So I have a hypothesis test $H_0: \theta = \theta_0$, $H_1: \theta \neq \theta_0$ and I have setup the following test
$$0.05 = \alpha = \Pr_{\theta_0}(T \leq c)$$
for some statistic $T(X)\sim N(5,5)$. I have data that says that $T=9$
Now what I did was
$$0.05 = \alpha = \Pr_{\theta_0}\left(Z \leq \frac{c-5}{\sqrt 5}\right)$$ $$\frac{c-5}{\sqrt 5} = -1.645 \implies c = 1.32$$
Well according I do not reject the null hypothesis.
But what if I had taken the following test instead?
$$0.05 = \alpha = \Pr_{\theta_0}(T \geq c)$$
$$0.05 = \alpha = \Pr_{\theta_0}\left(Z \geq \frac{c-5}{\sqrt 5}\right)$$ $$\frac{c-5}{\sqrt 5} = 1.645 \implies c = 8.68$$
Well now I am supposed to reject the null hypothesis?
Why this ambiguity? It seems like I will reject or accept $H_0$ depending on which test I choose. Is this supposed to be the case? Are both approaches correct?
BTW: the way I came across this issue, was, I was looking at some examples and the authors would switch from $\leq$ to $\geq$ seemingly without consequence (they would change the constant) but it seems like the test depends on which inequality you take?!
The form of the rejection region should mirror the alternative hypothesis: reject if there is evidence in favor of the alternative. If the alternative hypothesis is really $H_1: \theta \ne \theta_0$, then both high and low values for $T$ are evidence that $\theta$ is different from $\theta_0$. So the rejection region should be two-sided: Reject if $T>c_{\text {high}}$ or if $T<c_{\text {low}}$.
The two rejection regions you've proposed are both wrong for the given alternative hypothesis; the first is appropriate if $H_1$ is $\theta<\theta_0$, while the second is appropriate if $H_1$ is $\theta>\theta_0$.