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 Leigh Roeger posted on Tuesday, November 23, 1999 - 10:15 pm
Can anyone provide an interpretation of the mean estimates produced from a multigroup meanstructure analysis for the non-reference groups?

Is there an easy way to convert these estimates to effect sizes (ala Cohen)?
 Bengt O. Muthen posted on Thursday, December 02, 1999 - 3:07 pm
Mean estimates for latent variables in non-reference groups are given in
comparison to the zero latent variable means for the reference group. Mean
estimates for the observed variables should simply be seen as regular
means, hopefully closely fitting the observed sample means. The Cohen-type
effect sizes typically refer to differences in means from independent
samples divided by the SD of the mean. This would seem to carry over
directly to estimated means and their estimated SD's from different groups.
 Markku Niemivirta posted on Monday, September 09, 2002 - 12:03 am
Hi,

I'm testing a multigroup mediational model with structured means. My data are slightly nonnormal, so I would be using the MLM estimator. In the first step, I wanted to test the degree of factorial/structural invariance across my two groups, and then include mean structures in the next step. However, since the estimation procedure with MLM incorporates the mean structure, it seems I won't be able to do it this way. Am I correct?
 Linda K. Muthen posted on Monday, September 09, 2002 - 8:41 am
You can relax the equalities of the intercepts. A model with unstructured means is the same as a model without means so this should accomplish your purpose.
 Markku Niemivirta posted on Tuesday, September 10, 2002 - 12:25 am
If I understood this correctly, I should free the fixed latent means for the first group, and relax the equality constraints on the item intercepts for the second group. Well, if i do this, I run into identification problems (as one would expect). The error message I receive reports problems with the first latent mean parameter in the model. I guess I'm not doing exactly what you suggested. Any advice?
 Linda K. Muthen posted on Tuesday, September 10, 2002 - 8:13 am
You should do nothing to the latent variable means. You should free the intercepts across groups, for example in a two group analysis,

MODEL: f1 BY y1-y4;
MODEL males: [y1-y4];
 Markku Niemivirta posted on Tuesday, September 10, 2002 - 11:23 pm
If I free all intercepts, there will be problems with identification again. It rather seems that for identification purposes, I have to have at least one intercept per scale fixed. In fact, if I do this (fix one intercept for each scale to be equal across the two groups) I get an identified model with degrees of freedom equal to a model with no means at all. And that is exactly what I wanted. Thanks, Linda.
 Linda K. Muthen posted on Wednesday, September 11, 2002 - 7:24 am
All intercepts can be free across groups if the factor mean is zero in the first group and free in the others. If you get identification problems when you try to free all intercepts, then you must have the factor mean in your first group free. In this case, you need to hold one intercept equal across groups. This is simply an alternative parameterization.
 niemivirta posted on Wednesday, September 11, 2002 - 9:30 pm
The means in my first group are fixed. Check the syntax below. Here is my model for testing factorial invariance across the two groups (with MLM estimator).

MODEL:
lo by lo2 lo1 lo3 lo5;
po by po1 po3 po4;
meff by meff1-meff4;
mabi by mabi1-mabi4;
prior by pend;
int by pre9 pre3 pre7 pre4;
eff by pre5 pre8 pre4;
sh by pre6 pre2 pre10;
ross by r_1-r_3;
pend@2.41;

MODEL Group 2:
[lo1-r_3];

With this specification I get 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 176

The parameter 176 refers to the first latent mean for the second group.

This same happens even if I try to reproduce the simple one-factor model you described in your message.
 Linda K. Muthen posted on Thursday, September 12, 2002 - 7:04 am
Im sorry. When the intercepts are free in each group, you must fix the factor means to zero in all groups. So add,

[lo-ross@0] to the group specific model command, for example,

MODEL Group 2:
[lo1-r_3 lo-ross@0];
 Markku Niemivirta posted on Friday, September 13, 2002 - 4:00 am
But that adds fixed constraints into the model, and thus shoots up the dfs, right?
 Linda K. Muthen posted on Friday, September 13, 2002 - 8:01 am
No. Let's take as an example a model with one factor, three observed continuous factor indicators, and two groups.

