-----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...)
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