Date:         Thu, 14 Sep 2000 14:03:22 -0400
Reply-To:     Mark.K.Moran@CENSUS.GOV
Sender:       "SAS(r) Discussion" <SAS-L@LISTSERV.UGA.EDU>
Comments:     To: z <gzuckier@MY-DEJA.COM>
From:         Mark.K.Moran@CENSUS.GOV
Subject:      Re: Why offset in confidence interval for proportions?
Comments: To: SAS-L@LISTSERV.VT.EDU
Content-type: text/plain; charset=us-ascii

Mr./Ms. Zuckier,

You are talking about the Joint Commission on the Accreditation of Healthcare
Organizations?
Most Joint Commission events reporting that I've heard of (keep in mind that I
have only heard of
a *small part* of what they do) are for *rare* events--adverse outcomes,
sentinel events, deaths etc.
These events would not be distributed Gaussian or even Z, but they would have
some skewed distribution.
I think that accounts for the off-centering of the 99% CI.  More than that, I am
not sure about the exact
formula.  Recommend you ask someone with a PhD.

Mark







z <gzuckier@MY-DEJA.COM> on 09/14/2000 11:37:26 AM

Please respond to z <gzuckier@MY-DEJA.COM>








 To:      SAS-L@LISTSERV.VT.EDU

 cc:      (bcc: Mark K Moran/CSD/HQ/BOC)



 Subject: Why offset in confidence interval for proportions?








Hi. This is more a stats question than a SAS one, but.... I'm trying to
wade through the JCAHO ORYX reporting requirements. Their formulae for
upper and lower limits for a 99% confidence interval for a proportion
are
U=((p+(Z^2)/(2n))+Z(root))/(1+(Z^2)/n)
L=((p+(Z^2)/(2n))-Z(root))/(1+(Z^2)/n)
where root=((z^2)/(4(n^2))+p(1-p)/n)^.5
Z=2.576
n=Number of patients
p=observed proportion

I can't exactly see where this comes from (I would have come up with
p(+or-)Z(p(1-p)/n)^.5 myself), and more to the point, this gives me a
confidence interval whose center is offset from p by the (Z^2)/2n term.
Waiting for the JCAHO to clarify is a slow process, in the meantime,
can anyone explain the theory behind this calculation? Thanks.


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