Assignment 3
Albert Uriegas
6/21/2020
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.
library(ISLR)
attach(Weekly)(a) Produce some numerical and graphical summaries of the Weekly data. Do there appear to be any patterns?
Volume and Year have a high correlation (0.84) but there seem to be no other relationships.
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
##
##
##
## pairs(Weekly)
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 1.000000000(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?
Lag2 is the only statistically significant predictor with a p value of 0.0296.
glm.fit <- glm(Direction ~ Lag1 + Lag2 + Lag3 + Lag4 + Lag5 + Volume,
data = Weekly, family = binomial)
summary(glm.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(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.
The overall accuracy of predictions was 56.1% based on the confusion matrix below. Most of the errors were in predicting a down week. The model basically predicted an up week most of the time which resulted in an error rate of 88.8% for actual down weeks.
glm.probs <- predict(glm.fit, type = "response")
glm.pred <- rep("Down", length(glm.probs))
glm.pred[glm.probs > 0.5] = "Up"
table(glm.pred, Direction)## Direction
## glm.pred Down Up
## Down 54 48
## Up 430 557mean(glm.pred == Direction)## [1] 0.5610652(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)
Weeklytest <- Weekly[!train, ]
glm.fit <- glm(Direction ~ Lag2, data = Weekly, family = binomial, subset = train)
glm.probs <- predict(glm.fit, Weeklytest, type = "response")
glm.pred <- rep("Down", length(glm.probs))
glm.pred[glm.probs > 0.5] = "Up"
Direction.test <- Direction[!train]
table(glm.pred, Direction.test)## Direction.test
## glm.pred Down Up
## Down 9 5
## Up 34 56mean(glm.pred == Direction.test)## [1] 0.625(e) Repeat (d) using LDA.
library(MASS)
lda.fit <- lda(Direction ~ Lag2, data = Weekly, subset = train)
lda.pred <- predict(lda.fit, Weeklytest)
lda.class <- lda.pred$class
table(lda.class, Direction.test)## Direction.test
## lda.class Down Up
## Down 9 5
## Up 34 56mean(lda.class == Direction.test)## [1] 0.625(f) Repeat (d) using QDA.
qda.fit <- qda(Direction ~ Lag2, data = Weekly, ssubset = train)
qda.class <- predict(qda.fit, Weeklytest)$class
table(qda.class, Direction.test)## Direction.test
## qda.class Down Up
## Down 0 0
## Up 43 61mean(qda.class == Direction.test)## [1] 0.5865385(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)
knn.pred <- knn(train.X, test.X, train.Direction, k = 1)
table(knn.pred, Direction.test)## Direction.test
## knn.pred Down Up
## Down 21 30
## Up 22 31mean(knn.pred == Direction.test)## [1] 0.5(h) Which of these methods appears to provide the best results on this data?
The logistic regression model and the LDA model both had the best and identical results with a 62.5% accuracy rate.
(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.
The logistic regression model and the LDA model from (d) still had the highest accuracy rate of 62.5%.
# Logistic Regression with interaction of Lag1 & Lag 2
glm.int <- glm(Direction ~ Lag1*Lag2, data = Weekly, family = binomial,
subset = train)
glm.probs <- predict(glm.int, Weeklytest, type = "response")
glm.pred <- rep("Down", length(glm.probs))
glm.pred[glm.probs > 0.5] = "Up"
Direction.test <- Direction[!train]
table(glm.pred, Direction.test)## Direction.test
## glm.pred Down Up
## Down 7 8
## Up 36 53mean(glm.pred == Direction.test)## [1] 0.5769231# LDA with interaction of Lag1 & Lag2
lda.int <- lda(Direction ~ Lag1*Lag2, data = Weekly, subset = train)
lda.pred <- predict(lda.int, Weeklytest)
lda.class <- lda.pred$class
table(lda.class, Direction.test)## Direction.test
## lda.class Down Up
## Down 7 8
## Up 36 53mean(lda.class == Direction.test)## [1] 0.5769231# QDA with interaction of Lag1 & Lag2
qda.int <- qda(Direction ~ Lag1*Lag2, data = Weekly, ssubset = train)
qda.class <- predict(qda.int, Weeklytest)$class
table(qda.class, Direction.test)## Direction.test
## qda.class Down Up
## Down 17 30
## Up 26 31mean(qda.class == Direction.test)## [1] 0.4615385# KNN with k = 2(no interaction)
knn.pred <- knn(train.X, test.X, train.Direction, k = 2)
table(knn.pred, Direction.test)## Direction.test
## knn.pred Down Up
## Down 18 25
## Up 25 36mean(knn.pred == Direction.test)## [1] 0.5192308# KNN with k = 3(no interaction)
knn.pred <- knn(train.X, test.X, train.Direction, k = 3)
table(knn.pred, Direction.test)## Direction.test
## knn.pred Down Up
## Down 16 20
## Up 27 41mean(knn.pred == Direction.test)## [1] 0.5480769# KNN with k = 4(no interaction)
knn.pred <- knn(train.X, test.X, train.Direction, k = 4)
table(knn.pred, Direction.test)## Direction.test
## knn.pred Down Up
## Down 20 19
## Up 23 42mean(knn.pred == Direction.test)## [1] 0.5961538# KNN with k = 5(no interaction)
knn.pred <- knn(train.X, test.X, train.Direction, k = 5)
table(knn.pred, Direction.test)## Direction.test
## knn.pred Down Up
## Down 15 22
## Up 28 39mean(knn.pred == Direction.test)## [1] 0.5192308# KNN with k = 6(no interaction)
knn.pred <- knn(train.X, test.X, train.Direction, k = 6)
table(knn.pred, Direction.test)## Direction.test
## knn.pred Down Up
## Down 16 21
## Up 27 40mean(knn.pred == Direction.test)## [1] 0.5384615Problem 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)
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) :365attach(Auto)(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
Auto1 <- 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.
Cylinders, displacement, horsepower, weight, and possibly origin would be useful in predicting mpg01. 4 of the predictors are highly correlated with each other: cylinders, displacement, horsepower and weight.
pairs(Auto1)
cor(Auto1[, -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 1.0000000(c) Split the data into a training set and a test set.
train <- (year%%2 == 0) # even year
test <- !train
Auto1.train = Auto1[train, ]
Auto1.test = Auto1[test, ]
mpg01.test = mpg01[test](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?
The test error rate of the model is 12.64%.
lda.Auto1 <- lda(mpg01 ~ cylinders + weight + displacement + horsepower, data = Auto1,
subset = train)
lda.Auto1pred <- predict(lda.Auto1, Auto1.test)
mean(lda.Auto1pred$class != mpg01.test)## [1] 0.1263736(e) Perform QDA on the training data in order to predict mpg01 using the variables that seemed most associated with mpg01 in (b). What is the test error of the model obtained?
The test error rate is 13.19%.
qda.Auto1 <- qda(mpg01 ~ cylinders + weight + displacement + horsepower, data = Auto1,
subset = train)
qda.Auto1pred <- predict(qda.Auto1, Auto1.test)
mean(qda.Auto1pred$class != mpg01.test)## [1] 0.1318681(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?
The test error rate is 12.64%.
Auto1.log <- glm(mpg01 ~ cylinders + weight + displacement + horsepower, data = Auto1,
subset = train)
Auto1.probs <- predict(Auto1.log, Auto1.test, type = "response")
Auto1.logpred <- rep(0, length(Auto1.probs))
Auto1.logpred[Auto1.probs > 0.5] = 1
mean(Auto1.logpred != mpg01.test)## [1] 0.1263736(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?
The following test error rates were obtained: 15.38% for k = 1, 15.38% for k = 2, 13.74% for k = 3, 15.93% for k = 4, 14.84% for k = 5 and 15.93% for k = 10. The test error rate was lowest or k = 3.
train.X <- cbind(cylinders, weight, displacement, horsepower)[train, ]
test.X <- cbind(cylinders, weight, displacement, horsepower)[test, ]
train.mpg01 <- mpg01[train]
set.seed(1)
knn.pred = knn(train.X, test.X, train.mpg01, k = 1)
mean(knn.pred != mpg01.test)## [1] 0.1538462knn.pred = knn(train.X, test.X, train.mpg01, k = 2)
mean(knn.pred != mpg01.test)## [1] 0.1538462knn.pred = knn(train.X, test.X, train.mpg01, k = 3)
mean(knn.pred != mpg01.test)## [1] 0.1373626knn.pred = knn(train.X, test.X, train.mpg01, k = 4)
mean(knn.pred != mpg01.test)## [1] 0.1593407knn.pred = knn(train.X, test.X, train.mpg01, k = 5)
mean(knn.pred != mpg01.test)## [1] 0.1483516knn.pred = knn(train.X, test.X, train.mpg01, k = 10)
mean(knn.pred != mpg01.test)## [1] 0.1593407Problem 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.
