Let $X_1,\ldots,X_n\sim\text{Poisson}(\lambda)$, $H_0:\lambda=\lambda_0$, and $H_1:\lambda\neq\lambda_0$ for $\lambda_0>0$. Compute the size $\alpha$ Wald test, estimating $\lambda$ by $\overline{X}_n$.
The size $\alpha$ Wald test rejects $H_0$ if and only if $\left|\tfrac{\hat{\lambda}-\lambda_0}{\operatorname{se}(\hat{\lambda})}\right| > \Phi^{-1}\left(1-\dfrac{\alpha}{2}\right)$. $\hat{\lambda}=\overline{X}_n$, so $\mathbb{V}(\hat{\lambda})=\tfrac{\mathbb{V}(X_i)}{n}=\tfrac{\lambda}{n}$, so $\operatorname{se}(\hat{\lambda})=\sqrt{\tfrac{\lambda}{n}}$, which we can estimate by $\sqrt{\tfrac{\hat{\lambda}}{n}}=\sqrt{\tfrac{\overline{X}_n}{n}}$. I notice in The Wald test with Poisson distribution the second answer uses $\sqrt{\tfrac{\lambda_0}{n}}$ instead, which seems fine because we're assuming $H_0$. Of course if we assume $H_0$ completely then we'd know $\lambda=\lambda_0$, but the second answer only assumes $H_0$ partially, if that makes sense. Are both of our methods valid?