Reliability demonstration test for a finite population

QUALITY AND RELIABILITY ENGINEERING INTERNATIONAL

Qual. Reliab. Engng. Int.2001;17: 33–38

RELIABILITY DEMONSTRATION TEST FOR A FINITEPOPULATION

MING-WEI LU∗AND RICHARD J. RUDY

DaimlerChrysler Corporation, 800 Chrysler Drive, Auburn Hills, MI 48326, USA

SUMMARYIn many situations, we want to accept or reject a population with small or finite population size. In this paper, wewill describe Bayesian and non-Bayesian approaches for the reliability demonstration test based on the samplesfrom a finite population. The Bayesian method is an approach that combines prior experience with newer testdata in the application of statistical tools for reliability quantification. When test time and/or sample quantity islimited, the Bayesian approach should be considered. In this paper, a non-Bayesian reliability demonstration testis considered for both finite and large population cases. The Bayesian approach with ‘uniform’ prior distributions,Polya prior distributions, and sequential sampling is also presented. Copyright 2001 John Wiley & Sons, Ltd.

KEY WORDS: Bayesian approach; binomial distribution; hypergeometric distribution; Polya distribution;posterior distribution; prior distribution; reliability demonstration; sequential sampling

1. INTRODUCTION

In the product development and testing environmentfor automotive systems or components, the designer,drawing on a wealth of past experience, strives tomeet or exceed given reliability goals. However, thetraditional approach to statistical inference does notaccount for this past experience. Thus, data obtainedby testing a large number of samples is necessary todemonstrate a high reliability level with a high degreeof confidence. In practice, the number of systems orcomponents in actual life testing is usually very small.One answer to this dilemma is provided by Bayesianstatistics, which combines subjective judgement orprior experience with test data to provide estimatessimilar to those obtained from the traditional statisticalinference approach. That is, the Bayesian methodis an approach that combines prior experience withnewer test data in the application of statistical toolsfor reliability quantification. When test time and/orsample quantities are limited, the Bayesian approachshould be considered.

In many situations, we want to accept or reject apopulation of small or finite size. In this paper, we willdescribe both a Bayesian approach and non-Bayesianapproach for the reliability demonstration test basedon the samples from a finite population. A non-Bayesian reliability demonstration test is considered

∗Correspondence to: Ming-Wei Lu, DaimlerChrysler Corporation,800 Chrysler Drive, Auburn Hills, MI 48326, USA.

for both finite and large population cases. TheBayesian approach with ‘uniform’ prior distributions,Polya prior distributions, and sequential sampling isalso presented.

2. METHODS

In this section, we will describe a method fordetermining the sample size from a given finitepopulation size.

Notation:

N = Population sizeX = Number of defective items in a populationn = Sample size selected from a population of

sizeN

k = Number of defective items from a sampleof sizen

E = Event with{k defective items out ofnsamples}

q = X/N fraction of defective items in a population

By Bayes’ theorem [1],

P(X = x|E) = P(X = x)P(E|X = x)∑x P(X = x)P(E|X = x)

,

x = k, k + 1, . . . , N − n + k (1)

Sincek is the number of defective items in a sample

Received 3 April 2000Copyright 2001 John Wiley & Sons, Ltd. Revised 25 September 2000

34 M.-W. LU AND R. J. RUDY

Figure 1. 90% reliability with 90% confidence decision criteria

of sizen, then by the hypergeometric distribution [2]

P(E|X = x) =(xk

)(N−xn−k

)(Nn

) (2)

Equation (2) together with (1) allow us to updateour ‘prior’ distribution P(X = x) to obtain ournew ‘posterior’ distribution P(x|E). In this paper, the‘90% reliability with 90% confidence’ type decisioncriterion will be used for a population acceptanceor rejection. We will stop sampling and accept thepopulation when we have 90% confidence that 90% ofthe population is good. For use of this type of criterion,it is convenient to replaceX (the number of defectiveitems) by the variableq = X/N , which is the fractionof defective items in the population. If the area underthe curve (see Figure1) to the left ofq = 10% (0.1) is90%, then we have 90% confident that the fraction ofdefective items in the population is less than 10%. Inother words, we are 90% confident that the populationis 90% reliable.

For products that are highly critical, it may be usefulto describe a criterion of at least, say, 90% confidencethat there are zero defective items in the population.Such criteria can only possibly be met, if our samplefound zero defective items(k = 0). A stringentcriterion of this ‘zero defective’ type would requirealmost a 100% sample.

