## Application of Statistics in Gauge Quantification

Statistics is used in many disciplines today for effective analysis of data. Among these fields include engineering where it is used for gauge quantification. For instance, it can be used to quantify uncertainties in a gauge study. In such a study, engineers gather the appropriate data based on the specific variable (s) being observed and estimate standard deviations for the various sources that contribute to uncertainty. They apply powerful statistical techniques to arrive at the most accurate solutions.

### Attribute measurement analysis

Assuming a centered process what are the process control limits for the part

given USL = 0.95 in, LSL = 0.7 in

[email protected] = 0.0252 [email protected] = 0.0231

As we know for a centered process Xbar = (USL + LSL)/2 = 0.825 in

P_(pk ) = min( [USL - Xbar]/3*σ , [Xbar - LSL] / 3*σ ) = [USL - LSL] /6*σ

therefore σ = [0.95 - 0.7]/6*0.8

which gives σ = 0.05208 and total variance = σ = 0.05208

also d = [[email protected] + [email protected]] /2

therefore d = 0.02415

Now covering 99% of the variance

GRR = d/(5.15*TV)

which gives GRR = 0.02415/(5.15*0.05208)

= 0.09004 or 9%

process control limits

UCL =Xbar + 3*σ = 0.825 + 3*0.05208 = 0.98 in

LCL =Xbar - 3*σ = 0.825 - 3*0.05208 = 0.67 in

Assuming a centered process what are the process control limits for the part

given USL = 0.95 in, LSL = 0.7 in

[email protected] = 0.0252 [email protected] = 0.0231

As we know for a centered process Xbar = (USL + LSL)/2 = 0.825 in

P_(pk ) = min( [USL - Xbar]/3*σ , [Xbar - LSL] / 3*σ ) = [USL - LSL] /6*σ

therefore σ = [0.95 - 0.7]/6*1.3

which gives σ = 0.03205 and total variance = σ = 0.03205

also d = [[email protected] + [email protected]] /2

therefore d = 0.02415

Now covering 99% of the variance

GRR = d/(5.15*TV)

which gives GRR = 0.02415/(5.15*0.03205)

= 0.1463 or 14.63%

process control limits

UCL =Xbar + 3*σ = 0.825 + 3*0.03205 = 0.921 in

LCL =Xbar - 3*σ = 0.825 - 3*0.03205 = 0.73 in

Since the part is gaged by the ring gage for the maximum material condition, therefore,it's a go gage.

The diameter of the shaft =〖2.000〗_(-.005)^(+.008)in , range of diameter = 1.995 - 2.008 in

total tolerance = 0.013 in

generally gage tolerance is 10% of total tolerance therefore gage tolerance = 0.0013 in

for bilateral tolerance

we will divide gage tolerance by 2

we get 0.0013/2 = 0.00065 in

therefore size of go ring gage - 〖2.008〗_(-.00065)^(+.00065)

and the mean size of the ring gauge would be 2.008 in.

Part size | Number of acceptances in 20 measurements | |

2.004 | 20 | 1 |

2.0045 | 20 | 1 |

2.005 | 18 | 0.9 |

2.0055 | 13 | 0.65 |

2.006 | 8 | 0.4 |

2.0065 | 7 | 0.35 |

2.007 | 3 | 0.15 |

2.0085 | 1 | 0.05 |

2.009 | 0 | 0 |

2.0095 | 0 | 0 |

Part size | Number of acceptances in 20 measurements | Bias = reference size - measured size |

2.004 | 20 | 0.0040 |

2.0045 | 20 | 0.0045 |

2.005 | 18 | 0.0050 |

2.0055 | 13 | 0.0055 |

2.006 | 8 | 0.0060 |

2.0065 | 7 | 0.0065 |

2.007 | 3 | 0.0070 |

2.0085 | 1 | 0.0085 |

2.009 | 0 | 0.0090 |

2.0095 | 0 | 0.0095 |

No, more number of data would provide more accurate results.

