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.

  1. Generate a data set with n = 500 and p = 2, such that the observations belong to two classes with a quadratic decision boundary between them.
set.seed(1)
x1 <- runif(500) - 0.5
x2 <- runif(500) - 0.5
y <- 1 * (x1^2 - x2^2 > 0)
  1. Plot the observations, colored according to their class labels. Your plot should display X1 on the x-axis, and X2 on the yaxis.
plot(x1, x2, xlab = "x1", ylab = "x2", col = (4 - y), pch = (3 - y))

  1. Fit a logistic regression model to the data, using X1 and X2 as predictors.
set.seed(1)
glm.fit <- glm(y ~ x1 + x2, family = "binomial")
summary(glm.fit)
## 
## Call:
## glm(formula = y ~ x1 + x2, family = "binomial")
## 
## Deviance Residuals: 
##    Min      1Q  Median      3Q     Max  
## -1.179  -1.139  -1.112   1.206   1.257  
## 
## Coefficients:
##              Estimate Std. Error z value Pr(>|z|)
## (Intercept) -0.087260   0.089579  -0.974    0.330
## x1           0.196199   0.316864   0.619    0.536
## x2          -0.002854   0.305712  -0.009    0.993
## 
## (Dispersion parameter for binomial family taken to be 1)
## 
##     Null deviance: 692.18  on 499  degrees of freedom
## Residual deviance: 691.79  on 497  degrees of freedom
## AIC: 697.79
## 
## Number of Fisher Scoring iterations: 3
  1. 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.
data <- data.frame(x1 = x1, x2 = x2, y = y)
glm.probs <- predict(glm.fit, data, type = "response")
glm.preds <- rep(0, 500)
glm.preds[glm.probs > 0.48] <- 1
plot(data[glm.preds == 1, ]$x1, data[glm.preds == 1, ]$x2, col = (4 - 1), pch = (3 - 1), xlab = "X1", ylab = "X2")
points(data[glm.preds == 0, ]$x1, data[glm.preds == 0, ]$x2, col = (4 - 0), pch = (3 - 0))

  1. Now fit a logistic regression model to the data using non-linear functions of X1 and X2 as predictors (e.g. X21 , X1×X2, log(X2), and so forth).
glm.fit1 <- glm(y ~ poly(x1, 2) + poly(x2, 2) + I(x1 * x2), family = "binomial")
## Warning: glm.fit: algorithm did not converge
## Warning: glm.fit: fitted probabilities numerically 0 or 1 occurred
summary(glm.fit1)
## 
## Call:
## glm(formula = y ~ poly(x1, 2) + poly(x2, 2) + I(x1 * x2), family = "binomial")
## 
## Deviance Residuals: 
##        Min          1Q      Median          3Q         Max  
## -8.240e-04  -2.000e-08  -2.000e-08   2.000e-08   1.163e-03  
## 
## Coefficients:
##              Estimate Std. Error z value Pr(>|z|)
## (Intercept)    -102.2     4302.0  -0.024    0.981
## poly(x1, 2)1   2715.3   141109.5   0.019    0.985
## poly(x1, 2)2  27218.5   842987.2   0.032    0.974
## poly(x2, 2)1   -279.7    97160.4  -0.003    0.998
## poly(x2, 2)2 -28693.0   875451.3  -0.033    0.974
## I(x1 * x2)     -206.4    41802.8  -0.005    0.996
## 
## (Dispersion parameter for binomial family taken to be 1)
## 
##     Null deviance: 6.9218e+02  on 499  degrees of freedom
## Residual deviance: 3.5810e-06  on 494  degrees of freedom
## AIC: 12
## 
## Number of Fisher Scoring iterations: 25
  1. 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.
glm.probs1 <- predict(glm.fit1, data, type = "response")
glm.preds1 <- rep(0, 500)
glm.preds1[glm.probs1 > 0.48] <- 1
plot(data[glm.preds1 == 1, ]$x1, data[glm.preds1 == 1, ]$x2, col = (4 - 1), pch = (3 - 1), xlab = "X1", ylab = "X2")
points(data[glm.preds1 == 0, ]$x1, data[glm.preds1 == 0, ]$x2, col = (4 - 0), pch = (3 - 0))

