## 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
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
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
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
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
Taking reference value of part size = 2.000, the following table can be obtained for the bias of the gage
 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
average of bias value = 0.00655
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 t t 1 8 t t t 1 1 t t t 1 8 t t t 1 1 t t t 1 8 t t t 1 2 t t t 0 9 t t t 1 2 t t t 0 9 t t t 1 2 t t t 0 9 t t t 1 3 t t t 0 10 t t t 1 3 t t t 0 10 t f f 1 3 t t t 0 10 t t t 1 4 t t t 0 11 t t t 1 4 t t t 0 11 t t t 1 4 t t t 0 11 t t t 1 5 t t t 1 12 t t t 1 5 t f f 1 12 t f f 1 5 t t t 1 12 f t f 1 6 t t t 1 13 t t t 1 6 t f f 1 13 t t t 1 6 t t t 1 13 t t t 1 7 t t t 1 14 t t t 1 7 t t t 1 14 t t t 1 7 t t t 1 14 t 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 t t 1 22 t t t 1 15 t t t 1 22 t t t 1 15 t t t 1 22 t t t 1 16 t t t 1 23 t t t 1 16 t t t 1 23 f f t 1 16 t t t 1 23 t f f 1 17 t t t 1 24 t t t 1 17 t t t 1 24 t t t 1 17 t t t 1 24 t t t 1 18 t t t 1 25 t t t 1 18 t t t 1 25 t t t 1 18 t t t 1 25 t t t 1 19 t f f 1 26 t t t 1 19 f t f 1 26 t t t 1 19 f t f 1 26 t t t 1 20 t f f 1 27 t t t 0 20 f f t 1 27 t t t 0 20 f f t 1 27 f t f 0 21 t t t 1 28 t t t 1 21 t t t 1 28 t t t 1 21 t t t 1 28 t 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 t t 1 45 t t t 1 43 t t t 1 45 t t t 1 43 t t t 1 45 t t t 1 44 t t t 0 44 t t t 0 44 t t t 0
We will further create a cross-tabulation table for A/B, A/C, and B/C Agreement table for A/B kappa = (po -pe)/(1 - pe) where po = the sum of the actual counts in the diagonal cells/overall total pe = the sum of the expected counts in the diagonal cells/over total so                     k(A/B)   =  (126/135 -  79.08/135)/(1- 79.08/135)   =  83.91% for A/C
 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
Kappa(A/C)                        =   (120 - 77.52)/(135 - 77.52)   =  73.90% for B/C
 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
Kappa(B/C)                         =   (121 - 78.64)/(135 - 78.64)   =  75.16% 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% If you would like academic support on this topic, take our sampling assignment help.

### Probability

probability of non conforming part being called out as non conforming =  1 -  probability of  non conforming part being called out as conforming =  1  - probability( miss ) =   1  - P(miss by A) - P(miss by B) - P(miss by C ) =   1  -   0/24 - 1/24 - 0/24 =   23/24