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Solutions from Montgomery, D. C. (2017) Design and Analysis of Experiments, Wiley, NY

Chapter 5 Introduction to Factorial Designs Solutions 5.1. An interaction effect in the model from a factorial experiment involving quantitative factors is a way of incorporating curvature into the response surface model representation of the results. True

False

5.2. A factorial experiment may be conducted as a RCBD by running each replicate of the experiment in a unique block. True

False

5.3. If an interaction effect in a factorial experiment is significant the main effects of the factors involved in that interaction are difficult to interpret individually. True

False

5.4. A biomedical researcher has conducted a two-factor factorial experiment as part of the research to develop a new product. She performed the statistical analysis using a computer software package. A portion of the output is shown below: ANOVA for Selected Factorial Model Analysis of variance table [Partial sum of Squares] Source SS DF MS F Model 874.00 5 174.80 3.28 A 776.00 ? 388.00 7.27 B 5.33 1 5.33 0.10 AB 92.67 2 46.33 0.87 Error Total

320.00 1194.00

? 11

P 0.0904 0.0249 0.7625 0.4663

53.33

The complete ANOVA with the missing degrees of freedom is ANOVA for Selected Factorial Model Analysis of variance table [Partial sum of Squares] Source SS DF MS F Model 874.00 5 174.80 3.28 A 776.00 2 388.00 7.27 B 5.33 1 5.33 0.10 AB 92.67 2 46.33 0.87

P 0.0904 0.0249 0.7625 0.4663

Solutions from Montgomery, D. C. (2017) Design and Analysis of Experiments, Wiley, NY (a)

Interpret the F-statistic in the “Model” row of the ANOVA. Specifically, what hypotheses are being tested?

H0: The model is not significant; none of the parameters in the model is likely significant H1: The model is significant; at least one of the parameters in the model is likely significant With an F value of 3.28, and a P-value of 0.0904, marginally significant, at least one of the parameters in the model might be significant. (b)

What conclusions should be drawn regarding the individual model effects?

The individual parameter F values and corresponding P-values show that factor A is significant. B and the AB interaction are not significant. (c)

How many levels of factor A were used in this experiment?

With 2 degrees of freedom, there are three levels for factor A. (d)

How many replicates were run?

Based on the degrees of freedom for factors A and B, this is a 3x2 factorial experiment for a total of six runs. With 11 total degrees of freedom, a total of 12 runs were made. Therefore, two replicates were made for each experimental run.

5.5. A biomedical researcher has conducted a two-factor factorial experiment as part of the research to develop a new product. She performed the statistical analysis using a computer software package. A portion of the output is shown below: Source A B AB Error Total (a)

DF 1 ? 2 12 17

MS 50.00 40.00 15.00 ?

DF

MS 50.00 40.00 15.00 1.00

F ? ? ?

Complete the ANOVA calculations. Source A B AB Error Total

(b)

SS ? 80.00 30.00 ? 172.00

SS 50.00 80.00 30.00 12.00 172.00

1 2 2 12 17

F 50.00 40.00 15.00

Provide an interpretation of this experiment.

Both main effects A and B, as well as the AB interaction are significant. (c)

The pure error estimate of the standard deviation of the sample observations is 1.

Solutions from Montgomery, D. C. (2017) Design and Analysis of Experiments, Wiley, NY 5.6. The following output was obtained from a computer program that performed a two-factor ANOVA on a factorial experiment. Two-way ANOVA: y versus A, B Source DF SS MS A 1 0.322 ? B ? 80.554 40.2771 Interaction ? ? ? Error 12 105.327 8.7773 Total 17 231.551 (a)

F ? 4.59 ?

Fill in the blanks in the ANOVA table. You can use bounds on the P-values. Two-way ANOVA: y versus A, B Source DF SS MS A 1 0.322 0.3220 B 2 80.554 40.2771 Interaction 2 45.348 22.6740 Error 12 105.327 8.7773 Total 17 231.551

(b)

P ? ? ?

F 0.04 4.59 2.58

P 0.8513 0.0331 0.1167

How many levels were used for factor B?

3 levels. (c)

How many replicates of the experiment were performed?

3 replicates. (d)

What conclusions would you draw about this experiment?

Only factor B is significant; factor A and the two-factor interaction are not significant.

5.7. The following output was obtained from a computer program that performed a two-factor ANOVA on a factorial experiment. Two-way ANOVA: y versus A, B Source DF SS MS A 1 ? 0.0002 B ? 180.378 ? Interaction 3 8.479 ? Error 8 158.797 ? Total 15 347.653

F ? ? ?

