I'm an undergrad working on a biochemistry experiment where I take a bunch of fish fin samples (each of which is composed of n individual cut fins) and run the samples through a mass spec. It is critical that the group fin samples have the same total mass within about 10% of each other, but I can't directly weigh the samples in real time for multiple reasons (mainly that taking the time to weigh them would alter the integrity of the metabolic chemicals I'm trying to measure using the mass spec). However, I can estimate the population variance through sample variances obtained by weighing individual dehydrated fins from the same population of fish (such fins can't be used for mass spec but can be accurately weighed).
My thought is that given a large enough n (fins/sample), I can have a 95% or 99% confidence that the samples will weigh within 10% of each other without actually weighing any of them directly. How do I calculate n based on the data I have available (a sample variance from the population but for a different sample)?
I hope this isn't too confusing. If you're able to explain the overall mathematical reasoning behind the solution, that would be great. Thanks so much!
You state that you have a sample variance based on measurements of dried fins and you'd like to use this information to estimate an adequate sample size, $n$, so there is at most an $\alpha$% chance that at at least two samples will differ by more than 10%.
The variability of the dried fish fin massess will be less than the variability of the fresh fins, since the latter have entrained water. You will need to know, or be able to estimate, the distribution of FRESH fin masses, $m_f$. Either using water content and the dry fin masses or directly weighing fresh fins.
Now, you will need access to a computer, because there is no tidy equation to solve your problem, but you can get the answer via simulation (bootstrapping):
You will want to try a number of different $n$ for a given $m$ to see which one has a fraction less than, say 1%.