Message/Author
 Daniel Rodriguez posted on Tuesday, December 03, 2002 - 1:16 pm
Hi, I fit a model to my data that includes a significant quadratic trend factor. I am writing my results and want to make sure I am making appropriate interpretations. Is the log odds the sum of the Betas? Regarding my covariate, it's only significant for the linear trend. Would that be interpreted the same as the effect on a simple linear model? I already have the latest Hosmer and Lemeshow book, as well the Agresti text. I'd just like your valuable inisght
 bmuthen posted on Tuesday, December 03, 2002 - 7:57 pm
Because the odds refers to the dependent variable given a set of covariates (x's), the log odds is the whole right-hand-side of the equation,

beta_0 + beta_1*x_1 + beta_2*x_2 etc

- see page 339 of the Mplus User's Guide. So, it is not the sum of beta's alone.

 Daniel Rodriguez posted on Wednesday, December 04, 2002 - 5:11 am
Bengt, I'd like to thank you for being so helpful to all of us who strive to understand structural equation modeling. You are an exellent teacher. The rest of us can only hope to treat others as well as you and Linda do.
 Lange Gasse 20; 90403 nuremberg, Ge posted on Tuesday, May 19, 2009 - 5:17 am
Does anybody know a good source how to interpret the means of the growth factors with ordered categorical variables as dependent variables over time? How can I calculate the mean of the intercept out of the thresholds?
 Linda K. Muthen posted on Wednesday, May 20, 2009 - 8:27 am
See the following book:

Fitzmaurice, G., Davidian, M., Verbeke, G. & Molenberghs, G.
(2008). Longitudinal Data Analysis. Chapman & Hall/CRC Press.

A good way to interpret this is to look at plots. See the PLOT command.
 Lange Gasse 20; 90403 nuremberg, Ge posted on Monday, May 25, 2009 - 7:29 am
Thank you for this help. I ordered the book and had a look at the plots, but I want to make my problem more concrete. I modeled a growth curve with 5 time points and categorical indicators (0,1,2). The means of the intercept and linear slope are -1.303 and .103. Of course, there is an increase in the log odds of not answering category 1 (=0) over time. But how do you interpret the negative mean of the intercept?

SUMMARY OF CATEGORICAL DATA PROPORTIONS

S1D_N_T2
Category 1 0.907
Category 2 0.090
Category 3 0.003
S1D_N_T3
Category 1 0.876
Category 2 0.119
Category 3 0.006
S1D_N_T4
Category 1 0.867
Category 2 0.130
Category 3 0.003
S1D_N_T5
Category 1 0.806
Category 2 0.193
Category 3 0.001
S1D_N_T6
Category 1 0.821
Category 2 0.175
Category 3 0.004
 Linda K. Muthen posted on Monday, May 25, 2009 - 10:23 am
 Merril Silverstein posted on Monday, June 20, 2011 - 5:01 am
I am running a two-level growth curve model using four latent variables each indicated by three ordered categorical variables (3 categories). I assume the link function here is probit. The time basis variables are linear and quadratic terms for age.

I am at a bit at a loss trying to figure out how to convert the coefficients (and variable thresholds?) into probabilities (or some other more meaningful metric) that show the pattern of change in the latent variables by age.

Is this a meaningful question or does one just calculate and plot the growth curve using the mean estimates as in the continuous variable case?

Thank you.
 Bengt O. Muthen posted on Monday, June 20, 2011 - 10:27 am
I assume that when you talk about four latent variables, you mean one latent variable at four time points. You can compute the estimated means for this latent variable at each time point using the estimated growth factor means and the time scores.

To compute the probabilties of the categorical outcomes you need to do numerical integration over the latent variables so that's hard to do. Mplus does not currently provide that.
 Lisa M. Yarnell posted on Friday, October 28, 2011 - 5:12 pm
Hello, I set up my quadratic growth model as shown below. However, I got this warning (though could not identify any obvious problems using TECH4): WARNING: THE LATENT VARIABLE COVARIANCE MATRIX (PSI) IN GROUP CHINESE 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 THAN TWO LATENT VARIABLES. CHECK THE TECH4 OUTPUT FOR MORE INFORMATION. PROBLEM INVOLVING VARIABLE Q.

