Chi-square in V5 for 2PL model PreviousNext
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 Roger E. Millsap posted on Monday, February 11, 2008 - 9:12 am
I am now using Version 5, and I fit a model to 10 binary indicators using MLR, which the manual says will give 2PL output. In that output, two chi-square statistics appear as

Chi-Square Test of Model Fit for the Binary and Ordered Categorical
(Ordinal) Outcomes**


What null hypothesis do these chi-squares
test? One is a Pearson, and one is an LR chi-square. They do not appear to be
testing the fit of the 2PL model per se.

Sorry if this question has been asked before, but I couldn't find it on the list.

Roger Millsap
 Linda K. Muthen posted on Monday, February 11, 2008 - 9:21 am
These are tests of the observed versus expected multiway frequency tables for the categorical indicators.
 Roger E. Millsap posted on Monday, February 11, 2008 - 10:24 am
Hi Linda,
If I could follow up on this: when you
say "expected", under what model are the expected frequencies generated?

They do not seem to be generated under
a single-factor model for binary
indicators. Or are they?

The chi-squares are much different from
the fit of the single-factor model under
WLSMV, for example. Thanks,

Roger
 Bengt O. Muthen posted on Monday, February 11, 2008 - 10:52 am
In both cases expected is the H0 model such as a 1-factor model. The frequency table chi-squares (Pearson and LR) have higher df since they test against the full distribution whereas WLS tests against only bivariates - so df's are different (WLSMV df, as opposed to WLS and WLSM, is of course totally different since it is estimated). The p values should be comparable. As usual, freq table Pearson/LR can suffer from low cell count problems.
 Roger E. Millsap posted on Tuesday, February 12, 2008 - 9:04 am
Hi Bengt,
The p-values in my example are not even close to comparable.
I'm fitting a single-factor to 10 binary items with N=937. No missing data.
For the MLR run, the two chi-squares are:
Pearson chi=1039.751
df=1002
p=.1983
LR chi=589.487
df=1002
p=1.0000

For the single-factor run with WLSMV:

WLSMV chi=54.598
df=31
p=.0055

Clearly, the IRT-MLR results lead to a vastly different conclusion about fit.
What is the explanation? If you want
the data and programs, I can send them.

Roger
 Bengt O. Muthen posted on Tuesday, February 12, 2008 - 9:24 am
Typically when the Pearson and LR chi-squares for a frequency table are as different as in this MLR run, neither can be trusted most likely due to many freq table cells with (close to) zero cell counts. Adding Tech10 to the MLR run should expose this. The WLSMV chi-square can work better in such situations since it collapses to bivariate tables. If that doesn't explain it, please send materials to support.
 Roger E. Millsap posted on Tuesday, February 12, 2008 - 9:43 am
Hi Bengt,
I think that your conjecture is correct. When I cut the number of items down to 5-7, the p-values are much more comparable. Once I get up to 10 items, the divergence is there and there are many cells with zero's.
I guess that the lesson is to use the single-factor WLSMV to test the model if the number of items exceeds what can be supported with the N. Thanks,

Roger
 krisitne amlund hagen posted on Wednesday, February 13, 2008 - 5:18 am
Dear Drs. Muthen,
I am testing a model, in which the independent variable (IV) is categorical with 4 groups (4 groups of children with different social profiles). The grouping variable is gender.
I use version 5 now.
1. How do I specify that the independent (w1_socgr) variable is categorical?
When I run this model command:
W3_anti BY W3_devfA W3_police W3_drugA;
W2_soci BY W2_coopA W2_assrA W2_contA;
W2_soci ON w1_socgr;
W3_anti ON w1_socgr W2_soci;
it looks like w1_socgr is being correlated with w2_soci (a latent variable of social competence at w2)? If I choose w1_socgr as a grouping variable, how do I test for gender differences and how do I retain the three wave nature of the design?

2. One of the indicators for the outcome variable (antisocial at wave 3) is actually a count variable treated as a continuous variable (# of times the adolescent in contact with police). It has about 50% missing + many zeroís. Could I recode the variable into a dichotomous variable? What else could I do?

3. The results indicate that w2_soci is a mediator for girls, but not for boys, in that the path from w2_soci to w3_anti is significant only for girls. Do I have to run a test constraining that path to be equal across groups to test for significance? How do I specify that in the command?

Thank you for a wonderful forum,
 Linda K. Muthen posted on Wednesday, February 13, 2008 - 8:46 am
You need to represent a four-category nominal variable by three dummy variables and include these dummy variables as covariates in your model.

You can treat the variable as a count or dichotomize it. You could try both.

It depends on what question you are trying to answer. The z-test given for each group tests whether the coefficient is different from zero. A difference test tests whether the coefficients are different from each other.
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