1
$\begingroup$

I have a zero-inflated negative binomial model to a dataset (n = 47) with a over-dispersed dependent variable (mean = 6.70, sd = 17.68). The DV counts the number of fatalities in protest events, so the 0 value can be a result of the non-existence of protest, or of the lack of fatalities during the protests.

Thus, I estimate a ZINB model with the formula y ~ x1 + x2 + x3 + x4 | x1 + x2 + x3 + x5 + x6, and an offset.

The results in the inflation model show that no predictors are statistically significant (some of them are significant in the count model). Yet, a Vuong test between the ZINB model and a negative binomial (GLM with link negbin) model, with either all variables or only those in the count model, shows that the ZINB model is superior to the NB models. Neither the Log(theta) coefficient in the count part of the ZINB model or the (Intercept) in the inflation part are significant.

I don't understand this. If the inflation model is no significant, shouldn't a "plain" negative binomial model be better, or at least, indistinguishable? Is the ZINB model really a better fit?

PD. There are no important correlations between predictors that could alter significance values.

$\endgroup$
1
$\begingroup$

There is no contradiction here, at least from what you've said, althp

It appears that

  1. The NB model doesn't work well because of an excess of 0s
  2. The ZINB model worked better because it modeled these excess 0s
  3. That better modeling of the 0s reduced the significance of the predictors.

There are at least 2 reasons why the significance might go down: 1) The ZINB model has more parameters, so df is affected. 2) The significance in the NB model might be due to the violations of assumptions.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.