Lab 5
jace vayhinger
2025-03-07
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.
data(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
##
##
##
## a. Produce some numerical and graphical summaries of the Weekly data. Do there appear to be any patterns?
pairs(Weekly)
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 1.000000000It appears that there is only coorelation between year and volume.
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?
log_model <- glm(Direction ~ Lag1 + Lag2 + Lag3 + Lag4 + Lag5 + Volume,
data = Weekly, family = binomial)
summary(log_model)##
## 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: 4Based on the summary, lag2 is the only predictor that is statistically significant.
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.
pred_probs <- predict(log_model, type = "response")
pred_labels <- ifelse(pred_probs > 0.5, "Up", "Down")
table(Predicted = pred_labels, Actual = Weekly$Direction)## Actual
## Predicted Down Up
## Down 54 48
## Up 430 557mean(pred_labels == Weekly$Direction)## [1] 0.5610652(557+54)/1089## [1] 0.5610652The model predicted the trend correctly 56.1% of the time.
557/(557+48)## [1] 0.9206612The model correctly predicted the trend going up 92.1% of the time.
(54)/(54+430)## [1] 0.1115702The model correctly predicted the trend going down only 11.2% of the time. As we can see the model performed great when predicting an upwards trend, but extremely poorly when predicting downwards trends. The model makes most of its mistakes when predicting downward trends.
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$Year < 2009
test <- !train
log_model_train <- glm(Direction ~ Lag2, data = Weekly, family = binomial, subset = train)
test_probs <- predict(log_model_train, Weekly[test, ], type = "response")
test_preds <- ifelse(test_probs > 0.5, "Up", "Down")
table(Predicted = test_preds, Actual = Weekly$Direction[test])## Actual
## Predicted Down Up
## Down 9 5
## Up 34 56mean(test_preds == Weekly$Direction[test])## [1] 0.625The model correctly predicted the outcome 62.5% of the time.
56/(56+5)## [1] 0.9180328The model predicted the upward trend correctly 91.8% of the time.
9/(9+34)## [1] 0.2093023The moel predicted the trend would go down correctly 20.9% of the time.
e. Repeat (d) using LDA.
library(MASS)##
## Attaching package: 'MASS'## The following object is masked from 'package:dplyr':
##
## select## The following object is masked from 'package:ISLR2':
##
## Bostonlda_model <- lda(Direction ~ Lag2, data = Weekly, subset = train)
lda_preds <- predict(lda_model, Weekly[test, ])
lda_class <- lda_preds$class
table(Predicted = lda_class, Actual = Weekly$Direction[test])## Actual
## Predicted Down Up
## Down 9 5
## Up 34 56mean(lda_class == Weekly$Direction[test])## [1] 0.625Using LDA gave the same results as the logistic regression model. 62.5% accuracy.
f. Repeat (d) using QDA.
qda_model <- qda(Direction ~ Lag2, data = Weekly, subset = train)
qda_preds <- predict(qda_model, Weekly[test, ])
qda_class <- qda_preds$class
table(Predicted = qda_class, Actual = Weekly$Direction[test])## Actual
## Predicted Down Up
## Down 0 0
## Up 43 61mean(qda_class == Weekly$Direction[test])## [1] 0.5865385Using QDA we got 58.7% accuracy. It looks like there were no predictions for downward trend.
g. Repeat (d) using KNN with K =1.
library(class)
train_X <- as.matrix(Weekly[train, "Lag2"])
test_X <- as.matrix(Weekly[test, "Lag2"])
train_Y <- Weekly$Direction[train]
knn_preds <- knn(train_X, test_X, train_Y, k = 1)
table(Predicted = knn_preds, Actual = Weekly$Direction[test])## Actual
## Predicted Down Up
## Down 21 30
## Up 22 31mean(knn_preds == Weekly$Direction[test])## [1] 0.5Using KNN with K = 1 the model got a 51% accuracy.
h. Repeat (d) using naive Bayes.
library(e1071)
nb_model <- naiveBayes(Direction ~ Lag2, data = Weekly, subset = train)
nb_preds <- predict(nb_model, Weekly[test, ])
table(Predicted = nb_preds, Actual = Weekly$Direction[test])## Actual
## Predicted Down Up
## Down 0 0
## Up 43 61mean(nb_preds == Weekly$Direction[test])## [1] 0.5865385The Naive Bayes got an accuracy of 58.7%. It had the same results as part f QDA model.
i. Which of these methods appears to provide the best results on this data?
