Assume the population follows $\operatorname{Poi}(\lambda)$ distribution, I want to test $H_0: \lambda \geq \lambda_0$ versus $H_1:\lambda<\lambda_0$. The variable we usually use in this type of test is $$\frac{\bar X - \lambda}{\left(\frac{\lambda}{n}\right)^{1/2}}\sim^aN(0,1).$$
I would compute the p-value with the correction as such:
$$\text{p-value}\approx P\left(N(0,1) \leq \frac{\bar x - \lambda_0 + \frac{1}{2n}}{\left(\frac{\lambda_0}{n}\right)^{1/2}}\right)$$
This computation, we get it from $P(\sum X_i\leq\sum x_i)=P(\sum X_i\leq\sum x_i+1/2)$ and then subtracting and dividing by the obvious terms.
However, I think I get a bit confused when computing the rejection region. For dimension $\alpha$, the rejection region would be usually obtained from $\alpha=P(\sum X_i < c)$, where $c$ is a constant to be determined. From here we would proceed as $P(\sum X_i < c-1/2)$ giving $\alpha=P\left(\frac{\bar X - \lambda_0}{\left(\frac{\lambda_0}{n}\right)^{1/2}}\leq \frac{\bar x - \lambda_0-\frac{1}{2n}}{\left(\frac{\lambda_0}{n}\right)^{1/2}}\right)\approx P(N(0,1)\leq q_{\alpha})$
But now, it seems that the continuity correction has not changed anything, since I still get the same condition ($z_0<q_{\alpha}$) with or without... (although we get different constants $c$'s).
Thus, when computing with the p-value I my get it bigger (adding a $1/(2n)$ term) than $\alpha$, in which case I wouldn't reject, but I could still get $z_0<q_{\alpha}$ telling me to reject...
What am I doing wrong?