This is not a homework or similar, this is the real world, I am preparing some stats for a meeting I will attend.

I am a member of an association that has 26 members. There are two types of members, A (14 out of 26) and B (12).

The board consists of 5-6 members elected for one year at the time. The board members are split into two categories, "regular" board members and deputies. During the last 10 years, there have been 56 "man-year" on the board (that is, the board has had on average 5,6 members per year).

Out of the 56 man-years, 44 have been regular members and 12 for the deputies.

Out of the regular members' 44 man-years, 40 have been of member type A while 4 member type B.

OTOH just 1 out of the 12 man yeardeputies was A, while 11 was B.

Membership of the board should be randomly distributed. Everyone has one vote and everyone is eligible.

I would like to know how (un)likely it is that this distribution is random, for all three cases. How do I calculate that? 20 years since I studied econometrics and probability and I don't know where to start (nor finish...)

  • 1
    $\begingroup$ Why should this be random? Are position not assigned based on some kind of skill or experience that might correlate (or be perceived to correlate) with characteristic A or B (oh, and is the worry A/B vs. ordinary/deputy or A/B vs. on board/not on board?)? What makes this harder is that person-years are not uncorrelated, if people that are in a function in one year are more likely to run/volunteer/be elected in the next year. Finally, distributions that describe this process (even without the correlation) is not obvious to me, in case of doubt a permutation test might be an idea?! $\endgroup$
    – Björn
    May 6, 2020 at 19:23
  • $\begingroup$ @björn A and B live in identical flats, except for one detail, shower/bath tub. I think a reasonable null hypothesis is that skills for sitting on a board does not correlate to that in any significant manner. $\endgroup$
    – d-b
    May 6, 2020 at 21:27
  • $\begingroup$ @Björn Regarding recurrence, yes it is probably more likely that someone that is elected gets re-elected than that some random person is elected, in the short run. But in the long run, and in this case, 10 years is a pretty long run, this should even out. $\endgroup$
    – d-b
    May 6, 2020 at 21:52

1 Answer 1


If you really think that membership of the board should be randomly distributed (see Bjorn's comment -- I agree with him), and that each personXyear is iid from a distribution (importantly, you're assuming that me last year is a completely different person as me this year, and so on), and you are interested in tests of the form A/B vs ordinary/deputy, or A/B vs on board/not on board, or A/B vs ordinary/not ordinary, etc, then I'd probably just set up a bunch of 2x2 contingency tables. Of course, there are many caveats here, but if this is for a meeting and to make some points clear in that meeting, I think this approach is fine (if you were seeking to publish a peer-reviewed paper on lack of diversity of A and B people in the workplace, then certainly this is not good enough). So for example, let's say I want to test A/B vs regular/not regular. Then my table, using your numbers, is as follows:

\begin{array}{|c|c|c|c|} \hline & \text{reg} & \text{not reg} & \text{total} \\ \hline \text{A} & 40 & 100 & 140\\ \hline \text{B}& 4 & 116 & 120 \\ \hline \text{total} & 44 & 216 & 260 \\ \hline \end{array}

Then, your null hypothesis is that being in A or B has no effect on the probability of being reg vs not reg, so intuitively, you'd expect $44/216 \approx .2$ of both groups to be reg (i.e. $140*.2 = 28$ group A individuals and $120*.2 = 24$ group B individuals to be reg), but clearly, we don't observe that. The way to formalize that is to do a chi squared test. Check out the wikipedia link on chi squared tests to see how to implement them, and all statistical programming languages should easily allow you to perform these. You can also find online tools to quickly perform this analysis.

  • $\begingroup$ I don't understand your second column. Where do the numbers com from? $\endgroup$
    – d-b
    May 6, 2020 at 21:37
  • $\begingroup$ Could you also explain the totalts? 40 + 4 = 44 and 4 + 116 makes sense, but I don't understand 140 + 116 = 216 and 40 + 140 = 140. $\endgroup$
    – d-b
    May 6, 2020 at 21:39
  • $\begingroup$ I'm using your numbers. In this example, we are specifically interested in regular vs not regular, where not regular means anything else (including deputy). you could easily do board vs not board or anything else. $\endgroup$
    – doubled
    May 6, 2020 at 21:41
  • $\begingroup$ @d-b I had a typo for the first row of the second column, I fixed it. You said that of the 44 reg, 40 are A, 4 are B, so that's the first column. You also said 26 individuals over 10 years, 14 A, 12 B, and so that's 260 individualXyear total, with 140 A, 120 B observations (recall my assumption of individuals being independent across years for this analysis). Then if 40 A are reg, that means 140-40 A are not reg, and same for B. $\endgroup$
    – doubled
    May 6, 2020 at 21:46
  • $\begingroup$ I am interested in all three "types", that is total, regular and deputies. 26 is the pool the board, in itself 5-6 people, is elected from. $\endgroup$
    – d-b
    May 6, 2020 at 21:57

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