  1. 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.
library(e1071)
svm.fit <- svm(as.factor(y) ~ x1 + x2,data, kernel = "linear", cost = 0.01)
svm.preds <- predict(svm.fit, data)
plot(data[svm.preds == 0, ]$x1, data[svm.preds == 0, ]$x2, col = (4 - 0), pch = (3 - 0), xlab = "X1", ylab = "X2")
points(data[svm.preds == 1, ]$x1, data[svm.preds == 1, ]$x2, col = (4 - 1), pch = (3 - 1))

  1. 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.fit1 <- svm(as.factor(y) ~ x1 + x2, data, kernel = "radial", gamma = 1)
svm.preds1 <- predict(svm.fit1, data)
plot(data[svm.preds1 == 0, ]$x1, data[svm.preds1 == 0, ]$x2, col = (4 - 0), pch = (3 - 0), xlab = "X1", ylab = "X2")
points(data[svm.preds1 == 1, ]$x1, data[svm.preds1 == 1, ]$x2, col = (4 - 1), pch = (3 - 1))

  1. Comment on your results.

logistic regression w/ interaction terms and SVM non linear are good at identifying non linear decision boundaries. SVM with linear kernel and logistic without interactions are not effective in identifying non linear decision boundaries. we only tuned gamma for radial kernel so factors such as best interactions and tuning should be taken into consideration

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.

library(ISLR)
## Warning: package 'ISLR' was built under R version 4.0.3
library(e1071)
attach(Auto)
  1. 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.
newvar <- ifelse(Auto$mpg > median(Auto$mpg), 1, 0)
Auto$mpglevel <- as.factor(newvar)
  1. 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.
set.seed(1)
tuneout <- tune(svm, mpglevel ~ ., data = Auto, kernel = "linear", ranges = list(cost = c(0.01, 0.1, 1, 5, 10, 100)))
summary(tuneout)
## 
## Parameter tuning of 'svm':
## 
## - sampling method: 10-fold cross validation 
## 
## - best parameters:
##  cost
##     1
## 
## - best performance: 0.01025641 
## 
## - Detailed performance results:
##    cost      error dispersion
## 1 1e-02 0.07653846 0.03617137
## 2 1e-01 0.04596154 0.03378238
## 3 1e+00 0.01025641 0.01792836
## 4 5e+00 0.02051282 0.02648194
## 5 1e+01 0.02051282 0.02648194
## 6 1e+02 0.03076923 0.03151981

cross-validation error is lowest when cost = 1.

  1. 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(1)
tuneout1 <- tune(svm, mpglevel ~ ., data = Auto, kernel = "polynomial", ranges = list(cost = c(0.01, 0.1, 1, 5, 10, 100), degree = c(2, 3, 4)))
summary(tuneout1)
## 
## Parameter tuning of 'svm':
## 
## - sampling method: 10-fold cross validation 
## 
## - best parameters:
##  cost degree
##   100      2
## 
## - best performance: 0.3013462 
## 
## - Detailed performance results:
##     cost degree     error dispersion
## 1  1e-02      2 0.5511538 0.04366593
## 2  1e-01      2 0.5511538 0.04366593
## 3  1e+00      2 0.5511538 0.04366593
## 4  5e+00      2 0.5511538 0.04366593
## 5  1e+01      2 0.5130128 0.08963366
## 6  1e+02      2 0.3013462 0.09961961
## 7  1e-02      3 0.5511538 0.04366593
## 8  1e-01      3 0.5511538 0.04366593
## 9  1e+00      3 0.5511538 0.04366593
## 10 5e+00      3 0.5511538 0.04366593
## 11 1e+01      3 0.5511538 0.04366593
## 12 1e+02      3 0.3446154 0.09821588
## 13 1e-02      4 0.5511538 0.04366593
## 14 1e-01      4 0.5511538 0.04366593
## 15 1e+00      4 0.5511538 0.04366593
## 16 5e+00      4 0.5511538 0.04366593
## 17 1e+01      4 0.5511538 0.04366593
## 18 1e+02      4 0.5511538 0.04366593

