HW3
Nathaniel Sattler
3/3/2021
Q10. 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?
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 ...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 1.000000000anyNA(Weekly)## [1] FALSEplot(Weekly)
multi.hist(Weekly[,2:8])
There is no missing data and all numeric entries are normal except volume, which is right skewed. Also Volume appears to be correlated to the year resulting in a logarithmic function between the the two.
(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.weekly.fit = glm(Direction ~ . - Year - Today, data = Weekly, family = binomial)
summary(glm.weekly.fit)##
## Call:
## glm(formula = Direction ~ . - Year - Today, 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: 4The only statistically significant predictor in this set is lag2, with a p-value of .0296, which is less than the .05 cut off.
(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.probs = predict(glm.weekly.fit, type = "response")
glm.preds = rep("Down", length(glm.probs))
glm.preds[glm.probs > 0.5] = "Up"
table(glm.preds, Weekly$Direction)##
## glm.preds Down Up
## Down 54 48
## Up 430 557mean(glm.preds == Weekly$Direction) #accuracy## [1] 0.5610652557/(557+430) #precision## [1] 0.5643364557/(557+48) #recall## [1] 0.920661254/(54+430) #false positive rate## [1] 0.1115702The confusion matrix shows that we have an accuracy score of .5610652, meaning that the logistic regression is correct 56.11% of the time. This only a slight improvement over random guessing. The model is very precise when it comes to predicting Up with a Recall score of 92.07%. When it comes to properly classifying Down though, the model is only correct 11.16% of the time. Also the Precision score is .5643364 which means that out of all the Up predictions, about 56.43% were correct.
(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 = (Year < 2009)
weekly.train = Weekly[train,]
weekly.test = Weekly[!train,]
glm.weekly.train = glm(Direction~Lag2, data=Weekly,family=binomial, subset=train)
summary(glm.weekly.train)##
## Call:
## glm(formula = Direction ~ Lag2, family = binomial, data = Weekly,
## subset = train)
##
## Deviance Residuals:
## Min 1Q Median 3Q Max
## -1.536 -1.264 1.021 1.091 1.368
##
## 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.probsd = predict(glm.weekly.train, weekly.test, type = "response")
glm.predsd = rep("Down", length(glm.probsd))
glm.predsd[glm.probsd > 0.5] = "Up"
table(glm.predsd, weekly.test$Direction)##
## glm.predsd Down Up
## Down 9 5
## Up 34 56mean(glm.predsd == weekly.test$Direction) #accuracy## [1] 0.625The model has a 62.5% accuracy score, meaning that using lag 2 as the sole predictor of Direction and then training it on a large portion of the data nets this score for the test section of the data. This is another small improvement over the previous model and better than random guessing.
(e) Repeat (d) using LDA.
direction.test = Direction[!train]
lda.weekly.fit = lda(Direction~Lag2, data=Weekly, subset=train)
plot(lda.weekly.fit)
lda.pred = predict(lda.weekly.fit, weekly.test)
lda.class=lda.pred$class
table(lda.class, direction.test)## direction.test
## lda.class Down Up
## Down 9 5
## Up 34 56mean(lda.class==direction.test)## [1] 0.625The LDA model returns results that match those from the logistic model of part (d).
(f) Repeat (d) using QDA.
qda.weekly.fit = qda(Direction~Lag2, data=Weekly, subset=train)
qda.weekly.fit## Call:
## qda(Direction ~ Lag2, data = Weekly, subset = train)
##
## Prior probabilities of groups:
## Down Up
## 0.4477157 0.5522843
##
## Group means:
## Lag2
## Down -0.03568254
## Up 0.26036581qda.pred = predict(qda.weekly.fit, weekly.test)
qda.class=qda.pred$class
table(qda.class, direction.test)## direction.test
## qda.class Down Up
## Down 0 0
## Up 43 61mean(qda.class==direction.test)## [1] 0.5865385The model’s accuracy has decrease from the .625 of LDA and logistic regression to only .5865385. Also the model only predicted up in its results, similar to random guessing.
