[R] Shapiro-Wilk test

[Bill.Venables@CMIS.CSIRO.AU]

>  -----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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