國 立 交 通 大 學
統 計 學 研 究 所
碩 士 論 文
貨批中有
γ比例良品其信心之統計推論
γ
Statistical Inferences for Confidence of Percentage
Acceptable Products in Lot
研
究
生：蕭夙吟 (Su-Yin Hsiao)
指導教授：陳鄰安 博士 (Dr. Lin-An Chen)
貨批中有
γ比例良品其信心之統計推論
γ
Statistical Inferences for Confidence of Percentage
Acceptable Products in Lot
研
究 生：蕭夙吟
Student：Su-Yin Hsiao
指導教授：陳鄰安 博士 Advisor：Dr. Lin-An Chen
國 立 交 通 大 學
統計學研究所
碩 士 論 文
A Thesis
Submitted to Institute of Statistics
College of Science
National Chiao Tung University
In Partial Fulfillment of the Requirements
For the Degree of
Master
In
Statistics
June 2008
Hsinchu, Taiwan, Republic of China
中華民國九十七年六月
貨批中有
γ比例良品其信心之統計推論
研
究 生：蕭夙吟
指導教授：陳鄰安 博士
國立交通大學統計學研究所 碩士班
摘 要
容忍區間經常被用來研究一批貨物中具有γ比例良品的信心是
否達到特定水準。此篇文章說明此信心實際上是一個未知的參數，並
且證明一般所使用的 Eisenhart et al.(1941)所提出之最短容忍區
間，其所宣稱之信心是不適切的。我們將會提出一個更適切的檢定方
法來檢定實際的信心水準。最後將會根據新方法建立或決定樣本數的
大小，以期當實際信心高於或是低於預期的時候，能夠平衡生產者的
風險及利益。
Statistical Inferences for Confidence of Percentage γ
Acceptable Products in Lot
Student:Su-Yin Hsiao
Advisor:Dr. Lin-An Chen
Institute of Statistics
National Chiao Tung University
Abstract
The tolerance interval is often used to investigate if there is γ per-centage of acceptable products in a lot at some desired confidence. This paper shows that this confidence, with percentage γ fixed, is actually an unknown parameter and shows the popularly used shortest version of tolerance interval by Eisenhart et al. (1941) is not capable to serve as a test statistic for hypothesis assuming the unknown confidence to be a desired constant q0. A new test is shown to be more capable in this purpose. The sample size determination based on this new test ensuring to protect the manufacturer’s benefits and risks when the specification limits indicate true confidence well, respectively, above and below q0has been studied.
誌 謝
兩年的研究生生活在老師、同學還有家人的陪伴下即將邁入尾聲了，很開心 自己能夠成為統研所的一份子，在這一個大家庭裡，受到老師的指導而成長，認 識了互相扶持的好朋友，一路走來充滿歡樂、關愛與祝福，讓我的生命充滿感動 與幸福的樂章。 首先要謝謝我的指導教授陳鄰安博士，非常感謝老師對我的指導與包容，謝 謝老師帶領我作研究，耐心的指導與鼓勵，讓我論文能夠如期完成。老師是如此 的慈愛和藹，總是關心著我們的生活，對於茫茫的未來，給予我很多的方向；而 與老師每次的談話，也體驗到老師傳遞給我們人生的態度與價值觀。還要感謝所 辦郭姐與小陳哥給我的協助，以及口試委員許文郁教授、黃榮臣教授、彭南夫教 授給我的指導與建議。還要感謝我的母親、父親以及妹妹，謝謝父母的支持與栽 培，讓我能夠順利完成學業，而每當面對挫折的時候，母親與妹妹的打氣是我面 對挑戰的支持動力。 最後感謝我的同學、朋友們，佩芳、姿蒨、珮琦，謝謝妳們這兩年的陪伴， 一起努力討論功課，一起努力作論文，一起吃喝玩樂、一起抱怨，給我最大的支 持與鼓勵。鳳如、香菱、采玲、鈺婷、郃嵐、瑜達，謝謝這些日子以來給我的加 油打氣與幫忙。文廷、泰佐、翁賢、小賴學長在就算在自己也很忙碌的時候，總 是願意陪我一起討論問題，給我很大的幫助。感謝昱緯、仲竹、政言、重耕、鈺 玲、筱娟、怡玲、佩蓁一起奮鬥拼畢業，對我的包容與照顧。宗晟、宏瀚、勁葦、 泓毅，曉怡、小方、佩玲、琳婷、大雄、博全、旻原、心瑩妹妹及心瑩一家人、 瀅竹、瓊緯、育珊，謝謝你們一路上給我加油打氣。衷心感謝曾經幫助過我的人， 此篇文章謹獻給我親愛的家人、師長與同學，並致上我最深的謝意。 蕭夙吟 謹誌于 國立交通大學統計研究所 中華民國九十六年六月Contents
abstract in Chinese i abstract in English ii Acknowledgements iii Contents iv 1. Introduction 12. Specification Settings for Achieving Percentage γ Acceptable
Products at a Fixed Confidence 4
3. Power of the Classical Tolerance Interval 8
4. A New Test Based on Tolerance Interval 10
5. Sample Size Determination for Tolerance Interval 12
References 16
1. Introduction
The tolerance interval is popularly used by manufacturer and consumer for judgement of production lots. In mass-production, the manufacturer is interesting in an interval that contains a specified (usually large) percentage of the product and he knows that unless a fixed proportion (say γ) of the production is acceptable in the sense that the items’ characteristics conform to specification limits LSL and USL, he will lose money in this production. On the other hand, if this claim is not true, it will entail the consumer a loss of money. With this interest, the manufacturer and consumer want to know the following:
Whether there is percentage γ of acceptable measurements in a production lot?
