I have a series of 7 pairs with each pair adding up to 100%.

I now need to average column a and b to end up with a final pair totalling 100%.

Where do I start?

|       |   A   |   B   |   C
| Asus  | 74.38%| 25.62%| 100%
| HTC   | 0%    | 100%  | 100%
| LG    | 0%    | 100%  | 100%
| Moto  | 25%   | 75%   | 100%
|       |       |       |
|Android|  ?    |   ?   | 100%

Any linear combination of the rows will have this property if all row weights add up to one. The most sensible weighting scheme is one that takes into account the number of elements in each row (if there are twice as many Asus as LG elements in the total population, ideally the Asus row will count for twice as much), and the second most sensible is $1/n$ (treating every row the same).

For example, if we average each column we get $\frac{74.38+0+0+25}{4}=24.845$ for A and $\frac{25.62+100+100+75}{4}=75.155$ for B, which add up to 100.

To prove the first sentence, suppose we know that $a_i+b_i=1 \forall\ i$. We then want to show that:

$\sum_i (\alpha_i a_i+\alpha_i b_i)=1$ (the combination of percentages sums to 1)

$\sum_i \alpha_i(a_i+b_i)=1$ (using distributive property)

$\sum_i \alpha_i=1$ (using our assumption on sums for each $i$)

  • $\begingroup$ So sorry would you mind giving an example on the data set? $\endgroup$ – David Finder Dec 3 '15 at 16:07
  • $\begingroup$ @DavidFinder added in example. $\endgroup$ – Matthew Graves Dec 3 '15 at 16:16

Are you interested in proportion of A among all Android phones? Or are you interested in proportion of A among all Android models? The difference between these two is that each model has many actual physical phones.

If your answer is "among models", then you do a simple average. If your answer is "among phones", then a weighted average, where the weight is the number of phones for each model.


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