Assignment 3
Austin Thrash
2024-02-26
Question 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.
13.(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
##
##
##
## 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 1.000000000Off the bat, we don’t notice any distinct patterns. However there does seem to be an exponential relationship between volume and years. We can see that as time does on, the volume only increases at an exponential rate. This can be further exemplified by taking a look at the correlation between these two variables(0.84194162).
13.(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?
glm.direct = glm(Direction ~ . - Year - Today, data = Weekly, family = "binomial")
summary(glm.direct)##
## Call:
## glm(formula = Direction ~ . - Year - Today, 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: 4Lag2 seems to be the only statistically significant predictor to the response variable Direction. Lag2 is the only predictor with a P-Value less than 0.05 (0.0296). Because of its significance, it might be advised to take a closer look at a models with just Lag2 as the predictor.
13.(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.
glm.direct.probs = predict(glm.direct, type = "response")
glm.direct.pred = rep("Down", 1089)
glm.direct.pred[glm.direct.probs > 0.5] = "Up"
table(glm.direct.pred, Weekly$Direction)##
## glm.direct.pred Down Up
## Down 54 48
## Up 430 557From the confusion matrix we can see that the model correctly predicted the direction for 611 weeks (54 + 557) out of the 1089, giving us an accuracy of 0.561065. Lets take a closer look at the confusion matrix:
- True Positive: 557/605 = 0.92
- False Positive: 430/484 = 0.89
While our true positive value (i.e. recall) is fairly high, the number of false positives is also high. This may not be a good model depending on how much loss could occur when incorrectly predicting a down week.
13.(d) Now fit the logistic regression model using a training data period from 1990 to 2008, with Lag2 as the only predictor. Compute the confusion matrix and the overall fraction of correct predictions for the held out data (that is, the data from 2009 and 2010).
train = (Weekly$Year < 2009)
glm.fit = glm(Direction ~ Lag2, data = Weekly, subset = train, family = "binomial")
summary(glm.fit)##
## Call:
## glm(formula = Direction ~ Lag2, family = "binomial", data = Weekly,
## subset = train)
##
## 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: 4glm.probs = predict(glm.fit, Weekly[!train, ], type = "response")
glm.pred = rep("Down", dim(Weekly[!train, ])[1])
glm.pred[glm.probs > 0.5] = "Up"
table(glm.pred, Weekly[!train, ]$Direction)##
## glm.pred Down Up
## Down 9 5
## Up 34 56- Accuracy: 65/104 = 0.625
- True Positive: 56/61 = 0.91
- False Positive: 34/43 = 0.79
13.(e) Repeat (d) using LDA.
lda.fit = lda(Direction ~ Lag2, data = Weekly, subset = train)
lda.fit## Call:
## lda(Direction ~ Lag2, data = Weekly, subset = train)
##
## Prior probabilities of groups:
## Down Up
## 0.4477157 0.5522843
##
## Group means:
## Lag2
## Down -0.03568254
## Up 0.26036581
##
## Coefficients of linear discriminants:
## LD1
## Lag2 0.4414162lda.pred = predict(lda.fit, Weekly[!train, ])
table(lda.pred$class, Weekly[!train, ]$Direction)##
## Down Up
## Down 9 5
## Up 34 56mean(lda.pred$class == Weekly[!train, ]$Direction)## [1] 0.625- Accuracy: 65/104 = 0.625
- True Positive: 56/61 = 0.91
- False Positive: 34/43 = 0.79
13.(f) Repeat (d) using QDA.
qda.fit = qda(Direction ~ Lag2, data = Weekly, subset = train)
qda.fit## Call:
## qda(Direction ~ Lag2, data = Weekly, subset = train)
##
## Prior probabilities of groups:
## Down Up
## 0.4477157 0.5522843
##
## Group means:
## Lag2
## Down -0.03568254
## Up 0.26036581qda.pred = predict(qda.fit, Weekly[!train, ])
table(qda.pred$class, Weekly[!train, ]$Direction)##
## Down Up
## Down 0 0
## Up 43 61mean(qda.pred$class == Weekly[!train, ]$Direction)## [1] 0.5865385- Accuracy: 65/104 = 0.625
- True Positive: 61/61 = 1
- False Positive: 43/43 = 1
13.(g) Repeat (d) using KNN with K = 1.
