library(tidyverse)
library(ISLR)
library(modelr)
library(caret)(1) Using basic statistical properties of the variance, as well as single variable calculus, derive (5.6). In other words, prove that (α) given by (5.6) does indeed minimize (Var(αX +(1 − α)Y)).
(2) We will now derive the probability that a given observation is part of a bootstrap sample. Suppose that we obtain a bootstrap sample from a set of (n) observations.
(a) What is the probability that the first bootstrap observation is not the (j)th observation from the original sample? Justify your answer.
((n-1)/n). Because each observation in the sample has the same probability to be chosen, and the probability to pick any given observation is (1/n).
(b) What is the probability that the second bootstrap observation is not the (j)th observation from the original sample?
The same as in (a), because we’re sampling with replacement.
(c) Argue that the probability that the (j)th observation is not in the bootstrap sample is ((1 − 1/n)^n).
Because the probability that an event (of probability (p)) happens (n) times in a row is (p^n), and having the (j)th observation completely left out of the bootstrap sample is the same as having the event described in (a) happening (n) times in a row.
(d) When (n = 5), what is the probability that the (j)th observation is in the bootstrap sample?
1-(1-1/5)^5## [1] 0.67232
(e) When n = 100, what is the probability that the jth observation is in the bootstrap sample?
1-(1-1/100)^100## [1] 0.6339677
(f) When (n = 10,000), what is the probability that the (j)th observation is in the bootstrap sample?
1-(1-1/10000)^10000## [1] 0.632139
(g) Create a plot that displays, for each integer value of n from 1 to 100,000, the probability that the (j)th observation is in the bootstrap sample. Comment on what you observe.
plot_data <-
tibble(
n = 1:100000,
prob = 1-(1-1/n)^n
)
ggplot(plot_data, aes(n, prob)) +
geom_line() +
expand_limits(y = 0) +
scale_x_log10()
It starts at 1, but quickly converges to 0.632.
(h) We will now investigate numerically the probability that a bootstrap sample of size n = 100 contains the jth observation. Here j = 4. We repeatedly create bootstrap samples, and each time we record whether or not the fourth observation is contained in the bootstrap sample.
store <- rep (NA , 100000)
for (i in 1:100000) {
store[i] <- sum(sample(1:100, rep =TRUE) == 4) > 0
}
mean(store)## [1] 0.63404
Comment on the results obtained.
The probability of having the (j)th observation in the sample is very close to the value to which the plot in (g) converges.
(3) We now review k-fold cross-validation.
(a) Explain how k-fold cross-validation is implemented.
The available data is randomly splited in k samples (or “folds”) of equal size. Then the model is trained k times, leaving out a different fold each time, and using it after as test o validation data.
(b) What are the advantages and disadvantages of k-fold cross validation relative to:
i. The validation set approach?
K-fold cross validation greatly reduces the variability of the test error rate estimate, compared to using the validation set approach. Also it has less bias because we’re using a higher fraction of the available data to train the model. A minor disavantage is that it slightly increases the computation cost of obtaining error rate estimates (we need to fit the model k times, not just once).
ii. LOOCV?
K-fold CV requieres less computation cost than LOOCV, because we need to fit the model k times instead of n (and most of the times k < n). Also, it has less variance in the estimation, because there is less overlap in the training data used in each iteration.
However, k-fold CV has more bias than LOOCV, since we’re using less training observations for each model.
(4) Suppose that we use some statistical learning method to make a prediction for the response Y for a particular value of the predictor X. Carefully describe how we might estimate the standard deviation of our prediction.
Some statistical learning methods, like linear or logistic regression,
provide us with the standard deviations of the coefficient estimates and
the predictions. But other methods doesn’t provide this information. In
this case, we could apply a resampling method like the bootstrap. For
this we need to create a function that computes the statistical of
interest (in this case, the prediction) using a sample of the data. Then
we pass that function to the function boot() (along with the original
data and the number B of iterations) and then we get an estimation of
the standard error of the predictions, obtained through repeteadly
fitting the model in B samples (with replacement.)
