Assignment 4

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

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

The k-fold CV approach is implemented by randomly dividing the set of observations into k groups, or folds, of approximately equal size. The first fold is treated as a validation set, and a model is trained using the remaining k − 1 folds. The mean squared error, MSE₁, is then computed on the observations in the held-out fold. This procedure is repeated k times; each time, a different group of observations is treated as a validation set. This process results in k estimates of the test error, MSE₁,MSE₂, . . . ,MSE_k. The k-fold CV estimate is computed by averaging these values,

\[CV_{(k)} = \frac{1}{k}\sum_{i=1}^{k} MSE_i\]

Normally, the k-fold CV is performed with k = 5 or k = 10.

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

i. The validation set approach?

The disadvantages of Validation Set over k-fold is that the 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 as they are randomly divided from the available set of observations. Also, since only subset of observations are used in the training set than the validation set, the statistical method tends to perform worst with the validation set error rate tend to overestimate the test error rate as compare to the model that fits entire data set (which is the case in k-fold). On the other hand, the Validation Set is conceptually simple and easy to implement as compared to k-fold which is more computational than the Validation Set.

ii. LOOCV?

LOOCV is a special case of k-fold CV when k = n, where n is number of observations and in k-fold CV, the k is either 5 or 10. Because of this, k-fold has computational advantage than the LOOCV where LOOCV has to fit the statistical models n times (especially when n is extremely large). The other advantage of k-fold is that it has bias-variance trade off, that is, k-fold gives more accurate estimates of the test error rate. On the other hand, since the training are performed fewer sets than the LOOCV, it will lead to intermediate level of bias. So, from the bias reduction perspective, LOOCV is preferred than k-fold. In contrary to this, since k-fold CV are performed with k < n, the averaging outputs of the k fitted models are less correlated with each other as the overlap of training sets with each model is smaller. Because of this, the test error estimate resulting from LOOCV tends to have higher variance than the test error estimate resulting from 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`.

attach(Default)
set.seed(100)
glm.default.fit <- glm(default ~ income + balance, data = Default, family = "binomial")
summary(glm.default.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

Since the p-value for both predictors income and balance is much smaller than the significance value (0.05), their association with default variable is significant.

(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.

Using 75-25% split of the dataset for training and validation.

set.seed(120)
sample_split_75 <- sample.split(Default$default, SplitRatio=3/4)
train75 <- subset(Default, sample_split_75 == TRUE)
validation25 <-subset(Default, sample_split_75 == FALSE)

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

glm.train75.fit <- glm(default ~ income + balance, data = train75, family = "binomial")
summary(glm.train75.fit)

## 
## Call:
## glm(formula = default ~ income + balance, family = "binomial", 
##     data = train75)
## 
## Deviance Residuals: 
##     Min       1Q   Median       3Q      Max  
## -2.2077  -0.1451  -0.0575  -0.0210   3.7239  
## 
## Coefficients:
##               Estimate Std. Error z value Pr(>|z|)    
## (Intercept) -1.167e+01  5.024e-01 -23.219  < 2e-16 ***
## income       2.455e-05  5.713e-06   4.296 1.74e-05 ***
## balance      5.660e-03  2.635e-04  21.478  < 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: 2192.2  on 7499  degrees of freedom
## Residual deviance: 1188.6  on 7497  degrees of freedom
## AIC: 1194.6
## 
## 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.

default25.probs <- predict(glm.train75.fit, validation25, type="response")
default25.pred <- rep("No", length(validation25$default))
default25.pred[default25.probs > .5] <- "Yes"
table(default25.pred, validation25$default)

##               
## default25.pred   No  Yes
##            No  2407   50
##            Yes   10   33

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

mean(default25.pred != validation25$default)

## [1] 0.024

From the above mean calculation, the misclassification rate is 2.4% (validation set error rate). This can be be calculated from (iii) table as well, i.e., ((50+10)/2500) * 100 = 2.4%.

(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.

Split 1 (1/2 = 0.50)

#Split 50%
set.seed(200)
sample_split_50 <- sample.split(Default$default, SplitRatio=1/2)
train50 <- subset(Default, sample_split_50 == TRUE)
validation50 <-subset(Default, sample_split_50 == FALSE)
#Run the regression
glm.train50.fit <- glm(default ~ income + balance, data = train50, family = "binomial")
#Run the prediction
default50.probs <- predict(glm.train50.fit, validation50, type="response")
default50.pred <- rep("No", length(validation50$default))
default50.pred[default50.probs > .5] <- "Yes"
table(default50.pred, validation50$default)

##               
## default50.pred   No  Yes
##            No  4827  120
##            Yes    6   47

mean(default50.pred != validation50$default)

## [1] 0.0252

From the above mean calculation, the misclassification rate for the Split 1 (50-50 split) is 2.52% (validation set error rate). This can be be calculated from prediction table as well, i.e., ((120+6)/5000) * 100 = 2.52%.

