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 Daisy Chang posted on Tuesday, November 08, 2005 - 10:21 am

I am trying to do a second order factor model, but I constantly receiving an error message saying that THE STANDARD ERRORS OF THE MODEL PARAMETER ESTIMATES COULD NOT BE COMPUTED. THE MODEL MAY NOT BE IDENTIFIED. CHECK YOUR MODEL.

I have tried different ways including changing the starting value and fixing the loading of bpress to be 1, but none of it worked. I would really appreicate your input. Thanks. Here is my program.
title: project
data: file is u:\bio.dat;
BP_S BP_D WAISTHIP TOTHDL age control edu;
usevariables are lncor lnnor lntot sqep sbp_s sbp_d lnhb sqdh sqep ;
missing are all (9999);
define: lncor= log(cortis1) if cortis1 > 0 then lncor = log(cortis1);
lntot= log(tothdl) if tothdl > 0 then lntot= log(tothdl);
lnnor= log(norepi1) if norepi1 > 0 then lnnor = log(norepi1);
lnhb = log(hb_a1c) if hb_a1c > 0 then lnhb = log(hb_a1c);
sqep = sqrt(epinep1) if epinep1 >=0 then sqep = sqrt(epinep1);
sqdh = sqrt(dhea_s) if dhea_s >= 0 then sqdh = sqrt(dhea_s);
sbp_s = (bp_s)/50 if bp_s > 0 then sbp_s = (bp_s)/50;
sbp_d = (bp_d)/50 if bp_d > 0 then sbp_d = (bp_d)/50;

analysis: type = missing h1;

model:bpress by sbp_s sbp_d;
hpa by sqdh lncor;
metabo by lnhb lntot waisthip;
nerv by sqep lnnor;
age by hpa* bpress metabo nerv;
 Linda K. Muthen posted on Saturday, November 12, 2005 - 5:53 pm
You would need to send your input, data, output, and license number to to get an answer to this question.
 Maren Winkler posted on Monday, August 10, 2009 - 12:22 am

we are establishing a 2nd order model. We've done CFA for seven tests separately and now try to model them in one SEM.
The structure is as follows:

f1 BY x1 x2 x3;
f2 BY y1 y2 y3;
f7 BY z1 z2 z3;

g BY f1 f2 ...f7;

The model fits our data well. However, loadings of all items from the last test on its latent factor are negative (all z are negative). Hence, f7 loads negatively on the second order factor g when all other loadings (f1 to f6)are positive.
What could be the reason for this result? Does it have implications for further analyses?
Thank you very much for your help.
 Linda K. Muthen posted on Monday, August 10, 2009 - 6:42 am
If all are negative, they can all be changed to positive. I would try positive starting values.
 Sarah-Geneviève Trépanier posted on Tuesday, January 24, 2012 - 4:56 pm

I am new to Mplus (and SEM in general) and I was wondering if it is possible to integrate second-order factors in a SEM:


variables v1 v2 v3 v4 v5 v6 v7 v8 v9 v10 v11 v12 v13 v14 v15 v16 v17 v18 v19 v20;

f1 by v1 v2 v3;
f2 by v4 v5 v6;
f3 by v7 v8 v9;
f4 by v10 v11;
f5 by v12 v13 v14;
f6 by v15 v16 17;
f7 by v18 v19 v20;

f8 by f1* f2 f3;
f8 @1;

f9 by f4* f5 f6;
f9 @1;

F9 on F8;
F8 on F6 F7;

Thank you very much for your help!

 Linda K. Muthen posted on Tuesday, January 24, 2012 - 5:53 pm
Yes. See Example 5.6 in the user's guide.
 Sarah-Geneviève Trépanier posted on Tuesday, January 24, 2012 - 6:47 pm
Hi again,

I am sorry to bother you again. I just want to make sure my question was clear. As you suggested, I looked at chapiter 5 in the user's guide. It has a section on CFA (including second order factors: 5.6) and a section regarding SEM (structural paths). I do not see anything that combines both second-factor AND structural models like the syntax I gave in example in the previous post.

Sorry and thank you again.

 Linda K. Muthen posted on Wednesday, January 25, 2012 - 6:39 am
You can combine them. Run the analysis. If you have a problem, send it along with your license number to
 Patric Kirchner posted on Tuesday, March 31, 2015 - 12:26 pm
Dear Mrs. Muthen,

I'm new to Mplus an SEM, and hope that you can help reagarding the following question:

I did a second-order factor analysis with the following structure

f1 by x1-x3;
f2 by x4-x6;
f4 ..x10-x12;
f5 by f1-f4;

...and I have a qestion regarding the standardized model results: the estimate value for one of the first order factor (f2) on the second order factor (f5) is bigger than one. since I expected all loadings/estimates on the second order factor to be <1, I don't know how to interpret the result. what does this mean?

Thank you very much!

 Bengt O. Muthen posted on Tuesday, March 31, 2015 - 6:37 pm
See our FAQ on this: Stand'd greater than 1.
 Patric Kirchner posted on Tuesday, March 31, 2015 - 11:43 pm
Thanks a lot for your quick reply! That helped!
 Paulo Alexandre Ferreira Martins posted on Sunday, October 09, 2016 - 6:30 am
Dear Linda and Bengt Muthén,

Practicing a 2nd Order Factor i ran two models that thought would bring the same results:

1st Model: “Reference indicator method” fixing each fist indicator to 1. Tfist indicator to 1. To build a 2nd Order Factor i constrained each 1st Order Factors variance to 1 and their means to zero plus 2nd Order Factor to 1.

2nd Model: “Fixed factor method” by freeing first indicators. As MPlus would fix their related Factors to 1, i thought i just needed to write the remaining syntax like the previous model, i.e., 1st factor means to zero plus 2nd Order Factor to 1…

However, my second model popped up a warning message that this model may not be identified…

fist Model:
F1 BY var1 var4 var7 var10 var14;
F2 BY var 2 var5 var8 var11 var12 var15 var16;
F3 BY var3 var6 var9 var13 var17;

F1@1 F2@1 F3@1;

NeedSup by F1 F2 F3;

second MODEL:
F1 BY var1* var4 var7 var10 var14;
F2 BY var 2* var5 var8 var11 var12 var15 var16;
F3 BY var3* var6 var9 var13 var17;

NeedSup by F1 F2 F3;

Thank you!
 Bengt O. Muthen posted on Sunday, October 09, 2016 - 5:23 pm
The first model is needlessly restricted on both the first and second level because Mplus by default sets the factor metric by fixing the first loading and you set the metric also by fixing factor (residual) variances at one.

The second model is not identified because you don't set the factor metric on the first level. And it is needlessly restricted on the second level because the metric is again set twice.
 Paulo Alexandre Ferreira Martins posted on Monday, October 10, 2016 - 9:43 am

(fist model) - The problem is:

If i do not fix all 3 Factors variances of to 1, immediately pops up a warning message on "psi matrix", reporting a negative residual in each factor variance who was not constrained to 1....

Thank you for you attention!
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