Stata FAQ
How can I do EFA within a CFA framework? (Stata 12)

EFA within a CFA framework, as the name implies, combines aspects of both EFA and CFA. It produces a factor solution that is close to an EFA solution while providing features found in CFA, such as, standard errors, statistical tests and modification indices.

The trick to doing EFA within a CFA framework is to use the same number of constraints in CFA as are used in EFA. If m is the number of factors in an EFA model then the number of constraints is m². By using m² constraints in CFA we get a model with the same fit as an EFA model.

We will demonstrate EFA within a CFA framework using an artificial dataset with 500 observations on six variables. The data wee constructed to give a two factor solution.

We begin by loading the data and computing a correlation matrix.

use http://www.ats.ucla.edu/stat/data/efa_cfa, clear

correlate

(obs=500)

             |       y1       y2       y3       y4       y5       y6
-------------+------------------------------------------------------
          y1 |   1.0000
          y2 |   0.5212   1.0000
          y3 |   0.4719   0.5339   1.0000
          y4 |  -0.0008  -0.0345  -0.0101   1.0000
          y5 |  -0.0256  -0.0365  -0.0400   0.4297   1.0000
          y6 |   0.0176  -0.0159   0.0390   0.3688   0.4268   1.0000

Now we are ready to run a maximum likelihood EFA model followed by an oblique rotation using the quartimen normalize option. We will also use estat common to get the correlation between the oblique facts.

factor y1 y2 y3 y4 y5 y6, ml

(obs=500)
number of factors adjusted to 3
Iteration 0:   log likelihood =  -31.62904
Iteration 1:   log likelihood = -3.1615516
Iteration 2:   log likelihood = -.34164269
Iteration 3:   log likelihood = -.34066861  (backed up)
[output omitted]
Iteration 25:  log likelihood = -.05290372

number of factors adjusted to 2
Iteration 0:   log likelihood = -35.679836
Iteration 1:   log likelihood =  -1.103664
Iteration 2:   log likelihood = -1.1024515

Factor analysis/correlation                        Number of obs    =      500
    Method: maximum likelihood                     Retained factors =        2
    Rotation: (unrotated)                          Number of params =       11
                                                   Schwarz's BIC    =  70.5656
    Log likelihood = -1.102451                     (Akaike's) AIC   =  24.2049

    --------------------------------------------------------------------------
         Factor  |   Eigenvalue   Difference        Proportion   Cumulative
    -------------+------------------------------------------------------------
        Factor1  |      1.54078      0.30931            0.5558       0.5558
        Factor2  |      1.23147            .            0.4442       1.0000
    --------------------------------------------------------------------------
    LR test: independent vs. saturated:  chi2(15) =  593.54 Prob>chi2 = 0.0000
    LR test:   2 factors vs. saturated:  chi2(4)  =    2.19 Prob>chi2 = 0.7015

Factor loadings (pattern matrix) and unique variances

    -------------------------------------------------
        Variable |  Factor1   Factor2 |   Uniqueness 
    -------------+--------------------+--------------
              y1 |   0.6763    0.0631 |      0.5387  
              y2 |   0.7657    0.0354 |      0.4124  
              y3 |   0.6944    0.0612 |      0.5141  
              y4 |  -0.0675    0.6060 |      0.6282  
              y5 |  -0.0975    0.6969 |      0.5048  
              y6 |  -0.0290    0.6080 |      0.6295  
    -------------------------------------------------

rotate, oblique quartimin normalize

Factor analysis/correlation                        Number of obs    =      500
    Method: maximum likelihood                     Retained factors =        2
    Rotation: oblique quartimin (Kaiser on)        Number of params =       11
                                                   Schwarz's BIC    =  70.5656
    Log likelihood = -1.102451                     (Akaike's) AIC   =  24.2049

    --------------------------------------------------------------------------
         Factor  |     Variance   Proportion    Rotated factors are correlated
    -------------+------------------------------------------------------------
        Factor1  |      1.53686       0.5544
        Factor2  |      1.23729       0.4463
    --------------------------------------------------------------------------
    LR test: independent vs. saturated:  chi2(15) =  593.54 Prob>chi2 = 0.0000
    LR test:   2 factors vs. saturated:  chi2(4)  =    2.19 Prob>chi2 = 0.7015

