From: <Matthew.Findley_at_ch2m.com>

Date: Thu 13 Jul 2006 - 07:13:42 EST

Matthew C. Findley, CPSSc

Environmental Scientist

R-help@stat.math.ethz.ch mailing list

https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html Received on Thu Jul 13 07:33:36 2006

Date: Thu 13 Jul 2006 - 07:13:42 EST

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.test(As)

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.

Matthew C. Findley, CPSSc

Environmental Scientist

**CH2M HILL
**

mfindley@ch2m.com

R-help@stat.math.ethz.ch mailing list

https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html Received on Thu Jul 13 07:33:36 2006

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