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 Linda K. Muthen posted on Saturday, January 28, 2006 - 9:50 am
Formative model with an observed dependent variable ("friends")

TITLE: Hodge-Treiman social status modeling
DATA: FILE = htmimicn1.dat;
TYPE = COVARIANCE;
NOBS = 530;
VARIABLE:
NAMES = church member friends income occup educ;
USEV = friends-educ;
MODEL:
f BY friends*; ! defining the factor; same
! as regressing friends on
! f
f@0;
f ON income@1 occup educ;
OUTPUT:
TECH1 STANDARDIZED;


Formative model with a latent dependent variable ("fy")

TITLE: Hodge-Treiman social status modeling
DATA: FILE = htmimicn1.dat;
TYPE = COVARIANCE;
NOBS = 530;
VARIABLE:
NAMES = church members friends income occup educ;
USEV = church-educ;
MODEL:
fy BY church-friends;
f BY fy*;
f@0;
f ON income@1 occup educ;

1.
.3607 1.
.2104 .2655 1.
.1002 .2845 .1763 1.
.1563 .1924 .1363 .3046 1.
.1583 .3246 .2264 .3056 .3447 1.
 Cam McIntosh posted on Thursday, February 16, 2006 - 5:22 pm
Hi Linda,
Just a quick and maybe very silly question for now - what is the purpose of the f@0 command in these programs?
Thanks,
Cam
 Linda K. Muthen posted on Friday, February 17, 2006 - 6:27 am
Formative factors have no residual variance. This would not be identified.
 Robin posted on Friday, July 21, 2006 - 8:22 am
I followed the second code segment (for latent DV "fy") and drew the directions of the variable and indicators.

Can I clarify that this is a 2nd-order model, with "fy" as 1st-order factor having income, occup and educ as formative indicators, and "fy" being a reflective latent variable (indicator) of the 2nd-order factor "f"?

Thanks.
 Linda K. Muthen posted on Saturday, July 22, 2006 - 11:34 am
I would not call this a second-order factor.
 Son K. Lam posted on Saturday, April 07, 2007 - 8:33 am
Dear Linda,

How do I test interaction when I have
- 2 formative constructs (each with multiple indicators)?
- 1 formative and 1 reflective constructs?

Thanks
 Linda K. Muthen posted on Saturday, April 07, 2007 - 12:41 pm
You can try the XWITH command. I'm not sure if it will work.
 Anil Akpinar posted on Sunday, May 27, 2007 - 11:41 am
Dear Linda,

Just a quick question. In the Formative model with a latent dependent variable model above you set the path between income and f to one. But if that path is set to one, we cannot conclude anything about the significance of that specific path. My question is; since we cannot say anything about the significance of the path between income and f, how can we conclude that income is a formative indicator of f?

I saw the same model under Indicator Arrows Pointing to a Factor discussion but setting one of the paths between formative indicators and factor to one is not explained there neither.

Thank you...
 Linda K. Muthen posted on Monday, June 04, 2007 - 11:04 am
One of the paths needs to be fixed to a non-zero number for model identification. It does not necessarily need to be income.
 Gareth posted on Tuesday, September 04, 2007 - 3:50 am
In the example "Formative model with a latent dependent variable ("fy")", how would I include a covariate (e.g. age) that I hypothesize should be related to either or both of the formative and latent variables?

For example, f ON age implies that age is part of the formative variable, which I don't want:

f ON income@1 occup educ age;

Any suggestions would be most welcome.
 Linda K. Muthen posted on Tuesday, September 04, 2007 - 8:50 am
With a formative model, this cannot be statistically distinguished.
 Gareth posted on Monday, December 10, 2007 - 9:22 am
In the example above (Formative model with an observed dependent variable), what is the purpose of fixing the residual variance of f to zero (f@0)?
 Bengt O. Muthen posted on Monday, December 10, 2007 - 6:30 pm
The residual variance cannot be identified. The formative approach essentially is like forming a factor by a weighted sum of the indicators where the weights are estimated, but measurement error is not parsed out.
 Heiko Schimmelpfennig posted on Wednesday, April 30, 2008 - 4:38 pm
In the example 'Formative model with a latent dependent variable' the results depend on the path which is restricted to 1:

income@1: estimate for F BY FY: 0,108 (Est./S.E.: 3,825)
occup@1: estimate for F BY FY: 0,045 (Est./S.E.: 1,726)
educ@1: estimate for F BY FY: 0,156 (Est./S.E.: 4,945)

I would like to know why the results are ambiguous and how one can select the 'right' path setting to 1.

