Looking for an easy proof/intuition for the fact* that (assuming $E_i$):
$\sqrt{\frac{({O_i-E_i})^2}{E_i}} = \frac{{O_i-E_i}}{\sqrt{E_i}} \sim N(0,1)$
A similar question exits but is not as concrete as this one.
*The $\chi^2$-distribution is defined as the distribution of a sum of the squares of k independent standard normal random variables.
By exploring connections among the binomial, normal, and chi-squared distributions, one can answer.
$Y∼Bin(n,p)$ i.e. random variable $Y$ is number of successes in $n$ trials, with probability of success in a trial being $p$.
Assume $n$ and $p$ are such that $Y$ is well approximated by a normal distribution with mean = $np$ and variance = $np(1-p)$
Let's say $Y$ to be number of successes then,
$(Y−E(Y))^2/Var(Y)$ (i.e. square of standard normal.) will be approximately $\sim\chi^2_1$, where $E(Y)=np$ and $Var(Y)=np(1−p)$.
So $(Y−np)^2/np(1−p)$ will be approximately $\sim\chi^2_1$.
Using facts that $(Y−np)^2=[(n−Y)−n(1−p)]^2$ and the algebraic relation $1/p+1/(1−p)=1/p(1−p)$.
So,
$\frac{(Y-np)^2}{np(1-p)}$=$\frac{(Y-np)^2}{np}+\frac{(Y-np)^2}{n(1-p)}$ $\quad= \frac{(Y-np)^2}{np}+\frac{[(n-Y)-n(1-p)]^2}{n(1-p)} \\ \quad= \frac{(O_S-E_S)^2}{E_S}+\frac{(O_F-E_F)^2}{E_F}$
So, chi-square statistic $\frac{(O_S-E_S)^2}{E_S}+\frac{(O_F-E_F)^2}{E_F}$will have the distribution of the square of an approximately standard-normal random variable ex - $(Y-np)/\sqrt{np(1-p)}$
