STA 4723 Assignment 3
Tim Harrison
2023-02-24
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? There appears to be a corolation bewtween year and volume.
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 1.000000000pairs(Weekly)
- 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?
Lag 2 is statistically significant.
direct <- glm(Direction ~ Lag1 + Lag2 + Lag3 + Lag4 + Lag5 + Volume,
data = Weekly,
family = "binomial")
summary(direct)##
## 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- 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(direct, 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$Direction)## Up
## Down 0
## Up 1summary(Weekly$Direction)## Down Up
## 484 605dim(Weekly)## [1] 1089 9glm.pred = rep("Down", nrow(Weekly)) # uses the #of rows on weekly
glm.pred[glm.probs > .5] = "Up"
table(glm.pred, Weekly$Direction)##
## glm.pred Down Up
## Down 54 48
## Up 430 557- Now fit the logistic regression model using a training data period from 1990 to 2008, with Lag2 as the only predictor. Compute the confusion matrix and the overall fraction of correct predictions for the held out data (that is, the data from 2009 and 2010).
train = (Weekly$Year<2009)
Weekly.2009 = Weekly[!train,]
Direction.2009 = Weekly$Direction[!train]
dim(Weekly.2009)## [1] 104 9direct = glm(Direction ~ Lag2, data = Weekly, family =binomial, subset = train)
glm.probs = predict(direct, Weekly.2009, type = "response")
glm.pred = rep("Down", nrow(Weekly.2009))
glm.pred[glm.probs > .5] = "Up"
table(glm.pred, Direction.2009)## Direction.2009
## glm.pred Down Up
## Down 9 5
## Up 34 56- Repeat (d) using LDA.
lda.fit = lda(Direction~Lag2, data = Weekly, subset = train)
lda.fit## Call:
## lda(Direction ~ Lag2, data = Weekly, subset = train)
##
## Prior probabilities of groups:
## Down Up
## 0.4477157 0.5522843
##
## Group means:
## Lag2
## Down -0.03568254
## Up 0.26036581
##
## Coefficients of linear discriminants:
## LD1
## Lag2 0.4414162lda.pred = predict(lda.fit, Weekly.2009)
names(lda.pred)## [1] "class" "posterior" "x"lda.class = lda.pred$class
table(lda.class, Direction.2009)## Direction.2009
## lda.class Down Up
## Down 9 5
## Up 34 56- Repeat (d) using QDA.
qda.fit = qda(Direction~Lag2, data = Weekly, subset = train)
qda.fit## 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 0.26036581qda.class = predict(qda.fit, Weekly.2009)$class
table(qda.class, Direction.2009)## Direction.2009
## qda.class Down Up
## Down 0 0
## Up 43 61- Repeat (d) using KNN with K = 1.
attach(Weekly)
library(class)
train.X = cbind(Lag1, Lag2)[train,]
test.X = cbind(Lag1, Lag2)[!train,]
train.Direction = Weekly$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 32detach(Weekly)- Repeat (d) using naive Bayes.
library(e1071)
nb.fit = naiveBayes(Direction~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:
## 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 0 0
## Up 43 61- Which of these methods appears to provide the best results on this data? Logistic
- 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.
set.seed(1)
knn.pred <- knn(train.X,test.X,train.Direction ,k=4)
table(knn.pred, Direction.2009)## Direction.2009
## knn.pred Down Up
## Down 23 32
## Up 20 29mean(knn.pred==Direction.2009)## [1] 0.5nb_model <- naiveBayes(Direction~Lag1+Lag2+Lag3,data=Weekly, subset = train)
pred <- predict(nb_model,Weekly)
table(Direction.2009)## Direction.2009
## Down Up
## 43 61mean(pred==Direction.2009)## [1] 0.5629017qda.fit=qda(Direction~Lag1+Lag2+Lag3 ,data=Weekly ,subset=train)
qda.fit## Call:
## qda(Direction ~ Lag1 + Lag2 + Lag3, data = Weekly, subset = train)
##
## Prior probabilities of groups:
## Down Up
## 0.4477157 0.5522843
##
## Group means:
## Lag1 Lag2 Lag3
## Down 0.289444444 -0.03568254 0.17080045
## Up -0.009213235 0.26036581 0.08404044qda.class <- predict(qda.fit, Weekly)$class
mean(qda.class==Direction.2009)## [1] 0.5546373lda.fit <- lda(Direction~Lag1+Lag2+Lag3, data=Weekly, subset=train)
lda.fit## Call:
## lda(Direction ~ Lag1 + Lag2 + Lag3, data = Weekly, subset = train)
##
## Prior probabilities of groups:
## Down Up
## 0.4477157 0.5522843
##
## Group means:
## Lag1 Lag2 Lag3
## Down 0.289444444 -0.03568254 0.17080045
## Up -0.009213235 0.26036581 0.08404044
##
## Coefficients of linear discriminants:
## LD1
## Lag1 -0.29658609
## Lag2 0.29258490
## Lag3 -0.04766747lda.pred <- predict(lda.fit, Weekly)
names(lda.pred)## [1] "class" "posterior" "x"lda.class <- lda.pred$class
mean(lda.class==Direction.2009)## [1] 0.5775941Problem 14
- 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 <- ifelse(Auto$mpg > median(Auto$mpg), 1, 0)
Auto_V <- 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 predicting mpg01? Scatterplots and boxplots may be useful tools to answer this question. Describe your findings.
