Question 3 of 6: Confidence Intervals vs. Natural Tolerance Limits, and Process Capability Indices

Question 3: Confidence Intervals vs. Natural Tolerance Limits, and Process Capability Indices (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.

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

A confidence interval for $\mu$, $\bar x\pm Z_{\alpha/2}\,\sigma/\sqrt n$, is a statement about how precisely the SAMPLE has located the unknown POPULATION MEAN — it answers "where is $\mu$ itself." Its width shrinks toward zero as the sample size $n$ grows, because more data always lets us pin down the location of a fixed parameter more precisely; in the limit of an infinite sample, the interval collapses to a point (we would know $\mu$ exactly).

The natural tolerance limits, $\mu\pm Z\sigma$ (most commonly $\mu\pm3\sigma$, covering 99.73% of output), are a statement about where INDIVIDUAL, FUTURE measurements from the process are expected to fall — they answer "where will the next unit measure." Their width is governed entirely by the process's inherent standard deviation $\sigma$ and does not shrink as $n$ grows; collecting more data lets us estimate $\sigma$ (and $\mu$) more precisely, but it can never shrink the physical spread of the process itself. Even at "the same confidence level" the two intervals answer fundamentally different questions — one about the precision of an estimated parameter, the other about the inherent variability of the process — and confusing them (e.g., quoting a narrow confidence interval as though it bounded where individual parts will measure) is a common and serious capability-analysis error.

(b) Statistical control as a prerequisite for capability analysis; $C_p$, $C_{pk}$, and their relation

Yes, the process should be in statistical control before a capability analysis is performed. Capability indices are only meaningful if the estimated $\hat\sigma$ is a stable, repeatable property of the process that will persist into the future; if special causes are still active, $\hat\sigma$ reflects a mixture of common-cause variation plus whatever assignable causes happened to be present during data collection, so it is not a reliable predictor of tomorrow's output and any $C_p$/$C_{pk}$ computed from it is essentially fiction — the process could look "capable" today purely because a favourable transient assignable cause was in effect, and look very different once removed (or once a different assignable cause appears).

$C_p=\dfrac{USL-LSL}{6\sigma}$ measures POTENTIAL capability: the ratio of the specification width to the process's natural (6-sigma) spread, with no regard for where the process is centered. $C_{pk}=\min\!\left[\dfrac{USL-\mu}{3\sigma},\ \dfrac{\mu-LSL}{3\sigma}\right]$ measures ACTUAL capability: the distance from the process mean to the NEARER specification limit, in units of $3\sigma$, so it penalizes off-centering. Always $C_{pk}\le C_p$, with equality if and only if the process is perfectly centered on the midpoint of the specification ($\mu=T=(USL+LSL)/2$); any off-centering makes $C_{pk}$ strictly smaller than $C_p$. Examples where the two indexes differ: a machining process whose spread easily fits within the tolerance band (good $C_p$) but whose mean has drifted toward one limit because of tool wear (poor $C_{pk}$); or a chemical batch process capable of tight control in principle but run deliberately off-target to avoid a costly failure mode on one side, sacrificing $C_{pk}$ for safety margin on that side.

The three indexes are related by $\boxed{C_{pk}=C_p(1-k)}$, where $k=\dfrac{|\mu-T|}{(USL-LSL)/2}$ is the fractional off-centering of the mean relative to the specification target $T$ (with $0\le k\le1$ for a mean between the limits). A perfectly centered process has $k=0$ and $C_{pk}=C_p$; a process whose mean sits exactly on one specification limit has $k=1$ and $C_{pk}=0$, regardless of how large $C_p$ is.

(c) Fraction nonconforming, $C_{pm}$, and the centered-mean case

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

Find. Estimated fraction nonconforming $\hat p$; $C_{pm}$; and how $\hat p$ would change if the mean were centered.

Approach. Work entirely in units of $\sigma$: recover the off-centering fraction $k$ from $C_{pk}=C_p(1-k)$, convert to the distance (in $\sigma$) from the mean to each specification limit, then sum the two tail probabilities; recompute $C_{pm}$ from the same off-centering.

1. Off-centering fraction. $$k=1-\frac{C_{pk}}{C_p}=1-\frac{1.05}{1.33}=\boxed{0.211}$$
2. Distances to each limit, in units of $\sigma$. The half-specification width is $d=(USL-LSL)/2=3\sigma C_p=3.99\sigma$; the distance from $\mu$ to the NEARER limit is $3\sigma C_{pk}=3.15\sigma$, so the distance to the FARTHER limit is $2d-3.15\sigma=(7.98-3.15)\sigma=4.83\sigma$.
3. Fraction nonconforming. $$\hat p=\Phi(-3.15)+[1-\Phi(4.83)]=0.000816+0.0000007=\boxed{0.000817}\ (\approx817\text{ ppm, }0.0817\%)$$
4. $C_{pm}$. The Taguchi-style index $C_{pm}=\dfrac{USL-LSL}{6\sqrt{\sigma^2+(\mu-T)^2}}=\dfrac{C_p}{\sqrt{1+\left(\frac{\mu-T}{\sigma}\right)^2}}$; from step 1, $\mu-T=k\,d=0.211(3.99\sigma)=0.840\sigma$, so $$C_{pm}=\frac{1.33}{\sqrt{1+0.840^2}}=\frac{1.33}{\sqrt{1.706}}=\frac{1.33}{1.306}=\boxed{1.018}$$
5. If the mean is centered ($\mu=T$, so $C_{pk}=C_p=1.33$). Both tails become equidistant at $3\sigma C_p=3.99\sigma$: $$\hat p_{\text{centered}}=2\,[1-\Phi(3.99)]=2(0.0000330)=\boxed{0.0000661}\ (\approx66\text{ ppm})$$ Centering the process (with $\sigma$ unchanged) would cut the fraction nonconforming by more than a factor of 12, from about 817 ppm to about 66 ppm — the single largest, cheapest capability improvement available here, since it requires only re-targeting the process mean, not reducing its variance.