Let's say I have a sample mean $\bar X$ computed on $s$ random variables among $X_1, X_2,...,X_n$ where $n>s$:
\begin{align} \bar X = \frac{1}{s}\sum_{i=1}^{s}X_i \end{align}
For simplicity, let's assume the $X_i$ are i.i.d. $\text{Bernoulli}(p)$. My question is: what would be variance of $\bar X$?
By using the definition of variance, I have
\begin{align} \operatorname{Var}(\bar X) = \operatorname{Var} \left(\frac{1}{s}\sum_{i=1}^{s}X_i \right) = \frac{1}{s^2}\sum_{i=1}^{s}\operatorname{Var}(X_i) = \frac{1}{s^2}\sum_{i=1}^{s}p(1-p) = \frac{p(1-p)}{s} \end{align}
But this doesn't feel right to me since it ignores how many random variables there are, i.e. $n$.