What is $n$ in the Sampling Distribution of the Sample mean My question is: what is the correct value to use for "$n$" in formulas pertaining to the mean and variance of the sampling distribution of the sample mean.
Let's say we are taking $25$ samples from a population $64$ times.
And we want to calculate $\mu_{\bar{X}}$
I would think we would estimate for each sample of $25$ elements $\bar{X}_k=\frac{\Sigma_{i=1}^{25}X_i}{25}$.
And then $\mu_{\bar{X}}=\frac{\Sigma_{k=1}^{64}\bar{X}_k}{64}$
Similarly, for the variance, $\sigma_{\bar{X}}= \sqrt{\frac{\Sigma_{k=1}^{64}(\bar{X}_k- \mu_{\bar{X}})^2}{64}}$
What makes me skeptical that this is correct is that all the formulas for these parameter estimates have an "$n$" which seems to refer to the sample size. And In the context of what I'm asking, is that $25$ or $64$.
In particular in the relation: $\sigma_{\bar{X}}=\frac{\sigma}{\sqrt{n}}$ is $n=25$ or $64$, or maybe even $25\times 64= 1600$.
Thanks
 A: First of all, you are right with you answer. I think your confusion Côme from the fact you have two distinct situation.
First situation: you take sample of 25 induviduals from a population. Here $n=25$ and you used it to find $\bar X_k$.
Second situation: you have 64 of these d'amples. Here $n=64$. You used it to evalutate $\mu_{\bar X}$.
For your standard deviation $\sigma_{\bar X}$, you are dealing with values from the second situation, so $n=64$.
It is always a lot of fun to deal with those multileveled questions. It is important to label your variable to distinguish them.
EDIT OP added

In particular relating the actual variance of the population to the
  sample error of the mean: $\sigma_{\bar{X}}=\frac{\sigma}{\sqrt{n}}$
  is $n=25$ or $64$?

Once again, it is a matter te define variable properly.
Level $0$: the population has a mean $\mu$ and a standard deviation of $\sigma$.
Level $1$: sample of size $25$. For each sample, you will have a mean of $\bar X_k$ and a standard deviation of $\sigma_k$ (the standard deviation of the sample).
The expected mean distribution has a mean of $E[\bar X_k]=\mu$ (same has population) and a standard deviation of $$\sqrt{\mathrm{Var}[\bar X_k]}=\frac{\sigma}{\sqrt{25}}$$
Level $2$: $64$ samples. You took $64$ means of samples evaluate the mean.
The expected mean of the value of the level $2$ mean is
$$E[\bar X]=E[\bar X_k]=\mu$$
And the variance will be
$$\sqrt{\mathrm{Var}[\bar X]}=\frac{\sqrt{\mathrm{Var}[\bar X_k]}}{\sqrt{64}}=\frac{\sigma}{\sqrt{25}\sqrt{64}}$$

TL;DR Now to answer about your 
$$\sigma_{\bar X}=\sqrt{\frac{\sum_{k=1}^{64}(\bar X_k -\mu_{\bar X})^2}{64}}$$
It is the standard deviation of your $64$ sample means, it refer to the expected distribution of the level $1$. So it should be around $\frac{\sigma}{\sqrt{25}}$. 
A: Be careful. (This is not quite an answer, because it's not clear what your exact question is.)
The number $n$ represents sample size, but your question involves samples of different sizes from different distributions. Specifically, you talk about


*

*Samples of size $25$ from a distribution.

*A sample of size $64$ from a different distribution (the distribution of sample means for samples of size $25$ taken from your first distribution).


To ask “What is $n$?” when you are talking about two distributions and two sample sizes is impossible to answer.
Always be careful with your language. Since $\mu$ represents the population mean, you can never “calculate $\mu_{\bar X}$” from sample data, you can only estimate it. I think you know this, but you still shouldn't say “calculate $\mu_{\bar X}$”
You also mention “the sample error of the mean.” I don't know what that is. It isn’t a clear description of a number. If you are extremely careful about language, things are sometimes clearer.
A: I think notation is key here so hopefully I've understood and not messed up myself.
Let $X$ be a random variable with $\mathbb{E}(X)=\mu$ and $Var(X)=\sigma^2$.
To deal with the sample of size 25:
Let $\bar{X}_{25}=\frac{X_1+...+X_{25}}{25}$  then $\mathbb{E}(\bar{X}_{25})=\mu$ and $Var(\bar{X}_{25})=\frac{\sigma^2}{25}$ 
Here each of the $X_i$ are identical and independently distributed random variables. For example $X_3$ is the 3rd observation in the sample of 25 etc. 
Note that each $X_i$ is a random variable that is just $X$ in disguise. Hence $\bar{X}_{25}$ is a random variable.
The expectation and variance of $\bar{X}_{25}$ can be easily found using the algebra rules for  expectation and variance.
Now for the second part. Repeating the sample 64 times.
Let $Y=\bar{X}_{25}$ then $Y$ is a random variable  with $\mathbb{E}(Y)=\mu$ and $Var(Y)=\frac{\sigma^2}{25}$
Let $\bar{Y}_{64}=\frac{Y_1+..+Y_{64}}{64}$ and so $\bar{Y}_{64}$ is a random variable in a similar way to before.  
We now have $\mathbb{E}(\bar{Y}_{64})=\mu$ and $Var(\bar{Y}_{64})=\frac{\sigma^2}{25\times64}$
This deals with repeating 64 times.
Hope this helps.