The full logistic regression model was fitted to the training data, which resulted in an error rate of 18.18%. This was a better rate than the logistic regression model without chas and tax, which were not statistically significant in the full model. This model had an error rate of 18.58%. The LDA model had a 13.44% error rate. Out of the KNN models, the one with k = 10 had the lowest error rate of 11.46%.
attach(Boston)
summary(Boston)## crim zn indus chas
## Min. : 0.00632 Min. : 0.00 Min. : 0.46 Min. :0.00000
## 1st Qu.: 0.08205 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. :50.00pairs(Boston)
cor(Boston)## crim zn indus chas nox
## crim 1.00000000 -0.20046922 0.40658341 -0.055891582 0.42097171
## zn -0.20046922 1.00000000 -0.53382819 -0.042696719 -0.51660371
## indus 0.40658341 -0.53382819 1.00000000 0.062938027 0.76365145
## chas -0.05589158 -0.04269672 0.06293803 1.000000000 0.09120281
## nox 0.42097171 -0.51660371 0.76365145 0.091202807 1.00000000
## rm -0.21924670 0.31199059 -0.39167585 0.091251225 -0.30218819
## age 0.35273425 -0.56953734 0.64477851 0.086517774 0.73147010
## dis -0.37967009 0.66440822 -0.70802699 -0.099175780 -0.76923011
## rad 0.62550515 -0.31194783 0.59512927 -0.007368241 0.61144056
## tax 0.58276431 -0.31456332 0.72076018 -0.035586518 0.66802320
## ptratio 0.28994558 -0.39167855 0.38324756 -0.121515174 0.18893268
## black -0.38506394 0.17552032 -0.35697654 0.048788485 -0.38005064
## lstat 0.45562148 -0.41299457 0.60379972 -0.053929298 0.59087892
## medv -0.38830461 0.36044534 -0.48372516 0.175260177 -0.42732077
## rm age dis rad tax ptratio
## crim -0.21924670 0.35273425 -0.37967009 0.625505145 0.58276431 0.2899456
## zn 0.31199059 -0.56953734 0.66440822 -0.311947826 -0.31456332 -0.3916785
## indus -0.39167585 0.64477851 -0.70802699 0.595129275 0.72076018 0.3832476
## chas 0.09125123 0.08651777 -0.09917578 -0.007368241 -0.03558652 -0.1215152
## nox -0.30218819 0.73147010 -0.76923011 0.611440563 0.66802320 0.1889327
## rm 1.00000000 -0.24026493 0.20524621 -0.209846668 -0.29204783 -0.3555015
## age -0.24026493 1.00000000 -0.74788054 0.456022452 0.50645559 0.2615150
## dis 0.20524621 -0.74788054 1.00000000 -0.494587930 -0.53443158 -0.2324705
## rad -0.20984667 0.45602245 -0.49458793 1.000000000 0.91022819 0.4647412
## tax -0.29204783 0.50645559 -0.53443158 0.910228189 1.00000000 0.4608530
## ptratio -0.35550149 0.26151501 -0.23247054 0.464741179 0.46085304 1.0000000
## black 0.12806864 -0.27353398 0.29151167 -0.444412816 -0.44180801 -0.1773833
## lstat -0.61380827 0.60233853 -0.49699583 0.488676335 0.54399341 0.3740443
## medv 0.69535995 -0.37695457 0.24992873 -0.381626231 -0.46853593 -0.5077867
## black lstat medv
## crim -0.38506394 0.4556215 -0.3883046
## zn 0.17552032 -0.4129946 0.3604453
## indus -0.35697654 0.6037997 -0.4837252
## chas 0.04878848 -0.0539293 0.1752602
## nox -0.38005064 0.5908789 -0.4273208
## rm 0.12806864 -0.6138083 0.6953599
## age -0.27353398 0.6023385 -0.3769546
## dis 0.29151167 -0.4969958 0.2499287
## rad -0.44441282 0.4886763 -0.3816262
## tax -0.44180801 0.5439934 -0.4685359
## ptratio -0.17738330 0.3740443 -0.5077867
## black 1.00000000 -0.3660869 0.3334608
## lstat -0.36608690 1.0000000 -0.7376627
## medv 0.33346082 -0.7376627 1.0000000crime01 <- rep(0, length(crim))
crime01[crim > median(crim)] = 1
Boston <- data.frame(Boston, crime01)
names(Boston)## [1] "crim" "zn" "indus" "chas" "nox" "rm" "age"
## [8] "dis" "rad" "tax" "ptratio" "black" "lstat" "medv"
## [15] "crime01"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]Full Logistic Regression Model
glm.fit <- glm(crime01 ~ . - crim - crime01, data = Boston, family = binomial,
subset = train)
summary(glm.fit)##
## Call:
## glm(formula = crime01 ~ . - crim - crime01, family = binomial,
## data = Boston, subset = train)
##
## Deviance Residuals:
## Min 1Q Median 3Q Max
## -2.83229 -0.06593 0.00000 0.06181 2.61513
##
## Coefficients:
## Estimate Std. Error z value Pr(>|z|)
## (Intercept) -91.319906 19.490273 -4.685 2.79e-06 ***
## zn -0.815573 0.193373 -4.218 2.47e-05 ***
## indus 0.354172 0.173862 2.037 0.04164 *
## chas 0.167396 0.991922 0.169 0.86599
## nox 93.706326 21.202008 4.420 9.88e-06 ***
## rm -4.719108 1.788765 -2.638 0.00833 **
## age 0.048634 0.024199 2.010 0.04446 *
## dis 4.301493 0.979996 4.389 1.14e-05 ***
## rad 3.039983 0.719592 4.225 2.39e-05 ***
## tax -0.006546 0.007855 -0.833 0.40461
## ptratio 1.430877 0.359572 3.979 6.91e-05 ***
## black -0.017552 0.006734 -2.606 0.00915 **
## lstat 0.190439 0.086722 2.196 0.02809 *
## medv 0.598533 0.185514 3.226 0.00125 **
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## (Dispersion parameter for binomial family taken to be 1)
##
## Null deviance: 329.367 on 252 degrees of freedom
## Residual deviance: 69.568 on 239 degrees of freedom
## AIC: 97.568
##
## Number of Fisher Scoring iterations: 10glm.probs <- predict(glm.fit, Boston.test, type = "response")
glm.pred <- rep(0, length(glm.probs))
glm.pred[glm.probs > 0.5] = 1
table(glm.pred, crime01.test)## crime01.test
## glm.pred 0 1
## 0 68 24
## 1 22 139mean(glm.pred != crime01.test)## [1] 0.1818182Logistic Regression Model without chas and tax
glm.fit <- glm(crime01 ~ . - crim - crime01 - chas - tax, data = Boston,
family = binomial, subset = train)
glm.probs <- predict(glm.fit, Boston.test, type = "response")
glm.pred <- rep(0, length(glm.probs))
glm.pred[glm.probs > 0.5] = 1
table(glm.pred, crime01.test)## crime01.test
## glm.pred 0 1
## 0 67 24
## 1 23 139mean(glm.pred != crime01.test)## [1] 0.1857708LDA
lda.fit <- lda(crime01 ~ . - crim - crime01, data = Boston, subset = train)
lda.pred <- predict(lda.fit, Boston.test)
table(lda.pred$class, crime01.test)## crime01.test
## 0 1
## 0 80 24
## 1 10 139mean(lda.pred$class != crime01.test)## [1] 0.1343874KNN, k = 1
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.crime01 <- crime01[train]
set.seed(1)
knn.pred <- knn(train.X, test.X, train.crime01, k = 1)
table(knn.pred, crime01.test)## crime01.test
## knn.pred 0 1
## 0 85 111
## 1 5 52mean(knn.pred != crime01.test)## [1] 0.458498KNN, k = 2
knn.pred <- knn(train.X, test.X, train.crime01, k = 2)
table(knn.pred, crime01.test)## crime01.test
## knn.pred 0 1
## 0 86 79
## 1 4 84mean(knn.pred != crime01.test)## [1] 0.3280632KNN, k = 3
knn.pred <- knn(train.X, test.X, train.crime01, k = 3)
table(knn.pred, crime01.test)## crime01.test
## knn.pred 0 1
## 0 84 61
## 1 6 102mean(knn.pred != crime01.test)## [1] 0.2648221KNN, k = 4
knn.pred <- knn(train.X, test.X, train.crime01, k = 4)
table(knn.pred, crime01.test)## crime01.test
## knn.pred 0 1
## 0 84 48
## 1 6 115mean(knn.pred != crime01.test)## [1] 0.2134387KNN, k = 5
knn.pred <- knn(train.X, test.X, train.crime01, k = 5)
table(knn.pred, crime01.test)## crime01.test
## knn.pred 0 1
## 0 84 37
## 1 6 126mean(knn.pred != crime01.test)## [1] 0.1699605KNN, k = 10
knn.pred <- knn(train.X, test.X, train.crime01, k = 10)
table(knn.pred, crime01.test)## crime01.test
## knn.pred 0 1
## 0 83 22
## 1 7 141mean(knn.pred != crime01.test)## [1] 0.1146245KNN, k = 15
knn.pred <- knn(train.X, test.X, train.crime01, k = 15)
table(knn.pred, crime01.test)## crime01.test
## knn.pred 0 1
## 0 83 23
## 1 7 140mean(knn.pred != crime01.test)## [1] 0.1185771KNN, k = 20
knn.pred <- knn(train.X, test.X, train.crime01, k = 20)
table(knn.pred, crime01.test)## crime01.test
## knn.pred 0 1
## 0 83 24
## 1 7 139mean(knn.pred != crime01.test)## [1] 0.1225296