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| Carlijn C posted on Tuesday, July 29, 2014 - 1:42 am
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I'm trying to conduct a two-part latent growth model with two groups, but I get an error. I used this syntax:
DATA TWOPART:
NAMES = WODCT1-WODCT3;
BINARY = bin1-bin3;
CONTINUOUS = cont1-cont3;
variable: NAMES ARE Group WODCT1 WODCT2 WODCT3;
USEVAR = cont1-cont3 bin1-bin3 ;
CATEGORICAL = bin1-bin3;
CLASSES = c (2);
KNOWNCLASS = c (Group);
ANALYSIS: ESTIMATOR = MLR;
ALGORITHM = INTEGRATION;
TYPE=MIXTURE
MODEL:
iu su | bin1@0 bin2@1 bin3@2;
iy sy | cont1@0 cont2@1 cont3@2;
su@0; iu WITH sy@0;
I get this error: ERROR in MODEL command
No OVERALL or class label for the following MODEL statement(s):
IU SU | BIN1@0 BIN2@1 BIN3@2;
What did I wrong? |
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| You are using TYPE=MIXTURE. You need %OVERALL% after MODEL: See the examples in Chapter 8 for correct syntax. Please note that you need a semicolon after TYPE=MIXTURE. |
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| Carlijn C posted on Tuesday, July 29, 2014 - 6:55 am
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| Thank you very much for your help! |
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| Carlijn C posted on Thursday, July 31, 2014 - 5:12 am
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I have another question. The model works, but I want to test the difference between the slope means of the two groups. I used this syntax:
MODEL:
%OVERALL%
iu su | bin1@0 bin2@1 bin3@2;
iy sy | cont1@0 cont2@1 cont3@2;
su@0; iu WITH sy@0;
sy@0;
%c#1%
[iu] (p1);
iu (p2);
[su] (p3);
[iy] (p4);
iy (p5);
[sy] (p6);
%c#2%
[iu] (p1);
iu (p2);
[su] (p3);
[iy] (p4);
iy (p5);
[sy] (p60);
Model test:
p6=p60;
However, I get this error:
ONE OR MORE PARAMETERS WERE FIXED TO AVOID SINGULARITY OF THE INFORMATION MATRIX. THE SINGULARITY IS MOST LIKELY BECAUSE THE MODEL IS NOT IDENTIFIED, OR BECAUSE OF EMPTY CELLS IN THE JOINT DISTRIBUTION OF THE CATEGORICAL VARIABLES IN THE MODEL.
THE FOLLOWING PARAMETERS WERE FIXED: 12.
What does this mean? |
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| The growth modeling parametrization for categorical outcomes holds thresholds equal across time and fixes the intercept growth factor mean at zero. |
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| Carlijn C posted on Thursday, July 31, 2014 - 12:07 pm
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I'm sorry, but I'm not sure if I understand it. Does the fixed parameter 12 means that the intercept growth factor mean is set at zero? The intercept regression weights on the three time points are set to one, so, automatically, there is no intercept growth factor? Why is it then fixed to zero?
Despite of this error, can I use this model for my data? I don't see any other options to improve the model.
I've tried this model with the same data, but now with some missingness. Then I get this error:
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 0.305D-10. PROBLEM INVOLVING PARAMETER 12.
Can I conclude that the model doesn't fit for my data with missingness? |
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| We can't know which parameter number 12 refers to without looking at your output, but it is likely the iu factor mean (check TECH1 or use version 7.2). You should fix this mean to zero because it cannot be identified at the same time as the threshold which is held equal across time. This fixing does not mean imposing a restriction on the data fitting. It does not mean that there is not an intercept growth factor; we just standardize its mean to zero so that we don't have an indeterminacy. To better understand this, see handouts and videos for Topics 3 and 4. |
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| Carlijn C posted on Thursday, July 31, 2014 - 2:01 pm
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Thank you for your answer. So should I fix the iu factor mean to zero? Or does Mplus this automatically (as it says in the error: 'the following parameters were fixed: 12')? However, the iu factor mean is estimated in the output.
If I fix the iu factor mean to zero [iu@0] in both groups, the problem is solved. So, is this the right solution for the problem? |
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| You should fix [iu] in only one class. |
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