I'd like to analyse data of 11 plots with 15 plant individuals on each plot.

A variable was measured on each plant in 9 different years.

So 11 plots with 15 plants on each plot and measurements in 9 years = 1485 observations.

  year   id plot  val
1 2000 A_01    A 0.70
2 2000 A_02    A 0.90
3 2000 A_03    A 0.79
4 2000 A_04    A 1.04
5 2000 A_05    A 0.84
6 2000 A_06    A 0.84
     year   id plot  val
1480 2008 N_10    N 0.35
1481 2008 N_11    N 0.72
1482 2008 N_12    N 0.36
1483 2008 N_13    N 0.20
1484 2008 N_14    N 0.41
1485 2008 N_15    N 0.51

My goal is to find differences on each plot between years as well as differences in each year between the plots.

Thus my analysis in R looks like this so far:

mod <- lmer(val ~ year * plot + (1|id), data = dat)

> summary(mod)
Linear mixed model fit by REML ['lmerMod']
Formula: val ~ year * plot + (1 | id)
   Data: dat

REML criterion at convergence: 299.6

Scaled residuals: 
    Min      1Q  Median      3Q     Max 
-3.0167 -0.5385 -0.0967  0.4548 14.6926 

Random effects:
 Groups   Name        Variance  Std.Dev. 
 id       (Intercept) 5.596e-17 7.481e-09
 Residual             5.992e-02 2.448e-01

As I want to calculate p-values to find significant differences between years/plots I think I need to make sure I have normally distributed residuals of the random effects (?).

So I'm looking at the residuals ..

mod.res <- ranef(mod)$id$`(Intercept)`

.. and find that they are right-skewed:

enter image description here

So I found that there is the package robustlmm that - I think - can take care of that.

rmod <- rlmer(val ~ year * plot + (1|id), data = dat)

However, now the variance of the random effect is 0

Robust linear mixed model fit by DAStau 
Formula: val ~ year * plot + (1 | id) 
   Data: dat 

Scaled residuals: 
    Min      1Q  Median      3Q     Max 
-3.5596 -0.6193 -0.0762  0.6120 19.8078 

Random effects:
 Groups   Name        Variance Std.Dev.
 id       (Intercept) 0.00000  0.000   
 Residual             0.03766  0.194 

I am stuck now. What does that mean? How can I proceed?


1 Answer 1


These models are both telling you that there is no variation between plants. The second model estimates the random intercept variance as zero, while the lmer model estimates it as 5.596e-17, which is $0.000000000000000005596$

So you don't need to fit random intercepts to this data. You can simply use lm instead.

As for the residuals, these are not particularly skewed. There are 2, possibly 3, outliers and I would not be overly concerned about this. These residuals are plausibly normally distributed. But to be on the "safe side", you could use robust standard errors - see the sandwich package for more details.

  • $\begingroup$ Thanks a lot for your answer. I had also already assumed that I don't necessarily need random effects. However, I'd essentially like to carry out a repeated-measures ANOVA and run a Tukey test for group differences, hence I can't simply use lm or aov, can I? I'll have a look at the sandwich package! $\endgroup$
    – user45065
    Commented Aug 9, 2016 at 12:04
  • $\begingroup$ @beetroot there is no point running repeated measures ANOVA because there is no between-group variation. Why don't you just work with the linear model ? If you want custom contrasts then use lsmeans $\endgroup$ Commented Aug 9, 2016 at 12:18
  • $\begingroup$ Thanks, I am working with aov/lm and lsmeans now, however, I think there's quite some heteroskedasticity in my data, is that a problem when using lsmeans? $\endgroup$
    – user45065
    Commented Aug 11, 2016 at 9:47
  • $\begingroup$ Please post that as a separate question. $\endgroup$ Commented Aug 11, 2016 at 9:50
  • $\begingroup$ I asked a new question, maybe you have some advice ;) $\endgroup$
    – user45065
    Commented Aug 12, 2016 at 7:10

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