[R-sig-ME] back log transformation
espesser
robert.espesser at lpl-aix.fr
Thu Mar 24 11:37:39 CET 2011
Dear all,
This subject has been previously discussed, but I am not sure I proceed
the right way with the use of the variances.
Here is the summary of my lmer model :
Linear mixed model fit by REML
Formula: log(TIME) ~ LONG + ACCO + (1 | SUJET)
Data: dssPUISS
AIC BIC logLik deviance REMLdev
899.6 934.1 -442.8 856.7 885.6
Random effects:
Groups Name Variance Std.Dev.
SUJET (Intercept) 0.019090 0.13817
Residual 0.130297 0.36097
Number of obs: 1018, groups: SUJET, 24
Fixed effects:
Estimate Std. Error t value
(Intercept) 5.77423 0.04462 129.42
LONG 0.02883 0.01129 2.55
ACCO1 -0.05722 0.02272 -2.52
LONG is continuous .
ACCO is a 2 levels factor .
I would proceed so:
1) To compute TIME at this specific point :
sujet== "s3"
long == 6
acco == "1"
TIME = exp( intercept + 6*LONG + ACCO1
+ estimate_of_s3_intercept + 0.5*var(Residual) )
with var( Residual) == 0.130297
Is it correct ?
2) I am mainly interested to back-transform the fixed effects, at the
same point.
2.1) I would use:
TIME = exp( intercept + 6*LONG + ACCO1
+ 0.5*var(SUJET) +0.5*var(Residual) )
with var(SUJET) == 0.019090
2.2) Suppose there was a second random intercept (say b) in my model,
I would use:
TIME = exp( intercept + 6*LONG + ACCO1
+ 0.5*var(SUJET) + 0.5*var(b) + 0.5*var(Residual) )
Are these 2 expressions correct ?
2.3)
Suppose there was a random slope in the model, something like:
log(TIME) ~ LONG + ACCO + (LONG | SUJET)
How can I get TIME on the original scale ?
3) Related question :
To extract the stddev of the SUJET random intercept , I use:
attr(VarCorr(MyModel.lmer)$SUJET,"stddev")
Is it the right way ?
Thank you very much for your help .
Robert
--
Robert Espesser
CNRS UMR 6057 - Université de Provence
5 Avenue Pasteur
13100 AIX-EN-PROVENCE
Tel: +33 (0)442 95 36 26
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