Can you help me identify the problem?

MODEL: !This is the Chinese group.
i s q | YEAR1@0 YEAR2@0.385 YEAR3@0.692 YEAR4@1;
i with s;
i with q;
s with q;
!i on gender;
!s on gender;
!q on gender;

MODEL KOREAN:
i s q | YEAR1@0 YEAR2@0.385 YEAR3@0.692 YEAR4@1;
i with s;
i with q;
s with q;
!i on gender;
!s on gender;
!q on gender;
 Bengt O. Muthen posted on Friday, October 28, 2011 - 5:51 pm
You would have to send your input, output, and data to Support for us to be able to tell.
 Lisa M. Yarnell posted on Monday, October 31, 2011 - 10:14 pm
Hi again, I spotted the error in my output. There was a correlation between latent variables greater than |1.00|, which is illogical. In the Chinese group, the correlations between Intercept, Linear Growth and Quadratic growth (in the standardized solution) were:

I WITH
S -0.277 0.245 -1.128 0.259
Q 0.144 0.301 0.478 0.633

S WITH
Q -1.009 0.053 -18.900 0.000

The -1.009 is the illogical value.

However, setting S with Q @ -1.00 in the MODEL command doesn't yield a logical solution because that refers to setting the covariance (the estimate in the unstandardized solution) to -1.00, not the correlation (the standardized solution) to -1.00. When I set S with Q @ -1.00, the illogical correlation in the standardized solution remains.

How do I set S with Q @ -1.00 in the STANDARDIZED solution, or set the negative bound for this estimate to be -1.00? Is it possible to use the "@" command to set standardized estimates rather than unstandardized estimates?

Thank you!
 Lisa M. Yarnell posted on Monday, October 31, 2011 - 10:31 pm
Hi again, in my question, I meant to write: Is it possible to set the lower bound for this estimate to be -1.00?
 Linda K. Muthen posted on Tuesday, November 01, 2011 - 1:52 pm
You can set bounds for parameters using MODEL CONSTRAINT. However, in your situation I don't believe this is appropriate since the model for Chinese is inadmissible. You should fit a growth model for each group separately. If the same growth model does not fit well for each groups, across group comparisons are not relevant. It may be that the Chinese group cannot be included because they require a different growth model.
 Lisa M. Yarnell posted on Tuesday, November 01, 2011 - 2:18 pm
Actually, as a follow-up, I got the model to converge by centering the loadings for slope (subtracting -.5 from all loadings). Thanks! This issue is resolved.
 Meghan Cain posted on Thursday, January 18, 2018 - 6:38 pm
In reply to Bengt's comment, "To compute the probabilties of the categorical outcomes you need to do numerical integration over the latent variables so that's hard to do. Mplus does not currently provide that."

Is this still true?
 Bengt O. Muthen posted on Friday, January 19, 2018 - 11:42 am
I need to see the exact model to say.
 Meghan Cain posted on Friday, January 19, 2018 - 11:48 am
USEVARIABLES are l1almo l2almo l3almo l4almo l5almo l6almo l7almo
dum1intfm dum2intfm dummale dum1eth dum2eth;
CATEGORICAL are l1almo l2almo l3almo l4almo l5almo l6almo l7almo (*);
MISSING are all .;
ANALYSIS:
STARTS = 75 15;
ALGORITHM = integration;
MODEL:
int slope quad | l1almo@-3 l2almo@-2 l3almo@-1 l4almo@0 l5almo@1 l6almo@2 l7almo@3;
int ON dum1intfm dum2intfm dummale dum1eth dum2eth;
slope ON dum1intfm dum2intfm dummale dum1eth dum2eth;
quad ON dum1intfm dum2intfm dummale dum1eth dum2eth;
 Bengt O. Muthen posted on Friday, January 19, 2018 - 1:14 pm
The estimated probabilities are still obtained only for models without x's.