Question 5 of 6: Variables vs. Attributes Charts, and a $u$-Chart for Gas Water Heater Inspection

Question 5: Variables vs. Attributes Charts, and a $u$-Chart for Gas Water Heater Inspection (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) