Category: PROC GLIMMIX

The following is a proc glimmix example syntax.  I ran it using a fake dataset, so the results are also fake.  The outcome is an interval variable and the model is a linear model (not a non-linear model like the logistic regression model).  The random statement makes this model "multilevel."  Level-1 units are students and level-2 units are schools (IDs are nces_school_name).  The intercept is being estimated as the grand average of school-specific intercepts.

This model is a multilevel model and different from the OLS model because it estimates intercept as random effects (school-specific intercepts are derived as random effects and the grand average of those are reported as the intercept).

proc glimmix data=temp1 ;
class nces_school_name ;
model y=x1 x2 x3/
solution ddfm=kr dist=normal link=identity ;
random intercept /subject=nces_school_name;
run;

The main result table you will need to look at is the following.  You can interpret this just like you would interpret the OLS regression result.

Solutions for Fixed Effects
Effect	Estimate	Standard Error	DF	t Value	Pr > |t|
Intercept	0.5062	0.01600	1040	31.64	<.0001
x1	0.005776	0.01717	3483	0.34	0.7365
x2	0.000465	0.01684	3486	0.03	0.9780
x3	-0.01117	0.01708	3487	-0.65	0.5133

Another table that you want to look at is this.  Residual (0.54) is Level-1 variance (student-level variance).  Intercept/nces_school_name (0.65) is Level-2 variance (school-level variance).  I made up these numbers.

Covariance Parameter Estimates
Cov Parm	Subject	Estimate	Standard Error
Intercept	nces_school_name	0.65	0.0181
Residual		0.54	0.2002

I will focus on the first coefficient table when discussing results; however, I would also report information related to the second table.  I would report the following information:

ICC (intraclass correlation):

level2 variance / (level1+level2 variance)

That is:

0.65 / (0.65 + 0.54) =0.546218

This shows the degree to which outcome variance is located between groups (schools, clusters) as opposed to within individuals.

I would also report "variance explained."

I need to run the additional model, which is the analysis-of-variance model where I have no covariate in the model:

proc glimmix data=temp1 ;
class nces_school_name ;
model y=/
solution ddfm=kr dist=normal link=identity ;
random intercept /subject=nces_school_name;
run;

Let's take a look at the covariance table (I made up these values).

Covariance Parameter Estimates
Cov Parm	Subject	Estimate	Standard Error
Intercept	nces_school_name	0.92	0.181
Residual		0.83	0.200

 

Let's combine this table with the other one in this way.

	Analysis of variance model	Final model	Variance explained
Level 1 variance	0.83	0.54	0.34939759
Level 2 variance	0.92	0.65	0.293478261

 

 

Variance explained was calculated as:

(0.83-0.54)/0.83

and

(0.92-0.65)/0.92

 

I would make the final table look look like this.   I didn't round numbers, but you should.

	Estimates	Standard error	p-value	statistical test
Intercept	0.5062	0.016	<.0001	***
x1	0.00578	0.01717	0.7365
x2	0.00047	0.01684	0.978
x3	-0.0112	0.01708	0.5133
level-1 variance	0.83
level-2 variance	0.92
ICC	0.55
Level-1 variance explained	0.35
Level-2 variance explained	0.3
Notes: *** if p < 0.001, ** if p < 0.01, * if p < 0.05.

 

How to interpret coefficients in the table

In the table above, x1's coefficient is 0.00578.  This means that one unit increase in X1 will lead to an increase of 0.00578 in Y.   The p-value associated with this is 0.74.  So the coefficient here is not statistically significant at alpha=0.05.

One unit increase in X1 means ... if X is about height in meters (e.g., 1.5 meter, 1.7 meter), then 1 meter is 1 unit increase.  If X is a binary variable (0 or 1), then one unit increase means "0 to 1 increase".

For my work, I almost always have a variable called TREATMENT which is 1 if subjects received treatment/intervention and 0 if the subjects did not.  The coefficient for this is called "program impact effect."  If, for example, the program impact effect is 0.25, I just say that and I also mention that other covariates are in the model and the program impact effect is adjusted for these factors.  If the estimated program effect is 0.25, it means that the difference between the two groups (treatment vs. control) is 0.25 in outcome.

