There is no single sample size formula. The right method is decided by what your test has to prove — pass a reliability gate, characterise a life distribution, measure margin, discover limits, or screen production. Pick the test type below; each is an independent calculator with its own inputs, its own assumptions, and its own statement of what the answer does not mean.
You already know the target. The question is how much test it takes to defend it.
Use when units are cheap relative to test time, or when you have no defensible β.
The classic reliability demonstration. Test n units, allow at most f failures, and read off what that proves. Pick which quantity to solve for; the other three are inputs.
n = sample size, f = allowed failures, R = reliability, CL= confidence level. At f = 0 this reduces to n = ln(1−CL) / ln(R).
Sample size is not one calculation. Asking “how many units do I need?” without saying what the test must prove is what produces the familiar answer of 22 units for everything. These five test types each answer a different question, and they disagree with each other by design.
The demonstration test. You have a target such as R90C90 — 90% reliability at 90% confidence — and you need to show the design meets it. With zero allowed failures, the sample size is n = ln(1 − C) / ln(R), which gives 22 units for R90C90 and 230 units for R99C90. That jump is the central fact of reliability demonstration: high targets are punishingly expensive to prove by counting.
Allowing failures raises the count — at R90C90, one allowed failure needs 38 units, two needs 52, three needs 65 — because a test that can survive a failure must be bigger to prove the same thing.
Extended bogeytrades units for time. If failures follow a Weibull with shape β, running each unit to a multiple of one life earns credit: n = ln(1 − C) / [(t/T)β · ln R]. At β = 2, testing to two lives cuts R90C90 from 22 units to 6. But the saving lives entirely in β: at β = 1 — random failures — doubling the test only halves the units, and below 1 it buys almost nothing. A β borrowed from a handbook rather than measured from failure data is the most common way this method produces a confident wrong answer.
MTBF demonstration answers in time, not units. The total test time is T = MTBFreq · χ²(1−C, 2r+2) / 2, where r is the number of failures allowed. At 90% confidence with zero failures the multiplier is 2.3026, so demonstrating a 5,000 hour MTBF takes 11,513 unit-hours — splittable across any number of units. It assumes a constant failure rate, which is exactly the assumption the Weibull tool exists to test.
If you want to know when things fail rather than whether they pass, there is no closed-form sample size, because what you need is failures, not units. Two or three failures give an engineering feel for the mode; around seven support a usable β; fifteen to twenty tighten the confidence bounds enough to quote a B10 life. Units follow from failures once you assume a censoring fraction. Accelerated life testing needs at least three stress levels with failures at everylevel — a level that produces none contributes nothing to the model — and degradation testing needs far fewer units, typically five to ten, because each unit contributes a whole curve rather than a single data point.
Counting pass/fail outcomes throws away most of the information in a measurement. If the characteristic is continuous and roughly normal, a tolerance intervalproves the same claim with far fewer units: 90/90 by the attribute route takes 22 units, by the variables route about 10. The saving is not free. It holds only if the design carries the margin — at n = 10 the one-sided factor is k = 2.0657, so you need the spec limit to sit at least 2.07 standard deviations from the mean. A design running close to its limit gets no discount, and the method requires normality.
A comparative test(A versus B) is sized from effect size and statistical power, not from R and C at all — it answers whether two designs differ, not whether either is reliable. Detecting a one standard deviation difference at 80% power needs 17 units per arm. Weibayescredits existing test evidence from a comparable design against the new requirement, which can cut the remaining test dramatically — and is only as defensible as the claim that the two designs are comparable.
HALT is discovery, not proof. It finds operating and destruct limits by stepping stress past specification until things break, so no reliability or confidence figure can be claimed from it, no matter how many units you run. The unit count is a budget question: how many stress axes you explore, whether you push to destruct limits, and how many units each axis consumes. It usually lands at three to six. Build the stress profile itself in the HALT/HASS Wizard.
HASS is a production screen, a different problem from HALT: it removes latent defects from units you intend to ship. Sampling is a lot-acceptance question — how many parts per lot must be screened to catch a lot sitting at the target escape rate — and the honest answer is often a large fraction of the lot, which is why screening starts at 100% and reduces only after a proof-of-screen. Screen strength must stay well inside the destruct limits found in HALT; a screen that consumes useful life ships weakened product.
Every method here trades units, time, and assumptions. When schedule is the binding constraint rather than unit cost, combine with an acceleration model — Arrhenius for temperature-driven mechanisms, Coffin-Manson for thermal cycling — and check the resulting claim against your field target with the FIT calculator.