# A Study on Overestimating a Given Fraction Defective by an Imperfect Inspector

Authors: Moon Hee Yang; Kyung Chang
Journal: Mathematical Problems in Engineering
 (2014)

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## Abstract

It has been believed that even an imperfect inspector with nonzero inspection errors could either overestimate or underestimate a given FD (fraction defective) with a 50 : 50 chance. What happens to the existing inspection plans, if an imperfect inspector overestimates a known FD, when it is very low? We deal with this fundamental question, by constructing four mathematical models, under the assumptions that an infinite sequence of items with a known FD is given to an imperfect inspector with nonzero inspection errors, which can be constant and/or randomly distributed with a uniform distribution. We derive four analytical formulas for computing the probability of overestimation (POE) and prove that an imperfect inspector overestimates a given FD with more than 50%, if the FD is less than a value termed as a critical FD. Our mathematical proof indicates that the POE approaches one when FD approaches zero under our assumptions. Hence, if a given FD is very low, commercial inspection plans should be revised with the POE concept in the near future, for the fairness of commercial trades.


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## Body

## 1. Introduction

Our research started from a BLU (backlight unit) company in Korea. Inspections of a BLU, which is one of the major components attached to the back of a thin film transistor liquid crystal display unit, can be divided into several functional inspections and external appearance inspections. In the Korean BLU industry, inspection operation is usually done, like other production lines, at the end of a line, due to related costs, and the reworkability of a BLU. The inspection decision is made only by a single attribute: conforming (or C) or nonconforming (or NC). In reality, items will be misclassified, even if only a few. A C (or NC) item may be classified as NC (or C), and this probability is typically termed as type I (or type II) error. Correctly or falsely accepted items at the end of each BLU line are packaged into lots and transported to a clean storage area, where an acceptance sampling plan by attribute is performed, by a source inspector affiliated to a buyer. Even if only one NC item is found in a lot, the lot is rejected by the source inspector. The rejected items, even if there are very few NC items, must be 100% reinspected later, in another clean room (Yang and Cho [1]).As the fraction defective (FD) of BLU items waiting for source inspections has been gradually lowered to either the thousands or hundreds PPM level, most of the quality control managers have continually raised the possibility that the FD judged by a source inspector has “always” been overestimated, because his inspection could not be perfect, but also his inspection severity would be advantageous to his company. In addition, various questions about unknown dependencies between FD and inspection errors have been raised. In fact, the FD judged by an inspector should be underestimated or overestimated with a 50 : 50 chance, regardless of a given FD and/or inspections errors. In order to verify their presumptions, after having formed lots with a very low FD comprised of hundreds of BLU items randomly mixed up with C and NC items, they carried out significant field experiments controlled by an expert, where the lots were tested by an inspector, with type I error 0.86% and type II error 4.50%, as estimated. They concluded that the possibility of overestimation by the source inspector seemed to be at least significantly larger than 50% but they could not prove it mathematically and that there might exist some relationships between overestimation, very low FD, and inspection errors. From the above facts, many basic questions may be raised, but above all, here we are interested only in the following fundamental and theoretical questions:“Does an imperfect inspector overestimate a given fraction defective, when it is very low?”In other words, “Could the FDA (fraction defective after inspection) always be larger than the FDB (fraction defective before inspection), if the FDB is very or extremely low?” In order to answer the above question, we need to find a way to compute the probability of overestimation (POE), when an FD and inspection errors are given.As far as we know, there have been no papers directly related to our problem. However, some studies dealing with nonzero inspection errors have appeared in the literature, since the 1970s. Collins et al. [2] considered the effects of inspection error on the probability of a lot of acceptance, average outgoing quality, and average total inspection, under both replacement and nonreplacement assumptions, and suggested that an acceptance sampling plan may be designed, based on inspection error. Dorris and Foote [3] surveyed the state of knowledge on inspection and measurement errors and suggested future lines of investigation about inspection errors. Raz and Thomas [4] presented a branch-and-bound method for determining an optimum sequencing inspection plan, for a group of inspectors operating at different skill and cost levels. Tang [5] provided a rule for determining the optimal sequence of multiple quality characteristics, for minimizing the cost of inspection within each inspection stage. Lee [6] developed the stop rule, for seeking the optimal number of inspection stages. Sylla and Drury [7] dealt with the apparent fraction nonconforming qe=(1-q)α+q(1-β), where q is an FD, α is the probability of rejecting a C item, and β is the probability of accepting an NC item. They found n′, the sample size and c′, the cut-off value for single sampling by attributes, considering fraction defective, types I and II errors, and error-related payoffs, and proposed the concept of liability which is an inspector’s ability to respond to information, like payoffs, fraction defectives, and discriminability between noise and signal distributions. Burk et al. [8] derived a relation q=(qe-α)(1-α-β)-1 and showed a table about the relation. They noted that as the type I error approaches qe, q approaches zero and that for very good process, qe is actually a type I error. They suggested a procedure for estimating the types I and II errors and gave an industrial example. These were shown as major variables, in the last contribution related to our current research.Several models that are partially related to our problem have appeared in the literature. In order to attain a prespecified quality rate at the end of an assembly line, Yang [9] suggested a K-stage inspection-rework (K-IR) system, which was composed of a series of K stages, each of which included an inspection process and a rework process. He suspected the effectiveness of the K-IR system and proved mathematically that FDA is always larger than FDB if FDB is less than a value that depends on a FD of rework and inspection errors. Based on his assumptions, he suggested a necessary condition for inspection effectiveness that the sum of two errors must be less than one. However, the necessary condition is so rough that it cannot be used practically.In this paper, we deal with the above fundamental question. In Section2, we describe our problem in detail. Assuming that an imperfect inspector classifies an infinite sequence of items with a known FD and that each inspection error of type I or type II is either a constant or a uniform random variable on an interval, we provide four mathematical models: Model I (C, C) with both type I error and type II error being constant, Model II (R, C) with type I error and type II error being random and constant, respectively, Model III (C, R) with type I error and type II error being constant and random, respectively, and Model IV (R, R) with type I error and type II error being random and random, respectively. In Sections 3 through 6, for each model, we derive formulas for computing the probability of overestimation (POE) and a critical FD satisfying POE = 50%. In Section 7, in order to extract some relation between the results of the previous sections, we make a reasonable assumption and prove a theorem that answers our question.