2.1. No prior distribution ofP(X = x) is assumed

Let K be the number of defective items in a sampleof sizen. Hence (by the hypergeometric distribution),

P(K = k) =(Xk

)(N−Xn−k

)(Nn

) ,

k = a, a + 1, . . . , b − 1, b (3)

wherea = max{0, n − N + X} andb = min{X,n}.The relationship between lot size(N), sample size

(n), number of defective items(k) in the sample of

sizen, confidence level(C), and fraction of defectiveitems(X0/N) is given by

∑ki=0

(X0i

)(N−X0n−i

)(Nn

) ≤ 1 − C (4)

where X0 is the largestX value satisfying theequation (4).

For the special case where no defective itemsoccurred(k = 0), equation (4) becomes

(N−X0

n

)(Nn

) ≤ 1 − C (5)

AsN is large, letq = X/N be the fraction of defectiveitems in the lot. Then, from equation (3),

P(K = k) =(

n

k

)qk(1 − q)n−k (6)

That is, when the sampling fractionn/N is smallenough, binomial probabilities can be used as anapproximation to the hypergeometric probabilities.Hence, we are back to the usual binomial distribution.Different prior distributions on the Binomial distri-bution testing situation can be used. The relationshipbetween sample size(n), number of defective items(k) in the sample of sizen, confidence level(C), andfraction defective items(q) is given by

k∑i=0

(n

i

)qi(1 − q)n−i ≤ 1 − C (7)

For the special case wherek = 0, then

(1 − q)n ≤ 1 − C (8)

From equation (5), the minimum sample sizes(n)

required with no failures by different lot sizes(N) andquality targets (reliability: 90%, 95% with confidence:90%, 95%) are given in Table1.

Copyright 2001 John Wiley & Sons, Ltd. Qual. Reliab. Engng. Int.2001;17: 33–38

RELIABILITY DEMONSTRATION TEST 35

Table 1. Minimum sample size requirements with no failures by lot sizes and quality targets

90% reliability with 95% reliability with 90% reliability with 95% reliability with90% confidence 90% confidence 95% confidence 95% confidence

N n N n N n N n

20 14 20 18 20 16 20 1940 17 40 27 40 21 40 3160 19 60 32 60 23 60 3880 20 80 35 80 24 80 42

100 21 100 37 100 25 100 45230 22 120 38 150 26 150 48

140 39 270 27 200 51160 40 290 28 250 52200 41 300 29 300 54260 42 400 55340 43 500 56460 45 600 59

Example 2.1.

(a) For a lot sizeN = 40 with 95% reliabilitywith 90% confidence target requirement, byequation (5), n = 27. That is,n = 27 samplestested without failures will demonstrate 95%reliability with 90% confidence target.

(b) For a lot sizeN = 284,C = 90% confidence,and the number of defective itemsk = 1 ina sample of sizen = 30, by equation (4), thefraction of defective items is 12%.

2.2. When P(X = x) = 1/(1 + N) forx = 0, 1, 2, . . . , N

If we have no ‘strong’ information on our priordistribution P(X = x), we can start with a ‘uniform’prior distribution. That is, P(X = x) = 1/(1 + N) forx = 0, 1, 2, . . . , N . Since

N−n+k∑x=k

(x

k

)(N − x

n − k

)=

(N + 1

n + 1

)(9)

then, from equations (1), (2), and (9), we have

P(X = x|E) =(xk

)(N−xn−k

)(N+1n+1

) ,

x = k, k + 1, . . . , N − n + k (10)

The relationship between lot size(N), sample size(n),number of defective items(k) in the sample of sizen,confidence level(C), and fraction of defective items(X0/N) is given by∑X0

x=k

(xk

)(N−xn−k

)(N+1n+1

) ≥ C (11)

where X0 is the largestX value satisfying equa-tion (11).

For the special case wherek = 0, fromequation (10),

P(X = x|E) =(N−x

n

)(N+1n+1

) , x = 0, 1, . . . , N − n (12)

The relationship between lot size(N), sample size(n),confidence level(C), and fraction of defective items(X0/N) is given by∑X0

x=0

(N−x

n

)(N+1n+1

) ≥ C (13)

where X0 is the largestX value satisfying equa-tion (13).

From equation (13), the minimum sample sizes(n)

required with no failures by different lot sizes(N) andquality targets (reliability: 90%, 95% with confidence:90%, 95%) are given in Table2.

Example 2.2.A production total of 40 parts of agiven component (Battery Information Center) forthe Electric Minivan were built in a given year (lotsize N = 40). The reliability demonstration targetwas a reliability of 95% with 90% confidence. Fromequation (13) with X0 = 2, we haven = 21.Hence, a total of 21 samples tested without failureswill demonstrate 95% reliability with 90% confidencerequirement.