I also want to provide a standardized version of the program effect.  I would run the same statistical model with the z-score version of the outcome variable.  To do this, I usually use proc standard:

data abc2; set abc1;

Z_Y=Y;

run;

proc standard data=abc2 out=abc3 mean=0 std=1;

var Z_Y;

run;

Another approach would be to code this by hand in a datastep.  if the mean of Y is -0.42 and SD is 0.5:

data abc2; set abc1;

Z_Y= (Y - 0.42)/0.5 ;

run;

 

 

 

The goal of this exercise is to find out what happens if I intentionally use a level2 variable at level 1 in HLM.   I found that the coefficients and standard errors remain about the same.  The parameter that differed was just the degree of freedom, which was consistent with my expectation.

Using my old NELS dataset, I ran two different HLM models using Bryk and Raudenbush’s software (See model 1 and model 2 equations in the table below).

• The URBAN as level 1 covariate model (I entered the level2 variable URBAN wrongly at level 1)
• The URBAN as level 2 covariate model (I entered the level2 variable URBAN correctly at level 2)

The outcome variable is the achievement composite (POSTTEST), students are level 1 and schools are level 2.  When expressed as mixed models, the two models are identical, which is why I expected most parameters to come out the same.

POSTTEST_ij = γ₀₀
+ γ₁₀*URBAN_ij  + u_0j+ r_ij

The first model (MODEL 1; see below) included URBAN (students are in urban school) as a level 1 predictor.  Of course this is a wrong specification because urban is a school characteristic.  In the second model (MODEL 2), I used it at the expected level, which is at level 2 (school level).

These models look different, but AGAIN when expressed as mixed models, they are identical.  As the third model (MODEL 3), I replicated the same HLM model using SAS PROC GLIMMIX.  SAS requires that the equation be expressed as a mixed model.

Results showed that coefficients and standard errors are more or less the same across three models.  The only one thing that was different was degree of freedom.

Conclusion: As long as variables enter the model as fixed effects as done here, there is nothing magical about the HLM model.  HLM software or SAS PROC GLIMMIX (option ddfm=kr) adjust degree of freedom values, accounting for the fact that URBAN is a school-level variable and thus should not be awarded a value that is too large.  Notice that under the correct specification (MODEL 2 and MODEL 3), the degree of freedom for URBAN is close to the number of schools, not to the number of students.

Thanks for any comments you may have.

MODEL 1	MODEL 2	MODEL 3
URBAN as LEVEL 1 covariate	URBAN as level 2 covariate	SAS PROC GLIMMIX
Level-1 Model POSTTESTij = β0j + β1j*(URBANij) + rij Level-2 Model β0j = γ00 + u0j β1j = γ10 Mixed Model POSTTESTij = γ00 + γ10*URBANij  + u0j+ rij	Level-1 Model POSTTESTij = β0j + rij Level-2 Model β0j = γ00 + γ01*(URBAN_LEj) + u0j Mixed Model POSTTESTij = γ00 + γ01*URBAN_LEj  + u0j+ rij	proc glimmix data=kaz.level1; class schoolID; model posttest = urban /solution ddfm=kr dist=normal link=identity s ; random schoolID; run;
Final estimation of fixed effects (with robust standard errors) Fixed Effect  Coefficient  Standard error  t-ratio  Approx. d.f.  p-value For INTRCPT1, β0     INTRCPT2, γ00 52.643432 0.526139 100.056 125 <0.001 For URBAN slope, β1     INTRCPT2, γ10 -0.450022 1.157924 -0.389 692 0.698   Final estimation of variance components Random Effect Standard Deviation Variance Component   d.f. χ2 p-value INTRCPT1, u0 3.76951 14.20923 125 292.48369 <0.001 level-1, r 8.39004 70.39271 Statistics for current covariance components model	Final estimation of fixed effects (with robust standard errors) Fixed Effect  Coefficient  Standard error  t-ratio  Approx. d.f.  p-value For INTRCPT1, β0     INTRCPT2, γ00 52.643459 0.526140 100.056 124 <0.001     URBAN_LE, γ01 -0.449983 1.157920 -0.389 124 0.698   Final estimation of variance components Random Effect Standard Deviation Variance Component   d.f. χ2 p-value INTRCPT1, u0 3.76919 14.20678 124 292.48068 <0.001 level-1, r 8.39007 70.39334 Statistics for current covariance components model	Solutions for Fixed Effects Effect Estimate Standard Error DF t Value Pr > |t| Intercept 52.6434 0.5460 95.01 96.41 <.0001 urban -0.4501 1.1095 132.4 -0.41 0.6856     Covariance Parameter Estimates Cov Parm Estimate Standard Error schoolID 14.2150 3.4610 Residual 70.3913 3.7455