## 2. Problem Statement

Suppose that an imperfect inspector with nonzero inspection errors classifies one-by-one an infinite sequence of items with a known fraction defectiveq, but which is unknown to the inspector. Then the sequence can be considered as an infinite Bernoulli process {Xi,i=1,2,3,…} such that, for each i, the value of Xi is either zero, representing a C item, or one, representing an NC item; for all values of i, the probability that Xi=1, Pr⁡⁡{Xi=1}, is the same number q. Let Ai be the probability (the type I error) that the ith C item is misclassified as NC and falsely rejected by the inspector; let Bi be the probability (the type II error) that the ith NC item is misclassified as C and falsely accepted by the inspector. Let Yi be zero, if the ith item is judged as a C item by the inspector, and one, otherwise. That is, Ai=Pr⁡⁡{Yi=1∣Xi=0} and Bi=Pr⁡⁡{Yi=0∣Xi=1}. Then, Pr⁡⁡{Yi=1} and E[Yi] can be obtained as
(1)E[Yi]=Pr⁡{Yi=1}=∑k=01Pr⁡⁡{Yi=1∣Xi=k}Pr⁡{Xi=k}=Ai(1-q)+(1-Bi)q.Since the expected number of rejected items after thenth inspections is ∑i=1nE[Yi], the FD of an infinite number of items judged by the inspector, denoted by Q, becomes lim⁡n→∞⁡(1/n)∑i=1nE[Yi] and can be reduced to
(2)Q=q+lim⁡n→∞⁡1n{(1-q)∑i=1nAi-q∑i=1nBi}.The above equation implies that the value ofQ becomes q if all Ai and Bi are zeros since the value of the limit term becomes zero. We assume that two types of inspection errors are nonzero and less than or equal to one unless specially mentioned. That is, 0<Ai,Bi≤1 for all i. Hence, the value of the limit terms can be positive, zero, or negative, corresponding to overestimation, correct estimation, or underestimation, respectively. Either overestimation or underestimation does not raise any problems by themselves, as long as their probabilities are exactly the same. In fact, we are likely to believe that all inspectors are expected to either overestimate or underestimate a given FD with a 50 : 50 chance. Otherwise, either buyer or supplier must face economic loss due to an unfair inspection game. However, unfortunately, it turns out in this paper that the 50 : 50 chance is not always true and that it depends upon q and {(Ai,Bi),i=1,2,3,…}. Let POE be the probability of overestimation by the inspector, which can be reduced to
(3)POE=Pr⁡{Q>q}=Pr⁡{lim⁡n→∞⁡1n{(1-q)∑i=1nAi-q∑i=1nBi}>0}.Ifq is an input variable, POE can be expressed as POE(q). If there exists a unique FD q* such that POE(q*)=0.5, then q* is termed as “the critical fraction defective” or CFD. This definition implies that E[Q∣q=q*]=q*; that is, Q is an conditional unbiased estimator of CFD when Q is a random variable. In the case that such a CFD does not exist, q will be called as CFD only if an inspector estimate q correctly, that is, Q=q, and POE(q) is defined to be 0.5.Since inspection errorAi as well as Bi can be assumed to be either a constant or a random variable, we need four kinds of analyses of POE and CFD, as shown in Table 1. Note that the subscript “c” (or “r”) under the right sides of POE and CFD in the table represents that the type I or type II error is assumed to be constant (or random). In order to obtain some fundamental properties, we assume that each random variable follows a uniform distribution with an interval on (0, an upper value]. That is, fA(a)=(1/αu)I(0,αu](a), and fB(b)=(1/βu)I(0,βu](b) where I(0,xu](x) is an indicator function with one for 0<x≤xu, and zero otherwise. This uniformity assumption with zero/an upper value may be justified, since it has been hoped that inspection errors would become smaller and smaller, and there has been almost no information on the distribution of inspection error, up to now. It is well known that uniform distribution gives maximum uncertainty. Our problem can be summarized as follows: derive both POE(q) and CFD for each model and answer the fundamental question:“Does an imperfect inspector overestimate a given FD when it is very low?”Table 1
Four types of POE and CFD analysis.
Models
Model I (C, C)
Model II (R, C)
Model III (C, R)
Model IV (R, R)
Input
Known FD
Constant,q
Constant,q
Constant,q
Constant,q
Type I error
Constant,α
Random variable,A~fA(a)
Constant,α
Random variable,A~fA(a)
Type II error
Constant,β
Constant,β
Random variable,B~fB(b)
Random variable,B~fB(b)
Output
POE
POE
cc
(
q
)
POE
rc
(
q
)
POE
cr
(
q
)
POE
rr
(
q
)
CFD
CFDcc
CFDrc
CFDcr
CFDrrThroughout this paper, the first and second derivatives of a functionf(x) will be expressed as f′(x) and f′′(x), respectively, and the expectation of a random variable X will be denoted by E[X].


