In article <[EMAIL PROTECTED]>, sleszyk <[EMAIL PROTECTED]> wrote:
<Snip>
> I have another question about the proportion confidence interval:
> With the above formula one can calcute the symmetrical interval, which
> is ok as long as the proportion is not too close to 0 or 1. However,
> from what I've heard the "real" CI is always asymmetrical (well, maybe
> except for p=0,5).
> Does anybody know how to calculate an asymmetrical CI for a
> proportion? I've been looking for an answer on the web some time ago
> and couldn't find it.
This is nicely and simply given in the latest edition of "Statistics with
Confidence", by Gardner and Altman, pages 45 to 56. The techniques and
worked examples are given, so that you can check your own implementation!
It is published (in the UK) by the BMJ. Sorry, I don't have the ISBN.
Here's a precis of the method :-
Let n = sample size and r = number of "events". Then p = r/n,
and let q = (1-p). Let A = required probability, say 0.05 for example.
Then 100*(1-A)% would represent the 95% interval.
Let Z = 100(1-A/2) percentile from the normal distribution, which is
readily available in many softwares and tables.
Now calculate B = 2r + Z*Z, C = Z*SQRT(Z*Z + 4rq) and D= 2(n + Z*Z)
The confidence interval for the population proportion is then
(B - C)/D to (B + C)/D
The examples they give are for 1 event in a sample of 29, (p = 0.03448)
for which the 95% CI evaluates to 0.006 to 0.172, and
zero events in a sample of 20, clearly a proportion of 0.0 giving a 95%
CI of 0 to 0.161
In the book they give other related methods, such as confidence
intervals for differences in proportions for two samples, for both the
paired case and unpaired case.
--
Robin Edwards ZFC W Serious Statistical Software
REAL Statistics with Graphics for RISC OS machines
Please email [EMAIL PROTECTED] for details of our loan software.
.
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