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 Christopher Desjardins posted on Tuesday, June 30, 2009 - 8:33 am
Hi,
Is there a way to get standardized coefficients and fit indices (RMSEA, CLI, etc.) with MLR? I am comparing models and selecting the best with BIC but because I have an interaction model using the xwith command I don't seem to have these available. Thanks.
 Linda K. Muthen posted on Tuesday, June 30, 2009 - 9:05 am
These are not available for TYPE=RANDOM.
 Christopher Desjardins posted on Tuesday, June 30, 2009 - 10:22 am
Thanks Linda. Do I have any other kind of options? I can't do a multi group because of a small sample size.
 Linda K. Muthen posted on Wednesday, July 01, 2009 - 8:35 am
With TYPE=RANDOM; the variance of y varies with the values of x. This precludes the calculation of standardized coefficients and chi-square and related fit statistics. In this case, nested models are tested using -2 times the loglikelihood difference which is distributed as chi-square.
 Syd posted on Thursday, July 22, 2010 - 1:30 pm
Hi,
I am testing the direct and interaction effects of two second-order continuous latent variables (f1 and f2) on another continous latent variable (f3), and the mediation effect of this variable on a binary (f4) dependent variable with a single indicator.
My model is roughly as follows (I have left out BY commands for f5 through f10 for simplicity):

f4 BY y1;
f1 BY f5 f6 f7;
f2 BY f8 f9 f10;
f3 BY x1-x6;
f3 ON f1 f2;
f1xf2 | f1 xwith f2;
f3 ON f1xf2;
f4 ON f3 f1 f2 f1xf2

I have two questions:

1. I understand that I need to use either MLR or WLSMV estimators in a random effects model with numerical integration. Is this correct?

2. I also understand that no fit indices are provided when testing continuous latent variable interactions. If my understanding is correct, then how can I understand whether my model is good or ill fitting for the full model?

I will greatly appreciate any suggestions and guidance.
 Syd posted on Thursday, July 22, 2010 - 1:37 pm
Just to clarify my second question, above, I am wondering if there is a way to assess absolute fit of my full model.

Thank you.
 Linda K. Muthen posted on Thursday, July 22, 2010 - 2:40 pm
1. Only MLR is available with XWITH.

2. No absolute fit statistic is available for this model. I would make sure the fit is good for the model without the interaction and then check that the interaction is significant. See a recent paper in Psychometrika on fit for interaction models by Moojart and Satorra.
 Syd posted on Thursday, July 22, 2010 - 7:50 pm
Thank you for the very prompt explanation and the reference. It was very helpful.
 Syd posted on Saturday, July 24, 2010 - 4:25 am
Hi Linda,

While testing a model with a single-indicator, binary dependent variable, I have run into an issue that I can't figure out.

My model is as follows:
VARIABLE: CATEGORICAL ARE y1;
ANALYSIS: ESTIMATOR=MLR;
MODEL:
f1 BY x1-x3;
f2 BY x4-x6;
f3 BY x7-x9;
f4 BY x10-x12;
f5 BY x13-x15;
f6 BY x16-x18;
f7 BY x19-x25;
f8 BY f1-f3;
f9 BY f4-f6;
f10 BY y1;
f7 ON f5 f6;
f10 ON f7;

The model runs fine with a single indicator continuous dependent variable, I get the following message when I use the binary DV:

THE MODEL ESTIMATION DID NOT TERMINATE NORMALLY DUE TO A NON-ZERO DERIVATIVE OF THE OBSERVED-DATA LOGLIKELIHOOD.

THE MCONVERGENCE CRITERION OF THE EM ALGORITHM IS NOT FULFILLED. CHECK YOUR STARTING VALUES OR INCREASE THE NUMBER OF MITERATIONS. ESTIMATES CANNOT BE TRUSTED. THE LOGLIKELIHOOD DERIVATIVE FOR PARAMETER 72 IS 0.16653809D+00.

I have tried increasing the MITERATIONS up to 1000, but the result remained the same. I would greatly appreciate some insight regarding what I might be doing wrong.

Second, for such a model, is there a way to get the indirect effect of f8 (via f7) on f10?

Thank you,
 Linda K. Muthen posted on Saturday, July 24, 2010 - 5:57 am
Please send the output and your license number to support@statmodel.com.
 Syd posted on Monday, July 26, 2010 - 4:21 am
Hi,
Is there any way to calculate standardized coefficients when using MLR and TYPE=RANDOM?
Thank you,
 Linda K. Muthen posted on Monday, July 26, 2010 - 7:40 am
No, we don't give them automatically. You can use MODEL CONSTRAINT to do this.
 yao lu posted on Tuesday, March 13, 2012 - 7:37 pm
Hi Linda,

Following up your post above (July 26, 2010-7:40am),how can Model Constraint calculate standardized coefficients when using Type = Random? I read the Model Constraint section in the manual, but still not quite get it.

