Assignment3
Lisa Smith
2022-09-25
Problem #13
This question should be answered using the Weekly data set, which is part of the ISLR2 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.
- Produce some numerical and graphical summaries of the Weekly
data.
Do there appear to be any patterns?
## Warning: package 'ISLR2' was built under R version 4.1.3## 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
##
##
##
## ## 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
There seems to be little to no correlation between today’s returns and previous day returns. However, there seems to be a substantial correlation between Year and Volume, meaning that the number of shares traded daily increased (see plot).
- 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?
glm.fits <- glm(
Direction ~ Lag1 + Lag2 + Lag3 + Lag4 + Lag5 + Volume ,
data = Weekly, family = binomial
)
summary(glm.fits)##
## 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: 4The smallest p-value is associated with Lag2 with a p-value of 0.0296. This value is statistically significant as it’s 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.
glm.probs <- predict(glm.fits, type = "response")
glm.probs [1:10]## 1 2 3 4 5 6 7 8
## 0.6086249 0.6010314 0.5875699 0.4816416 0.6169013 0.5684190 0.5786097 0.5151972
## 9 10
## 0.5715200 0.5554287contrasts(Weekly$Direction)## Up
## Down 0
## Up 1glm.pred <- rep("Down", 1089)
glm.pred[glm.probs > .5] = "Up"
table(glm.pred, Weekly$Direction)##
## glm.pred Down Up
## Down 54 48
## Up 430 557mean(glm.pred == Weekly$Direction)## [1] 0.5610652The logistic regression correctly predicted the movement of the market 56.10% of the time. However, since this is the training set. The training error rate (100% - 56.10% = 43.90%) is often too optimistic and tends to underestimate the test error rate.
- Now fit the logistic regression model using a training data period from 1990 to 2008, with Lag2 as the only predictor. Compute the confusion matrix and the overall fraction of correct predictions for the held out data (that is, the data from 2009 and 2010)
train = (Weekly$Year <2009)
Weekly.2009 <- Weekly[!train, ]
dim(Weekly.2009)## [1] 104 9Direction.2009 <- Weekly$Direction[!train]glm.fits <- glm(
Direction ~ Lag2, data = Weekly , family = binomial , subset = train
)
glm.probs <- predict (glm.fits, Weekly.2009, type = "response")
glm.pred <- rep("Down", 104)
glm.pred[glm.probs > .5] <- "Up"
table (glm.pred, Direction.2009)## Direction.2009
## glm.pred Down Up
## Down 9 5
## Up 34 56mean (glm.pred == Direction.2009)## [1] 0.625mean (glm.pred != Direction.2009)## [1] 0.375- Repeat (d) using LDA.
library(MASS)##
## Attaching package: 'MASS'## The following object is masked from 'package:ISLR2':
##
## Bostonlda.fit <- lda(Direction ~ Lag1 + Lag2, data = Weekly, subset = train)
lda.fit## Call:
## lda(Direction ~ Lag1 + Lag2, data = Weekly, subset = train)
##
## Prior probabilities of groups:
## Down Up
## 0.4477157 0.5522843
##
## Group means:
## Lag1 Lag2
## Down 0.289444444 -0.03568254
## Up -0.009213235 0.26036581
##
## Coefficients of linear discriminants:
## LD1
## Lag1 -0.3013148
## Lag2 0.2982579plot(lda.fit)
lda.pred <- predict(lda.fit, Weekly.2009)
names(lda.pred)## [1] "class" "posterior" "x"lda.class <- lda.pred$class
table(lda.class, Direction.2009)## Direction.2009
## lda.class Down Up
## Down 7 8
## Up 36 53mean(lda.class == Direction.2009)## [1] 0.5769231sum(lda.class == Direction.2009)## [1] 60- Repeat (d) using QDA.
qda.fit <- qda(Weekly$Direction ~ Lag1 + Lag2, data = Weekly, subset = train)
qda.fit## Call:
