Statistical Quality Control, IE 3255 March Homework #6 Due: Fri, April points

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1 Statistical Quality Control, IE 355 March Homework #6 Due: Fri, April points Use Ecel, Minitab and a word processor to present quality answers to the following statistical process control problems. Solve the following problems from your tetbook (Introduction to Statistical Quality Control, 5 th Edition): 1. Eercise 7-4 (Word) 7-4. n= 5; = ; R = ; = ; tolerances: 0 ± 0.01 ˆ USL LSL C p = = = (0.0073) The process produces product that uses approimately 8% of the total specification band. ˆ USL ˆ µ Cpu = = = (0.0073) ˆ ˆ µ LSL ( 0.01) Cpl = = = (0.0073) Cˆ = min( Cˆ, Cˆ ) = 1.09 pk pl pu This process is not considered capable, failing to meet the minimally acceptable definition of capable C pk 1.33 T V = = = S Cˆ ˆ p 1. C pm = = = V 1 + ( 0.399) Since C pm is greater than 1, the mean µ lies within approimately the middle third of the specification band. ˆ ˆ µ T ξ = = = Cˆ ˆ pk 1.09 Cpkm = = = 1.01 ˆ 1+ ξ

2 . Eercise 7-6 (Word) 7-6. n= 4; ˆ µ = = 199; R = 3.5; = R d = = 1.70 USL = = 08; LSL = 00 8 = 19 (a) USL LSL Potential: ˆ C p = = = (1.70) The process produces product that uses approimately 64% of the total specification band. (b) Actual: ˆ USL ˆ µ Cpu = = = ˆ σ 3(1.70) ˆ ˆ µ LSL Cpl = = = (1.70) Cˆ = min( Cˆ, Cˆ ) = 1.37 pk pl pu (c) The current fraction nonconforming is: pˆ = Pr{ < LSL} + Pr{ > USL} Actual [ ] = Pr{ < LSL} + 1 Pr{ USL} LSL ˆ µ USL ˆ µ = Pr z< + 1 Pr z = Pr z< + 1 Pr z =Φ( ) + 1 Φ(5.941) = [ ] [ ] = If the process mean could be centered at the specification target, the fraction nonconforming would be: pˆ Potential = Pr z < 1.70 = =

3 3. Eercise 7-10 (Word) 7-10 (7-8). Process A: Process B: ˆ µ = = = = = A 0(100) 000; A 0 0(3.191) ˆ µ = = = = = B 0(105) 100; B 0 0(1.064) Process B will result in fewer defective assemblies. For the parts ( C ˆ, 1.045) ( ˆ pk A= < =C pkb, ) indicates that more parts from Process B are within specification than from Process A.

4 4. Eercise 7-0 (Minitab or Ecel) 7-0 (7-14). n= 30; = 97; S = 1.6; USL = 100;LSL = 90 (a) ˆ USL ˆ µ Cpu = = = ˆ σ 3(1.6) ˆ ˆ µ LSL Cpl = = = ˆ σ 3(1.6) Cˆ = min( Cˆ, Cˆ ) = 0.63 pk pl pu (b) α = 0.05 z = z = α / Cˆ 1 ˆ pk zα/ + C 1 pk Cpk + zα/ + ˆ ˆ 9nC ( n 1) 9 ( n 1) pk nc pk C C pk 9(30)(0.63) (30 1) 9(30)(0.63) (30 1) pk

5 5. Eercise 7-8 (Minitab) 7-8. MTB > Stat > ANOVA > Balanced ANOVA In Results, select Display epected mean squares and variance components ANOVA: E7-8Reading versus E7-8Part, E7-8Op Factor Type Levels E7-8Part random 0 E7-8Op random 3 Factor Values E7-8Part 1,, 3, 4, 5, 6, 7, 8, 9, 10, 11, 1, 13, 14, 15, 16, 17, 18, 19, 0 E7-8Op 1,, 3 Analysis of Variance for E7-8Reading Source DF SS MS F P E7-8Part E7-8Op E7-8Part*E7-8Op Error Total S = R-Sq = 95.33% R-Sq(adj) = 90.74% Epected Mean Square Variance Error for Each Term (using Source component term unrestricted model) 1 E7-8Part (4) + (3) + 6 (1) E7-8Op (4) + (3) + 40 () 3 E7-8Part*E7-8Op (4) + (3) 4 Error (4) = MS = 0.99 Repeatability Error MSP O MSE Part Operator = = = n MSO MSP O Operator = = = pn 0() MSP MSP O Part = = = on 3() The manual calculations match the MINITAB results. Note the Part Operator variance component is negative. Since the Part Operator term is not significant (α = 0.10), we can fit a reduced model without that term. For the reduced model: ANOVA: E7-8Reading versus E7-8Part, E7-8Op Epected Mean Square for Each Term (using Variance Error unrestricted

6 Source component term model) 1 E7-8Part (3) + 6 (1) E7-8Op (3) + 40 () 3 Error (3)

7 (a) = = Reproducibility Repeatability Error Operator = = (b) = + = = Gauge Reproducibility Repeatability = Gauge (c) 6 ˆGauge σ P/ T = = = USL-LSL 60 6 This gauge is borderline capable since the estimate of P/T ratio just eceeds Estimates of variance components, reproducibility, repeatability, and total gauge variability may also be found using: MTB > Stat > Quality Tools > Gage Study > Gage R&R Study (Crossed) Gage R&R Study - ANOVA Method Two-Way ANOVA Table With Interaction Source DF SS MS F P E7-8Part E7-8Op E7-8Part * E7-8Op Repeatability Total Two-Way ANOVA Table Without Interaction Source DF SS MS F P E7-8Part E7-8Op Repeatability Total Gage R&R %Contribution Source VarComp (of VarComp) Total Gage R&R Repeatability Reproducibility E7-8Op Part-To-Part Total Variation Study Var %Study Var Source StdDev (SD) (6 * SD) (%SV) Total Gage R&R Repeatability Reproducibility E7-8Op Part-To-Part Total Variation Number of Distinct Categories = 4

8 7-8 continued Visual representations of variability and stability are also provided: Gage R&R (ANOVA) for E7-8Reading Gage name: Date of study: Reported by: Tolerance: Misc: 100 Components of Variation % Contribution % Study Var 30 E7-8Reading by E7-8Part Percent Sample Range Sample Mean Gage R&R Repeat Reprod R Chart by E7-8Op 1 3 Xbar Chart by E7-8Op 1 3 Part-to-Part UCL=3.757 _ R=1.15 LCL=0 _ UCL=4.55 X=.39 LCL= Average E7-8Part E7-8Reading by E7-8Op E7-8Op E7-8Op * E7-8Part Interaction 3 E7-8Op E7-8Part

9 6. Eercise 7-31 (Word) 7-31 (7-3). 1~ N(0, 0.3 ); ~ N (19.6, 0.4 ) Nonconformities will occur if y = 1 < 0.1 or y = 1 > 0.9 µ = µ µ = = 0.4 y 1 σ = σ + σ = = 0.5 y 1 σ = 0.50 y Pr{Nonconformities} = Pr{ y< LSL} + Pr{ y > USL} = Pr{ y< 0.1} + Pr{ y > 0.9} = Pr{ y< 0.1} + 1 Pr{ y 0.9} =Φ + 1 Φ =Φ( 0.6) + 1 Φ(1.00) = =