В Sat, 25 Oct 2025 11:45:42 +0200 Christophe Dutang <[email protected]> пишет:
> 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) ______________________________________________ [email protected] mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide https://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
