Hypothesis testing: can we model the critical regions as pockets, rather than a designated "extreme region"? Say using the sampling distribution of the same mean, we standardize
and obtain the standard normal $N(0,1)$ value:
$$
Z=\frac{\bar{X}-\mu}{\sqrt{\frac{\sigma}{n}^{2}}}
$$
We wish to test the hypothesis $H_{0}\mu=a;$ $H_{1}:$$\mu>a.$ After
computing the statistic, we find the critical value corresponding
to a specific chosen significance level $\alpha$, $Z_{\alpha},$
such that if $Z>Z_{\alpha}$, we reject. One potential issue here
is that if we reduce $\alpha,$ we would increase $\beta$ (Type II
error). My question is this: $Z_{\alpha}$ here is computed as the
smallest value such $Z_{\alpha}$ such that
$$
\mathbb{P}\left(Z>Z_{\alpha}\right)=0.05
$$
In other words, we choose $Z_{\alpha}$ such that:
$$
\int_{Z_{\alpha}}^{\infty}f(Z,\mu_{0},\sigma_{0}^{2})dZ=0.05
$$
where $f\left(.\right)$ is the standard normal pdf, and $\mu_{0}$
and $\sigma_{0}^{2}$ are the distribution of the mean and variance
parameters under the null. Imagine an alternative definition,
$$
\text{New Critical Region=}\int_{a}^{x_{1}}f(Z,\mu,\sigma^{2})dZ+\int_{x_{2}}^{x_{3}}f(Z,\mu,\sigma^{2})dZ+...+\int_{x_{n-1}}^{x_{n}}f(Z,\mu,\sigma^{2})dZ=0.05
$$
where the $x_{1},x_{2},x_{3}$ each represent points on the real
line, greater than $\alpha,$ such that this new critical region can
be envisioned as the sum of separate ``pockets'' of critical regions.
The advantage, it seems to me, would perhaps be that we could increase
our power (decrease probability of Type II error), while maintaining
current significance levels. In other words, we could perhaps define
a constrained mimization problem, such that, for a given alternative
(say we have some knowledge a priori), we define
$$
\text{min}_{x_{1},...x_{n}}\text{Type II error }
$$
such that
$$
\int_{a}^{x_{1}}f(Z,\mu,\sigma^{2})dZ+\int_{x_{2}}^{x_{3}}f(Z,\mu,\sigma^{2})dZ+...+\int_{x_{n-1}}^{x_{n}}f(Z,\mu,\sigma^{2})dZ=0.05
$$
This seems “mechanical” and counter-intuitive, but I was wondering if similar tests have been proposed.
 A: The standard Gauss-test with the standard rejection region is a uniformly most powerful test, as was shown originally by Neyman and Pearson, see for example Lehmann and Romano's "Testing Statistical Hypotheses". So what you propose here will not increase the power, but in fact will make it worse, unless you choose your $x_1,\ldots,x_n$ in such a way that it's equivalent to the standard test, i.e. all your intervals together are just $[Z_\alpha,\infty]$ . For this reason such tests are not considered.
In other words, solving your minimisation problem will just reproduce the standard test (as long as you allow $x_n=\infty$).
Some words on why that holds: The likelihood ratio of the distribution of the test statistic between $\mu=a$ and $\mu=b$ for any $b>a$ is monotonic, meaning that generally the higher the value of the test statistic, the stronger evidence it provides against $\mu=a$. If there were a "hole" in the rejection region, this would mean to not reject $\mu=a$ at a certain larger value of the test statistic when it is rejected at a smaller value. This is inefficient trying to make the type II error probability as small as possible, because according to the likelihood (directly related to the type II error probability, which integrates the likelihoods) that larger value is more likely under the alternative compared to the null (quotient of likelihoods) than the smaller value.
