|Date: ||Wed, 28 Jan 2004 06:43:06 -0800|
|Reply-To: ||Natasha <nak23@MOLE.BIO.CAM.AC.UK>|
|Sender: ||"SAS(r) Discussion" <SAS-L@LISTSERV.UGA.EDU>|
|From: ||Natasha <nak23@MOLE.BIO.CAM.AC.UK>|
|Subject: ||Re: nested ANOVA and F ratio calculation|
|Content-Type: ||text/plain; charset=ISO-8859-1|
Paige Miller advice was really helpful and clarified that the second
model was the most appropriate where the effect of status would be
compared to the biological replicate variability. I am, however, new
to statistics and SPSS. Does any one have any advice on how to alter
the syntax to ensure the F ratio is calculated as discussed?
The current syntax is:
spotvol BY status replicat
/METHOD = SSTYPE(3)
/INTERCEPT = INCLUDE
/PRINT = HOMOGENEITY LOF
/PLOT = SPREADLEVEL RESIDUALS
/SAVE = SRESID
/CRITERIA = ALPHA(.05)
/DESIGN = status replicat (status).
Paige Miller <email@example.com> wrote in message >
> Based upon your description of the problem, let me try to re-state
> the problem in my own words, and then pontificate on the proper
> Paige re-states the problem: You have a fixed factor (not a random
> factor) named status (two levels). Nested within status, there are
> biological replicates, which is considered a random term in the
> ANOVA model; and then nested within each biological replicate, you
> have several technical replicates, also random. The technical
> replicates are what most statistical packages label as "error".
> Paige's answer: The F-test for status should use the mean square for
> biological replicates in the denominator; the F-test for biological
> replicates should use the mean square for technical replicates (or
> the mean square for error) in the denominator.
> When you work out the expected mean squares for these ANOVA model
> terms (and I'm not going to do that here), you get the answer I
> proposed assuming that biological replicates are a random factor.
> Nested variables usually are random. But as a sort of non-technical
> explanation, for the effect of status to be statistically
> significant, you want it to be larger than the biological replicate
> variability combined with the technical replicate variability; thus
> you use the biological replicate mean square in the denominator
> because the biological replicate mean square contains elements of
> both the biological replicate and technical replicate variability.
> (If you do the F-test for status as described in the next paragraph
> with technical replicates in the denominator, you are only comparing
> the the effect of status to the technical replicate variability and
> NOT to the biological replicate variability)
> Had biological replicates been fixed (hard to see how that can be
> with them also being nested), then you get the first solution you
> mentioned, where both F-tests have the error mean square in the
> denominator. Without additional explanation, I can't see this being
> Paige Miller
> Eastman Kodak Company
> paige dot miller at kodak dot com
> "It's nothing until I call it!" -- Bill Klem, NL Umpire
> "When you get the choice to sit it out or dance, I hope you dance"
> -- Lee Ann Womack