Summing "black-box" confidence I have a table with observed and estimated values for political districts like this
BEZIRK  EDU_OBS EDU_WEI EDU_CI
101     157     5129    15

BEZIRK is an ID for the district, EDU_OBS is the number of actually observed people in a sample (lets say, inhabitants), EDU_WEI is the estimation based on some unknown, black-box, extrapolation, and EDU_CI is the 95% confidence interval in percentage of the estimated value, e.g. for district 101, the estimation would be 5129 $\pm$ 770 with 95% confidentiality. 
Now, let's say I have another table with the same districts but other values. 
BEZIRK  EDU_OBS EDU_WEI EDU_CI
101     180     5987    14

I would like to add the values (the estimates) of those with the first table, per district.
What happens to the CI? I.e. how do I recalculate the new CI of the combined values without knowing how it was computed in the first place? Can I just average it? I don't think so. 
 A: Based on the answer to my comment above, I think that there are two different parameters being estimated.  The first is $N_{CW}$, number of work-related commuters.  The second is $N_{CE}$, the number of education-related commuters.  These are estimated by estimators, $\hat{N}_{CW}, \hat{N}_{CE}$.  Also given are 95% confidence intervals for these two quantities: $\hat{N}_{CW} \pm W_{CW}$ and $\hat{N}_{CE} \pm W_{CE}$.  What is desired is an estimate and confidence interval for $N_{CW}+N_{CE}$.
Assuming that the given confidence intervals were calculated using the normal approximation, they were calculated as:
\begin{align}
\hat{N}_{CW} &\pm 1.96\sqrt{\hat{V}(\hat{N}_{CW})}\\
\hat{N}_{CE} &\pm 1.96\sqrt{\hat{V}(\hat{N}_{CE})}
\end{align}
So, it is straightforward to "back out," at least approximately, the variances of the two estimators:
\begin{align}
\hat{V}(\hat{N}_{CW}) &= (W_{CW}/1.96)^2 = 154,076\\
\hat{V}(\hat{N}_{CE}) &= (W_{CE}/1.96)^2 = 182,878
\end{align}
Now, again presuming that the normal approximation is appropriate, the 95% confidence interval for the sum, $N_{CW}+N_{CE}$ is:
\begin{align}
\hat{N}_{CW}+\hat{N}_{CE} &\pm 1.96 \sqrt{\hat{V}(\hat{N}_{CW}+\hat{N}_{CE})}\\
\hat{N}_{CW}+\hat{N}_{CE} &\pm 1.96 
    \sqrt{\hat{V}(\hat{N}_{CW})+\hat{V}(\hat{N}_{CE})+2\widehat{Cov}(\hat{N}_{CW},\hat{N}_{CE})}
\end{align}
We have everything to calculate this except the covariance.  I asked in the comment about how the two tables were related to try to get some idea about the covariance.  If the two quantities were estimated from two different samples, then it is very reasonable to assume that the covariance is zero.  I'm not sure that the answer was, in fact, that they were estimated from different samples, and I suspect they were estimated from the same sample.  However, let's first do the calculation, assuming that they were estimated from different samples, so that they have covariance of zero:
\begin{align}
\hat{N}_{CW}+\hat{N}_{CE} &\pm 1.96 
    \sqrt{\hat{V}(\hat{N}_{CW})+\hat{V}(\hat{N}_{CE})}\\
11,116 &\pm 1.96 \sqrt{154,076+182,878}\\
11,116 &\pm 1,138
\end{align}
Assuming that they were estimated on the same sample, then there is not enough information to calculate the confidence interval properly.  We must get information on the covariance in order to do the calculation properly.  We can do a worst-case scenario, however.  The widest the confidence interval could be is if the correlation between the two estimators were to be 1.  This would make the covariance equal to the geometric mean of the two variances, or 167,860.  Then we would calculate:
\begin{align}
\hat{N}_{CW}+\hat{N}_{CE} &\pm 1.96 
    \sqrt{\hat{V}(\hat{N}_{CW})+\hat{V}(\hat{N}_{CE})+2\widehat{Cov}(\hat{N}_{CW},\hat{N}_{CE})}\\
11,116 &\pm 1.96 \sqrt{154,076+182,878+2\cdot167,860}\\
11,116 &\pm 1,607
\end{align}
