Second order factors
Message/Author
 Anonymous posted on Thursday, June 10, 2004 - 6:23 am
HI Linda,

I have a second order factor within a larger sem model.The second order factor is measured by 2 first order factors and one measured variable. which of the following methods should I use for this part of my model. Assuming I have 7 measured variables (x1-x7) for this part of the model

method 1:
model:
f1 by x1-x3;
f2 by x4-x6;
f3 by x7@1;
f4 by f1-f3;

Method2:
model:
f1 by x1-x3;
f2 by x4-x6;
f4 by f1 f2 x7;

Thanks
 Linda K. Muthen posted on Thursday, June 10, 2004 - 8:07 am
I believe they would be identical if you add x7@0; to the MODEL command of Method 1. I would use Method 2 because there is less room for error in setting up the model.
 ldumenci posted on Tuesday, June 29, 2004 - 6:14 am
Hi Linda,

Do you see any identification problems below?

f1 by x1-x5;
f2 by x6-x10;
f3 by x11-x15
f by f1 f2 f3;
f f1 f2 f3 on y1 y2;

Thanks
 Linda K. Muthen posted on Tuesday, June 29, 2004 - 6:44 am
If you are receiving an error about identification, send the output including TECH1 to support@statmodel.com and I will be happy to take a look at it.
 anonymous posted on Wednesday, October 26, 2005 - 12:10 pm
I am trying to test a model where there are 4 first-order factors subsumed by a single higher-order factor, and there is also a fifth first-order factor that covaries with the higher-order factor.

Is this model meaningfully different than a model that has all 5 first-order factors subsumed by the higher-order factor?
 Linda K. Muthen posted on Wednesday, October 26, 2005 - 5:37 pm
To answer you , I used Example 5.6 in the Mplus User's Guide and ran your two models. The model fit and the number of parameters are the same. The difference is that in one you estimate a factor loading and a residual variance and in the other you estimate a variance and a covariace.
 Anonymous posted on Thursday, October 27, 2005 - 10:46 am
Thank you. So is this difference (factor loading and residual variance versus variance and covariance) make the models meaningfully distinct, given that the # of parameters and the fit statistics are the same? That is, is there any reason to test both these models?
 Linda K. Muthen posted on Thursday, October 27, 2005 - 11:35 am
These models are not statistically distinguihable.
 anonymous posted on Thursday, January 19, 2006 - 9:19 am
I am trying to ceate a CFA model as a basis for nested models I want to compare. However, with my input, my output does not show any numbers for CFI, TLI and RSMEA. How get I get the numbers for them?

INPUT INSTRUCTIONS

TITLE: Fieke

DATA:
FILE IS "D:\dataspss.dat";
FORMAT is 12f1;

VARIABLE:
NAMES ARE country inno1 inno2 inno3 inno4 quality inno5 custsat
custloy emplsat emplret emplloc;
USEVARIABLES ARE country inno1 inno2 inno3 inno4 quality inno5
custsat custloy emplsat emplret emplloc;
GROUPING IS country (1=UnitedKingdom 2=Austria 3=Ireland
5=NewZealand 6=Australia);

ANALYSIS:
TYPE IS MEANSTRUCTURE;
ESTIMATOR IS ML;
ITERATIONS = 100000;
CONVERGENCE = 0.00001;

MODEL:
f1 BY inno1 inno2 inno3 inno4 inno5;
f2 BY quality;
f3 BY custsat custloy;
f4 BY emplsat emplret emplloc;
f1 ON f2 f3 f4;
f1 BY inno1*;
f2 BY quality*;
f3 BY custsat*;
f4 BY emplsat*;

MODEL UnitedKingdom:
f1@0.0;
f2@0.0;
f3@0.0;
f4@0.0;

Fieke

SUMMARY OF ANALYSIS

Number of groups 5
Number of observations
Group UNITEDKINGDOM 487
Group AUSTRIA 657
Group IRELAND 249
Group NEWZEALAND 472
Group AUSTRALIA 250

Number of y-variables 11
Number of x-variables 0
Number of continuous latent variables 4

Observed variables in the analysis
INNO1 INNO2 INNO3 INNO4 QUALITY INNO5
CUSTSAT CUSTLOY EMPLSAT EMPLRET EMPLLOC

Grouping variable COUNTRY

Continuous latent variables in the analysis
F1 F2 F3 F4

Estimator ML
Information matrix EXPECTED
Maximum number of iterations 100000
Convergence criterion 0.100D-04
Maximum number of steepest descent iterations 20

Input data file(s)
D:\dataspss.dat

Input data format
(12F1)

THE MODEL ESTIMATION TERMINATED NORMALLY

THE STANDARD ERRORS OF THE MODEL PARAMETER ESTIMATES COULD NOT BE
COMPUTED. THE MODEL MAY NOT BE IDENTIFIED. CHECK YOUR MODEL.
PROBLEM INVOLVING PARAMETER 59.

MODEL RESULTS

Estimates

Group UNITEDKINGDOM

F1 BY
INNO1 0.908
INNO2 0.940
INNO3 0.908
INNO4 0.695
INNO5 0.459

F2 BY
QUALITY 0.338

F3 BY
CUSTSAT 0.737
CUSTLOY 0.816

F4 BY
EMPLSAT 0.725
EMPLRET 0.811
EMPLLOC 0.439

F1 ON
F2 -0.016
F3 1.845
F4 1.883

F3 WITH
F2 0.228

F4 WITH
F2 0.078
F3 0.118

Means
F2 0.000
F3 0.000
F4 0.000

Intercepts
INNO1 3.468
INNO2 3.389
INNO3 3.420
INNO4 3.310
QUALITY 3.700
INNO5 3.448
CUSTSAT 3.715
CUSTLOY 3.639
EMPLSAT 3.313
EMPLRET 3.346
EMPLLOC 3.420
F1 0.000

Variances
F2 0.000
F3 0.000
F4 0.000

Residual Variances
INNO1 0.265
INNO2 0.211
INNO3 0.164
INNO4 0.445
QUALITY 0.505
INNO5 0.526
CUSTSAT 0.603
CUSTLOY 0.643
EMPLSAT 0.724
EMPLRET 0.964
EMPLLOC 0.694
F1 0.000