(g) Repeat (d) using KNN with K = 1.
train.lag2 = as.matrix(Lag2[train])
test.lag2 = as.matrix(Lag2[!train])
direction.train = Direction[train]
dim(train.lag2)## [1] 985 1dim(test.lag2)## [1] 104 1set.seed(1)
knn.pred=knn(train.lag2, test.lag2, direction.train, k=1)
table(knn.pred,direction.test)## direction.test
## knn.pred Down Up
## Down 21 30
## Up 22 31mean(knn.pred == direction.test)## [1] 0.5The model results in a 50% accuracy score, meaning that model acts like random guessing.
(h) Which of these methods appears to provide the best results on this data?
Logistic Regression and LDA as they both have 62.5% model accuracy for this scenario.
(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 lag1 and lag2 with all terms
glm.weekly.int <-glm(Direction~. - Today -Year + Lag1:Lag2, data=Weekly,family=binomial, subset=train)
summary(glm.weekly.int)##
## Call:
## glm(formula = Direction ~ . - Today - Year + Lag1:Lag2, family = binomial,
## data = Weekly, subset = train)
##
## Deviance Residuals:
## Min 1Q Median 3Q Max
## -1.650 -1.249 0.977 1.084 1.567
##
## Coefficients:
## Estimate Std. Error z value Pr(>|z|)
## (Intercept) 0.334016 0.094312 3.542 0.000398 ***
## Lag1 -0.057932 0.031247 -1.854 0.063736 .
## Lag2 0.044264 0.029978 1.477 0.139802
## Lag3 -0.016404 0.029600 -0.554 0.579463
## Lag4 -0.032939 0.029582 -1.113 0.265496
## Lag5 -0.037754 0.029246 -1.291 0.196729
## Volume -0.090234 0.054202 -1.665 0.095960 .
## Lag1:Lag2 0.003122 0.007615 0.410 0.681791
## ---
## 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: 1342.2 on 977 degrees of freedom
## AIC: 1358.2
##
## Number of Fisher Scoring iterations: 4glm.probsi = predict(glm.weekly.train, weekly.test, type = "response")
glm.predsi = rep("Down", length(glm.probsi))
glm.predsi[glm.probsi > 0.5] = "Up"
table(glm.predsi, weekly.test$Direction)##
## glm.predsi Down Up
## Down 9 5
## Up 34 56mean(glm.predsi == weekly.test$Direction) #accuracy## [1] 0.625This model returns the same accuracy score as the best models from before, but also none of the terms are significant within a .05 margin. Lag2 and Volume come close if we were using a .10 cutoff they would be significant.
glm.weekly.int <-glm(Direction~. - Today -Year - Volume + log(Volume), data=Weekly,family=binomial, subset=train)
summary(glm.weekly.int)##
## Call:
## glm(formula = Direction ~ . - Today - Year - Volume + log(Volume),
## family = binomial, data = Weekly, subset = train)
##
## Deviance Residuals:
## Min 1Q Median 3Q Max
## -1.7794 -1.2475 0.9776 1.0881 1.5187
##
## Coefficients:
## Estimate Std. Error z value Pr(>|z|)
## (Intercept) 0.19325 0.06731 2.871 0.00409 **
## Lag1 -0.05918 0.02913 -2.031 0.04222 *
## Lag2 0.04837 0.02950 1.639 0.10111
## Lag3 -0.01267 0.02931 -0.432 0.66547
## Lag4 -0.02827 0.02903 -0.974 0.33011
## Lag5 -0.03497 0.02907 -1.203 0.22889
## log(Volume) -0.10227 0.06593 -1.551 0.12084
## ---
## 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: 1342.7 on 978 degrees of freedom
## AIC: 1356.7
##
## Number of Fisher Scoring iterations: 4glm.probsi = predict(glm.weekly.train, weekly.test, type = "response")
glm.predsi = rep("Down", length(glm.probsi))
glm.predsi[glm.probsi > 0.5] = "Up"
table(glm.predsi, weekly.test$Direction)##
## glm.predsi Down Up
## Down 9 5
## Up 34 56mean(glm.predsi == weekly.test$Direction) #accuracy## [1] 0.625This model uses a log transformation on the Volume variable in hopes of correcting the right skew. In this case, the transformed variable is still not significant, but the significant variable is now Lag1. Also the model predicts with the same accuracy as the other best performing models.
direction.test = Direction[!train]
lda.weekly.fit = lda(Direction~. - Today -Year - Volume + log(Volume), data=Weekly, subset=train)
plot(lda.weekly.fit)
lda.pred = predict(lda.weekly.fit, weekly.test)
lda.class=lda.pred$class
table(lda.class, direction.test)## direction.test
## lda.class Down Up
## Down 23 30
## Up 20 31mean(lda.class==direction.test)## [1] 0.5192308Using the transformation on the LDA model does not result in a better accuracy, most likely because the transformed variable still has no significant effect on the data.
train.lag2 = as.matrix(Lag2[train])
test.lag2 = as.matrix(Lag2[!train])
direction.train = Direction[train]
dim(train.lag2)## [1] 985 1dim(test.lag2)## [1] 104 1set.seed(1)
knn.pred=knn(train.lag2, test.lag2, direction.train, k=4)
table(knn.pred,direction.test)## direction.test
## knn.pred Down Up
## Down 20 17
## Up 23 44mean(knn.pred == direction.test)## [1] 0.6153846Using the KNN method, this time with k=4 returns a result of .6153846 model accuracy, which is very close to the best performing models. Also its no longer only predicting Up.
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.
detach(Weekly)
attach(Auto)## The following object is masked from package:ggplot2:
##
## mpgsummary(Auto)## mpg cylinders displacement horsepower weight
## Min. : 9.00 Min. :3.000 Min. : 68.0 Min. : 46.0 Min. :1613
## 1st Qu.:17.00 1st Qu.:4.000 1st Qu.:105.0 1st Qu.: 75.0 1st Qu.:2225
## Median :22.75 Median :4.000 Median :151.0 Median : 93.5 Median :2804
## Mean :23.45 Mean :5.472 Mean :194.4 Mean :104.5 Mean :2978
## 3rd Qu.:29.00 3rd Qu.:8.000 3rd Qu.:275.8 3rd Qu.:126.0 3rd Qu.:3615
## Max. :46.60 Max. :8.000 Max. :455.0 Max. :230.0 Max. :5140
##
## acceleration year origin name
## Min. : 8.00 Min. :70.00 Min. :1.000 amc matador : 5
## 1st Qu.:13.78 1st Qu.:73.00 1st Qu.:1.000 ford pinto : 5
## Median :15.50 Median :76.00 Median :1.000 toyota corolla : 5
## Mean :15.54 Mean :75.98 Mean :1.577 amc gremlin : 4
## 3rd Qu.:17.02 3rd Qu.:79.00 3rd Qu.:2.000 amc hornet : 4
## Max. :24.80 Max. :82.00 Max. :3.000 chevrolet chevette: 4
## (Other) :365(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.
mpg01 = rep(0, length(mpg))
mpg01[mpg > median(mpg)] = 1
Auto = data.frame(Auto, mpg01)(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.
library(corrplot)## Warning: package 'corrplot' was built under R version 4.0.3## corrplot 0.84 loadedcorrplot(cor(Auto[,-9]), method="square")
multi.hist(Auto[c(1,3:6)])
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 1.0000000Using Correlation plot, we can see that cylinders, displacement, horsepower, and weight are all strongly negatively correlated with mpg01, while origin seems to have a moderate positive correlation with mpg01.
Also, the numeric data distributions appear skewed to the right, with the exception of acceleration which has a normal bell shaped curved.
