sample comes from the reference distribution). Ideally, I would try the
procedure suggested by Marsiglia to compute p-values but this is far beyond my
coding skills.
> On 5 Sep 2019, at 21:21, Rui Barradas wrote:
>
> Hello,
>
> Inline.
>
> Às 20:09 de 05/09/19, Boo G. esc
ogorov-smirnov distance
d_stat[i, 2] <- ht$statistic
d_stat[i, 3] <- ht$p.value
}
hist(d_stat[, 2])
hist(d_stat[, 3])
Note that d_2 is not sorted, but the results are equal in the sense of function
identical(), meaning they are *exactly* the same. Why shouldn't they?
Hope this helps,
_stat[i, 3] <- ht$p.value
> }
>
> hist(d_stat[, 2])
> hist(d_stat[, 3])
>
>
> Note that d_2 is not sorted, but the results are equal in the sense of
> function identical(), meaning they are *exactly* the same. Why shouldn't they?
>
> Hope this he
Hello,
I am trying to perform a Kolmogorov–Smirnov test to assess the difference
between a distribution and samples drawn proportionally to size of different
sizes. I managed to compute the Kolmogorov–Smirnov distance but I am lost with
the p-value. I have looked into the ks.test function unsuc
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