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First, I have read this post, this post and this post. All have very useful information. I have three other more specific questions.

I have estimated a negative binomial model using the glm.nb function of MASS and discovered the following parameters Theta: 9.0487, S.E: 0.444

  1. Is it correct to assume that dispersion parameter has a standard deviation of 20.38?
  2. Does this value correspond to the Poisson overdispersion that is corrected by the negative binomial model or is my model still overdispersed?
  3. Joseph Hilbe states in his book that R's glm.nb function employs an inverted relationship of the dispersion parameter, theta. Thus a Poisson model results when theta approaches infinity. Suppose now that my second glm.nb model had estimates of Theta: 19.0487, S.E: 0.444. Would this model be less overdispersed than the first model?
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    $\begingroup$ Can you explain where the value of 20.38 in part 1 is coming from? $\endgroup$ – Aniko Dec 16 '15 at 15:21
  • $\begingroup$ I am dividing the value of Theta with the standard error $\endgroup$ – user3218416 Dec 16 '15 at 18:45
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  1. Standard error is the standard deviation of an estimate of a parameter (see eg Wikipedia). So the standard deviation of the estimate $\hat\theta$ of the dispersion parameter $\theta$ is $0.444$. The ratio of an estimate to its standard error $\hat\theta/SE(\hat\theta)=20.38$ is often used as a test statistic for the null-hypothesis $\theta=0$. A large value here suggests that this null-hypothesis would probably be rejected. However,

    • as you noted, $\theta\rightarrow\infty$ corresponds to no overdispersion, so testing $\theta=0$ is not very meaningful

    • your third link discusses that testing for overdispersion has to be done carefully, especially in the context of the negative binomial distribution as the null-hypothesis is at the edge of the parameter space (whatever way it is parameterized).

  2. The function fits a negative binomial distribution, and $\theta$ is one of its parameters. The negative binomial distribution is always overdispersed compared to the Poisson distribution (unless $\theta=\infty$). Since by modifying $\theta$ for a fixed $\mu$ the negative binomial distribution can achieve any variance ($Var(NB) = \mu +\frac{\mu^2}{\theta}$), there is no such thing as "overdispersion relative to the negative binomial distribution". Of course, it is possible that the negative binomial does not provide a good fit to the data, but the concept of overdispersion does not apply.

  3. If the means are the same, then yes, if you have a larger $\theta$, then that model is less overdispersed compared to the Poisson distribution. If you change $\mu$ as well, then you have to think about quantifying overdispersion. For the Poisson distribution $E(X)=Var(X)$, but do you want to quantify deviations as $Var(X)/E(X)$? or $Var(X)-E(X)$? or some other way? Your question will have different answers for different ways to quantify overdispersion.

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