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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!! |
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| The graph plots the means. In your model with a covariate, the intercepts are estimated not the means. |
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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 |
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| 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. |
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| TJ posted on Wednesday, January 01, 2020 - 12:33 pm
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| 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? |
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| 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. |
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| TJ posted on Thursday, January 02, 2020 - 7:51 pm
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| Thanks. And how do we obtain the Intercept and Means for the non-reference group because the intercept and mean reflected in the output file are for the reference group. |
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Perhaps you are asking how you get the mean/intercept for the non-reference 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. |
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| TJ posted on Sunday, January 05, 2020 - 9:06 pm
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I dont think i have b in my model command infront of gender. I only want to find out the intercept for the non-reference 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? |
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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. |
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| TJ posted on Tuesday, January 07, 2020 - 4:30 pm
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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? |
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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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