## qda(Weekly$Direction ~ Lag1 + Lag2, data = Weekly, subset = train)
##
## Prior probabilities of groups:
## Down Up
## 0.4477157 0.5522843
##
## Group means:
## Lag1 Lag2
## Down 0.289444444 -0.03568254
## Up -0.009213235 0.26036581qda.class <- predict (qda.fit, Weekly.2009)$class
table(qda.class, Direction.2009)## Direction.2009
## qda.class Down Up
## Down 7 10
## Up 36 51mean(qda.class == Direction.2009)## [1] 0.5576923- Repeat (d) using KNN with K = 1.
library(class)
train.X <- cbind (Weekly$Lag1 , Weekly$Lag2)[train , ]
test.X <- cbind (Weekly$Lag1 , Weekly$Lag2)[!train , ]
train.Direction <- Weekly$Direction[train]
set.seed(1)
knn.pred <- knn(train.X, test.X, train.Direction, k = 1)
table(knn.pred, Direction.2009)## Direction.2009
## knn.pred Down Up
## Down 18 29
## Up 25 32knn.pred <- knn(train.X, test.X, train.Direction, k = 3)
table(knn.pred, Direction.2009)## Direction.2009
## knn.pred Down Up
## Down 22 29
## Up 21 32mean(knn.pred == Direction.2009)## [1] 0.5192308- Repeat (d) using naive Bayes.
library(e1071)## Warning: package 'e1071' was built under R version 4.1.3nb.fit <- naiveBayes(Direction ~ Lag1 + Lag2, data = Weekly, subset = train)
nb.fit##
## Naive Bayes Classifier for Discrete Predictors
##
## Call:
## naiveBayes.default(x = X, y = Y, laplace = laplace)
##
## A-priori probabilities:
## Y
## Down Up
## 0.4477157 0.5522843
##
## Conditional probabilities:
## Lag1
## Y [,1] [,2]
## Down 0.289444444 2.211721
## Up -0.009213235 2.308387
##
## Lag2
## Y [,1] [,2]
## Down -0.03568254 2.199504
## Up 0.26036581 2.317485nb.class <- predict(nb.fit, Weekly.2009)
table(nb.class, Direction.2009)## Direction.2009
## nb.class Down Up
## Down 3 8
## Up 40 53mean(nb.class == Direction.2009)## [1] 0.5384615- Which of these methods appears to provide the best results on this data? It appears that, out of these methods, the Logistic Regression and the LDA provide the best results.
- 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.
Problem #14
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)
head(Auto)## mpg cylinders displacement horsepower weight acceleration year origin
## 1 18 8 307 130 3504 12.0 70 1
## 2 15 8 350 165 3693 11.5 70 1
## 3 18 8 318 150 3436 11.0 70 1
## 4 16 8 304 150 3433 12.0 70 1
## 5 17 8 302 140 3449 10.5 70 1
## 6 15 8 429 198 4341 10.0 70 1
## name
## 1 chevrolet chevelle malibu
## 2 buick skylark 320
## 3 plymouth satellite
## 4 amc rebel sst
## 5 ford torino
## 6 ford galaxie 500mpg01 <- rep(0, length(mpg))
mpg01[mpg > median(mpg)] <- 1
Auto = data.frame(Auto, mpg01)
median(mpg01)## [1] 0.5- 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.
par(mfrow = c(2, 2))
plot(factor(Auto$mpg01), Auto$year, ylab = "Manufacture year", xlab = "Cars")
plot(factor(Auto$mpg01), Auto$weight, ylab = "Weight (lbs)", xlab = "Cars")
plot(factor(Auto$mpg01), Auto$horsepower, ylab = "Horsepower", xlab = "Cars")
plot(factor(Auto$mpg01), Auto$acceleration, ylab = "Time to reach 60mpg in seconds", xlab = "Cars")
The boxplots suggest that there is some overlap between manufacture year and time to reach 60mpg in seconds. It appears that there is a greater difference in the weight and the horsepower between the cars.
- Split the data into a training set and a test set.
train <- (year %% 2 == 0)
train_auto <- Auto[train,]
test_auto <- Auto[!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?
autolda_fit <- lda(mpg01~displacement+horsepower+weight+year+cylinders+origin, data=train_auto, subset = train)
autolda_pred <- predict(autolda_fit, Auto[-train, ])
table(autolda_pred$class, Auto[-train, "mpg01"])##
## 0 1
## 0 171 14
## 1 24 1821 - mean(autolda_pred$class == Auto[-train, "mpg01"])## [1] 0.0971867The test error of the model obtained is 9.71%
- 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 ~ cylinders + displacement + horsepower + weight + year + origin, data = train_auto, subset = train)
autoqda_pred = predict(autoqda_fit, Auto[-train, ])
table(autoqda_pred$class, Auto[-train, "mpg01"])##
## 0 1
## 0 175 21
## 1 20 1751 - mean(autoqda_pred$class == Auto[-train, "mpg01"])## [1] 0.1048593The test error of the model obtained is 10.49%
- 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?