Rotated factor loadings (pattern matrix) and unique variances

    -------------------------------------------------
        Variable |  Factor1   Factor2 |   Uniqueness 
    -------------+--------------------+--------------
              y1 |   0.6794    0.0117 |      0.5387  
              y2 |   0.7657   -0.0228 |      0.4124  
              y3 |   0.6972    0.0084 |      0.5141  
              y4 |  -0.0073    0.6095 |      0.6282  
              y5 |  -0.0281    0.7025 |      0.5048  
              y6 |   0.0313    0.6086 |      0.6295  
    -------------------------------------------------

Factor rotation matrix

    --------------------------------
                 | Factor1  Factor2 
    -------------+------------------
         Factor1 |  0.9971  -0.0990 
         Factor2 |  0.0758   0.9951 
    --------------------------------

estat common

Correlation matrix of the quartimin rotated common factors

    ----------------------------------
         Factors |  Factor1   Factor2 
    -------------+--------------------
         Factor1 |        1           
         Factor2 |  -.02325         1 
    ----------------------------------

The rotated factor solution gives us a rather clean two factor model. We note that the model fit versus a saturated model has a chi-square of 2.19 with four degrees of freedom. This is a very good fit for an EFA and reflects the synthetic nature of the data. We also note the the two factors have a small correlation of -.02325.

Since the EFA solution has two factors we will need to make a total of four constraints in the CFA model. The easiest ones are to set the factor variances to one (2 factors, 2 constraints). The other constraints revolve around identifying anchor variables for each factor and constraining the cross loadings for the anchor variables to be zero (2 cross loadings, 2 more constraints for a total of 4).

An anchor item has a high loading on one factor and low loadings on the remaining factors. From the rotated solution above we see that y2 has the highest loading on Factor1 and has low loading on Factor2. We will make y2 the anchor item for Factor1 and will constrain the loading to equal zero in Factor2. We will use y5 as the anchor item for Factor2.

Here is the sem command for the EFA within a CF framework.

sem (F1 -> y1 y2   y3 y4 y5@0 y6)    ///
    (F2 -> y1 y2@0 y3 y4 y5   y6) ,  ///
    variance(F1@1 F2@1) standardized 

Endogenous variables

Measurement:  y1 y2 y3 y4 y5 y6

Exogenous variables

Latent:       F1 F2

Fitting target model:

Iteration 0:   log likelihood = -5064.3487  (not concave)
Iteration 1:   log likelihood = -5005.6323  (not concave)
Iteration 2:   log likelihood = -4997.4943  (not concave)
Iteration 3:   log likelihood = -4982.0445  (not concave)
Iteration 4:   log likelihood = -4978.5317  (not concave)
[output omitted]
Iteration 15:  log likelihood = -4975.0886  (not concave)
Iteration 16:  log likelihood = -4975.0731  
Iteration 17:  log likelihood =    2163798  (not concave)
[output omitted]
Iteration 40:  log likelihood =    2163798  (not concave)
Iteration 41:  log likelihood =    2163798  (not concave)
Iteration 42:  log likelihood =    2163798  (not concave)
--Break--
r(1);

The sem would have run forever if we had let it. However, since the log likelihood did not change from the 17th iteration on, we broke out of the program. There are two possible reasons for the endless iterations: 1) Either the model is not identified or 2) the starting values did not allow sem to converge on a solution. I'm betting that it is the second reason, bad starting values.

To investigate this we will run the sem command again but limit the iterations to 16 and see what the results look like.

sem (F1 -> y1 y2   y3 y4 y5@0 y6)                 ///
    (F2 -> y1 y2@0 y3 y4 y5   y6) ,               ///
    variance(F1@1 F2@1) standardized iterate(16) 

Endogenous variables

Measurement:  y1 y2 y3 y4 y5 y6

Exogenous variables

Latent:       F1 F2

Fitting target model:

Iteration 0:   log likelihood = -5064.3487  (not concave)
Iteration 1:   log likelihood = -5005.6323  (not concave)
Iteration 2:   log likelihood = -4997.4943  (not concave)
[output omitted]
Iteration 15:  log likelihood = -4975.0886  (not concave)
Iteration 16:  log likelihood = -4975.0731  
convergence not achieved

Structural equation model                       Number of obs      =       500
Estimation method  = ml
Log likelihood     = -4975.0731