First, let's consider the model without means. We have six sample statistics for each group for a total of 12. We are estimating four factor loadings, six residual variances, and two factor variances for a total of 12 parameters and zero degrees of freedom.

Now, let's consider the same model with means. This adds six sample statistics for a total of 18. With the factor means fixed to zero in each group and the intercepts free across groups, this adds six parameters for a toal of 18 and zero degrees of freedom.

Note that in this example there are only six observed means and they are represented by six parameters so that the estimated parameters will be the observed means and the residuals for the means will be zero not adding to the chi-square.

You can try this using ML where you can run the model without means and with means by adding TYPE=MEANSTRUCTURE.
 Markku Niemivirta posted on Sunday, September 15, 2002 - 9:41 pm
Of course, thanks!
 Anonymous posted on Monday, October 21, 2002 - 8:12 am
I'm following-up on the posting from November 23, 1999 concerning estimated means and variances for latent variables. I've completed a multi-group, multi-level analysis with MPLUS where I examined model differences by race (achievement is the outcome variable). Can I interpret the estimated between (criterion) latent variable means as "adjusted means," and can I compare these for the two race groups using the estimated latent variable variances (i.e., SDs)?

Thanks!
 bmuthen posted on Monday, October 21, 2002 - 8:44 am
When you say adjusted means, it sounds like you refer to intercepts in endogenous latent variables regressed on background variables. If that's the case, I think you are right if the regression slopes are the same across groups - so that the ANCOVA analogy holds. If I am reading you correctly, then the estimated latent variable SDs would be relevant to use in the comparison.
 Anonymous posted on Tuesday, October 22, 2002 - 12:25 pm
In this case we have one covariate (income) and 3 endogenous latent variables, the outcome latent variable being test score. Between and within factor loadings are invariant across race groups; paths are not (as expected). We're using random intercepts across schools (level 2), but not random slopes. Can the estimated latent variable SDs still be used to develop CIs around the estimated latent variable means (or calculate effect sizes)?
 bmuthen posted on Tuesday, October 22, 2002 - 1:50 pm
If the factor loadings are the same for Within and Between so that you referring to the same factor on those two levels, I would use the sum of the between and within variance in the factor to get the denominator SD for the ES.
 Anonymous posted on Tuesday, March 15, 2005 - 6:07 pm
Hello. I need to clarify in my own head some things previously discussed in this thread (i.e., estimating effect sizes in standard-deviation units for mean differences). The Mplus output provides StdY and StdYX values (which are equal) for the means of each group. Are these values effect sizes for the non-reference groups? If not, would you mind explicitly telling me how to calculate an effect size of mean difference in standard-deviation units using data from the output? Thank you!
 Linda K. Muthen posted on Wednesday, March 16, 2005 - 7:19 am
A standardized mean is not an effect size. An effect size is the difference between two means divided by the appropriate standard error, for example:

x1 - x2 / sqrt (var(x1) + var (x2))
 Anonymous posted on Wednesday, March 16, 2005 - 8:02 am
Perfect, thank you. I got a little confused with the factor mean of a non-reference group being in fact a difference between the reference group and the non-reference group. Then, I thought that the standardized value of that mean (a difference value in and of itself) might be an effect size. But, you clarified things for me well. Thank you, again.
 Anonymous posted on Tuesday, July 19, 2005 - 8:33 am
Hello!
I am working on a multi-group SEM and am a little confused about estimating means and SDs. I have four latent variables and MPlus provides the mean and variance for one of them (there is only one variable that only serves an independent variable). The others operate as both dependent and independent variables and I do not get means for these on my output. How would I go about estimating the means and variances for the other variables in the model for each group? Thank you for your help!
 bmuthen posted on Tuesday, July 19, 2005 - 11:17 am
Request Tech4 in the Output command.
 Anonymous posted on Thursday, August 11, 2005 - 5:49 pm
Hi, I have a question regarding the mean estimate:

...
CATEGORICAL IS Y1 Y2;
ANALYSIS: PARAMETERIZATION=THETA;
TYPE=MEANSTRUCTURE;
MODEL: Y1 ON X1 X2;
Y2 ON Y1 Y3 X2;
Y3 ON X3 X4 X5;
...

