Interpretation of poisson regression using random effects

[Jessica Paiva]

I am working on a research that seeks to explain the influence of rains on the occurrence of traffic accidents. I need help to interpret the result displayed by STATA.

-----------------------------------------------------------------------------------------------------------------------------
acc | Coef. Std. Err. z P>|z| [95% Conf. Interval]
-------------+---------------------------------------------------------------------------------------------------------------
precipitacao | .0079662 .0005433 14.66 0.000 .0069015 .009031
tempo | 8.26e-06 9.97e-07 8.29 0.000 6.31e-06 .0000102
temposq | -1.39e-10 1.04e-11 -13.41 0.000 -1.60e-10 -1.19e-10
_cons | -1.539297 .4425398 -3.48 0.001 -2.406659 -.6719348
-------------+------------------------------------------------------------------------------------------------ ---------------
/lnalpha | 2.217552 .2220223 1.782396 2.652708
-------------+--------------------------------------------------------------------------------------------------------------
alpha | 9.18482 2.039235 5.944084 14.19242
------------------------------------------------------------------------------------------------------------------------------
Likelihood-ratio test of alpha=0: chibar2(01) = 4.4e+04 Prob>=chibar2 = 0.000


I'm totally lost in the interpretation. The variable depends on ac (accidents) and independent variables are "precipitacao" ("precipitation"), "tempo" (time) and "temposq" (time to square). The time variables were added because over time the accidents tend to increase.

What does / lnalpha and alpha mean?
Does the fact that the coefficient of time have an inverse relationship with the variable accidents proves that with the progression of time the accidents will decrease?
Please, somebody help me!

[Carlo Lazzaro]

Jessica:
-as this is not your first post, please use CODE delimiters (as per FAQ) to make the output of -xtpoisson- readable. Thanks.
-I fail to get -time- unit, as it seems really weird.
-Why creating squared term by hand when -fvvarlist- can do it for you?:

xtpoisson acc precipitacao c.tempo##c.tempo, re

-Have you investigated if your regression suffers from overdispersion?
- as per -xtpoisson- entry, Stata (not STATA, please) .pdf manual, page 340:

likelihood-ratio test of [alfa]= 0, [...] compares the panel estimator with the pooled (Poisson) estimator

[Clyde Schechter]

Alpha is a statistic that assesses the extent to which the use of random effects enhances the fit of your model (compared to just using pooled estimation with -poisson-.) In your case, it seems clear that there is enough within-panel clustering of your data that you do need the two-level model.

As for the negative coefficient of temposq, that tells you that the quadratic relationship between tempo and acc is an upside down (inverted-U) shaped relationship. So it says that as tempo increases from a very low value, the number of accidents will first increase, then reach a maximum value, and then further increases in tempo will be associated with a decrease in acc. From standard algebra of quadratics, the peak value of acc will occur when tempo = -coefficient of tempo/(2*coefficient of temposq), which in your case calculates to about 29700. So for values of tempo < 29,700 acc increases as tempo increases, but above 29,700 acc decreases as tempo increases farther. I don't know what values tempo takes on in your data. But if 29,700 is within the range of your observed values of data, you truly have an inverted-U relationship. If 29,700 is larger than any value of tempo in your data, then you just have acc increasing as tempo increases, albeit at a declining rate.

Really the best way to see this is to literally see it. So you should make some graphs of your results. Unfortunately, you didn't use factor-variable notation in your regression, so you can't do that with these results. Here's a better way. Pick some values of precipitacao and tempo that are fairly representative of your data. Let's say, for the sake of illustration, that precipitacao runs from 0 to 10 in your data, and that tempo runs from 0 to 30000.

xtpoisson acc precipitacao c.tempo##c.tempo, re
margins, at(tempo =(0(1000)30000) precipitacao(0(2)10)) predict(nu0)
marginsplot

One other thing, you might consider rescaling the variable tempo. The coefficients you are getting for it, and its square are very small numbers that are not easy to grasp intuitively. So consider scaling them down by, say, 1000, or 10000 to get coefficients that are more like "normal" numbers. Or, if, for example, tempo is currently measured in days, consider switching to years or even decades.

Added: Crossed with #2.