The difference between $\frac{\bar X-\mu}{S´/\sqrt{n}}$ and $\frac{\bar X-\mu}{s´/\sqrt{n}}$ Imagine we're dealing with a normal population, with known mean $\mu$, but unknown variance $\sigma^2$.
We know that $\frac{\bar X-\mu}{S'/\sqrt{n}}\sim t(n-1)$. (T-Student distribution)
Now, suppose someone asks us to calculate $P(\bar X >c)$ and give us $s'=3$.
I've seen people write the following: $P(\bar X >c)=P(\frac{\bar X-\mu}{s'/\sqrt{n}}<\frac{c-\mu}{s'/\sqrt{n}})=P(T>\frac{c-\mu}{s'/\sqrt{n}})$, where $T\sim t(n-1)$. This sequence of equalities doesn't seem to be correct, since a draw of $S'$, i.e. $s'$, is not the same as the r.v.  $S'$. If it's correct, why is that?
Any help would be appreciated. 
 A: You are right, these people are mistaken. 
Let us try to follow their "reasoning". Since $T = \frac{\overline X - \mu}{S'/\sqrt{n}}$ has the $t(n-1)$ distribution, then for any $x\in \mathbb{R}$
$$
P_{\sigma^2}(T>x) =  P(t(n-1) >x).
$$
So in order to compute $P(\overline X> c)$, we plug in $x = \frac{c - \mu}{s'/\sqrt{n}}$ in there and get the answer. Right? Wrong!  
Essentially, this line of reasoning relies on equality 
$$
P_{\sigma^2}\left(T>\frac{c - \mu}{S'/\sqrt{n}}\right) = P\big(t(n-1) > x\big)\big|_{x=\frac{c - \mu}{S'/\sqrt{n}}},
$$
which is absurd (the left-hand side is non-random, while the right-hand side is random).

Here is a much simpler example, which I think is more illustrative. Let $X,Y$ be independent having the distribution $P(X= a) = 1-P(X=- a) = 0.1$ with unknown $a>0$. You observe $X=x$. What is the probability that $Y>0$? 
Using a similar "reasoning", you might say that, no matter what $a$ is, $Y/X= 1$ with probability $0.82$ and $-1$ with probability $0.18$. "Therefore", $P(Y>0) = 0.82$ when $x>0$ and $P(Y>0)=0.18$ when $x<0$. Right? No way! 
Of course, $P(Y>0) = 0.1$ always, independently of $x$.
