I have some data which contains the mean values from some distribution. Each value also has its standard error.

Now I want to re-average those values. Which it will be just the sum of the mean values divided by the number of them, but I wonder how to treat the standard error associated?

For example, if I have the following data:

$\bar{x} = 3.2, 3.1, 3.2, 3.0, 3.4, 3.1$

$\sigma_{\bar{x}} = 0.001, 0.002, 0.005, 0.001, 0.002, 0.001$

The new average will be 3.16, but what about the standard error of the mean?

I suppose I have to use the values given by $\sigma_{\bar{x}}$ rather than computing the standard deviation for the new average and divide it by $\sqrt{N}$, where $N$ is the length of $\bar{x}$, but I don't know how.


1 Answer 1


First, let's write this down theoretically. Suppose you have $N$ sample means, obtained from iid samples of size $n_1, n_2 ,\dots, n_N$, respectively. That is,

$$\bar{X}_1 = \dfrac{1}{n_1} \sum_j X_{1j}, \quad \text{ where } \quad X_{11}, X_{12}, \dots, X_{1n_1} \overset{iid}{\sim} F $$ $$\vdots $$ $$\bar{X}_N = \dfrac{1}{n_N} \sum_j X_{Nj}, \quad \text{ where } \quad X_{N1}, X_{N2}, \dots, X_{Nn_N} \overset{iid}{\sim} F $$

Let $F$ have mean $\mu$ and variance $\sigma^2$. Then by the CLT on each individual sample mean $\bar{X}_i$

$$\sqrt{n_{i}}(\bar{X}_i - \mu) \overset{d}{\to} N(0, \sigma^2) $$

Your values of $\sigma_{\bar{x}}=0.001,0.002,0.005,0.001,0.002,0.001$ are $$\dfrac{\sigma}{\sqrt{n_1}}, \dfrac{\sigma}{\sqrt{n_2}}, \dots, \dfrac{\sigma}{\sqrt{n_6}}\,. $$

Now, when combining all the means in the following way: $$\bar{\bar{X}} = \dfrac{\bar{X}_1 + \bar{X}_2 + \dots + \bar{X}_N}{N}\,, $$

each of the $\bar{X}_i$ are approximately Normally distributed (as long as $n_i$ are not too small). This this case

\begin{align*} \text{Var}(\bar{\bar{X}}) & = \text{Var} \left( \dfrac{\bar{X}_1 + \bar{X}_2 + \dots + \bar{X}_N}{N} \right) \\ & = \dfrac{1}{N^2}\sum_{i=1}^{N} \text{Var}(\bar{X}_i)\\ & = \dfrac{1}{N^2} \sum_{i=1}^{N} \dfrac{\sigma^2}{n_i} \,. \end{align*}

So, the overall standard error for the overall sample mean is $$\sigma_{\bar{\bar{X}}} = \sqrt{\dfrac{1}{N^2} \sum_{i=1}^{N}\sigma^2_{\bar{X}_i} } = \dfrac{\sqrt{\sum_{i=1}^{N}\sigma^2_{\bar{X}_i}}}{N}$$

In your case, it will be .001 I think.

An important assumption made in this theory is that all the samples obtained are independent and come from the same distribution.

  • $\begingroup$ Independence is important, but being equidistributed is not. The introduction of asymptotically Normal distributions is irrelevant. $\endgroup$
    – whuber
    Jan 11, 2021 at 17:18

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