Auto_NQ <- Auto_V %>% select_if(is.numeric)
names(Auto_V)## [1] "mpg" "cylinders" "displacement" "horsepower" "weight"
## [6] "acceleration" "year" "origin" "name" "mpg01"names(Auto_NQ)## [1] "mpg" "cylinders" "displacement" "horsepower" "weight"
## [6] "acceleration" "year" "origin" "mpg01"cor(Auto_NQ)## 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 1.0000000plot(Auto_NQ)
glm.fits <- glm(mpg01~displacement+horsepower+weight+acceleration+year, data=Auto_NQ,family = binomial)
summary(glm.fits)##
## Call:
## glm(formula = mpg01 ~ displacement + horsepower + weight + acceleration +
## year, family = binomial, data = Auto_NQ)
##
## Deviance Residuals:
## Min 1Q Median 3Q Max
## -2.1982 -0.1129 0.0114 0.2243 3.3045
##
## Coefficients:
## Estimate Std. Error z value Pr(>|z|)
## (Intercept) -15.844178 5.636740 -2.811 0.004941 **
## displacement -0.007024 0.006519 -1.078 0.281237
## horsepower -0.035762 0.023176 -1.543 0.122826
## weight -0.003984 0.001084 -3.676 0.000237 ***
## acceleration 0.008144 0.141248 0.058 0.954019
## year 0.414101 0.072623 5.702 1.18e-08 ***
## ---
## 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: 159.34 on 386 degrees of freedom
## AIC: 171.34
##
## Number of Fisher Scoring iterations: 8After using logistic regression weight and year have significant impacts on fuel effieciency.
- Split the data into a training set and a test set.
set.seed(1)
sample1 <- sample(c(TRUE, FALSE), nrow(Auto_NQ), replace=TRUE, prob = c(.6,.4))
train <- Auto_NQ[sample1, ]
test <- Auto_NQ[!sample1, ]
dim(train)## [1] 246 9dim(test)## [1] 146 9- 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?
lda.fit.auto <- lda(mpg01 ~ weight+year, data=train)
lda.fit.auto## Call:
## lda(mpg01 ~ weight + year, data = train)
##
## Prior probabilities of groups:
## 0 1
## 0.504065 0.495935
##
## Group means:
## weight year
## 0 3620.040 74.70968
## 1 2315.344 77.63115
##
## Coefficients of linear discriminants:
## LD1
## weight -0.00176319
## year 0.11400180lda.class.auto <- predict(lda.fit.auto,test)$class
table(lda.class.auto,test$mpg01)##
## lda.class.auto 0 1
## 0 62 5
## 1 10 69mean(lda.class.auto==test$mpg01)## [1] 0.89726031-mean(lda.class.auto==test$mpg01)## [1] 0.1027397The test error is 10.27%.
- 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?
qda.fit.auto <- qda(mpg01 ~ weight+year, data=train)
qda.fit.auto## Call:
## qda(mpg01 ~ weight + year, data = train)
##
## Prior probabilities of groups:
## 0 1
## 0.504065 0.495935
##
## Group means:
## weight year
## 0 3620.040 74.70968
## 1 2315.344 77.63115qda.class.auto <- predict(qda.fit.auto,test)$class
table(qda.class.auto,test$mpg01)##
## qda.class.auto 0 1
## 0 61 6
## 1 11 68mean(qda.class.auto==test$mpg01)## [1] 0.88356161-mean(qda.class.auto==test$mpg01)## [1] 0.1164384The test error is 11.64%.
