Q10 This question should be answered using the Weekly data set, which is part of the ISRL package. This data is similar in nature to 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)
data("Weekly")
dim(Weekly)
## [1] 1089    9
**(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  
##            
##            
##            
## 
cor(Weekly[,-9])
##               Year         Lag1        Lag2        Lag3         Lag4
## Year    1.00000000 -0.032289274 -0.03339001 -0.03000649 -0.031127923
## Lag1   -0.03228927  1.000000000 -0.07485305  0.05863568 -0.071273876
## Lag2   -0.03339001 -0.074853051  1.00000000 -0.07572091  0.058381535
## Lag3   -0.03000649  0.058635682 -0.07572091  1.00000000 -0.075395865
## Lag4   -0.03112792 -0.071273876  0.05838153 -0.07539587  1.000000000
## Lag5   -0.03051910 -0.008183096 -0.07249948  0.06065717 -0.075675027
## Volume  0.84194162 -0.064951313 -0.08551314 -0.06928771 -0.061074617
## Today  -0.03245989 -0.075031842  0.05916672 -0.07124364 -0.007825873
##                Lag5      Volume        Today
## Year   -0.030519101  0.84194162 -0.032459894
## Lag1   -0.008183096 -0.06495131 -0.075031842
## Lag2   -0.072499482 -0.08551314  0.059166717
## Lag3    0.060657175 -0.06928771 -0.071243639
## Lag4   -0.075675027 -0.06107462 -0.007825873
## Lag5    1.000000000 -0.05851741  0.011012698
## Volume -0.058517414  1.00000000 -0.033077783
## Today   0.011012698 -0.03307778  1.000000000
attach(Weekly)
plot(Volume)

The correlations between the “lag” variables and today’s returns are close to zero. The only substantial correlation is between “Year” and “Volume”. When we plot “Volume”, we see that it is increasing over time.

(b) Use the full data set to perform a logistic regressino 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?

fit.glm <- glm(Direction ~ Lag1 + Lag2 + Lag3 + Lag4 + Lag5 + Volume, data = Weekly, family = binomial)
summary(fit.glm)
## 
## 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

“Lag2” is the only predictor statistically significant as its p-value is less than 0.05.

(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.

probs <- predict(fit.glm, type = "response")
pred.glm <- rep("Down", length(probs))
pred.glm[probs > 0.5] <- "Up"
table(pred.glm, Direction)
##         Direction
## pred.glm Down  Up
##     Down   54  48
##     Up    430 557

The percentage of correct predictions on the training data is (54+557)/1089 wich is equal to 56.1065197%. In other words 43.8934803% is the training error rate, which is often overly optimistic. We could also say that for weeks when the market goes up, the model is right 92.0661157% of the time (557/(48+557)). For weeks when the market goes down, the model is right only 11.1570248% of the time (54/(54+430)).

(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).

train <- (Year < 2009)
Weekly.20092010 <- Weekly[!train, ]
Direction.20092010 <- Direction[!train]
fit.glm2 <- glm(Direction ~ Lag2, data = Weekly, family = binomial, subset = train)
summary(fit.glm2)
## 
## Call:
## glm(formula = Direction ~ Lag2, family = binomial, data = Weekly, 
##     subset = train)
## 
## Deviance Residuals: 
##    Min      1Q  Median      3Q     Max  
## -1.536  -1.264   1.021   1.091   1.368  
## 
## Coefficients:
##             Estimate Std. Error z value Pr(>|z|)   
## (Intercept)  0.20326    0.06428   3.162  0.00157 **
## Lag2         0.05810    0.02870   2.024  0.04298 * 
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## (Dispersion parameter for binomial family taken to be 1)
## 
##     Null deviance: 1354.7  on 984  degrees of freedom
## Residual deviance: 1350.5  on 983  degrees of freedom
## AIC: 1354.5
## 
## Number of Fisher Scoring iterations: 4
probs2 <- predict(fit.glm2, Weekly.20092010, type = "response")
pred.glm2 <- rep("Down", length(probs2))
pred.glm2[probs2 > 0.5] <- "Up"
table(pred.glm2, Direction.20092010)
##          Direction.20092010
## pred.glm2 Down Up
##      Down    9  5
##      Up     34 56

