[R] shapiro.test() output
robert-mcfadden at o2.pl
Thu Jul 13 09:03:03 CEST 2006
> -----Original Message-----
> From: r-help-bounces w stat.math.ethz.ch [mailto:r-help-
> bounces w stat.math.ethz.ch] On Behalf Of Matthew.Findley w ch2m.com
> Sent: Wednesday, July 12, 2006 11:14 PM
> To: r-help w stat.math.ethz.ch
> Subject: [R] shapiro.test() output
> R Users:
> My question is probably more about elementary statistics than the
> mechanics of using R, but I've been dabbling in R (version 2.2.0) and
> used it recently to test some data .
> I have a relatively small set of observations (n = 12) of arsenic
> concentrations in background groundwater and wanted to test my
> assumption of normality. I used the Shapiro-Wilk test (by calling
> shapiro.test() in R) and I'm not sure how to interpret the output.
> Here's the input/output from the R console:
> >As = c(13, 17, 23, 9.5, 20, 15, 11, 17, 21, 14, 22, 13)
> Shapiro-Wilk normality test
> data: As
> W = 0.9513, p-value = 0.6555
> How do I interpret this? I understand, from poking around the internet,
> that the higher the W statistic the "more normal" the data.
> What is the null hypothesis - that the data is normally distributed?
> What does the p-value tell me? 65.55% chance of what - getting
> W-statistic greater than or equal to 0.9513 (I picked this up from the
> Dalgaard book, Introductory Statistics with R, but its not really
> sinking in with respect to how it applies to a Shipiro Wilk test).?
> The method description - retrieved using ?shapiro.test() - is a bit
> light on details.
> Thanks much.
The null hypothesis: the data is normally distributed.
If p-value > \alpha (significance level) it means that there is no evidence
to reject null hypothesis. Otherwise you reject - your data is not normally
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