In the modeling results:
...
Means
Y3=-2 ...

My question: Y3 is a continuous variable and its sample mean is positive. Why its mean here is negative?

Thanks!
 Linda K. Muthen posted on Friday, August 12, 2005 - 6:39 am
You may be reading the data incorrectly. If you send your input, data, output, and license number to support@statmodel.com, I can take a look at it.
 kberon posted on Thursday, December 15, 2005 - 7:11 am
Hi,
I'm running a multigroup, missing data CFA and have used the general approach, described earlier in this discussion, to "remove" the mean structure by freeing both groups intercepts and fixing the latent means to zero. I then allow the mean structure to come back in. I believe that the "covariance model," with removed means, is nested in the mean structure model and so I can use a chi-square test to see the improvement (or not) from the mean structure. However I have not been able to find any confirmation of this. Does this seem reasonable?
By the way, thanks for a great program - I've been using it since Liscomp days!
 bmuthen posted on Thursday, December 15, 2005 - 7:50 am
Yes, if you have free intercepts and factor means fixed at zero, then that is the same as a analyzing a covariance matrix because the mean vector is unstructured, so you have zero mean residuals giving zero contribution to the chi-square.

Amazing, a former Liscomp user! Good to hear from that niche group.
 kberon posted on Thursday, December 15, 2005 - 2:00 pm
Thanks for the info. When you go on your next worldwide tour I'll have to have you autograph my original Liscomp manual (and disk!) that I still have.
 bmuthen posted on Friday, December 16, 2005 - 8:12 am
Be happy to.
 Madeline Hogan posted on Sunday, July 16, 2006 - 1:18 pm
The default in Mplus sets means of latent variables to zero. I have two continuous latent variables with categorical indicators, and I want to estimate the mean of both of the latent variables. How do I do this? Would it make sense to fix the thresholds of the factor indicators to zero and free the variances of the latent variables? When I do that, I get a negative value for my mean - which I don't understand.
 Bengt O. Muthen posted on Sunday, July 16, 2006 - 1:58 pm
You can fix one threshold and thereby identify (be able to estimate) a factor mean. But there is nothing gained by this since the two models are equivalent. The factor mean simply takes the value of the threshold with negative sign (if you restrict the threshold for the indicator with unit loading). Factor means really only need to be free with multiple-group or growth analysis.

Note that the factor variance is free as the default.
 Madeline Hogan posted on Thursday, July 27, 2006 - 2:32 pm
Thanks for your help. I am working on a project using a matched pairs design. Though I am not using the "multigroup" option in Mplus because I want to take advantage of the matched pairs, I am essentially trying to conduct a multigroup analysis. I need to estimate the mean difference between CDGIRLS and CDBOYS. I think I have done this (see below)? Also, I found measurement invariance for factor loadings for CDGIRLS and CDBOYS, but not for the thresholds. Is there a way to estimate the means for CDBOYS and CDGIRLS without holding the thresholds equal?

