I have made a Linear mixed model in R, but I'm having trouble interpreting the results.
The model is Y ~ Group + TP + (1+TP| Subjectnum
, where Y is volume in %, Group is a factor of 3 groups, TP is continous time points and subjectnum is factor of 305 subjects. I want to see the difference between the groups, but I'm having trouble interpreting the intercept
. Is this the value for group 1? Or how do I find the estimate for group 1? I have read other posts, which mentioned that intercept is the reference, but couldn't quite figure out what the implifications for the groups.
I'm new to lme and any help would be greatly appreciated!
Linear mixed model fit by REML. t-tests use Satterthwaite's method ['lmerModLmerTest']
Formula: Y ~ Group + TP + (1 + TP | Subjectnum)
Data: Data
REML criterion at convergence: 2841
Scaled residuals:
Min 1Q Median 3Q Max
-3.2312 -0.2925 0.0100 0.3206 3.8045
Random effects:
Groups Name Variance Std.Dev. Corr
Subjectnum (Intercept) 0.2948 0.5430
TP 0.4419 0.6648 -1.00
Residual 0.7822 0.8844
Number of obs: 915, groups: Subjectnum, 305
Fixed effects:
Estimate Std. Error df t value Pr(>|t|)
(Intercept) 0.30604 0.10270 832.29413 2.980 0.00297 **
Group2 0.05221 0.10371 434.59132 0.503 0.61489
Group3 -0.06066 0.10448 434.59132 -0.581 0.56181
TP -0.30305 0.05226 346.02747 -5.799 1.51e-08 ***
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
Correlation of Fixed Effects:
(Intr) Group2 Group3
Group2 -0.507
Group3 -0.504 0.499
TP -0.698 0.000 0.000
convergence code: 0
boundary (singular) fit: see ?isSingular
Best Answer
Group
is a factor. So, so called 'reference level' is chosen for it (by default this would be its first level:Group1
).All the coefficients of the model can be interpreted as a difference between 'modelled' and 'reference' level.
So, in you case:
(Intercept)
is parameter forGroup1
, soY
is on average equal 0.30604 for subject inGroup1
inTP
=0Y
is larger by, on average, 0.05211 inGroup2
than inGroup1
givenTP
is equal in both groups (= in the same timepoint)Y
is samller by, on average, 0.06066 inGroup3
than inGroup1
givenTP
is equal in both groups