# For Chi-Squared test on contingency tables there is a proof to get from: sum((O_i−E_i)^2)/(E_i) which equals (N(ad−bc)2)/((a+b)(c+d)(a+c)(b+d)) Can anyone explain the steps in the proof i know how to get from one to other but not sure why certain steps happen!

For Chi-Squared test on contingency tables there is a proof to get from: $\sum \frac{\left({O}_{i}-{E}_{i}{\right)}^{2}}{{E}_{i}}$ which equals $\frac{N\left(ad-bc{\right)}^{2}}{\left(a+b\right)\left(c+d\right)\left(a+c\right)\left(b+d\right)}$Can anyone explain the steps in the proof i know how to get from one to other but not sure why certain steps happen!
Below ill put the proof if anyone wants to see it or can explain it?
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domino671v
Same way, but done by human:
Starting with ${\chi }^{2}=\sum _{i=1}^{4}\frac{\left({o}_{i}-{e}_{i}{\right)}^{2}}{{e}_{i}}$, and expand the square:
${\chi }^{2}=\sum _{i=1}^{4}\frac{\left({o}_{i}-{e}_{i}{\right)}^{2}}{{e}_{i}}=\sum _{i=1}^{4}\left(\frac{{o}_{i}^{2}}{{e}_{i}}-2{o}_{i}+{e}_{i}\right)=\sum _{i=1}^{4}\frac{{o}_{i}^{2}}{{e}_{i}}-n$
(since $\sum {o}_{i}=\sum {e}_{i}=n$)
Substitute the values in:
${\chi }^{2}=\frac{n{a}^{2}}{\left(a+c\right)\left(a+b\right)}-a+\frac{n{b}^{2}}{\left(b+d\right)\left(a+b\right)}-b+\frac{n{c}^{2}}{\left(a+c\right)\left(c+d\right)}-c+\frac{n{d}^{2}}{\left(c+d\right)\left(b+d\right)}-d$
Notice:
$an-\left(a+c\right)\left(a+b\right)=a\left(a+b+c+d\right)-\left(a+c\right)\left(a+b\right)=ad-bc$
So,
$\frac{a}{\left(a+c\right)\left(a+b\right)}\left[na-\left(a+c\right)\left(a+b\right)\right]=\frac{a\left(ad-bc\right)}{\left(a+c\right)\left(a+b\right)}$
Do the same thing for b,c,d.