Multigroup alignment method
Message/Author
 emmanuel bofah posted on Monday, August 12, 2013 - 9:48 am
on page 2 Mplus LANGUAGE ADDENDUM 7.1 is said that the
The alignment optimization method consists of three steps:
1. Analysis of a configural model with the same number of factors and same pattern of zero factor loadings in all groups.
2. Alignment optimization of the measurement parameters, factor loadings and intercepts/thresholds according to a
simplicity criterion that favors few non-invariant measurement parameters.
3. Adjustment of the factor means and variances in line with the optimal alignment.
my question is why is that a model with ALIGNMENT = FREE (CONFIGURAL);
give the same estimates as ALIGNMENT = FREE; because i was thinking you need to be able to compare the configural and the scalar but their estimates are the same.
 Linda K. Muthen posted on Monday, August 12, 2013 - 9:51 am
The default is CONFIGURAL so the two specifications you show are the same.
 emmanuel bofah posted on Monday, August 12, 2013 - 10:49 am
why not possible to specify:
ALIGNMENT = FREE (METRIC);
ALIGNMENT = FREE (SCALAR);
How can i specific scalar model with the alignment.
 Linda K. Muthen posted on Monday, August 12, 2013 - 11:34 am
The alignment method avoids using a metric or scalar model. The definition of the alignment method is that it is based on the configural model.
 Peter Halpin posted on Friday, February 14, 2014 - 9:40 am
Hello,

Is alignment implemented for ordered categorical data?
 Bengt O. Muthen posted on Friday, February 14, 2014 - 11:16 am
No, not yet. Only binary.
 Tait Medina posted on Tuesday, February 25, 2014 - 6:19 am
Is it possible to test the equality of regression coefficients across groups under the alignment method? For example, I am interested in regressing the factor on age and testing whether the coefficient for age is different across groups?

Thank you!
 Bengt O. Muthen posted on Tuesday, February 25, 2014 - 12:15 pm
The alignment method currently doesn't handle covariates. But you can divide people into age groups and then you get a*b groups in the alignment run, where a is the number of age groups and b is the number of original groups.
 Tait Medina posted on Tuesday, February 25, 2014 - 3:11 pm

I have a question about the output from Alignment=Fixed. How are the estimates given under MODEL RESULTS connected to the numbers given under “Item Parameters In The Alignment Optimization Metric”? I have read Web Note 18, but am having a hard time mapping the output in these two sections onto the equations.

Also, the R2 measures given in Table 7 in Web Note 18 need to be calculated by hand using Eq 13 and 14. Correct?

Thank you.
 Tihomir Asparouhov posted on Wednesday, February 26, 2014 - 5:16 pm
The “Item Parameters In The Alignment Optimization Metric” section contains the alignment results in the metric in which the alignment optimization is performed, i.e., after all indicator variables are standardized and also under constraint (10) from web note 18, also including the factor mean fixed to 0 in the corresponding group. These parameters are scale reversed back to the original metric of the variables and also to factor variance fixed to 1 in the corresponding group if that is the requested parameterization. There is a way to get these parameters as your final model estimates. You will need to standardize your variables using the standardized option of the define command as well as the analysis option METRIC=PRODUCT;

R2 is computed with the upcoming Mplus 7.2 but you can compute it by hand as well.
 Tait Medina posted on Monday, April 07, 2014 - 1:21 pm
Thank you for your response. I have another question about the alignment approach.

I have noticed that when I use ALIGNMENT=FREE I receive a warning that I should switch to ALIGNMENT=FIXED and a reference group (or baseline group) is suggested. How is the suggestion for a baseline group determined? I have played around with using different baseline groups trying to get a feel for this new approach and have noticed that the choice of group impacts the results under the APPROXIMATE MEASUREMENT INVARIANCE (NONINVARIANCE) FOR GROUPS section. Could you provide a bit more insight into this? Thank you.
 Tihomir Asparouhov posted on Monday, April 07, 2014 - 2:50 pm
> I have noticed that when I use ALIGNMENT=FREE I receive a warning that I should switch to ALIGNMENT=FIXED and a reference group (or baseline group) is suggested. How is the suggestion for a baseline group determined?

It is the group with the smallest absolute factor mean value. Presumably fixing that parameter to 0 would lead to the smallest misspecification.

> I have played around with using different baseline groups trying to get a feel for this new approach and have noticed that the choice of group impacts the results under the APPROXIMATE MEASUREMENT INVARIANCE (NONINVARIANCE) FOR GROUPS section. Could you provide a bit more insight into this?

