13-1. a) Because factor df = total df – error df = 19 − 16 = 3 (and the degrees of freedom equals the number of levels minus one), 4 levels of the factor were used.
b) Because the total df = 19, there were 20 trials in the experiment. Because there are 4 levels for the factor, there were 5 replicates of each level.
c) From part (a), the factor df = 3
MS(Error) = 396.8/16 = 24.8, f = MS(Factor)/MS(Error) = 39.1/24.8 = 1.58.
From Appendix Table VI, 0.1 < P-value < 0.25
d) We fail to reject H0. There are not significance differences in the factor level means at α = 0.05.
Analysis of Variance for STRENGTH
Source DF SS MS F P
COTTON 4 475.76 118.94 14.76 0.000
Error 20 161.20 8.06
Total 24 636.96
Reject H0 and conclude that cotton percentage affects mean breaking strength.
b) Tensile strength seems to increase up to 30% cotton and declines at 35% cotton.
c) The normal probability plot and the residual plots show that the model assumptions are reasonable.