Group AUSTRIA

F1 BY
INNO1 0.908
INNO2 0.940
INNO3 0.908
INNO4 0.695
INNO5 0.459

F2 BY
QUALITY 0.338

F3 BY
CUSTSAT 0.737
CUSTLOY 0.816

F4 BY
EMPLSAT 0.725
EMPLRET 0.811
EMPLLOC 0.439

F1 ON
F2 1.127
F3 -0.268
F4 -0.283

F3 WITH
F2 0.349

F4 WITH
F2 0.466
F3 0.368

Means
F2 -0.207
F3 0.013
F4 0.192

Intercepts
INNO1 3.468
INNO2 3.389
INNO3 3.420
INNO4 3.310
QUALITY 3.700
INNO5 3.448
CUSTSAT 3.715
CUSTLOY 3.639
EMPLSAT 3.313
EMPLRET 3.346
EMPLLOC 3.420
F1 0.475

Variances
F2 0.671
F3 0.481
F4 0.747

Residual Variances
INNO1 0.304
INNO2 0.230
INNO3 0.174
INNO4 0.415
QUALITY 0.455
INNO5 0.459
CUSTSAT 0.235
CUSTLOY 0.196
EMPLSAT 0.252
EMPLRET 0.227
EMPLLOC 0.401
F1 0.299

Group IRELAND

F1 BY
INNO1 0.908
INNO2 0.940
INNO3 0.908
INNO4 0.695
INNO5 0.459

F2 BY
QUALITY 0.338

F3 BY
CUSTSAT 0.737
CUSTLOY 0.816

F4 BY
EMPLSAT 0.725
EMPLRET 0.811
EMPLLOC 0.439

F1 ON
F2 0.676
F3 -0.291
F4 0.505

F3 WITH
F2 0.360

F4 WITH
F2 0.119
F3 0.397

Means
F2 0.138
F3 0.015
F4 0.173

Intercepts
INNO1 3.468
INNO2 3.389
INNO3 3.420
INNO4 3.310
QUALITY 3.700
INNO5 3.448
CUSTSAT 3.715
CUSTLOY 3.639
EMPLSAT 3.313
EMPLRET 3.346
EMPLLOC 3.420
F1 -0.011

Variances
F2 0.588
F3 0.531
F4 0.649

Residual Variances
INNO1 0.387
INNO2 0.259
INNO3 0.174
INNO4 0.395
QUALITY 0.363
INNO5 0.468
CUSTSAT 0.213
CUSTLOY 0.175
EMPLSAT 0.397
EMPLRET 0.502
EMPLLOC 0.560
F1 0.429

Group NEWZEALAND

F1 BY
INNO1 0.908
INNO2 0.940
INNO3 0.908
INNO4 0.695
INNO5 0.459

F2 BY
QUALITY 0.338

F3 BY
CUSTSAT 0.737
CUSTLOY 0.816

F4 BY
EMPLSAT 0.725
EMPLRET 0.811
EMPLLOC 0.439

F1 ON
F2 1.012
F3 -0.640
F4 0.105

F3 WITH
F2 0.430

F4 WITH
F2 0.243
F3 0.289

Means
F2 0.160
F3 0.149
F4 0.403

Intercepts
INNO1 3.468
INNO2 3.389
INNO3 3.420
INNO4 3.310
QUALITY 3.700
INNO5 3.448
CUSTSAT 3.715
CUSTLOY 3.639
EMPLSAT 3.313
EMPLRET 3.346
EMPLLOC 3.420
F1 0.360

Variances
F2 0.707
F3 0.486
F4 0.718

Residual Variances
INNO1 0.219
INNO2 0.237
INNO3 0.167
INNO4 0.488
QUALITY 0.422
INNO5 0.498
CUSTSAT 0.233
CUSTLOY 0.213
EMPLSAT 0.302
EMPLRET 0.289
EMPLLOC 0.560
F1 0.230

Group AUSTRALIA

F1 BY
INNO1 0.908
INNO2 0.940
INNO3 0.908
INNO4 0.695
INNO5 0.459

F2 BY
QUALITY 0.338

F3 BY
CUSTSAT 0.737
CUSTLOY 0.816

F4 BY
EMPLSAT 0.725
EMPLRET 0.811
EMPLLOC 0.439

F1 ON
F2 0.607
F3 -0.015
F4 0.029

F3 WITH
F2 0.405

F4 WITH
F2 0.226
F3 0.388

Means
F2 0.177
F3 -0.101
F4 0.300

Intercepts
INNO1 3.468
INNO2 3.389
INNO3 3.420
INNO4 3.310
QUALITY 3.700
INNO5 3.448
CUSTSAT 3.715
CUSTLOY 3.639
EMPLSAT 3.313
EMPLRET 3.346
EMPLLOC 3.420
F1 0.134

Variances
F2 0.847
F3 0.623
F4 1.061

Residual Variances
INNO1 0.199
INNO2 0.248
INNO3 0.121
INNO4 0.321
QUALITY 0.469
INNO5 0.568
CUSTSAT 0.383
CUSTLOY 0.332
EMPLSAT 0.408
EMPLRET 0.262
EMPLLOC 0.791
F1 0.45
 Linda K. Muthen posted on Thursday, January 19, 2006 - 9:54 am
Please do not post outputs on Mplus Discussion. It takes too much room. Please send your input, data, output, and license number to support@statmodel.com and we will look into your problem.
 Raheem Paxton posted on Thursday, March 16, 2006 - 9:06 am
I'm testing the measurement of 2 second order model by five factors each. If the hypothesized model do not fit the data, is it appropriate to add correlations between the first order factors, or are the 2nd order factors suppose to represent those correlation.
 Linda K. Muthen posted on Friday, March 17, 2006 - 6:59 am
Yes, this would be similar to adding residual covarariances to the first order factors.
 john posted on Wednesday, August 09, 2006 - 9:28 am
hi

sorry, i have two questions.

1. do you know how to figure out the discriminant validity Fornell Larker test? i am stuck with this as i do not know what the avarage variance extracted is...any help will be a massive reief.

