INTRODUCTION TO STATISCAL LEARNING
CHAPTER 5: RESAMPLING METHODS
Applied Exercises
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.
- Fit a logistic regression model that uses income and balance to predict default.
library(ISLR)
set.seed(1)
fit.glm = glm(default ~ income + balance, data = Default, family = "binomial")
summary(fit.glm)##
## 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- Using the validation set approach, estimate the test error of this model. In order to do this, you must perform the following steps:
- Split the sample set into a training set and a validation set.
train = sample(dim(Default)[1], dim(Default)[1] / 2)- Fit a multiple logistic regression model using only the train- ing observations.
fit.glm = glm(default ~ income + balance, data = Default[train,], family = "binomial")
fit.glm = glm(default ~ income + balance, data = Default, family = "binomial", subset = train)
#Both above formulas have the same outcome.
summary(fit.glm)##
## Call:
## glm(formula = default ~ income + balance, family = "binomial",
## data = Default, subset = train)
##
## Deviance Residuals:
## Min 1Q Median 3Q Max
## -2.3583 -0.1268 -0.0475 -0.0165 3.8116
##
## Coefficients:
## Estimate Std. Error z value Pr(>|z|)
## (Intercept) -1.208e+01 6.658e-01 -18.148 <2e-16 ***
## income 1.858e-05 7.573e-06 2.454 0.0141 *
## balance 6.053e-03 3.467e-04 17.457 <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: 1457.0 on 4999 degrees of freedom
## Residual deviance: 734.4 on 4997 degrees of freedom
## AIC: 740.4
##
## Number of Fisher Scoring iterations: 8- 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(fit.glm, newdata = Default[-train, ], type="response")
glm.pred=rep("No",5000)
glm.pred[glm.probs>0.5] = "Yes"- Compute the validation set error, which is the fraction of the observations in the validation set that are misclassified.
mean(glm.pred != Default[-train, ]$default)## [1] 0.0286- Repeat the process in (b) three times, using three different splits of the observations into a training set and a validation set. Comment on the results obtained.
train <- sample(dim(Default)[1], dim(Default)[1] / 2)
fit.glm <- glm(default ~ income + balance, data = Default, family = "binomial", subset = train)
probs <- predict(fit.glm, newdata = Default[-train, ], type = "response")
pred.glm <- rep("No", length(probs))
pred.glm[probs > 0.5] <- "Yes"
mean(pred.glm != Default[-train, ]$default)## [1] 0.0236train <- sample(dim(Default)[1], dim(Default)[1] / 2)
fit.glm <- glm(default ~ income + balance, data = Default, family = "binomial", subset = train)
probs <- predict(fit.glm, newdata = Default[-train, ], type = "response")
pred.glm <- rep("No", length(probs))
pred.glm[probs > 0.5] <- "Yes"
mean(pred.glm != Default[-train, ]$default)## [1] 0.028train <- sample(dim(Default)[1], dim(Default)[1] / 2)
fit.glm <- glm(default ~ income + balance, data = Default, family = "binomial", subset = train)
probs <- predict(fit.glm, newdata = Default[-train, ], type = "response")
pred.glm <- rep("No", length(probs))
pred.glm[probs > 0.5] <- "Yes"
mean(pred.glm != Default[-train, ]$default)## [1] 0.0268We see that the validation estimate of the test error rate can be variable, depending on precisely which observations are included in the training set and which observations are included in the validation set.
- 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.
train <- sample(dim(Default)[1], dim(Default)[1] / 2)
fit.glm <- glm(default ~ income + balance + student, data = Default, family = "binomial", subset = train)
pred.glm <- rep("No", length(probs))
probs <- predict(fit.glm, newdata = Default[-train, ], type = "response")
pred.glm[probs > 0.5] <- "Yes"
mean(pred.glm != Default[-train, ]$default)## [1] 0.0264It doesn’t seem that adding the “student” dummy variable leads to a reduction in the validation set estimate of 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.
