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?
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: firstname.lastname@example.org