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In the answer to this question

How to Calculate a Confidence Interval for Estimating a Proportion When the Population is Very Small

the method for constructing an exact 95% confidence interval for the number of successes in the population is given. This interval is exact in the sense that it uses the hypergeometric distribution to construct the interval. However, the interval may not have a coverage of exactly 95%. Can someone suggest a reference that discusses how to calculate the achieved coverage for one of these exact confidence intervals? Thanks.

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  • $\begingroup$ Isn't the point that you can't? Hence you have to resort to 'randomized p-values'. $\endgroup$ – conjugateprior Dec 18 '15 at 15:39
  • $\begingroup$ I am not asking to get 95%, I am asking what I got. Is it 98.6% or what? $\endgroup$ – Thomas Dec 18 '15 at 18:43
  • $\begingroup$ Ah, I see. Well, the realized coverage is very choppy for almost all types of interval, depending on the true proportion : See graphs in Brown, Cai and Dasgupta 2001 and 2002. $\endgroup$ – conjugateprior Dec 18 '15 at 19:32
  • $\begingroup$ Those papers appear to discuss confidence intervals for the number of successes for an infinite population (the binomial distribution), not for a finite population (hypergeometric distribution). $\endgroup$ – Thomas Dec 19 '15 at 11:56
  • $\begingroup$ True. However, the hypergeometric intervals seem to look rather the same - presumably for the same reasons - at least if I'm interpreting Fig. 23 of this. I'm well out of area, as you can tell, but it seems like Section 6 might possibly be relevant to your question. The author has R code too, I believe. $\endgroup$ – conjugateprior Dec 19 '15 at 22:48
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Getting good Binomial intervals is hard and coverage is very variable (see Brown, Kai, Dasgupta 2001) due to the discreteness of the observations. While that is not the question here, it is related. Scholz provides a technical note that also deals with hypergeometric intervals (section 6) and shows that these intervals have similar behaviour, presumably for similar reasons.

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