Difference between K-S manual test and K-S test with R?

Question

On the Internet there is an example of k-s test being applied relative to distribution of number of bird varieties over different five hour periods. The observed distribution was:

a=c(0,1,1,9,4)

The expected distribution (if there is no difference between the five hours) could be:

b=c(3,3,3,3,3)

After I found the two cumulate distributions, I calculated manually, D = 0,4667 (a value that is similar with the internet). But if I try to use R, I find a different value of D:

> a=c(0,1,1,9,4)

> b=c(3,3,3,3,3)

> ks.test(a,b)

Two-sample Kolmogorov-Smirnov test

data: a and b 

D = 0.6, p-value = 0.3291

alternative hypothesis: two-sided .......

• What is leading to the difference between my manual calculation and result that R gives?

Answer 1

You are testing a different thing.

While you think c(0,1,1,9,4) means you are looking at 0 values of one, 1 value of two, 1 value of three, 9 values of four, and 4 values of five, R thinks you are looking at one value of 0, two values of 1, one value of 9, and one value of 4.

To get D = 0.4667..., try the rather verbose

ks.test( c(2,3,4,4,4,4,4,4,4,4,4,5,5,5,5), 
         c(1,1,1,2,2,2,3,3,3,4,4,4,5,5,5) ) 

giving

    Two-sample Kolmogorov-Smirnov test 

D = 0.4667, p-value = 0.07626 
alternative hypothesis: two-sided  

Answer 2

First of all, the Kolmogorov-Smirnoff test is probably incorrect for your situation. I assume you want to test whether the distribution of the number bird sightings is uniform (constant accross time). A chi-square goodness-of-fit test would be the simplest solution for that.

Second, the R-produced value of 0.6 seems correct to me: at $x=2.5$, $F_a(x)=0.6$, while $F_b(x)=0$.