Question 2 of 6: Attribute vs. Variable Control Charts and an $\bar X$–$R$ Chart with Two-Point ARL Design

Question 2: Attribute vs. Variable Control Charts and an $\bar X$–$R$ Chart with Two-Point ARL 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.

Part (a) — One chart for attributes, a pair for variables; R chart vs. S chart

A variables characteristic (a continuous measurement such as a weight or a length) carries two independent pieces of information per sample — where the process is centred (location) and how spread out it is (dispersion) — and these two moments can shift independently of one another; a process can drift off target while its variability stays constant, or become erratic while its average stays put. A single chart cannot see both failure modes at once, so variables monitoring always uses a pair of charts: one for location ($\bar X$ or individuals) and one for dispersion ($R$ or $S$), evaluated together. An attributes characteristic (conforming/nonconforming, or a defect count) collapses to a single number per sample — a fraction or a count — whose own sampling distribution (binomial or Poisson) has a variance that is a fixed function of its mean ($np(1-p)$ or $\lambda$); there is no independent second "spread" parameter to track, so one chart (p, np, c, or u) is both necessary and sufficient.

Both the $R$ chart (subgroup range) and the $S$ chart (subgroup standard deviation) estimate process dispersion, but they differ in statistical efficiency. The range uses only the two extreme observations in a subgroup and discards the interior data, so its efficiency as an estimator of $\sigma$ falls off rapidly once subgroup size $n$ exceeds about 8–10; $S$ uses every observation and remains efficient at any $n$. The $S$ chart is generally preferable, especially at larger $n$ or whenever the extra arithmetic is not a burden (routine with modern data-collection software), because it makes full use of the sample and gives a lower-variance estimate of $\sigma$ for the same sample size. The $R$ chart remains attractive mainly for small, manually plotted subgroups ($n\le5$, e.g. hand-charted shop-floor data), where the range is trivial to compute by eye and its slightly lower efficiency is an acceptable trade for simplicity.

The 3-sigma limits for the $S$ chart are $UCL_S=B_4\bar S$, $LCL_S=B_3\bar S$, $CL_S=\bar S$, where $B_3=1-3\sqrt{1-c_4^2}/c_4$ and $B_4=1+3\sqrt{1-c_4^2}/c_4$, and $c_4$ is the usual unbiasing constant for the sample standard deviation at subgroup size $n$. Because $c_4$ increases toward 1 as $n$ grows, $B_3$ and $B_4$ both move toward 1 (the limits tighten in relative terms) as $n$ increases — larger subgroups estimate $\sigma$ more precisely, so the chart can afford narrower relative limits around $\bar S$ while keeping the same in-control false-alarm rate. This is a structurally different behaviour from the $R$ chart's $D_3,D_4$ factors, which do not converge to 1 in the same way, another reason the $S$ chart scales better to larger $n$.

Part (b) — $\bar X$–$R$ chart for injection-molding part weight ($n=5$, 15 samples)

Given. Five parts are measured every 20 minutes; 15 samples of $n=5$, with the sample average $\bar X_i$ and range $R_i$ reported for each (grams).

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

Approach. Control the $R$ chart first (it validates that within-sample dispersion is stable); once $R$ is in control, use its centre line to set the $\bar X$ chart limits and revise those in turn if any sample average signals.

