one sample t-test ${\rm H_0}:\mu=1$ vs ${\rm H_0}:\mu<1$ Suppose I want to test
\begin{align}
&{\rm H_0}: \mu = 1,\notag\\
&{\rm H_A}: \mu < 1,
\end{align}
and I get a p-value of 0.3, meaning I don't reject the null for the alternative.
In this post
One sided test $H_0:\mu=0$ or $H_0:\mu\leq 0$?
can write this test this way:
\begin{align}
{\rm H_0}: \mu < 1,\notag\\
\end{align}
I think the p-value for this test is just 1 minus the p-value of the above test, so I get a p-value of $0.7$. However I am a little confused about the interpretation. A p-value of 0.7 (>0.05) means that we don't reject the null. But the null here is $\mu < 1$. It seems to me that this contradicts my previous conclusion.
Can anyone help me explain this? Thanks!
 A: In this post, we'll write down rejection regions associated with both tests in OP's question. We'll show that the rejection regions for both tests do not partition space: it's possible that the realized data is in neither rejection region, leading to no rejections at all.
The model
Suppose that $X_i \stackrel{\textrm{ind}}{\sim} \mathcal{N}(\mu, 1)$ are independent draws from a normal distribution with common mean and variance. For convenience, we'll assume that the variance is known and so we're working with a $Z$ test rather than a $t$ test.
Getting the rejection regions
In this section, we will write down the rejection regions for both of the tests defined below. The displayed equations are the only important parts. Notationally, we use "primes" on all quantities associated with the second test.
For the first hypothesis test
$\begin{align*}
 H_0: \mu = 1 \\
 H_1: \mu < 1
\end{align*}$
an optimal rejection region is given by $R=\{\bar{X} < c\}$ for some rejection cutoff $c$. Whenever the sample mean is small, we reject the null hypothesis $H_0$ and conclude that $\mu < 1$. If we wish the rejection region $R$ to have an $\alpha$ type I error rate, then we can choose the cutoff so that: $$R_\alpha = \{\bar{X} < 1+\frac{\Phi^{-1}(\alpha)}{\sqrt{n}}\},$$ where $\Phi$ is the standard normal distribution function. Note, we found this rejection region $R_\alpha$ by solving for $\mathbb{P}_{\mu=0} \left[ \bar{X} \in R_\alpha \right] = \alpha.$
For the second hypothesis test
$\begin{align*}
 H_0': \mu < 1 \\
 H_1': \mu = 1
\end{align*}$
an optimal rejection region is given by $R'=\{\bar{X} > c'\}$ for some rejection cutoff $c'$. Whenever the sample mean is large, we reject the null hypothesis $H_0'$ and conclude that $\mu = 1$. If we wish the rejection region $R'$ to have an $\alpha$ type I error rate, then we can choose the cutoff so that: $${R'}_\alpha = \{\bar{X} > 1-\frac{\Phi^{-1}(\alpha)}{\sqrt{n}}\},$$ where $\Phi$ is the standard normal distribution function. Note, we found this rejection region ${R'}_\alpha$ by solving for $\sup_{\mu < 1} \mathbb{P}_{\mu} \left[ \bar{X} \in {R'}_\alpha \right] = \alpha.$
Interpreting the rejection regions
Here's where we are now: by writing down the rejection regions, we've found the sample means which would lead us to reject either test. Now we will interpret the tests for a particular choice of the sample size $n$ and the chosen Type I error rate $\alpha$.
Suppose that $\alpha = 0.05$ and $n = 100$. Then $$R_{0.05} = \{\bar{X} < 0.8355\}$$ and $${R'}_{0.05} = \{ \bar{X} > 1.1645 \}.$$ Therefore, when the sample mean is smaller than $0.8355$, we would reject $H_0$ and conclude that $H_1$ is true, i.e. $\mu < 1$. Further, when the sample mean is larger than $1.1645$, we would reject $H_0'$ and conclude that $H_1'$ is true, i.e. $\mu=1$.
What happens if $\bar{X}$ is not "small" or "large"? In this case, neither $H_0$ nor $H_0'$ would be rejected, and we cannot conclude anything from these particular tests. One way to make some conclusion would be making $\alpha$ larger until one of the tests reject---then concluding the alternative of that test. In some sense, this is choosing the "most likely" parameter region. This would correspond to choosing the alternative of the test with the smallest $p$ value.