## 3. Analysis of Model I (C, C)

Suppose thatAi=α and Bi=β, for all i, and both α and β are real-valued constants. Then, since Pr⁡⁡{Yi=1∣Xi=0}=α and Pr⁡⁡{Yi=1∣Xi=1}=1-β, the process {Yi,i=1,2,3,…} becomes an infinite Bernoulli process with Pr⁡{Yi=1} being the same number {α(1-q)+(1-β)q}. The following proposition indicates that POEcc(q) depends on q and ρcc where ρcc=β/α and that CFDcc depends only on ρcc, not q. Let qcc=1/(1+ρcc) and assume that ρcc is a rational number.Proposition 1.
Under the assumption thatAi=α and Bi=β, for all i, and both α and β are real-valued constants with 0<α,β≤1, (1)
(4)POEcc(q)={1,for0≤q<qcc,0.5,forq=qcc,0,forqcc<q≤1,(2)
CFDcc=qcc,(3)
an inspector with(α,β) always overestimates q for 0≤q<qcc, estimates q correctly for q=qcc with POEcc(q)=0.5, and always underestimates q for qcc<q≤1.Proof.
Equations (2) and (3) can be derived, respectively, as
(5)Qcc=q+(α+β)(qcc-q),POEcc(q)=Pr⁡{Qcc>q}=Pr⁡{(α+β)(qcc-q)>0}.
For0≤q<qcc, since the inequality (α+β)(qcc-q)>0 is “always” true, (i.e., an inspector with ρcc “always” overestimates q), POEcc(q) can be defined to be one. For qcc<q≤1, since an inspector with ρcc “always” underestimates q, POEcc(q) can be defined to be zero. In the case of q=qcc, since Qcc=q, that is, an inspector estimate q correctly, by our definition in Section 2, it follows that CFDcc=qcc and POEcc(qcc)=0.5. However, q is a rational number (note that q can be expressed as lim⁡n→∞⁡(1/n)∑i=1nXi, which can be expressed as the fraction m/n of two integers, with n>0), while qcc can be an irrational number depending on the values of α and β. Hence, q cannot be equal to qcc if ρcc (or qcc) is an irrational number. Since we assume that ρcc (or qcc) is a rational number, we have, Qcc=q if and only if q=qcc and Proposition 1-(3) holds true.From Proposition1, POEcc(q) can be drawn as shown in Figure 1(a). Suppose that both q and ρcc are input variables. Then, since an inspector with ρcc overestimates if and only if 0≤q<qcc=1/(1+ρcc), every point (q,ρcc) satisfying q(1+ρcc)<1 gives overestimation. For q≠0, every point (q,ρcc) satisfying 0<ρcc<(1-q)/q gives overestimation. For q=0, every point (0,ρcc) for ρcc>0 gives overestimation. Let Rcc be the overestimation region as shown in Figure 1(b). Then, an inspector with ρcc overestimates q if (q,ρcc)∈Rcc, estimates q correctly if ρcc=(1-q)/q, and underestimates q otherwise. Note that the point (0,0) cannot be included in Rcc since ρcc≠0 due to the assumption that α≠0.The graph ofPOEcc(q) and the overestimation region in two different planes.
(a)
The graph ofPOEcc(q)
(b)
The region of overestimation in(q,ρcc) plane
(c)
The region of overestimation in(α,β) plane when qs>β/αSuppose that bothα and β are input variables and that q is given as a constant. Then, since an inspector with ρcc overestimates if and only if q<qcc=α/(α+β), every point (α,β) satisfying q(α+β)<α gives overestimation. For q≠0, every point (α,β) satisfying β<qsα gives overestimation where qs=(1-q)/q. For q=0, every point (α,β) satisfying α>0 gives overestimation. Let Rcc′ be the overestimation region bounded by 0<β<qsα, 0<α≤1, and 0<β≤1 as shown in Figure 1(c) where qs>β/α. That is, an inspector with (α,β) overestimates q if (α,β)∈Rcc′, estimates q correctly if β=qsα, and underestimates q otherwise.Ifq=0, then Qcc=α from (2) and the inspector estimates correctly if and only if α=0. Hence, POEcc(0) has two values either 0.5 (when α=0) or one (when 0<α≤1) depending on α. If q=1, then Qcc=1-β from (2), and the inspector estimates correctly if and only if β=0. Hence, POEcc(1) has two values either 0.5 (when β=0) or zero (when 0<β≤1) depending on β. These special cases of (q,α)=(0,0) and (q,β)=(1,1) will be discussed only if necessary.Since CFDcc depends only on ρcc, CFDcc remains the same even if both α and β are multiplied by the same number. For example, the CFDcc of (α,β)=(1ppm,3ppm) is 0.25, the same as that of (α,β)=(10%,30%). Typical CFDcc values for 0≤α,β≤1 are summarized in Table 2. As shown in the table, the value of CFDcc is 50% on the diagonal, more than 50% above the diagonal, and less than 50% below the diagonal. It can be proved that (1)CFDcc is zero if and only if α=0 and is one if and only if β=0 except the case that (α,β)=(0,0), and (2) for a given β, CFDcc increases strictly and forms a concave shape as α increases, while for a given α,  CFDcc decreases strictly and forms a convex shape as β increases.Table 2
Critical fraction defective for0≤α,β≤1.
β
α
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
0
—
1.00
1.00
1.00
1.00
1.00
1.00
1.00
1.00
1.00
1.00
0.1
0.00
0.50
0.67
0.75
0.80
0.83
0.86
0.88
0.89
0.90
0.91
0.2
0.00
0.33
0.50
0.60
0.67
0.71
0.75
0.78
0.80
0.82
0.83
0.3
0.00
0.25
0.40
0.50
0.57
0.63
0.67
0.70
0.73
0.75
0.77
0.4
0.00
0.20
0.33
0.43
0.50
0.56
0.60
0.64
0.67
0.69
0.71
0.5
0.00
0.17
0.29
0.38
0.44
0.50
0.55
0.58
0.62
0.64
0.67
0.6
0.00
0.14
0.25
0.33
0.40
0.45
0.50
0.54
0.57
0.60
0.63
0.7
0.00
0.13
0.22
0.30
0.36
0.42
0.46
0.50
0.53
0.56
0.59
0.8
0.00
0.11
0.20
0.27
0.33
0.38
0.43
0.47
0.50
0.53
0.56
0.9
0.00
0.10
0.18
0.25
0.31
0.36
0.40
0.44
0.47
0.50
0.53
1
0.00
0.09
0.17
0.23
0.29
0.33
0.38
0.41
0.44
0.47
0.50Example 2.
Suppose thatq=5% and in order to estimate the FD correctly, we must select one inspector among three inspectors with (αi,βi) for i=1,2,3 as shown in Table 3. In order to reflect actual inspection errors of a back-light unit manufacturer, the value of αi is set to be smaller than that of βi. At first glance, the inspector with either (α1,β1)=(0.1%,2%) or (α2,β2)=(0.2%,2.5%) is more likely to be selected than the inspector with (α3,β3)=(0.3%,5.7%) since α1<α2<α3 and β1<β2<β3. However, it is a wrong decision since inspector 1 underestimates q as 4.995% and inspector 2 overestimates q as 5.065%, while the right decision is to select the worst inspector 3 since only inspector 3 can estimate q correctly. This decision seems to be very strange at first time but gives a new concept to quality control managers, who could utilize intentionally this perspective in special situations.Table 3
POE
cc
(
q
), CFDcc, and related results for n=10,000, q=5%, and (αj,βj) for j=1,2,3.
Inspector 1
Inspector 2
Inspector 3
α
j
0.1%
0.2%
0.3%
β
j
2.0%
2.5%
5.7%
ρ
cc
20.0
12.5
19.0
Q
cc
4.995%
5.065%
5.000%
POE
cc
(
q
)
0
1
0.5
CFD
cc
(=qcc)
4.7619%
7.4074%
5.0000%


## 4. Analysis of Model II (R, C)

Suppose thatAi=A and Bi=β for all i where β is a constant with 0<β≤1 and A is a random variable distributed with the probability density function fA(a) for 0<a≤αu and 0<αu≤1. Then, since Pr⁡⁡{Yi=1∣Xi=0}=A and Pr⁡⁡{Yi=0∣Xi=1}=β, the process {Yi,i=1,2,3,…} becomes an infinite “random-Bernoulli” process with Pr⁡⁡{Yi=1} being not a constant but the random variable {(1-q)A+(1-β)q}. From (2), Qrc can be obtained as (1-q)A+(1-β)q. From (3), we have
(6)POErc(q)=Pr⁡{Qrc>q}=Pr⁡{(1-q)A>βq}.Forq≠1, (6) can be further reduced to(7)POErc(q)=Pr⁡{A>βq1-q}={1,forβq1-q=0,1-FA(βq1-q),for0<βq1-q≤αu,0,forαu<βq1-q,
where FA(q)=∫0qfA(a)da. For q=1, POErc(1) becomes zero since Qrc=(1-β) is always smaller than q=1. If q=1 and β=0, then since Qrc=q=1, every inspector with β=0 always estimates correctly regardless of any distribution of fA(a), and by our definition of POE, POErc(1)=0.5. Hence, POErc(1) has two values either 0.5 (when β=0) or zero (when 0<β≤1) depending on β. Suppose that fA(a)=(1/αu)I(0,αu](a). Then, we have the following proposition indicating that POErc(q) can be expressed as a function of two input variables ρrc, and q where ρrc=2β/αu.Proposition 3.
Under the assumptions thatAi=A and Bi=β for all i, where β is a constant with 0<β≤1 and A is a random variable distributed with fA(a)=(1/αu)I(0,αu](a) for 0<a≤αu and 0<αu≤1,(1)
(8)POErc(q)={1-ρrcq2(1-q),for0≤q≤qrc,0,forqrc≤q≤1,whereqrc=2/(2+ρrc),(2)
POErc(q) is a strictly decreasing concave function of qfor0≤q≤qrc,(3)
CFDrc=1/(1+ρrc),(4)
the inspector with(αu,β) overestimates q with POE>0.5 for 0≤q<CFDrc, estimates q correctly with POE=0.5 for q=CFDrc, and underestimates q with POE<0.5 for CFDrc<q≤1.Proof.
From (7), since FA(βq/(1-q))=βq/αu(1-q) for 0≤q≤qrc, POErc(q) can be derived as above. Since POErc′(q)=-ρrc/2(1-q)2<0 and POErc′′(q)=-ρrc/(1-q)3<0, POErc(q) is a strictly decreasing concave function of q for 0≤q≤qrc. Since 0≤POErc(q)≤1 for 0≤q≤qrc, solving POErc(q*)=0.5 for 0≤q*≤qrc gives CFDrc=1/(1+ρrc). Hence, (4) holds true.Note thatPOErc(q) and CFDrc for any density function fA(a) can be generally derived as shown in Proposition A.1 in Appendix. Also note that E[Qrc∣q=CFDrc]=(1-CFDrc)E[A]+(1-β)CFDrc=CFDrc where E[A]=αu/2.A representative graph ofPOErc(q) is shown in Figure 2(a). Suppose that both q and ρrc are input variables. Then, solving POErc(q)=1-ρrcq/2(1-q)>0.5 for 0≤q≤qrc gives 0<ρrc<(1-q)/q and Proposition 3 implies that every point (q,ρrc) in the shaded region Rrc as shown in Figure 2(b) gives overestimation with POE>0.5, where Rrc can be represented by {(q,ρrc)∣0<ρrc<(1-q)/q,0≤q<1}. That is, an inspector with ρrc overestimates q if (q,ρrc)∈Rrc, estimates q with POE=0.5 if ρrc=(1-q)/q, and underestimates q otherwise. Note that if β=0, then every point (q,0) for 0<q<1 on the line ρrc=0 gives overestimation with POE>0.5 since Qrc=(1-q)A+q is always greater than q.The graph ofPOErc(q) and the overestimation region in two different planes.
(a)
The graph ofPOErc(q)
(b)
The region of overestimation in(q,ρrc) plane
(c)
The region of overestimation in(αu,β) plane given q when 0.5qs>β/αuSuppose that bothαu and β are input variables and q is given as a constant. Then, solving POErc(q)=1-βq/αu(1-q)>0.5 for 0≤q≤qrc gives 0<β<0.5qsαu and Proposition 3 implies that every point (αu,β) in the shaded region Rrc′ as shown in Figure 2(c) gives overestimation with POE>0.5 where Rrc′={(αu,β)∣0<β<0.5qsαu,0<αu≤1,0<β≤1}. That is, an inspector with (αu,β) overestimates with POE>0.5 if (αu,β)∈Rrc′, estimates q with POE=0.5 if β=0.5qsαu, and underestimates q with POE<0.5 otherwise.Example 4.
Forq=5% and three inspectors given in Table 4, POErc(q) and CFDrc for each inspector can be computed and summarized in the table. If we would like to select the inspector satisfying POErc(q)=50%, then inspector 3 will be selected again.Table 4
POE
rc
(
q
) and CFDrc for q=5%
and different inspectors.
Inspector 1
Inspector 2
Inspector 3
α
u (%)
0.2
0.4
0.6
β (%)
2.0
2.5
5.7
ρ
rc
=
2
β
/
α
u
20.0
12.5
19.0
CFD
rc
=
q
rc
4.76%
7.41%
5.00%
POE
rc
(
q
)
47.37%
67.11%
50.00%


