# Metrology_ExercisesAndExamples

GR&R Study Example

A GR&R study was performed using 3 operators and a gauge to measure the height of a sample of 7 nails whose specification is from 2.4 to 2.6 cms. (USL=2.6, LSL=2.4). Each of these nails was measured 3 times by each operator.

 GAUGE R&R STUDY SHEET UNIT Oper REPLICA 1 2 3 4 5 6 7 8 9 10 AVERAGE A 1 2.67 2.45 2.5 2.61 2.35 2.55 2.4 2.5043 2 2.67 2.45 2.5 2.61 2.35 2.56 2.39 2.5043 3 2.66 2.44 2.49 2.61 2.35 2.56 2.39 2.5 AVG. 2.6667 2.4467 2.4967 2.61 2.35 2.5567 2.3933 2.5029 Xb1 RANGE 0.01 0.01 0.01 0 0 0.01 0.01 0.0071 Rb1 Xb1=(2.5043+2.5043+2. 2.5)/3=2.5029 Rb1=(sum of ranges)/7=0.0071 B 1 2.65 2.45 2.51 2.6 2.35 2.56 2.39 2.5014 2 2.64 2.46 2.49 2.6 2.33 2.55 2.39 2.4943 3 2.64 2.46 2.51 2.61 2.34 2.54 2.41 2.5014 AVG. 2.6433 2.4567 2.5033 2.6033 2.34 2.55 2.3967 2.499 Xb2 RANGE 0.01 0.01 0.02 0.01 0.02 0.02 0.02 0.0157 Rb2 Xb2=(2.5014+2.4943+2.5014)/3=2.4990 Rb2=(sum of ranges)/7=0.0157 C 1 2.65 2.44 2.5 2.6 2.34 2.54 2.4 2.4957 2 2.67 2.44 2.5 2.6 2.34 2.55 2.4 2.5 3 2.66 2.45 2.5 2.6 2.34 2.55 2.4 2.5 AVG. 2.66 2.4433 2.5 2.6 2.34 2.5467 2.4 2.4986 Xb3 RANGE 0.02 0.01 0 0 0 0.01 0 0.0057 Rb3 Xb3=(2.4957+2.5+2.5)/3=2.4986 Rb3=(sum of ranges)/7=0.0057 UNITS AVG. 2.65667 2.44889 2.5 2.60444 2.34333 2.55111 2.39667 Rp (Xbp) 0.3133 Xbp1=(2.6667+2.6433+2.66)/3=2.65667 Rp=Max(Xbp)-Min(Xbp)=2.65667-2.34333=0.3133

 Rb(OP) 0.00952 r=No. repl.= 3 p=No. op.= 3 Xbb= 2.5002 Xbdiff= MAX(Xb - Min(Xb))= 0.0043 D4= 2.58 A2= 1.023 n=No. units = 7 xb out 18 UCL(R) = D4(Rb)= 0.02457 D4=3.27 y A2=1.88 for 2 replicates No. pts 21 LCL(R) = 0 D4=2.58 y A2=1.023 for 3 replicates LCL(Xb)=Xbb-A2Rb= 2.49042 UCL(Xb)=Xbb+A2Rb= 2.5099 85.71 % of the observations are outside of the control limits of averages. More than half indicate the ability of the gauge to detect variation. UCL(R) represents the upper limit for individual ranges Detect the ranges above this limit and measure them again Unit number Gauge number Unit name Nail Gauge name Characteristic Height Gauge type Specification 2.4-2.6 cms Tol/6= 0.03333

Xbb = (Xb1+Xb2+Xb3)/3 = (2.5029+2.499+2.4986)/3 = 2.5002
Rb(OP) = (Rb1+Rb2+Rb3)/3 = (0.0071+0.0157+0.0057)/3 = 0.0095
Xbdiff = max(Xb1,Xb2,Xb3) - min(Xb1,Xb2,Xb3) = 2.5029 - 2.4986 = 0.0043

UCL(R) = D4(Rb) = 2.58 * (0.00952) = 0.02456 LCL(R)=0 (for 2 & 3 replicates)
LCL(Xb) = Xbb - A2(Rb) = 2.5002-1.023(0.0095)=2.4904
UCL(Xb) = Xbb - A2(Rb) = 2.5002+1.023(0.0095)=2.5099

