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# A quick introduction to interpretation of Exploratory Factor Analysis: Mplus Example

May 22, 2013 | 1 Comment

Last week I wrote a bit about how to get an exploratory factor analysis using Mplus. The question now, is what does that output MEAN ?

First, you just get some information on the programming statements or defaults that produced your output:

INPUT READING TERMINATED NORMALLY

Exploratory Factor Analysis ;

SUMMARY OF ANALYSIS

Number of groups 1

Number of observations 730

Number of dependent variables 6

Number of independent variables 0

Number of continuous latent variables 0

Observed dependent variables

Continuous

Q1F1 Q2F1 Q3F1 Q1F2 Q2F2 Q3F2

Estimator ML

Rotation GEOMIN

Row standardization CORRELATION

Type of rotation OBLIQUE

This tells us we our analyzing all of the data as one group, and not, for example, separate analyses for males and females. We have 730 records, six variables, all of which are continuous and listed above. The maximum likelihood method (ML) of estimation is used and the default rotation, GEOMIN, which is an oblique method, that is it allows the factors to be correlated.

Here we have a list of our eigenvalues

RESULTS FOR EXPLORATORY FACTOR ANALYSIS

EIGENVALUES FOR SAMPLE CORRELATION MATRIX

1 ……… 2 ……… 3 4 5

________ ________ _____ ________ ________

1.866 1.262 0.866 0.750 0.716

EIGENVALUES FOR SAMPLE CORRELATION MATRIX

6

________

0.539

In this case, you could go ahead with the eigenvalue greater than one rule, but let’s take a look at a couple of other statistics. First, we have the results from the one factor solution. Here we have the chi-square testing the goodness of fit of the model

Chi-Square Test of Model Fit

Value 96.228

Degrees of Freedom 9

P-Value 0.0000

We want this test to be non-significant because our null hypothesis is there is no difference between the observed data and our hypothesized one-factor model. This null is soundly rejected.

Let’s take a look at the Chi-square for our two-factor solution

Chi-Square Test of Model Fit

Value 3.016

Degrees of Freedom 4

P-Value 0.5552

You can clearly see that the chi-square is much smaller and non-significant.

Let’s take a look at two other tests. The Root Mean Square Error of Approximation (RMSEA) for the one-factor solution is .115, as shown below. We would like to see an RMSEA less than .05 which is clearly not the case here.

RMSEA (Root Mean Square Error Of Approximation)

Estimate 0.115

90 Percent C.I. 0.095 0.137

Probability RMSEA <= .05 0.000

For the two factor solution, our RMSEA rounds to zero, as shown below

RMSEA (Root Mean Square Error Of Approximation)

Estimate 0.000

90 Percent C.I. 0.000 0.049

Probability RMSEA <= .05 0.954

Clearly, we are liking the two-factor solution here, yes? The eigenvalue > 1 rule (which should not be TOO emphasized) points there, as does the model fit chi-square and the RMSEA.

In their course on factor analysis, Muthen & Muthen give this very nice example of a table comparing different factor solutions using the data

They also like the scree plot, which I do, too. I also agree with them that one should never blindly follow some rule but rather have some theory or expectation about how the factors should fall out. I also agree with them in looking at multiple indicators, for example, scree plot, chi-square, RMSEA and eigen-values.

# Comments

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i would like to continue on the same topic of data output interpretation by EFA<CFA and SEM model and if you have such example on interpretation i would love to have it urgently