Question 2 of 6: Traditional vs. Special Control Charts, X̄/R Chart for Part Weight, and Chart Design for Two ARL Targets

Question 2: Traditional vs. Special Control Charts, X̄/R Chart for Part Weight, and Chart Design for Two ARL Targets (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) Traditional vs. special (EWMA/CUSUM) control charts

The traditional Shewhart chart ($\bar X$, $R$, $S$, $p$, $c$, $u$) bases its decision entirely on the CURRENT sample: a point signals only if it falls outside the $\pm3\sigma$ limits, so the chart has no memory of prior points. It is simple to construct and interpret, and it is very effective at detecting LARGE shifts (roughly $\ge1.5\sigma$) quickly — e.g. an $\bar X$ chart catching a tool breakage that suddenly shifts a dimension by $2\sigma$. It is comparatively slow, however, at detecting SMALL, sustained shifts (e.g. $0.5$–$1\sigma$), because a small shift only modestly increases the chance any single point falls outside the wide 3-sigma band. EWMA (Exponentially Weighted Moving Average) and CUSUM (Cumulative Sum) charts instead plot a statistic that accumulates information across MANY past samples — EWMA as a geometrically-weighted running average $z_i=\lambda x_i+(1-\lambda)z_{i-1}$, CUSUM as a running sum of deviations from target, $C_i^+=\max[0,\,x_i-(\mu_0+K)+C_{i-1}^+]$ (and the mirror-image $C_i^-$) — so a persistent small shift steadily builds up evidence across samples rather than relying on any one sample being extreme enough to cross a fixed limit. A typical EWMA/CUSUM application is monitoring a chemical process's concentration or a machining dimension where slow tool wear or catalyst depletion causes a small, gradual drift that a Shewhart chart would miss for many samples; a typical Shewhart application is a high-volume assembly line where an abrupt, large-magnitude assignable cause (broken tool, wrong material lot) is the dominant failure mode.

EWMA/CUSUM are more sensitive to small mean shifts because their plotted statistic is a WEIGHTED ACCUMULATION of the shift's effect across many samples rather than a single-sample average: even if one sample's deviation from target is too small to trigger a Shewhart 3-sigma signal by itself, that same small deviation, repeated shift after shift, steadily pushes the EWMA statistic or the CUSUM running total toward its own control limit, so the SIGNAL-TO-NOISE of a persistent shift accumulates over time while pure sampling noise (independent, zero-mean deviations) tends to average out. This is exactly analogous to why a larger sample size $n$ makes an $\bar X$ chart more sensitive (averaging reduces noise) — EWMA/CUSUM instead average/accumulate ACROSS TIME rather than across a single sample's units.

The V-mask is a graphical CUSUM decision procedure: a V-shaped template (parameters lead distance $d$ and half-angle $\theta$, related to the CUSUM's reference value $k$ and decision interval $h$) is overlaid on the plotted cumulative-sum chart with its vertex a fixed lead distance ahead of the most recent point; an out-of-control signal occurs if any earlier plotted point falls outside the two arms of the V. Its main limitations are: it is graphically cumbersome and error-prone to apply consistently by hand or by different operators; it does not naturally give a numerical estimate of the current cumulative sum's distance-to-signal (harder to build automated alarms/software rules around); and, most importantly, its ARL performance is not as easily or precisely calibrated as the tabular form. The tabular CUSUM instead computes the two one-sided cumulative sums $C_i^+$ and $C_i^-$ numerically each period and signals when either exceeds a numeric decision interval $H$ (typically $H=5\sigma$) — it is mathematically equivalent to the V-mask (the same reference value $k$ and decision interval, related by $h=H$ and $d=H/\tan\theta$), but is far easier to compute, automate, tabulate, and to design/verify precisely against a target ARL, which is why the tabular form is standard in modern SPC software while the V-mask survives mainly as a teaching/visualization device.

(b) X̄/R chart: control limits, revision, and in-control process estimates

Given. Injection-molding part weight, $n=4$ parts/sample, $m=15$ shifts (samples). From Appendix VI (this paper) for $n=4$: $A_2=0.729$, $D_3=0$, $D_4=2.282$, $d_2=2.059$.

Find. Trial and (if needed) revised $\bar X$- and $R$-chart control limits; the in-control process mean $\hat\mu$ and standard deviation $\hat\sigma$.

