= Solution
The variance $\mu+\mu^2/\theta$ approaches the <Poisson distribution> variance $\mu$ when $\theta\to\infty$. Equivalently, with $\alpha=1/\theta\geq0$, the Poisson model is the boundary value $\alpha=0$. The usual <Wilks theorem> for a <likelihood-ratio test> assumes that the null parameter is an interior point of a smooth parameter space, so comparing the statistic with an ordinary $\chi_1^2$ law is invalid here.
A valid <parametric bootstrap> proceeds as follows. First fit the null Poisson model and retain its fitted means $\widehat\mu_i$. For each bootstrap repetition $b=1,\ldots,B$, independently simulate
$$
Y_i^{(b)}\sim\operatorname{Pois}(\widehat\mu_i),
$$
using the original doses, refit both the Poisson and negative-binomial models to that simulated data, and calculate
$$
T_b=2\left\{\ell_{NB}^{(b)}-\ell_P^{(b)}\right\}.
$$
For the observations calculate the analogous $T_{obs}$. The bootstrap $p$-value
$$
\widehat p=\frac{1+\sum_{b=1}^B\mathbf1_{\{T_b\geq T_{obs}\}}}{B+1}
$$
uses the null distribution with its boundary and finite-sample fitting behaviour automatically reproduced. A large $B$ controls the <Monte Carlo error> of this estimate.
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