Homework3
Brittany Blankenship
2023-02-19
- This question should be answered using the Weekly data set, which is part of the ISLR package. This data contains 1,089 weekly stock 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?
library(ISLR)
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
##
##
##
## pairs(Weekly)
cor(subset(Weekly, select = -Direction))## 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 1.000000000Baseup upon the visuals above, there appears to be a relationship between Year and Volume. There seems to be no other patterns in the data.
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?
logit_model <- glm(Direction ~ Lag1 + Lag2 + Lag3 + Lag4 + Lag5 + Volume,
data = Weekly,
family = binomial)
summary(logit_model)##
## 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: 4Based upond the outcome above there appears to be a statistical signifigance with a p-value of less than 0.05 in “Lag2”.
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.
show_model_performance <- function(predicted_status, observed_status) {
confusion_matrix <- table(predicted_status,
observed_status,
dnn = c("Predicted Status", "Observed Status"))
print(confusion_matrix)
error_rate <- mean(predicted_status != observed_status)
cat("\n") # \n means newline so it just prints a blank line
cat(" Error Rate:", 100 * error_rate, "%\n")
cat("Correctly Predicted:", 100 * (1-error_rate), "%\n")
cat("False Positive Rate:", 100 * confusion_matrix[2,1] / sum(confusion_matrix[,1]), "%\n")
cat("False Negative Rate:", 100 * confusion_matrix[1,2] / sum(confusion_matrix[,2]), "%\n")
}
# get prediction as Up/Down direction - only needed for GLM models
predict_glm_direction <- function(model, newdata = NULL) {
predictions <- predict(model, newdata, type="response")
return(as.factor(ifelse(predictions < 0.5, "Down", "Up")))
}predicted_direction <- predict_glm_direction(logit_model)
show_model_performance(predicted_direction, Weekly$Direction)## Observed Status
## Predicted Status Down Up
## Down 54 48
## Up 430 557
##
## Error Rate: 43.89348 %
## Correctly Predicted: 56.10652 %
## False Positive Rate: 88.84298 %
## False Negative Rate: 7.933884 %D.) Now fit the logistic regression model using a training data period from 1990 to 2008, with Lag2 as the only predictor. Compute the confusion matrix and the overall fraction of correct predictions for the held out data (that is, the data from 2009 and 2010).
train <- (Weekly$Year < 2009)
train_set <- Weekly[train, ]
test_set <- Weekly[!train, ]
logit_model <- glm(Direction ~ Lag2, data = Weekly, family = binomial, subset = train)
predicted_direction <- predict_glm_direction(logit_model, test_set)
show_model_performance(predicted_direction, test_set$Direction)## Observed Status
## Predicted Status Down Up
## Down 9 5
## Up 34 56
##
## Error Rate: 37.5 %
## Correctly Predicted: 62.5 %
## False Positive Rate: 79.06977 %
## False Negative Rate: 8.196721 %E.) Repeat (d) using LDA.
library(MASS)
lda_model <- lda(Direction ~ Lag2, data = Weekly, subset = train)
predictions <- predict(lda_model, test_set, type="response")
show_model_performance(predictions$class, test_set$Direction)## Observed Status
## Predicted Status Down Up
## Down 9 5
## Up 34 56
##
## Error Rate: 37.5 %
## Correctly Predicted: 62.5 %
## False Positive Rate: 79.06977 %
## False Negative Rate: 8.196721 %F.) Repeat (d) using QDA
qda_model <- qda(Direction ~ Lag2, data = Weekly, subset = train)
predictions <- predict(qda_model, test_set, type="response")
show_model_performance(predictions$class, test_set$Direction)## Observed Status
## Predicted Status Down Up
## Down 0 0
## Up 43 61
##
## Error Rate: 41.34615 %
## Correctly Predicted: 58.65385 %
## False Positive Rate: 100 %
## False Negative Rate: 0 %G.) Repeat (d) using KNN with K = 1.
library(class)
run_knn <- function(train, test, train_class, test_class, k) {
set.seed(12345)
predictions <- knn(train, test, train_class, k)
cat("KNN: k =", k, "\n")
show_model_performance(predictions, test_class)
}
train_matrix <- as.matrix(train_set$Lag2)
test_matrix <- as.matrix(test_set$Lag2)
run_knn(train_matrix, test_matrix, train_set$Direction, test_set$Direction, k = 1)## KNN: k = 1
## Observed Status
## Predicted Status Down Up
## Down 21 29
## Up 22 32
##
## Error Rate: 49.03846 %
## Correctly Predicted: 50.96154 %
## False Positive Rate: 51.16279 %
## False Negative Rate: 47.54098 %H.) Which of these methods appears to provide the best results on this data? The methods that appear to provide the best results would be LDA and Logistic regression based upon the above data/outsomes.
