There are multiple methods to generate a bootstrap confidence interval. The simplest in terms of implementation is a percentile bootstrap interval: generate B bootstrap samples of the original data and calculate B statistics from these samples. Then take the alpha/2 and 1-alpha/2 quantiles of the statistic to form a confidence interval of level alpha. However, this interval does not behave as well as other intervals, as we will show:

Take 20 random exponential variables with mean 3. In R this looks like:

`x = rexp(20,rate=1/3)`

Then generate B=1000 bootstrap samples of x, and calculate the mean for each bootstrap sample.

`s = numeric(B)`

for (j in 1:B) {

boot = sample(n,replace=TRUE)

s[j] = mean(x[boot])

}

Then, for an alpha = .05 / 95% confidence interval, look at the .025 and .975 quantiles of the bootstrap statistics in the vector s:

`simple.ci = quantile(s,c(.025,.975))`

If I repeat this process from the start (including drawing a new x of 20 random exponential variables of mean 3) I can see how often the intervals actually contain the true mean. 11 of the intervals do not contain the true mean 3. On average we would expect 5 if these are 95% confidence intervals. If I repeat this 1000 times, I get 88.4% of the intervals containing the true mean.

In* An Introduction to the Bootstrap* by Efron and Tibshirani, the authors write,

The quantity (theta-hat – theta)/se-hat is called an approximate pivot: this means that its distribution is approximately the same for each value of theta….

Some elaborate theory shows that in large samples the coverage of the bootstrap-t interval tends to be closer to the desired level than the coverage of the standard interval or the interval based on the t table….

Notice also that the normal and t percentage points are symmetric about zero, and as a consequence the resulting intervals are symmetric about the point estimate theta-hat. In contrast, the bootstrap-t percentiles can be asymmetric about 0, leading to intervals which are longer on the left or right. This asymmetry represents an important part of the improvement in coverage it enjoys.

Note the authors here are not comparing the bootstrap-t interval to the bootstrap percentile method. Also, they add a caveat: “The bootstrap-t method, at least in its simple form, cannot be trusted for more general problems, like setting a confidence interval for a correlation coefficient.”

But it works well for calculating the mean, as in this case. This time we calculate a pivotal quantity as the bootstrapped statistic. For the vector x of random exponential variables of mean 3, we first calculate the mean and standard deviation of the original dataset:

`x = rexp(n,rate=1/true.mean)`

mean.x = mean(x)

sd.x = sd(x)

Then for each bootstrap sample, calculate the difference between the mean of the bootstrap data and the original data, divided by the standard deviation of the bootstrap (note that there may be better estimators for the denominator here).

`z = numeric(B)`

for (j in 1:B) {

boot = sample(n,replace=TRUE)

z[j] = (mean(x[boot]) - mean.x)/sd(x[boot])

}

Then to form a confidence interval, take quantiles of our bootstrapped statistic, multiply by the standard deviation of the original data and substract this from the mean of the original data:

`pivot.ci = mean.x - sd.x * quantile(z,c(.975,.025))`

If I do this 1000 times, I get 95.1% of the intervals containing the true mean. The mean interval size has increased, from 2.4 for the simple intervals to 3.2 for the bootstrap-t intervals.

Here is an R script for this simulation.

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Good article, question below — you are using, SD — but it should be SE?

se.x = sd(x) / sqrt(len(n))

rather than ?

sd.x = sd(x)

Hi Aida,

see here: http://en.wikipedia.org/wiki/Pivotal_quantity

it’s like a z-score, in that we want to scale the differences down by the sample standard deviation and then rescale afterward.

Hey mike, I’m wondering how long such a script takes you? Bootstrap T with 1000 bootstraps and 1000 nested bootstraps. I’m running a Bootstrap T to get a confidence interval for a logistic regression. Its been running for 2 weeks!!

inteffect<-rep(NA,1000)

intstd<-rep(NA,1000)

T<-rep(NA,1000)

for(i in 1:1000)

{

boot<-forbootframe[sample(297, 297,replace=TRUE), ]

bootlogit<-glm(Case~gender+allele+Ever+allele*Ever, family=binomial(logit),data=boot)

inteffect[i]<-summary(bootlogit)$coef[5,1]

nesteffect<-rep(NA,1000)

for(u in 1:1000)

{

nestboot<-boot[297, 297,replace=TRUE), ]

bootlogit<-glm(Case~gender+allele+Ever+allele*Ever, family=binomial(logit),data=nestboot)

nesteffect[u]<-summary(bootlogit)$coef[5,1]

}

intstd[i]<-sqrt(var(nesteffect))

T[i]<-(inteffect[i]+2.0469301)/intstd[i]

}

bootstd<-sqrt(var(inteffect))

-2.0469301-(quantile(T,0.025)*(bootstd))

-2.0469301-(quantile(T,0.975)*(bootstd))

sorry no suggestion for you. the bootstrap is very amenable to parallelization.