2.3. WhenX is a Polya distribution

The Polya distribution is a very rich distributionthat may take on a variety of shapes, including the



Table 2. Minimum sample size requirements with no failures by lot sizes and quality targets

90% reliability with 95% reliability with 90% reliability with 95% reliability with90% confidence 90% confidence 95% confidence 95% confidence

N n N n N n N n

20 10 20 13 20 12 20 1540 14 40 21 40 17 40 2560 16 60 26 60 20 60 3180 17 80 29 80 21 80 35

100 18 100 31 100 22 100 38150 19 150 36 150 24 150 46200 19 200 37 200 25 200 46250 20 250 39 250 25 250 50300 21 300 39 300 27 300 50400 23 400 40 400 31 400 51500 26 500 41 500 35 500 53

Table 3. Effect of prior parameter selection from the Polyadistribution

s 0.5 1 3 5t 4.5 9 27 45n 6 8 10 11σ2X 15.75 9.82 5.23 4.24

uniform distribution (Section2.2) and the binomialdistribution. WhenX has the binomial distribution,there is no correlation between the number of defectiveitems in the sample and the number of defective itemsin the unsampled portion of the lot [3,4]. Hence, thebinomial distribution is not a realistic one for use inany lot-sampling problem.

The Polya distribution is given by the following:

P(X = x) =(

N

x

)0(s + x)0(t + N − x)0(s + t)

0(s)0(t)0(s + t + N),

x = 0, 1, . . . , N (14)

For this distribution

q̄ = E(X/N) = s/(s + t) (15)

σ 2X = Nst

(s + t)2

s + t + N

s + t + 1(16)

The relationship between lot size(N), sample size(n), number of defective items(k) in the sampleof size n, confidence level(C), and percentage ofdefective items(X0/N) is given by

X0∑x=k

P(X = x|E) ≥ C (17)

For the special case wherek = 0, equation (17)becomes

X0∑x=0

P(X = x|E) ≥ C (18)

Example 2.3.Given N = 30, s = 1, and t = 9with a 95% reliability with 90% confidence targetrequirement, from equation (18), n can be determined(with no failures occurring) by the following equation:

P(X = 0|E) + P(X = 1|E) ≥ 0.9

Hence,n = 8 samples tested without failures willdemonstrate 95% reliability with 90% confidencetarget.

The selection of the Polya prior distribution isnot an easy task and this selection has a directeffect on the subsequent sample size requirement. InExample 2.3,̄q = s/(s + t) = 1/10 = 10% might beviewed as the subjective estimate of the percentage ofdefectives prior to testing. It can be seen that manycombinations ofs and t exist that would providea mean of 10%. Table3 provides different samplesize (n) requirements with no failure at differentcombinations ofs andt (such thats/(s + t) = 10%)with a 95% reliability with 90% confidence targetrequirement from a lot size of 30.

It can be seen that the larger the values ofs andt , the larger the sample sizen required (with nofailure) for a 95% reliability with 90% confidencetarget requirement from a lot size of 30. The strongprior can be attributed to the fact that the variance of aPolya distribution is given byσ 2

X (equation (16)) andas boths andt increase, the variance decreases. Thissuggests that if one wants to use a smaller test samplesize, then small values fors andt are in order.


RELIABILITY DEMONSTRATION TEST 37

Figure 2. Trend of the state of knowledge curves

2.4. Sequential sampling

Using equation (2) together with equation (1), onecan update the ‘prior’ distribution P(X = x) toobtain a new ‘posterior’ distribution P(X = x|E).Thus, supposeE represents the results of today’ssampling work and suppose the distribution P(X =x) represents our state of knowledge aboutx at theend of yesterday’s work. One can update the ‘prior’distribution P(X = x) to obtain a new ‘posterior’distribution P(X = x|E) which is our state ofknowledge at the end of today’s sampling. In this way,we can update our knowledge at the end of each day.

Suppose one started the sampling program with thestate of knowledge P0(X = x). After the first sampleof sizen1, our knowledge would be expressed by thedistribution P1(X = x); after the second sample ofsize n2, the knowledge would be expressed by thedistribution P2(X = x) and so on. Pictorially, thesecurves would appear as in Figure2, which shows ourknowledge sharpening as the sample gets bigger.

For eachi, let Ei be the event with{ki defectiveitems out of sample sizeni}. For i = 1,

P(E1|X = x) =(

xk1

)(N−xn1−k1

)(Nn1

) (19)

For i = 2, 3, 4, . . . ,

P(Ei |X = x) =(xki

)(Ni−xni−ki

)(Ni

ni

) ,

whereNi = N −i−1∑j=1

(nj − kj ) (20)

From equation (1), the new ‘posterior’ distribution isgiven by, fori = 1,

P1(X = x|Ei) = P0(X = x)P(E1|X = x)∑x P0(X = x)P(E1|X = x)

,

x = k1, . . . , N − n1 + k1 (21)

and, fori = 2, 3, 4, . . . ,

Pi (X = x|Ei) = Pi−1(X = x)P(Ei|X = x)∑x Pi−1(X = x)P(Ei |X = x)

,

x =i−1∑j=1

kj , . . . , Ni (22)

The relationship between lot size(N), sample size(n),number of defective items(ki) in the sample of sizeni ,confidence level(C), and fraction of defective items(X0/N) is given by

X0∑x=∑I−1

j=1 kj

P1(X = x|E1) ≥ C (23)

whereX0 is the largestX value andI (I > 1) is theminimumI th sampling satisfying equation (23).

