Assignment

3. We now review k-fold cross-validation

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

Answer:

An alternate process to LOOCV is k-fold cross-validation. This process involves randomly selecting and dividing observations into k groups or k folds of equal size approximately.

The first fold is treated as a validation set and the method is fitted on the remaining k-1 folds. The mean square error(MSE) is calculated on this observations in the remaining folds(held out). And, this procedure is repeated for k times, and each time the different observations group is considered for validation set. Thus, the process results in k estimates of the test error such that MSE1,MSE2..MSEk. The k-fold cross validation is compuated or implemented by averaging these values of MSEs(MSE1,MSE2…MSEK).

The most obvious advantage of using the k-fold cross-validation is computational and this is very general approach and which can be applied to most statistical learning approach.

b) What are the advantages and disadvantages of k-fold crossvalidation relative to:

i. The validation set approach?

Answer:

Advantages:

1.Compared to the validation set approach, the k-fold CV has substantially less variability, especially for smaller datasets. Because the data in a validation set is partitioned only once, a model may appear more or less favorable “by chance” depending on the observations that were included in the train and test datasets. For this reason, decisions about model selection and tuning may be heavily influenced by these observations.

2.Model performance is tested and trained using all of the data.

3.Since most models improve with more data and a major fraction is removed entirely from training, the validation set strategy may overestimate the test error when compared to a model that is trained on the entire dataset.

Less bias when compared to the validation set approach.

Disadvantages:

A model is trained and evaluated only once, which is a computational advantage of the validation set approach. For conventional values of k such that 5 and 10, these training datasets will also typically be larger than those used in the validation procedure. In a k-fold CV, k models will be trained. All of this indicates that huge data sets and high values of k may require significantly longer k-fold CV processing times.

Here validation set approach has computational advantage over k-fold CV usually when folds are higher.

ii. LOOCV?

Answer:

Advantages:

At typical values of k, k-fold CV scales well and requires significantly less computing power. An example is 5 and 10 and When k = n, k-fold CV and LOOCV are the same. However, in the case of k = 10 and n = 10,000, k-fold CV will fit 10 models while LOOCV will fit 10,000.
The bias-variance trade-off (LOOCV has larger variance but lower bias), there is some evidence to suggest that k-fold CV can provide a more accurate estimate of the test error rate than LOOCV.
Besides the computational concerns, k-fold CV more accurate estimates of the test error rate than does LOOCV, which is more beneficial when related to trade-offs between bias and variation.

Disadvantages:

k-fold has high bias.
Similar to validation set approach k-fold CV has element of randomness while selecting the k-fold.
In few cases, LOOCV require less computational than k-fold.

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.

Answer:

Required Library:

library(ISLR)

## Warning: package 'ISLR' was built under R version 4.3.2

library(MASS)

## Warning: package 'MASS' was built under R version 4.3.2

library(caret)

## Warning: package 'caret' was built under R version 4.3.2

## Loading required package: ggplot2

## Warning: package 'ggplot2' was built under R version 4.3.2

## Loading required package: lattice

library(boot)

## 
## Attaching package: 'boot'

## The following object is masked from 'package:lattice':
## 
##     melanoma

attach(Default) # from ISLR package.
set.seed(1)

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

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

Interpretation: From the summary of logistic regression model that uses income and balance to predict default shows that predictors(income and balance) are significant at 0.05.

Predict Default:

logreg_pred<-predict(logreg,Default$default,type="response")
logreg_pred.class<-ifelse(logreg_pred>0.5,"Yes","No")
Misclassification_Rate=round(mean(Default$default!=logreg_pred.class),4)
print(Misclassification_Rate)

## [1] 0.0263

print(paste("Misclassification Rate:", round(Misclassification_Rate * 100, 2), "%"))

## [1] "Misclassification Rate: 2.63 %"

Interpretation: 0.0263 suggests that the misclassification rate of the logistic regression model on the Default dataset is approximately 2.63%.