1. Trial $R$ chart. $\bar R=\dfrac{\sum R_i}{15}=\dfrac{264}{15}=17.6$ g. With $D_3=0$, $D_4=2.114$ ($n=5$): $UCL_R=D_4\bar R=2.114(17.6)=\boxed{37.21\ \text{g}}$, $LCL_R=0$. Sample 12 ($R=38>37.21$) plots above $UCL_R$ — an assignable-cause spread event — so it is removed.
2. Revise $R$ chart. Dropping sample 12 leaves 14 samples: $\bar R=\dfrac{264-38}{14}=16.14$ g, $UCL_R=2.114(16.14)=\boxed{34.13\ \text{g}}$. All 14 remaining ranges (max 32 g) are below this revised limit, so the $R$ chart is now in control.
3. Trial $\bar X$ chart (on the 14 $R$-clean samples). $\bar{\bar X}=\dfrac{\sum_{i\ne12}\bar X_i}{14}=2033.64$ g. With $A_2=0.577$: $UCL_{\bar X}=2033.64+0.577(16.14)=\boxed{2042.96\ \text{g}}$, $LCL_{\bar X}=2033.64-0.577(16.14)=\boxed{2024.33\ \text{g}}$. Checking each remaining $\bar X_i$: samples 1 (2024), 2 (2008), 4 (2058) and 5 (2047) all plot outside — four assignable-cause shifts in the process average.
4. Revise $\bar X$ chart. Dropping samples 1, 2, 4, 5 (in addition to 12) leaves the 10 samples 3, 6–11, 13–15: $$\bar{\bar X}_{\text{rev}}=\boxed{2033.4\ \text{g}},\qquad \bar R_{\text{rev}}=\boxed{16.7\ \text{g}}.$$ $$UCL_{\bar X}=2033.4+0.577(16.7)=\boxed{2043.04\ \text{g}},\qquad LCL_{\bar X}=2033.4-0.577(16.7)=\boxed{2023.76\ \text{g}}.$$ All 10 remaining averages fall inside these revised limits, and re-checking the $R$ chart on the same final 10 samples ($UCL_R=2.114(16.7)=35.30$ g, max remaining range 32 g) confirms it is still in control — no further revision is needed.
5. In-control process-parameter estimates. Using the final revised $\bar R=16.7$ g and $d_2=2.326$ ($n=5$): $$\hat\mu_0=\bar{\bar X}_{\text{rev}}=\boxed{2033.4\ \text{g}},\qquad \hat\sigma_0=\dfrac{\bar R_{\text{rev}}}{d_2}=\dfrac{16.7}{2.326}=\boxed{7.18\ \text{g}}.$$

Part (c) — Designing an $\bar X$ chart to meet a two-point ARL specification

Given. $k=3$ (3-sigma limits); required $ARL\ge100$ at a shift of $\delta_1=0.25\sigma$; required $ARL\le10$ at a shift of $\delta_2=1.20\sigma$.

Find. The sample size $n$ (using $\hat\mu_0,\hat\sigma_0$ from part (b)) that satisfies both requirements.

Approach. For an $\bar X$ chart with $k$-sigma limits, a mean shift of $\delta\sigma$ is detected on a given sample with probability $1-\beta$, where $\beta=\Phi(k-\delta\sqrt n)-\Phi(-k-\delta\sqrt n)$ and $ARL=1/(1-\beta)$. Since $ARL$ falls as $n$ grows (for a fixed $\delta$), the $\delta_1=0.25\sigma$ requirement bounds $n$ from above and the $\delta_2=1.20\sigma$ requirement bounds it from below; evaluate $ARL(\delta,n)$ over a small range of integers to find the feasible window.

1. Scan $n$ against both constraints. Computing $ARL(0.25,n)$ and $ARL(1.20,n)$ for $n=1,2,\dots$: $ARL(0.25,n)\ge100$ holds up to $n=7$ ($ARL=101.99$; $n=8$ gives only 90.65), and $ARL(1.20,n)\le10$ first holds at $n=3$ ($ARL=5.61$; $n=2$ still gives 10.38). The feasible window is therefore $\boxed{3\le n\le7}$.
2. Select the design point. Taking the smallest (most economical) feasible sample size, $n=3$, and checking both requirements directly with $k=3$: $$ARL(0.25,3)=\dfrac{1}{1-[\Phi(3-0.25\sqrt3)-\Phi(-3-0.25\sqrt3)]}=\boxed{184.2}\ \ (\ge100,\ \checkmark),$$ $$ARL(1.20,3)=\dfrac{1}{1-[\Phi(3-1.20\sqrt3)-\Phi(-3-1.20\sqrt3)]}=\boxed{5.61}\ \ (\le10,\ \checkmark).$$