ks.gof or not?

[<Andre.Mery@aventis.com>]

Hi,

 

I have a problem with the ks.gof function. Basically I have 2 distributions like that:

 

First:

Class              0    1     2    3     4     5     6 

Observed counts    0    1    12    4    17    25    41

 

Second:

Class                0        1      2       3       4       5       6

Expected counts    1.5625  9.3750 23.4375 31.2500 23.4375  9.3750  1.5625

 

Using ks.gof for testing the difference between these 2 distributions leads to:

 

> ks.gof(Observed, Expected)

Two-Sample Kolmogorov-Smirnov Test

data:  Observed and Expected
ks = 0.2857, p-value = 0.9627
alternative hypothesis:
  cdf of Observed does not equal the
              cdf of Observed for at least one sample point.

 

H0 is that the distributions are identical, true ? Then as p=0.96, I cannot reject H0. But the observed distribution is clearly J-shaped and the expected distribution is from a binomial law with p=0.5, n=100, and is completely symmetrical. Plotting the 2 shows a clear difference.

 

What did I miss in the way to use ks.gof ? Is it inappropriate in that case or do I use it wrongly ?

 

Thanks.

 

André Méry
Aventis Pharma
20, avenue Raymond Aron - 92160 Antony
[Courrier / Mail : Tri B2/13]
[Tél. / Tel. : 01 55 71 68 69]

 

André Méry
Aventis Pharma
20, avenue Raymond Aron - 92160 Antony
[Courrier / Mail : Tri B2/13]
[Tél. / Tel. : 01 55 71 68 69]