Bootstrap t

• if underlying population is normal
• and we know \( \sigma \)

• if the underlying population is normal,
• or at least symmteric and we have a large sample size
• we use the sample standard deviation (aka standard error) instead of \( \sigma \)

• we need a sampling distribution to use instead of the t-distribution
• bootstrap
• previously we saw a percentage bootstrap confidence interval
• now we'll study a version called the bootstrap t

• Load the Bangladesh data set
• make a vector of the chlorine levels
• make sure to omit the NA values
• Look at and comment on the sample data

• Find a 95% bootstrap confidence interval
• This is what we did in Chapter 5.
• Remember the idea:
  • do a bootstrap resample
  • compute its mean
  • store that mean in a vector
  • repeat 10⁵ times
  • find the middle 95% of the bootstrap distribution
• time the bootstrap loop

start_time <- Sys.time()
N <- 10^5
percboot <- numeric(N)
for (i in 1:N)
{samp <- sample(chlor, length(chlor), replace=T)
percboot[i] <- mean(samp)
}
end_time <- Sys.time()
end_time - start_time

Time difference of 5.656472 secs

• let's think a little more about how t confidence intervals work
• an estimate of the true mean, \( \mu \) is given by the confidence interval: \[ \begin{align} \overline{X}-q\frac{S}{\sqrt{n}} &\leq \mu \leq \overline{X}+q\frac{S}{\sqrt{n}}\\ -q\frac{S}{\sqrt{n}} &\leq \mu-\overline{X} \leq +q\frac{S}{\sqrt{n}}\\ -q &\leq \frac{\mu-\overline{X}}{\frac{S}{\sqrt{n}}} \leq +q\\ \end{align} \]

• The statistic

\[ \frac{\mu-\overline{X}}{\frac{S}{\sqrt{n}}} \] is usually written

\[ \frac{\overline{X}-\mu}{\frac{S}{\sqrt{n}}} \]

• My way will avoid an small complication with the inequalities

• The idea is to mimic what we do with a \( t \)-distribution
• We will replace \( \mu \) with the sample mean \( \overline{x} \)
• We will use a bootstrap mean for \( \overline{X} \): \( \overline{X^*} \)
• We will use the standard deviation of the bootstrap resample: \( S^* \)
• We will generate a distribution for \[ \frac{\overline{x}-\overline{X^*}}{\frac{S^*}{\sqrt{n}}} \] by running the resample loop many times
• We will find bootstrap quantiles, \( q_1 \) and \( q_2 \) such that \[ P\left(q_1 \leq \frac{\overline{x}-\overline{X^*}}{\frac{S^*}{\sqrt{n}}} \leq q_2\right)=1-\alpha \] where \( 1-\alpha \) is the confidence level (eg. 0.95)
• Unlike quantiles from a normal or \( t \) distribution \( q_1 \) and \( q_2 \) will not be symmetric

• Let \( \overline{X^*} \) be the mean of a bootstrap resample
• Let \( S^* \) be the standard deviation of the bootstrap resample
• \( \overline{x} \) is the mean of the sample
• \( n \) is the size of the sample
• Compute \( \frac{\overline{x}-\overline{X^*}}{\frac{S^*}{\sqrt{n}}} \) for many (\( 10^5 \)) bootstrap resamples

library(resampledata)
chlor <- na.omit(Bangladesh$Chlorine)

xbar=mean(chlor)
n=length(chlor)

start_time <- Sys.time()
N <- 10^5
bootT <- numeric(N)
for (i in 1:N)
{Xstar <- sample(chlor, n, replace=T)
Sstar <- sd(Xstar)
bootT[i] <- (xbar-mean(Xstar))/(Sstar/sqrt(n))
}
end_time <- Sys.time()
end_time - start_time

Time difference of 13.89275 secs

• Compute quantiles, \( q_1 \) and \( q_2 \), from this bootstrap \( t \) distribution

• compute the confidence interval
• you'll need the MOE:
  • use the sample standard deviation
  • use \( \overline{x} \) from the sample

samplesd <- sd(chlor)
xbar+quantile(bootT, 0.025)*samplesd/sqrt(length(chlor))

    2.5% 
56.82553 

xbar+quantile(bootT, 0.975)*samplesd/sqrt(length(chlor))

   97.5% 
112.1318 

• The \( t \) confidence interval is almost certainly bad because the distribution of the population is not normal
• However, the \( t \)-test confidence interval is fast
• The bootstrap-\( t \) confidence interval is by far the most computationally expensive
• Is this critical? I honestly don't know. It took about 13 seconds to run for 269 data points. In the “real world” data sets might be much larger.
• Which is best? The bootstrap-\( t \) is best.
• Why? Running simulations from known populations shows this to be the case. (See section 7.5.3 for this discussion.)