P ? ? 0.932

Solutions from Montgomery, D. C. (2017) Design and Analysis of Experiments, Wiley, NY (a)

Fill in the blanks in the ANOVA table. You can use bounds on the P-values. Two-way ANOVA: y versus A, B Source DF SS MS A 1 0.0002 0.0002 B 3 180.378 60.1260 Interaction 3 8.479 2.8263 Error 8 158.797 19.8496 Total 15 347.653

(b)

F P 0.00001 0.998 3.02907 0.093 0.14239 0.932

How many levels were used for factor B?

4 levels. (c)

How many replicates of the experiment were performed?

2 replicates. (d)

What conclusions would you draw about this experiment?

Factor B is moderately significant with a P-value of 0.93. Factor A and the two-factor interaction are not significant.

5.8. The yield of a chemical process is being studied. The two most important variables are thought to be the pressure and the temperature. Three levels of each factor are selected, and a factorial experiment with two replicates is performed. The yield data follow:

Temperature 150 160 170

(a)

200 90.4 90.2 90.1 90.3 90.5 90.7

Pressure 215 90.7 90.6 90.5 90.6 90.8 90.9

230 90.2 90.4 89.9 90.1 90.4 90.1

Analyze the data and draw conclusions. Use α = 0.05.

Both pressure (A) and temperature (B) are significant, the interaction is not. Design Expert Output Response:Surface Finish ANOVA for Selected Factorial Model Analysis of variance table [Partial sum of squares] Sum of Source Squares DF Model 1.14 8 A 0.77 2 B 0.30 2 AB 0.069 4 Residual 0.16 9 Lack of Fit 0.000 0 Pure Error 0 16 9

Mean Square 0.14 0.38 0.15 0.017 0.018 0 018

F Value 8.00 21.59 8.47 0.97

Prob > F 0.0026 0.0004 0.0085 0.4700

significant

Solutions from Montgomery, D. C. (2017) Design and Analysis of Experiments, Wiley, NY The Model F-value of 8.00 implies the model is significant. There is only a 0.26% chance that a "Model F-Value" this large could occur due to noise. Values of "Prob > F" less than 0.0500 indicate model terms are significant. In this case A, B are significant model terms. Values greater than 0.1000 indicate the model terms are not significant. If there are many insignificant model terms (not counting those required to support hierarchy), model reduction may improve your model.

(b)

Prepare appropriate residual plots and comment on the model’s adequacy.

The residual plots show no serious deviations from the assumptions.

(c)

Under what conditions would you operate this process?

Solutions from Montgomery, D. C. (2017) Design and Analysis of Experiments, Wiley, NY

Set pressure at 215 and Temperature at the high level, 170 degrees C, as this gives the highest yield. The standard analysis of variance treats all design factors as if they were qualitative. In this case, both factors are quantitative, so some further analysis can be performed. In Section 5.5, we show how response curves and surfaces can be fit to the data from a factorial experiment with at least one quantative factor. Since both factors in this problem are quantitative and have three levels, we can fit linear and quadratic effects of both temperature and pressure. The Design-Expert output, including the response surface plots, now follows. Design Expert Output Response:Surface Finish ANOVA for Selected Factorial Model Analysis of variance table [Partial sum of squares] Sum of Source Squares DF Model 1.13 5 A 0.10 1 B 0.067 1 A2 0.67 1 B2 0.23 1 AB 0.061 1 Residual 0.17 12 Lack of Fit 7.639E-003 3 Pure Error 0.16 9 Cor Total1.30 17

Mean Square 0.23 0.10 0.067 0.67 0.23 0.061 0.014 2.546E-003 0.018

F Value 16.18 7.22 4.83 47.74 16.72 4.38

Prob > F < 0.0001 0.0198 0.0483 < 0.0001 0.0015 0.0582

0.14

0.9314

The Model F-value of 16.18 implies the model is significant. There is only a 0.01% chance that a "Model F-Value" this large could occur due to noise. Values of "Prob > F" less than 0.0500 indicate model terms are significant. In this case A, B, A2, B2 are significant model terms. Values greater than 0.1000 indicate the model terms are not significant. If there are many insignificant model terms (not counting those required to support hierarchy), model reduction may improve your model.

Std. Dev. Mean C.V.

