IE575_Assignment-7_MaulikPatel
Maulik Patel
October 9, 2016
1. Using a little bit of algebra, prove that the equation (4.2): p(X)=(e(B0+B1X))/(1+e(B0+B1X)) is equivalent to the equation (4.3): p(X)/(1-p(X))=e(B0+B1X). In other words, the logistic function representation and logit representation for the logistic regression model are equivalent.
Answer:1
p(X)=(e(B0+B1X))/(1+e(B0+B1X)) ……………..(1)
let’s say e(B0+B1X) = Y …………………………(2)
now, p(X) = Y / 1+Y ……………….as per (1) & (2)
so p(X) / 1-p(X) = (Y/1+Y) / 1-(Y/1+Y)
p(X) / 1-p(X) = (Y/1+Y) / (1-Y+Y/1+Y)
p(X) / 1-p(X) = (Y/1+Y) / (1/1+Y)
p(X) / 1-p(X) = (Y/1+Y) * (1+Y/1)
p(X) / 1-p(X) = Y
p(X) / 1-p(X) = e(B0+B1X) ……………..as per (2)
2. 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 1089 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?
suppressMessages(library(ISLR))
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 Downweekly = Weekly
names(weekly) = tolower(names(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
## Min. :-18.1950 Min. :-18.1950 Min. :0.08747
## 1st Qu.: -1.1580 1st Qu.: -1.1660 1st Qu.:0.33202
## Median : 0.2380 Median : 0.2340 Median :1.00268
## Mean : 0.1458 Mean : 0.1399 Mean :1.57462
## 3rd Qu.: 1.4090 3rd Qu.: 1.4050 3rd Qu.:2.05373
## Max. : 12.0260 Max. : 12.0260 Max. :9.32821
## today direction
## Min. :-18.1950 Down:484
## 1st Qu.: -1.1540 Up :605
## Median : 0.2410
## Mean : 0.1499
## 3rd Qu.: 1.4050
## Max. : 12.0260cor(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 1.000000000pairs(weekly[,-9])
My Observations:
- year : appears to have a strong correlation with volume and volume has been increasing over the years and has flattened out in recent years. year variable has no correlation with any other rate of return variables as indicated by pairwise correlation plots as well as correlation coefficient numbers close to 0.
- lag 1-5 : previous 5 weeks’ rate of returns have no correlation with any other variables as indicated by pairwise correlation plots as well as correlation coefficient numbers close to 0.
- today : same as previous weeks’ rate of return, today’s rate of return also has no correlation with any other variables as indicated by pairwise correlation plots as well as correlation coefficient numbers close to 0.
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?
(Hint: Use function glm() and pass in the argument family=binomial to run logistic regression. Use Pr(>|z|) to find out which variables are significant.)
glm.fit = glm(direction ~ lag1+lag2+lag3+lag4+lag5+volume, data=weekly, family=binomial)
summary(glm.fit)##
## 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: 4Significant Predictors:
based on p-values of individual predictor variables in summary output of the logistic regression model, lag2 is the only significant predictor of the “direction” as it has p-value less than significance level of 0.05.
c. Compute the confusion matrix and overall fraction of correct predictions. Regard value “Down” as the positive class. Calculate the sensitivity and specificity according to the confusion matrix.
(Hint: Use function predict() to make the predictions on the same data set you use to train a logistic regression model. Remember to pass in the argument type=response to obtain prediction probabilities. Let a predicted probability > 0.5 be “Up” and <=0.5 be “Down”. Then use function table() to obtain the confusion matrix. Calculate the overall fraction of correct predictions, sensitivity and specificity according to confusion matrix.)
glm.probs = predict(glm.fit, weekly, type="response")
contrasts(weekly$direction)## Up
## Down 0
## Up 1glm.pred = ifelse(glm.probs > 0.5, "Up","Down")
glm.confusion = table(glm.pred, weekly$direction)
glm.confusion##
## glm.pred Down Up
## Down 54 48
## Up 430 557# Regard value "Down" as the positive class.
glm.accuracy.percentage = round(mean(glm.pred == weekly$direction)*100,2)
glm.sensitivity.percentage = round(length(which(glm.pred == "Down" & weekly$direction == "Down"))/length(which(weekly$direction == "Down"))*100,2)
glm.specificity.percentage = round(length(which(glm.pred == "Up" & weekly$direction == "Up"))/length(which(weekly$direction == "Up"))*100,2)
glm.evaluation = data.frame(Accuracy = glm.accuracy.percentage, Sensitivity = glm.sensitivity.percentage, Specificity = glm.specificity.percentage)
glm.evaluation## Accuracy Sensitivity Specificity
## 1 56.11 11.16 92.07d. Now fit the logistic regression model using a training data period from 1990 to 2008, with Lag2 as the only predictor. Compute the confusion matrix and the overall fraction of correct predictions for the held out data (that is, the data from 2009 and 2010).
