Assignment 3
CJ Patteson
7/3/2021
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 1089 weekly returns for 21 years, from the beginning of 1990 to the end of 2010.
- Produce some numerical and graphical summaries of the “Weekly” data. Do there appear to be any patterns ?
library(ISLR)## Warning: package 'ISLR' was built under R version 4.0.5summary(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 1.000000000attach(Weekly)
plot(Volume)
There is hardly any correlation between the “lag” variables and today’s return. “Year” and “Volume” are the only ones that seem to have any correlation, and you can see “Volume” increasing as time goes on.
- 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 ?
fit.glm <- glm(Direction ~ Lag1 + Lag2 + Lag3 + Lag4 + Lag5 + Volume, data = Weekly, family = binomial)
summary(fit.glm)##
## 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: 4It can see that “Lag2” looks like the only predictor statistically significant due to its p-value to be less than 0.05.
- 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.
probs <- predict(fit.glm, type = "response")
pred.glm <- rep("Down", length(probs))
pred.glm[probs > 0.5] <- "Up"
table(pred.glm, Direction)## Direction
## pred.glm Down Up
## Down 54 48
## Up 430 557The confusion matrix is telling us that the percentage of correct predictions on the training data is equal to 56.1065197%. Which means 43.8934803% is the training error rate. It also tells us that when the market goes up, the model is often right 92.0661157% of the time. When the market goes down, the model is only right 11.1570248% of the time.
- 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 to 2010).
train <- (Year < 2009)
Weekly.20092010 <- Weekly[!train, ]
Direction.20092010 <- Direction[!train]
fit.glm2 <- glm(Direction ~ Lag2, data = Weekly, family = binomial, subset = train)
summary(fit.glm2)##
## Call:
## glm(formula = Direction ~ Lag2, family = binomial, data = Weekly,
## subset = train)
##
## Deviance Residuals:
## Min 1Q Median 3Q Max
## -1.536 -1.264 1.021 1.091 1.368
##
## Coefficients:
## Estimate Std. Error z value Pr(>|z|)
## (Intercept) 0.20326 0.06428 3.162 0.00157 **
## Lag2 0.05810 0.02870 2.024 0.04298 *
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## (Dispersion parameter for binomial family taken to be 1)
##
## Null deviance: 1354.7 on 984 degrees of freedom
## Residual deviance: 1350.5 on 983 degrees of freedom
## AIC: 1354.5
##
## Number of Fisher Scoring iterations: 4probs2 <- predict(fit.glm2, Weekly.20092010, type = "response")
pred.glm2 <- rep("Down", length(probs2))
pred.glm2[probs2 > 0.5] <- "Up"
table(pred.glm2, Direction.20092010)## Direction.20092010
## pred.glm2 Down Up
## Down 9 5
## Up 34 56This shows the percentage of correct predictions on the test data is equal to 62.5%. Meaning the test error rate is 37.5%. This also shows that when the market goes up, the model is right 91.8032787%. When the market goes down, the model is right only 20.9302326%.
- Repeat (d) using LDA.
library(MASS)
fit.lda <- lda(Direction ~ Lag2, data = Weekly, subset = train)
fit.lda## Call:
## lda(Direction ~ Lag2, data = Weekly, subset = train)
##
## Prior probabilities of groups:
## Down Up
## 0.4477157 0.5522843
##
## Group means:
## Lag2
## Down -0.03568254
## Up 0.26036581
##
## Coefficients of linear discriminants:
## LD1
## Lag2 0.4414162pred.lda <- predict(fit.lda, Weekly.20092010)
table(pred.lda$class, Direction.20092010)## Direction.20092010
## Down Up
## Down 9 5
## Up 34 56This shows that the percentage of correct predictions on the test data is 62.5%. Meaning 37.5% is the test error rate. Also telling us when the market goes up, the model is right 91.8032787%, and when the market goes down, the model is only right 20.9302326%. This shows us that the results are pretty similar to the results of the logistic regression model.
