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Shapiro-Wilk test

2 messages · Ralf Goertz, Brian Ripley

Original
Ralf Goertz
# 
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.

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,

Ralf
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Brian Ripley
#
On Mon, 2 Jul 2001, Ralf Goertz wrote:

            

      
      
      
    

        
Well, both I and the Shapiro-Wilk test can.  That sample has too short
tails:

            

      
      
      
    

        
Shapiro-Wilk normality test

data:  qnorm((a + 0:8)/(8 + 2 * a))
W = 0.9925, p-value = 0.9986

            

      
      
      
    

        
Shapiro-Wilk normality test

data:  qnorm((a + 0:8)/(8 + 2 * a))
W = 0.9998, p-value = 1

(Values less than 1 are normally used in Q-Q plots. For example

            

      
      
      
    

        
[1] 0.06756757 0.17567568 0.28378378 0.39189189 0.50000000 0.60810811 0.71621622
[8] 0.82432432 0.93243243

corresponds to a = 5/8.)

            

      
      
      
    

        
Actually

            

      
      
      
    

        
[1] 0.9999907

            

      
      
      
    

        
[1] 0.9999999

            

      
      
      
    

        
Tail sensitivity varies between tests, and  K-S is notoriously
insensitive.

            

      
      
      
    

        
If a command does the wrong thing, that is a bug.  But be sure you know
for certain what it ought to have done.  If you aren't familiar with the
command, or don't know for certain how the command is supposed to work,
then it might actually be working right.  Rather than jumping to
conclusions, show the problem to someone who knows for certain.