library(ISLR)

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

library(corrplot)

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

## corrplot 0.84 loaded

library(MASS)

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

library(class)

Question 3: We now review k-fold cross-validation.

Part A: Explain how k-fold cross-validation is implemented.

Answer: The k-fold cross validation is done by taking the amount of observations (n) and randomly splitting them into k, non-overlapping groups of length of n/k. These groups acts as a validation set, and the remainder acts as a training set. The test error is then estimated by averaging the k resulting MSE estimates.

Part B: What are the advantages and disadvantages of k-fold crossvalidation relative to:

i. The validation set approach?

Answer: Advantage of k-fold cross validation relative to the validation set: The validation estimate of the test error rate can be highly variable, depending on precisely which observations are included in the training set and which observations are included in the validation set. Moreover, validation set error rate may tend to overestimate the test error rate for the model fit on the entire data set.

Disadvantage of k-fold cross validation relative to the validation set: Validation set approach is conceptually simple and easy to implement.

ii. LOOCV?

Answer: Advantage of k-fold cross validation relative to LOOCV: LOOCV requires fitting the statistical learning method n times. This has the potential to be computationally expensive.

Disadvantage of k-fold cross validation relative to LOOCV: If the main purpose bias reduction, LOOCV should be preffered to k-fold CV since it tends to has less bias.

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.

Part A: Fit a logistic regression model that uses income and balance to predict default.

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

## 
## 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

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

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

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

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

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

## 
## Call:
## glm(formula = default ~ income + balance, family = "binomial", 
##     data = Default, subset = train)
## 
## Deviance Residuals: 
##     Min       1Q   Median       3Q      Max  
## -2.5830  -0.1428  -0.0573  -0.0213   3.3395  
## 
## Coefficients:
##               Estimate Std. Error z value Pr(>|z|)    
## (Intercept) -1.194e+01  6.178e-01 -19.333  < 2e-16 ***
## income       3.262e-05  7.024e-06   4.644 3.41e-06 ***
## balance      5.689e-03  3.158e-04  18.014  < 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: 1523.8  on 4999  degrees of freedom
## Residual deviance:  803.3  on 4997  degrees of freedom
## AIC: 809.3
## 
## Number of Fisher Scoring iterations: 8

iii. 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.

probs <- predict(fit.glm, newdata = Default[-train, ], type = "response")
pred.glm <- rep("No", length(probs))
pred.glm[probs > 0.5] <- "Yes"

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

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

## [1] 0.0254

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

## [1] 0.0274

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

## [1] 0.0244

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

## [1] 0.0244

Answer: The three different splits each produce a different resulting error rate which shows that the rate varies by which observations are in the training/validation sets.

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

## [1] 0.0278

Answer: Including the dummy variable for student leads did not result in a reduction in 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.

Part A: 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)
attach(Default)

## The following objects are masked from Default (pos = 3):
## 
##     balance, default, income, student

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

## 
## 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

Answer: The glm() estimates of the standard errors for the coefficients intercept, income and balance are respectively 4.348e-01, 4.985e-06 and 2.274e-04.

Part B: 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) {
    fit <- glm(default ~ income + balance, data = data, family = "binomial", subset = index)
    return (coef(fit))
}

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

library(boot)
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 -3.945460e-02 4.344722e-01
## t2*  2.080898e-05  1.680317e-07 4.866284e-06
## t3*  5.647103e-03  1.855765e-05 2.298949e-04

Answer: The bootstrap estimates of the standard errors for the coefficients B0, B1 and B2 are respectively 0.4239, 4.583 x 10^(-6) and 2.268 x 10^(-4).

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

Answer: The estimated standard errors obtained by each of the two methods are very similar

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

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

library(MASS)
attach(Boston)
mu.hat <- mean(medv)
mu.hat

## [1] 22.53281

Part B: Provide an estimate of the standard error of (mu-hat). 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.

se.hat <- sd(medv) / sqrt(dim(Boston)[1])
se.hat

## [1] 0.4088611

Part C: Now estimate the standard error of (mu-hat) using the bootstrap. How does this compare to your answer from (b)?

set.seed(1)
boot.fn <- function(data, index) {
    mu <- mean(data[index])
    return (mu)
}
boot(medv, boot.fn, 1000)

## 
## ORDINARY NONPARAMETRIC BOOTSTRAP
## 
## 
## Call:
## boot(data = medv, statistic = boot.fn, R = 1000)
## 
## 
## Bootstrap Statistics :
##     original      bias    std. error
## t1* 22.53281 0.007650791   0.4106622

Part D: 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 [(mu-hat) 2SE(mu-hat), (mu-hat) + 2SE(mu-hat)].

