

Effectivity intervention:comparison b... 

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Carlijn C posted on Monday, June 23, 2014  1:42 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  1:44 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? 


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