Process Capability, Natural Tolerance Limits, and Detection-Speed Chart Design: 23-Ind-A5 Quality Planning, Control, and Assurance, December 2015, Q3

Question 3: Process Capability, Natural Tolerance Limits, and Detection-Speed Chart Design (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) Proportion nonconforming, in control and after a mean shift

Given. From Question 2(d): $\hat\mu_0=175.6$ lb, $\hat\sigma=11.15$ lb. Specification $175\pm35$, i.e. $LSL=140$, $USL=210$ lb.

Find. $p_{noncon}$ when the process runs at $\mu_0$, and again after a shift to $\mu_1=185$.

In-control proportion nonconforming. $$Z_U=\frac{USL-\mu_0}{\hat\sigma}=\frac{210-175.6}{11.15}=3.085,\qquad Z_L=\frac{LSL-\mu_0}{\hat\sigma}=\frac{140-175.6}{11.15}=-3.192.$$ $$p_{noncon}=P(Z>3.085)+P(Z<-3.192)=(1-\Phi(3.085))+\Phi(-3.192)=0.00102+0.00071=\boxed{0.00172\ (0.172\%)}.$$
Shifted proportion nonconforming ($\mu_1=185$). $$Z_U=\frac{210-185}{11.15}=2.242,\qquad Z_L=\frac{140-185}{11.15}=-4.035.$$ $$p_{noncon}=(1-\Phi(2.242))+\Phi(-4.035)=0.01249+0.00003=\boxed{0.01251\ (1.251\%)}.$$

Shifting the mean 9.4 lb (0.84$\hat\sigma$) toward the upper specification more than sevenfolds the expected nonconforming fraction (0.17% → 1.25%) even though the process is still well inside spec on average, illustrating why early detection of small mean shifts (the subject of part (c)) matters economically.

(b) $C_p$, $C_{pk}$, and 99%/95% natural tolerance limits

Given. $\hat\mu_0=175.6$ lb, $\hat\sigma=11.15$ lb (from Question 2(d), individuals-chart estimate), sample standard deviation of the raw 20 weights $s=11.99$ lb, $N=20$, $USL=210$, $LSL=140$. Required tolerance-limit coverage: 99% of the population between limits, at 95% confidence.

Find. $C_p$, $C_{pk}$; the two-sided natural tolerance limits; and what they mean for this process.

Capability indexes. $$C_p=\frac{USL-LSL}{6\hat\sigma}=\frac{210-140}{6(11.15)}=\boxed{1.046}.$$ $$C_{pu}=\frac{USL-\mu_0}{3\hat\sigma}=\frac{210-175.6}{3(11.15)}=1.028,\qquad C_{pl}=\frac{\mu_0-LSL}{3\hat\sigma}=\frac{175.6-140}{3(11.15)}=1.064.$$ $$C_{pk}=\min(C_{pu},C_{pl})=\boxed{1.028}.$$
Natural tolerance limits (Appendix VII, $N=20$). A two-sided normal tolerance interval is $\bar X\pm Ks$, where $K$ is read from the tolerance-factor table at $N=20$, 95% confidence, 99% population coverage: $K=3.615$. Using the sample standard deviation of the 20 individual weights, $s=11.99$ lb, $$TL=\bar X\pm Ks=175.6\pm 3.615(11.99)=175.6\pm43.35=\boxed{[132.25,\ 218.95]\ \text{lb}}.$$ (Cross-check: the closed-form $K=Z_{0.995}\sqrt{(N-1)/\chi^2_{0.05,19}}=2.576\sqrt{19/10.12}=3.53$ agrees with the tabulated value to within table-rounding.)
Interpretation. We can state with 95% confidence that at least 99% of the individual output of this process falls between 132.25 lb and 218.95 lb — a statement about the SPREAD of individual product, in contrast to a confidence interval, which would describe uncertainty about the unknown MEAN. Comparing to the specification band $[140,\,210]$: the tolerance interval extends slightly outside the spec on both sides, consistent with $C_p=1.046$ being only marginally above the minimum-acceptable value of 1.0 — the process is barely capable when centred, and $C_{pk}=1.028

Check: the tolerance-limit factor is read directly from this paper's own Appendix VII (two-sided normal tolerance limits, $N=20$ row) rather than derived from an assumed table lookup elsewhere — using the paper's own attached table is the intended method for this question.

