STA Assignment 3
Marina Allen
3/4/2021
library(ISLR) ## Warning: package 'ISLR' was built under R version 4.0.3library(corrplot)## Warning: package 'corrplot' was built under R version 4.0.4## corrplot 0.84 loadedlibrary(MASS)## Warning: package 'MASS' was built under R version 4.0.3library(class)Question 10: 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.
Part A: Produce some numerical and graphical summaries of the Weeklydata. 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
##
##
##
## corrplot(cor(Weekly[,-9]), method="square")
Answer: Year and Volume are the only variables that seem to have a linear relationship.
Part 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: 4Answer: The only variable that was significant at 0.05 (alpha) is Lag2. Otherwise the other variables fail to reject the null hypothesis.
Part 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 557Answer: The model predicted the weekly market trend correctly 56.11% of the time.
Part 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] 0.625Answer: When spliting up the whole Weekly dataset into a training and test dataset, the model correctly predicted weekly trends at rate of 62.5%. 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.
Part E: Repeat (d) using LDA.
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] 0.625Part 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] 0.5865385Answer: Quadratic Linear Analysis created a model with an accuracy of 58.65%, which is lower than the other methods. Also this model only considered predicting the correctness of weekly upward trends.
Part G: Repeat (d) using KNN with K = 1.
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] 0.5Answer: The K-Nearest neighbor resulted in a classifying model with an accuracy rate of 50% which is equal to random chance.
Part H: Which of these methods appears to provide the best results on this data?
Answer: The methods that have the highest accuracy rates are the Logistic Regression and Linear Discriminant Analysis (62.5%).
Question 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.
library(ISLR)
attach(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) :365Part 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)Part 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 1.0000000pairs(Auto)
boxplot(year ~ mpg01, data = Auto, main = "Year vs mpg01")
boxplot(weight ~ mpg01, data = Auto, main = "Weight vs mpg01")
boxplot(acceleration ~ mpg01, data = Auto, main = "Acceleration vs mpg01")
boxplot(horsepower ~ mpg01, data = Auto, main = "Horsepower vs mpg01")
### Answer: There may be some association between “mpg01” and “cylinders”, “weight”, “displacement” and “horsepower”.
Part C: Split the data into a training set and a test set.
train <- (year %% 2 == 0)
train.auto <- Auto[train,]
test.auto <- Auto[-train,]Part 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?
fit.lda <- lda(mpg01 ~ cylinders + weight + displacement + horsepower, data = Auto, subset = train)
fit.lda## Call:
## lda(mpg01 ~ cylinders + weight + displacement + horsepower, data = Auto,
## subset = train)
##
## Prior probabilities of groups:
## 0 1
## 0.4571429 0.5428571
##
## Group means:
## cylinders weight displacement horsepower
## 0 6.812500 3604.823 271.7396 133.14583
## 1 4.070175 2314.763 111.6623 77.92105
##
## Coefficients of linear discriminants:
## LD1
## cylinders -0.6741402638
## weight -0.0011465750
## displacement 0.0004481325
## horsepower 0.0059035377autolda.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$mpg01)## [1] 0.08439898Answer: Using LDA method to create a classifying model resulted in a test error rate of 8.44%.
Part E: Perform QDA on the training data in order to predict mpg01 using the variables that seemed most associated with mpg01 in (b). What is the test error of the model obtained?
autoqda.fit <- qda(mpg01~displacement+horsepower+weight+year+cylinders+origin, data=train.auto)
autoqda.pred <- predict(autoqda.fit, test.auto)
table(autoqda.pred$class, test.auto$mpg01)##
## 0 1
## 0 176 20
## 1 19 176mean(autoqda.pred$class != test.auto$mpg01)## [1] 0.09974425Answer: Using QDA method for classification resulted in a model with a test error rate of 9.97%
Part F: Perform logistic regression on the training data in order to predict mpg01 using the variables that seemed most associated with mpg01 in (b). What is the test error of the model obtained?
auto.fit<-glm(mpg01~displacement+horsepower+weight+year+cylinders+origin, data=train.auto,family=binomial)
auto.probs = predict(auto.fit, test.auto, type = "response")
auto.pred = rep(0, length(auto.probs))
auto.pred[auto.probs > 0.5] = 1
table(auto.pred, test.auto$mpg01)##
## auto.pred 0 1
## 0 174 12
## 1 21 184mean(auto.pred != test.auto$mpg01)## [1] 0.08439898Answer: The logistic regression method resulted in a model with a test error rate of 8.44%.
