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!

fofopausiomiava

fofopausiomiava

Answered question

2022-10-11

For Chi-Squared test on contingency tables there is a proof to get from: ( O i E i ) 2 E i which equals N ( a d b c ) 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?

Answer & Explanation

domino671v

domino671v

Beginner2022-10-12Added 8 answers

Same way, but done by human:
Starting with χ 2 = i = 1 4 ( o i e i ) 2 e i , and expand the square:
χ 2 = i = 1 4 ( o i e i ) 2 e i = i = 1 4 ( o i 2 e i 2 o i + e i ) = i = 1 4 o i 2 e i n
(since o i = e i = n)
Substitute the values in:
χ 2 = n a 2 ( a + c ) ( a + b ) a + n b 2 ( b + d ) ( a + b ) b + n c 2 ( a + c ) ( c + d ) c + n d 2 ( c + d ) ( b + d ) d
Notice:
a n ( a + c ) ( a + b ) = a ( a + b + c + d ) ( a + c ) ( a + b ) = a d b c
So,
a ( a + c ) ( a + b ) [ n a ( a + c ) ( a + b ) ] = a ( a d b c ) ( a + c ) ( a + b )
Do the same thing for b,c,d.

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