I can't answer this without seeing exactly what you are doing. You should get the same results using the Wald test versus testing two nested models. I have seen examples where these are the same. So perhaps there are some defaults that change. If you send the inputs, data, outputs, and your license number to firstname.lastname@example.org, I can take a look at it.
Kihan Kim posted on Wednesday, September 22, 2010 - 9:42 am
I was going to run Wald chi-square difference test with one degree of freedom by constrainting one path (i.e., cc on itv) to 0.
I used the following syntax, but I get the same chi-square value as the model without the MODEL TEST command.
Is there anything wrong in my syntax?
Model: dv3 on dc cc;
dc on itv ts pd cent form12;
cc on itv(p1) ts pd cent form12;
dc with cc;
itv with ts cent form12 pd; ts with cent form12 pd; cent with form12 pd; form12 with pd;
Hello, I would like to know if I can use the MODEL TEST command to compare the strength of two different path coefficients predicting the same outcome. I am worried that if the two predictors have a different scale of measurement, then constraining them to equality may not be meaningful, if the constraint applies to unstandardized path coefficients. But if the test allows to compare the strength of the standardized coefficients for the two paths, then this strategy seems OK. Thanks for your help.