IIona-

I think you may be misinterpreting the t-test.

In model 1, consider the speciesH coefficient.  A test that speciesH = 0,
essentially asks: Is speciesH the same as speciesB?  The test statistic for
this hypothesis is the t-value reported in the table.  (t-value= -4.2,
p-value=0.0001)

In model 2, the corresponding coefficient, t-value, and p-value do not
correspond to the same hypothesis test from model 1.

If your goal is to test the overall species effect, then the test you want
is:

model0<-lme(sprouts~moisture,random=~time|ID,method="ML")
model1<-lme(sprouts~species+moisture,random=~time|ID,method="ML")
model2<-lme(sprouts~species*moisture,random=~time|ID,method="ML")

anova(model0,model2)  #TEST OF SPECIES EFFECT IN MODEL WITH INTERACTION
anova(model0,model1)  #TEST OF SPECIES EFFECT IN MODEL WITHOUT INTERACTION

And as an added bonus, (which should probably be done before the test of
Species Effect)

anova(model1,model2)  #TEST OF INTERACTION.  That is, do I really need the
more complex model?

Hope that helps.
-tgs


On Tue, Jun 29, 2010 at 5:09 AM, Ilona Leyer <ileyer@yahoo.de> wrote:

> Dear all,
> In a greenhouse experiment we tested performance of 4 different species
> (B,H,P,R) under 3 different water levels in 10 replications. As response
> variable e.g. the number of emerging sprouts were measured on three dates. A
> simple Anova considering every measurement date separately shows a higly
> significant effect of species and moisture (and partly the interaction of
> both). The mixed-effects model with species and moisture shows a highly
> significant effect of species and moisture as well. However, when I included
> the interaction the t-values of the species dropped strongly and the SE
> increase and the results for the species are not significant anymore. For me
> this does not seem plausible. Has anybody an idea, how this can be
> interpreted and if I have done a mistake in calculating the data?
>
> Thanks in advance for any help!
> Ilona
>
>
> model1<-lme(sprouts~species+moisture,random=~time|ID)
> model2<-lme(sprouts~species*moisture,random=~time|ID)
>
>
> Fixed effects: sprouts ~ species + moisture
>                         Value Std.Error  DF   t-value p-value
> (Intercept)           7.971267  1.330500 240  5.991180  0.0000
> speciesH             -6.459344  1.536329 114 -4.204400  0.0001
> speciesP            -10.063604  1.536329 114 -6.550421  0.0000
> speciesR             -5.051894  1.536329 114 -3.288288  0.0013
> moisturemoist         2.228835  1.330500 114  1.675185  0.0966
> moisturewaterlogged  17.111149  1.330500 114 12.860688  0.0000
>
>
> Fixed effects: sprouts ~ species * moisture
>                                  Value Std.Error  DF   t-value p-value
> (Intercept)                    4.831965  1.750970 240  2.759594  0.0062
> speciesH                      -4.464197  2.476245 108 -1.802809  0.0742
> speciesP                      -3.986787  2.476245 108 -1.610013  0.1103
> speciesR                      -0.809376  2.476245 108 -0.326856  0.7444
> moisturemoist                  3.505506  2.476245 108  1.415654  0.1598
> moisturewaterlogged           24.766934  2.476245 108 10.001811  0.0000
> speciesH:moisturemoist        -0.457291  3.501939 108 -0.130582  0.8963
> speciesP:moisturemoist        -2.458125  3.501939 108 -0.701932  0.4842
> speciesR:moisturemoist        -2.555356  3.501939 108 -0.729697  0.4672
> speciesH:moisturewaterlogged  -5.597498  3.501939 108 -1.598400  0.1129
> speciesP:moisturewaterlogged -15.538272  3.501939 108 -4.437048  0.0000
> speciesR:moisturewaterlogged -10.206874  3.501939 108 -2.914635  0.0043
>
>
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>

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