Assignment 8

library(ggplot2)
library(e1071)
library(ISLR2)
library(caret)

## Loading required package: lattice

Question 5

We have seen that we can fit an SVM with a non-linear kernel in order to perform classification using a non-linear decision boundary. We will now see that we can also obtain a non-linear decision boundary by performing logistic regression using non-linear transformations of the features.

(a) Generate a data set with n = 500 and p = 2, such that the obser- vations belong to two classes with a quadratic decision boundary between them. For instance, you can do this as follows:

set.seed(42)
x1 <- runif (500) - 0.5
x2 <- runif (500) - 0.5
y <- 1 * (x1^2 - x2^2 > 0)

(b) Plot the observations, colored according to their class labels. Your plot should display X1 on the x-axis, and X2 on the y- axis.

plot(x1[y == 0], x2[y == 0], col = "red", xlab = "X1", ylab = "X2")
points(x1[y == 1], x2[y == 1], col = "blue")

(c) Fit a logistic regression model to the data, using X1 and X2 as predictors.

logit.fit <- glm(y ~ x1 + x2, family = "binomial")
summary(logit.fit)

## 
## Call:
## glm(formula = y ~ x1 + x2, family = "binomial")
## 
## Coefficients:
##             Estimate Std. Error z value Pr(>|z|)
## (Intercept)  0.09978    0.08976   1.112    0.266
## x1          -0.17659    0.30658  -0.576    0.565
## x2          -0.20067    0.30978  -0.648    0.517
## 
## (Dispersion parameter for binomial family taken to be 1)
## 
##     Null deviance: 691.79  on 499  degrees of freedom
## Residual deviance: 691.08  on 497  degrees of freedom
## AIC: 697.08
## 
## Number of Fisher Scoring iterations: 3

(d) Apply this model to the training data in order to obtain a predicted class label for each training observation. Plot the observations, colored according to the predicted class labels. The decision boundary should be linear.

df=data.frame(x1 = x1, x2 = x2, y = as.factor(y))

logit.prob=predict(logit.fit,newdata=df,type='response')
logit.pred=ifelse(logit.prob>0.5,1,0)
plot(x1, x2, col = c('blue', 'red')[logit.pred+1])

(e) Now fit a logistic regression model to the data using non-linear functions of X1 and X2 as predictors (e.g. X12, X1 ×X2, log(X2), and so forth).

logit.fit2 <- glm(y ~  poly(x1,2) + poly(x2,2), family = "binomial")

## Warning: glm.fit: algorithm did not converge

## Warning: glm.fit: fitted probabilities numerically 0 or 1 occurred

summary(logit.fit2)

## 
## Call:
## glm(formula = y ~ poly(x1, 2) + poly(x2, 2), family = "binomial")
## 
## Coefficients:
##               Estimate Std. Error z value Pr(>|z|)
## (Intercept)      241.2     2464.3   0.098    0.922
## poly(x1, 2)1   -1560.4    43432.8  -0.036    0.971
## poly(x1, 2)2  150754.9  1452847.9   0.104    0.917
## poly(x2, 2)1    3829.4    54613.8   0.070    0.944
## poly(x2, 2)2 -145721.8  1403130.1  -0.104    0.917
## 
## (Dispersion parameter for binomial family taken to be 1)
## 
##     Null deviance: 6.9179e+02  on 499  degrees of freedom
## Residual deviance: 6.5043e-05  on 495  degrees of freedom
## AIC: 10
## 
## Number of Fisher Scoring iterations: 25

(f) Apply this model to the training data in order to obtain a predicted class label for each training observation. Plot the observations, colored according to the predicted class labels. The decision boundary should be obviously non-linear. If it is not, then repeat (a)-(e) until you come up with an example in which the predicted class labels are obviously non-linear.

logit.prob=predict(logit.fit2,newdata=df,type='response')
logit.pred=ifelse(logit.prob>0.5,1,0)
plot(x1, x2, col = c('blue', 'red')[logit.pred+1])

(g) Fit a support vector classifier to the data with X1 and X2 as predictors. Obtain a class prediction for each training observation. Plot the observations, colored according to the predicted class labels.