Part G: Perform KNN on the training data, with several values of K, in order to predict mpg01. Use only the variables that seemed most associated with mpg01 in (b). What test errors do you obtain? Which value of K seems to perform the best on this data set?
train.K= cbind(displacement,horsepower,weight,cylinders,year, origin)[train,]
test.K=cbind(displacement,horsepower,weight,cylinders, year, origin)[-train,]
set.seed(1)
autok.pred=knn(train.K,test.K,train.auto$mpg01,k=1)
mean(autok.pred != test.auto$mpg01)## [1] 0.07161125autok.pred=knn(train.K,test.K,train.auto$mpg01,k=5)
mean(autok.pred != test.auto$mpg01)## [1] 0.112532autok.pred=knn(train.K,test.K,train.auto$mpg01,k=10)
mean(autok.pred != test.auto$mpg01)## [1] 0.1253197Answer: After reviewing the results of the K nearest neighbors,it appears that K=1 has the lowest error rate of 7.16%. As the value of K increases so does the error rate for this particular model.
Question 13: Using the Boston data set, fit classification models in order to predict whether a given suburb has a crime rate above or below the median.Explore logistic regression, LDA, and KNN models using various subsets of the predictors. Describe your findings.
attach(Boston)
crim01 <- rep(0, length(crim))
crim01[crim > median(crim)] <- 1
Boston <- data.frame(Boston, crim01)
train <- 1:(length(crim) / 2)
test <- (length(crim) / 2 + 1):length(crim)
Boston.train <- Boston[train, ]
Boston.test <- Boston[test, ]
crim01.test <- crim01[test]
fit.glm <- glm(crim01 ~ . - crim01 - crim, data = Boston, family = binomial, subset = train)## Warning: glm.fit: fitted probabilities numerically 0 or 1 occurredprobs <- predict(fit.glm, Boston.test, type = "response")
pred.glm <- rep(0, length(probs))
pred.glm[probs > 0.5] <- 1
table(pred.glm, crim01.test)## crim01.test
## pred.glm 0 1
## 0 68 24
## 1 22 139mean(pred.glm != crim01.test)## [1] 0.1818182Answer: We have a test error rate of 18.1818182%.
fit.glm <- glm(crim01 ~ . - crim01 - crim - chas - nox, data = Boston, family = binomial, subset = train)## Warning: glm.fit: fitted probabilities numerically 0 or 1 occurredprobs <- predict(fit.glm, Boston.test, type = "response")
pred.glm <- rep(0, length(probs))
pred.glm[probs > 0.5] <- 1
table(pred.glm, crim01.test)## crim01.test
## pred.glm 0 1
## 0 78 28
## 1 12 135mean(pred.glm != crim01.test)## [1] 0.1581028Answer: We have a test error rate of 15.8102767%.
fit.lda <- lda(crim01 ~ . - crim01 - crim, data = Boston, subset = train)
pred.lda <- predict(fit.lda, Boston.test)
table(pred.lda$class, crim01.test)## crim01.test
## 0 1
## 0 80 24
## 1 10 139mean(pred.lda$class != crim01.test)## [1] 0.1343874Answer: For this LDA, we have a test error rate of 13.4387352%.
fit.lda <- lda(crim01 ~ . - crim01 - crim - chas - nox, data = Boston, subset = train)
pred.lda <- predict(fit.lda, Boston.test)
table(pred.lda$class, crim01.test)## crim01.test
## 0 1
## 0 82 30
## 1 8 133mean(pred.lda$class != crim01.test)## [1] 0.1501976Answer: For this LDA, we have a test error rate of 15.0197628%.
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)
pred.knn <- knn(train.X, test.X, train.crim01, k = 1)
table(pred.knn, crim01.test)## crim01.test
## pred.knn 0 1
## 0 85 111
## 1 5 52mean(pred.knn != crim01.test)## [1] 0.458498Answer: For this KNN (k=1), we have a test error rate of 45.8498024%.
pred.knn <- knn(train.X, test.X, train.crim01, k = 10)
table(pred.knn, crim01.test)## crim01.test
## pred.knn 0 1
## 0 83 23
## 1 7 140mean(pred.knn != crim01.test)## [1] 0.1185771Answer: For this KNN (k=10), we have a test error rate of 11.8577075%.
pred.knn <- knn(train.X, test.X, train.crim01, k = 100)
table(pred.knn, crim01.test)## crim01.test
## pred.knn 0 1
## 0 86 120
## 1 4 43mean(pred.knn != crim01.test)## [1] 0.4901186