Multiple group analysis
Message/Author
 Alyson Zalta posted on Friday, March 09, 2007 - 6:45 am
Hello,
I am trying to run a multiple group analysis and test for differences between the groups (men & women). The model is comprised of all observed, continuous variables. I understand how to use the grouping function and constrain the paths (see syntax below). However I'm having trouble understanding the output. The approach I learned was to go step by step: First constrain path coefficients to be equal and test for differences. Then add on constraint for intercepts and test for differences. Then add on constrain for disturbances and test for differences. This way, you know where the difference lies. I'm confused what MPlus is doing with the syntax below and how to go about examining these differences. Any help would be much appreciated!

MODEL:
ib ON smsr (1)
icsrle (2);
cisst ON smsr (3);
pswq rrs ON smsr (4)
icsrle (5)
ib (6)
cisst (7)
sxc (8);
rrs WITH pswq (9);

ANALYSIS:
TYPE IS GENERAL;
ESTIMATOR IS MLM;
ITERATIONS = 1000;
CONVERGENCE = 0.00005;

OUTPUT: SAMPSTAT MODINDICES(3.84) RESIDUAL STANDARDIZED TECH2;
 Linda K. Muthen posted on Friday, March 09, 2007 - 8:25 am
See TECH1 or the results to see the effect of the equality constraints that you have placed. I think you may not mean some of them, for example,

pswq rrs ON smsr (4)

holds the regression coefficients of pswq on smsr and rrs on smsr equal to each other and equal across groups.

To test the equality across groups, you can use chi-square difference testing of the model with constraints versus the model without constraints. You can also use the MODEL TEST command to obtain the Wald chi-square test.
 Alyson Zalta posted on Friday, March 09, 2007 - 11:46 am
Thanks for your help. I am now trying to calculate the Chi-Square difference test using the Satorra-Bentler Scale Chi-Square. I have followed steps 1-4 using the following data and come up with a TRd = 21.68. What should I use for degrees of freedom to get the significance value?

T0 = 32.561, C0 = 1.125, D0 = 27
T1 = 7.31, C1 = 0.986, D1 = 6
 Linda K. Muthen posted on Friday, March 09, 2007 - 2:48 pm
The degrees of freedom is the difference in the number of free parameters.
 Lluis Coromina posted on Thursday, May 17, 2007 - 12:34 pm
Hello,
I am doing a CFA multigroup analysis for testing invariance (scalar).
Mplus by default fixes the mean of the latent variable to 0 in the first group. The intercepts and loadings are equal across groups.
I would like to use something different from the Mplus default. I am interested in the means of the latent variable. So, I fixed an intercept for one observed variable (tau)to 0 and I want to have all latent variable means free.
I would like to know how to get a latent variable mean also in the first group?
I hope my question is clear.

An example is below:
....
analysis: type= meanstructure;

model:
MEDIA by TV* RD* NWSP@1;
TV* RD* NWSP* ;
MEDIA* ;
[TV* RD* NWSP@0] ;
[MEDIA*];

model ES:
TV* RD* NWSP*;
MEDIA*;
[MEDIA* ];

Lluís Coromina
 Linda K. Muthen posted on Thursday, May 17, 2007 - 2:26 pm
You have to free the mean in the first group. From what you show, I can't tell which is the first group. If you can't figure this out, please send the input, data, output, and your license number to support@statmodel.com.
 Anjali Gupta posted on Thursday, August 27, 2009 - 9:32 am
Hello,

I'd like to use multigroup analysis to determine if moderation exists.

I have a path model and would like to check for differences by subgroup.

Can I run a multiple group model when
my base unconstrained model is fully saturated?

Thank you.
 Linda K. Muthen posted on Thursday, August 27, 2009 - 3:17 pm
If your path model is saturated in each group, you can test if a regression coefficient is different across groups.
 Anjali Gupta posted on Monday, August 31, 2009 - 2:13 pm
Hello,

To confirm - is this via multiple group analysis, where I constrain the predictor path (and/or covariate paths)? And then compare the non-constrained model and constrained model by analyzing the x2 and df?

How, in this situation, would multiple group analysis differ from adding an interaction term? Would I arrive at the same conclusion of moderation or no moderation?

Thank you.
 Linda K. Muthen posted on Monday, August 31, 2009 - 6:46 pm
This is an alternative way of testing for an interaction where you use the DEFINE command to create the interaction variable.
 Anjali Gupta posted on Wednesday, September 02, 2009 - 11:15 am
Hello,

Thank you for the response. To clarify - could using multiple groups analysis or interactions be 'interchanged'. That is, are there advantages/disadvantages to using either technique?

Or times one should be used over the other?

Also - to replicate running a regression on subgroups, say in SAS, would one, using Multiple Group Analysis, ideally constrain all paths (predictors, covariates, autoregressive paths)? Or focus on the predictor to Dep. Var path?

It's my guess that to replicate regression models with subgroups - one would compare unconstrained versus all paths constrained - including the covariates.

Thank you.
 Linda K. Muthen posted on Wednesday, September 02, 2009 - 1:37 pm
Multiple group analysis allows more parameters to be compared.

I am not familiar with SAS but I would imagine subgroups in SAS would be groups in Mplus.