(1.1)
Statisticians seek solution to verify this problem through two steps. We start
with a random sample X = (X1, ..., Xn)0 from a distribution with probability
density function fθ(x) representing observations from the same production
pro-cess. For the first step, the pioneer article Wilks (1941) introduced a γ-content tolerance interval with confidence 1 − α which is defined as a random interval
(T1, T2) = (t1(X), t2(X)) satisfying
Pθ{Pθ[X0 ∈ (T1, T2)|X] ≥ γ} ≥ 1 − α for θ ∈ Θ (1.2)
where X0 represents the future observation from the same production process.
Let (t1, t2) be the observed of this tolerance interval. The general rule for
verifying a manufacturer’s problem using the tolerance interval is as follows:
If (t1, t2) ⊂ (LSL, U SL), the lot of product is acceptable,
because we have confidence 1 − α that at least 100γ% of the products
conform to specification limits.
(1.3)
Much attention has been paid for developing tolerance intervals, for examples Wilks (1941), Wald (1943), Paulson (1943), Guttman (1970) and, for a review,
Patel (1986). In general, a common effort been made in the literature is to investigate the version with minimum width, for which Eisenhart et al. (1947) constructed an approximate minimum width tolerance interval for normal ran-dom variable. This normal tolerance interval is now popularly implemented in manufacturing industries and is presented in text books of engineering statis-tics. The interest of this paper is to study if the tolerance interval is appropriate to deal with problem in (1.1) for the manufacturer and consumer.
To study the appropriateness of tolerance interval in this engineering prob-lem, we need to clarify its role with classical hypothesis testing problem. The classical hypothesis testing problem set a simple hypothesis and derive a test with a specified significance level, usually a small value such as 0.05 or 0.01. However, varied philosophies set null hypothesis in different ways that may lead to completely different conclusions. The most commonly accepted rule for setting null hypothesis through the following philosophy:
If one wishes to prove that a hypothesis A is true, one first assumes that it isn’t true.
(1.4)
This philosophy favors right of the consumer (buyer). In a clinical trial, one wish to see if a new drug has a different effect. In general, the null hypothesis might be that the new drug is no better, on average, than the current drug. Why is this philosophy in these problems? Switching a new product (drug) or technique usually requires large initial expenditures, and a decision maker should not do so unless the new product is significantly better than the old one.
There is an approach of philosophy considers a situation employed in some engineering problems. The quality of product is distributed from background noise and abnormal noise. If the process operator adjust the process based on the tests performed periodically for hypothesis defined from rule of (1.4), it will often overacted to the background noise and to deteriorate the performance of the process. Hence, to prevent this unnecessary adjustment, a rule guiding the
null hypothesis is as follows:
Unless it is broken, we do not fix it. (1.5)
This rule follows based on the philosophy from the aspect of the manufacturer (producer).
In the on line quality control, without evidence of existing assignable cause, the manufacturing process is considered to be in control. For example, for
detecting if there is a mean shift, one construct the ¯X chart as
U CL = µ0+ 3√σ0n
LCL = µ0 − 3√σ0n
(1.6)
where µ0 and σ0 are, respectively, the mean and standard deviation for in
con-trol process. Probability 0.9973 of acceptable region does in favor of hypothesis
H0. Why use this philosophy for process control? Suppose that the process
operators adjust the manufacturing process based on tests obeying philosophy one. Only strong evidence showing in data for supporting the hypothesis of in control process will make the operators often overreacted to the random cause for unnecessary process adjustments. These unnecessary process adjust-ments can actually result in a deterioration of process performance. From the above discussion, the process control using the control chart does follows the philosophy of (1.5) assuming that the process is in control as the hypothesis.