Accuracy:

Accuracy=1−MisclassificationRate

accuracy <- 1 - Misclassification_Rate
print(paste("Accuracy:", round(accuracy * 100, 2), "%"))

## [1] "Accuracy: 97.37 %"

Interpretation: The accuracy of the logistic regression model that uses income and balance to predict default on the Default dataset is approximately 97.37%.

(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.

set.seed(123)
index <- createDataPartition(y = Default$default, p = 0.5, list = F) # I chose 50-50 split approach

train_set <- Default[index, ]
valid_set <- Default[-index, ]

nrow(train_set) / nrow(Default)

## [1] 0.5001

nrow(valid_set) / nrow(Default)

## [1] 0.4999

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

log_reg_splt<-glm(default~income+balance,family = binomial,data=train_set)
summary(log_reg_splt)

## 
## Call:
## glm(formula = default ~ income + balance, family = binomial, 
##     data = train_set)
## 
## Coefficients:
##               Estimate Std. Error z value Pr(>|z|)    
## (Intercept) -1.158e+01  6.209e-01 -18.656  < 2e-16 ***
## income       2.296e-05  7.206e-06   3.187  0.00144 ** 
## balance      5.617e-03  3.200e-04  17.555  < 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: 1463.76  on 5000  degrees of freedom
## Residual deviance:  791.05  on 4998  degrees of freedom
## AIC: 797.05
## 
## Number of Fisher Scoring iterations: 8

Interpretation: From the summary of logistic regression model that uses income and balance to predict default shows that predictors(income and balance) are significant at 0.05.

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.

logreg_splt_pred<-predict(log_reg_splt,valid_set,type="response")
logreg_splt_pred.class<-ifelse(logreg_splt_pred>0.5,"Yes","No")
table(valid_set$default,logreg_splt_pred.class,dnn=c("Actual","Predicted"))

##       Predicted
## Actual   No  Yes
##    No  4813   20
##    Yes  117   49

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

Misclassification_Rate2=round(mean(logreg_splt_pred.class!=valid_set$default),4)
print(Misclassification_Rate2)

## [1] 0.0274

print(paste("Misclassification Rate:", round(Misclassification_Rate2 * 100, 2), "%"))

## [1] "Misclassification Rate: 2.74 %"

Interpretation: 0.0274 suggests that the misclassification rate of the logistic regression model on the Default dataset is approximately 2.74%.

Accuracy:

Accuracy=1−MisclassificationRate

accuracy2 <- 1 - Misclassification_Rate2
print(paste("Accuracy:", round(accuracy2 * 100, 2), "%"))

## [1] "Accuracy: 97.26 %"

(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.

I chose 80-20 split approach

set.seed(123)
index_80 <- createDataPartition(y = Default$default, p = 0.8, list = F) 

train_set_80 <- Default[index_80, ]
valid_set_80 <- Default[-index_80, ]

nrow(train_set_80) / nrow(Default)

## [1] 0.8001

nrow(valid_set_80) / nrow(Default)

## [1] 0.1999

log_reg_splt_80<-glm(default~income+balance,family = binomial,data=train_set_80)
summary(log_reg_splt_80)

## 
## Call:
## glm(formula = default ~ income + balance, family = binomial, 
##     data = train_set_80)
## 
## Coefficients:
##               Estimate Std. Error z value Pr(>|z|)    
## (Intercept) -1.186e+01  5.007e-01 -23.691  < 2e-16 ***
## income       2.708e-05  5.670e-06   4.777 1.78e-06 ***
## balance      5.715e-03  2.574e-04  22.201  < 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: 2340.6  on 8000  degrees of freedom
## Residual deviance: 1252.2  on 7998  degrees of freedom
## AIC: 1258.2
## 
## Number of Fisher Scoring iterations: 8

logreg_splt_pred_80<-predict(log_reg_splt_80,valid_set_80,type="response")
logreg_splt_pred_80.class<-ifelse(logreg_splt_pred_80>0.5,"Yes","No")
table(valid_set_80$default,logreg_splt_pred_80.class,dnn=c("Actual","Predicted"))