(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.
glm_default <- glm(default ~ balance + income,
data = Default,
family = "binomial")
summary(glm_default)##
## Call:
## glm(formula = default ~ balance + income, 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 ***
## balance 5.647e-03 2.274e-04 24.836 < 2e-16 ***
## income 2.081e-05 4.985e-06 4.174 2.99e-05 ***
## ---
## 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.
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.
set.seed(1989)
train_default <-
Default %>%
sample_frac(size = 0.5)
test_default <-
Default %>%
anti_join(train_default)
glm_train_default <- glm(default ~ balance + income,
data = train_default,
family = "binomial")
test_default <- test_default %>%
add_predictions(glm_train_default, type = "response") %>%
mutate(pred_class = ifelse(
pred > 0.5, "Yes", "No"),
pred_class = factor(pred_class, levels = c("No", "Yes")))
caret::confusionMatrix(test_default[["pred_class"]],
reference = test_default[["default"]])## Confusion Matrix and Statistics
##
## Reference
## Prediction No Yes
## No 4813 111
## Yes 19 57
##
## Accuracy : 0.974
## 95% CI : (0.9692, 0.9782)
## No Information Rate : 0.9664
## P-Value [Acc > NIR] : 0.001154
##
## Kappa : 0.4558
##
## Mcnemar's Test P-Value : 1.449e-15
##
## Sensitivity : 0.9961
## Specificity : 0.3393
## Pos Pred Value : 0.9775
## Neg Pred Value : 0.7500
## Prevalence : 0.9664
## Detection Rate : 0.9626
## Detection Prevalence : 0.9848
## Balanced Accuracy : 0.6677
##
## 'Positive' Class : No
##
Validation set error is as follows:
1-0.974## [1] 0.026
(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.
estimate_error_glm_default <- function(...) {
train_default <-
Default %>%
sample_frac(size = 0.5)
test_default <-
Default %>%
anti_join(train_default,
by = c("default", "student", "balance", "income"))
glm_train_default <- glm(default ~ balance + income,
data = train_default,
family = "binomial")
test_default <- test_default %>%
add_predictions(glm_train_default, type = "response") %>%
mutate(
pred_class = ifelse(pred > 0.5, "Yes", "No"),
pred_class = factor(pred_class, levels = c("No", "Yes"))
)
conf_matrix <-
caret::confusionMatrix(test_default[["pred_class"]],
reference = test_default[["default"]])
1 - conf_matrix[["overall"]][["Accuracy"]]
}
(test_error_rates <-
map_dbl(1:3, estimate_error_glm_default))## [1] 0.0222 0.0278 0.0246
mean(test_error_rates)## [1] 0.02486667
sd(test_error_rates)## [1] 0.002809508
The average test error rate is slightly higher than in (b), and we also see a standard eviation of 0.0011 in these three cases.
(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(1991)
train_default <-
Default %>%
sample_frac(size = 0.5)
test_default <-
Default %>%
anti_join(train_default)## Joining, by = c("default", "student", "balance", "income")
glm_train_default_2 <- glm(default ~ balance + income + student,
data = train_default,
family = "binomial")
test_default <- test_default %>%
add_predictions(glm_train_default, type = "response") %>%
mutate(pred_class = ifelse(
pred > 0.5, "Yes", "No"),
pred_class = factor(pred_class, levels = c("No", "Yes")))
conf_matrix <-
caret::confusionMatrix(test_default[["pred_class"]],
reference = test_default[["default"]])
1-conf_matrix[["overall"]][["Accuracy"]]## [1] 0.0272
The test error rate obtained is almost the same as with the simpler
model, so I wouldn’t say that incorporing the variable student
improves the Accuracy.
(7) 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.
summary(glm_default)##
## Call:
## glm(formula = default ~ balance + income, 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 ***
## balance 5.647e-03 2.274e-04 24.836 < 2e-16 ***
## income 2.081e-05 4.985e-06 4.174 2.99e-05 ***
## ---
## 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) 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, indexes) {
data <- data[indexes,]
glm(default ~ balance + income,
data = data,
family = "binomial") %>%
coefficients()
}(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::boot(Default, boot.fn, 1000)##
## ORDINARY NONPARAMETRIC BOOTSTRAP
##
##
## Call:
## boot::boot(data = Default, statistic = boot.fn, R = 1000)
##
##
## Bootstrap Statistics :
## original bias std. error
## t1* -1.154047e+01 1.567597e-04 4.359111e-01
## t2* 5.647103e-03 1.510398e-06 2.279440e-04
## t3* 2.080898e-05 -1.694973e-07 4.876856e-06
(d) Comment on the estimated standard errors obtained using the glm()
function and using your bootstrap function.
