|
|
 |
| DF in ESEM and target rotation question |
 
|
| Message/Author |
|
|
|
|
Hi,
I have a question about target rotation in ESEM which led me to think about the df in in ESEM, so I'll ask that first.
Where do the df come from in ESEM? If it was possible to run a CFA for, for example, a two factor model with six indicators with all indicators loading on both factors (I know it would not be identified), the df would (hypothetically) be 2 (21 observed variances/covariances minus 19 parameters to be estimated). For a corresponding ESEM, the df are 4, so where do they come from?
Second, when I run an ESEM with target rotation so that all non-intended factor loadings are set to zero, I get exactly the same chi sq/df as for an ESEM without target rotation. Is this correct? If so target rotation imposes no more model constraints, clearly. Here's the syntax I'm using, reduced to just a six item example for simplicity:
VARIABLE:
NAMES ARE BB1 BB2 BB3 NB1 NB2 NB3;
ANALYSIS:
ROTATION=TARGET(OBLIQUE);
MODEL:
F1 BY BB1-NB3 NB1~0 NB2~0 NB3~0(*t);
F2 BY BB1-NB3 BB1~0 BB2~0 BB3~0(*t);
OUTPUT: stand;
Am I doing something wrong?
Thank you,
David. |
|
|
|
|
In an ESEM model with only the measurement part, the df is the same as in EFA.
The targets only affect the rotation, not the ESEM fit. |
|
|
|
|
Thank you Bengt. I'm still wondering how the df are calculated in ESEM.
Regards,
David. |
|
|
|
|
In an ESEM model with only the measurement part, the df is the same as in EFA. Our Topic 1 handout gives the formula on slide 104 as df = b-a:
a = p*m + m*(m+1)/2 + p – m*m
b = p*(p + 1)/2
where p = number of observed variables
m = number of factors |
|
|
|
|
Dear Bengt,
Thank you for this. It makes sense to me now.
David. |
|
|
| Back to top |
|
|