train.X = data.frame(Weekly[train, ]$Lag2)
test.X = data.frame(Weekly[!train, ]$Lag2)
train.Direction = Weekly[train, ]$Direction
set.seed(1)
knn.pred = knn(train.X, test.X, train.Direction, k = 1)
table(knn.pred, Weekly[!train, ]$Direction)##
## knn.pred Down Up
## Down 21 30
## Up 22 31mean(knn.pred == Weekly[!train, ]$Direction)## [1] 0.5- True Positive: 31/61 = 0.508
- False Positive: 22/43 = 0.512
13.(h) Repeat (d) using naive Bayes.
train.X <- data.frame(Weekly[train, ]$Lag2)
test.X <- data.frame(Weekly[!train, ]$Lag2)
train.Direction <- Weekly[train, ]$Direction
naive_model <- naiveBayes(train.X, train.Direction)
naive_pred <- predict(naive_model, test.X)## Warning in predict.naiveBayes(naive_model, test.X): Type mismatch between
## training and new data for variable 'Weekly.train....Lag2'. Did you use factors
## with numeric labels for training, and numeric values for new data?table(naive_pred, Weekly[!train, ]$Direction)##
## naive_pred Down Up
## Down 0 0
## Up 43 61accuracy <- mean(naive_pred == Weekly[!train, ]$Direction)- Accuracy: 65/104 = 0.625
- True Positive: 61/61 = 1
- False Positive: 43/43 = 1
13.(i) Which of these methods appears to provide the best results on this data?
It seems as if Logistic regression and LDA are the best models to use on this data. Both these models performed equally well when trained and tested on the data
13.(j) Experiment with different combinations of predictors, including possible transformations and interactions, for each of the methods. Report the variables, method, and associated confusion matrix that appears to provide the best results on the held out data. Note that you should also experiment with values for K in the KNN classifier.
weighted.lag.avg = 0.4*Weekly$Lag1 + 0.35*Weekly$Lag2 + 0.15*Weekly$Lag3 + 0.05*Weekly$Lag4 + 0.05*Weekly$Lag5
Weekly = data.frame(Weekly, weighted.lag.avg)
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 Down
## weighted.lag.avg
## 1 0.10055
## 2 0.20515
## 3 -1.12070
## 4 0.58290
## 5 1.15560
## 6 1.10520cor(Weekly$Today, Weekly$weighted.lag.avg)## [1] -0.03724141glm.fit = glm(Direction ~ weighted.lag.avg, data = Weekly, subset = train, family = "binomial")
summary(glm.fit)##
## Call:
## glm(formula = Direction ~ weighted.lag.avg, family = "binomial",
## data = Weekly, subset = train)
##
## Coefficients:
## Estimate Std. Error z value Pr(>|z|)
## (Intercept) 0.21349 0.06446 3.312 0.000926 ***
## weighted.lag.avg -0.02816 0.05347 -0.527 0.598508
## ---
## 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: 1354.4 on 983 degrees of freedom
## AIC: 1358.4
##
## Number of Fisher Scoring iterations: 3glm.probs = predict(glm.fit, Weekly[!train, ], type = "response")
glm.pred = rep("Down", dim(Weekly[!train, ])[1])
glm.pred[glm.probs > 0.5] = "Up"
table(glm.pred, Weekly[!train, ]$Direction)##
## glm.pred Down Up
## Up 43 61mean(glm.pred == Weekly[!train, ]$Direction)## [1] 0.5865385lda.fit = lda(Direction ~ weighted.lag.avg, data = Weekly, subset = train)
lda.pred = predict(lda.fit, Weekly[!train, ])
table(lda.pred$class, Weekly[!train, ]$Direction)##
## Down Up
## Down 0 0
## Up 43 61mean(lda.pred$class == Weekly[!train, ]$Direction)## [1] 0.5865385qda.fit = qda(Direction ~ weighted.lag.avg, data = Weekly, subset = train)
qda.pred = predict(qda.fit, Weekly[!train, ])
table(qda.pred$class, Weekly[!train, ]$Direction)##
## Down Up
## Down 0 0
## Up 43 61mean(qda.pred$class == Weekly[!train, ]$Direction)## [1] 0.5865385train.X = data.frame(Weekly[train, "weighted.lag.avg"])
test.X = data.frame(Weekly[!train, "weighted.lag.avg"])
train.Direction = Weekly[train, "Direction"]
set.seed(1)
knn.pred = knn(train.X, test.X, train.Direction, k = 1)
table(knn.pred, Weekly[!train, ]$Direction)##
## knn.pred Down Up
## Down 14 28
## Up 29 33mean(knn.pred == Weekly[!train, ]$Direction)## [1] 0.4519231Question 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.