svm.fit <- svm(y ~ x1 + x2, df, kernal = 'linear', cost = .01)

svm.pred = predict(svm.fit, df)
data.pos = df[svm.pred == 1, ]
data.neg = df[svm.pred == 0, ]
plot(data.pos$x1, data.pos$x2, col = "blue", xlab = "X1", ylab = "X2", pch = "+")
points(data.neg$x1, data.neg$x2, col = "red", pch = 4)

(h) Fit a SVM using a non-linear kernel to the data. Obtain a class prediction for each training observation. Plot the observations, colored according to the predicted class labels.

svm.fit <- svm(y ~ x1 + x2, df, kernel = 'radial', gamma=1)

svm.pred = predict(svm.fit, df)
data.pos = df[svm.pred == 1, ]
data.neg = df[svm.pred == 0, ]
plot(data.pos$x1, data.pos$x2, col = "blue", xlab = "X1", ylab = "X2", pch = "+")
points(data.neg$x1, data.neg$x2, col = "red", pch = 4)

(i) Comment on your results.

The Support Vector Classifier from part (g) was unable to detect any linear boundaries in the data, but in part (h), the SVM was able to detect the non-linear boundary present in the data.

Question 7

In this problem, you will use support vector approaches in order to predict whether a given car gets high or low gas mileage based on the Auto data set.

(a) Create a binary variable that takes on a 1 for cars with gas mileage above the median, and a 0 for cars with gas mileage below the median.

auto.df<-Auto
auto.df<-na.omit(auto.df)
auto.df$MPG_Abv_Med <- as.factor(ifelse(auto.df$mpg > median(auto.df$mpg),1,0))
head(auto.df)

##   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 MPG_Abv_Med
## 1 chevrolet chevelle malibu           0
## 2         buick skylark 320           0
## 3        plymouth satellite           0
## 4             amc rebel sst           0
## 5               ford torino           0
## 6          ford galaxie 500           0

(b) Fit a support vector classifier to the data with various values of cost, in order to predict whether a car gets high or low gas mileage. Report the cross-validation errors associated with different values of this parameter. Comment on your results. Note you will need to fit the classifier without the gas mileage variable to produce sensible results.

set.seed(42)

tune.fit <- tune(svm,MPG_Abv_Med ~ . - mpg ,data=auto.df,kernel="linear", ranges=list(cost=c(0.001, 0.01, 0.1, 1,5,10,100)))
summary(tune.fit)

## 
## Parameter tuning of 'svm':
## 
## - sampling method: 10-fold cross validation 
## 
## - best parameters:
##  cost
##  0.01
## 
## - best performance: 0.08916667 
## 
## - Detailed performance results:
##    cost      error dispersion
## 1 1e-03 0.12775641 0.06746999
## 2 1e-02 0.08916667 0.05258186
## 3 1e-01 0.09160256 0.05869690
## 4 1e+00 0.09173077 0.04357345
## 5 5e+00 0.10942308 0.04734731
## 6 1e+01 0.11705128 0.05314992
## 7 1e+02 0.12993590 0.05797340

The results above show that the best value for the cost parameter is .01 as it give the least error of 0.08916667.

(c) Now repeat (b), this time using SVMs with radial and polynomial basis kernels, with different values of gamma and degree and cost. Comment on your results.

set.seed(42)

tune.fit2 <- tune(svm,MPG_Abv_Med ~ . - mpg ,data=auto.df,kernel="radial", ranges=list(cost=c(0.001, 0.01, 0.1, 1,5,10,100), gamma = c(0.01, 0.1, 1)))
summary(tune.fit2)

