Date:         Mon, 3 Aug 1998 07:16:59 -0400
Reply-To:     "William B. Ware" <wbware@EMAIL.UNC.EDU>
Sender:       "SPSSX(r) Discussion" <SPSSX-L@UGA.CC.UGA.EDU>
From:         "William B. Ware" <wbware@EMAIL.UNC.EDU>
Subject:      Re: t-test's df when variances not equal?
Comments: To: Gonzalo Kmaid <gkmaid@INTERNET.COM.UY>
In-Reply-To:  <3.0.16.19980802194635.3faf1b58@mail.internet.com.uy>
Content-Type: TEXT/PLAIN; charset=US-ASCII

On Sun, 2 Aug 1998, Gonzalo Kmaid wrote:
 
> Hi! Just a minor question on how to compute df for a t-test (testing
> differences between two means) when is not possible to assume equal
> variances between groups.
...
> My question is, how df are computed in this case?
>
> In the mentioned manual's example (groups sizes are 50 and 48)
>
> Df Equal 96 (clearly N1 + N2 -2)
> Df Unequal 89.43 ????
 
The value for the df is calculated using a formula that is a relative
complex function of the sample variances and sample sizes.  I believe that
the original goes back to Satterwaite (sp?).
 
It appears in several introductory statistics books... for example,
Hinkle, Wiersma, & Jurs.
 
__________________________________________________________________________
William B. Ware, Professor and Chair                 Psychological Studies
CB# 3500                                                      in Education
University of North Carolina                         PHONE  (919)-962-7848
Chapel Hill, NC      27599-3500                      FAX:   (919)-962-1533
http://www.unc.edu/~wbware/                          EMAIL: wbware@unc.edu
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