Engineering & Technology · Quality Engineering, Experiments & Process Evidence

Capability, sampling, acceptance and quality decisions

A supplier reports a capability number, while a receiving team examines a sample from one lot. These activities ask related but different questions. Capability concerns a process distribution relative to requirements; acceptance sampling applies a specified rule to a particular lot.

Review measurement and uncertainty first. Means, ranges, square roots and percentages are used directly. The inference lesson defines its paired-difference model and supplies its Student-t quantile; mathematics develops probability and inference further. These examples teach interpretation rather than prescribe production sampling, monitoring or release rules.

Back to the chapter

After reading, you can

Distinguish a conditional process-capability description from a sample-based product disposition decision.

A ratio needs a stable distribution

For a stable normal population model with mean μ and standard deviation σ, the familiar index Cp=(USL−LSL)/(6σ) compares specification width with a six-sigma spread. Cpk uses the nearer specification distance: the minimum of (USL−μ)/(3σ) and (μ−LSL)/(3σ). The limits are supplied requirements, not values selected to flatter the process.

Our original population model supplies limits 9 and 11 mm, mean 10.6 mm and σ=0.2 mm. Cp=2/1.2≈1.667, while Cpk=min (0.4/0.6, 1.6/0.6)≈0.667. The process is offset toward the upper bound. The figure shows the location problem that Cp alone would conceal. No universal pass threshold or actual defect percentage is inferred.

An estimated index also has uncertainty

A real capability study must examine stability, measurement suitability, the population and the distributional model. Estimated parameters and indices have sampling uncertainty. A few convenient values cannot establish a long-term process distribution, and a normal-model index should not silently imply normal-tail behaviour for a skewed or mixed population.

The mean±3σ span in the figure is a model spread, not a guarantee that all items fall within it. Normal distributions have tails beyond that span. A changed mean, a new operating state or a measurement problem can invalidate the original comparison. Report the assumptions and scope rather than promoting an index to a complete quality certificate.

Specification width 11−9=2 mm; six-sigma width 6(0.2)=1.2 mm, giving Cp=1.667. Distances from mean 10.6 to bounds are 0.4 and 1.6 mm. Dividing the smaller by 3σ=0.6 mm gives Cpk=0.667. The 10.0 to 11.2 span is mean±3σ, not a guaranteed range or acceptance limit.

A lot decision can miss nonconforming items

An acceptance plan defines the lot, random selection method, sample size, defect definition, acceptance count and disposition rules. Producer and consumer risks depend on the plan and lot quality. Passing a sample rule does not prove every untested item conforms, and it does not automatically establish a stable process.

Consider a separate original finite lot with 100 items, of which 4 are nonconforming in the teaching model. Select 10 distinct items uniformly without replacement and accept only if none is nonconforming. The probability of missing all 4 is (96/100)×(95/99)×…×(87/91), about 0.652. This is a constructed calculation illustrating sampling risk, not a recommended plan or a claim that an observed clean sample reveals the hidden count.

Follow the conditional probabilities one draw at a time

In the finite-lot example, 96 of the 100 items conform. Assume the classification is perfect for this calculation. The first randomly selected item therefore has probability 96/100 of conforming. Conditional on that first item conforming and being removed, 95 conforming items remain among 99 total. The probability that the first two both conform is (96/100)×(95/99)≈0.9212: the second fraction describes only the cases in which the first draw was conforming.

If those first two conform, the next conditional fraction is 94/98. Continuing to ten draws gives the stated product ending at 87/91. Multiplying these conditional chances gives the probability of the complete all-conforming path; it does not assume that the draws are independent. Both numerator and denominator decrease because no sampled item is returned. Using the unchanged first-draw probability 0.96 ten times would describe a different replacement model.

Under the zero-nonconforming acceptance rule, that all-conforming path is exactly acceptance. Its probability is about 0.6516 when four nonconforming items are actually present. The complementary probability of finding at least one is 1−0.6516≈0.3484. For a sample of 20, continue the same conditional product to 77/81: acceptance probability is about 0.4033 and finding-at-least-one probability about 0.5967. More inspection changes the conditional risk and consumes more items/time; the calculation alone does not select an appropriate plan.

The finite-without-replacement count model is called hypergeometric. Its hidden lot composition is an input here. Calculating the probability of a clean sample given four nonconforming items is not the reverse probability that four exist given a clean sample. An observed clean sample does not establish the assumed composition. The plotted zero-sample value is only the mathematical all-conforming-path reference; no real release rule is being recommended.

Under the stated lot and perfect-classification model, multiply the successive conforming-item fractions, starting at 96/100 and reducing both numbers by one at each subsequent draw, to obtain the probability of a clean sample. At sample size 0 it is 1; at 10 approximately 0.6516305491; at 20 approximately 0.4033382425; at 30 approximately 0.2338287142; at 40 approximately 0.1243578219. Probability of finding at least one is one minus this value. These probabilities assume the hidden count four; an observed clean sample does not establish that hidden count or a defect-free lot. The plot is not a recommended release plan.

Interpret two conditional calculations

In the capability model, moving the supplied mean to 10 mm while retaining σ=0.2 mm would make both boundary distances 1 mm and Cpk=1/0.6≈1.667. This arithmetic describes a changed hypothetical model. It does not establish that an actual adjustment preserves the same variation or produces those parameters.

In the separate finite-lot account, a clean sample remains quite possible even though four nonconforming items exist. The sample rule and risk need justification for the intended decision and consequences. Keep lot acceptance, process monitoring and process capability as distinct evidence paths that can inform one another without substituting for one another.

Try the reasoning

Why can Cp be large while Cpk is much smaller?
Show the answer and reasoning
Cp ignores centring; Cpk reflects the nearer boundary, so an offset mean can reduce it.
Does a clean sample prove a lot has no defects?
Show the answer and reasoning
No. A sampling plan can miss nonconforming items, with a probability that depends on the model and plan.
Can the supplied normal-population index be treated as a real study result?
Show the answer and reasoning
No. Its parameters are invented assumptions, and no empirical process or measurement study has been supplied.
For the same 100-item lot with four nonconforming items and perfect classification, change the distinct-item sample from 10 to 20 while retaining acceptance only with zero found. What changes, and what does a clean result establish?
Show the answer and reasoning
Multiply twenty conditional fractions, starting at 96/100 and ending at 77/81, to obtain approximately 0.4033. This is lower than the ten-item acceptance probability 0.6516; probability of finding at least one rises from about 0.3484 to 0.5967. These are probabilities conditional on the stipulated hidden count. A clean result still does not prove the lot is defect-free or reveal that four are present. Plan suitability also depends on consequences and inspection conditions.
In the separate stable normal population model, move the mean from 10.6 to 10.2 mm while keeping limits 9/11 and σ=0.2 mm. Compare Cp and Cpk without inventing a pass threshold.
Show the answer and reasoning
Cp remains (11−9)/(6×0.2)=5/3≈1.667 because width and spread did not change. Cpk becomes min((11−10.2)/0.6,(10.2−9)/0.6)=min(4/3,2)=4/3≈1.333, compared with 0.667 before. The nearer-boundary distance increased in this hypothetical model. It does not demonstrate an actual centring intervention, unchanged variation or compliance with a real release criterion.

What to carry forward

State the distribution and evidence behind capability, and state the sampling rule and risks behind a lot decision.

References

Teaching resources and research supporting this lesson’s explanations and examples.

National Institute of Standards and Technology (NIST)5 sources