Hypothesis testing: Two-tailed tests and deciding $H_0$ I'm kind of studying Hypothesis testing by myself and was looking for a clarification on the next topic:
In order to confirm my $H_1$ I basically need to reject $H_0$.
What do I do in case I cas assume either way?
Let's assume there's a packing maching in a factory and I need to decide whether it's well-calibrated so the packings have the average weight of: $\mu_x=80g$ and suppose I also have the S.D. $\sigma_x=2g$.
Given a certain $\alpha$, let's suppose $\alpha = 0.4$. since it's a two-tailed test I need to devide 0.2 worth of area in either side.
My question is: Do I need to hypothesize my assumptions in such a way that the area of rejection is always on the edges of the Gaussian curve?
Let us suppose: $$H_0: \mu_x=0.8$$ $$H_1: \mu_x\neq 0.8$$ Then the rejection area is necessarily on the edges. But since im not asked to assume anything about this machine I can assume in advance it's not calibrated and then: $$H_0: \mu_x\not=0.8$$ $$H_1: \mu_x= 0.8$$
And now I supposedly have a limited rejection area symetrically around $\mu_x$.
I tryed to look around for a thumb-rule. All of the examples I've seen used the first method I described but never justified it or explained why. Is it a thumb-rule that the rejection area has to be infinite, or under the circumstances of the problem the latter way can be equally fine?
If there is just one right way I'd like to know.

 A: 
Do I need to hypothesize my assumptions in such a way that the area of rejection is always on the edges of the Gaussian curve?

Well, this is because of the definition of $\alpha$, which is the level of significance, which is defined as follows: 
$$\alpha = \Pr(\mathcal{H}_1 \mid \mathcal{H}_0)$$
which reads as follows: Decide $\mathcal{H}_1$ is true, given that, in fact, $\mathcal{H}_0$ is true. In your case, we can write (in a very sloppy way)
$$\alpha = \Pr(\hat{\mu}_x \neq 0.8 \mid \mu_x = 0.8)$$
which means we have decided that our estimator $\hat{\mu}_x$ is not $0.8$, when the true one is. The estimator is random, which is modelled in this test as Normal. The reason I say sloppy is because the above happens with probability $1$. To avoid this, we build an interval
$$\alpha = \Pr(\hat{\mu}_x \notin  [\mu_x \pm K] \mid \mu_x = 0.8)$$
where  the interval is the complement of the so called "confidence interval" (corresponds to the shaded part in the left figure) Said differently, $[\mu_x \pm K]$ corresponds to the interval to the right side. Note that I say "corresponds" as the areas are the same and not the interval values, because $[\mu_x \pm K]$  is not necessarily centered. Usually, you will find in the literature a centered version of $[\mu_x \pm K]$, which makes it easier to read off z-tables, and so.
A: The basic concept of a null hypothesis is that you're seeing what the probability of seeing the observed evidence given the null hypothesis is. Thus, the point of the null hypothesis is to choose a hypothesis such that the probability is well defined. If you choose as your null hypothesis that weights are normally distributed with a mean of 80$^{[1]}$ and a sd of 2, then the probability distribution under this hypothesis is well defined. If you choose your null hypothesis as being "the weights are normally distributed with a sd of 2, and the mean isn't 80", then how do you calculate probabilities? "Probability of seeing such-and-such given that the mean isn't 80" is not well defined.
[1] You have 80 at one point and .8 at another, so that's a bit confusing.
