2 Inspection’s role and type Although the modern QM emphasizes the principle of prevention in advance, sampling inspection has still its special role, especially in inspecting the quality of raw material, WIP and finished goods.Inspection: a. no inspectionb. Acceptance samplingc. 100% inspection
3 When to use acceptance sampling? When testing is destructive.When the cost of 100% inspection is high.When 100% inspection is not technologically feasible or would require so much calendar time and/or expenses.When there are many items to be inspected and the inspection error rate is high.When the vendor has an excellent quality history, and some reduction in inspection from 100% is desired.When there are potentially serious product liability risks.
4 Acceptance Sampling Purposes Determine quality level Ensure quality is within predetermined level4
5 Advantages of Acceptance Sampling Less expensive because of less inspection.There is less handling damage of the product.It is applicable to destructive testing.Fewer personnel are involved in inspection activities.It often greatly reduces the amount of inspection error.The rejection of entire lots often provides a stronger motivation to the vendor for quality improvement.
6 Disadvantages of Acceptance Sampling Risks of accepting ”bad” lots and rejecting “good” lots. In the “good” lot, there might be nonconformities.Sample provides less information than 100-percent inspectionAcceptance sampling requires more time on planning and documentation of the acceptance sampling procedure.5
7 Data Type Attribute Data：Only conformity and nonconformity Variable Data ：length, width etc.-- Liberman, G. J., and G. J. Resnikiff(1955).”Sampling plans for Inspection by variables”Journal of Quality Technology, Vol. 29, No.2.
8 Types of Acceptance sampling Plans Single-sampling planDouble-sampling planMultiple-sampling planSequential-sampling plan
9 Single-sampling plan (N, p) (n,c) Total number ：N Acc the lot The proportion of defects :PAcc the lotSn≦C(n,c)(N, p)Reject the lotSn>CWhere Sn is the number of the actual defects in the sample.
10 Double-sampling plan (N, p) (n,c1 ,r1) Acc the lot (n1+n2, c2) Reject the lotSn1≦c1Sn1>c1(n1+n2, c2)c1<Sn1<r1S(n1+n2) ≦c2S(n1+n2) >c2(N, p)
14 Risk Acceptable Quality Level (AQL) a (Producer’s risk) Max. acceptable percentage of defectives agreed by the producer and user.a (Producer’s risk)The probability of rejecting a good lot.Lot Tolerance Percent Defective (LTPD)Percentage of defectives that defines consumer’s rejection point. (Consumer’s risk)The probability of accepting a bad lot.8
15 Decision Law Accept the lot Reject the lot Acceptable Producer’s Risk sentencequalityAccept the lotReject the lotAcceptableProducer’s RiskUn-acceptable(Rejected)Consumer’s Risk
16 Operating Characteristic (OC) Curve Probability of acceptance for the lot to be inspected under different percentage of defectives.OC (p)=Pr(Accept the lot |p) ≡P (Sn ≦C |p)
22 Acceptance Sampling: Single Sampling Plan A simple goalDetermine (1) how many units, n, to sample from a lot, and (2) the maximum number of defective items, c, that can be found in the sample before the lot is rejected.7
23 Example: Acceptance Sampling Problem Zypercom, a manufacturer of video interfaces, purchases printed wiring boards from an outside vender, Procard. Procard has set an acceptable quality level of 1% and accepts a 5% risk of rejecting lots at or below this level. Zypercom considers lots with 3% defectives to be unacceptable and will assume a 10% risk of accepting a defective lot.Develop a sampling plan for Zypercom and determine a rule to be followed by the receiving inspection personnel.10
24 Example: Step 1. What is given and what is not? In this problem, AQL is given to be 0.01 and LTDP is given to be We are also given an alpha of 0.05 and a beta of 0.10.What you need to determine your sampling plan is “c” and “n.”11
25 Example: Step 2. Determine “c” First divide LTPD by AQL.Then find the value for “c” by selecting the value in the TN7.10 “n(AQL)”column that is equal to or just greater than the ratio above.Exhibit TN 7.10cLTPD/AQLn AQL44.8900.05253.5492.613110.9460.35563.2063.28626.5090.81872.9573.98134.8901.36682.7684.69544.0571.97092.6185.426So, c = 6.11
26 Example: Step 3. Determine Sample Size Now given the information below, compute the sample size in units to generate your sampling plan.c = 6, from Tablen (AQL) = 3.286, from TableAQL = .01, given in problemn(AQL/AQL) = 3.286/.01 = 328.6, or 329 (always round up)Sampling Plan:Take a random sample of 329 units from a lot.Reject the lot if more than 6 units are defective.13