The constrained versus unconstrained models would be driven by the test you want to do.
 D C posted on Tuesday, August 03, 2010 - 5:35 pm
Hello,

I am trying to do a two-group CFA analysis with categorical indicators. Hence, I have to perform the difftest:

ANALYSIS:
type = mgroup;
estimator = wlsmv;
difftest is deriv.dat;

I did this test by firstly estimating the less restrictive model (letting all loadings be free under 'MODEL group:' command with 'SAVEDATA: DIFFTEST IS deriv.dat;') then I estimated the more restrictive model (restricting equality of parameters under the 'MODEL:' command and 'ANALYSIS: difftest is deriv.dat;').

Hence, I got the following result indicating that constraining the parameters of the nested model significantly worsens the fit of the model:

Chi-Square Test for Difference Testing
Value 42.291
Degrees of Freedom 14**
P-Value 0.0001

However, I am afraid this result may still be driven by the large sample size of my data (group 1 N=2000 and group 2 N=1600). Are there other test I can use to show that the theorized factor structure fits the data well in both groups? Or, rather are there other tests to conclude that the loadings in the two groups are sufficiently similar to proceed with a full group analysis?

Thank you!
Dasha
 Linda K. Muthen posted on Wednesday, August 04, 2010 - 9:16 am
You might consider using an ESEM model as shown in Example 5.27.
 D C posted on Thursday, August 12, 2010 - 7:25 pm
Hello,

Is it possible to free certain loadings across say, race, in a MIMIC model while regressing the latent factors on race? I would want to do this if the mentioned loadings actually fit statically significantly differently across race in a multi-group analysis.

My second question: what does the direct regression effect from a factor indicator to race mean in a MIMIC model that also simultaneously regresses the latent factors on race?

Thank you!
 Linda K. Muthen posted on Friday, August 13, 2010 - 9:44 am
The most straightforward way to see if factor loadings are different for race is to do a multiple group analysis.

The regression of a factor indicator on a dummy variable is a direct effect and describes the intercept difference in the factor indicator across the levels of the dummy variable.

See the Topic 1 course video and handout where MIMIC, multiple group, and measurement invariance are discussed in detail.
 J.D. Haltigan posted on Saturday, August 28, 2010 - 12:12 am
Hello,
Much like the first post in this thread, I am interested in running an MGA in a path model with all observed indicators. This a moderation model whereby the dichotomous groups are thought to moderate the direct effect of a predictor on 5 outcomes. I understand that individual or summary data can be used. Is it okay to use a correlation matrix if one chooses to use summary data? I am unclear if this is an appropriate strategy in that all variables except the moderator (in the model) are continuous.
 Linda K. Muthen posted on Saturday, August 28, 2010 - 7:27 am
I would only use summary data when individual-level data were not available. You can use a correlation is you have a scale free model and all continuous outcomes so that would be okay for your model.
 J.D. Haltigan posted on Saturday, August 28, 2010 - 1:42 pm
Thank you. Two further questions. If individual level data in our analyses are mean composite variables does this render the data as mean structure? Secondly, in testing the models at various levels of the moderator, we are interested only in a certain subset of paths that might differ as a function of moderator level. Thus, the overall chi-square test of model fit (which would test all paths in the model) would seem inappropriate. Is there a specific way to compare only a subset of paths in the model at a given level of the moderator? In both models (i.e., at both levels of the moderator) we will constrain these paths to equality for theoretical reasons.
 Linda K. Muthen posted on Sunday, August 29, 2010 - 9:49 am
Meanstructure means that the means are estimated as model parameters.

You can test nested models using chi-square or loglikelihood difference testing.
 J.D. Haltigan posted on Sunday, August 29, 2010 - 12:54 pm
Thanks Linda. One last question: it indicates in the user's guide that for summary data in the case of MGA that the first group is represented by the first set of summary data found in the summary data set. Thus, if we are opting to use a correlation matrix, would this mean then that the grouping variable (i.e., moderator) should appear as the first variable in a lower correlation matrix?
 Linda K. Muthen posted on Sunday, August 29, 2010 - 2:22 pm
There is no grouping variable with summary data. See the rest of the writeup where NOBOSERVATIONS and NGROUPS are explained. A correlation matrix is not allowed with multiple group analysis. A multiple group model is not a scale free model.
 J.D. Haltigan posted on Sunday, August 29, 2010 - 2:51 pm
Thanks. I did read the instructions for summary data and NOBSERVATIONS and NGROUPS after I initially posted. I am having trouble in that when I use my summary data file (a covariance matrix) and lay it out as is indicated with the covariances for g1 on top and g2 on the bottom, I get the warning that there are more NOBSERVATIONS than NGROUPS (NOBSERVATIONS being 138 and 387 respectively). Am I missing something?
 Linda K. Muthen posted on Sunday, August 29, 2010 - 2:58 pm
 J.D. Haltigan posted on Sunday, August 29, 2010 - 3:48 pm
Hi: I did detect an error in my input file command statement that was responsible for the warning and was able to run the model successfully!
 Caroline Vancraeyveldt posted on Wednesday, December 14, 2011 - 5:53 am
Hello,

Currently I am testing the effect of an intervention on (amongst others) the growth of problem behavior in MPLUS. Therefore I have tested latent growth curve models, and multiple group models. I have noticed that the latent growth curve models sometimes give bad results for the effect of the intervention on problem behavior, whereas multiple group models mostly lead to good results. Can I choose for multiple group models to represent my data, or am I bound to use latent growth modeling?

