Assignment #7 STA6543 SMR 2025 UTSA

install.packages("pacman")

Error in install.packages : Updating loaded packages

library("pacman")

p_load(ggplot2, dplyr, tidyr, tidyverse, ISLR, ISLR2, tibble, readr, purrr,stringr, forcats, lubridate, glmnet, leaps, caret, klaR, pls, mlbench, randomForest, tree, gbm, BART, MASS, rattle,rpart)

3. Consider the Gini index, classification error, and entropy in a simple classification setting with two classes. Create a single plot that displays each of these quantities as a function of ˆpm1. The x-axis should display ˆpm1, ranging from 0 to 1, and the y-axis should display the value of the Gini index, classification error, and entropy. Hint: In a setting with two classes, ˆpm1 = 1 − ˆpm2. You could make this plot by hand, but it will be much easier to make in R.

# set the axis range from 0 to 1
p <- seq(0, 1, 0.01)

# Calculate each error rate. Since 2 classes, use pˆm1 = 1 − pˆm2
classification.error <- 1 - pmax(p, 1 - p)

gini.index <- 2 * p * (1 - p)

entropy <- - (p * log(p) + (1 - p) * log(1 - p))

# Build the plot
matplot(p, cbind(classification.error, gini.index, entropy), col = c("blue", "green", "red" ), pch=16,main = "Classification Tree Measures with 2 classes", xlab = "Proportion of training observations", ylab = "Measure Values",ylim=c(0,1.1), lty=1)
legend(x='top', pch = 16,title = "Measures", col=c("blue", "green", "red"), legend=c("Classification Error Rate", "Gini Index", "Entropy"), box.lty=1)

8. In the lab, a classification tree was applied to the Carseats data set after converting Sales into a qualitative response variable. Now we will seek to predict Sales using regression trees and related approaches, treating the response as a quantitative variable.

1. Split the data set into a training set and a test set.
2. Fit a regression tree to the training set. Plot the tree, and interpret the results. What test MSE do you obtain?
3. Use cross-validation in order to determine the optimal level of tree complexity. Does pruning the tree improve the test MSE?
4. Use the bagging approach in order to analyze this data. What test MSE do you obtain? Use the importance() function to determine which variables are most important.
5. Use random forests to analyze this data. What test MSE do you obtain? Use the importance() function to determine which variables are most important. Describe the effect of m, the number of variables considered at each split, on the error rate obtained.
6. Now analyze the data using BART, and report your results.

set.seed(997)

train_index <- createDataPartition(Carseats$Sales, p = 0.6, list = FALSE, times = 2)

train <- Carseats[train_index, ]

test <- Carseats[-train_index, ]

dim(train)

[1] 482  11

2. Fit a regression tree to the training set. Plot the tree, and interpret the results. What test MSE do you obtain? 4.284189

tree.Carseats <- tree(Sales ~ ., data = train)

summary(tree.Carseats)

Regression tree:
tree(formula = Sales ~ ., data = train)
Variables actually used in tree construction:
[1] "ShelveLoc"   "Price"       "Age"         "Income"      "CompPrice"  
[6] "Advertising"
Number of terminal nodes:  17 
Residual mean deviance:  2.372 = 1103 / 465 
Distribution of residuals:
    Min.  1st Qu.   Median     Mean  3rd Qu.     Max. 
-5.20700 -1.14500 -0.01706  0.00000  1.05200  4.45200 

plot(tree.Carseats)

text(tree.Carseats, pretty = 0, cex = .55)

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NA

pred.tree <- predict(tree.Carseats, newdata = test)

mse.tree <- mean((pred.tree - test$Sales)^2)

mse.tree

[1] 4.284189

3. Use cross-validation in order to determine the optimal level of tree complexity. Does pruning the tree improve the test MSE? No.It goes up to 4.923

set.seed(997)

cv.carseats <- cv.tree(tree.Carseats, FUN = prune.tree)

cv.carseats

$size
 [1] 17 16 15 14 13 12 11  9  8  7  5  4  3  2  1

$dev
 [1] 1656.432 1734.245 1764.190 1774.177 1790.883 1790.883 1824.284 1831.200
 [9] 2185.604 2185.604 2416.917 2412.573 2543.957 2815.409 3924.927

