I am new to statistics and got following confusion:
Suppose there is a population (with variance $ \sigma^2 $) and we draw 10 samples each of size 7 from this population.
For example: \begin{align*} S_1 &= \{2,3,4,5,6,3,5\}\\ S_2 &= \{1,3,4,5,6,4,5\}\\ &\ \ \vdots\\ S_{10} &= \{1,4,4,5,1,4,5\} \end{align*}
From these 10 samples, we get 10 means respectively. And if we draw the distribution out of these 10 means, it would be the "sampling distribution of sample means".
According to this post: https://math.stackexchange.com/a/2114833 , variance is given as $ \frac1 n\sigma^2 $
What would be the variance of the "sampling distribution of sample means" in this case?
$ \frac1 {10}\sigma^2 $ or $ \frac1 7\sigma^2 $ ?
Formula/proof given in the post: $ \text{var}\left(\sum X_i\right) =\sum\text{var}\left( X_i\right)= \sum \sigma^2 = n\sigma^2 $, the n looks to be the number of samples each having variance $ \sigma^2 $, if it is the case then it would be the number of samples 10 instead of 7. But in the following link, the sample size is the size of each sample: https://www.khanacademy.org/math/ap-statistics/sampling-distribution-ap/sampling-distribution-mean/v/sampling-distribution-of-the-sample-mean which is 7 in our case.
Edit 1:
I tried a simple example:
$ X_1 = {1,2,4,3} $
$ X_2 = {2,5,1,4} $
$ Var(\overline X) = Var(\frac{X_1+X_2}{2}) $
$ = Var(\frac{1+2}{2},\frac{1+5}{2},\frac{1+1}{2},\frac{1+4}{2},\frac{2+2}{2},\frac{2+5}{2},\frac{2+1}{2},\frac{2+4}{2},\frac{4+2}{2},\frac{4+5}{2},\frac{4+1}{2},\frac{4+4}{2},\frac{3+2}{2},\frac{3+5}{2},\frac{3+1}{2},\frac{3+4}{2}) $
$ = Var(1.5,3.0,1.0,2.5,2.0,3.5,1.5,3.0,3.0,4.5,2.5,4.0,2.5,4.0,2.0,3.5)$
$ = 0.9375 $
By the variance formula:
$ Var(\overline X) = Var(\frac{X_1+X_2}{2}) $
$ = Var(\frac{1}{2} (X_1+X_2)) $
$ = \frac{1}{4} Var(X_1+X_2) $
$ = \frac{1}{4} [Var(X_1) + Var(X_2)] $
$ = \frac{1}{4} [Var(1,2,4,3) + Var(2,5,1,4)] $
$ = \frac{1}{4} [1.25 + 2.5] $
$ = \frac{1}{4} (3.75) $
$ = 0.9375 $
Here we can see that n=2 (number of samples) gives proper variance and not n=4 (sample size).
Edit2: I think I got what I was missing. I think I was missing the basic understanding about the samples, i.e. each individual value in the sample is given by a random variable, therefore the sample size is decided by the no. of random variables used and thus if the random variables are equi-variant, having the same variance, the variance gets added n times, i.e. number of random variables or sample size. The sample size in Edit 1 is actually 2 (which I was mistaking to be 4).