Hi,Stepping briefly outside the R context, I noticed a statistical point in the text you linked that, in my opinion, isn't quite right. I believe there's a key misunderstanding here: The statement that the z-test does not depend on the number of cases is incorrect. The p-value of the z-test is —just like other tests— very much dependent on the sample size, assuming the same mean difference and standard deviation. The text you linked is actually calculating an Effect Size, which is (largely) independent of the sample size. Effect Size answers the question of how "relevant" or "large" the difference between groups is. This is fundamentally different from testing for "significant" differences. Specifically, the crucial 1/\sqrt{n} term, which is necessary for calculating the standard error of the mean difference, seems to be missing from the presented formula for the z-score. I just wanted to quickly point this out.
Best regards Am 27.10.2025 um 14:12 schrieb Petr Pikal:
Hallo The t test is probably not the best option in your case. With 95 observations your data behave more like a population and you may get better insight using z-test. See https://toxictruthblog.com/avoiding-little-known-problems-with-the-t-test/ Best regards. Petr <https://www.avast.com/sig-email?utm_medium=email&utm_source=link&utm_campaign=sig-email&utm_content=webmail> Neobsahuje žádné viry.www.avast.com <https://www.avast.com/sig-email?utm_medium=email&utm_source=link&utm_campaign=sig-email&utm_content=webmail> <#DAB4FAD8-2DD7-40BB-A1B8-4E2AA1F9FDF2> so 25. 10. 2025 v 11:46 odesílatel Christophe Dutang <[email protected]> napsal:Dear list, I'm computing a p-value for the Student test and discover some inconsistencies with the cdf pt(). The observed statistic is 11.23995 for 95 observations, so the p-value is very smallt_score <- 11.23995 n <- 95 print(pt(t_score, df = n-2, lower=FALSE), digits=22)[1] 2.539746620181247991746e-19print(integrate(dt, lower=t_score, upper=Inf, df=n-2)$value, digits = 22)[1] 2.539746631161970791961e-19 But if I compute with pt(lower=TRUE), I got 0print(1-pt(t_score, df = n-2, lower=TRUE), digits=22)[1] 0 Indeed, the p-value is lower than the epsilon machinept(t_score, df = n-2, lower=FALSE) < .Machine$double.eps[1] TRUE Using the square of t statistic which follows a Fisher distribution, I got the same issue:print(pf(z, 1, n-2, lower=FALSE), digits=22)[1] 5.079493240362495983491e-19print(integrate(df, lower=z, upper=Inf, df1=1, df2=n-2)$value, digits =22) [1] 5.079015231299358486828e-19print(1-pf(z, 1, n-2, lower=TRUE), digits=22)[1] 0 When using the t.test() function, the p-value is naturally printed : p-value < 2.2e-16. Any comment is welcome. ChristopheR.version_ platform aarch64-apple-darwin20 arch aarch64 os darwin20 system aarch64, darwin20 status major 4 minor 5.1 year 2025 month 06 day 13 svn rev 88306 language R version.string R version 4.5.1 (2025-06-13) nickname Great Square Root ------------------------------------------------- Christophe DUTANG LJK, Ensimag, Grenoble INP, UGA, France ILB research fellow Web: http://dutangc.free.fr ------------------------------------------------- [[alternative HTML version deleted]] ______________________________________________ [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.[[alternative HTML version deleted]] ______________________________________________ [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.
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