Assignment 4, Ch 5, Q3,5,6,9

Question 3

We now review k-fold cross-validation.

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

KFCV is similar to LOOCV, except instead of having the data run through each row of data to compare it to the rest of the data, we instead choose smaller chunks (K=5 or 10). When we split the data in this manner, each set consists of a random set of training/testing data of approximately equal size (either 1/5 or 1/10 depending on K chosen), and then run through a linear model. Each iteration of this, the MSE is found and reported to compare and take an average.

(b) What are the advantages and disadvantages of k-fold cross-validation relative to:

i. The validation set approach?

The KFCV has less variability in test error estimates compared to the validation set approach. This is due to the KFCV utilizing all of its data, and the validation set forcing to leave test data completely separate from the training data.

ii. LOOCV?

KFCV is less intensive in regards to computations, so it is able to run much faster. This is due to KFCV using less “chunks” of computations (n = 5 or 10) compared to LOOCV having n = rows. Also, by using KFCV you are able to see multiple outputs and its average. That being said, LOOCV has less bias and higher variance than KFCV when k < n.

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.

library(ISLR)
attach(Default)

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

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

(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(1)
train <- sample(nrow(Default), nrow(Default)/2)
str(train)

##  int [1:5000] 1017 8004 4775 9725 8462 4050 8789 1301 8522 1799 ...

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

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

## 
## 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.

glm.probs = predict(glm.fit2, newdata = Default[-train, ], type="response")
glm.pred=rep("No", length(train))
glm.pred[glm.probs>0.5] = "Yes"
table(glm.pred)

## glm.pred
##   No  Yes 
## 4932   68

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

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

## [1] 0.0254

(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.

set.seed(2)
train <- sample(nrow(Default), nrow(Default)/2)
str(train)

##  int [1:5000] 4806 8465 5469 3276 3453 690 6787 4914 9015 5695 ...

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

## 
## Call:
## glm(formula = default ~ income + balance, family = "binomial", 
##     data = Default, subset = train)
## 
## Deviance Residuals: 
##     Min       1Q   Median       3Q      Max  
## -2.3702  -0.1628  -0.0673  -0.0259   3.6470  
## 
## Coefficients:
##               Estimate Std. Error z value Pr(>|z|)    
## (Intercept) -1.090e+01  5.749e-01 -18.955   <2e-16 ***
## income       1.622e-05  6.891e-06   2.354   0.0186 *  
## balance      5.365e-03  3.049e-04  17.598   <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: 1483.83  on 4999  degrees of freedom
## Residual deviance:  854.49  on 4997  degrees of freedom
## AIC: 860.49
## 
## Number of Fisher Scoring iterations: 8

glm.probs = predict(glm.fit2, newdata = Default[-train, ], type="response")
glm.pred=rep("No", length(train))
glm.pred[glm.probs>0.5] = "Yes"
table(glm.pred)

## glm.pred
##   No  Yes 
## 4920   80

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

## [1] 0.0238

set.seed(3)
train <- sample(nrow(Default), nrow(Default)/2)
str(train)

##  int [1:5000] 3770 8844 5095 6692 6842 8168 2923 5087 8584 788 ...

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

## 
## Call:
## glm(formula = default ~ income + balance, family = "binomial", 
##     data = Default, subset = train)
## 
## Deviance Residuals: 
##     Min       1Q   Median       3Q      Max  
## -2.5804  -0.1390  -0.0524  -0.0180   3.7789  
## 
## Coefficients:
##               Estimate Std. Error z value Pr(>|z|)    
## (Intercept) -1.204e+01  6.326e-01 -19.031  < 2e-16 ***
## income       2.462e-05  6.941e-06   3.547  0.00039 ***
## balance      5.894e-03  3.287e-04  17.929  < 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: 1536.99  on 4999  degrees of freedom
## Residual deviance:  793.34  on 4997  degrees of freedom
## AIC: 799.34
## 
## Number of Fisher Scoring iterations: 8

glm.probs = predict(glm.fit2, newdata = Default[-train, ], type="response")
glm.pred=rep("No", length(train))
glm.pred[glm.probs>0.5] = "Yes"
table(glm.pred)

## glm.pred
##   No  Yes 
## 4931   69

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

## [1] 0.0264

set.seed(4)
train <- sample(nrow(Default), nrow(Default)/2)
str(train)

