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### ANOVA [Two-Factor Factorial Experiment, Fixed Effects Model]

Here we present a program to compute analysis of variance for 2 levels using 3 settings for each and to the confident interval and P-value requirement.

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

INPUT

H = [   H1       H2       H3       H4       H5

H6       H7       H8       H9       H10

H11     H12     H13     H14     H15 ]

= [H]

I = [     I1         I2         I3         I4         I5

I6         I7         I8         I9         I10

I11       I12       I13       I14       I15 ]

= [I]

J = [    J1        J2        J3        J4        J5

J6        J7        J8        J9        J10

J11      J12      J13      J14      J15 ]

= [J]

Assuming [H], [I], and [J] are three settings where each setting contains three levels and each level contains 5 replications. Levels and settings could represent two factors.

Null hypothesis: there is no difference between the means of the different levels

Alternative: at least two of the means are different

OUTPUT

 source of variation SS df MS F Sample A X L Columns B Y J Interaction C Z K Within D W Total E

Level of significance, confidence interval

Hypothesis testing

`2 → DIM(L6)`
`DIM([E]) → L6`
`L6(1) → S`
`L6(2) → T`
`{1,1} → DIM([G])`
`0 → [G](1,1)`
`{S,S} → DIM([G])`

`For(P,1,S,1)`
`For(M,1,T,1)`
`[H](P,M) + [G](P,1) → [G](P,1)`
`[I](P,M) + [G](P,2) → [G](P,2)`
`[J](P,M) + [G](P,3) → [G](P,3)`
`END`
`END`

`FOR(P,1,S,1)`
`FOR(M,1,S,1)`
`[G](P,M) / T → [G](P,M)`
`END`
`END`

`0 → DIM(L4)`
`0 → DIM(L5)`
`S → DIM(L4)`
`S → DIM(L5)`

`FOR(P,1,S,1)`
`FOR(M,1,S,1)`
`L4(P) +[G](P,M) → L4(P)`
`L5(M) +[G](P,M) → L5(M)`
`END`
`END`
`L4/S → L4`
`L5/S → L5`
`0 → Z`
`FOR(M,1,S,1)`
`Z+L5(M) → Z`
`END`

`Z/S → Z`
`0 → A`
`0 → B`

`FOR(P,1,S,1)`
`A+(L4(P)-Z)^2 → A`
`B+(L5(P)-Z)^2 → B`
`END`

`A*S*T → A`
`B*S*T → B`

`FOR(P,1,S,1)`
`FOR(M,1,S,1)`
`([G](P,M)-L4(P)-L5(M)+Z)^2→[G](P,M)`
`END`
`END`

`0 → C`
`FOR(M,1,S,1)`
`FOR(P,1,S,1)`
`C + [G](M,P) → C`
`END`
`END`

`T*C → C`
`AUGMENT([H],[I]) → [E]`
`AUGMENT([J],[E]) → [E]`
`S*T → R`

`FOR(M,1,S,1)`
`FOR(P,1,R,1)`
`([E](M,P)-Z)^2 → [E](M,P)`
`END`
`END`
`0 → D`

`FOR(M,1,S,1)`
`FOR(P,1,R,1)`
`D+[E](M,P) → D`
`END`
`END`
`D-A-B-C → E`
`(S-1) → X`
`(S-1) → Y`
`(S-1) * (S-1) → V`
`S * S * (T-1) → W`

`DISP “USING F DISTRIBUTION”`
`((X + W)/2-1)! * ((X/W)^(X/2))→ K`
`K / ((X/2-1)! * (W/2-1)!) → K`
`(X/2)-1 → M`
`(X/W) → R`
`(X+W)/2 → Q`

`“K * (X^M) / ((R*x+1)^Q)→ Y1`

`((Y+W)/2-1)! * ((Y/W)^(Y/2))→ K`
`K / ((Y/2-1)! * (W/2-1)!) → K`
`(Y/2)-1 → M`
`(Y/W) → R`
`(Y+W)/2 → Q`

`“K * (X^M) / ((R*x+1)^Q)→ Y2`

`((V+W)/2-1)! * ((V/W)^(V/2))→ K`
`K / ((V/2-1)! * (W/2-1)!) → K`
`(V/2)-1 → M`
`(V/W) → R`
`(V+W)/2 → Q`

`“K * (X^M) / ((R*x+1)^Q)→ Y3`

`A/X → M`
`B/Y → N`
`C/V → P`
`E/W → R`

`M/R → L`
`N/R → J`
`P/R → K`

`fnInt(Y1,x,0,L) → H`
`1-H → Z`
`H*100 → H`

`DISP “for each setting or sample, required confidence interval to accept null hypothesis”, H`
`DISP “for each setting or sample, required level of significance to accept null hypothesis”, Z`

`fnInt(Y2,x,0,J) → H`
`1-H → Z`
`H*100 → H`

`DISP “for each level or column, required confidence interval to accept null hypothesis”, H`
`DISP “for each level or column, required level of significance to accept null hypothesis”, Z`

`fnInt(Y3,x,0,K) → H`
`1-H → Z`
`H*100 → H`

`DISP “for interaction between settings or factors, required confidence interval to accept null hypothesis”, H`
`DISP “for interaction between settings or factors, required level of significance to accept null hypothesis”, Z`