Understanding formulation of hypotheses in difference between two sample means (z test) Introduction
When testing a sample against a population, one has one of the following sets of hypotheses:



*

*Two-tailed test: $H_0: \mu=\ldots$, i.e. "no difference", and $H_1: \mu\neq\ldots$

*Right-tailed test: $H_0: \mu\leq\ldots$, and $H_1: \mu>\ldots$

*Left-tailed test: $H_0: \mu\geq\ldots$, and $H_1: \mu<\ldots$

I know that some textbooks write $H_0: \mu=\ldots$, and $H_1: \mu>\ldots$ for one-tailed tests, i.e. leave out the other side. However, I always considered that bad practice.
Now suppose testing the difference between two sample means. Here, the formula states:
$z=\frac{(\bar{X}_1-\bar{X}_2)-(\mu_1-\mu2)}{\sqrt{\frac{\sigma_1^2}{n_1}+\frac{\sigma_2^2}{n_2}}}$
Then, it is assumed that $H_0:\mu_1=\mu_2$. Thus, we get:
$z=\frac{(\bar{X}_1-\bar{X}_2)}{\sqrt{\frac{\sigma_1^2}{n_1}+\frac{\sigma_2^2}{n_2}}}$
Problem
This is all fine to me when considering two-tailed tests. However, I encountered an exercise that asks whether there is a significant increase between two samples (sales of first and second year).
Now I face two questions:


*

*Isn't it correct to state $H_0: \mu_1\geq\mu_2$, and $H_1: \mu_1<\mu_2$ for a left-tailed test comparing two sample means? But then the $\mu$s would not sum to zero. I assume that this does not matter since if $\mu_1-\mu_2$ increases, in the formula, we would get a smaller $z$ which would lead us to rejecting $H_0$ with a higher probability anyway. Thus, every case is covered with $H_0:\mu_1=\mu_2$ already.

*Since the two samples have "equal rights", referring to the symmetry between the samples, can we formulate the hypotheses either way, i.e. measure an increase of one or, alternatively, swap $H_0$ and $H_1$ and switch the equality signs? This is, of course, a no-go for "normal" tests, but facing two sample means, it does not matter which way to look at them, or does it?
 A: 1) If you want your hypotheses to partition the universe, where do you stop? Imagine you have a dataset drawn from a normal distribution. You might consider the hypotheses:
$H_1$: $\mu>0$
$H_0$: $\mu\leq0$
$H_{-1}$: The data weren't Normal after all but followed some other distribution
$H_{-2}$: The data didn't even have to be real valued, we just happened to observe nothing but real values
...
In practice, if the sample mean of population 2 is lower than that of population 1 we will always be rejecting an alternative hypothesis of $\mu_2 > \mu_1$ whether the null hypothesis contained an equals or a $"\leq"$.
It is perfectly possible to perform a likelihood ratio test for an "$=$" hypothesis against a "$<$" hypothesis.
2) Yes, you can swap the signs but you don't swap which hypothesis is the null. The two formulations are equivalent:
$H_0: \mu_1 \leq \mu_2\\H_1: \mu_1 > \mu_2$
$H_0: \mu_2 \geq \mu_1\\H_1: \mu_2 < \mu_1$
The reason for this is that the datasets were symmetric in a sense, but the null hypothesis enjoys a special privilege (it is "innocent until proven guilty", or at least 95%-so.)
A: *

*Your assumption is correct and you explained it nicely yourself.

*If you swap the hypotheses, then you must keep in mind that level of significance $\alpha$ and 1-power will swap places.
