Multigroup tests in multiple growth c...
Message/Author
 Anonymous  posted on Friday, January 27, 2006 - 2:30 am
Hi,
I have specified a multiple growth model as reflected in the syntax below.

What is the correct way to test for gender differences in the intercept and slope? I'm use to conducting multigroup tests in the Mplus environment, but here the mean of the intercept is fixed a zero by default - what kind of syntax should I use to examine whether the overall level or trend in the measures differ among genders (i.e., how should I fixe intercept and/or slope to be equal among the two groups) ?

Anonymous

Syntax:

i s | T1@0 T2@1 T3@3;
 Anonymous  posted on Friday, January 27, 2006 - 5:33 am
Just to add to my previous message, the measurement invariance has also been specified for two items per mesurement occasion, although not included in the syntax description.
-Anonymous
 Linda K. Muthen posted on Friday, January 27, 2006 - 8:54 am
For the intercept growth factor, you would compare the default model of zero in the first group and free in the other groups to the model where the mean is zero in all groups. For the slope growth factor, you would compare the model where the means are free across groups to the model where the means are constrained to be equal across groups.
 Alexandru Cernat posted on Tuesday, November 19, 2013 - 4:24 am
Hello,

I am trying to compare a categorical latent growth model across two groups (Single vs. MM) using parametrization = THETA. I want to see if the variance, means and correlations for the slope and intercept are equal. Here I try to model a baseline model from which I can start to constrain. Could you please let me know if I'm doing anything wrong.

Model:
i s | sf1_1@0 sf1_2@1 sf1_3@2 sf1_4@3;

Model Single:
[i@0];
[s@0];

Model MM:

[i* s*];
i* s*;

Thank you,

Alex
 Bengt O. Muthen posted on Tuesday, November 19, 2013 - 9:51 am
With categorical outcomes and a single group the mean of i is fixed at zero and the mean of s is free. So don't fix the mean of s to zero in the second group. Also, in the second group you want to free the covariance between i and s.
 Alexandru Cernat posted on Wednesday, November 20, 2013 - 2:05 am
Thank you for the answer but it does not really solve the problem when I use THETA. I have applied all the restrictions you mentioned and those recommended by Millsap & Yun-Tein (2004) and I get the same errors.

Also, the model seems to be working with Delta parametrization. I don't manage to understand why this is happening. Any thoughts?
Thank you,
 Alexandru Cernat posted on Wednesday, November 20, 2013 - 2:05 am
The new syntax:
Model:
i s | sf1_1@0 sf1_2@1 sf1_3@2 sf1_4@3;
sf1_1@1;
sf1_2@1;
sf1_3@1;
sf1_4@1;
Model Single:
[i@0];
[s@0];
[sf1_1\$1] (a);
[sf1_1\$2] (b);
[sf1_2\$1] (c);
[sf1_2\$2] (d);
[sf1_3\$1] (e);
[sf1_3\$2] (f);
[sf1_4\$1] (g);
[sf1_4\$2] (h);
[sf1_1\$3] (i);
[sf1_1\$4] (j);
[sf1_2\$3] (k);
[sf1_2\$4] (l);
[sf1_3\$3] (m);
[sf1_3\$4] (n);
[sf1_4\$3] (o);
[sf1_4\$4] (p);
Model MM:
[i@0];
[s*];
i with s*;
[sf1_1\$1] (a);
[sf1_1\$2] (b);
[sf1_2\$1] (c);
[sf1_2\$2] (d);
[sf1_3\$1] (e);
[sf1_3\$2] (f);
[sf1_4\$1] (g);
[sf1_4\$2] (h);
[sf1_1\$3] (i);
[sf1_1\$4] (j);
[sf1_2\$3] (k);
[sf1_2\$4] (l);
[sf1_3\$3] (m);
[sf1_3\$4] (n);
[sf1_4\$3] (o);
[sf1_4\$4] (p);
 Bengt O. Muthen posted on Wednesday, November 20, 2013 - 6:31 pm
You should hold thresholds equal across time in a growth model, not just equal across groups. Also, you should not have [s@0] in the first group - it should be free. See also page 678 of the V7 UG.
 Alexandru Cernat posted on Thursday, November 21, 2013 - 1:56 am
Dear Bengt, thank you for the answer but unfortunately it's still not working. The new version of the syntax:

Model:
i s | sf1_1@0 sf1_2@1 sf1_3@2 sf1_4@3;
sf1_1@1;
sf1_2@1;
sf1_3@1;
sf1_4@1;

Model Single:
[i@0 s];
[sf1_1\$1-sf1_4\$1] (1);
[sf1_1\$2-sf1_4\$2] (2);
[sf1_1\$3-sf1_4\$3] (3);
[sf1_1\$4-sf1_4\$4] (4);

Model MM:
[i@0 s];
i with s*;
[sf1_1\$1-sf1_4\$1] (1);
[sf1_1\$2-sf1_4\$2] (2);
[sf1_1\$3-sf1_4\$3] (3);
[sf1_1\$4-sf1_4\$4] (4);
 Linda K. Muthen posted on Thursday, November 21, 2013 - 6:09 am
 Catherine Drane posted on Tuesday, June 09, 2015 - 10:58 pm
Hi Linda

Just hoping you might clarify something for me. I ran 2 parallel process models in 2 completely separate data sets, then realized that I should have run them as a nested multigroup parallel process model instead, using a grouping variable. Both data sets were grouped the same. The results for both approaches are different, can you explain to me why this would be so?
Thank you very much
 Linda K. Muthen posted on Wednesday, June 10, 2015 - 6:23 am