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 CMP posted on Friday, January 09, 2015 - 3:32 am
I am planning a study where I would use latent class profiling to identify different classes (preferably two) and include these classes as a covariate in a CFA with two factors (each having 3 indicators). I want to calculate the minimum required sample size using the not-close-fit hypothesis approach.
Question:
I would like to know if I should include the indicators of the latent class when counting my free parameters for determining the df of the model.
 Bengt O. Muthen posted on Friday, January 09, 2015 - 3:59 pm
Seems like you need to do a Monte Carlo study to get this power so I don't know what you would use the df for.
 CMP posted on Monday, January 12, 2015 - 1:16 pm
Thank you for your response.
Following your advice, I attempted to do a Monte Carlo study. However, I got error messages which I don't understand.
Below are the error messages received:


-------------

*** ERROR in MODEL command
One or more pairs of ordered thresholds are not increasing in Class 2.
Check your starting values. Problem with the following pairs:
Y3$2 (1.000) and Y3$3 (-0.120)
Y5$1 (0.307) and Y5$2 (0.000)
Y6$1 (0.307) and Y6$2 (0.150)
*** ERROR in MODEL POPULATION command
One or more pairs of ordered thresholds are not increasing in Class 1.
Check your population values. Problem with the following pairs:
Y3$1 (0.000) and Y3$2 (0.000)
Y3$2 (0.000) and Y3$3 (-0.250)
Y5$1 (0.000) and Y5$2 (0.000)
*** ERROR in MODEL POPULATION command
One or more pairs of ordered thresholds are not increasing in Class 2.
Check your population values. Problem with the following pairs:
Y3$1 (0.000) and Y3$2 (0.000)
Y3$2 (0.000) and Y3$3 (-0.120)
Y5$1 (0.000) and Y5$2 (0.000)
 CMP posted on Monday, January 12, 2015 - 1:17 pm
The syntax used was:
MONTECARLO:
NAMES ARE u y1-y14;
NOBSERVATIONS = 500;
NREPS = 500;
SEED = 4533;
GENERATE = y1 (1) y2 (1) y3(3) y4 (1) y5(2) y6 (2);
CATEGORICAL = y1-y6;
GENCLASSES = c (2);
CLASSES = c (2);
PATMISS = y7(.2) y8(.1) y9(.2) y10(.1) y11(.3) y12(.1) y13(.2)
y14(.1) y11| y13(.2) y14(.2);
PATPROBS = .4 | .6;
MODEL POPULATION:
%OVERALL%
f1 BY y7-y12*.6;
y13 ON u*.10;
y14 ON u*.15;
%c#1%
[y1$1*.26 y2$1*.-30 y3$3*-.25 y4$1*.26 y5$2*.-30 y6$2*.30];
%c#2%
[y1$1*.12 y2$1*.15 y3$3*-.12 y4$1*.13 y5$2*.-15 y6$2*.15];

ANALYSIS: TYPE = MIXTURE;
MODEL:
%OVERALL%
f1 BY y7-y12*.6;
y13 ON u*.10;
y14 ON u*.15;
%c#1%
[y1$1*.26 y2$1*.-30 y3$3*-.25 y4$1*.26 y5$2*.-30 y6$2*.30];
%c#2%
[y1$1*.12 y2$1*.15 y3$3*-.12 y4$1*.13 y5$2*.-15 y6$2*.15];
OUTPUT: TECH9;
 Bengt O. Muthen posted on Monday, January 12, 2015 - 5:19 pm
Please don't post in more than one window as requested.

When you have a variable such as y3 with 3 thresholds you need to give population values for all 3 thresholds, not just the highest one. And the values need to be ordered from low to high (so not the same).
 CMP posted on Wednesday, January 14, 2015 - 4:37 am
My sincere aplogies. Thank you for your response. I took it into account but got another error message:

*** FATAL ERROR
THE POPULATION COVARIANCE MATRIX THAT YOU GAVE
AS INPUT IS NOT POSITIVE DEFINITE AS IT SHOULD BE.

How can I address this problem? Thanking you in advance for your help.
 Linda K. Muthen posted on Wednesday, January 14, 2015 - 6:19 am
Check that you have given population parameter values for all parameters. If you don't, they are zero. Zero variances is one cause of this message.
 CMP posted on Wednesday, January 14, 2015 - 6:50 am
Thank so much for this helpful information. I can't identify where I got it wrong. Please could you point me to the omission? Many thanks!

MODEL POPULATION:
%OVERALL%
f1 BY y7@1 y8-y12*.6;
f1*.5;! f1 variance is .5
y8-y12*.4;
f1 WITH y13*.3;
f1 WITH y14*.3;
!y13 ON c*.10;
!y14 ON c*.15;
%c#1%
[y1$1*.26 y2$1*-.3 y3$1*-.25 y3$2*.5 y3$3*1 y4$1*.26 y5$1*-.3 y5$2*.8 y6$1*.3 y6$2*.5];

%c#2%
[y1$1*.26 y2$1*-.3 y3$1*-.25 y3$2*.5 y3$3*1 y4$1*.26 y5$1*-.3 y5$2*.8 y6$1*.3 y6$2*.5];
 Bengt O. Muthen posted on Wednesday, January 14, 2015 - 10:27 am
Please send input, output, and license number to Support.
 CMP posted on Wednesday, January 14, 2015 - 11:19 am
I will. I am so thankful for your help.
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