Q5
In Chapter 4, we used logisitc 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”.
library(ISLR)
attach(Default)
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: 8b.Using the validation set approach, estimate the test error of this model. In order to do this, you must perform the following steps:
1.Split the sample set into a training set and a validation set.
train = sample(dim(Default)[1],dim(Default)[1]/2)2.Fit a multiple logistic regression model using only the training observations.
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
## -2.5830 -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: 83.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.
probs = predict(fit.glm,newdata=Default[-train,],type="response")
pred.glm = rep("No",length(probs))
pred.glm[probs>0.5] = "Yes"4.Compute the validation set error, which is the fraction of the observations in the validation set that are misclassified.
mean(pred.glm!=Default[-train,]$default)## [1] 0.0254We have a 2.48% test error rate with the validation set approach.
c.Repeat the process in (b) three times, using three different splits of the observations into a training set and a validation set. Comment on the results obtained.
train = sample(dim(Default)[1], dim(Default)[1] / 2)
fit.glm = glm(default~income+balance,data=Default,family="binomial",subset=train)
probs = predict(fit.glm, newdata = Default[-train, ], type = "response")
pred.glm = rep("No", length(probs))
pred.glm[probs > 0.5] <- "Yes"
mean(pred.glm != Default[-train, ]$default)## [1] 0.0274train = 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, ]$default)## [1] 0.0244train = sample(dim(Default)[1], dim(Default)[1] / 2)
fit.glm = glm(default~income+balance,data=Default,family="binomial",subset=train)
probs = predict(fit.glm, newdata = Default[-train, ], type = "response")
pred.glm = rep("No", length(probs))
pred.glm[probs > 0.5] <- "Yes"
mean(pred.glm != Default[-train, ]$default)## [1] 0.0244We see that the validatoin 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.
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.
train = sample(dim(Default)[1],dim(Default)[1]/2)
fig.glm = glm(default~income+balance+student,data=Default,family="binomial",
subset=train)
probs = predict(fig.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] 0.0278It doesn’t seem that adding the “student” dummy variable leads to a reducation in the validation set estimate of the test error rate.
Q6
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 computes 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(1)
attach(Default)## The following objects are masked from Default (pos = 3):
##
## balance, default, income, studentfit.glm = glm(default~income+balance,family="binomial")
summary(fit.glm)##
## Call:
## glm(formula = default ~ income + balance, family = "binomial")
##
## 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: 8The glm() estimates of the standard errors for the coefficients \(\beta_0,\beta_1,\beta_2\) are repectively \(0.4348,4.985×10^{-6}\) and \(2.274×10^{-4}\)
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){
fit = glm(default~income+balance,data=data,family="binomial",subset=index)
return(coef(fit))
}c.Use the boot() function together with your boot.fn() function to estimate the standard errors of the logistic regression coefficients for “income” and “balance”.
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 -3.945460e-02 4.344722e-01
## t2* 2.080898e-05 1.680317e-07 4.866284e-06
## t3* 5.647103e-03 1.855765e-05 2.298949e-04The bootstrap estimates of the standard errors for the coefficients \(\beta_0,\beta_1\) and \(\beta_2\) are respectively \(0.4335,4.866284×10^{-6}\) and \(2.298949×10^{-4}\)
d.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.
Q7
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”.
set.seed(1)
attach(Weekly)
fit.glm = glm(Direction~Lag1+Lag2,family="binomial",data=Weekly)
summary(fig.glm)##
## Call:
## glm(formula = default ~ income + balance + student, family = "binomial",
## data = Default, subset = train)
##
## Deviance Residuals:
## Min 1Q Median 3Q Max
## -2.3850 -0.1413 -0.0568 -0.0212 3.7409
##
## Coefficients:
## Estimate Std. Error z value Pr(>|z|)
## (Intercept) -1.042e+01 6.921e-01 -15.058 <2e-16 ***
## income -5.472e-06 1.205e-05 -0.454 0.6498
## balance 5.638e-03 3.276e-04 17.212 <2e-16 ***
## studentYes -8.286e-01 3.468e-01 -2.389 0.0169 *
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## (Dispersion parameter for binomial family taken to be 1)
##
## Null deviance: 1409.45 on 4999 degrees of freedom
## Residual deviance: 767.12 on 4996 degrees of freedom
## AIC: 775.12
##
## Number of Fisher Scoring iterations: 8b.Fit a logistic regression model that predicts “Direction” using “Lag1” and “Lag2” using all but the first observation.
fit.glm.1 = glm(Direction~Lag1+Lag2,data=Weekly[-1,],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: 4c.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.glm(fit.glm.1,Weekly[1,],type="response")>0.5## 1
## TRUEWe may conclude that the prediction for the first observation is “Up”. This observation was not correctly classified as the true direction is “Down”.
d.Write a 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 :
1.Fit a logistic regression model using all but the ith observation to predict “Direction” using “Lag1” and “Lag2”.
