Question 3 of 6: Type I/II Errors and the OC Curve, a p-Chart for PC Inspection, and Minimum Sample Size for Shift Detection

Question 3: Type I/II Errors and the OC Curve, a p-Chart for PC Inspection, and Minimum Sample Size for Shift 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 $\bar X$ chart's OC curve

On a control chart, a Type I error (false alarm, probability $\alpha$) occurs when a point plots outside the control limits while the process is actually still IN control — the operator stops the process, hunts for an assignable cause that does not exist, and wastes time/production capacity on an unnecessary investigation. A Type II error (probability $\beta$) occurs when the process has genuinely shifted out of control but the plotted point still falls INSIDE the limits, so the chart fails to signal — the process continues running out of control, producing nonconforming or higher-variability output until a later sample (if any) finally detects it. The practical trade-off is direct: WIDENING the control limits (e.g. moving from 3-sigma to a wider multiple) reduces $\alpha$ (fewer false alarms, less wasted investigation cost) but increases $\beta$ (real shifts go undetected longer, producing more scrap/rework before correction); NARROWING the limits does the reverse. The economically optimal limit width therefore balances the cost of unnecessary process investigations against the cost of running out of control undetected, rather than defaulting blindly to 3-sigma in every application.

The operating-characteristic (OC) curve for an $\bar X$ chart plots $\beta$ (probability of NOT signalling, i.e. missing the shift) as a function of the true process shift size, usually expressed in standardized units $k=|\mu_1-\mu_0|/\sigma$: $\beta(k)=\Phi(3-k\sqrt n)-\Phi(-3-k\sqrt n)$. At $k=0$ (process still on target) $\beta$ is highest by construction (the chart should rarely signal); as $k$ grows, $\beta$ falls toward zero (larger shifts are essentially always caught). The OC curve's STEEPNESS depends directly on the sample size $n$: because $n$ enters through $k\sqrt n$, a larger $n$ makes $\beta$ fall off faster as $k$ increases — the curve becomes steeper and hugs the axes more tightly, meaning the chart discriminates more sharply between "on target" and "shifted," and even small shifts are caught reliably. A smaller $n$ produces a flatter, more gradually-declining OC curve: only larger shifts are reliably detected, because a smaller sample averages out less of the sampling noise that would otherwise mask a modest true shift in the mean.

(b) p-chart for PC final inspection: control limits, revision, and $\hat p$

Given. 10 days of 100% inspection, sample size varies by day.

Find. Trial (and, if needed, revised) p-chart control limits; the in-control estimate $\hat p$; the smallest sample size giving a strictly positive $LCL$.

1. Pooled estimate of the fraction nonconforming. Total nonconforming $=4+7+5+8+6+6+4+3+9+1=53$; total inspected $=4(80)+6(100)=920$. $$\hat p=\bar p=\frac{53}{920}=\boxed{0.05761\ (5.76\%)}.$$
2. Control limits (vary with $n_i$). $UCL_i=\bar p+3\sqrt{\bar p(1-\bar p)/n_i}$, $LCL_i=\max\!\big[0,\,\bar p-3\sqrt{\bar p(1-\bar p)/n_i}\big]$: $$n_i=80:\ UCL=0.05761+3\sqrt{\tfrac{0.05761(0.94239)}{80}}=\boxed{0.1358},\quad LCL=\boxed{0\ (\text{negative, clipped})};$$ $$n_i=100:\ UCL=0.05761+3\sqrt{\tfrac{0.05761(0.94239)}{100}}=\boxed{0.1275},\quad LCL=\boxed{0}.$$
3. Check each day and revise if necessary. Daily $\hat p_i=d_i/n_i$ ranges from $1/80=0.0125$ (day 10) to $8/80=0.100$ (day 4) and $9/100=0.09$ (day 9) — every value is below its day's $UCL$ (0.1358 for $n=80$, 0.1275 for $n=100$) and above $LCL=0$. No point is out of control, so no revision is needed; $\bar p=0.05761$ stands as the in-control estimate.
4. Smallest sample size giving a positive $LCL$. $LCL>0$ requires $\bar p-3\sqrt{\bar p(1-\bar p)/n}>0$, i.e. $n>9(1-\bar p)/\bar p$: $$n>\frac{9(1-0.05761)}{0.05761}=\frac{9(0.94239)}{0.05761}=147.2\ \Longrightarrow\ \boxed{n=148\ (\text{smallest integer giving }LCL>0)}.$$

(c) Minimum fixed sample size to detect a shift to $p_1=0.12$ within two samples

Given. $p_0=\bar p=0.05761$ (part b); shift target $p_1=0.12$; requirement: probability of a signal on the first or second post-shift sample $\ge0.6$.

Find. The minimum fixed sample size $n$.

1. Translate the "detect within 2 samples" requirement into a per-sample detection probability. If $\beta(n)$ is the probability a single post-shift sample fails to signal, then the probability of NOT detecting in either of the first two samples is $\beta(n)^2$, so $$P(\text{detect within 2 samples})=1-\beta(n)^2\ge0.6\ \Longrightarrow\ \beta(n)\le\sqrt{0.4}=0.6325.$$
2. Express $\beta(n)$ using the FIXED $p_0$-based limits evaluated at the shifted true fraction $p_1$. Control limits stay fixed at $UCL=p_0+3\sqrt{p_0(1-p_0)/n}$, $LCL=\max[0,\,p_0-3\sqrt{p_0(1-p_0)/n}]$, while the sample proportion after the shift is (normal approximation) $\hat p\sim N\!\big(p_1,\,p_1(1-p_1)/n\big)$: $$\beta(n)=\Phi\!\left(\frac{UCL-p_1}{\sqrt{p_1(1-p_1)/n}}\right)-\Phi\!\left(\frac{LCL-p_1}{\sqrt{p_1(1-p_1)/n}}\right).$$
3. Solve for the minimum integer $n$ with $\beta(n)\le0.6325$. $\beta(n)$ decreases smoothly with $n$ in the region of interest; evaluating: $$n=87:\ \beta=0.6404\ (\text{fails}),\qquad n=89:\ \beta=0.6327\ (\text{fails, marginally}),$$ $$n=90:\ \beta=0.6289\ \Longrightarrow\ 1-\beta^2=0.6045\ge0.6\ (\text{passes}).$$ So the minimum fixed sample size is $\boxed{n=90}$.
4. Sanity check. $n=90$ gives $n p_0=90(0.05761)=5.18\ge5$, comfortably inside the usual rule-of-thumb range for the normal approximation to a p-chart's binomial sampling distribution to be reasonable — the numeric answer is not an artefact of an under-sized sample where the approximation itself would be suspect.