MIMIC LGM Model PreviousNext
Mplus Discussion > Growth Modeling of Longitudinal Data >
Message/Author
 Michael Moore posted on Thursday, February 16, 2012 - 3:26 pm
My question involves a MIMIC model where I intend to compare a baseline LGM to an LGM nested within it (wherein intercept and slope paths are freed) using the traditional Chi-square difference test. My (abbreviated) syntax is included below:

BASELINE MODEL:
GROUPING IS PRCATO15_2cat (1 = secure 2 = insecure);
MODEL:
I S | STANXX1F@0 STANXXG3@.25 STANXXG5@.5 STANXXG6@.6 STANXXX5@1;
I S on L08X5MEAN_R;
STANXX1F STANXXG3 STANXXG5 STANXXG6 STANXXX5(1);
OUTPUT:
SAMPSTAT;
STANDARDIZED;

NESTED MODEL:
GROUPING IS PRCATO15_2cat (1 = secure 2 = insecure);
MODEL:
I S | STANXX1F@0 STANXXG3@.25 STANXXG5@.5 STANXXG6@.6 STANXXX5@1;
I S on L08X5MEAN_R;
STANXX1F STANXXG3 STANXXG5 STANXXG6 STANXXX5(1);
MODEL secure:
I S on L08X5MEAN_R;
OUTPUT:
SAMPSTAT;
STANDARDIZED;

When I run this syntax, the degrees of freedom between the baseline and nested models are identical. How is this possible if I'm specifying the freeing of I S on L08X5MEAN_R; for the secure group? I appreciate any help you can provide.
 Linda K. Muthen posted on Thursday, February 16, 2012 - 8:54 pm
There is no difference between the two models. The ON statement you mention in the group-specific MODEL command is not constrained to be equal as the default.
 Michael Moore posted on Thursday, February 23, 2012 - 6:25 pm
To make the statement "I S on L08X5MEAN_R" equal between groups the default in the baseline model, would this work:

BASELINE MODEL:
GROUPING IS PRCATO15_2cat (1 = secure 2 = insecure);
MODEL:
I S | STANXX1F@0 STANXXG3@.25 STANXXG5@.5 STANXXG6@.6 STANXXX5@1;
I S on L08X5MEAN_R;
STANXX1F STANXXG3 STANXXG5 STANXXG6 STANXXX5(1);
MODEL secure:
I S | STANXX1F@0 STANXXG3@.25 STANXXG5@.5 STANXXG6@.6 STANXXX5@1(p1);
I S on L08X5MEAN_R(p2);
STANXX1F STANXXG3 STANXXG5 STANXXG6 STANXXX5(1);
MODEL insecure:
I S | STANXX1F@0 STANXXG3@.25 STANXXG5@.5 STANXXG6@.6 STANXXX5@1(p3);
I S on L08X5MEAN_R(p4);
STANXX1F STANXXG3 STANXXG5 STANXXG6 STANXXX5(1);
MODEL CONSTRAINT:
0 = p1 - p3
0 = p2 - p4
OUTPUT:
SAMPSTAT;
STANDARDIZED;

Thanks again for your help!
 Linda K. Muthen posted on Friday, February 24, 2012 - 5:20 pm
You should place the equality in the general MODEL command:

MODEL:
I S | STANXX1F@0 STANXXG3@.25 STANXXG5@.5 STANXXG6@.6 STANXXX5@1;
I on L08X5MEAN_R (10);
S on L08X5MEAN_R (11);
 Michael Moore posted on Monday, February 27, 2012 - 5:10 pm
So, just to be clear, the syntax would read:

BASELINE MODEL:
GROUPING IS PRCATO15_2cat (1 = secure 2 = insecure);
MODEL:
I S | STANXX1F@0 STANXXG3@.25 STANXXG5@.5 STANXXG6@.6 STANXXX5@1;
I on L08X5MEAN_R (10);
s on L08X5MEAN_R (11);
STANXX1F STANXXG3 STANXXG5 STANXXG6 STANXXX5(1);
MODEL secure:
I S | STANXX1F@0 STANXXG3@.25 STANXXG5@.5 STANXXG6@.6 STANXXX5@1;
I on L08X5MEAN_R (10);
s on L08X5MEAN_R (11);
STANXX1F STANXXG3 STANXXG5 STANXXG6 STANXXX5(1);
MODEL insecure:
I S | STANXX1F@0 STANXXG3@.25 STANXXG5@.5 STANXXG6@.6 STANXXX5@1;
I on L08X5MEAN_R (10);
s on L08X5MEAN_R (11);
STANXX1F STANXXG3 STANXXG5 STANXXG6 STANXXX5(1);
OUTPUT:
SAMPSTAT;
STANDARDIZED;

Thanks again!
 Linda K. Muthen posted on Monday, February 27, 2012 - 5:30 pm
You don't need the group-specific MODEL commands. They are redundant. There is a section in Chapter 14 of the user's guide that discusses multiple group analysis.
Back to top
Add Your Message Here
Post:
Username: Posting Information:
This is a private posting area. Only registered users and moderators may post messages here.
Password:
Options: Enable HTML code in message
Automatically activate URLs in message
Action: