Testing a Process Against a Spec

Testing against a spec is simply asking whether your data give solid evidence that a process meets a requirement, rather than just a hopeful average. It's the one-sample t-test from the intro course, aimed at an engineering question. The statistic $t_0 = \frac{\bar x - \mu_0}{s/\sqrt n}$ is three familiar pieces: $\bar x - \mu_0$ (how far the sample mean sits from the spec value), $s/\sqrt n$ (the standard error) and their ratio (how many standard errors that is). The engineering judgment is in choosing which side has to prove itself, and that's what this lesson is really about.

Who has to prove what. A fastener supplier says the mean clamp load of its bolts is at least 12 kN. If you'll accept the lot only on evidence, the claim goes in the alternative: $H_0: \mu \le 12$ against $H_1: \mu > 12$, and the supplier's data must push past it. If you'll accept unless there's evidence of a shortfall, it's $H_0: \mu \ge 12$ against $H_1: \mu < 12$, and the benefit of the doubt flips. Same data, possibly different conclusions, so decide before looking. The test is run at the boundary $\mu = \mu_0$: a test that holds false alarms to α there holds them below α everywhere else in $H_0$.

The statistic, and why it's t. $t_0 = (\bar x - \mu_0)/(s/\sqrt n)$ with $n - 1$ degrees of freedom (t tests). For normal data, $Z = (\bar x - \mu)/(\sigma/\sqrt n)$ is standard normal, $V = (n-1)s^2/\sigma^2$ is $\chi^2_{n-1}$ (intervals for variability) and independent of $\bar x$, and $Z/\sqrt{V/(n-1)}$ is, by definition, the t distribution with $n - 1$ degrees of freedom. The unknown σ cancels, leaving $t_0$.

p-values from a table. A t table lists only a few upper-tail points, so you bracket. For ν = 14:

A one-sided $t_0 = 2.40$ falls between 2.145 and 2.624, so $0.01 < p < 0.025$; software gives 0.015. Two-sided, double both ends: $0.02 < p < 0.05$.

Statistical vs practical significance. With 400 bolts, a mean 0.06 kN below a 12 kN spec can be "highly significant" and still irrelevant if the design margin is 1 kN. With 5 bolts and $s = 1.5$ kN, even a 1 kN shortfall can hide in the noise ($t_0 \approx -1.5$). So report the confidence bound next to the p-value: it says how big the effect plausibly is. The two always agree, because a level-α one-sided test rejects exactly when the $1 - \alpha$ confidence bound clears $\mu_0$. To show that a change made no practical difference, flip the logic and test equivalence: two one-sided tests against the edges of the band you care about (the challenge).

Mistakes to avoid: choosing the direction after seeing which way the data went (that doubles the real α); reading $p = 0.03$ as "a 3% chance that $H_0$ is true"; and treating "not significant" as "meets spec", when a small sample may simply have had little power (planning sample sizes).

Incoming inspection, lot acceptance and design verification all test against a spec: is the mean fill at least the labeled volume, does the mean bond strength exceed the requirement, has a supplier's mean drifted? Regulated industries add equivalence tests when a process changes, so that "no difference" has to be demonstrated rather than assumed, and quality engineers report confidence bounds so managers see the size of an effect, not just its p-value.