a) Construct a normal probability plot of the effect estimates. Which effects appear to be large?

Transcripción

1 Problema 6- n experiment was run in a semiconductor fabrication plant in an effort to increase yield. Five factors each at two levels were studied. The factors and levels were = aperture setting (small, large) = exposure time (% bellow nominal, % above nominal) = development time (3s, 5s) D = mask dimension (small, large) and E = etch time (.5 min, 5.5 min) the unreplicated 5 design show bellow was run: () = 7 d = 8 e = 8 de = 6 a = 9 ad = ae = ade = b = 3 bd = 3 be = 35 bde = 3 ab = 55 abd = 5 abe = 5 abde = 53 c = 6 cd = 8 ce = 5 cde = 5 ac = acd = ace = acde = bc = bcd = bce = 5 bcde = abc = 6 abcd = 6 abce = 65 abcde = 63 a) onstruct a normal probability plot of the effect estimates. Which effects appear to be large? Normal Probability Plot of the Effects (response is Yield, lpha =.5) Percent Effect Type Not Significant Significant Factor Name D D E E Effect Lenth's PSE =.6565

2 Según la grafica los factores,, y la interacción son los únicos significativos en el modelo. b) onduct an analysis of variance to confirm your findings to part a. Factorial Fit: Yield versus,,, D, E Estimated Effects and oefficients for Yield (coded units) Term Effect oef onstant D E * * *D *E * *D *E *D *E D*E ** **D **D **E **E *D*E.85.6 **D **E *D*E *D*E ***D ***E **D*E **D*E **D*E ***D*E S = * nalysis of Variance for Yield (coded units) Source DF Seq SS dj SS dj MS F P Main Effects * * -Way Interactions * * 3-Way Interactions * * -Way Interactions * * 5-Way Interactions * * Error * * * Total 3 66.

3 Los efectos más grandes se estiman para los factores,, y la interacción. l hacer el nova teniendo en cuenta solo los factores, y se obtiene: NOV: Yield versus,, Factor Type Levels Values fixed -, fixed -, fixed -, nalysis of Variance for Yield Source DF SS MS F P * * *....9 ** Error Total S = R-Sq = 99.35% R-Sq(adj) = 99.6% c) Write down the regression model relating yield to the significant process variables: Y = X X +.8X X ijk d) Plot the residuals on normal probability paper. Is the plot satisfactory?

4 Plots for Yield Normal Probability Plot of the s s Versus the Fitted Values Percent Fitted Value 5 6 Frequency Histogram of the s s Versus the Order of the Data Observation Order Se ve que los residuales se distribuyen normalmente y que no hay patrones ni inconsistencia con la variabiliad. e) Plot the residuals versus the predicted yields and versus each of the five factors. 3 s Versus Yield (response is Yield) Yield 5 6 7

5 3 s Versus (response is Yield) s Versus (response is Yield) s Versus (response is Yield)

6 f) Interpret any significant interactions Se encontró que la única interacción significativa es la de los factores y por lo tanto al variar los niveles de ambos factores a la vez, la respuesta se ve afectada. g) What are your recomendations regarding process operating conditions? Main Effects Plot (data means) for Yield 5 3 Mean of Yield 5 - D - - E El Yield se incrementa cuando los factores, y estan en su nivel alto de manera que estos deben mantenerse en este nivel. Los factores D y E se pueden mantener en cualquiera de los dos niveles según lo que más convenga al productor porque la respuesta no se afecta significativamente con su variación. h) Project the 5 design in this problem into a k design in the important factors. Sketch the design and show the average and range of yields at each run. Does this sketch aid in interpreting the results of this experiment?

7 ube Plot (data means) for Yield Problema 6.6 n experiment was conducted on a chemical process that produces a polymer. The four factors studied al two levels where: temperature (), catalysis concentration (), time (), and pressure (D). Two responses, molecular weight and viscosity, were observed. The design matrix and response data are shown in the table: Run ctual Molecular D number run order Weight Viscosity

8 The factor levels are: Factors Low (-) High (+) (elsius) (%) 8 (min) 3 D (psi) 6 75 a) onsider only the molecular weight response. Plot of the effect estimates on a normal probability scale. What effects appear to be important?