This is explained in Section 5.3 in
http://statmodel.com/examples/webnotes/webnote18.pdf
Fixing the factor mean to 0 in one group can lead to biased results if that mean is not 0.

You can also try using TOLERANCE=0.01. This option seems to yield more robust results and will be Mplus default in the upcoming Mplus 7.2
 Tait Medina posted on Tuesday, April 15, 2014 - 9:24 am
I am a little confused by this output (below). The p-values seem to suggest that the item intercept for CHILD is noninvariant in group 3 as compared to groups 1 and 2.

APPROXIMATE MEASUREMENT INVARIANCE (NONINVARIANCE) FOR GROUPS

Intercepts
NEIGHB 1 2 3
FRIEND 1 2 (3)
SOCIAL 1 2 (3)
WORK 1 2 3
MARRY 1 2 3
CHILD 1 2 3

Intercept for CHILD
Group Group Value Value Difference SE P-value
2 1 3.378 3.385 -0.006 0.027 0.817
3 1 3.501 3.385 0.117 0.039 0.003
3 2 3.501 3.378 0.123 0.042 0.003
Approximate Measurement Invariance Holds For Groups:
1 2 3
Weighted Average Value Across Invariant Groups: 3.430

Invariant Group Values, Difference to Average and Significance
Group Value Difference SE P-value
1 3.385 -0.046 0.019 0.019
2 3.378 -0.052 0.022 0.017
3 3.501 0.071 0.023 0.002
 Tihomir Asparouhov posted on Tuesday, April 15, 2014 - 10:05 am
The process is explained in Section 4
http://statmodel.com/examples/webnotes/webnote18.pdf
but to summarize the invariance is not determined by pairwise comparison but rather by this: compare group 3 against the average of group 1,2,3. Also due to multiple testing we use smaller p-value 0.001 as the cutoff value.
 Tait Medina posted on Tuesday, April 15, 2014 - 11:58 am
Is the pairwise comparison portion of the output related to the "first step" of the algorithm used to determine a starting set of invariant groups that is described in Section 4? "We conduct a pairwise test for each pair of groups and we "connect" two groups if the p-value obtained by the pairwise comparison test is bigger than 0.01." (pg. 15).

Finally, when dichotomous outcome variables are used, how are scale factors/residual variances handled? Are they fixed to 1 in all groups?

Thank you.
 Tihomir Asparouhov posted on Tuesday, April 15, 2014 - 3:06 pm
Yes on the first question.

The second question also yes - we use the theta parameterization where all residual variances are fixed to 1 during the configural model estimation. After that ... the alignment is done without any consideration for the residual variances, i.e., the alignment is for the intercepts and loadings only and it does not use residual variances in the computations.

I have to also correct my message from Feb 26. To get the
"Item Parameters In The Alignment Optimization Metric"
as your final parameter estimates you have to use a linear scale transformation for each indicator variable Y like this
define:Y=(Y-a)/b;
where a and b are obtained from the configural model estimates as follows

a=average Y intercept across the groups
 Tait Medina posted on Friday, April 18, 2014 - 12:51 pm
Thank you, that makes sense.

I have a follow-up question about Eq. 9 in Webnote 18. I am trying to make sure I understand Eq. 9 by plugging in the estimates taken from the output (using ML, Alignment=Fixed) using my own data. For now, I am using 2 groups. The loading for item 1 is .585 in group 1 and .588 in group 2. Taking the difference of these loadings gives me -.003. Scaling this by the CLF (using the small number .0001) gives me f(x)= .103. The Contribution to the Fit Function for this item, given in the output under Loadings, is -.316. I am not sure how to arrive at that number. The sample size for group 1 is 698 and for group 2 it is 949. The sqrt(N1*N2) is therefore 813.881. Weighting f(x) by 813.881 gives me 83.512. What am I misunderstanding about Eq. 9?
 Tihomir Asparouhov posted on Monday, April 21, 2014 - 8:38 am
The loss function that is reported in the output has a negative sign. See footnote 2 on page 10. You are also using 0.01 not 0.0001. Also the weight is standardized: scaled so the total weight is equal to the total number of cross group comparisons NG*(NG-1)/2 which is 1 in your case. So the actual weight that we use for the tech8 output is
w=((NG-1)*NG/2)*w0/sum(w0)
where
w0=sqrt(N1*N2)
and NG is the number of groups.
The weight standardization of course doesn't affect the optimization since it is a constant multiple. It is done so that all weights are 1 when the groups are of equal sizes. In your case the weight is 1 because there is just one cross group comparision. Thus the loss function for that loading is
-sqrt(sqrt(0.003^2+0.01))=-.316
 Tait Medina posted on Monday, April 28, 2014 - 6:54 am
Dr. Asparouhov, thank you so much for taking the time to address my questions. It has been tremendously helpful!

