Assignment 4

Load libraries

library(ISLR)

## Warning: package 'ISLR' was built under R version 4.0.3

library(boot)

## Warning: package 'boot' was built under R version 4.0.4

library(MASS)

Question 3

We now review k-fold cross-validation. (a) Explain how k-fold cross-validation is implemented.

The data is first divided into k parts. The first part is treated as a validation set, and the model is fit on k-1 parts This is repeated k times, using different parts as the validation set each time. The Mean Squared Error is calculated for each cycle which is then averaged to find the k-fold cross-validation estimate

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

1. The validation set approach? Advantages: low variability for k-fold cross validation compared to validation set approach Disadvantage: k-fold cross validation takes more computational power and expense compared to validation set approach

2. LOOCV? Advantages: k-fold cross validation has less variance compared to LOOCV k-fold cross validation less computationally expensive compared to LOOCV Disadvantages: k-fold cross validation can have higher bias than LOOCV k-fold cross validation has more bias compared to LOOCV

Question 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. (a) Fit a logistic regression model that uses income and balance to predict default.

data("Default")
set.seed(1)
glm.fit <- glm(default ~ income + balance, data = Default, family = "binomial")

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.

set.seed(1)
train2 <- sample(1:nrow(Default), 0.5*nrow(Default))

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

glm.fit2 <- glm(default ~ income + balance, data = Default, family = binomial, subset = train2)

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.

glm.fit2.prob <- predict(glm.fit2, Default[-train2,], type = "response")
pred_class2 = ifelse(glm.fit2.prob > .5, "Yes", "No")

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

mean(pred_class2 != Default[-train2, ]$default)

## [1] 0.0254

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.

set.seed(1)
#70-30 split 
train3 <- sample(1:nrow(Default), 0.7*nrow(Default))
glm.fit3 <- glm(default ~ income + balance, data = Default, family = binomial, subset = train3)
glm.fit3.probs <- predict(glm.fit3, Default[-train3, ], type = "response")
pred_class3 <- ifelse(glm.fit3.probs > 0.5, "Yes", "No")
mean(pred_class3 != Default[-train3, ]$default)

## [1] 0.02666667

set.seed(1)
#80-20 split 
train4 <- sample(1:nrow(Default), 0.8*nrow(Default))
glm.fit4 <- glm(default ~ income + balance, data = Default, family = binomial, subset = train4)
glm.fit4.probs <- predict(glm.fit4, Default[-train4, ], type = "response")
pred_class4 <- ifelse(glm.fit4.probs > 0.5, "Yes", "No")
mean(pred_class4 != Default[-train4, ]$default)

## [1] 0.026

set.seed(1)
#90-10 split 
train5 <- sample(1:nrow(Default), 0.9*nrow(Default))
glm.fit5 <- glm(default ~ income + balance, data = Default, family = binomial, subset = train5)
glm.fit5.probs <- predict(glm.fit5, Default[-train5, ], type = "response")
pred_class5 <- ifelse(glm.fit5.probs > 0.5, "Yes", "No")
mean(pred_class5 != Default[-train5, ]$default)

## [1] 0.029

The three sets, with a 70-30 split, a 80-20 split, and a 90-10 split all had different test error rates of 0.02666667, 0.026, and 0.029 respectively. They are not significantly different and have a small range.

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.

train6 <- sample(nrow(Default), 0.5*nrow(Default))
glm.fit6 <- glm(default ~ income + balance + student, data = Default, family = binomial, subset = train6)
summary(glm.fit6)

## 
## Call:
## glm(formula = default ~ income + balance + student, family = binomial, 
##     data = Default, subset = train6)
## 
## Deviance Residuals: 
##     Min       1Q   Median       3Q      Max  
## -2.1148  -0.1404  -0.0558  -0.0214   3.6412  
## 
## Coefficients:
##               Estimate Std. Error z value Pr(>|z|)    
## (Intercept) -1.036e+01  6.830e-01 -15.165  < 2e-16 ***
## income      -7.013e-06  1.160e-05  -0.605  0.54534    
## balance      5.657e-03  3.259e-04  17.358  < 2e-16 ***
## studentYes  -9.754e-01  3.366e-01  -2.897  0.00376 ** 
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## (Dispersion parameter for binomial family taken to be 1)
## 
##     Null deviance: 1429.88  on 4999  degrees of freedom
## Residual deviance:  777.79  on 4996  degrees of freedom
## AIC: 785.79
## 
## Number of Fisher Scoring iterations: 8

glm.fit6.probs <- predict(glm.fit6, Default[-train6, ], type = "response")
pred_class6 <- ifelse(glm.fit6.probs > 0.5, "Yes", "No")
mean(pred_class6 != Default[-train6, ]$default)