$k
 [1]       -Inf   46.46058   49.46359   50.39076   52.98134   53.58919
 [7]   56.04805   64.42033  102.95100  102.97797  141.39331  154.54278
[13]  193.94041  418.38086 1118.65084

$method
[1] "deviance"

attr(,"class")
[1] "prune"         "tree.sequence"

plot(cv.carseats$size , cv.carseats$dev, type = "b")

NA
NA

prune.carseats <- prune.tree(tree.Carseats , best = 5)

plot(prune.carseats) 

text(prune.carseats , pretty = 0)

NA
NA

pred.tree.prune <- predict(prune.carseats, newdata = test)

mse.tree <- mean((pred.tree.prune - test$Sales)^2)

mse.tree

[1] 4.923

4. Use the bagging approach in order to analyze this data. What test MSE do you obtain? Use the importance() function to determine which variables are most important. Price and Shelf location are the most important

set.seed(997)

bag.Carseats <- randomForest(Sales ~., data = train, ntree = 500, mtry = ncol(train)-1, importance= TRUE) 

bag.Carseats

Call:
 randomForest(formula = Sales ~ ., data = train, ntree = 500,      mtry = ncol(train) - 1, importance = TRUE) 
               Type of random forest: regression
                     Number of trees: 500
No. of variables tried at each split: 10

          Mean of squared residuals: 1.245393
                    % Var explained: 84.67

install.packages("pacman")

Warning in install.packages :
  package ‘pacman’ is in use and will not be installed

pred.carseats <- predict(bag.Carseats, newdata = test)

mse.carseats <- mean((pred.carseats - test$Sales)^2)

mse.carseats

[1] 2.255516

randomForest::importance(bag.Carseats)

               %IncMSE IncNodePurity
CompPrice    55.264271     416.03406
Income       35.365201     228.65812
Advertising  44.294243     281.95318
Population   20.722038     104.81350
Price       111.812343    1093.21565
ShelveLoc   111.655615    1257.93758
Age          43.386355     307.35694
Education    27.637200     122.34371
Urban         7.459482      17.89376
US           11.484217      23.91682

5. Use random forests to analyze this data. What test MSE do you obtain? Use the importance() function to determine which variables are most important. Describe the effect of m, the number of variables considered at each split, on the error rate obtained. m is reduced to 3 instead of 10. The error rate is slightly increased. Price and Shelf location are still the most influential factors

set.seed(997)

rf.carseats <- randomForest(Sales~., data = train, importance = TRUE)

rf.carseats

Call:
 randomForest(formula = Sales ~ ., data = train, importance = TRUE) 
               Type of random forest: regression
                     Number of trees: 500
No. of variables tried at each split: 3

          Mean of squared residuals: 1.525753
                    % Var explained: 81.22

pred.rfCarseats <- predict(rf.carseats, mtry=3, newdata = test)

mse.rf <- mean((pred.rfCarseats - test$Sales)^2)

mse.rf

[1] 2.329565

randomForest::importance(rf.carseats)

              %IncMSE IncNodePurity
CompPrice   35.060236     351.43146
Income      28.020843     277.23553
Advertising 36.421151     313.80800
Population  21.151835     233.73415
Price       62.483767     951.51825
ShelveLoc   61.262379     990.05196
Age         36.129786     406.01624
Education   23.340310     181.89627
Urban        8.790943      33.74796
US          13.333512      55.10219

6. Now analyze the data using BART, and report your results. BART performed the best.

x_train <- train[, -which(names(train) == "Sales")]

x_test <- test[, -which(names(test) == "Sales")]

y_train <- train$Sales


set.seed(997)

bart_fit <- wbart(x.train = x_train, y.train = y_train, x.test = x_test)

*****Into main of wbart
*****Data:
data:n,p,np: 482, 14, 64
y1,yn: 1.948402, -1.611598
x1,x[n*p]: 138.000000, 1.000000
xp1,xp[np*p]: 113.000000, 1.000000
*****Number of Trees: 200
*****Number of Cut Points: 71 ... 1
*****burn and ndpost: 100, 1000
*****Prior:beta,alpha,tau,nu,lambda: 2.000000,0.950000,0.287616,3.000000,0.210871
*****sigma: 1.040455
*****w (weights): 1.000000 ... 1.000000
*****Dirichlet:sparse,theta,omega,a,b,rho,augment: 0,0,1,0.5,1,14,0
*****nkeeptrain,nkeeptest,nkeeptestme,nkeeptreedraws: 1000,1000,1000,1000
*****printevery: 100
*****skiptr,skipte,skipteme,skiptreedraws: 1,1,1,1