(c) Split the data into a training set and a test set.
train = (Auto$year %% 2 == 0)
train.auto = Auto[train,]
test.auto = 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.auto.fit = lda(mpg01~displacement+horsepower+weight+cylinders+origin, data=train.auto)
lda.pred = predict(lda.auto.fit, test.auto)
table(lda.pred$class, test.auto$mpg01)##
## 0 1
## 0 86 11
## 1 14 71mean(lda.pred$class == test.auto$mpg01)## [1] 0.86263741 - mean(lda.pred$class == test.auto$mpg01)## [1] 0.1373626The test error of the LDA model is 0.1373626.
(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.fit = qda(mpg01~displacement+horsepower+weight+cylinders+origin, data=train.auto)
qda.auto.fit## Call:
## qda(mpg01 ~ displacement + horsepower + weight + cylinders +
## origin, data = train.auto)
##
## Prior probabilities of groups:
## 0 1
## 0.4571429 0.5428571
##
## Group means:
## displacement horsepower weight cylinders origin
## 0 271.7396 133.14583 3604.823 6.812500 1.166667
## 1 111.6623 77.92105 2314.763 4.070175 2.035088qda.pred = predict(qda.auto.fit, test.auto)
qda.class=qda.pred$class
table(qda.class, test.auto$mpg01)##
## qda.class 0 1
## 0 88 13
## 1 12 69mean(qda.class==test.auto$mpg01)## [1] 0.86263741 - mean(qda.class==test.auto$mpg01)## [1] 0.1373626The test error of the QDA model is 0.1373626.
(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.auto.fit = glm(mpg01~displacement+horsepower+weight+cylinders+origin, data=train.auto, family=binomial)
glm.probs = predict(glm.auto.fit, test.auto, type = "response")
glm.preds = rep(0, length(glm.probs))
glm.preds[glm.probs > 0.5] = 1
table(glm.preds, test.auto$mpg01)##
## glm.preds 0 1
## 0 89 11
## 1 11 71mean(glm.preds == test.auto$mpg01) #accuracy## [1] 0.87912091 - mean(glm.preds == test.auto$mpg01)## [1] 0.1208791The test error of the logistic regression model is 0.1208791.
(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?
train.X = cbind(displacement,horsepower, weight, cylinders, origin)[train,]
test.X = cbind(displacement,horsepower, weight, cylinders, origin)[!train,]
train.mpg01 = mpg01[train]
test.mpg01 = mpg01[!train]
set.seed(1)
knn.pred=knn(train.X, test.X, train.mpg01, k=1)
table(knn.pred,test.mpg01)## test.mpg01
## knn.pred 0 1
## 0 83 11
## 1 17 711 - mean(knn.pred == test.mpg01)## [1] 0.1538462knn.pred=knn(train.X, test.X, train.mpg01, k=3)
table(knn.pred,test.mpg01)## test.mpg01
## knn.pred 0 1
## 0 84 9
## 1 16 731 - mean(knn.pred == test.mpg01)## [1] 0.1373626knn.pred=knn(train.X, test.X, train.mpg01, k=5)
table(knn.pred,test.mpg01)## test.mpg01
## knn.pred 0 1
## 0 82 9
## 1 18 731 - mean(knn.pred == test.mpg01)## [1] 0.1483516knn.pred=knn(train.X, test.X, train.mpg01, k=10)
table(knn.pred,test.mpg01)## test.mpg01
## knn.pred 0 1
## 0 79 7
## 1 21 751 - mean(knn.pred == test.mpg01)## [1] 0.1538462After testing out the K Nearest Neighbors method on different fold sizes (1,3,5 and 10), 3 performed the best with an error rate of 0.1373626. Also K=1 and K=10 had the same error rate for this model.