### Gage R&R (Xbar/R) for Height

 EQUIPMENT VARIATION (REPEATABILITY) EV=Rb*k1= 0.00563 k1= 0.5908 k2= 0.5231 k1=0.8862, r=2 k3= 0.3534 k1=0.5908, r=3 OPERATORS' VARIATION (REPRODUCIBILITY) \begin{align} OV = \sqrt{ (X_{bdiff} * k_2)^2 - \frac{EV^2}{nr} } \\ \end{align} 0.00188 If OV is negative inside the squrare root, set it up to zero k2=0.7071 p=2 operators k2=0.5231 p=3 operators REPEATABIILTY AND REPRODUCUBILITY \begin{align} OV = \sqrt{ EV^2 + OV^2 } \\ \end{align} 0.00593 n k3 UNIT'S VARIATION 2 0.7071 UV=R3k3 0.11073 3 0.5231 4 0.4467 5 0.4030 TOTAL VARIATION 6 0.3742 7 0.3534 \begin{align} TOT = \sqrt{RR^2 + UV^2}\\ \end{align} 0.11089 8 0.3375 9 0.3249 10 0.3146

\begin{align} OV &= \sqrt{ (X_{bdiff} * k_2)^2 - \frac{EV^2}{nr} } \\ &= \sqrt{ (0.0043 * 0.5231)^2 - \frac{0.0056^2}{7(3)} } = 0.001888 \\ \end{align} n=No. of units=7
r=No. of replicates=3
\begin{align} RR &= \sqrt{ EV^2 + OV^2 } \\ &= \sqrt{ 0.00563^2 + 0.00188^2 } = 0.00593 \\ TOT &= \sqrt{ RR^2 + OV^2 } \\ &= \sqrt{ 0.00593^2 + 0.11073^2 } = 0.11089 \\ \end{align}

 % EQUIPMENT VARIATION(REPEATBILITY) EV(%)=100(EV/TOT)= 5.074 TOL=Tolerence/6 Tolerence = USL - LSL EV(%)=100(EV/TOL)= 16.882 % OPERATORS' VARIATION(REPRODUCIBILITY) OV(%)=100(OV/TOT)= 1.692 OV(%)=100(OV/TOL)= 5.628 % REPEATBILITY AND REPRODUCIBILITY RR(%)=100(RR/TOT)= 5.349 RR(%)=100(RR/TOL)= 17.795 % UNITS' VARIATION UV(%)=100(UV/TOT)= 99.857 UV(%)=100(UV/TOL)= 332.229 NOTES: RR(%)<=10 is excellent. Between 10-30% depends on the characteristic. Greater than 30 needs calibration. DISCRIMINATION 1.41*(UV/RR)= 26.324 Greater or equal to 5 is acceptable (if r=2, 4 or more is acceptable)

EV(%) = 100 (EV/TOT) = 100*(0.00563/0.11089) = 5.07

EV(%) = 100 (EV/TOL) = 100*(0.00563/0.03333) = 16.89

OV(%) = 100 (OV/TOT) = 100*(0.00188/0.11089) = 1.69

OV(%) = 100 (OV/TOL) = 100*(0.00188/0.03333) = 0.564

RR(%) = 100 (RR/TOT) = 100*(0.00593/0.11089) = 5.348

RR(%) = 100 (RR/TOL) = 100*(0.00593/0.03333) = 17.79

UV(%) = 100 (UV/TOT) = 100*(0.11073/0.11089) = 99.86

UV(%) = 100 (UV/TOL) = 100*(0.11073/0.03333) = 332.22

### Gauge R&R Study -Xbar/R Method(Using Minitab)

 Source VarComp %Contribution (of VarComp) Total gauge R&R 0.0000352 0.29 Repeatability 0.0000316 0.26 Reproducibility 0.0000035 0.03 Part-To-Part 0.0122586 99.71 Total Variation 0.0122938 100.00
• If “Total gauge R&R” is less or equal 10% the measurement system is acceptable
• If “Total gauge R&R” is greater or equal 30% the measurement system is unacceptable
• If “Total gauge R&R” is more than 10% but less than 30% the measurement system is marginally acceptable (It’s acceptable if the measured characteristic is not critical, otherwise it’s unacceptable)
 Source StdDev(SD) Study Var (6*SD) %Study Var(%SV) %Tolerance(SV/Toler) Total gauage R&R 0.005931 0.035584 5.35 17.79 Repeatability 0.005625 0.033752 5.07 16.88 Reproducibility 0.001878 0.011270 1.69 5.63 Part-To-Part 0.110718 0.664311 99.86 332.16 Total Variation 0.110877 0.665263 100.00 332.63

Number of Distinct Categories = 26 (This is the discriminating capability of the measurement instrument. It’s acceptable if greater or equal to 5)

Conclusion: The measurement system is marginally acceptable because “Total gauge R&R”=17.79

Classic GR&R Exercise
 UNIT OP Replica 1 2 3 4 5 Average A 1 21 24 20 27 24 23.2 2 20 23 21 27 23 22.8 AVG 20.5 23.5 20.5 27.0 23.5 23.0 RANGE 1.0 1.0 1.0 0.0 1.0 0.8 B 1 20 22 24 28 19 22.6 2 20 22 23 26 18 21.8 AVG 20.0 22.0 23.5 27.0 18.5 22.2 RANGE 0.0 0.0 1.0 2.0 1.0 0.8 C 1 19 23 20 25 18 21.0 2 18 22 19 24 18 20.2 AVG RANGE

ANOVA Exercise

Apply the ANOVA method to the information of the classic GR&R exercise and compare both analyses.