MCMC
done 0 (out of 1100)
done 100 (out of 1100)
done 200 (out of 1100)
done 300 (out of 1100)
done 400 (out of 1100)
done 500 (out of 1100)
done 600 (out of 1100)
done 700 (out of 1100)
done 800 (out of 1100)
done 900 (out of 1100)
done 1000 (out of 1100)
time: 6s
check counts
trcnt,tecnt,temecnt,treedrawscnt: 1000,1000,1000,1000

pred_bart <- colMeans(bart_fit$yhat.test)

mse_bart <- mean((pred_bart - test$Sales)^2)

mse_bart

[1] 1.649856

9. This problem involves the OJ data set which is part of the ISLR2 package.

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

data(OJ)

str(OJ)

'data.frame':   1070 obs. of  18 variables:
 $ Purchase      : Factor w/ 2 levels "CH","MM": 1 1 1 2 1 1 1 1 1 1 ...
 $ WeekofPurchase: num  237 239 245 227 228 230 232 234 235 238 ...
 $ StoreID       : num  1 1 1 1 7 7 7 7 7 7 ...
 $ PriceCH       : num  1.75 1.75 1.86 1.69 1.69 1.69 1.69 1.75 1.75 1.75 ...
 $ PriceMM       : num  1.99 1.99 2.09 1.69 1.69 1.99 1.99 1.99 1.99 1.99 ...
 $ DiscCH        : num  0 0 0.17 0 0 0 0 0 0 0 ...
 $ DiscMM        : num  0 0.3 0 0 0 0 0.4 0.4 0.4 0.4 ...
 $ SpecialCH     : num  0 0 0 0 0 0 1 1 0 0 ...
 $ SpecialMM     : num  0 1 0 0 0 1 1 0 0 0 ...
 $ LoyalCH       : num  0.5 0.6 0.68 0.4 0.957 ...
 $ SalePriceMM   : num  1.99 1.69 2.09 1.69 1.69 1.99 1.59 1.59 1.59 1.59 ...
 $ SalePriceCH   : num  1.75 1.75 1.69 1.69 1.69 1.69 1.69 1.75 1.75 1.75 ...
 $ PriceDiff     : num  0.24 -0.06 0.4 0 0 0.3 -0.1 -0.16 -0.16 -0.16 ...
 $ Store7        : Factor w/ 2 levels "No","Yes": 1 1 1 1 2 2 2 2 2 2 ...
 $ PctDiscMM     : num  0 0.151 0 0 0 ...
 $ PctDiscCH     : num  0 0 0.0914 0 0 ...
 $ ListPriceDiff : num  0.24 0.24 0.23 0 0 0.3 0.3 0.24 0.24 0.24 ...
 $ STORE         : num  1 1 1 1 0 0 0 0 0 0 ...

set.seed(997)



train_index <- sample(1:nrow(OJ), 800)

train <- OJ[train_index, ]

test <- OJ[-train_index, ]

2. Fit a tree to the training data, with Purchase as the response and the other variables as predictors. Use the summary() function to produce summary statistics about the tree, and describe the results obtained. What is the training error rate? How many terminal nodes does the tree have?

tree_OJ = rpart(Purchase~., data = train, method = "class", control = rpart.control(minsplit = 10, cp = 0.01))

summary(tree_OJ)

Call:
rpart(formula = Purchase ~ ., data = train, method = "class", 
    control = rpart.control(minsplit = 10, cp = 0.01))
  n= 800 

          CP nsplit rel error    xerror       xstd
1 0.47648903      0 1.0000000 1.0000000 0.04341424
2 0.03134796      1 0.5235110 0.5611285 0.03695189
3 0.02507837      3 0.4608150 0.4952978 0.03529884
4 0.01567398      4 0.4357367 0.4576803 0.03424756
5 0.01000000      5 0.4200627 0.4733542 0.03469566

Variable importance
       LoyalCH      PriceDiff        StoreID    SalePriceMM WeekofPurchase 
            47              8              8              6              5 
       PriceMM         DiscMM      PctDiscMM        PriceCH  ListPriceDiff 
             5              5              5              4              4 
         STORE    SalePriceCH 
             1              1 

Node number 1: 800 observations,    complexity param=0.476489
  predicted class=CH  expected loss=0.39875  P(node) =1
    class counts:   481   319
   probabilities: 0.601 0.399 
  left son=2 (448 obs) right son=3 (352 obs)
  Primary splits:
      LoyalCH     < 0.5036    to the right, improve=126.45590, (0 missing)
      StoreID     < 3.5       to the right, improve= 39.41901, (0 missing)
      PriceDiff   < 0.015     to the right, improve= 22.68647, (0 missing)
      SalePriceMM < 2.04      to the right, improve= 19.96728, (0 missing)
      Store7      splits as  RL, improve= 17.61501, (0 missing)
  Surrogate splits:
      StoreID        < 3.5       to the right, agree=0.639, adj=0.179, (0 split)
      WeekofPurchase < 237.5     to the right, agree=0.614, adj=0.122, (0 split)
      PriceCH        < 1.825     to the right, agree=0.600, adj=0.091, (0 split)
      PriceMM        < 2.04      to the right, agree=0.596, adj=0.082, (0 split)
      ListPriceDiff  < 0.085     to the right, agree=0.586, adj=0.060, (0 split)

Node number 2: 448 observations,    complexity param=0.02507837
  predicted class=CH  expected loss=0.1495536  P(node) =0.56
    class counts:   381    67
   probabilities: 0.850 0.150 
  left son=4 (420 obs) right son=5 (28 obs)
  Primary splits:
      PriceDiff   < -0.39     to the right, improve=14.53601, (0 missing)
      LoyalCH     < 0.7645725 to the right, improve=13.57180, (0 missing)
      DiscMM      < 0.72      to the left,  improve=10.57814, (0 missing)
      SalePriceMM < 1.435     to the right, improve=10.57814, (0 missing)
      PctDiscMM   < 0.3342595 to the left,  improve=10.57814, (0 missing)
  Surrogate splits:
      DiscMM      < 0.72      to the left,  agree=0.971, adj=0.536, (0 split)
      SalePriceMM < 1.435     to the right, agree=0.971, adj=0.536, (0 split)
      PctDiscMM   < 0.3342595 to the left,  agree=0.971, adj=0.536, (0 split)
      SalePriceCH < 2.075     to the left,  agree=0.944, adj=0.107, (0 split)

Node number 3: 352 observations,    complexity param=0.03134796
  predicted class=MM  expected loss=0.2840909  P(node) =0.44
    class counts:   100   252
   probabilities: 0.284 0.716 
  left son=6 (178 obs) right son=7 (174 obs)
  Primary splits:
      LoyalCH   < 0.2761415 to the right, improve=17.104590, (0 missing)
      STORE     < 1.5       to the left,  improve= 6.014799, (0 missing)
      StoreID   < 3.5       to the right, improve= 5.962806, (0 missing)
      PriceDiff < 0.215     to the right, improve= 5.514752, (0 missing)
      Store7    splits as  RL, improve= 4.357584, (0 missing)
  Surrogate splits:
      STORE       < 1.5       to the left,  agree=0.634, adj=0.259, (0 split)
      StoreID     < 3.5       to the right, agree=0.599, adj=0.190, (0 split)
      PriceCH     < 1.875     to the left,  agree=0.585, adj=0.161, (0 split)
      SalePriceCH < 1.775     to the left,  agree=0.585, adj=0.161, (0 split)
      SalePriceMM < 2.04      to the left,  agree=0.577, adj=0.144, (0 split)

Node number 4: 420 observations
  predicted class=CH  expected loss=0.1166667  P(node) =0.525
    class counts:   371    49
   probabilities: 0.883 0.117 

Node number 5: 28 observations,    complexity param=0.01567398
  predicted class=MM  expected loss=0.3571429  P(node) =0.035
    class counts:    10    18
   probabilities: 0.357 0.643 
  left son=10 (5 obs) right son=11 (23 obs)
  Primary splits:
      LoyalCH     < 0.9748025 to the right, improve=5.031056, (0 missing)
      PriceCH     < 2.04      to the right, improve=2.777143, (0 missing)
      DiscMM      < 0.47      to the left,  improve=2.777143, (0 missing)
      SalePriceMM < 1.64      to the right, improve=2.777143, (0 missing)
      SalePriceCH < 2.04      to the right, improve=2.777143, (0 missing)