dim(weekly)## [1] 1089 9train = weekly[weekly$year %in% c(1990:2008),]
dim(train)## [1] 985 9levels(as.factor(train$year)) # sanity check## [1] "1990" "1991" "1992" "1993" "1994" "1995" "1996" "1997" "1998" "1999"
## [11] "2000" "2001" "2002" "2003" "2004" "2005" "2006" "2007" "2008"test = weekly[weekly$year %in% c(2009:2010),]
dim(test)## [1] 104 9levels(as.factor(test$year)) # sanity check## [1] "2009" "2010"glm.fit2 = glm(direction ~ lag2, data=train, family=binomial)
summary(glm.fit2)##
## Call:
## glm(formula = direction ~ lag2, family = binomial, data = train)
##
## Deviance Residuals:
## Min 1Q Median 3Q Max
## -1.536 -1.264 1.021 1.091 1.368
##
## Coefficients:
## Estimate Std. Error z value Pr(>|z|)
## (Intercept) 0.20326 0.06428 3.162 0.00157 **
## lag2 0.05810 0.02870 2.024 0.04298 *
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## (Dispersion parameter for binomial family taken to be 1)
##
## Null deviance: 1354.7 on 984 degrees of freedom
## Residual deviance: 1350.5 on 983 degrees of freedom
## AIC: 1354.5
##
## Number of Fisher Scoring iterations: 4glm.probs2 = predict(glm.fit2, test, type="response")
glm.pred2 = ifelse(glm.probs2 > 0.5, "Up","Down")
glm.confusion2 = table(glm.pred2, test$direction)
glm.confusion2##
## glm.pred2 Down Up
## Down 9 5
## Up 34 56# Regard value "Down" as the positive class.
glm.accuracy.percentage2 = round(mean(glm.pred2 == test$direction)*100,2)
glm.sensitivity.percentage2 = round(length(which(glm.pred2 == "Down" & test$direction == "Down"))/length(which(test$direction == "Down"))*100,2)
glm.specificity.percentage2 = round(length(which(glm.pred2 == "Up" & test$direction == "Up"))/length(which(test$direction == "Up"))*100,2)
glm.evaluation2 = data.frame(Accuracy = glm.accuracy.percentage2, Sensitivity = glm.sensitivity.percentage2, Specificity = glm.specificity.percentage2)
glm.evaluation2## Accuracy Sensitivity Specificity
## 1 62.5 20.93 91.8Accuracy of predicting stock market’s direction in terms of rate of return improved from 56.11% to 62.5% by using “lag2” only as a predictor in logistoc regression model.
sensitivity improved from 11.16% to 20.93%.
specificity decreased slightly from 92.07% to 91.8%.
e. Repeat (d) using LDA.
(Hint: Use function lda() to train a LDA model, which is part of the MASS package.)
suppressMessages(library(MASS))
lda.fit = lda(direction ~ lag2, data=train)
lda.fit## Call:
## lda(direction ~ lag2, data = train)
##
## Prior probabilities of groups:
## Down Up
## 0.4477157 0.5522843
##
## Group means:
## lag2
## Down -0.03568254
## Up 0.26036581
##
## Coefficients of linear discriminants:
## LD1
## lag2 0.4414162summary(lda.fit)## Length Class Mode
## prior 2 -none- numeric
## counts 2 -none- numeric
## means 2 -none- numeric
## scaling 1 -none- numeric
## lev 2 -none- character
## svd 1 -none- numeric
## N 1 -none- numeric
## call 3 -none- call
## terms 3 terms call
## xlevels 0 -none- listlda.pred = predict(lda.fit, test)
names(lda.pred)## [1] "class" "posterior" "x"lda.pred$class## [1] Up Up Down Down Up Up Up Down Down Down Down Up Up Up