(e) Repeat (d) using LDA.

library(MASS)
fit.lda <- lda(Direction ~ Lag2, data = Weekly, subset = train)
fit.lda
## 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
pred.lda <- predict(fit.lda, Weekly.20092010)
table(pred.lda$class, Direction.20092010)
##       Direction.20092010
##        Down Up
##   Down    9  5
##   Up     34 56

In this case, we may conclude that the percentage of correct predictions on the test data is 62.5%. We could also say that for weeks when the market goes up, the model is right 91.8032787% of the time. For weeks when the market goes down, the model is right only 20.9302326% of the time. These results are very close to those obtained with the logistic regression model which is not surpising.

(f) Repeat (d) using QDA.

fit.qda <- qda(Direction ~ Lag2, data = Weekly, subset = train)
fit.qda
## 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
pred.qda <- predict(fit.qda, Weekly.20092010)
table(pred.qda$class, Direction.20092010)
##       Direction.20092010
##        Down Up
##   Down    0  0
##   Up     43 61

In this case, we may conclude that the percentage of correct predictions on the test data is 58.6538462%. We could say that for weeks when the market goes up, the model is right 100% of the time. For weeks when the market goes down, the model is right only 0% of the time. We may note, that QDA achieves a correctness of 58.6538462% even though the model chooses “Up” the whole time !

(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]
set.seed(1)
pred.knn <- knn(train.X, test.X, train.Direction, k = 1)
table(pred.knn, Direction.20092010)
##         Direction.20092010
## pred.knn Down Up
##     Down   21 30
##     Up     22 31

In this case, we may conclude that the percentage of correct predictions on the test data is 50%. In other words 50% is the test error rate. We could also say that for weeks when the market goes up, the model is right 50.8196721% of the time. For weeks when the market goes down, the model is right only 48.8372093% of the time.

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

If we compare the test error rates, we see that logistic regression and LDA have the minimum error rates, followed by QDA and KNN. The LDA model with Lag2 as its only predictor did the best. Furthermore it was the only model that beat the baseline of just guessing Direction == Test, which would give a score og 0.5865385

(i) Experiment with different combinations of predictors, includ- ing possible transformations and interactions, for each of the methods. Report the variables, method, and associated confu- sion 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.

fit.glm3 <- glm(Direction ~ Lag2:Lag1, data = Weekly, family = binomial, subset = train)
probs3 <- predict(fit.glm3, Weekly.20092010, type = "response")
pred.glm3 <- rep("Down", length(probs3))
pred.glm3[probs3 > 0.5] = "Up"
table(pred.glm3, Direction.20092010)
##          Direction.20092010
## pred.glm3 Down Up
##      Down    1  1
##      Up     42 60
mean(pred.glm3 == Direction.20092010)
## [1] 0.5865385
fit.lda2 <- lda(Direction ~ Lag2:Lag1, data = Weekly, subset = train)
pred.lda2 <- predict(fit.lda2, Weekly.20092010)
mean(pred.lda2$class == Direction.20092010)
## [1] 0.5769231
fit.qda2 <- qda(Direction ~ Lag2 + sqrt(abs(Lag2)), data = Weekly, subset = train)
pred.qda2 <- predict(fit.qda2, Weekly.20092010)
table(pred.qda2$class, Direction.20092010)
##       Direction.20092010
##        Down Up
##   Down   12 13
##   Up     31 48
mean(pred.qda2$class == Direction.20092010)
## [1] 0.5769231
pred.knn2 <- knn(train.X, test.X, train.Direction, k = 10)
table(pred.knn2, Direction.20092010)
##          Direction.20092010
## pred.knn2 Down Up
##      Down   17 18
##      Up     26 43
mean(pred.knn2 == Direction.20092010)
## [1] 0.5769231
pred.knn3 <- knn(train.X, test.X, train.Direction, k = 100)
table(pred.knn3, Direction.20092010)
##          Direction.20092010
## pred.knn3 Down Up
##      Down    9 12
##      Up     34 49
mean(pred.knn3 == Direction.20092010)
## [1] 0.5576923

Out of these combinations, the original logistic regression and LDA have the best performance in terms of test error rates.

