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 Carlijn C posted on Monday, June 23, 2014 - 7:42 pm
I'm studying the effectiveness of a intervention (with a RCT).
First, I study the normative development by a separate analysis of the control group. The four growth parameters (intercept/slope mean and variance) found in the model of the control group are repeated in the model for the experimental group, and I test equality of parameters.
I've used this syntax to test the difference in intercept variance:
model: i s | BVLT1@0 BVLT2@1 BVLT3@2;
[i@2.447]; [s@-0.086]; i@0.172 ; s@0.006 ;
Model control: i s | BVLT1@0 BVLT2@1 BVLT3@2;
[i@2.447] (p1); [s@-0.086] (p2); i@0.172 (p3); s@0.006 (p4);
Model experimental: i s | BVLT1@0 BVLT2@1 BVLT3@2;
[i] (p10); [s@-0.086] (p2); i@0.172 (p3); s@0.006 (p4);
Model test:
p1=p10;
But this doesn't work, I got an error.
 Carlijn C posted on Monday, June 23, 2014 - 7:44 pm
I've also used another syntax:
model: i s | BVLT1@0 BVLT2@1 BVLT3@2;
[i] (p1); [s] (p2); i (p3); s (p4);
Model control: i s | BVLT1@0 BVLT2@1 BVLT3@2;
[i] (p1); [s] (p2); i (p3); s (p4);
Model experimental: i s | BVLT1@0 BVLT2@1 BVLT3@2;
[i] (p1); [s] (p2); i (p3); s (p4);
Model test;
p1=p10;

Then I get also an error because this model doesn't fit anymore (slope variance is negative).
So now I'm using two different models for the control group (i s | BVLT1@0 BVLT2@1 BVLT3@2) and experimental group (i s | BVLT1@0 BVLT2@1 BVLT3@2; s@0). Is this the right way?
 Linda K. Muthen posted on Monday, June 23, 2014 - 10:09 pm
Please send the outputs showing the errors and your license number to support@statmodel.com.

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