The logistic regression model had the best accuracy at 62.5%. LDA had the same results as well.
j. (j) 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.
train <- Weekly$Year < 2009
test <- !train
train_data <- Weekly[train, ]
test_data <- Weekly[test, ]Logistic Regression Model:
log_model_interact <- glm(Direction ~ Lag1 * Lag2 + Volume, data = train_data, family = binomial)
log_probs_interact <- predict(log_model_interact, test_data, type = "response")
log_preds_interact <- ifelse(log_probs_interact > 0.5, "Up", "Down")
table(Predicted = log_preds_interact, Actual = test_data$Direction)## Actual
## Predicted Down Up
## Down 27 32
## Up 16 29mean(log_preds_interact == test_data$Direction)## [1] 0.5384615LDA with Multiple Predictors
lda_model_multi <- lda(Direction ~ Lag1 + Lag2 + Lag3 + Volume, data = train_data)
lda_preds_multi <- predict(lda_model_multi, test_data)$class
table(Predicted = lda_preds_multi, Actual = test_data$Direction)## Actual
## Predicted Down Up
## Down 30 37
## Up 13 24mean(lda_preds_multi == test_data$Direction)## [1] 0.5192308QDA with Polynomial Terms
qda_model_poly <- qda(Direction ~ poly(Lag2, 2) + Volume, data = train_data)
qda_preds_poly <- predict(qda_model_poly, test_data)$class
table(Predicted = qda_preds_poly, Actual = test_data$Direction)## Actual
## Predicted Down Up
## Down 31 42
## Up 12 19mean(qda_preds_poly == test_data$Direction)## [1] 0.4807692Naive Bayes with Different Predictors
nb_model_multi <- naiveBayes(Direction ~ Lag1 + Lag2 + Volume, data = train_data)
nb_preds_multi <- predict(nb_model_multi, test_data)
table(Predicted = nb_preds_multi, Actual = test_data$Direction)## Actual
## Predicted Down Up
## Down 41 58
## Up 2 3mean(nb_preds_multi == test_data$Direction)## [1] 0.4230769KNN with Different K Values
train_X <- as.matrix(train_data[, c("Lag1", "Lag2", "Volume")])
test_X <- as.matrix(test_data[, c("Lag1", "Lag2", "Volume")])
train_Y <- train_data$Direction
k_values <- c(1, 3, 5, 7, 10)
for (k in k_values) {
knn_preds <- knn(train_X, test_X, train_Y, k = k)
accuracy <- mean(knn_preds == test_data$Direction)
print(paste("K =", k, "Accuracy =", accuracy))
}## [1] "K = 1 Accuracy = 0.5"
## [1] "K = 3 Accuracy = 0.480769230769231"
## [1] "K = 5 Accuracy = 0.557692307692308"
## [1] "K = 7 Accuracy = 0.548076923076923"
## [1] "K = 10 Accuracy = 0.509615384615385"I think I would choose the KNN model with K = 10 because it had the best accuracy percentage.
14. In this problem, you will develop a model to predict whether a given car gets high or low gas mileage based on the Auto data set
a. Create a binary variable, mpg01, that contains a 1 if mpg contains a value above its median, and a 0 if mpg contains a value below its median. You can compute the median using the median() function. Note you may find it helpful to use the data.frame() function to create a single data set containing both mpg01 and the other Auto variables.
data(Auto)
Auto$mpg01 <- ifelse(Auto$mpg > median(Auto$mpg), 1, 0)
summary(Auto$mpg01)## Min. 1st Qu. Median Mean 3rd Qu. Max.