USEVARIABLES ARE wagf ranf lief stlf dpf ftf weaf injf
bulf litf brof harsh1f harsh2f harsh3f harsh4f harsh5f
harsh6f wagm ranm liem stlm dpm ftm weam injm bulm
litm brom harsh1m harsh2m harsh3m harsh4m harsh5m harsh6m;

CATEGORICAL ARE
wagf ranf lief stlf dpf ftf weaf injf bulf litf
brof harsh1f harsh2f harsh6f wagm ranm liem
stlm dpm ftm weam injm bulm litm brom harsh1m
harsh2m harsh6m;


Missing = .;

ANALYSIS: Type = Meanstructure;
MODEL:
CDGIRLS by
brof
wagf (2)
ranf (3)
lief (4)
stlf (5)
dpf (6)
ftf (7)
weaf (8)
injf (9)
bulf (10)
litf (11);

[CDGIRLS *];

CDBOYS by
brom
wagm (2)
ranm (3)
liem (4)
stlm (5)
dpm (6)
ftm (7)
weam (8)
injm (9)
bulm (10)
litm (11);

[CDBOYS@0];

[brof$1](18);
[wagf$1] (19);
[ranf$1] (20);
[lief$1] (21);
[stlf$1] (22);
[dpf$1] (23);
[ftf$1] (24);
[weaf$1] (25);
[injf$1] (26);
[bulf$1] (27);
[litf$1] (28);

[brom$1] (18);
[wagm$1] (19);
[ranm$1] (20);
[liem$1] (21);
[stlm$1] (22);
[dpm$1] (23);
[ftm$1] (24);
[weam$1] (25);
[injm$1] (26);
[bulm$1] (27);
[litm$1] (28);

HARSHG by
harsh6f
harsh2f (13)
harsh3f (14)
harsh1f (15)
harsh5f (16)
harsh4f (17);


HARSHB by
harsh6m
harsh2m (13)
harsh3m (14)
harsh1m (15)
harsh5m (16)
harsh4m (17);

[harsh1f$1] (29);
[harsh2f$1] (30);
[harsh3f] (31);
[harsh4f] (32);
[harsh5f] (33);
[harsh6f$1] (34);

[harsh1m$1] (29);
[harsh2m$1] (30);
[harsh3m] (31);
[harsh4m] (32);
[harsh5m] (33);
[harsh6m$1] (34);

[HARSHG *];
[HARSHB@0];

CDGIRLS ON HARSHG;
CDBOYS ON HARSHB;
 Bengt O. Muthen posted on Thursday, July 27, 2006 - 4:44 pm
You have to have at least some items with invariant thresholds in order to identify group differences in factor means.
 J.W. posted on Wednesday, February 25, 2009 - 2:51 pm
I would like to confirm that the following interpretation of SEM results is correct:
¡°The estimated means of the endogenous latent variables produced by Tech4 are in fact the intercepts of regressing the latent variables on covariates.¡±
Thanks a lot for your help!
 Linda K. Muthen posted on Wednesday, February 25, 2009 - 4:15 pm
If your model has covariates, your endogenous variable will have an intercept estimated for it in Model Results. TECH4 gives the model estimated mean not the intercept.
 J.W. posted on Wednesday, February 25, 2009 - 7:45 pm
Thank you so much for your quick reply to my question. This was a single group MIMIC model with three factors (each factor has 6 indicators); and four covariates (individual background variables) were used to predict each of the factors. In the Mplus output I found that intercept estimates were only for the 18 factor indicators. That was why I was wondering whether the factor mean estimates by Tech4 would be the intercepts for the factors.
 Linda K. Muthen posted on Thursday, February 26, 2009 - 9:42 am
In a single-group analysis, factor means and factor intercepts are fixed at zero. The factor means in TECH4 for the conditional model take information about the covariates into account and are not zero. With an unconditional model, the factor means in TECH4 are zero.
 Holly Ketterer posted on Saturday, April 03, 2010 - 10:51 am
Can you please provide some insight regarding the interpretation of unstandardized and standardized latent factor means? I understand that mean estimates for latent variables in non-reference groups are given in
comparison to the zero latent variable means for the reference group.

