Date:         Mon, 27 Oct 2003 11:57:51 -0500
Reply-To:     Peter Flom <flom@NDRI.ORG>
Sender:       "SAS(r) Discussion" <SAS-L@LISTSERV.UGA.EDU>
From:         Peter Flom <flom@NDRI.ORG>
Subject:      How to calculate the 95% CI (was RE: chi square analysis
              toidentify the outliers)
Comments: To: sassysaser@YAHOO.COM
Content-Type: text/plain; charset=US-ASCII

Peter L. Flom, PhD
Assistant Director, Statistics and Data Analysis Core
Center for Drug Use and HIV Research
National Development and Research Institutes
71 W. 23rd St
www.peterflom.com
New York, NY 10010
(212) 845-4485 (voice)
(917) 438-0894 (fax)



>>> Karriere Sucher <sassysaser@YAHOO.COM> 10/27/2003 10:34:19 AM >>>
<<<
I am still thinking over how to calculate the 95% CI. Just to refresh
your memory, my original question is I have ten scores: 8, 25, 35, 41,
50, 75, 75, 79, 92, 99 and I want to construct a 95% CI for those ten
scores.
>>>

First, a definition (quoting from The Cambridge Dictionary of
Statistics, by BS Everitt)

Confidence Interval: "A range of values, calculated from the sample
observations, that are believed, with a particular probability, to
contain the true parameter value.  A 95% CI, for example, implies that,
were the estimation process repeated again and again, then 95% of the
calculated intervals would be expected to contain the true parameter
value" (p. 74)

Second, calculation.  David is correct (surprise).  To see why, in a
general way, there needs to be some adjustment for N, suppose you took
two samples, one with 10 people, one with 1,000  people.  from a
normally distributed population of 100,000 people with a mean of 0 and
an sd of 1.  Both of these would have means approximately = 0, and sd
approximately 1.  Using YOUR method, the CIs would be equal, using
David's, they would be different.  Clearly, a larger sample makes for a
better estimate.



Peter