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For Chi-Squared test on contingency tables there is a proof to get from: $\sum\frac{(O_i - E_i)^2}{E_i}$ which equals $\frac{N(ad-bc)^2}{(a+b)(c+d)(a+c)(b+d)}$
Can anyone explain the steps in the proof i know how to get from one to other but not sure why certain steps happen!
Below ill put the proof if anyone wants to see it or can explain it?
Thanks
Same way, but done by human:
Starting with $\chi^2=\sum^4_{i=1}\frac{(o_i-e_i)^2}{e_i}$, and expand the square: $$\chi^2=\sum^4_{i=1}\frac{(o_i-e_i)^2}{e_i}=\sum^4_{i=1}\bigl(\frac{o_i^2}{e_i}-2o_i+e_i\bigr)=\sum^4_{i=1}\frac{o_i^2}{e_i}-n$$ (since $\sum o_i=\sum e_i =n$)
Substitute the values in: $$\chi^2=\frac{na^2}{(a+c)(a+b)}-a+\frac{nb^2}{(b+d)(a+b)}-b+\frac{nc^2}{(a+c)(c+d)}-c+\frac{nd^2}{(c+d)(b+d)}-d$$
Notice: $$an-(a+c)(a+b)=a(a+b+c+d)-(a+c)(a+b)=ad-bc$$
So, $$\frac{a}{(a+c)(a+b)}[na-(a+c)(a+b)]=\frac{a(ad-bc)}{(a+c)(a+b)}$$
Do the same thing for b,c,d.
It now gets fairly staight forward
I went through the same problem as yours. I used Mathematica to ease the computations.
$\chi ^2=\sum _{i=1}^4 \frac{(O_i-E_i)^2}{E_i}=\frac{n(a d-b c)^2}{(a+c)(b+d)(a+b)(c+d)}$
$O_i$ are the observed values, $O_1=a, O_2=b,O_3=c$ and $O_4=d$
$E_i$ are the expected values, $E_1=\frac{(a+b)(a+c)}{n}, E_2=\frac{(a+b)(b+d)}{n},E_3=\frac{(a+c)(c+d)}{n}$ and $E_4=\frac{(b+d)(c+d)}{n}$
Then by expanding the summation we have: $\sum _{i=1}^4 \frac{(O_i-E_i)^2}{E_i}=\frac{(O_1-E_1)^2}{E_1}+\frac{(O_2-E_2)^2}{E_2}+\frac{(O_3-E_3)^2}{E_3}+\frac{(O_4-E_4)^2}{E_4}$
Substituting in the summation we have:
$\frac{n\left(a-\frac{(a + b)(a+c)}{n}\right)^2}{(a + b)(a+c)}+\frac{n\left(b-\frac{(a + b)(b+d)}{n}\right)^2}{(a + b)(b+d)}+\frac{n\left(c-\frac{(a + c)(c+d)}{n}\right)^2}{(a + c)(c+d)}+\frac{n\left(d-\frac{(b + d)(c+d)}{n}\right)^2}{(b + d)(c+d)}$
$\text{Simplify}\left[\frac{n\left(a-\frac{(a + b)(a+c)}{n}\right)^2}{(a + b)(a+c)}+\frac{n\left(b-\frac{(a + b)(b+d)}{n}\right)^2}{(a + b)(b+d)}+\frac{n\left(c-\frac{(a + c)(c+d)}{n}\right)^2}{(a + c)(c+d)}+\frac{n\left(d-\frac{(b + d)(c+d)}{n}\right)^2}{(b + d)(c+d)}\right]$
$\frac{((b+d) (c+d) ((a+b) (a+c)-a n)^2+(a+c) (c+d) ((a+b) (b+d)-b n)^2+(a+b) (b+d) ((a+c) (c+d)-c n)^2+(a+b) (a+c) ((b+d) (c+d)-d n)^2}{((a+b) (a+c) (b+d) (c+d) n)}$
Then expanding the numerator and distributing the denominator all over we have:
$\text{Expand}[\frac{((b+d) (c+d) ((a+b) (a+c)-a n)^2+(a+c) (c+d) ((a+b) (b+d)-b n)^2+(a+b) (b+d) ((a+c) (c+d)-c n)^2+(a+b) (a+c) ((b+d) (c+d)-d n)^2}{((a+b) (a+c) (b+d) (c+d) n)}]$
