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