WARNING: Could not compute standard errors!

[car...@ucn.cl]

Dear all,

I am starting with lavaan, and I really appreciate if someone could help me with this warning.

I am developing a study about the effect of formalization of task on team exploration capacity.  I am calculating "team exploration" using two sources of information: team member’s perception and outside observer’s ratings. On one side, a team member’s perception is measured using five questions assessed with a Likert scale (1-5). On the other side, outside observer rating the teams' formalization comparing a given team with other teams answering two questions which are scaled from 1 for “far below to average” to 7 “far above average”. Thus, I created a second order variable averaging these two standardized variables. However, when I run the CFA, the standard error can’t be calculated. Is it any form to overcome this warning?

WARNING: Could not compute standard errors! The information matrix could not be inverted. This may be a symptom that the model is not identified. 

Here is the R code I have used to run the model, complete with resulting summary output:

#SECOND ORDER “EXPLORATION”

EXR <- ' 

EXRJ =~ EXJ1 + EXJ2 

EXRG =~ EXRG1 + EXRG2 + EXRG3 +EXRG4 + EXRG5 

GLOBAL=~ EXRJ + EXRG'

#CFA

EXR.fit = cfa(EXR, data=DATAPROM, std.lv = TRUE)

summary(EXR.fit, fit.measures=TRUE, standardized = TRUE, rsquare = TRUE)

fitMeasures(EXR.fit)

lavaan 0.6-2 ended normally after 70 iterations

  Optimization method                           NLMINB

  Number of free parameters                         16

  Number of observations                               62

  Estimator                                         ML

  Model Fit Test Statistic                       5.691

  Degrees of freedom                                12

  P-value (Chi-square)                           0.931

Model test baseline model:

  Minimum Function Test Statistic              563.305

  Degrees of freedom                                    21

  P-value                                                          0.000

User model versus baseline model:

  Comparative Fit Index (CFI)                    1.000

  Tucker-Lewis Index (TLI)                          1.020

Loglikelihood and Information Criteria:

  Loglikelihood user model (H0)                     -144.421

  Loglikelihood unrestricted model (H1)       -141.576

  Number of free parameters                         16

  Akaike (AIC)                                                     320.843

  Bayesian (BIC)                                                 354.877

  Sample-size adjusted Bayesian (BIC)           304.536

Root Mean Square Error of Approximation:

  RMSEA                                          0.000

  90 Percent Confidence Interval          0.000  0.037

  P-value RMSEA <= 0.05                          0.961

Standardized Root Mean Square Residual:

  SRMR                                           0.012

Parameter Estimates:

  Information                                                Expected

  Information saturated (h1) model          Structured

  Standard Errors                                          Standard

Latent Variables:

                   Estimate  Std.Err  z-value  P(>|z|)   Std.lv  Std.all

  EXRJ =~                                                               

    EXJ1              0.449       NA                      0.821    0.943

    EXJ2              0.455       NA                      0.831    0.982

  EXRG =~                                                               

    EXRG1             0.271       NA                      0.495    0.872

    EXRG2             0.297       NA                      0.542    0.883

    EXRG3             0.297       NA                      0.542    0.948

    EXRG4             0.287       NA                      0.524    0.948

    EXRG5             0.290       NA                      0.529    0.958

  GLOBAL =~                                                             

    EXRJ              1.529       NA                      0.837    0.837

    EXRG              1.527       NA                      0.837    0.837

Variances:

                   Estimate  Std.Err  z-value  P(>|z|)   Std.lv  Std.all

   .EXJ1              0.084       NA                      0.084    0.110

   .EXJ2              0.026       NA                      0.026    0.036

   .EXRG1          0.077       NA                      0.077    0.239

   .EXRG2          0.083       NA                      0.083    0.220

   .EXRG3          0.033       NA                      0.033    0.102

   .EXRG4          0.031       NA                      0.031    0.101

   .EXRG5          0.025       NA                      0.025    0.082

    EXRJ              1.000                                  0.299    0.299

    EXRG             1.000                                  0.300    0.300

    GLOBAL        1.000                                  1.000    1.000

R-Square:

