This seems like it should be a pretty common problem. I have four estimates of fishing effort, each with its own variance. For subsequent calculations, I want the mean of the four estimates, and a variance that captures variability of the numbers being averaged, and also propagates the measurement error.
In general terms, I have a series of estimates $(X_1, X_2, \dots X_n)$, each with a variance $(\sigma^2_1, \sigma^2_2, \dots \sigma^2_n)$. I know how to calculate the overall mean, $\bar{X}$ (sum of the $X$s over n). I am unhappy with the formulas I have found for the variance of $\bar{X}$.
It seems that that variance can come in two forms.
First, there is the 'standard' variance that measures the spread of the observed values around the mean. i.e.,
$$ \text{Var}_1 = \frac{\sum((X_i - \bar{X})^2)}{(n-1)}$$
But this variance ignores the fact that each of the $X$ values was measured with error.
Second, I could propagate the error, and calculate the variance as:
$$\text{Var}_2 = \frac{1}{n^2} (\sigma^2_1 + \sigma^2_2 + ... + \sigma^2_n)$$
But this variance ignores the fact that each of the X values differed from each other.
I have also seen:
$$\text{variance} = \frac{\sigma^2}{n}$$
but that appears to be for the situation in which the original $X$s all have the same variance, which doesn't apply here. If anything, I am looking for a 'final variance' that is bigger than any of the original $\sigma^2_i$ values, and also maybe bigger than the thing I called '$\text{Var}_1$', above.
How can I calculate a variance that captures variability of the numbers being averaged, and also propagates the measurement error?