# Confidence intervals for averages of averages

Suppose we have an experiment involving $$N$$ independent samples of single variable functions, $$y_1(t),...,y_N(t)$$ where $$y_k(t) = \dfrac{1}{M}\sum_{j = 1}^{M} x_j(t); \ \ k = 1,...,N.$$ I am interested in the average over independent samples, $$\bar{y}(t) = \dfrac{1}{N} \sum_{k = 1}^N y_k(t)$$

How should I combine the uncertainty in the measurements $$x_j(t)$$, $$y_k(t)$$ and the final average $$\bar{y}(t)$$? For instance, if I produce a plot of each $$y_i(t)$$, I could draw error bars at each time point equal to the standard deviation $$\tilde{y}_i(t) = y_i(t) \pm \sqrt{\text{Var}[y(t)]}$$ to give some indication of uncertainty. The same could be done for $$\bar{y}$$, $$\hat{y}_i(t) = \bar{y}(t) \pm \sqrt{\text{Var}[\bar{y}(t)]}.$$ Clearly if for all $$t$$, $$\sqrt{\text{Var}[y(t)]} \le \sqrt{\text{Var}[\bar{y}(t)]},$$ we could just consider the error in the final average; under what conditions would that be true? If the $$x_j(t)$$'s are i.i.d with finite variance and the CLT applies, then $$\sqrt{\text{Var}[y(t)]} \sim \mathcal{O}(M^{-1/2})(?)$$ Then $$N \ll M$$ would justify us to indicate only the width (SD) of $$\bar{y}(t)$$ as a simple error estimate. I imagine this situation is very common, any standard texts or references discussing it? This post is similar, but I am hoping for a conceptual explanation.

If $$X$$ has standard deviation of $$\sigma$$, the distribution of the means of $$n_1$$ samples has standard deviation $$\dfrac{\sigma}{\sqrt{n_1}}$$.
Now you can apply the same idea. The means of $$n_2$$ samples of the distribution of the mean of $$n_1$$ samples will have a standard deviation of $$\dfrac{\sigma}{\sqrt{n_1} \cdot \sqrt{n_2}}$$.
• Thanks! This combines the sample sizes in the final result of SD in $\bar{y}$ but should there not be a $\sigma_1$ and $\sigma_2$ for each averaged quantity? To clarify my question, I would like to have some measure of the error of the $\bar{y}$ object which accounts for errors in the determination of the $y$'s. In your answer, the relationship between SD's of each object $\bar{y}$ and $y$ is independent of $\sigma$. Apr 29, 2020 at 9:06