The estimated standard errors for all the coefficients are very close in
the glm() and bootstrap functions.
(7) In Sections 5.3.2 and 5.3.3, we saw that the cv.glm() function can
be used in order to compute the LOOCV test error estimate.
Alternatively, one could compute those quantities using just the glm()
and predict.glm() functions, and a for loop. You will now take this
approach in order to compute the LOOCV error for a simple logistic
regression model on the Weekly data set. Recall that in the context of
classification problems, the LOOCV error is given in (5.4).
(a) Fit a logistic regression model that predicts Direction using
Lag1 and Lag2.
glm(Direction ~ Lag1 + Lag2,
data = Weekly,
family = "binomial") %>%
summary()##
## Call:
## glm(formula = Direction ~ Lag1 + Lag2, family = "binomial", data = Weekly)
##
## Deviance Residuals:
## Min 1Q Median 3Q Max
## -1.623 -1.261 1.001 1.083 1.506
##
## Coefficients:
## Estimate Std. Error z value Pr(>|z|)
## (Intercept) 0.22122 0.06147 3.599 0.000319 ***
## Lag1 -0.03872 0.02622 -1.477 0.139672
## Lag2 0.06025 0.02655 2.270 0.023232 *
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## (Dispersion parameter for binomial family taken to be 1)
##
## Null deviance: 1496.2 on 1088 degrees of freedom
## Residual deviance: 1488.2 on 1086 degrees of freedom
## AIC: 1494.2
##
## Number of Fisher Scoring iterations: 4
(b) Fit a logistic regression model that predicts Direction using
Lag1 and Lag2 using all but the first observation.
glm_lfo <-
glm(Direction ~ Lag1 + Lag2,
data = Weekly[-1, ],
family = "binomial") (c) Use the model from (b) to predict the direction of the first observation. You can do this by predicting that the first observation will go up if (P(Direction ="Up"|Lag1, Lag2) > 0.5). Was this observation correctly classified?
Weekly %>%
filter(row_number() == 1) %>%
add_predictions(glm_lfo, type = "response") %>%
select(Direction, pred)## Direction pred
## 1 Down 0.5713923
It is incorrectly classified.
(d) Write a for loop from (i = 1) to (i = n), where (n) is the number of observations in the data set, that performs each of the following steps:
i. Fit a logistic regression model using all but the (ith) observation
to predict Direction using Lag1 and Lag2.
ii. Compute the posterior probability of the market moving up for the ith observation.
iii. Use the posterior probability for the ith observation in order to predict whether or not the market moves up.
iv. Determine whether or not an error was made in predicting the direction for the ith observation. If an error was made, then indicate this as a 1, and otherwise indicate it as a 0.
is_error <- vector("integer", nrow(Weekly))
for(i in 1:nrow(Weekly)) {
fit_wo_i <-
glm(Direction ~ Lag1 + Lag2,
data = Weekly[-i, ],
family = "binomial")
prediction_i <-
predict(fit_wo_i, newdata = Weekly[i,], type = "response") %>%
as.numeric()
prediction_class <- ifelse(prediction_i > 0.5,
"Up",
"Down")
error <-
ifelse(
prediction_class == as.character(Weekly[i, "Direction"]),
0,
1)
is_error[[i]] <- error
}(e) Take the average of the n numbers obtained in (d)iv in order to obtain the LOOCV estimate for the test error. Comment on the results.
mean(is_error)## [1] 0.4499541
We get 44.9% as estimate for the error rate using the LOOCV method.
(8) We will now perform cross-validation on a simulated data set.
(a) Generate a simulated data set as follows:
set.seed(1)
simulated <-
tibble(
x = rnorm(100),
y= x - 2*x^2 + rnorm (100)
)
simulated## # A tibble: 100 x 2
## x y
## <dbl> <dbl>
## 1 -0.626 -2.03
## 2 0.184 0.158
## 3 -0.836 -3.14
## 4 1.60 -3.34
## 5 0.330 -0.542
## 6 -0.820 -0.400
## 7 0.487 0.729
## 8 0.738 0.558
## 9 0.576 0.297
## 10 -0.305 1.19
## # ... with 90 more rows
In this data set, what is n and what is p? Write out the model used to generate the data in equation form.
n is 100 and p is 2. The model used to generate the data is as follows:
(y = x - 2x^2 + e) with (e) as an error term with mean 0 and standard deviation 1.