Split 2 (70-30 ratio)

#Split 70%
set.seed(300)
sample_split_70 <- sample.split(Default$default, SplitRatio=0.70)
train70 <- subset(Default, sample_split_70 == TRUE)
validation30 <-subset(Default, sample_split_70 == FALSE)
#Run the regression
glm.train70.fit <- glm(default ~ income + balance, data = train70, family = "binomial")
#Run the prediction
default30.probs <- predict(glm.train70.fit, validation30, type="response")
default30.pred <- rep("No", length(validation30$default))
default30.pred[default30.probs > .5] <- "Yes"
table(default30.pred, validation30$default)

##               
## default30.pred   No  Yes
##            No  2887   67
##            Yes   13   33

mean(default30.pred != validation30$default)

## [1] 0.02666667

From the above mean calculation, the misclassification rate for the Split 2 (70-30 split) is 2.67% (validation set error rate). This can be be calculated from prediction table as well, i.e., ((13+67)/3000) * 100 = 2.67%.

Split 3 (4/5 = 0.80)

#Split 80%
set.seed(350)
sample_split_80 <- sample.split(Default$default, SplitRatio=4/5)
train80 <- subset(Default, sample_split_80 == TRUE)
validation20 <-subset(Default, sample_split_80 == FALSE)
#Run the regression
glm.train80.fit <- glm(default ~ income + balance, data = train80, family = "binomial")
#Run the prediction
default20.probs <- predict(glm.train80.fit, validation20, type="response")
default20.pred <- rep("No", length(validation20$default))
default20.pred[default20.probs > .5] <- "Yes"
table(default20.pred, validation20$default)

##               
## default20.pred   No  Yes
##            No  1926   46
##            Yes    7   21

mean(default20.pred != validation20$default)

## [1] 0.0265

From the above mean calculation, the misclassification rate for the Split 3 (80-20 split) is 2.65% (validation set error rate). This can be be calculated from prediction table as well, i.e., ((7+46)/2000) * 100 = 2.65%.

In all the 3 splits, we see that the 50-50 split has the lowest misclassification rate of 2.52%. But, when this is compared to the 75-25 split (2.4%) in (b), its marginally higher.

(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.

For this logistic regression model we are using 75-25% split as we found out this is the lowest so far from (b) and (c).

#Split 75-25% ratio
set.seed(120)
sample_dummy_75 <- sample.split(Default$default, SplitRatio=0.75)
train.dummy.75 <- subset(Default, sample_dummy_75 == TRUE)
val.dummy.25 <-subset(Default, sample_dummy_75 == FALSE)
#Run the regression
glm.train.dummy.fit <- glm(default ~ income + balance + student, data = train.dummy.75, family = "binomial")
#Run the prediction
def.dummy.probs <- predict(glm.train.dummy.fit, val.dummy.25, type="response")
def.dummy.pred <- rep("No", length(val.dummy.25$default))
def.dummy.pred[def.dummy.probs > .5] <- "Yes"
table(def.dummy.pred, val.dummy.25$default)

##               
## def.dummy.pred   No  Yes
##            No  2405   54
##            Yes   12   29

mean(def.dummy.pred != val.dummy.25$default)

## [1] 0.0264

From the above mean calculation, the misclassification rate is 2.64% (validation set error rate). This can be be calculated from prediction table as well, i.e., ((12+54)/2500) * 100 = 2.64%. This is higer than the same split without the addition of the dummy varaible student in the logistic regression model in (b). So, the additional variable DID NOT reduce the test error rate.

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(120)
#Fit the logistic regression model for `income` and `balance`
glm.default.fit <- glm(default ~ income + balance, data = Default, family = "binomial")
summary(glm.default.fit)$coefficients[2:3,]

##             Estimate   Std. Error   z value      Pr(>|z|)
## income  2.080898e-05 4.985167e-06  4.174178  2.990638e-05
## balance 5.647103e-03 2.273731e-04 24.836280 3.638120e-136

The estimated standard errors for the coefficients associated with income and balance are 4.985167e-06 and 2.273731e-04 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) {
  glm.fit <- glm(default ~ income + balance, data = data, family = "binomial", subset = index)
  return(summary(glm.fit)$coefficients[2:3, 2])
}

(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)
set.seed(600)
boot(Default, boot.fn, R=500)

## 
## ORDINARY NONPARAMETRIC BOOTSTRAP
## 
## 
## Call:
## boot(data = Default, statistic = boot.fn, R = 500)
## 
## 
## Bootstrap Statistics :
##         original       bias     std. error
## t1* 4.985167e-06 5.416004e-09 1.386681e-07
## t2* 2.273731e-04 8.407107e-07 1.065559e-05

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

The estimated standard errors of the coefficients of the income and balance predictors from the logistic regression function glm() are SE($\hat\beta_{income}$) = 4.985167e-06 and SE($\hat\beta_{balance}$) = 2.273731e-04 respectively.

The estimate standard errors of the coefficients of the income and balance predictors from the bootstrap function are SE($\hat\beta_{income}$) = 1.386681e-07 and SE($\hat\beta_{balance}$) = 1.065559e-05 respectively.

The Standard Errors of the coefficients of the income and balance from the bootstrap function is lesser than that of the logistic regression. This is because the bootstrap function does not rely on any assumptions that the logistic regression function are.