Thanks in advance.
 Linda K. Muthen posted on Thursday, May 01, 2008 - 9:27 am
Please send the input, data, output including the STANDARDIZED option of the OUTPUT command, and your license number to support@statmodel.com.
 Heiko Schimmelpfennig posted on Thursday, May 01, 2008 - 12:12 pm
Dear Linda,

I used the input and data from the example above. I only changed the path which is restricted to 1. The standardized estimates for F BY FY are the same in the three cases, but I cannot assess definitely, if the coefficient is significant.
 Linda K. Muthen posted on Thursday, May 01, 2008 - 6:16 pm
Please send the files and your license number so I don't have to put them together.
 Esperanza Camargo posted on Saturday, February 14, 2009 - 4:41 pm
Can someone help me out with this? Please

I'm trying to define Family's SES based on family's poverty level (POV), female's SES (FSES), and male's SES (MSES).

Can I define the following formative factor?

POV is nominal (richest - poorest)
education occupation ---> MSES
education occupation employment ----> FSES

POV MSES FSES ----> Family's SES

Where can I find literature and/or annotated syntax of formative factors

Thanks a lot
 Linda K. Muthen posted on Sunday, February 15, 2009 - 10:50 am
For the nominal variable, you need to create dummy variables. See the Topic 1 course handout for inputs for formative factors.
 Esperanza Camargo posted on Sunday, February 15, 2009 - 1:48 pm
Thank you very much.
 Esperanza Camargo posted on Monday, February 16, 2009 - 8:54 am
Sorry, I need to be sure of what you're saying.

for my model

POV MSES FSES ---> FSES

Not only POV is nominal. Education and occupation are nominal too. Employment is dichotomous.

So, should I create dummy variables for all my nominal variables or just for POV? and what is the reason to do that.

thank you so much for your help
 Linda K. Muthen posted on Monday, February 16, 2009 - 9:05 am
Any nominal variable with more than two categories must be turned into a set of dummy variables. Formative factors are specified using ON. Variables on the right-hand side of ON are covariates in a regression and must be binary (dummy) or continuous as in regular regression.
 Esperanza Camargo posted on Monday, February 16, 2009 - 11:36 am
Understood. Thank you so much.
 Yu-Shan Chang posted on Thursday, April 02, 2009 - 3:48 am
I want to do CFA with a latent variable(kks) and four formative indicators(x1 x2 x3 x4).
so I write my program...

DATA: FILE IS 2.dat;
VARIABLE: NAMES ARE x1-x4;
MODEL: kks ON x1 x2 x3 x4;

but it doesn't work.
The results show...
*** ERROR in MODEL command
Unknown variable(s) in an ON statement: KKS

I don't know how to correct my mistake.
please help me.
Thank you so much.
 Linda K. Muthen posted on Thursday, April 02, 2009 - 5:45 am
Please see short course Topic 1 starting with Slide 238 for the proper specification of formative indicators.
 Yu-Shan Chang posted on Thursday, April 02, 2009 - 7:01 am
Thank you so much....
 Alexandru Agache posted on Tuesday, August 11, 2009 - 9:54 am
Hi,

is it possible in Mplus to standardize the variance of the formatively measured latent by fixing itís variance to unity (instead of fixing one of the slopes)? If yes... how?

Thanks,
Alex
 Linda K. Muthen posted on Wednesday, August 12, 2009 - 8:49 am
I don't think this model is identified.
 Alexandru Agache posted on Wednesday, August 12, 2009 - 11:49 am
Hi Linda and thanks for your thought.
Letís assume an indentified model where the formative measured latent has 3 cause indicators and is directly connected to 2+ endogenous variables . In order to set a scale for the formative latent one has following options (e.g., Edwards, 2001, p.161):
1. either fix the paths leading to or from the formative construct to 1 (like the examples above with SES - but not necessarily with zeta set to 0)
or
2. fix the variance of the construct to unity, thereby standardizing the construct.

I would prefer the second option because I want to test the S.E. for all paths. Frankeet al. (2008) show that the effects in the model change depending on the scaling method (They used lisrel and I have read somewhere that it is also possible with ramon but I was curios about mplus).


Edwards (2001): http://orm.sagepub.com/cgi/content/abstract/4/2/144
Franke, Rigdon, Preacher, 2008, in the Jou of Business Research- spec. issue on formative constructs
 Bengt O. Muthen posted on Wednesday, August 12, 2009 - 2:18 pm
You can constrain the variance of the construct to unity using Model Constraint, where you express the variance of the construct in terms of model parameters and the formative indicators' sample covariance matrix and set the construct variance at 1.

Note that in Mplus you can test all paths even if you fix a slope at 1 - this is because Mplus gives you SEs also for the standardized coefficients.