If I = 1, then the summation ofx starts fromk1 toX0 in equation (23). For the special case whenni = 1andki = 0, i = 1, 2, 3, . . . , then

P(Ei |X = x) = (Ni − x)/N (24)

whereNi = N − i + 1.

Example 2.4.For a population size ofN = 10 andconsecutive samples of sizeni = 1 with no failurefor each i = 1, 2, . . . , we will stop sampling andaccept the population, when we have 90% confidencethat 90% of the population is free of defective items.The maximum number of defective items allowed ina lot of sizeN = 10 is X0 = 1. If we start withP0(X = x) = 1/11 = 9.09% forx = 0, 1, . . . , 10, thePi (X = x) distribution in per cent(i = 1, 2, , . . . 7) isshown in Table4.

Since P7(X = 0) + P7(X = 1) = 94.55%> 90%,then a total of seven consecutive samples is requiredto meet the 90% reliability with 90% confidencerequirement.

If we start with P0(X = 0) = 50%, P0(X =1) = 30%, and P0(X = 2) = 20%, then a total of



Table 4. Probability distributions in per cent fori = 1, 2, . . . , 7

x P0(X = x) P1(X = x) P2(X = x) P3(X = x) P4(X = x) P5(X = x) P6(X = x) P7(X = x)

0 9.09 18.18 27.27 36.36 45.45 54.55 63.64 72.731 9.09 16.36 21.82 25.45 27.27 27.27 25.45 21.822 9.09 14.55 16.97 16.97 15.15 12.12 8.48 4.853 9.09 12.73 12.73 10.61 7.58 4.55 2.12 0.614 9.09 10.91 9.10 6.06 3.25 1.30 0.03 0.005 9.09 9.09 6.06 3.03 1.08 0.22 0.00 0.006 9.09 7.27 3.64 1.21 0.22 0.00 0.00 0.007 9.09 5.45 1.82 0.30 0.00 0.00 0.00 0.008 9.09 3.64 0.61 0.00 0.00 0.00 0.00 0.009 9.09 1.82 0.00 0.00 0.00 0.00 0.00 0.00

10 9.09 0.00 0.00 0.00 0.00 0.00 0.00 0.00

four consecutive samples is required to meet the 90%reliability with 90% confidence requirement.

3. CONCLUSIONS

The Bayesian method is an approach that combinesprior experience with newer test data in the applicationof statistical tools for reliability quantification. Whentest time and/or sample quantity is limited, theBayesian approach should be considered. It is evidentthat the prior distribution is an essential element in theapplication of Bayesian methods. In many situations,we want to accept or reject a population of smallor finite size. In this paper, the Bayesian approachwith different prior distributions and non-Bayesianapproach for the reliability demonstration test basedon the samples from a finite population size arepresented.

REFERENCES

1. Kapur KC, Lamberson LR.Reliability in Engineering Design.John Wiley and Sons: 1977.

2. Gibra IN. Probability and Statistical Inference for Scientistsand Engineers. Prentice-Hall, Inc.: 1973.

3. Case KE, Keats JB. On the selection of a prior distribution inBayesian acceptance sampling.Journal of Quality Technology1982;14(1):10–18.

4. Mood KE. On the dependence of sampling selection plans uponpopulation distribution.Annals of Mathematical Statistics1943;14(1):415–425.

Authors’ biographies:

Ming-Wei Lu is a Senior Quality and Reliability Specialistin the Reliability Process Management Department, Daim-lerChrysler Corporation. He received his PhD in Statisticswith reliability application from Purdue University. He is amember of the American Society of Quality (ASQ) and aCertified Quality Engineer (CQE) and Certified ReliabilityEngineer (CRE).

Richard J. Rudy is Senior Manager of the ReliabilityProcess Management Department, DaimlerChrysler Corpo-ration. He received his degree in Physics from MichiganState University. He is a member of the American Societyof Quality (ASQ), Society of Automotive Engineers (SAE),and Society of Reliability Engineers (SRE). He is a Certi-fied Reliability Engineer (CRE), Certified Quality Engineer(CQE), and Certified Quality Manager (CQMgr).


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Reliability demonstration test for a finite population