Why is it called the "sampling distribution of the mean"? Is there a good (or even a bad) reason why it's called the "sampling distribution of the mean" and not the "distribution of the sample mean"?
If we take multiple samples all of the same size, $n$, we get a distribution of sample means, $\bar{X}$. If I get this right, this is called the "sampling distribution of the mean". But that seems like an overly confusing name. I can be fussy about names of things. But then sometimes there's a good reason for a "bad" name. So why did we give a distribution of sample means this unwieldy name? Is it wrong to call it a "distribution of sample means"?
 A: In order to keep often-used terminology brief statisticians and textbook authors often resort to abbreviation. That works OK if
the abbreviations are more or less consistent and everyone
understands the reason for the abbreviation.
Suppose you're trying to estimate the mean of a normal
population with unknown $\mu$ and $\sigma.$ You take a random sample of size $n$ from the population to get a 95% confidence interval
for $\mu.$ [That should really be a 95% confidence for estimating $\mu,$ but if you're making a CI everyone knows estimation is
involved, so we often drop the word estimating.]
The confidence interval may be of the form $\bar X \pm t^*S/\sqrt{n},$ where $t^*$ cuts off probability from the
upper tail of Student's t distribution with $n-1$ degrees of
freedom. The distribution of $\bar X$ is $\mathsf{Norm}(\mu, \sigma/\sqrt{n}).$ This is called the sampling distribution of the (sample) mean.=--or, in context, just sampling distribution.
The full story is that we have taken
a random sample of size $n$ for the population so that
each observation is $X_i \sim \mathsf{Norm}(\mu,\sigma),$ and these observations are independent so that $\bar X \sim
\mathsf{Norm}(\mu,\sigma/\sqrt{n}).$ That is, $E(\bar X) = \mu$
and $SD(\bar X) = \sigma/\sqrt{n}.$ But people who know what
is going on are happy to refer to the sampling distribution of $\bar X.$
Also, we can say that $\sigma/\sqrt{n}$ is the standard deviation of $\bar X.$ We often say the standard error (of the mean) is $\sigma/\sqrt{n}.$ It is understood that 'standard error' is
used only for estimators.
Moreover, because $\sigma$ is unknown, we estimate it by the
sample standard deviation $S.$ Then the estimated standard deviation of $\bar X$ is $S/\sqrt{n},$ but we're happy to
shorten this to saying $S/\sqrt{n}$ is the standard error. Maybe it ought to be the estimated standard deviation of the sampling distribution of the sample mean, but the estimated is omitted because seeing $S$
we know it's estimated and we also say it's a standard error for short.
The terminology standard error may also be used for the estimated standard deviation $\sqrt{\frac{\hat p(1-\hat p)}{n}}$ for the point estimate $\hat p$ of a binomial proportion, which is sometimes used as the standard deviation
of a normal approximation to the distribution of $\hat p:$
a similar story, but with a few extra details.
Because of such abbreviated language your stat book may weigh
only 5 lbs instead of 7. Watch out for bold face type often
printed in blue ink. That's often code for the first use of
abbreviated terminology. Pay attention, you'll see it again.
