We then considered whether this number of significant hits exceeded the number of hits that one could expect solely from chance. Given, for example, a p-value threshold of p = .05 and N = 1000 SNPs, one would expect N × p = 50 “hits” by chance, if each test were independent (an assumption to which we will return shortly). This expectation of 50 hits is actually the mean of a distribution of expected outcomes, where the standard deviation of that distribution is (N∗p∗(1−p). Consequently, the number of observed hits can be converted to a Z score, and a one-tailed Z test can provide what is essentially a p-value for the set of p-values; in other words, it is a determination of the probability of observing this number of hits if there were no true association between the phenotype and the SNPs in this gene set.