lowest cv error obtained for degree = 2 and cost = 100

set.seed(1)
tuneout2 <- tune(svm, mpglevel ~ ., data = Auto, kernel = "radial", ranges = list(cost = c(0.01, 0.1, 1, 5, 10, 100), gamma = c(0.01, 0.1, 1, 5, 10, 100)))
summary(tuneout2)
## 
## Parameter tuning of 'svm':
## 
## - sampling method: 10-fold cross validation 
## 
## - best parameters:
##  cost gamma
##   100  0.01
## 
## - best performance: 0.01282051 
## 
## - Detailed performance results:
##     cost gamma      error dispersion
## 1  1e-02 1e-02 0.55115385 0.04366593
## 2  1e-01 1e-02 0.08929487 0.04382379
## 3  1e+00 1e-02 0.07403846 0.03522110
## 4  5e+00 1e-02 0.04852564 0.03303346
## 5  1e+01 1e-02 0.02557692 0.02093679
## 6  1e+02 1e-02 0.01282051 0.01813094
## 7  1e-02 1e-01 0.21711538 0.09865227
## 8  1e-01 1e-01 0.07903846 0.03874545
## 9  1e+00 1e-01 0.05371795 0.03525162
## 10 5e+00 1e-01 0.02820513 0.03299190
## 11 1e+01 1e-01 0.03076923 0.03375798
## 12 1e+02 1e-01 0.03583333 0.02759051
## 13 1e-02 1e+00 0.55115385 0.04366593
## 14 1e-01 1e+00 0.55115385 0.04366593
## 15 1e+00 1e+00 0.06384615 0.04375618
## 16 5e+00 1e+00 0.05884615 0.04020934
## 17 1e+01 1e+00 0.05884615 0.04020934
## 18 1e+02 1e+00 0.05884615 0.04020934
## 19 1e-02 5e+00 0.55115385 0.04366593
## 20 1e-01 5e+00 0.55115385 0.04366593
## 21 1e+00 5e+00 0.49493590 0.04724924
## 22 5e+00 5e+00 0.48217949 0.05470903
## 23 1e+01 5e+00 0.48217949 0.05470903
## 24 1e+02 5e+00 0.48217949 0.05470903
## 25 1e-02 1e+01 0.55115385 0.04366593
## 26 1e-01 1e+01 0.55115385 0.04366593
## 27 1e+00 1e+01 0.51794872 0.05063697
## 28 5e+00 1e+01 0.51794872 0.04917316
## 29 1e+01 1e+01 0.51794872 0.04917316
## 30 1e+02 1e+01 0.51794872 0.04917316
## 31 1e-02 1e+02 0.55115385 0.04366593
## 32 1e-01 1e+02 0.55115385 0.04366593
## 33 1e+00 1e+02 0.55115385 0.04366593
## 34 5e+00 1e+02 0.55115385 0.04366593
## 35 1e+01 1e+02 0.55115385 0.04366593
## 36 1e+02 1e+02 0.55115385 0.04366593

lowest cv error obtained for gamma = 0.01 and cost = 100

  1. Make some plots to back up your assertions in (b) and (c).
svm.linear <- svm(mpglevel ~ ., data = Auto, kernel = "linear", cost = 1)
svm.poly <- svm(mpglevel ~ ., data = Auto, kernel = "polynomial", cost = 100, degree = 2)
svm.radial <- svm(mpglevel ~ ., data = Auto, kernel = "radial", cost = 100, gamma = 0.01)
plotpairs = function(fit) {
    for (name in names(Auto)[!(names(Auto) %in% c("mpg", "mpglevel", "name"))]) {
        plot(fit, Auto, as.formula(paste("mpg~", name, sep = "")))
    }
}
plotpairs(svm.linear)

plotpairs(svm.poly)

plotpairs(svm.radial)

QUESTION 8: This problem involves the OJ data set which is part of the ISLR package.

attach(OJ)
  1. Create a training set containing a random sample of 800 observations, and a test set containing the remaining observations.
set.seed(1)
row.number <- sample(nrow(OJ), 800)
OJ_train <- OJ[row.number, ]
OJ_test <- OJ[-row.number, ]
  1. 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.
set.seed(1)
ojsvmlinear = svm(Purchase ~ ., kernel = "linear", data = OJ_train, cost = 0.01)
summary(ojsvmlinear)
## 
## Call:
## svm(formula = Purchase ~ ., data = OJ_train, kernel = "linear", cost = 0.01)
## 
## 
## Parameters:
##    SVM-Type:  C-classification 
##  SVM-Kernel:  linear 
##        cost:  0.01 
## 
## Number of Support Vectors:  435
## 
##  ( 219 216 )
## 
## 
## Number of Classes:  2 
## 
## Levels: 
##  CH MM

SVM creates 435 support vectors. 219 belong to the CH level 216 belong to the MM level

  1. What are the training and test error rates?
ojsvm.train.pred <- predict(ojsvmlinear, OJ_train)
table(OJ_train$Purchase, ojsvm.train.pred)
##     ojsvm.train.pred
##       CH  MM
##   CH 420  65
##   MM  75 240
(75+65)/(420+65+75+240)
## [1] 0.175

The training error rate is 17.5%

ojsvm.test.pred <- predict(ojsvmlinear, OJ_test)
table(OJ_test$Purchase, ojsvm.test.pred)
##     ojsvm.test.pred
##       CH  MM
##   CH 153  15
##   MM  33  69
(33+15)/(153+15+33+69)
## [1] 0.1777778

The training error rate is 17.78%

  1. Use the tune() function to select an optimal cost. Consider values in the range 0.01 to 10.
set.seed(1)
ojtuneout <- tune(svm, Purchase ~ ., data = OJ_train, kernel = "linear", ranges = list(cost = 10^seq(-2, 1, by = 0.25)))
summary(ojtuneout)
## 
## Parameter tuning of 'svm':
## 
## - sampling method: 10-fold cross validation 
## 
## - best parameters:
##      cost
##  3.162278
## 
## - best performance: 0.16875 
## 
## - Detailed performance results:
##           cost   error dispersion
## 1   0.01000000 0.17625 0.02853482
## 2   0.01778279 0.17625 0.03143004
## 3   0.03162278 0.17125 0.02829041
## 4   0.05623413 0.17625 0.02853482
## 5   0.10000000 0.17250 0.03162278
## 6   0.17782794 0.17125 0.02829041
## 7   0.31622777 0.17125 0.02889757
## 8   0.56234133 0.17125 0.02703521
## 9   1.00000000 0.17500 0.02946278
## 10  1.77827941 0.17375 0.02729087
## 11  3.16227766 0.16875 0.03019037
## 12  5.62341325 0.17375 0.03304563
## 13 10.00000000 0.17375 0.03197764

optimal cost = 3.162278.

  1. Compute the training and test error rates using this new value for cost.
ojsvmlinear1 <- svm(Purchase ~ ., kernel = "linear", data = OJ_train, cost = ojtuneout$best.parameter$cost)
ojsvm.train.pred1 <- predict(ojsvmlinear1, OJ_train)
table(OJ_train$Purchase, ojsvm.train.pred1)
##     ojsvm.train.pred1
##       CH  MM
##   CH 423  62
##   MM  70 245
(70+62)/(423+62+70+245)
## [1] 0.165

The train error rate decreased to 16.5%

ojsvm.test.pred1 <- predict(ojsvmlinear1, OJ_test)
table(OJ_test$Purchase, ojsvm.test.pred1)
##     ojsvm.test.pred1
##       CH  MM
##   CH 156  12
##   MM  29  73
(29+12)/(156+12+29+73)
## [1] 0.1518519