Node number 6: 178 observations,    complexity param=0.03134796
  predicted class=MM  expected loss=0.4382022  P(node) =0.2225
    class counts:    78   100
   probabilities: 0.438 0.562 
  left son=12 (102 obs) right son=13 (76 obs)
  Primary splits:
      PriceDiff      < 0.05      to the right, improve=12.206500, (0 missing)
      SalePriceMM    < 2.04      to the right, improve=11.423010, (0 missing)
      WeekofPurchase < 230.5     to the right, improve= 5.154373, (0 missing)
      ListPriceDiff  < 0.035     to the right, improve= 4.904148, (0 missing)
      PriceMM        < 2.04      to the right, improve= 4.791957, (0 missing)
  Surrogate splits:
      SalePriceMM   < 1.94      to the right, agree=0.938, adj=0.855, (0 split)
      DiscMM        < 0.08      to the left,  agree=0.826, adj=0.592, (0 split)
      PctDiscMM     < 0.038887  to the left,  agree=0.826, adj=0.592, (0 split)
      PriceMM       < 2.04      to the right, agree=0.747, adj=0.408, (0 split)
      ListPriceDiff < 0.135     to the right, agree=0.747, adj=0.408, (0 split)

Node number 7: 174 observations
  predicted class=MM  expected loss=0.1264368  P(node) =0.2175
    class counts:    22   152
   probabilities: 0.126 0.874 

Node number 10: 5 observations
  predicted class=CH  expected loss=0  P(node) =0.00625
    class counts:     5     0
   probabilities: 1.000 0.000 

Node number 11: 23 observations
  predicted class=MM  expected loss=0.2173913  P(node) =0.02875
    class counts:     5    18
   probabilities: 0.217 0.783 

Node number 12: 102 observations
  predicted class=CH  expected loss=0.4019608  P(node) =0.1275
    class counts:    61    41
   probabilities: 0.598 0.402 

Node number 13: 76 observations
  predicted class=MM  expected loss=0.2236842  P(node) =0.095
    class counts:    17    59
   probabilities: 0.224 0.776 

fancyRpartPlot(tree_OJ)

set.seed(997)

tree.OJ <- tree(Purchase ~., data = train)

summary(tree.OJ)

Classification tree:
tree(formula = Purchase ~ ., data = train)
Variables actually used in tree construction:
[1] "LoyalCH"     "SalePriceMM" "PriceDiff"  
Number of terminal nodes:  8 
Residual mean deviance:  0.7751 = 613.9 / 792 
Misclassification error rate: 0.1675 = 134 / 800 

Classification tree built both with growing tree with rpart and classification. Both models showed that Brand Loyalty was the most influential factor, followed by prices. Store location showed as equal weight in the rpart model, which points more towards the customer demographics and store display. Would need to be further investigated.

3. Type in the name of the tree object in order to get a detailed text output. Pick one of the terminal nodes, and interpret the information displayed. Brand Loyalty followed by Sale Price, and Price Differential were still the most influential factors

tree.OJ

node), split, n, deviance, yval, (yprob)
      * denotes terminal node

 1) root 800 1076.000 CH ( 0.60125 0.39875 )  
   2) LoyalCH < 0.574494 383  473.100 MM ( 0.30809 0.69191 )  
     4) LoyalCH < 0.0616725 76   10.650 MM ( 0.01316 0.98684 ) *
     5) LoyalCH > 0.0616725 307  408.100 MM ( 0.38111 0.61889 )  
      10) SalePriceMM < 2.04 174  203.000 MM ( 0.27011 0.72989 ) *
      11) SalePriceMM > 2.04 133  184.000 CH ( 0.52632 0.47368 )  
        22) LoyalCH < 0.277977 45   52.190 MM ( 0.26667 0.73333 ) *
        23) LoyalCH > 0.277977 88  112.900 CH ( 0.65909 0.34091 ) *
   3) LoyalCH > 0.574494 417  321.400 CH ( 0.87050 0.12950 )  
     6) LoyalCH < 0.764572 161  186.900 CH ( 0.73292 0.26708 )  
      12) PriceDiff < -0.065 38   50.020 MM ( 0.36842 0.63158 ) *
      13) PriceDiff > -0.065 123  105.900 CH ( 0.84553 0.15447 )  
        26) PriceDiff < 0.31 79   84.790 CH ( 0.77215 0.22785 ) *
        27) PriceDiff > 0.31 44    9.545 CH ( 0.97727 0.02273 ) *
     7) LoyalCH > 0.764572 256   90.760 CH ( 0.95703 0.04297 ) *