## 5. Analysis of Model III (C, R)

Suppose thatAi=α and Bi=B for all i where α is a constant with 0<α≤1 and B is a random variable distributed with the probability density function fB(b) for 0<b≤βu and 0<βu≤1. Then, since Pr⁡⁡{Yi=1∣Xi=0}=α, Pr⁡⁡{Yi=0∣Xi=1}=B, the process {Yi,i=1,2,3,…} becomes an infinite random-Bernoulli process with Pr⁡{Yi=1} being the random variable {α(1-q)+(1-B)q}. From (2), Qcr can be obtained as α(1-q)+(1-B)q. From (3), we have
(9)POEcr(q)=Pr⁡{Qcr>q}=Pr⁡{qB<α(1-q)}.Forq≠0, (9) can be further reduced to
(10)POEcr(q)=Pr⁡{B<α(1-q)q}={1,forα(1-q)q≥βu,FB(α(1-q)q),forα(1-q)q<βu,
where FB(q)=∫0qfB(b)db.Forq=0, POEcr(0) becomes one since Qcr=α is always greater than q=0 from (9). If q=α=0, then since Qcr(=0) is always equal to q(=0), every inspector with α=0 always estimates correctly regardless of any distribution of fB(b), and, by our definition of POE, POEcr(0) is not one but 50% in this case. This result implies that POEcr(0) can be either one or 50% depending on the value of α. Since we assume that 0<α≤1, we have POEcr(0)=1. Suppose that fB(b)=(1/βu)I(0,βu](b). Then, we have the following proposition indicating that POEcr(q) can be expressed as a function of two input variables ρcr and q where ρcr=βu/2α.Proposition 5.
Under the assumption thatAi=α and Bi=B for all i where α is a constant with 0<α≤1 and B is a random variable distributed with fB(b) for 0<b≤βu and 0<βu≤1, (1)
(11)POEcr(q)={1,for0≤q≤qcr,(1-q)2ρcrq,forqcr≤q≤1,whereqcr=1/(1+2ρcr),(2)
POEcr(q) is a strictly decreasing convex function of q for qcr≤q≤1,(3)
CFDcr=1/(1+ρcr),(4)
the inspector with(α,βu) overestimates q with POE>0.5 for 0≤q<CFDcr, estimates q with POE=0.5 for q=CFDcr, and underestimates q with POE<0.5 for CFDcr<q≤1.Proof.
From (10), since FB(α(1-q)/q)=α(1-q)/βuq for qcr≤q≤1, POEcr(q) can be derived as above. Since POEcr′(q)=-1/2ρcrq2<0 and POEcr′′(q)=1/ρcrq3>0 for qcr≤q≤1, POEcr(q) is a strictly decreasing convex function of q for qcr≤q≤1. Since 0≤POEcr(q)≤1 for qcr≤q≤1, solving POEcr(q*)=0.5 for qcr≤q*≤1 gives CFDcr=1/(1+ρcr). Hence, (4) holds true.Note that generallyPOEcr(q) and CFDcr for any density function fB(b) can be derived as shown in Proposition A.2 in Appendix. Also note that E[Qcr∣q=CFDcr]=(1-CFDcr)α+(1-E[B])CFDcr=CFDcr where E[B]=βu/2.A representative graph ofPOEcr(q) is shown in Figure 3(a). Suppose that both q and ρcr are input variables. Then, solving POEcr(q)=(1-q)/2ρcrq>0.5 for qcr≤q≤1 gives 0<ρcr<(1-q)/q and Proposition 5 implies that every point (q,ρcr) in the shaded region Rcr as shown in Figure 3(b) gives overestimation with POE>0.5 where Rcr={(q,ρcr)0<ρcr<(1-q)/q,0≤q<1}. That is, an inspector with ρcr overestimates q if (q,ρcr)∈Rcr, estimates q with POE=0.5 if ρcr=(1-q)/q, and underestimates q with POE<0.5 otherwise. Note that if α=0, then every point (q,0) for 0<q<1 gives underestimation since Pr⁡{Qcr>q}=Pr⁡{B<0}=0.The graph ofPOEcr(q) and the overestimation region in two different planes.
(a)
The graph ofPOEcr(q)
(b)
The region of overestimation in(q,ρcr) plane
(c)
The region of overestimation in(α,βu) plane given q when 2qs>βu/αSuppose that bothα and βu are input variables and q is given as a constant. Then, solving POEcr(q)=(1-q)α/qβu>0.5 for qcr≤q≤1 gives 0<βu<2qsα and Proposition 5 implies that every point (α,βu) in the shaded region Rcr′ as shown in Figure 1(c) which gives overestimation with POE>0.5 where Rcr′={(α,βu)∣0<βu<2qsα,0<α≤1,0<βu≤1}. That is, an inspector with (α,βu) overestimates with POE>0.5 if (α,βu)∈Rcr′, estimates q with POE=0.5 if β=2qsαu, and underestimates q with POE<0.5 otherwise.Example 6.
Forq=5% and three inspectors given in Table 5, POEcr(q) and CFDcr for each inspector can be computed and summarized in the table. If we would like to select the inspector satisfying POEcr(q)=50%, then inspector 3 will be selected again.Table 5
POE
cr
(
q
) and CFD for q=5% and different inspectors.
Inspector 1
Inspector 2
Inspector 3
α(%)
0.1
0.2
0.3
β
u (%)
4.0
5.0
11.4
ρ
cr
=
β
u
/
2
α
20.0
12.5
19.0
CFD
cr
=
q
cr
4.76%
7.41%
5.00%
POE
cr
(
q
)
47.50%
76.00%
50.00%