Q13. 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.
detach(Auto)
attach(Boston)
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. :50.00crim01 = rep(0, length(crim))
crim01[crim > median(crim)] = 1
Boston = data.frame(Boston, crim01)corrplot(cor(Boston), method="square")
multi.hist(Boston[c(1:3, 5:14)])
cor(Boston)## crim zn indus chas nox
## crim 1.00000000 -0.20046922 0.40658341 -0.055891582 0.42097171
## zn -0.20046922 1.00000000 -0.53382819 -0.042696719 -0.51660371
## indus 0.40658341 -0.53382819 1.00000000 0.062938027 0.76365145
## chas -0.05589158 -0.04269672 0.06293803 1.000000000 0.09120281
## nox 0.42097171 -0.51660371 0.76365145 0.091202807 1.00000000
## rm -0.21924670 0.31199059 -0.39167585 0.091251225 -0.30218819
## age 0.35273425 -0.56953734 0.64477851 0.086517774 0.73147010
## dis -0.37967009 0.66440822 -0.70802699 -0.099175780 -0.76923011
## rad 0.62550515 -0.31194783 0.59512927 -0.007368241 0.61144056
## tax 0.58276431 -0.31456332 0.72076018 -0.035586518 0.66802320
## ptratio 0.28994558 -0.39167855 0.38324756 -0.121515174 0.18893268
## black -0.38506394 0.17552032 -0.35697654 0.048788485 -0.38005064
## lstat 0.45562148 -0.41299457 0.60379972 -0.053929298 0.59087892
## medv -0.38830461 0.36044534 -0.48372516 0.175260177 -0.42732077
## crim01 0.40939545 -0.43615103 0.60326017 0.070096774 0.72323480
## rm age dis rad tax ptratio
## crim -0.21924670 0.35273425 -0.37967009 0.625505145 0.58276431 0.2899456
## zn 0.31199059 -0.56953734 0.66440822 -0.311947826 -0.31456332 -0.3916785
## indus -0.39167585 0.64477851 -0.70802699 0.595129275 0.72076018 0.3832476
## chas 0.09125123 0.08651777 -0.09917578 -0.007368241 -0.03558652 -0.1215152
## nox -0.30218819 0.73147010 -0.76923011 0.611440563 0.66802320 0.1889327
## rm 1.00000000 -0.24026493 0.20524621 -0.209846668 -0.29204783 -0.3555015
## age -0.24026493 1.00000000 -0.74788054 0.456022452 0.50645559 0.2615150
## dis 0.20524621 -0.74788054 1.00000000 -0.494587930 -0.53443158 -0.2324705
## rad -0.20984667 0.45602245 -0.49458793 1.000000000 0.91022819 0.4647412
## tax -0.29204783 0.50645559 -0.53443158 0.910228189 1.00000000 0.4608530
## ptratio -0.35550149 0.26151501 -0.23247054 0.464741179 0.46085304 1.0000000
## black 0.12806864 -0.27353398 0.29151167 -0.444412816 -0.44180801 -0.1773833
## lstat -0.61380827 0.60233853 -0.49699583 0.488676335 0.54399341 0.3740443
## medv 0.69535995 -0.37695457 0.24992873 -0.381626231 -0.46853593 -0.5077867
## crim01 -0.15637178 0.61393992 -0.61634164 0.619786249 0.60874128 0.2535684
## black lstat medv crim01
## crim -0.38506394 0.4556215 -0.3883046 0.40939545
## zn 0.17552032 -0.4129946 0.3604453 -0.43615103
## indus -0.35697654 0.6037997 -0.4837252 0.60326017
## chas 0.04878848 -0.0539293 0.1752602 0.07009677
## nox -0.38005064 0.5908789 -0.4273208 0.72323480
## rm 0.12806864 -0.6138083 0.6953599 -0.15637178
## age -0.27353398 0.6023385 -0.3769546 0.61393992
## dis 0.29151167 -0.4969958 0.2499287 -0.61634164
## rad -0.44441282 0.4886763 -0.3816262 0.61978625
## tax -0.44180801 0.5439934 -0.4685359 0.60874128
## ptratio -0.17738330 0.3740443 -0.5077867 0.25356836
## black 1.00000000 -0.3660869 0.3334608 -0.35121093
## lstat -0.36608690 1.0000000 -0.7376627 0.45326273
## medv 0.33346082 -0.7376627 1.0000000 -0.26301673
## crim01 -0.35121093 0.4532627 -0.2630167 1.00000000Note that indus, nox, age, rad,and tax are all positively correlated with crim01 as well as dis being negatively correlated with crim01.