##  int [1:5000] 5624 587 2867 1795 4167 684 4794 307 2360 6558 ...

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

## 
## Call:
## glm(formula = default ~ income + balance, family = "binomial", 
##     data = Default, subset = train)
## 
## Deviance Residuals: 
##     Min       1Q   Median       3Q      Max  
## -1.9240  -0.1468  -0.0558  -0.0191   3.5650  
## 
## Coefficients:
##               Estimate Std. Error z value Pr(>|z|)    
## (Intercept) -1.135e+01  6.096e-01 -18.613   <2e-16 ***
## income       9.504e-06  7.000e-06   1.358    0.175    
## balance      5.772e-03  3.281e-04  17.591   <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: 1477.13  on 4999  degrees of freedom
## Residual deviance:  792.65  on 4997  degrees of freedom
## AIC: 798.65
## 
## Number of Fisher Scoring iterations: 8

glm.probs = predict(glm.fit2, newdata = Default[-train, ], type="response")
glm.pred=rep("No", length(train))
glm.pred[glm.probs>0.5] = "Yes"
table(glm.pred)

## glm.pred
##   No  Yes 
## 4922   78

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

## [1] 0.0256

Of the four splits that we have run through the model, the results were: 0.0254, 0.0238, 0.0264, and 0.0256. While these numbers may not vary much, there is still some variance occurring due to each split consisting of different train/test data.

(d) Now consider a logistic regression model that predicts the prob- ability 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(1)
train <- sample(nrow(Default), nrow(Default)/2)
str(train)

##  int [1:5000] 1017 8004 4775 9725 8462 4050 8789 1301 8522 1799 ...

glm.fit3 <- glm(default~income+balance+student, data=Default, family="binomial", subset = train)
summary(glm.fit3)

## 
## Call:
## glm(formula = default ~ income + balance + student, family = "binomial", 
##     data = Default, subset = train)
## 
## Deviance Residuals: 
##     Min       1Q   Median       3Q      Max  
## -2.5823  -0.1419  -0.0554  -0.0210   3.3961  
## 
## Coefficients:
##               Estimate Std. Error z value Pr(>|z|)    
## (Intercept) -1.134e+01  6.937e-01 -16.346   <2e-16 ***
## income       1.686e-05  1.122e-05   1.502   0.1331    
## balance      5.767e-03  3.213e-04  17.947   <2e-16 ***
## studentYes  -5.992e-01  3.324e-01  -1.803   0.0715 .  
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## (Dispersion parameter for binomial family taken to be 1)
## 
##     Null deviance: 1523.77  on 4999  degrees of freedom
## Residual deviance:  800.07  on 4996  degrees of freedom
## AIC: 808.07
## 
## Number of Fisher Scoring iterations: 8

glm.probs3 = predict(glm.fit3, newdata = Default[-train, ], type="response")
glm.pred3=rep("No", length(train))
glm.pred3[glm.probs3>0.5] = "Yes"
table(glm.pred3)

## glm.pred3
##   No  Yes 
## 4937   63

mean(glm.pred3 != Default[-train, ]$default)

## [1] 0.026

There does not appear to be any difference (better or worse) to add student into the model. The test error rate remains roughly the same as previous versions of the model.

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 co- efficients 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

library(boot)

(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)
train <- sample(nrow(Default), nrow(Default)/2)
str(train)

##  int [1:5000] 1017 8004 4775 9725 8462 4050 8789 1301 8522 1799 ...

glm.fit.Q6 <- glm(default~income+balance, data=Default, family="binomial")
summary(glm.fit.Q6)$coef

##                  Estimate   Std. Error    z value      Pr(>|z|)
## (Intercept) -1.154047e+01 4.347564e-01 -26.544680 2.958355e-155
## income       2.080898e-05 4.985167e-06   4.174178  2.990638e-05
## balance      5.647103e-03 2.273731e-04  24.836280 3.638120e-136

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

(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 -3.912114e-02 4.347403e-01
## t2*  2.080898e-05  1.585717e-07 4.858722e-06
## t3*  5.647103e-03  1.856917e-05 2.300758e-04

detach(Default)

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

The Standard Errors for income and balance from glm() are 4.98e-6 and 2.27e-4, and the bootstrap functions shows them to be 4.94e-6, and 2.27e-4. These results are very similar which shows that the functions themselves are similar as well

Question 9

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

library(ISLR2)

## 
## Attaching package: 'ISLR2'

## The following objects are masked from 'package:ISLR':
## 
##     Auto, Credit

attach(Boston)
set.seed(1)

(a) Based on this data set, provide an estimate for the population mean of `medv`. Call this estimate $\hat{\mu}$ (`mu.hat`).

