In Bayes success-run theorem (below), total "population" size is not a parameter. Why is this?
$$R = (1-C) ^{ (1/n)}$$ where:
- $R =$ Reliability (or probability of success),
- $C =$ confidence level,
- $n =$ sample size.
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$\begingroup$ We could use more contextual information: in particular, exactly how is the sample obtained? $\endgroup$– whuber ♦Feb 7, 2017 at 16:18
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$\begingroup$ Randomized sample $\endgroup$– Justin BorromeoFeb 7, 2017 at 16:23
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$\begingroup$ That doesn't help much, because there are many ways of obtaining random samples. You need to be specific (because the answer depends on it). $\endgroup$– whuber ♦Feb 7, 2017 at 16:41
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$\begingroup$ Simple random sampling. 500 units (each with unique IDs) are being sampled and a computer will generate a randomized list of n units from the set of 500 units. $\endgroup$– Justin BorromeoFeb 7, 2017 at 16:46
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$\begingroup$ See stats.stackexchange.com/a/80851/919 for a good discussion. That answer makes it clear that the theorem applies to sampling from a process, not a finite population. In your case--assuming you are limiting your conclusions to just those 500 units--a finite-population correction is certainly possible. $\endgroup$– whuber ♦Feb 7, 2017 at 18:10
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