9 Normal Probability Plot of the Standardized Effects (response is Molecular Weight, lpha =.5) Effect Type Not Significant Significant Percent Factor D Name D 5 - Standardized Effect 6 8 Los factores importantes son, y según la grafica, pero como la interacción de esta incluida, se debe considerar el factor como significativo también. b) onduct an analysis of variance using the normal probability plot in part (a) is there indication of curvature? Factorial Fit: Molecular Weight versus,, Estimated Effects and oefficients for Molecular Weight (coded units) Term Effect oef SE oef T P onstant * t Pt S = R-Sq = 93.83% R-Sq(adj) = 9.63% nalysis of Variance for Molecular Weight (coded units) Source DF Seq SS dj SS dj MS F P Main Effects Way Interactions urvature Error Lack of Fit Pure Error Total Unusual Observations for Molecular Weight

10 Molecular Obs StdOrder Weight Fit SE Fit St Resid R R R denotes an observation with a large standardized residual. El p value para la curvature es mayor a.5 por lo tanto no hay indicios de la misma. c) Write down the regression model relating to the molecular weight to the significant process variables Y = X 5.6X +.6X + 6X ijk d) nalyze the residuals from this experiment Plots for Molecular Weight Normal Probability Plot of the s s Versus the Fitted Values 99 5 Percent Fitted Value 7 8 Histogram of the s 5 s Versus the Order of the Data Frequency Observation Order 6 8 Se observa que los residuales tienen un comportamiento aproximadamente normal, no hay un orden que siga un patrón. e) Repeat parts a-d using the viscosity response

11 Normal Probability Plot of the Standardized Effects (response is Viscosity, lpha =.5) Effect Type Not Significant Significant Percent Factor D Name D Standardized Effect omo se nota que los únicos factores significativos son y se realiza entonces el NOV teniendo en cuenta solo estos factores: Factorial Fit: Viscosity versus, Estimated Effects and oefficients for Viscosity (coded units) Term Effect oef SE oef T P onstant t Pt S = 3.75 R-Sq = 8.8% R-Sq(adj) = 78.% nalysis of Variance for Viscosity (coded units) Source DF Seq SS dj SS dj MS F P Main Effects urvature Error Lack of Fit Pure Error Total Unusual Observations for Viscosity Obs StdOrder Viscosity Fit SE Fit St Resid R

12 Plots for Viscosity 99 Normal Probability Plot of the s 5 s Versus the Fitted Values Percent Fitted Value 55 6 Histogram of the s s Versus the Order of the Data 5 8 Frequency Observation Order 6 8 No se observa curvatura y los residuales se notan aproximandamente normal aunque se ve un patrón en la varianza de los mismos, a mayor viscosidad, la variabilidad aumenta. Problem 7.9 onsider the data from the 5 design in problem 6.. Suppose that it was necessary to run this design in four blocks with DE and D (and consequently E) confounded. nalyze the data for this design. Teniendo en cuenta que hay bloques se realize el analisis que produjo los siguientes factores principales:

13 Normal Probability Plot of the Effects (response is YIELD, lpha =.5) Percent Effect Type Not Significant Significant Factor Name D D E E Effect Lenth's PSE =.6565 on los factores,, y la interaccion, se hizo el anova para entonces darle grados de libertad al error: Factorial Fit: YIELD versus lock,,, Estimated Effects and oefficients for YIELD (coded units) Term Effect oef SE oef T P onstant lock lock lock * S =.78 R-Sq = 99.35% R-Sq(adj) = 99.6% nalysis of Variance for YIELD (coded units) Source DF Seq SS dj SS dj MS F P locks Main Effects Way Interactions Error Total 3 66.

14 Percent Normal Probability Plot of the s Plots for YIELD s Versus the Fitted Values Fitted Value 6.8 Histogram of the s 3. s Versus the Order of the Data Frequency Observation Order Los residuales se ven aproximadamente normales y no tienen un patrón definido, al igual que se ve una varianza aproximadamente constante. Problem 7- Design an experiment for confounding a 6 factorial in four blocks. Suggest an appropriate confounding scheme. Un experimento de este tipo tendrían 6 factores:,,, D, E y F con dos niveles cada uno. l tener bloques se necesitan entonces 3 generadores. ada bloque tendría 6 tratamientos que corresponde entonces a un total de 6 tratamientos para todo el experimento. Se tomaron los generadores D y DEF que conducen a un tercer generador EF. Los mismos se pueden tomar a discreción de la persona que haga el experimento. l hacer el procedimiento por Minitab se obtuvieron los generadores F, DE Y DEF Problem 7- onsider the design in two blocks with confounded. Prove algebraically that SS = SS locks.

15 loque ab loque a b SS bloques = ( + ab) + ( a + b) ( + ab + a + b ) + ab + a + = (() + ab a b) = = SS b + ()ab () a () b a( ab) b( ab) Problem Los siguientes son los tratamientos correspondientes a uno de los niveles de la fuente bloqueada:

16 Teniendo en cuenta un experimento factorial 5 con bloques encuentre los generadores correspondientes al diseño y proponga los tratamientos correspondientes a los 3 bloques restantes. Los generadores para estos tratamientos (teniendo en cuenta que los puntos negros representan las proyecciones) serian:

17 Y el tercero seria DE si se multiplican los dos anteriormente hallados. llí se presenta el bloque, los demás bloques se hallan a partir del primero y los generadores, también se halla con las proyecciones hechas para cada generador: loque loque loque 3 loque abc cd abd be ace ade bcde ac ad e cde b bcd abce abde c abcd bce acde ab d ae bde a bc bd ce de abe acd abcde