0.12 90.41 0.13

R-Squared Adj R-Squared Pred R-Squared

0.8708 0.8170 0.6794

significant

not significant

Solutions from Montgomery, D. C. (2017) Design and Analysis of Experiments, Wiley, NY

Factor Intercept A-Pressure B-Temperature A2 B2 AB

Estimate 90.52 -0.092 0.075

DF 1 1 1

Error 0.062 0.034 0.034

Low 90.39 -0.17 6.594E-004

High 90.66 -0.017 0.15

VIF 1.00 1.00

-0.41

1

0.059

-0.54

-0.28

1.00

0.24 -0.087

1 1

0.059 0.042

0.11 -0.18

0.37 3.548E-003

1.00 1.00

Final Equation in Terms of Coded Factors: Yield +90.52 -0.092 +0.075 -0.41 +0.24 -0.087

= *A *B * A2 * B2 *A*B

Final Equation in Terms of Actual Factors: Yield +48.54630 +0.86759 -0.64042 -1.81481E-003 +2.41667E-003 -5.83333E-004

= * Pressure * Temperature * Pressure2 * Temperature2 * Pressure * Temperature

Solutions from Montgomery, D. C. (2017) Design and Analysis of Experiments, Wiley, NY 5.9. An engineer suspects that the surface finish of a metal part is influenced by the feed rate and the depth of cut. She selects three feed rates and four depths of cut. She then conducts a factorial experiment and obtains the following data:

0.15

Depth of 0.18

Cut (in) 0.20

0.25

0.20

74 64 60

79 68 73

82 88 92

99 104 96

0.25

92 86 88

98 104 88

99 108 95

104 110 99

0.30

99 98 102

104 99 95

108 110 99

114 111 107

Feed Rate (in/min)

(a)

Analyze the data and draw conclusions. Use α = 0.05.

The depth (A) and feed rate (B) are significant, as is the interaction (AB). Design Expert Output Response: Surface Finish ANOVA for Selected Factorial Model Analysis of variance table [Partial sum of squares]

Source Model A-Depth B-Feed AB Residual Lack of Fit Pure Error Cor Total

Sum of Squares 5842.67 2125.11 3160.50 557.06 689.33 0.000 689.33 6532.00

DF 11 3 2 6 24 0 24 35

Mean Square 531.15 708.37 1580.25 92.84 28.72

F Value 18.49 24.66 55.02 3.23

Prob > F < 0.0001 < 0.0001 < 0.0001 0.0180

28.72

The Model F-value of 18.49 implies the model is significant. There is only a 0.01% chance that a "Model F-Value" this large could occur due to noise. Values of "Prob > F" less than 0.0500 indicate model terms are significant. In this case A, B, AB are significant model terms.

significant

Solutions from Montgomery, D. C. (2017) Design and Analysis of Experiments, Wiley, NY (b)

Prepare appropriate residual plots and comment on the model’s adequacy.

The residual plots shown indicate nothing unusual.

(c)

Obtain point estimates of the mean surface finish at each feed rate. Feed Rate 0.20 0.25 0.30

Average 81.58 97.58 103.83

Solutions from Montgomery, D. C. (2017) Design and Analysis of Experiments, Wiley, NY

(d)

Find P-values for the tests in part (a).

The P-values are given in the computer output in part (a).

5.10. For the data in Problem 5.9, compute a 95 percent interval estimate of the mean difference in response for feed rates of 0.20 and 0.25 in/min. We wish to find a confidence interval on, where µ is the mean surface finish for 0.20 in/min and µ2 is the mean surface finish for 0.25 in/min.

2 MSE 2 MSE ≤ µ1 − µ2 ≤ y1.. − y2.. + tα 2,ab( n− 1) ) n n 2(28.7222) = −16 ±9.032 (81.5833 −97.5833) ±(2.064) 3

y1.. − y2.. − tα 2,ab( n− 1)

Therefore, the 95% confidence interval for µ1 – µ2 is –16.000 ± 9.032.

5.11. An article in Industrial Quality Control (1956, pp. 5-8) describes an experiment to investigate the effect of the type of glass and the type of phosphor on the brightness of a television tube. The response variable is the current necessary (in microamps) to obtain a specified brightness level. The data are as follows: Glass Type 1

2

1 280 290 285

Phosphor Type 2 300 310 295

3 290 285 290

230 235 240

260 240 235

220 225 230

Solutions from Montgomery, D. C. (2017) Design and Analysis of Experiments, Wiley, NY

Both factors, phosphor type (A) and Glass type (B) influence brightness. Design Expert Output Response: Current in microamps ANOVA for Selected Factorial Model Analysis of variance table [Partial sum of squares]

Source Model A B AB Residual Lack of Fit Pure Error Cor Total

Sum of Squares 15516.67 933.33 14450.00 133.33 633.33 0.000 633.33 16150.00

DF 5 2 1 2 12 0 12 17

Mean Square 3103.33 466.67 14450.00 66.67 52.78

F Value 58.80 8.84 273.79 1.26

Prob > F < 0.0001 0.0044 < 0.0001 0.3178

significant

52.78

The Model F-value of 58.80 implies the model is significant. There is only a 0.01% chance that a "Model F-Value" this large could occur due to noise. Values of "Prob > F" less than 0.0500 indicate model terms are significant. In this case A, B are significant model terms.

