Assignment 4

3 We now review k-fold cross-validation.

1. Explain how k-fold cross-validation is implemented.

K- fold cross validation is made up of k groups similar in size. The first k-fold group is used as a validation set that is then used to fit the remaining k-1 groups. This process will occur k number of times and will have varying amounts for the values for the mean square error(MSE).These MSE values are then averaged to result in the k-fold cross-validation estimate (CVk).

2. What are the advantages and disadvantages of k-fold cross validation relative to:

1. The validation set approach? This approach creates gives a test estimate that is very fluctuating when you compare it to the k-fold cross validation test estimate. The randomization of the separation of observations put into the training and testing data sets are what cause this fluctuation.

2. LOOCV? LOOCV can cause problems when the value of n is large because it requires the fitting of the statistical learning method n times. Since the test error estimate is the average of the n fitted models with similar observations this causes the estimate to have high fluctuation as well. The K-fold validation however could possibly have a value of k resulting in an unbiased error test rate with a lower variance.

5 In Chapter 4, we used logistic regression to predict the probability of default using income and balance on the Default data set. We will now estimate the test error of this logistic regression model using the validation set approach. Do not forget to set a random seed before beginning your analysis.

library(ISLR)

## Warning: package 'ISLR' was built under R version 4.0.5

attach(Default)
set.seed(8)

1. Fit a logistic regression model that uses income and balance to predict default.

default.fit<-glm(default~income+balance, data=Default, family="binomial")
summary(default.fit)

## 
## Call:
## glm(formula = default ~ income + balance, family = "binomial", 
##     data = Default)
## 
## Deviance Residuals: 
##     Min       1Q   Median       3Q      Max  
## -2.4725  -0.1444  -0.0574  -0.0211   3.7245  
## 
## Coefficients:
##               Estimate Std. Error z value Pr(>|z|)    
## (Intercept) -1.154e+01  4.348e-01 -26.545  < 2e-16 ***
## income       2.081e-05  4.985e-06   4.174 2.99e-05 ***
## balance      5.647e-03  2.274e-04  24.836  < 2e-16 ***
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## (Dispersion parameter for binomial family taken to be 1)
## 
##     Null deviance: 2920.6  on 9999  degrees of freedom
## Residual deviance: 1579.0  on 9997  degrees of freedom
## AIC: 1585
## 
## Number of Fisher Scoring iterations: 8

2. Using the validation set approach, estimate the test error of this model. In order to do this, you must perform the following steps:

1. Split the sample set into a training set and a validation set.

train.default<-sample(dim(Default)[1],dim(Default)[1]/2)

2. Fit a multiple logistic regression model using only the training observations.

spl.default<-glm(default~income+balance, data=Default, family="binomial", subset = train.default)

3. Obtain a prediction of default status for each individual in the validation set by computing the posterior probability of default for that individual, and classifying the individual to the default category if the posterior probability is greater than 0.5.

default.probs=predict(spl.default,Default[-train.default,], type="response")
spldefault.pred=rep("No",length(default.probs))
spldefault.pred[default.probs>0.5]="Yes"

4. Compute the validation set error, which is the fraction of the observations in the validation set that are misclassified.

mean(spldefault.pred != Default[-train.default, ]$default)

## [1] 0.0272

Validation set approach test error rate = 2.72%.

3. Repeat the process in (b) three times, using three different splits of the observations into a training set and a validation set. Comment on the results obtained.

train.default<-sample(dim(Default)[1],dim(Default)[1]/2)
spl.default<-glm(default~income+balance, data=Default, family="binomial", subset = train.default)
default.probs=predict(spl.default,Default[-train.default,], type="response")
spldefault.pred=rep("No",length(default.probs))
spldefault.pred[default.probs>0.5]="Yes"
mean(spldefault.pred != Default[-train.default, ]$default)

## [1] 0.0258

train.default<-sample(dim(Default)[1],dim(Default)[1]/2)
spl.default<-glm(default~income+balance, data=Default, family="binomial", subset = train.default)
default.probs=predict(spl.default,Default[-train.default,], type="response")
spldefault.pred=rep("No",length(default.probs))
spldefault.pred[default.probs>0.5]="Yes"
mean(spldefault.pred != Default[-train.default, ]$default)

## [1] 0.0288

train.default<-sample(dim(Default)[1],dim(Default)[1]/2)
spl.default<-glm(default~income+balance, data=Default, family="binomial", subset = train.default)
default.probs=predict(spl.default,Default[-train.default,], type="response")
spldefault.pred=rep("No",length(default.probs))
spldefault.pred[default.probs>0.5]="Yes"
mean(spldefault.pred != Default[-train.default, ]$default)

## [1] 0.0268

Validation set approach test error rate = (2.58% 2.88%, 2.68%) all being slightly different from the randomization.

