Tolerance Intervals

A tolerance interval is simply a range that, with high confidence, contains almost every part the process will ever make, not just the average or the next part. The formula $\bar x \pm k s$ looks just like the others: center (the sample mean) plus or minus a multiple $k$ of the spread (the sample SD). The only new thing is where $k$ comes from: a table, indexed by how much of the population you want covered ($p$), how sure you want to be ($\gamma$) and how many parts you measured ($n$). Two percentages and a lookup, nothing scarier.

The spec question. A drawing says $12.000 \pm 0.015$ mm. The customer doesn't care where the mean is or where the next shaft lands: they care whether essentially all shafts are inside. That's a claim about the population's middle 99%, and it needs a tolerance interval.

Two probabilities, two jobs. $p$ is how much of the population the interval should contain; $\gamma$ is how sure we are that it does. A "95%/99% tolerance interval" means: with 95% confidence, at least 99% of all parts lie inside.

Building k. If $\mu$ and $\sigma$ were known, $\mu \pm z_{(1-p)/2}\sigma$ would contain exactly $p$. We only have estimates, so pad each one:

• the center $\bar x$ misses $\mu$ by about $\sigma/\sqrt n$, so widen by $\sqrt{1 + 1/n}$, the prediction interval factor;
• $s$ might be smaller than $\sigma$, so replace $\sigma$ with its upper confidence bound $s\sqrt{(n-1)/\chi^2_{\gamma,n-1}}$ from intervals for variability.

Multiplying the pads gives Howe's approximation, within about 1% of the exact factor even at $n = 5$. Exact tables come from integrating over the joint distribution of $\bar x$ and $s$; in practice you read $k$ from a table or software.

k shrinks toward z. As $n \to \infty$ both pads go to 1 and $k \to z_{(1-p)/2}$: 1.960 for $p = 0.95$, 2.576 for $p = 0.99$. At $n = 10$ the 95%/95% factor is 3.393, so small samples pay heavily.

CI, PI or TI? Same data ($n = 10$, $\bar x = 50.0$, $s = 2.0$), all at 95% confidence:

Only a tolerance interval answers "do the parts meet spec?": put it next to LSL and USL. If it fits inside, you can claim at least $p$ of the parts are in spec, with confidence $\gamma$.

One-sided bounds such as a minimum strength $\bar x - k_1 s$ use their own table of $k_1$, smaller than the two-sided $k$ because all the uncovered parts sit on one side.

Mistakes to avoid: reading $k$ for the wrong $(p, \gamma)$ pair; using a two-sided $k$ for a one-sided spec; thinking the interval holds exactly $p$ (usually it holds more, the price of confidence); and forgetting normality. Tolerance intervals are about the tails of the population, so they're only as good as the normal model out there: check a probability plot, or use a distribution-free bound like the one in the challenge.

Where it leads: comparing a tolerance interval with the spec is a capability statement with honest uncertainty (process capability).

Aerospace and automotive design allowables are one-sided tolerance bounds (99% or 90% of material above the value, with 95% confidence). Medical-device and pharmaceutical validation use two-sided tolerance intervals to show that nearly every unit meets spec, and supplier quality agreements often ask for 95%/99% intervals inside the drawing limits. Each one answers the question a customer actually asks: not "what's your average?" but "will my parts be good?"