 Caroline Vancraeyveldt posted on Wednesday, December 14, 2011 - 5:56 am
PS: Maybe it is important to know that I have a relatively small sample size (89 persons in the treatment group, and 86 persons in the control gropu)
 Linda K. Muthen posted on Wednesday, December 14, 2011 - 11:46 am
I can't quite understand what you are doing. Please send the growth and multiple group outputs along with your license number to support@statmodel.com.
 Lily Wang posted on Monday, July 02, 2012 - 6:50 pm
Hi Drs. Muthen,

I am working on a multi group analysis and have a question:

I did two sets of analyses, one based on gender and the other on race and I found that a path is significant for both males and females in the gender based multigroup analysis, but not significant for either the white or the minority group in the race based multigroup analysis. This doesn't make intuitive sense, but maybe I miss something important here? Could you provide some guidance as to why this could happen?
 Bengt O. Muthen posted on Monday, July 02, 2012 - 9:30 pm
Sounds like you need to do a multi-group analysis with gender crossed with race groups.
 Lily Wang posted on Tuesday, July 03, 2012 - 6:46 am
Dr. Muthen,

Thanks so much for your advice. Could I follow up on that and ask if you could provide any additional information on when to make such a decision--e.g. instead of two sets of analyses respectively based on gender and race, running one analysis with four groups (white female, white male, minority female and minority male)? That is, only after the analyses and I see the weird pattern of that one path, or are there other considerations? Do you know of any references or examples that deal with similar problems? I really appreciate your time!
 Bengt O. Muthen posted on Tuesday, July 03, 2012 - 5:08 pm
Given the generality of the question, you might enjoy posing it on SEMNET.
 Esperanza Camargo posted on Monday, November 26, 2012 - 11:16 am
Hello,
I want to measure differences/similarities between men's and women's behavior.
My dataset does not allow me to run a group analysis. Instead of have a variable such as gender, it does have different variables for each item, e.g. X= husband hit by mother and Y= wife hit by mother.
I ran exploratory and confirmatory analyses, independently, for men and women. The models are exactly the same. I am making my conclusions based on the correlations between four factors as well as their correlations with a few correlates (independent variables). All of the indicators of fit are excellent for both models (CTI, TLI, and RMSEA).
Is this a legitimate way to compare these behaviors? Is there any way to run them in the same model?

Thank you so much in advance,

Esperanza
 Esperanza Camargo posted on Monday, November 26, 2012 - 12:13 pm
Sorry another question.

Could I use the 'test of model fit for the baseline model' from each model to claim difference between these populations?
 Linda K. Muthen posted on Tuesday, November 27, 2012 - 10:16 am
What is the unit of analysis in your data set. Is it couple?
 Esperanza Camargo posted on Friday, November 30, 2012 - 11:38 am
No. The individual.
 Linda K. Muthen posted on Friday, November 30, 2012 - 3:27 pm
If the unit of observation is the individual, how can you have variables such as husband hit by mother and wife hit by mother? It seems the unit is either the husband or the wife.
 Sara Geven posted on Friday, September 27, 2013 - 12:23 am
Hello,

I am conducting a multiple group analysis in which I am comparing paths across different groups. Because I am not interested in comparing (latent) means, I checked for metric invariance. In my structural model I now freely estimated the intercepts in each group. This seemed right to me, since I did not test for scalar invariance. Is this the correct approach?

Kind regards,
Sara
 Linda K. Muthen posted on Friday, September 27, 2013 - 7:26 am
This seems correct.
 Sara Geven posted on Sunday, September 29, 2013 - 9:57 am
Thanks a lot. It makes me feel more certain about my results. Sara
 baozhenzhou posted on Tuesday, February 25, 2014 - 1:27 am
Hello,
I am trying to run a multiple group analysis and test for differences between two groups (male & female). The model is comprised of all observed, continuous variables(y is dependent variable,x is independent variable, w is moderator ).

VARIABLE:
names are y x w gender;
grouping is gender (0 = female, 1 = male);
usevariables are x y w gender xw;
DEFINE:
xw = x*w;
ANALYSIS:
bootstrap = 1000;
MODEL:
y on x
w
xw;
MODEL male:
y on x
w
xw;
OUTPUT: STDYX CINTERVAL(BCBOOTSTRAP);
The output only present the coefficients of the moderation for male and female.My question is how can i decide whether these moderation effects have gender differences? Thank you very much!
 Linda K. Muthen posted on Tuesday, February 25, 2014 - 6:25 am
You can use MODEL TEST or you can do a chi-square difference test between your model above and the model with regression coefficients held equal:

MODEL:
y on x (1)
w
xw;
 baozhenzhou posted on Tuesday, February 25, 2014 - 5:41 pm
VARIABLE:
names are y x w gender;
grouping is gender (0 = female, 1 = male);
usevariables are x y w gender xw;
DEFINE:
xw = x*w;
ANALYSIS:
bootstrap = 1000;
MODEL female:
y on x (fe_b1)
w (fe_b2)
xw (fe_b3);
MODEL male:
y on x (ma_b1)
w (ma_b2)
xw (ma_b3);
MODET TEST:
fe_b3=ma_b3;
OUTPUT: STDYX CINTERVAL(BCBOOTSTRAP);
I am interested in whether the moderation effect has gender difference, so I set the path of moderation to be equal and check the result of Wald test. Is it correct ? Thank you very much!
 Linda K. Muthen posted on Wednesday, February 26, 2014 - 11:26 am
MODEL TEST should be

MODET TEST:
0 = fe_b3 - ma_b3;

Note that the parameters have different labels. They are not held equal.
 Nicholas Bishop posted on Monday, March 17, 2014 - 2:04 pm
Hello,
To use the Model Test option to test for invariance in a parameter across 4 groups, would Form A or Form B of the model statement below be correct?

