Odp: Three sigma rule
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 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.
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.int
function (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]]
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