Solutions Montgomery

Solutions from Montgomery, D. C. (2004) Design and Analysis of Experiments, Wiley, NY

The 2 Factorial Design

k

Chapter 6 Solutions

6-5 A router is used to cut locating notches on a printed circuit board. The vibration level at the surface of the board as it is cut is considered to be a major source of dimensional variation in the notches. Two factors are thought to influencevibration: bit size (A) and cutting speed (B). Two bit sizes (1/16 and 1/8 inch) and two speeds (40 and 90 rpm) are selected, and four boards are cut at each set of conditions shown below. The response variable is vibration measured as a resultant vector of three accelerometers (x, y, and z) on each test circuit board.
Treatment A + + B + + Combination (1) a b ab I 18.2 27.2 15.9 41.0 Replicate II 18.924.0 14.5 43.9 III 12.9 22.4 15.1 36.3 IV 14.4 22.5 14.2 39.9

(a) Analyze the data from this experiment.
Design Expert Output Response: Vibration ANOVA for Selected Factorial Model Analysis of variance table [Partial sum of squares] Sum of Mean Source Squares DF Square Model 1638.11 3 546.04 A 1107.23 1 1107.23 B 227.26 1 227.26 AB 303.63 1 303.63 Residual 71.72 12 5.98 Lack of Fit 0.000 0 PureError 71.72 12 5.98 Cor Total 1709.83 15

F Value 91.36 185.25 38.02 50.80

Prob > F < 0.0001 < 0.0001 < 0.0001 < 0.0001

significant

The Model F-value of 91.36 implies the model is significant. There is only a 0.01% chance that a "Model F-Value" this large could occur due to noise.

(b) Construct a normal probability plot of the residuals, and plot the residuals versus the predictedvibration level. Interpret these plots.

6-1

Solutions from Montgomery, D. C. (2004) Design and Analysis of Experiments, Wiley, NY

Normal plot of residuals
3 .6 2 5 99 95

Residuals vs. P redicted

N orm al % probability

90

1 .7 2 5

50 30 20 10 5 1

R es iduals
-3 .9 7 5 -2 .0 7 5 -0 .1 7 5 1 .7 2 5 3 .6 2 5

80 70

-0 .1 7 5

-2 .0 7 5

-3 .9 7 5 1 4 .9 2 2 1 .2 62 7 .6 0 3 3 .9 4 4 0 .2 7

R es idual

Predicted

There is nothing unusual about the residual plots. (c) Draw the AB interaction plot. Interpret this plot. What levels of bit size and speed would you recommend for routine operation? To reduce the vibration, use the smaller bit. Once the small bit is specified, either speed will work equally well, because the slope of the curve relatingvibration to speed for the small tip is approximately zero. The process is robust to speed changes if the small bit is used.
DE S IG N-E X P E RT P l o t V i b ra ti o n 4 3 .9

Interaction Graph
C utting Speed

X = A : B i t S i ze Y = B : Cu tti n g S p e e d De si g n P o i n ts B - -1 .0 0 0 B + 1 .0 0 0 3 6 .1 5

Vibration

2 8 .4

2 0 .6 5

1 2 .9 -1 .0 0 -0 .5 0 0 .0 0 0 .5 0 1 .00

Bit Size

6-7 An experiment was performed to improve the yield of a chemical process. Four factors were selected, and two replicates of a completely randomized experiment were run. The results are shown in the following table:
Treatment Combination (1) Replicate I 90 Replicate II 93 Treatment Combination d Replicate I 98 Replicate II 95

6-2

Solutions from Montgomery, D. C. (2004)Design and Analysis of Experiments, Wiley, NY
a b ab c ac bc abc 74 81 83 77 81 88 73 78 85 80 78 80 82 70 ad bd abd cd acd bcd abcd 72 87 85 99 79 87 80 76 83 86 90 75 84 80

(a) Estimate the factor effects.
Design Expert Output Term Model Intercept Error A Error B Error C Error D Error AB Error AC Error AD Error BC Error BD Error CD Error ABC Error ABD Error ACD Error BCD Error ABCD Effect-9.0625 -1.3125 -2.6875 3.9375 4.0625 0.6875 -2.1875 -0.5625 -0.1875 1.6875 -5.1875 4.6875 -0.9375 -0.9375 2.4375 SumSqr % Contribtn 40.3714 0.84679 3.55038 7.62111 8.11267 0.232339 2.3522 0.155533 0.0172814 1.3998 13.228 10.8009 0.432036 0.432036 2.92056

657.031 13.7812 57.7813 124.031 132.031 3.78125 38.2813 2.53125 0.28125 22.7812 215.281 175.781 7.03125 7.03125 47.5313

(b) Prepare an...