- Repeat (d) using QDA.
fit.qda <- qda(Direction ~ Lag2, data = Weekly, subset = train)
fit.qda## Call:
## qda(Direction ~ Lag2, data = Weekly, subset = train)
##
## Prior probabilities of groups:
## Down Up
## 0.4477157 0.5522843
##
## Group means:
## Lag2
## Down -0.03568254
## Up 0.26036581pred.qda <- predict(fit.qda, Weekly.20092010)
table(pred.qda$class, Direction.20092010)## Direction.20092010
## Down Up
## Down 0 0
## Up 43 61This shows us that the percentage of correct predictions on the test data is 58.6538462%. Meaning 41.3461538% is the test error rate. Showing us that when the market goes up, the model is right 100%, and when the market goes down, the model is not right ever at a rate of 0%.
- Repeat (d) using KNN with K=1.
library(class)
train.X <- as.matrix(Lag2[train])
test.X <- as.matrix(Lag2[!train])
train.Direction <- Direction[train]
set.seed(1)
pred.knn <- knn(train.X, test.X, train.Direction, k = 1)
table(pred.knn, Direction.20092010)## Direction.20092010
## pred.knn Down Up
## Down 21 30
## Up 22 31This shows us that the percentage of correct predictions on the test data is 50%. Meaning 50% is the test error rate. When the market goes up the model is right 50.8196721%, and when it goes down, the model is only right 48.8372093%.
- Which of these methods appears to provide the best results on this data?
When we compare the test error rates, we can see that logistic regression and LDA have the lowest error rates, then followed by QDA and KNN.
- 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.
fit.glm3 <- glm(Direction ~ Lag2:Lag1, data = Weekly, family = binomial, subset = train)
probs3 <- predict(fit.glm3, Weekly.20092010, type = "response")
pred.glm3 <- rep("Down", length(probs3))
pred.glm3[probs3 > 0.5] = "Up"
table(pred.glm3, Direction.20092010)## Direction.20092010
## pred.glm3 Down Up
## Down 1 1
## Up 42 60mean(pred.glm3 == Direction.20092010)## [1] 0.5865385fit.lda2 <- lda(Direction ~ Lag2:Lag1, data = Weekly, subset = train)
pred.lda2 <- predict(fit.lda2, Weekly.20092010)
mean(pred.lda2$class == Direction.20092010)## [1] 0.5769231fit.qda2 <- qda(Direction ~ Lag2 + sqrt(abs(Lag2)), data = Weekly, subset = train)
pred.qda2 <- predict(fit.qda2, Weekly.20092010)
table(pred.qda2$class, Direction.20092010)## Direction.20092010
## Down Up
## Down 12 13
## Up 31 48mean(pred.qda2$class == Direction.20092010)## [1] 0.5769231pred.knn2 <- knn(train.X, test.X, train.Direction, k = 10)
table(pred.knn2, Direction.20092010)## Direction.20092010
## pred.knn2 Down Up
## Down 17 18
## Up 26 43mean(pred.knn2 == Direction.20092010)## [1] 0.5769231pred.knn3 <- knn(train.X, test.X, train.Direction, k = 100)
table(pred.knn3, Direction.20092010)## Direction.20092010
## pred.knn3 Down Up
## Down 9 12
## Up 34 49mean(pred.knn3 == Direction.20092010)## [1] 0.5576923We can conclude that the original logistic regression and LDA have the best test error rates out of them all.
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.
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.
attach(Auto)
mpg01 <- rep(0, length(mpg))
mpg01[mpg > median(mpg)] <- 1
Auto <- data.frame(Auto, mpg01)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 predictiong “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(cylinders ~ mpg01, data = Auto, main = "Cylinders vs mpg01")
boxplot(displacement ~ mpg01, data = Auto, main = "Displacement vs mpg01")
boxplot(horsepower ~ mpg01, data = Auto, main = "Horsepower vs mpg01")
boxplot(weight ~ mpg01, data = Auto, main = "Weight vs mpg01")
boxplot(acceleration ~ mpg01, data = Auto, main = "Acceleration vs mpg01")
boxplot(year ~ mpg01, data = Auto, main = "Year vs mpg01")
We can gather that there is some correlation between “mpg01” and “cylinders”, “weight”, “displacement” and “horsepower”.
