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Let's say I studied the shoulders of 10 individuals, and I scored the presence/absence of a criteria (ex: arthrosis, presence=0;absence=1) on the left side and the right side.

I need to test the bilateral asymmetry, to evaluate if there is a statistically significant difference between the left and right side.

So I am dealing with categorical/binary data, which are paired, and there are 4 different possibilities for one individual : no arthrosis at all (0 0); arthrosis on both sides (1 1); arthrosis on the left but not on the right (1 0); and the contrary (0 1).

Considering the nature of data, I assumed the McNemar test was the solution, but if I have for example five pairs with presence on the left and absence on the right (1 0), and the five others with absence on the left and presence on the right (0 1), then McNemar test gives a p value of 1.

If I'm correct, this would mean that there is no significant difference between both sides, while the fact is that there is 100% of bilateral asymmetry since no pair share the same value (no ties).

So maybe the test was not appropriate and I would need to use a Chi square/Fisher's exact test instead? But then, how to make these tests compare the pairs and not globally both sides?(a contingency table from the same example also gives a p value of 1, because there are five 0s and five 1s for each side). Or should I consider the data as ordinal and use Wilcoxon signed-ranks test instead?

I have been looking for a solution for a very long time but have not succeeded. Thank you very much for your help.

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    $\begingroup$ If I understand what you're trying to test, --- and I doubt I do understand --- you actually have two conditions. 1. Asymmetry, which includes left only or right only. 2. Symmetry, which includes both sides or neither side. In this case, you might use a binomial test testing against a probability of 0.5. $\endgroup$
    – Sal Mangiafico
    Commented Oct 7, 2019 at 21:56
  • $\begingroup$ Thank you very much for your answer! Indeed, even if there are 4 possible combinations for a pair, there are actually 2 situations, asymmetry or symmetry, and I need to test if there is any asymmetry in the sample and if it is significant. I am not familiar with this test but as asymmetry is the most uncommon (and interesting) situation, is a probability of 0.5 appropriate? And if I understand, I must transform all symmetry cases to a 0 for example, and all asymmetry cases to a 1? If so, is there any way I could use a Chi-square/Fisher's exact test instead? $\endgroup$
    – wburton
    Commented Oct 9, 2019 at 3:55
  • $\begingroup$ @wburton Fisher's exact test concerns a relationship between 2 variables. There is only 1 variable (yes, you can arbitrarily code symmetric cases = 0 and asymmetric cases = 1). With such a small $N$, a $\chi^2$ goodness-of-fit statistic will not follow a $\chi^2$ distribution, but a binomial test works if you have a reasonable $H_0$ (it does not have to be $\pi=50\%$): en.wikipedia.org/wiki/Binomial_test $\endgroup$
    – Terrence
    Commented Mar 29, 2022 at 8:59

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If I'm correct, this would mean that there is no significant difference between both sides, while the fact is that there is 100% of bilateral asymmetry since no pair share the same value (no ties).

This is not the correct interpretation. The correct interpretation would be that your data bring no proof to say that the proportion of unilateral right arthrosis and the proportion of unilateral left arthrosis are different in the whole population. And as you indicate, this is the case: you have exactly the same proportions here.

This is not the same question as symmetry at the individual-level. To answer this question, you have a bunch of classical tools, such as Cohen's Kappa, and other friends.

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  • $\begingroup$ Could you edit your answer to explain how Cohen's kappa would be used in this situation? The binomial test Sal Mangiafico referred to above might be what @wburton is looking for, but perhaps it would also be informative to quantify how related the left- and right-hand criteria are (e.g., higher kappa indicates the criterion is more commonly met on both/neither side than asymmetric). $\endgroup$
    – Terrence
    Commented Mar 29, 2022 at 9:06

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