2. if my total variance extracted is 30% how big a problem is this for me?

thankyou
john
 john posted on Wednesday, August 09, 2006 - 9:56 am
hi

or what is the best way to calculate discriminent validity using SPSS EFA?

john
 Bengt O. Muthen posted on Wednesday, August 09, 2006 - 2:08 pm
I am have not heard of the Fornell Larker test. Maximizing the percent variance extracted is not the goal of factor analysis. Since you seem to be using SPSS, they might be able to guide you further.
 Nyankomo Wambura Marwa posted on Thursday, August 10, 2006 - 7:55 am
Dear Prof Muthen

I am fitting categorical LCA models for four categorical manifest variables with two levels in each.Actually they are diagnostic tests.Also i have one covariate which is the age of the patient.Previous my model did not fit well.Through diagnostics for local dependence by using CONDEP program i detected that two test are locally dependent.I tried to fit the model with adjustment of local dependence in LEM software the model was ok.

I real want to fit the same model in Mplus but now accounting for local dependence.I tried but i don't seem to get it right.I tried example 7.16 for Qu model which is some what close to my situation it does not seem to work well either in my situation.

I guess may be i don get the coding right.Please will you kindly advice me accordingly.

Here are my mplus syntax for the model:{the local depenedence is bewteen A and D variables).

DATA:file is D:\sum\zlon.txt ;
variable:names are id age A D B H cou;
usevar age A D B H;
categorical are A D B H;
classes=cl(2);
analysis:
type = mixture ;
starts = 0;
model:
%overall%
cl#1 on age;
[A$1*10 D$1*10 B$1*10 H$1*10];
%cl#1%
[A$1*-10 D$1*-10 B$1*-10 H$1*-10];

output:tech10 tech11;
 Linda K. Muthen posted on Thursday, August 10, 2006 - 3:45 pm
The conditional independence in Example 7.16 is modeled by the f BY statement. I don't see this in your input. Please look at the example again.
 Nyankomo Wambura Marwa posted on Thursday, August 10, 2006 - 11:34 pm
Dear linda.
Actually i am afraid if was able to communicate well my problem to you.Here is the original code i used for adjusting local dependence.But it did not seem to work well.That is why i gave you the naive code i.e. one without local dependence to see how we adjust it.

I will appreciate your help once again.

DATA:file is D:\sum\zlon.txt ;
variable:names are id age u1-u4 cou;
USEVARIABLES ARE age u1-u4;
CATEGORICAL = u1-u4;
CLASSES = c(2);
ANALYSIS: TYPE = MIXTURE;
ALGORITHM = INTEGRATION;
MODEL:
%OVERALL%
c#1 on age;
f by u1-u2@0;
f@1; [f@0];
%c#1%
[u1$1-u4$1*-1];
f by u1@1 u2;
OUTPUT: TECH1 TECH8 tech10;
 Linda K. Muthen posted on Friday, August 11, 2006 - 6:06 am
I don't know what you mean by it did not seem to work well. Perhaps you should send your input, data, output, license number, and explanation of what you mean to support@statmodel.com.
 Lars Penke posted on Wednesday, September 19, 2007 - 2:19 pm
Hello,

From prior analyses, I have a well-fitting MPlus solution to a hierarchical CFA with 3 latent factors and a higher-order factor. Now I fixed all loadings to the solution I found and then wanted to estimate the correlations between all 4 factors and an external variable:

VARIABLES ARE rv1r rv2r rv5r re_1 re_2 re_4 ra_1 ra_2 ra_5 extvar;
ANALYSIS: TYPE = general;
MODEL:
rv BY rv1r@1, rv2r@0.975, rv5r@0.898;
re BY re_1@1, re_2@2.063, re_4@2.581;
ra BY ra_1@1, ra_2@0.967, ra_5@0.819;
rho BY rv@1 re@0.653 ra@4.315;
rv WITH extvar;
re WITH extvar;
ra WITH extvar;
rho WITH extvar;

However, I receive the following error message:

THE STANDARD ERRORS OF THE MODEL PARAMETER ESTIMATES COULD NOT BE
COMPUTED. THE MODEL MAY NOT BE IDENTIFIED. CHECK YOUR MODEL.
PROBLEM INVOLVING PARAMETER 27.

TECH1 output: Parameter 27 is "rho WITH extvar".

I don't see why this model is not identified. And how could I fix it?
 Linda K. Muthen posted on Thursday, September 20, 2007 - 11:12 am
You cannot identify all of the direct and indirect effects in the model as you are trying to do.
 Michael W Myers posted on Thursday, August 07, 2008 - 5:38 pm
Hi,
I'm a new user of mplus, so I apologize if this is an obvious question. I tried to create a second-level factor based on three latent factors and one observed variable. This was my model syntax:
alc by C81AL71 P7AlcD P7AlcA;
pot by C81CL71 P7CanD P7CanA ;
tob by C81TL71 P7TobD P7TobW ;
anti by alc pot tob C5F7ASR;
While the model fit the data well, I was surprised to see that the output provided two set of estimates for my second-level factor (anti) -- one for just my observed variable (C5F7ASR) and another for my three latent factors. Additionally, since I had only one observed variable that was measuring my second-level factor, it looks like mplus set this variable's factor loading at 1. However, it also did this to one of my latent factors as well.
Is there something wrong with my syntax, or can I not define a second-level factor that is measured by a combination of latent factors and observed variables? If it is possible, can you explain how I interpret the output for this second-level factor?
 Linda K. Muthen posted on Thursday, August 07, 2008 - 5:56 pm
 Carla Bann posted on Friday, August 15, 2008 - 7:36 am
I am fitting the following second order factor model with two groups of respondents.

