Inconsistency in two-sided hypothesis testing Suppose you have two sets of data with known population variances and want to test the null hypothesis that two means are equal, ie. $H_{0}: \mu_{1} = \mu_{2}$ against $H_{1}: \mu_{1} > \mu_{2}$. There's a certain way I want to think about it, which is the following:
\begin{align}
P(\mu_{1} > \mu_{2}) &= P(-(\mu_{1} - \mu_{2}) < 0) \\
&= P\left(\frac{\bar{x}_{1} - \bar{x}_{2} - (\mu_{1} - \mu_{2})}{\sigma_{\delta \bar{x}}}<\frac{\bar{x}_{1} - \bar{x}_{2}}{\sigma_{\delta \bar{x}}} \right) \\
&=P\left(z < \frac{\bar{x}_{1} - \bar{x}_{2}}{\sigma_{\delta \bar{x}}}  \right)
\end{align}
To me, this 'derivation' makes it perfectly clear what's actually going on. You're actually calculating the probability that $H_{1}$ is true and not just blindly looking up some $z$-score. However, now suppose that $H_{1}: \mu_{1} \neq \mu_{2}$. The problem with this is that the method I just described doesn't seem to work. If I write
$$
P(\mu_{1} \neq \mu_{2}) =  P(\mu_{1} < \mu_{2}) + P(\mu_{1} > \mu_{2})
$$
Then all that happens is $P(\mu_{1} \neq \mu_{2}) = 1$. I think I'm probably not interpreting the above equation correctly.

If is probably that $\mu_1<\mu_2$, then  $\mu_1\geq\mu_2$ is rejected, then not is possible that holds $\mu_1<\mu_2$ and $\mu_1>\mu_2$ simultaneusly.