train = 1:(length(Boston$crim)[1]/2)
test = (length(Boston$crim)[1]/2 + 1):length(Boston$crim)[1]
train.Boston = Boston[train, ]
test.Boston = Boston[test, ]
crim01.test = crim01[test]
crim01.train = crim01[train]glm.Boston.fit = glm(crim01~ indus+nox+age+dis+rad+tax, data=train.Boston,family=binomial)
summary(glm.Boston.fit)##
## Call:
## glm(formula = crim01 ~ indus + nox + age + dis + rad + tax, family = binomial,
## data = train.Boston)
##
## 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: 8glm.probs = predict(glm.Boston.fit, test.Boston, type = "response")
glm.preds = rep(0, length(glm.probs))
glm.preds[glm.probs > 0.5] = 1
table(glm.preds, crim01.test)## crim01.test
## glm.preds 0 1
## 0 75 8
## 1 15 1551 - mean(glm.preds == crim01.test)## [1] 0.09090909lda.Boston.fit <-lda(crim01~ indus+nox+age+dis+rad+tax, data=train.Boston, family=binomial)
lda.preds = predict(lda.Boston.fit, test.Boston)
lda.class=lda.preds$class
table(lda.class, crim01.test)## crim01.test
## lda.class 0 1
## 0 81 18
## 1 9 1451 - mean(lda.class == crim01.test)## [1] 0.1067194qda.Boston.fit = qda(crim01~ indus+nox+age+dis+rad+tax, data=train.Boston)
qda.Boston.fit## Call:
## qda(crim01 ~ indus + nox + age + dis + rad + tax, data = train.Boston)
##
## Prior probabilities of groups:
## 0 1
## 0.6442688 0.3557312
##
## Group means:
## indus nox age dis rad tax
## 0 7.042454 0.4702902 54.68957 4.843857 4.239264 300.4417
## 1 13.959778 0.6119889 85.67889 2.895154 5.155556 358.0111qda.preds = predict(qda.Boston.fit, train.Boston)
qda.class=qda.preds$class
table(qda.class, crim01.test)## crim01.test
## qda.class 0 1
## 0 71 99
## 1 19 641 - mean(qda.class==crim01.test)## [1] 0.4664032train.X = cbind(indus,nox,age,dis,rad,tax)[train,]
test.X = cbind(indus,nox,age,dis,rad,tax)[test,]
set.seed(1)
knn.pred=knn(train.X, test.X, crim01.train, k=1)
table(knn.pred,crim01.test)## crim01.test
## knn.pred 0 1
## 0 82 151
## 1 8 121 - mean(knn.pred == crim01.test)## [1] 0.6284585knn.pred=knn(train.X, test.X, crim01.train, k=5)
table(knn.pred,crim01.test)## crim01.test
## knn.pred 0 1
## 0 80 22
## 1 10 1411 - mean(knn.pred == crim01.test)## [1] 0.1264822knn.pred=knn(train.X, test.X, crim01.train, k=10)
table(knn.pred,crim01.test)## crim01.test
## knn.pred 0 1
## 0 83 23
## 1 7 1401 - mean(knn.pred == crim01.test)## [1] 0.11857714 methods of modeling were used to classify if where the data had a higher or lower crime rate than the median crime rate in Boston suburbs. Using Logistic Regression, LDA, QDA, and K Nearest Neighbors, the best model came from the implementation of the logistic regression. It had a mean error rate of 0.09090909. The worst model was the QDA (0.4664032 mean error rate) and KNN with k=1 (0.6284585 mean error rate). When the amount of folds for the KNN was increased to 10, the model performed a much better (0.1185771), almost on par with the second best model, LDA (0.1067194).