CHI-SQUARE TEST

Here we present a program to compute the chi square test for 3 groups and 7 parameters.

Using material from
Montgomery C. Douglas and Runger C. George, Applied Statistics and Probability for Engineers, Wiley, 2007.

INPUT
F = [ F1 F2 F3 F4 F5 F6 F7
          F8 F9 F10 F11 F12 F13 F14
          F15 F16 F17 F18 F19 F20 F21 ]
= [F]

Assuming these are three separate samples and each sample contains 7 parameters

OUTPUT
Null hypothesis: there is no difference between the means of the different levels
Alternative: at least two of the means are different

2 → DIM(L6)
DIM([F]) → L6
L6(1) → S
L6(2) → T
0 → DIM (L4)
S → DIM (L4)
0 → A

FOR(P,1,S,1)
FOR(M,1,T,1)
L4(P) + [F](P,M) → L4(P)
END
A + L4(P) → A
END

0 → DIM(L5)
T → DIM(L5)

FOR(M,1,T,1)
FOR(P,1,S,1)
L5(M) + [F](P,M) → L5(M)
END
END

DIM([F]) → DIM([G])

FOR(M,1,T,1)
FOR(P,1,S,1)
(L4(P) * L5(M))/A) → [G](P,M)
END
END

0 → B

FOR(M,1,7,1)
FOR(P,1,3,1)
B+(([G](P,M)-[F](P,M) )^2)/[G](P,M)) → B
END
END

(S-1)*(T-1) → K
K/2 → K

“(x^(K-1))*e^(-x/2))/ ((2^K)*(K-1)!)→ Y1

fnInt(Y1,x,0,B) → H
1-H → Z
H*100 → H

DISP “required confidence interval to accept null hypothesis”, H
DISP “required level of significance to accept null hypothesis”, Z


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