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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 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.


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