Question 4 of 6: Attributes Charts — p vs. u, the c Chart's Inspection Unit, and a u Chart Build

Question 4: Attributes Charts — p vs. u, the c Chart's Inspection Unit, and a u Chart Build (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) One chart for attributes, two for variables; $p$ vs. $u$ charts

A variables measurement (e.g. tensile strength) can drift in two statistically independent ways — its central tendency ($\mu$) and its dispersion ($\sigma$) — and a normal distribution needs both parameters to be fully specified; an X̄ chart alone can miss a pure variance increase (mean unchanged), so a companion R or S chart is required to track spread. An attributes characteristic (fraction nonconforming $p$, or count of nonconformities per unit $u$) instead comes from a single-parameter distribution (binomial or Poisson) in which the variance is a fixed function of that one parameter ($\sigma^2=np(1-p)$ for binomial, $\sigma^2=u$ for Poisson) — there is no independent second parameter to track, so a single chart on the parameter itself fully characterizes the process.

$p$ chart: plots the fraction of nonconforming units in a sample (each unit is simply classified conforming/nonconforming — go/no-go), modeled as Binomial$(n,p)$; centre line $\bar p$, limits $\bar p\pm3\sqrt{\bar p(1-\bar p)/n}$. $u$ chart: plots the average number of nonconformities per inspection unit — a single physical unit can have several distinct nonconformities (e.g. 3 scratches and a dent on one panel) which a $p$ chart cannot represent — modeled as Poisson$(u)$; centre line $\bar u$, limits $\bar u\pm3\sqrt{\bar u/n}$, where $n$ is the number of inspection units in the sample (naturally handling variable sample sizes, unlike the closely related $c$ chart which requires a constant inspection-unit count).

(b) Why a $c$ chart's inspection unit aggregates several physical units; the $c_i=0$ policy

If the true nonconformity rate per single physical unit is very low, defining 1 inspection unit = 1 physical part gives a Poisson mean $c$ close to zero; the resulting $c$ chart would have an LCL clamped at 0 and a very "sparse" count (mostly 0's and occasional 1's), which is statistically inefficient (poor power to detect a real increase) and hard to interpret visually. Aggregating several physical units into one inspection unit (e.g. 5 parts) inflates the expected count $c$ to a more workable magnitude (rule of thumb: $\bar c\gtrsim$ about 1–2, ideally higher) so the chart has meaningful resolution and the normal approximation used for the 3-sigma limits is reasonable.

When $LCL=0$ and $c_i=0$: a single observation of $c_i=0$ is not below the LCL (it sits exactly at the boundary, which is not a violation) and, by itself, is not a statistical signal — under a Poisson process with a moderate mean, an occasional zero count is a normal, expected outcome, so the process should not be stopped for a single zero. The situation calls for investigation only when zero counts occur far more often than the Poisson model predicts — e.g. a run of several consecutive $c_i=0$ points (a classic Western Electric run-type rule), which is itself an assignable-cause signal (possibly a genuine, desirable improvement, but also possibly that inspection is not actually being performed, or a change in the inspection/measurement system) and should be investigated even though no individual point violated a control limit.

(c) u chart for nonconformities per inspection unit (variable daily units)

Given. 10 days of data; 1 inspection unit = 2 assemblies.

Find. The appropriate control chart with 3-sigma limits, revised if necessary, and the in-control mean number of nonconformities per assembly.

Approach. The number of inspection units inspected varies day to day (1 or 2), so a plain $c$ chart (which assumes a constant inspection-unit count) is not appropriate — a $u$ chart (nonconformities per inspection unit, with sample-size–dependent control limits) is required instead.

1. Convert to inspection units and compute $u_i$. $n_i=(\text{assemblies inspected})/2$; $u_i=c_i/n_i$.

   Day	1	2	3	4	5	6	7	8	9	10
   $n_i$ (insp. units)	1	2	1	2	1	1	2	2	1	2
   $u_i$	8.0	13.0	10.0	15.0	18.0	9.0	10.0	12.0	6.0	14.0

2. Centre line. $$\bar u=\frac{\sum c_i}{\sum n_i}=\frac{179}{15}=11.93$$
3. Variable 3-sigma limits, $UCL_i/LCL_i=\bar u\pm3\sqrt{\bar u/n_i}$. For $n_i=1$: $UCL=11.93+3\sqrt{11.93/1}=22.30$, $LCL=11.93-3\sqrt{11.93}=1.57$. For $n_i=2$: $UCL=11.93+3\sqrt{11.93/2}=19.26$, $LCL=11.93-3\sqrt{11.93/2}=4.61$.
4. Check for out-of-control points. Comparing every $u_i$ above against its day's own limits (1.57–22.30 for the 1-unit days, 4.61–19.26 for the 2-unit days), all 10 points fall inside their control limits — the largest, $u_5=18.0<22.30$, and the smallest, $u_9=6.0>1.57$. $$\boxed{\text{No point is out of control} \Rightarrow \text{no revision of the limits is needed}}$$
5. In-control mean nonconformities per assembly. $\bar u=11.93$ is per inspection unit (= 2 assemblies), so per assembly: $$\hat c_{\text{assembly}}=\frac{\bar u}{2}=\frac{11.93}{2}=5.97$$ $$\boxed{\hat c_{\text{assembly}}=5.97\text{ nonconformities/assembly}}$$