F1 by var1-var4;
F2 by var5-var8;
F3 by var9-var12;
F4 by F1-F3;

The model runs fine when I analyze the two groups separately; however, I get an error about identification when I try to include both groups and use the grouping statement. The identification error does not occur when I remove the second order factor (i.e., just have the 3 first-order factors). I'd really appreciate any suggestions. Thank you!
 Linda K. Muthen posted on Friday, August 15, 2008 - 8:19 am
You need to fix the intercepts of the first-order factors to zero in all groups.
 Carla Bann posted on Monday, August 18, 2008 - 9:38 am
This may be a dumb question. However, I am running two models, one with 3 factors and the other with the 3 factors loading on one second order factor. The fit indices for the two models are exactly the same. Is that to be expected?
 Linda K. Muthen posted on Monday, August 18, 2008 - 10:14 am
The second-order factor is just-identified. This is why it does not change model fit.
 Andrea Hildebrandt posted on Tuesday, October 28, 2008 - 2:00 pm
how can I set residual variances of first-order factors to 1?

model:
f1 BY y1 y2 y3;
f2 BY y4 y5 y6;
f3 BY f1 f2;

with f1@1 I would fix the total variance of this first order factor to 1, wouldn't I?

Thank you!
 Linda K. Muthen posted on Tuesday, October 28, 2008 - 3:02 pm
In a situation where f1 is a dependent variable as in your example where it is a factor indicator, f1@1 fixed the residual variance at one.
 Liu-Qin Yang posted on Tuesday, December 30, 2008 - 1:01 pm
Hi, all,
I have a quick question about second-order factor in a measurement model.
Specifically, I am trying to run a measurement model with a second-order factor built upon 2 first-order underlying variables. See below.

MODEL:
F1 by burdn9 burdn2
burdn6 burdn10;
F2 by burdn3 burdn4 burdn5
burdn7;
F3 by F1 F2;
F1 @ 0;

F1 with F2;

But Mplus did not like it.

Any suggestions?
Thanks a lot and happy new year!
 Linda K. Muthen posted on Tuesday, December 30, 2008 - 1:55 pm
A second-order factor with two first-order factor indicators is not identified.
 JPower posted on Wednesday, March 18, 2009 - 11:13 am
Hello,
Can you provide some guidance as to how to choose between a model with correlated factors and a model with those factors loading on a second order factor? As the second order factor is just identified, it does not change model fit. What else should be considered? R-square for the first order latent factors? Anything else? Thanks.
 Bengt O. Muthen posted on Friday, March 20, 2009 - 10:59 am
Statistically, you can choose between those models only if you have more than 3 first-order factors.
 Anja Tausch posted on Friday, May 15, 2009 - 2:56 am
Hi,
I want to test a model with 8 first-order factors and 3 second-order factors. One of the second-order factors is indicated by only two first-order factors. Are there any 'tricks' (like setting parameters to 1 or 0) to get this model identified?
 Linda K. Muthen posted on Friday, May 15, 2009 - 8:00 am
I think this model is identified because the factor with two factor indicators will borrow information from the other factors. If you run this and have a problem, please send the full output and your license number to support@statmodel.com.
 Alexandre Morin posted on Friday, May 15, 2009 - 10:33 am
The overall model will be identified. What you have there is a factor that will be locally non-identified. This should not pose problem in the estimation process. However, if you do want to have it locally identified, you may either want to fix both loadings to 1 or to fix them to equality and to fix the higher order variance to 1.
Little et al. discuss this in the context of first order models.
Little, T.D., Lindenberger, U., & Nesselroade, J.R. (1999). On selecting indicators for multivariate measurement and modeling with latent variables: When “good” indicators are bad and “bad” indicators are good. Psychological Methods, 4, 192-211.
 Lisa M. Yarnell posted on Monday, August 22, 2011 - 7:24 pm
Hi Drs. Muthen,

A professor told me that when you have latent variables loading onto a higher-order latent variable, it is not possible to estimate means for the higher-order factor and ALL of the lower-order factors. He said that you'd either need to NOT estimate the mean for the higher-order factor, or NOT estimate the mean for one of the lower-order factors--elsewise your model would not be identified.

If this is true, how do I choose to NOT estimate either the mean of the higher order factor or the mean of one of the lower order factors?

Do I set the mean to zero using "@0" for one of these? Or, do I just comment out the portion of my code where I had previously been asking for a mean?

I apologize for my oblivion on how to do this; I tried several things in my Mplus code just now, but none of them worked, and so I could use some advice.

Lisa
 Linda K. Muthen posted on Tuesday, August 23, 2011 - 11:47 am
In a cross-sectional study, factor means cannot be identified at all. It is only with multiple groups or multiple time points that factor means can be estimated in all but one group and at all but one time point.
 Lisa M. Yarnell posted on Tuesday, August 23, 2011 - 2:20 pm
Thanks, Dr. Muthen. Sometimes these theoretical points about SEM are challenging to translate into Mplus code. Thanks again!
 Linda K. Muthen posted on Wednesday, August 24, 2011 - 7:16 am
This is the default in Mplus so no code is necessary.
 Jasmina Memetovic posted on Friday, September 23, 2011 - 4:59 pm
Dear. Dr Muthén,

I am testing a second-order CFA of a measure with 28 items, using 5 first-order and 2 second-order latent variables. All items use a 5 point Likert scale – however some items show skewness of > +1.5 and the skewness varies from item to item (some are positive, some negative, and quite a few are skewed). Here is the second-order LV portion of my code:

F6 BY F1* F2;
F7 BY F3* F4 F5;

F6@1;
F7@1;

I first used an ML algorithm and got satisfactory fit (CFI =0.907, SRMR=0.060). However, I tried re-running the model using the WLSMV (and specifying all items as categorical), and received the following warning:

WARNING: THE LATENT VARIABLE COVARIANCE MATRIX (PSI) IS NOT POSITIVE
DEFINITE….

I checked to see whether any latent variables had negative residual variances, and one of the first-order residual variances was -1.337, which seems too large to fix the variance to zero. I am not sure what else I could try…

 Linda K. Muthen posted on Saturday, September 24, 2011 - 9:03 am
I would suggest doing an EFA on the first-order factor indicators and using the correlation matrix of the five factors as data in an EFA of the five factors. It may be that the second-order factor structure is not correct.
 Lisa M. Yarnell posted on Friday, January 06, 2012 - 3:39 pm
Hi Linda and Bengt,

Normally, a factor model with three indicators is just-identified; a factor model with two indicators is underidentified; and a factor model with 4+ indicators is overidentified.