- Split the data into a training set and a test set.
train <- (year %% 2 == 0)
Auto.train <- Auto[train, ]
Auto.test <- Auto[!train, ]
mpg01.test <- mpg01[!train]- 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 0.0059035377pred.lda <- predict(fit.lda, Auto.test)
table(pred.lda$class, mpg01.test)## mpg01.test
## 0 1
## 0 86 9
## 1 14 73mean(pred.lda$class != mpg01.test)## [1] 0.1263736We can see that the test error rate is 12.6373626%.
- Perform QDA on the training data in order to predict “mpg01” using the variables that seemed most associated with “mpg01” in (b). What is the test error of the model obtained?
fit.qda <- qda(mpg01 ~ cylinders + weight + displacement + horsepower, data = Auto, subset = train)
fit.qda## Call:
## qda(mpg01 ~ cylinders + weight + displacement + horsepower, data = Auto,
## subset = train)
##
## Prior probabilities of groups:
## 0 1
## 0.4571429 0.5428571
##
## Group means:
## cylinders weight displacement horsepower
## 0 6.812500 3604.823 271.7396 133.14583
## 1 4.070175 2314.763 111.6623 77.92105pred.qda <- predict(fit.qda, Auto.test)
table(pred.qda$class, mpg01.test)## mpg01.test
## 0 1
## 0 89 13
## 1 11 69mean(pred.qda$class != mpg01.test)## [1] 0.1318681We can see that the test error rate is 13.1868132%.
- 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?
fit.glm <- glm(mpg01 ~ cylinders + weight + displacement + horsepower, data = Auto, family = binomial, subset = train)
summary(fit.glm)##
## Call:
## glm(formula = mpg01 ~ cylinders + weight + displacement + horsepower,
## family = binomial, data = Auto, subset = train)
##
## Deviance Residuals:
## Min 1Q Median 3Q Max
## -2.48027 -0.03413 0.10583 0.29634 2.57584
##
## Coefficients:
## Estimate Std. Error z value Pr(>|z|)
## (Intercept) 17.658730 3.409012 5.180 2.22e-07 ***
## cylinders -1.028032 0.653607 -1.573 0.1158
## weight -0.002922 0.001137 -2.569 0.0102 *
## displacement 0.002462 0.015030 0.164 0.8699
## horsepower -0.050611 0.025209 -2.008 0.0447 *
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## (Dispersion parameter for binomial family taken to be 1)
##
## Null deviance: 289.58 on 209 degrees of freedom
## Residual deviance: 83.24 on 205 degrees of freedom
## AIC: 93.24
##
## Number of Fisher Scoring iterations: 7probs <- predict(fit.glm, Auto.test, type = "response")
pred.glm <- rep(0, length(probs))
pred.glm[probs > 0.5] <- 1
table(pred.glm, mpg01.test)## mpg01.test
## pred.glm 0 1
## 0 89 11
## 1 11 71mean(pred.glm != mpg01.test)## [1] 0.1208791We can see that the test error rate is 12.0879121%.
- Perform KNN on the training data, with several values of K, in order to predict “mpg01” using 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, weight, displacement, horsepower)[train, ]
test.X <- cbind(cylinders, weight, displacement, horsepower)[!train, ]
train.mpg01 <- mpg01[train]
set.seed(1)
pred.knn <- knn(train.X, test.X, train.mpg01, k = 1)
table(pred.knn, mpg01.test)## mpg01.test
## pred.knn 0 1
## 0 83 11
## 1 17 71mean(pred.knn != mpg01.test)## [1] 0.1538462We can see that the test error rate is 15.3846154% for K=1.
pred.knn <- knn(train.X, test.X, train.mpg01, k = 10)
table(pred.knn, mpg01.test)## mpg01.test
## pred.knn 0 1
## 0 77 7
## 1 23 75mean(pred.knn != mpg01.test)## [1] 0.1648352We can see that the test error rate is 16.4835165% for K=10.
pred.knn <- knn(train.X, test.X, train.mpg01, k = 100)
table(pred.knn, mpg01.test)## mpg01.test
## pred.knn 0 1
## 0 81 7
## 1 19 75mean(pred.knn != mpg01.test)## [1] 0.1428571We can see that the test error rate is 14.2857143% for K=100. Showing us that a K value of 100 performs the best.
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 the logistic regression, LDA, and KNN models using various subsets of the predictors. Describe your findings.
library(MASS)
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] 0.1818182We can see that the test error rate for this logistic regression is 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] 0.1581028fit.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] 0.1343874We can see that the test error rate for LDA is 13.4387352%.
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] 0.458498We can see that the test error rate for KNN (k=1) is 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] 0.1185771We can see that the test error rate for KNN (k=10) is 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] 0.4901186We can see that the test error rate for KNN (k=100) is 49.0118577%.