Exploratory Factor Analysis: Why the propotion (share of variance explained) by the only factor retained using ipf is equal to 1.0000?

[Junior Boumsong]

I am conducting an exploratory factor analysis (EFA) from a polychoric matrix using the iterated principal factor method (ipf). After specifying the minimum of eigenvalues to be retained, only one factor was retained (please see code and output below),

Why the proportion (the % of total variance explained) of the only factor retained is 1.0000?

Is it normal that the proportion of this factor retained is equal to 1.0000?

Is this possible to calculate the proportion explained by retained factors using ipf? If so, what is the Stata command?

[Erik Ruzek]

Junior,

This is due to the fact that you are using the mineigen(1) syntax in factormat, which specifies that the minimum value of eigenvalues to be retained is one. If you were to drop this option, you would see that, very likely, your first factor has a proportion < 1.0. See below, which is a slightly altered version of what is provided in -help factormat-:

matrix D =  ( 1.000, 0.943,  0.901  \ ///
              0.943, 1.000,  0.805  \ ///
              0.901, 0.805,  1.000  )

** With mineigen(1)
factormat D, n(331) names(satis1 satis2 satis3)  mineigen(1) ipf
(obs=331)

Factor analysis/correlation                      Number of obs    =        331
    Method: iterated principal factors           Retained factors =          1
    Rotation: (unrotated)                        Number of params =          3

    Beware: solution is a Heywood case
            (i.e., invalid or boundary values of uniqueness)

    --------------------------------------------------------------------------
         Factor  |   Eigenvalue   Difference        Proportion   Cumulative
    -------------+------------------------------------------------------------
        Factor1  |      2.66709      2.66698            1.0000       1.0000
        Factor2  |      0.00011      0.00023            0.0000       1.0000
        Factor3  |     -0.00012            .           -0.0000       1.0000
    --------------------------------------------------------------------------
    LR test: independent vs. saturated:  chi2(3)  = 1307.20 Prob>chi2 = 0.0000

** Dropping mineigen(1) from the code
factormat D, n(331) names(satis1 satis2 satis3)  ipf // mineigen(1)
(obs=331)

Factor analysis/correlation                      Number of obs    =        331
    Method: iterated principal factors           Retained factors =          2
    Rotation: (unrotated)                        Number of params =          3

    Beware: solution is a Heywood case
            (i.e., invalid or boundary values of uniqueness)

    --------------------------------------------------------------------------
         Factor  |   Eigenvalue   Difference        Proportion   Cumulative
    -------------+------------------------------------------------------------
        Factor1  |      2.69585      2.61905            0.9723       0.9723
        Factor2  |      0.07680      0.07689            0.0277       1.0000
        Factor3  |     -0.00009            .           -0.0000       1.0000
    --------------------------------------------------------------------------
    LR test: independent vs. saturated:  chi2(3)  = 1307.20 Prob>chi2 = 0.0000

[Junior Boumsong]

Eric, thank you for your reply!

I have tried to follow your advice (Please see the results below).

Step 1: Here we can retain one factor because it has an eigenvalue over 1 (Kaiser criterion),

Step 2: However, after rotating the factor loadings to get a clearer pattern and a final solution, it seems that two factors should be retained if we consider that it is common to arbitrarily consider factor of 0.32 or 0.40 as salient (please see the rotation results below). In this case, does it always make sense to retain one factor according to the results in the step 1.


Step 1 results

Step 2 results (rotation)

[Bruce Weaver]

Hello Junior Boumsong. Why are you using the Eigenvalues < 1 criterion? I ask because nowadays, many authors argue that it is not a good method. See this article, for example:

• https://quantpsy.org/pubs/preacher_maccallum_2003.pdf

Cheers,
Bruce

[Junior Boumsong]

Hello Bruce Weaver thank you for your reply

The results of the scree test also indicate that only one factor should be retained