The formula for the Chi-Square test statistic is the following:
$$\chi^2=\sum_{i=1}^n\frac{({O_i-E_i})^2}{E_i}$$
where $O_i$ is observed data, and $E_i$ is expected.
I am just curious why this follows the $\chi^2$ distribution?
[Math] Why the chi-squared statistic follows chi-squared distribution
chi squaredprobability distributionsstatistics
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In the version of this test that I am familiar with, individual data is categorical, not quantitative like your examples. And the expected and observed values should be frequencies of some category (a count of how many times it occurs), not some individual's quantitative measurement. The numbers that go in to the $E_i$ and $O_i$ positions are unitless, as they are just counts.
So for example, in a box with mixed fruit, maybe 12 pieces were bananas, but you were expecting 15 to be bananas. You will have the term $$\frac{(12-15)^2}{15}$$ and there is no way to rescale units as you did. Writing $$\frac{(12000-15000)^2}{15000}$$ would correspond to a very different scenario. There you would have seen 12000 bananas when you were expecting 15000. And the corresponding $P$ value should be a lot smaller, because it should be a lot less likely to be off by 3000 out of 15000 than 3 out of 15, when you consider the variance from one piece of fruit to the next on its chances to be a banana. So $\chi^2$ should be a lot larger in the latter case.
The general idea is that we want the test statistic $\chi^{2}$ to be independent of the scale of the data (or more precisely, the spread) as the actual probability distribution is independent of the scale. That is, given a distribution with variance $\sigma^{2}$, we can always divide out by $\sigma^{2}$ to get a unit-variance distribution, and it is both valid and simpler to always just work with a unit-variance distribution (we just have one fewer parameter to think about).
More specifically to $\chi^{2}$, the reason that we divide by $E_{i}$ has to do with the underlying distribution. First, it is important to realize that what $\chi^{2}$ random variables are. Generally, a $\chi^{2}$ variable with $n$ degrees of freedom is the sum of squares of $n$ unit-variance normal random variables. So when we define $$ \chi^{2} = \sum \frac{(O_{i} - E_{i})^{2}}{E_{i}} $$ we are using the fact that each $\frac{(O_{i} - E_{i})}{\sqrt{E_{i}}}$ is normally distributed with unit-variance. More precisely, we can treat $O_{i}$ as being normally distributed with mean $E_{i}$ and variance $E_{i}$. The value $\frac{(O_{i} - E_{i})}{\sqrt{E_{i}}}$ is the standardized residual.
There is a twist that explains why $E_{i}$ is the variance: the value $O_{i}$ is actually poisson distributed. This makes sense since we are working with counts and not something continuous. In the poisson distribution, the variance is precisely the same as the mean $E_{i}$. The question then is why we can treat $O_{i}$ as normally distributed. The reason is that when the expected counts $E_{i}$ is `large enough', the poisson distribution is reasonably similar to the normal distribution (very, very roughly).
On a final note, the variance of a $\chi^{2}$ variable is actually not one but is dependent on the degrees of freedom. Regardless, the distribution only has the one parameter (the degrees of freedom) and no variance parameter, so it's the same idea as using a unit-variance distribution.
Best Answer
It is $\frac{O_i-E_i}{E_i}$ that follows a normal distribution not its square root. We are just assuming that relative errors are Gaussian. It is just an assumption. The square of the gaussian variable ~ Gamma Distribution (Chi-square). The sum of the these squared variables follow a Chi-square with $n$ degrees of freedom. Now if we were to look at absolute values , namely $\left|\frac{O_i-E_i}{E_i}\right|$ , we would have a half-normal distribution and the sum $\sum_{i=1}^n\left|\frac{O_i-E_i}{E_i}\right|$would end up converging to a Gaussian, though unlike Chi-Square, without a clearly standard distribution for $n$ small.