ME.6.MC:5.1 - Compare centralized calculation with combined local summaries

Two teams must report the mean of all their measurements. The proposed arrangements are to send every measurement to one calculator, or to compute a summary in each team and combine the summaries. All values have the same meaning and units, and every included measurement has equal weight.

Represent a finite list (x=(x_1,\ldots,x_n)) by

[ T(x)=\left(\sum_{i=1}^{n}x_i,\ n\right). ]

Combine pairs by ((s,n)\oplus(t,m)=(s+t,n+m)). For concatenation of lists (x) and (y),

[ T(x,y)=T(x)\oplus T(y). ]

Both components follow from adding the sums and counts. Pair addition is associative, so repeated regrouping preserves the summary. For a nonempty combined list, recover the mean as total sum divided by total count. This supplies the composition and recovery needed by the parallel arrangement.

With lists ((0,4)) and ((10)), the summaries are ((4,2)) and ((10,1)). Their combination is ((14,3)), giving (14/3), the same mean as central calculation. Averaging the two local means instead gives ((2+10)/2=6), answering a different weighting question.

The mathematical comparison supports using sum-and-count summaries when this mean is the receiving result. Application requires consistent inclusion and weighting rules, with every included record assigned to one team and counted once. The algebra uses exact addition; a rounded implementation needs the relevant numerical error comparison. Whether the two teams use those rules needs grounds from their work.

Now the recipient asks how many measurements exceed (3). Replacing ((0,4)) by ((2,2)) leaves the local sum-and-count pair unchanged, and leaves the combined pair ((14,3)) unchanged, but changes the requested count from two to one. No rule using only that pair can distinguish the cases. The earlier mean-equivalence remains valid; the new question needs another retained quantity or a return to the measurements. The method decision must not treat the old summary as a substitute for all uses of the records.