Not all statistical hypotheses problems design a test with null hypothesis following a philosophy only to protect the risk of either the manufacturer or the consumer. Acceptance sampling, popularized by Dodge and Roming and was originally applied by the U.S. military, provides a rule of decision of accepting or rejecting a production lot. With a sample of size n for items selected at random from a lot, a plan as denoted as {n, c} is conducted with lot accepted when there are defectives of number less than or equal to c. By calling the percentage defective as average quality level (AQL), this plan is generally designed to have high probability of lot acceptance when there is low AQL value and to have low probability of lot acceptance when there is large AQL value.
On the other hand, a test based on tolerance interval expects to have high probability of hypothesis acceptance when lots of production are with γ per-centage of acceptable products at confidence remarkably larger than 1 − α and to have low probability of hypothesis acceptance when lots of production are with γ percentage of acceptable products at confidence remarkably lower than 1 − α. Hence, this kind of test should not accompany with hypothesis following the philosophy only to protect the risk of either the manufacturer or the con-sumer. We will see that for lots of production, the confidence for γ percentage of acceptable products may be formulated as an unknown parameter. The test presented in (1.3) in fact deals with a hypothesis on value of this unknown confidence. Hence there are two problems we may concern that we want to deal in this paper. (a) Is the test presented in (1.3) appropriate for protecting both the risks of manufacturer and consumer? (b) Is there alternative test, also formulated based on tolerance interval, that performs better to protect the risks of manufacturer and consumer.
This paper formulate the unknown confidence in Section 2, for a γ percent-age acceptable products, in an explicit expression. We then study in Section 3 the power of the γ content tolerance interval of Eisenhart et al. (1947) with test of (1.3). We will propose in Section 4 a new test for the same purpose of studying if there is γ percentage of acceptable products at confidence 1 − α. A sample size determination problem will also be studied in Section 5 that guarantees a low probability of acceptance of the hypothesis when the true specification limits are moderately shorter than the desired ones and a large probability of acceptance of the hypothesis when the true specification limits are moderately wider than the desired ones.
2. Specification Settings for Achieving Percentage γ Acceptable
Products at a Fixed Confidence
Let X be a random variable having distribution function Fθwith probability
is {LSL, U SL}. We call the probability that a product to be acceptable
pitem(θ) =
Z U SL
LSL
fθ(x) dx = Fθ(U SL) − Fθ(LSL) (2.1)
the item reliability where Fθ is the distribution function. Suppose that the lot
size is known as constant k (usually a large number). For this production lot,
the number of acceptable products is with binomial distribution b(k, pitem(θ)).
Then the true confidence for having proportion γ of production lot conforming to specification limits is q =Pk i=[kγ] k i
pitem(θ)i(1 − pitem(θ))k−i. (2.2)
This expression shows that the confidence of a γ percentage of acceptable prod-ucts in lots interesting for the manufacturer actually is an unknown parameter. Hence, the interest for a manufacturer is to test the following hypothesis:
H∗ : q ≥ q0 (2.3)
for some specified (large) value q0. We wouldn’t call H∗ a null or alternative
hypothesis since it is not appropriate to consider a classical test for it.
The classical approach to test H∗ is rule (1.3) through a γ-content tolerance
interval (T1, T2) at confidence q0 that may be re-written as
Accept H∗ if (t1, t2) ⊂ (LSL, U SL) (2.4)
where (t1, t2) is the observation of (T1, T2) (see Bowker and Goode (1952) and
Papp (1992) for this application). With this observation that there is an un-known true confidence, it raises the question that what we may expect a test
for its power function representing the probability that H∗ is accepted.
Recall that the manufacturer expects to have proportion γ or more
accept-able products at confidence q0. Let’s define the minimum item reliability that
guarantees proportion γ acceptable products at confidence q0 as pq0 satisfying
Pk i=[kγ] k i (pq0) i (1 − pq0) k−i = q 0. (2.5)
With product’s characteristic variable having a distribution function Fθ, the
manufacturer is desired to have item reliability pitem(θ) with
pitem(θ) = Fθ(U SL) − Fθ(LSL) ≥ pq0. (2.6)
From (2.5), the manufacturer will loss money most of the times when pitem(θ) is
moderately smaller than pq0. For a given pairs (γ, q0), we list the item
reliabil-ities that achieve exactly proportion γ of acceptable products with confidence
q0 in the following table.
