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SMU EMIS 7364

SMU EMIS 7364. NTU TO-570-N. Statistical Quality Control Dr. Jerrell T. Stracener, SAE Fellow. Acceptance Sampling for Attributes Updated: 4.4.02. Decision Risk Producer’s Risk – conforming lots are rejected Consumer’s Risk – nonconforming lots are accepted. Decision Risk

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SMU EMIS 7364

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  1. SMU EMIS 7364 NTU TO-570-N Statistical Quality Control Dr. Jerrell T. Stracener, SAE Fellow Acceptance Sampling for Attributes Updated: 4.4.02

  2. Decision Risk • Producer’s Risk – conforming lots are rejected • Consumer’s Risk – nonconforming lots are accepted.

  3. Decision Risk True Situation Lot meets Lot does not Decision requirements meet requirements Accept Lot no error Type II error Reject Lot Type I error no error

  4. Lot-by-Lot Acceptance Sampling by Attributes – Single Sampling

  5. Single Sampling N = lot size = the number of items in the lot from which the sample is to be drawn. n = sample size = the number of items drawn at random from the lot c = the maximum allowable number of defective items in the sample. More than c defectives in the sample will cause rejection of the lot.

  6. Type A OC Function for Single Sampling Plan • Sampling Plan Specification N = Lot Size n = Sample Size c = Acceptance number • D = true number of defectives in the lot • X = number of defective items in the random sample.

  7. Type A OC Function for Single Sampling Plan • Probability Distribution of X • X ~ H(N, D, n) • Probability Mass Function of X

  8. Type A OC Function for Single Sampling Plan • OC Function

  9. Example – Single Sampling Plan A • Determine and plot the OC Function for a single sampling plan specified by

  10. Example Solution – Single Sampling Plan OC(D) D

  11. Concept Suppose that the lot size N is large (theoretically infinite). Under this condition, the distribution of the number of defectives d in a random sample of n items is binomial with parameters n and p, where p is the fraction of defective items in the lot. An equivalent way to conceptualize this is to draw lots of N items at random from a theoretically infinite process, and then to draw random samples of n from these lots. Sampling from the lot in this manner is the equivalent of sampling directly from the process.

  12. Type B OC Function for Single Sampling Plan • Sampling Plan Specification N = Lot Size n = Sample Size c = Acceptance number • N is infinite, or at least much larger than n • D = true number of defective items in the lot • p = proportion of the population that is defective, i.e., • X = number of defective items in the random sample.

  13. Type B OC Function for Single Sampling Plan • Probability Distribution of X • X ~ B(n, p) • Probability Mass Function of X

  14. Acceptance Sampling The probability of observing exactly x defective items is: for x = 0, 1, . . ., n

  15. OC Function The probability of acceptance is the probability that d is less than or equal to c, or:

  16. Example – Single Sampling Plan B • Determine and plot the OC Function for a single sampling plan specified by • Compare the OC curve for N=500, n=98, c=2.

  17. Example Solution – Single Sampling Plan OC(p) p

  18. Example If the lot fraction defective is p = 0.02, n = 89 and c = 2, then

  19. Example The OC curve is developed by evaluating PA(p) for various values of p. The following table displays the calculated value of several points on the curve. The OC curve shows the discriminatory power of the sampling plan. For example, in the sampling plan n = 98, c = 2, if the lots are 2% defective, the probability of acceptance is approximately 0.74. This means that if 100 lots from a process that manufactures 2% defective product are submitted to this sampling plan, we will expect, in the long run, to accept 74 of the lots and reject 26 of them.

  20. Example Probabilities of Acceptance for the Single-Sampling plan n = 89, c = 2 Fraction Probability of Defective, p Acceptance, Pa (p) 0.005 0.9897 0.010 0.9397 0.020 0.7366 0.030 0.4985 0.040 0.3042 0.050 0.1721 0.060 0.0919 0.070 0.0468 0.080 0.0230 0.090 0.0109

  21. If N = 500, n =98 & c=2, For comparison, select p=0.02. Then

  22. Example -p 0 10 20 30 40 -D

  23. Single Sample Test Plan Design • Probability Distribution of X P0 = specified fraction defective P1 = minimum acceptable fraction defective a = producer’s risk b = consumer’s risk

  24. Test Procedure • To test • H0: p = p0 • vs H1: p = p1 • at the   100% level of significance, • Obtain a random sample of size n • Inspect the n items and determine the number, X, that are defective • Reject H0 if x > c, otherwise accept H0

  25. Single Sampling Plan x number of defective items reject c . . . accept 2 1 0 0 n n = sample size x = number of defective items c = maximum number of defective items for acceptance

  26. Operating Characteristic (OC) Function Note that: and

  27. p0 OC Curve 1 1 -   p1 0 1 p

  28. Test Plan Design To determine a single sample test plan for testing H0: p = p0, specify values of p0, p1, , and  such that PA(p0) = 1 -  and PA(p1) = , then find the values of n and c that satisfy the following equations. and

  29. Lot-by-Lot Acceptance Sampling by Attributes – Double Sampling

  30. Double Sampling A double-sampling plan is a procedure in which, under certain circumstances, a second sample is required before the lot can be sentenced. A double- sampling plan is defined by four parameters: n1 = sample size on the first sample c1 = acceptance number of the first sample n2 = sample size on the second sample c2 = acceptance number for both samples

  31. Single Sampling Plan x number of defective items Reject Reject c2 . . . Continue Accept c1+1 c1 . . . . . . Accept 2 1 0 0 2 n1 n1+n2 1 . . . . . .

