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)
library(corrplot)
## corrplot 0.92 loaded

(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            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)

Well, I do not see much correlation between the variables, except for ‘year’ and ‘volume’. Trying corrplot to cross verify.

corrplot(cor(Weekly[,-9]), method="square")

The corrplot shows that ‘Volume’ and ‘year’ are correlated.

(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?

attach(Weekly)
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
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    430 557

‘Lag2’ is significantly associated with ‘Direction’. The p-value is 0.0296, which is lower than 0.5. The other variables fail to reject the NULL hypothesis.

(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 percentage of correct prediction is (54+557)/(54+48+430+557) = 0.5610. i.e., the model predicted up and down trends correctly 56% of the time.

(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).

#Creating a vector index
attach(Weekly)
## The following objects are masked from Weekly (pos = 3):
## 
##     Direction, Lag1, Lag2, Lag3, Lag4, Lag5, Today, Volume, Year
#creaing train dataset
train=(Year<2009)

#Creating a test dataset
Weekly.0910=  Weekly[!train,]
glm.fits=glm(Direction~Lag2, data=Weekly, subset=train, family=binomial)
Direction.0910=Direction[!train]
glm.probs=predict(glm.fits, Weekly.0910, type='response')
glm.pred=rep('Down', 104)
glm.pred[glm.probs>0.5]='Up'
table(glm.pred,Direction.0910)
##         Direction.0910
## glm.pred Down Up
##     Down    9  5
##     Up     34 56
#How many times did logistic regression predict right
mean(glm.pred==Direction.0910)
## [1] 0.625

After splitting the whole data into training and test data, the model correctly predicted 62.5% of the time. This model is better than the previous one, where the whole dataset was used.

(e) Repeat (d) using LDA.

library(MASS)
lda.fits<-lda(Direction~Lag2, data=Weekly, subset=train)
lda.fits
## 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 0.4414162
#plot(lda.fits)
lda.pred=predict(lda.fits,Weekly.0910)
lda.class=lda.pred$class
table(lda.class,Direction.0910)
##          Direction.0910
## lda.class Down Up
##      Down    9  5
##      Up     34 56
mean(lda.class==Direction.0910)
## [1] 0.625

The Linear Discriminant analysis yielded the similar results as in logistic regression.

(f) Repeat (d) using QDA.

qda.fits<-qda(Direction~Lag2, data=Weekly, subset=train)
qda.fits
## 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    0.26036581
#Create class
qda.class=predict(qda.fits,Weekly.0910)$class
table(qda.class,Direction.0910)
##          Direction.0910
## qda.class Down Up
##      Down    0  0
##      Up     43 61
mean(qda.class == Direction.0910)
## [1] 0.5865385

The quadratic discriminant analysis model accuracy is only 58.65%, which is lower than the last two models.

(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]
#dim(train.X)
#dim(test.X)
#length(train.Direction)
#Set the seed to ensure the reproducibility of the results
set.seed(1)
knn.pred=knn(train.X,test.X,train.Direction,k=1)
table(knn.pred,Direction.0910)
##         Direction.0910
## knn.pred Down Up
##     Down   21 30
##     Up     22 31
mean(knn.pred==Direction.0910)
## [1] 0.5

The KNN model resulted in classifying model with an accuracy of only 50%.

(h) Which of these methods appears to provide the best results on this data?

The logistic regression and LDA provide the best results. Both the models have accuracy 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.

An interaction term of Lag2 and Lag4 is added to the model.

attach(Weekly)
## The following objects are masked from Weekly (pos = 5):
## 
##     Direction, Lag1, Lag2, Lag3, Lag4, Lag5, Today, Volume, Year
## The following objects are masked from Weekly (pos = 6):
## 
##     Direction, Lag1, Lag2, Lag3, Lag4, Lag5, Today, Volume, Year
Weekly.fit1<-glm(Direction ~ Lag2:Lag4 + Lag2, data=Weekly,family=binomial, subset=train)
Weekly.prob1= predict(Weekly.fit1, Weekly.0910, type = "response")
Weekly.pred1 = rep("Down", length(Weekly.prob1))
Weekly.pred1[Weekly.prob1 > 0.5] = "Up"
Direction.0910 = Direction[!train]
table(Weekly.pred1, Direction.0910)
##             Direction.0910
## Weekly.pred1 Down Up
##         Down    3  4
##         Up     40 57

The logistic regression predicted correctly, (57+3)/(57+3+40+4) = 57.69% of the time.

