# [R] Shapiro-Wilk test

Bill.Venables@CMIS.CSIRO.AU Bill.Venables at CMIS.CSIRO.AU
Mon Jul 2 15:06:41 CEST 2001

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>  -----Original Message-----
> From: 	Ralf Goertz [mailto:R.Goertz at psych.uni-frankfurt.de]
> Sent:	Monday, 2 July 2001 10:48 PM
> To:	r-help at stat.math.ethz.ch
> Subject:	[R] Shapiro-Wilk test
>
> Hi,
>
> does the shapiro wilk test in R-1.3.0 work correctly? Maybe it does, but
can
> anybody tell me why the following sample doesn't give "W = 1" and
> "p-value = 1":
>
> R> x<-1:9/10;x
> [1] 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9
> R> shapiro.test(qnorm(x))
>
>         Shapiro-Wilk normality test
>
> data:  qnorm(x)
> W = 0.9925, p-value = 0.9986
>
> I can't imagine a sample being more "normal" than this. Furthermore, the
> Kolmogorov-Smirnov test gives a p-value of 1.

Even though the expected probability content between any two consecutive
order statistics is 1/(n+1), the *expected* values of the order statistics
themselves are not the percentiles.  You can get a more "normal" looking
sample by choosing one where the values are the normal scores.
qnorm(ppoints(n)) gives a pretty good approximation to these:

> shapiro.test(qnorm(1:9/10))

Shapiro-Wilk normality test

data:  qnorm(1:9/10)
W = 0.9925, p-value = 0.9986

> shapiro.test(qnorm(ppoints(9)))

Shapiro-Wilk normality test

data:  qnorm(ppoints(9))
W = 0.9965, p-value = 1

>

So the Shapiro-Wilk test is really a check to see if the order statistics
depart significantly from *their* expectation under the normality
hypothesis.

> R> ks.test(qnorm(x),"pnorm",mean=0,sd=1)
>
>         One-sample Kolmogorov-Smirnov test
>
> data:  qnorm(x)
> D = 0.1, p-value = 1
> alternative hypothesis: two.sided
>
> Can someone explain this please,

The two tests are testing slightly different aspects of normality.  (At
least that's my guess...)

> Ralf
>
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