I have a general question about the Alignment approach. In many applications of multiple group factor analysis when the outcome variables are continuous, you will see a sequence of progressively more restrictive invariance tests performed, and distinction made between metric invariance (invariance of factor loadings) and scalar invariance (invariance of factor loadings, and item intercepts). It is only once metric invariance is found as a tenable hypothesis, that the hypothesis of scalar invariance is considered. In some ways the focus on testing for metric invariance first and then scalar invariance seems unnecessary to me when the goal is to compare factor means across groups. I wonder if this is perhaps a bit of a historical artifact stemming from the fact that EFA was based on correlation matrices, and then CFA expanded this to covariances matrices, and then Joreskog (1971) expanded CFA to multiple groups, and then Sorbom (1974) expanded multiple group CFA to include a mean structure. I am wondering if the Alignment approach makes this hierarchical distinction between metric and scalar invariance?
 Tihomir Asparouhov posted on Monday, April 28, 2014 - 7:39 pm
With alignment you can compute factor means and compare the factor means across groups even when full loading invariance is not fulfilled. As long as loading invariance and intercept invariance is violated to some minor extent factor means will be estimated well, see the simulation studies in Section 5
http://statmodel.com/examples/webnotes/webnote18.pdf
 Bengt O. Muthen posted on Tuesday, April 29, 2014 - 6:12 am
I agree that testing metric first is a bit of a historical artifact. Alignment does not make the distinction between metric and scalar invariance.
 Jessica Kay Flake posted on Wednesday, June 04, 2014 - 9:04 am
I know this has been asked before, but given there is a recent, new version out, I wanted to confirm. Can ordered categorical data be handled by the alignment in version 7.2?
 Tihomir Asparouhov posted on Wednesday, June 04, 2014 - 6:50 pm
Not yet. You can use BSEM with the Diff priors as an alternative.
 Joana posted on Thursday, July 03, 2014 - 7:33 am
Hello,

I’m testing measurement invariance of a scale in two groups because I need to compare means between both. For this and based on webonte 18 and on article from schoot et al (2013), I decided to use: alignment = fixed (bsem) approach.
But I have some difficulties to understand the output. More specifically on the following issues:

1 I got this message USE THE FBITERATIONS OPTION TO INCREASE THE NUMBER OF ITERATIONS BY A FACTOR OF AT LEAST TWO TO CHECK CONVERGENCE AND THAT THE PSR VALUE DOES NOT INCREASE. I added to the model (in 3 different times) FBITERATIONS= 1000, 5000, 20000 and I always get the same message. This means that the model doesn’t converge and I can’t continue with the invariance analysis?

2 The results from the alignment output indicate that the intercepts of 3 items (in 14) are variant between the 2 groups. How do I know the model has a good fit? Can I say that the measure is approximately invariant between the 2 groups? How can I calculate the factor scores to compare factor means between the 2 groups?

Thank you for help.

Best Regards

Joana Carvalho
 Bengt O. Muthen posted on Thursday, July 03, 2014 - 4:12 pm
1. That is an automatic message that always comes out and does not reflect on the quality of your run. You should check that the PSR is 1 for the different FBITER runs and if the results are approx the same.

2. When a minority of the measurement parameters are non-invariant the factor means and variances for the different groups are typically trustworthy. As our website handout for the May UCONN M3 workshop shows you can do a Monte Carlo study to check that the factor means and variances are dependable. No need to compute factor scores to compare the factor means.
 Joana posted on Wednesday, July 09, 2014 - 2:05 pm
Thanks very much for your help!

Before the means comparison I tried to study the quality of the alignment results and run a monte carlo simulation according to the following papers: IRT studies of many groups: the alignment method (version 2 - july 2014) and New Methods for the Study of Measurement Invariance with Many Group (october 2013), and I have one more doubt:

- After running the model I get the following warning message: "All variables are uncorrelated with all other variables within class" and I can't figure out what I did wrong on input specification... Is this the reason why I don't get the correlations results to evaluate the quality of alignment results?