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.mu.hat <- c(22.53 - 2 * 0.4119, 22.53 + 2 * 0.4119)
CI.mu.hat

## [1] 21.7062 23.3538

Part E: Based on this data set, provide an estimate, (mu-hat)med, for the median value of medv in the population.

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

## [1] 21.2

Part F: We now would like to estimate the standard error of (mu-hat)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.

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

## 
## ORDINARY NONPARAMETRIC BOOTSTRAP
## 
## 
## Call:
## boot(data = medv, statistic = boot.fn, R = 1000)
## 
## 
## Bootstrap Statistics :
##     original  bias    std. error
## t1*     21.2 -0.0386   0.3770241

Answer: We get an estimated median value of 21.2 which is equal to the value obtained in (e), with a standard error of 0.3874 which is relatively small compared to median value.

Part G: Based on this data set, provide an estimate for the tenth percentile of medv in Boston suburbs. Call this quantity (mu-hat)0.1. (You can use the quantile() function.)

percent.hat <- quantile(medv, c(0.1))
percent.hat

##   10% 
## 12.75

Part H: Use the bootstrap to estimate the standard error of (mu-hat)0.1. Comment on your findings.

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

## 
## ORDINARY NONPARAMETRIC BOOTSTRAP
## 
## 
## Call:
## boot(data = medv, statistic = boot.fn, R = 1000)
## 
## 
## Bootstrap Statistics :
##     original  bias    std. error
## t1*    12.75  0.0186   0.4925766

Answer: We get an estimated tenth percentile value of 12.75 which is again equal to the value obtained in (g), with a standard error of 0.5113 which is relatively small compared to percentile value.

STA Assignment #4

Marina Allen

3/26/2021

Question 3: We now review k-fold cross-validation.

Part A: Explain how k-fold cross-validation is implemented.

Part B: What are the advantages and disadvantages of k-fold crossvalidation relative to:

i. The validation set approach?

Disadvantage of k-fold cross validation relative to the validation set: Validation set approach is conceptually simple and easy to implement.

ii. LOOCV?

Answer: Advantage of k-fold cross validation relative to LOOCV: LOOCV requires fitting the statistical learning method n times. This has the potential to be computationally expensive.

Disadvantage of k-fold cross validation relative to LOOCV: If the main purpose bias reduction, LOOCV should be preffered to k-fold CV since it tends to has less bias.

Part A: Fit a logistic regression model that uses income and balance to predict default.

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

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

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

iii. 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.

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

Part C: 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.

Answer: The three different splits each produce a different resulting error rate which shows that the rate varies by which observations are in the training/validation sets.

Answer: Including the dummy variable for student leads did not result in a reduction in the test error rate.

Part A: 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.

Answer: The glm() estimates of the standard errors for the coefficients intercept, income and balance are respectively 4.348e-01, 4.985e-06 and 2.274e-04.

Part B: 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.

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

Answer: The bootstrap estimates of the standard errors for the coefficients B0, B1 and B2 are respectively 0.4239, 4.583 x 10^(-6) and 2.268 x 10^(-4).

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

Answer: The estimated standard errors obtained by each of the two methods are very similar

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

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

Part B: Provide an estimate of the standard error of (mu-hat). 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.

Part C: Now estimate the standard error of (mu-hat) using the bootstrap. How does this compare to your answer from (b)?

Part D: 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 [(mu-hat) 2SE(mu-hat), (mu-hat) + 2SE(mu-hat)].

Part E: Based on this data set, provide an estimate, (mu-hat)med, for the median value of medv in the population.

Part F: We now would like to estimate the standard error of (mu-hat)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.

Answer: We get an estimated median value of 21.2 which is equal to the value obtained in (e), with a standard error of 0.3874 which is relatively small compared to median value.

Part G: Based on this data set, provide an estimate for the tenth percentile of medv in Boston suburbs. Call this quantity (mu-hat)0.1. (You can use the quantile() function.)

Part H: Use the bootstrap to estimate the standard error of (mu-hat)0.1. Comment on your findings.

Answer: We get an estimated tenth percentile value of 12.75 which is again equal to the value obtained in (g), with a standard error of 0.5113 which is relatively small compared to percentile value.