Help me choose R and C (optional)
Example only. Set per your risk management procedure. The table below is an illustration, not a recommendation. Edit it to match your organization's risk-to-sample-size table. Your edits are stored only in this browser.
| Use | Level | Description | C (%) | R (%) |
|---|
Zero-failure attribute test (success-run)
How many units must all pass to show reliability R with confidence C?
Formula used
n = ceil( ln(1 − C) / ln(R) ). If the true reliability were only R, the chance that n units in a row all pass is Rn. n is the smallest size for which that chance is at most 1 − C. This is the exact binomial bound for zero failures.
Attribute test allowing c failures
The smallest n such that observing c or fewer failures still demonstrates reliability R with confidence C (exact binomial).
Formula used
n is the smallest integer with P(X ≤ c | n, p = 1 − R) ≤ 1 − C, where X is binomial. The binomial CDF is computed exactly through the regularized incomplete beta function. With c = 0 this reduces to the success-run formula.
Normal tolerance interval: k-factor and pass/fail
Paste your measurements or enter summary statistics, then enter the specification limits. One limit gives a one-sided check, two limits give a two-sided check.
Which two-sided rule?
Whole interval inside spec (default). This uses the two-sided factor k2. PASS means that, with confidence C, at least R of all units lie between LSL and USL. It is the statistical tolerance interval of ISO 16269-6 and NIST 7.2.6.3, and it is the conservative choice.
One-sided bound at each limit. This uses the one-sided factor k1 separately at each limit. PASS means that, with confidence C for each limit separately, at most 1 − R of units fall below LSL and at most 1 − R fall above USL. Up to 2(1 − R) of units could be out of spec in total, and the joint confidence is lower than C. This rule can be argued when each limit protects against a different failure mode with its own risk. Document that argument. We found no verified standard or handbook reference that prescribes this rule, so none is cited.
Data checks
Paste data above to run normality and outlier checks.
Normality assumption. These k-factors are valid only if the data come from a normal distribution. Check normality (e.g., Anderson-Darling) before using. With summary statistics only, the page cannot check this for you.
Plan the test: n and the probability of passing
A future sample can fail even if the process is good, because the sample mean and SD vary. Enter what you expect the true mean and SD to be. You get the probability that a test of size n passes, computed exactly with the noncentral t distribution, and the n needed to reach a target probability.
Formula used
One-sided (exact): k = t′C(n − 1, zR√n) / √n, where t′ is the noncentral t quantile with noncentrality zR√n. It is computed by numerical integration over the sampling distribution of the SD.
Two-sided (exact): k solves 2∫0∞ φ(z) · P(χ²n−1 > (n − 1) r²(z/√n) / k²) dz = C, where r(x) solves Φ(x + r) − Φ(x − r) = R. This is the exact Odeh-type integral, not the Howe or Wald-Wolfowitz approximation.
Decision: observed k = (mean − LSL)/SD and/or (USL − mean)/SD. PASS if the observed k (the smaller one for two-sided) is at least the required k.
Probability of passing: with KL = (μ − LSL)/σ and KU = (USL − μ)/σ, P = E[ max(0, Φ(√n(KU − kV)) − Φ(√n(kV − KL))) ], where V = s/σ. With one limit this is 1 − Ft′(k√n; n − 1, K√n). The SD upper bound is s·√((m − 1)/χ²1−conf, m−1) for m pilot units (NIST 1.3.5.8).
Checks: Anderson-Darling with the modified statistic A* = A²(1 + 0.75/n + 2.25/n²) and the Stephens (1986) p-value, as in R nortest. Shapiro-Wilk by Royston's algorithm AS R94, as in R shapiro.test. Two-sided Grubbs test with the t-distribution critical value (NIST 1.3.5.17.1). Box-Cox λ is chosen to maximize the normal probability plot correlation (NIST 1.3.3.6).
Nonparametric (distribution-free) tolerance
How many units are needed so that the sample minimum (one-sided), or the minimum and maximum (two-sided), bound proportion R with confidence C, for any continuous distribution?