## 
## Parameter tuning of 'svm':
## 
## - sampling method: 10-fold cross validation 
## 
## - best parameters:
##  cost gamma
##     1     1
## 
## - best performance: 0.07891026 
## 
## - Detailed performance results:
##     cost gamma      error dispersion
## 1  1e-03  0.01 0.59679487 0.05312225
## 2  1e-02  0.01 0.59679487 0.05312225
## 3  1e-01  0.01 0.11493590 0.05837054
## 4  1e+00  0.01 0.08660256 0.05519479
## 5  5e+00  0.01 0.09166667 0.05754492
## 6  1e+01  0.01 0.08660256 0.06001684
## 7  1e+02  0.01 0.10442308 0.05805399
## 8  1e-03  0.10 0.59679487 0.05312225
## 9  1e-02  0.10 0.29326923 0.08606908
## 10 1e-01  0.10 0.08666667 0.05390785
## 11 1e+00  0.10 0.08660256 0.05519479
## 12 5e+00  0.10 0.08647436 0.03940028
## 13 1e+01  0.10 0.08647436 0.04445518
## 14 1e+02  0.10 0.10173077 0.04696584
## 15 1e-03  1.00 0.59679487 0.05312225
## 16 1e-02  1.00 0.59679487 0.05312225
## 17 1e-01  1.00 0.59679487 0.05312225
## 18 1e+00  1.00 0.07891026 0.03633038
## 19 5e+00  1.00 0.08910256 0.04132724
## 20 1e+01  1.00 0.08910256 0.04132724
## 21 1e+02  1.00 0.08910256 0.04132724

Looking at the results above it can be seen that the best value for cost is 1, and the best value for gamma is 1, as that combination of those parameters result in an error of 0.07891026. This is using a radial based kernel.

set.seed(42)

tune.fit3 <- tune(svm,MPG_Abv_Med ~ . - mpg ,data=auto.df, kernel="polynomial", ranges=list(cost=c(0.001, 0.01, 0.1, 1,5,10,100), gamma = c(0.01, 0.1, 1), degree = c(2, 3, 4)))
summary(tune.fit3)

## 
## Parameter tuning of 'svm':
## 
## - sampling method: 10-fold cross validation 
## 
## - best parameters:
##  cost gamma degree
##   100   0.1      3
## 
## - best performance: 0.08641026 
## 
## - Detailed performance results:
##     cost gamma degree      error dispersion
## 1  1e-03  0.01      2 0.59679487 0.05312225
## 2  1e-02  0.01      2 0.59679487 0.05312225
## 3  1e-01  0.01      2 0.59679487 0.05312225
## 4  1e+00  0.01      2 0.58653846 0.05488085
## 5  5e+00  0.01      2 0.39064103 0.10386602
## 6  1e+01  0.01      2 0.31698718 0.09209637
## 7  1e+02  0.01      2 0.28083333 0.06885413
## 8  1e-03  0.10      2 0.59679487 0.05312225
## 9  1e-02  0.10      2 0.58653846 0.05488085
## 10 1e-01  0.10      2 0.31698718 0.09209637
## 11 1e+00  0.10      2 0.28083333 0.06885413
## 12 5e+00  0.10      2 0.16775641 0.10359086
## 13 1e+01  0.10      2 0.17294872 0.08470264
## 14 1e+02  0.10      2 0.19839744 0.07771274
## 15 1e-03  1.00      2 0.31698718 0.09209637
## 16 1e-02  1.00      2 0.28083333 0.06885413
## 17 1e-01  1.00      2 0.17294872 0.08470264
## 18 1e+00  1.00      2 0.19839744 0.07771274
## 19 5e+00  1.00      2 0.21121795 0.08538310
## 20 1e+01  1.00      2 0.22391026 0.09572557
## 21 1e+02  1.00      2 0.25717949 0.09485533
## 22 1e-03  0.01      3 0.59679487 0.05312225
## 23 1e-02  0.01      3 0.59679487 0.05312225
## 24 1e-01  0.01      3 0.59679487 0.05312225
## 25 1e+00  0.01      3 0.59679487 0.05312225
## 26 5e+00  0.01      3 0.36064103 0.14593226
## 27 1e+01  0.01      3 0.30166667 0.10288776
## 28 1e+02  0.01      3 0.25301282 0.07977961
## 29 1e-03  0.10      3 0.59679487 0.05312225
## 30 1e-02  0.10      3 0.30166667 0.10288776
## 31 1e-01  0.10      3 0.25301282 0.07977961
## 32 1e+00  0.10      3 0.09679487 0.05029680
## 33 5e+00  0.10      3 0.09929487 0.05393973
## 34 1e+01  0.10      3 0.09166667 0.05101515
## 35 1e+02  0.10      3 0.08641026 0.05301971
## 36 1e-03  1.00      3 0.09679487 0.05029680
## 37 1e-02  1.00      3 0.09166667 0.05101515
## 38 1e-01  1.00      3 0.08641026 0.05301971
## 39 1e+00  1.00      3 0.10179487 0.04394941
## 40 5e+00  1.00      3 0.10442308 0.03816415
## 41 1e+01  1.00      3 0.10442308 0.03816415
## 42 1e+02  1.00      3 0.10442308 0.03816415
## 43 1e-03  0.01      4 0.59679487 0.05312225
## 44 1e-02  0.01      4 0.59679487 0.05312225
## 45 1e-01  0.01      4 0.59679487 0.05312225
## 46 1e+00  0.01      4 0.59679487 0.05312225
## 47 5e+00  0.01      4 0.59679487 0.05312225
## 48 1e+01  0.01      4 0.59679487 0.05312225
## 49 1e+02  0.01      4 0.41666667 0.13285341
## 50 1e-03  0.10      4 0.59679487 0.05312225
## 51 1e-02  0.10      4 0.41666667 0.13285341
## 52 1e-01  0.10      4 0.34256410 0.11428217
## 53 1e+00  0.10      4 0.27314103 0.06981031
## 54 5e+00  0.10      4 0.19602564 0.07480024
## 55 1e+01  0.10      4 0.21929487 0.08090842
## 56 1e+02  0.10      4 0.20416667 0.05433005
## 57 1e-03  1.00      4 0.21929487 0.08090842
## 58 1e-02  1.00      4 0.20416667 0.05433005
## 59 1e-01  1.00      4 0.21410256 0.06210638
## 60 1e+00  1.00      4 0.22166667 0.07114891
## 61 5e+00  1.00      4 0.22166667 0.07114891
## 62 1e+01  1.00      4 0.22166667 0.07114891
## 63 1e+02  1.00      4 0.22166667 0.07114891

