Design And Analysis Of Experiments Chapter 8 Solutions Apr 2026

Order: (1), a, b, ab, c, ac, bc, abc.

:

y = [25, 22, 20, 30, 24, 28, 32, 35]

AC: (+1,-1,+1,-1,-1,+1,-1,+1) = 25-22+20-30-24+28-32+35 = (25-22=3; 3+20=23; 23-30=-7; -7-24=-31; -31+28=-3; -3-32=-35; -35+35=0) ✅

Compute: 25 1 +22 (-1)+20*(-1)+30 1 +24 (-1)+28 1 +32 1 +35*(-1) = 25 -22 -20 +30 -24 +28 +32 -35 = (25-22=3; 3-20=-17; -17+30=13; 13-24=-11; -11+28=17; 17+32=49; 49-35=14). design and analysis of experiments chapter 8 solutions

: A (2^3) design with 2 replicates, each in 2 blocks. In replicate I, confound ABC; in replicate II, confound AB. Estimate all effects.

A: -25+22-20+30-24+28-32+35 = (-25+22=-3; -3-20=-23; -23+30=7; 7-24=-17; -17+28=11; 11-32=-21; -21+35=14) ✅ Order: (1), a, b, ab, c, ac, bc, abc

ABC: confounded with block — contrast is the block difference. ABC contrast = (+1,-1,-1,+1,-1,+1,+1,-1)?? Wait, sign pattern for ABC = A B C = (1): +++ → +1; a: +-- → -1; b: -+- → -1; ab: --+ → +1; c: -++ → -1; ac: +-+ → +1; bc: ++- → +1; abc: --- → -1. So ABC: +1, -1, -1, +1, -1, +1, +1, -1.

Thus, in this design, we cannot estimate ABC, ABD, or CD separately from block differences. When a design is replicated in blocks but different effects are confounded in different replicates, we have partial confounding . This allows estimation of all effects, but with reduced precision for the confounded ones. In replicate I, confound ABC; in replicate II, confound AB