Question 3 of 6: X̄/S Charts for Sheet-Metal Thickness and Detection-Speed Chart Design

Question 3: X̄/S Charts for Sheet-Metal Thickness 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) Effect of sample size on the $\bar X$ chart

The $\bar X$ chart's control limits are $\mu_0\pm3\sigma/\sqrt n$: as the sample size $n$ increases, the standard error of the plotted statistic $\sigma/\sqrt n$ shrinks, so the limits pull in TOWARD the center line even though the process's own inherent variability $\sigma$ is unchanged. A tighter band around the centre line means a smaller shift in the true mean is needed to push the sample average outside the limits, which is exactly why the chart becomes more sensitive as $n$ grows — the $\bar X$ chart never reacts to the raw measurement's spread, only to the spread of the SAMPLE AVERAGE, and averaging over more units always reduces that spread by the familiar $1/\sqrt n$ factor. The OC (operating-characteristic) function, $\beta=P(\text{no signal}\mid\text{shift }k\sigma)=\Phi(3-k\sqrt n)-\Phi(-3-k\sqrt n)$, falls faster (steeper OC curve) as $n$ increases for any fixed shift size $k\sigma$, because the $k\sqrt n$ term in the argument grows with $n$ — the probability of missing a given-size shift on any one sample drops. The ARL (average run length), $ARL=1/(1-\beta)$ off-target and $1/\alpha$ on-target, falls (fewer samples needed, on average, to detect a real shift) as $n$ increases, again because $\beta$ falls; the in-control ARL stays fixed at $1/\alpha=1/0.0027\approx370$ for standard 3-sigma limits since $\alpha$ does not depend on $n$. In short: larger samples do not change how quickly the process itself moves, but they average out sampling noise faster, so a genuine shift stands out against a narrower "normal" band sooner — this is the single mechanism behind all three effects (limits, OC curve, ARL).

(b) $\bar X$/S chart limits, process estimates, and proportion conforming

Given. Sheet-metal thickness, $n=4$ per sample, $m=20$ samples: $\sum\bar X_i=200$, $\sum S_i=1.635$. Specification $9.95\pm0.3$ mm, i.e. $LSL=9.65$, $USL=10.25$ mm. For $n=4$: $A_3=1.628$, $B_3=0$, $B_4=2.266$, $c_4=0.9213$ (Appendix VI, this paper).

Find. $\bar X$- and $S$-chart control limits; in-control $\hat\mu$, $\hat\sigma$; estimated proportion conforming; and the mean location that maximizes that proportion.

1. Centre lines from the 20 samples. $\bar{\bar X}=\sum\bar X_i/m=200/20=10.000$ mm; $\bar S=\sum S_i/m=1.635/20=0.08175$ mm.
2. $\bar X$-chart limits ($n=4$, $A_3=1.628$): $$UCL_{\bar X}=\bar{\bar X}+A_3\bar S=10.000+1.628(0.08175)=\boxed{10.1331\ \text{mm}},\quad LCL_{\bar X}=10.000-1.628(0.08175)=\boxed{9.8669\ \text{mm}}.$$
3. $S$-chart limits ($B_3=0$, $B_4=2.266$): $$UCL_S=B_4\bar S=2.266(0.08175)=\boxed{0.1852\ \text{mm}},\qquad LCL_S=B_3\bar S=\boxed{0\ \text{mm}}.$$
4. In-control process mean and standard deviation. $\hat\mu=\bar{\bar X}=10.000$ mm. $\bar S$ is a biased estimator of $\sigma$ (bias factor $c_4$), so $$\hat\sigma=\bar S/c_4=0.08175/0.9213=\boxed{0.08873\ \text{mm}}.$$
5. Proportion meeting specification. Standardizing the spec limits against $\hat\mu,\hat\sigma$: $$Z_{USL}=\frac{10.25-10.000}{0.08873}=2.817,\qquad Z_{LSL}=\frac{9.65-10.000}{0.08873}=-3.944.$$ From Appendix II, $\Phi(2.82)\approx0.99760$ and $\Phi(-3.94)\approx0.00004$, so $$P(\text{conforming})=\Phi(Z_{USL})-\Phi(Z_{LSL})=0.99760-0.00004=\boxed{0.9975\ (99.75\%)}.$$
6. Optimal mean location. With $\sigma$ fixed, the conforming proportion for a symmetric two-sided spec is maximized by centring the process exactly midway between $LSL$ and $USL$ (this splits the fixed tolerance band symmetrically about the normal distribution's own axis of symmetry, giving the smallest possible combined tail area for that $\sigma$): $$\mu^\ast=\frac{LSL+USL}{2}=\frac{9.65+10.25}{2}=\boxed{9.95\ \text{mm}}\ (\text{the nominal target}).$$ The current estimate $\hat\mu=10.000$ is $0.05$ mm above this optimum, which is why the upper tail ($Z_{USL}=2.82$) is tighter than the lower tail ($Z_{LSL}=-3.94$) above — a small recentring toward $9.95$ would further reduce the (already small) nonconforming fraction.

(c) Minimum sample size for a 1.4σ shift, $P(RL\ge5)\le0.2$

Given. $\hat\mu_0=10.000$, $\hat\sigma=0.08873$ mm from part (b); shift of interest $\mu_0-1.4\sigma$; requirement $P(\text{run length}\ge5)\le0.2$ after the shift.

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

1. Translate the run-length requirement into a per-sample detection probability. "Run length $\ge5$" means the first four post-shift samples all fail to signal. If $p$ is the probability that a single post-shift sample signals (falls outside the 3-sigma limits), then $P(RL\ge5)=(1-p)^4$, so the requirement is $$(1-p)^4\le0.2\ \Longrightarrow\ 1-p\le0.2^{1/4}=0.6687\ \Longrightarrow\ p\ge0.3313.$$
2. Express $p$ as a function of $n$. After a downward shift of $1.4\sigma$, $\bar X\sim N(\mu_0-1.4\sigma,\ \sigma^2/n)$. The chance of signalling low dominates (the chance of signalling high is negligible for a downward shift with 3-sigma limits): $$p(n)\approx P\!\left(\bar X<\mu_0-3\sigma/\sqrt n\right)=\Phi\!\left(\frac{\mu_0-3\sigma/\sqrt n-(\mu_0-1.4\sigma)}{\sigma/\sqrt n}\right)=\Phi\!\left(-3+1.4\sqrt n\right).$$
3. Solve for the minimum integer $n$ with $p(n)\ge0.3313$. Trying successive integers: $$n=3:\ p=\Phi(-0.575)=0.283\ (\text{fails, }P(RL\ge5)=0.265>0.2),$$ $$n=4:\ p=\Phi(-0.200)=0.421\ (\text{passes, }P(RL\ge5)=(1-0.421)^4=0.113\le0.2).$$ So the minimum sample size is $\boxed{n=4}$.
4. Control limits at $n=4$. Using $\hat\mu_0=10.000$, $\hat\sigma=0.08873$: $$UCL=10.000+\frac{3(0.08873)}{\sqrt4}=\boxed{10.1331\ \text{mm}},\qquad LCL=10.000-\frac{3(0.08873)}{\sqrt4}=\boxed{9.8669\ \text{mm}}.$$ These match the limits already in use from part (b) — the sample size already adopted for routine monitoring happens to be exactly the minimum needed to meet this detection-speed requirement, so no change to the sampling scheme is required.