Looking at the results above it can be seen that the best value for cost is 100, the best value for gamma is .1, and the best value for degree is 3 as that combination of those parameters result in an error of 0.08641026 This is using a polynomial based kernel.

(d) Make some plots to back up your assertions in (b) and (c).

auto.df.subset<- auto.df[, c("horsepower", "weight", "MPG_Abv_Med")]

svc.fit<- svm(MPG_Abv_Med ~ ., data=auto.df.subset, kernel="linear", cost = .01)
svm.fit.rad<- svm(MPG_Abv_Med ~ ., data=auto.df.subset, kernel="radial", cost = 1, gamma = 1)
svm.fit.poly<- svm(MPG_Abv_Med ~ ., data=auto.df.subset, kernel="polynomial", cost = 100, gamma = .1, degree = 3)

plot(svc.fit, data = auto.df.subset)

plot(svm.fit.rad, data = auto.df.subset)

plot(svm.fit.poly, data = auto.df.subset)

Question 8

(a) Create a training set containing a random sample of 800 observations, and a test set containing the remaining observations.

oj.df = OJ

set.seed(42)

oj_train_index <- sample(1:nrow(oj.df), 800)
oj_train_data <- oj.df[oj_train_index, ]
oj_test_data <- oj.df[-oj_train_index, ]

(b) Fit a support vector classifier to the training data using cost = 0.01, with Purchase as the response and the other variables as predictors. Use the summary() function to produce summary statistics, and describe the results obtained.

svc.fit<- svm(Purchase ~ ., data=oj_train_data, kernel="linear", cost = .01)
summary(svc.fit)

## 
## Call:
## svm(formula = Purchase ~ ., data = oj_train_data, kernel = "linear", 
##     cost = 0.01)
## 
## 
## Parameters:
##    SVM-Type:  C-classification 
##  SVM-Kernel:  linear 
##        cost:  0.01 
## 
## Number of Support Vectors:  432
## 
##  ( 215 217 )
## 
## 
## Number of Classes:  2 
## 
## Levels: 
##  CH MM

The support vector classifier resulted in a total of 432 vectors. 215 vectors belong to the Level CH, and 217 vectors belong to the Level MM.