I.) Experiment with different combinations of predictors, including possible transformations and interactions for each of the methods. Report the variables, method, and associated confusion matrix that appears to provide the best results on the held out data. Note that you should also experiment with values for K in the KNN classifier.
#interaction of Lag2
logit_model <- glm(Direction ~ Lag1 * Lag2, data = Weekly, family = binomial, subset = train)
predicted_direction <- predict_glm_direction(logit_model, test_set)
show_model_performance(predicted_direction, test_set$Direction)## Observed Status
## Predicted Status Down Up
## Down 7 8
## Up 36 53
##
## Error Rate: 42.30769 %
## Correctly Predicted: 57.69231 %
## False Positive Rate: 83.72093 %
## False Negative Rate: 13.11475 %#interaction of LDA between Lag1 * Lag2
lda_model <- lda(Direction ~ Lag1 * Lag2, data = Weekly, subset = train)
predictions <- predict(lda_model, test_set, type="response")
show_model_performance(predictions$class, test_set$Direction)## Observed Status
## Predicted Status Down Up
## Down 7 8
## Up 36 53
##
## Error Rate: 42.30769 %
## Correctly Predicted: 57.69231 %
## False Positive Rate: 83.72093 %
## False Negative Rate: 13.11475 %#the QDA model with the sqrt(abs(lag2))
qda_model <- qda(Direction ~ Lag2 + sqrt(abs(Lag2)), data = Weekly, subset = train)
predictions <- predict(qda_model, test_set, type="response")
show_model_performance(predictions$class, test_set$Direction)## Observed Status
## Predicted Status Down Up
## Down 12 13
## Up 31 48
##
## Error Rate: 42.30769 %
## Correctly Predicted: 57.69231 %
## False Positive Rate: 72.09302 %
## False Negative Rate: 21.31148 %# The KNN with k =10
run_knn(train_matrix, test_matrix, train_set$Direction, test_set$Direction, k = 10)## KNN: k = 10
## Observed Status
## Predicted Status Down Up
## Down 18 21
## Up 25 40
##
## Error Rate: 44.23077 %
## Correctly Predicted: 55.76923 %
## False Positive Rate: 58.13953 %
## False Negative Rate: 34.42623 %# The KNN with k = 100
run_knn(train_matrix, test_matrix, train_set$Direction, test_set$Direction, k = 100)## KNN: k = 100
## Observed Status
## Predicted Status Down Up
## Down 10 13
## Up 33 48
##
## Error Rate: 44.23077 %
## Correctly Predicted: 55.76923 %
## False Positive Rate: 76.74419 %
## False Negative Rate: 21.31148 %By looking at the above computations, LDA and Logistic regression appear to have a better outcome/preformance based upon test error rate.
- In this problem, you will develop a model to predict whether a given car gets high or low gas mileage based on the Auto dataset from the ISLR package.
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. We can use ifelse() to create a variable with values of 0 or 1, or we could just use the logical TRUE/FALSE which is even easier to create.
Auto$mpg01 <- Auto$mpg > median(Auto$mpg)head(Auto$mpg01, n = 20)## [1] FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE
## [13] FALSE FALSE TRUE FALSE FALSE FALSE TRUE TRUEB.) 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.
cor(subset(Auto, select = -name))## 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 1.0000000pairs(Auto)
Based upon the above graphics, displacement,cylinders,weight and
horsepower appear to be likely used in the prediction of mpg01
C.) Split the data into a training set and a test set.
train <- sample(nrow(Auto) * 0.7)
train_set <- Auto[train, ]
test_set <- Auto[-train, ]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?