autoglm_fit = glm(mpg01 ~cylinders + displacement + horsepower + weight + year + origin, data = Auto, family = "binomial")
summary(autoglm_fit)##
## Call:
## glm(formula = mpg01 ~ cylinders + displacement + horsepower +
## weight + year + origin, family = "binomial", data = Auto)
##
## Deviance Residuals:
## Min 1Q Median 3Q Max
## -2.4289 -0.1044 0.0081 0.2125 3.1750
##
## Coefficients:
## Estimate Std. Error z value Pr(>|z|)
## (Intercept) -1.680e+01 4.831e+00 -3.477 0.000506 ***
## cylinders -1.636e-01 4.235e-01 -0.386 0.699274
## displacement 1.963e-03 1.199e-02 0.164 0.869905
## horsepower -4.297e-02 1.664e-02 -2.581 0.009839 **
## weight -4.251e-03 9.889e-04 -4.299 1.72e-05 ***
## year 4.286e-01 7.485e-02 5.726 1.03e-08 ***
## origin 4.755e-01 3.615e-01 1.315 0.188440
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## (Dispersion parameter for binomial family taken to be 1)
##
## Null deviance: 543.43 on 391 degrees of freedom
## Residual deviance: 157.55 on 385 degrees of freedom
## AIC: 171.55
##
## Number of Fisher Scoring iterations: 8autoglm_probs = predict(autoglm_fit, Auto[-train, ], type = "response")
autoglm_pred = rep(0, dim(Auto[-train, ])[1])
autoglm_pred[autoglm_probs > 0.5] = 1
table(autoglm_pred, Auto[-train, "mpg01"], dnn = c("Predicted", "Actual"))## Actual
## Predicted 0 1
## 0 172 13
## 1 23 1831- mean(autoglm_pred == Auto[-train, "mpg01"])## [1] 0.09207161The test error rate is 8.43%
- Perform naive Bayes 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?
autonb_fit <- naiveBayes(mpg01 ~cylinders + displacement + horsepower + weight + year + origin, data = Auto, subset = train)
nb.fit##
## Naive Bayes Classifier for Discrete Predictors
##
## Call:
## naiveBayes.default(x = X, y = Y, laplace = laplace)
##
## A-priori probabilities:
## Y
## Down Up
## 0.4477157 0.5522843
##
## Conditional probabilities:
## Lag1
## Y [,1] [,2]
## Down 0.289444444 2.211721
## Up -0.009213235 2.308387
##
## Lag2
## Y [,1] [,2]
## Down -0.03568254 2.199504
## Up 0.26036581 2.317485autonb_pred <- predict(autonb_fit, Auto)- 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?
autoknn_train <- cbind( cylinders, displacement, horsepower, weight
)[train,]
autoknn_test <- cbind( cylinders, displacement, horsepower, weight
)[!train,]
set.seed(1)
autoknn_1 <- knn(autoknn_train, autoknn_test, train_auto$mpg01, k = 1)
table(autoknn_1, test_auto$mpg01)##
## autoknn_1 0 1
## 0 83 11
## 1 17 711 - mean(autoknn_1 == test_auto$mpg01)## [1] 0.1538462The test error rate is 15.38% with K=1.
autoknn_10 <- knn(autoknn_train, autoknn_test, train_auto$mpg01, k = 10)
table(autoknn_10, test_auto$mpg01)##
## autoknn_10 0 1
## 0 77 7
## 1 23 751 - mean(autoknn_10 == test_auto$mpg01)## [1] 0.1648352The test error rate is 16.48% with K=10.
autoknn_5 <- knn(autoknn_train, autoknn_test, train_auto$mpg01, k = 5)
table(autoknn_5, test_auto$mpg01)##
## autoknn_5 0 1
## 0 82 9
## 1 18 731 - mean(autoknn_5 == test_auto$mpg01)## [1] 0.1483516The test error rate is 14.83% with K=5.
autoknn_100 <- knn(autoknn_train, autoknn_test, train_auto$mpg01, k = 100)
table(autoknn_100, test_auto$mpg01)##
## autoknn_100 0 1
## 0 81 7
## 1 19 751 - mean(autoknn_100 == test_auto$mpg01)## [1] 0.1428571The test error rate is 14.29% with K=100.
Problem #16
Using the Boston data set, fit classification models in order to predict whether a given census tract has a crime rate above or below the median. Explore logistic regression, LDA, naive Bayes, and KNN models using various subsets of the predictors. Describe your findings.
Hint: You will have to create the response variable yourself, using the variables that are contained in the Boston data set.
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. :50.00#crime is "crim" in the data set
attach(Boston)Boston_crime <- rep(0, length(crim))
Boston_crime[crim > median(crim)] <- 1
Boston = data.frame(Boston, Boston_crime)
train = 1:(dim(Boston)[1]/2)
test = (dim(Boston)[1]/2 + 1):dim(Boston)[1]
Boston_train = Boston[train, ]
Boston_test = Boston[test, ]
Boston_crime_test = Boston_crime[test]pairs(Boston)
The paired correlation plots indicate that the variables “indus” (proportion of non-retail business acres per town.), “nox” (nitrogen oxides concentration (parts per 10 million), “age” (proportion of owner-occupied units built prior to 1940.), “dis” (weighted mean of distances to five Boston employment centres), “rad” (index of accessibility to radial highways.), and “tax” full-value property-tax rate per $10,000.) could be correlated with the crime rate.
