Question 4 of 6: Attributes Charts, a $u$-Chart for Disk-Drive Nonconformities, and Minimum Sample Size

Question 4: Attributes Charts, a $u$-Chart for Disk-Drive Nonconformities, and Minimum Sample Size (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) p vs. np, c vs. u, and the demerit chart

The p-chart plots the fraction nonconforming ($\hat p=x/n$) in each sample and, because it is a proportion, naturally accommodates a varying sample size $n$ from period to period (its control limits, $\bar p\pm3\sqrt{\bar p(1-\bar p)/n}$, simply widen or narrow with $n$). The np-chart plots the raw count of nonconforming units and requires a constant sample size, because a count is only comparable across samples if the opportunity to accumulate that count (the sample size) does not change; it is otherwise the same underlying binomial model as the p-chart and is often preferred on the shop floor because operators find a raw count more intuitive than a fraction.

The c-chart plots the count of nonconformities (not nonconforming units — a single unit can have several nonconformities) found in a constant inspection unit, modelled as Poisson with limits $\bar c\pm3\sqrt{\bar c}$. The u-chart plots nonconformities per inspection unit, $u=c/n$, and like the p-chart accommodates a varying number of inspection units per sample, with limits $\bar u\pm3\sqrt{\bar u/n}$. Both are used when a product can exhibit multiple, independent defects of the same general type (solder joints, cosmetic blemishes, workmanship nonconformities) rather than being simply good or bad.

A demerit chart extends the c/u-chart idea to account for the fact that not all nonconformities are equally serious: each defect is classified into a severity class (e.g., very serious, serious, moderate, minor) and assigned a demerit weight (a common convention is 100/50/10/1), and the chart plots a single weighted demerit score per unit rather than a plain count. This is more informative than a c- or u-chart whenever defect severity varies materially, because it prevents a large number of trivial cosmetic nonconformities from masking, or a single critical nonconformity from being under-weighted relative to, the count-only picture.

(b) $u$-chart for daily workmanship nonconformities

Given. Ten days of inspection; inspection unit $=$ 1 assembly; sample size (assemblies inspected) and total nonconformities found vary by day:

Find. $u$-chart control limits (varying, since $n_i$ varies), whether the process is in control, and the in-control estimate of nonconformities per assembly, $\hat\lambda=\bar u$.

Approach. Pool all ten days to estimate the process average nonconformity rate $\bar u=\sum c_i/\sum n_i$, then compute a separate control limit pair for each day from its own $n_i$ (the varying-sample-size u-chart), and check every plotted $u_i=c_i/n_i$ against its own day-specific limits.

1. Centre line. $\sum c_i=52$ nonconformities over $\sum n_i=26$ assemblies, so $$\bar u=\frac{\sum c_i}{\sum n_i}=\frac{52}{26}=\boxed{2.00\ \text{nonconformities/assembly}}.$$
2. Day-specific limits. $UCL_i=\bar u+3\sqrt{\bar u/n_i}$, $LCL_i=\max\!\big(0,\ \bar u-3\sqrt{\bar u/n_i}\big)$. For $n_i=1$ (days 4, 10): $UCL=2+3\sqrt2=\boxed{6.24}$; for $n_i=2$ (days 1, 3, 7): $UCL=2+3\sqrt{1}=\boxed{5.00}$; for $n_i=3$ (days 5, 9): $UCL=2+3\sqrt{2/3}=\boxed{4.45}$; for $n_i=4$ (days 2, 6, 8): $UCL=2+3\sqrt{0.5}=\boxed{4.12}$. Every $LCL_i$ evaluates to a negative number and is therefore set to 0 in every case (the smallest $\bar u-3\sqrt{\bar u/n_i}$, at $n_i=1$, is $2-4.24=-2.24$).
3. Check every point. $u_i=c_i/n_i$ gives $0,\ 2.50,\ 4.00,\ 5.00,\ 1.67,\ 1.25,\ 2.00,\ 1.50,\ 2.00,\ 3.00$; comparing each to its own day's limits above, every value is comfortably inside $[0,UCL_i]$ — the largest, day 4 at $u=5.00$ against $UCL=6.24$, is still under its limit. No point signals; the trial limits need no revision.
4. In-control nonconformity rate. Since no revision is required, the estimate of the in-control expected number of nonconformities per assembly is simply the pooled centre line, $\hat\lambda=\boxed{2.00}$.

(c) Minimum sample size to detect a $\lambda\to1.5\lambda$ shift within two samples, at probability $\geq0.7$

Given. $\hat\lambda=2.00$ nonconformities/assembly from part (b); the new inspection unit is redefined as 2 assemblies; the shift of interest is from $\lambda$ to $1.5\lambda$ per assembly; target: detect on the 1st or 2nd sample after the shift with probability $\geq0.7$.

Find. The minimum constant number of (2-assembly) inspection units per sample, $k$, that meets the target.

Approach. Re-express the in-control and shifted rates on the new (2-assembly) inspection unit, fix the control limit at the pre-shift rate (the chart doesn't know a shift has happened), then find the smallest sample size $k$ (number of new inspection units per sample) for which the single-sample detection probability $p_1$ satisfies $1-(1-p_1)^2\geq0.7$, i.e. $p_1\geq1-\sqrt{0.3}=0.4523$.

1. Rates per new (2-assembly) inspection unit. Pre-shift: $m_0=2\hat\lambda=2(2.00)=\boxed{4.00}$. Post-shift ($1.5\times$ the per-assembly rate): $m_1=2(1.5\times2.00)=\boxed{6.00}$.
2. Required single-sample detection probability. $1-(1-p_1)^2\geq0.7\ \Rightarrow\ (1-p_1)^2\leq0.3\ \Rightarrow\ p_1\geq1-\sqrt{0.3}=\boxed{0.4523}.$
3. Search over sample size $k$ (inspection units). With control limit fixed at the pre-shift rate, $UCL(k)=m_0+3\sqrt{m_0/k}$, and using the normal approximation to the post-shift Poisson count at rate $m_1$ per unit over $k$ units ($U\sim N(m_1,\,m_1/k)$), the single-sample detection probability is $p_1(k)=P\!\big(U>UCL(k)\big)=1-\Phi\!\Big(\dfrac{UCL(k)-m_1}{\sqrt{m_1/k}}\Big)$:

   $k$	6	7	8	9	10
   $UCL(k)$	6.449	6.268	6.061	6.000	5.897
   $p_1(k)$	0.327	0.386	0.444	0.500	0.553
   $1-(1-p_1)^2$	0.547	0.623	0.691	0.750	0.800

   $k=8$ still falls short (0.691 < 0.7); $k=9$ is the first value that clears the target (0.750 $\geq$ 0.7), and at $k=9$ the fixed $UCL(9)=4+3\sqrt{4/9}=6.00$ lands exactly on $m_1$, giving $p_1=0.500$ by symmetry.
4. Minimum sample size. $k=\boxed{9}$ inspection units of 2 assemblies each, i.e. $\boxed{18\ \text{assemblies per sample}}$.