LDA

library(MASS)
lda.fiti = lda(Direction~Lag2:Lag4+Lag2, data=Weekly, subset=train)
lda.fiti
## Call:
## lda(Direction ~ Lag2:Lag4 + Lag2, data = Weekly, subset = train)
## 
## Prior probabilities of groups:
##      Down        Up 
## 0.4477157 0.5522843 
## 
## Group means:
##             Lag2  Lag2:Lag4
## Down -0.03568254 0.78824608
## Up    0.26036581 0.04407141
## 
## Coefficients of linear discriminants:
##                  LD1
## Lag2       0.3442725
## Lag2:Lag4 -0.0608715
#plot(lda.fits)
lda.pred=predict(lda.fiti,Weekly.0910)
lda.class=lda.pred$class
table(lda.class,Direction.0910)
##          Direction.0910
## lda.class Down Up
##      Down    3  3
##      Up     40 58
mean(lda.class==Direction.0910)
## [1] 0.5865385

LDA predicted 58.65% of the time.

QDA

qda.fits<-qda(Direction~Lag2 + Lag4:Lag2, data=Weekly, subset=train)
qda.fits
## Call:
## qda(Direction ~ Lag2 + Lag4:Lag2, data = Weekly, subset = train)
## 
## Prior probabilities of groups:
##      Down        Up 
## 0.4477157 0.5522843 
## 
## Group means:
##             Lag2  Lag2:Lag4
## Down -0.03568254 0.78824608
## Up    0.26036581 0.04407141
#Create class
qda.class=predict(qda.fits,Weekly.0910)$class
table(qda.class,Direction.0910)
##          Direction.0910
## qda.class Down Up
##      Down    7  6
##      Up     36 55
mean(qda.class == Direction.0910)
## [1] 0.5961538

The QDA predicted correctly 59.6% of the time.

KNN

Trying Knn with different values of k for the earlier model - which had only 50% of accuracy.

library(class)
train.X=as.matrix(Lag2[train])
test.X=as.matrix(Lag2[!train])
train.Direction=Direction[train]
dim(train.X)
## [1] 985   1
dim(test.X)
## [1] 104   1
length(train.Direction)
## [1] 985
#Set the seed to ensure the reproducibility of the results
set.seed(1)
knn.pred=knn(train.X,test.X,train.Direction,k=10)
table(knn.pred,Direction.0910)
##         Direction.0910
## knn.pred Down Up
##     Down   17 21
##     Up     26 40
mean(knn.pred==Direction.0910)
## [1] 0.5480769

The KNN model with k=10 predicted correctly 54.8% of the time.Increasing k=100.

library(class)
train.X=as.matrix(Lag2[train])
test.X=as.matrix(Lag2[!train])
train.Direction=Direction[train]
dim(train.X)
## [1] 985   1
dim(test.X)
## [1] 104   1
length(train.Direction)
## [1] 985
#Set the seed to ensure the reproducibility of the results
set.seed(1)
knn.pred=knn(train.X,test.X,train.Direction,k=100)
table(knn.pred,Direction.0910)
##         Direction.0910
## knn.pred Down Up
##     Down   10 11
##     Up     33 50
mean(knn.pred==Direction.0910)
## [1] 0.5769231

Now, the KNN predicted correctly by 57.69% of the time when k=100 for this model.

