Assignment 3 STA 6543
AC Band
2025-07-09
13. This question should be answered using the Weekly data set, which is part of the ISLR2 package. This data is similar in nature to the Smarket data from this chapter’s lab, except that it contains 1, 089 weekly returns for 21 years, from the beginning of 1990 to the end of 2010. (a) Produce some numerical and graphical summaries of the Weekly data. Do there appear to be any patterns?
Yes, between Volume, Year, and Today
library(ISLR)
str(Weekly)## 'data.frame': 1089 obs. of 9 variables:
## $ Year : num 1990 1990 1990 1990 1990 1990 1990 1990 1990 1990 ...
## $ Lag1 : num 0.816 -0.27 -2.576 3.514 0.712 ...
## $ Lag2 : num 1.572 0.816 -0.27 -2.576 3.514 ...
## $ Lag3 : num -3.936 1.572 0.816 -0.27 -2.576 ...
## $ Lag4 : num -0.229 -3.936 1.572 0.816 -0.27 ...
## $ Lag5 : num -3.484 -0.229 -3.936 1.572 0.816 ...
## $ Volume : num 0.155 0.149 0.16 0.162 0.154 ...
## $ Today : num -0.27 -2.576 3.514 0.712 1.178 ...
## $ Direction: Factor w/ 2 levels "Down","Up": 1 1 2 2 2 1 2 2 2 1 ...install.packages("pacman")## Installing package into 'C:/Users/aliso/AppData/Local/R/win-library/4.5'
## (as 'lib' is unspecified)## package 'pacman' successfully unpacked and MD5 sums checked
##
## The downloaded binary packages are in
## C:\Users\aliso\AppData\Local\Temp\Rtmpya75V2\downloaded_packageslibrary(pacman)
p_load(ggplot2, dplyr, stringr, GGally, ggstats, DataExplorer, class, MASS, ISLR)
p_update(update = FALSE)## [1] "ggstats" "Rcpp"head(Weekly)## Year Lag1 Lag2 Lag3 Lag4 Lag5 Volume Today Direction
## 1 1990 0.816 1.572 -3.936 -0.229 -3.484 0.1549760 -0.270 Down
## 2 1990 -0.270 0.816 1.572 -3.936 -0.229 0.1485740 -2.576 Down
## 3 1990 -2.576 -0.270 0.816 1.572 -3.936 0.1598375 3.514 Up
## 4 1990 3.514 -2.576 -0.270 0.816 1.572 0.1616300 0.712 Up
## 5 1990 0.712 3.514 -2.576 -0.270 0.816 0.1537280 1.178 Up
## 6 1990 1.178 0.712 3.514 -2.576 -0.270 0.1544440 -1.372 Downplot(Weekly)
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
##
##
##
## create_report(Weekly)##
##
## processing file: report.rmd## | | | 0% | |. | 2% | |.. | 5% [global_options] | |... | 7% | |.... | 10% [introduce] | |.... | 12% | |..... | 14% [plot_intro]## | |...... | 17% | |....... | 19% [data_structure] | |........ | 21% | |......... | 24% [missing_profile]## | |.......... | 26% | |........... | 29% [univariate_distribution_header] | |........... | 31% | |............ | 33% [plot_histogram]## | |............. | 36% | |.............. | 38% [plot_density] | |............... | 40% | |................ | 43% [plot_frequency_bar]## | |................. | 45% | |.................. | 48% [plot_response_bar] | |.................. | 50% | |................... | 52% [plot_with_bar] | |.................... | 55% | |..................... | 57% [plot_normal_qq]## | |...................... | 60% | |....................... | 62% [plot_response_qq] | |........................ | 64% | |......................... | 67% [plot_by_qq] | |.......................... | 69% | |.......................... | 71% [correlation_analysis]## | |........................... | 74% | |............................ | 76% [principal_component_analysis]## | |............................. | 79% | |.............................. | 81% [bivariate_distribution_header] | |............................... | 83% | |................................ | 86% [plot_response_boxplot] | |................................. | 88% | |................................. | 90% [plot_by_boxplot] | |.................................. | 93% | |................................... | 95% [plot_response_scatterplot] | |.................................... | 98% | |.....................................| 100% [plot_by_scatterplot] ## output file: C:/Users/aliso/Documents/UTSA/Machine Learning 101/report.knit.md## "C:/Program Files/RStudio/resources/app/bin/quarto/bin/tools/pandoc" +RTS -K512m -RTS "C:\Users\aliso\DOCUME~1\UTSA\MACHIN~1\REPORT~1.MD" --to html4 --from markdown+autolink_bare_uris+tex_math_single_backslash --output pandoc6aac37a46c4e.html --lua-filter "C:\Users\aliso\AppData\Local\R\win-library\4.5\rmarkdown\rmarkdown\lua\pagebreak.lua" --lua-filter "C:\Users\aliso\AppData\Local\R\win-library\4.5\rmarkdown\rmarkdown\lua\latex-div.lua" --lua-filter "C:\Users\aliso\AppData\Local\R\win-library\4.5\rmarkdown\rmarkdown\lua\table-classes.lua" --embed-resources --standalone --variable bs3=TRUE --section-divs --table-of-contents --toc-depth 6 --template "C:\Users\aliso\AppData\Local\R\win-library\4.5\rmarkdown\rmd\h\default.html" --no-highlight --variable highlightjs=1 --variable theme=yeti --mathjax --variable "mathjax-url=https://mathjax.rstudio.com/latest/MathJax.js?config=TeX-AMS-MML_HTMLorMML" --include-in-header "C:\Users\aliso\AppData\Local\Temp\Rtmpya75V2\rmarkdown-str6aac23792d50.html"##
## Output created: report.htmlplot_missing(Weekly)
plot_histogram(Weekly)
plot_correlation(Weekly)
matrixThisIs <- cor(Weekly [, -9])
matrixThisIs## 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 1.000000000max_correlation <- max(matrixThisIs[upper.tri(matrixThisIs)])
max_correlation## [1] 0.8419416attach(Weekly)
plot(Volume, type = "l")
(b) Use the full data set to perform a logistic regression with Direction as the response and the five lag variables plus Volume as predictors. Use the summary function to print the results. Do any of the predictors appear to be statistically significant? If so, which ones?
logRegresWeek = glm(Direction ~ Lag1+Lag2+Lag3+Lag4+Lag5+Volume, data = Weekly, family = binomial)
summary(logRegresWeek)##
## Call:
## glm(formula = Direction ~ Lag1 + Lag2 + Lag3 + Lag4 + Lag5 +
## Volume, family = binomial, data = Weekly)
##
## 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: 4coef(logRegresWeek)## (Intercept) Lag1 Lag2 Lag3 Lag4 Lag5
## 0.26686414 -0.04126894 0.05844168 -0.01606114 -0.02779021 -0.01447206
## Volume
## -0.02274153**(b) cont. Do any of the predictors appear to be statistically significant? If so, which ones?
Lag2 has a positive correlation with direction.
- 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.**
logRegresWeek_probs <- predict(logRegresWeek, type = "response")
logRegresWeek_probs [1:10]## 1 2 3 4 5 6 7 8
## 0.6086249 0.6010314 0.5875699 0.4816416 0.6169013 0.5684190 0.5786097 0.5151972
## 9 10
## 0.5715200 0.5554287contrasts(Direction)## Up
## Down 0
## Up 1logRegresWeek_pred = rep("Down", 1089)
logRegresWeek_pred[logRegresWeek_probs > 0.5] = "Up"table(logRegresWeek_pred, Direction)## Direction
## logRegresWeek_pred Down Up
## Down 54 48
## Up 430 557(54+557)/1089## [1] 0.5610652mean(logRegresWeek_pred == Direction)## [1] 0.5610652So the current model predicts the next day’s volume 56% of the time.
