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:
Then, it is assumed that $H_0:\mu_1=\mu_2$. Thus, we get:
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?