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st: How to apply and interpret lincom function as an additional analysis to xtmixed


From   Jakob Nielsen <[email protected]>
To   "[email protected]" <[email protected]>
Subject   st: How to apply and interpret lincom function as an additional analysis to xtmixed
Date   Fri, 3 Jan 2014 16:02:11 +0000

Im using tmixed and need to make a post estimation analysis (lincom) in
 addition to the information I get from the xtmixed analysis. However, I
 have doubt how to write the command correctly. 

I am using Stata 12.1 for windows. 

----- 

Firstly,
 my dataset consists of two groups performing two different 
interventions. There is an uneven number of measurement points between 
the groups (4 vs. 3), but the first three measurements have been 
performed at the same time between the groups.  

Group 1        Group 2
id    Trial 1 Trial 2 Trial 3 Trial 4       Id    Trial 1 Trial 2 Trial 3
FP1    x.xx    x.xx    x.xx    x.xx        FP13    x.xx    x.xx    x.xx
Fp2    x.xx    x.xx    x.xx    x.xx        Fp14    x.xx    x.xx    x.xx
FP3    x.xx    x.xx    x.xx    x.xx        FP15    x.xx    x.xx    x.xx
FP4    x.xx    x.xx    x.xx    x.xx        FP16    x.xx    x.xx    x.xx
FP5    x.xx    x.xx    x.xx    x.xx        FP19    x.xx    x.xx    x.xx
FP6    x.xx    x.xx    x.xx    x.xx        FP20    x.xx    x.xx    x.xx
FP7    x.xx    x.xx    x.xx    x.xx        FP21    x.xx    x.xx    x.xx
FP9    x.xx    x.xx    x.xx    x.xx                    
FP10   x.xx    x.xx    x.xx    x.xx                    
FP11   x.xx    x.xx    x.xx    x.xx                    

When I compute the xtmixed analysis, I have the above data on the long format, and I get an output looking like this:

----
. xi: xtmixed measurement i.time*i.group || id: ,mle
i.time            _Itime_1-4          (naturally coded; _Itime_1 omitted)
i.group           _Igroup_1-2         (naturally coded; _Igroup_1 omitted)
i.time*i.group    _ItimXgro_#_#       (coded as above)
note: _ItimXgro_4_2 omitted because of collinearity

Performing EM optimization: 

Performing gradient-based optimization: 

Iteration 0:   log likelihood =  68.201963  
Iteration 1:   log likelihood =  68.202082  
Iteration 2:   log likelihood =  68.202082  

Computing standard errors:

Mixed-effects ML regression                     Number of obs      =        60
Group variable: id                              Number of groups   =        17
                                               
                                                Obs per group: min =         3
                                                avg                =       3.5
                                                max                =         4


                                                Wald chi2(6)       =    164.59
Log likelihood =  68.202082                     Prob> chi2        =    0.0000

-------------------------------------------------------------------------------
  measurement |      Coef.   Std. Err.    z       P>|z|     [95% Conf. Interval]
--------------+----------------------------------------------------------------
     _Itime_2 |  .3080587    .032767     9.40     0.000     .2438365    .3722809
     _Itime_3 |  .257        .0317744    8.09     0.000     .1947233    .3192767
     _Itime_4 |  .238        .0317744    7.49     0.000     .1757233    .3002767
    _Igroup_2 |  .0041429    .0391134    0.11     0.916    -.0725181    .0808038
_ItimXgro_2_2 | -.3109158    .0501596   -6.20     0.000    -.4092268   -.2126049
_ItimXgro_3_2 | -.2541429    .0495169   -5.13     0.000    -.3511941   -.1570916
_ItimXgro_4_2 |    0  (omitted)
        _cons |  .103        .0250987    4.10     0.000     .0538075    .1521925
-------------------------------------------------------------------------------

------------------------------------------------------------------------------
  Random-effects Parameters  |   Estimate   Std. Err.     [95% Conf. Interval]
-----------------------------+------------------------------------------------
id: Identity                 |
                   sd(_cons) |   .0353746   .0129137      .0172964    .0723481
-----------------------------+------------------------------------------------
                sd(Residual) |   .0710498   .0075191      .0577405    .0874268
------------------------------------------------------------------------------
LR test vs. linear regression: chibar2(01) =     3.15 Prob>= chibar2 = 0.0379

.----

This
 is all fine. However in addition to this, I would like to get a 
comparison within group comparison of the timepoints, which is not 
provided with the xtmixed analysis. 

That being: 

Group 1: time 2 and 3
Group 1: time 2 and 4
Group 1: time 3 and 4 

And 

Group 2: time 1 and 2
Group 2: time 1 and 3
Group 2: time 2 and 3 

I have been told that the command to use is the following lincom commands: 

Group 1: 

lincom _Itime_2 + _Itime_3

 ( 1)  [measurement]_Itime_2 + [measurement]_Itime_3 = 0

------------------------------------------------------------------------------
 measurement |      Coef.   Std. Err.      z    P>|z|     [95% Conf. Interval]
-------------+----------------------------------------------------------------
         (1) |   .5650587   .0556139    10.16   0.000     .4560574    .6740599
------------------------------------------------------------------------------

This should compare time 2 with time 3 in group 1. 

Group 2: 

lincom _Itime_2 + _ItimXgro_3_2

 ( 1)  [measurement]_Itime_2 + [measurement]_ItimXgro_3_2 = 0

------------------------------------------------------------------------------
 measurement |      Coef.   Std. Err.      z    P>|z|     [95% Conf. Interval]
-------------+----------------------------------------------------------------
         (1) |   .0539158   .0501596     1.07   0.282    -.0443951    .1522268
------------------------------------------------------------------------------

This should compare time 2 with time 3 in group 2. 

However,
 I am not entirely sure, whether I have setup the command correctly. 
Thus, the question is, whether the above lincom command is correct setup
 in regards to the purpose (within group comparisons of time 2 and 3)? 

Thanks in advance. 

Best, Jakob Nielsen 		 	   		  
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