(c) What are the training and test error rates?

# Confusion Matrix for prediction on train data
oj.preds = predict(svc.fit, oj_train_data)
confusionMatrix(oj.preds, oj_train_data$Purchase, positive = 'MM')

## Confusion Matrix and Statistics
## 
##           Reference
## Prediction  CH  MM
##         CH 432  77
##         MM  60 231
##                                           
##                Accuracy : 0.8288          
##                  95% CI : (0.8008, 0.8542)
##     No Information Rate : 0.615           
##     P-Value [Acc > NIR] : <2e-16          
##                                           
##                   Kappa : 0.6346          
##                                           
##  Mcnemar's Test P-Value : 0.1716          
##                                           
##             Sensitivity : 0.7500          
##             Specificity : 0.8780          
##          Pos Pred Value : 0.7938          
##          Neg Pred Value : 0.8487          
##              Prevalence : 0.3850          
##          Detection Rate : 0.2888          
##    Detection Prevalence : 0.3638          
##       Balanced Accuracy : 0.8140          
##                                           
##        'Positive' Class : MM              
##

# Confusion Matrix for prediction on test data
oj.preds = predict(svc.fit, oj_test_data)
confusionMatrix(oj.preds, oj_test_data$Purchase, positive = 'MM')

## Confusion Matrix and Statistics
## 
##           Reference
## Prediction  CH  MM
##         CH 142  25
##         MM  19  84
##                                          
##                Accuracy : 0.837          
##                  95% CI : (0.7875, 0.879)
##     No Information Rate : 0.5963         
##     P-Value [Acc > NIR] : <2e-16         
##                                          
##                   Kappa : 0.6585         
##                                          
##  Mcnemar's Test P-Value : 0.451          
##                                          
##             Sensitivity : 0.7706         
##             Specificity : 0.8820         
##          Pos Pred Value : 0.8155         
##          Neg Pred Value : 0.8503         
##              Prevalence : 0.4037         
##          Detection Rate : 0.3111         
##    Detection Prevalence : 0.3815         
##       Balanced Accuracy : 0.8263         
##                                          
##        'Positive' Class : MM             
##

The train error is 0.1712, and the test error is 0.163.

(d) Use the tune() function to select an optimal cost. Consider values in the range 0.01 to 10.

set.seed(42)

tune.fit <- tune(svm,Purchase ~ ., data=oj_train_data, kernel="linear", ranges=list(cost=c(0.01, 0.1, 1,5,10)))
summary(tune.fit)

## 
## Parameter tuning of 'svm':
## 
## - sampling method: 10-fold cross validation 
## 
## - best parameters:
##  cost
##     1
## 
## - best performance: 0.175 
## 
## - Detailed performance results:
##    cost   error dispersion
## 1  0.01 0.17750 0.02415229
## 2  0.10 0.17625 0.03356689
## 3  1.00 0.17500 0.02886751
## 4  5.00 0.18375 0.02703521
## 5 10.00 0.18625 0.02729087

(e) Compute the training and test error rates using this new value for cost.

best.svc<- svm(Purchase ~ ., data=oj_train_data, kernel="linear", cost = 1)

# Confusion Matrix for prediction on train data
oj.preds = predict(best.svc, oj_train_data)
confusionMatrix(oj.preds, oj_train_data$Purchase, positive = 'MM')

## Confusion Matrix and Statistics
## 
##           Reference
## Prediction  CH  MM
##         CH 434  76
##         MM  58 232
##                                           
##                Accuracy : 0.8325          
##                  95% CI : (0.8048, 0.8577)
##     No Information Rate : 0.615           
##     P-Value [Acc > NIR] : <2e-16          
##                                           
##                   Kappa : 0.6424          
##                                           
##  Mcnemar's Test P-Value : 0.1419          
##                                           
##             Sensitivity : 0.7532          
##             Specificity : 0.8821          
##          Pos Pred Value : 0.8000          
##          Neg Pred Value : 0.8510          
##              Prevalence : 0.3850          
##          Detection Rate : 0.2900          
##    Detection Prevalence : 0.3625          
##       Balanced Accuracy : 0.8177          
##                                           
##        'Positive' Class : MM              
##

