### CHI-SQUARE TEST

 Here we present a program to compute the chi square test for 3 groups and 7 parameters.Using material fromMontgomery C. Douglas and Runger C. George, Applied Statistics and Probability for Engineers, Wiley, 2007. INPUTF = [ 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 parametersOUTPUTNull hypothesis: there is no difference between the means of the different levelsAlternative: 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`