xtlogit (random effects) interpretation help

[Jamal Nasir]

Hi all,

I hope you are all well. I had a quick question with regards to interpreting a xtlogit stata output (random effects logistic regression). In particular, interpreting the dummy variable outcome.

If someone could give an example of one year and the age variable that will be very helpful. (would it be easier to interpret using the OR function?)

For year the base group is 1

xtlogit quit i.year age if lastsmoke==1, i(pid) re


quit | Coef. Std. Err. z P>|z| [95% Conf. Interval]
-------------+----------------------------------------------------------------
year |
3 | -.0035738 .1019349 -0.04 0.972 -.2033625 .196215
4 | -.0384937 .1073351 -0.36 0.720 -.2488665 .1718792
5 | .0289347 .1076174 0.27 0.788 -.1819916 .2398609
6 | -.2405922 .1120875 -2.15 0.032 -.4602798 -.0209047
7 | -.2317875 .1060899 -2.18 0.029 -.4397198 -.0238552
8 | -.0694481 .1050903 -0.66 0.509 -.2754213 .136525
9 | .1774066 .0937444 1.89 0.058 -.0063291 .3611423
10 | -.0270623 .0978131 -0.28 0.782 -.2187725 .1646478
age | -.0001354 .0016902 -0.08 0.936 -.003448 .0031773


Regards

Jamal

[yasirzuberi]

If someone could give some light on this i'd much appreciate it also!

[Joseph Coveney]

How about the following?

For year, among respondents whose last smoke is coded as 1, the log odds ratio of quitting for year == 3 versus year == 1 (your baseline) is -0.0036 in a model that includes age as a continuous explanatory variable. Likewise, the log odds ratio of quitting for year == 10 versus year == 1 is -0.027 in the same model among the same respondents.

For age, among respondents whose last smoke is coded as 1, the log odds ratio of quitting for each increment of one year of age (quit of age versus quit of the previous year's age) is -0.00014 in a model that includes year as an unordered categorical explanatory variable.

[Amir Sariaslan]

Jamal,

You may benefit from reading the following paper:
Larsen, K., & Merlo, J. (2005). Appropriate assessment of neighborhood effects on individual health: integrating random and fixed effects in multilevel logistic regression. Am J Epidemiol, 161(1), 81-88.

[David Crow]

As Joseph said, the interpretation of the coefficients (i.e., the fixed effects part of the model) in a random effects logit model is exactly the same as for a normal logit model: the natural log of the odds ratio (LOR) for quitting in year==t vs. year==1, that is:

log[(quit==1 | year== t / quit==0 | year==t) / (quit==1 | year==1 / quit==0 | year==0)]

Exponentiating (with the "or" option) definitely helps interpretation. Now we have the simple odds ratio:

(quit==1 | year== t / quit==0 | year==t) /
(quit==1 | year==1 / quit==0 | year==0)

So, for example, exp(B_9) = exp(.1774066) = 1.19412, which means that the odds of quitting in year nine are nearly 20% higher than in year 1.

Two comments, though:

1) It's not clear why you're using a random effects logit model. Your output doesn't show the intraclass correlation coefficient ("rho"), and it's not clear to me how your data are structured. Can an individual quit smoking more than once, for example? Is an observation in year==t dependent on an observation in year==t-1?

One of the products of a random effects model is teasing out how much overall variation in the outcome of interest is due to within-subjects variability and how much to between-subjects variability (the intra-class correlation coefficient). Is this an important part of your research question?

2) It looks like what you're really modelling is duration data, or "time to failure"--that is, the time dependency of quitting smoking. In other words, is someone more or less likely to give up smoking as time goes on (along with the effects of other covariates). If this is the case, why not consider a proportion hazards (Cox) model ("stcox") or other type of survival model, such as a discrete-time hazard model?

Hope this helps,
David

[Marko Ledic]

t [/I]/ quit==0 | year==t) / (quit==1 | year==1 / quit==0 | year==0)]

There's a small typo I guess so the odds ratio for quitting should be: log[(quit==1 | year== t / quit==0 | year==t) / (quit==1 | year==1 / quit==0 | year==1)]

Best,
Marko