Best regards
Joana

Here is an excerpt of the input:

Montecarlo: NAMES ARE mhc1-mhc14;
ngroups=2;
nobservations=2(2000);
Nreps=50;

ANALYSIS:
type= mixture;
ESTIMATOR=ml;
alignment=fixed(1);
processors=8;

Model population:

%OVERALL%

f1 BY mhc1-mhc3*1;
f2 BY mhc4-mhc9*1;
f3 BY mhc10-mhc14*1;

%g#1%

f1 BY mhc1*0.67816 ;
f1 BY mhc2*0.67552 ;
 Bengt O. Muthen posted on Wednesday, July 09, 2014 - 3:58 pm
Perhaps you are not specifying factor variances.
 Julia Higdon posted on Saturday, August 30, 2014 - 8:55 pm
I have used the alignment optimization method and want to use the results in a subsequent SEM analysis.

Is it possible in an alignment optimization analysis to save the factor scores and then use those factor scores in a subsequent analysis? As in:

Analysis:
Type = mixture;
estimator = bayes;
alignment = fixed(1);
thin=50;
fbiterations = (5000);

model: ...

savedata:
file is fscores.txt;
save = fscores(10);

Thank you
 Bengt O. Muthen posted on Sunday, August 31, 2014 - 10:58 am
Yes, this is possible.
 Tait Medina posted on Sunday, December 14, 2014 - 10:37 am
I have noticed that when I have few groups (<10) that ALIGNEMENT=FREE tends not to work and I have to move to ALIGNEMNENT=FIXED. However, when I have more groups (15 or more), ALIGNMENT=FREE does tend to work. Are there any characteristics of the data that you would say tend to support ALIGNMENT=FREE? Have you seen this in regards to increases in group number?
 Tihomir Asparouhov posted on Thursday, December 18, 2014 - 10:19 am
For ALIGNMENT=FREE to work well you need a certain level of non-invariance. The more non-invariance there is the better ALIGNMENT=FREE will be compared to ALIGNMENT=FIXED. The more groups you have the more likely it is that enough non-invariance will be accumulated to warrant ALIGNMENT=FREE.
 Stephus Daus posted on Friday, January 09, 2015 - 4:16 am
With 7.3 and the ordinal alignment method: I have two CFA factors specified, and in one of them there are a few (necessary) items with 3 categories, the rest have 4 categories. I notice that Mplus suddenly considers these 3-categorical items as 4-categorical items (with the last category empty), judging by the "Univariate proportions and counts" and the non-identification errors. Is this a bug/not yet implemented? Or is it yet impossible for this method? Any legitimate work-arounds? Collapsing all items is here problematic...
Cheers
 Linda K. Muthen posted on Friday, January 09, 2015 - 9:32 am
Please send the input, data, and output to support@statmodel.com.
 Tait Medina posted on Tuesday, February 10, 2015 - 8:30 am
Do you have any current recommendations for how to test if the factor variances estimated using the alignment method are significantly different across any two groups?

Thank you.
 Tihomir Asparouhov posted on Tuesday, February 10, 2015 - 12:18 pm
You can use MODEL TEST or MODEL CONSTRAINTS to look at the difference between the variances.
 Jessica Kay Flake posted on Tuesday, March 03, 2015 - 4:34 pm
Hello--
A few questions about ordered categorical alignment.

In the SEM paper you state the parameters are, by default, reported in a standardized metric in the MODEL RESULTS part of the output. Then, is it appropriate to interpret thresholds from a binary or ordered categorical alignment as zscore units and the loading as standardized loadings?

Related--Above Dr. Asparouhov said the theta parameterization was used for specifying the configural model, but not when minimizing the loss function with respect to the intercepts and loadings-- How does the theta parameterization in the earlier part of the estimation impact the results reported in the MODEL RESULTS section?
 Tihomir Asparouhov posted on Wednesday, March 04, 2015 - 10:00 am
The standardized results can be obtained using OUTPUT:STANDARDIZED command.