## 6. Analysis of Model IV (R, R)

Suppose thatAi=A and Bi=B for all i where A and B are random variables distributed with the probability density functions fA(a) and fB(b), respectively, for 0<a≤αu, 0<b≤βu and 0<αu,βu≤1. Then, since Pr⁡⁡{Yi=1∣Xi=0}=A and Pr⁡⁡{Yi=0∣Xi=1}=B, the process {Yi,i=1,2,3,…} becomes an infinite random-Bernoulli process with Pr⁡{Yi=1} being the random variable {A(1-q)+(1-B)q}. From (2), Qrr can be obtained as A(1-q)+(1-B)q. From (3), POE can be expressed as
(12)POErr(q)=Pr⁡{Qrr>q}=Pr⁡{(1-q)A>qB}.Ifq=1, then POErr(q)=0 since POErr(q)=Pr⁡{B<0}=0; that is, even perfect inspector always underestimates q regardless of any distribution of (fA(a),fB(b)). On the other hand, if q=0, then POErr(0)=1 since POErr(0)=Pr⁡{A>0}=1; that is, even perfect inspector always overestimates q regardless of any distribution of (fA(a),fB(b)). For 0<q<1, (12) can be reduced to
(13)POErr(q)=∫0βuPr⁡{(1-q)A>qB∣B=b}fB(b)db=∫0βuPr⁡{qb1-q<A≤αu}fB(b)db=∫0βu{∫qb/(1-q)αufA(a)da}fB(b)db=∬(a,b)∈SrrfA(a)fB(b)dadb,
where Srr={(a,b)|b<qsa,0≤a≤αu,0≤b≤βu,and0<αu,βu≤1}.Suppose thatfA(a)=(1/αu)I(0,αu](a) and fB(b)=(1/βu)I(0,βu](b). Then, we have the following proposition indicating that POErr(q) can be expressed as a function of two input variables ρrr and q where ρrr=βu/αu.Proposition 7.
Under the assumption thatAi=A and Bi=B for all i where A and B are random variables distributed with fA(a)=(1/αu)I(0,αu](a) and fB(b)=(1/βu)I(0,βu](b), respectively, for 0<a≤αu, 0<b≤βu, and 0<αu,βu≤1, (1)
(14)POErr(q)={POErr1(q)=1-ρrrq2(1-q),for0≤q≤qrr(orqs≥ρrr),POErr2(q)=1-q2ρrrq,forqrr≤q≤1(or0≤qs≤ρrr),whereqrr=1/(1+ρrr),(2)
POErr1(q) is a strictly decreasing concave function of q with POErr1(0)=1, and POErr2(q) is a strictly decreasing convex function of q with POErr2(1)=0,(3)
CFDrr=1/(1+ρrr),(4)
the inspector withρrr overestimates q with POE>0.5 for 0≤q<CFDrr, estimates q with POE=0.5 for q=CFDrr, and underestimates q with POE<0.5 for CFDrr<q≤1.Proof.
It is proved from (12) that POErr(0)=1 and POErr(1)=0. For 0<q<1, since the shape of Srr depends upon qs and ρrr, for qs≥ρrr as shown in Figure 4(a), (13) can be reduced to
(15)POErr1(q)=1αuβu∫0βu∫qb/(1-q)αudadb=1-ρrrq2(1-q);
and for 0<qs≤ρrr as shown in Figure 4(b), (13) can be reduced to
(16)POErr2(q)=1αuβu∫0((1-q)/q)αu∫qb/(1-q)αudadb=1-q2ρrrq.
Equation (15) holds when q=0 (equivalently, qs=∞) since replacing q in (15) with zero gives POErr1(0)=1 and (16) holds when q=1 (equivalently, qs=0) since replacing q in (16) with one gives POErr2(1)=0. For 0≤q≤qrr, since POErr1′(q)=-ρrr/2(1-q)2<0 and POErr1′′(q)=-ρrr/(1-q)3<0, POErr1(q) is a strictly decreasing concave function of q. For qrr≤q≤1, since POErr2′(q)=-1/2ρrrq2<0 and POErr2′′(q)=1/ρrrq3>0, POErr2(q) is a strictly decreasing convex function of q. Note that POErr(q) is differentiable at q=qrr. Hence, POErr(q) is a strictly decreasing function of q with POErr(0)=POErr1(0)=1 and POErr(1)=POErr2(1)=0. Since POErr1(qrr)=POErr2(qrr)=0.5, CFDrr is qrr. Since POErr1(q)> 0.5 for 0≤q<CFDrr and POErr2(q)< 0.5 for CFDrr<q≤1, (4) holds true.The region ofSrr depending on qs and ρrr.
