Presented at March 2009 meeting of ASQ Fort Worth Section

1. The Application of Statistical ToleranceThe Application of Statistical Tolerance
Limits in Process QualificationLimits in Process Qualification
SpeakerSpeaker
David A. Goodrich, P.E., CQE, CQADavid A. Goodrich, P.E., CQE, CQA

2. What are Tolerance Limits?What are Tolerance Limits?
 Tolerance limits define an interval that coversTolerance limits define an interval that covers
a proportion of the overall population with aa proportion of the overall population with a
given confidence levelgiven confidence level
 In 1931 tolerance limits were introduced byIn 1931 tolerance limits were introduced by
Shewhart in his book “ Economic Control ofShewhart in his book “ Economic Control of
Quality of Manufactured Product.”Quality of Manufactured Product.”

3. Why Use Tolerance Limits for ProcessWhy Use Tolerance Limits for Process
Qualification?Qualification?
 Provides a statement of confidence and reliabilityProvides a statement of confidence and reliability
about the processabout the process
 Accounts for uncertainty due to sample sizeAccounts for uncertainty due to sample size
 Allows comparison of the practical processAllows comparison of the practical process
boundaries to the design specificationboundaries to the design specification
 Using Cpk for process qualification is flawed due toUsing Cpk for process qualification is flawed due to
the relatively small sample sizes used and the lack ofthe relatively small sample sizes used and the lack of
sufficient process data to establish that the processsufficient process data to establish that the process
data is in statistical controldata is in statistical control

4. Formula for Tolerance LimitsFormula for Tolerance Limits
 Standard deviation method (normality assumed)Standard deviation method (normality assumed)
 For two-sided limits:For two-sided limits:
Tolerance limitsTolerance limits = X + K= X + K22ss
 For one-sided limit:For one-sided limit:
Tolerance limitTolerance limit = X= X ++ KK11s or Xs or X -- KK11ss
Where:Where: X = sample averageX = sample average
KK11 = tolerance limit factor (one-sided)= tolerance limit factor (one-sided)
KK22 = tolerance limit factor (two-sided)= tolerance limit factor (two-sided)
S = sample standard deviationS = sample standard deviation

5. Calculating the Tolerance LimitsCalculating the Tolerance Limits
 Test for a normal distributionTest for a normal distribution
 Two-sided or one-sided specificationTwo-sided or one-sided specification
 Sample Mean and Standard DeviationSample Mean and Standard Deviation
 Confidence and reliability for process acceptanceConfidence and reliability for process acceptance
 Based on risk (product safety)Based on risk (product safety)
 Based on economics (cost of poor quality)Based on economics (cost of poor quality)
 Compute using K-factor tables (Juran, etc.)Compute using K-factor tables (Juran, etc.)
 Sample Size used to determine Mean and Standard DeviationSample Size used to determine Mean and Standard Deviation
 Confidence and ReliabilityConfidence and Reliability
 Computer applicationsComputer applications
 Minitab (Minitab free macro available)Minitab (Minitab free macro available)
 ““Home grown” Excel spreadsheets availableHome grown” Excel spreadsheets available

6. Example 1- Container filling operationExample 1- Container filling operation
 The fill specification is 6.95 to 7.05 gThe fill specification is 6.95 to 7.05 g
 A sample of 66 containers were weighed whichA sample of 66 containers were weighed which
resulted in a mean weight of 7.0113 g and aresulted in a mean weight of 7.0113 g and a
standard deviation of 0.0066 gstandard deviation of 0.0066 g
 What are the 95% upper and lower tolerance limitsWhat are the 95% upper and lower tolerance limits
for 95% of the population?for 95% of the population?

