hw5 sol.pdf - Q1 1 There are 6 factors(k=6 only 8=23...

  • No School
  • AA 1
  • 4

This preview shows page 1 - 3 out of 4 pages.

1 Q1. 1) There are 6 factors (k=6), only 8=2 3 runs (2 k-p =2 3 ). This is a 2 6-3 design. 2) Next, find generators. The fractional design contains a full factorial design from factor 1,2,3 The possible generators from the 3 factors include: 1, 2, 3, 12, 13, 23, 123. We can compare the column 4, 5, 6 against those candidates to find the 3 generators: 4=12, 5=-13, 6=23 1 2 3 4 5 6 12 13 23 123 y 1 -1 -1 -1 1 -1 1 1 1 1 -1 265 2 1 -1 -1 -1 1 1 -1 -1 1 1 155 3 -1 1 -1 -1 -1 -1 -1 1 -1 1 135 4 1 1 -1 1 1 -1 1 -1 -1 -1 205 5 -1 -1 1 1 1 -1 1 -1 -1 1 195 6 1 -1 1 -1 -1 -1 -1 1 -1 -1 205 7 -1 1 1 -1 1 1 -1 -1 1 -1 125 8 1 1 1 1 -1 1 1 1 1 1 315 divisor 4 4 4 4 4 4 Put the 3 generators in standard form: I=124, I=-135, I=236 Adding 2-way combination and 3-way combination, the final defining relation is I = 124 = -135 = 236 = -2345 = 1346 = -1256 = -456 3) The resolution of this design is III. 4) Confounding pattern for main effects 1 = 24 = -35 = 1236 = -12345 = 346 = -256 = -1456 2 = 14 = -1235 = 36 = -345 = 12346 = -156 = -2456 3 = 1234 = -15 = 26 = -245 = 146 = -12356 = -3456 4 = 12 = -1345 = 2346 = -235 = 136 = -12456 = -56 5 = 1245 = -13 = 2356 = -234 = 13456 = -126 = -46 6 = 1246 = -1356 = 23 = -23456 = 134 = -125 = -45 5) main effect estimation using contrast coefficient method l 1 =(-265+155-135+205-195+205-125+315)/4=40 l 2 =(-265-155+135+205-195-205+125+315)/4=-10 l 3 =(-265-155-135-205+195+205+125+315)/4=20 l 4 =(265-155-135+205+195-205-125-315)/4=90 l 5 =(-265+155-135+205-195+205-125+315)/4=-60 l 6 =(265+155-135-205-195-205+125+315)/4=30 the assumption is that i- factor interaction effects (i≥2) is negligible.
Image of page 1

Subscribe to view the full document.