Quantity	Value
$C_p$	1.046
$C_{pu}$, $C_{pl}$	1.028, 1.064
$C_{pk}$	1.028
$K$ (Appendix VII, $N=20$, 95%/99%)	3.615
Natural tolerance limits	[132.25, 218.95] lb

(c) Minimum $\bar X$-chart sample size to detect the shift within two samples

Given. $\mu_0=175.6$ lb, $\hat\sigma=11.15$ lb (in-control estimates), shift target $\mu_1=185$ lb, required $P(\text{detect on sample 1 or 2})>0.7$.

Find. The minimum future $\bar X$-chart sample size $n$, and the resulting control limits.

Approach. Express the standardized shift as $\delta=|\mu_1-\mu_0|/\hat\sigma$; a single sample's detection probability after the shift is $p=1-\beta=P(Z>3-\delta\sqrt n)+P(Z<-3-\delta\sqrt n)$; require $1-(1-p)^2>0.7$ (miss on both of the first two samples has probability under 0.3), and search over integer $n$.

Standardized shift. $\delta=\dfrac{|185-175.6|}{11.15}=0.843$.
Search over $n$. For each $n$, $p(n)=P(Z>3-0.843\sqrt n)$ (the lower-tail term is negligible here) and the two-sample detection probability is $1-(1-p(n))^2$: $$n=11:\ p=0.419,\ 1-(1-p)^2=0.663<0.7\ (\text{fails}).$$ $$n=12:\ p=0.468,\ 1-(1-p)^2=\boxed{0.717>0.7}\ (\text{passes}).$$ So the minimum sample size is $\boxed{n=12}$.
Control limits for the designed chart ($n=12$). $$UCL_{\bar X}=\mu_0+\frac{3\hat\sigma}{\sqrt n}=175.6+\frac{3(11.15)}{\sqrt{12}}=\boxed{185.26\ \text{lb}},\qquad LCL_{\bar X}=175.6-\frac{3(11.15)}{\sqrt{12}}=\boxed{165.94\ \text{lb}}.$$

Quantity	Value
Standardized shift $\delta$	0.843
Minimum sample size $n$	12
$P(\text{detect by 2nd sample})$ at $n=12$	0.717
$UCL_{\bar X},\ LCL_{\bar X}$ ($n=12$)	185.26 lb, 165.94 lb

Concept & Theory

Topic: Process capability indexes, statistical tolerance intervals, and $\bar X$-chart sample-size design for a target detection speed.

Key relations: $C_p=(USL-LSL)/6\sigma$, $C_{pk}=\min(C_{pu},C_{pl})$; two-sided normal tolerance interval $\bar X\pm Ks$; OC-function-based detection probability $1-\beta=\Phi(3-\delta\sqrt n)$ combined across consecutive samples as $1-(1-p)^k$.

Why this works: $C_p$ measures potential capability (spread vs. spec, ignoring centring) while $C_{pk}$ penalizes off-centre processes; the gap $C_p-C_{pk}$ is a direct measure of how much capability is being lost purely to a poorly centred mean. The two-sample detection-probability search is the same OC-curve machinery used for single-sample ARL design, just composed over 2 independent trials via $1-(1-p)^2$.

Common pitfall: Confusing a confidence interval for the mean with a tolerance interval for individual values — the tolerance interval in part (b) is deliberately wider because it must bound 99% of individual products, not just locate the unknown mean.

Source: Montgomery, Introduction to Statistical Quality Control (8th ed.), Ch. 8 (process capability, natural tolerance limits), Ch. 5 (OC curves and $\bar X$-chart sample-size design).

Question 3 of 6: Process Capability, Natural Tolerance Limits, and Detection-Speed Chart Design

Question 3: Process Capability, Natural Tolerance Limits, and Detection-Speed Chart Design (20 marks)

(a) Proportion nonconforming, in control and after a mean shift

(b) $C_p$, $C_{pk}$, and 99%/95% natural tolerance limits

(c) Minimum $\bar X$-chart sample size to detect the shift within two samples