Assignment 3
Arwa N
6/25/2020
Chapter 04 (page 168): 10, 11, 13)
- 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
library(ISLR)## Warning: package 'ISLR' was built under R version 3.6.3data("Weekly")
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
##
##
##
## - Produce some numerical and graphical summaries of the Weekly data. Do there appear to be any patterns?
Year and volume have a distinct pattern
pairs(Weekly)
- 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?
Lag 2 is statistically significant
LOG.model = glm(Direction ~ Lag1 + Lag2 + Lag3 + Lag4 + Lag5 + Volume, data = Weekly, family = binomial)
summary(LOG.model)##
## 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- 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.
Accuracy = 56.1% 92% accuracy when the markets are up and 11% error when the markets are down.
LOG.probs = predict(LOG.model, type = "response")
LOGPRED = rep("Down", length(LOG.probs))
LOGPRED[LOG.probs > 0.5] = "Up"
table(LOGPRED, Weekly$Direction)##
## LOGPRED Down Up
## Down 54 48
## Up 430 557- Now fit the logistic regression model using a training data period from 1990 to 2008, with Lag2 as the only predictor. Compute the confusion matrix and the overall fraction of correct predictions for the held out data (that is, the data from 2009 and 2010).
Accuracy = 62.5% = 65/104
trainset = Weekly[Weekly$Year<2009,]
testset = Weekly[Weekly$Year>2008,]
LOG.model2 = glm(Direction ~ Lag2, data= trainset, family = "binomial")
testprob = predict(LOG.model2, type = "response", newdata = testset)
testDir = Weekly$Direction[Weekly$Year>2008]
testPredic = rep("Down", length(testprob))
testPredic[testprob > 0.5] = "Up"
table(testPredic, testDir)## testDir
## testPredic Down Up
## Down 9 5
## Up 34 56- Repeat (d) using LDA.
Accuracy = 62.5% = 65/104
library(MASS)LDAModel = lda(Direction ~ Lag2, data= trainset)
LDAPRED = predict(LDAModel, newdata=testset, type="response")
LDACLA = LDAPRED$class
table(LDACLA, testset$Direction)##
## LDACLA Down Up
## Down 9 5
## Up 34 56- Repeat (d) using QDA.
Accuracy = 58.6% = 61/104
QDAModel = qda(Direction ~ Lag2, data= trainset)
QDAPRED = predict(QDAModel, newdata=testset, type="response")
QDACLA = QDAPRED$class
table(QDACLA, testset$Direction)##
## QDACLA Down Up
## Down 0 0
## Up 43 61- Repeat (d) using KNN with K = 1.
Accuracy = 50% = 61/104
library(class)set.seed(1)
train.1 = cbind(trainset$Lag2)
test.1 = cbind(testset$Lag2)
train.P = cbind(trainset$Direction)
KNNPRED = knn(train.1, test.1, train.P, k=1)
table(KNNPRED, testset$Direction)##
## KNNPRED Down Up
## 1 21 30
## 2 22 31- Which of these methods appears to provide the best results on this data?
Logistic and LDA provide the best out of the others
- 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.
- 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.
- 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.
data(Auto)attach(Auto)
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
## - 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.
pairs(Auto)
Cylinders, Displacement, horsepower, weight
- Split the data into a training set and a test set.
datacut = sample(1:dim(Auto)[1], size=dim(Auto)[1]*0.80)
trainauto = Auto[datacut,]
testauto = Auto[-datacut,]- 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?
Test error = 10.1%
LDAAUTO = lda(mpg01 ~ cylinders + weight + displacement + year, data = trainauto)
LDAPREDAUTO = predict(LDAAUTO, newdata=testauto, type="response")$class
table(LDAPREDAUTO, testauto$mpg01)##
## LDAPREDAUTO 0 1
## 0 37 0
## 1 8 34- Perform QDA on the training data in order to predict mpg01 using the variables that seemed most associated with mpg01 in (b). What is the test error of the model obtained?
Test error = 12.6%
QDAAUTO = qda(mpg01 ~ cylinders + weight + displacement + year, data = trainauto)
QDAPREDAUTO = predict(QDAAUTO, newdata=testauto, type="response")$class
table(QDAPREDAUTO, testauto$mpg01)##
## QDAPREDAUTO 0 1
## 0 39 4
## 1 6 30- Perform logistic regression on the training data in order to predict mpg01 using the variables that seemed most associated with mpg01 in (b). What is the test error of the model obtained?
Test error = 10.2%
LOGAUTO = glm(mpg01 ~ cylinders + weight + displacement + year, family="binomial", data=trainauto)
LOGPROBAUTO = predict(LOGAUTO,testauto, type="response")
LOGPREDAUTO = rep(0, length(LOGPROBAUTO))
LOGPREDAUTO[LOGPROBAUTO>0.5]=1
table(LOGPREDAUTO, testauto$mpg01)##
## LOGPREDAUTO 0 1
## 0 41 1
## 1 4 33- 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?
Test error = 15.1%
trainXYZ = cbind(trainauto$cylinders, trainauto$weight, trainauto$displacement, trainauto$year)
testXYZ = cbind(testauto$cylinders, testauto$weight, testauto$displacement, testauto$year)
train.P1 = cbind(trainauto$mpg01)
set.seed(1)
KNNPREDAUTO = knn(trainXYZ, testXYZ, train.P1, k=1)
table(KNNPREDAUTO, testauto$mpg01)##
## KNNPREDAUTO 0 1
## 0 37 4
## 1 8 30- 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.
Lowest error rate is for the logistic Reg which is 9.09%. KNN had the highest error rate with 62.84%
data("Boston")attach(Boston)
crim01 = rep(0, length(crim))
crim01[crim > median(crim)] = 1
Boston = data.frame(Boston, crim01)traincut = 1:(dim(Boston)[1]/2)
testcut = (dim(Boston)[1]/2 + 1):dim(Boston)[1]
trainBOS = Boston[traincut, ]
testBOS = Boston[testcut, ]LOGBOS = glm(crim01 ~ indus+nox+age+dis+rad+tax, family="binomial", data=trainBOS)
LOGPROBBOS = predict(LOGBOS,testBOS, type="response")
LOGPREDBOS = rep(0, length(LOGPROBBOS))
LOGPREDBOS[LOGPROBBOS>0.5]=1
table(LOGPREDBOS, testBOS$crim01)##
## LOGPREDBOS 0 1
## 0 75 8
## 1 15 155mean(LOGPREDBOS !=testBOS$crim01)## [1] 0.09090909LDABOS = lda(crim01 ~ indus+nox+age+dis+rad+tax, data = trainBOS)
LDAPREDBOS = predict(LDABOS, newdata=testBOS, type="response")$class
table(LDAPREDBOS, testBOS$crim01)##
## LDAPREDBOS 0 1
## 0 81 18
## 1 9 145mean(LDAPREDBOS !=testBOS$crim01)## [1] 0.1067194trainXXX = cbind(trainBOS$indus, trainBOS$nox, trainBOS$age, trainBOS$dis, trainBOS$rad,trainBOS$tax)
testXXX = cbind(testBOS$indus, testBOS$nox, testBOS$age, testBOS$dis, testBOS$rad,testBOS$tax)
train.P2 = cbind(trainBOS$crim01)
set.seed(1)
KNNPREDBOS = knn(trainXXX, testXXX, train.P2, k=1)
table(KNNPREDBOS, testBOS$crim01)##
## KNNPREDBOS 0 1
## 0 82 151
## 1 8 12mean(KNNPREDBOS !=testBOS$crim01)## [1] 0.6284585