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Chapter 8 Exercise Solutions

Chapter 8 Exercise Solutions 8.3. ö  x  10.375; Rx  6.25; öx  R d 2  6.25 2.059  3.04

USL x  [(350  5)  350]  10  50; LSL x  [(350  5)  350] 10  50 xi  (obsi  350) 10 USL x  LSL x 50  (50) Cöp    5.48 6öx 6(3.04) The process produces product that uses approximately 18% of the total specification band. USL x  ö 50  10.375   4.34 Cöpu  3öx 3(3.04)

ö  LSL x 10.375  (50)   6.62 Cöpl  3öx 3(3.04) Cöpk  min(Cöpu , Cöpl )  4.34 This is an extremely capable process, with an estimated percent defective much less than 1 ppb. Note that the Cpk is less than Cp, indicating that the process is not centered and is not achieving potential capability. However, this PCR does not tell where the mean is located within the specification band. 8.6. n  4; ö  x  199; R  3.5; öx  R d 2  3.5 2.059  1.70 USL = 200 + 8 = 208; LSL = 200 – 8 = 192 (a)

USL  LSL 208  192 Potential: Cöp    1.57 6ö 6(1.70) The process produces product that uses approximately 64% of the total specification band.

(b)

USL  ö 208  199   1.76 Cöpu  3ö 3(1.70) ö  LSL 199  192   1.37 Actual: Cöpl  3ö 3(1.70) Cö  min(Cö , Cö )  1.37 pk

pl

pu

(c) The current fraction nonconforming is:

Chapter 8 Exercise Solutions pöActual  Pr{x  LSL}  Pr{x  USL}

 Pr{x  LSL}  1  Pr{x  USL}

LSL  ö   USL  ö     Pr  z    1  Pr  z   ö   ö    208  199   192  199     Pr  z    1  Pr  z   1.70   1.70      (4.1176)  1   (5.2941)  0.0000191  1  1  0.0000191 If the process mean could be centered at the specification target, the fraction nonconforming would be:  192  200  pöPotential  2  Pr  z   1.70    2  0.0000013

 0.0000026

8.25. MTB > Stat > Control Charts > Variables Charts for Subgroups > X-bar R

Test Results for Xbar Chart of Ex8.25All TEST 1. One point more than 3.00 standard deviations from center line. Test Failed at points: 2, 3

Chapter 8 Exercise Solutions

The x chart has a couple out-of-control points, and the R chart is in control. This indicates that the operator is not having difficulty making consistent measurements. (b) x  98.2; R  2.3; öGauge  R d 2  2.3 1.693  1.359 2 öTotal  4.717 2 2 2 öProduct  öTotal  öGauge  4.717  1.3592  2.872

öProduct  1.695 (c) öGauge

öTotal

 100 

1.359 100  62.5% 4.717

(d) USL = 100 + 15 = 115; LSL = 100 – 15 = 85 6öGauge P 6(1.359)    0.272 T USL  LSL 115  85

Chapter 8 Exercise Solutions

8.28. MTB > Stat > ANOVA > Balanced ANOVA In Results, select “Display expected mean squares and variance components” ANOVA: Ex8.28Reading versus Ex8.28Part, Ex8.28Op Factor Ex8.28Part Ex8.28Op

Type random random

Factor Ex8.28Part 18, 19, 20 Ex8.28Op

Values 1, 2,

Levels 20 3

3,

4,

5,

6,

7,

8,

1, 2, 3

Analysis of Variance for Ex8.28Reading Source DF SS MS Ex8.28Part 19 1185.425 62.391 Ex8.28Op 2 2.617 1.308 Ex8.28Part*Ex8.28Op 38 27.050 0.712 Error 60 59.500 0.992 Total 119 1274.592 S = 0.995825

1 2 3 4

9, 10, 11, 12, 13, 14, 15, 16, 17,

R-Sq = 95.33%

Source Ex8.28Part Ex8.28Op Ex8.28Part*Ex8.28Op Error

F 87.65 1.84 0.72

P 0.000 0.173 0.861

R-Sq(adj) = 90.74%

Variance component 10.2798 0.0149 -0.1399 0.9917

Error term 3 3 4

Expected Mean Square for Each Term (using unrestricted model) (4) + 2 (3) + 6 (1) (4) + 2 (3) + 40 (2) (4) + 2 (3) (4)

2 öRepeatability  MS Error  0.992

MS P? O  MS E 0.712  0.992   0.1400  0 n 2 MS O  MS P? O 1.308  0.712 2  =  0.0149 öOperator pn 20(2) MS P  MS P? O 62.391  0.712 2  =  10.2798 öPart on 3(2) 2 öPart? Operator 

The manual calculations match the MINITAB results. Note the Part  Operator variance component is negative. Since the Part  Operator term is not significant ( = 0.10), we can fit a reduced model without that term. For the reduced model: ANOVA: Ex8.28Reading versus Ex8.28Part, Ex8.28Op …

1 2 3

Source Ex8.28Part Ex8.28Op Error

Variance component 10.2513 0.0106 0.8832

Error term 3 3

Expected Mean Square for Each Term (using unrestricted model) (3) + 6 (1) (3) + 40 (2) (3)

Chapter 8 Exercise Solutions

(a) 2 2 öReproducibility  öOperator  0.0106 2 2 öRepeatability  öError  0.8832

(b) 2 2 2 öGauge  öReproducibility  öRepeatability  0.0106  0.8832  0.8938

öGauge  0.9454 (c)

/ T  6  öGauge  6  0.9454  0.1050 P USL-LSL 60  6 This gauge is borderline capable since the estimate of P/T ratio just exceeds 0.10.

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 Ex8.28Part Ex8.28Op Ex8.28Part * Ex8.28Op Repeatability Total

DF 19 2 38 60 119

SS 1185.43 2.62 27.05 59.50 1274.59

MS 62.3908 1.3083 0.7118 0.9917

F 87.6470 1.8380 0.7178

Two-Way ANOVA Table Without Interaction Source Ex8.28Part Ex8.28Op Repeatability Total

DF 19 2 98 119

SS 1185.43 2.62 86.55 1274.59

MS 62.3908 1.3083 0.8832

F 70.6447 1.4814

P 0.000 0.232

Gage R&R Source Total Gage R&R Repeatability Reproducibility Ex8.28Op Part-To-Part Total Variation

VarComp 0.8938 0.8832 0.0106 0.0106 10.2513 11.1451

%Contribution (of VarComp) 8.02 7.92 0.10 0.10 91.98 100.00

Study Var Source StdDev (SD) (6 * SD) Total Gage R&R 0.94541 5.6724 Repeatability 0.93977 5.6386 Reproducibility 0.10310 0.6186 Ex8.28Op 0.10310 0.6186 Part-To-Part 3.20176 19.2106 Total Variation 3.33842 20.0305 Number of Distinct Categories = 4

%Study Var (%SV) 28.32 28.15 3.09 3.09 95.91 100.00

P 0.000 0.173 0.861

Chapter 8 Exercise Solutions 8.28 continued

Visual representations of variability and stability are also provided: