Assignment 4 Resampling Methods

library(tidyverse)
library(openintro)

Question 3

We now review k-fold cross-validation.

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

The k-fold cross-validation is implemented by taking, n, number of observations and splitting it into to k sets, each set being approximately n/k. The rest then become the training set. THe training set is then used to fit the data to the model and obtain the k MSE estimates. The testing error is then computed by averaging the k MSE estimates.

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

The advantages of a k-fold cross validation is that it can avoid overestimating test error.(Something the validation set can lead to) However it is more costly and time consuming since you would need to repeat the process many times to gain your results.

2. LOOVC

The LOOVC has less bias than the k-fold. However it is more expensive and has lower accuracy than the k-fold

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.

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

library(ISLR2)

## Warning: package 'ISLR2' was built under R version 4.1.3

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

bdata = sample(dim(Default)[1], dim(Default)[1] / 2)

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

glm.fit2 = glm(default ~ income + balance, data = Default, family = "binomial", subset = bdata)
summary(glm.fit2)

## 
## Call:
## glm(formula = default ~ income + balance, family = "binomial", 
##     data = Default, subset = bdata)
## 
## 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

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.

prob = predict(glm.fit2, newdata = Default[-bdata, ], type = "response")
glm.pred2 = rep("No", length(prob))
glm.pred2[prob > 0.5] = "Yes"

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

mean(glm.pred2 != Default[-bdata, ]$default)

## [1] 0.0254

From the model above we have a test error of 2.74%.

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.

bdata = sample(dim(Default)[1], dim(Default)[1] / 2)
glm.fit2 = glm(default ~ income + balance, data = Default, family = "binomial", subset = bdata)
prob = predict(glm.fit2, newdata = Default[-bdata, ], type = "response")
glm.pred2 = rep("No", length(prob))
glm.pred2[prob > 0.5] = "Yes"
mean(glm.pred2 != Default[-bdata, ]$default)

## [1] 0.0274

According to this fitted model, we have a test error of 2.44%.

bdata = sample(dim(Default)[1], dim(Default)[1] / 2)
glm.fit2 = glm(default ~ income + balance, data = Default, family = "binomial", subset = bdata)
prob = predict(glm.fit2, newdata = Default[-bdata, ], type = "response")
glm.pred2 = rep("No", length(prob))
glm.pred2[prob > 0.5] = "Yes"
mean(glm.pred2 != Default[-bdata, ]$default)

## [1] 0.0244

This model having the same test error as the previous on of 2.44%.

bdata = sample(dim(Default)[1], dim(Default)[1] / 2)
glm.fit2 = glm(default ~ income + balance, data = Default, family = "binomial", subset = bdata)
prob = predict(glm.fit2, newdata = Default[-bdata, ], type = "response")
glm.pred2 = rep("No", length(prob))
glm.pred2[prob > 0.5] = "Yes"
mean(glm.pred2 != Default[-bdata, ]$default)

## [1] 0.0244

This final model’s test error is 2.7%. Overall the test errors of the model are similar.

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.

bdata = sample(dim(Default)[1], dim(Default)[1] / 2)
glm.fit2 = glm(default ~ income + balance + student, data = Default, family = "binomial", subset = bdata)
prob = predict(glm.fit2, newdata = Default[-bdata, ], type = "response")
glm.pred2 = rep("No", length(prob))
glm.pred2[prob > 0.5] = "Yes"
mean(glm.pred2 != Default[-bdata, ]$default)

## [1] 0.0278

With the dummy variable “Student” we have a test error of 2.64% for this model. The dummy variable didn’t seem to reduce the test error of this model and is similar to the other models that did not have the dummy variable, “Student”.

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)
attach(Default)
glm.fit2 = glm(default ~ income + balance, data = Default, family = "binomial")
summary(glm.fit2)

## 
## 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. 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.

library(boot)

## 
## Attaching package: 'boot'

## The following object is masked from 'package:openintro':
## 
##     salinity

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

The standard deviation for this 4.0769 and it’s balance is 2.08089.

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

The standard errors are relatively similar when using the glm() function.

Question 9

We will now consider the Boston housing data set, from the ISLR2 library.

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

mean(Boston$medv)

## [1] 22.53281

2. Provide an estimate of the standard error of ˆµ. Interpret this result.

sd(Boston$medv)/sqrt(dim(Boston)[1]-1)

## [1] 0.4092658

The standard error of the mean of medv is 0.40927. This indicates that we have variability in our model. However it appears we do not have much variability in this particular model.

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

meanfunc=function(data, index){return(mean(Boston$medv[index]))}
boot(Boston, meanfunc, 100)

## 
## ORDINARY NONPARAMETRIC BOOTSTRAP
## 
## 
## Call:
## boot(data = Boston, statistic = meanfunc, R = 100)
## 
## 
## Bootstrap Statistics :
##     original     bias    std. error
## t1* 22.53281 0.02923123   0.3997709

The standard error above is 0.4304 which is similar to oour previous standard deviation of 0.40927.

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).

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

The 95% confidence interval is from [21.722, 23.34] for the mean of medv.

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

median(Boston$medv)

## [1] 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.

medianfunc=function(data, index){return(median(Boston$medv[index]))}
boot(Boston, medianfunc, 100)

## 
## ORDINARY NONPARAMETRIC BOOTSTRAP
## 
## 
## Call:
## boot(data = Boston, statistic = medianfunc, R = 100)
## 
## 
## Bootstrap Statistics :
##     original  bias    std. error
## t1*     21.2 -0.0135   0.4046357

The median has a standard error 0.3325 when using the bootstrap method. This close to when we found the standard error for the mean of medv.

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

quantile(Boston$medv, 0.1)

##   10% 
## 12.75

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

bootfunc=function(data, index){return(quantile(Boston$medv[index], 0.1))}
boot(Boston, bootfunc, 100)

## 
## ORDINARY NONPARAMETRIC BOOTSTRAP
## 
## 
## Call:
## boot(data = Boston, statistic = bootfunc, R = 100)
## 
## 
## Bootstrap Statistics :
##     original  bias    std. error
## t1*    12.75  0.0345   0.5122586

In the tenth percentile the standard error is 0.4821. This is close to the other standard error we found.