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Figure 7 Fractional Factorials This first example (C=AB) is in the book c1 c2 a1b1 b2 a2b1 b2 ConditionGMABCABACBCABC a 1 b 1 c 1 1-1-1-1 1 1 1-1 a 1 b 1 c 2 1-1-1 1 1-1-1 1 a 1 b 2 c 1 1-1 1-1-1 1-1 1 a 1 b 2 c 2 1-1 1 1-1-1 1-1 a 2 b 1 c 1 1 1-1-1-1-1 1 1 a 2 b 1 c 2 1 1-1 1-1 1-1-1 a 2 b 2 c 1 1 1 1-1 1-1-1-1 a 2 b 2 c 2 1 1 1 1 1 1 1 1 ConditionGMABCABACBCABC a 1 b 1 c 2 1-1-1 1 1-1-1 1 a 1 b 2 c 1 1-1 1-1-1 1-1 1 a 2 b 1 c 1 1 1-1-1-1-1 1 1 a 2 b 2 c 2 1 1 1 1 1 1 1 1

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Figure 7, continued "confounding" or "aliases": start with confounding grand mean and highest order interaction effect in full 2 3 in 1/2 2 3 where GM = ABC GMGM = ABC AA(ABC) = A 2 BC = BCA & BC are aliases BB(ABC) = AB 2 C = ACB & AC are aliases CC(ABC) = ABC 2 = ABC & AB are aliases ABAB(ABC) = A 2 B 2 C = C ACAC(ABC) = A 2 BC 2 = B BCBC(ABC) = AB 2 C 2 = A ABCABC(ABC) = A 2 B 2 C 2 = GM This first example (C=AB) is in the book

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Figure 8 Fractional Factorial: Planning ConditionGMABCABACBCABC a 1 b 1 c 1 1-1-1-1 1 1 1-1 a 1 b 2 c 1 1-1 1-1-1 1-1 1 a 2 b 1 c 2 1 1-1 1-1 1-1-1 a 2 b 2 c 2 1 1 1 1 1 1 1 1 c1 c2 a1b1 b2 a2b1 b2 GMAC GM & AC are aliases AA(AC) = A 2 C = CA & C are aliases BB(AC) = ABC = ABCB & ABC are aliases CC(AC) = AC 2 = A ABAB(AC) = A 2 BC = BCAB & BC are aliases ACAC(AC) = A 2 C 2 = GM BCBC(AC) = ABC 2 = AB ABCABC(AC) = A 2 BC 2 = B

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so ANOVA table:sourcedf A (or BC)1 B (or AC)1 C (or AB)1 need error term need to assume higher order interactions are not significant or interesting Fractional Factorials

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2nd ex/ 2 4 factorial, fraction confound GM = ABCD Source in 2 4 in fractional factorial AA(ABCD) = A 2 BCD = BCD BB(ABCD) = AB 2 CD = ACD CABC 2 D = ABD DABCD 2 = ABC ABA 2 B 2 CD = CD ACA 2 BC 2 D = BD ADA 2 BCD 2 = BC BCAB 2 C 2 D = AD BDAB 2 CD 2 = AC CDABC 2 D 2 = AB ABCA 2 B 2 C 2 D = D ABDA 2 B 2 CD 2 = C ACDA 2 BC 2 D 2 = B BCD AB 2 C 2 D 2 = A ABCDGM Fractional Factorials So, assuming you expect main effects A,B,C, & D, couldn't untangle them from BCD, ACD, ABD, ABC. Furthermore, aliasing/confounding of AB = CD, AC = BD, AD = BC. Clearly if you expect some higher-order interactions, better think hard before using this design. (used in conjoint…)

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