(a)
Forqs≥ρrr (or 0<q≤qrr)
(b)
For0<qs≤ρrr (or qrr≤q<1)SinceA and B are uniform random variables, respectively, (15) and (16) can be derived by a different method as follows. Equation (13) can be expressed as
(17)POErr(q)=∬(a,b)∈SrrfA(a)fB(b)dadb=1αuβu∬(a,b)∈Srrdadb.Since the value of the integral is equivalent to the size of the areaSrr as shown in Figure 4, POErr(q) can be interpreted as the ratio of the size of Srr to the size of the rectangle αuβu. The shape of Srr varies depending on the straight line b=qsa, which cuts the rectangle into two parts, that is, an upper part and a lower part as shown in Figure 4. The lower part of the rectangle under the line corresponds to Srr, a right-angled triangle when 0<qs≤ρrr, and a trapezoid when qs>ρrr. Since the size of the trapezoid and the triangle can be derived as (2qsαu-βu)βu/2qs and qsαu2/2, respectively, POErr(q) can be easily obtained.It seems to be mathematically hard to derivePOErr(q) and CFDrr generally for any density functions. From Propositions A.1 and A.2 in Appendix, our conjecture is that CFDrr=FA-1(0.5)/(FA-1(0.5)+FB-1(0.5)). Note that E[Qrr∣q=CFDrr]=(1-CFDrr)E[A]+(1-E[B])CFDrr=CFDrr.From Proposition7, a graph of POErr(q) can be drawn as shown in Figure 5(a). Suppose that q and ρrr are input variables. Then, since POErr1(q)> 0.5, solving POErr1(q)=1-ρrrq/2(1-q)>0.5 for 0≤q≤qrr=1/(1+ρrr) gives 0<ρrr<(1-q)/q and Proposition 7 implies that every point (q,ρrr) in the shaded region Rrr as shown in Figure 5(b) gives overestimation with POE>0.5 where Rrr={(q,ρrr)∣0<ρrr<(1-q)/q,0≤q<1}. That is, an inspector with ρrr overestimates q with POE>0.5 if (q,ρrr)∈Rrr, estimates q with POE=0.5 if ρrr=qs, and underestimates q with POE<0.5 otherwise.The graph ofPOEcr(q) and the overestimation region in two different planes.
(a)
The graph ofPOErr(q)
(b)
The region of overestimation in(q,ρrr) plane
(c)
The region of overestimation in(αu,βu) plane for given q when qs>βu/αuSuppose thatq is a given constant and both αu and βu are input variables. Then, similarly solving POErr1(q)=1-(q/2(1-q))·(βu/αu)>0.5 for 0≤q≤qrr=αu/(αu+βu) gives 0<βu<((1-q)/q)αu and Proposition 7 implies that every point (αu,βu) in the shaded region Rrr′ as shown in Figure 5(c) gives overestimation POE>0.5 where Rrr′={(αu,βu)∣0<βu<qsαu,0<αu≤1,0<βu≤1}. That is, an inspector with (αu,βu) overestimates q with POE>0.5 if (αu,βu)∈Rrr′, estimates q with POE=0.5 if βu=qsαu, and underestimates q with POE<0.5 otherwise.Example 8.
Forq=5% and three inspectors given in Table 6, POErr(q) and CFDrr for each inspector can be computed and summarized in the table. If we would like to select the inspector satisfying POErr(q)=50%, then inspector 3 will be selected again as before.Table 6
POE
rr
(
q
) and CFD for q=5% and different inspectors.
Inspector 1
Inspector 2
Inspector 3
α
u (%)
0.2
0.4
0.6
β
u (%)
4.0
5.0
11.4
ρ
rr
=
β
u
/
α
u
20.0
12.5
19.0
CFD
rr
=
q
rr
4.76%
7.41%
5.00%
POE
rr
(
q
)
47.50%
67.11%
50.00%


