10. This question should be answered using the Weekly data set, which is part of the ISLR 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?

library(ISLR)
pairs(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  
##            
##            
##            
## 
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

Outside of Year and Volume, there doesn’t seem to be any obvious patterns. (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 <- glm(Direction ~ Lag1 + Lag2 + Lag3 + Lag4 + Lag5 + Volume,  data = Weekly,  family = "binomial")
summary(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

The predictor with the most significance is 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.

glm.pred <- rep("Down", 1089)
glm.probs=predict(glm,type="response")
glm.pred[glm.probs>.5]="Up"
table(glm.pred, Weekly$Direction)
##         
## glm.pred Down  Up
##     Down   54  48
##     Up    430 557

The confusion matrix is telling us that the models has (54+557)/1089 = .56 or a 56% accuracy rate, which is not too great. (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[Weekly$Year <= 2008, ]
test <- Weekly[Weekly$Year > 2008, ]
glm.lag2 <- glm(Direction ~ Lag2, data = train, family = "binomial")
glm.lag2.probs <- predict(glm.lag2, newdata = test, type="response")
glm.pred.lag2 <- rep("Down", 1089)
glm.pred.lag2[glm.lag2.probs>.5]="Up"
table(glm.pred.lag2, Weekly$Direction)
##              
## glm.pred.lag2 Down  Up
##          Down   63  85
##          Up    421 520

For this scenario, the fraction of correct predictions is \((63 + 520)/1089 = .53\) or 53% (e) Repeat (d) using LDA.

library(MASS)
train <- Weekly[Weekly$Year <= 2008, ]
test <- Weekly[Weekly$Year > 2008, ]
lda <- lda(Direction ~ Lag2, data = train)
lda.pred = predict(lda, newdata = test)
table(lda.pred$class, test$Direction)
##       
##        Down Up
##   Down    9  5
##   Up     34 56

11. 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.

Auto$mpg01 <- factor(as.numeric(Auto$mpg > median(Auto$mpg)))

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

rand_obs <- sample.int(nrow(Auto), nrow(Auto)*.7)
train_set <- Auto[rand_obs,]
test_set <- Auto[-rand_obs,]

(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? I decided on weight, horsepower, acceleration.

lda.fit=lda(mpg01~weight+acceleration+horsepower,data=train_set)
lda.fit
## Call:
## lda(mpg01 ~ weight + acceleration + horsepower, data = train_set)
## 
## Prior probabilities of groups:
##         0         1 
## 0.4927007 0.5072993 
## 
## Group means:
##     weight acceleration horsepower
## 0 3634.504     14.44370  132.17037
## 1 2372.619     16.45324   80.05036
## 
## Coefficients of linear discriminants:
##                       LD1
## weight       -0.001823084
## acceleration  0.057643547
## horsepower    0.003441345

(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.fit=qda(mpg01~weight+acceleration+horsepower,data=train_set)
qda.fit
## Call:
## qda(mpg01 ~ weight + acceleration + horsepower, data = train_set)
## 
## Prior probabilities of groups:
##         0         1 
## 0.4927007 0.5072993 
## 
## Group means:
##     weight acceleration horsepower
## 0 3634.504     14.44370  132.17037
## 1 2372.619     16.45324   80.05036

(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? 

glm.fit=glm(mpg01~weight+acceleration+horsepower,data=train_set, family = binomial())
summary(glm.fit)
## 
## Call:
## glm(formula = mpg01 ~ weight + acceleration + horsepower, family = binomial(), 
##     data = train_set)
## 
## Deviance Residuals: 
##     Min       1Q   Median       3Q      Max  
## -2.2796  -0.2380   0.1136   0.3904   2.8851  
## 
## Coefficients:
##                Estimate Std. Error z value Pr(>|z|)    
## (Intercept)  12.0550136  3.0063344   4.010 6.08e-05 ***
## weight       -0.0030545  0.0007067  -4.322 1.54e-05 ***
## acceleration  0.0284943  0.1358897   0.210   0.8339    
## horsepower   -0.0382763  0.0226309  -1.691   0.0908 .  
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## (Dispersion parameter for binomial family taken to be 1)
## 
##     Null deviance: 379.79  on 273  degrees of freedom
## Residual deviance: 160.57  on 270  degrees of freedom
## AIC: 168.57
## 
## Number of Fisher Scoring iterations: 7

(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? ### 13. 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. Change the Crime column into 1s and 0s based on if it’s above or below the median:

Boston$crim <- factor(ifelse(Boston$crim > median(Boston$crim), 1, 0))

Create a sets for training and test

rand_obs <- sample.int(nrow(Boston), nrow(Boston)*.7)
train_set <- Boston[rand_obs,]
test_set <- Boston[-rand_obs,]