Intuition for inverting one sided hypothesis tests Is there an intuitive way to remember if we get an upper bound or lower bound for a parameter when inverting a one-sided hypothesis test? In Casella & Berger, it is written:
If we have $H_0: \mu = \mu_0, H_1: \mu < \mu_0$ then "$H_1$ specifies 'large' values of $\mu_0$, so the confidence set will contain 'small' values." I don't really understand this. I interpret $H_1$ as saying that the real value of $\mu$ is smaller than $\mu_0$, I don't really interpret it as saying that $\mu_0$ is large. Can anyone help me build intuition on this?

The authors are not ambivalent here.
If $\mathcal H_1: \mu<\mu_0, $ what should the test function look like? It must reject the null hypothesis $\mathcal H_0: \mu=\mu_0 $ if $\bar x$ of a certain $\mathbf X$ falls shorter than $\mu_0$ by some specific value ie. $$\varphi(\mathbf X) =\begin{cases}1,&\bar x<\mu_0-\textrm{value}\\0,&\textrm{otherwise}\end{cases}\tag 1\label 1.$$ Here, in $\eqref{1}, $ that $\textrm{value}=t_{n-1, ~\alpha}\frac{S}{\sqrt n}.$
So, what should the acceptance region look like? It's simple.
$$A(\mu_0) =\left\{\mathbf X: \bar x \geq \mu_0-t_{n-1, ~\alpha}\frac{S}{\sqrt n}\right\}.\tag 2\label 2$$ It admittedly provides an upper bound when inverting in that
$$C(\mathbf X) =\left\{\mu_0: \bar x +t_{n-1, ~\alpha}\frac{S}{\sqrt n}\geq \mu_0\right\};\tag 3$$ in this sense $\mu_0$ will be "small": smaller than a bound.
