Coefficients:
Estimate t value Std.Err. p raw p Bonf p adj
Al800-Al100 -2.253 -10.467 0.213 0.000 0.000 0.000
Al600-Al100 -2.185 -10.389 0.207 0.000 0.000 0.000
Al400-Al100 -2.036 -9.850 0.210 0.000 0.000 0.000
Al200-Al100 -1.712 -8.051 0.215 0.000 0.000 0.000
control-Al100 -1.487 -7.243 0.205 0.000 0.000 0.000
control-Al800 0.767 -5.282 0.143 0.000 0.000 0.000
control-Al600 0.698 -5.072 0.148 0.000 0.000 0.000
control-Al400 0.550 -4.160 0.155 0.000 0.001 0.001
Al800-Al200 -0.542 -3.488 0.141 0.001 0.006 0.006
Al600-Al200 -0.473 -3.191 0.140 0.002 0.014 0.012
Al400-Al200 -0.325 -2.267 0.147 0.027 0.135 0.110
control-Al200 0.225 -1.593 0.132 0.116 0.466 0.341
Al800-Al400 -0.217 -1.475 0.152 0.145 0.466 0.341
Al600-Al400 -0.149 -1.064 0.138 0.292 0.583 0.466
Al800-Al600 -0.068 -0.449 0.145 0.655 0.655 0.655
> #a friend told me that it is possible to do multiple comparisons for lme
in a simplest way, i.e. :
> anova(lm1,L=c("treatmentcontrol"=1,"treatmentAl200"=-1))
F-test for linear combination(s)
treatmentAl200 treatmentcontrol
-1 1
numDF denDF F-value p-value
1 1 12 2.538813 0.1371
> anova(lm1,L=c("treatmentcontrol"=1,"treatmentAl400"=-1))
F-test for linear combination(s)
treatmentAl400 treatmentcontrol
-1 1
numDF denDF F-value p-value
1 1 12 17.30181 0.0013
> anova(lm1,L=c("treatmentcontrol"=1,"treatmentAl600"=-1))
F-test for linear combination(s)
treatmentAl600 treatmentcontrol
-1 1
numDF denDF F-value p-value
1 1 12 25.72466 3e-04
> anova(lm1,L=c("treatmentcontrol"=1,"treatmentAl800"=-1))
F-test for linear combination(s)
treatmentAl800 treatmentcontrol
-1 1
numDF denDF F-value p-value
1 1 12 27.9043 2e-04
> # however, p values are different that those obtained above. Is this way