What Do We Mean by “Zero Defects”?
Part 2 of 3: Non-Conformities When Sampling a Continuum
Q: How do we handle zero defects or non-conformities when sampling a continuum?
A. In our first article on this topic, we found that while “zero defects” is a standard in quality excellence, the occurrence of x = 0 in a sample does not mean that defective items can never occur. We developed formulas for the upper confidence bound on the true process fraction non-conforming when sampling a process or a very large lot. The data are binary, of the “go-no go” type, and we observe x = 0 non-conforming items in a sample of size n.
In this article we develop a similar upper confidence bound for a rate of occurrence when sampling a continuum, and we observe zero defects or non-conformities within the inspection region. Before this is taken up, we briefly revisit the binary data scenario and consider how an inspection error probability may be accounted for in the upper confidence bound. By inspection error we mean the error of misclassifying non-conforming items as conforming. This is also called “consumer risk,” or the risk to a customer. We also assume that the second type of error, the “producer’s risk,”1 is zero for this application. If the error of misclassification is θ and if the upper confidence bound without considering the error is pu, then the corrected upper bound, considering the misclassification error, is equal to pu/(1 - θ).
The Poisson Distribution
In a sample n = 50 objects, we find x = 0 non-conforming. At 95 percent confidence, using the formulas developed in part 1, the upper confidence bound ispu= 0.05816. If the error of misclassifying a non-conforming item as conforming is, say, 10 percent, then the corrected upper bound would be pu/(1 - 0.1) = 0.05816/(1 - 0.1) = 0.06462. Often, the quantity (1 - θ) is referred to as the probability of detection for the specific attribute.
When sampling a continuum we call the events that are found either defects or non-conformities. By a continuum, we mean any type of inspection region defined by space, time, area, volume or other similar metric. An event within the inspection region is the occurrence of a defect or non-conformity. Thus, for this type of sampling, we are not classifying objects, rather we are counting events. We further assume that the inspection region being observed is homogeneous and that events occur randomly throughout the region. This means that the rate of event occurrence remains constant throughout the inspection region. Some examples include: a) paper rolling off a mill, where events are defined as any abnormality on the paper’s surface, e.g., the rate might be defined as abnormalities per square yard; b) surface pits or other abnormality per square foot for sheet metal stock; or c) blemishes on a painted surface per square foot. There are numerous applications for this type of inspection process. Wherever there is an event of interest within some defined interval we can use this method.
The statistical model that governs this type of sampling is the Poisson distribution. In this distribution there is a single parameter, λ, interpreted as the unknown rate of occurrence for the event in question. Sometimes there is an auxiliary parameter, t, attached to the rate as λt. If the number t is the size of the inspection region, then λt is interpreted as the mean number of events in a region of size t where the rate of event occurrence is λ. In this scenario, the units of t and λ must be the same. For example, if we are looking at blemishes on a metal surface and the rate is in the units of blemishes per square yard, then the inspection region would have to be specified in square yards. If the inspection region were, say, 3 square feet, then t would be 1/3 since 3 ft2 is a third of a square yard. The mean within the inspection region would then be λ/3. When the rate is events per unit and the inspection interval is 1 unit then λ is the mean number of events within the inspection interval.
In the Poisson distribution with rate λ, the probability of observing x = 0 is:
Using an argument similar to that in part 1, when we observe zero events we require that this probability be at least some small value such as 0.05 pr 0.01. This is cast as the quantity 1 - C where C is confidence. Applying this to (1) gives:
Then solving the inequality for the rate λ gives the upper confidence bound when x = 0 is observed in a region of size t:
We can also take (3) and solve for either C or t. For example, solving (3) for C gives:
The interpretation of (4) is that the confidence is at least C that the rate is equal to a prespecified value, λ0, when we observe zero events in an inspection region of size t. When there is the risk of missing events, say with probability θ (consumer’s risk), this enters into the upper bound calculation in the same way as previously shown for the binary case. The upper bound with misclassification error is:
1) Suppose the inspection region is 20 board feet of processed timber used in high end furniture manufacturing. The event is defined as a certain size of abnormality in wood products. At confidence C = 0.95, what is the upper bound on the rate λ when zero events are observed within the inspection region? Assume no inspection error
Use (3) with C = 0.95 and t = 20. The answer is, approximately: λ ≤ 0.15 abnormalities per board foot of timber.
2) Suppose in illustration 1 that there is a potential for an inspection error of approximately 10 percent. This means that the inspector may miss a real event about 10 percent of the time.
Use (5) with C = 0.95, t = 20 and θ = 0.1. The answer is, approximately: λ ≤ 0.17 abnormalities per board foot of timber.
3) A business unit manager of a large manufacturing cell is interested in tracking the number of returns or “turnbacks” of cell product per thousand units produced. Last week, a typical week for this operation, the cell produced 17,500 units with 0 turnbacks observed. Under similar controlled conditions, what rate per thousand units is being demonstrated with 90 percent confidence? Assume that there is a risk of approximately 3 percent of missing a real turnback.
Use (5) with C = 0.9, t = 17.5 and θ = 0.03. The answer is: 0.1356 turnbacks per thousand units produced.
Stephen N. Luko, Pratt & Whitney Aircraft, is the immediate past chairman of Committee E11 on Quality and Statistics and a fellow of ASTM International.
Dean V. Neubauer, Corning Inc., is vice chairman of Committee E11 on Quality and Statistics, chairman of Subcommittee E11.30 on Statistical Quality Control and chairman of E11.90.03 on Publications; he also coordinates the DataPoints column and is an ASTM International fellow.
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