 UNIT OPER REPL 1 2 3 4 5 SUM A 1 21 24 20 27 24 230 2 20 23 21 27 23 SUM 41 47 41 54 47 B 1 20 22 24 28 19 222 2 20 22 23 26 18 SUM 40 44 47 54 37 C 1 19 23 20 25 18 2 18 22 19 24 18 SUM Total Sum 658

The calculations of the ANOVA table are presented below. Please click on the images to see the answers.

 Sources of Variation SS df MS F Units Operators Units x Operators Repeatability TOTAL

\begin{align} F_{\alpha, df_1,df_2} (units) &= F \\ F_{\alpha, df_1,df_2} (op) &= F \\ F_{\alpha, df_1,df_2} (units,op) &= F \\ MS = \frac{SS} {df}\\ \end{align}

### Conclusions

Statistically significant factors?

### Study of Variation

r=number of replicates=3 | o=number of operators=3 | n=number of units=7 | SS=sum of squares

 Sources of Variation Variance Std. Dev. Study Var. %Study Var. %Contribution Total GR%R *Repeatability *Reproducibility **Operator **Unit Oper. Unit Total Variation

\begin{align} Variance &= \hat{\sigma}^2_i \text{, } \%Contrib=\frac{(\%Study \text{ } Var.)^2}{100} \text{, } Std. \text{ } dev. = \hat{\sigma} = \sqrt{\hat{\sigma}^2_i}\\ Study \text{ } Var. &= 6\hat{\sigma} \text{, } \%Study \text{ } Var.=\frac{(100(Study \text{ } Var.))}{Total \text{ } Var._{(Study \text{ } Var.)} } \text{, } \%Tol. = \frac{100(Study \text{ } Var.)}{Tolerance}\\ \end{align}

 %Study Var. Source of Variation ANOVA GR&R Total GR&R *Repeatability *Reproducibility **Operator **Unit Oper. Unit Total

Classic GR&R Exercise Solutions

### Gauge R&R Study - XBar/R Method

 Source VarComp %Contributuion Total GR&R 2.01444 22.68 Repeatability 0.48393 5.45 Reproducibility 1.53051 17.23 Part to Part 6.86947 77.32 Total Variation 8.88391 100.00

 Study Var %Study Var %Tolerance Source SD 6 * SD (%SV) (SV/Toler) Total GR&R 1.41931 8.5159 47.62 85.16 Repeatability 0.69565 4.1739 23.34 41.74 Reproducibility 1.23714 7.4228 41.51 74.23 Part to Part 2.62097 15.7258 87.93 157.26 Total Variation 2.98059 17.8835 100.00 178.84

Number of Distinct Categories = 2

### Gage R&R (Xbar/R) Value

Since Total GR&R > 30 with respect to %Study var. and %Tolerance, the measurement system is unacceptable and must be improved.

ANOVA Exercise Solutions

### Two-Way ANOVA Table With Interaction

 Source DF SS MS F P Part 4 167.533 41.8833 8.4899 0.006 Operator 2 29.867 14.9333 3.0270 0.105 Part * Operator 8 39.467 4.9333 10.5714 0.000 Repeatability 15 7.000 0.4667 Total 29 243.867

### Gauge R&R ANOVA Method

 Source VarComp %Contribution(of VarComp) Total GR&R 3.70000 37.53 Repeatability 0.46667 4.73 Reproducibility 3.23333 32.80 Operator 1.0000 10.14 Operator * Part 2.23333 22.65 Part to Part 6.15833 62.47 Total Variation 9.85833 100.00

 Study Var %Study Var %Tolerance Source SD 6 * SD (%SV) (SV/Toler) Total GR&R 1.92354 11.5412 61.26 115.41 Repeatability 0.68313 4.0988 21.76 40.99 Reproducibility 1.79815 10.7889 57.27 107.89 Operator 1.0000 6.000 31.85 60.00 Operator * Part 1.49943 8.9666 47.60 89.67 Part to Part 2.48160 14.8896 79.04 148.90 Total Variation 3.13980 18.8388 100.00 188.39

Number of Distinct Categories = 1

### Comparison of Two Methods

 %Study Var. Source of Variation ANOVA GR&R classic Total GR&R 61.26 47.62 *Repeatability 21.76 23.34 *Reproducibility 57.27 41.51 **Operator 31.85 41.51 **Unit Oper. 47.6 NA Unit 79.04 87.94 Total Variation 100.00 100.00

The components of variation are more precisely estimated. Since Total GR&R > 30 with respect to %Study var. and %Tolerance, the measurement system is unacceptable and must be improved.