## 0.0 0.0 0.5 0.5 1.0 1.0b. 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(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 1.0000000pairs(Auto)
The features: displacement, horsepower, weight would be useful in predicting mpg01.
ggplot(Auto, aes(x = factor(mpg01), y = horsepower)) +
geom_boxplot() +
labs(title = "Horsepower vs MPG Class")
ggplot(Auto, aes(x = factor(mpg01), y = weight)) +
geom_boxplot() +
labs(title = "Weight vs MPG Class")
c. Split the data into a training set and a test set.
set.seed(1)
train_indices <- sample(1:nrow(Auto), nrow(Auto) * 0.7)
train_data <- Auto[train_indices, ]
test_data <- Auto[-train_indices, ]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_auto <- lda(mpg01 ~ horsepower + weight + displacement, data = train_data)
lda_preds <- predict(lda_auto, test_data)$class
table(Predicted = lda_preds, Actual = test_data$mpg01)## Actual
## Predicted 0 1
## 0 47 1
## 1 14 56mean(lda_preds == test_data$mpg01)## [1] 0.8728814accuracy <- mean(lda_preds == test_data$mpg01)
print(paste("Test Accuracy:", accuracy))## [1] "Test Accuracy: 0.872881355932203"test_error <- 1 - accuracy
print(paste("Test Error:", test_error))## [1] "Test Error: 0.127118644067797"The test error is 12.7%.
e. Perform QDA on the training data in order to predict mpg01 using the variables that seemed most associated with mpg01 in (b). What is the test error of the model obtained?
qda_auto <- qda(mpg01 ~ horsepower + weight + displacement, data = train_data)
qda_preds <- predict(qda_auto, test_data)$class
table(Predicted = qda_preds, Actual = test_data$mpg01)## Actual
## Predicted 0 1
## 0 51 3
## 1 10 54mean(qda_preds == test_data$mpg01)## [1] 0.8898305The test error is 11%.
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?
log_auto <- glm(mpg01 ~ horsepower + weight + displacement, data = train_data, family = binomial)
log_probs <- predict(log_auto, test_data, type = "response")
log_preds <- ifelse(log_probs > 0.5, 1, 0)
table(Predicted = log_preds, Actual = test_data$mpg01)## Actual
## Predicted 0 1
## 0 53 3
## 1 8 54mean(log_preds == test_data$mpg01)## [1] 0.9067797The test error is 9.3%.
g. Perform naive Bayes 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?
nb_auto <- naiveBayes(mpg01 ~ horsepower + weight + displacement, data = train_data)
nb_preds <- predict(nb_auto, test_data)
table(Predicted = nb_preds, Actual = test_data$mpg01)## Actual
## Predicted 0 1
## 0 50 2
## 1 11 55mean(nb_preds == test_data$mpg01)## [1] 0.8898305The test error is the same as QDA at 11%.
h. 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?
train_X <- as.matrix(train_data[, c("horsepower", "weight", "displacement")])
test_X <- as.matrix(test_data[, c("horsepower", "weight", "displacement")])
train_Y <- train_data$mpg01
k_values <- c(1, 3, 5, 7, 10)
for (k in k_values) {
knn_preds <- knn(train_X, test_X, train_Y, k = k)
accuracy <- mean(knn_preds == test_data$mpg01)
print(paste("K =", k, "Accuracy =", accuracy))
}## [1] "K = 1 Accuracy = 0.864406779661017"
## [1] "K = 3 Accuracy = 0.889830508474576"
## [1] "K = 5 Accuracy = 0.872881355932203"
## [1] "K = 7 Accuracy = 0.872881355932203"
## [1] "K = 10 Accuracy = 0.855932203389831"Test Errors: K = 1 - 13.6% K = 3 - 11% K = 5 - 12.7% K = 7 - 12.7% K = 10 - 14.4%
The KNN model that performs the best is the K = 3 model.