lda_model <- lda(mpg01 ~ cylinders + weight + displacement + horsepower,
data = Auto,
subset = train)
predictions <- predict(lda_model, test_set)
show_model_performance(predictions$class, test_set$mpg01)## Observed Status
## Predicted Status FALSE TRUE
## FALSE 18 16
## TRUE 2 82
##
## Error Rate: 15.25424 %
## Correctly Predicted: 84.74576 %
## False Positive Rate: 10 %
## False Negative Rate: 16.32653 %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 + weight + displacement + horsepower,
data = Auto,
subset = train)
predictions <- predict(qda_model, test_set)
show_model_performance(predictions$class, test_set$mpg01)## Observed Status
## Predicted Status FALSE TRUE
## FALSE 18 19
## TRUE 2 79
##
## Error Rate: 17.79661 %
## Correctly Predicted: 82.20339 %
## False Positive Rate: 10 %
## False Negative Rate: 19.38776 %F.) 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?
logit_model <- glm(mpg01 ~ cylinders + weight + displacement + horsepower,
data = Auto,
family = binomial,
subset = train)
predictions <- predict(logit_model, test_set, type = "response")
show_model_performance(predictions > 0.5, test_set$mpg01)## Observed Status
## Predicted Status FALSE TRUE
## FALSE 20 21
## TRUE 0 77
##
## Error Rate: 17.79661 %
## Correctly Predicted: 82.20339 %
## False Positive Rate: 0 %
## False Negative Rate: 21.42857 %G.) Perform KNN on the training data, with several values of K, in order to predict mpg01. Use only the variables that seemed most associated with mpg01 in (b). What test errors do you obtain? Which value of K seems to perform the best on this data set?
vars <- c("cylinders", "weight", "displacement", "horsepower")
train_matrix <- as.matrix(train_set[, vars])
test_matrix <- as.matrix(test_set[, vars])
predictions <- knn(train_matrix, test_matrix, train_set$mpg01, 1)
run_knn(train_matrix, test_matrix, train_set$mpg01, test_set$mpg01, k = 1)## KNN: k = 1
## Observed Status
## Predicted Status FALSE TRUE
## FALSE 18 24
## TRUE 2 74
##
## Error Rate: 22.0339 %
## Correctly Predicted: 77.9661 %
## False Positive Rate: 10 %
## False Negative Rate: 24.4898 %run_knn(train_matrix, test_matrix, train_set$mpg01, test_set$mpg01, k = 10)## KNN: k = 10
## Observed Status
## Predicted Status FALSE TRUE
## FALSE 20 22
## TRUE 0 76
##
## Error Rate: 18.64407 %
## Correctly Predicted: 81.35593 %
## False Positive Rate: 0 %
## False Negative Rate: 22.44898 %run_knn(train_matrix, test_matrix, train_set$mpg01, test_set$mpg01, k = 100)## KNN: k = 100
## Observed Status
## Predicted Status FALSE TRUE
## FALSE 20 25
## TRUE 0 73
##
## Error Rate: 21.18644 %
## Correctly Predicted: 78.81356 %
## False Positive Rate: 0 %
## False Negative Rate: 25.5102 %16.) 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 subsets of the predictors. Describe your findings.
summary(Boston)## crim zn indus chas
## Min. : 0.00632 Min. : 0.00 Min. : 0.46 Min. :0.00000
## 1st Qu.: 0.08205 1st Qu.: 0.00 1st Qu.: 5.19 1st Qu.:0.00000
## Median : 0.25651 Median : 0.00 Median : 9.69 Median :0.00000
## Mean : 3.61352 Mean : 11.36 Mean :11.14 Mean :0.06917
## 3rd Qu.: 3.67708 3rd Qu.: 12.50 3rd Qu.:18.10 3rd Qu.:0.00000
## Max. :88.97620 Max. :100.00 Max. :27.74 Max. :1.00000
## nox rm age dis
## Min. :0.3850 Min. :3.561 Min. : 2.90 Min. : 1.130
## 1st Qu.:0.4490 1st Qu.:5.886 1st Qu.: 45.02 1st Qu.: 2.100
## Median :0.5380 Median :6.208 Median : 77.50 Median : 3.207
## Mean :0.5547 Mean :6.285 Mean : 68.57 Mean : 3.795
## 3rd Qu.:0.6240 3rd Qu.:6.623 3rd Qu.: 94.08 3rd Qu.: 5.188
## Max. :0.8710 Max. :8.780 Max. :100.00 Max. :12.127
## rad tax ptratio black
