Two neighborhoods, unequal populations
In a hypothetical example, 20 of 100 residents in Area A have a characteristic, while 90 of 900 in Area B do. The percentages are 20% and 10%.
Their simple average is 15%. But the combined proportion is 110 out of 1,000, or 11%.
The 15% figure answers a different question: the unweighted average of the two area percentages.
Check compatibility before adding
- Are the measures defined the same way?
- Do the populations and periods match?
- Are the areas nonoverlapping?
- Are the values counts, weighted estimates, or model-based predictions?
Those distinctions affect whether and how aggregation is valid.
Survey uncertainty needs its own calculation
For estimated counts, the uncertainty of a combined value cannot generally be found by averaging margins of error. Use the source's appropriate formulas or replicate methods and account for dependencies where relevant.
Medians are different again
You cannot recover the median income of a combined population by population-weighting the areas' median incomes. Medians do not combine like counts.
Keep the arithmetic tied to the statistic. If the needed underlying information is unavailable, report the limitation rather than producing a precise-looking combined number from an invalid shortcut.
