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I'm measuring binding constants my system and I appreciate the usual methods of using replicates to measure standard errors and using those errors to calculate propagation of error. I'm curious if bootstrapping is a reasonable alternative to calculate uncertainty?

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    $\begingroup$ This is probably more suitable for stats.stackexchange.com, or do you think biological assumptions will influence the answer (I'm not familiar with binding constants)? $\endgroup$
    – fileunderwater
    Jan 23, 2015 at 11:04
  • $\begingroup$ I approve this move. @fileunderwater Please let me know how to make that transition. I was under the impression that this was a fairly regular question that was relevant to the biochemistry community? $\endgroup$
    – bobthejoe
    Jan 28, 2015 at 21:37
  • $\begingroup$ I've flagged it asking for migration, but I don't know if it is possible. This is not my topic, so I cannot say anything about methods/problems in biochemistry. To me, it feels like a clearcut statistical issue though. $\endgroup$
    – fileunderwater
    Jan 28, 2015 at 23:08

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My understanding of bootstrapping is that you may estimate variance (and thus standard error of the population mean) iff your measurements are independent and have the same population distribution, in which case a number of sampling-with-replacement calculations can be done. I suspect that this method of estimation would be less desirable with small n values because of the effect that outliers or large standard deviations may have on the calculation. If you have a large dataset then it shouldn't be an issue.

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    $\begingroup$ What is the definition of independent in this case? Does running a dilution series maintain independence? Now I'm even more curious how legitimate this is. $\endgroup$
    – bobthejoe
    Mar 4, 2012 at 7:29
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    $\begingroup$ Independent really depends on the context of the experiment. If you are doing a serial dilution then I would consider those to be dependent on each other since the lower concentrations derive from the larger. This would be better analyzed as a replicate. $\endgroup$
    – user560
    Mar 4, 2012 at 12:43

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