## [15] Up Up Up Up Up Up Down Up Up Up Up Up Up Up
## [29] Up Up Up Up Up Up Up Up Up Up Up Up Up Up
## [43] Up Up Down Up Up Up Up Up Up Up Up Up Up Up
## [57] Down Up Up Up Up Up Up Up Up Up Up Up Up Up
## [71] Up Down Up Down Up Up Up Up Down Down Up Up Up Up
## [85] Up Down Up Up Up Up Up Up Up Up Up Up Up Up
## [99] Up Up Up Up Up Up
## Levels: Down Uplda.confusion = table(lda.pred$class, test$direction)
lda.confusion##
## Down Up
## Down 9 5
## Up 34 56# Regard value "Down" as the positive class.
lda.accuracy.percentage = round(mean(lda.pred$class == test$direction)*100,2)
lda.sensitivity.percentage = round(length(which(lda.pred$class == "Down" & test$direction == "Down"))/length(which(test$direction == "Down"))*100,2)
lda.specificity.percentage = round(length(which(lda.pred$class == "Up" & test$direction == "Up"))/length(which(test$direction == "Up"))*100,2)
lda.evaluation = data.frame(Accuracy = lda.accuracy.percentage, Sensitivity = lda.sensitivity.percentage, Specificity = lda.specificity.percentage)
lda.evaluation## Accuracy Sensitivity Specificity
## 1 62.5 20.93 91.8f. Repeat (d) using QDA.
(Hint: Use function qda() to train a QDA model, which is part of the MASS package.)
qda.fit = qda(direction ~ lag2, data=train)
qda.fit## Call:
## qda(direction ~ lag2, data = train)
##
## Prior probabilities of groups:
## Down Up
## 0.4477157 0.5522843
##
## Group means:
## lag2
## Down -0.03568254
## Up 0.26036581summary(qda.fit)## Length Class Mode
## prior 2 -none- numeric
## counts 2 -none- numeric
## means 2 -none- numeric
## scaling 2 -none- numeric
## ldet 2 -none- numeric
## lev 2 -none- character
## N 1 -none- numeric
## call 3 -none- call
## terms 3 terms call
## xlevels 0 -none- listqda.pred = predict(qda.fit, test)
names(qda.pred)## [1] "class" "posterior"qda.pred$class## [1] Up Up Up Up Up Up Up Up Up Up Up Up Up Up Up Up Up Up Up Up Up Up Up
## [24] Up Up Up Up Up Up Up Up Up Up Up Up Up Up Up Up Up Up Up Up Up Up Up
## [47] Up Up Up Up Up Up Up Up Up Up Up Up Up Up Up Up Up Up Up Up Up Up Up
## [70] Up Up Up Up Up Up Up Up Up Up Up Up Up Up Up Up Up Up Up Up Up Up Up
## [93] Up Up Up Up Up Up Up Up Up Up Up Up
## Levels: Down Upqda.confusion = table(qda.pred$class, test$direction)
qda.confusion##
## Down Up
## Down 0 0
## Up 43 61# Regard value "Down" as the positive class.
qda.accuracy.percentage = round(mean(qda.pred$class == test$direction)*100,2)
qda.sensitivity.percentage = round(length(which(qda.pred$class == "Down" & test$direction == "Down"))/length(which(test$direction == "Down"))*100,2)
qda.specificity.percentage = round(length(which(qda.pred$class == "Up" & test$direction == "Up"))/length(which(test$direction == "Up"))*100,2)
qda.evaluation = data.frame(Accuracy =qda.accuracy.percentage, Sensitivity = qda.sensitivity.percentage, Specificity = qda.specificity.percentage)
qda.evaluation## Accuracy Sensitivity Specificity
## 1 58.65 0 100g. Which of these methods appears to provide the best results on this data?
(Hint: Use the overall fraction of correct predictions to determine which method(s) perform the best.)
model_comparison = rbind(glm.evaluation2, lda.evaluation, qda.evaluation)
model_comparison = cbind(Model = c("Logistic_Regression","LDA","QDA"), model_comparison)
model_comparison[order(model_comparison$Accuracy, decreasing = TRUE),]## Model Accuracy Sensitivity Specificity
## 1 Logistic_Regression 62.50 20.93 91.8
## 2 LDA 62.50 20.93 91.8
## 3 QDA 58.65 0.00 100.0based on all 3 model’s accuracy% in above table, we can say that “logistic regression” and “LDA” provide best result for this dataset for predicting direction of the stock market in terms rate of return.
3. 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.