The theta parametrization is the only one available at this time with the ordered categorical variables. The MODEL RESULTS are reported in the theta parametrization.
 Natalia Dmitrieva posted on Wednesday, March 25, 2015 - 4:35 pm
I have a dataset with 835 participants and 45 continuous variables. Participants came from 7 different studies, which range from 40 to 346 participants. To examine invariance (in factor loadings, intercepts, etc) across the 7 samples, I am using multiple group factor analysis with alignment=fixed. My model terminates normally, however, I get the following warning messages:

WARNING: THE MODEL ESTIMATION HAS REACHED A SADDLE POINT OR A POINT WHERE THE
OBSERVED AND THE EXPECTED INFORMATION MATRICES DO NOT MATCH.
AN ADJUSTMENT TO THE ESTIMATION OF THE INFORMATION MATRIX HAS BEEN MADE.
THE CONDITION NUMBER IS -0.158D+02.
THE PROBLEM MAY ALSO BE RESOLVED BY DECREASING THE VALUE OF THE
MCONVERGENCE OR LOGCRITERION OPTIONS OR BY CHANGING THE STARTING VALUES
OR BY USING THE MLF ESTIMATOR.

THE STANDARD ERRORS OF THE MODEL PARAMETER ESTIMATES MAY NOT BE
TRUSTWORTHY FOR SOME PARAMETERS DUE TO A NON-POSITIVE DEFINITE
FIRST-ORDER DERIVATIVE PRODUCT MATRIX. THIS MAY BE DUE TO THE STARTING
VALUES BUT MAY ALSO BE AN INDICATION OF MODEL NONIDENTIFICATION. THE
CONDITION NUMBER IS -0.315D-17. PROBLEM INVOLVING PARAMETER 117.

NOTE THAT THE NUMBER OF PARAMETERS IS GREATER THAN THE SAMPLE SIZE.

An earlier reply stated that the first warning message may be disregarded. May the other 2 warning messages be disregarded as well?
 Bengt O. Muthen posted on Wednesday, March 25, 2015 - 4:44 pm
We need to see the full output - please send to support@statmodel.com along with your license number.
 Jessica Kay Flake posted on Friday, March 27, 2015 - 10:20 am
Hello Dr. Asparohouv,

I am doing a simulation study on the polytomous alignment for my dissertation. For my model population, I am specifying starting values that use results from a real, single group, data analysis. I want to make sure that there is not a mismatch between the metric of the results from the real data analysis and what I input for the alignment simulation. To ensure this, should I complete the real data analysis using PARAMETERIZATION=THETA, then input those results in my model population for simulating in the alignment framework? I am just trying to simulate data that are characteristic of the real factor analysis I did with one group. Previously I have completed a single group polytomous factor model with the default, DELTA. Does this create a mismatch? Or, in other words, when I input the results from the real single group, polyomous factor model using DELTA into model population using the alignment, is it interpreting those as being THETA parameterized starting values?

I apologize for my long question, I hope this is clear.
 Tihomir Asparouhov posted on Friday, March 27, 2015 - 4:45 pm
It is a mismatch. You should analyze the real data using PARAMETERIZATION=THETA and use those values in model population. Alternatively you can redo the real data analysis using the ML estimator (which is based on the theta parameterization as well).
 Jessica Kay Flake posted on Saturday, March 28, 2015 - 12:03 pm
Dr. Asparouhov,
Thank you for getting back to me so quickly. Very helpful to have confirmation on this issue, will rerun!
 Johannes Bauer posted on Wednesday, May 20, 2015 - 12:46 pm
I have two questions on the "FACTOR MEAN COMPARISON" part of the output. This output reports results using a 5% significance level.

1) I am wondering whether there is a way to apply a correction for multiple testing. Can e.g. the exact p-values be requested to apply Bonferroni correction?

2) Since the output is labeled "Groups With Significantly Smaller Factor Mean", is the applied test one-tailed or two-tailed?

Many thanks
Johannes
 Tihomir Asparouhov posted on Wednesday, May 20, 2015 - 5:07 pm
1) You can use code like this to get the exact p-value

model:
f by y1-y5;
%C2%
[f] (m2);
%C3%
[f] (m3);

model constraints:
new(md); md=m2-m3;

The p-value reported for md is the exact value.

2) It is two-tailed.
 Rafael Goldszmidt posted on Tuesday, May 26, 2015 - 2:41 pm
I am running a multigroup alignment analysis. Is it possible to obtain fit indexes (RMSEA, etc) for the final model?
 Tihomir Asparouhov posted on Tuesday, May 26, 2015 - 3:13 pm
You can get that using model=configural command. This is a sample input.

VARIABLE: NAMES=y1-y5;
ANALYSIS: MODEL = CONFIGURAL;
MODEL: f by y1-y5;
 Rafael Goldszmidt posted on Tuesday, June 02, 2015 - 5:09 am
Dear Prof Asparouhov,

Thanks for your fast reply! With this model, I get the fit indexes for the model with no equality constraints on loadings or intercepts.

