Dear Sir, I have data, coming from tests, consisting of 300 values. Is there a way in R with which I can confirm this data to 68-95-99.8 rule or three-sigma rule? I need to look around percentile ranks and prediction intervals for this data. I, however, used SixSigma package and used ss.ci() function, which produced 95% confidence intervals. I still am not certain about percentile ranks conforming to 68-95-99.7 rule for this data. Thanks and regards, Salil Sharma [[alternative HTML version deleted]]
On May 28, 2011, at 2:12 PM, Salil Sharma wrote:> Dear Sir, > > > > I have data, coming from tests, consisting of 300 values. Is there a > way in > R with which I can confirm this data to 68-95-99.8 rule or three- > sigma rule?Can you describe this rule? I get the idea that it might be "private language" adopted by the SigxSigma sect.> > I need to look around percentile ranks and prediction intervals for > this > data. I, however, used SixSigma package and used ss.ci() function, > which > produced 95% confidence intervals. I still am not certain about > percentile > ranks conforming to 68-95-99.7 rule for this data.The quantile function is pretty much "standard operating procedure". fivenum will return the values that would appear in a box-and-whiskers plot. -- David Winsemius, MD West Hartford, CT
Hi r-help-bounces at r-project.org napsal dne 28.05.2011 20:12:33:> "Salil Sharma" <salil31 at gmail.com> > Odeslal: r-help-bounces at r-project.org > Dear Sir, > > > > I have data, coming from tests, consisting of 300 values. Is there a wayin> R with which I can confirm this data to 68-95-99.8 rule or three-sigmarule?> > I need to look around percentile ranks and prediction intervals for this > data. I, however, used SixSigma package and used ss.ci() function, which > produced 95% confidence intervals. I still am not certain aboutpercentile> ranks conforming to 68-95-99.7 rule for this data. >Not sure what you exactly want but you could look at function quantile. Or you could compute confidence interval for mean by e.g.> mean.intfunction (x, p = 0.95) { x.na <- na.omit(x) mu <- mean(x.na) odch <- sd(x.na) l <- length(x.na) alfa <- (1 - p)/2 mu.d <- mu - qt(1 - alfa, l - 1) * odch/sqrt(l) mu.h <- mu + qt(1 - alfa, l - 1) * odch/sqrt(l) return(data.frame(mu.d, mu, mu.h)) } Regards Petr> > > Thanks and regards, > Salil Sharma > > > [[alternative HTML version deleted]] > > ______________________________________________ > R-help at r-project.org mailing list > https://stat.ethz.ch/mailman/listinfo/r-help > PLEASE do read the posting guidehttp://www.R-project.org/posting-guide.html> and provide commented, minimal, self-contained, reproducible code.
On May 28, 2011, at 5:12 PM, David Winsemius wrote:> > On May 28, 2011, at 2:12 PM, Salil Sharma wrote: > >> Dear Sir, >> >> >> >> I have data, coming from tests, consisting of 300 values. Is there >> a way in >> R with which I can confirm this data to 68-95-99.8 rule or three- >> sigma rule? > > Can you describe this rule? I get the idea that it might be "private > language" adopted by the SigxSigma sect.Given the mention of the SixSigma package I can perhaps be forgiven for jumping to the conclusion that it might be "private language" and I still cannot be sure that a corruption of standard statistical theory has not been adopted by the SSers. Looking at Wikipedia I get a different "answer" to the question what is the "three sigma rule" than I do by looking at "The American Statistician". My hierarchy for probity assigns a higher level of confidence to TAS. The Three Sigma Rule Author(s): Friedrich Pukelsheim Source: The American Statistician, Vol. 48, No. 2 (May, 1994), pp. 88-91 Published by: American Statistical Association Stable URL: http://www.jstor.org/stable/2684253 . For a distribution whose density is unimodal (and notice _not_ assuming symmetry): Pr( abs( X-mean(X) ) > 3*sd(X) ) < 4/18 < 0.05 It seemed trivial to test this with a normal distribution, so I illustrate it with a skewed distribution: > X <- rexp(300) > sum( abs( X-mean(X) ) > 3*sd(X) )/300 [1] 0.02>> >> I need to look around percentile ranks and prediction intervals for >> this >> data. I, however, used SixSigma package and used ss.ci() function, >> which >> produced 95% confidence intervals. I still am not certain about >> percentile >> ranks conforming to 68-95-99.7 rule for this data.Would those percentiles be: > 50 -c(68, 95, 99.7)/2 [1] 16.00 2.50 0.15 > 50 + c(68, 95, 99.7)/2 [1] 84.00 97.50 99.85> > The quantile function is pretty much "standard operating procedure". > > fivenum will return the values that would appear in a box-and- > whiskers plot. > > > -- > David Winsemius, MD > West Hartford, CT-- David Winsemius, MD West Hartford, CT