Question 2 of 6: SPC Tools, an $\bar X$-$R$ Chart for Injection-Molded Part Weight, and Chart Design for Two Simultaneous ARL Targets

Question 2: SPC Tools, an $\bar X$-$R$ Chart for Injection-Molded Part Weight, and Chart Design for Two Simultaneous 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) SPC tools and process variability; why attributes need one chart and variables need two

SPC tools (control charts together with the other "magnificent seven" — Pareto diagrams, cause-and-effect diagrams, check sheets, histograms, scatter diagrams and defect-concentration diagrams) reduce process variability and instability by separating common-cause variation, which is the process's own stable, irreducible spread, from special-cause variation, which is an assignable, removable disturbance. A control chart makes that separation visible in real time: every point beyond the control limits, or every non-random pattern flagged by the zone rules, is a signal to investigate and permanently remove an assignable cause. Repeating this find-and-remove cycle steadily strips special causes out of the process, so both its instability (fewer shifts, trends and drifts) and, because each removed special cause was itself adding variance, its underlying variability fall over time — the process converges toward its natural, common-cause-only capability.

An attribute chart needs only one chart because the discrete distribution behind it (binomial for $p$/$np$, Poisson for $c$/$u$) has a single parameter that fixes both its mean and its variance at once — e.g. for the binomial, mean $=np$ and variance $=np(1-p)$ are both determined by $p$ alone. Monitoring the mean (the proportion or count) therefore automatically monitors the spread; there is nothing independent left for a second chart to track. A variable (a continuous measurement) has its location (mean) and its spread (variance) as two separately estimated parameters of, typically, a normal distribution — the process can shift in mean while its spread stays fixed, or vice versa — so two charts are required: one to track central tendency ($\bar X$, or individuals for $n=1$) and one to track dispersion ($R$, $S$, or moving range).

Between $R$ and $S$: the range uses only the largest and smallest of the $n$ sample values, ignoring everything in between, so its efficiency as an estimator of $\sigma$ falls off quickly as $n$ grows past about 10; $S$ (the sample standard deviation) uses every observation and remains a statistically efficient estimator at any $n$, including when $n$ varies from sample to sample (where $R$'s $d_2$-based conversion becomes awkward). $R$'s only real advantage is that it is trivial to compute by hand, which is why it was historically preferred for small, constant-$n$, shop-floor charts; once data collection and charting are automated, that advantage disappears and $S$ — being the more efficient, more generally applicable estimator — is the preferable chart.

(b) $\bar X$-$R$ chart for injection-molded part weight, $n=5$, 15 shifts

Given. Fifteen consecutive shift samples of $n=5$ parts each; the sample average and range are recorded per shift:

For $n=5$: $A_2=0.577$, $D_3=0$, $D_4=2.114$, $d_2=2.326$.

Find. Trial and (if needed) revised control limits for both charts, and the resulting in-control process mean $\hat\mu_0$ and standard deviation $\hat\sigma_0$.

Approach. Standard two-stage procedure: establish the $R$-chart's own trial limits first and remove any assignable-cause range before the $\bar X$ limits are computed from it (an inflated $R$ would otherwise widen the $\bar X$ limits and mask a real mean shift), then trial the $\bar X$ chart on the $R$-clean data and revise it in turn.

