Chi-square test vs K-S test of uniformity, is it necessary to normalize values between 0-1?

In summary, the conversation discusses the need to normalize two digit numbers between 0 and 1 for the K-S test, but it is not clear if this is necessary for the chi-square test. The idea of normalization is mentioned, but it is unclear how it should be done and if it is necessary for the chi-square test. The concept of normalization for this calculation is ambiguous and may depend on personal standards.f
  • #1
Say two digits numbers are given like 10,20,30,55,95,85,12,13,52...etc. Is it necessary to normalize them to numbers between 0 to 1? i.e 0.10 for 10, 0.20 for 20 and so on? I've read this to be the case for K-S test. But I'm not sure for chi-square test. I'm not 100% sure on this information as I've not seen it everywhere.
It looks like it's the case for K-S test as x needs to be between 0 to 1, not sure for Chi-square test?
  • #2
It doesn't seem as though it's necessary according to
(also 10 doesn't normalize to .10 unless the total is 100)
They define:
##\chi^2 = \Sigma \frac{(O-E)^2}{E}## which isn't quite a normalization. O is observed E is expected.
You also need some standard, which they call a critical value.
I am not sure if it matters, though.

If we look at a specific example for 1 data point (using your "normalization", i.e. divide by 100):
O = 15, E =10 -> ##\chi^2 = \frac{5^2}{10} = 2.5## vs ## \frac{.05^2}{.1} =.025##

Thinking about it, it's not even clear how you should normalize O and E. Is it always necessary that Sum(O) = Sum(E)? I think it depends.
If your list is randomly generated, you should expect 50 from each, say we have 10 data points, that's 500 total expected units. It could be the case that in some niche trial the computer generates all 1's or all 99's, in which case there would be more observed units than expected units. The only way to actually normalize both O and E would be to normalize them separately, like ##\frac{O_i}{\Sigma_k O_k}## and the same for E. I'm not sure if that's reasonable or not.
Looking at the case where the numbers are all random, we generate 10 points from 1 to 99, the average is 50, which is what we should expect from each. The computer generates 10-1's. Our normalized O values would be 1/10, and our normalized E values would be 50/500 = 1/10. This would give chi^2 = 0, which would imply that our model was right on the money. This is clearly not the case.

If we normalize using ##\frac{O_i}{\Sigma_k E_k}## we get ##\bar{O_i} = \frac{1}{500} \to \chi^2 = \Sigma \frac{(.002 - .1)^2}{.1} = .9604##
Compare that without "normalization" and we have
##\chi^2 = \Sigma \frac{(1-50)^2}{50} = 480.2##
2 very different numbers, but interestingly enough the 2nd one is 500 times larger.
The first one seems to be something to the effect of %error, and at the end of they day, I think it's all going to come down to standards. The website I read didn't do it, nor did they mention it, and in the wikipedia, they give a specific example and don't do it either.

I haven't seen it written, but I think the concept of "normalization" is ambiguous for this calculation.

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