# Confusion Matrix for prediction on test data
oj.preds = predict(best.svc, oj_test_data)
confusionMatrix(oj.preds, oj_test_data$Purchase, positive = 'MM')

## Confusion Matrix and Statistics
## 
##           Reference
## Prediction  CH  MM
##         CH 140  23
##         MM  21  86
##                                          
##                Accuracy : 0.837          
##                  95% CI : (0.7875, 0.879)
##     No Information Rate : 0.5963         
##     P-Value [Acc > NIR] : <2e-16         
##                                          
##                   Kappa : 0.6605         
##                                          
##  Mcnemar's Test P-Value : 0.8802         
##                                          
##             Sensitivity : 0.7890         
##             Specificity : 0.8696         
##          Pos Pred Value : 0.8037         
##          Neg Pred Value : 0.8589         
##              Prevalence : 0.4037         
##          Detection Rate : 0.3185         
##    Detection Prevalence : 0.3963         
##       Balanced Accuracy : 0.8293         
##                                          
##        'Positive' Class : MM             
##

The train error is 0.1675, and the test error is 0.163. The training error got slightly better.

(f) Repeat parts (b) through (e) using a support vector machine with a radial kernel. Use the default value for gamma.

svm.fit.rad<- svm(Purchase ~ ., data=oj_train_data, kernel="radial", cost = .01)
summary(svm.fit.rad)

## 
## Call:
## svm(formula = Purchase ~ ., data = oj_train_data, kernel = "radial", 
##     cost = 0.01)
## 
## 
## Parameters:
##    SVM-Type:  C-classification 
##  SVM-Kernel:  radial 
##        cost:  0.01 
## 
## Number of Support Vectors:  621
## 
##  ( 308 313 )
## 
## 
## Number of Classes:  2 
## 
## Levels: 
##  CH MM

The support vector classifier resulted in a total of 621 vectors. 308 vectors belong to the Level CH, and 313 vectors belong to the Level MM.

# Confusion Matrix for prediction on train data
oj.preds = predict(svm.fit.rad, oj_train_data)
confusionMatrix(oj.preds, oj_train_data$Purchase, positive = 'MM')

## Confusion Matrix and Statistics
## 
##           Reference
## Prediction  CH  MM
##         CH 492 308
##         MM   0   0
##                                           
##                Accuracy : 0.615           
##                  95% CI : (0.5803, 0.6489)
##     No Information Rate : 0.615           
##     P-Value [Acc > NIR] : 0.5156          
##                                           
##                   Kappa : 0               
##                                           
##  Mcnemar's Test P-Value : <2e-16          
##                                           
##             Sensitivity : 0.000           
##             Specificity : 1.000           
##          Pos Pred Value :   NaN           
##          Neg Pred Value : 0.615           
##              Prevalence : 0.385           
##          Detection Rate : 0.000           
##    Detection Prevalence : 0.000           
##       Balanced Accuracy : 0.500           
##                                           
##        'Positive' Class : MM              
##

# Confusion Matrix for prediction on test data
oj.preds = predict(svm.fit.rad, oj_test_data)
confusionMatrix(oj.preds, oj_test_data$Purchase, positive = 'MM')

## Confusion Matrix and Statistics
## 
##           Reference
## Prediction  CH  MM
##         CH 161 109
##         MM   0   0
##                                           
##                Accuracy : 0.5963          
##                  95% CI : (0.5351, 0.6553)
##     No Information Rate : 0.5963          
##     P-Value [Acc > NIR] : 0.5263          
##                                           
##                   Kappa : 0               
##                                           
##  Mcnemar's Test P-Value : <2e-16          
##                                           
##             Sensitivity : 0.0000          
##             Specificity : 1.0000          
##          Pos Pred Value :    NaN          
##          Neg Pred Value : 0.5963          
##              Prevalence : 0.4037          
##          Detection Rate : 0.0000          
##    Detection Prevalence : 0.0000          
##       Balanced Accuracy : 0.5000          
##                                           
##        'Positive' Class : MM              
##

