Message/Author 

anonymous posted on Monday, January 23, 2006  5:36 am



Hello, I'm working on a project where I have interactions. My interactions were generated using the "define" code. I would like to plot the mean, one sd above mean, and one sd below the mean. How may I do this most efficiently in Mplus? Thanks 


This cannot currently be done in Mplus. 


Dear Drs. Muthen I am trying to plot a latent interaction based on Mplus output. All the variables are continuous. By reading the Mplus discussion list, Mplus handout 3(slide 170), and your comment below, I am guessing that one can actually plot an interaction. But I don't understand which part of a Mplus interaction outout tells the intercept value ("a" as in Y = a + b1*x1 + b2*x2 + b3*x1x2). Do you have an example of plotting a latent interaction for continuous variables? Also, you mentioned in the message below that one can calculate low and high latent variable using factor mean and standard errors. How do you do that? Thank you in advance A post on April 2008 "concerning interpretation of latent interactions, it seems that following the Aiken and West approach is recommended...I would plot lines for equations with the path coefficients (b1, b2, and b3) for low and high values of the predictor and moderator, as factor means are 0. What is the best way to obtain high and low factor valuesis calculating the standard deviation from the variance obtained in the measurement model acceptable?" B. Muthen replied; "high and low factor values can be computed using the factor mean and variance estimates. Use e.g. + and  1 SD from the mean. For an example in a growth model context, see handout for Topic 2 on our web site, slides 125132." 


Please see the Topic 3 course handout starting on slide 162 and in particular slides 170171. 


Dear Drs.Muthen, Based on Mplus manual p.71 (latent interaction), I tried the Mplus input below. The output doesn't seem to produce the intercept value necessary to construct an interaction plot. I studied slides 162171 from topic 3 just like you said. I see that on page 169, you are using the intercept on S (0.417) into the unstandardized equation. But my output doesn't have the intercept on the latent criterion variables. I'm stuck and need help. Thank you in advance. ANALYSIS: TYPE = RANDOM; ALGORITHM=INTEGRATION model: AFFECT BY Aper1Aper3; JS BY JS1JS3; CWB BY CWB1CWB4; NA BY NAper1NAper3; IJ BY SIJ1SIJ6; AxIJ  AFFECT XWITH IJ; CWB ON AxIJ; JS ON AxIJ; JS ON AFFECT; CWB ON AFFECT; output: SAMPSTAT STANDARDIZED Tech1 TECH8; 


It sounds like you may be using an older version of Mplus where TYPE=MEANSTRUCTURE is not the default. 


Hello Drs. Muthen, Actually, I have the latest Mplus version, which is Mplus6. The user's guide that I refer to is supposed to be the latest version also. The only missing information for me to plot 2 latent interactions are an intercepts for a latent criterion (which is intercept ON JS and CWB in my case). Do you see any problem with my Mplus Input? Should I type in Type = MEANSTRUCTURE? Thank you for your help. 


Please send the full output and your license number to support@statmodel.com. 


I would also like to plot a latent factor interaction. I'm using the unconditional approach by Marsh so that I may fit a two group model by gender (apparently, not possible using XWITH). Unfortunately, I do not see an estimate for the intercept of the dependent latent factor. How do I obtain this estimate? 


An intercept for the dependent latent variable cannot be identified separately from the latent variable indicator intercept parameters. You can use XWITH with gender. 

Luisa Walls posted on Saturday, June 09, 2012  10:38 am



Dr Muthen, I am plotting a three way interaction. I have followed the procedure described by Aiken and West. However, I do not how to request or where to find the intercept of the equation. I have seen the Topic 3 course handout. In that example the intercept of the independent variable is provided, I have the same model commands but it is not showed in my output, only the intercepts of the residual variances. could you provide me guidance in how to request or how to find the intercept? 


I am confused by your questions. A dependent variable has an intercept. An independent variable does not. I don't know what you mean by the intercepts of the residual variances. 


Sorry my mistake. I need to know how to request the intercept of the dependent variable when I perform an Interaction effect (xwith) because the output does not show it. 


Please send the output and your license number to support@statmodel.com. 

Helen Zhao posted on Sunday, February 03, 2013  5:26 am



Hi, Linda, I read the handouts and all relevant threads but still have exactly the same problem as people above. I would like to plot interaction but couldn't find the intercept for my dependent variable. That is, under the section of "intercept" I can only find intercepts of IVs but not DVs. Can you help? Thanks! 


Latent variables in a single group analysis have means and intercepts of zero. 


Hello I ran a simple moderation model with X, M, and Y variables and 2 binary categorical covariates; X and M are also binary categorical variables. In the Mplus output, where can I find the estimated mean values of the Y variable at each level of the X and M variables? I will use these estimated values to plot the interaction effect in Excel. Under the Output command, I wrote: cinterval (bcbootstrap) STAND SAMPSTAT; Thank you. 


It's like regression of Y on binary M and binary x (regression on dummy variables). That gives you 1 intercept and 2 slopes (or 3 slopes if you interact X and M). This translates into the Y means as in regular regression with dummy variables (the Y mean for X=0, M=0 is the intercept, etc). 

Ye Liu posted on Friday, October 09, 2015  8:34 am



I am trying to incorporate an interaction in a SEM model between two latent exogenous variables by using the XWITH option in Mplus. Because the means of latent variables are 0, which means that there is only slopes and no intercepts between x and the y, I have learned above that y will have zero as their intercepts.How to plot the interaction effects like this ​in the case of structural equation modelling? 


See our FAQ Latent variable interaction LOOP plot. 


I’m trying to plot the quadratic effect of my (standardized) latent variable on continuoustime survival response using the code below with the LOOPcommand. The plot looks rather ok to me but I’m still wondering whether the code used is correct? MODEL: LAT = x1* x2 x3; LAT@1; LAT2  LAT XWITH LAT; Y = LAT (b1) LAT2 (b2); MODEL CONSTRAINT: PLOT (predY); LOOP (LAT,3,3,0.1); predY = EXP((b1*LAT)+b2*(LAT*LAT)); PLOT: TYPE =PLOT3; Thanks! 


The variable predY is the hazard ratio (not the expected value of Y). The effect of the covariates estimated by a proportional hazards model is reported as that hazard ratios. So the formulas are correct except in the model there is no = sign. Rather LAT by x1* x2 x3; Y on LAT (b1) LAT2 (b2); 


Thanks Tihomir. And yes, there was a typo in my code. 


If you would have a path model (only observed variables) with multiple dependent variables (y1, y2, and y3), and also multiple independent variables (a and b), and you would find a signficant interaction effect (e.g., a*b on y2 is signficant), would it be acceptable to plot this interaction using this equation? y2=intc + B*a + B*b + B*ab (where you get the intercept of y2 and the unstandardized regression coefficients from Mplus and you just fill in meaningful values for a and b (like 1 and 1 if these variables are standardized)) Or in other words, would it be acceptable to ignore the other variables in the path model and plot the interaction just like in a normal/simple regression analysis? 


Check our Mediation page on our website and the description of ex 318 plotting. 


I may be wrong but isn't the example about a model with only one dependent variable? I was wondering about plotting interactions in a model with more than one dependent variable? In such a model, do you plot an interaction on a specific DV like you do in this example thus ignoring the other DVs in the plot? 


Yes, you do one DV at a time. 


OK, thanks a lot! 

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