# Why is an average of an average usually incorrect?

Can someone explain why taking an average of an average usually results in a wrong answer? Is there ever a case where the average of the average can be used correctly?

As an example, let's say that an assessment is given to three schools and I want to find out the average score for all three schools combined and the average score per school. When I attempt to add the three individual scores and divide by three I get a number that is very close (+/- 1 percent) to the actual overall average.

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It's not wrong, it needs interpretation. If school A has 10000 students, with average marks in chemistry, say 2.56, and school B has one pupil (ok, bit extreme, but why not), with average mark in chemistry 1, then you may either want to know the average marks w.r.t. the number of schools (2.56 + 1)/2 or w.r.t to pupils (10000*2.56 +1)/10001. Both number make sense but have, of course, completely different implications and interpretations. – user20266 Jan 2 '12 at 21:20
Say two of the schools are small and very good, and the third school is huge, and pretty bad. If you average over students, the good results from the small schools hardly make a dent. If you first find the average per school, and average the averages, this will tend to hide the poor results from the huge school. In the real world, disparities of size or performance are often not very dramatic, so the two procedures can end up giving fairly similar numbers. One thing we can be sure of: whoever is trying to make a point will choose the procedure that supports his/her point. – André Nicolas Jan 2 '12 at 21:22
@AndréNicolas - You may want to post that as your answer. I feel as though you are implying that neither approach is wrong, is this true? – O.O Jan 2 '12 at 21:32
An interesting real-world case. When a university reports the "average class size", the result is quite different if they report the class size for the average student, or the class size for the average instructor. – GEdgar Jan 2 '12 at 21:38
@subt13: In case I do not get around to posting, either approach may be useful. Doing a school by school analysis, but using more than simple averages, and including variance, can help to highlight regional disparities, and may motivate remedial action. Averaging over students produces information of a different kind. – André Nicolas Jan 2 '12 at 21:45

If there are $n_1$, $n_2$, and $n_3$ scores in the three schools, and the average for each school is $a_1$,$a_2$,$a_3$, respectively, the correct average is a "weighted average:"

$$\frac{n_1}{n_1+n_2+n_3}a_1+\frac{n_2}{n_1+n_2+n_3}a_2+\frac{n_3}{n_1+n_2+n_3}a_3$$

The average of the averages is:

$$\frac{1}{3}a_1 + \frac{1}{3}a_2 + \frac{1}{3}a_3$$

These two values will be exactly the same if each school has exactly the same number of students, and will tend to be "close" if the schools are relatively close in size and/or the scores for the three schools are close.

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I changed some $n_1$'s to $n_2$ and $n_3$... – David Mitra Jan 3 '12 at 0:52
I don't know if i have to open new question. How would you weight average potions of impresions & clicks keywords? – jrosell Mar 4 at 7:03

For example: the average of $\{ 2,2,2,2,2,2,2,2,2,2,2,2,2\}$ is 2, (N=13) and the average of $\{4\}$ is 4, (N=1). The average of the averages is 3. But the average of all numbers is $30/14$ ≃ $2.14$.

I hope this is enough to explain what goes wrong (you're giving equal weights to the "first averages" when you take their average, which isn't the correct thing to do if you want the average of all the numbers).

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A B C D
6 5 4 8
7 6 3 6
8 7 2 5
9 5 5 7
4 8 6 6
= = = =
6.8 6.2 4 6.4 average section (A;B;C;D) = 5.85

average all number = 5.85

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Yes, if you have the same number of items in each sample, then the average of the averages is the same as the overall average. The discrepancies arise where the samples are of different sizes as the average for each sample needs to be weighted appropriately. – IanF1 Nov 24 '14 at 7:50

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