%overall%
ia sa| a1 a2 a3 a4 a5 a6 at t1-t6;

[ia];

%cg#1%
[ia] (p1);
%cg#2%
[ia] (p2);
%cg#3%
[ia] (p3);
%cg#4%
[ia] (p4);

!Form A
model test:
0 = p1-p2;
0 = p1-p3;
0 = p1-p4;

!Form B
model test:
0 = p1-p2;
0 = p1-p3;
0 = p1-p4;
0 = p2-p3;
0 = p2-p4;
0 = p3-p4;
 Linda K. Muthen posted on Monday, March 17, 2014 - 3:06 pm
Form A is correct.
 Nicholas Bishop posted on Tuesday, March 18, 2014 - 8:56 am
Thank you Linda. I am a still confused as to why Form B would not be the test I would want to use. For example, Form A would tell me if P2, P3, and P4 were equal to P1, but not if P2 was equal to P3 and P4, or if P3 and P4 were equal.

Is Form B not a more complete test of invariance across all four groups than Form A?
 Linda K. Muthen posted on Tuesday, March 18, 2014 - 1:29 pm
MODEL TEST runs a joint test not individual tests. The tests from Form B that are not in Form A are implied by Form A.
 SY Khan posted on Wednesday, April 30, 2014 - 6:46 am
Hi Dr. Muthen,

I am doing a path analysis with multigroup differnces for gender. I have specified two models i.e paths estimated freely (Model 1)and paths constrained (Model 2). Kindly advise if my syntax is correct:

MODEL 1:

grouping= gender (0 = female, 1 = male);

MODEL:

YA ON ABL MOT OPP COM;
YD ON ABL MOT OPP COM;
YJ ON ABL MOT OPP COM;
YO ON ABL MOT OPP COM;

J1 ON ABL MOT OPP COM;

YA ON J1;
YD ON J1;
YJ ON J1;
YO ON J1;

MODEL 2:

MODEL:

YA ON ABL MOT OPP COM;
YD ON ABL MOT OPP COM;
YJ ON ABL MOT OPP COM;
YO ON ABL MOT OPP COM;

J1 ON ABL MOT OPP COM;

YA ON J1;
YD ON J1;
YJ ON J1;
YO ON J1;

MODEL FEMALE:
YA ON ABL MOT OPP COM (a);
YD ON ABL MOT OPP COM (b);
YJ ON ABL MOT OPP COM (c);
YO ON ABL MOT OPP COM (d);

J1 ON ABL MOT OPP COM (e);

YA ON J1 (f);
YD ON J1 (g);
YJ ON J1 (h);
YO ON J1(i);

MODEL MALE:

YA ON ABL MOT OPP COM (a);
YD ON ABL MOT OPP COM (b);
YJ ON ABL MOT OPP COM (c);
YO ON ABL MOT OPP COM (d);

J1 ON ABL MOT OPP COM (e);

YA ON J1 (f);
YD ON J1 (g);
YJ ON J1 (h);
YO ON J1(i);

Many thanks.
 Linda K. Muthen posted on Wednesday, April 30, 2014 - 8:26 am
The best way to tell if a syntax is correct is to run the analysis and look at your results.

The following statement holds all four regression coefficients equal to each other. Is that what you want?

YA ON ABL MOT OPP COM (a);
 SY Khan posted on Wednesday, April 30, 2014 - 9:18 am
Dr. Muthen thank you for the prompt reply. I realized as well that holding the four regression co-efficient is not what I intended.So modified the syntax as follows:

YA ON ABL (a);
YA ON MOT(b);
YA ON OPP (c);
YA ON COM(d);

and the rest of the syntax is changed accordingly as well for both FEMALE and MALE Models.

When I run the analysis I get the following Results:

UNCONSTRAINED MODEL:
Chi-Sq(0)= 0.000
RMSEA= 0.000
CFI=1.000
TLI=1.000

CONSTRAINED: (ALL SIMILAR PATHS held Equal)
Chi-Sq(24)=146.423 p-value=0.0000
RMSEA=0.023
CFI=0.991
TLI=0.972

Do these model fit results mean that gender does not have an effect on the regression relationships?

I am not sure how to interpret these results. Some of the insignificant regression results in the unconstrained model become significant in the constrained model and vice versa. But most stay the same.

Thanks.
 Linda K. Muthen posted on Wednesday, April 30, 2014 - 3:48 pm
The question you answer when you compare a model where the regression coefficients are free across groups to the model where the regression coefficients are equal across groups is whether holding them equal worsens model fit. If the chi-square difference is significant, it indicates that constraining the parameters to be equal worsens the fit of the model. This means they are not the same for males and females.
 Angela Wolff posted on Thursday, May 08, 2014 - 1:48 pm
Hi
I am having an error message (using v 5.2) with this syntax and have checked the variable,
*** ERROR Categorical variable ET10 contains less than 2 categories.