2.Compute the posterior probability of the market moving up for the ith observation.
3.Use the posterior probability for the ith observation in order to predict whether or not the market moves up.
4.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 1 0 0
## [38] 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 1 0 1 1 0 0
## [75] 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 1 0 0 1 0 1 0 0 1
## [112] 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 1 1 1 0 0 0 1 0 0 0 0 0
## [149] 0 1 1 1 0 1 0 0 1 1 0 1 0 0 1 1 0 0 1 0 0 1 0 0 1 1 1 0 1 0 1 0 0 0 0 0 0
## [186] 0 0 1 1 0 1 0 1 0 1 0 1 0 0 1 0 0 1 0 0 1 0 1 0 1 1 1 0 0 1 1 0 1 0 0 1 1
## [223] 0 0 0 1 1 1 0 1 0 1 0 1 0 0 0 1 1 0 1 0 1 0 1 0 1 0 1 1 0 1 0 0 1 0 0 1 0
## [260] 0 0 0 0 1 0 0 0 1 0 0 1 0 0 0 1 0 0 1 0 0 1 0 0 1 0 1 1 0 0 0 0 0 1 0 1 0
## [297] 0 1 0 0 0 1 0 0 1 1 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
## [334] 1 1 1 1 0 1 0 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
## [371] 1 1 1 1 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
## [408] 0 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 1 1
## [445] 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 0 0 1 0 0
## [482] 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 1 0 1 0 0 0 1 0
## [519] 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 1 1 0 1 0 0 0 1 1 1 1
## [556] 1 1 0 1 0 1 0 0 1 0 0 1 1 1 0 0 0 1 1 1 1 1 1 1 1 1 1 0 1 0 0 1 0 0 1 0 1
## [593] 0 0 1 0 0 1 0 1 1 0 1 1 1 0 1 0 1 0 1 0 0 0 1 0 1 0 1 0 1 1 0 1 1 0 1 0 1
## [630] 0 1 1 1 1 0 1 1 0 0 0 1 1 1 1 0 1 1 1 0 1 0 0 0 1 1 1 1 1 1 0 1 0 0 1 0 0
## [667] 0 1 1 0 1 0 1 1 1 1 0 0 0 1 1 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 1 0 0 1 0 1 1
## [704] 0 0 0 0 1 0 1 0 1 0 1 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
## [741] 1 1 1 1 0 0 0 1 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
## [778] 1 1 0 0 0 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
## [815] 0 1 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 1
## [852] 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 0 1 0 1
## [889] 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 0 0 0 1 0 0 1
## [926] 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 1 0 0 1 1 1 1 1 0 1
## [963] 0 0 0 1 1 1 0 1 1 1 1 0 0 0 0 1 1 0 0 0 0 1 0 0 1 1 1 0 0 1 1 1 0 1 0 0 0
## [1000] 0 1 0 0 1 0 1 0 0 1 1 1 1 0 1 0 0 1 0 0 1 0 0 1 1 0 1 1 1 0 1 1 0 0 0 1 0
## [1037] 1 0 1 1 1 1 0 0 1 0 0 0 0 0 0 1 0 1 1 0 0 0 0 0 1 1 1 0 0 0 1 0 1 1 1 0 0
## [1074] 0 0 1 0 0 0 0 0 1 0 1 0 0 0 0 0mean(error)## [1] 0.4499541The LOOCV estimate for the test error rate is 93.2%!
Q8
We will now perform cross-validation on a simulated data set.
a.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 \[Y=X−2X^2+\epsilon\]
b.Create a scatterplot of X against Y. Comment on what you find.
plot(x,y,pch=19)
The data obviously suggests a curved relationship.