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.

library(ISLR)
attach(Auto)
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)           :365
median(mpg)
## [1] 22.75
Auto$mpg01 <- ifelse(Auto$mpg>22.75, 1, 0)

(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.

library(dplyr)
## 
## Attaching package: 'dplyr'
## The following object is masked from 'package:MASS':
## 
##     select
## The following objects are masked from 'package:stats':
## 
##     filter, lag
## The following objects are masked from 'package:base':
## 
##     intersect, setdiff, setequal, union
library(ggplot2)
## 
## Attaching package: 'ggplot2'
## The following object is masked from 'Auto':
## 
##     mpg
#Calling another plotting library
library(GGally)
## Registered S3 method overwritten by 'GGally':
##   method from   
##   +.gg   ggplot2
#For easy plotting removing the columns - mpg and name from Auto

Auto1 <- Auto[,-c(1,9)]


#We use the function "ifelse" to assess the categorical based on mpg
Auto1$mpg01 <- ifelse(Auto1$mpg01 >= 1, "above", "below")


ggpairs(Auto1, aes(colour = mpg01))
## `stat_bin()` using `bins = 30`. Pick better value with `binwidth`.
## `stat_bin()` using `bins = 30`. Pick better value with `binwidth`.
## `stat_bin()` using `bins = 30`. Pick better value with `binwidth`.
## `stat_bin()` using `bins = 30`. Pick better value with `binwidth`.
## `stat_bin()` using `bins = 30`. Pick better value with `binwidth`.
## `stat_bin()` using `bins = 30`. Pick better value with `binwidth`.
## `stat_bin()` using `bins = 30`. Pick better value with `binwidth`.

The box plot shows that ‘weight’ and ‘year’ are clearly showing the separation between ‘above’ and ‘below’ median. For ‘cylinder’, ‘displacement’, ‘horsepower’ the decision boundary is clear. However, the distribution of observations are far from normal. It is difficult to classify ‘acceleration’ as well as ‘origin’.

(c) Split the data into a training set and a test set.

#Creating a vector index
attach(Auto)
## The following object is masked from package:ggplot2:
## 
##     mpg
## The following objects are masked from Auto (pos = 6):
## 
##     acceleration, cylinders, displacement, horsepower, mpg, name,
##     origin, weight, year
train=(year<81)

#Creating a test dataset
Auto.8182=  Auto[!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?

##Checking what variables are most associated with mpg01
glm.fita=glm(mpg01~cylinders+displacement+horsepower+weight+acceleration+year, data = Auto,family=binomial)
summary(glm.fita)
## 
## Call:
## glm(formula = mpg01 ~ cylinders + displacement + horsepower + 
##     weight + acceleration + year, family = binomial, data = Auto)
## 
## Deviance Residuals: 
##     Min       1Q   Median       3Q      Max  
## -2.1999  -0.1126   0.0115   0.2249   3.3019  
## 
## Coefficients:
##                Estimate Std. Error z value Pr(>|z|)    
## (Intercept)  -15.828787   5.653426  -2.800  0.00511 ** 
## cylinders     -0.015009   0.405220  -0.037  0.97045    
## displacement  -0.006745   0.009961  -0.677  0.49831    
## horsepower    -0.035608   0.023543  -1.512  0.13042    
## weight        -0.003986   0.001085  -3.673  0.00024 ***
## acceleration   0.007983   0.141357   0.056  0.95497    
## year           0.414204   0.072700   5.697 1.22e-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 385  degrees of freedom
## AIC: 173.34
## 
## Number of Fisher Scoring iterations: 8

‘weight’ and ‘year’ are significantly associated with mpg01. Hence, the LDA is performed to predict mpg01 using weight and year.

LDA

lda.fita<-lda(mpg01~ weight+year, data=Auto, subset=train)
lda.fita
## Call:
## lda(mpg01 ~ weight + year, data = Auto, subset = train)
## 
## Prior probabilities of groups:
##         0         1 
## 0.5748503 0.4251497 
## 
## Group means:
##     weight     year
## 0 3629.292 74.25521
## 1 2299.430 76.04930
## 
## Coefficients of linear discriminants:
##                 LD1
## weight -0.001685936
## year    0.087475780
#plot(lda.fits)
lda.pred=predict(lda.fita,Auto.8182)
lda.class=lda.pred$class
mpg01.8182=mpg01[!train]
table(lda.class,mpg01.8182)
##          mpg01.8182
## lda.class  0  1
##         0  2  2
##         1  2 52
mean(lda.class==mpg01.8182)
## [1] 0.9310345
mean(lda.class!=mpg01.8182)
## [1] 0.06896552

The test error rate of LDA model is 6.89%.The LDA correctly predicted 93.1% of the time.

