Thanks for your answers.
I was not aware of the R function expm1().
I?m completely aware that 1 == 1 - 5e-19. But I was wondering why pt() returns
something smaller than double.eps.
For students who will use this exercise, it is disturbing to find 0 or 5e-19 :
yet it will be a good exercise to find that these quantities are equalled.
Regards, Christophe
> Le 25 oct. 2025 ? 12:14, Ivan Krylov <ikrylov at disroot.org> a ?crit
:
>
> ? Sat, 25 Oct 2025 11:45:42 +0200
> Christophe Dutang <dutangc at gmail.com> ?????:
>
>> Indeed, the p-value is lower than the epsilon machine
>>
>>> pt(t_score, df = n-2, lower=FALSE) < .Machine$double.eps
>> [1] TRUE
>
> Which means that for lower=TRUE, there will not be enough digits in R's
> numeric() type to represent the 5*10^-19 subtracted from 1 and
> approximately 16 zeroes.
>
> Instead, you can verify your answer by asking for the logarithm of the
> number that is too close to 1, thus retaining more significant digits:
>
> print(
> -expm1(pt(t_score, df = n-2, lower=TRUE, log.p = TRUE)),
> digits=16
> )
> # [1] 2.539746620181249e-19
> print(pt(t_score, df = n-2, lower=FALSE), digits=16)
> # [1] 2.539746620181248e-19
>
> expm1(.) computes exp(.)-1 while retaining precision for numbers that
> are too close to 0, for which exp() would otherwise return 1.
>
> See the links in
>
https://cran.r-project.org/doc/FAQ/R-FAQ.html#Why-doesn_0027t-R-think-these-numbers-are-equal_003f
> for a more detailed explanation.
>
> --
> Best regards,
> Ivan
> (flipping the "days since referring to R FAQ 7.31" sign back to
0)