c.Set a random seed, and then compute the LOOCV errors that result from fitting the following four models using least squares : 1.\(Y=β_0+β_1X+ε\)
library(boot)
set.seed(1)
Data = data.frame(x, y)
fit.glm.1 = glm(y ~ x)
cv.glm(Data, fit.glm.1)$delta[1]## [1] 5.8909792.\(Y=β_0+β_1X+β_2X^2+ε\)
fit.glm.2 = glm(y~poly(x,2))
cv.glm(Data,fit.glm.2)$delta[1]## [1] 1.0865963.\(Y=β_0+β_1X+β_2X^2+β_3X^3+ε\)
fit.glm.3 = glm(y~poly(x,3))
cv.glm(Data,fit.glm.3)$delta[1]## [1] 1.1025854.\(Y=β_0+β_1X+β_2X^2+β_3X^3+β_4X^4+ε\)
fit.glm.4 = glm(y~poly(x,4))
cv.glm(Data,fit.glm.4)$delta[1]## [1] 1.114772d.Repeat (c) using another random seed, and report your results. Are your results the same as what you got in (c) ? Why ?
set.seed(10)
fit.glm.1 = glm(y~x)
cv.glm(Data,fit.glm.1)$delta[1]## [1] 5.890979fit.glm.2 = glm(y~poly(x,2))
cv.glm(Data,fit.glm.2)$delta[1]## [1] 1.086596fig.glm.3 = glm(y~poly(x,3))
cv.glm(Data,fit.glm.3)$delta[1]## [1] 1.102585fig.glm.4 = glm(y~poly(x,4))
cv.glm(Data,fit.glm.4)$delta[1]## [1] 1.114772The results above are identitical to the results in(c) since LOOCV evaluates n folds of a single observation.
e.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.
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 ?
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: 2The p-value show that the linear and quardratic 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.
Q9
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 \(\hat{μ}\).
library(MASS)
attach(Boston)
mu.hat = mean(medv)
mu.hat## [1] 22.53281b.Provide an estimate of the standard error of \(\hatμ\). Interpret this result.
se.hat = sd(medv)/sqrt(dim(Boston)[1])
se.hat## [1] 0.4088611c.Now estimate the standard error of \(\hatμ\) using the bootstrap. How does this compare to your answer from (b) ?
set.seed(1)
boot.fn = function(data,index){
mu = mean(data[index])
return (mu)
}
boot(medv,boot.fn,1000)##
## ORDINARY NONPARAMETRIC BOOTSTRAP
##
##
## Call:
## boot(data = medv, statistic = boot.fn, R = 1000)
##
##
## Bootstrap Statistics :
## original bias std. error
## t1* 22.53281 0.007650791 0.4106622The bootstrap estimated stadard error of \(\hatμ\) of 0.4106622 is very close to the estimate found in(b) of 0.4088611.
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).
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
## 22.53281CI.mu.hat = c(22.53-2*0.4106622,22.53+2*0.4106622)
CI.mu.hat## [1] 21.70868 23.35132The bootstrap confidence interval is very close to the one provided by the t.test() function.
e.Based on this data set, provide an estimate, \(\hatμ_{med}\), for the median value of “medv” in the population.
med.hat = median(medv)
med.hat## [1] 21.2f.We now would like to estimate the standard error of \(\hatμ_{med}\). 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 = function(data,index){
mu = median(data[index])
return (mu)
}
boot(medv,boot.fn,1000)##
## ORDINARY NONPARAMETRIC BOOTSTRAP
##
##
## Call:
## boot(data = medv, statistic = boot.fn, R = 1000)
##
##
## Bootstrap Statistics :
## original bias std. error
## t1* 21.2 -0.0386 0.3770241We get an estimated median value of 21.2 which is equal to the value obtained in (e), with a standard error of 0.3770241 which is relatively small compared to median value.
g.Based on this data set, provide an estimate for the tenth percentile of “medv” in Boston suburbs. Call this quantity \(\hatμ_{0.1}\).
percent.hat = quantile(medv,c(0.1))
percent.hat## 10%
## 12.75h.Use the bootstrap to estimate the standard error of \(\hatμ_{0.1}\). Comment on your findings.
boot.fn = function(data,index){
mu = quantile(data[index],c(0.1))
return (mu)
}
boot(medv,boot.fn,1000)##
## ORDINARY NONPARAMETRIC BOOTSTRAP
##
##
## Call:
## boot(data = medv, statistic = boot.fn, R = 1000)
##
##
## Bootstrap Statistics :
## original bias std. error
## t1* 12.75 0.0186 0.4925766We get an estimated tenth percentile value of 12.75 which is again equal to the value obtained in(g), with a standard error of 0.4925766 is relatively small compared to percentile value.