 

Datasets:

www.nippondream.com/file/datafiles_HLM.zip

Results from Model 1

Results from Model 2

Results from PROC GLIMMIX

 

I encountered a situation where somehow PROC GLIMMIX can't produce results while PROC MIXED can.  In this WORD document, I showed you how I edited the PROC GLIMMIX syntax to be a PROC MIXED syntax. Please ignore my macro %w.

https://drive.google.com/open?id=0B7AoA5fyqX_sVXQ1YzRlSlFlaW8

While running PROC GLIMMIX with a large dataset with categorical variables explicitly treated as CLASS variables AND with weight, I got the following error message:

WARNING: Obtaining minimum variance quadratic unbiased estimates as starting values for the         covariance parameters failed.

The weight and the use of categorical variables were the cause of the problem as I found out that without weight the model produces results.

SAS support staff spent time trying to figure out what the issue is/was.  I sent them a dataset (though I replaced the variables with randomly generated values) and the exact syntax.  They determined that (thank you so much for going extra miles to figure this out!):

"the computation of MIVQUE fails because the default singular value (1e-12) leads to the detection of singularity while sweeping of the mixed model equation"  -- May 12, 2016.

They proposed that singular value should be specified as a very small value to let the conversion happen:

proc glimmix data=example singular=1e-9;
weight psweight;
class race_cate level2 ;
model outcome= race_cate / s dist=normal link=identity ddfm=kr;
random intercept/subject=level2;
run;

****

Before hearing this final response from SAS, I wrote this and said that I chose to use PROC MIXED because it produces results.

 

SAS Support person suggested that I use of PARMS statement after Random, so I feed in the initial values for variance-covariance values manually/directly.  I did something like this:

PARMS (0.2) (0.4);

Then I get:

“ERROR: Values given in PARMS statement are not feasible.”

Out of curiosity, I used PROC MIXED to run the same model without PARMS but with a weight statement.  This time, the model produced results without encountering any errors.

PROC MIXED and PROC GLIMMIX produce identical results (or almost or essentially identical results) when the same models are specified.

I *think* my solution for now is to use PROC MIXED for this specific model.

The following two produced the identical results.
proc mixed data=analysis noclprint ;
where random_number > 0.8;
weight psweight;
class level2 race_cate2 loc_cate;
model zposttest=
<I put variable names here -- including categorical variables, level2 race_cate2 loc_cate>
/ s ddfm=kr ;
random int/subject=level2;
ParameterEstimates=kaz5
Diffs=DIF_GROUP5
LSMeans=LS5
Tests3=jointtest;
run;

proc glimmix data=analysis noclprint ;
where random_number > 0.8;
weight psweight;
class level2 race_cate2 loc_cate;
model zposttest=
<I put variable names here -- including categorical variables, race_cate loc_cate>
/ s dist=normal link=identity ddfm=kr STDCOEF ;
lsmeans
race_cate2 loc_cate
/ ilink diff
at (
treat
Zpretest
male
LEP
SPECED
DISADV
)=(0 0 0 0 0 0);
random int/subject=level2;
*parms (.03 ) (.45) /* /hold=1,2*/ ;
*ods output
ParameterEstimates=kaz5
Diffs=DIF_GROUP5
LSMeans=LS5
Tests3=jointtest;
run;