- 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.2009 = Weekly[!train ,]
Direction.2009 = Direction[!train]
dim(Weekly.2009)## [1] 104 9logRegresWeek_L2 = glm(Direction ~ Lag2, data = Weekly, subset = train, family = binomial)
logRegresWeek_probs2 = predict(logRegresWeek_L2, Weekly.2009, type = "response")
logRegresWeek_pred2 = rep("Down", 104)
logRegresWeek_pred2[logRegresWeek_probs2 > 0.5] ="Up"
table(logRegresWeek_pred2, Direction.2009)## Direction.2009
## logRegresWeek_pred2 Down Up
## Down 9 5
## Up 34 56mean(logRegresWeek_pred2 == Direction.2009)## [1] 0.625(9+56)/104## [1] 0.625- Repeat (d) using LDA.
library(MASS)
lda.fit = lda(Direction ~ Lag2, data = Weekly, subset = train)
lda.fit## 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 0.4414162lda.pred = predict(lda.fit, Weekly.2009)
names(lda.pred)## [1] "class" "posterior" "x"lda.class = lda.pred$class
table(lda.class, Direction.2009)## Direction.2009
## lda.class Down Up
## Down 9 5
## Up 34 56mean(lda.class == Direction.2009)## [1] 0.625- Repeat (d) using QDA.
qda.fit = qda(Direction~Lag2, data = Weekly, subset = train)
qda.fit## Call:
## qda(Direction ~ Lag2, data = Weekly, subset = train)
##
## Prior probabilities of groups:
## Down Up
## 0.4477157 0.5522843
##
## Group means:
## Lag2
## Down -0.03568254
## Up 0.26036581qda.class = predict(qda.fit, Weekly.2009)$class
table(qda.class, Direction.2009)## Direction.2009
## qda.class Down Up
## Down 0 0
## Up 43 61mean(qda.class == Direction.2009)## [1] 0.5865385- Repeat (d) using KNN with K = 1.
train.X <- data.frame(Lag2 = Weekly$Lag2[train])
test.X <- data.frame(Lag2 = Weekly$Lag2[!train])
train.Direction = Direction[train]set.seed(997)
knn.pred <- knn(train.X, test.X, train.Direction, k =2)
table(knn.pred , Direction.2009)## Direction.2009
## knn.pred Down Up
## Down 21 28
## Up 22 33mean(knn.pred == Weekly$Direction[!train])## [1] 0.5192308set.seed(997)
knn.pred <- knn(train.X, test.X, train.Direction, k =3)
table(knn.pred , Direction.2009)## Direction.2009
## knn.pred Down Up
## Down 15 19
## Up 28 42mean(knn.pred == Weekly$Direction[!train])## [1] 0.5480769set.seed(997)
knn.pred <- knn(train.X, test.X, train.Direction, k =4)
table(knn.pred , Direction.2009)## Direction.2009
## knn.pred Down Up
## Down 21 21
## Up 22 40mean(knn.pred == Weekly$Direction[!train])## [1] 0.5865385- Repeat (d) using naive Bayes.
**(i) Which of these methods appears to provide the best results on this data?
Logistic Regression with Lag 2 only. And Linear Discriminant Analysis with Lag 2 only.**
- Experiment with different combinations of predictors, including possible transformations and interactions, for each of the methods. Report the variables, method, and associated confusion matrix that appears to provide the best results on the held out data. Note that you should also experiment with values for K in the KNN classifier.
p_load(MASS)
library(MASS)
lda.fit2 = lda(Direction ~ Lag1 + Lag2, data = Weekly, subset = train)
lda.fit2## Call:
## lda(Direction ~ Lag1 + Lag2, data = Weekly, subset = train)
##
## Prior probabilities of groups:
## Down Up
## 0.4477157 0.5522843
##
## Group means:
## Lag1 Lag2
## Down 0.289444444 -0.03568254
## Up -0.009213235 0.26036581
##
## Coefficients of linear discriminants:
## LD1
## Lag1 -0.3013148
## Lag2 0.2982579lda.pred2 = predict(lda.fit2, Weekly.2009)
names(lda.pred2)## [1] "class" "posterior" "x"lda.class2 = lda.pred2$class
table(lda.class2, Direction.2009)## Direction.2009
## lda.class2 Down Up
## Down 7 8
## Up 36 53mean(lda.class2 == Direction.2009)## [1] 0.5769231- 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.
- Create a binary variable, mpg01, that contains a 1 if mpg contains a value above its median, and a 0 if mpg contains a value below its median. You can compute the median using the median() function. Note you may find it helpful to use the data.frame() function to create a single data set containing both mpg01 and the other Auto variables.
