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It appears to me that the mean of four numbers is equivalent to the mean of the means of two pairs of those numbers:

$\text{mean}(a,b,c,d) = \frac{a+b+c+d}{4}$

$\text{mean}(\text{mean}(a,b),\text{mean}(c,d)) = \frac{\text{mean}(a,b)+\text{mean}(c,d)}{2} = \frac{\frac{a+b}{2}+\frac{c+d}{2}}{2} \cdot \frac{2}{2} = \frac{a+b+c+d}{4}$

I am confused after reading Simpson's paradox though. It seems like this shouldn't necessarily be the case? This leads be to a number of questions:

  1. Is this derivation correct for 4 numbers?
  2. How does this differ from Simpson's paradox?
  3. If it only differs from Simpson's paradox under certain conditions, do those conditions change if we only use integers?
  4. If it only differs from Simpson's paradox under certain conditions, do those conditions change if we specify the exact number (i.e., 4) of numbers we use?
sdasdadas
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    Sure, if you take the average of $n$ numbers, each of which is the average of $m$ numbers, then that is the same as the average of the full set of $nm$ numbers. Simpson’s paradox happens becaus the sets of numbers are not all the same size. – Thomas Andrews Jul 15 '21 at 21:29
  • Related, maybe duplicate: https://math.stackexchange.com/q/95909/7933 – Thomas Andrews Jul 15 '21 at 21:29
  • Thank you! I believe it is a duplicate. As a side note, does this hold for median as well? – sdasdadas Jul 15 '21 at 22:40
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    Median doesn’t even work with $4.$ – Thomas Andrews Jul 15 '21 at 22:43
  • If you had two different groups of different sizes, then thee mean of the mean of the two groups would not necessarily be the same as the mean of all the individual observations. Mean of 1,2,3 is $2.$' mean of 3.5 is $4,$ mean of 2,4 is 3. However mean of 1,2,3,3,5 is $14/5 \ne 3.$ – BruceET Jul 16 '21 at 02:11

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