(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?

qda.fita<-qda(mpg01~ weight+year, data=Auto, subset=train)
qda.fita
## Call:
## qda(mpg01 ~ weight + year, data = Auto, subset = train)
## 
## Prior probabilities of groups:
##         0         1 
## 0.5748503 0.4251497 
## 
## Group means:
##     weight     year
## 0 3629.292 74.25521
## 1 2299.430 76.04930
#Create class
qda.class=predict(qda.fita,Auto.8182)$class
table(qda.class,mpg01.8182)
##          mpg01.8182
## qda.class  0  1
##         0  2  2
##         1  2 52
mean(qda.class == mpg01.8182)
## [1] 0.9310345
mean(qda.class != mpg01.8182)
## [1] 0.06896552

Using QDA, we got the similar results. QDA model correctly predicted 93.1% of the time.The test error rate is 6.89%.

(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? 

glm.fits=glm(mpg01~weight+year, data=Auto, subset=train, family=binomial)
mpg01.8182=mpg01[!train]
glm.probs=predict(glm.fits, Auto.8182, type='response')
glm.pred=rep('Down', 52)
glm.pred[glm.probs>0.5]='Up'
table(glm.pred,mpg01.8182)
##         mpg01.8182
## glm.pred  0  1
##     Down  2  2
##     Up    2 52

The percentage of correct prediction is (52+2)/(52+2+2+2) = 0.9310. i.e., the model predicted up and down trends correctly 93% of the time. The test error rate is (2+2)/(52+2+2+2)=6.89%.

(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? 

library(class)
train.X=cbind(weight,year)[train,]
test.X=cbind(weight,year)[!train,]
train.mpg01=mpg01[train]
dim(train.X)
## [1] 334   2
dim(test.X)
## [1] 58  2
length(train.mpg01)
## [1] 334
#Set the seed to ensure the reproducibility of the results
set.seed(1)
knn.pred=knn(train.X,test.X,train.mpg01,k=1)
table(knn.pred,mpg01.8182)
##         mpg01.8182
## knn.pred  0  1
##        0  3 13
##        1  1 41
mean(knn.pred==mpg01.8182)
## [1] 0.7586207

The KNN model resulted in classifying model with an accuracy of only 75.86% for k=1. In this dataset, logistic regression, LDA and QDA performed better than KNN.Trying for k=10.

library(class)
train.X=cbind(weight,year)[train,]
test.X=cbind(weight,year)[!train,]
train.mpg01=mpg01[train]
dim(train.X)
## [1] 334   2
dim(test.X)
## [1] 58  2
length(train.mpg01)
## [1] 334
#Set the seed to ensure the reproducibility of the results
set.seed(1)
knn.pred=knn(train.X,test.X,train.mpg01,k=10)
table(knn.pred,mpg01.8182)
##         mpg01.8182
## knn.pred  0  1
##        0  4 13
##        1  0 41
mean(knn.pred==mpg01.8182)
## [1] 0.7758621

For K=10, KNN predicted correctly 77.59%. As k increases, KNN model is getting better.

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.

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
median(mpg)
## [1] 22.75
attach(Boston)

median(crim) #Median is 0.25651
## [1] 0.25651
Boston1<-Boston
Boston1$crim01 <- ifelse(Boston1$crim>0.25651, 1, 0)
View(Boston1)
attach(Boston1)
## The following objects are masked from Boston:
## 
##     age, black, chas, crim, dis, indus, lstat, medv, nox, ptratio, rad,
##     rm, tax, zn
#The data is split such that 80% is training dataset and 20% is the test dataset

## 80% of the sample size
smp_size <- floor(0.80 * nrow(Boston1))

## set the seed to make your partition reproducible
set.seed(123)
train_ind <- sample(seq_len(nrow(Boston1)), size = smp_size)

Boston.train <- Boston1[train_ind, ]
Boston.test <- Boston1[-train_ind, ]

#test1<-dim(test)
#Check the dimensions
dim(Boston.train)
## [1] 404  15
dim(Boston.test)
## [1] 102  15
library(corrplot)
corrplot(cor(Boston1), method="square")

The variable, ‘indus’, ‘nox’, ‘age’, ‘rad’and ’tax’ are correlated with ‘crim01’.