  32. Double Sampling - Advantages The principal advantage of a double-sampling plan with respect to single sampling is that it may reduce the total amount of required inspection. Suppose that the first sample taken under a double-sampling plan is smaller than the sample that would be required using a single-sampling plan that offers the consumer the same protection. In all cases, then, in which a lot is accepted or rejected on the first sample, the cost of inspection will be lower for double sampling than it would be for single sampling. It is also possible to reject a lot without complete inspection of the second sample (called curtailment of the second sample).

  33. Double Sampling - Disadvantages Double sampling has two potential disadvantages: 1. Unless curtailment is used on the second sample, under some circumstances double sampling may require more total inspection than would be required in a single-sampling plan that offers the same protection. 2. Double-sampling is administratively more complex, which may increase the opportunity for the occurrence of inspection errors. Furthermore, there may be problems in storing and handling raw materials or component parts for which one sample has been taken, but that are awaiting a second sample before a final lot dispositioning decision can be made.

  34. Double Sampling - The OC Curve The performance of a double-sampling plan can be conveniently summarized by means of its operating- characteristic (OC) curve. A double-sampling plan has a primary OC curve that gives the probability of acceptance as a function of lot of process quality. It also has supplementary OC curves that show the probability of acceptance as a function of lot acceptance and rejection on the first sample.

  35. Double Sampling Operation of the double-sampling plan with n1 = 50, c1 = 1, n2 = 100, c2 = 3 Inspect a random sample of n1 = 50 from the lot d1 = number of observed defectives Accept the lot Reject the lot d1> c1 = 3 d1 c1 = 1 Inspect a random sample of n2 = 100 from the lot d2 = number of observed defectives Accept the lot Reject the lot d1 + d2 c2 = 3 d1 + d2> c2 = 3

  36. Example If denotes the probability of acceptance on the combined samples, and and denote the probability of acceptance on the first and second samples, respectively, then is just the probability that we will observe d1  c1 = 1 defectives out of a random sample of n1 = 50 items. Thus

  37. Example If p = 0.05 is the fraction defective in the incoming lot, then To obtain the probability of acceptance on the second sample, we must list the number of ways the second sample can be obtained. A second sample is drawn only if there are two or three defectives on the first sample - that is, if c1 < d1  c2.

  38. Example 1. d1 = 2 and d2 = 0 or 1; that is, we find two defectives on the first sample and one or less defectives on the second sample. The probability of this is: P(d1 = 2, d2  1) = P(d1 = 2) x P(d2  1) = (0.261)(0.037) = 0.009

  39. Example 2. d1 = 3 and d2 = 0; that is, we find three defectives on the first sample and no defectives on the second sample. The probability of this is: P(d1 = 3, d2  0) = P(d1 = 3) x P(d2 = 0) = (0.220)(0.0059) = 0.001

  40. Example Thus, the probability of acceptance on the second sample is The probability of acceptance of a lot that has fraction defective p = 0.05 is therefore

  41. Double Sampling - The OC Curve OC Curves for the double-sampling plan with n1 = 50, c1 = 1, n2 = 100, c2 = 3 Probability of acceptance on combined sample Probability, P Probability of rejection on first sample Probability of acceptance on first sample Lot fraction defective, p

  42. Rectifying Inspection Programs

  43. Rectifying Inspection Programs • Acceptance-sampling programs require corrective action when lots are rejected. • Generally takes the form of 100% inspection or screening of rejected lots, with all discovered defective items either removed for subsequent rework or return to the vendor, or replaced from a stock of known good items. Such sampling programs are called rectifying inspection programs.

  44. Rectifying Inspection Programs (continued) • The inspection activity affects the final quality of the outgoing product. Suppose that the incoming lots to the inspection activity have fraction defective, po. Some of these lots will be accepted, and others will be rejected. The rejected lots will be screened, and their final fraction defective will be zero. However, accepted lots have fraction defective p0. Consequently, the outgoing lots from the inspection activity are a mixture of lots with fraction defective p0 and fraction defective zero, so the average fraction defective in the stream of outgoing lots is p1, which is less that p0, and serves to “correct” lot quality.

  45. Inspectionactivity Rectifying Inspection Rejected lots Fraction defective0 Incoming lotsFraction defectivep0 Outgoing lotsFraction defectivep1 Fraction defectivep0 Accepted lots

  46. Rectifying Inspection Programs (continued) • Used in situations where the manufacturer wishes to know the average level of quality that is likely to result at a given stage of the manufacturing operations. • Used either at receiving inspection, in-process inspection of semi-finished products, or at final inspection of finished goods • The objective of in-plant usage is to give assurance regarding the average quality of material used in the next stage of the manufacturing operations.

  47. Handling of Rejected Lots • The best approach is to return rejected lots to the vendor, and require it to perform the screening and rework activities. • Has the psychological effect of making the vendor responsible for poor quality • May exert pressure on the vendor to improve its manufacturing processes or to install better process controls. • Screening and rework take place at the consumer level because the components or raw materials are required in order to meet production schedules.

  48. Average Outgoing Quality • Widely used for the evaluation of a rectifying sampling plan. • Is the quality in the lot that results from the application of rectifying inspection • Is the average value of lot quality that would be obtained over a long sequence of lots from a process with fraction defective p.

  49. Average Outgoing Quality (AOQ) The average fraction defective, called average outgoing quality is Where the lot size is N and that all defectives are replaces with good units. Then in lots of size N, we have • N items in the lot that, after inspection, contain no defectives, because all discovered defectives are replaced • N – n items that, if the lots is rejected, contain no defectives • N – n items that, if accepted, contain (N-n)p defectives

  50. Example Suppose that N = 10,000, n = 89, and c = 2, and that the incoming lots are of quality p = 0.01. Now at p = 0.01, we have Pa = 0.9397, and the AOQ is That is, the average outgoing quality is at 0.93% defective.

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