Would it be more appropriate to use negative Binomial regression instead of Poisson regression if my sample variance is greater than my sample mean? My response variable $y_i$ denotes the number of articles produced by journalist $i$ in the  last two years. It is a count variable with fixed exposure hence why I chose to use Poisson regression. 
I have a dataset containing the number of articles produced by numerous different journalists and I also have values of other variables (my predictor variables). 
I have calculated the sample mean and sample variance of the variable $y$, and the sample variance is nearly double (over-dispersion). Is this a good enough reason for me to use negative Binomial regression? Should I consider anything else? The deviance of the model also drops by a noticeable amount when I use NB over Poisson.  
This is my first time considering using a negative Binomial regression so I apologise if the question is very basic. 
 A: Formally, i don't think you can go off your raw sample variance and mean, because the distribution is conditional i.e. $Y|X \thicksim P(\lambda)$
I can be totally wrong on the first point, so someone correct me. But generally as a rule of thumb test for overdispersion is to fit your model and divide the residual deviance by the residual degrees of freedom; if it's much larger than 1 then it's probably overdispersed. 
A more formal hypothesis test is to simply test if the negative binomial fits better. That is, the Poisson is the Negative Binomial when $k \rightarrow 0$. So, we are actually testing if 
$H_0: k = 0$ and $H_1: k > 0$ 
This can be tested using the log likelihood ratio test:
$LR=2(\ell_{NB}-\ell_p)$ which is distributed $\chi_p^2$ where $p$ is the number of parameters. Only tricky thing is the cutoff is at $\alpha=0.1$ instead of $0.05$ because the distribution of LR has a mass of $0.5$ at $k=0$ and a half-$\chi_p^2$ distribution above zero. Basically just fit two models and use the log likelihood reatio test to compare their deviances. 
In R, you can either compute it by hand using the logLik to pull out the loglikelihoods and substituting into the formmula and then looking up the chi squared criitical value at 0.1. Or, the pscl package has a odTest command to test this for you. See more here, but I haven't tried that one myself. 
