Question 3 of 6: Type I/II Errors, a $p$-Chart for Personal-Computer Inspection, and Minimum Sample Size for Detection

Question 3: Type I/II Errors, a $p$-Chart for Personal-Computer Inspection, and Minimum Sample Size for Detection (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) Type I/II errors and the zone rules

A Type I error ($\alpha$, a false alarm) is concluding the process is out of control when it is actually still in control. Its practical cost is an unnecessary process stoppage and investigation — wasted engineering/operator time chasing a cause that does not exist, and, if someone "corrects" an in-control process anyway, tampering that can actually increase variability (Deming's funnel experiment). A Type II error ($\beta$, a missed signal) is concluding the process is in control when it has actually shifted. Its practical cost is that out-of-specification product keeps being produced and shipped undetected, so the cost of nonconformance keeps accumulating — often escalating from an internal to a far more expensive external failure cost — until the shift is eventually caught by the chart or by a customer complaint. The two errors trade off directly: narrowing the control limits (e.g. $2\sigma$ instead of $3\sigma$) lowers $\beta$ (shifts are caught sooner) but raises $\alpha$ (more false alarms); the conventional $3\sigma$ limit is the standard compromise between the two costs.

The Western Electric zone rules (one point beyond $3\sigma$; two of three consecutive points beyond $2\sigma$ on the same side; four of five beyond $1\sigma$ on the same side; eight consecutive points on one side of the centre line) supplement the basic $3\sigma$ test by treating non-random patterns, not just single extreme points, as signals. This makes the chart substantially more sensitive to small, sustained shifts — it lowers $\beta$ for shifts that a lone $3\sigma$ test would take a long time to catch — but it does so at the cost of a higher combined false-alarm rate: the nominal single-point rate of about 0.27% rises to roughly 2% once all four rules are applied simultaneously to a genuinely in-control process, because there are now many more ways for the chart to signal by chance alone.

(b) $p$-chart for PC inspection, 10 days, variable daily sample size

Given. All units produced are inspected each day; both the units inspected and the nonconforming count vary day to day:

Find. The $p$-chart (individual limits at each day's own $n_i$), the estimated in-control fraction nonconforming, whether revision is needed, and the smallest constant $n$ that gives a strictly positive LCL.

Approach. With sample size varying day to day, the centre line uses the pooled estimate $\bar p=\sum D_i/\sum n_i$, and the control limits must be recomputed at each day's own $n_i$: $UCL_i=\bar p+3\sqrt{\bar p(1-\bar p)/n_i}$, $LCL_i=\max\!\big(0,\ \bar p-3\sqrt{\bar p(1-\bar p)/n_i}\big)$.

1. Pooled fraction nonconforming. $\sum n_i=880$, $\sum D_i=39$, so $\bar p=\dfrac{39}{880}=\boxed{0.04432}$ (4.43%).
2. Control limits at each sample size. For $n_i=70$: $\sqrt{\bar p(1-\bar p)/70}=0.02461$, so $UCL=0.04432+3(0.02461)=\boxed{0.1181}$, $LCL=0.04432-0.0738<0\Rightarrow\boxed{0}$. For $n_i=100$: $\sqrt{\bar p(1-\bar p)/100}=0.02059$, so $UCL=0.04432+3(0.02059)=\boxed{0.1061}$, $LCL=0$ likewise.
3. Check every day against its own limit. The daily fractions are $0.043,\,0,\,0.060,\,0.114,\,0.050,\,0.020,\,0,\,0.050,\,0.100,\,0$; the closest approaches to a limit are day 4 ($8/70=0.1143$, against $UCL=0.1181$) and day 9 ($10/100=0.100$, against $UCL=0.1061$) — both stay just inside their limits. No day exceeds its UCL and no day is below its (zero) LCL, so the chart is already in statistical control — no revision of $\bar p$ is necessary.
4. Smallest sample size for a positive LCL. $LCL>0$ requires $\bar p-3\sqrt{\bar p(1-\bar p)/n}>0 \iff n>\dfrac{9(1-\bar p)}{\bar p}=\dfrac{9(0.9557)}{0.04432}=194.1$, so the smallest usable integer sample size is $\boxed{n=195}$.

Whether to delete the zero-count days (2, 7, 10) because they coincide with $LCL=0$: no. When $\bar p$ is small enough that the $3\sigma$ lower limit computes to a negative number, the limit is clamped at $LCL=0$ because a fraction cannot be negative — a plotted point of exactly zero nonconforming units then lies at or inside that clamped floor, not below it. Zero defectives is a legitimate, desirable outcome of an in-control (or better-than-typical) day, not evidence of an assignable cause; deleting those good days would bias $\bar p$ upward and understate the process's true capability. (This is the mirror image of a genuine point above the UCL, which IS removed as an assignable cause during a revision pass.)

(c) Minimum fixed sample size to detect a shift to $p=0.12$ within 3 samples with probability $\ge0.6$

Given. Baseline (in-control) $p_0=\bar p=0.04432$ from part (b); a future shift to $p_1=0.12$; require $P(\text{detected by the 1st, 2nd, or 3rd sample})\ge0.6$, using a fixed sample size $n$ for every future sample.

Find. The minimum $n$.

Approach. With control limits fixed at $UCL(n)=p_0+3\sqrt{p_0(1-p_0)/n}$ (and $LCL(n)=0$ for the $n$ values in play here), each future sample independently has some probability $p_d(n)$ of signalling once the true fraction has shifted to $p_1$; the probability of detecting the shift within the first 3 samples is $1-(1-p_d)^3$. Using the standard normal (OC-curve) approximation for the shifted sample proportion's own sampling distribution, $p_d(n)=1-\Big[\Phi\big(\tfrac{UCL(n)-p_1}{\sigma_1(n)}\big)-\Phi\big(\tfrac{LCL(n)-p_1}{\sigma_1(n)}\big)\Big]$ with $\sigma_1(n)=\sqrt{p_1(1-p_1)/n}$, this is monotonic in $n$ and can be scanned directly.

1. Required single-sample detection probability. $1-(1-p_d)^3\ge0.6 \iff p_d\ge1-0.4^{1/3}=\boxed{0.2632}$.
2. Scan $n$. At $n=24$: $UCL=0.04432+3\sqrt{0.04432(0.9557)/24}=0.1703$, $\sigma_1=\sqrt{0.12(0.88)/24}=0.06633$, giving $p_d=1-\Phi\!\big(\tfrac{0.1703-0.12}{0.06633}\big)=0.2592$ — just short. At $n=25$: $UCL=0.1678$, $\sigma_1=0.06498$, $p_d=1-\Phi\!\big(\tfrac{0.1678-0.12}{0.06498}\big)=0.2635\ge0.2632$ — meets the target, with $1-(1-0.2635)^3=0.6004\ge0.6$.

$$\boxed{n=25}$$