Question 4 of 6: Variables vs. Attributes Charts, the $p=0$ Decision Rule, and a $u$-Chart for Disk-Drive Nonconformities

Question 4: Variables vs. Attributes Charts, the $p=0$ Decision Rule, and a $u$-Chart for Disk-Drive Nonconformities (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) Two charts for variables vs. one for attributes; $p$ vs. $np$; $c$ vs. $u$; multi-part inspection units

Continuous (variable) measurements carry information about two STATISTICALLY INDEPENDENT properties of a normal population — its central tendency (mean) and its dispersion (variance) — and a process can drift out of control in either one without affecting the other (the mean can shift with spread unchanged, or the spread can inflate with the mean still on target). A single chart cannot reliably detect both kinds of shift at once, so variable data are always monitored with a PAIR of charts: one for location ($\bar X$ or individuals) and one for spread ($R$ or $s$). Attribute data (counts or proportions of nonconforming units/nonconformities) instead follow a single-parameter distribution — binomial with parameter $p$, or Poisson with parameter $\lambda$ — in which the VARIANCE IS A FIXED, KNOWN FUNCTION of that one parameter ($np(1-p)$ for the binomial count, $\lambda$ for the Poisson count). There is no separate, independently-estimated "spread" parameter: once the chart's centre-line parameter is estimated, the control limits' width is theoretically fixed by the same distribution, so one chart fully characterizes the process.

The $p$-chart plots the FRACTION nonconforming in a sample, $\hat p=x/n$; because it is a proportion, its limits ($\bar p\pm3\sqrt{\bar p(1-\bar p)/n}$) automatically rescale for a sample size $n$ that VARIES from period to period. The $np$-chart plots the raw COUNT of nonconforming units instead; it is the same underlying binomial model but requires a CONSTANT $n$, since a raw count is comparable across samples only if the opportunity to accumulate it (the sample size) does not change — operators sometimes prefer it because a plain count is more intuitive on the shop floor than a fraction. Analogously, the $c$-chart plots the COUNT of nonconformities (not nonconforming units — a single unit can carry several) found within an inspection unit of CONSTANT size, modelled as Poisson, while the $u$-chart plots nonconformities PER UNIT, $u=c/n$, the Poisson analogue of the $p$-chart, and likewise accommodates a VARYING inspection-unit size or count — the case needed in part (c), where the number of assemblies inspected changes from day to day.

An inspection unit is frequently DEFINED as several physical parts (e.g. 1 inspection unit = 4 parts) because the Poisson $c$-chart behaves poorly, and its lower control limit is uninformative, whenever the average count per inspection unit is close to zero: if a single physical part rarely shows more than 0 or 1 nonconformity, the resulting chart sits at a very low mean with $LCL$ pinned at zero and a $UCL$ so close to the centre line that the chart has almost no power to discriminate a real increase from ordinary Poisson noise. Bundling several physical units into one inspection unit raises the expected count to a workable magnitude (typically several nonconformities per inspection unit) while still keeping a CONSTANT "area of opportunity" for the count to be checked against sample to sample, which is exactly what the $c$-chart's Poisson model requires.

(b) $p$-chart with $LCL=0$: when a sample $\hat p_i=0$ warrants investigation, and when it does not

A single sample landing exactly at $\hat p_i=0$, sitting ON the lower control limit rather than beyond it, is not by itself a signal to stop the process and search for an assignable cause. Under the binomial model, zero nonconforming units in a sample is entirely consistent with in-control operation at the estimated rate $\bar p$ — indeed, when $\bar p$ is small it is typically the single MOST LIKELY outcome for any one sample — and because it sits on the boundary rather than crossing it, there is no statistical evidence of a genuine change. The situation that DOES call for investigation is a long, statistically improbable RUN of consecutive zero (or near-zero) samples: under the fitted $\bar p$, the probability of, say, 7–8 samples in a row all showing zero defectives can itself be vanishingly small, and such a run is a legitimate nonrandom-pattern signal (one of the standard supplementary Western Electric-type run rules) even though every individual point sits inside (on) the limits. When that pattern appears, the appropriate response is to investigate WHY the count has stayed at zero — genuine, sustained process improvement is one possible (welcome) explanation, but so are inspectors failing to detect/record real nonconformities, a change in inspection method that under-counts, or gaming of the reported figures — 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 inspection; assemblies inspected $n_i$ and total nonconformities $c_i$ per day vary day to day:

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 assemblies inspected per day varies, the $c$-chart's constant-area-of-opportunity requirement is violated; the correct chart is a $u$-chart ($u_i=c_i/n_i$), whose control limits vary with $n_i$: $UCL_i=\bar u+3\sqrt{\bar u/n_i}$, $LCL_i=\max\!\big(0,\ \bar u-3\sqrt{\bar u/n_i}\big)$.

1. Centre line. $$\bar u=\frac{\sum c_i}{\sum n_i}=\frac{8+21+10+30+18+10+20+24+15+26}{1+3+2+4+2+1+3+4+2+4}=\frac{182}{26}=\boxed{7.00\ \text{nonconformities/assembly}}.$$
2. Day-by-day $u_i$ and limits. (values in nonconformities/assembly) $$u_i:\ 8.00,\ 7.00,\ 5.00,\ 7.50,\ 9.00,\ 10.00,\ 6.67,\ 6.00,\ 7.50,\ 6.50.$$ $$UCL_i\ (n_i{=}1,2,3,4):\ 14.94,\ 12.61,\ 11.58,\ 10.97;\qquad LCL_i\ (n_i{=}1,2,3,4):\ 0,\ 1.39,\ 2.42,\ 3.03.$$
3. Check each point against its own day's limits. Comparing every $u_i$ against the $UCL_i$/$LCL_i$ for that day's $n_i$ (e.g. day 6: $n_6=1$, $u_6=10.00<UCL_6=14.94$; day 5: $n_5=2$, $u_5=9.00<UCL_5=12.61$), every one of the 10 points falls strictly between its own $UCL_i$ and $LCL_i$ — no point signals.
4. Revision. Since no point is out of control, $\boxed{\text{no revision is necessary}}$; $\bar u=7.00$ nonconformities per assembly is retained as the in-control estimate.