Inspection is a mandatory but nonvalue-adding activity, and our objective is to do as little as possible, provided that we continue to fulfill the customer’s requirements. The zero acceptance number (c = 0) sampling plan requires far less inspection than the corresponding ANSI/ASQ Z1.4 (formerly MIL-STD 105) plan, and becomes viable when the supplier is extremely confident in its level of quality.1
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An ANSI/ASQ Z1.4 plan consists of a sample size n, and an acceptance number c. The inspector checks n items, and accepts the lot if c or fewer defects or nonconformances are found. These plans are designed to give (roughly) a 95-percent chance of acceptance at the acceptable quality level (AQL), which is one of the parameters for the plan’s selection.
The c = 0 plan, on the other hand, rejects the lot if any defects or nonconformances are found, but it requires a considerably smaller sample size. The drawback is that the producer’s risk (α) of rejecting a lot at the AQL is usually far greater than the textbook 5 percent, so the c = 0 plan should be used only when quality is much better than the AQL. This reinforces a basic principle of industrial statistics: We can have low risks or small sample sizes, but we can’t have both at the same time.
How to design a c = 0 plan
Any ANSI/ASQ Z1.4 plan can be converted into a c = 0 plan as follows.
1. Define the ANSI/ASQ Z1.4 plan on the basis of a) the lot size N; b) the inspection level; and c) the acceptable quality level (AQL).
2. ANSI/AZQ Z1.4 plans don’t have formal rejectable quality levels (RQLs), but pretend that the RQL is the nonconforming fraction (p) for which the acceptance probability, or consumer’s risk ß, is 10 percent.
3. Compute the minimum sample size necessary to get an acceptance probability of 10 percent or less at the RQL for an acceptance number of zero.
The formula is the same as for discovery sampling, where we want a 90 percent or better chance of finding at least one nonconformance when quality is at the RQL.
Example:
• Given a lot of N = 600 items
• Inspection level II
• AQL = 1.0 percent
We will use the tables from MIL-STD 105E, as these are in the public domain. ANSI/ASQ Z1.4 is, however, little if at all different. In this case, a lot of 600 and a general inspection level II gives us code letter J.
Now go into the normal inspection table, and cross-reference code letter J with the column for AQL = 1 percent. This is important because of the “chutes and ladders” nature of this table, which can give us a different code letter than the one we started with. (Always pay attention to this for any ASQ certification exam.) If, for example, the AQL was 0.40 percent instead of 1.0 percent, we would have to go down one row to letter K and a sample size of 125. In this case, there is no arrow, so the sample size is 80, and the acceptance number is 2.
The adjacent rejection number may seem redundant because, for a single sample plan with an acceptance number of 2, it is obvious that three or more defects or nonconformances will reject the lot. Double and multiple sampling plans, as well as reduced inspection plans, often have nonadjacent acceptance and rejection numbers. If we get a defect or nonconformance count between the acceptance and rejection numbers for double or multiple sampling plans, it means we must draw another sample. In the case of reduced inspection, we accept the lot but go back to normal sampling according to the standard’s switching rules.
Now find the RQL, which is in this case 6.52 percent. The probability of acceptance for an ANSI/ASQ Z1.4 plan is the cumulative binomial probability of getting c or fewer defects or nonconformances, given a sample of n and a nonconforming fraction of p. =BINOM.DIST(2,80,0.0652,1) in Excel returns 0.09977, which rounds off to 0.100.
Now compute the necessary sample size for a c = 0 plan. The operating characteristic curve for the c = 0 plan sets the acceptance probability equal to (1–p)n for a nonconforming fraction of p. Rearrangement of this equation yields the same equation as for discovery sampling, where we want (in this case) 90-percent confidence of finding at least one defect or nonconformance among n parts at the RQL.
This should technically be rounded up rather than off, although Nicholas Squeglia uses n = 34. If the nonconforming fraction is 0.0652, there is a (1–0.0652)34 = 0.101 (10.1%) chance of acceptance, which is technically too large. It is also important, regardless of whether we round the sample size up or off, to note the acceptance probability when the nonconforming fraction is 1 percent. This is (1–0.01)34 = 0.711, which means the producer’s risk of rejecting a lot at the AQL is 28.9 percent. This underscores the key takeaway that we should use a c = 0 plan only when quality is considerably better than the AQL.
Squeglia adds that the sample size for AQL = 4% is the same under ANSI/ASQ Z1.4 as it is for AQL = 1%. That is, we still need a sample of 80, even though the acceptance number is now seven rather than two, and a smaller sample should be adequate for protection against a much poorer level of quality. Table X-J-1 shows that, for AQL = 4%, the RQL is 14.3 percent. A sample of only 15 (or 16, if we round up) is adequate to ensure comparable protection for the customer at the RQL.
On the other hand, the chance of accepting a lot at the AQL is (1–0.04)15 = 0.542, so the producer’s risk of rejecting such a lot is 0.458 rather than the target of roughly 5 percent. This again underscores the need to have very good quality before an attempt is made to use a c = 0 plan.
