Question 3 of 6: X̄ vs. CUSUM/EWMA, and the X̄-S Control Chart

Question 3: X̄ vs. CUSUM/EWMA, and the X̄-S Control Chart (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) X̄ chart vs. tabular CUSUM; sensitivity to small shifts; EWMA's dual role in EPC and SPC

An X̄ chart is memoryless in the strict sense that its signal depends only on the current sample: each point is compared to the same fixed 3-sigma limits, independent of every prior point. A tabular CUSUM chart instead accumulates deviations from a target value over time, keeping running one-sided sums $C_i^+=\max\!\left[0,\ C_{i-1}^++(\bar x_i-\mu_0-K)\right]$ and $C_i^-=\max\!\left[0,\ C_{i-1}^-+(\mu_0-K-\bar x_i)\right]$, signalling when either sum exceeds a decision interval $H$. Because it integrates evidence across MANY samples rather than judging each in isolation, the CUSUM is substantially more sensitive to small, sustained shifts (typically in the 0.5–2$\sigma$ range) — a shift too small to push any single sample outside the X̄ limits still steadily accumulates in $C_i^+$ or $C_i^-$ until it crosses $H$. The X̄ chart, conversely, reacts fastest to LARGE, abrupt shifts (a big jump immediately exceeds the fixed limits) but is comparatively slow to detect small drifts, since each sample is judged with no memory of the ones before it.

The EWMA statistic $z_i=\lambda\bar x_i+(1-\lambda)z_{i-1}$ sits between these two extremes: choosing $\lambda$ close to 1 recovers X̄-like behaviour (fast on large shifts), while small $\lambda$ gives CUSUM-like accumulated memory (sensitive to small shifts) — $\lambda$ is a single tuning knob that lets one chart be matched to the shift size of interest, and it is also robust to non-normal data. The SAME recursive form is exploited for two different purposes: in SPC, $z_i$ is plotted against control limits to DETECT a special cause so it can be investigated and removed; in EPC (engineering process control), the identical exponentially-weighted-average recursion is the minimum-mean-square-error one-step-ahead forecast for an IMA(1,1) disturbance model, and is used to compute a compensatory ADJUSTMENT that is fed back into the process automatically. The rationale for using EWMA in both roles is that its statistic is simultaneously the best available forecast of where the process is drifting to (useful for automatic compensation) and a sensitive detector of when that drift has changed character (useful for stopping the process and investigating) — it naturally bridges the "adjust vs. investigate" decision that separates EPC from SPC.

(b) X̄-S chart: control limits, process parameters, and $P(RL\ge10)$ after a shift

Given. $m=20$ samples of size $n=6$; $\bar{\bar X}=40$; $\bar S=5.26$. From the paper's Appendix VI, for $n=6$: $A_3=1.287$, $B_3=0.030$, $B_4=1.970$, $c_4=0.9515$.

Find. The X̄ and S chart 3-sigma limits, the process $\hat\mu$/$\hat\sigma$, and $P(RL\ge10)$ on the X̄ chart if the mean shifts to 48.

Approach. Apply the $A_3$/$B_3$/$B_4$ factor relations for the limits and $\bar S/c_4$ for $\hat\sigma$; then convert the post-shift mean to a standardized shift, compute the per-sample miss probability $\beta$, and use $P(RL\ge10)=\beta^9$ for the geometric run-length distribution.

1. X̄ chart limits. $$UCL_{\bar X}=40+1.287(5.26)=\boxed{46.77},\qquad LCL_{\bar X}=40-1.287(5.26)=\boxed{33.23}.$$
2. S chart limits. $$UCL_S=1.970(5.26)=\boxed{10.36},\qquad LCL_S=0.030(5.26)=\boxed{0.158}.$$
3. Process parameters. $$\hat\mu=\bar{\bar X}=\boxed{40},\qquad \hat\sigma=\frac{\bar S}{c_4}=\frac{5.26}{0.9515}=\boxed{5.528}.$$
4. $P(RL\ge10)$ after a shift to 48. Standardized shift $\delta=(48-40)/5.528=1.447\sigma$; per-sample miss probability $$\beta=\Phi\!\left(3-1.447\sqrt6\right)-\Phi\!\left(-3-1.447\sqrt6\right)=\Phi(-0.545)-\Phi(-6.545)=\boxed{0.2930}.$$ Since $RL$ is geometric with parameter $1-\beta$, $P(RL\ge10)=\beta^{9}$: $$P(RL\ge10)=0.2930^{9}=\boxed{1.59\times10^{-5}}.$$

(c) Designing the X̄ chart for a target run-length performance

Given. From (b): $\hat\mu=40$, $\hat\sigma=5.528$. The mean is to shift from 40 to 45 (an absolute shift of 5), and the design requirement is $P(RL\ge10)\le0.2$.

Find. The minimum sample size $n$ and the corresponding 3-sigma X̄ control limits.

Approach. Express the target shift in standardized units, then scan integer $n$ for the smallest value at which $\beta(n)^9\le0.2$, i.e. $\beta(n)\le0.2^{1/9}=0.8248$.

1. Standardized shift. $$\delta=\frac{45-40}{5.528}=\boxed{0.9045}\ \sigma.$$
2. Scan for the minimum $n$. With $\beta(n)=\Phi(3-\delta\sqrt n)-\Phi(-3-\delta\sqrt n)$: $$n=4:\ \beta=0.8832,\ \beta^9=0.327>0.2;\qquad n=5:\ \beta=0.8359,\ \beta^9=0.199\le0.2.$$ The minimum sample size is $\boxed{n=5}$.
3. Control limits at $n=5$. $$UCL_{\bar X}=40+\frac{3(5.528)}{\sqrt5}=\boxed{47.42},\qquad LCL_{\bar X}=40-\frac{3(5.528)}{\sqrt5}=\boxed{32.58}.$$