(b) Create a scatterplot of X against Y. Comment on what you find.
qplot(x, y, data = simulated)
The data follows a quadratic pattern (as expected, since y was
generated from x^2). As x moves away from zero, the value of y
goes down.
(c) Set a random seed, and then compute the LOOCV errors that result from fitting the following four models using least squares:
i. (Y = β_0 + β_1X + e) ii. (Y = β_0 + β_1X + β_2X_2 + e) iii. (Y = β_0 + β_1X + β_2X_2 + β_3X_3 + e) iv. (Y = β_0 + β_1X + β_2X_2 + β_3X_3 + β_4X_4 + e).
models <-
list(
glm(y ~ poly(x, 1), data = simulated),
glm(y ~ poly(x, 2), data = simulated),
glm(y ~ poly(x, 3), data = simulated),
glm(y ~ poly(x, 4), data = simulated)
)set.seed(1989)
loocv_error <-
map(models, ~boot::cv.glm(data = simulated, glmfit = .))
loocv_error %>%
map("delta")## [[1]]
## [1] 7.288162 7.284744
##
## [[2]]
## [1] 0.9374236 0.9371789
##
## [[3]]
## [1] 0.9566218 0.9562538
##
## [[4]]
## [1] 0.9539049 0.9534453
(d) Repeat (c) using another random seed, and report your results. Are your results the same as what you got in (c)? Why?
set.seed(1991)
loocv_error <-
map(models, ~boot::cv.glm(data = simulated, glmfit = .))
loocv_error %>%
map("delta")## [[1]]
## [1] 7.288162 7.284744
##
## [[2]]
## [1] 0.9374236 0.9371789
##
## [[3]]
## [1] 0.9566218 0.9562538
##
## [[4]]
## [1] 0.9539049 0.9534453
The results are the same because randomness doesn’t influence LOOCV results (it’s always the same iteration using all the observations in the data).
(e) Which of the models in (c) had the smallest LOOCV error? Is this what you expected? Explain your answer.
The second model (with x as a quadratic term) is the one with lowest
LOOCV error. This is as expected, since y was generated using a
quadratic function of x.
(f) Comment on the statistical significance of the coefficient estimates that results from fitting each of the models in (c) using least squares. Do these results agree with the conclusions drawn based on the cross-validation results?
map(models, summary)## [[1]]
##
## Call:
## glm(formula = y ~ poly(x, 1), data = simulated)
##
## Deviance Residuals:
## Min 1Q Median 3Q Max
## -9.5161 -0.6800 0.6812 1.5491 3.8183
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) -1.550 0.260 -5.961 3.95e-08 ***
## poly(x, 1) 6.189 2.600 2.380 0.0192 *
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## (Dispersion parameter for gaussian family taken to be 6.760719)
##
## Null deviance: 700.85 on 99 degrees of freedom
## Residual deviance: 662.55 on 98 degrees of freedom
## AIC: 478.88
##
## Number of Fisher Scoring iterations: 2
##
##
## [[2]]
##
## Call:
## glm(formula = y ~ poly(x, 2), data = simulated)
##
## Deviance Residuals:
## Min 1Q Median 3Q Max
## -1.9650 -0.6254 -0.1288 0.5803 2.2700
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) -1.5500 0.0958 -16.18 < 2e-16 ***
## poly(x, 2)1 6.1888 0.9580 6.46 4.18e-09 ***
## poly(x, 2)2 -23.9483 0.9580 -25.00 < 2e-16 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## (Dispersion parameter for gaussian family taken to be 0.9178258)
##
## Null deviance: 700.852 on 99 degrees of freedom
## Residual deviance: 89.029 on 97 degrees of freedom
## AIC: 280.17
##
## Number of Fisher Scoring iterations: 2
##
##
## [[3]]
##
## Call:
## glm(formula = y ~ poly(x, 3), data = simulated)
##
## Deviance Residuals:
## Min 1Q Median 3Q Max
## -1.9765 -0.6302 -0.1227 0.5545 2.2843
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) -1.55002 0.09626 -16.102 < 2e-16 ***
## poly(x, 3)1 6.18883 0.96263 6.429 4.97e-09 ***
## poly(x, 3)2 -23.94830 0.96263 -24.878 < 2e-16 ***
## poly(x, 3)3 0.26411 0.96263 0.274 0.784
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## (Dispersion parameter for gaussian family taken to be 0.9266599)