- 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. (Hint: Use function data.frame() to make the format of Auto data set as data frame. Use function ifelse() to create the binary variable mpg01 according to variable mpg.)
suppressMessages(library(MASS))
auto = Auto
names(auto) = tolower(names(auto))
head(auto)## mpg cylinders displacement horsepower weight acceleration year origin
## 1 18 8 307 130 3504 12.0 70 1
## 2 15 8 350 165 3693 11.5 70 1
## 3 18 8 318 150 3436 11.0 70 1
## 4 16 8 304 150 3433 12.0 70 1
## 5 17 8 302 140 3449 10.5 70 1
## 6 15 8 429 198 4341 10.0 70 1
## name
## 1 chevrolet chevelle malibu
## 2 buick skylark 320
## 3 plymouth satellite
## 4 amc rebel sst
## 5 ford torino
## 6 ford galaxie 500dim(auto)## [1] 392 9med = median(auto$mpg)
auto$mpg01 = ifelse(auto$mpg > med, 1,0)
table(auto$mpg01)##
## 0 1
## 196 196b. 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.
(Hint: Use function plot() to make scatterplot and use function boxplot() to make boxplot.)
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[,-9])
par(mfrow = c(1,2))
for(i in 1:(ncol(auto)-2)){
boxplot(auto[,i] ~ as.factor(auto$mpg01),
xlab = "Gas Mileage (0:Low, 1:High)",
ylab = colnames(auto)[i],
col = c("red","green"))
}
par(default_par)Observation of useful Predictors:
Based on, correlation plots and correlation-coefficient numbers and boxplot of each predictor with respect to gas mileage and looking at the whether it has clear distinction between low and high gas mileage:
Strong and Negative correlation:
- cylinders
- displacement
- horsepower
- weight
average to weak and Positive correlation:
- acceleration
- year
- origin
“mpg01” was directly derived by filtering “mpg” variable so as expeced they have positive strong correlation.
c. Split the data into a training set and a test set. Use function set.seed(13) and function createDataPartition in caret package to split the data into 67% for training and 33% for testing. That is, in R, code:
set.seed(13) trainIndex = createDataPartition(Auto$mpg01, p=.67, list=FALSE, times=1)
Then Auto[trainIndex,] is your training set and Auto[-trainIndex,] is your test set.
suppressMessages(library(caret))
set.seed(13)
auto$mpg01 = as.factor(as.character(auto$mpg01))
auto = auto[,-9] # removing "name" column to avoid the error i am getting during predict.glm()
trainIndex = createDataPartition(auto$mpg01, p=.67, list=FALSE, times=1)
trainauto = auto[trainIndex,]
testauto = auto[-trainIndex,]
dim(auto)## [1] 392 9dim(trainauto)## [1] 264 9dim(testauto)## [1] 128 9d. Perform LDA on the training data in order to predict mpg01 using all the other variables except of “name” and “mpg” in (b). What is the test error rate of the model obtained?
(Hint: Calculate the test error rate using the confusion matrix on the test set.)
lda.fit = lda(mpg01 ~ .-mpg, data=trainauto)
lda.fit## Call:
## lda(mpg01 ~ . - mpg, data = trainauto)
##
## Prior probabilities of groups:
## 0 1
## 0.5 0.5
##
## Group means:
## cylinders displacement horsepower weight acceleration year
## 0 6.666667 266.8864 128.56061 3584.735 14.57879 74.19697
## 1 4.143939 113.5189 79.44697 2317.992 16.28939 77.52273
## origin
## 0 1.189394
## 1 2.015152
##
## Coefficients of linear discriminants:
## LD1
## cylinders -0.533671238
## displacement 0.002693468
## horsepower 0.005944805
## weight -0.001117793
## acceleration -0.016153715
## year 0.134896097
## origin 0.239014920summary(lda.fit)## Length Class Mode
## prior 2 -none- numeric
## counts 2 -none- numeric
## means 14 -none- numeric
## scaling 7 -none- numeric
## lev 2 -none- character
## svd 1 -none- numeric
## N 1 -none- numeric
## call 3 -none- call
## terms 3 terms call
## xlevels 0 -none- listlda.pred = predict(lda.fit, testauto)
names(lda.pred)## [1] "class" "posterior" "x"lda.pred$class## [1] 0 0 0 1 0 0 1 0 0 0 0 0 0 1 1 0 0 0 0 0 1 1 0 0 0 0 0 1 0 1 0 0 1 0 0