Check measured data against the limits (optional)
Formula used
One-sided: the smallest n with 1 − Rn ≥ C (the same arithmetic as the success-run).
Two-sided: the smallest n with 1 − nRn−1 + (n − 1)Rn ≥ C (NIST/SEMATECH e-Handbook 7.2.6.4).
Data check: PASS when every measured value is inside the limit(s) and n is at least the required n.
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Frequently asked questions
Why 59 samples?
59 is the smallest number of units that, if all pass, demonstrates at least 95% reliability with 95% confidence. If the true reliability were only 95%, the chance that 59 units in a row all pass is 0.95^59, about 4.8%, which is below 5%. With 58 units it is about 5.1%. The general zero-failure (success-run) formula is n = ln(1 - C) / ln(R), rounded up. The same 59 is also the distribution-free sample size for a one-sided 95%/95% tolerance limit, because the arithmetic is identical.
What does 95/95 mean?
95/95 means 95% confidence that at least 95% of units (the reliability, or coverage) meet the requirement. Authors order the two numbers differently: 95/90 can mean 95% confidence with 90% reliability (29 units with zero failures) or 95% reliability with 90% confidence (45 units). Always write both numbers with their names in a protocol. The levels are usually tied to the severity of harm in the risk management file (ISO 14971). This calculator does not choose them for you.
What is the difference between a tolerance interval and a confidence interval?
A confidence interval bounds a population parameter such as the mean. A tolerance interval bounds a stated proportion of individual units, such as 95% of all units, at a stated confidence. Design verification asks whether individual units meet the specification, so a tolerance interval or an attribute test is usually the right tool. A confidence interval on the mean can sit inside the specification while many individual units fall outside it.
Why does this calculator give 3.393 for a two-sided 95/95 interval with n = 10 when some tables show 3.379?
3.393 is the exact factor, matching Odeh and Owen (1980) and the R EnvStats package. 3.379 comes from an older approximation (Wald-Wolfowitz) that many legacy tables reproduce. The difference is small, but the exact value is the defensible one. The reference-value check report lists both.
When should I use an attribute test instead of a tolerance interval?
Use an attribute (pass/fail) test when the result can only be pass or fail. Use a variables method (tolerance interval) when you measure a continuous value such as seal strength or force. It usually needs far fewer units, but the normal method assumes normally distributed data, so check normality (for example with an Anderson-Darling test) first. The distribution-free method makes no such assumption but needs more units.
What if my data are not normal?
Paste your data and the page runs the Anderson-Darling and Shapiro-Wilk tests and draws a normal probability plot. If normality is rejected, you can transform the data and the specification limits together (log or Box-Cox), which leaves the proportion inside the specification unchanged. You can also use the distribution-free method, which needs more units but makes no distribution assumption. If you use the normal method anyway, justify and document that choice.
Do ISO 2859-1 or ISO 3951-1 apply to design verification?
ISO 2859-1 (attributes) and ISO 3951-1 (variables) are acceptance-sampling schemes indexed by an acceptable quality limit (AQL) for lot-by-lot inspection. They are built to control lot acceptance in production and incoming inspection, not to demonstrate a reliability at a stated confidence, so design verification rationales more often use confidence/reliability methods like the ones on this page. ISO 16269-6 covers statistical tolerance intervals.
Which requirement asks for a sample size rationale?
ISO 13485:2016 clause 7.3.6 expects design verification plans to state the statistical techniques used and the basis for the sample size, where appropriate. Since 2 February 2026, the FDA Quality Management System Regulation (21 CFR Part 820) incorporates ISO 13485:2016 by reference.
Validation
Every method is checked against published reference values: the NIST/SEMATECH e-Handbook, worked examples from ISO 16269-6, the Odeh & Owen tables, and R output for the normality tests. The checks also include independent numerical cross-checks and Monte Carlo simulation. Open the reference-value check report, or run the self-test in your own browser and print the results for your tool validation file.