- 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(dim(Default)[1], dim(Default)[1] / 2)
fit.glm <- glm(default ~ income + balance, data = Default, family = "binomial", subset = train)
summary(fit.glm)##
## Call:
## glm(formula = default ~ income + balance, family = "binomial",
## data = Default, subset = train)
##
## Deviance Residuals:
## Min 1Q Median 3Q Max
## -2.3583 -0.1268 -0.0475 -0.0165 3.8116
##
## Coefficients:
## Estimate Std. Error z value Pr(>|z|)
## (Intercept) -1.208e+01 6.658e-01 -18.148 <2e-16 ***
## income 1.858e-05 7.573e-06 2.454 0.0141 *
## balance 6.053e-03 3.467e-04 17.457 <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: 1457.0 on 4999 degrees of freedom
## Residual deviance: 734.4 on 4997 degrees of freedom
## AIC: 740.4
##
## Number of Fisher Scoring iterations: 8The glm() estimates of the standard errors for the coefficients β0, β1 and β2 are respectively 0.4347564, 4.9851672 x 10-6 and 2.2737314 x 10-4.
- 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) {
fit <- glm(default ~ income + balance, data = data, family = "binomial", subset = index)
return (coef(fit))
}- 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)
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 -8.331926e-03 4.240329e-01
## t2* 2.080898e-05 5.792741e-08 4.590086e-06
## t3* 5.647103e-03 2.526993e-06 2.268457e-04The bootstrap estimates of the standard errors for the coefficients β0, β1 and β2 are respectively 0.4239, 4.583 x 10(-6) and 2.268 x 10(-4).
- Comment on the estimated standard errors obtained using the glm() function and using your bootstrap function.
The estimated standard errors obtained by the two methods are pretty close.
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).
- Fit a logistic regression model that predicts Direction using Lag1 and Lag2.
fit.glm = glm(Direction ~ Lag1 + Lag2, data= Weekly, family = binomial)
summary(fit.glm)##
## 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- Fit a logistic regression model that predicts Direction using Lag1 and Lag2 using all but the first observation.
fit.glm = glm(Direction ~ Lag1 + Lag2, data= Weekly[-1,], family = binomial)
summary(fit.glm)##
## Call:
## glm(formula = Direction ~ Lag1 + Lag2, family = binomial, data = Weekly[-1,
## ])
##
## Deviance Residuals:
## Min 1Q Median 3Q Max
## -1.6258 -1.2617 0.9999 1.0819 1.5071
##
## Coefficients:
## Estimate Std. Error z value Pr(>|z|)
## (Intercept) 0.22324 0.06150 3.630 0.000283 ***
## Lag1 -0.03843 0.02622 -1.466 0.142683
## Lag2 0.06085 0.02656 2.291 0.021971 *
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## (Dispersion parameter for binomial family taken to be 1)
##
## Null deviance: 1494.6 on 1087 degrees of freedom
## Residual deviance: 1486.5 on 1085 degrees of freedom
## AIC: 1492.5
##
## Number of Fisher Scoring iterations: 4- 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?
predict(fit.glm, newdata = Weekly[1,], type = "response") > 0.5## 1
## TRUEWeekly[1,]$Direction## [1] Down
## Levels: Down UpWe may conclude that the prediction for the first observation is “Up”. This observation was not correctly classified as the true direction is “Down”.
- 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:
Fit a logistic regression model using all but the ith observation to predict Direction using Lag1 and Lag2.
Compute the posterior probability of the market moving up for the ith observation.
Use the posterior probability for the ith observation in order to predict whether or not the market moves up.
