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Hello, I got the following problem with an LCGA (4 classes and one covariate): The output is saying that my slope for class 1 is negative (2.627), but the graph (estimated means)for this class is showing a positive slope. I don't understand how this can be? I am an absolute beginner working with Mplus and it would be wonderful if you could help me. Thank you very much!! 


The graph plots the means. In your model with a covariate, the intercepts are estimated not the means. 


Ok, thank you so much. Just to make sure I got it right: This output shows the intercepts (NOT the means as in a model without covariate): Intercepts I 22.007 2.332 9.436 0.000 S 2.421 0.781 3.101 0.002 And this one the effects for the regression of the intercept and slope on the covariate: I ON BAGE 0.009 0.019 0.466 0.641 S ON BAGE 0.027 0.008 3.446 0.001 


In a conditional model, a model with covariates, the intercepts of the intercept and slope growth factors are estimated. In an unconditional model, a model without covariates, the means of the intercept and slope growth factors are estimated. 

TJ posted on Wednesday, January 01, 2020  12:33 pm



I would like to clarify what is the function of "specify value for variable" when plotting graphs with covariates? I am trying to plot seperate graphs for Males and Females. I checked the User Guide but it does not give detailed explaination about the function of each button. Are there resources that can help me better understand what I should be entering into those boxes when trying to plot my graphs with covariates? 


If you have a variable scored 0/1 for male/female, you simply enter 0 when you want the male plot and 1 when you want the female plot. 

TJ posted on Thursday, January 02, 2020  7:51 pm



Thanks. And how do we obtain the Intercept and Means for the nonreference group because the intercept and mean reflected in the output file are for the reference group. 


Perhaps you are asking how you get the mean/intercept for the nonreference group. If so, you can use Model constraint to express that and get the estimate and its SE. For instance, in the regression Y = a + b*X + ... X where X is scored 0/1 as above, the intercept for females is expressed in Model Constraint as intfem = a + b; where a and b are parameter labels defined in the Model command. See the UG for more details. 

TJ posted on Sunday, January 05, 2020  9:06 pm



I dont think i have b in my model command infront of gender. I only want to find out the intercept for the nonreference group male. 0=female, 1=male Model: I S  get@1 get@2 get@3; S1@0; I S ON gender; Based on your response above, I tried adding a model constraint command: Model: I S  get@1 get@2 get@3; S1@0; I S ON gender; Model Constraint: new(intmale); I S on intmale; I was prompted with an error message saying that I need a parameter before intmale which I don't have one. does it have to do with me constraining my slope to 0? 


When you say e.g. I ON Gender; this estimates the linear regression I = a + b*Gender + e so you can give the label b as I ON Gender (b); and then use that b in Model Constraint. 

TJ posted on Tuesday, January 07, 2020  4:30 pm



I managed to generate an output with the following command: MODEL: I1 S1  get1@0 get2@1 get3@2; I1 S1 ON gender (b); MODEL CONSTRAINT: NEW (intfemale); intfemale = b; The output gave me the following: New/Additional Parameters INTFEMAL 0.054 0.566 0.096 0.924 I was given the estimate .054 and am wondering what is this estimate? 


Remember my earlier message where I said if you write the regression as Y = a + b*X + ... X where X is scored 0/1 as above, the intercept for females is expressed in Model Constraint as intfem = a + b; The label a in the Model command is given as the intercept [y] (a); You may want to study our RMA book where regression matters like this are discussed with Mplus inputs. 

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