4. Now consider a logistic regression model that predicts the probability of default using income, balance, and a dummy variable for student. Estimate the test error for this model using the validation set approach. Comment on whether or not including a dummy variable for student leads to a reduction in the test error rate.

train.default<-sample(dim(Default)[1],dim(Default)[1]/2)
spl.default<-glm(default~income+balance+student, data=Default, family="binomial", subset = train.default)
default.probs=predict(spl.default,Default[-train.default,], type="response")
spldefault.pred=rep("No",length(default.probs))
spldefault.pred[default.probs>0.5]="Yes"
mean(spldefault.pred != Default[-train.default, ]$default)

## [1] 0.027

There is no need to create another variable because the student was already a binary categorical variable. Adding a student to the model wouldn’t cause a decrease in the test error rate because 2.7% falls in the range of the previous test error rates.

6 We continue to consider the use of a logistic regression model to predict the probability of default using income and balance on the Default data set. In particular, we will now compute estimates for the standard errors of the income and balance logistic regression coefficients in two different ways:(1) using the bootstrap, and (2) using the standard formula for computing the standard errors in the glm() function. Do not forget to set a random seed before beginning your analysis.

1. Using the summary() and glm() functions, determine the estimated standard errors for the coefficients associated with income and balance in a multiple logistic regression model that uses both predictors.

library(boot)
set.seed(12)
default.fit<-glm(default~income+balance, data=Default, family="binomial")
summary(default.fit)

## 
## Call:
## glm(formula = default ~ income + balance, family = "binomial", 
##     data = Default)
## 
## Deviance Residuals: 
##     Min       1Q   Median       3Q      Max  
## -2.4725  -0.1444  -0.0574  -0.0211   3.7245  
## 
## Coefficients:
##               Estimate Std. Error z value Pr(>|z|)    
## (Intercept) -1.154e+01  4.348e-01 -26.545  < 2e-16 ***
## income       2.081e-05  4.985e-06   4.174 2.99e-05 ***
## balance      5.647e-03  2.274e-04  24.836  < 2e-16 ***
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## (Dispersion parameter for binomial family taken to be 1)
## 
##     Null deviance: 2920.6  on 9999  degrees of freedom
## Residual deviance: 1579.0  on 9997  degrees of freedom
## AIC: 1585
## 
## Number of Fisher Scoring iterations: 8

Estimated standard error for income (β1) = 4.985e−06; balance(β2) = 2.274e−4; intercept (β0) = 0.4348

2. Write a function, boot.fn(), that takes as input the Default data set as well as an index of the observations, and that outputs the coefficient estimates for income and balance in the multiple logistic regression model.

boot.fn <- function(data, index){
  def.fit<-glm(default~income+balance, data=data, family="binomial", subset=index)
  return(coef(def.fit))}

3. Use the boot() function together with your boot.fn() function to estimate the standard errors of the logistic regression coefficients for income and balance.

boot(Default,boot.fn,1000)

## 
## ORDINARY NONPARAMETRIC BOOTSTRAP
## 
## 
## Call:
## boot(data = Default, statistic = boot.fn, R = 1000)
## 
## 
## Bootstrap Statistics :
##          original        bias     std. error
## t1* -1.154047e+01 -8.387469e-03 4.484914e-01
## t2*  2.080898e-05 -1.550111e-07 5.010512e-06
## t3*  5.647103e-03  6.338097e-06 2.313711e-04

detach(Default)

Estimated coefficients standard error for intercept = 0.4485; income = 5.0105e−6; balance = 2.3137e−4