Is it appropriate to interpret the standardized latent mean for the reference group as a Cohen-type standardized mean difference?
 Linda K. Muthen posted on Sunday, April 04, 2010 - 8:32 am
If you change reference to non-reference in the last sentence above, I think this is true.
 QianLi Xue posted on Monday, April 04, 2011 - 7:41 am
suppose I have two sets of items measured at two timepoints t1 and t2, and the items are the indicators of two factors f1 and f2 correspoinding to t1 and t2. In MPLUS, can I model the difference between f1 and f2 as a function of covariates?
 Linda K. Muthen posted on Monday, April 04, 2011 - 8:52 am
Yes, you would specify:

MODEL:
f1 BY ...
f2 BY ...
f BY f1@-1 f2@1;
f ON covariates
 QianLi Xue posted on Monday, April 04, 2011 - 9:45 am
Thanks so much!
 Andrea Vocino posted on Wednesday, March 14, 2012 - 7:40 pm
I was able to establish partial scalar invariance across four different groups. I now want to test latent means invariance. Do I need to fix factor means to zero only to one group or to them all? See below

MODEL:


suc by ms1 ms2 ms3 ms4;
hap by mh1 mh2 mh3;
conc by c1 c2 c3 c4 c5 c8;
intn by in2 in3 in4 in5 in6;
indr by pb1 pb2 pb3 ab1 ab3 ab5;
dir by hb2 hb3 hb4 sb4 sb5;


MODEL chi:


conc by c8;
intn by in4 in6;
[c8 in4 in6];
[c1 ms1 hb3 in5 c3 c4 hb2];
[suc-dir@0]

MODEL hk:


dir by sb4 ;
[sb4];
[c3 pb3];


MODEL tai:

intn by in5;
dir by hb4 sb4;
[in5 hb4 sb4];
[c3 in6 c1 hb3];


MODEL sin:

suc by ms2 ms3 ms4;
hap by mh2 mh3;
conc by c2 c4 c8;
indr by ab1;
dir by hb4 ;
[ms2 ms3 ms4 mh2 mh3 c2 c8 ab1 hb4];
[pb2 ab5 sb5]
 Linda K. Muthen posted on Wednesday, March 14, 2012 - 8:01 pm
The test of the invariance of means is one model with means zero in all groups versus another model with means zero in one group and free in the others. This is shown in the Topic 1 course handout toward the end of the multiple group discussion.
 Andrea Vocino posted on Wednesday, March 14, 2012 - 8:28 pm
I have seen that but the example refers only to two groups. So if I were to test equality of latent means across the four different groups after having established partial scalar invariance do I need to add [suc-dir@0] to each group? Thank you!
 Linda K. Muthen posted on Thursday, March 15, 2012 - 6:50 am
Yes.
 Andrea Vocino posted on Thursday, March 15, 2012 - 11:50 am
Thank you!
 UMDEDMS posted on Thursday, April 26, 2012 - 1:53 am
Hi Dr. Muthen,

I have problem to interpret the standardized indicator intercept from a multigroup means analysis, as shown in below.
I thought the latent variable and indicators are standardized to have mean of 0 and variance of 1. Why did I got standardized indicator intercept much larger than 0?

Your clarification is highly appreciated!


STDYX Standardization
READPROF BY
SATVOC 0.883 0.011 80.825 0.000
SATCOMP 0.884 0.011 81.005 0.000
SATLANG 0.737 0.015 50.418 0.000

Means
READPROF 0.000 0.000 999.000 999.000

Intercepts
SATVOC 15.483 0.337 46.003 0.000
SATCOMP 17.558 0.381 46.032 0.000
SATLANG 17.421 0.352 49.558 0.000

Variances
READPROF 1.000 0.000 999.000 999.000

Residual Variances
SATVOC 0.220 0.019 11.427 0.000
SATCOMP 0.219 0.019 11.377 0.000
SATLANG 0.457 0.022 21.228 0.000
 Linda K. Muthen posted on Thursday, April 26, 2012 - 1:40 pm
You are thinking about standardized variables that have a mean of zero and variance of one. When parameters are standardized, they do not have means of zero. They have variances of one.
 Min Liu posted on Thursday, April 26, 2012 - 1:58 pm
Hi Linda,

Thanks for pointing out this to me.
I figure out how to compute those values for standardized loadings. But I still don't know how to generate standardized indicator intercept.Could you show me using just one example?Here is my output for unstandardized results. Or refer some formula for computing it? Many thanks in advance.

