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Student Solutions Manual Applied Statistics and Probability for Engineers, Fifth Edition by Douglas C. Montgomery, George C. Runger

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CHAPTER 13

Section 13-2

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.

13-3.    a)   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.

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b)   Tensile strength seems to increase up to 30% cotton and declines at 35% cotton.

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c)   The normal probability plot and the residual plots show that the model assumptions are reasonable.

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13-5.    a)   Analysis ...

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