Updated 10-14-2002
Confirmatory Factor Analysis - CFA
Same form of the model as Exploratory Factor analysis
but in CFA, the
is not
necessarily constrained to be the identity matrix, AND constraints
can be placed on the elements of
, AND
is not necessarily
constrained to be diagonal (i.e. the errors may be correlated).
As with EFA, we are interested in fitting the model
covariance matrix
to the sample covariance
matrix.
where
represents all the free parameters in
,
,
and
.
Maximum likelihood assuming normality for the
data will be used to estimate
.
The degrees of freedom for the confirmatory factor analysis model equals p*(p+1)/2 - number of free parameters in
.
Comparison of EFA to CFA - EXAMPLE
Exploratory Factor analysis as a precursor to Confirmatory factor
analysis, E-mail from Bengt Muthen on SEMNET here.
More on the subject can be found at HTTP://bama.ua.edu/cgi-bin/wa?A1=ind0010&L=semnet
Requirements for CFA Model Identification.
(pp. 188-192 of Maruyama talks about identification for SEM in general)
Necessary conditions:
- Parameters
observations, in other words d.f
0.
- Every factor must have a scale - achieved in one of two ways
- Fix the variance of the factor to be 1 or
- Fix one of the factor loadings to be 1
If the CFA model has all measurement errors uncorrelated,
(i.e.
is diagonal) and satisfies both of the necessary
conditions above, then the model is identified if
- each factor has
two indicators, when the model has
2
factors
- the factor has
three indicators when the model has only
one factor.
If the CFA model has some measurement errors correlated,
defining rules for identifiability is difficult. One common
recommendation is to the let the software crank it out. AMOS can
give a message that the model is not identified (based on empirical
considerations). This use is not fool proof though, it is possible
the software can miss. (see Bollen 1989 and Rigdon 1995 for more
rules).
Model fitting
(pp.196-201 in Maruyama)
Let
represent the true covariance structure
of the data. We want to find a model
that describes
.
- Given a particular model for the covariance structure, the
chi-square test of model fit is testing
Ho:
(your model)
HA:
The saturated model (just-identified model)
- Since the degrees of freedom for the saturated model are 0,
this means it fits the data perfectly.
- So we are comparing
to a model
that fits the data perfectly. THUS IF IT IS NOT SIGNIFICANTLY
DIFFERENT THAN THE MODEL THAT FITS PERFECTLY, IT MEANS IT IS PRETTY GOOD.
- Look for p-value larger than .05
AMOS LAB - using the visual perception/ verbal ability data fit a two factor confirmatory factor analysis model.
degrees of freedom should be 21 - 13 = 8. There are 13 parameters being estimated.....
NESTED Models
(pp. 235-238 in Maruyama)
- Models are nested whenever one model is the same as the other model
except that it has some parameters constrained. In other words, if
you unconstrained some parameters in one model, you would get the
other.
- Examples
- The CFA model with factor loadings constrained to be equal is
nested within the CFA model where factor loadings are allowed to be
estimated freely (from Lab)
- A two factor CFA model with simple structure is nested within
a two factor EFA model (constrains multiple loadings
to equal zero)
- A one factor model is nested within a two factor simple
structure CFA (constrains correlation between factors to equal 1)
- When models are nested, we can perform a Chi-square difference
test to compare them.
Ho:
More
restrictive model (the one with some parameters constrained)
HA:
Less restrictive model
Chi-square test to compare these models =
Chi-square test value for More restrictive model -
Chi-square test value for Less restrictive model
Compare this test stat to a chi-squared distribution with d.f. =
(d.f. for More restrictive model) - (d.f. for Less restrictive
model)
- Example from Lab
- Ho:
Tau-equivalent model
HA:
Congeneric model
- Chi-square test value for Congeneric model = 238.5 with 9
degrees of freedom
- Chi-square test value for Tau-equivalent model = 263.8 with 14
degrees of freedom
- Chi-square difference test for hypothesis above = 263.8-238.5
= 25.3 with 14-9 = 5 d.f.
- The p-value associated with this test is <.0001, so we reject
the Null hypothesis in favor of the alternative.
- Thus the Tau-equivalent model does not fit as well as the
Congeneric model.
- NOTE: This does not mean the Congeneric model fits well, for
this we should consider how it compares to the saturated model or
look at other Goodness of Fit tests.
Modification Indices
- Each path that has not been included in the model has an associated
"modification index".
- A modification index tells you how much the
overall chi-square test statistic would change if that particular
path was added to the model.
- Adding one new path will decrease
the degrees of freedom by one. So, since the model without the path
can be considered to be "nested" within the model with the path we
can consider the difference in the chi-square test as a test for
whether the path is needed or not.
- Since 3.84 is the .05 cut-off value
for a chi-square with 1 degree of freedom, modification indices
greater than 3.84 suggest that a path should be added.
- AMOS has a default threshold of 4 for outputting modification
indices. This can be changed in the output tab window under
analysis properties.
- The ParChange found in the AMOS modification index output tells you
what the unstandardized estimate of the path would be if it was
added to the model.
- NOTE: modification indices should only be considered one at a time. If
you add more than one path the change in overall chi-square will not
necessarily equal the sum of the two modification indices.
Sample Size and Normality
Chi-squared test of model fit and standard errors for factor loadings are based on asymptotic theory
- How big should n be before we trust these numbers????
- Rule of thumb: n > 15*(number of free parameters)
- But, we should consider at least 2 other factors
- to what extent multivariate normality is violated
- validity/reliability of measures
- How non-normal is non-normal??
- Rules of thumb - examine univariate skew and kurtosis.
- West, Finch and Curran (1995) "Structural equation models with non-normal variables: Problems and remedies". In R.H. Hoyle (Ed.) Structural equation modeling (pp. 56-75) Thousand Oaks, CA: Sage.
- This paper gives rule of thumb of skew > 2 and kurtosis > 7 to indicate problematic non-normal data.
- Departures from normality cause the chi-square test to be larger than it should be and standard errors to be smaller than they should be.
- If n is very large (i.e. rule of thumb n>1000), do not need to worry about non-normality. Amemiya and Anderson (1988,1990)
For another good reference for non-normality, check out http://www.utexas.edu/cc/fa
qs/stat/general/gen33.html
Handling non-normal data in SEM and http://www.utexas.edu/cc/faqs/
stat/amos/amos7.html Handling non-normal data using AMOS
Here is a .pdf file containing notes I made outlining an article by Hu, Bentler (1995) ``Evaluating Model
Fit'' in SEM: Concepts, Issues, and Applications, edited by
Hoyle, RH, Thousand Oaks, Calif., Sage Pub, p. 76-99.
A newer reference by Hu and Bentler is: Hu, L-T., & Bentler,
P.M. (1999). "Cutoff criteria for fit indexes in covariance structure
analysis: Conventional criteria versus new alternatives", Structural
Equation Modeling, 6, 1-55.
Also check out: West S., Finch, J., and Curran, P.J. "Structural Equation Models with nonnormal variables: Problems and remedies" in SEM: Concepts, Issues, and Applications. Edited by Hoyle RH. Thousand Oaks, CA, Sage Pub. pp 56-75. Note to MW: make notes from this article and put here
Other Fit Indices and Testing procedures
(We will touch on these topics later)
- Cross-validation
- Bootstrapping
- Bollen-Stine Chi-Square test - Bootstrapped version of the
Chi-square test
- Bootstrapping standard errors
- Parametric bootstrap
- A wide array of goodness of fit indices
- discussed in Chpt 10 of Maruyama
- We will focus only on RMSEA
- Many of the fit indices make every model look good (Read
e-mail from SEMNET)