Because different scaling settings lead to different results I think it might be more straightforward to set the construct residual variance at zero, acknowledging that we don't have information on it.
 Alexandru Agache posted on Thursday, August 13, 2009 - 1:24 pm
Many thanks!
(I will try to model the variance using the model constraint. I was using version 4.1 where the stdxy; option is not implemented; it seems that the consequences of different scalings could be a good case for a new monte carlo study )
 Lewina Lee posted on Monday, August 27, 2012 - 3:33 pm
Dear Drs. Muthen,

I'm building an SEM involving a formative indicator (CMAT1) predicting a latent outcome variable (LPOS). When I ran the model specifying one of the composite paths to 1 (CMAT1 on AGE @1 MARRY EDU;) and freely estimating the link between the outcome variable and the composite variable (LPOS on CMAT1;), as in METHOD 1 below, the model would not converge.

However, when I slightly changed the model so that I freely estimated all the composite paths (CMAT1 on AGE MARRY EDU;) but set the link between the composite indicator and the outcome variable to 1 (LPOS ON CMAT1 @1;), as in METHOD 2 below, the model was identified.

I see that METHOD 2 is slightly different than what you have adviced here and in your handout, am I doing anything wrong?

(MARRY is a dichotomous variable, if that may make any difference).

Thank you,
Lewina

***METHOD 1***
LPOS by LSAT MCS;
CMAT1 by;
CMAT1 on AGE @1 MARRY EDU;
CMAT1 @0;
LPOS on CMAT1;

***METHOD 2***
LPOS by LSAT MCS;
CMAT1 by;
CMAT1 on AGE MARRY EDU;
CMAT1 @0;
LPOS on CMAT1 @1;
 Linda K. Muthen posted on Monday, August 27, 2012 - 5:05 pm
I wonder if the problem is that lpos is not identified. Try playing with that.
 Lewina Lee posted on Tuesday, August 28, 2012 - 8:35 am
Thanks, Linda. Would you consider METHOD 1 & METHOD 2 above equivalent? If METHOD 2 allows the model to converge (despite the slight departure from your suggested approach in the handouts & on this forum), can I go along with the results?

Thank you,
Lewina
 Linda K. Muthen posted on Tuesday, August 28, 2012 - 10:52 am
No, the two methods are not the same. You need to fix on indicator to one.
 Kathleen Van Benthem posted on Tuesday, February 18, 2014 - 2:17 pm
Hello,
To evaluate formative constructs the P values for the weights are one criteria for relevance of the composite indicators. How can I determine the weights (and p values) used to calculate each of the composite indicators? The manual demos setting weights, which is not what I need to do.
I have modeled a reflective-formative construct as below:
Analysis: Estimator = mlr;
!for possible skew or non-normality in raw data
MODEL:
A by X1@1 X2* X3* ; ! create the reflective lv "A"

B by Y1@1 Y2*; ! create the reflective lv "B"

C by Z1@1 Z2* Z3*;

Q by ; !create the form. lv "Q"

!D is a single indicator construct for Q

Q ON A@.42 B@.34 C@.38 D@.35 ; ! specify the form. lv "Q" use pop values- or make equal
A with B; ! check r between 1st order lvs
A with C; !etc

Q@0; ! set Q residual var to 0
[Q@0]; ! set Q mean at 0
thanks!
 Bengt O. Muthen posted on Tuesday, February 18, 2014 - 6:05 pm
You fix one of the weights at 1 and free the rest. You fix the residual variance of Q at 0 as you have done. Note that

[q@0];

doesn't fix the mean of Q at zero, but its intercept (since Q is a DV).
 Elizabeth Salib posted on Monday, April 07, 2014 - 8:46 am
Hello. I have a model where a latent variable predicts the 2 latent cause indicators of a formative model and the formative variable predicts 2 latent variables.

SL -> U
SL -> B
U & B -> I (Formative model)
I -> C
I -> T

How can I tell if this model is identified? Mplus is able to estimate this model but I'm not sure it is identified.
 Elizabeth Salib posted on Monday, April 07, 2014 - 8:48 am
Syntax is:

SL BY SL2 SL25;
UNIQ BY Unique4 Unique6;
PIS BY PIS1 PIS5;
UNIQ with PIS;
Inc by;
Inc@0;
Inc on Uniq@1 PIS@1;
CRT by Crt2 Crt4;
TC by TeamCit1 TeamCit3;
TC ON INC;
CRT ON INC;
UNIQ ON SL;
PIS ON SL;
 Bengt O. Muthen posted on Monday, April 07, 2014 - 4:41 pm
I assume "PIS" plays the role of "B" in your first notation. The model is identified and you should free the coefficients for PIS in

Inc on Uniq@1 PIS@1;

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