Table 1. Minimum item reliability pq0
q0 γ = 0.8 0.85 0.9 0.95 0.99 (k = 1, 000) 0.8 0.8099 0.8587 0.9071 0.9549 0.9918 0.85 0.8123 0.8608 0.9089 0.9562 0.9923 0.9 0.8152 0.8635 0.9111 0.9577 0.9929 0.95 0.8196 0.8673 0.9142 0.9599 0.9938 0.99 0.8277 0.8744 0.9200 0.9638 0.9952 k = (10, 000) 0.8 0.8033 0.8529 0.9025 0.9518 0.9907 0.85 0.8040 0.8536 0.9030 0.9522 0.9909 0.9 0.8050 0.8545 0.9038 0.9527 0.9912 0.95 0.8064 0.8557 0.9048 0.9534 0.9915 0.99 0.8091 0.8581 0.9068 0.9549 0.9921 k = (100, 000) 0.8 0.8010 0.8509 0.9008 0.9505 0.9902 0.85 0.8013 0.8512 0.9010 0.9507 0.9903 0.9 0.8016 0.8514 0.9012 0.9509 0.9904 0.95 0.8021 0.8518 0.9016 0.9511 0.9905 0.99 0.8029 0.8526 0.9022 0.9516 0.9907
We will derive the specification limits that achieve the minimum item reli-ability for advanced study later in this paper . Suppose that the characteristic
variable of interest obeys a normal distribution N (µ, σ2). Then item reliability, the probability that an item conforming to specifications, is
pitem(µ, σ) = Z U SL LSL 1 √ 2πσe −(x−µ)2 2σ2 dx
and then the true confidence to have proportion γ acceptable products is
q =Pk i=[kγ] k i
pitem(µ, σ)i(1 − pitem(µ, σ))k−i
=Pk i=[kγ] k i Φ U SL−µ σ − Φ LSL−µ σ i 1 − Φ U SL−µσ − Φ LSL−µσ k−i.
A production lot to have proportion γ acceptable products with confidence q0
requires that pitem(µ, σ) = Φ U SL − µ σ − Φ LSL − µ σ ≥ pq0 (2.7)
where pq0 may be found in Table 1.
One interesting question is that how short the specification limits should be to achieve the minimum item reliability. Let the specification limits be
{LSL, U SL} = {µ − lσ, µ + lσ} and we denote lq0 as the l so that the item
reliability is
pq0 = P (µ − lq0σ ≤ X ≤ µ + lq0σ).
Table 2. Specification limits (LSL, U SL) = (µ − lq0σ, µ + lq0σ) to achieve item
reliability exactly equal to pq0
q γ = 0.8 0.85 0.9 0.95 0.99 (k = 1, 000) 0.8 1.3102 1.4711 1.6807 2.0043 2.6456 0.85 1.3174 1.4790 1.6898 2.0161 2.6682 0.9 1.3265 1.4890 1.7013 2.0310 2.6947 0.95 1.3397 1.5037 1.7184 2.0532 2.7372 0.99 1.3650 1.5317 1.7508 2.0955 2.8210 (k = 10, 000) 0.8 1.2910 1.4499 1.6569 1.9752 2.6020 0.85 1.2931 1.4525 1.6597 1.9787 2.6096 0.9 1.2961 1.4556 1.6633 1.9833 2.6187 0.95 1.3001 1.4601 1.6686 1.9901 2.6324 0.99 1.3080 1.4689 1.6786 2.0033 2.6561 k = (100, 000) 0.8 1.2845 1.4428 1.6488 1.9647 2.5841 0.85 1.2853 1.4436 1.6497 1.9662 2.5879 0.9 1.2861 1.4446 1.6509 1.9647 2.5906 0.95 1.2875 1.4460 1.6525 1.9698 2.5928 0.99 1.2900 1.4489 1.6554 1.9737 2.6007
This table will be used in next section to study the power of the classical tol-erance interval in detection of manufacturer’s confidence.
