Problema 8.6: R.D. Snee ( Experimenting with a large number of variables, in experiments in Industry: Design, Analysis and Interpretation of Results, by R. D. Snee, L.B. Hare, and J. B. Trout, Editors, ASQC, 985) describes an experiment in which a 5- design with I = ABCDE was used to investigate the effects of five factors on the color of a chemical product. The factors are A= solvent/reactant, B = catalyst/reactant, C = temperature, D = reactant purity, and E = reactant ph. The results obtained were as follows: Tratamientos Resultados e -.6 a.5 b -.68 abe.66 c.6 ace. bce -.9 abc.9 d 6.79 ade 5.7 bde.5 abd 5.68 cde 5. acd.8 bcd. abcde.5 a) Prepare a normal probability plot of the effects. Which effects seem active?
Normal Probability Plot of the Effects (response is Results, Alpha =.5) 95 9 8 7 6 5 D Effect Type Not Significant Significant Factor Name A A B B C C D D E E 5 - - Effect 5 Lenth's PSE =.765 El efecto significativo es la D = pureza del reactante. b) Calculate the residuals. Construct a normal probability plot of the residuals and plot the residuals versus the fitted values. Comment on the plots. Normal Probability Plot of the s 9 5 Plots for Results s Versus the Fitted Values - - -... Fitted Value.6.8 Histogram of the s s Versus the Order of the Data Frequency - - - - 5 6 7 8 9 5 6 Observation Order Los residuales muestran un comportamiento aproximadamente normal y al compararlos con los fitted values no se ve un cambio en la varianza poco significativo.
c) If any factors are negligible, collapse the 5- design into a full factorial in the active factors. Comment on the resulting design, and interpret the results. Factorial Fit: Respuesta versus D Estimated Effects and Coefficients for Respuesta (coded units) Term Effect Coef SE Coef T P Constant.78.9 6.7. D...9 5.7. S =.6556 R-Sq = 68.% R-Sq(adj) = 65.86% Analysis of Variance for Respuesta (coded units) Source DF Seq SS Adj SS Adj MS F P Main Effects 78.6 78.56 78.6 9.9. Error 6.5 6.57.6 Pure Error 6.5 6.57.6 Total 5.686 Unusual Observations for Respuesta Obs StdOrder Respuesta Fit SE Fit St Resid -.68.975.579 -.775 -.R R denotes an observation with a large standardized residual. Alias Structure I D Normal Probability Plot of the s 9 5 Plots for Respuesta s Versus the Fitted Values - - -... Fitted Value.6.8 Histogram of the s s Versus the Order of the Data Frequency - - - - 5 6 7 8 9 5 6 Observation Order
Problema 8.8: An article in Industrial and Engineering Chemistry uses a 5- design to investigate the effect of A = condensation temperature, B = amount of material, C = solvent volume, D = condensation time, and E = amount of material on yield. The results obtained are as follows: Tratamientos e. ab 5.5 ad 6.9 bc 6. cd.8 ace. bde 6.8 abcde 8. Resultados a) Verify that the design generators used were I = ACE and I = BDE. Factors: 5 Base Design:, 8 Resolution: III Runs: 8 Replicates: Fraction: / Blocks: Center pts (total): * NOTE * Some main effects are confounded with two-way interactions. Design Generators: D = ABC, E = AC Alias Structure (up to order ) I + ACE + BDE A + CE + BCD + ABDE B + DE + ACD + ABCE C + AE + ABD + BCDE D + BE + ABC + ACDE E + AC + BD + ABCDE AB + CD + ADE + BCE AD + BC + ABE + CDE ABCD Los generadores si son ACE y BDE.