##       Predicted
## Actual   No  Yes
##    No  1922   11
##    Yes   49   17

Misclassification_Rate2_80=round(mean(logreg_splt_pred_80.class!=valid_set_80$default),4)
print(Misclassification_Rate2_80)

## [1] 0.03

print(paste("Misclassification Rate:", round(Misclassification_Rate2_80 * 100, 2), "%"))

## [1] "Misclassification Rate: 3 %"

Accuracy:

Accuracy=1−MisclassificationRate

accuracy2_80 <- 1 - Misclassification_Rate2_80
print(paste("Accuracy:", round(accuracy2_80 * 100, 2), "%"))

## [1] "Accuracy: 97 %"

I chose 70-30 split approach

set.seed(123)
index_70 <- createDataPartition(y = Default$default, p = 0.7, list = F) 

train_set_70 <- Default[index_70, ]
valid_set_70 <- Default[-index_70, ]

nrow(train_set_70) / nrow(Default)

## [1] 0.7001

nrow(valid_set_70) / nrow(Default)

## [1] 0.2999

log_reg_splt_70<-glm(default~income+balance,family = binomial,data=train_set_70)
summary(log_reg_splt_70)

## 
## Call:
## glm(formula = default ~ income + balance, family = binomial, 
##     data = train_set_70)
## 
## Coefficients:
##               Estimate Std. Error z value Pr(>|z|)    
## (Intercept) -1.142e+01  5.131e-01 -22.250  < 2e-16 ***
## income       2.383e-05  5.973e-06   3.989 6.64e-05 ***
## balance      5.533e-03  2.681e-04  20.639  < 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: 2050.6  on 7000  degrees of freedom
## Residual deviance: 1139.1  on 6998  degrees of freedom
## AIC: 1145.1
## 
## Number of Fisher Scoring iterations: 8

logreg_splt_pred_70<-predict(log_reg_splt_70,valid_set_70,type="response")
logreg_splt_pred_70.class<-ifelse(logreg_splt_pred_70>0.5,"Yes","No")
table(valid_set_70$default,logreg_splt_pred_70.class,dnn=c("Actual","Predicted"))

##       Predicted
## Actual   No  Yes
##    No  2885   15
##    Yes   61   38

Misclassification_Rate2_70=round(mean(logreg_splt_pred_70.class!=valid_set_70$default),4)
print(Misclassification_Rate2_70)

## [1] 0.0253

print(paste("Misclassification Rate:", round(Misclassification_Rate2_70 * 100, 2), "%"))

## [1] "Misclassification Rate: 2.53 %"

Accuracy:

Accuracy=1−MisclassificationRate

accuracy2_70 <- 1 - Misclassification_Rate2_70
print(paste("Accuracy:", round(accuracy2_70 * 100, 2), "%"))

## [1] "Accuracy: 97.47 %"

I chose 90-10 split approach

set.seed(123)
index_90 <- createDataPartition(y = Default$default, p = 0.9, list = F) 

train_set_90 <- Default[index_90, ]
valid_set_90 <- Default[-index_90, ]

nrow(train_set_90) / nrow(Default)

## [1] 0.9001

nrow(valid_set_90) / nrow(Default)

## [1] 0.0999

log_reg_splt_90<-glm(default~income+balance,family = binomial,data=train_set_90)
summary(log_reg_splt_90)

## 
## Call:
## glm(formula = default ~ income + balance, family = binomial, 
##     data = train_set_90)
## 
## Coefficients:
##               Estimate Std. Error z value Pr(>|z|)    
## (Intercept) -1.133e+01  4.495e-01 -25.210  < 2e-16 ***
## income       1.895e-05  5.259e-06   3.603 0.000315 ***
## balance      5.552e-03  2.353e-04  23.596  < 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: 2630.7  on 9000  degrees of freedom
## Residual deviance: 1436.8  on 8998  degrees of freedom
## AIC: 1442.8
## 
## Number of Fisher Scoring iterations: 8

logreg_splt_pred_90<-predict(log_reg_splt_90,valid_set_90,type="response")
logreg_splt_pred_90.class<-ifelse(logreg_splt_pred_90>0.5,"Yes","No")
table(valid_set_90$default,logreg_splt_pred_90.class,dnn=c("Actual","Predicted"))

