Question 3 of 6: Tolerance Limits, Process Capability, and Fraction Nonconforming from $C_p$, $C_{pk}$

Question 3: Tolerance Limits, Process Capability, and Fraction Nonconforming from $C_p$, $C_{pk}$ (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Part (a) — Confidence interval for $\mu$ vs. natural tolerance limits

A confidence interval for $\mu$ is a statement about the parameter: given a sample of size $n$, the interval $\bar X\pm t_{\alpha/2,n-1}s/\sqrt n$ is constructed so that, over repeated sampling, it captures the true (unknown, fixed) population mean $\mu$ in $100(1-\alpha)\%$ of samples. It shrinks toward zero width as $n\to\infty$, because more data pins down the location of $\mu$ ever more precisely, and it says nothing directly about where an individual unit's measurement will fall — only about the accuracy of the estimate $\bar X$ as a stand-in for $\mu$.

The natural tolerance limits, by contrast, are a statement about the population of individual units: $\mu\pm3\sigma$ (or, when estimated from a sample, $\bar X\pm K s$ for a tolerance factor $K$) is constructed to contain a specified proportion (conventionally 99.73% for $\pm3\sigma$, or any other stated percentage) of the individual measurements the process will produce, at a stated confidence level. Because it describes the spread of individual units rather than the precision of an estimated mean, this interval does not shrink to zero as $n$ grows — it converges to the fixed population interval $\mu\pm3\sigma$, whose width is set by the process's inherent variability, not by how much data was collected.

The confidence level means something correspondingly different in each case. For the CI on $\mu$, it is the long-run fraction of such intervals (across repeated samples) that would bracket the true, single-valued $\mu$. For the tolerance limits, the confidence level is the long-run fraction of times the computed interval actually contains at least the stated proportion (e.g. 99%) of the population — it is a statement about how reliably the interval achieves its coverage promise about individuals, which is inherently a harder (two-part) claim than pinning down a single parameter, and is why tolerance-interval factors $K$ are always noticeably wider than the corresponding $t$/$z$ multiplier used for a CI on $\mu$ at the same sample size.

Part (b) — Statistical control as a precondition for capability analysis; $C_p$, $C_{pk}$, and $k$

Yes, the process should be in statistical control before performing capability analysis. Capability indices are computed from an estimated process mean and standard deviation, and both quantities are only meaningful as stable, predictable descriptors of the process's future output if the process is currently free of assignable causes. If the process is not in control, $\hat\mu$ and $\hat\sigma$ describe only the particular unstable mixture of in-control and out-of-control behaviour that happened to occur during the sampled period — a capability index computed on that basis is not a reliable forecast of what the process will actually produce going forward, because an assignable cause could appear, disappear, or recur at any time, and the "process" being characterized is not a single, fixed statistical entity. Practically, one should first bring the process into control via control charts (as in Questions 2, 4 and 5), and only then use the in-control estimates $\hat\mu_0,\hat\sigma_0$ for capability analysis.

$C_p$ (the "potential" capability index) compares only the width of the specification band to the process spread, ignoring where the process is centred: $C_p=(USL-LSL)/(6\sigma)$. It answers "could this process, if perfectly centred, fit inside the tolerance?" and is unaffected by any off-target mean. $C_{pk}$ (the "actual" capability index) additionally accounts for centring: $C_{pk}=\min\!\left(\dfrac{USL-\mu}{3\sigma},\dfrac{\mu-LSL}{3\sigma}\right)$, i.e. it uses the nearer specification limit and therefore penalizes an off-centre mean. The two are equal ($C_{pk}=C_p$) only when the process is perfectly centred on the midpoint of the tolerance; whenever the mean drifts toward one limit, $C_{pk}\lt C_p$, and the gap between them is a direct measure of how much of the process's potential capability is being lost to poor centring rather than to excess variability.

The $k$ index quantifies exactly that off-centring, as a fraction of the tolerance half-width: $k=\dfrac{|\mu-M|}{(USL-LSL)/2}$, where $M=(USL+LSL)/2$ is the specification midpoint. The three indices are tied together by $C_{pk}=C_p(1-k)$ — $C_{pk}$ is always $C_p$ reduced by the factor $(1-k)$, so $k=0$ (mean exactly on target) recovers $C_{pk}=C_p$, and $k\to1$ (mean approaching a specification limit) drives $C_{pk}\to0$ regardless of how tight $\sigma$ is. This relation is the working tool behind part (c): given only $C_p$ and $C_{pk}$, $k$ can be recovered directly as $k=1-C_{pk}/C_p$, without needing the actual $\mu$, $\sigma$, or specification limits.

Part (c) — Fraction nonconforming, $C_{pm}$, and the effect of centering, from $C_p=1.33$, $C_{pk}=1.05$

Given. $C_p=1.33$, $C_{pk}=1.05$; process normal; two-sided specification limits $LSL$, $USL$; target $T=(USL+LSL)/2=M$ (the specification midpoint).

Find. The process fraction nonconforming; $C_{pm}$; and how the fraction nonconforming changes if the mean is re-centred.

Approach. $C_p$ and $C_{pk}$ alone are enough to recover the nearer- and farther-limit distances in $\sigma$-units (via $k=1-C_{pk}/C_p$), without needing $\mu$, $\sigma$, or the spec limits individually.

1. Off-centring fraction $k$. $$k=1-\dfrac{C_{pk}}{C_p}=1-\dfrac{1.05}{1.33}=\boxed{0.2105}.$$
2. Distances to the near and far specification limits, in $\sigma$-units. The near limit is exactly $C_{pk}$'s definition: $Z_{\text{near}}=3C_{pk}=3(1.05)=\boxed{3.15}$. The far limit is the mirror distance, $Z_{\text{far}}=3C_p(1+k)=3(2C_p-C_{pk})=3(2(1.33)-1.05)=\boxed{4.83}$.
3. Fraction nonconforming. $$\hat p=P(Z>Z_{\text{near}})+P(Z<-Z_{\text{far}})=[1-\Phi(3.15)]+\Phi(-4.83)=0.000816+0.0000007=\boxed{0.000817\ (0.0817\%)}.$$
4. $C_{pm}$. With $T=M$, the mean's offset from target in $\sigma$-units is $|\mu-T|/\sigma=3(C_p-C_{pk})=3(1.33-1.05)=0.84$, so $$C_{pm}=\dfrac{C_p}{\sqrt{1+\left(\frac{\mu-T}{\sigma}\right)^2}}=\dfrac{1.33}{\sqrt{1+0.84^2}}=\dfrac{1.33}{1.306}=\boxed{1.018}.$$
5. Effect of centering the mean. If the mean is moved onto target ($\mu=T$, so $k=0$ and $C_{pk}=C_p=1.33$), both tails become equidistant at $Z=3C_p=3.99$: $$\hat p_{\text{centered}}=2[1-\Phi(3.99)]=2(0.0000331)=\boxed{0.0000661\ (0.00661\%)}.$$ Centering the mean, with $\sigma$ unchanged, cuts the fraction nonconforming by more than an order of magnitude (from 0.0817% to 0.0066%) — the process's inherent spread ($C_p=1.33$) was already comfortably capable, and essentially all of the shortfall between $C_p$ and $C_{pk}$ was a centring problem, not a variability problem.