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.
error <- rep(0, dim(Weekly)[1])
for (i in 1:dim(Weekly)[1]) {
fit.glm <- glm(Direction ~ Lag1 + Lag2, data = Weekly[-i, ], family = "binomial")
pred.up <- predict.glm(fit.glm, Weekly[i, ], type = "response") > 0.5
true.up <- Weekly[i, ]$Direction == "Up"
if (pred.up != true.up)
error[i] <- 1
}
error## [1] 1 1 0 1 0 1 0 0 0 1 1 0 1 0 1 0 1 0 1 0 0 0 1 1 1 1 1 1 0 1 1 1 1 0
## [35] 1 0 0 0 1 0 1 0 0 1 0 1 1 1 0 1 0 0 0 1 0 0 1 1 0 0 0 0 1 0 1 1 0 0
## [69] 1 0 1 1 0 0 0 1 0 1 1 0 0 1 1 0 1 1 0 0 1 0 0 1 1 1 0 0 0 0 0 1 0 1
## [103] 1 0 0 1 0 1 0 0 1 1 0 0 1 0 0 1 0 0 1 1 1 1 0 0 0 1 0 1 0 1 1 0 0 0
## [137] 1 1 1 0 0 0 1 0 0 0 0 0 0 1 1 1 0 1 0 0 1 1 0 1 0 0 1 1 0 0 1 0 0 1
## [171] 0 0 1 1 1 0 1 0 1 0 0 0 0 0 0 0 0 1 1 0 1 0 1 0 1 0 1 0 0 1 0 0 1 0
## [205] 0 1 0 1 0 1 1 1 0 0 1 1 0 1 0 0 1 1 0 0 0 1 1 1 0 1 0 1 0 1 0 0 0 1
## [239] 1 0 1 0 1 0 1 0 1 0 1 1 0 1 0 0 1 0 0 1 0 0 0 0 0 1 0 0 0 1 0 0 1 0
## [273] 0 0 1 0 0 1 0 0 1 0 0 1 0 1 1 0 0 0 0 0 1 0 1 0 0 1 0 0 0 1 0 0 1 1
## [307] 0 0 1 0 0 0 0 1 0 1 1 0 0 1 0 1 0 1 1 0 0 0 1 0 1 0 0 1 1 1 1 0 1 0
## [341] 0 1 0 0 0 1 0 1 0 1 0 0 0 0 0 1 1 0 0 1 0 0 1 0 0 0 1 1 0 1 1 1 1 1
## [375] 0 0 0 1 0 0 0 0 0 0 1 0 1 1 0 0 1 1 0 0 0 0 0 1 0 0 1 1 1 0 1 0 1 0
## [409] 1 1 1 0 1 0 0 0 0 0 0 0 0 0 0 1 0 1 0 1 0 1 0 0 1 0 1 0 0 0 0 0 1 1
## [443] 1 1 0 1 1 0 1 0 1 1 0 1 0 0 1 0 0 1 1 0 0 0 0 1 1 0 0 1 0 1 0 0 0 1
## [477] 0 0 1 0 0 0 1 1 1 0 1 0 0 0 1 0 1 1 1 0 0 0 0 1 1 1 0 1 1 0 1 0 0 0
## [511] 1 0 1 0 0 0 1 0 1 1 0 0 1 1 0 0 0 1 1 0 1 0 1 1 1 1 1 0 0 0 1 0 0 0
## [545] 1 1 0 1 0 0 0 1 1 1 1 1 1 0 1 0 1 0 0 1 0 0 1 1 1 0 0 0 1 1 1 1 1 1
## [579] 1 1 1 1 0 1 0 0 1 0 0 1 0 1 0 0 1 0 0 1 0 1 1 0 1 1 1 0 1 0 1 0 1 0
## [613] 0 0 1 0 1 0 1 0 1 1 0 1 1 0 1 0 1 0 1 1 1 1 0 1 1 0 0 0 1 1 1 1 0 1
## [647] 1 1 0 1 0 0 0 1 1 1 1 1 1 0 1 0 0 1 0 0 0 1 1 0 1 0 1 1 1 1 0 0 0 1
## [681] 1 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 1 0 0 1 0 1 1 0 0 0 0 1 0 1 0 1 0 1
## [715] 0 0 1 1 0 0 0 0 0 0 0 0 0 1 0 0 1 0 0 1 1 1 0 1 1 0 1 1 1 1 0 0 0 1
## [749] 1 1 1 1 1 0 1 0 0 0 0 0 0 1 0 1 1 1 0 0 0 1 0 0 1 0 0 0 1 1 1 0 0 0
## [783] 1 0 0 1 1 1 0 0 1 0 1 0 1 0 0 1 0 0 1 0 0 0 0 0 1 0 1 1 0 0 1 1 0 1
## [817] 1 1 0 0 0 0 0 1 1 0 0 1 0 0 1 0 1 0 0 0 1 1 0 1 1 0 1 0 1 0 1 1 0 0
## [851] 1 1 1 0 1 1 0 0 0 1 0 1 0 1 