# Bootstrap on samples with different precision

Imagine the following setup:

• I have N specimens that come from the same population (e.g., they are from the dame material)
• For each specimen i, I have a single estimate of a given property in terms of its mean value $$x_i$$ and a confidence interval $$[x_i-CI_i, x+CI_i]$$
• So I have N pairs of slightly different ($$x_i,CI_i$$), for $$i=1,...,N$$, due to the random nature of producing specimens
• Now, I want to estimate my confidence on the mean of the mean properties of the N specimens, i.e., $$E(x_i)$$, the mean of the group of N specimens.

I thought to simply bootstrap on the N mean values of each specimen and build a confidence interval from that. But then I feel it will not be taking into consideration the information contained on the confidence interval of each of their individual mean estimates, which is an information I have.

I started considering making a weighted bootstrap, weighting each original sample by its confidence interval range (e.g. basically divide each $$x_i$$ by a normalized weight computed from their $$CI_i$$).

But I could not find a solid reference (paper, book) on that and since I am not a statistician I am afraid of getting into some pitfall.

What are your thoughts (and references) on that?

• @Ggjj11 thanks for the insights! Regarding the inverse variance weighting, I did not know it, thanks for that. So from your inputs I envisioned two possible approaches: 1) instead of a bootstrap sample being made from random sampling with replacement from the set of N values of $x_i$, I would take one random sample out of each distribution parametrized by each pair of $(x_i,CI_i)$; 2) make a traditional bootstrap, using the original set of $x_i$, but then study the inverse-variance weighted average instead of the simple average. Would you know if one is theorectically sounder?
• I think it is important to note that you are trying to find an estimator $\hat{E}[y]=f(x_1[y],...x_N[y])$ where the $x_i=\hat{E}[y]$ are estimated mean values already. Now there is different choices for $f$ and they may result in a higher or lower variance $Var[\hat{E}[y]]$. The least variance of the mean estimator will be given by inverse variance weighting (which also gives you a formula for the variance of the estimated mean). You will directly be able to see it in simulations as well e.g. (x_1=20, sigma(x_1) =60), (x_2=70, sigma(x_2)=1) should intuitively have a better estimate around 70 Commented Apr 3 at 18:11