

Negative Mean for NonNegative Distal... 

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I'm attempting to use the (manual) BCH approach to fit a model for the mean of a (manifest) continuous variable (DTSCORE) as a function of a fourcategory latent category variable (C), first without adjustment (for other covariates) and then with adjustment for other covariates. The continuous variable has only nonnegative integer values (eg., 0, 1, . . . , 22). However, for the unadjusted model, the estimated mean for DTSCORE is negative for category C = 2. I was wondering how the mean can be negative? I'm using the following syntax for the unadjusted model: ANALYSIS: TYPE = COMPLEX MIXTURE; Starts = 0; Estimator= MLR; MODEL: %OVERALL% DTSCORE; %c#1% [DTSCORE] (mn1); %c#2% [DTSCORE] (mn2); %c#3% [DTSCORE] (mn3); %c#4% [DTSCORE] (mn4); MODEL TEST: mn1 = mn2; And here are the model results for Class 2, showing the negative mean (0.586): MODEL RESULTS TwoTailed Estimate S.E. Est./S.E. PValue Latent Class 2 Means DTSCORE 0.586 0.667 0.879 0.379 Variances DTSCORE 2.998 0.681 4.401 0.000 


To follow up, I'm guessing that the problem is that, in the BCH method, the weights can take on negative values if the entropy is not high, resulting in admissible estimates for the auxiliary model. However, I just want to make sure the problem is not due to incorrect syntax, or to my interpretation of results. 


It seems possible to get a negative estimated mean if your variable has a large percentage at 0. Your model assumes a normally distributed variable so the left tail may go into the negative. If you have a large percentage at 0 and integer outcomes you might want to declare your variable as a count variable. 


Thank you so much for the suggestion. I tried treating the distal outcome as a count variable by adding the following option under VARIABLE: COUNT is DTSCORE (nb); However, in the results, the means of DTSCORE are negative for three of the classes (classes 1 and 4, in addition to 2). 


The modeling of counts uses log(mean), so if you want to get back to means you have to exponentiate the value. See chapter 6 of our new book. 

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