##
## Null deviance: 700.852 on 99 degrees of freedom
## Residual deviance: 88.959 on 96 degrees of freedom
## AIC: 282.09
##
## Number of Fisher Scoring iterations: 2
##
##
## [[4]]
##
## Call:
## glm(formula = y ~ poly(x, 4), data = simulated)
##
## Deviance Residuals:
## Min 1Q Median 3Q Max
## -2.0550 -0.6212 -0.1567 0.5952 2.2267
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) -1.55002 0.09591 -16.162 < 2e-16 ***
## poly(x, 4)1 6.18883 0.95905 6.453 4.59e-09 ***
## poly(x, 4)2 -23.94830 0.95905 -24.971 < 2e-16 ***
## poly(x, 4)3 0.26411 0.95905 0.275 0.784
## poly(x, 4)4 1.25710 0.95905 1.311 0.193
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## (Dispersion parameter for gaussian family taken to be 0.9197797)
##
## Null deviance: 700.852 on 99 degrees of freedom
## Residual deviance: 87.379 on 95 degrees of freedom
## AIC: 282.3
##
## Number of Fisher Scoring iterations: 2
Yes, the statistical significance results match the conclusions from the
LOOCV. The terms x^3 and x^4 are not statistically significant in
the models iii and iv, and also adding those terms doesn’t reduce the
LOOCV error below the one in the model ii.
- We will now consider the
Bostonhousing data set, from theMASSlibrary.
(a) Based on this data set, provide an estimate for the population mean
of medv. Call this estimate (\hat{\mu}).
Boston <- MASS::Boston
Boston[["medv"]] %>% mean()## [1] 22.53281
(b) Provide an estimate of the standard error of (\hat{\mu}). Interpret this result.
sd_medv <- Boston[["medv"]] %>% sd()
n_obs <- nrow(Boston)
sd_medv/sqrt(n_obs)## [1] 0.4088611
(c) Now estimate the standard error of (\hat{\mu}) using the bootstrap. How does this compare to your answer from (b)?
get_mean_boot <- function(data, indexes) {
data[indexes] %>% mean()
}
(mean_by_boot <-
boot::boot(Boston[["medv"]], get_mean_boot, 1000))##
## ORDINARY NONPARAMETRIC BOOTSTRAP
##
##
## Call:
## boot::boot(data = Boston[["medv"]], statistic = get_mean_boot,
## R = 1000)
##
##
## Bootstrap Statistics :
## original bias std. error
## t1* 22.53281 0.003395652 0.4081243
The bootstrap estimate for the standard error is a bit higher than the one calculed in (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).
sd_boot <- sd(mean_by_boot[["t"]])
mean_boot <- mean_by_boot[["t0"]]
c(mean_boot - 2*sd_boot, mean_boot + 2*sd_boot)## [1] 21.71656 23.34905
(e) Based on this data set, provide an estimate, (\hat{\mu}_{med}),
for the median value of medv in the population.
Boston[["medv"]] %>% median()## [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.
get_median_boot <- function(data, indexes) {
data[indexes] %>% median()
}
(median_by_boot <-
boot::boot(Boston[["medv"]], get_median_boot, 1000))##
## ORDINARY NONPARAMETRIC BOOTSTRAP
##
##
## Call:
## boot::boot(data = Boston[["medv"]], statistic = get_median_boot,
## R = 1000)
##
##
## Bootstrap Statistics :
## original bias std. error
## t1* 21.2 -0.02035 0.3908339
The bootstrap estimate for the median itself is the same as in (e), and the estimate for standard error is lower than the estimate for the mean.
(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
(h) Use the bootstrap to estimate the standard error of (\hat{\mu}_{0.1}).Comment on your findings.
get_quantile10_by_boot <- function(data, indexes) {
data[indexes] %>% quantile(., 0.1)
}
boot::boot(Boston[["medv"]], get_quantile10_by_boot, 1000)##
## ORDINARY NONPARAMETRIC BOOTSTRAP
##
##
## Call:
## boot::boot(data = Boston[["medv"]], statistic = get_quantile10_by_boot,
## R = 1000)
##
##
## Bootstrap Statistics :
## original bias std. error
## t1* 12.75 0.015 0.4885936