##       Predicted
## Actual  No Yes
##    No  962   4
##    Yes  24   9

Misclassification_Rate2_90=round(mean(logreg_splt_pred_90.class!=valid_set_90$default),4)
print(Misclassification_Rate2_90)

## [1] 0.028

print(paste("Misclassification Rate:", round(Misclassification_Rate2_90 * 100, 2), "%"))

## [1] "Misclassification Rate: 2.8 %"

Accuracy:

Accuracy=1−MisclassificationRate

accuracy2_90 <- 1 - Misclassification_Rate2_90
print(paste("Accuracy:", round(accuracy2_90 * 100, 2), "%"))

## [1] "Accuracy: 97.2 %"

Interpretation:

Based on the dataset split into three different ratios splits and the model is fitted on training set and the misclassification rate is calculated on validation set.

Missclassification rate(validation estimate of the test error rate ) (Train:Test–80:20): 0.03

Missclassification rate(validation estimate of the test error rate ) (Train:Test–70:30): 0.0253

Missclassification rate(validation estimate of the test error rate ) (Train:Test–90:10): 0.028

From the above metric is understood that validation estimate error rate is nearly different based on the split ratio.

(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.

set.seed(123)
index_70_d <- createDataPartition(y = Default$default, p = 0.7, list = F) 

train_set_70_d <- Default[index_70_d, ]
valid_set_70_d <- Default[-index_70_d, ]

nrow(train_set_70_d) / nrow(Default)

## [1] 0.7001

nrow(valid_set_70_d) / nrow(Default)

## [1] 0.2999

log_reg_splt_70_d<-glm(default~income+balance+student,family = binomial,data=train_set_70_d)
summary(log_reg_splt_70_d)

## 
## Call:
## glm(formula = default ~ income + balance + student, family = binomial, 
##     data = train_set_70_d)
## 
## Coefficients:
##               Estimate Std. Error z value Pr(>|z|)    
## (Intercept) -1.048e+01  5.782e-01 -18.122  < 2e-16 ***
## income      -5.519e-07  9.660e-06  -0.057  0.95444    
## balance      5.638e-03  2.733e-04  20.628  < 2e-16 ***
## studentYes  -8.937e-01  2.774e-01  -3.221  0.00128 ** 
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## (Dispersion parameter for binomial family taken to be 1)
## 
##     Null deviance: 2050.6  on 7000  degrees of freedom
## Residual deviance: 1128.8  on 6997  degrees of freedom
## AIC: 1136.8
## 
## Number of Fisher Scoring iterations: 8

logreg_splt_pred_70_d<-predict(log_reg_splt_70_d,valid_set_70_d,type="response")
logreg_splt_pred_70_d.class<-ifelse(logreg_splt_pred_70_d>0.5,"Yes","No")
table(valid_set_70_d$default,logreg_splt_pred_70_d.class,dnn=c("Actual","Predicted"))

##       Predicted
## Actual   No  Yes
##    No  2884   16
##    Yes   63   36

Misclassification_Rate2_70_d=round(mean(logreg_splt_pred_70_d.class!=valid_set_70_d$default),4)
print(Misclassification_Rate2_70_d)

## [1] 0.0263

print(paste("Misclassification Rate:", round(Misclassification_Rate2_70_d * 100, 2), "%"))

## [1] "Misclassification Rate: 2.63 %"

Accuracy:

Accuracy=1−Misclassification_Rate2_70_d

accuracy2_70_d <- 1 - Misclassification_Rate2_70_d
print(paste("Accuracy:", round(accuracy2_70_d * 100, 2), "%"))

## [1] "Accuracy: 97.37 %"

Interpretation: Except for 70-30 ratio split of the dataset(slight increase in the error rate), all other ratios has slightly reduced(decreased) in test error rate when added a dummy variable student to the model, however, their isn’t much variation in the test error rate reduction.

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.