ET10 does have more than 2 categories. What am I doing wrong?
Thanks
Angela

IDVARIABLE IS ID;
MISSING ARE ALL (99);
USEVARIABLES ARE ET1 ET2 ET3 ET4 ET5 ET6 ET7 ET8 ET9
ET10 ET11 ET12 ET13 ET14 REL1 REL2 REL3
REL4 REL5 REL6 REL7 ENV1 ENV2 ENV3 ENV4
ENV5 ENV6 CLX1 CLX2 CLX3 CLX4 PRAC1 PRAC2 PRAC3 PRAC4 PRAC5 PRAC6 LS1 LS2 LS3 LS4 LS5 Level;

CATEGORICAL ARE ET1 ET2 ET3 ET4 ET5 ET6 ET7 ET8 ET9
ET10 ET11 ET12 ET13 ET14 REL1 REL2 REL3
REL4 REL5 REL6 REL7 ENV1 ENV2 ENV3 ENV4
ENV5 ENV6 CLX1 CLX2 CLX3 CLX4 PRAC1 PRAC2
PRAC3 PRAC4 PRAC5 PRAC6 LS1 LS2 LS3 LS4 LS5;
GROUPING IS Level (1=L1 2=L2 3=L3 4=L4);
ANALYSIS:
ESTIMATOR = WLSMV;
MODEL: FAC1 BY ET1 ET2 ET3 ET4 ET5 ET6 ET7 ET8 ET9
ET10 ET11 ET12 ET13 ET14 REL1 REL2 REL3
REL4 REL5 REL6 REL7 ENV1 ENV2 ENV3 ENV4 ENV5 ENV6
CLX1 CLX2 CLX3 CLX4 PRAC1 PRAC2 PRAC3 PRAC4 PRAC5 PRAC6
LS1 LS2 LS3 LS4 LS5;

MODEL L1: FAC1 BY ET1 ET2 ET3 ET4 ET5 ET6 ET7 ET8 ET9
ET11 ET12 ET13 ET14 REL1 REL2 REL3
REL4 REL5 REL6 REL7 ENV1 ENV2 ENV3 ENV4 ENV5 ENV6
CLX1 CLX2 CLX3 CLX4 PRAC1 PRAC2 PRAC3 PRAC4 PRAC5 PRAC6
LS1 LS2 LS3 LS4 LS5;
 Linda K. Muthen posted on Thursday, May 08, 2014 - 2:08 pm
 Melissa Lopez posted on Friday, August 15, 2014 - 2:26 pm
Hello,

I'm attempting a two group mean comparison with a sample of 9,932. I have 23 variables and am using 6 of those as Usevariables.

I am receiving the error message: Data file contains more data than necessary. Data contents may be incorrectly supplied.

The data works fine for a path analysis. This is my first time conducting a means difference on a path analysis so I may be doing something completely wrong. I've spend a long time searching before asking, but can't find an answer.

Any help is greatly appreciated.
 Bengt O. Muthen posted on Friday, August 15, 2014 - 4:21 pm
Please send output and data to support so you don't have to keep searching.
 Jon H posted on Thursday, February 26, 2015 - 9:22 am
I am wondering if there is a way to use multiple group analysis (or something similar) to look at nested models with a lot of constraints. For example, let's say I'm looking at the following code for a composite variable (Model 1):
f1 by x1(a1) x2(a2) x3(a3);
f2 by x4(a4) x5(a5) x6(a6)
f3 by x7(a7) x8(a8) x9(a9)
fy1 by y1(b1) y2(b2) y3(b3);
c1 by;
c1 on f1@1 f2(q2) f3(q3);
fy1 on c1;

And then for Model 2, I'm looking at:
f1 by x1(a1) x2(a2) x3(a3);
f2 by x4(a4) x5(a5) x6(a6)
f3 by x7(a7) x8(a8) x9(a9)
fy1 by y1(b1) y2(b2) y3(b3);
c1 by;
c1 on f1@1 f2(q2) f3(q3);
fy1 on c1;
fy1 on c1 race sex income;

What I'm looking for is a way to constrain many of these paths to be the same from model 1 to model 2, as indicted by the labels above. I'm trying to figure out what happens to the relationship between fy1 and c1 if I introduce a bunch of additional covariates, but otherwise everything else remains the same.

My understanding is constraining variables across models only works in multiple group analysis, but I can't see how it would work. Can you think of a way to do this in Mplus?
 Bengt O. Muthen posted on Thursday, February 26, 2015 - 7:01 pm
Mutiple-group won't do it since it is the same people. It sounds somewhat akin to 3-step. I guess you don't simply want to fix the equal parameters when you go to Model 2. No bright ideas come to mind.
 Luo Wenshu posted on Wednesday, March 25, 2015 - 11:47 pm
Dear Dr. Muthen,

In two-group CFA analysis, if there is evidence to show
1) metric and scalar equivalence
2) equivalence of factor variance & covariance matrix
3) but non-equivalence of indicator residual variances,
can we then combine the two groups to run full SEM to examine structural relationships among factors?

Thank you very much!
 Bengt O. Muthen posted on Thursday, March 26, 2015 - 8:21 am
I would not combine the groups into a single-group analysis because it may distort the estimates relative to a two-group analysis. I would just continue the two-group analysis under scalar invariance. See also my 1989 Psychometrika article "Heterogeneous...".
 YH, Son posted on Thursday, November 26, 2015 - 3:16 am
Dear Dr. Muthen,

In two-group LGM, i want to check the difference between growth factors.

I S1| a1@0 a2@1 a3@2 a4@3 a5@4;
I S2| a1@0 a2@0 a3@1 a4@1 a5@1;
s1@0;(because of only two time point)

I know i can't test the difference of s1 variance becasue s1 fixed to 0.
But, I want to test the difference of s1 "means" between two groups.

Model a: [s1] (a_s1);
MODEL B: [S1] (B_S1);
MODEL TEST: A_S1 B_S1

BUT, WARNING SAYS " All continuous latent variable covariances involving S2 have been fixed to 0
because the variance of S2 is fixed at 0.".

CAN'T I test the differec of means?
 Bengt O. Muthen posted on Thursday, November 26, 2015 - 6:34 pm
It is s2@0 that is needed.

You can test the s2 means. That message is just a warning not an error. But you should write

Model Test:

0 = A_s1 - B_s1;
 Irene Dias posted on Friday, April 01, 2016 - 12:34 pm
Dear Dr Muthen,

I am using multigroup analysis to test if mediation exists (all variables in the model are observed). The unconstrained model is fully saturated and I am trying to test if differences in the regression paths exist between students in grade 4 and in grade 8, but I am working with a longitudinal design, meaning that the group of students is the same but assessed in both grades.
Can I use multigroup analysis and use the define command to test for differences? Given that all variables are observed and the model fully saturated, but I have a longitudinal design, should I use the command type=meanstructure? Should I control in any way for the covariance of the scores, given that they have been measured in the same participants?

Thank you so much.
 Bengt O. Muthen posted on Sunday, April 03, 2016 - 5:17 pm
If the grade 4 students are the same as the grade 8 students you can't use multi-group analysis because it assumes independent observations. Instead, formulate a model for both grades (wide format) and how the variables relate across grades and then use Model Test or Model constraint to check for differences of any parameters or product of parameters that you wish.
 M.Y posted on Monday, March 06, 2017 - 1:43 pm
Hi Professor, thanks for your response to my previous question.

Another question about multiple group modeling. My model involves several covariates. Some covariates and all my exogenous variables are observed variables (measured by single indicator). So I am wondering when I assess measurement invariance across groups, do I need to include these observed variables into the model? If yes, should I include them as covariates and use multi-group MIMIC model or should I treat them as one-indicator latent variable with zero measurement error?
 Bengt O. Muthen posted on Monday, March 06, 2017 - 5:52 pm
Multi-group MIMIC.
 M.Y posted on Saturday, March 11, 2017 - 3:22 pm
Hi Professor Muthen,

I am conducting multiple group modeling and received the follow error message.

THE STANDARD ERRORS OF THE MODEL PARAMETER ESTIMATES MAY NOT BE
TRUSTWORTHY FOR SOME PARAMETERS DUE TO A NON-POSITIVE DEFINITE
FIRST-ORDER DERIVATIVE PRODUCT MATRIX. THIS MAY BE DUE TO THE STARTING
VALUES BUT MAY ALSO BE AN INDICATION OF MODEL NONIDENTIFICATION. THE
CONDITION NUMBER IS 0.149D-15. PROBLEM INVOLVING THE FOLLOWING PARAMETER:
Parameter 123, V1

THIS IS MOST LIKELY DUE TO HAVING MORE PARAMETERS THAN THE SAMPLE SIZE.

However, the sample size of each group is actually greater than the number of estimated parameters when the model was estimated separately for each group. But for a configure model where parameters are not allowed to be equal across groups (certainly there will be more parameters to be estimated), the number of total parameters are greater than the size of one of the groups.

So my question is for multiple group modeling, when it says the number of cases in each group should be greater than the number of parameters, do we mean the no. of parameters in a configural model or the model estimated separately for each group?
 M.Y posted on Sunday, March 19, 2017 - 10:19 pm
Hi Professor, could you please respond to my previous post? Thanks!

Hi Professor Muthen,

I am conducting multiple group modeling and received the follow error message.

THE STANDARD ERRORS OF THE MODEL PARAMETER ESTIMATES MAY NOT BE
TRUSTWORTHY FOR SOME PARAMETERS DUE TO A NON-POSITIVE DEFINITE
FIRST-ORDER DERIVATIVE PRODUCT MATRIX. THIS MAY BE DUE TO THE STARTING
VALUES BUT MAY ALSO BE AN INDICATION OF MODEL NONIDENTIFICATION. THE
CONDITION NUMBER IS 0.149D-15. PROBLEM INVOLVING THE FOLLOWING PARAMETER:
Parameter 123, V1

THIS IS MOST LIKELY DUE TO HAVING MORE PARAMETERS THAN THE SAMPLE SIZE.

However, the sample size of each group is actually greater than the number of estimated parameters when the model was estimated separately for each group. But for a configure model where parameters are not allowed to be equal across groups (certainly there will be more parameters to be estimated), the number of total parameters are greater than the size of one of the groups.

So my question is for multiple group modeling, when it says the number of cases in each group should be greater than the number of parameters, do we mean the no. of parameters in a configural model or the model estimated separately for each group?
 Bengt O. Muthen posted on Monday, March 20, 2017 - 5:25 pm
We mean the number of parameters in the configural model (so for all groups).

If you have further questions on this, send your output to Support along with your license number.
 Moe Machida-Kosuga posted on Tuesday, February 20, 2018 - 5:04 pm
I am trying to do a multiple group analysis with only observed variables (one IV and one DV) and four groups.

Like the post above, I am comparing two models: paths estimated freely (Model 1)and paths constrained (Model 2).

I have two questions about interpreting the results:

Model 1 (unconstrained)
ANALYSIS:
TYPE=GENERAL;
ITERATIONS = 10000;
ESTIMATOR IS MLM;
MODEL:
LC2 ON MR_C;
Chi-Square (df 0) =0.000
P = 0.0000
RMSEA
Estimate 0.000 (90CI 0.000, 0.000)
CFI 1.000
TLI 1.000

Model 2 (constrained)
MODEL:
LC2 ON MR_C (1);
LC2 (2);
MR_C (3);
Chi-Square (df 9) =25.541
P= 0.0024
RMSEA
Estimate 0.176 (90CI 0.098, 0.258)
CFI 0.000
TLI 0.328

Q1:
Can I interpret this results that constraining the paths equal across four groups worsens the model fit and the relationships between IV and DV are different across four groups?

