still wasn’t as low as logistic regression or LDA.