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Feb 9, 2016 at 23:19 comment added Glen_b @Guy I wouldn't use the normal approximation for that; with 100 intervals, even if all the assumptions held, the expected number outside is a small count (binomial but well approximated by a Poisson(5) ... so pretty discrete)...
Feb 9, 2016 at 20:33 comment added Guy Davidson Which, if I wanted a 95% CI, would be 0.95 +/- 1.96 * sqrt(0.95 * 0.05 / 100) => 0.95 +/- 0.0427, which essential means about 90.5-100% of the time. Interesting!
Feb 9, 2016 at 20:23 comment added Guy Davidson That makes sense. Would I be justified in using something like this? en.wikipedia.org/wiki/Binomial_proportion_confidence_interval (looks like I can't line break in comments, which is a little awkward).
Feb 8, 2016 at 2:22 history answered Glen_b CC BY-SA 3.0