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 2022-10-11 Answered
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?
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Answers (1)

domino671v
Answered 2022-10-12 Author has 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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