0 1 0 0 0 0 0 1 0 0 1 1 0 0 1 0 1 0 1 1
## [885] 0 1 0 1 1 0 0 0 1 0 0 0 0 0 0 0 1 0 1 0 1 0 0 0 1 1 1 1 1 0 1 1 0 0
## [919] 0 0 0 1 0 0 1 0 0 0 0 1 0 1 1 1 0 0 1 1 0 1 1 1 1 0 1 0 1 0 1 0 1 0
## [953] 1 0 0 1 1 1 1 1 0 1 0 0 0 1 1 1 0 1 1 1 1 0 0 0 0 1 1 0 0 0 0 1 0 0
## [987] 1 1 1 0 0 1 1 1 0 1 0 0 0 0 1 0 0 1 0 1 0 0 1 1 1 1 0 1 0 0 1 0 0 1
## [1021] 0 0 1 1 0 1 1 1 0 1 1 0 0 0 1 0 1 0 1 1 1 1 0 0 1 0 0 0 0 0 0 1 0 1
## [1055] 1 0 0 0 0 0 1 1 1 0 0 0 1 0 1 1 1 0 0 0 0 1 0 0 0 0 0 1 0 1 0 0 0 0
## [1089] 0- 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(error)## [1] 0.4499541The LOOCV estimate for the test error rate is 44.9954086%.
8. We will now perform cross-validation on a simulated data set.
- Generate a simulated data set as follows:
set.seed(1)
y <- rnorm(100)
x=rnorm(100)
y=x-2*x^2+rnorm(100)In this data set, what is n and what is p? Write out the model used to generate the data in equation form.
Here we have that n=100 and p=2, the model used is \(Y=X−2X^2+ε.\)
- Create a scatterplot of X against Y . Comment on what you find.
plot(x,y)
(c) Set a random seed, and then compute the LOOCV errors that result from fitting the following four models using least squares:
- Y = β0 + β1X + ε
set.seed(1)
Data <- data.frame(x, y)
fit.glm.1 <- glm(y ~ x)
cv.glm(Data, fit.glm.1)$delta[1]## [1] 5.890979- Y = β0 + β1X + β2X2 + ε
fit.glm.2 <- glm(y ~ poly(x, 2))
cv.glm(Data, fit.glm.2)$delta[1]## [1] 1.086596- Y = β0 +β1X +β2X2 +β3X3 +ε
fit.glm.3 <- glm(y ~ poly(x, 3))
cv.glm(Data, fit.glm.3)$delta[1]## [1] 1.102585- Y = β0 +β1X +β2X2 +β3X3 +β4X4 +ε.
fit.glm.4 <- glm(y ~ poly(x, 4))
cv.glm(Data, fit.glm.4)$delta[1]## [1] 1.114772Note you may find it helpful to use the data.frame() function to create a single data set containing both X and Y .
- Repeat (c) using another random seed, and report your results. Are your results the same as what you got in (c)? Why?
set.seed(10)
fit.glm.1 <- glm(y ~ x)
cv.glm(Data, fit.glm.1)$delta[1]## [1] 5.890979fit.glm.2 <- glm(y ~ poly(x, 2))
cv.glm(Data, fit.glm.2)$delta[1]## [1] 1.086596fit.glm.3 <- glm(y ~ poly(x, 3))
cv.glm(Data, fit.glm.3)$delta[1]## [1] 1.102585fit.glm.4 <- glm(y ~ poly(x, 4))
cv.glm(Data, fit.glm.4)$delta[1]## [1] 1.114772The results above are identical to the results obtained in (c) since LOOCV evaluates n folds of a single observation.