3. Power of the Classical Tolerance Interval
Suppose that we have X1, ..., Xn a random sample for the
characteris-tic variable of interest and the specification limits for the characterischaracteris-tic are
{LSL, U SL}. Let (T1, T2) be a tolerance interval of Wilks (1941) constructed
is to test hypothesis H∗ based on rule of (2.4). With the fact that the true confidence is a parameter, it is interesting to evaluate power function of this tolerance interval, in terms of specification limits, is
π(LSL, U SL) = Pθ((T1, T2) ⊂ (LSL, U SL)). (3.1)
It provides the probability that we should conclude that there is proportion γ or more acceptable products in a lot at confidence 1 − α. The optimal tolerance
interval, if there is, should have power value 1 for q ≥ q0 and value zero for
q < q0. This is generally not attainable. Hence, there are two properties that
we expect a tolerance interval to be satisfies:
(a) The power function is nondecreasing in terms of item reliability pitem.
(b) For a balance of the manufacturer’s benefits and risks, the power when true
confidence q is equal to q0 is close to 0.5.
We want to simulate the powers for the Eisenhart et al.’s tolerance interval for several combinations of specification limits. Let’s set replication number m and specification limits (LSL, U SL) = (−b, b). The simulated power of a
tolerance interval (T1, T2) is defined as
ˆ π = 1 m m X j=1 I((tj1, tj2) ⊂ (−b, b)) (3.2)
where (tj1, tj2) is the observation of (T1, T2) from the jth sample. The power
of (3.2) simulates the chance of (3.1) that the tolerance interval (T1, T2) may
conclude that the production lot includes a proportion γ of acceptable products with confidence 1 − α.
To study (3.2), suppose that the random sample X1, ..., Xn is drawn from
normal distribution N (µ, σ2) where both µ and σ are unknown. The general
form of a prediction interval for a future normal random variable is of the form
( ¯X − m∗s, ¯X + m∗s) (3.3)
where the 100(1 − α)% confidence interval (prediction interval) is the form
with m∗ = t1−α2(n − 1)
q
α
2th quantile of the central t-distribution with degrees of freedom. For the
Wilks’ tolerance interval, Eisenhart et al. (1947) developed the shortest one which is now the most popular version of tolerance interval to deal with the manufacturer’s problem when the characteristic variable does obey a normal
distribution. We select values m∗ corresponding with γ = 0.9, 1 − α = 0.95
from the table developed in Eisenhart et al. (1947).
With replication m = 100, 000, we generate random sample of size n from
distribution N (0, 1). Let ¯Xj and Sj2 be the sample mean and sample variance
for jth sample. We compute this tolerance interval and study its powers of (3.2) with several sample size n = 20, 30, 50 and various values b where b = 1.7184, 1.6686 and 1.6525 corresponds, respectively, to specification limits such that their true confidences are identical to 1 − α = 0.95. The simulated results are listed in Table 3,4,5.
We have several comments drawn from the these tables:
(a) As expected, the power of the tolerance interval is increasing when the
specification limits are wider indicating increasing in pitem. For b ≥ 1.7184
with k = 1, 000, the corresponding confidence q ≥ q0 = 0.95, we see that the
larger the sample size the more the chance (probability) to accept H∗.
(b) When b = 1.7184 for k = 1, 000, the process does guarantee confidence 0.95 with percentage 0.9 of acceptable products. However, the simulated power values are 0.0257, 0.0289, 0.0466, respectively, for sample sizes n = 20, 30, 50. These revealed little chance to observe that the lots are already γ = 0.9 per-centage of acceptable products at confidence 0.95. Hence, the test of (2.3) is not satisfactory in losing benefits for the manufacturer.
4. A New Test Based on Tolerance Interval
A test for hypothesis H∗ is expected to have power not too far from 0.5
when q = q0 is true. The classical test based on tolerance interval does not
meet this requirement. We then introduce a new test.
Suppose that we have an appropriate estimate, denoted by ˆθ of parameter
variable X as fθˆ(x). The rule of the new test for hypothesis H∗ of (2.3) is:
Accepting H∗ if
Z
(t1,t2)∩(LSL,U SL)
fθˆ(x)dx ≥ pq0. (4.1)
We have two comments for setting the above rule to test hypothesis H∗:
(a) For given a γ-content tolerance interval (T1, T2) at confidence 1 − α, if
its observation (t1, t2) does contain percentage of products conforming with
specification limits pq0 or more, we conclude with confidence 1 − α that there
is percentage γ acceptable products in a lot.
(b) This test sets test statistic R(T
1,T2)∩(LSL,U SL)fθˆ(x)dx. However, the critical
point pq0 is not the cut off point based on distribution of the test statistic.
Hence, this test does not follows the classical hypothesis testing to ensures a specified significance level.
