# Question on G-Test

1. Apr 15, 2014

### mnb96

Hello,

it is claimed that the so called G-Test can be used as a replacement for the well-known Chi-squared test. The G-test is defined as: $$G = 2\sum_i O_i \cdot \log \left( \frac{O_i}{E_i}\right)$$where Oi and Ei are the observed and expected counts in the cell i of a contingency table.

I see a big problem with this.
Namely, the value G is directly proportional to the total amount N of observations!

This is easily seen even with the most trivial example of a coin toss. Suppose we want to test wheter a coin is fair or not. We collect N=10 samples and we obtain {1 head, 9 tails}. Thus, according to the above formula G≈7.36.
Now suppose we collect N=100 samples and we obtain {10 heads, 90 tails}. Well, according to the above formula we now get G≈73.6, exactly ten times more.

So, what is the threshold value for G above which we reject the null-hypothesis that the coin is fair?

2. Apr 16, 2014

### Stephen Tashi

What's bad about that? Intuitively, more trials provide more evidence of a trend toward tails.

What $\alpha$ do you want to use?

"The" chi-square distribution is actually a family of distributions. You have to specify the "degrees of freedom" to specify a particular distribution.

3. Apr 16, 2014

### mnb96

Well, let's say I want to set α=0.005.
If we stick with the example of the coin-toss in my previous post, we have only 1 degree of freedom, to which it correspond a P-value of 7.879. Thus in the first case, where we had only 10 tosses, we won't yet reject the hypothesis that the coin is fair (G was ~7.36).
In the second case when we have 100 tosses (more evidence), we obtained G≈73.6 which is more than enough to reject the hypothesis that the coin is fair.

Yes, now it makes sense. I was just missing the correct interpretation.
I believe that what confused me is that in both scenarios we had 90% tails and 10% heads, and I wrongly expected to get the same G-value.

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