Logistic regression

set.seed(1)
glm.fitb = glm(crim01 ~ indus+nox+age+rad+tax,data = Boston.train, family=binomial)
#crim01test=crim01[test1]
summary(glm.fitb)
## 
## Call:
## glm(formula = crim01 ~ indus + nox + age + rad + tax, family = binomial, 
##     data = Boston.train)
## 
## Deviance Residuals: 
##      Min        1Q    Median        3Q       Max  
## -1.90833  -0.27994  -0.02743   0.01473   2.74633  
## 
## Coefficients:
##               Estimate Std. Error z value Pr(>|z|)    
## (Intercept) -19.819313   3.230436  -6.135 8.51e-10 ***
## indus        -0.013366   0.053069  -0.252 0.801152    
## nox          35.379066   6.954376   5.087 3.63e-07 ***
## age           0.015777   0.009910   1.592 0.111359    
## rad           0.554555   0.130690   4.243 2.20e-05 ***
## tax          -0.009433   0.002815  -3.351 0.000805 ***
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## (Dispersion parameter for binomial family taken to be 1)
## 
##     Null deviance: 560.02  on 403  degrees of freedom
## Residual deviance: 199.63  on 398  degrees of freedom
## AIC: 211.63
## 
## Number of Fisher Scoring iterations: 8
glm.probb=predict(glm.fitb, Boston.test, type='response')
glm.predb=rep('Down', length(glm.probb))
glm.predb[glm.probb>0.5]='Up'

View(Boston.test)

#dim(Boston.train$crim01)
table(glm.predb,Boston.test$crim01)
##          
## glm.predb  0  1
##      Down 44  6
##      Up    5 47

The logistic regression model correctly predicted (44+47)/(44+47+6+5), i.e., 81.22% of the time.

LDA

lda.fitb<-lda(crim01 ~ indus+nox+age+rad+tax,data = Boston.train, family=binomial)
lda.fitb
## Call:
## lda(crim01 ~ indus + nox + age + rad + tax, data = Boston.train, 
##     family = binomial)
## 
## Prior probabilities of groups:
##         0         1 
## 0.5049505 0.4950495 
## 
## Group means:
##       indus       nox      age      rad      tax
## 0  6.831618 0.4699054 50.94363  4.22549 309.6765
## 1 15.332850 0.6361000 86.59250 14.79000 508.6700
## 
## Coefficients of linear discriminants:
##                LD1
## indus  0.039468916
## nox    6.599459223
## age    0.014213119
## rad    0.106187045
## tax   -0.003217974
#plot(lda.fits)
lda.predb=predict(lda.fitb,Boston.test)
lda.classb=lda.predb$class
table(lda.classb,Boston.test$crim01)
##           
## lda.classb  0  1
##          0 44 13
##          1  5 40
mean(lda.classb==Boston.test$crim01)
## [1] 0.8235294

LDA predicted 82.35% of the time. The previous model is better than LDA.

KNN

library(class)
train.X=cbind(indus,nox,age,rad,tax)[Boston.train$crim01,]
test.X=cbind(indus,nox,age,rad,tax)[Boston.test$crim01,]
train.crim01=crim01[Boston.train$crim01]
#Set the seed to ensure the reproducibility of the results
set.seed(1)
knn.pred=knn(train.X,test.X,train.crim01,k=1)
#table(knn.pred,Boston.test$crim01)
mean(knn.pred==Boston.test$crim01)
## Warning in `==.default`(knn.pred, Boston.test$crim01): longer object length is
## not a multiple of shorter object length
## Warning in is.na(e1) | is.na(e2): longer object length is not a multiple of
## shorter object length
## [1] 0.4803922
#table(knn.pred,mpg01.8182)
#mean(knn.pred==mpg01.8182)

KNN predicted only 48.03% of the time only. Thus, LDA is a better model.