Is there any potential problems in doing that?

Q2:
Because the model is with with 0 degree of freedom 0 in the unconstrained model (Model 1), model fit (CFI and TLI) cannot be assessed, correct? Do you advise to still report these values in the results of a research?
 Bengt O. Muthen posted on Wednesday, February 21, 2018 - 4:55 pm
Q1: Yes.
No.

Q2: Right.
No, just say that the model has df=0.
 Moe Machida-Kosuga posted on Saturday, March 03, 2018 - 3:16 pm
Dr. Muthen:

Thank you for the fast reply to my post. I have been contemplating the models that I have run on mplus.

To explain my research question more fully, I am interested in seeing if the relationship between DV (LC2) and IV(MR_C) are different across these four groups.

Q1: In this case, in the constrained model (Model 2), should I only constrain the regression of LC2 ON MR_C to be held constant across the four groups? (and not constrain residual variances and intercepts)

Model 1 (unconstrained)
ANALYSIS:
TYPE=GENERAL;
ITERATIONS = 10000;

MODEL:
LC2 ON MR_C;

Chi-Square (df 0) =0.000
P = 0.0000

Model 2 (paths constrained)
MODEL:
LC2 ON MR_C (1);

Chi-Square (df 3) =11.739
P= 0.0083

Q2: It seems that constraining the regression significantly worsens the model fit. And can I conclude that the regression is different across four groups? (and proceed to relax the regression in each group to be freed one by one and see where the difference lies)

Thank you again for your help!
 Bengt O. Muthen posted on Sunday, March 04, 2018 - 2:18 pm
Q1: Yes.

Q2: Yes.

For general analysis strategy questions you may want to use SEMNET.
 Moe Machida-Kosuga posted on Monday, March 05, 2018 - 4:00 pm
Dr. Muthen:

Thank you so much for your response and introducing SEMNET!

I will check that out as well.

Thank you again for all your help.
 Jason Jabbari posted on Friday, May 18, 2018 - 4:26 pm
Dear Dr. Muthen,

I am trying to test the moderating role that gender (male and female) plays in a SEM model that has latent construct X predicting latent construct Y.

By testing the difference between constrained and unconstrained latent constructs for males and females, I find that the latent constructs within the model are NOT invariant to gender.

Here are my questions:

1. Does this mean that gender moderates the relationship between X and Y? Or, does group moderation only occur when models are invariant to groups?

2. Having read previous posts, i understand that it then makes little sense for me to compare the structural paths between males and females because their constructs are different. However, is it possible--and does it make sense--to add model constraints in order to allow comparisons between the groups to be adequately made? In other words, are group comparisons possible in models that are not invariant to the group?

I would GREATLY appreciate any help that you can offer.

Thank you,

Jason
 Bengt O. Muthen posted on Saturday, May 19, 2018 - 3:50 pm
You have to make a distinction between measurement invariance and structural invariance. You can't test the latter until the former has been established. This means that analyzing moderation of structural relationships doesn't make sense if measurement invariance is not present.
 Jason Jabbari posted on Sunday, May 20, 2018 - 4:20 pm
Thank you very much for your response. I have two follow up questions.

I have altered my latent constructs (x and y), so that they are now invariant to gender (males and females).

1. If my main interest is to see how the the structural relationship between x and y are different for males and females, do I also need to provide evidence of structural invariance?

2. If structural invariance is evident in my model, and I want to make comparisons for males and females, do I make the comparisons with the measurement and structural constraints in the model OR without the measurement and structural constraints in the model?

Thank you for your time. It is much appreciated.
 Bengt O. Muthen posted on Sunday, May 20, 2018 - 5:55 pm
1. You should show that the gender difference is significant.

2. If you have structural invariance there is nothing to compare across gender - they are the same.

These general analysis strategy questions are better asked on SEMNET.
 Jason Jabbari posted on Monday, May 21, 2018 - 9:33 am
Unfortunately, I am now having problems with running a model where each group is estimated freely. My understanding is that to estimate groups freely, i should fix factor means to zero in the general model and estimate loadings and intercepts separately. However, when i run the model, i get the following error message:"the standard errors of the model parameter estimates could not be computed. the model may not be identified. check your model. problem involving parameter 49, Group FEMALE: Math1"

USEVAR = X1MTHID
X1TXMTH
zHiMath09
X2MTHID
X2TXMTH
zHiMath12
X3DROPSTAT_recoded;
Categorical is X3DROPSTAT_recoded;
GROUPING IS X1SEX( 0 = male 1 = female);

Analysis:
Type = COMPLEX ;
MODEL:
Math1 by X1MTHID
X1TXMTH
zHiMath09;
Math2 by X2MTHID
X2TXMTH
zHiMath12;

Math2 on Math1;
X3DROPSTAT_recoded on Math2;

X2MTHID WITH X1MTHID;
X2TXMTH WITH X1TXMTH;
zHiMath12 WITH zHiMath09;

[Math1@0 Math2@0];

Model male:

Model female:
Math1 by X1MTHID
X1TXMTH
zHiMath09;
Math2 by X2MTHID
X2TXMTH
zHiMath12;

[X1MTHID X1TXMTH zHiMath09];
[X2MTHID X2TXMTH zHiMath12];
 Jason Jabbari posted on Monday, May 21, 2018 - 12:24 pm

Thanks!
 Johanna Ziemes posted on Tuesday, June 12, 2018 - 4:46 am
Dear all,

I would like to do a MG-CFA with an attitude scale on gender equality.
I have: Students in classes in countries.

