Question 5 of 6: Attribute Charts, a $p$ Chart, and Detection Speed after a Proportion Shift

Question 5: Attribute Charts, a $p$ Chart, and Detection Speed after a Proportion Shift (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) — $p$, $c$, $u$ charts; when a zero-defective sample should (or should not) trigger investigation

The $p$ chart monitors the fraction nonconforming in samples of size $n$ (constant or varying), where each unit is classified simply conforming/nonconforming — used whenever the count of interest is naturally a proportion of inspected units, e.g. the fraction of parts failing a go/no-go gauge. The $c$ chart monitors the count of nonconformities (defects, not defective units) found in an inspection unit of constant size or opportunity — a single unit can carry more than one nonconformity, e.g. the number of paint blemishes on one identical car-body panel. The $u$ chart generalizes the $c$ chart to a varying inspection-unit size, plotting the nonconformities per unit of opportunity (e.g. defects per square metre of fabric, or per 100 circuit boards), so that samples of different size remain comparable on the same chart.

When the $LCL$ for a $p$ chart computes to a negative number, it is conventionally set to zero, since a proportion cannot be negative — but this means the chart can never signal a decrease in the defective rate through the ordinary lower-limit test, because zero defectives is always inside (in fact, sitting exactly on) the boundary. Whether a sample with zero defectives should trigger an investigation therefore depends on context, not on the chart alone. It should NOT be stopped/investigated when an occasional zero-defective sample is simply consistent with ordinary sampling variation around a low in-control $\bar p$ — e.g. if $\bar p=0.01$ and $n=50$, the expected count is only 0.5 defectives, so a sample of exactly zero is unremarkable and occurs routinely by chance; treating every such sample as a signal would be tampering with a process that has not actually changed. It SHOULD be investigated when zero defectives occurs as part of a longer non-random pattern that points to a genuine (and, in this direction, desirable) process improvement — e.g. a run of several consecutive zero-defective samples where $\bar p$ and $n$ would ordinarily predict at least a few defectives in that many samples combined. Such a run is itself an assignable-cause signal (by the standard runs/zone-rule logic, just applied to an improvement rather than a degradation), and it is worth investigating and, if a real cause is found (a maintenance fix, a supplier change, a process adjustment), formally revising $\bar p$ downward and locking in the gain, rather than dismissing the run as luck.

Part (b) — $p$ chart for 15 samples of $n=150$ parts

Given. 15 samples, $n=150$ parts each, number of nonconforming parts per sample:

Find. Trial and (if needed, revised) $p$-chart limits, the in-control proportion $\hat p_0$, and $P(\text{detect on sample 1 or 2})$ after a shift to $p_1=0.035$.

Approach. Pool all 15 samples for the trial centre line, form 3-sigma limits, remove any assignable-cause point, and re-check; then use the normal approximation to the binomial for the shifted-process detection probability.

1. Trial limits. Total nonconforming $=\sum d_i=61$ over $15\times150=2250$ inspected, so $\bar p=\dfrac{61}{2250}=0.02711$. With $\hat\sigma_p=\sqrt{\bar p(1-\bar p)/n}=\sqrt{0.02711(0.97289)/150}=0.01326$: $$UCL_p=\bar p+3\hat\sigma_p=0.02711+3(0.01326)=\boxed{0.0669},\qquad LCL_p=\max(0,\,0.02711-3(0.01326))=\boxed{0}.$$
2. Check and revise. Sample 9 has $\hat p_9=12/150=0.080\gt UCL_p=0.0669$ — an assignable-cause point, removed. Recomputing over the remaining 14 samples: $\bar p_{\text{rev}}=\dfrac{61-12}{14(150)}=\dfrac{49}{2100}=\boxed{0.02333}$, $\hat\sigma_{p,\text{rev}}=\sqrt{0.02333(0.97667)/150}=0.01233$, giving $$UCL_{p,\text{rev}}=0.02333+3(0.01233)=\boxed{0.0603},\qquad LCL_{p,\text{rev}}=0.$$ All 14 remaining sample proportions (max $7/150=0.047$) fall inside these revised limits — no further revision is needed.
3. In-control proportion. $\hat p_0=\bar p_{\text{rev}}=\boxed{0.0233\ (2.33\%)}$.
4. Detection probability within the first two samples after a shift to $p_1=0.035$. Using the shifted-process standard error $\sigma_{p_1}=\sqrt{p_1(1-p_1)/n}=\sqrt{0.035(0.965)/150}=0.01502$, the probability the shifted proportion plots inside the (unchanged) revised limits on a single sample is $$\beta=\Phi\!\left(\dfrac{UCL_{p,\text{rev}}-p_1}{\sigma_{p_1}}\right)-\Phi\!\left(\dfrac{LCL_{p,\text{rev}}-p_1}{\sigma_{p_1}}\right)=\Phi(1.658)-\Phi(-2.330)=0.9515-0.0099=\boxed{0.944}.$$ The probability the shift goes undetected on both of the first two samples is $\beta^2$, so $$P(\text{detect on sample 1 or 2})=1-\beta^2=1-0.944^2=\boxed{0.108\ (10.8\%)}.$$

Part (c) — $ARL$ and $ATS$ after a shift of $1.5\hat\sigma$

Given. $\hat p_0=0.02333$, $\hat\sigma_{p_0}=0.01233$ (part (b)); shift to $p_1=p_0+1.5\hat\sigma_{p_0}$; sampling every half hour.

Find. $ARL$ and $ATS$ for $p_1$.

1. Shifted proportion. $$p_1=0.02333+1.5(0.01233)=\boxed{0.0418\ (4.18\%)}.$$
2. Detection probability per sample at $p_1$. With $\sigma_{p_1}=\sqrt{p_1(1-p_1)/150}=0.01631$: $$\beta=\Phi\!\left(\dfrac{0.0603-0.0418}{0.01631}\right)-\Phi\!\left(\dfrac{0-0.0418}{0.01631}\right)=\Phi(1.135)-\Phi(-2.564)=0.8718-0.0052=\boxed{0.8666}.$$
3. $ARL$ and $ATS$. $$ARL=\dfrac{1}{1-\beta}=\dfrac{1}{1-0.8666}=\boxed{7.45\ \text{samples}},\qquad ATS=ARL\times0.5\ \text{hr}=\boxed{3.72\ \text{hr}}.$$

A $1.5\sigma$ shift (a much larger relative jump than Question 5(b)'s $p_1=0.035$, which was only about $0.94\hat\sigma_{p_0}$ above $\hat p_0$) is detected roughly four times faster — about 7–8 samples (under 4 hours) on average, versus the 0.94$\sigma$ shift's much slower single-sample detection probability of only 5.6% ($1-\beta=1-0.944$) computed in part (b), consistent with the general rule that $p$-chart detection speed improves quickly as the shift size grows relative to $\sigma_{p_0}$.