- 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?
glm.auto <- glm(mpg01~weight+year, family=binomial, data=train)
summary(glm.auto)##
## Call:
## glm(formula = mpg01 ~ weight + year, family = binomial, data = train)
##
## Deviance Residuals:
## Min 1Q Median 3Q Max
## -1.9209 -0.1360 -0.0007 0.2159 3.1868
##
## Coefficients:
## Estimate Std. Error z value Pr(>|z|)
## (Intercept) -2.082e+01 6.708e+00 -3.104 0.00191 **
## weight -5.946e-03 8.835e-04 -6.730 1.69e-11 ***
## year 4.929e-01 1.063e-01 4.636 3.55e-06 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## (Dispersion parameter for binomial family taken to be 1)
##
## Null deviance: 341.01 on 245 degrees of freedom
## Residual deviance: 93.95 on 243 degrees of freedom
## AIC: 99.95
##
## Number of Fisher Scoring iterations: 7glm.probs.auto <- predict(glm.auto,test,type="response")
glm.pred.auto <- rep(0,nrow(test))
glm.pred.auto[glm.probs.auto > 0.50]=1
table(glm.pred.auto,test$mpg01)##
## glm.pred.auto 0 1
## 0 64 9
## 1 8 65mean(glm.pred.auto==test$mpg01)## [1] 0.88356161-mean(glm.pred.auto==test$mpg01)## [1] 0.1164384The test error is 11.64%.
- Perform naive Bayes on the training data in order to predict mpg01 using the variables that seemed most associated with mpg01 in (b). What is the test error of the model obtained?
nb_model.auto <- naiveBayes(mpg01~weight+year,data=train)
summary(nb_model.auto)## Length Class Mode
## apriori 2 table numeric
## tables 2 -none- list
## levels 2 -none- character
## isnumeric 2 -none- logical
## call 4 -none- callprob.auto <- predict(nb_model.auto,test)
pred.auto <- rep(0,nrow(test))
pred.auto[prob.auto > 0.50]=1
table(pred.auto,test$mpg01)##
## pred.auto 0 1
## 0 72 74mean(pred.auto==test$mpg01)## [1] 0.49315071-mean(pred.auto==test$mpg01)## [1] 0.5068493The test error is 50.68%.
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.
crim01 <- ifelse(Boston$crim > median(Boston$crim), 1, 0)
Boston_V <- data.frame(Boston, crim01)
cor(Boston_V)## 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
## crim01 0.40939545 -0.43615103 0.60326017 0.070096774 0.72323480
## 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
## crim01 -0.15637178 0.61393992 -0.61634164 0.619786249 0.60874128 0.2535684
## black lstat medv crim01
## crim -0.38506394 0.4556215 -0.3883046 0.40939545
## zn 0.17552032 -0.4129946 0.3604453 -0.43615103
## indus -0.35697654 0.6037997 -0.4837252 0.60326017
## chas 0.04878848 -0.0539293 0.1752602 0.07009677
## nox -0.38005064 0.5908789 -0.4273208 0.72323480
## rm 0.12806864 -0.6138083 0.6953599 -0.15637178
## age -0.27353398 0.6023385 -0.3769546 0.61393992
## dis 0.29151167 -0.4969958 0.2499287 -0.61634164
## rad -0.44441282 0.4886763 -0.3816262 0.61978625
## tax -0.44180801 0.5439934 -0.4685359 0.60874128
## ptratio -0.17738330 0.3740443 -0.5077867 0.25356836
## black 1.00000000 -0.3660869 0.3334608 -0.35121093
## lstat -0.36608690 1.0000000 -0.7376627 0.45326273
## medv 0.33346082 -0.7376627 1.0000000 -0.26301673
## crim01 -0.35121093 0.4532627 -0.2630167 1.00000000set.seed(1)
sample_b <- sample(c(TRUE, FALSE), nrow(Boston_V), replace=TRUE, prob = c(.6,.4))
trainb <- Boston_V[sample_b, ]
testb <- Boston_V[!sample_b, ]
dim(trainb)## [1] 314 15dim(testb)## [1] 192 15glm.bos <- glm(crim01~indus+nox+age+rad+tax+lstat, family=binomial, data=trainb)
