STA-4233 Data Mining HW 3 CH 4
Savanna Munoz
3/6/2020
Problem 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.
require(ISLR)## Loading required package: ISLR## Warning: package 'ISLR' was built under R version 3.6.2require(MASS)## Loading required package: MASSrequire(class)## Loading required package: class(a) Produce some numerical and graphical summaries of the Weekly data. 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
## 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. : 12.0260cor(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)
The only substantial correlation is between “Year” and “Volume”. When we plot “Volume”, it is evident that it is increasing 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?
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: 4“Lag2” is the only predictor statistically significant as its p-value is less than 0.05.
(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.
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 557We can conclude that the percentage of correct predictions on the training data is (54+557)/1089 wich is equal to 56.11%. That means 43.89% is the training error rate. We could say that for weeks when the market goes up, the model is right 92.07% of the time (557/(48+557)). For weeks when the market goes down, the model is right only 11.16% of the time (54/(54+430)).
(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.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 56When spliting up the 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. This model also 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)
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$class, Direction.20092010)## Direction.20092010
## Down Up
## Down 9 5
## Up 34 56Using Linear Discriminant Analysis to develop a classifying model yielded similar results as the logistic regression model created in part D.
(f) Repeat (d) using QDA.
fit.qda <- qda(Direction ~ Lag2, data = Weekly, subset = train)
fit.qda## Call:
## qda(Direction ~ Lag2, data = Weekly, subset = train)
##
## Prior probabilities of groups:
## Down Up
## 0.4477157 0.5522843
##
## Group means:
## Lag2
## Down -0.03568254
## Up 0.26036581pred.qda <- predict(fit.qda, Weekly.20092010)
table(pred.qda$class, Direction.20092010)## Direction.20092010
## Down Up
## Down 0 0
## Up 43 61Quadratic Linear Analysis created a model with an accuracy of 58.65%. This 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)
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 31The K-Nearest neighbors resulted in a classifying model with an accuracy rate of 50%.
(h) Which of these methods appears to provide the best results on this data?
When comparing the test error rates, we find that logistic regression and linear discriminant analysis have the minimum error rates with accuracy 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 Lag2:Lag1
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] 0.5865385# LDA with Lag2 interaction with Lag1
fit.lda2 <- lda(Direction ~ Lag2:Lag1, data = Weekly, subset = train)
pred.lda2 <- predict(fit.lda2, Weekly.20092010)
mean(pred.lda2$class == Direction.20092010)## [1] 0.5769231# QDA with sqrt(abs(Lag2))
fit.qda2 <- qda(Direction ~ Lag2 + sqrt(abs(Lag2)), data = Weekly, subset = train)
pred.qda2 <- predict(fit.qda2, Weekly.20092010)
table(pred.qda2$class, Direction.20092010)## Direction.20092010
## Down Up
## Down 12 13
## Up 31 48mean(pred.qda2$class == Direction.20092010)## [1] 0.5769231# KNN k =10
pred.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] 0.5769231# KNN k = 100
pred.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] 0.5576923Still, the original logistic regression and LDA have the best performance when comparing test error rates.
Problem 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.
(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.
attach(Auto)
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.
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")
Some association between “mpg01” and “cylinders”, “weight”, “displacement” and “horsepower” exists.
(c) 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](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 0.0059035377pred.lda <- predict(fit.lda, Auto.test)
table(pred.lda$class, mpg01.test)## mpg01.test
## 0 1
## 0 86 9
## 1 14 73mean(pred.lda$class != mpg01.test)## [1] 0.1263736Using LDA method to create a classifying model resulted in a test error rate is 12.64%.
(e) Perform QDA on the training data in order to predict mpg01 using the variables that seemed most associated with mpg01 in (b). What is the test error of the model obtained?
fit.qda <- qda(mpg01 ~ cylinders + weight + displacement + horsepower, data = Auto, subset = train)
fit.qda## Call:
## qda(mpg01 ~ cylinders + weight + displacement + horsepower, data = Auto,
## subset = train)
##
## Prior probabilities of groups:
## 0 1
## 0.4571429 0.5428571
##
## Group means:
## cylinders weight displacement horsepower
## 0 6.812500 3604.823 271.7396 133.14583
## 1 4.070175 2314.763 111.6623 77.92105pred.qda <- predict(fit.qda, Auto.test)
table(pred.qda$class, mpg01.test)## mpg01.test
## 0 1
## 0 89 13
## 1 11 69mean(pred.qda$class != mpg01.test)## [1] 0.1318681Using QDA method for classification resulted with a test error rate of 13.19%.
(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?
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] 0.1208791The logistic regression method resulted in a model with a test error rate is 12.09%.
(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.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] 0.1538462The test error rate for K = 1 is 15.38%.
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] 0.1648352The test error rate for K = 10 is 16.48%.
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] 0.1428571The test error rate for K = 100 seems to perform the best at 14.29%.
Problem 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.
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] 0.1818182The test error rate for this logistic regression is 18.18%.
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] 0.1581028For this logistic regression, the test error rate is 15.81%.
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$class != crim01.test)## [1] 0.1343874For this LDA, the test error rate is 13.44%.
fit.lda <- lda(crim01 ~ . - crim01 - crim - chas - nox, data = Boston, subset = train)
pred.lda <- predict(fit.lda, Boston.test)
table(pred.lda$class, crim01.test)## crim01.test
## 0 1
## 0 82 30
## 1 8 133mean(pred.lda$class != crim01.test)## [1] 0.1501976The test error rate for this LDA is 15.02%.
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] 0.458498For this KNN at K = 1, the test error rate is 45.85%.
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] 0.1185771For this KNN at K = 10, the test error rate is 11.86%.
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] 0.4901186For this KNN at K = 100, the test error rate is 49.01%.
After reviewing the results of each classification method, the K-Nearest neighbors where K = 10 had the lowest test error rate of 11.86%.