7. Example 1
7.0507.0357.0207.0056.9906.9756.960
14
12
10
8
6
4
2
0
Fill Weight (grams)
Frequency
6.95 7.05
Mean 7.011
StDev 0.006628
N 66
Histogram (with Normal Curve) of Fill Weight

8. Test For NormalityTest For Normality
 Assessing normality using the Ryan-Joiner test.Assessing normality using the Ryan-Joiner test.
 Null hypothesis: the data {Null hypothesis: the data {xx1, ...,1, ..., xnxn} are a random} are a random
sample of sizesample of size nn from a normal distribution.from a normal distribution.
 Alternative hypothesis: the data are a random sampleAlternative hypothesis: the data are a random sample
from some other distribution.from some other distribution.
 Desired confidence level 95%Desired confidence level 95%

9.  P > 0.10 Cannot reject the null hypothesis. The data appearP > 0.10 Cannot reject the null hypothesis. The data appear
to be consistent with a sample from a normal distribution.to be consistent with a sample from a normal distribution.
P-Value (approx): > 0.1000
R: 0.9860
W-test for Normality
N: 66
StDev: 0.0066281
Average: 7.01129
7.0257.0157.0056.995
.999
.99
.95
.80
.50
.20
.05
.01
.001
Probability
Fill Weight
Normal Probability PlotExample 1

10. Tolerance Limits CalculatedTolerance Limits Calculated
 From K-factor table (two sided)From K-factor table (two sided)
 For N=60 (closest to 66)For N=60 (closest to 66)
 Confidence = .95Confidence = .95
 Reliability (population) = .95Reliability (population) = .95
 KK22 = 2.333= 2.333
Tolerance limitsTolerance limits = X= X ++ KK22ss
= 7.0113 g= 7.0113 g ++ 2.333(0.0066)2.333(0.0066)
== 6.99596.9959 to 7.0267 gto 7.0267 g
Compared to specification of 6.95 to 7.05 g, this process meetsCompared to specification of 6.95 to 7.05 g, this process meets
the 95%/95% confidence/reliability for acceptancethe 95%/95% confidence/reliability for acceptance
Example 1

11. Example 2 – Container Seal BurstExample 2 – Container Seal Burst
StrengthStrength
 Minimum burst specification is 26.3 psiMinimum burst specification is 26.3 psi
 A sample size of 40 container seals were burst toA sample size of 40 container seals were burst to
failure with a mean failure pressure of 48.175 psifailure with a mean failure pressure of 48.175 psi
and a standard deviation of 4.590 psiand a standard deviation of 4.590 psi
 Will the 95% lower tolerance limit for 95% of theWill the 95% lower tolerance limit for 95% of the
population meet the specification?population meet the specification?

12. 555045403020100
10
8
6
4
2
0
Burst Pressure (psi)
Frequency
26.3
Mean 48.17
StDev 4.590
N 40
Histogram (with Normal Curve) of Burst Pressure (psi)
Example 2

13. Average: 48.175
StDev: 4.59033
N: 40
W-test for Normality
R: 0.9913
P-Value (approx): > 0.1000
40 45 50 55
.001
.01
.05
.20
.50
.80
.95
.99
.999
Probability
Pressure
Burst
Normal Probability Plot
Example 2

14. Lower Tolerance Limit CalculatedLower Tolerance Limit Calculated
 From K-factor table (one sided)From K-factor table (one sided)
 For N=40For N=40
 Confidence = .95Confidence = .95
 Reliability (population) = .95Reliability (population) = .95
 KK11 == 2.1262.126
Tolerance limitTolerance limit = X – K= X – K11ss
= 48.175 - 2.126(4.590 )= 48.175 - 2.126(4.590 )
== 38.4238.42 psipsi
Compared to the specification of 26.3 psi, this process meets theCompared to the specification of 26.3 psi, this process meets the
95%/95% confidence/reliability for acceptance95%/95% confidence/reliability for acceptance
Example 2

