I am looking at several dependant variables for which I created LMMs of the following kind:

DV ~ Group + (1|Subject) + (1|Time)

Now I am struggling with how to interpret the output concerning the random effects, e.g. for DV1:

Random effects:
 Groups   Name        Variance Std.Dev.
 Subject   (Intercept)  6.2     2.49    
 Time      (Intercept) 67.7     8.23    
 Residual             50.2     7.08    

and for DV2:

 Groups   Name        Variance Std.Dev.
 Subject   (Intercept)  27.2     5.22   
 Time      (Intercept)  17.7     4.21   
 Residual             125.2    11.19   

Are conclusions such as the following valid?

  1. Variability between subjects was about twice as high for DV2 than DV1.
  2. Variability between dates (time) was higher for DV1 than DV2.
  3. For DV1 time accounted for a higher variability than subject
  4. For DV2 a similar amount of variability can be attributed to time and subject.
  5. For DV1 there was a variability of about 8 [units] between years.
  6. For DV2 variability between subjects was about 5 [units].

Thanks!

  • 1
    Keep in mind what you are modelling. Each subject has its own intercept, and your model assumes that these intercepts follow a N(0,v) distribution. The variances reported are simply the "v" parameter here. Similarly for the time variable. – ahfoss Aug 28 '14 at 14:25
up vote 8 down vote accepted
+100

I'd suggest centering the interpretation around intraclass correlations:

  1. For DV1, the total random variation not explained by the group differences (the fixed factor) is $\hat\sigma^2_{tot} = 6.2 + 67.7 + 50.2 = 124.1$, so that the total SD is $\hat\sigma_{tot}=11.14$. For DV2, these figures are $\hat\sigma^2_{tot} = 170.1$ and $\hat\sigma_{tot}=13.04$
  2. For DV1, the correlation between two observations on the same subject but different times is $r_S = 6.2/124.1=.050$, while for DV2 it is $27.2/170.1 = .160$
  3. For DV1, the correlation between two observations at the same time but different subjects is $r_T = 67.7/124.1=.546$, while for DV2 it is $17.7/170.1=.104$
  4. For DV1, the correlation between two observations on the same subject and the same time is $r_{ST} = (6.2 + 67.7)/124.1 = 1-50.2/124.1=.595$, while for DV2 it is $1-125.2/170.1=.264$

Of course, point (4) may or may not be meaningful in your experiment. The largest correlation, by far is the Time one for DV1: Time effects contribute hugely to the variations in DV1.

I agree with ahfoss on avoiding the word "variability"

A bit more on intraclass correlation

Most experimental design books discuss this topic under "model II" or "random-effects models". Suppose that a measurement $Y$ follows $Y = \mu + S + E$ where $S$ is random variation among subjects and $E$ is error variance. Then $Var(Y) = Var(S)+Var(E) = \sigma^2_S + \sigma^2_E$. Now let $Y_1$ and $Y_2$ be two such measurements. Then in general, $$Var(Y_1-Y_2)=Var(Y_1)+Var(Y_2)-2Cov(Y_1,Y_2) = 2(\sigma^2_S+\sigma^2_E)-2Cov(Y_1,Y_2)$$ If the measurements are on the same subject, then the $S$ term cancels out and we have $$Var(Y_1-Y_2)=2\sigma^2_E$$ Equating these, we have $$Cov(Y_1,Y_2) = (\sigma^2_S+\sigma^2_E) - \sigma^2_E = \sigma^2_S$$ and thus that $Corr(Y_1,Y_2) = \sigma^2_S / (\sigma^2_S+\sigma^2_E)$.

This is called the intraclass correlation because it is the correlation of two measurements in the same class. It is useful because having random effects in a model implies a correlation structure in the data, and this quantifies it. The above ideas apply when there are more random effects; you just add the variances of the random effects you are conditioning upon.

  • Thanks! Could you explain a bit more what the idea behind these "interclass correlations" is and how I interpret those correlation coefficients? – beetroot Aug 31 '14 at 7:55
  • OK, I added some discussion to my answer. – rvl Aug 31 '14 at 14:49
  • Although this is not exactly what the original question asked, I would say this is a better answer than mine since ICC is an ideal framework in which to present the results. One suggestion: in your point #1, you might clarify that "group differences" refers to a fixed effect in the model called "group", not the "Groups" listed in the random effects results. That confused me for a sec... – ahfoss Sep 3 '14 at 22:01
  • @ahfoss makes a good point. I edited the answer to clarify what I meant there. – rvl Sep 4 '14 at 17:36
  • Many thanks, I think that's just what I need! By any chance, do you know if there is a package/function in R with which it is quick and easy to calculate the interclass correlations for a LMM? – beetroot Sep 10 '14 at 11:21

A few things. First, be careful with the general term "variability." In a general context it is fine, but if you are referring to variance or standard deviation specifically, use the correct term. Also, as per my comment to your statement, remember that these are variances of the distributions of subject- and time-specific intercepts.

Your statements (1) and (2) are not valid. If your goal in this analysis is to make an inference about absolute differences in variances between two variables with respect to a grouping, you would need some formal hypothesis test, which you have not implemented. You can make relative statements, however. You might modify (1) as "between-subject variability accounted for a greater proportion of the variance in DV2 compared to DV1"

Modify your remaining statements like this:

(3) Time accounted for a larger proportion of the variance in DV1 than subject

(4) modify as in (3), but how you define "similar" should be carefully considered based on your goals. In some contexts, a difference of 1 might be very meaningful...

(5) For DV1, time-specific means were drawn from a normal distribution with standard deviation 8 (assuming you have checked that normality assumptions hold, approximately)

  • Many thanks for your answer, one more question concerning (5), which normality are you referring to? I looked at the QQplot of the model residuals, is that sufficient or do I have to do further tests? – beetroot Aug 29 '14 at 6:52
  • Looking at the residuals is good. What I was referring to is verifying that the distribution of the random intercepts are also approximately normal, which you should also check. – ahfoss Sep 3 '14 at 21:48

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.