Solved – Does the birth date of professional boxers matter? Prove/disprove what an astrologer might predict

If your hypothesis was formulated a priori, then the data are quite strongly significant.

Your null hypothesis is that astrology does not predict anything. This would mean that the probability of a boxer to be born under an "earth" sign is $0.25$, and the same is true for the "ethereal" signs. I assume that you selected these signs a priori, before looking at the actual birth dates.

You want to disprove the null hypothesis, and you are interested in deviations in one particular direction: more boxers born under earth signs, and less under the ethereal signs (not vice versa). This means you can conduct one-sided tests. Here is how.

Consider earth signs. Under the null hypothesis, the most probable number of boxers born under these signs out of total $67$ is $67/4$. But for any integer number $x$ between $0$ and $67$ one can compute a probability that exactly so many boxers were born under these signs. This gives a function $p(x)$, known as binomial probability density function. You can then ask, what is the probability that $27$ or more boxers would be born under an earth sign? The answer is given by a sum $sum_{x=27}^{67} p(x)$. Computing it gives $0.004$.

This is known as a p-value: a probability that you could have observed your result, or an even more extreme result, under the null hypothesis. P-value of $p=0.004$ is pretty low, and most people would call it "significant", i.e. the data seem to speak against the null hypothesis.

We can do the same with the ethereal signs, arriving at the p-value of $p=0.03$ that $10$ or less boxers were born under them. This is also quite low.

Note, however, that these two p-values are not independent: certainly, if more boxers were born under the earth signs, it would automatically mean that less boxers could have been born under the ethereal signs. I don't know how to compute a probability of observing $67$ or more and $10$ or less at the same time, but it is easy to simulate. Let's generate $1:000:000$ parallel worlds where null hypothesis is true. We can then count the number of worlds where number of earth-born boxers is $27$ or more; where number of ethereal-born boxers is $10$ or less; and where both is true. This is called a Monte Carlo simulation.

Dividing the counts by $1:000:000$, I obtain: $0.004$, $0.03$, and $0.0008$. First two numbers are identical to the ones obtained above. The last number is the most relevant one.

I would argue that $p=0.0008$ is low enough to think that maybe there is something interesting here! If one has some strong a priori reasons to doubt astrology, one would want to use a much more stringent criterium than a conventional threshold of $p<0.05$: "extraordinary claims require extraordinary evidence". But $p=0.0008$ looks quite convincing (even though can still be a fluke).

Finally, let me remind you that all of the above crucially depends on the fact that you selected your Zodiac signs before looking at the data and that you selected your boxers without looking at their birth dates. If that is not true, then p-value can easily change to about $sim 0.05$, as nicely shown here by @whuber.

Matlab code:

N = 1e+6; counts = [0 0 0];  n1 = binornd(67, 0.25, [N 1]); n2 = binornd(67-n1, 1/3);  counts(1) = length(find(n1>=27)); counts(2) = length(find(n2<=10)); counts(3) = length(find(n1>=27 & n2<=10));  display(['Monte Carlo results: ' num2str(counts/N, 2)]) display(['Analytical solution: ' num2str(1-binocdf(26,67,0.25), 2)]) display(['Analytical solution: ' num2str(binocdf(10,67,0.25), 2)]) 

Running this takes 4.5 seconds on my laptop and results in

Monte Carlo results: 0.0043    0.034   0.00082 Analytical solution: 0.0042 Analytical solution: 0.034