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

## [1] 22.53281

(b) Provide an estimate of the standard error of $\hat{\mu}$. 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(length(medv))
se

## [1] 0.4088611

(c) Now estimate the standard error of $\hat{\mu}$ using the bootstrap. How does this compare to your answer from (b)?

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

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

The difference in standard errors from question 9b (0.4088), and 9c (0.4002) are minimal - they both calculated standard error of $\hat{\mu}$ similarly.

(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$medv)`. Hint You can approximate a 95% confidence interval using the formula [$\hat{\mu}$ − 2SE($\hat{\mu}$), $\hat{\mu}$ + 2SE($\hat{\mu}$)].

medv.CI.high <- mu.hat-2*se
medv.CI.low <- mu.hat+2*se
medv.CI.high

## [1] 21.71508

medv.CI.low

## [1] 23.35053

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

The 5% and 95% confidence intervals are very similar between the t.test() and the calculated boundaries.

(e) Based on this data set, provide an estimate, $\hat{\mu}$med, for the median value of `medv` in the population.

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

## [1] 21.2

(f) We now would like to estimate the standard error of $\hat{\mu}$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.med <- function(data, index){
  mu.med1 <- median(data[index])
  return(mu.med1)
}
mu.med.boot <- boot(medv, boot.fn.med, 1000)
mu.med.boot

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

Both the bootstrap and the calculated median of medv result in a median of 21.2 (which is great because it means it works!). The standard error from the bootstrap is shown to be 0.379, which is small by comparison.

(g) Based on this data set, provide an estimate for the tenth percentile of `medv` in Boston census tracts. Call this quantity $$0.1. (You can use the `quantile()` function.)

mu.tenth <- quantile(medv, probs = 0.1)
mu.tenth

##   10% 
## 12.75

(h) Use the bootstrap to estimate the standard error of $$0.1. Comment on your findings

boot.fn.tenth <- function(data, index){
  mu.tenth1 <- quantile(data[index], probs=0.1)
  return(mu.tenth1)
}
mu.tenth.boot <- boot(medv, boot.fn.tenth, 1000)
mu.tenth.boot

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

Both g and h have a value of 12.75, and the bootstrap revealed a standard error of 0.5114 - this is still pretty small by comparison to the value

detach(Boston)

Assignment 4, Ch 5, Q3,5,6,9

Garrett Garceau

2023-02-22

Question 3

We now review k-fold cross-validation.

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

(b) What are the advantages and disadvantages of k-fold cross-validation relative to:

i. The validation set approach?

ii. LOOCV?

Question 5

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

(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.

(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.

Question 6

(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.

Question 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 \(\hat{\mu}\) (`mu.hat`).

(b) Provide an estimate of the standard error of \(\hat{\mu}\). 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 \(\hat{\mu}\) using the bootstrap. How does this compare to your answer from (b)?

(e) Based on this data set, provide an estimate, \(\hat{\mu}\)med, for the median value of `medv` in the population.

(f) We now would like to estimate the standard error of \(\hat{\mu}\)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.

(g) Based on this data set, provide an estimate for the tenth percentile of `medv` in Boston census tracts. Call this quantity $$0.1. (You can use the `quantile()` function.)

(h) Use the bootstrap to estimate the standard error of $$0.1. Comment on your findings

Assignment 4, Ch 5, Q3,5,6,9

Garrett Garceau

2023-02-22

Question 3

We now review k-fold cross-validation.

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

(b) What are the advantages and disadvantages of k-fold cross-validation relative to:

i. The validation set approach?

ii. LOOCV?

Question 5

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

(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.

(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.

Question 6

(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.

Question 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 \(\hat{\mu}\) (mu.hat).

(b) Provide an estimate of the standard error of \(\hat{\mu}\). 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 \(\hat{\mu}\) using the bootstrap. How does this compare to your answer from (b)?

(e) Based on this data set, provide an estimate, \(\hat{\mu}\)med, for the median value of medv in the population.

(f) We now would like to estimate the standard error of \(\hat{\mu}\)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.

(g) Based on this data set, provide an estimate for the tenth percentile of medv in Boston census tracts. Call this quantity $$0.1. (You can use the quantile() function.)

(h) Use the bootstrap to estimate the standard error of $$0.1. Comment on your findings

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

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.

(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.

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 \(\hat{\mu}\) (`mu.hat`).

(e) Based on this data set, provide an estimate, \(\hat{\mu}\)med, for the median value of `medv` in the population.

(g) Based on this data set, provide an estimate for the tenth percentile of `medv` in Boston census tracts. Call this quantity $$0.1. (You can use the `quantile()` function.)