With this test, the power function is
Pθ{
Z
(T1,T2)∩(LSL,U SL)
fθˆ(x)dx ≥ pq0}. (4.2)
Consider that we have a random sample X1, ..., Xn drawn from the
distri-bution N (µ, σ2). Let ˆµ = ¯x and ˆσ2 = 1
n−1
Pn
i=1(xi− ˆµ)2. The rule for testing
hypothesis H0 is:
Accepting H0 if
Z
(t1,t2)∩(LSL,U SL)
φµ,ˆˆσ(x)dx ≥ pq0, (4.3)
and the empirical power of tolerance interval (T1, T2) is
ˆ πSpe= 1 m m X j=1 I( Z (tj1,tj2)∩(LSL,U SL) φµ,ˆˆσ(x)dx ≥ pq0). (4.4)
We conduct a simulation with the same design set in Section 3 to study the power function of this new test. The simulated results for lot sizes, k = 1, 000, 10, 000, 100, 000 are listed in Tables 7,8,9 (see these tables in the end of this paper). However, the tolerance intervals considered including the shortest version (STL) and the version (CITL) developed by Huang, Chen and Welsh (2007).
(a) Monotone power values are as our expectation. However, the power values are relatively higher than them based on test of (2.4). Hence, there are larger probabilities in all settings of specification limits and sample sizes for accepting
H∗.
(b) There is no significant differences in the performance between the shortest tolerance interval and the version of Huang, Chen and Welsh.
(c) When b is the value (1.7184 for k = 1, 000, 1.6686 for k = 10, 000 and 1.6525 for k = 100, 000) that the true confidence is identical to 0.95 the simulated power values are all close to 0.48. This is interesting indicating that this new test is more capable in our purpose.
5. Sample Size Determination for Tolerance Interval
The most popular technique to judge if a production lot is accepted is the acceptance sampling plan which specifies a sample size n and an acceptance number c and the lot is accepted if the number of defective items in this sample is less than or equal to c. An important question in acceptance sampling plan approach is to determine the sample size that the probabilities of acceptance
when fraction of defectives are p1 and p2 with p1 < p2, respectively, achieve two
specified values. This is to protect the manufacturer having large probability
of acceptance when fraction defective is as low as p1 and protect the consumer
having small probability of acceptance when fraction defective is as large as
p2. This idea may be extended to the acceptance of lots at some confidence by
tolerance intervals.
Suppose that we have a random sample X1, ..., Xndrawn from a distribution
with distribution function Fθ. The probability that a product to be acceptable
expressed in (2.1) and the confidence for having proportion γ of products in a lot conforming to specification limits expressed in (2.2) are both dependent on parameter θ, specification limits LSL, U SL and lot size k. However, its power of (4.2) actually relies on the efficiency of the estimator of unknown parameter θ. Generally, efficiency may be improved when sample size n is increased. Hence, the sample size n should be set to satisfy the consumer with a small
power of (4.2) when the interval of specification limits is shorter and to satisfy the manufacturer with a large power when the interval is wilder.
We conduct a simulation to generate the power function (4.4) for normal distribution and display the simulated power functions for several sample sizes. The resulted power functions are shown in Figure 1.
● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● 1.0 1.5 2.0 2.5 0.0 0.2 0.4 0.6 0.8 1.0
Confidence Curves Varying Sampling Size
Spec Acceptance Probability n=50 n=20 n=10 n=5
The graph that plots the confidence q(Spec) versus the specification limits may be called the OC graph. It is interesting to design a tolerance interval requiring that the confidence is a desired large value when the interval of spec-ification limits is wilder and it is a desired small value when the interval is
q1 < q2, find sample size n such that the followings Pθ{ Z (T1,T2)∩(LSL1,U SL1) fθˆ(x)dx ≥ pq0} ≤ q1 (5.1) Pθ{ Z (T1,T2)∩(LSL2,U SL2) fθˆ(x)dx ≥ pq0} ≥ q2.
We consider an empirical solution of sample size n as minimum n satisfying the following equations
1 m m X j=1 I( Z (tj1,tj2)∩(LSL1,U SL1) φµ,ˆˆσ(x)dx ≥ pq0) ≤ q1 (5.2) 1 m m X j=1 I( Z (tj1,tj2)∩(LSL2,U SL2) φµ,ˆˆσ(x)dx ≥ pq0) ≥ q2.
For the purpose of sample size determination, we further make the following assumptions:
(a) The lot size k and percentage value γ are both fixed.
(b) The parameter(s) is assumed to be known as θ0. For case that θ is not
known, we assume that there is a training sample for us to estimate it.
Let’s use the normal distribution as an example for explaining the technique of sample size determination. The sample size is determined from (5.2) with item reliability pitem(Spec) = Φ U SL − µ0 σ0 − Φ LSL − µ0 σ0 .
