chapter 1

Chapter 11: Multifactor Analysis of Variance15.a. Σ 2 2 ⎛ 3 ⎞⎛24 ⎞α i= 24,so Φ = ⎜ ⎟⎜⎟ = 1. 125 , Φ =1. 06 , ν1= 3,2= 6,⎝ 4 ⎠⎝16 ⎠figure 10.5, power ≈ . 2 . For the second alternative, Φ =1. 59 , and power . 43ν and from≈ .2⎛ 1 ⎞ β j⎜ ⎟∑ 2⎝ J ⎠ σ≈ . .⎛ 4 ⎞⎛20 ⎞⎜ ⎟⎜⎟⎝ 5 ⎠⎝16⎠2b. Φ = = = 1. 00power 3, so =1. 00Φ , ν1= 4,2=12,ν andSection 11.216.a.Source Df SS MS fA 2 30,763.0 15,381.50 3.79B 3 34,185.6 11,395.20 2.81AB 6 43,581.2 7263.53 1.79Error 24 97,436.8 4059.87Total 35 205,966.6b. =1. 79f which is not ≥ F 2. 51AB. 05,6,24=conclude that no interaction is present.c. = 3. 79A, so H oAB cannot be rejected, and wef which is ≥ F 3. 40 , so H oA is rejected at level .05.d. = 2. 81B. 05,2,24=f which is not ≥ F 3. 01e. 3. 53. 05,3,24=4059.87Q. 05,3,24= , = 3 .53 = 64. 9312w ., so H oB is not rejected.3 1 23960.02 4010.88 4029.10Only times 2 and 3 yield significantly different strengths.319