Dear Ana
This really depends on your scientific question. The two techniques you
have shown do different things and there must be many more which could
be applied.
Michael
On 17/06/2020 20:57, Ana Marija wrote:> Hello,
>
> I have p values from two distributions, Pold and Pnew
>> head(m)
> CHR POS MARKER Pnew Pold
> 1: 1 785989 rs2980300 0.1419 0.9521
> 2: 1 1130727 rs10907175 0.1022 0.4750
> 3: 1 1156131 rs2887286 0.3698 0.5289
> 4: 1 1158631 rs6603781 0.1929 0.2554
> 5: 1 1211292 rs6685064 0.6054 0.2954
> 6: 1 1478153 rs3766180 0.6511 0.5542
> ...
>
> In order to compare those two distributions (QQ plots shown in attach)
> does it make sense to use:
>
> var.test(m$Pold, m$Pnew, alternative = "two.sided")
>
> F test to compare two variances
>
> data: m$Pold and m$Pnew
> F = 0.99937, num df = 1454159, denom df = 1454159, p-value = 0.7057
> alternative hypothesis: true ratio of variances is not equal to 1
> 95 percent confidence interval:
> 0.9970808 1.0016750
> sample estimates:
> ratio of variances
> 0.9993739
>
>
> Or some other test makes more sense?
>
> Thanks
> Ana
>
>
>
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--
Michael
http://www.dewey.myzen.co.uk/home.html