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Latent curve model on the basis of cl... |
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Drs. Muthen, My question concerns the manual 3-step estimation. Is it possible to put any model one desires into that third step, even a latent curve? Specifically, I’ve estimated a latent class model on the basis of some indicators (say, X1-X10). Then I want to see if the resultant latent classes are linked to different latent curves based on different variables (say, Z1-Z5). I have tried to run this in a way similar to Input File 16 in Web Note 15. So for instance, given that modal class is treated as a categorical variable N, the overall statement and the first class’ model look like this: %OVERALL% I S | Z1@0 Z2@1 Z3@2 Z4@3 Z5@4; S@0; %C#1% [N#1@3.789]; [N#2@-.546]; [N#3@1.345]; [I] (i1); [S] (s1); ...etc. I set the overall within-class variance of the slope to zero to ease estimation burden. The model estimated just fine, and I got results that make sense. My question, though: is this a tenable way of doing things? The classes do appear to change slightly, but I will be checking allocation consistency in the way outlined in the Web Note. Thank you very much! |
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This is fine. |
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Hi Dr. Muthen, Thanks so much for your response. I have (what I hope is) a quick follow-up question. When I ran the model as written above, allocation consistency from Step 1 to Step 3 wasn't great. However, I decided to try ALGORITHM = INTEGRATION and for some reason when I do that the class assignment barely changes at all from Step 1 to Step 3. It was pretty much perfect. However, it doesn't seem to say anything about using ALGORITHM = INTEGRATION in Web Note 15, so I was concerned that this wasn't the right way to go about it. Do you have any thoughts about why ALGORITHM = INTEGRATION changes things so much for the better? And, absent any red flags in the output, is it OK to proceed having used ALGORITHM = INTEGRATION? Thank you very much! |
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Hi Dr. Muthen, Sorry, I just realized that I made an error here -- ALGORITHM = INTEGRATION obviously doesn't make things better, since STARTS = 0. My mistake. I retract my most recent question. Thank you, and apologies. |
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