The alignment model, on the other hand establishes some level of invariance. Would it be possible to get the fit indexes for the model considering the constraints defined by the alignment model?
 Tihomir Asparouhov posted on Tuesday, June 02, 2015 - 8:23 am
Rafael - the fit of the alignment model is the same as the fit of the configural model. You can see that they have the same log-likelihood value. There is no penalty for the level of invariance that the alignment provides - it is kind of the maximum invariance you can get for "free"/with no penalty in fit.
 Mircea Comsa posted on Thursday, September 24, 2015 - 11:41 pm
Hi,
Can I use alignment in conjunction with a bifactorial model? According to your paper (2014, MG factor analysis alignment) is not possible yet. There is an alternative? Thank you.
 Tihomir Asparouhov posted on Friday, September 25, 2015 - 9:03 am
It is not available yet.
 DC posted on Wednesday, October 07, 2015 - 1:14 pm
In running a model with the alignment method, I am getting the following message:
"THE CHI-SQUARE TEST CANNOT BE COMPUTED BECAUSE THE FREQUENCY TABLE FOR THE
LATENT CLASS INDICATOR MODEL PART IS TOO LARGE."

Could you please explain what this means?

Does this affect the interpretation of the results?

Is there a way to fix the issue and obtain the chi-square test?
 Tihomir Asparouhov posted on Wednesday, October 07, 2015 - 5:53 pm
The frequency table of the joint distribution of the categorical variables is too large and the Pearson chi-square
(https://en.wikipedia.org/wiki/Pearson%27s_chi-squared_test)
can not be computed. When the table is so large that goes over the computational limits the chi-square asymptotics break down anyway so I would not worry about it. With 30 binary variables for example the joint distribution of the binary variables has over a billion cells. So the chi-square will have over a billion degrees of freedom. This is not the chi-square that the WLSMV estimator computes.

There is no problem with the reported results or their interpretation.
 Alvin  posted on Sunday, October 11, 2015 - 3:54 pm
Hi Bengt, as mentioned previously, it is not yet possible to look at covariates using the alignment method. As I understand, it is possible to use factor scores derived from the analysis for further analysis, but does factor score indeterminancy impact on this type of analysis. And if so, how? Thanks
 Tihomir Asparouhov posted on Monday, October 12, 2015 - 8:01 pm
There is a paper by Herbet Marsh that I would recommend. Please contact him regarding the AwC (alignment with covariates) paper.

If you use the factor scores approach you should use the plausible values. See Section 4.2

 Alvin  posted on Monday, October 12, 2015 - 8:35 pm
Many thanks Tihomir. can you send me the link to the paper please? Best, Alvin
 Tihomir Asparouhov posted on Wednesday, October 14, 2015 - 3:32 pm
I don't think it is online yet. It is under review.
 Grace Icenogle posted on Monday, December 07, 2015 - 2:35 pm
Hello Drs. Muthen and Asparouhov,

I am hoping for some guidance in determining the quality of my results from an 11-group (11 countries) model (n=3000). I have a self-report measure with 6 binary indicators. I am struggling to integrate the results.

Under APPROXIMATE MEASUREMENT INVARIANCE, less than 25% of parameters are non-invariant:
Intercepts/Thresholds
I1\$1 (1) 3 5 6 7 8 9 12 13 16 17
I2\$1 1 3 5 6 (7) (8) (9) 12 13 16 17
I3\$1 (1) 3 5 6 7 8 9 12 13 16 17
I4\$1 (1) 3 5 6 7 8 9 12 (13) 16 17
I5\$1 1 3 5 6 7 (8) 9 12 13 16 17
I6\$1 1 3 5 6 7 8 9 12 13 16 17
I1 thru I6 indicate invariance for all groups.

However, the R-square values seem low compared to the example in Asparouhov & Muthen (2014).

Intercepts:
Fit R-Sq
I1 -31.66 0.08
I2 -35.41 0.18
I3 -31.20 0.00
I4 -43.43 0.07
I5 -38.88 0.27
I6 -49.92 0.00
Fit R-Sq
I1 -32.75 0.10
I2 -36.84 0.25
I3 -35.97 0.00
I4 -32.18 0.42
I5 -30.61 0.09
I6 -46.05 0.16

"Fit" = fit function contribution.

Could you assist in interpreting these findings (which seem contradictory)?

Thank you so much for your time.