- Which of the models in (c) had the smallest LOOCV error ? Is this what you expected ? Explain your answer.
We may see that the LOOCV estimate for the test MSE is minimum for “fit.glm.2”, this is not surprising since we saw clearly in (b) that the relation between “x” and “y” is quadratic.
- 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 crossvalidation results?
summary(fit.glm.4)##
## Call:
## glm(formula = y ~ poly(x, 4))
##
## Deviance Residuals:
## Min 1Q Median 3Q Max
## -2.8914 -0.5244 0.0749 0.5932 2.7796
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) -1.8277 0.1041 -17.549 <2e-16 ***
## poly(x, 4)1 2.3164 1.0415 2.224 0.0285 *
## poly(x, 4)2 -21.0586 1.0415 -20.220 <2e-16 ***
## poly(x, 4)3 -0.3048 1.0415 -0.293 0.7704
## poly(x, 4)4 -0.4926 1.0415 -0.473 0.6373
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## (Dispersion parameter for gaussian family taken to be 1.084654)
##
## Null deviance: 552.21 on 99 degrees of freedom
## Residual deviance: 103.04 on 95 degrees of freedom
## AIC: 298.78
##
## Number of Fisher Scoring iterations: 2he p-values show that the linear and quadratic terms are statistically significants and that the cubic and 4th degree terms are not statistically significants. This agree strongly with our cross-validation results which were minimum for the quadratic model.
9. We will now consider the Boston housing data set, from the MASS library.
- Based on this data set, provide an estimate for the population mean of medv. Call this estimate μˆ.
library(MASS)
mu.hat = mean(Boston$medv)
mu.hat## [1] 22.53281- Provide an estimate of the standard error of μˆ. Interpret this result. Hint: We can compute the standard error of the sample mean by dividing the sample standard deviation by the square root of the number of observations.
se.hat = sd(Boston$medv) /sqrt(dim(Boston)[1])
se.hat## [1] 0.4088611- Now estimate the standard error of μˆ 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(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 0.008517589 0.4119374- Based on your bootstrap estimate from (c), provide a 95 % con- fidence interval for the mean of medv. Compare it to the results obtained using t.test(Boston$medv). Hint: You can approximate a 95 % confidence interval using the formula [μˆ − 2SE(μˆ), μˆ + 2SE(μˆ)].
t.test(Boston$medv)##
## One Sample t-test
##
## data: Boston$medv
## t = 55.111, df = 505, p-value < 2.2e-16
## alternative hypothesis: true mean is not equal to 0
## 95 percent confidence interval:
## 21.72953 23.33608
## sample estimates:
## mean of x
## 22.53281CI.mu.hat <- c(22.53 - 2 * 0.4119, 22.53 + 2 * 0.4119)
CI.mu.hat## [1] 21.7062 23.3538The bootstrap confidence interval is very close to the one provided by the t.test() function
- Based on this dataset, provide an estimate, μˆmed, for the median value of medv in the population.
med.hat <- median(Boston$medv)
med.hat## [1] 21.2- We now would like to estimate the standard error of μˆmed. 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 <- function(data, index) {
mu <- median(data[index])
return (mu)
}
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 -0.0098 0.3874004We get an estimated median value of 21.2 which is equal to the value obtained in (e), with a standard error of 0.3874 which is relatively small compared to median value.
- Based on this data set, provide an estimate for the tenth percentile of medv in Boston suburbs. Call this quantity μˆ0.1. (You can use the quantile() function.)
percent.hat <- quantile(Boston$medv, c(0.1))
percent.hat## 10%
## 12.75- Use the bootstrap to estimate the standard error of μˆ0.1. Comment on your findings.
boot.fn <- function(data, index) {
mu <- quantile(data[index], c(0.1))
return (mu)
}
boot(Boston$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 0.00515 0.5113487We get an estimated tenth percentile value of 12.75 which is again equal to the value obtained in (g), with a standard error of 0.5113 which is relatively small compared to percentile value.