Conclusion to the classic GR&R
The ANOVA variances are estimated with greater precision. The interaction between the units and the operators are also taken into account. Observation of the F-values showed that the statistically significant factors are the units, and the interaction between them and the operators. which means that there exists a statistically significant difference between the sampled units and the interaction between them and the operators. This difference can also be seen in the %Study Var. shown in the table above. According to this example, the classic GR&R is underestimating the variation because it doesn’t include the interaction between units and operators therefore the ANOVA GR&R is providing a better solution.

Attributes Exercise

A measurement study was performed to assess a pass/fail measurement system. The requirements of number of operators and minimum samples is not met on purpose to limit the amount of data to be analyzed.

 Unit Operator 1 Operator 2 Expert 1 D ND ND ND ND 2 ND ND ND D ND 3 D D D D D 4 D D ND ND D 5 ND ND ND ND ND 6 D D D D ND 7 ND ND ND ND ND 8 D D D D D

\begin{align} LL &= \frac{v_{1i} F_{0.025,v_{1i},v_{2i}}} {v_{2i} +v_{1i}F_{0.025,v_{1i},v_{2i}} }\\ UL &= \frac{v_{1s} F_{0.975,v_{1s},v_{2s}}} {v_{2s} +v_{1s} F_{0.975,v_{1s},v_{2s}}} \\ v_{1i} &= 2m \text{, } v_{2i} = 2(N-m-1) \\ v_{1s} &= 2(m+1) \text{, } v_{2s} = 2(N-m) \\ \end{align}

m=No. of successes, N=Total number of tests

1. Internal agreement (within operators):

-Operator 1: CI:
-Operator 2: CI:

2. Each operator versus the expert (%AOE):

-Operator 1: CI:
-Operator 2: CI:

Detailed errors’ analysis:
-Operator 1 ( mistakes):
Mixture =
D-ND =
ND-D =

-Operator 2 ( mistakes):
Mixture =
D-ND =
ND-D =

 Op D-D ND-D Total D-ND ND-ND Total 1 2

(Table based on the 16 individual evaluations from each operator)
D-ND means the operator said the unit was D when it actually was ND (expert)

3. Agreement between operators

4. All operators versus the expert

Decision table (MSA, 2002):

 Decision %AOE %ND-D %D-ND Acceptable ≥90 ≤2 ≤5 Marginal ≥80 ≤5 ≤10 Unacceptable <80 >5 >10

For this exercise,

 OP %AOE %ND-D %D-ND Conclusion 1 2

Attributes Exercise Solution

Within Appraisers
Assessment Agreement

 Appraiser #Inspected #Matched Percent 95 %CI 1 8 7 87.50 (47.35, 99.68) 2 8 7 87.50 (47.35, 99.68)

# Matched: Appraiser agrees with him/herself across trials.

Each Appraiser vs Standard
Assessment Agreement

 Appraiser #Inspected #Matched Percent 95 %CI 1 8 6 72.00 (34.91, 96.81) 2 8 5 62.50 (24.49, 91.48)

# Matched: Appraiser's assessment across trials agrees with the known standard.

Assessment Disagreement

 Appraiser #ND / D Percent #D / ND Percent #Mixed Percent 1 0 0.00 1 20.00 1 12.50 2 1 33.33 1 20.00 1 12.50

# ND / D: Assessments across trials = ND / standard = D.
# D / ND: Assessments across trials = D / standard = ND.
# Mixed: Assessments across trials are not identical.

Between Appraisers
Assessment Agreement

 #Inspected #Matched Percent 95 %CI 8 5 62.50 (24.49, 91.48)

# Matched: All appraisers' assessments agree with each other.

All Appraisers vs Standard
Assessment Agreement

 #Inspected #Matched Percent 95 %CI 8 4 40.00 (15.70, 84.30)

# Matched: All appraisers' assessments agree with the known standard.

For this exercise,

 Op D-D ND-D Total D-ND ND-D Total 1 6 0 6 3 7 10 0.00% 30.0% 2 4 2 6 3 7 10 33.33% 30.0%

 Op %AOE %ND-D %D-ND Conclusion 1 75 0 30 Unacceptable 2 62.5 33.33 30 Unacceptable