## [36] 1 0 0 0 1 1 1 1 0 0 0 0 1 0 1 1 1 1 1 1 1 0 0 0 0 1 1 0 0 1 1 0 1 0 1
## [71] 1 0 0 0 0 1 1 1 1 0 0 0 1 0 0 0 0 0 1 1 1 1 0 0 0 0 1 1 1 1 0 1 1 1 1
## [106] 1 1 1 1 1 1 1 1 1 1 1 0 1 1 1 1 1 1 1 1 1 1 1
## Levels: 0 1lda.confusion = table(lda.pred$class, testauto$mpg01)
lda.confusion##
## 0 1
## 0 57 3
## 1 7 61# Regarding value "1:high" as the positive class.
lda.accuracy.percentage = round(mean(lda.pred$class == testauto$mpg01)*100,2)
lda.error.percentage = round(mean(lda.pred$class != testauto$mpg01)*100,2)
lda.error.percentage+lda.accuracy.percentage## [1] 100lda.sensitivity.percentage = round(length(which(lda.pred$class == "1" & testauto$mpg01 == "1"))/length(which(testauto$mpg01 == "1"))*100,2)
lda.specificity.percentage = round(length(which(lda.pred$class == "0" & testauto$mpg01 == "0"))/length(which(testauto$mpg01 == "0"))*100,2)
lda.evaluation = data.frame(Test_Error_rate = lda.error.percentage, Accuracy = lda.accuracy.percentage, Sensitivity = lda.sensitivity.percentage, Specificity = lda.specificity.percentage)
lda.evaluation## Test_Error_rate Accuracy Sensitivity Specificity
## 1 7.81 92.19 95.31 89.06Test error-rate of “LDA” model = 7.81 %
e. Perform QDA on the training data in order to predict mpg01 using all the other variables except of “name” and “mpg” in (b). What is the test error rate of the model obtained?
qda.fit = qda(mpg01 ~ .-mpg, data=trainauto)
qda.fit## Call:
## qda(mpg01 ~ . - mpg, data = trainauto)
##
## Prior probabilities of groups:
## 0 1
## 0.5 0.5
##
## Group means:
## cylinders displacement horsepower weight acceleration year
## 0 6.666667 266.8864 128.56061 3584.735 14.57879 74.19697
## 1 4.143939 113.5189 79.44697 2317.992 16.28939 77.52273
## origin
## 0 1.189394
## 1 2.015152summary(qda.fit)## Length Class Mode
## prior 2 -none- numeric
## counts 2 -none- numeric
## means 14 -none- numeric
## scaling 98 -none- numeric
## ldet 2 -none- numeric
## lev 2 -none- character
## N 1 -none- numeric
## call 3 -none- call
## terms 3 terms call
## xlevels 0 -none- listqda.pred = predict(qda.fit, testauto)
names(qda.pred)## [1] "class" "posterior"qda.pred$class## [1] 0 0 0 1 0 0 1 0 0 0 0 0 0 1 1 0 0 0 0 0 1 1 0 0 0 0 0 1 0 1 0 0 1 0 0
## [36] 1 0 0 0 1 1 1 1 0 0 0 0 1 0 1 1 1 1 0 1 1 0 0 0 0 1 1 0 0 1 1 0 0 0 1
## [71] 1 0 0 0 0 1 1 1 1 0 0 0 1 0 0 0 0 0 1 1 1 1 0 0 0 0 1 1 1 1 0 1 1 1 1
## [106] 1 1 1 1 1 1 1 1 1 1 0 0 0 1 1 1 0 1 0 1 1 1 1
## Levels: 0 1qda.confusion = table(qda.pred$class, testauto$mpg01)
qda.confusion##
## 0 1
## 0 61 5
## 1 3 59# Regarding value "1:high" as the positive class.
qda.accuracy.percentage = round(mean(qda.pred$class == testauto$mpg01)*100,2)
qda.error.percentage = round(mean(qda.pred$class != testauto$mpg01)*100,2)
qda.error.percentage+qda.accuracy.percentage## [1] 100qda.sensitivity.percentage = round(length(which(qda.pred$class == "1" & testauto$mpg01 == "1"))/length(which(testauto$mpg01 == "1"))*100,2)
qda.specificity.percentage = round(length(which(qda.pred$class == "0" & testauto$mpg01 == "0"))/length(which(testauto$mpg01 == "0"))*100,2)
qda.evaluation = data.frame(Test_Error_rate = qda.error.percentage, Accuracy = qda.accuracy.percentage, Sensitivity = qda.sensitivity.percentage, Specificity = qda.specificity.percentage)
qda.evaluation## Test_Error_rate Accuracy Sensitivity Specificity
## 1 6.25 93.75 92.19 95.31Test error-rate of “QDA” model = 6.25 %
f. Perform logistic regression on the training data in order to predict mpg01 using all the other variables except of “name” and “mpg” in (b). What is the test error rate of the model obtained?
glm.fit3 = glm(mpg01 ~ .-mpg, data=trainauto, family = binomial)
summary(glm.fit3)##
## Call:
## glm(formula = mpg01 ~ . - mpg, family = binomial, data = trainauto)
##
## Deviance Residuals:
## Min 1Q Median 3Q Max
## -2.37410 -0.10718 0.00796 0.22286 3.07994
##
## Coefficients:
## Estimate Std. Error z value Pr(>|z|)
## (Intercept) -22.134093 7.400266 -2.991 0.00278 **
## cylinders -0.389773 0.512952 -0.760 0.44734
## displacement 0.006032 0.015974 0.378 0.70573
## horsepower -0.021597 0.026927 -0.802 0.42252
## weight -0.004735 0.001480 -3.198 0.00138 **
## acceleration 0.056806 0.164796 0.345 0.73031
## year 0.480997 0.098479 4.884 1.04e-06 ***
## origin 0.586447 0.453602 1.293 0.19606
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## (Dispersion parameter for binomial family taken to be 1)
##
## Null deviance: 365.98 on 263 degrees of freedom
## Residual deviance: 107.00 on 256 degrees of freedom
## AIC: 123
##
## Number of Fisher Scoring iterations: 8# I had to remove "name" columns as during predict.glm() i was getting error of "Error in model.frame.default..." as levels of variable "name" were not same in both training and test sets. this might be due to the suggested method of splitting the dataset. I understand there might be a better way to split the datasets to avoid this error, for example using stratified from the splitstackshape package. but at this point to continue with the exercise i have removed the column "name" explicitly instead of during modeling and i didn't get the error after that.
# in presence of "name" column, i would have used below formula to fit the glm model to data:
# glm.fit3 = glm(mpg01 ~ .-mpg-name, data=trainauto, family = binomial)
contrasts(trainauto$mpg01)## 1
## 0 0
## 1 1glm.probs3 = predict(glm.fit3, testauto, type="response")
dim(testauto)## [1] 128 9length(glm.probs3)## [1] 128glm.pred3 = ifelse(glm.probs3 > 0.5, "1","0")
glm.confusion3 = table(glm.pred3, testauto$mpg01)
glm.confusion3##
## glm.pred3 0 1
## 0 59 4
## 1 5 60# Regarding value "1:High" as the positive class.
glm.error.percentage3 = round(mean(glm.pred3 != testauto$mpg01)*100,2)
glm.accuracy.percentage3 = round(mean(glm.pred3 == testauto$mpg01)*100,2)
glm.accuracy.percentage3 + glm.error.percentage3## [1] 100glm.sensitivity.percentage3 = round(length(which(glm.pred3 == "1" & testauto$mpg01 == "1"))/length(which(testauto$mpg01 == "1"))*100,2)
glm.specificity.percentage3 = round(length(which(glm.pred3 == "0" & testauto$mpg01 == "0"))/length(which(testauto$mpg01 == "0"))*100,2)
glm.evaluation3 = data.frame(Test_Error_rate = glm.error.percentage3, Accuracy = glm.accuracy.percentage3, Sensitivity = glm.sensitivity.percentage3, Specificity = glm.specificity.percentage3)
glm.evaluation3## Test_Error_rate Accuracy Sensitivity Specificity
## 1 7.03 92.97 93.75 92.19Test error-rate of “Linear regression” model = 7.03 %
model_comparison = rbind(glm.evaluation3, lda.evaluation, qda.evaluation)
model_comparison = cbind(Model = c("Logistic_Regression","LDA","QDA"), model_comparison)
model_comparison[order(model_comparison$Accuracy, decreasing = TRUE),]## Model Test_Error_rate Accuracy Sensitivity Specificity
## 3 QDA 6.25 93.75 92.19 95.31
## 1 Logistic_Regression 7.03 92.97 93.75 92.19
## 2 LDA 7.81 92.19 95.31 89.06