15. Example 3 – Minimum Sample SizeExample 3 – Minimum Sample Size
 15 new sealing machines must be validated (all exact make and model)15 new sealing machines must be validated (all exact make and model)
for pouch a sealing processfor pouch a sealing process
 A minimum of five runs required per validation (2 at max/min, 3 atA minimum of five runs required per validation (2 at max/min, 3 at
nominal process parameters)nominal process parameters)
 For a C = 0 failures attribute sampling plan, a 95% confidence level of
95% reliability would require 59 samples per run, for a total of 4425
samples to validate 15 machines
 N = ln (1 – confidence) / ln reliabilityN = ln (1 – confidence) / ln reliability
 A thorough process study of one existing (identical) machine wasA thorough process study of one existing (identical) machine was
conducted with the following results:conducted with the following results:
 Data distribution is normalData distribution is normal
 Process mean = 6.788 lbfProcess mean = 6.788 lbf
 Process standard deviation = 1.378 lbfProcess standard deviation = 1.378 lbf
 Based on the available process data, for a specification of 1.0 lbfBased on the available process data, for a specification of 1.0 lbf
minimum peel strength, what is the minimum practical sample size perminimum peel strength, what is the minimum practical sample size per
run that can be used for the remaining machines?run that can be used for the remaining machines?

16. Minimum Sample Size CalculatedMinimum Sample Size Calculated
 Specification Limit = 1.0 lbfSpecification Limit = 1.0 lbf
 Process Mean = 6.788 lbfProcess Mean = 6.788 lbf
 Process Standard Deviation = 1.378 lbfProcess Standard Deviation = 1.378 lbf
1) Solve for one-sided K-factor1) Solve for one-sided K-factor
6.788 - 1.0 = K ( 1.378)6.788 - 1.0 = K ( 1.378)
K = (6.788 – 1.0)/1.378K = (6.788 – 1.0)/1.378
K = 4.20K = 4.20
2) Reverse look-up in K-factor Table2) Reverse look-up in K-factor Table
Confidence = .95Confidence = .95
Reliability (population) = .95Reliability (population) = .95
For KFor K11 = 4.20= 4.20 Therefore:Therefore: NN = 5 samples= 5 samples
Therefore, the total number of samples required to validate 15 machinesTherefore, the total number of samples required to validate 15 machines
based onbased on variables data is reduced from 4425 tois reduced from 4425 to 375 samples (5x15x5)(5x15x5)
Example 3

17. Data with Unknown DistributionData with Unknown Distribution
- Requires a non-parametric estimate of the toleranceRequires a non-parametric estimate of the tolerance
intervalinterval
- Two-Sided Intervals (Juran Table W)Two-Sided Intervals (Juran Table W)
- Based on sample size only, Juran Table W shows for a giveBased on sample size only, Juran Table W shows for a give
confidence, the proportion of the process that liesconfidence, the proportion of the process that lies
between the extremes of the sample measurementsbetween the extremes of the sample measurements
- One-Sided IntervalsOne-Sided Intervals
- Based on sample size only, an upper or lower toleranceBased on sample size only, an upper or lower tolerance
limit can be estimated based on the one-sided extreme oflimit can be estimated based on the one-sided extreme of
the measurements:the measurements:
N = ln (1 – C) / ln PN = ln (1 – C) / ln P
where:where: C = confidence levelC = confidence level
P = population proportion (reliability)P = population proportion (reliability)

18. Data with Unknown DistributionData with Unknown Distribution
(Example)(Example)
- One-Sided Lower Limit Example:One-Sided Lower Limit Example:
- What is the sample size required to determine with 95%What is the sample size required to determine with 95%
confidence that a process yields results with 90% ofconfidence that a process yields results with 90% of
population equal to or above the lowest measurement inpopulation equal to or above the lowest measurement in
the samplethe sample
N = ln (1 – C) / ln PN = ln (1 – C) / ln P
wherewhere C = confidence levelC = confidence level
P = population proportionP = population proportion
N = ln (1 - .95) / ln .90N = ln (1 - .95) / ln .90
N = ln .05 / ln .90N = ln .05 / ln .90
N = 28.4 (rounded up to nearest whole number N = 29)N = 28.4 (rounded up to nearest whole number N = 29)

19. Questions ?Questions ?
 Helpful Web addressesHelpful Web addresses
 Mintab (http://www.Minitab.Com)Mintab (http://www.Minitab.Com)
 Excel spreadsheet for tolerance limitsExcel spreadsheet for tolerance limits
(http://statpages.Org/tolintvl.Html)(http://statpages.Org/tolintvl.Html)
 NIST engineering statistics handbookNIST engineering statistics handbook
(http://www.itl.nist.gov/div898/handbook/index.htm)(http://www.itl.nist.gov/div898/handbook/index.htm)