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(123)
logreg_6a<-glm(default~income+balance,family = binomial,data=Default)
summary(logreg_6a)

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

Interpretation: From the above logistic regression summary the Std.Error co-efficients are obtained as 4.348e-01, 4.985e-06, and 2.274e-04 for beta0,beta1 and beta2 respectively.

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

(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.

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 -2.754771e-02 4.204817e-01
## t2*  2.080898e-05  1.582518e-07 4.729534e-06
## t3*  5.647103e-03  1.296980e-05 2.217214e-04

Interpretation: The bootstrap estimates using the boot function of the standard errors for the coefficients of beta0,beta1,beta2 are respectively 4.204817e-01, 4.729534e-06, and 2.217214e-04.

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

Answer:

From the logistic regression summaries coefficients (beta0, beta1, and beta2), the predicted standard errors using the bootstrap method and the `glm()} function closely match, suggesting the bootstrap’s dependability in capturing coefficient fluctuation. By using a non-parametric resampling strategy, this consistency highlights the bootstrap approach’s robustness, particularly in small or complicated datasets(such as Default) where standard methods may not be as effective. As a result, confidence in coefficient estimates is increased.

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

attach(Boston)

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

mu<-mean(medv)
mu

## [1] 22.53281

Interpretation:

The estimated population mean (mu) of medv in the dataset is 22.53281.

(b) 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.

SE<-sd(medv)/sqrt(nrow(Boston))
SE

## [1] 0.4088611

Interpretation: From the above , the sample mean estimate of medv in the Boston dataset is probably an approximation of the population mean, with relatively low variability around this estimate, according to the standard error of 0.4088611.

(c) Now estimate the standard error of ˆμ 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

Interpretation: The estimate of 0.4088611 obtained in (b) and the bootstrap estimated standard error of (mu) of 0.4106622 are both extremely very close. Demonstrating the consistency of the direct calculation method utilized in part (b) with the bootstrapped standard error estimation method.

(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 [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_mu<-c(mu-2*0.4106622,mu+2*0.4106622)
CI_mu

## [1] 21.71148 23.35413

Interpretation: The confidence interval offered by the t.test() function and the bootstrapped one are very close intervals.

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

mu_med<-median(medv)
mu_med

## [1] 21.2

Interpretation: Median value of medv in the population is 21.2.

(f) 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(123)
boot.fn<-function(data,index){
  med_mu<-median(data[index])
  return(med_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 -0.0203   0.3676453

Interpretation: From the above median value is 21.2 which is equal to the estimated value obtained in question (e), with a standard error of 0.3676453 which is small when relatively compared to median value of 21.2.

(g) 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.)

quntl_mu<-quantile(medv,c(0.1))
quntl_mu

##   10% 
## 12.75

Interpretation: The boston census tracts wth 10(tenth) percentile of medv is 12.75.

(h) Use the bootstrap to estimate the standard error of ˆμ0.1. Comment on your findings.

boot.fn <- function(data, index) {
  quantile_mu <- quantile(data[index], c(0.1))
  return (quantile_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  0.0071   0.5138566

Interpretation: From the above quantile value is 12.75 which is equal to the estimated value obtained in question (g), with a standard error of 0.5138566 which is small when relatively compared to median value of 12.75.

Assignment_4

Sai

2024-03-08

3. We now review k-fold cross-validation

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

Answer:

b) What are the advantages and disadvantages of k-fold crossvalidation relative to:

i. The validation set approach?

Answer:

Advantages:

Disadvantages:

ii. LOOCV?

Answer:

Advantages:

Disadvantages:

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

Answer:

Required Library:

Predict Default:

Accuracy:

(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.

Accuracy:

(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.

I chose 80-20 split approach

Accuracy:

I chose 70-30 split approach

Accuracy:

I chose 90-10 split approach

Accuracy:

Accuracy:

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.

(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

(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.

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

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

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

(b) 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.

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

(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 [ˆμ − 2SE(ˆμ), ˆμ + 2SE(ˆμ)].

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

(f) 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.

(g) 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.)

(h) Use the bootstrap to estimate the standard error of ˆμ0.1. Comment on your findings.