The test error rate decreased to 15.19%

  1. Repeat parts (b) through (e) using a support vector machine with a radial kernel. Use the default value for gamma.
set.seed(1)
ojsvmradial <- svm(Purchase ~ ., kernel = "radial", data = OJ_train)
summary(ojsvmradial)
## 
## Call:
## svm(formula = Purchase ~ ., data = OJ_train, kernel = "radial")
## 
## 
## Parameters:
##    SVM-Type:  C-classification 
##  SVM-Kernel:  radial 
##        cost:  1 
## 
## Number of Support Vectors:  373
## 
##  ( 188 185 )
## 
## 
## Number of Classes:  2 
## 
## Levels: 
##  CH MM
ojsvm.train.pred2 <- predict(ojsvmradial, OJ_train)
table(OJ_train$Purchase, ojsvm.train.pred2)
##     ojsvm.train.pred2
##       CH  MM
##   CH 441  44
##   MM  77 238
(77+44)/(441+44+77+238)
## [1] 0.15125
ojsvm.test.pred2 <- predict(ojsvmradial, OJ_test)
table(OJ_test$Purchase, ojsvm.test.pred2)
##     ojsvm.test.pred2
##       CH  MM
##   CH 151  17
##   MM  33  69
(33+17)/(151+17+33+69)
## [1] 0.1851852

The radial kernel with default gamma creates 373 support vectors 188 belong to level CH 185 belong to level MM

The classifier has a training error = 15.13%, test error = 18.52%. Train error lower than linear kernel Test error higher than linear kernel

set.seed(1)
ojtuneout1 <- tune(svm, Purchase ~ ., data = OJ_train, kernel = "radial", ranges = list(cost = 10^seq(-2, 
    1, by = 0.25)))
summary(ojtuneout1)
## 
## Parameter tuning of 'svm':
## 
## - sampling method: 10-fold cross validation 
## 
## - best parameters:
##       cost
##  0.5623413
## 
## - best performance: 0.16875 
## 
## - Detailed performance results:
##           cost   error dispersion
## 1   0.01000000 0.39375 0.04007372
## 2   0.01778279 0.39375 0.04007372
## 3   0.03162278 0.35750 0.05927806
## 4   0.05623413 0.19500 0.02443813
## 5   0.10000000 0.18625 0.02853482
## 6   0.17782794 0.18250 0.03291403
## 7   0.31622777 0.17875 0.03230175
## 8   0.56234133 0.16875 0.02651650
## 9   1.00000000 0.17125 0.02128673
## 10  1.77827941 0.17625 0.02079162
## 11  3.16227766 0.17750 0.02266912
## 12  5.62341325 0.18000 0.02220485
## 13 10.00000000 0.18625 0.02853482
ojsvmradial1 <- svm(Purchase ~ ., kernel = "radial", data = OJ_train, cost = ojtuneout1$best.parameter$cost)
summary(ojsvmradial1)
## 
## Call:
## svm(formula = Purchase ~ ., data = OJ_train, kernel = "radial", cost = ojtuneout1$best.parameter$cost)
## 
## 
## Parameters:
##    SVM-Type:  C-classification 
##  SVM-Kernel:  radial 
##        cost:  0.5623413 
## 
## Number of Support Vectors:  397
## 
##  ( 200 197 )
## 
## 
## Number of Classes:  2 
## 
## Levels: 
##  CH MM
ojsvm.train.pred3 <- predict(ojsvmradial1, OJ_train)
table(OJ_train$Purchase, ojsvm.train.pred3)
##     ojsvm.train.pred3
##       CH  MM
##   CH 437  48
##   MM  71 244
(71+48)/(437+48+71+244)
## [1] 0.14875

The train error decreased to 14.88%

ojsvm.test.pred3 <- predict(ojsvmradial1, OJ_test)
table(OJ_test$Purchase, ojsvm.test.pred3)
##     ojsvm.test.pred3
##       CH  MM
##   CH 150  18
##   MM  30  72
(30+18)/(150+18+30+72)
## [1] 0.1777778

The train error decreased to 17.78%.

