# [R] Tukey Test, lme, error: less than two groups

Lilith sweelith at gmx.de
Thu Dec 2 18:39:10 CET 2010

```Dear R-group,

I am trying desperately to get this Tukey test working. Its my first time
here so I hope my question is not too stupid, but I couldn't find anything
helpful in the help or in the forum.

I am analysing a dataset of treated grasses. I want to find out, if the
grasses from different Provenances react differently.
In the aov test I found a significance for the combination Treatment and
provenance:

summary(PAMaov<-aov(PAMval~Treatmentf*Pretreatmentf*Provenancef+Error(Datef/Code)))

Treatmentf:Provenancef               p-value:  0.008023 **

In the Linear fixed effects model lme, I can see that there is a
significance for two provenances (HU and ES)

summary(PAM.lme<-lme(PAMval~Treatmentf*Provenancef*Pretreatmentf, random=
~1|Datef/Code,na.action=na.omit))

Value  Std.Error  DF   t-value
p-value
(Intercept)                               0.6890317 0.06117401 994 11.263473
0.0000
TreatmentfF                              -0.2897619 0.05484590 467 -5.283201
0.0000
ProvenancefDE                             0.0105873 0.05484590 467  0.193037
0.8470

TreatmentfF:ProvenancefES                 0.1647302 0.08226884 467  2.002340
0.0458
TreatmentfF:ProvenancefHU                 0.1569524 0.07756381 467  2.023526
0.0436

No the big mystery is the Tukey test. I just can't find the mistake, it
keeps telling me, that there are " less than two groups"

summary(glht(PAM.lme, linfct = mcp(Provenancef = "Tukey")))

Fehler in contrMat(table(mf[[nm]]), type = types[pm]) :
less than two groups

I guess its important to know that I made factors out of some of the data.
Here is the code:

PAMdata\$provenance[PAMdata\$provenance == "5"] = "ES"
PAMdata\$provenance[PAMdata\$provenance == "6"] = "HU"
# etc.

Treatmentf <- factor(PAMdata\$treatment, levels=c("C","F"))
Datef <- factor(PAMdata\$Date, levels=c( "25.05.10 14:00","26.05.10
19:00","27.05.2010 7:30","27.05.10 14:00","01.06.10 14:00","02.06.10
19:00","23.06.10 12:30"),ordered=TRUE)

Pretreatmentf <- as.factor(PAMdata\$pretreatment)
Provenancef <- as.factor(PAMdata\$provenance)
Greenhousef <- as.factor(PAMdata\$greenhouse)
Individualf <- as.factor(PAMdata\$individual)

PAMval <- (PAMdata\$DataPAM)
Code<-(PAMdata\$code)

Thank you for any hint! That Tukey test seems so easy, I just can't find the
mistake....
Thank you very much fpr your help and greetings from Tanzania,
Lilith

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