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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
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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 for Group1, so Y is on average equal 0.30604 for subject in Group1 in TP=0
  • Y is larger by, on average, 0.05211 in Group2 than in Group1 given TP is equal in both groups (= in the same timepoint)
  • Y is samller by, on average, 0.06066 in Group3 than in Group1 given TP is equal in both groups
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  • $\begingroup$ Thank you so much! that was exactly what I needed. $\endgroup$ – Hjalte Quist May 7 '20 at 12:54

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