Q11 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. 

data("Auto")

(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.

attach(Auto)
mpg01 <- rep(0, length(mpg))
mpg01[mpg > median(mpg)] <- 1
Auto <- data.frame(Auto, mpg01)

(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 predictiong “mpg01” ? Scatterplots and boxplots may be useful tools to answer this question. Describe your findings.

cor(Auto[, -9])
##                     mpg  cylinders displacement horsepower     weight
## mpg           1.0000000 -0.7776175   -0.8051269 -0.7784268 -0.8322442
## cylinders    -0.7776175  1.0000000    0.9508233  0.8429834  0.8975273
## displacement -0.8051269  0.9508233    1.0000000  0.8972570  0.9329944
## horsepower   -0.7784268  0.8429834    0.8972570  1.0000000  0.8645377
## weight       -0.8322442  0.8975273    0.9329944  0.8645377  1.0000000
## acceleration  0.4233285 -0.5046834   -0.5438005 -0.6891955 -0.4168392
## year          0.5805410 -0.3456474   -0.3698552 -0.4163615 -0.3091199
## origin        0.5652088 -0.5689316   -0.6145351 -0.4551715 -0.5850054
## mpg01         0.8369392 -0.7591939   -0.7534766 -0.6670526 -0.7577566
##              acceleration       year     origin      mpg01
## mpg             0.4233285  0.5805410  0.5652088  0.8369392
## cylinders      -0.5046834 -0.3456474 -0.5689316 -0.7591939
## displacement   -0.5438005 -0.3698552 -0.6145351 -0.7534766
## horsepower     -0.6891955 -0.4163615 -0.4551715 -0.6670526
## weight         -0.4168392 -0.3091199 -0.5850054 -0.7577566
## acceleration    1.0000000  0.2903161  0.2127458  0.3468215
## year            0.2903161  1.0000000  0.1815277  0.4299042
## origin          0.2127458  0.1815277  1.0000000  0.5136984
## mpg01           0.3468215  0.4299042  0.5136984  1.0000000
boxplot(cylinders ~ mpg01, data = Auto, main = "Cylinders vs mpg01")

par(mfrow=c(2,3))
boxplot(cylinders ~ mpg01, data = Auto, main = "Cylinders vs mpg01")
boxplot(displacement ~ mpg01, data = Auto, main = "Displacement vs mpg01")
boxplot(horsepower ~ mpg01, data = Auto, main = "Horsepower vs mpg01")
boxplot(weight ~ mpg01, data = Auto, main = "Weight vs mpg01")
boxplot(acceleration ~ mpg01, data = Auto, main = "Acceleration vs mpg01")
boxplot(year ~ mpg01, data = Auto, main = "Year vs mpg01")

There appears to be some association between “mpg01” and “cylinders”, “weight”, “displacement” and “horsepower”.

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

set.seed(123)
train <- sample(1:dim(Auto)[1], dim(Auto)[1]*.7, rep=FALSE)
test <- -train
training_data<- Auto[train, ]
testing_data= Auto[test, ]
mpg01.test <- mpg01[test]

(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?

library(MASS)
lda_model <- lda(mpg01 ~ cylinders + weight + displacement + horsepower, data = training_data)
lda_model
## Call:
## lda(mpg01 ~ cylinders + weight + displacement + horsepower, data = training_data)
## 
## Prior probabilities of groups:
##         0         1 
## 0.4963504 0.5036496 
## 
## Group means:
##   cylinders   weight displacement horsepower
## 0  6.786765 3641.022     275.2941  130.96324
## 1  4.188406 2314.000     114.5290   78.00725
## 
## Coefficients of linear discriminants:
##                        LD1
## cylinders    -0.3974647924
## weight       -0.0009670704
## displacement -0.0029615583
## horsepower    0.0049004106
lda_pred = predict(lda_model, testing_data)
names(lda_pred)
## [1] "class"     "posterior" "x"
pred.lda <- predict(lda_model, testing_data)
table(pred.lda$class, mpg01.test)
##    mpg01.test
##      0  1
##   0 50  3
##   1 10 55
mean(pred.lda$class != mpg01.test)
## [1] 0.1101695