library(MASS)
library(ISLR)
library(class)
library(dplyr)data("Auto")
head(Auto)## 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
## 1 chevrolet chevelle malibu
## 2 buick skylark 320
## 3 plymouth satellite
## 4 amc rebel sst
## 5 ford torino
## 6 ford galaxie 500data("Auto")med_mpg <-median(Auto$mpg, na.rm = TRUE)
med_mpg## [1] 22.75mpg_lvl = ifelse(Auto$mpg > med_mpg, 1, 0)
Auto$mpg_lvl <- as.factor(mpg_lvl)
table(mpg_lvl)## mpg_lvl
## 0 1
## 196 196- Explore the data graphically in order to investigate the association between mpg01 and the other features. Which of the other features seem most likely to be useful in predicting mpg01? Scatterplots and boxplots may be useful tools to answer this question. Describe your findings.
boxplot(weight ~ mpg_lvl, data = Auto, main = "Weight")
boxplot(horsepower ~ mpg_lvl, data = Auto, main = "Horsepower")
boxplot(year ~ mpg_lvl, data = Auto, main = "Year")
boxplot(cylinders ~ mpg_lvl, data = Auto, main = "Cylinders")
boxplot(displacement ~ mpg_lvl, data = Auto, main = "Displacement")
(c) Split the data into a training set and a test set.
set.seed(445)
train_idx <- sample(1:dim(Auto)[1], dim(Auto)[1] * 0.6, rep=FALSE)
testX <- -train_idx
trainX3 <- Auto[train_idx, ]
testX3 <- Auto[testX, ]
mpg_lvl_test <- mpg_lvl[testX]- Perform LDA on the training data in order to predict mpg01 using the variables that seemed most associated with mpg01 in (b). What is the test error of the model obtained?
lda.fit3 <- lda(mpg_lvl ~ weight + horsepower + year + displacement, data = trainX3)
lda.fit3## Call:
## lda(mpg_lvl ~ weight + horsepower + year + displacement, data = trainX3)
##
## Prior probabilities of groups:
## 0 1
## 0.5148936 0.4851064
##
## Group means:
## weight horsepower year displacement
## 0 3596.471 128.5785 74.21488 269.9091
## 1 2353.298 78.7807 77.64035 114.2588
##
## Coefficients of linear discriminants:
## LD1
## weight -0.0008853028
## horsepower 0.0140164551
## year 0.1220238325
## displacement -0.0105400125lda_pred = predict(lda.fit3, testX3)
names(lda_pred)## [1] "class" "posterior" "x"pred_lda3 <- predict(lda.fit3, testX3)
table(pred_lda3$class, mpg_lvl_test)## mpg_lvl_test
## 0 1
## 0 66 3
## 1 9 79mean(pred_lda3$class != mpg_lvl_test)## [1] 0.07643312Test error rate = 7.64%
- Perform QDA on the training data in order to predict mpg01 using the variables that seemed most associated with mpg01 in (b). What is the test error of the model obtained?
qda_model = qda(mpg_lvl ~ weight + horsepower + year + displacement, data = trainX3)
qda_model## Call:
## qda(mpg_lvl ~ weight + horsepower + year + displacement, data = trainX3)
##
## Prior probabilities of groups:
## 0 1
## 0.5148936 0.4851064
##
## Group means:
## weight horsepower year displacement
## 0 3596.471 128.5785 74.21488 269.9091
## 1 2353.298 78.7807 77.64035 114.2588qda.class = predict(qda_model, testX3)$class
table(qda.class,mpg_lvl_test)## mpg_lvl_test
## qda.class 0 1
## 0 70 5
## 1 5 77mean(qda.class != mpg_lvl_test)## [1] 0.06369427Test error rate = 6.3%
- Perform logistic regression on the training data in order to predict mpg01 using the variables that seemed most associated with mpg01 in (b). What is the test error of the model obtained?
