Deviation from Hardy Weinberg equilibrium (HWE)
A significance test (i.e., p-value) allows us to reject the null hypothesis, but not to confirm it. In other words, when $p&lt;0.05$ one can say that the conditions for HWE are not met, however $p&gt;0.05$ 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 $p&lt;0.05$ 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 $\alpha=0.05$:
$$
\alpha\longrightarrow\frac{\alpha}{N}=\frac{0.05}{150}\approx 0.0003
$$