[R] comparing variances/distributions

Ana Marija @okov|c@@n@m@r|j@ @end|ng |rom gm@||@com
Wed Jun 17 22:20:22 CEST 2020


would using Kolmogorov-Smirnov test make more sense here?

> x=m$Pold
> y=m$Pnew
> ks.test(x,y)

    Two-sample Kolmogorov-Smirnov test

data:  x and y
D = 0.0049066, p-value = 1.221e-15
alternative hypothesis: two-sided

Warning message:
In ks.test(x, y) : p-value will be approximate in the presence of ties

as I understand high p-values here say I cannot claim statistical
support for a difference, but low p-values are not evidence of
sameness?
D should be the maximum difference between the two probability distributions?

On Wed, Jun 17, 2020 at 3:06 PM Ana Marija <sokovic.anamarija using gmail.com> 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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