We conclude that we have a test error rate of 11.86%

(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_model = qda(mpg01 ~ cylinders + horsepower + weight + acceleration, data=training_data)
qda_model
## Call:
## qda(mpg01 ~ cylinders + horsepower + weight + acceleration, data = training_data)
## 
## Prior probabilities of groups:
##         0         1 
## 0.4963504 0.5036496 
## 
## Group means:
##   cylinders horsepower   weight acceleration
## 0  6.786765  130.96324 3641.022     14.55588
## 1  4.188406   78.00725 2314.000     16.55072
qda.class=predict(qda_model, testing_data)$class
table(qda.class, testing_data$mpg01)
##          
## qda.class  0  1
##         0 53  4
##         1  7 54
mean(qda.class != testing_data$mpg01)
## [1] 0.09322034

We may conclude that we have a test error rate of 0.093%.

(f) Perform logistic regression on the training data in order to pre- dict mpg01 using the variables that seemed most associated with mpg01 in (b). What is the test error of the model obtained?

glm_model <- glm(mpg01 ~ cylinders + weight + displacement + horsepower, data = training_data, family = binomial)
summary(glm_model)
## 
## Call:
## glm(formula = mpg01 ~ cylinders + weight + displacement + horsepower, 
##     family = binomial, data = training_data)
## 
## Deviance Residuals: 
##      Min        1Q    Median        3Q       Max  
## -2.44120  -0.17870   0.08712   0.31147   3.05303  
## 
## Coefficients:
##                Estimate Std. Error z value Pr(>|z|)    
## (Intercept)  11.8103006  2.0819718   5.673 1.41e-08 ***
## cylinders     0.1869071  0.3972245   0.471  0.63797    
## weight       -0.0020251  0.0008573  -2.362  0.01817 *  
## displacement -0.0164493  0.0095899  -1.715  0.08629 .  
## horsepower   -0.0443408  0.0172072  -2.577  0.00997 ** 
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## (Dispersion parameter for binomial family taken to be 1)
## 
##     Null deviance: 379.83  on 273  degrees of freedom
## Residual deviance: 138.27  on 269  degrees of freedom
## AIC: 148.27
## 
## Number of Fisher Scoring iterations: 7
probs <- predict(glm_model, testing_data, type = "response")
pred.glm <- rep(0, length(probs))
pred.glm[probs > 0.5] <- 1
table(pred.glm, mpg01.test)
##         mpg01.test
## pred.glm  0  1
##        0 53  6
##        1  7 52
mean(pred.glm != mpg01.test)
## [1] 0.1101695

Test error rate in logistic regression is 11.01695%.

(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?

str(Auto)
## 'data.frame':    392 obs. of  10 variables:
##  $ mpg         : num  18 15 18 16 17 15 14 14 14 15 ...
##  $ cylinders   : num  8 8 8 8 8 8 8 8 8 8 ...
##  $ displacement: num  307 350 318 304 302 429 454 440 455 390 ...
##  $ horsepower  : num  130 165 150 150 140 198 220 215 225 190 ...
##  $ weight      : num  3504 3693 3436 3433 3449 ...
##  $ acceleration: num  12 11.5 11 12 10.5 10 9 8.5 10 8.5 ...
##  $ year        : num  70 70 70 70 70 70 70 70 70 70 ...
##  $ origin      : num  1 1 1 1 1 1 1 1 1 1 ...
##  $ name        : Factor w/ 304 levels "amc ambassador brougham",..: 49 36 231 14 161 141 54 223 241 2 ...
##  $ mpg01       : num  0 0 0 0 0 0 0 0 0 0 ...
data = scale(Auto[,-c(9,10)])
set.seed(1234)
train <- sample(1:dim(Auto)[1], 392*.7, rep=FALSE)
#train <- sample(1:dim(Auto)[1], dim(Auto)[1]*.7, rep=FALSE)
test <- -train
training_data = data[train,c("cylinders","horsepower","weight","acceleration")]
testing_data = data[test, c("cylinders", "horsepower","weight","acceleration")]
## KNN take the training response variable seperately
train.mpg01 = Auto$mpg01[train]

## we also need the have the testing_y seperately for assesing the model later on
test.mpg01= Auto$mpg01[test]
library(class)
set.seed(1234)
knn_pred_y = knn(training_data, testing_data, train.mpg01, k = 1)
table(knn_pred_y, test.mpg01)
##           test.mpg01
## knn_pred_y  0  1
##          0 57  5
##          1  7 49
mean(knn_pred_y != test.mpg01)
## [1] 0.1016949

