## Williams design 3-way [Design Issues]

Hi vezz,

» » Unlikely. If you have true period effects, your estimates will be biased and you have no means to correct it.
»
» This is not to support the design proposed by Brus, but it seems to me that with those sequences we should still be able to obtain unbiased estimates of treatment effect also in presence of a period effect (based on the model suggested by the EMA Guideline on BE).

Nope. Let’s consider a simple example. A = B = C = 1, no carryover, additive model.
• Latin Square
No period effects
     p1  p2  p3 s1  A=1 B=1 C=1 s2  B=1 C=1 A=1 s3  C=1 A=1 B=1 A/C = (s1p1+s2p3+s3p2)/(s1p3+s2p2+s3p1)=(1+1+1)/(1+1+1) = 1 B/C = (s1p2+s2p1+s3p3)/(s1p3+s2p2+s3p1)=(1+1+1)/(1+1+1) = 1
Period effects (p2 = p1–0.5, p3 = p2+1)
      p1   p2   p3 s1  A=1 B=0.5 C=1.5 s2  B=1 C=0.5 A=1.5 s3  C=1 A=0.5 B=1.5 A/C = (s1p1+s2p3+s3p2)/(s1p3+s2p2+s3p1)=(1+1.5+0.5)/(1.5+0.5+1) = 1 B/C = (s1p2+s2p1+s3p3)/(s1p3+s2p2+s3p1)=(0.5+1+1.5)/(1.5+0.5+1) = 1
• Brus’ design
No period effects
      p1  p2  p3 s1  A=1 B=1 C=1 s2  B=1 A=1 C=1 s3  C=1 A=1 B=1 s4  C=1 B=1 A=1 A/C = (s1p1+s2p2+s3p2+s4p3)/(s1p3+s2p3+s3p1+s4p1)=(1+1+1+1)/(1+1+1+1) = 1 B/C = (s1p2+s2p1+s3p3+s4p2)/(s1p3+s2p3+s3p1+s4p1)=(1+1+1+1)/(1+1+1+1) = 1
Period effects (p2 = p1–0.5, p3 = p2+1)
     p1   p2   p3 s1  A=1 B=0.5 C=1.5 s2  B=1 A=0.5 C=1.5 s3  C=1 A=0.5 B=1.5 s4  C=1 B=0.5 A=1.5 A/C = (s1p1+s2p2+s3p2+s4p3)/(s1p3+s2p3+s3p1+s4p1)=(1+0.5+0.5+1.5)/(1.5+1.5+1+1) = 0.7 B/C = (s1p2+s2p1+s3p3+s4p2)/(s1p3+s2p3+s3p1+s4p1)=(0.5+1+1.5+0.5)/(1.5+1.5+1+1) = 0.7

Dif-tor heh smusma 🖖
Helmut Schütz

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