Compare LCGA through chiČ
Message/Author
 Sofie Danneel posted on Tuesday, December 11, 2018 - 12:26 am
I want to do LCGA with a cohortsequential design. I ran LCGA's with slope and intercept constrained and unconstrained, but I don't get a chiČ value.
classes=C(6);
knownclass = C(cohort=1 cohort=2 cohort=3 cohort=4 cohort=5 cohort=6);
model: %overall%
I S | LAC1@0 LAC2@1 LAC3@2 LAC4@3;
I;
S;
I with S;
%c#1%
[I] (1);
[S] (2);
%c#2%
I S | LAC2@1 LAC3@2 LAC4@3;
[I] (1);
[S] (2);
%c#3%
I S | LAC1@1 LAC2@2 LAC3@3 LAC4@4;
[I] (1);
[S] (2);
%c#4%
I S | LAC1@2 LAC2@3 LAC3@4;
[I] (1);
[S] (2);
%c#5%
I S | LAC1@3 LAC2@4;
[I] (1);
[S] (2);
%c#6%
I S | LAC1@4;
[I] (1);
[S] (2);

How do I get a chiČ to compare the models through chiČ difference test?
 Bengt O. Muthen posted on Tuesday, December 11, 2018 - 12:28 pm
You use the loglikelihood values for the 2 models and compute a chi-square test from that as twice the loglikelihood difference.
 Sofie Danneel posted on Friday, December 14, 2018 - 6:31 am
So from this part of the output of each model?
Loglikelihood

H0 Value -15474.940
H0 Scaling Correction Factor 1.8329
for MLR

I also found this: https://www.statmodel.com/chidiff.shtml
But I don't get a loglikelihood for H1 with LCGA, only when I do LCGM. In that formula, would L0 be the logelikelihood of the constrained model and L1 of the unconstrained model? Or do I need to use another formula?
 Bengt O. Muthen posted on Friday, December 14, 2018 - 11:47 am
You use the H0 logL value for each of the two models and use the link you give but go to the heading

Difference Testing Using the Loglikelihood
 Sofie Danneel posted on Monday, December 17, 2018 - 6:57 am
Thanks!

And how do I define the number of parameters in an LCGA? And specifically in my study with 6 cohorts?
 Bengt O. Muthen posted on Monday, December 17, 2018 - 4:46 pm
The number of parameters is printed in your output.
 Sofie Danneel posted on Tuesday, December 18, 2018 - 1:18 am
Do I need to specify something in the output command to see the number of parameters?
All I find now is the number of freely estimated parameters: I assume that is not what I need?
 Bengt O. Muthen posted on Tuesday, December 18, 2018 - 8:11 am
Q1: No.

Q2: That is what you need.