### Hypothesis testing

repeatability = variation due to measurement environment

therefore repeatability = 0.001921

To determine if the bias is significantly different from 0, we need to do a t test to test that

Null Hypothesis: Mean of bias is zero

Alternate Hypothesis: Mean of bias is not zero

We need to perform a two-tailed t test to test our hypothesis

Mean of bias= .00655

Sample SD = 0.001921

Degrees of freedom = n-1 = 9

Test value of t-statistic = (Mean of bias - 0)/(Sample SD/SQRT(10)) = 10.78

At a level of significance = 0.01 and df = 9, the critical values of t are 土3.2498

As the test value of t falls outside the range of critical values of t, we reject the null hypothesis

In conclusion, at 99% confidence, we can say that bias is not zero

To plot a gage performance curve we require following inputs

USL = 2.008, LSL = 1.995

Accuracy ( = -bias) = -0.00655 GRR(SD) = 0.001921

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### Gauge tolerance

given USL = 0.56 , LSL = 0.44

P_pk = 0.6

gage tolerance = 10% of part tolerance

Assuming a centered process

P_(pk ) = min( [USL - Xbar]/3*σ , [Xbar - LSL] / 3*σ ) = [USL - LSL] /6*σ

therefore σ = [0.56 - 0.44]/6*0.6

which gives σ = 0.0333

mean = 0.50

Agreement calculation between A and B

(heret represent agreement and f represent disagreement.)

part | Agreement (A/B)[1 or 0 by A] | Agreement (A/C) | Agreement (B/C) | Pass or fail (based on reference dimension (0.44-0.56) | part | Agreement (A/B) [1 or 0 by A] | Agreement (A/C) | Agreement (B/C) | Pass or fail (based on reference dimension (0.44-0.56) |

1 | t[1] | t | t | 1 | 8 | t[1] | t | t | 1 |

1 | t[1] | t | t | 1 | 8 | t[1] | t | t | 1 |

1 | t[1] | t | t | 1 | 8 | t[1] | t | t | 1 |

2 | t[0] | t | t | 0 | 9 | t[1] | t | t | 1 |

2 | t[0] | t | t | 0 | 9 | t[1] | t | t | 1 |

2 | t[0] | t | t | 0 | 9 | t[1] | t | t | 1 |

3 | t[0] | t | t | 0 | 10 | t[0] | t | t | 1 |

3 | t[0] | t | t | 0 | 10 |
t[0] | f |
f |
1 |

3 | t[0] | t | t | 0 | 10 | t[0] | t | t | 1 |

4 | t[0] | t | t | 0 | 11 | t[1] | t | t | 1 |

4 | t[0] | t | t | 0 | 11 | t[1] | t | t | 1 |

4 | t[0] | t | t | 0 | 11 | t[1] | t | t | 1 |

5 | t[1] | t | t | 1 | 12 | t[1] | t | t | 1 |

5 |
t[1] | f |
f |
1 |
12 |
t[1] | f |
f |
1 |

5 | t[0] | t | t | 1 | 12 |
f[0] |
t | f |
1 |

6 | t[1] | t | t | 1 | 13 | t[1] | t | t | 1 |

6 |
t[1] | f |
f |
1 |
13 | t[1] | t | t | 1 |

6 | t[1] | t | t | 1 | 13 | t[1] | t | t | 1 |

7 | t[1] | t | t | 1 | 14 | t[1] | t | t | 1 |

7 | t[1] | t | t | 1 | 14 | t[1] | t | t | 1 |

7 | t[1] | t | t | 1 | 14 | t[1] | t | t | 1 |

part | Agreement (A/B) [1 or 0 by A] | Agreement (A/C) | Agreement (B/C) | Pass or fail (based on reference dimension (0.44-0.56) | part | Agreement (A/B) [1 or 0 by A] | Agreement (A/C) | Agreement (B/C) | Pass or fail (based on reference dimension (0.44-0.56) |

15 | t[1] | t | t | 1 | 22 | t[1] | t | t | 1 |

15 | t[1] | t | t | 1 | 22 | t[1] | t | t | 1 |

15 | t[1] | t | t | 1 | 22 | t[1] | t | t | 1 |

16 | t[1] | t | t | 1 | 23 | t[0] | t | t | 1 |

16 | t[1] | t | t | 1 | 23 | f[1] | f | t | 1 |

16 | t[1] | t | t | 1 | 23 | t[0] | f | f | 1 |

17 | t[1] | t | t | 1 | 24 | t[1] | t | t | 1 |

17 | t[1] | t | t | 1 | 24 | t[1] | t | t | 1 |

17 | t[1] | t | t | 1 | 24 | t[1] | t | t | 1 |

18 | t[1] | t | t | 1 | 25 | t[1] | t | t | 1 |

18 | t[1] | t | t | 1 | 25 | t[1] | t | t | 1 |

18 | t[1] | t | t | 1 | 25 | t[1] | t | t | 1 |

19 | t[1] | f | f | 1 | 26 | t[1] | t | t | 1 |

19 | f[1] | t | f | 1 | 26 | t[1] | t | t | 1 |

19 | f[0] | t | f | 1 | 26 | t[1] | t | t | 1 |

20 | t[0] | f | f | 1 | 27 | t[0] | t | t | 0 |

20 | f[0] | f | t | 1 | 27 | t[0] | t | t | 0 |

20 | f[1] | f | t | 1 | 27 | f[0] | t | f | 0 |

21 | t[1] | t | t | 1 | 28 | t[1] | t | t | 1 |

21 | t[1] | t | t | 1 | 28 | t[1] | t | t | 1 |

21 | t[1] | t | t | 1 | 28 | t[1] | t | t | 1 |

part | Agreement (A/B) | Agreement (A/C) | Agreement (B/C) | Pass or fail (based on reference dimension (0.44-0.56) | part | Agreement (A/B) | Agreement (A/C) | Agreement (B/C) | Pass or fail (based on reference dimension (0.44-0.56) |

43 | t[1] | t | t | 1 | 45 | t[0] | t | t | 1 |

43 | t[1] | t | t | 1 | 45 | t[0] | t | t | 1 |

43 | t[1] | t | t | 1 | 45 | t[0] | t | t | 1 |

44 | t[0] | t | t | 0 | |||||

44 | t[0] | t | t | 0 | |||||

44 | t[0] | t | t | 0 |

operator | C | C | C | C | |||

1 | 1 | 0 | 0 | total | |||

actual count | expected | actual count | expected | ||||

A | 1 | actual count | 86 | 8 | 94 | ||

A | 1 | expected | 64.76 | 29.24 | 94 | ||

A | 0 | actual count | 7 | 34 | 41 | ||

A | 0 | expected | 28.24 | 12.76 | 41 | ||

total | 93 | 93 | 42 | 42 | 135 |

operator | C | C | C | C | |||

1 | 1 | 0 | 0 | total | |||

actual count | expected | actual count | expected | ||||

B | 1 | actual count | 88 | 9 | 97 | ||

B | 1 | expected | 66.82 | 30.18 | 97 | ||

B | 0 | actual count | 5 | 33 | 38 | ||

B | 0 | expected | 26.18 | 11.82 | 38 | ||

total | 93 | 93 | 42 | 42 | 135 |

*A general rule of thumb is that values of kappa greater than 0.75 indicate good to the excellent agreement (with a maximum kappa = 1); values less than 0.40 indicate poor agreement."*Thus we can say that A/B and B/C has good agreement while A/C agreement is slightly poorer.

**Sampling **

Miss rate of an operator = total of accept the bad/(total sample that are bad in study*number of trials per operator)
miss rate of A = 0/(8*3) = 0%
false alarm rate = total of reject the good/(total sample that are good in study*number of trials per operator)
false alarm rate = 17/ (37*3) = 15.32%
(5)
effectiveness of operator A for agreement with reference value
= [ number of good correct + number of bad correct]/number of samples
= [ 27 + 8] /45 = 35/45 = 77.78%
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