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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 SSdfMS  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
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