$ -\frac{2 a^3 b c}{(a+b) (a+c) (b+d) (c+d)}-\frac{4 a^2 b^2 c}{(a+b) (a+c) (b+d) (c+d)}-\frac{2 a b^3 c}{(a+b) (a+c) (b+d) (c+d)}-\frac{4 a^2 b c^2}{(a+b) (a+c) (b+d) (c+d)}-\frac{6 a b^2 c^2}{(a+b) (a+c) (b+d) (c+d)}-\frac{2 b^3 c^2}{(a+b) (a+c) (b+d) (c+d)}-\frac{2 a b c^3}{(a+b) (a+c) (b+d) (c+d)}-\frac{2 b^2 c^3}{(a+b) (a+c) (b+d) (c+d)}-\frac{2 a^3 b d}{(a+b) (a+c) (b+d) (c+d)}-\frac{4 a^2 b^2 d}{(a+b) (a+c) (b+d) (c+d)}-\frac{2 a b^3 d}{(a+b) (a+c) (b+d) (c+d)}-\frac{2 a^3 c d}{(a+b) (a+c) (b+d) (c+d)}-\frac{10 a^2 b c d}{(a+b) (a+c) (b+d) (c+d)}-\frac{10 a b^2 c d}{(a+b) (a+c) (b+d) (c+d)}-\frac{2 b^3 c d}{(a+b) (a+c) (b+d) (c+d)}-\frac{4 a^2 c^2 d}{(a+b) (a+c) (b+d) (c+d)}-\frac{10 a b c^2 d}{(a+b) (a+c) (b+d) (c+d)}-\frac{6 b^2 c^2 d}{(a+b) (a+c) (b+d) (c+d)}-\frac{2 a c^3 d}{(a+b) (a+c) (b+d) (c+d)}-\frac{2 b c^3 d}{(a+b) (a+c) (b+d) (c+d)}-\frac{2 a^3 d^2}{(a+b) (a+c) (b+d) (c+d)}-\frac{6 a^2 b d^2}{(a+b) (a+c) (b+d) (c+d)}-\frac{4 a b^2 d^2}{(a+b) (a+c) (b+d) (c+d)}-\frac{6 a^2 c d^2}{(a+b) (a+c) (b+d) (c+d)}-\frac{10 a b c d^2}{(a+b) (a+c) (b+d) (c+d)}-\frac{4 b^2 c d^2}{(a+b) (a+c) (b+d) (c+d)}-\frac{4 a c^2 d^2}{(a+b) (a+c) (b+d) (c+d)}-\frac{4 b c^2 d^2}{(a+b) (a+c) (b+d) (c+d)}-\frac{2 a^2 d^3}{(a+b) (a+c) (b+d) (c+d)}-\frac{2 a b d^3}{(a+b) (a+c) (b+d) (c+d)}-\frac{2 a c d^3}{(a+b) (a+c) (b+d) (c+d)}-\frac{2 b c d^3}{(a+b) (a+c) (b+d) (c+d)}+\frac{a^4 b c}{(a+b) (a+c) (b+d) (c+d) n}+\frac{3 a^3 b^2 c}{(a+b) (a+c) (b+d) (c+d) n}+\frac{3 a^2 b^3 c}{(a+b) (a+c) (b+d) (c+d) n}+\frac{a b^4 c}{(a+b) (a+c) (b+d) (c+d) n}+\frac{3 a^3 b c^2}{(a+b) (a+c) (b+d) (c+d) n}+\frac{7 a^2 b^2 c^2}{(a+b) (a+c) (b+d) (c+d) n}+\frac{5 a b^3 c^2}{(a+b) (a+c) (b+d) (c+d) n}+\frac{b^4 c^2}{(a+b) (a+c) (b+d) (c+d) n}+\frac{3 a^2 b c^3}{(a+b) (a+c) (b+d) (c+d) n}+\frac{5 a b^2 c^3}{(a+b) (a+c) (b+d) (c+d) n}+\frac{2 b^3 c^3}{(a+b) (a+c) (b+d) (c+d) n}+\frac{a b c^4}{(a+b) (a+c) (b+d) (c+d) n}+\frac{b^2 c^4}{(a+b) (a+c) (b+d) (c+d) n}+\frac{a^4 b d}{(a+b) (a+c) (b+d) (c+d) n}+\frac{3 a^3 b^2 d}{(a+b) (a+c) (b+d) (c+d) n}+\frac{3 a^2 b^3 d}{(a+b) (a+c) (b+d) (c+d) n}+\frac{a b^4 d}{(a+b) (a+c) (b+d) (c+d) n}+\frac{a^4 c d}{(a+b) (a+c) (b+d) (c+d) n}+\frac{8 a^3 b c d}{(a+b) (a+c) (b+d) (c+d) n}+\frac{14 a^2 b^2 c d}{(a+b) (a+c) (b+d) (c+d) n}+\frac{8 a b^3 c d}{(a+b) (a+c) (b+d) (c+d) n}+\frac{b^4 c d}{(a+b) (a+c) (b+d) (c+d) n}+\frac{3 a^3 c^2 d}{(a+b) (a+c) (b+d) (c+d) n}+\frac{14 a^2 b c^2 d}{(a+b) (a+c) (b+d) (c+d) n}+\frac{16 a b^2 c^2 d}{(a+b) (a+c) (b+d) (c+d) n}+\frac{5 b^3 c^2 d}{(a+b) (a+c) (b+d) (c+d) n}+\frac{3 a^2 c^3 d}{(a+b) (a+c) (b+d) (c+d) n}+\frac{8 a b c^3 d}{(a+b) (a+c) (b+d) (c+d) n}+\frac{5 