However, in a higher-order model with two higher-order correlated factors, must each higher-order factor have at least 3 first-order factors for the model to be identified? If one of the two higher-order factors has only two first-order factors, will this present estimation problems?

To provide specifics, I have Aggressiveness and Rule-Breaking first-order factors loading onto an Externalizing second-order factor; and I have Somatic Complaints, Withdrawl, and Anxious/Depressed first-order factors loading onto an Internalizing second-order factor. Is the Externalizing portion of the model OK, or will this setup present estimation problems?
 Linda K. Muthen posted on Friday, January 06, 2012 - 4:44 pm
A second-order factor must have at least three first-order factor indicators to be identified.
 Lisa M. Yarnell posted on Friday, January 06, 2012 - 5:07 pm
Thank you so much!
 Tarcia Davoglio posted on Tuesday, February 07, 2012 - 12:46 pm
I need a very basic information. I'm not sure I'm properly observing the output of the second order factor analysis. Where to visualize the correlations between the presence of second and first order? are these?

BY F5
F1 0,734 0,068 10,813 0,000
F2 0,961 0,035 27,219 0,000
F3 1,053 0,053 19,840 0,000
F4 0,980 0,044 22,270 0,000

The errors of the four factors? These are:

residual variances
F1 0,462 0,100 4,636 0,000
F2 0,076 0,068 1,118 0,264
F3 999,000 999,000 999,000 999,000
F4 0,039 0,086 0,453 0,651

Thank you!
 Linda K. Muthen posted on Tuesday, February 07, 2012 - 5:26 pm
These parameters are not identified in a second-order factor model.
 Heidi Kjogx posted on Thursday, March 29, 2012 - 1:16 am
Hi.

I can see that my problem has been seen before, but I still do not understand how to fix it. I am running a model with 3 factors loading on one second order factor. The fit indices for the two models are exactly the same. You have previously written that this is because the model is "just-identified". I am still not sure what this means and what can be done to fix it?

What can I do to make it work?

Here is my model:

ANALYSIS: type=general; estimator=mlm;

MODEL: f1 BY PCS1 PCS2 PCS3 PCS4 PCS5 PCS12;
f2 BY PCS8 PCS9 PCS10 PCS11;
f3 BY PCS6 PCS7 PCS13;
f4 BY f1-f3;

OUTPUT: standardized;
 Linda K. Muthen posted on Thursday, March 29, 2012 - 5:42 am
There is nothing wrong with the model being just-identified. It means that there are zero degrees of freedom and the fit of the model cannot be tested. You would need four or more first-order factors to change this.
 Terry Ng-Knight posted on Tuesday, September 18, 2012 - 1:53 am
Hi Linda and Bengt,
I have run a first-order factor model (5 factors and 1 observed variable) which runs ok and fits ok. I have now run a second order model where one of the first order factors (SCSOC) now has a negative residual variance. I have quite a small sample size (approx 230) and realise that I can set the error for SCSOC to 0 or drop the factor (i think this would make my second order model unidentified) but neither of these is ideal. Is there anything else I can do to remedy this problem?

Here is my model input:

MODEL:
ATTAIN BY english2 maths2 science2;
attndr;
LIKING BY sl2_1r sl2_3r sl2_4r sl2_6r sl2_7r;
BEHAV BY disrup2r cooper2r;
SCSOC BY SC2frnds SC2older SC2bulld;
SCSCH BY SC2class SC2ntchr SC2trvl SC2ltppl SC2size SC2break SC2dnnr;
attndr WITH LIKING;
SL2_6R WITH SL2_3R;

SCFUNC BY ATTAIN BEHAV attndr;
SCAFF BY SCSCH SCSOC;
 Linda K. Muthen posted on Tuesday, September 18, 2012 - 10:33 am
This points to the model being misspecified. You might want to look at modification indices to see if cross-loadings or residual covariances are needed. Or you might want to go back to an EFA to see if the first-order factors fit are well-specified.
 caroline masquillier posted on Thursday, October 25, 2012 - 3:20 am
Dear Prof. Muthen,
My measurement model consists of

- three first order latent constructs
- and a second order latent construct with five first order factors. One of those five first order factors (i.e. general mood) has 7 items of which 3 items are negative worded. Adding a negative wording factor (method factor) to the second-order factor solution improved the loadings of these three negatively worded items.

However, when we estimate the structural model between those four latent factors, the loadings of the three negative worded items on the first order construct (i.e. general mood) diminish far below 0.40 boundary line.
Can you help me to clarify this problem?
 Linda K. Muthen posted on Thursday, October 25, 2012 - 9:54 am
I would ask for MODINDICES (ALL) in the OUTPUT command. The need for cross-loadings and residual covariances may be the cause of the misfit.
 caroline masquillier posted on Friday, October 26, 2012 - 7:14 am
Dear Prof. Muthen,
Thank you very much for your quick answer. Unfortunately the suggested modification indices are not possible with respect to content. Are these suggestions of the modification indices the only solution for this problem? Or could there be other underlying problems which can be addressed?
Thank you very much in advance.
 Linda K. Muthen posted on Friday, October 26, 2012 - 1:39 pm
If the suggested modification indices do not make sense, you may need to rethink your model. The modification indices show where the model and the data do not agree. Perhaps you should try testing parts of your model to see where the misfit is.
 Terry Ng-Knight posted on Tuesday, October 30, 2012 - 6:21 am
Hi Linda
I understand that a 2nd-order factor needs 3 1st-order factors to be just-identified. However, Mplus seems happy to run my model below, where I have only 2 1st-order factors and 1 observed variable. Is it ok to do this?

