How can I calculate uncertainty of the mean of a set of samples with different uncertainties? I have a set of independent measurements with associated measurement errors. However, each of the errors is substantially different. For example:
10 ± 0.8
12 ± 1.2
7 ± 0.5
8 ± 1.5
I want to calculate the mean of these samples along with the error of this mean. Calculating the mean is straightforward, but I'm having trouble figuring out how to properly propagate the measurement errors. Any clues?
 A: Using straightforward quadrature sum, the standard deviation (error) for the average, $\bar{x}$, of your 10, 12, 7, and 8 would be
\begin{equation}
\sigma_{\bar{x}} = \sqrt{0.8^2 + 1.2^2 + 0.5^2 + 1.5^2},
\end{equation}
so the result is $\bar{x} \pm \sigma_{\bar{x}}$. 
Calculating this out, the result is (7.1,11.4).
Another approach for propagation, called weighted means, would suggest that the weights are the inverse squares of the corresponding uncertainties, $w_j=1/\sigma_j^2$, and the best estimate, $x_{best}$, is 
\begin{equation}
x_{best} = \frac{\sum_{j=1}^p w_jx_j}{\sum_{j=1}^p w_j},
\end{equation}
and the overall uncertainty is
\begin{equation}
\sigma_{x,best} = \left(\sum_{j=1}^p w_j \right) ^{-1/2}.
\end{equation}
Upon substitution with results for $p=4$ measurements, we get 
\begin{align}
x_{best} &= \frac{1.56*10 + 0.694*12 + 4*7  + 0.444*8}{1.56 + 0.694 + 4 + 0.444} \\
     &= 8.28,\\
\end{align}
and the uncertainty is
\begin{align}
\sigma_{x,best} &= \frac{1}{\sqrt{1.56 + 0.694 + 4 + 0.444}}\\
                &=\frac{1}{\sqrt{6.7}}\\
                &=0.386,
\end{align}
yielding a result of 8.3(7.9,8.7).
For Monte Carlo simulation, since no distributional information is available, I would employ triangle ("lack-of-knowledge") distributions (5000 realizations, zero correlation).  The resulting mean $y$ is normally distributed, and the sigma (0.217) is smaller leading to (9,9.5), while normal distributions resulted in (8.7,9.8), where sigma was 0.531.  Below are the histograms for the triangle distribution and the normally-distributed mean distribution (Shapiro-Wilk $P$=0.2382).

A: Ideally, you need the original measures. For example, for your 7 +- 0.5, you should have something like. [6.5, 7, 7.5, 7 , 6.5, 7.5 etc] 
If you do not have them then try to "Unroll"  them yourself in a similar way, for example take 1.000 random samples with mean 7 and variance 0.5. 
Then you will have one  bigger sample, which can be analyzed, further. Find its distribution, get the mean and the variance, or do a One Sample T-test which is more statistically correct for the mean.
From this you can easily derive the range of uncertainty of the calculated sampled mean for a given level of uncertainty (usually you take the 0.95)