Kindly,
Grace
 Tihomir Asparouhov posted on Monday, December 07, 2015 - 4:16 pm
You can compute the R2 by hand using the results in tech8. Most likely the issue is due to empty cells in certain groups or very small variation in the factor mean and variance across groups or very dis-balanced group design. If you still don't see why this happens send the example to support@statmodel.com.
 seulki jang posted on Friday, February 12, 2016 - 8:02 am
Hi Dr. Muthen and Dr. Asparouhov,

Hope you are having a good day. I have a question about the alignment optimization variance calculation. In the alignment optimization fit table, there are three indices (fit function contribution, r-squared, and variance). In the output, I see fit function contribution values and r-squred values, but not variance values. How do we calculate variances for each item factor loading and factor intercept?
Thank you!

Best regards,
Seulki
 Tihomir Asparouhov posted on Friday, February 12, 2016 - 11:11 pm
It is the sample variance for the aligned parameter across the groups. It is not in the output yet but can be computed manually.
 Marc Wigley posted on Thursday, April 14, 2016 - 12:11 pm
Hi
Running multiple group alignment for 27 countries I get the error
'The number of variable/value pairs do not match the number of classes
for class variable C'.

I've specified
classes = c(27);
knownclass =c(country);

Is that correct?

Many thanks
 Bengt O. Muthen posted on Thursday, April 14, 2016 - 6:16 pm
 aaron k Haslam posted on Tuesday, June 28, 2016 - 6:31 am
On the posting dated May 20, 2015 - 5:07 pm. Dr. Asparouhov, responded to a question regarding exact p-values. I would like to know how the standard errors are calculated for these pairwise comparisons?

Thank you,

Aaron
 Linda K. Muthen posted on Tuesday, June 28, 2016 - 1:46 pm
The standard errors are calculated using the Delta method.
 Jan-Benedict Steenkamp posted on Monday, August 01, 2016 - 2:18 pm
I am trying out the alignment method and am a little uncertain how to write the code. Could you make the code for the 26-country study published in Struct Eq Modeling available? That may be sufficient to figure out what I do wrong.

JB Steenkamp
 Bengt O. Muthen posted on Monday, August 01, 2016 - 4:22 pm
Will send them to you.
 John Gelissen posted on Thursday, December 22, 2016 - 3:40 am
I have run the alignment method on a set of 7 ordinal indicators (each having 4 categories) for 1 latent variable for 35 countries. When I look at the alignment output, I see, for example, that it is reported that approximate measurement invariance holds for the first item threshold for 34 countries, whereas approximate measurement invariance in this threshold does not hold for only one country. Upon inspection of the reported R-squared measurement invariance index, I find a value of only .019 for this threshold. I am trying to understand how the latter value can be so extremely low, given the other result. Any suggestion?

thanks, John
 Tihomir Asparouhov posted on Friday, December 23, 2016 - 9:55 am
There could be several different reasons.

1. The one threshold that is non-invariant is large (due to non-occurrence of a particular category in one group) and that accounts for the majority of the variability in the threshold.

2. The factor mean variability is small

4. It can also be a combination of the above and large standard errors that lean to not being able to establish significant non-invariance

It is quite straight forward to compute this by hand and figure out exactly why this happens: R2= (computation across groups)

--------------------------
 Alvin Tay posted on Friday, January 06, 2017 - 4:01 am
The residuals of the ordinal items in my 2-factor alignment model based on a sample of 8000+across 8 districts seem a bit odd. I understand that standardised residuals are at best indicative of model misspecifications but in the case of alignment analysis, what is the best approach to compare the fit of different models. Thanks Alvin
 Tihomir Asparouhov posted on Friday, January 06, 2017 - 5:39 pm
Alignment does not have an effect on fit. The fit of the alignment is the same as that of the configural model. Any model fit issue should be addressed prior to alignment by running a configural model and verifying that 2 factors are enough in each group.

Substantial model misfit can be addressed using an additional factor or by switching to Bayes where residual correlations for categorical variables can be included in the model.
 Jian-Bin Li posted on Monday, March 06, 2017 - 10:57 am
Hello,

As stated in the seminal paper about alignment analysis (Asparouhov & Muthen, 2014), this analysis starts from configural model. However, I am not sure if I need to test and report the model fit of each group as well as the configural model before using alignment analysis? If the model fit of the configural model doesn't go well, can I still use the alignment to compare the factor means across groups? Thank you.
 Bengt O. Muthen posted on Monday, March 06, 2017 - 5:51 pm
Q1. That would be good.

