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:

UNIANOVA spotvol BY status replicat /METHOD = SSTYPE(3) /INTERCEPT = INCLUDE /PRINT = HOMOGENEITY LOF /PLOT = SPREADLEVEL RESIDUALS /SAVE = SRESID /CRITERIA = ALPHA(.05) /DESIGN = status replicat (status).

Thanks, Natasha

Paige Miller <paige.miller@kodak.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 > solution. > > 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 > correct. > > -- > Paige Miller > Eastman Kodak Company > paige dot miller at kodak dot com > http://www.kodak.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