## Min. : 1.000 Min. :187.0 Min. :12.60 Min. : 0.32
## 1st Qu.: 4.000 1st Qu.:279.0 1st Qu.:17.40 1st Qu.:375.38
## Median : 5.000 Median :330.0 Median :19.05 Median :391.44
## Mean : 9.549 Mean :408.2 Mean :18.46 Mean :356.67
## 3rd Qu.:24.000 3rd Qu.:666.0 3rd Qu.:20.20 3rd Qu.:396.23
## Max. :24.000 Max. :711.0 Max. :22.00 Max. :396.90
## lstat medv
## Min. : 1.73 Min. : 5.00
## 1st Qu.: 6.95 1st Qu.:17.02
## Median :11.36 Median :21.20
## Mean :12.65 Mean :22.53
## 3rd Qu.:16.95 3rd Qu.:25.00
## Max. :37.97 Max. :50.00attach(Boston)the code below will create a binary “crim” variable
crime01 <- rep(0, length(crim))
crime01[crim > median(crim)] <- 1
Boston= data.frame(Boston,crime01)the code below will be splitting the first half of the dataset
train = 1:(dim(Boston)[1]/2)
test = (dim(Boston)[1]/2 + 1):dim(Boston)[1]
Boston.train = Boston[train, ]
Boston.test = Boston[test, ]
crime01.test = crime01[test]Logistic regression below
set.seed(1)
Boston.fit <-glm(crime01~ indus+nox+age+dis+rad+tax, data=Boston.train,family=binomial)
Boston.probs = predict(Boston.fit, Boston.test, type = "response")
Boston.pred = rep(0, length(Boston.probs))
Boston.pred[Boston.probs > 0.5] = 1
table(Boston.pred, crime01.test)## crime01.test
## Boston.pred 0 1
## 0 75 8
## 1 15 155mean(Boston.pred != crime01.test)## [1] 0.09090909summary(Boston.fit)##
## Call:
## glm(formula = crime01 ~ indus + nox + age + dis + rad + tax,
## family = binomial, data = Boston.train)
##
## Deviance Residuals:
## Min 1Q Median 3Q Max
## -1.97810 -0.21406 -0.03454 0.47107 3.04502
##
## Coefficients:
## Estimate Std. Error z value Pr(>|z|)
## (Intercept) -42.214032 7.617440 -5.542 2.99e-08 ***
## indus -0.213126 0.073236 -2.910 0.00361 **
## nox 80.868029 16.066473 5.033 4.82e-07 ***
## age 0.003397 0.012032 0.282 0.77772
## dis 0.307145 0.190502 1.612 0.10690
## rad 0.847236 0.183767 4.610 4.02e-06 ***
## tax -0.013760 0.004956 -2.777 0.00549 **
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## (Dispersion parameter for binomial family taken to be 1)
##
## Null deviance: 329.37 on 252 degrees of freedom
## Residual deviance: 144.44 on 246 degrees of freedom
## AIC: 158.44
##
## Number of Fisher Scoring iterations: 8LDA below
Boston.ldafit <-lda(crime01~ indus+nox+age+dis+rad+tax, data=Boston.train,family=binomial)
Bostonlda.pred = predict(Boston.ldafit, Boston.test)
table(Bostonlda.pred$class, crime01.test)## crime01.test
## 0 1
## 0 81 18
## 1 9 145mean(Bostonlda.pred$class != crime01.test)## [1] 0.1067194KNN Below
#K=1
train.K=cbind(indus,nox,age,dis,rad,tax)[train,]
test.K=cbind(indus,nox,age,dis,rad,tax)[test,]
Bosknn.pred=knn(train.K, test.K, crime01.test, k=1)
table(Bosknn.pred,crime01.test)## crime01.test
## Bosknn.pred 0 1
## 0 31 155
## 1 59 8mean(Bosknn.pred !=crime01.test)## [1] 0.8458498#K=100
train.K=cbind(indus,nox,age,dis,rad,tax)[train,]
test.K=cbind(indus,nox,age,dis,rad,tax)[test,]
Bosknn.pred=knn(train.K, test.K, crime01.test, k=100)
table(Bosknn.pred,crime01.test)## crime01.test
## Bosknn.pred 0 1
## 0 21 6
## 1 69 157mean(Bosknn.pred !=crime01.test)## [1] 0.2964427Based upon the outcomes above,the classification method logistic regression appeared to have the lowest test error rate of 9.09%. the only variables that were statistically significant were inud, nox, rad, and tax. Each model consisted of the same variables which help determine the best association with crime01. k=1 had the highest error rate which made the model ineffective, as the k value increased the error rate did improve but was not as low as LDA or the logistic regression.