Itraclass correlation for multilevel models, gllamm

[Alyssa Beavers]

Hello,

I am trying to figure out how to calculate intraclass correlation (ICC) for a three level model using gllamm, a user-contributed command for multilevel modeling. gllamm does not accept the "estat icc" option. From what I am seeing online, I believe there is not a simple command for ICC with gllamm, and that I have to write out the formula for ICC. To start with, I am trying to find the unconditional ICC for the levels in my model.

Here is my unconditional model:

gllamm Garden_Active_, i(Garden_ID LGardenZip_) family(binomial) link(logit)  nip(30) adapt

And here is my output:

gllamm Garden_Active_, i(Garden_ID LGardenZip_) family(binomial) link(logit)  nip(30) adapt

Running adaptive quadrature
Iteration 0:    log likelihood = -2718.5586
Iteration 1:    log likelihood =  -2606.836
Iteration 2:    log likelihood = -2600.9528
Iteration 3:    log likelihood = -2600.8307
Iteration 4:    log likelihood = -2600.8307


Adaptive quadrature has converged, running Newton-Raphson
Iteration 0:   log likelihood = -2600.8307  
Iteration 1:   log likelihood = -2600.8307  (backed up)
Iteration 2:   log likelihood = -2600.7843  
Iteration 3:   log likelihood = -2600.7388  
Iteration 4:   log likelihood = -2600.7386  
 
number of level 1 units = 4035
number of level 2 units = 2432
number of level 3 units = 31
 
Condition Number = 2.1411013
 
gllamm model 
 
log likelihood = -2600.7386
 
--------------------------------------------------------------------------------
Garden_Active_ |      Coef.   Std. Err.      z    P>|z|     [95% Conf. Interval]
---------------+----------------------------------------------------------------
         _cons |  -.0779572   .0776656    -1.00   0.315     -.230179    .0742645
--------------------------------------------------------------------------------
 
 
Variances and covariances of random effects
------------------------------------------------------------------------------

 
***level 2 (Garden_ID)
 
    var(1): 4.6620217 (.59617179)
 
***level 3 (LGardenZip_)
 
    var(1): .01965498 (.03711558)
------------------------------------------------------------------------------

However, I'm not understanding the formula and what numbers I need to plug in to calculate ICC for my level 2 and 3. Could anyone please help?

Many thanks,
Alyssa

[Andrew Musau]

You can just use the formula

ICC1= var(_cons1)/(var(_cons1)+ var(_cons2) + (pi)^2/3)
ICC2= var(_cons2)/(var(_cons1)+ var(_cons2) + (pi)^2/3)

where var(_consj) is the variance of the random intercept j. I think you should also be able to estimate the above model with -xtmelogit- which allows you to use -estat icc- for comparison with your calculations.

[Alyssa Beavers]

Hi Andrew,

Thanks for your answer. Just to make sure I understand, to determine the ICC for my level 3, it would be .01965498/(.01965498+4.6620217+pi^2/3)
and for my level 2 it would be 4.6620217/(.01965498+4.6620217+pi^2/3) ?

Many thanks,
Alyssa

[Andrew Musau]

For the lowest level, it should be the sum of both variances, which I neglected to include in #2. See the following example

. webuse towerlondon
(Tower of London data)

. melogit dtlm difficulty i.group || family: || subject:, nolog

Mixed-effects logistic regression               Number of obs     =        677

-------------------------------------------------------------
                |     No. of       Observations per Group
 Group Variable |     Groups    Minimum    Average    Maximum
----------------+--------------------------------------------
         family |        118          2        5.7         27
        subject |        226          2        3.0          3
-------------------------------------------------------------

Integration method: mvaghermite                 Integration pts.  =          7

                                                Wald chi2(3)      =      74.90
Log likelihood = -305.12041                     Prob > chi2       =     0.0000
--------------------------------------------------------------------------------
          dtlm |      Coef.   Std. Err.      z    P>|z|     [95% Conf. Interval]
---------------+----------------------------------------------------------------
    difficulty |  -1.648505   .1932075    -8.53   0.000    -2.027185   -1.269826
               |
         group |
            2  |  -.2486841   .3544076    -0.70   0.483    -.9433102     .445942
            3  |  -1.052306   .3999921    -2.63   0.009    -1.836276   -.2683357
               |
         _cons |  -1.485863   .2848455    -5.22   0.000     -2.04415   -.9275762
---------------+----------------------------------------------------------------
family         |
     var(_cons)|   .5692105   .5215654                      .0944757    3.429459
---------------+----------------------------------------------------------------
family>subject |
     var(_cons)|   1.137917   .6854853                      .3494165    3.705762
--------------------------------------------------------------------------------
LR test vs. logistic model: chi2(2) = 17.54               Prob > chi2 = 0.0002

Note: LR test is conservative and provided only for reference.

. estat icc

Residual intraclass correlation

------------------------------------------------------------------------------
                       Level |        ICC   Std. Err.     [95% Conf. Interval]
-----------------------------+------------------------------------------------
                      family |   .1139105   .0997727      .0181851    .4715289
              subject|family |   .3416307   .0889471       .192923    .5297291
------------------------------------------------------------------------------



. di .5692105/ (.5692105 + 1.137917+ c(pi)^2/3)
.11391055

. di (.5692105+1.137917) / (.5692105 + 1.137917+ c(pi)^2/3)
.34163078