 ( 1)  [var(F1)]_cons = 1
 ( 2)  [var(F2)]_cons = 1
------------------------------------------------------------------------------
             |                 OIM
Standardized |      Coef.   Std. Err.      z    P>|z|     [95% Conf. Interval]
-------------+----------------------------------------------------------------
Measurement  |
  y1 <-      |
          F1 |   .6708773   .0557835    12.03   0.000     .5615436    .7802109
          F2 |   .0120947    .042074     0.29   0.774    -.0703689    .0945583
       _cons |  -.0157292   .0447242    -0.35   0.725     -.103387    .0719286
  -----------+----------------------------------------------------------------
  y2 <-      |
          F1 |   .7702398   .0338197    22.77   0.000     .7039545    .8365251
       _cons |   .0181829   .0447252     0.41   0.684    -.0694768    .1058426
  -----------+----------------------------------------------------------------
  y3 <-      |
          F1 |   .6545966   .0596644    10.97   0.000     .5376566    .7715366
          F2 |   .0421295    .045045     0.94   0.350    -.0461571    .1304162
       _cons |   .0246474   .0447283     0.55   0.582    -.0630185    .1123133
  -----------+----------------------------------------------------------------
  y4 <-      |
          F1 |  -.3645916   .2553101    -1.43   0.153    -.8649902    .1358071
          F2 |   .3503213   .2420369     1.45   0.148    -.1240623     .824705
       _cons |  -.0156629   .0447241    -0.35   0.726    -.1033205    .0719948
  -----------+----------------------------------------------------------------
  y5 <-      |
          F2 |   .1567238   .0419252     3.74   0.000     .0745519    .2388958
       _cons |  -.0121302    .044723    -0.27   0.786    -.0997857    .0755253
  -----------+----------------------------------------------------------------
  y6 <-      |
          F1 |   13.86547   4.399784     3.15   0.002     5.242055    22.48889
          F2 |  -13.37937    4.19254    -3.19   0.001     -21.5966   -5.162141
       _cons |   .0384571   .0447411     0.86   0.390    -.0492339    .1261481
-------------+----------------------------------------------------------------
Variance     |
        e.y1 |   .5330075   .0478677                      .4469816    .6355898
        e.y2 |   .4067307   .0520985                      .3164286     .522803
        e.y3 |   .5127309   .0484352                      .4260697    .6170188
        e.y4 |   1.008326   .0090961                      .9906548    1.026313
        e.y5 |   .9754376   .0131414                      .9500181    1.001537
        e.y6 |   13.15271   9.979148                      2.973009    58.18808
          F1 |          1  (constrained)
          F2 |          1  (constrained)
-------------+----------------------------------------------------------------
Covariance   |
  F1         |
          F2 |   1.033392   .0123008    84.01   0.000     1.009282    1.057501
------------------------------------------------------------------------------
LR test of model vs. saturated: chi2(4)   =    140.82, Prob > chi2 = 0.0000
Warning: convergence not achieved

In the results above we see that all of the loadings are between 0 and 1 except for variable y6. Also the residual variance for y6 is much higher than any of the other residual variances. I think we can fix the problem by setting the initial starting value for the loading on Factor1 to 0 and on Factor2 to 0.5. We do this by including init in the sem command as seen below,

sem (F1 -> y1 y2   y3 y4 y5@0 (y6, init(0.0)))    ///
    (F2 -> y1 y2@0 y3 y4 y5   (y6, init(0.5))) ,  ///
    variance(F1@1 F2@1) standardized 

Endogenous variables

Measurement:  y1 y2 y3 y4 y5 y6

Exogenous variables

Latent:       F1 F2

Fitting target model:

Iteration 0:   log likelihood = -5467.1006  (not concave)
Iteration 1:   log likelihood = -5087.7064  (not concave)
[output omitted]
Iteration 8:   log likelihood = -4905.7634  
Iteration 9:   log likelihood = -4905.7634  

Structural equation model                       Number of obs      =       500
Estimation method  = ml
Log likelihood     = -4905.7634

