Bootstrap Confidence Interval 90%

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Samuel Wray
Samuel Wray on 26 Oct 2019
Edited: Adam Danz on 27 Oct 2019
We were asked to calculate the 90% confidence interval for a given dataset using bootci function. This was my line in Matlab
Pbci = bootci(2000,{@mean,Pb},'alpha',.1)%90 confidence interval
Is this the correct way?
Next we were asked to use the bootstrap technique to estimate the 90% confidence interval for the probability that the mean of Pb exceeds the MCL (i.e., 50ppm). Would i use Pbci = bootci(2000,{@mean,Pb},'weights',) for this? If so what do i put as my constraints for the weights part?

Accepted Answer

Adam Danz
Adam Danz on 26 Oct 2019
"We were asked to calculate the 90% confidence interval [using bootci()]... Is this the correct way?"
To determine if it's the correct way, compare the results with a lower level computation of the confidence intervals. Not only will this confirm that you're using the bootci() function correctly but you'll get a better understanding of how those intervals are computed.
Let's say your data contain 1000 samples and you're bootstrapping the mean of your data 2000 times. bootci() resamples the data with replacement 2000 times and computes the mean on each iteration. That means on each of the 2000 iterations, it randomly chooses 1000 samples, many of which will be duplicates (it uses the randi() function), and computes the mean. It then uses the 2000 means that were computed to determine the CI. There are several methods of doing this (explained under "types" in the documentation) but with enough bootstraps, the distribution of means should be normal (thanks to the Central Limit Theorem) and the CI type shouldn't make too much of a difference (though I typically recommend the percentile method which is not dependent on the shape of the distribution).
Follow thie brief tutorial below where the CIs are computing for a random dataset using bootci() and using a lower-level, direct computation of the CIs. As you can see by the figure it produces, the results are nearly the same. The only differences are due to a different randomized resampling of the data between methods.
% Create random data from a normal distribution
% with mean 28.25 and sd 8.5.
data = (randn(1,100000)*8.5 + 28.25)';
% Run bootci (percentile method is chosen since that's how we're
% computing it below in the other method.
nBoot = 2000; %number of bootstraps
[bci,bmeans] = bootci(nBoot,{@mean,data},'alpha',.1,'type','per'); %90 confidence interval
% Compute bootstrap sample mean
bmu = mean(bmeans);
% Now repeat that process with lower-level bootstrapping
% using the same sampling proceedure and the same data.
bootMeans = nan(1,nBoot);
for i = 1:nBoot
bootMeans(i) = mean(data(randi(numel(data),size(data))));
end
CI = prctile(bootMeans,[5,95]);
mu = mean(bootMeans);
% Plot the bootci() results
figure()
ax1 = subplot(2,1,1);
histogram(bmeans);
hold on
xline(bmu, 'k-', sprintf('mu = %.2f',bmu),'LineWidth',2,'FontSize',12)
xline(bci(1),'k-',sprintf('%.1f',bci(1)),'LineWidth',2,'FontSize',12)
xline(bci(2),'k-',sprintf('%.1f',bci(2)),'LineWidth',2,'FontSize',12)
title('bootci()')
% plot the lower-level, direct computation results
ax2 = subplot(2,1,2);
histogram(bootMeans);
hold on
xline(mu, 'k-', sprintf('mu = %.2f',mu),'LineWidth',2,'FontSize',12)
xline(CI(1),'k-',sprintf('%.1f',CI(1)),'LineWidth',2,'FontSize',12)
xline(CI(2),'k-',sprintf('%.1f',CI(2)),'LineWidth',2,'FontSize',12)
title('Lower level')
linkaxes([ax1,ax2], 'xy')
191026 150605-Figure 1.png
"Next we were asked to use the bootstrap technique to estimate the 90% confidence interval for the probability that the mean of Pb exceeds the MCL (i.e., 50ppm)."
I'm not sure I follow this part. What is MCL? Is it a scalar value (like 50) or is is the mean of a 2nd distribution?
  2 Comments
Adam Danz
Adam Danz on 27 Oct 2019
That desciption differs from what's in your question. The phrasing used in your question is:
we were asked to use the bootstrap technique to estimate the 90% confidence interval for the probability that the mean of Pb exceeds the MCL (i.e., 50ppm).
Pb are your your raw data so in that phrasing of the quesiton, you're comparing some value MCL to the mean of your data mean(Pb).
But the phrasing used in your comment above:
We're supposed to find the probability that the bootstrap means from the first question exceed a number such as 50.
The bootstrap mean is the mean of all 2000 bootstrap means (which probably nearly matches your population mean, anyway).
So I'm wondering if you should be using your population distribution (which could take any form) or the bootstrap'd distribtion of means (which is very close to a normal distribution). Since this sounds like homework, I'm sure there is text of lecture materail that would clear this up.
This may be informative:

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