## 7. Summary

In order to find relations between four propositions, the constantsα and β can be assumed to be E[A] and E[B], respectively. This assumption may be justified since a constant can be interpreted as a representative value. Since A and B are uniform random variables on (0,αu] and (0,βu], respectively, we have, α=E[A]=αu/2 and β=E[B]=βu/2, respectively. Thus, the following theorem holds true.Theorem 9.
Assuming an infinite sequence of items with a known FDq and nonzero inspection errors with α=E[A]=αu/2 and β=E[B]=βu/2, (1)
an imperfect inspector with nonzero inspection errors(E[A],E[B]) has his/her own POE curve and CFD,(2)
POE is a function of two variablesq and ρ, denoted by POE(q,ρ),(3)
POE is a decreasing function ofq with POE(0,ρ)=1 and POE(1,ρ)=0,(4)
POE is a decreasing function ofρ with POE(q,0)=1 and POE(q,1)=0,(5)
there always exists a uniqueCFD=1/(1+ρ), which depends only on inspection errors and not q,(6)
the inspector overestimatesq with POE>0.5 for 0≤q<CFD, estimates q with POE=0.5 for q=CFD, and underestimates q with POE<0.5 for CFD<q≤1.Proof.
Ifα=E[A]=αu/2 and β=E[B]=βu/2, then we have ρcc=ρrc=ρcr=ρrr=E[B]/E[A] and CFDcc=CFDrc=CFDcr=CFDrr=E[A]/(E[A]+E[B])=1/(1+ρ), where ρ=E[B]/E[A]. Let CFD=1/(1+ρ). Then, from the results of four propositions, the theorem except (4) holds true and those results are summarized in Table 7. By using the similar method used for proofs of propositions, (4) can be proved.Table 7
Four types of POE and CFD analysis under the assumption ofα=E[A]=αu/2 and β=E[B]=βu/2.
Models
Model I (C, C)
Model II (R, C)
Model III (C, R)
Model IV (R, R)
FD
Constant,q
Constant,q
Constant,q
Constant,q
Type I error
Constant,α
Uniform random variable on(0,αu]
Constant,α
Uniform random variable on(0,αu]
Type II error
Constant,β
Constant,β
Uniform random variable on(0,βu]
Uniform random variable on(0,βu]
ρ
ρ
cc
=
β
α
=
ρ
ρ
rc
=
2
β
α
u
=
ρ
ρ
cr
=
β
u
2
α
=
ρ
ρ
rr
=
β
u
α
u
=
ρ
POE
POE
cc
(
q
,
ρ
)
=
1
for
0
≤
q
<
1
1
+
ρ
0
.
5
for
q
=
1
1
+
ρ
0
for
1
1
+
ρ
<
q
≤
1
POE
rc
(
q
,
ρ
)
=
1
-
ρ
q
2
(
1
-
q
)
for
0
≤
q
≤
2
2
+
ρ
0
for
2
2
+
ρ
≤
q
≤
1
POE
cr
(
q
,
ρ
)
=
0
for
0
≤
q
≤
1
1
+
2
ρ
(
1
-
q
)
2
ρ
q
for
1
1
+
2
ρ
≤
q
≤
1
POE
rr
(
q
,
ρ
)
=
1
-
ρ
q
2
(
1
-
q
)
for
0
≤
q
≤
1
1
+
ρ
1
-
q
2
ρ
q
for
1
1
+
ρ
<
q
≤
1
CFD
CFD
cc
=
α
α
+
β
=
1
1
+
ρ
CFD
rc
=
0.5
α
u
0.5
α
u
+
β
=
1
1
+
ρ
CFD
cr
=
α
α
+
0.5
β
u
=
1
1
+
ρ
CFD
rr
=
α
u
α
u
+
β
u
=
1
1
+
ρSince we have(18)qcc=E[A]E[A]+E[B]=11+ρ,qrc=2E[A]2E[A]+E[B]=22+ρ,qcr=E[A]E[A]+2E[B]=11+2ρ,qrr=E[A]E[A]+E[B]=11+ρ,POErc(q)=1-ρq2(1-q)for0≤q≤22+ρ,POEcr(q)=1-q2ρqfor11+2ρ≤q≤1,POErr(q)={1-ρq2(1-q),for0≤q≤11+ρ,1-q2ρq,for11+ρ≤q≤1,
it can be observed in Table 7 that POErr(q) has the same form as POErc(q) for 0≤q≤1/(1+ρ) and POEcr(q) for 1/(1+ρ)≤q≤1 even though their related domains are not the same. Now, based on our theorem, our answer to the fundamental question “Does an imperfect inspector overestimate a known fraction defective when it is very low?” could be “certainly yes at least under our assumptions” since POE≈1 when FD is very low.


## 8. Conclusion

Overestimation by an inspector may be explained not only by inspection errors but also by other factors such as psychological aspects of inspectors, incentive plans for inspectors, workload, conflicts among inspectors, and so on. However, our results and concepts are based on four assumptions: the assumption of an infinite sequence of items, the assumption of a fixed known FD, the assumption of nonzero inspection errors, and the assumption of a uniform distribution. Further research may be concentrated on relaxing these assumptions. We may obtain slightly different results, by assuming a finite sequence of items; or by assuming that FD is not a fixed constant, but a random variable; or by assuming other distributions, such as a skewed triangular, a truncated normal, or an empirical distribution for a lower/upper limit interval. However, our strong conjecture is that our theorem will still be true, regardless of any distribution, and even a finite number of items, as long as FD is a constant. Since our mathematical models do not consider any related costs, a cost-based optimization model with POE could be constructed, to determine a trade-off point between buyers and sellers.If we consider the fair trade between a seller and a buyer, and the trend that FD’s of manufacturers have been continuously approaching zero, Theorem9 implies that either the ratio of type I error to type II error must go to infinity, or the type I error must be zero in order for CFD to approach zero, and that all commercial inspection plans should be revised with the concept of POE in the near future, for the fairness of commercial trades, since the smaller (up to several hundreds ppm level) the FD of items sold by sellers is, the more their unfair loss is forced to be. We hope that the concept of POE should become one of the major criteria in the future. Our methodology used in this paper could, with slight modification, be applied and extended to the existing sampling plans.



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*Source: 619639-2014-09-15.xml*