Without lose of generality, we let µ0 = 0 and σ0 = 1 and for our convenience,
we let specification limits {LSL, U SL} = {−`, `}. The item reliability then is
We choose k = 1, 000. From Table 2, the specification limits (LSL, U SL) = (−1.7184, 1.7184) guarantees to meet percentage 0.9 acceptable products at confidence 0.95. However, suppose that the manufacturer asks for probability 0.3 or lesser of accepting the lots most of the time when the specification limits
are {LSL1, U SL1} = {−1.5, 1.5} and probability 0.7 or more of accepting
the lots most of the time when the specification limits are {LSL2, U SL2} =
{−1.9, 1.9}. These conditions require sample size n = 12. There are other two cases of conditions that the sample sizes are also listed in Table 10.
Table 10. Desired sample sizes meet the probabilities of acceptance of lots when the interval of specification limits is wilder or shorter.
Spec1 q1 Spec2 q2 n
(−1.5, 1.5) 0.3 (−1.9, 1.9) 0.7 12
(−1.5, 1.5) 0.2 (−1.9, 1.9) 0.8 32
(−1.5, 1.5) 0.1 (−1.9, 1.9) 0.9 78
This explanation of sample size determination involves a simpler design of specification limits. For purpose of general applications, further studies are needed.
References
Eisenhart, C., Hastay, M. W. and Wallis, W. A. (1947). Techniques of Statis-tical Analysis. McGraw-Hill Book Company: New York.
Guttman, I. (1970). Statistical tolerance regions: Classical and Baysian. Grif-fin: London.
Huang, J.-Y., Chen,L.-A. and Welsh, A. (2007). The mode and its applica-tion to control charts and tolerance intervals. Submitted to Statistics and Probability Letter for publication (In revision).
Patel, J.K. (1986). Tolerance limits - A review. Communications in statistics - Theory and Methods, 15, 2719-2762.
Paulson, E. (1943). A note on tolerance limits. Annals of Mathematical Statis-tics, 14, 90-93.
Poulsen, O.M., Holst, E. & Christensen, J.M. (1997). Calculation and appli cation of coverage intervals for biological reference values (technical repo rt). Pure and Applied Chemistry, 69, 1601-1611.
Wald, A. (1943). An extension of Wilks’ method for setting tolerance limits. Annals of Mathematical Statistics, 14, 45-55.
Wilks, S. S. (1941). Determination of sample sizes for setting tolerance limits. Annals of Mathematical Statistics, 12, 91-96.
Tables
Table 3. Powers of the minimum-width tolerance interval ((γ, 1 − α) =
(0.9, 0.95)) k = 1000 Limits n = 20 n = 30 n = 50 (m∗) (2.310) (2.140) (1.969) b = 1.4 q = 1.371e − 08 0.0019 0.0008 0.0004 b = 1.5 q = 0.00071 0.0043 0.0033 0.0025 b = 1.6 q = 0.17899 0.0104 0.0096 0.0111 b = 1.645 q = 0.52786 0.0157 0.0153 0.0204 b = 1.7184 q = 0.949878 0.0257 0.0289 0.0466 b = 1.8 q = 0.9995 0.0436 0.0556 0.1001 b = 2.0 q = 1.0 0.1213 0.1888 0.3777 b = 2.2 q = 1.0 0.2611 0.4224 0.7313 b = 2.5 q = 1.0 0.5528 0.7829 0.9730 b = 3.0 q = 1.0 0.9111 0.9898 1.0000 b = 3.5 q = 1.0 0.9934 0.9999 1.0000