The train error is 0.385, and the test error is 0.4037.

set.seed(42)

tune.fit <- tune(svm,Purchase ~ ., data=oj_train_data, kernel="radial", ranges=list(cost=c(0.01, 0.1, 1,5,10)))
summary(tune.fit)

## 
## Parameter tuning of 'svm':
## 
## - sampling method: 10-fold cross validation 
## 
## - best parameters:
##  cost
##     1
## 
## - best performance: 0.18 
## 
## - Detailed performance results:
##    cost   error dispersion
## 1  0.01 0.38500 0.04199868
## 2  0.10 0.18125 0.03784563
## 3  1.00 0.18000 0.03343734
## 4  5.00 0.18625 0.03701070
## 5 10.00 0.19375 0.03738408

best.svm<- svm(Purchase ~ ., data=oj_train_data, kernel="radial", cost = 1)

# Confusion Matrix for prediction on train data
oj.preds = predict(best.svm, oj_train_data)
confusionMatrix(oj.preds, oj_train_data$Purchase, positive = 'MM')

## Confusion Matrix and Statistics
## 
##           Reference
## Prediction  CH  MM
##         CH 453  81
##         MM  39 227
##                                          
##                Accuracy : 0.85           
##                  95% CI : (0.8233, 0.874)
##     No Information Rate : 0.615          
##     P-Value [Acc > NIR] : < 2.2e-16      
##                                          
##                   Kappa : 0.675          
##                                          
##  Mcnemar's Test P-Value : 0.000182       
##                                          
##             Sensitivity : 0.7370         
##             Specificity : 0.9207         
##          Pos Pred Value : 0.8534         
##          Neg Pred Value : 0.8483         
##              Prevalence : 0.3850         
##          Detection Rate : 0.2838         
##    Detection Prevalence : 0.3325         
##       Balanced Accuracy : 0.8289         
##                                          
##        'Positive' Class : MM             
##

# Confusion Matrix for prediction on test data
oj.preds = predict(best.svm, oj_test_data)
confusionMatrix(oj.preds, oj_test_data$Purchase, positive = 'MM')

## Confusion Matrix and Statistics
## 
##           Reference
## Prediction  CH  MM
##         CH 146  28
##         MM  15  81
##                                           
##                Accuracy : 0.8407          
##                  95% CI : (0.7915, 0.8823)
##     No Information Rate : 0.5963          
##     P-Value [Acc > NIR] : < 2e-16         
##                                           
##                   Kappa : 0.6627          
##                                           
##  Mcnemar's Test P-Value : 0.06725         
##                                           
##             Sensitivity : 0.7431          
##             Specificity : 0.9068          
##          Pos Pred Value : 0.8438          
##          Neg Pred Value : 0.8391          
##              Prevalence : 0.4037          
##          Detection Rate : 0.3000          
##    Detection Prevalence : 0.3556          
##       Balanced Accuracy : 0.8250          
##                                           
##        'Positive' Class : MM              
##

The train error is 0.15, and the test error is 0.1593. Both errors got marginally better than before.

(g) Repeat parts (b) through (e) using a support vector machine with a polynomial kernel. Set degree = 2.

svm.fit.poly<- svm(Purchase ~ ., data=oj_train_data, kernel="polynomial", cost = .01, degree = 2)
summary(svm.fit.poly)

## 
## Call:
## svm(formula = Purchase ~ ., data = oj_train_data, kernel = "polynomial", 
##     cost = 0.01, degree = 2)
## 
## 
## Parameters:
##    SVM-Type:  C-classification 
##  SVM-Kernel:  polynomial 
##        cost:  0.01 
##      degree:  2 
##      coef.0:  0 
## 
## Number of Support Vectors:  621
## 
##  ( 308 313 )
## 
## 
## Number of Classes:  2 
## 
## Levels: 
##  CH MM