Switching rules
Supplement to Zero Acceptance Number Sampling Plans, Fourth Edition by Nicholas Squeglia says, “The c = 0 sampling plans are not AQL plans and do not require or recommend a switching procedure, although one can be used.” This is, however, up to the customer. If we tell our customer that our sampling plan’s protection against poor quality is at least equal to that of the traditional plan, the customer may insist that the switching rules be used. If so, it is absolutely vital that quality be sufficiently good so as to not invoke the switching rules by rejecting lots that the traditional plan would have accepted.
Alternatives to reduce inspection
If quality is not sufficiently good to permit use of c = 0 plans, alternatives are available. These include ANSI/ASQ Z1.4’s double and multiple sampling plans. Their advantage is that they will reject very bad lots quickly, often on the first sample, and accept extremely good ones almost as quickly. A sequential sampling plan can meanwhile be developed whose operating characteristic curve is identical to that of the ANSI/ASQ Z1.4 plan at the two points (AQL, 1–α) and (RQL, ß), where α is the producer’s risk of rejecting a lot at the AQL, and ß is the consumer’s risk of accepting it at the RQL. (Remember again that the ANSI/ASQ Z1.4 plan does not have a formal RQL, but we pretend that the nonconforming fraction at which the chance of acceptance is 10 percent is the RQL for the purpose of specifying a c = 0 or sequential sampling plan.)
Figure 4 shows average sample numbers for the single and double sampling plans for code letter J and AQL = 1%, as well as for the corresponding sequential sampling plan. All compare unfavorably to 34 or 35 for the c = 0 plan, but their risk of rejecting a lot at the AQL is roughly 5 percent rather than 28.9 percent. This means we should use a double, multiple, or sequential sampling plan when quality is not sufficiently good to allow a c = 0 plan.
References
1. Squeglia, Nicholas L. Zero Acceptance Number Sampling Plans, Fourth Edition. The fifth edition is available from ASQ Quality Press.
Comments
Consumer Risk or Producer Risk?
Thanks to Mr. Levinson for his paper on C=0 sampling plans. The sample size formula presented in his paper is one of the most useful tools available. I would like to add that recently under AS9138 and ARP9013 the definitions have been standardized for Acceptable Quality Level (AQL), Equal Risk Point (ERP), Lot Tolerance Percent Defective (LTPD), and Rejectable Quality Level (RQL). They are in fact simply different points on the same operating characteristic curve having probability of acceptance 0.90-0.95, 0.50, 0.10, and 0.05 respectively. During world war two, the point-of-view of the consumer (LTPD) was abandoned in favor of the producer's point-of-view (AQL) because, as HR Bellinson stated; 20,000,000 identical items were being procured from more than 50 different suppliers, and a c=0 sampling plan was thought to be unfair to small suppliers; rejecting their product more often than large suppliers product having exactly the same quality level (ASA 105th annual meeting, January 27th 1946.). Of course AQL based sampling plans can have larger sample sizes than c=0 plans in order to "bend" the operating characteristic curve at the AQL point providing a high probability of acceptance, while still holding the same consumer's LTPD point. It is improper to say that a c=0 plan has an AQL because by design, its point-of-view is the consumer, not the producer. This is the reason for the 0.542 probability of acceptance in Mr. Levinson's example with n=15, c=0 for a lot is 4% nonconforming (4.0 AQL). We can not design a set of consistent c=0 sampling plans with the producer's risk in mind. This was the main thrust and birth of AQL based sampling plans.
With the consumer in mind - A sampling plan which is finding the nonconformities most of the time, is doing its intended job and therefore does not need to be tightened. It is only when the economics of the situation dictate that the expense of finding a few nonconformities later outweighs the expense of more inspection that we will tighten inspection.
There are many attribute sampling procedures such as MIL-STD-105, MIL-STD-1916, APR9013, and AS9138 to name a few. Most have differing poin-of-view and it all becomes a little confusing, but thankfully the underlying mathematics remains constant and reveals that it is simply a different color "lipstick" on the same OC Curve.
Why did you use binomial distribution?
Dear Mr. William A Levinson.
I was instroduced Squeglia's c=0 plan at March of 2018. I was wanted to know Squeaglia's calculation logic to define sample size.
I read his book, 5th edition but failed to find out. So I am very thank you for your article. And I have one question on your article.
On Squeaglia's book, He told that he used Hypergeometric distirbution for more exact calculation, but you used only Binomial distribution
to get sample size n. I hope to know why you didn't use Hypergeometric distribution.
The another question is the book or article which Squeaglia wrote his logic and claculations.
On the 5th edition, "Zero Acceptance Number Sampling Plans", I couldn't find out how he calculate sample size n.
If you give me smallest idea on your reply, it will be very appreciated.
I'm looking forward to reading of your reply.
Best regards,
Hong Sung Kim
IQC engineer of display device manufacturer
Republic of Korea (South Korea)
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