Could you please provide me the coding for calculating standardized coefficients?

Thank you very much.
 Linda K. Muthen posted on Wednesday, March 14, 2012 - 2:35 pm
There is no single y variance with TYPE=RANDOM. Because of this, you cannot standardize with respect to y.

See Example 5.20 to see how MODEL CONSTRAINT can be used for standardization.
 Zeynep Türktorun posted on Tuesday, March 08, 2016 - 8:05 am
Dear Dr. Muthen,

I am currently testing a cross-level interaction. According to the discussions absolute statistic fits for those sorts of models (Type = Random;) are not available. You proposed to make sure the fit is good for the model without the interaction.

Which part of the model do I need to change to test the same model without the interaction.

This is my model:

Variable:
usevar = code S1 cs bo MG;
cluster = code;
within = cs; !
between = S1 MG;


analysis: type = twolevel random;
model: %within%
beta1j | bo on cs;
%between%
bo Beta1j ON S1 MG;
bo with beta1j;
output: sampstat;


Would it look like this...

(1) (...)
model:
%within%
bo on cs;

%between%
bo ON S1 MG;


Or like this...

(2)(...)
model: %within%
beta1j | bo on cs;

%between%
bo ON S1 MG;


Thanks a lot for your help!
 Bengt O. Muthen posted on Wednesday, March 09, 2016 - 11:55 am
You would not have a random slope specification, that is, not betaj |...

You can create an interaction variable for the Within vbl times the Between vble and use that in the model.
 Zeynep Türktorun posted on Thursday, March 10, 2016 - 2:59 am
Dear Dr. Muthen,
Thanks a lot for that hint!

In this case the type of model would change from "twolevel random" into "twolevel", is that right?

I not sure where to use the created interaction variable in the model. On within or between level?

Thank you very much.
 Bengt O. Muthen posted on Thursday, March 10, 2016 - 2:51 pm
Q1 Right

Q2 in Define And it will vary on both levels.
 Sabrina Krys posted on Wednesday, November 01, 2017 - 6:35 am
Hi!
I know that for Type=Random there is no standardized output available. However, I ran a moderated mediation analysis with several mediators and I want to build standardized main, direct, indirect, and interaction effects. I read Example 5.20, but it did not help to me. Is there any possibility to standardize the coefficients?
Thanks!
 Bengt O. Muthen posted on Thursday, November 02, 2017 - 3:01 pm
It is available in Version 8 using two-level Bayes. See description in


Hamaker, E.L., Asparouhov, T., Brose, A., Schmiedek, F. & Muthén, B. (2017). At the frontiers of modeling intensive longitudinal data: Dynamic structural equation models for the affective measurements from the COGITO study. Submitted to Multivariate Behavioral Research.
 Andrea Maloof posted on Thursday, October 04, 2018 - 8:12 pm
I am trying to standardize my variables in a 3-way interaction. I received this error: What does this mean? and how do I fix it?

THE MODEL ESTIMATION DID NOT TERMINATE NORMALLY DUE TO A NON-ZERO DERIVATIVE OF THE OBSERVED-DATA LOGLIKELIHOOD. THE MCONVERGENCE CRITERION OF THE EM ALGORITHM IS NOT FULFILLED. CHECK YOUR STARTING VALUES OR INCREASE THE NUMBER OF MITERATIONS. ESTIMATES CANNOT BE TRUSTED. THE LOGLIKELIHOOD DERIVATIVE
FOR THE FOLLOWING PARAMETER IS -0.16947489D+00:
Parameter 62, E

This was my syntax:

usevariables are
OD_Mean ID_Mean
EL_1 EL_2
EL_3 EL_4 EL_5 EL_6 EL_7 EL_8
F1 F2 F3 F4 F5 F6 F7
Hogan;

ANALYSIS: TYPE = RANDOM;
ALGORITHM = INTEGRATION;
DEFINE: standardize OD_Mean ID_Mean
EL_1 EL_2
EL_3 EL_4 EL_5 EL_6 EL_7 EL_8
F1 F2 F3 F4 F5 F6 F7
Hogan;

MODEL: WD BY OD_Mean ID_Mean;
EL BY EL_1 EL_2 EL_3 EL_4 EL_5
EL_6 EL_7 EL_8;
EC BY F1 F2 F3 F4 F5 F6 F7;
E BY Hogan;
ExEL | E XWITH EL;
ExEC | E XWITH EC;
ExELxEC | ExEL XWITH EC;
WD ON E EL EC ExEL ExEC ExELxEC;

Hogan@3.54;

output: TECH1 sampstat;
 Bengt O. Muthen posted on Friday, October 05, 2018 - 2:40 pm
Increase the number of iterations as the message suggest. But also study th eFAQ on our website called "Tech8 - negative abs changes". You have many dimensions of integration (see you Summary of analysis output section and also tech8 screen printing) and this makes for difficult computations.
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