(b)

Do the two factors interact? Use α = 0.05.

There is no interaction effect. (c)

Analyze the residuals from this experiment.

The residual plot of residuals versus phosphor content indicates a very slight inequality of variance. It is not serious enough to be of concern, however.

Solutions from Montgomery, D. C. (2017) Design and Analysis of Experiments, Wiley, NY

5.12. Johnson and Leone (Statistics and Experimental Design in Engineering and the Physical Sciences, Wiley 1977) describe an experiment to investigate the warping of copper plates. The two factors studied were the temperature and the copper content of the plates. The response variable was a measure of the amount of warping. The data were as follows:

Temperature (°C) 50 75 100 125 (a)

40 17, 20 12, 9 16, 12 21, 17

Copper 60 16, 21 18, 13 18, 21 23, 21

Content (%) 80 24, 22 17, 12 25, 23 23, 22

100 28, 27 27, 31 30, 23 29, 31

Is there any indication that either factor affects the amount of warping? Is there any interaction between the factors? Use α = 0.05.

Both factors, copper content (A) and temperature (B) affect warping, the interaction does not. Design Expert Output Response: Warping ANOVA for Selected Factorial Model Analysis of variance table [Partial sum of squares] Sum of Source Squares DF Model 968.22 15 A 698.34 3 B 156.09 3 AB 113.78 9 Residual 108.50 16 Lack of Fit 0.000 0 Pure Error 108.50 16 Cor Total 1076.72 31

Mean Square 64.55 232.78 52.03 12.64 6.78

F Value 9.52 34.33 7.67 1.86

6.78

The Model F-value of 9.52 implies the model is significant. There is only a 0.01% chance that a "Model F-Value" this large could occur due to noise. Values of "Prob > F" less than 0.0500 indicate model terms are significant.

Prob > F < 0.0001 < 0.0001 0.0021 0.1327

significant

Solutions from Montgomery, D. C. (2017) Design and Analysis of Experiments, Wiley, NY (b)

Analyze the residuals from this experiment.

There is nothing unusual about the residual plots.

Solutions from Montgomery, D. C. (2017) Design and Analysis of Experiments, Wiley, NY (c)

Plot the average warping at each level of copper content and compare them to an appropriately scaled t distribution. Describe the differences in the effects of the different levels of copper content on warping. If low warping is desirable, what level of copper content would you specify?

Design Expert Output Factor Name A Copper Content B Temperature

Level 40 Average

Low Level 40 50

High Level 100 125

Warping

Prediction 15.5

SE Mean 0.92

95% CI low 13.55

95% CI high 17.45

Factor A B

Name Copper Content Temperature

Level 60 Average

Low Level 40 50

High Level 100 125

Warping

Prediction 18.875

SE Mean 0.92

95% CI low 16.92

95% CI high 20.83

Factor A B

Name Copper Content Temperature

Level 80 Average

Low Level 40 50

High Level 100 125

Warping

Prediction 21

SE Mean 0.92

95% CI low 19.05

95% CI high 22.95

Factor A B

Name Copper Content Temperature

Level 100 Average

Low Level 40 50

High Level 100 125

Warping

Prediction 28.25

SE Mean 0.92

95% CI low 26.30

95% CI high 30.20

Use a copper content of 40 for the lowest warping. =

=

=

SE Pred 2.76

95% PI low 9.64

95% PI high 21.36

SE Pred 2.76

95% PI low 13.02

95% PI high 24.73

SE Pred 2.76

95% PI low 15.14

95% PI high 26.86

SE Pred 2.76

95% PI low 22.39

95% PI high 34.11

Solutions from Montgomery, D. C. (2017) Design and Analysis of Experiments, Wiley, NY (d)

Suppose that temperature cannot be easily controlled in the environment in which the copper plates are to be used. Does this change your answer for part (c)?

Use a copper of content of 40. This is the same as for part (c).

5.13. The factors that influence the breaking strength of a synthetic fiber are being studied. Four production machines and three operators are chosen and a factorial experiment is run using fiber from the same production batch. The results are as follows: Operator 1 109 110

2 110 115

Machine 3 108 109

4 110 108

2...