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

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

mu_hat <- mean(Boston$medv)
mu_hat

## [1] 22.53281

$\hat\mu$ = 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_mu_hat <- sd(Boston$medv) / sqrt(length(Boston$medv))
se_mu_hat

## [1] 0.4088611

The standard error of $\hat\mu$ = 0.4088611

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

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

## 
## ORDINARY NONPARAMETRIC BOOTSTRAP
## 
## 
## Call:
## boot(data = Boston$medv, statistic = boot.fn, R = 1000)
## 
## 
## Bootstrap Statistics :
##     original     bias    std. error
## t1* 22.53281 0.01153162   0.4154223

The standard error of $\hat\mu$ from the bootstrap is 0.4154223, which is very close to the standard error of $\hat\mu$ 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 [$\hat\mu$ − 2SE($\hat\mu$), $\hat\mu$ + 2SE($\hat\mu$)].

mu_hat_lower <- (22.53281 - (2 * 0.4154223))
mu_hat_upper <- (22.53281 + (2 * 0.4154223))
sprintf("The lower bound of the confidence interval = %f", mu_hat_lower)

## [1] "The lower bound of the confidence interval = 21.701965"

sprintf("The upper bound of the confidence interval = %f", mu_hat_upper)

## [1] "The upper bound of the confidence interval = 23.363655"

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 confidence intervals that are calculated from the Bootstrap estimated standard error are very close to the One Sample t-test confidence intervals estimate (even we can say both are approximately same).

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

mu_hat_median <- median(Boston$medv)
mu_hat_median

## [1] 21.2

$\hat\mu_{median}$ = 21.2

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

## 
## ORDINARY NONPARAMETRIC BOOTSTRAP
## 
## 
## Call:
## boot(data = Boston$medv, statistic = boot.fn, R = 1000)
## 
## 
## Bootstrap Statistics :
##     original   bias    std. error
## t1*     21.2 -0.00225   0.3861691

The estimate of the standard error of the median using the bootstrap, $\hat\mu_{median}$ = 0.3861691.
The standard error of the median, $\hat\mu_{median}$, is smaller than the standard error of the mean, $\hat\mu$ = 0.4154223, which indicates that the median estimate of the entire population could be accurate.

(g) Based on this data set, provide an estimate for the tenth percentile of `medv` in Boston suburbs. Call this quantity $\hat\mu_{0.1}$. (You can use the `quantile()` function.)

quantile(Boston$medv, 0.1)

##   10% 
## 12.75

The estimate of the tenth percentile of medv, $\hat\mu_{0.1}$ = 12.75.

(h) Use the bootstrap to estimate the standard error of $\hat\mu_{0.1}$. Comment on your findings.

set.seed(1000)
boot.fn <- function(data, index) {
    mu.hat.quantile <- quantile(data[index], 0.1)
    return (mu.hat.quantile)
}
boot(Boston$medv, boot.fn, 1000)

## 
## ORDINARY NONPARAMETRIC BOOTSTRAP
## 
## 
## Call:
## boot(data = Boston$medv, statistic = boot.fn, R = 1000)
## 
## 
## Bootstrap Statistics :
##     original  bias    std. error
## t1*    12.75  0.0153   0.5081034

The estimated standard error of $\hat\mu_{0.1}$ = 0.5081034, which is very small compared to the percentile value in (g). Also, the estimated tenth percentile value 12.75 from bootstrap is same as the value in (g).

Assignment 4

Alexis Chinnappan (UTSA ID: sez126)

7/7/2021

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?

(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.

Split 1 (1/2 = 0.50)

Split 2 (70-30 ratio)

Split 3 (4/5 = 0.80)

(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 `MASS` library.

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

(b) Provide an estimate of the standard error of \(\hat\mu\). Interpret this result.

(c) Now estimate the standard error of \(\hat\mu\) 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)`.

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

(f) We now would like to estimate the standard error of \(\hat\mu_{median}\). 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 suburbs. Call this quantity \(\hat\mu_{0.1}\). (You can use the `quantile()` function.)

(h) Use the bootstrap to estimate the standard error of \(\hat\mu_{0.1}\). Comment on your findings.

Assignment 4

Alexis Chinnappan (UTSA ID: sez126)

7/7/2021

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?

(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.

Split 1 (1/2 = 0.50)

Split 2 (70-30 ratio)

Split 3 (4/5 = 0.80)

(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 MASS library.

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

(b) Provide an estimate of the standard error of \(\hat\mu\). Interpret this result.

(c) Now estimate the standard error of \(\hat\mu\) 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).

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

(f) We now would like to estimate the standard error of \(\hat\mu_{median}\). 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 suburbs. Call this quantity \(\hat\mu_{0.1}\). (You can use the quantile() function.)

(h) Use the bootstrap to estimate the standard error of \(\hat\mu_{0.1}\). 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.

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

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

(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)`.

(e) Based on this data set, provide an estimate, \(\hat\mu_{median}\), 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 suburbs. Call this quantity \(\hat\mu_{0.1}\). (You can use the `quantile()` function.)