Formulas
- Capability: smallest N with
Cpk − z · √(1/(9N) + Cpk²/(2(N−1))) ≥ required(Bissell, one-sided). - Mean:
n = (z · σ / E)² - Proportion:
n = z² · p(1 − p) / E² - Zero failures:
n = ln(1 − C) / ln(R)
For incoming inspection of lots, use the AQL sampling plan instead.
Frequently asked questions
›How many parts do I need for a capability study?
It depends on how close the expected Cpk is to the requirement. If you expect Cpk 1.67 and need to show 1.33 with 95 % confidence, about 40 parts are enough. If you expect only 1.45, you need more than 200. The calculator uses the Bissell confidence interval.
›What is the success-run theorem?
If n parts are tested without a failure, the reliability R is demonstrated with confidence C when 1 − R^n ≥ C, so n = ln(1 − C) / ln(R). For 90 % reliability at 90 % confidence you need 22 parts, for 99 % at 95 % confidence 299 parts.
›Why is the sample size for a proportion so large?
Rare events need many observations. To estimate a 1 % defect rate within ± 0.5 % you need about 1,500 parts. If a defect rate is very low, consider measuring the characteristic instead and using a capability study.