MODEL:
ATTAIN BY english2 maths2 science2;
BEHAV BY disrup2r cooper2r;
attndr;

SCFUNC BY ATTAIN BEHAV attndr;

Thanks, Terry
 Linda K. Muthen posted on Tuesday, October 30, 2012 - 1:22 pm
You need three factor indicators for the second-order factor. They can be first-order factors or observed variables.
 Marcus Crede posted on Wednesday, November 14, 2012 - 11:19 am
If I want to examine whether data (16 manifest variables, 4 theoretical first-order factors) is characterized by a single second-order factor onto which all four first-order factors load I believe that I need to examine the fit of such a model and compare it to the fit of an alternate model with four correlated first-order factors. My first question is this: do I base my decision on a comparison of chi-square values or on a comparison of other fit statistics (or do both)? My concern about using chi-square values is that I do not see how the more parsimonious higher-order model is nested within the model with correlated first-order factors. Hence my second question: Are these two models nested and if so, how are they nested?
Thank you for any guidance that you can offer.
Marcus
 Linda K. Muthen posted on Thursday, November 15, 2012 - 8:59 am
Yes, these models are nested. You can think of the model where the first-order factors as an unrestriced H1 model and the second-order factor model as a restrictive H0 model. You can compare them using a chi=square difference test.
 Claudia Pérez posted on Wednesday, November 28, 2012 - 4:17 pm
Dear Dr. Muthen,

I run a CFA with three first order factors (The first has 5 variables, the second 6, the third 5 variables), and then I conducted a second order analysis with 1 second order factor measured by the same 3 first order factors I just mentioned. The second order analysis gives me the exactly same results than the first order analysis. Am I doing something wrong? I did not obtain any warning in the output.
My syntax was
TITLE: CPSSinvestigation
DATA:
FILE IS "C:\Users\CP\Documents\AnalisisMPLUS\CPSS.dat";
FORMAT IS FREE ;
VARIABLE:
CATEGORICAL ARE CPSS1 CPSS2 CPSS3 CPSS4 CPSS5 CPSS6 CPSS7 CPSS9 CPSS10
CPSS11 CPSS12 CPSS13 CPSS14 CPSS15 CPSS16 CPSS17;
NAMES ARE ID CPSS1 CPSS2 CPSS3 CPSS4 CPSS5 CPSS6 CPSS7 CPSS9 CPSS10
CPSS11 CPSS12 CPSS13 CPSS14 CPSS15 CPSS16 CPSS17;
USEVARIABLES ARE CPSS1 CPSS2 CPSS3 CPSS4 CPSS5 CPSS6 CPSS7 CPSS9
CPSS10 CPSS11 CPSS12 CPSS13 CPSS14 CPSS15 CPSS16 CPSS17;

ANALYSIS:
TYPE IS GENERAL ;
ESTIMATOR IS WLSMV;
model:
f1 by CPSS1 - CPSS5;
f2 by CPSS6 - CPSS12;
f3 by CPSS13 - CPSS17;
f4 by f1 - f3;

OUTPUT: SAMPSTAT RESIDUAL STANDARDIZED;

SAVEDATA:
RESULTS IS second_orderCPSS;
 Linda K. Muthen posted on Wednesday, November 28, 2012 - 4:31 pm
A second-order factor with three-indicators is just-identified. It has zero degrees of freedom. It does not contribute to model fit. This is why the fit statistics are the same.
 Claudia Pérez posted on Thursday, November 29, 2012 - 5:46 am
Dear Dr. Muthen,

If my model is just identified, I could fix some parameters to gain some degrees of freedom, right? Could I specify at fixed levels some coeficients estimates whose magnitud was revealed in the first order analysis to make the model work? How can I do that? If not, what option do I have?
Sorry for asking you these basic questions.
 Bengt O. Muthen posted on Thursday, November 29, 2012 - 6:28 am
No, it is incorrect to fix parameters to gain degrees of freedom. Adding the second-order factor does not impose any further restrictions. You should just accept that and report the estimates.
 Claudia Pérez posted on Thursday, November 29, 2012 - 10:49 am
Dear Dr Muthen:

So, if adding the second-order factor does not impose any further restriction to the model, do this mean that both models are equivalents (the second order factor has the same model fit than the first order analysis)or just mean that the second order analysis could not be calculated?
Could the addition of a second order factor in an overidentified model make the fit of a model worse?
Thanks!
 Bengt O. Muthen posted on Thursday, November 29, 2012 - 11:12 am
It means that the models are equivalent. There is nothing wrong with your second-order model; the second-order part is just not testable.

With more than 3 indicators (3 first-order factors), a second-order model is overidentified and can make model fit worse.
 Louise Mewton posted on Tuesday, December 04, 2012 - 6:42 pm
Dear Dr Muthen,

I am unsure about how to interpret the r-square for a second order factor. I have 3 first order factors which all correlate quite highly (>.7) on a fourth higher order factor. The fit of this model is good and almost identical to the fit of a model with just 3 factors and no higher order factor. However, for theoretical reasons, I have chosen the model with the second order factor.

The r-square for the three first order factors ranges from .4 to .6 but the r-square for the second order factor is 0.008. Would I interpret this as indicating that the second order factor is not explaining much variance over and above that explained by the 3 first order factors? Does this argue against including the second order factor in my model?

Many thanks,
Louise
 Bengt O. Muthen posted on Tuesday, December 04, 2012 - 7:02 pm
I don't understand what you mean that the r-square for the second-order factor is 0.008. There should be one r-square for each dependent variable.

Also, note that the fit is exactly the same with our without a second-order factor when there are only 3 first-order factors.
 Louise Mewton posted on Tuesday, December 04, 2012 - 8:09 pm
Thanks for getting back to me so quickly. My apologies, I wasn't clear. I am running a second order factor model with covariates. After running an initial model and checking modindices I included specific paths in the model as follows:

f1 by GA1 GA2 GA3 GA4 GA5 GA6 GA7
GA8 GA9;
f2 by GD1 GD2 GD3 GD4 GD5 GD6 GD7 GD8 GD10
GD11 GD12 GD13 GD14 GD15 ;
f3 by NEU1 NEU2 NEU3 NEU4 NEU5 NEU6
NEU7 NEU8 NEU9 NEU10 NEU11 NEU12;
f4 by f1-f3;
F4 ON AGE SEX;
f1 on age;
f2 on sex age;
f3 on sex;

I guess I'm confused about how you interpret the r-square for the second order factor here?

Thanks,
Louise
 Linda K. Muthen posted on Wednesday, December 05, 2012 - 11:07 am
The R-square you get for the second-order factor is the variance explained in the second-order factor by the covariates.
 Maren Formazin posted on Thursday, March 28, 2013 - 9:09 am
Dear Bengt & Linda,

I have a second order factor model:

f12 BY f1 f2;
f13 BY f3 f4 f5 f6;
f14 by f7 f8 f9 f10 f11;

The three factors f12, f13 and f14 are correlated.