4. Create a plot of the tree, and interpret the results.

plot(tree.OJ)

text(tree.OJ, pretty = 3, cex = 0.7)

NA
NA

5. Predict the response on the test data, and produce a confusion matrix comparing the test labels to the predicted test labels. What is the test error rate? The test error rate is 19. The accuracy rate is 80%%

# use predict function to get the response on the test data
tree.preds <- predict(tree.OJ, test, type = "class")

# Build a confusion matrix
confusion <- table(tree.preds, test$Purchase)

confusion

          
tree.preds  CH  MM
        CH 147  28
        MM  25  70

misclassification_rate <- 1 - sum(diag(confusion)) / sum(confusion)

print(misclassification_rate)

[1] 0.1962963

accuracy_rate <- sum(diag(confusion)) / sum(confusion)

print(accuracy_rate)

[1] 0.8037037

6. Apply the cv.tree() function to the training set in order to determine the optimal tree size.

set.seed(997)

cv.oj <- cv.tree(tree.OJ, FUN = prune.misclass)

cv.oj

$size
[1] 8 7 5 2 1

$dev
[1] 158 158 173 179 319

$k
[1]       -Inf   0.000000   5.000000   9.333333 147.000000

$method
[1] "misclass"

attr(,"class")
[1] "prune"         "tree.sequence"

7. Produce a plot with tree size on the x-axis and cross-validated classification error rate on the y-axis.

plot(cv.oj$size, cv.oj$dev, type = "b",
     xlab = "Tree Size", ylab = "Cross-Validated Error",
     main = "Optimal Tree Size")

8. Which tree size corresponds to the lowest cross-validated classification error rate?

optimal_size <- cv.oj$size[which.min(cv.oj$dev)]

cat("Optimal Tree Size:", optimal_size, "\n")

Optimal Tree Size: 8 

1. Produce a pruned tree corresponding to the optimal tree size obtained using cross-validation. If cross-validation does not lead to selection of a pruned tree, then create a pruned tree with five terminal nodes.

prune.oj7 <- prune.misclass(tree.OJ, best = 7)

plot(prune.oj7)

text(prune.oj7, pretty = 0)

prune.oj8 <- prune.misclass(tree.OJ, best = 8)

plot(prune.oj8)

text(prune.oj8, pretty = 0)

prune.oj5 <- prune.misclass(tree.OJ, best = 5)

plot(prune.oj5)

text(prune.oj5, pretty = 0)

10. Compare the training error rates between the pruned and unpruned trees. Which is higher? the pruned tree to 5 nodes is slightly higher by 0.0125

summary(prune.oj8)

Classification tree:
tree(formula = Purchase ~ ., data = train)
Variables actually used in tree construction:
[1] "LoyalCH"     "SalePriceMM" "PriceDiff"  
Number of terminal nodes:  8 
Residual mean deviance:  0.7751 = 613.9 / 792 
Misclassification error rate: 0.1675 = 134 / 800 

summary(prune.oj5)

Classification tree:
snip.tree(tree = tree.OJ, nodes = 3L)
Variables actually used in tree construction:
[1] "LoyalCH"     "SalePriceMM"
Number of terminal nodes:  5 
Residual mean deviance:  0.8808 = 700.2 / 795 
Misclassification error rate: 0.18 = 144 / 800 

11. Compare the test error rates between the pruned and unpruned trees. Which is higher? The misclassification rate with the unpruned tree is slightly higher by 0.0074074 -repeated-

prune.oj.preds = predict(prune.oj5, newdata = test, type = "class")

confusion2 <- table(prune.oj.preds, test$Purchase)

confusion2

              
prune.oj.preds  CH  MM
            CH 153  32
            MM  19  66

misclassification_rate2 <- 1 - sum(diag(confusion2)) / sum(confusion2)

print(misclassification_rate2)

[1] 0.1888889

print(misclassification_rate)

[1] 0.1962963