summary(glm.bos)##
## Call:
## glm(formula = crim01 ~ indus + nox + age + rad + tax + lstat,
## family = binomial, data = trainb)
##
## Deviance Residuals:
## Min 1Q Median 3Q Max
## -2.08917 -0.19479 0.00026 0.01065 2.74919
##
## Coefficients:
## Estimate Std. Error z value Pr(>|z|)
## (Intercept) -22.535448 3.975270 -5.669 1.44e-08 ***
## indus -0.072520 0.056568 -1.282 0.199841
## nox 40.095414 8.392236 4.778 1.77e-06 ***
## age 0.022459 0.012367 1.816 0.069362 .
## rad 0.576305 0.151603 3.801 0.000144 ***
## tax -0.008295 0.003199 -2.593 0.009521 **
## lstat -0.011124 0.043571 -0.255 0.798480
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## (Dispersion parameter for binomial family taken to be 1)
##
## Null deviance: 435.25 on 313 degrees of freedom
## Residual deviance: 139.35 on 307 degrees of freedom
## AIC: 153.35
##
## Number of Fisher Scoring iterations: 8glm.probs.b <- predict(glm.bos,testb,type="response")
glm.pred.b <- rep(0,nrow(testb))
glm.pred.b[glm.probs.b > 0.50]=1
table(glm.pred.b,testb$crim01)##
## glm.pred.b 0 1
## 0 88 13
## 1 10 81mean(glm.pred.b==testb$crim01)## [1] 0.88020831-mean(glm.pred.b==testb$crim01)## [1] 0.1197917uccessfully predict 88.02% with a test error rate of 11.98%
lda.fit.b <- lda(crim01~indus+nox+age+rad+tax+lstat, data=trainb)
lda.fit.b## Call:
## lda(crim01 ~ indus + nox + age + rad + tax + lstat, data = trainb)
##
## Prior probabilities of groups:
## 0 1
## 0.4936306 0.5063694
##
## Group means:
## indus nox age rad tax lstat
## 0 6.663226 0.4648245 50.03871 4.096774 303.8065 9.240387
## 1 15.310189 0.6426730 86.79308 14.610063 506.4969 15.832264
##
## Coefficients of linear discriminants:
## LD1
## indus 0.019009728
## nox 7.126273863
## age 0.018582977
## rad 0.089442271
## tax -0.001811084
## lstat -0.022915042lda.class.b <- predict(lda.fit.b,testb)$class
table(lda.class.b,testb$crim01)##
## lda.class.b 0 1
## 0 90 23
## 1 8 71mean(lda.class.b==testb$crim01)## [1] 0.83854171-mean(lda.class.b==testb$crim01)## [1] 0.1614583LDA produces a lower success rate, with 83.85%. The test error rate is 16.15%.
nb_model.b <- naiveBayes(crim01~indus+nox+age+rad+tax+lstat, data=trainb)
summary(nb_model.b)## Length Class Mode
## apriori 2 table numeric
## tables 6 -none- list
## levels 2 -none- character
## isnumeric 6 -none- logical
## call 4 -none- callprob.b <- predict(nb_model.b,testb)
pred.b <- rep(0,nrow(testb))
pred.b[prob.b > 0.50]=1
table(pred.b,testb$crim01)##
## pred.b 0 1
## 0 98 94mean(pred.b==testb$crim01)## [1] 0.51041671-mean(pred.b==testb$crim01)## [1] 0.4895833Naive Bayes produces 51.04% success. 48.96% error rate.
set.seed(1)
train.Boston = trainb[,c("indus","nox","age","rad","tax","lstat")]
test.Boston = testb[,c("indus","nox","age","rad","tax","lstat")]
knn.pred=knn(train.Boston,test.Boston,trainb$crim01,k=1)
table(knn.pred,testb$crim01)##
## knn.pred 0 1
## 0 91 7
## 1 7 87mean(knn.pred==testb$crim01)## [1] 0.92708331-mean(knn.pred==testb$crim01)## [1] 0.07291667knn.pred=knn(train.Boston,test.Boston,trainb$crim01,k=2)
table(knn.pred,testb$crim01)##
## knn.pred 0 1
## 0 86 12
## 1 12 82mean(knn.pred==testb$crim01)## [1] 0.875knn.pred=knn(train.Boston,test.Boston,trainb$crim01,k=3)
table(knn.pred,testb$crim01)##
## knn.pred 0 1
## 0 91 10
## 1 7 84mean(knn.pred==testb$crim01)## [1] 0.9114583knn.pred=knn(train.Boston,test.Boston,trainb$crim01,k=4)
table(knn.pred,testb$crim01)##
## knn.pred 0 1
## 0 90 11
## 1 8 83mean(knn.pred==testb$crim01)## [1] 0.9010417KNN with K = 1 prodcies success rate of 92.71% with a test error rate of 7.29%.