14.(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.
Auto = na.omit(Auto)
# Creating factors of the origin variable
Auto$origin[Auto$origin == 1] = "American"
Auto$origin[Auto$origin == 2] = "European"
Auto$origin[Auto$origin == 3] = "Japanese"
Auto$origin = as.factor(Auto$origin)
# Creating binary variable
mpg01 = rep(0, dim(Auto)[1])
# If mpg is greater than median value, update value to 1
mpg01[Auto$mpg > median(Auto$mpg)] = 1
# Create new data frame with binary variable
Auto = data.frame(Auto, mpg01)
head(Auto)## mpg cylinders displacement horsepower weight acceleration year origin
## 1 18 8 307 130 3504 12.0 70 American
## 2 15 8 350 165 3693 11.5 70 American
## 3 18 8 318 150 3436 11.0 70 American
## 4 16 8 304 150 3433 12.0 70 American
## 5 17 8 302 140 3449 10.5 70 American
## 6 15 8 429 198 4341 10.0 70 American
## name mpg01
## 1 chevrolet chevelle malibu 0
## 2 buick skylark 320 0
## 3 plymouth satellite 0
## 4 amc rebel sst 0
## 5 ford torino 0
## 6 ford galaxie 500 014.(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 predicting mpg01? Scatterplots and boxplots may be useful tools to answer this question. Describe your findings.
par(mfrow = c(2, 3))
plot(factor(Auto$mpg01), Auto$cylinders, ylab = "cylinders")
plot(factor(Auto$mpg01), Auto$displacement, ylab = "displacement")
plot(factor(Auto$mpg01), Auto$horsepower, ylab = "horsepower")
plot(factor(Auto$mpg01), Auto$weight, ylab = "weight")
plot(factor(Auto$mpg01), Auto$acceleration, ylab = "acceleration")
plot(factor(Auto$mpg01), Auto$year, ylab = "year")
pairs(Auto)
Based on the box plots and scatter plots, horsepower, weight, displacement, and origin might be the best predictors to include in the model. All of these predictors appear to have clear clustering when compared against the response variable, mpg01.
14.(c) Split the data into a training set and a test set.
set.seed(47)
train = sample(dim(Auto)[1], size = 0.70*dim(Auto)[1])14.(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?
14.(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?
14.(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?
# Train logistic regression model
log.model <- glm(mpg01 ~ cylinders + displacement + horsepower + weight + year, subset = train, data = Auto, family = binomial)
summary(log.model)##
## Call:
## glm(formula = mpg01 ~ cylinders + displacement + horsepower +
## weight + year, family = binomial, data = Auto, subset = train)
##
## Coefficients:
## Estimate Std. Error z value Pr(>|z|)
## (Intercept) -13.864345 5.791219 -2.394 0.0167 *
## cylinders -0.203355 0.473310 -0.430 0.6675
## displacement 0.002807 0.011737 0.239 0.8110
## horsepower -0.038139 0.019361 -1.970 0.0489 *
## weight -0.004603 0.001171 -3.932 8.41e-05 ***
## year 0.408016 0.091124 4.478 7.55e-06 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## (Dispersion parameter for binomial family taken to be 1)
##
## Null deviance: 378.38 on 273 degrees of freedom
## Residual deviance: 106.36 on 268 degrees of freedom
## AIC: 118.36
##
## Number of Fisher Scoring iterations: 8