Simple Question about Chi-Squared Statistic I have a basic question about the chi-squared statistic. Perhaps I am misunderstanding, or perhaps I am looking for a different test. I am mathematical to a degree but am (becoming) self-educated in statistics. 
I understand how to calculate the statistic and do hypothesis testing, according to the textbooks. However, the statistic is very sensitive to the total frequency, in that, if X2(O,E) is the statistic, then increasing all the frequencies by some factor N increases the statistic by the same factor, i.e., 
X2(N*O,N*E)  = N * X2(O,E)          

where N*O is multiplying a vector by a scalar.  Therefore, the statistic is much more sensitive to the total frequency than to the degrees of freedom. When testing goodness-of-fit for a distribution, it seems like the total frequency should NOT be the most significant factor. 
This seems odd and I am wondering what I am missing, if anything. 
 A: This is a general problem with goodness of fit tests. 
If I fit a cancer risk model in 500,000 patients, I know I will have a nicely calibrated risk model. If I collect cancer outcomes in an independent set of 500,000 patients, I know that my previous model will fail most goodness of fit tests. 
The reason for the problem is over-reliance on the $p$-value: the more $N$ you have, the more likely you are to reject the null hypothesis even if the differences are small and unassuming. Therefore, I look to the $p$-value moreover as a measure of sample size than anything. Of course, in your example under the null, "goodness-of-fit" is only achieved if $O = E$ exactly. In that case, the chi-square statistic will have an expected value of the degrees of freedom of the test, and the $p$-value is 0.5 every time. 
However, you and I agree that small differences from $E$ and $O$ do not mean that the model fits poorly, and yet, if you were to sample ad infinitum, the test would ultimately reject and conclude that there is poor goodness of fit tests.
I summarily do not conduct goodness-of-fit tests without first visualizing the difference, if a picture cannot convince, neither should a statistical test.