b) Write down the complete defining relation and the aliases for this design. A = CE = BCD = ABDE B = DE = ACD = ABCE C = AE = ABD = BCDE D = BE = ABC = ACDE E = AC = BD = ABCDE AB = CD = ADE = BCE AD = BC = ABE = CDE ABCD c) Estimate the main effects. Estimated Effects and Coefficients for Respuesta (coded units) Term Effect Coef SE Coef T P Constant 9.8.787.. A -.55 -.76.787 -.97.5 B -5.75 -.587.787 -.9.8 C.75.8.787.5.85 D -.675 -.7.787 -..7 E.75.8.787.5.85 S =.66 R-Sq = 88.95% R-Sq(adj) = 6.% Normal Probability Plot of the Effects (response is Respuesta, Alpha =.5) 95 9 8 7 6 5 Effect Type Not Significant Significant Factor Name A A B B C C D D E E 5-7.5-5. -.5. Effect.5 5. Lenth's PSE =.775 No se ve ningun factor significativo sin embargo se verifico que factor tiene un efecto mayor sobre la respuesta con el siguiente grafico:
Main Effects Plot (data means) for Respuesta A B C Mean of Respuesta 8 6 - D - - E 8 6 - - Se observa que B es el que mas impacta la respuesta. d) Prepare an analysis of variance table. Verify that the AB and AD interactions are available to use as error. Analysis of Variance for Respuesta (coded units) Source DF Seq SS Adj SS Adj MS F P Main Effects 5 79.86 79.86 5.965 * * -Way Interactions 9.9 9.9.956 * * Error * * * Total 7 89.79 Effects Plot for Respuesta Alias Structure I + A*C*E + B*D*E + A*B*C*D A + C*E + B*C*D + A*B*D*E B + D*E + A*C*D + A*B*C*E C + A*E + A*B*D + B*C*D*E D + B*E + A*B*C + A*C*D*E E + A*C + B*D + A*B*C*D*E A*D + B*C + A*B*E + C*D*E A*B + C*D + A*D*E + B*C*E Dejando las interacciones mencionadas por fuera no se obtiene aun grados de libertad para el error. Sabiendo que B es el que mas impacta, se procedió a realizar una anova incluyendo solo el factor B:
Factorial Fit: Respuesta versus B Estimated Effects and Coefficients for Respuesta (coded units) Term Effect Coef SE Coef T P Constant 9.8.868.6. B -5.75 -.588.868 -.98.5 S =.555 R-Sq = 59.69% R-Sq(adj) = 5.97% Analysis of Variance for Respuesta (coded units) Source DF Seq SS Adj SS Adj MS F P Main Effects 5.56 5.56 5.56 8.88.5 Error 6 6.775 6.775 6. Pure Error 6 6.775 6.775 6. Total 7 89.788 Unusual Observations for Respuesta Obs StdOrder Respuesta Fit SE Fit St Resid 7 6.9.85.78 -.95 -.R R denotes an observation with a large standardized residual. * NOTE * Normal and Pareto effects plots require at least terms. Alias Structure I B e) Plot the residuals versus the fitted values. Also construct a normal probability plot of the residuals. Plots for Respuesta Normal Probability Plot of the s s Versus the Fitted Values 9 5. -.5-5. -.5..5 5. -5. 6 8 Fitted Value. Histogram of the s s Versus the Order of the Data Frequency.5..5. -.5. -5 - - - - -5. 5 6 Observation Order 7 8
Se observa un comportamiento aproximadamente normal, pero parece que a mayor valor en la respuesta mayor varianza en los residuales. Problema.9: An experiment was performed to investigate the capability of a measurement system. Ten parts were randomly selected, and two randomly selected operators measured each part three times. The tests were made in random order, and the data below resulted. Part number Operator Operator 5 9 5 5 8 5 5 5 5 5 5 5 5 5 5 5 5 5 9 5 5 8 5 5 5 8 9 8 8 9 8 6 5 5 5 5 5 5 7 5 5 5 5 5 5 8 5 5 9 5 8 5 9 5 5 5 5 8 9 7 6 9 6 7 8 General Linear Model: results versus operador, Part number Factor Type Levels Values operador random, Part number random,,,, 5, 6, 7, 8, 9, Analysis of Variance for results, using Adjusted SS for Tests Source DF Seq SS Adj SS Adj MS F P operador.7.7.7.69.7 Part number 9.7.7. 8.8. operador*part number 9 5.7 5.7.6..97 Error 6. 6..5 Total 59 6.85
Normal Probability Plot of the s (response is results).9 95 9 8 7 6 5 5. - - - Según lo anterior la interacción de las partes con el operador no es significativa. b) Find point estimates of the variance components using the analysis of variance method. σ = MS E =.5 σ τβ = (MS AB MS E ) / n = (.68 -.5) / <, asumiendo σ τβ = σ τ = (MS B MS AB ) / an = (.8 -.68) / * =.7 σ β = (MS A MS AB ) / bn <, asumiendo σ τ =