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Imagine a scenario where someone walks into a room with ten adults. Five of the people are in their twenties and are children of the other five people (who are in their fifties). The person who just came in is tasked with determining based on family resemblance who goes with who.

I'm trying to either find or develop a statistical test that can determine a p-value for getting a certain percentage of guesses correct. But I'm sure there's a test already out there. I just don't know what it would be called or how to go about searching for it.

I'm aware that technically, a chi-squared test can be used. A lot of nCr and nPr calculations would have to be made to determine the expected frequencies of getting matches right. But I would think that if this test has been developed, someone would have made a test to address this exact situation.

Again I'm not sure how you would even go about describing this test so I'm not sure what I would search for. Would we call this a "non-independent Bernoulli trial of matches without replacement"?

Simply put, my question is "is there a hypothesis test that was custom made to address this kind of situation. And what is that test called (if it exists)?"

Regards

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    $\begingroup$ This sounds like a classic counting problem. $\endgroup$
    – Galen
    Commented Jul 6, 2021 at 2:15
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    $\begingroup$ Are these pairs of one parent to one child, or can someone have multiple parents or children present? $\endgroup$
    – Henry
    Commented Jul 6, 2021 at 11:07
  • $\begingroup$ Ultimately I was going to look at both of those scenarios. $\endgroup$
    – the_photon
    Commented Jul 6, 2021 at 13:25

2 Answers 2

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I recommend a permutation test.

There are 5 ways to assign the first pair, 4 the second, 3 the third, 2 the fourth, and only 1 the last pair. There are 120=5 * 4 * 3 * 2 * 1 different random assignments of pairs. Only 1 of these is correct for all five pairs for a probability of 1/120. Note that if four are correct, so must the fifth match be correct. If three are correct, the remaining two pairs must be switched. There are 10 ways to choose the two pairs that are mismatched, for a probability of of 10/120 = 1/12 of exactly three matches. The total of 3 or more matches is 11/120 so this will happen randomly almost 10% of the time. The remaining numbers of 1 or 2 matches are not very interesting but could be enumerated. The last, no matches, is most easily found by subtraction.

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    $\begingroup$ What is the benefit of doing a permutation test, especially if we already know the probability distribution of guessing $x$ correct pairs? In my opinion, there is no parameter to estimate hence no hypothesis to test. Its simply a matter of computing the $\operatorname{Pr}(k \leq X)$ $\endgroup$
    – Demetri Pananos
    Commented Jul 6, 2021 at 3:07
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    $\begingroup$ A test does not always require a parameter. Here, a null hypothesis might be that assignment is random or that the person making the matches has no skill whatever. A permutation test addresses that question. The distribution of the number of correct pairs is based directly on the numbers that result from pure guessing. $\endgroup$
    – David Smith
    Commented Jul 6, 2021 at 13:26
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    $\begingroup$ YES! David! I think you understand my original question behind all this! I think I'm going to open a new question. I didn't really state my question completely. That distribution (that you speak of) of the number of correct pairs that result from pure guessing is the distribution I am seeking. $\endgroup$
    – the_photon
    Commented Jul 6, 2021 at 13:33
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Yes, Galen and Demetri are right. As stated, this is just a classic counting problem. A p-value can be established by mere combinatorics - there's no parameter to estimate and thus, no hypothesis test to be done really.

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