I am mot 100% sure how the CIs for random effects are computed---probably by exponentiating the ordinary CI on the logarithmic estimation metric---but because a random effect cannot be smaller than 0, 0 will never be in one. Two of the three are pretty small though how small depends on the scaling of the regressor.
That's why an explicit model comparison makes sense.
-----Original Message-----
From: "Vincenzo Carrieri" <[email protected]>
To: [email protected]
Sent: 2/26/2009 9:53 AM
Subject: Re: st: Re: xtmelogit post-estimation
Jay Verkuilen wrote:
This LR test is an omibus test of ALL random effects vs. none, i.e.,
ordinary logistic regression. Based on my cursory reading of what you
have here, you probably don't need quintp~1 and could probably do OK
without escluso but might want to keep it. This can be resolved with an
LR test between the random intercept model and the one with random
slopes, though because you are dealing with a parameter on the boundary
of the parameter space, you have to use the dreaded chibar distribution.
Also do you mean for these all to be independent? You may want to fit
several models and get AICs for each.
I'd suggest holding the fixed effects constant and explicitly compare:
Fixed effects only
Random intercept
Random intercept + slopes (independent)
Random intercept + slopes (dependent)
JV
1. How do you say that I don't need quintp~1? It seems that interval
confidence does not include 0.
2. There is a way to get per each region (my 2-level variables) the
estimated random coefficients and confidence interval. The predict
command gives me only the estimated parameters and estimated sd, but
not the confidence interval. Do you suggest something to obtain CI?
many thanks,
Vincenzo
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