1. $R$-chart trial limits. $\bar R=\dfrac{\sum R_i}{15}=\dfrac{265}{15}=\boxed{17.60\ \text{g}}$, so $UCL_R=D_4\bar R=2.114(17.60)=\boxed{37.21\ \text{g}}$ (LCL$_R=D_3\bar R=0$). Shift 12 ($R=38$) exceeds $UCL_R$ and is removed as an assignable-cause point.
2. $R$-chart revision. On the remaining 14 shifts, $\bar R_{rev}=\dfrac{265-38}{14}=16.14\ \text{g}$, giving $UCL_{R,rev}=2.114(16.14)=34.13\ \text{g}$. All 14 remaining ranges are now inside this limit — the $R$-chart needs no further revision.
3. $\bar X$-chart trial limits (on the 14 $R$-clean shifts). $\bar{\bar x}=\dfrac{\sum_{i\ne12}\bar x_i}{14}=2033.64\ \text{g}$, so $$UCL_{\bar X}=\bar{\bar x}+A_2\bar R_{rev}=2033.64+0.577(16.14)=\boxed{2042.96\ \text{g}},\qquad LCL_{\bar X}=2033.64-0.577(16.14)=\boxed{2024.33\ \text{g}}.$$ Four shifts fall outside these limits: shift 2 (2008, below LCL), shift 4 (2058, above UCL), shift 5 (2047, above UCL), and shift 13 (2024, below LCL).
4. $\bar X$-chart revision. Removing shifts 2, 4, 5 and 13 leaves 10 shifts (1, 3, 6, 7, 8, 9, 10, 11, 14, 15): $$\bar{\bar x}_{rev}=2033.40\ \text{g},\qquad \bar R_{rev,2}=16.70\ \text{g},$$ $$UCL_{\bar X,rev}=2033.40+0.577(16.70)=\boxed{2043.04\ \text{g}},\qquad LCL_{\bar X,rev}=2033.40-0.577(16.70)=\boxed{2023.76\ \text{g}}.$$ All 10 remaining shift averages now fall inside these limits, and re-checking the $R$-chart limit on this same final set ($UCL_R=2.114(16.70)=35.30\ \text{g}$) still clears all 10 ranges — no further revision is needed on either chart.
5. In-control process mean and standard deviation. Using the final 10-shift set, $$\hat\mu_0=\bar{\bar x}_{rev}=\boxed{2033.4\ \text{g}},\qquad \hat\sigma_0=\frac{\bar R_{rev,2}}{d_2}=\frac{16.70}{2.326}=\boxed{7.18\ \text{g}}.$$

(c) Designing the $\bar X$ chart for two simultaneous ARL/$\beta$ targets

Given. 3-sigma $\bar X$ limits ($k=3$); require $ARL_{\mu_1}\ge100$ for any shift with $|\mu_1-\mu_0|\le0.25\sigma$, and simultaneously $\beta(\mu_2)\le0.3$ for any shift with $|\mu_2-\mu_0|>1.2\sigma$; the only free design variable is the sample size $n$.

Find. The sample size $n$ that satisfies both requirements.

Approach. For 3-sigma limits, the probability that one sample signals when the true shift is $\delta\sigma$ is $p(\delta,n)=1-\big[\Phi(3-\delta\sqrt n)-\Phi(-3-\delta\sqrt n)\big]$, with $ARL=1/p$ and $\beta=1-p$. Because $p$ is monotonically increasing in $n$ for fixed $\delta$, the small-shift requirement ($ARL\ge100$ at $\delta=0.25$, i.e. $p\le0.01$ at the worst case $\delta=0.25$) bounds $n$ from above, while the large-shift requirement ($\beta\le0.3$ at $\delta=1.2$, i.e. $p\ge0.7$ at the worst case $\delta=1.2$) bounds $n$ from below — the two pull in opposite directions, since a bigger sample makes the chart more sensitive to every shift, small and large alike. Solve each boundary in closed form, then check whether the two bounds actually overlap.

1. Upper bound on $n$ from the small-shift ARL requirement. Ignoring the negligible lower-tail term, $p(0.25,n)\le0.01 \iff \Phi(0.25\sqrt n-3)\le0.01 \iff 0.25\sqrt n-3\le\Phi^{-1}(0.01)=-2.326$, so $$\sqrt n\le\frac{3-2.326}{0.25}=2.696 \;\Rightarrow\; n\le\boxed{7.26}.$$
2. Lower bound on $n$ from the large-shift $\beta$ requirement. Similarly $p(1.2,n)\ge0.7 \iff \Phi(1.2\sqrt n-3)\ge0.7 \iff 1.2\sqrt n-3\ge\Phi^{-1}(0.7)=0.524$, so $$\sqrt n\ge\frac{3+0.524}{1.2}=2.937 \;\Rightarrow\; n\ge\boxed{8.63}.$$
3. Check feasibility. The small-shift requirement needs $n\le7.26$; the large-shift requirement needs $n\ge8.63$. Since $7.26<8.63$, no sample size satisfies both requirements at once — the two design goals are fundamentally in tension for a fixed-$n$, 3-sigma $\bar X$ chart: any $n$ small enough to avoid over-reacting to a negligible $0.25\sigma$ wobble is too small to reliably catch a genuine $1.2\sigma$ shift, and any $n$ large enough to catch the real shift is already over-sensitive to the negligible one.