b^2 c^3 d}{(a+b) (a+c) (b+d) (c+d) n}+\frac{a c^4 d}{(a+b) (a+c) (b+d) (c+d) n}+\frac{b c^4 d}{(a+b) (a+c) (b+d) (c+d) n}+\frac{a^4 d^2}{(a+b) (a+c) (b+d) (c+d) n}+\frac{5 a^3 b d^2}{(a+b) (a+c) (b+d) (c+d) n}+\frac{7 a^2 b^2 d^2}{(a+b) (a+c) (b+d) (c+d) n}+\frac{3 a b^3 d^2}{(a+b) (a+c) (b+d) (c+d) n}+\frac{5 a^3 c d^2}{(a+b) (a+c) (b+d) (c+d) n}+\frac{16 a^2 b c d^2}{(a+b) (a+c) (b+d) (c+d) n}+\frac{14 a b^2 c d^2}{(a+b) (a+c) (b+d) (c+d) n}+\frac{3 b^3 c d^2}{(a+b) (a+c) (b+d) (c+d) n}+\frac{7 a^2 c^2 d^2}{(a+b) (a+c) (b+d) (c+d) n}+\frac{14 a b c^2 d^2}{(a+b) (a+c) (b+d) (c+d) n}+\frac{7 b^2 c^2 d^2}{(a+b) (a+c) (b+d) (c+d) n}+\frac{3 a c^3 d^2}{(a+b) (a+c) (b+d) (c+d) n}+\frac{3 b c^3 d^2}{(a+b) (a+c) (b+d) (c+d) n}+\frac{2 a^3 d^3}{(a+b) (a+c) (b+d) (c+d) n}+\frac{5 a^2 b d^3}{(a+b) (a+c) (b+d) (c+d) n}+\frac{3 a b^2 d^3}{(a+b) (a+c) (b+d) (c+d) n}+\frac{5 a^2 c d^3}{(a+b) (a+c) (b+d) (c+d) n}+\frac{8 a b c d^3}{(a+b) (a+c) (b+d) (c+d) n}+\frac{3 b^2 c d^3}{(a+b) (a+c) (b+d) (c+d) n}+\frac{3 a c^2 d^3}{(a+b) (a+c) (b+d) (c+d) n}+\frac{3 b c^2 d^3}{(a+b) (a+c) (b+d) (c+d) n}+\frac{a^2 d^4}{(a+b) (a+c) (b+d) (c+d) n}+\frac{a b d^4}{(a+b) (a+c) (b+d) (c+d) n}+\frac{a c d^4}{(a+b) (a+c) (b+d) (c+d) n}+\frac{b c d^4}{(a+b) (a+c) (b+d) (c+d) n}+\frac{a^2 b c n}{(a+b) (a+c) (b+d) (c+d)}+\frac{a b^2 c n}{(a+b) (a+c) (b+d) (c+d)}+\frac{a b c^2 n}{(a+b) (a+c) (b+d) (c+d)}+\frac{2 b^2 c^2 n}{(a+b) (a+c) (b+d) (c+d)}+\frac{a^2 b d n}{(a+b) (a+c) (b+d) (c+d)}+\frac{a b^2 d n}{(a+b) (a+c) (b+d) (c+d)}+\frac{a^2 c d n}{(a+b) (a+c) (b+d) (c+d)}+\frac{b^2 c d n}{(a+b) (a+c) (b+d) (c+d)}+\frac{a c^2 d n}{(a+b) (a+c) (b+d) (c+d)}+\frac{b c^2 d n}{(a+b) (a+c) (b+d) (c+d)}+\frac{2 a^2 d^2 n}{(a+b) (a+c) (b+d) (c+d)}+\frac{a b d^2 n}{(a+b) (a+c) (b+d) (c+d)}+\frac{a c d^2 n}{(a+b) (a+c) (b+d) (c+d)}+\frac{b c d^2 n}{(a+b) (a+c) (b+d) (c+d)} $
Then we fully simplify that messy horrible thing
FullSimplify[Messy horrible thing from above]
$ -2 (a+b+c+d)+\frac{(a+b+c+d)^2}{n}+\frac{\left(b c (d (c+d)+b (2 c+d))+a^2 (b (c+d)+d (c+2 d))+a \left(b^2 (c+d)+c d (c+d)+b \left(c^2+d^2\right)\right)\right) n}{(a+b) (a+c) (b+d) (c+d)} $
Up to this point we observe we have the term $(a + b + c + d)$ which we know is equivalent to $n$. We substitute that by $n$ in the following equation and simplify
$ \text{Simplify}[-2 (n)+\frac{(n)^2}{n}+\frac{\left(b c (d (c+d)+b (2 c+d))+a^2 (b (c+d)+d (c+2 d))+a \left(b^2 (c+d)+c d (c+d)+b \left(c^2+d^2\right)\right)\right) n}{(a+b) (a+c) (b+d) (c+d)}] $
$ \frac{(b c-a d)^2 n}{(a+b) (a+c) (b+d) (c+d)} $
There you go, hope this helps.