  1. Repeat parts (b) through (e) using a support vector machine with a polynomial kernel. Set degree=2.
set.seed(1)
ojsvm.poly <- svm(Purchase ~ ., kernel = "polynomial", data = OJ_train, degree = 2)
summary(ojsvm.poly)
## 
## Call:
## svm(formula = Purchase ~ ., data = OJ_train, kernel = "polynomial", 
##     degree = 2)
## 
## 
## Parameters:
##    SVM-Type:  C-classification 
##  SVM-Kernel:  polynomial 
##        cost:  1 
##      degree:  2 
##      coef.0:  0 
## 
## Number of Support Vectors:  447
## 
##  ( 225 222 )
## 
## 
## Number of Classes:  2 
## 
## Levels: 
##  CH MM

The polynomial kernel with default gamma creates 447 support vectors 225 belonging to level CH 222 belong to level MM

ojsvm.train.pred4 <- predict(ojsvm.poly, OJ_train)
table(OJ_train$Purchase, ojsvm.train.pred4)
##     ojsvm.train.pred4
##       CH  MM
##   CH 449  36
##   MM 110 205
(110+36)/(449+36+110+205)
## [1] 0.1825

The train error is 18.25%

ojsvm.test.pred4 <- predict(ojsvm.poly, OJ_test)
table(OJ_test$Purchase, ojsvm.test.pred4)
##     ojsvm.test.pred4
##       CH  MM
##   CH 153  15
##   MM  45  57
(45+15)/(153+15+45+57)
## [1] 0.2222222

The test error is 22.22%.

set.seed(1)
ojtune.out2 <- tune(svm, Purchase ~ ., data = OJ_train, kernel = "polynomial", degree = 2, ranges = list(cost = 10^seq(-2, 
    1, by = 0.25)))
summary(ojtune.out2)
## 
## Parameter tuning of 'svm':
## 
## - sampling method: 10-fold cross validation 
## 
## - best parameters:
##      cost
##  3.162278
## 
## - best performance: 0.1775 
## 
## - Detailed performance results:
##           cost   error dispersion
## 1   0.01000000 0.39125 0.04210189
## 2   0.01778279 0.37125 0.03537988
## 3   0.03162278 0.36500 0.03476109
## 4   0.05623413 0.33750 0.04714045
## 5   0.10000000 0.32125 0.05001736
## 6   0.17782794 0.24500 0.04758034
## 7   0.31622777 0.19875 0.03972562
## 8   0.56234133 0.20500 0.03961621
## 9   1.00000000 0.20250 0.04116363
## 10  1.77827941 0.18500 0.04199868
## 11  3.16227766 0.17750 0.03670453
## 12  5.62341325 0.18375 0.03064696
## 13 10.00000000 0.18125 0.02779513
ojsvm.poly1 <- svm(Purchase ~ ., kernel = "polynomial", degree = 2, data = OJ_train, cost = ojtune.out2$best.parameter$cost)
summary(ojsvm.poly1)
## 
## Call:
## svm(formula = Purchase ~ ., data = OJ_train, kernel = "polynomial", 
##     degree = 2, cost = ojtune.out2$best.parameter$cost)
## 
## 
## Parameters:
##    SVM-Type:  C-classification 
##  SVM-Kernel:  polynomial 
##        cost:  3.162278 
##      degree:  2 
##      coef.0:  0 
## 
## Number of Support Vectors:  385
## 
##  ( 197 188 )
## 
## 
## Number of Classes:  2 
## 
## Levels: 
##  CH MM
ojsvm.train.pred5 <- predict(ojsvm.poly1, OJ_train)
table(OJ_train$Purchase, ojsvm.train.pred5)
##     ojsvm.train.pred5
##       CH  MM
##   CH 451  34
##   MM  90 225
(90+34)/(451+34+90+225)
## [1] 0.155

The train error rate decreased to 15.5%

ojsvm.test.pred5 <- predict(ojsvm.poly1, OJ_test)
table(OJ_test$Purchase, ojsvm.test.pred5)
##     ojsvm.test.pred5
##       CH  MM
##   CH 154  14
##   MM  41  61
(41+14)/(154+14+41+61)
## [1] 0.2037037

The test error rate decreased to 20.37%

  1. Overall, which approach seems to give the best results on this data?

SVM model with the radial kernel and optimal cost seems to give the best results