# Hypothesis testing: decision is arbitrary?!

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.

$$0.05 = \alpha = \Pr_{\theta_0}(T \geq c)$$

$$0.05 = \alpha = \Pr_{\theta_0}\left(Z \geq \frac{c-5}{\sqrt 5}\right)$$ $$\frac{c-5}{\sqrt 5} = 1.645 \implies c = 8.68$$

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?!

• Please explain the used tests! I do not see why we do not reject in example $1$, but reject in example $2$. What is the critical value in both cases ? – Peter Oct 12 '17 at 20:57
• @Peter case1 $T = 9 > c = 1.32$ so it is outside the critical region $T\leq c$ ........ case2 $T = 9 > c = 8.68$ so it is inside the critical region $T \geq c$ so we reject H_0 in this case but not in the first case. It seems like the the critical region is defined arbitrary but it affects the decision you then end up making?! – MortyPython Oct 12 '17 at 21:01
• Personally I would have asked this question at cross validated. – clark Oct 12 '17 at 21:05
• @Peter Sorry just a typo – MortyPython Oct 12 '17 at 21:06

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 $\theta>\theta_0$.
• so we don't have a rejection region $T\in[0,c]$ but rather $T\notin [c_1, c_2]$ ?? Not sure what the correct formulation of the rejection region is in this case?! – MortyPython Oct 12 '17 at 21:28
• @MortyPython The rejection region should be $T\not\in [c_1,c_2]$, signifying either high or low values of $T$. More precisely, it has the form $|T-\theta_0|>c$. Standardizing, this has the form $|T-5|/\sqrt 5 > c/\sqrt 5$. – grand_chat Oct 12 '17 at 21:29