To test significance, we calculate a logistic regression model:
set.seed(1)
Boston_fit <-glm(Boston_crime~ indus+nox+age+dis+rad+tax, data=Boston_train,family=binomial)
Boston_probs = predict(Boston_fit, Boston_test, type = "response")
Boston_pred = rep(0, length(Boston_probs))
Boston_pred[Boston_probs > 0.5] = 1
table(Boston_pred, Boston_crime_test)## Boston_crime_test
## Boston_pred 0 1
## 0 75 8
## 1 15 155mean(Boston_pred != Boston_crime_test)## [1] 0.09090909summary(Boston_fit)##
## Call:
## glm(formula = Boston_crime ~ indus + nox + age + dis + rad +
## tax, family = binomial, data = Boston_train)
##
## Deviance Residuals:
## Min 1Q Median 3Q Max
## -1.97810 -0.21406 -0.03454 0.47107 3.04502
##
## Coefficients:
## Estimate Std. Error z value Pr(>|z|)
## (Intercept) -42.214032 7.617440 -5.542 2.99e-08 ***
## indus -0.213126 0.073236 -2.910 0.00361 **
## nox 80.868029 16.066473 5.033 4.82e-07 ***
## age 0.003397 0.012032 0.282 0.77772
## dis 0.307145 0.190502 1.612 0.10690
## rad 0.847236 0.183767 4.610 4.02e-06 ***
## tax -0.013760 0.004956 -2.777 0.00549 **
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## (Dispersion parameter for binomial family taken to be 1)
##
## Null deviance: 329.37 on 252 degrees of freedom
## Residual deviance: 144.44 on 246 degrees of freedom
## AIC: 158.44
##
## Number of Fisher Scoring iterations: 8Next we calculate the LDA:
Bostonlda_fit <-lda(Boston_crime~ indus+nox+age+dis+rad+tax, data=Boston_train,family=binomial)
Bostonlda_pred = predict(Bostonlda_fit, Boston_test)
table(Bostonlda_pred$class, Boston_crime_test)## Boston_crime_test
## 0 1
## 0 81 18
## 1 9 145mean(Bostonlda_pred$class != Boston_crime_test)## [1] 0.1067194Next, we calculate Naive Bayes:
Boston_nb_fit <- naiveBayes(Boston_crime ~ indus+nox+age+dis+rad+tax, data=Boston)
Boston_nb_fit##
## Naive Bayes Classifier for Discrete Predictors
##
## Call:
## naiveBayes.default(x = X, y = Y, laplace = laplace)
##
## A-priori probabilities:
## Y
## 0 1
## 0.5 0.5
##
## Conditional probabilities:
## indus
## Y [,1] [,2]
## 0 7.002292 5.514454
## 1 15.271265 5.439010
##
## nox
## Y [,1] [,2]
## 0 0.4709711 0.05559789
## 1 0.6384190 0.09870365
##
## age
## Y [,1] [,2]
## 0 51.31028 25.88190
## 1 85.83953 17.87423
##
## dis
## Y [,1] [,2]
## 0 5.091596 2.081304
## 1 2.498489 1.085521
##
## rad
## Y [,1] [,2]
## 0 4.158103 1.659121
## 1 14.940711 9.529843
##
## tax
## Y [,1] [,2]
## 0 305.7431 87.4837
## 1 510.7312 167.8553Next, we calculate the KNN. First, K=1
Bostonknn_train=cbind(indus,nox,age,dis,rad,tax)[train,]
Bostonknn_test=cbind(indus,nox,age,dis,rad,tax)[test,]
Bostonknn_pred=knn(Bostonknn_train, Bostonknn_test, Boston_crime_test, k=1)
table(Bostonknn_pred,Boston_crime_test)## Boston_crime_test
## Bostonknn_pred 0 1
## 0 31 155
## 1 59 8mean(Bostonknn_pred != Boston_crime_test)## [1] 0.8458498Next, K=5
Bostonknn_pred5=knn(Bostonknn_train, Bostonknn_test, Boston_crime_test, k=5)
table(Bostonknn_pred5,Boston_crime_test)## Boston_crime_test
## Bostonknn_pred5 0 1
## 0 42 20
## 1 48 143mean(Bostonknn_pred5 != Boston_crime_test)## [1] 0.2687747Next, K=100
Bostonknn_pred100=knn(Bostonknn_train, Bostonknn_test, Boston_crime_test, k=100)
table(Bostonknn_pred100,Boston_crime_test)## Boston_crime_test
## Bostonknn_pred100 0 1
## 0 21 7
## 1 69 156mean(Bostonknn_pred100 != Boston_crime_test)## [1] 0.3003953In summary, we can see that the logistic regression model had the lowest test error rate of 9.09%, followed by the LDA with 10.67%. The statistically significant variables turned out to be indus, nox, rad, and tax.