glm_model <- glm(mpg_lvl ~ weight + horsepower + year + displacement, data = trainX3, family = binomial)
summary(glm_model)##
## Call:
## glm(formula = mpg_lvl ~ weight + horsepower + year + displacement,
## family = binomial, data = trainX3)
##
## Coefficients:
## Estimate Std. Error z value Pr(>|z|)
## (Intercept) -12.619209 6.014234 -2.098 0.03589 *
## weight -0.002180 0.001161 -1.878 0.06037 .
## horsepower -0.055636 0.021473 -2.591 0.00957 **
## year 0.355223 0.090132 3.941 8.11e-05 ***
## displacement -0.019679 0.008905 -2.210 0.02712 *
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## (Dispersion parameter for binomial family taken to be 1)
##
## Null deviance: 325.57 on 234 degrees of freedom
## Residual deviance: 100.18 on 230 degrees of freedom
## AIC: 110.18
##
## Number of Fisher Scoring iterations: 8probs <- predict(glm_model, testX3, type = "response")
pred.glm <- rep(0, length(probs))
pred.glm[probs > 0.65] <- 1
table(pred.glm, mpg_lvl_test)## mpg_lvl_test
## pred.glm 0 1
## 0 75 11
## 1 0 71mean(pred.glm != mpg_lvl_test)## [1] 0.07006369Test error rate is 7.0%
- 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 ...
## $ mpg_lvl : Factor w/ 2 levels "0","1": 1 1 1 1 1 1 1 1 1 1 ...trainX4 <- scale(trainX3[, c("weight","horsepower","year","displacement")])
testX4 <- scale(testX3[, c("weight","horsepower","year","displacement")])
trainY4 <- trainX3$mpg_lvl
set.seed(997)
knn.pred = knn(trainX4, testX4, trainY4, k = 5)
table(knn.pred, mpg_lvl_test)## mpg_lvl_test
## knn.pred 0 1
## 0 73 6
## 1 2 76mean(knn.pred != mpg_lvl_test)## [1] 0.05095541- Using the Boston data set, fit classification models in order to predict whether a given census tract has a crime rate above or below the median. Explore logistic regression, LDA, naive Bayes, and KNN models using various subsets of the predictors. Describe your findings. Hint: You will have to create the response variable yourself, using the variables that are contained in the Boston data set.
library(MASS)
data("Boston")
attach(Boston)
med_crime <-median(Boston$crim, na.rm = TRUE)
med_crime## [1] 0.25651crime_lvl = ifelse(Boston$crim > med_crime, 1, 0)
Boston$crime_lvl <- as.factor(crime_lvl)
table(crime_lvl)## crime_lvl
## 0 1
## 253 253set.seed(997)
train_idx <- sample(1:nrow(Boston), nrow(Boston)*0.6)
train <- Boston[train_idx,]
test <- Boston[-train_idx,]glm_Boston <-glm(crime_lvl ~ ., data = train, family = binomial)## Warning: glm.fit: algorithm did not converge## Warning: glm.fit: fitted probabilities numerically 0 or 1 occurredglm_probs <- predict(glm_Boston, test, type = "response")
glm_pred <- ifelse(glm_probs > 0.5, 1, 0)
table(Predicted = glm_pred, Actual = test$crime_lvl)## Actual
## Predicted 0 1
## 0 103 1
## 1 3 96glm_Boston <-glm(crime_lvl ~ zn + indus + dis + ptratio + rad + nox, data = train, family = binomial)
glm_probs <- predict(glm_Boston, test, type = "response")
glm_pred <- ifelse(glm_probs > 0.5, 1, 0)
table(Predicted = glm_pred, Actual = test$crime_lvl)## Actual
## Predicted 0 1
## 0 87 10
## 1 19 87mean(glm_pred != test$crime_lvl)## [1] 0.1428571lda_Boston <- lda(crime_lvl ~ zn + indus + dis + ptratio + rad + nox, data = train)
lda_pred <- predict(lda_Boston, test)
table(Predicted = lda_pred$class, Actual = test$crime_lvl)## Actual
## Predicted 0 1
## 0 96 22
## 1 10 75mean(lda_pred$class != test$crime_lvl)## [1] 0.1576355train.X <- scale(train[, c("zn", "indus", "dis", "ptratio", "rad", "nox")])
test.X <- scale(test[, c("zn", "indus", "dis", "ptratio", "rad", "nox")])
train.Y <- train$crime_lvl
for (k in c(1, 3, 5, 10)) {
knn.pred <- knn(train.X, test.X, train.Y, k = k)
err <- mean(knn.pred != test$crime_lvl)
cat("Test error for k =", k, ":", round(err, 3), "\n")
}## Test error for k = 1 : 0.039
## Test error for k = 3 : 0.039
## Test error for k = 5 : 0.074
## Test error for k = 10 : 0.079