Odp: Three sigma rule
I think you really want a normality test. If that's what you want, you have more options than the three-sigma rule. http://en.wikipedia.org/wiki/Normality_test Tom
On Tue, May 31, 2011 at 12:31 PM, Bert Gunter <gunter.berton at gene.com> wrote:
Folks: On Tue, May 31, 2011 at 8:48 AM, Petr PIKAL <petr.pikal at precheza.cz> wrote:
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.
-- Nor am I, but ...
Or you could compute confidence interval for mean by e.g.
I'm pretty sure that this is NOT what he wants. -- Bert
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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