I have problems with the factor loading for f1: even if I fix the variance of f12 (f12@1) and allow the factor loadings to be freely estimated (f12 BY f1* f2), Mplus seems to fix the factor loading: in the output for STDYX, the factor loading is 1. Moreover, in Tech9 I get the information that there is a problem with this factor loading for all datasets (I use 10 imputed datasets) due to a non-positive definite first-order derivative product matrix.

Am I specifying the model in a wrong way?

 Linda K. Muthen posted on Thursday, March 28, 2013 - 10:39 am
 H Steen posted on Thursday, May 02, 2013 - 1:51 am
Dear dr. Muthen,

I doing a CFA analysis, with a model that is partly first and partly second order. (it is done with WLSMV as it concerns all categorical data)

f1 by x1 x2;
f2 by x3 x4;

f3 by f1 f2 x5 x6 x7 x8;

It shows a RMSEA of 0.044 and CFI and TLI above 0.95, so it seems to perform fine.

I have two questions:

1) is it indeed ok do do a CFA partly second order?
2) the output gives me loadings on f3 for the latents f1 and f2 seperately from x5 to x8, meaning both f1 and x5 have fixed loadings on f3 of 1. How should I interpret this? Do all items and factor contribute in the same way to f3?
 Linda K. Muthen posted on Thursday, May 02, 2013 - 6:11 am
1. Yes.
2. You should free one of the fixed loadings. All factor indicators of f3 are treated in the same way.
 H Steen posted on Wednesday, May 15, 2013 - 4:37 am
Thank you very much!

Just to be sure, the output on Mplus is:

F3 by
x5
x6
x7
x8

F3 by
F1
F2

Thank you so much in advance.
 Linda K. Muthen posted on Wednesday, May 15, 2013 - 6:25 am
This is the same as if the results were presented:

F3 by
x5
x6
x7
x8
F1
F2

The observed and latent factor indicators are simply present separately. They are estimated together.
 H Steen posted on Wednesday, May 15, 2013 - 7:41 am
Thank you very much!
 H Steen posted on Wednesday, November 13, 2013 - 8:53 am
I have a follow up question about the distribution of the resulting factor scores of this partly second order WLSMV CFA.

The distribution is skewed, but this is understandable as the items are generally easy.

The distribution has a range of -1,3 to 0.71, mean -0,02.

What I find strange is that between 0,55 and 0,7 there are no scores, and then a peak again at 0,71 of 5,3% of the respondents.This seems a ceiling effect, but I wonder why this gap?

Have you come across such a finding before, and do you know what can cause this?

The model is working fine in other respects.
 Linda K. Muthen posted on Wednesday, November 13, 2013 - 1:15 pm
This can happen depending on the location of the item thresholds and the subjects' responses.
 H Steen posted on Thursday, November 14, 2013 - 3:08 am
Thank you for your prompt answer. Could yyou recommand literature on this subject?
 Linda K. Muthen posted on Thursday, November 14, 2013 - 8:30 am
I don't know of any. This is based on experience. You might want to contact Steve Reise at UCLA.
 Loryana L. Vie, Ph.D. posted on Friday, March 14, 2014 - 8:06 am
Hello,

We are attempting to fit a second order CFA that we will ultimately examine in a parallel process growth model.

We are wondering whether the intercepts of the marker variables should be set to 0, whether the metric of the first-order factors should be set to 1.0 (is this automatic in Mplus), and how we should set the metric at the 2nd order level.

Also, should the intercepts of the first-order factors be set to zero, and should the factor loadings of the 2nd order factor (on the five primary factors) be constrained to equivalence across time if the 1st order factor loadings are constrained across time?

Any advice would be greatly appreciated.

Thank you,

Loryana
 Linda K. Muthen posted on Friday, March 14, 2014 - 10:13 am
See Example 5.6. The default in Mplus is to fix the first factor loading to one to set the metric of the factor. You can free that and fix the factor variance to you if you wish, for example,

f BY y1* y2 y3;
f@1;
 Loryana L. Vie, Ph.D. posted on Monday, March 17, 2014 - 5:40 pm
Thank you. We're having trouble running a parallel process growth model reflecting growth in a 2nd-order factor and growth in a 1st-order factor. (LV cov. matrix not positive definite). Any advice?

!Construct 1 (C1), 1st-order, T1(a)
F1a BY x1a x2a-x3a (1-2);
F2a BY x4a x5a-x6a (3-4);
F3a BY x7a x8a-x9a (5-6);
F4a BY x10a x11a-x12a (7-8);
F5a BY x13a x14a-x15a (9-10);

...

!C1, T4(d)
F1d BY x1d x2d-x3d (1-2);
...
F5d ...

!C1, 2nd-order, T1-4
F6a BY F1a F2a (11)
F3a (12)
F4a (13)
F5a (14);
...

F5d (14);

!C2, 1st-order, T1-4
F7a BY x16a x17a-x18a (15-16);
...
F7d BY x16d x17d-x18d (15-16);

!Growth, 2nd- and 1st-order factors
i1 s1 | F6a@0 F6b@1 F6c@2 F6d@3;
i2 s2 | F7a@0 F7b@1 F7c@2 F7d@3;

[i1@0 i2@0];

[F1a-F5d@0];

[x1a-x1d] (18);
...
[x18a-x18d] (35);
 Linda K. Muthen posted on Tuesday, March 18, 2014 - 1:41 pm
 Loryana L. Vie, Ph.D. posted on Thursday, March 20, 2014 - 1:57 pm
Is a parameter, like correlated slopes, that is out of bounds (greater than 1) an indication of model identification problems or the model not being able to find a single estimable solution? Might this be caused by collinearity between the intercept terms? Thank you.
 Linda K. Muthen posted on Thursday, March 20, 2014 - 3:01 pm
When slopes correlate greater than one, this is not an indication of model identification problems or that the model cannot find a single solution. It means the the model is inadmissible and needs to be changed in some way.
 Sarah  posted on Friday, March 21, 2014 - 4:06 am
Hello,

I am attempting to construct a measurement model of child well-being. I have constructed a four factor model with each factor representing a domain of child well-being. This model works well. However when I attempt to add in a second order factor of 'child well-being' I run into problems.
I receive warnings stating that the latent variable covariance matrix is not positive definite. There are no negative variances or residual variances, and no correlations greater than or equal to one between latent variables. However, the residual variance (unstandardized) for one of the first order factors is huge at 28552.928. The warning states that the problem is with the second order variable of child well-being and many of the estimates involving this factor are not calculated.