1. Trial centre lines and $R$-chart limits. $\bar{\bar X}=\sum\bar X_i/15=2033.93$ g, $\bar R=\sum R_i/15=17.60$ g. $$UCL_R=D_4\bar R=2.282(17.60)=\boxed{40.16\ \text{g}},\qquad LCL_R=D_3\bar R=\boxed{0\ \text{g}}.$$ Every $R_i$ (max $=38$, sample 12) is below $UCL_R=40.16$, so the R chart needs no revision — the process's short-term variability is stable across all 15 shifts, and $\bar R=17.60$ g is used unchanged for every subsequent step.
2. Trial $\bar X$-chart limits. $$UCL_{\bar X}=\bar{\bar X}+A_2\bar R=2033.93+0.729(17.60)=\boxed{2046.76\ \text{g}},\qquad LCL_{\bar X}=2033.93-0.729(17.60)=\boxed{2021.10\ \text{g}}.$$ Checking each $\bar X_i$ against these limits: sample 2 ($2008<2021.10$), sample 4 ($2058>2046.76$), and sample 5 ($2047>2046.76$) all plot outside — three assignable-cause shifts in the process AVERAGE (not the spread, since the R chart was already clean).
3. Revise the $\bar X$ centre line. Drop samples 2, 4, 5 and recompute the grand average from the remaining 12 in-control samples: $$\bar{\bar X}_{\text{rev}}=\frac{2024+2030+2025+2041+2025+2042+2030+2028+2038+2040+2035+2038}{12}=\boxed{2033.00\ \text{g}}.$$ $\bar R$ is unchanged at $17.60$ g since the $R$ chart itself required no revision.
4. Revised $\bar X$-chart limits. $$UCL_{\bar X,\text{rev}}=2033.00+0.729(17.60)=\boxed{2045.83\ \text{g}},\qquad LCL_{\bar X,\text{rev}}=2033.00-0.729(17.60)=\boxed{2020.17\ \text{g}}.$$ All 12 remaining averages (range 2024–2042 g) fall comfortably inside these revised limits, so no further revision is needed.
5. In-control process mean and standard deviation. The revised grand average estimates the mean directly; $\bar R/d_2$ (using the un-revised, already-clean $\bar R$) estimates $\sigma$: $$\hat\mu=\bar{\bar X}_{\text{rev}}=\boxed{2033.00\ \text{g}},\qquad \hat\sigma=\frac{\bar R}{d_2}=\frac{17.60}{2.059}=\boxed{8.548\ \text{g}}.$$

(c) Designing the future-production $\bar X$ chart: minimum $n$ for two ARL targets

Given. $\hat\mu_0=2033.00$ g, $\hat\sigma=8.548$ g from part (b); 3-sigma limits ($L=3$); requirement 1: $P(RL\ge100)\ge0.6$ for any shift up to $|\mu_1-\mu_0|\le0.2\sigma$; requirement 2: $ARL_{\mu_2}\le8$ at $\mu_2=\mu_0+1.2\sigma$.

Find. The minimum sample size $n$ satisfying both requirements, and the resulting control limits.

1. Express the "no-signal" probability $\beta$ as a function of shift size $\delta=|\mu_1-\mu_0|/\sigma$ and $n$. With 3-sigma limits, a single sample fails to signal with probability $$\beta(\delta,n)=\Phi(3-\delta\sqrt n)-\Phi(-3-\delta\sqrt n)\approx\Phi(3-\delta\sqrt n)$$ (the second term is negligible for every $n$ tried below). Requirement 1 uses the WORST case within the stated band, $\delta=0.2$ (the largest shift the process is allowed to have and still need $P(RL\ge100)\ge0.6$, since larger $\delta$ gives smaller $\beta$ and is therefore the binding case); requirement 2 uses $\delta=1.2$.
2. Requirement 1 as a bound on $n$. Run length is geometric with per-sample no-signal probability $\beta$, so $P(RL\ge100)=\beta^{99}$. Requiring $\beta(0.2,n)^{99}\ge0.6$ gives $\beta(0.2,n)\ge0.6^{1/99}=0.99485$. Evaluating $\beta(0.2,n)$ for successive $n$: $$n=4:\ \beta=0.99500,\ P(RL\ge100)=0.99500^{99}=0.609\ (\ge0.6,\ \text{passes}),$$ $$n=5:\ \beta=0.99437,\ P(RL\ge100)=0.99437^{99}=0.572\ (<0.6,\ \text{fails}).$$ So requirement 1 by itself limits the design to $\boxed{n\le4}$.
3. Requirement 2 as a bound on $n$. $ARL_{\mu_2}=1/(1-\beta(1.2,n))\le8\iff\beta(1.2,n)\le7/8=0.875$. Evaluating: $$n=2:\ \beta=0.9037,\ ARL=10.38\ (>8,\ \text{fails}),$$ $$n=3:\ \beta=0.8216,\ ARL=1/(1-0.8216)=5.61\ (\le8,\ \text{passes}).$$ So requirement 2 by itself needs $\boxed{n\ge3}$.
4. Combine. Both requirements must hold simultaneously, so $3\le n\le4$. The minimum (most economical) sample size meeting BOTH is $$\boxed{n=3}\quad(\beta(0.2,3)=0.9956,\ P(RL\ge100)=0.647\ge0.6;\ \ \beta(1.2,3)=0.8216,\ ARL_{\mu_2}=5.61\le8).$$ $n=4$ also satisfies both requirements, with a larger margin on requirement 2 ($ARL_{\mu_2}=3.65$) at the cost of one extra unit per sample; $n=3$ is adopted as the minimum sample size that meets the stated design targets.
5. Resulting control limits at $n=3$. $$UCL=\hat\mu_0+\frac{3\hat\sigma}{\sqrt3}=2033.00+\frac{3(8.548)}{\sqrt3}=\boxed{2047.80\ \text{g}},\qquad LCL=2033.00-\frac{3(8.548)}{\sqrt3}=\boxed{2018.20\ \text{g}}.$$