4. Comment on the estimated standard errors obtained using the glm() function and using your bootstrap function.

The estimated standard error for bootstrap was a little less compared to the estimated standard error for glm

9 We will now consider the Boston housing data set, from the MASS library.

library(MASS)
attach(Boston)

1. Based on this data set, provide an estimate for the population mean of medv. Call this estimate μ^.

mu.hat <- mean(medv)
mu.hat

## [1] 22.53281

Population mean for medv = 22.5328

2. Provide an estimate of the standard error of μ^. Interpret this result. Hint: We can compute the standard error of the sample mean by dividing the sample standard deviation by the square root of the number of observations.

mu.hat.err= sd(medv)/sqrt(length(medv))
mu.hat.err

## [1] 0.4088611

Estimated standard error of μ^ = 0.4089, which is pretty small compared to the mean of medv

3. Now estimate the standard error of μ^ using the bootstrap. How does this compare to your answer from (b)?

bootmu.fn<- function(data, index){
  mu<-mean(data[index])
  return(mu)}
boot(medv, bootmu.fn, 1000)

## 
## ORDINARY NONPARAMETRIC BOOTSTRAP
## 
## 
## Call:
## boot(data = medv, statistic = bootmu.fn, R = 1000)
## 
## 
## Bootstrap Statistics :
##     original      bias    std. error
## t1* 22.53281 0.005427866   0.3967152

The standard error estimate resulting at 0.3967

4. Based on your bootstrap estimate from (c), provide a 95% confidence interval for the mean of medv. Compare it to the results obtained using t.test(Boston$medv). Hint: You can approximate a 95% confidence interval using the formula: (μ^−2SE(μ),μ^+2SE(μ))

t.test(medv)

## 
##  One Sample t-test
## 
## data:  medv
## t = 55.111, df = 505, p-value < 2.2e-16
## alternative hypothesis: true mean is not equal to 0
## 95 percent confidence interval:
##  21.72953 23.33608
## sample estimates:
## mean of x 
##  22.53281

CI.muhat<-c(22.53281 - 2 * 0.4119,  22.53281 + 2 * 0.4119)
CI.muhat

## [1] 21.70901 23.35661

Using the standard error calculated from bootstrapping resulted in A 95% confidence interval was resulted for the standard error, also being relatively close to the confidence interval resulting in the t-test. The bootstrap value is about ~.02 less than the t-test value.

5. Based on this data set, provide an estimate, μ^med, for the median value of medv in the population.

mu.med<-median(medv)
mu.med

## [1] 21.2

6. We now would like to estimate the standard error of μ^med. Unfortunately, there is no simple formula for computing the standard error of the median. Instead, estimate the standard error of the median using the bootstrap. Comment on your findings.

bootmed.fn <-function(data,index){
  med.mu<-(median(data[index]))
  return(med.mu)}
boot(medv,bootmed.fn, 1000)

## 
## ORDINARY NONPARAMETRIC BOOTSTRAP
## 
## 
## Call:
## boot(data = medv, statistic = bootmed.fn, R = 1000)
## 
## 
## Bootstrap Statistics :
##     original   bias    std. error
## t1*     21.2 -0.02815   0.3773028

The median and the bootstrap estimate show to hold the same value of 21.2. Bootstraping’s median standard error = .3773 which is small compared to the mean.

7. Based on this data set, provide an estimate for the tenth percentile of medv in Boston suburbs. Call this quantity μ0.1^. (You can use the quantile() function.)

mu.01 <- quantile(medv, c(0.1))
mu.01

##   10% 
## 12.75

Results show to be in the tenth percentile of medv.

8. Use the bootstrap to estimate the standard error of μ0.1^. Comment on your findings.

boot01.fn<-function(data,index){
  mu01<-quantile(data[index],c(0.1))
  return(mu01)
}
boot(medv,boot01.fn,1000)

## 
## ORDINARY NONPARAMETRIC BOOTSTRAP
## 
## 
## Call:
## boot(data = medv, statistic = boot01.fn, R = 1000)
## 
## 
## Bootstrap Statistics :
##     original  bias    std. error
## t1*    12.75  0.0326   0.4849664

Bootstrap estimate = 12.75 (same results as part g), resulting in a standard error = 0.4850 making it small compared to the tenth percentile’s value.