Unstandardized Model result

READPROF BY
SATVOC 1.433 0.038 38.040 0.000
SATCOMP 1.240 0.033 38.153 0.000
SATLANG 1.000 0.000 999.000 999.000
Means
READPROF 0.000 0.000 999.000 999.000
Intercepts
SATVOC 710.029 1.410 503.478 0.000
SATCOMP 696.598 1.219 571.307 0.000
SATLANG 668.183 1.100 607.624 0.000
Variances
READPROF 798.586 51.498 15.507 0.000
Residual Variances
SATVOC 463.594 40.054 11.574 0.000
SATCOMP 345.173 29.958 11.522 0.000
SATLANG 672.464 35.065 19.177 0.000
 Linda K. Muthen posted on Thursday, April 26, 2012 - 2:13 pm
Divide the raw intercept by its standard deviation.
 Min Liu posted on Thursday, April 26, 2012 - 2:22 pm
Thanks so much for your quick response
 Sarah Phillips posted on Friday, June 29, 2012 - 5:47 pm
Hello,

I'm trying to follow some advice that you gave to Markku Niemivirta on September 12th, 2002 in this thread.

I am doing a multi-group analysis of a 2nd order engagement scale, and it seems as if I do not have intercept invariance on the second order factor itself. I'd like to relax this constraint and continue using the grouping function.
I followed your advice to Markku but want to make sure that I am allowing the intercept of my 2nd order latent factor to vary between groups. Is this the correct code:

Model:
BE by m_a3 m_a31 m_a28 m_b50;
EE by m_a62;
m_a62@0;
CE by m_a72 m_a74 m_a79 m_a80;
Engage by BE EE CE;

Model AfAm: [EE-CE Engage@0];
Model Latino: [EE-CE Engage@0];

Thank you,

Sarah
 Bengt O. Muthen posted on Friday, June 29, 2012 - 8:37 pm
Your code fixes all means at zero. If you want to study invariance you should hold

[ee-ce@0];

in both groups and fix

[engage@0];

in the first group while having it free in the second group. You can then use Modindices to see which of the ee-ce intercepts might not be invariant=0.
 Kyle Stephenson posted on Friday, January 25, 2013 - 3:45 pm
Hello!

I'm trying to run a higher order factor model with two groups (see input below).

Grouping is CSA (1=CSA 0=NSA);

Model: Desire BY fsfi1 fsfi2;
Arousal BY fsfi3 fsfi4 fsfi5 fsfi6;
Lub BY fsfi7 fsfi8 fsfi9 fsfi10;
Orgasm BY fsfi11 fsfi12 fsfi13;
Satisfaction BY fsfi14 fsfi15 fsfi16;
Pain BY fsfi17 fsfi18 fsfi19;
SexFunct BY Desire Arousal Lub Orgasm Satisfaction Pain;

Model NSA: Desire BY fsfi2;
Arousal BY fsfi4 fsfi5 fsfi6;
Lub BY fsfi8 fsfi9 fsfi10;
Orgasm BY fsfi12 fsfi13;
Satisfaction BY fsfi15 fsfi16;
Pain BY fsfi18 fsfi19;
SexFunct BY Arousal Lub Orgasm Satisfaction Pain;
[fsfi1-fsfi19];

I'm at the initial step of testing for configural invariance, but I keep getting the following error message in the output:

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 115.

Parameter 115 is the alpha for the first 1st-level factor: Desire. However, I've tried removing desire from the model and I still keeping getting the same message, just for the next 1st-order factor. Thank you so much for any help you can provide!
 Bengt O. Muthen posted on Friday, January 25, 2013 - 4:12 pm
It looks like you free the intercepts for all items in the second group, but you have to also fix at zero the factor means in the second group, both the 1st-order intercepts and the 2nd-order factor means.
 Kyle Stephenson posted on Saturday, January 26, 2013 - 8:49 am
Thanks so much for responding!