Test error rate is 8.47% for K=1

knn_pred_y = NULL
error_rate = NULL
for(i in 1:dim(testing_data)[1]){
set.seed(1234)
knn_pred_y = knn(training_data,testing_data,train.mpg01,k=i)
error_rate[i] = mean(test.mpg01 != knn_pred_y)
}

### find the minimum error rate
min_error_rate = min(error_rate)
print(min_error_rate)
## [1] 0.09322034
K = which(error_rate == min_error_rate)
print(K)
## [1] 4

When we train a KNN model with k=4, then we get the lowest misclassification error rate of 6.77%.

(Q13)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 sub- sets of the predictors. Describe your findings.

library(MASS)
attach(Boston)
crim01 <- rep(0, length(crim))
crim01[crim > median(crim)] <- 1
Boston <- data.frame(Boston, crim01)

train <- 1:(length(crim) / 2)
test <- (length(crim) / 2 + 1):length(crim)
Boston.train <- Boston[train, ]
Boston.test <- Boston[test, ]
crim01.test <- crim01[test]

fit.glm <- glm(crim01 ~ . - crim01 - crim, data = Boston, family = binomial, subset = train)
## Warning: glm.fit: fitted probabilities numerically 0 or 1 occurred
probs <- predict(fit.glm, Boston.test, type = "response")
pred.glm <- rep(0, length(probs))
pred.glm[probs > 0.5] <- 1
table(pred.glm, crim01.test)
##         crim01.test
## pred.glm   0   1
##        0  68  24
##        1  22 139
mean(pred.glm != crim01.test)
## [1] 0.1818182

We may conclude that, for this logistic regression, we have a test error rate of 18.1818182%.

fit.glm <- glm(crim01 ~ . - crim01 - crim - chas - nox, data = Boston, family = binomial, subset = train)
## Warning: glm.fit: fitted probabilities numerically 0 or 1 occurred
probs <- predict(fit.glm, Boston.test, type = "response")
pred.glm <- rep(0, length(probs))
pred.glm[probs > 0.5] <- 1
table(pred.glm, crim01.test)
##         crim01.test
## pred.glm   0   1
##        0  78  28
##        1  12 135
mean(pred.glm != crim01.test)
## [1] 0.1581028

We may conclude that, for this logistic regression, we have a test error rate of 15.8102767%.

fit.lda <- lda(crim01 ~ . - crim01 - crim, data = Boston, subset = train)
pred.lda <- predict(fit.lda, Boston.test)
table(pred.lda$class, crim01.test)
##    crim01.test
##       0   1
##   0  80  24
##   1  10 139
mean(pred.lda$class != crim01.test)
## [1] 0.1343874

We may conclude that, for this LDA, we have a test error rate of 15.0197628%.

train.X <- cbind(zn, indus, chas, nox, rm, age, dis, rad, tax, ptratio, black, lstat, medv)[train, ]
test.X <- cbind(zn, indus, chas, nox, rm, age, dis, rad, tax, ptratio, black, lstat, medv)[test, ]
train.crim01 <- crim01[train]
set.seed(1)
pred.knn <- knn(train.X, test.X, train.crim01, k = 1)
table(pred.knn, crim01.test)
##         crim01.test
## pred.knn   0   1
##        0  85 111
##        1   5  52
mean(pred.knn != crim01.test)
## [1] 0.458498

We may conclude that, for this KNN (k=1), we have a test error rate of 45.8498024%.

pred.knn <- knn(train.X, test.X, train.crim01, k = 10)
table(pred.knn, crim01.test)
##         crim01.test
## pred.knn   0   1
##        0  83  23
##        1   7 140
mean(pred.knn != crim01.test)
## [1] 0.1185771

We may conclude that, for this KNN (k=10), we have a test error rate of 11.8577075%

pred.knn <- knn(train.X, test.X, train.crim01, k = 100)
table(pred.knn, crim01.test)
##         crim01.test
## pred.knn   0   1
##        0  86 120
##        1   4  43
mean(pred.knn != crim01.test)
## [1] 0.4901186

We may conclude that, for this KNN (k=100), we have a test error rate of 49.0118577%.