I want to inspect the measuring invariance for boys and girls while controlling for the layered structure - is that possible?

If not: Is it possible to investigate measuring invariance at country and gender level simultaneously?
 Bengt O. Muthen posted on Tuesday, June 12, 2018 - 5:52 pm
This modeling of groups on the within level is a bit tricky to do correctly. See the Mplus Web Note 16 on our website:

Asparouhov, T. & Muthén, B. (2012). Multiple group multilevel analysis. Mplus Web Notes: No. 16. November 15, 2012.
 Christopher Bratt posted on Wednesday, December 19, 2018 - 2:49 am
Is there a simple way to free all parameters in a multigroup SEM analysis (no measurement invariance for factors in the SEM model)?

I have 20 groups, in my mind too many groups to specify the model for each group separatedly (not only because I'm lazy, but also because one command is less prone to errors than specifying the model for 20 groups separatedly).
 Bengt O. Muthen posted on Wednesday, December 19, 2018 - 3:37 pm
Model = Configural is set up to do this.
 Mengting Li posted on Tuesday, February 05, 2019 - 3:44 pm
Hi,I am trying to do a multiple group analysis and comparing the baseline model and the constrained model.It seems that the chi-square and CFI become worse when constraining the parameters to be equal. However, RMSEA become better.
So, I am wondering if this is possible? Does the comstrained model worse or better in such a case?
 Bengt O. Muthen posted on Tuesday, February 05, 2019 - 5:10 pm
Q1: Yes, this happens a lot - fit indices don't always agree.

Q2: It is becoming worse because chi-square says so. But it may be good enough according to RMSEA. BIC is also useful.
 SHANNA TRICHES LUCCHESI posted on Tuesday, April 09, 2019 - 5:56 am
Hi everyone.

We are conducting a multigroup analysis with two latent variables, a series of indicators and some background variables. When evaluating the full restricted model, the background variables don't seem to be fixed as their loads changed over groups. How can I restricted their loads to posteriorly compare their impact between groups?
 Bengt O. Muthen posted on Tuesday, April 09, 2019 - 5:34 pm
You can apply equality constraints in your Model command to hold the effects of the background variables equal across groups. And then check Modindices to see which of those equalities don't hold.
 SHANNA TRICHES LUCCHESI posted on Saturday, April 13, 2019 - 3:30 am
Thank you very much, Mr. Muther. It works! However, the same happened with the relation between the latent variables, with load change over groups. I tried to apply constraints but the model does not converge. What do you recommend to keep all the model restricted?
 Bengt O. Muthen posted on Saturday, April 13, 2019 - 4:20 pm
I would try to figure out why it doesn't converge with the equality constraints. Are the results from separate-group analyses so different? Is the model too complex. Have you set it up correctly?

 shonnslc posted on Tuesday, August 27, 2019 - 7:49 am
Hi,

I am doing a multiple group SEM with path coefficients freely estimated between two groups. However, I got this warning:

WARNING: THE LATENT VARIABLE COVARIANCE MATRIX (PSI) IN GROUP SENIOR
IS NOT POSITIVE DEFINITE. THIS COULD INDICATE A NEGATIVE VARIANCE/
RESIDUAL VARIANCE FOR A LATENT VARIABLE, A CORRELATION GREATER OR EQUAL
TO ONE BETWEEN TWO LATENT VARIABLES, OR A LINEAR DEPENDENCY AMONG MORE
PROBLEM INVOLVING VARIABLE C_1.

I think this is a Heywood case. I found the residual variance of C_1 is negative. So, I constrained the variance to be zero but I still got the same warning in my second run. Then, I tried to constrain the paths to be equal between two groups (without constraining the residual variance), Mplus terminated normally without any warning. What should I do for my case? Thanks.
 Bengt O. Muthen posted on Wednesday, August 28, 2019 - 2:49 pm
We need to see your two outputs - with the negative variance and with the variance fixed at zero.
 Carlos Sierra posted on Friday, September 27, 2019 - 8:20 am
HI Dr. Muthen,

I am trying to examine group differences (men vs. women) through the use of a multiple group conditional LGM. I have been successful at fitting the model, including adding covariates which predicts the intercept of the LGM. One of the covariates (child's gender) was significant in predicting the intercept but did so for women and not for men.

I then tried to use the MODEL CONSTRAINT and MODEL TEST commands to examine if my covariates deferentially affected the LGM's intercept for men and women. The results showed that neither the omnibus Wald test or any of the new parameters (resulting from the use of MODEL TEST command) were significant.

My question is how to interpret this. Is it accurate to say that, despite the significant estimate of child's gender on the intercept for women but not for men, there is not difference on this coefficient between men and women? or, is there a better way of interpreting these seemingly incongruous findings?

 Bengt O. Muthen posted on Friday, September 27, 2019 - 11:29 am
The "not significant" for males statement refers to a single coefficient whereas your Model Constraint and Model Test approaches may have tested more than one coefficient.

However, even with one coefficient, this outcome is quite common. A coefficient for one group may be too close to zero but not as far from that of the other group.
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