posted 17 years ago
Hmm, I would think an acceptable approximation is
&sigmap = sqrt(p*(1-p)/n)
Unfortunately we don't really know p here - measuring p is the point of the excercise. But it's usually acceptable to use the measured, approximate value of p in lieu of the real thing.
(What else could one do?)
So for case E, the variance of the mean would be sqrt((.00001)*(.99999)/1000) = .0001. That means, roughly, that each time you repeat the experiment 1000 times, you can expect that the measured value of p might vary by 0.01 %. (Bearing in mind that it can vary by more than one standard deviation, but this gets increasingly unlikely the farther out you get.) Since this is ten times as large as the measurement itself (at least, for the current best estimate), you would presumably want to repeat the measurement a few more times.
If you want the variance of the mean to be within, say, 0.0001% (a tenth of the presumed value) then solve for n:
n = p*(1-p)/σ2 = (.00001)(.99999)/(.000001)2 = 9999900
Yes, that's a lot of repetitions, unfortunately.
I may be misremembering some of the details on this, as it's been a long time - but it feels right to me, at least for a simple view of the distrbution. There were probably some simplifying assumptions that went into the formula, which I don't remember right now. Hopefully it's close enough for a rough idea though.
[ December 18, 2006: Message edited by: Jim Yingst ]
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