I am working with two data sets of unequal sample sizes of 998 and 857. Average of first (998 samples), come out to be higher than the other dataset. To my surprise, when I split my complete data into unequal halves; first into 803 and 195 samples and other also into 819 and 38 samples. then comparison of average values of 803 subset of first complete dataset with the average of 819 subset of second complete data set showed a reverse trend in their mean values. Same reverse trend was observed with other subset of both the data set.

My question is that is it possible that if mean of total items in A>B, their two subsets showed reverse trend in their means i.e. mean of A1

Is this because of unequal sample sizes???? or because of sample density distribution trend????? or both????

If this is possible also, then is there any way to explain this quantitatively??????

It would be really helpful, if anyone can help me on this..

  • 1
    $\begingroup$ In other words, you want to know if Simpson's paradox can appear with mean differences instead of odds ratios? $\endgroup$ – Michael M Nov 8 '13 at 13:45

The answer is "Yes". This is Simpson's paradox applied to mean differences instead of odds ratios. You can read Wiki's article (http://en.wikipedia.org/wiki/Simpson%27s_paradox) to understand the mechanisms behind it. It's a projection problem: If you only see a two dimensional projection of a three dimensional object, you can get quite a wrong impression about the whole picture. In balanced settings (equal group sizes), this is not possible.

Consider, for instance, the following simple setting:

  • $A_1$ consists of 99 times the value 1
  • $A_2$ consists of the value 100
  • $B_1$ consists of the value -9
  • $B_2$ consists of the value 99

The average of $A = A_1 \cup A_2$ is about 2 and thus much smaller than the average 45 of $B = B_1 \cup B_2$. On the other hand, the average 1 of $A_1$ is larger than the average -9 of $B_1$. Similarly, the average 100 of $A_2$ is larger than the average 99 of $B_2$.

  • $\begingroup$ Thanks for the reply. Now one more problem here is if I want to draw an inference regarding the trend of data in A and B. Then what will be the right approach??? Should I look at the whole data and should not split it???? we decided to split our data into two as the data in A and B were found to form two clusters above and below 60 value. Please direct me for the right approach here. $\endgroup$ – Suneyna Nov 9 '13 at 6:27
  • $\begingroup$ How exactly did you split the two samples? $\endgroup$ – Michael M Nov 9 '13 at 12:17
  • $\begingroup$ From A, I separate out values less than and equal to 60 in A1 and other values more than 60 was put in A2 so that number of samples in A1+A2=A. Similarily, splitting was done in case of dataset B. Then we compare mean of A to B, A1 to B1 and A2 to B2. $\endgroup$ – Suneyna Nov 12 '13 at 5:13
  • $\begingroup$ Is 60 a limit that was specified before looking at the data? If no, it's humbug. $\endgroup$ – Michael M Nov 12 '13 at 7:56
  • $\begingroup$ No, this 60 limit was set after looking at the data only. As the data was seen to distributed around 60 in such a way that all thousands of data appeared as two clusters above and below 60. So, After this discussion I think I must not split my data and rely on the whole data set only. $\endgroup$ – Suneyna Nov 13 '13 at 8:27

If you split your data into smaller samples, you can definitely have your means change. If you split your data by removing all the low points, then you immediately increase your mean.

The only time this will not be true is if all points are equal. Then removing points won't change the average.

Note that this does not have any significance! The fact that you took a subset does not prove anything about the overall distribution or trend. It just happened. If anything, it could imply that you incorrectly split your data.

  • $\begingroup$ Thanks for the reply. But I want to know if mean of A>B and A1, A2 are the two subsets of A. Similarily, B1, B2 are two subsets of B. Is the condition A1<B1 and A2<B2 mathematically possible????? As I think it is not. I think that either both or at least one subset must have shown the same relationship in their average values as that of whole data set A and B. $\endgroup$ – Suneyna Nov 8 '13 at 12:33

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