20.
Sample Size
90% Conf of Fraction
Meeting Specs
95% Conf of Fraction
Meeting Specs
99% Conf of Fraction
Meeting Specs
n 90% 95% 99% 90% 95% 99% 90% 95% 99%
5 3.49 4.15 5.42 4.28 5.08 6.63 6.61 7.86 10.3
6 3.13 3.72 4.87 3.71 4.41 5.78 5.34 6.35 8.30
7 2.90 3.45 4.52 3.37 4.01 5.25 4.61 5.45 7.19
8 2.74 3.26 4.28 3.14 3.73 4.89 4.15 4.94 6.47
9 2.63 3.13 4.10 2.97 3.53 4.63 3.82 4.55 5.97
10 2.54 3.02 3.96 2.83 3.38 4.43 3.58 4.27 5.60
12 2.40 2.86 3.76 2.65 3.16 4.15 3.25 3.87 5.08
14 2.31 2.76 3.62 2.53 3.01 3.95 3.03 3.61 4.74
16 2.25 2.68 3.51 2.44 2.90 3.81 2.87 3.42 4.49
18 2.19 2.61 3.43 2.37 2.82 3.70 2.75 3.28 4.31
20 2.15 2.56 3.37 2.31 2.75 3.62 2.66 3.17 4.16
25 2.08 2.47 3.25 2.21 2.63 3.46 2.49 2.97 3.90
30 2.03 2.41 3.17 2.14 2.55 3.35 2.39 2.84 3.73
40 1.96 2.33 3.07 2.05 2.44 3.21 2.25 2.68 3.52
50 1.92 2.28 3.00 2.00 2.38 3.13 2.16 2.58 3.39
60 1.89 2.25 2.95 1.96 2.33 3.07 2.10 2.51 3.29
K Factors for Two-Sided NormalK Factors for Two-Sided Normal
Tolerance Limits by Sample SizeTolerance Limits by Sample Size

21.
Sample Size
90% Conf of Fraction
Meeting Specs
95% Conf of Fraction
Meeting Specs
99% Conf of Fraction
Meeting Specs
n 90% 95% 99% 90% 95% 99% 90% 95% 99%
5 2.74 3.40 4.67 3.41 4.20 5.74
6 2.49 3.09 4.24 3.01 3.71 5.06 4.41 5.41 7.33
7 2.33 2.89 3.97 2.76 3.40 4.64 3.86 4.73 6.41
8 2.22 2.76 3.78 2.58 3.19 4.35 3.50 4.29 5.81
9 2.13 2.65 3.64 2.45 3.03 4.14 3.24 3.97 5.39
10 2.07 2.57 3.53 2.36 2.91 3.98 3.05 3.74 5.08
12 1.97 2.45 3.37 2.21 2.74 3.75 2.77 3.41 4.63
14 1.90 2.36 3.26 2.11 2.61 3.59 2.59 3.19 4.34
16 1.84 2.30 3.17 2.03 2.52 3.46 2.46 3.03 4.12
18 1.80 2.25 3.11 1.97 2.45 3.37 2.36 2.91 3.96
20 1.77 2.21 3.05 1.93 2.40 3.30 2.28 2.81 3.83
25 1.70 2.13 2.95 1.84 2.29 3.16 2.13 2.63 3.60
30 1.66 2.08 2.88 1.78 2.22 3.06 2.03 2.52 3.45
40 1.60 2.01 2.79 1.70 2.13 2.94 1.90 2.37 3.25
50 1.56 1.97 2.74 1.65 2.07 2.86 1.82 2.30 3.12
60 1.53 1.93 2.69 1.61 2.02 2.81 1.76 2.20 3.04
K Factors for One-Sided NormalK Factors for One-Sided Normal
Tolerance Limits by Sample SizeTolerance Limits by Sample Size