We have $X_i \sim ^{iid} N(\mu,\sigma^2)$with known mean $\mu$, and unknown $\sigma^2$. Let's suppose we're given $s^2$ (sample variance) and $n$ (sample size).
We know $\frac{\bar X-\mu}{S/\sqrt{n-1}} \sim t(n-1)$.
Is there a way to compute $P(\bar X>c)$?
Now, I've seen several times the following reasoning:
$$P(\bar X>c)=P\left(\frac{\bar X-\mu}{S/\sqrt{n-1}}>\frac{c-\mu}{s/\sqrt{n-1}}\right)$$
However, I find this odd, instead I think we should have
$$P(\bar X>c)=P\left(\frac{\bar X-\mu}{S/\sqrt{n-1}}>\frac{c-\mu}{S/\sqrt{n-1}}\right)$$
because we're supposed to divide by the same expression on both sides of the inequality.
(Note: By $s$ I mean the r.v $S$ evaluated at a given sample)
The motivation for this question is the following: Imagine I wanted to test the null $\mu \leq \mu_0$. If we knew the population variance, then we would expect to reject the null if $\bar X > c'>\mu_0$. So,
$$P(\bar X>c')=P\left(\frac{\bar X-\mu_0}{\sigma/\sqrt{n-1}}>\frac{c'-\mu_0}{\sigma/\sqrt{n-1}}\right)$$
and we would chose $\frac{c'-\mu_0}{\sigma/\sqrt{n-1}}:= q_{1-\alpha}$, the $1-\alpha$ quantile for $N(0,1)$.
However, this type of reasoning cannot be done for the case at the begining of this question. We can prove that a similar rejection interval does indeed come out from using the Likelihood Ratio Test(LRT). I was wondering if there was a way to reach the same rejection interval without having to use the LRT...
Any help would be appreciated.