I have a set of 10 genes, each gene contains around 15 SNPs. I have tested the deviation from Hardy Weinberg equilibrium (HWE) for all SNPs. All of the SNPs didn't deviate from HWE, except 5 SNPs in ABCG8, they had p<0.05. I have found different factors that can affect the HWE. My question: Why for the rest of the SNPs met the HWE, why only 5 SNPs in ABCG8 deviate largely from HWE?

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    $\begingroup$ Be careful with the claims based on p-values: a p-value allows to reject the null hypothesis, but not to confitm it. In other words, when $p<0.05$ one can say that HWE is broken, however $p>0.05$ does not mean that HWE holds - it could be still broken, but we cannot prove it. $\endgroup$
    – Roger V.
    Jan 5, 2021 at 13:33
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    $\begingroup$ I would quibble with @Vadim a bit- when you do 150 tests, you expect that $0.05 * 150 = 7.5$ SNPs to be $p < 0.05$, as under the null hypothesis, p-values from independent tests are uniformly distributed. P-values are a quite low standard of evidence except when they're very small. Having all 5 occur in one gene might be surprising, but nonetheless I would say that "rejecting" the null hypothesis based on it is a bit strong, as the observations are still pretty consistent with the null hypothesis (without seeing additional data). $\endgroup$ Jan 5, 2021 at 21:12
  • $\begingroup$ @MaximilianPress I was talking about more basic difference about p-value and power, objecting the claim that the rest of the SNPs met the HWE. However I do agree with you that, given the large number of tests, even the five SNPs with $p<0.05$ are not a proof that HWE is violated. $\endgroup$
    – Roger V.
    Jan 6, 2021 at 8:26
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    $\begingroup$ @Vadim ok, that makes sense. I agree with your overall point about p-values used to confirm the null hypothesis, I think I was just worried about the other side of the issue, which wasn't clear. I think that you actually addressed the question quite directly and accurately! I would suggest making it into an answer, actually. $\endgroup$ Jan 6, 2021 at 18:11

1 Answer 1


A significance test (i.e., p-value) allows us to reject the null hypothesis, but not to confirm it. In other words, when $p<0.05$ one can say that the conditions for HWE are not met, however $p>0.05$ does not mean that HWE holds - the conditions might still not be met, but we cannot prove it. Thus, by making statements such as All of the SNPs didn't deviate from HWE and the rest of the SNPs met the HWE one may be committing Type II statistical error.

As @MaximilianPress correctly pointed out, in this case one should not be making any statements about the SNPs with $p<0.05$ either, due to the high possibility of False discovery. Indeed, a p-value gives us the probability that the data would be at least as extreme as was observed, even though the null hypothesis is true. Given $N=150$ SNPs one expects about $$N\times\alpha = 150 \times 0.05= 7.5$$ statistically significant outcomes to occur by chance, even if the HWE holds. That is, one could be also committing the Type I error of rejecting the null hypothesis, when it is true. To exclude this error one needs to use a more stringent test, e.g., Bonferroni correction, i.e., testing against the significance level $N$ times smaller than the current $\alpha=0.05$: $$ \alpha\longrightarrow\frac{\alpha}{N}=\frac{0.05}{150}\approx 0.0003 $$


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