Q2. No; the configural model needs to fit ok.
 Jian-Bin Li posted on Thursday, March 23, 2017 - 7:15 am
Thank you Dr Muthen.

I have couple of follow-up questions:

(1) can version 7.3 deal with categorical data in the alignment analysis? I tried to run the model and there was no warning message.

(2) when I tried to run the model (i.e., a comparison on 21 items rated on a 5-point Likert scale across two countries), I treated the 21 items as categorical data. However the output stated that the estimator is MLR instead of WLSMV. Why's that?

(3) when reporting the results, do I need to report the (non)invariance of all thresholds of each item as follows?
LAY1\$1 1 2
LAY1\$2 1 2
LAY1\$3 1 2
LAY1\$4 1 2
LAY2\$1 (1) (2)
LAY2\$2 1 2
LAY2\$3 1 2
LAY2\$4 1 2

(4) my case concerns only two countries. That means if the loading or threshold shows non-invariance for one country, then it is also noninvariant for the other country. In this case, how to calculate the percentage of non-invariance? Take the 8 thresholds listed above as an example, do I calculate the percentage as 1/16 or 2/16?

I am Sorry for my long questions and hope I have made my points clearly.Thank you very much in advance.
 Jian-Bin Li posted on Friday, March 24, 2017 - 6:30 am
Thank you Prof. Muthen. I have two follow-up questions:

(1) I tried to compare the latent factor mean across 2 countries on v7.31(mac). The latent factor has 21 indicators(items) rated on a 5-point Likert scale. I treated all the items as categorical. The output showed the estimator is MLR instead of WLSMV which is used for categorical variable. I am wondering if I missed something in the syntax?

(2) Since the variables were treated categorical, each item has 4 thresholds. Does it mean that I need to report the (non-)invariance for all the 84 (4*21) thresholds?

Thank you.
 Tihomir Asparouhov posted on Monday, March 27, 2017 - 10:52 pm
(1) Yes
(2) see bottom of page 612 in the user's guide
(3) yes
(4) 2/16

(1) MLR is available for categorical
(2) yes
 Jian-Bin Li posted on Friday, March 31, 2017 - 4:58 am
Thank you. Now I am clear. Just one last question, I notice that a limit of 25% of non-invariance is a rough rule of thumb. I am wondering what if the number of non-invariance exceeds 25%? Is there any solution or guideline that help address this issue, or simply that the results should be abandoned? Thank you.
 Bengt O. Muthen posted on Saturday, April 01, 2017 - 4:39 pm
Alignment is probably better than alternatives also in that case. But you could do a Monte Carlo simulation study to find out.
 Philipp Sischka posted on Thursday, June 01, 2017 - 5:58 am
Hello,

I have a question regarding the FIXED alignment method.

I have run the alignment method on a set of 5 indicators (each having 6 categories) for 1 latent variable for 33 countries.

The initial FREE alignment model provide the warning that the model may be poorly identified and suggests to switch to the FIXED method with group 17 as baseline group.
But neither this group nor any other group has a mean close to 0 (as recommended in Asparouhov & Muthén, 2014). Group 17 in fact has a latent mean of -0.764.

My question is: Under this circumstances, does the FIXED method provide trustworthy parameter estimates?

Furthermore in the paper from Marsh et al. (2017) it is stated that
"For the present purposes we used the FIXED option available in the Mplus CFA-MI.AL model, in which the latent factor mean and variance of one arbitrarily selected group (in this case the first group, Australia) were fixed to 0 and 1, respectively".
However, as far as I understood, the choice of group selection is not arbitrary, but depends on the size of the latent mean which is closest to zero...

Could you provide a bit more insight into this? Thank you very much in advance.
 Tihomir Asparouhov posted on Thursday, June 01, 2017 - 6:29 pm
The recommended group is the one with mean closest to 0 by absolute value. I don't have any reason to doubt the conclusion of the estimation, i.e., that the FIXED method is better than the FREE method in overall terms of bias and standard error considerations.

The FREE method poor identifiability could be due to not enough noninvariance.
 Tino Nsenene posted on Wednesday, June 07, 2017 - 9:41 pm
Hello,

I am trying out a measurement invariance analysis for 20 groups with four continous indicators.

As one might expect, scalar invariance cannot be established. So I switched to the alignment method using ML.

My more general question is (approximately) how many parameters should be invariant in alignment analyses to consider the resultung factor means as trustworthy? I am bit scared that althought the alignment approach works straightforward, a large share of invariant parameters (say, 40%) certainly does not helpt to reach robust conclusions?