I tried fixing the item intercepts and factor means at zero in the second group (below):

[fsfi1-fsfi19];
fsfi1-fsfi19 @ 0;
Desire Arousal Lub Orgasm
Satisfaction Pain SexFunct @ 0;
[Desire Arousal Lub Orgasm
Satisfaction Pain SexFunct]

However, now I get a new error message:

NO CONVERGENCE. SERIOUS PROBLEMS IN ITERATIONS. ESTIMATED COVARIANCE MATRIX NON-INVERTIBLE.CHECK YOUR STARTING VALUES.

I've tried increasing the max iterations, using MPlus's suggested start values, and removing a number of estimated parameters to cut down on model complexity. None of these has worked. Any advice? Thanks so much!
 Linda K. Muthen posted on Sunday, January 27, 2013 - 6:27 am
Please send your output and license number to support@statmodel.com.
 Georg Datler posted on Thursday, June 13, 2013 - 7:50 am
In a Multiple Group CFA with Covariates and a Mean Structure (such as Ex. 5.15.):

How do we interpret the latent variable intercept and differences of latent variable intercepts across groups?
(as compared to latent variable mean differences in a Multiple Group CFA)

Thank you!
 Linda K. Muthen posted on Thursday, June 13, 2013 - 9:02 am
The mean difference for people with the same covariate values is the intercept difference.
 Leslie Rutkowski posted on Monday, July 29, 2013 - 10:51 pm
Hello,

I am looking at a multiple groups analysis with 8 groups, 8 indicators and I'm trying to estimate a model with and without mean structure, but I'm getting an identical model regardless (according to fit statistics).

My mean structure model is as follows:

MODEL:
scl1 by
v1 v2 v3 v4 v5 v6 v7 v8;
[v1 v2 v3 v4 v5 v6 v7 v8];

MODEL G2:
scl1 by
v2 v3 v4 v5 v6 v7 v8 ;
[ v2 v3 v4 v5 v6 v7 v8];

...


My "no mean structure" model is:

MODEL:
scl1 by
v1 v2 v3 v4 v5 v6 v7 v8;
[v1* v2 v3 v4 v5 v6 v7 v8];
[scl1@0];

MODEL G2:
scl1 by
v1 v2 v3 v4 v5 v6 v7 v8;
[v1* v2 v3 v4 v5 v6 v7 v8];

...

The models have identical loadings and unique variance estimates and differ in that in the second analysis, the LV means are 0 and all of the intercepts correspond to the empirical means. Whereas, in the first model, the LV mean in group 1 is 0, the intercepts correspond to the empirical mean for that group, and the first intercept of all of the other groups are equal to group 1.

According to the DF and number of free parameters, I am estimating a mean structure in both of these analyses.

How can I really remove the mean structure (and get back the associated DF)?

Thanks,
Leslie
 Bengt O. Muthen posted on Tuesday, July 30, 2013 - 5:47 am
The models are equivalent, just being reparameterizations of each other. Relative to the second model, the first model let's the mean of the first indicator vary across groups with the help of the varying factor mean.

If you want a mean structure you need to have more than one measurement intercept equal across groups in the first model. The second model is the correct "no mean structure" model.
 Leslie Rutkowski posted on Tuesday, July 30, 2013 - 6:22 am
Thanks, Bengt. This is a big help.

Leslie
 Yuyu Fan posted on Tuesday, February 25, 2014 - 3:29 pm
Dear Muthen,

Is it possible to get estimates of latent means in a ful SEM model in a single group study?

Thank you!
 Linda K. Muthen posted on Tuesday, February 25, 2014 - 5:16 pm
No, they are not identified in this case. Factor means are identified only if there are multiple groups or multiple time points.
 Yuyu Fan posted on Tuesday, February 25, 2014 - 5:46 pm
Got it. Thank you very much!
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