This is a repost of
https://stat.ethz.ch/pipermail/r-help/2007-September/141727.html
Please do study the posting guide to see why you did not get an answer and
what to do when you do not.
It is nothing to do with TukeyHSD, as the differences are there in the
means.
It is clear that your covariates are havng an effect on the fit.
Presumably you suspected that because you tried a fit with them included,
and the fit with covariates is therefore the one to use.
As the posting guide points out this is not a list for statistical advice,
so please make use of your local statistical advice service for help on
the substantive issues here.
On Thu, 27 Sep 2007, Mariana Botelho wrote:
Hello,
I have some doubts on TukeyHSD application.
I want to investigate the effects of depth, latitude and month variation on
the length of a fish. These are orthogonal and observational data.
For this, I have made an aov model (L~month+lat+prof+month*lat), after
applying drop1 and step functions. But when I applied TukeyHSD I had
unexpected results.
For instance, I have three levels for latitude and the mean and standard
deviation of lengths are:
aggregate(LtMm,list(FLat=FLat),mean)
FLat x
1 24.5 431.8745
2 25 415.9973
3 25.5 416.0420
aggregate(LtMm,list(FLat=FLat),sd)
FLat x
1 24.5 114.6516
2 25 108.9774
3 25.5 105.5219
So, it's expected to have 25 and 25.5 levels closer than 24.5, and we see
this making a simple aov model:
aov.LtArL <-aov(LtMm~FLat)
TukeyHSD(aov.LtArL, ordered = TRUE)
Tukey multiple comparisons of means
95% family-wise confidence level
factor levels have been ordered
Fit: aov(formula = LtMm ~ FLat)
$FLat
diff lwr upr p adj
25.5-25 0.04474535 -16.009079 16.09857 0.9999764
24.5-25 15.87715429 -5.371913 37.12622 0.1860347
24.5-25.5 15.83240894 -3.213078 34.87790 0.1251572
Nevertheless, the complete model indicates just the opposite:
aov.LtAr<-aov(LtMm~FMes+FLat+FProf+FMes*FLat)
TukeyHSD(aov.LtAr,"FLat",ordered=T)
Tukey multiple comparisons of means
95% family-wise confidence level
factor levels have been ordered
Fit: aov(formula = LtMm ~ FMes + FLat + FProf + FMes * FLat)
$FLat
diff lwr upr p adj
24.5-25.5 6.46322 -11.706623 24.63306 0.6815646
25-25.5 19.72066 4.404934 35.03639 0.0072350
25-24.5 13.25744 -7.014670 33.52955 0.2751153
Which should be the right interpretation?
Thanks in advance for any help!
Best regards,
Mariana L. L. A. Botelho
MSc candidate
São Paulo Fisheries Institute
Brazil
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--
Brian D. Ripley, [EMAIL PROTECTED]
Professor of Applied Statistics, http://www.stats.ox.ac.uk/~ripley/
University of Oxford, Tel: +44 1865 272861 (self)
1 South Parks Road, +44 1865 272866 (PA)
Oxford OX1 3TG, UK Fax: +44 1865 272595
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