Question 5 of 6: Attributes Control Charts and a u-Chart for Disk-Drive Assemblies

Question 5: Attributes Control Charts and a u-Chart for Disk-Drive Assemblies (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) Purpose of the p, np, c, u, and demerit charts

The $p$ chart tracks the FRACTION of nonconforming units in a sample (each unit classified simply good/bad) and, unlike the $np$ chart, naturally accommodates a sample size that VARIES from period to period. The $np$ chart tracks the raw COUNT of nonconforming units and requires a CONSTANT sample size (since the count's expected value and control limits both depend directly on $n$). The $c$ chart tracks the COUNT of individual nonconformities (a single unit can have several) found within an inspection unit of CONSTANT size/area of opportunity. The $u$ chart tracks nonconformities PER UNIT ($u=c/n$) and is the correct choice when the area of opportunity (number of units inspected, or units per inspection lot) VARIES, since its control limits are recomputed for each sample's own $n$. The demerit chart extends the $c$/$u$ family to nonconformities of DIFFERING severity (e.g. critical, major, minor) by assigning each class a weight and charting a single weighted "demerit" score per unit, so that one critical defect and several trivial ones are not treated as equally serious.

(b) A zero-count sample on a $u$-chart with $LCL=0$: when to investigate

A single sample showing zero nonconformities, with $\hat u_i=0$ landing exactly ON the lower control limit rather than crossing below it, is not by itself a signal to stop the process and search for an assignable cause. Under the Poisson model that underlies the $u$-chart, a count of zero is entirely consistent with in-control operation at the estimated rate $\bar u$ — indeed, for a modest $\bar u$, zero is often the SINGLE MOST LIKELY outcome for any one sample — and because the point sits on the boundary rather than beyond it, there is no statistical evidence of a genuine change.

What DOES call for investigation is a long, statistically improbable RUN of consecutive zero (or near-zero) samples: under the fitted Poisson rate $\bar u$, the probability of several samples in a row all showing zero nonconformities can itself become vanishingly small, making such a run a legitimate non-random-pattern signal (one of the standard supplementary run rules) even though every individual point sits inside the limits. When that pattern appears, the correct response is to investigate WHY the count has stayed at zero — genuine, sustained process improvement is one welcome explanation, but so are inspectors failing to detect/record real nonconformities, a change in inspection method that under-counts, or misreporting — and only once a genuine cause is confirmed should the centre line be revised downward.

(c) $u$-chart for the disk-drive nonconformity data

Given. 10 days of final inspection; assemblies inspected and total workmanship nonconformities per day vary day to day. One inspection unit is defined as two assemblies:

Find. An appropriate 3-sigma control chart; whether any point requires revision; and the in-control mean nonconformities per assembly.

Approach. Because the number of inspection units per day varies, the $c$-chart's constant-area-of-opportunity requirement is violated, so the correct chart is a $u$-chart ($u_i=c_i/n_i$), whose control limits vary with $n_i$: $UCL_i=\bar u+3\sqrt{\bar u/n_i}$, $LCL_i=\max\!\left(0,\ \bar u-3\sqrt{\bar u/n_i}\right)$.

1. Centre line. $$\bar u=\frac{\sum c_i}{\sum n_i}=\frac{15+40+18+50+36+28+42+52+12+60}{2+3+2+4+2+1+3+4+2+4}=\frac{353}{27}=\boxed{13.07\ \text{nonconformities/assembly}}.$$
2. Day-by-day $u_i$ and limits. (values in nonconformities/assembly) $$u_i:\ 7.50,\,13.33,\,9.00,\,12.50,\,18.00,\,28.00,\,14.00,\,13.00,\,6.00,\,15.00.$$ $$UCL_i\ (n_i{=}1,2,3,4):\ 23.92,\,20.74,\,19.34,\,18.50;\qquad LCL_i\ (n_i{=}1,2,3,4):\ 2.23,\,5.40,\,6.81,\,7.65.$$
3. Check each point against its own day's limits. Day 6 has $n_6=1$, so $UCL_6=23.92$; its observed $u_6=28.00\gt UCL_6$, an out-of-control signal. Every other day falls strictly between its own $UCL_i$/$LCL_i$ (e.g. day 5: $n_5=2$, $u_5=18.00\lt UCL_5=20.74$; day 10: $n_{10}=4$, $u_{10}=15.00\lt UCL_{10}=18.50$).
4. Revise. Since day 6 signals, remove it and recompute the centre line from the remaining 9 days: $$\bar u_{\text{rev}}=\frac{353-28}{27-1}=\frac{325}{26}=\boxed{12.50\ \text{nonconformities/assembly}}.$$ Recomputing $UCL_i$/$LCL_i$ at $\bar u_{\text{rev}}=12.50$ for the remaining 9 days, no further point signals — the revised chart is stable.