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Hi there. I was told that one of the ways to compare nonnested models was to scrutinize the standardized matrix of the residuals....whichever one has the most offdiagonal entries larger than .1 is the poorer model. Well, I ran my models in Mplus and almost ALL the standardized residuals (found under Standardized Residuals (zscores) for Covariances/ Correlations/Residual Corr) for both models were above .1! Does this mean there is something wrong with both my models, or am I interpreting the matrix incorrectly? Or perhaps I've been given bad advice? Thank you very much, Andrea 


I have not heard of doing this. If I were looking at zscores, I would use a value of about 2. 


Thanks so much for getting back to me so quickly  and on a long weekend! What would you recommend I look at when comparing nonnested models....AIC/BIC and RMSEA? Thanks very much, Andrea 


If the nonnested models have the same set of observed variables, you can look at BIC. If they do not have the same set of observed variables, I know of no way to compare them. 


Yes, the models have the same set of observed variables. Thanks for your help! Andrea 


I have recently encountered a statistically minded person who insists that BIC can be used to compare nonnested models which are NOT based on the same set of observed variables (e.g., M1 has items 1,2,3,4,5 & M2 has items 4,5,6,7,8). Any literature which supports or refutes this assertion would be most helpful. 


sorry, in my just recent post, I should have asked "can BIC be used to compare nonnested models which are NOT *completely* based on the same set of observed variables, but have some overlap (e.g., M1 has items 1,2,3,4,5 & M2 has items 4,5,6,7,8)? 


I can't see how BIC can be used here. The log likelihood that the BIC is based on is in a different metric if the set of DVs is not the same. 


Hello. This means that there is no way of comparing two nonnested models with only overlapping observable variables? To give a concrete example: A mediation analysis with one indipendent variable X1, two mediators M1 M2 and one outcome variable Y1, and I would like to show that adding M1 to X1>Y1 and then adding M2 to X1>M1>Y1 does not "make the model worst" with the help of some indices. 


You need to have the same set of observed variables to compare models. Use the full set in both analyses and fix the paths you don't want to zero. 


Hello. I want to compare a 4 with a 3 factor model, such that the 3 factor model has one full factor (and corresponding indicators) removed. It appears that there is no test for this nonnested model, is that correct? Thanks 


No, not even BIC can be used due to having different variables in the two models. 


Regarding the question above, I understand that BIC can only be used to compare nonnested models with the same set of observed variables. Linda mentions the possibility of using the full set of observed variables in both analyses and fixing certain paths to zero. Could this be used in a CFA framework? If I want to compare two versions of a measure with overlapping items, but not all the same items, these models are clearly not nested. Could I work around this by fixing certain paths from my latent variable to my indicators to be zero? For example, In one model, factor 1 is defined as: f1 by y1y5; in the 2nd model factor 1 is defined as: f1 by y1y3; items 4,5 are not included. Would it be plausible to use f1 by y1@1 y2* y3* y4@0 y5@0; in order to compare one model to the other with the BIC? Of course this is a simplification of my real model. I am just wondering if this could work in theory and if there are any other implications of fixing those paths to zero that I am not considering. 


I don't think this will work. 


Hello. I have two latent growth models, one for changes in sibling conflict and one for changes in friendship conflict across 4 years in adolescence. The same scale was used for both measures (we simply changed the relationship they needed to report on). Both LGM show significant linear decreases in conflict over time. I am trying to find out if there is a way for me to test whether the slopes in these two models are significantly different. That is, is the decrease in sibling conflict steeper than the decrease in friendship conflict, or vice versa. Would you be able advise me on this matter? Thank you, Dayuma 


You can test this difference using MODEL TEST. See the user's guide. 


I ran a parallel process model so that I could use MODEL TEST, but the slopes were too highly correlated, so I constrained all the covariances between the LGMs to zero. Syntax as follows: MODEL: !Friend iF sF  fC_W1@0 fC_W2@1 fC_W3@2 fC_W4@3 fC_W5@4; fC_W1 fC_W2 fC_W3 fC_W4 fC_W5 (v1); iF sF; iF WITH sF; [fC_W1@0 fC_W2@0 fC_W3@0 fC_W4@0 fC_W5@0]; [iF] (m1); [sF] (m2); !Sib iS sS  sC_W1@0 sC_W2@1 sC_W3@2 sC_W4@3 sC_W5@4; sC_W1 sC_W2 sC_W3 sC_W4 sC_W5 (v4); iS sS; iS WITH sS; [sC_W1@0 sC_W2@0 sC_W3@0 sC_W4@0 sC_W5@0]; [iS] (m3); [sS] (m4); !Covariances iF WITH iS sS@0; sF WITH iS sS@0; MODEL TEST: m2 = m4; Is this appropriate? Thank you 


I do not think this is a good solution. Sometimes in a parallel process model, when growth factors are highly correlated, there is a need to correlate residuals at each time point across the processes. I would try that. 


Thank you for the quick response. 

Sara Geven posted on Wednesday, May 15, 2013  8:31 am



Hello, I am trying to compare a model without a mediator to a model in which a mediator is included. In this thread I read that the BIC can only be used when the observed variables are the same across the two models. Hence, Prof Muthen suggested to fix some paths in the analysis without the mediation to zero. However, when I did so, I saw that the RMSEA and the CFI also went down compared to a model in which I do not include the mediator at all (maybe because the intercepts and variances are now estimated for the mediator, but there are no predictors for the mediator in the model?). Is it ok to use the BIC of the model in which the paths are fixed to zero and to rely on the RMSEA and the CFI of the model in which the mediator and its paths are not included? Thank you in advance. Kind regards, Sara Geven 


I would use only BIC for model comparisons. 

Elli posted on Friday, March 18, 2016  11:02 am



Hello, When comparing these models, I get the same model fit criteria. A > B > C B > A > C However, I get different model fit criteria if I compare the following A>B>C A>C>B Does this seem correct? Thanks 


Please send this general analysis question to SEMNET. 


Hello, I would like to ask about a similar method I was experimenting with. For now, Im concerned with a simple regression. I also wanted to compare models with different sets of variables like this... Model: x1 ON y1; Model: x2 ON y1; Model: x3 ON y1; I tried the suggestion of including all x variables and allowing only one x to regress on y1 at a time, disallowing unwanted covariances, and comparing the BIC. The other method I experimented with consisted on creating duplicates of y with the DEFINE command to run the alternative models at the same time, disallowing all unwanted covariances and then using the model test command to compare the parameters of x on y. DEFINE: y2 = y1 +0; y3 = y1 +0; MODEL: x1 ON y1 (p1); x2 ON y2 (p2); x3 ON y3 (p3); x1 with y2y3@0 x2x3@0; etc... MODEL TEST: p1=p2; The estimates, standard errors and pvalues for x ON y that I got with this method were same as with the previous suggested method. Thus, instead of using BIC, I used the size of the estimates and the pvalues of the various wald tests to determine the best model. My question would be: Is there a problem with using this method or is it ok? I think that this might be a similar method to the case of Dauyma Vargas, but Im not familiar with growth models enough to say it is. Thanks in advance, Anton. 


I trust that when you say x ON y; you mean that x is the DV and y is the IV (I ask because the naming convention in Mplus is the reverse). I don't know why you don't simply say x1x3 on y1 (p1p3); (It doesn't matter if you add x1x3 with x1x3@0;) And then do the Model Test you mention. 


Yes, I see. Thank you very much! 

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