Process Capability Indices, Tolerance Limits, and a Hypothesis Test on C p: 23-Ind-A5 Quality Planning, Control, and Assurance, May 2014, Q4

Question 4: Process Capability Indices, Tolerance Limits, and a Hypothesis Test on C_p (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) $C_p$, $C_{pk}$, and natural tolerance limits

Given. $USL=10.25$, $LSL=9.65$ mm; $N=200$, $\bar X=9.98$, $S=0.085$ mm; $\alpha=0.01$, confidence level $\gamma=0.95$.

Find. $C_p$, $C_{pk}$, and the two-sided natural tolerance limits at the stated $\alpha$/$\gamma$; comment on capability.

Potential capability, $C_p$. With $N=200$ large, $S$ is used directly as $\hat\sigma$: $$C_p=\frac{USL-LSL}{6S}=\frac{10.25-9.65}{6(0.085)}=\frac{0.60}{0.51}=\boxed{1.176}.$$
Actual capability, $C_{pk}$. The mean $\bar X=9.98$ sits above the spec midpoint (9.95), so the upper side is tighter: $$C_{pu}=\frac{USL-\bar X}{3S}=\frac{0.27}{0.255}=1.059,\qquad C_{pl}=\frac{\bar X-LSL}{3S}=\frac{0.33}{0.255}=1.294,$$ $$C_{pk}=\min(C_{pu},C_{pl})=\boxed{1.059}.$$
Natural tolerance limits (accounting for sampling uncertainty in $S$). A two-sided tolerance interval covering a proportion $1-\alpha$ of the population with confidence $\gamma$ is $\bar X\pm K S$, with the tolerance factor $K=Z_{1-\alpha/2}\sqrt{(N-1)/\chi^2_{1-\gamma,\,N-1}}$ (the lower-tail $\chi^2$ percentile widens $K$ to cover the chance that this particular sample's $S$ under-estimates $\sigma$). With $Z_{0.995}=2.576$ and $\chi^2_{0.05,199}=167.4$: $$K=2.576\sqrt{199/167.4}=\boxed{2.809}.$$ $$\text{Natural tolerance limits}=9.98\pm2.809(0.085)=9.98\pm0.239=\boxed{[9.741,\ 10.219]\ \text{mm}}.$$
Comment on capability. The natural tolerance limits $[9.741, 10.219]$ lie entirely inside the specification band $[9.65, 10.25]$, consistent with $C_p=1.176>1$ (the process's inherent spread fits within the tolerance) — but $C_{pk}=1.059$ is noticeably lower than $C_p$ because the process mean sits $0.03$ mm above the nominal target of $9.95$, toward the USL. The process is capable in a basic sense ($C_p>1$) but only marginally so on the $C_{pk}$ criterion most practitioners require ($C_{pk}\ge1.33$ for a "capable" process); a small recentring toward $9.95$ mm (the same conclusion reached in Question 3(b)) would raise $C_{pk}$ toward $C_p$ without any reduction in variability.

Quantity	Result
$C_p$	1.176
$C_{pk}$	1.059 (limiting side: upper)
Natural tolerance limits ($\alpha=0.01$, $\gamma=0.95$)	[9.741, 10.219] mm

Check: the paper does not supply a tolerance-factor ($K$) table, so $K$ is computed here from the standard closed-form large-sample tolerance-interval formula $K=Z_{1-\alpha/2}\sqrt{(N-1)/\chi^2_{1-\gamma,N-1}}$ (the same chi-square machinery Q4(b) uses for the $C_p$ hypothesis test). At $N=200$ this is close to the even-simpler large-$N$ approximation $K\to Z_{1-\alpha/2}=2.576$ (limits $[9.761,10.199]$) — both readings support the same "capable but off-target" conclusion.

(b) Hypothesis test of the claim $C_p>1.05$

Given. $\hat C_p=1.176$ (part a), $N=200$, claimed threshold $C_{p0}=1.05$, $\alpha=0.05$; approximate formula $\chi^2_{\alpha,m}\doteq m\big(1-\tfrac{2}{9m}+Z_\alpha\sqrt{\tfrac{2}{9m}}\big)^3$.

Find. Test $H_0:C_p=1.05$ vs. $H_1:C_p>1.05$ at $\alpha=0.05$; state whether the claim is accepted.

Test statistic. Since $(N-1)S^2/\sigma^2\sim\chi^2_{N-1}$, and $\hat C_p/C_{p0}=\sigma_0/S$ (where $\sigma_0=(USL-LSL)/6C_{p0}$ is the standard deviation implied by the claimed $C_{p0}$), it follows that under $H_0$ $$\chi^2_0=(N-1)\left(\frac{\hat C_p}{C_{p0}}\right)^2\sim\chi^2_{N-1}.$$ A larger true $C_p$ inflates $\hat C_p$ (smaller $S$ than the claim implies), pushing $\chi^2_0$ above a typical $\chi^2_{N-1}$ value — so large $\chi^2_0$ supports $H_1$. With $m=N-1=199$: $$\chi^2_0=199\left(\frac{1.176}{1.05}\right)^2=199(1.2549)=\boxed{249.8}.$$
Critical value. Using the given approximation with $Z_{0.95}=1.645$, $m=199$: $$\chi^2_{0.95,199}\doteq199\left(1-\frac{2}{9(199)}+1.645\sqrt{\frac{2}{9(199)}}\right)^{3}=199(1.0539)^3=\boxed{232.9}.$$
Decision. Since $\chi^2_0=249.8>\chi^2_{0.95,199}=232.9$, reject $H_0$ in favour of $H_1:C_p>1.05$ at the $5\%$ significance level — the data provide statistically significant evidence supporting the claim. $$\boxed{\text{Accept the claim: }C_p>1.05\text{ (}\alpha=0.05\text{)}.}$$

Quantity	Result
Test statistic $\chi^2_0$	249.8
Critical value $\chi^2_{0.95,199}$	232.9
Decision	Reject $H_0$ — accept the claim $C_p>1.05$

(c) Power of the test when the true $C_p=1.2$

Given. Rejection region $\chi^2_0>232.9$ from part (b); true $C_p=1.2$.

Find. $P(\text{reject }H_0\mid C_p=1.2)$.

Rescale the rejection region. Under a true value $C_p$, $\chi^2_0=(N-1)(\hat C_p/C_{p0})^2=W\cdot(C_p/C_{p0})^2$ where $W\sim\chi^2_{199}$ exactly (the genuine sampling chi-square of $S^2$). Rejecting when $\chi^2_0>232.9$ is therefore rejecting when $$W>\frac{232.9}{(C_p/C_{p0})^2}=\frac{232.9}{(1.2/1.05)^2}=\frac{232.9}{1.3061}=\boxed{178.3}.$$
Power. $$\text{Power}=P(W>178.3),\ W\sim\chi^2_{199}=\boxed{0.851\ (85.1\%)}.$$

Quantity	Result
Power at $C_p=1.2$	0.851 (85.1%)

Check: the power is evaluated numerically from the $\chi^2_{199}$ distribution (matching the exam's own Wilson–Hilferty approximation to within 0.01 of the critical value used); an $85\%$ probability of correctly detecting a true $C_p=1.2$ against a $C_{p0}=1.05$ claim is a reasonable, moderate-power result for $N=200$.

Concept & Theory

Topic. Process capability indices $C_p$/$C_{pk}$, statistical tolerance intervals, and a chi-square-based hypothesis test (and its power) on $C_p$.

Key relations: $C_p=(USL-LSL)/6\sigma$; $C_{pk}=\min[(USL-\mu)/3\sigma,(\mu-LSL)/3\sigma]$; $(N-1)S^2/\sigma^2\sim\chi^2_{N-1}$; test statistic $\chi^2_0=(N-1)(\hat C_p/C_{p0})^2$; power $=P\big(\chi^2_{N-1}>\chi^2_{crit}/(C_{p,true}/C_{p0})^2\big)$.

Why this works. $C_p$ is a simple deterministic rescaling of $\sigma$ (equivalently of $S$), so the entire inferential machinery for $C_p$ — confidence/tolerance intervals in part (a), the hypothesis test in part (b), and the power calculation in part (c) — is inherited directly from the well-known sampling distribution of $S^2$, namely $\chi^2_{N-1}$; nothing new needs to be derived once that substitution is recognized.

Common pitfall. Testing $C_p \gt C_{p0}$ using a two-sided or lower-tail rejection region (the correct region is upper-tail, since a larger $C_p$ makes $\hat C_p$, and hence $\chi^2_0$, larger); and forgetting that the natural tolerance limits use $S$ directly (uncorrected by $c_4$) because they describe individual measurements from a large sample, not a subgroup-averaged control-chart statistic.

Source. Montgomery, Introduction to Statistical Quality Control, Ch. 8 (process capability analysis, confidence intervals and hypothesis tests on $C_p$/$C_{pk}$).

Question 4 of 6: Process Capability Indices, Tolerance Limits, and a Hypothesis Test on C p

Question 4: Process Capability Indices, Tolerance Limits, and a Hypothesis Test on Cp (20 marks)

(a) $C_p$, $C_{pk}$, and natural tolerance limits

(b) Hypothesis test of the claim $C_p>1.05$

(c) Power of the test when the true $C_p=1.2$

Question 4: Process Capability Indices, Tolerance Limits, and a Hypothesis Test on C_p (20 marks)