Table 4. Powers of the minimum-width tolerance interval ((γ, 1 − α) = (0.9, 0.95)) k = 10000 Limits n = 20 n = 30 n = 50 (m∗) (2.310) (2.140) (1.969) b = 1.4 q = 9.71747e − 71 0.0015 0.0009 0.0005 b = 1.5 q = 5.569e − 25 0.0045 0.0028 0.0022 b = 1.6 q = 0.00099 0.0109 0.0093 0.0108 b = 1.645 q = 0.5124351 0.0155 0.0150 0.0205 b = 1.6686 q = 0.9501104 0.0181 0.0186 0.0270 b = 1.8 q = 1.0 0.0416 0.0571 0.1002 b = 2.0 q = 1.0 0.1227 0.1890 0.3758 b = 2.2 q = 1.0 0.2622 0.4231 0.7274 b = 2.5 q = 1.0 0.5528 0.7832 0.9719 b = 3.0 q = 1.0 0.9112 0.9898 1.0000 b = 3.5 q = 1.0 0.9932 1.0000 1.0000
Table 5. Powers of the minimum-width tolerance interval ((γ, 1 − α) = (0.9, 0.95)) k = 100000 Limits n = 20 n = 30 n = 50 (m∗) (2.310) (2.140) (1.969) b = 1.4 q = 0 0.0017 0.0009 0.0003 b = 1.5 q = 7.83579e − 232 0.0043 0.0031 0.0029 b = 1.6 q = 3.809744e − 23 0.0106 0.0097 0.0108 b = 1.645 q = 0.5153559 0.0154 0.0147 0.0206 b = 1.6525 q = 0.9521992 0.0160 0.0161 0.0223 b = 1.8 q = 1.0 0.0430 0.0558 0.1016 b = 2.0 q = 1.0 0.1215 0.1889 0.3762 b = 2.2 q = 1.0 0.2616 0.4222 0.7272 b = 2.5 q = 1.0 0.5512 0.7803 0.9719 b = 3.0 q = 1.0 0.9114 0.9904 1.0000 b = 3.5 q = 1.0 0.9941 1.0000 1.0000
Table 7. Powers of the minimum-width tolerance interval and coverage in-terval based tolerance inin-terval ((γ, 1 − α) = (0.9, 0.95), k = 1000)
Limits n = 20 STL CITL n = 30 STL CITL n = 50 STL CITL b = 1.4 0.1067 0.1074 0.0651 0.0659 0.0258 0.0269 b = 1.5 0.1954 0.1966 0.1496 0.1511 0.0892 0.0923 b = 1.6 0.3140 0.3154 0.2799 0.2826 0.2273 0.2339 b = 1.645 0.3746 0.3763 0.3529 0.3560 0.3132 0.3206 b = 1.7184 0.4781 0.4800 0.4813 0.4850 0.4773 0.4875 b = 1.8 0.5950 0.5969 0.6420 0.6273 0.6565 0.6661 b = 2 0.8245 0.8257 0.8781 0.8799 0.9357 0.9401 b = 2.2 0.9458 0.9465 0.9780 0.9786 0.9953 0.9959 b = 2.5 0.9956 0.9957 0.9993 0.9993 1 1 b = 3 1 1 1 1 1 1
Table 8. Powers of the minimum-width tolerance interval and coverage interval based tolerance interval ((γ, 1 − α) = (0.9, 0.95), k = 10000)
Limits n = 20 STL CITL n = 30 STL CITL n = 50 STL CITL b = 1.4 0.1413 0.1419 0.0942 0.0949 0.0459 0.0468 b = 1.5 0.2465 0.2475 0.2022 0.2035 0.1415 0.1411 b = 1.6 0.3814 0.3824 0.3586 0.3604 0.3251 0.3297 b = 1.645 0.4434 0.4448 0.4387 0.4408 0.4258 0.4308 b = 1.6686 0.4781 0.4793 0.4836 0.4856 0.4788 4841 b = 1.8 0.6675 0.6686 0.7096 0.7116 0.7673 0.7719 b = 2 0.8739 0.8747 0.9232 0.9243 0.9688 0.9701 b = 2.2 0.9665 0.9668 0.9887 0.9889 0.9986 0.9986 b = 2.5 0.9978 0.9978 0.9998 0.9998 1 1 b = 3 1 1 1 1 1 1
Table 9. Powers of the minimum-width tolerance interval and coverage interval based tolerance interval ((γ, 1 − α) = (0.9, 0.95), k = 100000)
Limits n = 20 STL CITL n = 30 STL CITL n = 50 STL CITL b = 1.4 0.1528 0.1534 0.1067 0.1074 0.0539 0.0548 b = 1.5 0.2642 0.2650 0.2227 0.2242 0.1671 0.1695 b = 1.6 0.3995 0.4003 0.3858 0.3876 0.3603 0.3639 b = 1.645 0.4688 0.4699 0.4695 0.4713 0.4626 0.4668 b = 1.6525 0.4786 0.4797 0.4813 0.4831 4835 0.4877 b = 1.8 0.6898 0.6909 0.7368 0.7385 0.7965 0.8003 b = 2 0.8875 0.8880 0.9348 0.9357 0.9757 0.9765 b = 2.2 0.9724 0.9726 0.9916 0.9918 0.9991 0.9992 b = 2.5 0.9983 0.9983 0.9999 0.9999 1 1 b = 3 1 1 1 1 1 1
