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