                   Estimate

    EXJ1              0.890

    EXJ2              0.964

    EXRG1          0.761

    EXRG2          0.780

    EXRG3          0.898

    EXRG4          0.899

    EXRG5          0.918

    EXRJ             0.701

    EXRG           0.700

Thank you in advance

Best regards,

Carolina Rojas

Phd (C ) in Engineering Sciences

Pontificia Universidad Católica de Chile 

[Edward Rigdon]

The model is not identified. For a stand-alone second order factor model to be identified, there must be 3 variables dependent on the second order factor, and you only have 2. You could establish identification by constraining the loadings on the second order factor to be equal to each other, and fixing the variance of the second order factor to be 1.

EXR <- ' 

EXRJ =~ EXJ1 + EXJ2 

EXRG =~ EXRG1 + EXRG2 + EXRG3 +EXRG4 + EXRG5 

GLOBAL=~ a*EXRJ + a*EXRG

GLOBAL~~1*GLOBAL'

However, the resulting second order factor will not be an average of the 2 first order factors, but rather the commonality between the 2 first order factors.

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[Jeremy Miles]

My first thought is that that is a pretty small sample, and a pretty large model. I'd be surprised if that's not the route of your problem.

Jeremy

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[Carolina Rojas Cordova]

Thank you very much Edwards for your answer!. 

I tried your proposal and it worked.

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[Carolina Rojas Cordova]

Dear Edward, 

One more question, the fact that the second order factor is not an average means that I can't use the average of the 2 first order factor as a measure of exploration in a regression model?

Thank you very much in advance 

[Edward Rigdon]

     You could attempt this if:

You perform all opf this analysis within the same model. If you try estimating your factor model first, and then using those results in a later, separste analysis, you face the problem of factor indeterminacy, which may be quite severe in your case.

But if you have another variable Z and you want the average of EXRG and EXRJ to predict Z, you could do that with constraints. You could, for example, set the variances of EXRG and EXRJ to 1 (while allowing loadings to be freely estimated), and then require the regression coefficients for EXRG and EXRJ to be equal:

' 

EXRJ =~ NA*EXJ1 + EXJ2 

EXRG =~ NA*EXRG1 + EXRG2 + EXRG3 +EXRG4 + EXRG5 

EXRG ~~ 1*EXRG

EXRJ ~~ 1*EXRJ

Z ~ b*EXRG + b*EXRJ

'

So then the two factors would have the same variances and the same regression weights on Z, hence you would have an average. The only slight complication is that the regression weight b would encompass both the averaging function and the regression of Z on EXRG and EXRJ.

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[Carolina Rojas Cordova]

 I am going to calculate Z as you indicate 

I want to use Z as dependent variable of a regression of three independents team level variables. 

 thank you very much!!!!! for your help

[Edward Rigdon]

In my example, Z was some observed variable. You cannot create an average in this way. Even if you could, you could not also regress that variable on other predictors within the same analysis, because you are already modeling it as the sum of two other variables.

If you want the average of the two factors to be your dependent variable, then I suggest:

fix the variances of the two factors as I suggested;

regress each of the two factors on the 3 predictors, but

constrain the regression coefficients to equality across the two factors:

' 

EXRJ =~ NA*EXJ1 + EXJ2 

EXRG =~ NA*EXRG1 + EXRG2 + EXRG3 +EXRG4 + EXRG5 

EXRG ~~ 1*EXRG

EXRJ ~~ 1*EXRJ

EXRG  ~ a*X + b*Y + c*Z

EXRJ  ~ a*X + b*Y + c*Z

'

where X, Y and Z are your 3 predictors.

[Carolina Rojas Cordova]

Thank you very much for your help!!!!