t-test: denominator as random variable? It is probably a simple question... But I am struggling with the following:
In the t-test, $\frac{\bar X - \mu_0}{SE}$, I understand that $\bar X$ is a random variable that represents the sampling distribution of the mean. When we usually see this formula, the denominator is not written as "$SE$" but as its estimated form $S/\sqrt{n}$. However, my understanding is that in this test, the ratio aims to "normalize" its distribution. Therefore, the denominator corresponds to the standard deviation of the sampling distribution of the mean, i.e. the standard error of the mean.
Is this correct? If yes, is the denominator also a random variable ?
Finally, I am asking this later question because if this is correct, then $SE$ should result from $\bar X$... which should not be considered as a random variable (well... I think this is the part where I am struggling the most).
I hope I made myself clear and thank you in advance your for help.

As you pointed out (and assuming that your random sample comes from a normal distribution), an estimator of the standard error $SE$ is $\sqrt{\frac{S^2}{n}}$. In the same way you see the $\bar{X}$ as a random variable that represents the sampling distribution of the mean, you can see $S^2$ as a random variable representing the the sampling distribution of the variance.