I also receive an error that the model may not be identified. However, my four factor first order model was identified and my second order factor includes four first order factors so I thought it was identified?

Do these difficulties probably mean that a second order factor is simply inappropriate for the model?

 Linda K. Muthen posted on Friday, March 21, 2014 - 6:56 am
 Heather Yang posted on Saturday, April 05, 2014 - 8:57 am
Hi Dr. Muthen,

I would like to compare a second-order factor model (one second order factor and two first order factors) and a first-order factor model (two first order factors). I know that a second-order factor must have at least three first-order factor indicators to be identified.

However, my data do not allow me to do so. I am wondering whether there are any other ways I could compare them?
 Linda K. Muthen posted on Saturday, April 05, 2014 - 11:09 am
The estimates of a non-identified model are not meaningful.
 Lisa M. Yarnell posted on Thursday, October 23, 2014 - 1:58 pm
Hello, Why is Mplus not recognizing the lower-order factors set up the following higher-order mixture model, where I have known classes? I get the message:
*** ERROR in MODEL command
Unknown variable(s) in a BY statement: SUBST1

My code is below.

ANALYSIS:
TYPE = MIXTURE;
MITERATIONS = 1000;
PROCESSORS = 4; !8
INTEGRATION=MONTECARLO (500);
MCONVERGENCE = 0.015;

MODEL: %OVERALL%

SUBST1 BY
alc30d1 !(1)
mar30d1 (2)
hdr30d1 (3);

SUBST3 BY
alc30d3 !(1)
mar30d3 (2)
hdr30d3 (3);

SUBST4 BY
alc30d4 !(1)
mar30d4 (2)
hdr30d4 (3);

!* higher-order Intercept and Slope factors *

INT_HORD BY
SUBST1@1
SUBST3@1
SUBST4@1;

SLP_HORD BY
SUBST1@-1.5
SUBST3@0
SUBST4@1.5;

SUBST1 WITH SUBST3@0 SUBST4@0;
SUBST3 WITH SUBST4@0;

INT_HORD with SLP_HORD;

INT_HORD;
SLP_HORD;

%VIOL#2%
INT_HORD with SLP_HORD;

INT_HORD;
SLP_HORD;
 Linda K. Muthen posted on Thursday, October 23, 2014 - 2:47 pm
 Johnson Song posted on Sunday, November 23, 2014 - 9:37 pm
Dear Dr. Muthen,

Is a model which is composed of two second-order latent factors with each being measured by two first-order latent factors identifiable?
The results of running with Mplus did not show any improper results.

!! examples
variable: names are a1-a4 b1-b4 c1-c4 d1-d5;
analysis: estimator = mlr;
model:
a by a1-a4;
b by b1-b4;
c by c1-c4;
d by d1-d5;
fab by a b;
fcd by c d;

If yes, this should be also applied to a multilevel CFA model which both the within and between models have the same second-order measurement structure as shown in the above example, right?

Thank you very much!

John
 Linda K. Muthen posted on Monday, November 24, 2014 - 7:57 am
A second-order factor with two indicators is not identified. When a model has two second-order factors with two indicators, the model is identified because information from other parts of the model are used for identification. This is not an ideal situation.
 Guillermo posted on Thursday, November 27, 2014 - 7:11 am
Dear Dr. Muthen,

I have a question similar to the one that opens this thread and to that posted by H Steen. I have a second-order CFA in which the second-order latent variable is derived from 4 first order latent variables, two of them measured with only one manifest variable.

f1 by x1-x5;
f2 by x6-x10;
f3 by x11@1; x11@0;
f4 by x12@1; x12@0;
f5 by f1-f4;

When I try to fit this model, the output displays the message: "NO CONVERGENCE. NUMBER OF ITERATIONS EXCEEDED".

I have also tried this other way:

f1 by x1-x5;
f2 by x6-x10;
f5 by f1 f2 x11 x12;

In this case the output displays the message: "The metric for the following factor cannot be determined because the factor has both observed and latent indicators".

Am I doing something wrong?

Thank you so much.
 Linda K. Muthen posted on Thursday, November 27, 2014 - 10:47 am
Try

f1 by x1-x5;
f2 by x6-x10;
f5 by f1@1 f2 x11 x12;
 Guillermo posted on Thursday, November 27, 2014 - 1:47 pm
It does not work. When I try it, the output displays the message:

WARNING: THE LATENT VARIABLE COVARIANCE MATRIX (PSI) IS NOT POSITIVE
DEFINITE. THIS COULD INDICATE A NEGATIVE VARIANCE/RESIDUAL VARIANCE FOR A
LATENT VARIABLE, A CORRELATION GREATER OR EQUAL TO ONE BETWEEN TWO LATENT
VARIABLES, OR A LINEAR DEPENDENCY AMONG MORE THAN TWO LATENT VARIABLES.
PROBLEM INVOLVING VARIABLE F1.

However, when I let f1 free for estimation and I fix the link of the first manifest variable to 1, it works perfectly.

f1 by x1-x5;
f2 by x6-x10;
f5 by f1* f2 x11@1 x12;

Luckily, this is exactly the link I needed to fix to 1, so my problem is completely solved. Nevertheless, I do not understand why other options do not work. When I fix f2 to 1 letting f1 free, the output also displays the previous warning message, and when I fix x12 to 1 instead of x11, the output displays the message "NO CONVERGENCE. NUMBER OF ITERATIONS EXCEEDED".

Thank you so much for your help.
 Linda K. Muthen posted on Thursday, November 27, 2014 - 3:02 pm
I would need to see the outputs to explain what is happening. Send them and your license number to support@statmodel.com.