 ( 1)  [var(F1)]_cons = 1
 ( 2)  [var(F2)]_cons = 1
------------------------------------------------------------------------------
             |                 OIM
Standardized |      Coef.   Std. Err.      z    P>|z|     [95% Conf. Interval]
-------------+----------------------------------------------------------------
Measurement  |
  y1 <-      |
          F1 |   .6814311   .0353949    19.25   0.000     .6120583     .750804
          F2 |   .0319918   .0506257     0.63   0.527    -.0672328    .1312163
       _cons |  -.0158254   .0447242    -0.35   0.723    -.1034832    .0718323
  -----------+----------------------------------------------------------------
  y2 <-      |
          F1 |   .7665428   .0340075    22.54   0.000     .6998894    .8331962
       _cons |   .0183253   .0447251     0.41   0.682    -.0693343     .105985
  -----------+----------------------------------------------------------------
  y3 <-      |
          F1 |   .6991976   .0352692    19.82   0.000     .6300711     .768324
          F2 |   .0292474    .050601     0.58   0.563    -.0699286    .1284235
       _cons |   .0248039   .0447282     0.55   0.579    -.0628618    .1124697
  -----------+----------------------------------------------------------------
  y4 <-      |
          F1 |   .0171268   .0517733     0.33   0.741     -.084347    .1186006
          F2 |   .6110693   .0447048    13.67   0.000     .5234494    .6986891
       _cons |  -.0156637   .0447241    -0.35   0.726    -.1033213    .0719939
  -----------+----------------------------------------------------------------
  y5 <-      |
          F2 |   .7036822   .0455108    15.46   0.000     .6144826    .7928817
       _cons |  -.0121345    .044723    -0.27   0.786      -.09979     .075521
  -----------+----------------------------------------------------------------
  y6 <-      |
          F1 |   .0557532   .0519712     1.07   0.283    -.0461084    .1576148
          F2 |   .6112825     .04504    13.57   0.000     .5230058    .6995593
       _cons |   .0368424   .0447365     0.82   0.410    -.0508395    .1245244
-------------+----------------------------------------------------------------
Variance     |
        e.y1 |   .5386683   .0475129                      .4531498    .6403259
        e.y2 |   .4124122   .0521364                      .3219024    .5283707
        e.y3 |   .5140572   .0485346                      .4272139    .6185539
        e.y4 |   .6282406   .0540083                      .5308237    .7435355
        e.y5 |   .5048314   .0640503                      .3936865    .6473546
        e.y6 |   .6295414   .0541293                      .5319077    .7450962
          F1 |          1  (constrained)
          F2 |          1  (constrained)
-------------+----------------------------------------------------------------
Covariance   |
  F1         |
          F2 |  -.0926641   .0801889    -1.16   0.248    -.2498313    .0645032
------------------------------------------------------------------------------
LR test of model vs. saturated: chi2(4)   =      2.20, Prob > chi2 = 0.6981

This time sem converged on a solution. All of the factor loadings and residual variances look reasonable. The model fit versus a saturated model has a chi-square of 2.20 with 4 df. This value is almost identical to the chi-square of 2.19 with 4 df from the maximum likelihood EFA model we ran at the beginning.

Before comparing the EFA within a CFA framework with the EFA solution, let's look at some of the various indicators of goodness of fit.

estat gof, stat(all)

----------------------------------------------------------------------------
Fit statistic        |      Value   Description
---------------------+------------------------------------------------------
Likelihood ratio     |
          chi2_ms(4) |      2.205   model vs. saturated
            p > chi2 |      0.698
         chi2_bs(15) |    596.921   baseline vs. saturated
            p > chi2 |      0.000
---------------------+------------------------------------------------------
Population error     |
               RMSEA |      0.000   Root mean squared error of approximation
 90% CI, lower bound |      0.000
         upper bound |      0.051
              pclose |      0.947   Probability RMSEA <= 0.05
---------------------+------------------------------------------------------
Information criteria |
                 AIC |   9857.527   Akaike's information criterion
                 BIC |   9954.463   Bayesian information criterion
---------------------+------------------------------------------------------
Baseline comparison  |
                 CFI |      1.000   Comparative fit index
                 TLI |      1.012   Tucker-Lewis index
---------------------+------------------------------------------------------
Size of residuals    |
                SRMR |      0.007   Standardized root mean squared residual
                  CD |      0.925   Coefficient of determination
----------------------------------------------------------------------------

It is not surprising that the gof values all look good considering the artificial nature of the data.

Now, let's compare the EFA within CFA solution to the EFA solution.

Summary of results

                             efa rotated maximum
efa within cfa loadings      likelihood loading
     F1         F2           Factor1   Factor2
y1  .6814311   .0319918       0.6794    0.0117
y2  .7665428    0             0.7657   -0.0228
y3  .6991976   .0292474       0.6972    0.0084
y4  .0171268   .6110693      -0.0073    0.6095
y5   0         .7036822      -0.0281    0.7025
y6  .0557532   .6112825       0.0313    0.6086 
    
factor: LR test: 2 factors vs. saturated: chi2(4) = 2.19 Prob>chi2 = 0.7015
   sem: LR test   of model vs. saturated: chi2(4) = 2.20 Prob>chi2 = 0.6981

The EFA within a CFA framework results appear very close to the rotated maximum likelihood EFA results while having the added benefits of standard errors, statistical tests and access to modification indices, if needed.