Are any pairs of these four structural models nested in another? I think the second-order single factor model is nested in the second-order two factor model (second order single factor model is when covariance between the two second order factors in second-order two factor model is 1) but I can't determine whether other pairs are nested or not.
Solved – Are any pairs of these four structural models nested in another
confirmatory-factornested-modelsstructural-equation-modeling
Related Solutions
Main Point
With only two observed variables per factor, the latent variable can not generally be estimated. Having correlations between factors presumably adds enough constraints to the model to allow estimation of the latent factors, but without those correlations, the latent factors are not estimatable.
What should you do?
- Add more observed variables per factor. If you have three or more observed variables, your latent factors will generally be estimatable.
- Constrain factor loadings to be equal. As a secondary option, you could constrain you factor loadings for the two items to be equal. Make sure they are on the same scale so that this makes sense.
I think the option of having at least three observed variables per factor is preferable.
Note also that setting the covariance between each factor to 1 will not create uncorrelated factors. If the factors have unit variance (i.e., like z-scores), then your factors will perfectly correlated (r=1). Thus, it would be equivalent to having a single factor. If the factors do not have unit variance, then it be constraining some other specific correlation structure that is related to the variance of each of the factors.
Comments below relate to the original post prior to the update.
Degrees of freedom
- If you constrain a parameter to be a constant, then you should get an additional degree of freedom in an SEM. Thus, when you specify $k$ latent factors to be uncorrelated, then that implies you are specifying $k(k-1)/2$ potential parameters to be zero and you should have that many additional degrees of freedom.
- If you are constraining $k$ latent factors to have a common covariance, then you should get $k(k-1)/2 - 1$ additional degrees of freedom, because you are still estimating one parameter.
So why might your degrees of freedom not be changing? Here are a few thoughts:
- Have you included the second order factor in your model testing the first-order model? If you have, remove the second-order factor.
- Have you specified constraints correctly in Amos?
Constraining factor correlations to be equal
- Note that in social sciences, it is often the correlation between factors that is conceptualised to be approximately the same from a theoretical perspective. Thus, you may want to ensure that your factors are all on the same scale before constraining covariances. For example, you could constrain the covariance of the factors to be one, rather than constraining one of the loadings, as is the default in Amos. By placing the factors on the same scale, constraining covariances to be equal is the same as constraining correlations to be equal.
Negative variances
- In terms of errors, it may be that the factors are so highly correlated that forcing the factors to be uncorrelated is causing estimation problems.
- You may want to just try removing the double headed arrows in Amos rather than specifying constraints to see whether that makes a difference to see whether that makes a difference
- You may have issues with number of items per factor.
There's also some discussion here
While your data and model don't fit the formal definition of nested, they are in essence nested because you can redefine the variables in such a way that they do fit the definition. For example change to 2 variables:
rab is 0 for rats and 1 for rabbits
breed is 0 for breed 1 (either species) and 1 for breed 2 (either species)
then fit the model:
$ y = a + b1(rab) + b2(breed) + b3(rab \times breed) + e$
Now you can see that your first model is clearly nested in this model and this model will give the exact same predictions and therefore the exact same sums of squares, likelihood, etc. so any test using these values will give the same results for your model and my model, so your 2 models can be treated as nested.
Best Answer
a) is the saturated model. All models are nested within the saturated model. That's the basis of the chi-square test. You're testing against the saturated model, which has 0 df and 0 chi-square.
c) and d) are equivalent models. d) isn't identified without a constraint on the loadings, so there is only one parameter to estimate.
All that we are left with is the relationship between b), and c)/d). These are nested, but not (quite) in the usual way. Constraing the correlation to 1 in model b) gives you nested models, but you can't test these nested models using the chi-square test, because the parameter is on the boundary of the permitted space. In a sense it's a one-tailed test - when you relax the correlation in model c) we know that the parameter will change, but it must get smaller, not larger. The nested model chi-square test assumes that the parameter can change in any direction. It's surprisingly tricky. Here's a paper: http://psycnet.apa.org/?&fa=main.doiLanding&doi=10.1037/1082-989X.13.2.150