## [1] 0.026

The test error rate for both models, with or without student, is 2.6% Adding student does not help or worsen the test error rate

Question 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.

set.seed(1)
glm.fit7 <- glm(default ~ income + balance, data = Default, family = binomial)
summary(glm.fit7)

## 
## 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

The standard error for income is 4.985e-06 and 2.274e-04 is balance.

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){
  return(coef(glm(default ~ income + balance, data=data, family="binomial", subset=index)))
}

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, 100)

## 
## ORDINARY NONPARAMETRIC BOOTSTRAP
## 
## 
## Call:
## boot(data = Default, statistic = boot.fn, R = 100)
## 
## 
## Bootstrap Statistics :
##          original        bias     std. error
## t1* -1.154047e+01  8.556378e-03 4.122015e-01
## t2*  2.080898e-05 -3.993598e-07 4.186088e-06
## t3*  5.647103e-03 -4.116657e-06 2.226242e-04

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

Using glm function, the standard error for income is 4.985e-06 and 2.274e-04 for balance. Using bootstrap, the coefficient for income is 4.186088e-06 and balance is 2.226242e-04 These are quite similar to each other

Question 9

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

data("Boston")

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

set.seed(1)
mu <- mean(Boston$medv)
mu

## [1] 22.53281

2. Provide an estimate of the standard error of ˆμ. 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.

stdev.mu <- sd(Boston$medv)/sqrt(length(Boston$medv))
stdev.mu

## [1] 0.4088611

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

set.seed(1)

boot.mu = function(data, index) return(mean(data[index]))

boot(Boston$medv, boot.mu, 500)

## 
## ORDINARY NONPARAMETRIC BOOTSTRAP
## 
## 
## Call:
## boot(data = Boston$medv, statistic = boot.mu, R = 500)
## 
## 
## Bootstrap Statistics :
##     original      bias    std. error
## t1* 22.53281 0.004967589   0.4265264

The standard error in part b is 0.4088611 The standard error using bootstrap is 0.4265264 The errors are similar

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 [ˆμ − 2SE(ˆμ), ˆμ + 2SE(ˆμ)].

t.test(Boston$medv)

## 
##  One Sample t-test
## 
## data:  Boston$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 = c(mu - 2 * stdev.mu, mu + 2 * stdev.mu)
ci

## [1] 21.71508 23.35053

The confidence interval using t-test is 21.72953, 23.33608 The confidence interval based on bootstrap is 21.71508, 23.35053 The confidence intervals are extremely similar

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

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

## [1] 21.2

the median is 21.2

6. We now would like to estimate the standard error of ˆμmed. 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.

set.seed(1)
boot.fn <- function(data, index) {
  return(median(data[index]))
}

boot(Boston$medv, boot.fn, 100)

## 
## ORDINARY NONPARAMETRIC BOOTSTRAP
## 
## 
## Call:
## boot(data = Boston$medv, statistic = boot.fn, R = 100)
## 
## 
## Bootstrap Statistics :
##     original  bias    std. error
## t1*     21.2  -0.029   0.3461316

the standard error, 0.3461, for the median, 21.2, is small.

7. Based on this data set, provide an estimate for the tenth percentile of medv in Boston suburbs. Call this quantity ˆμ0.1. (You can use the quantile() function.)

muhat.1 <- quantile(Boston$medv, c(0.1))
muhat.1

##   10% 
## 12.75

10th percentile is 12.75

8. Use the bootstrap to estimate the standard error of ˆμ0.1. Comment on your findings.

set.seed(1)
boot.fn2 <- function(data, index) return(quantile(data[index], c(0.1)))

boot(Boston$medv, boot.fn2, 100)

## 
## ORDINARY NONPARAMETRIC BOOTSTRAP
## 
## 
## Call:
## boot(data = Boston$medv, statistic = boot.fn2, R = 100)
## 
## 
## Bootstrap Statistics :
##     original  bias    std. error
## t1*    12.75   0.008   0.5370477

the standard error estimate is about 0.5370 using bootstrap method for the tenth percentile of medv in Boston compared to the other estimates, it is higher