standard errors from standard error 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. 
 A: 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.
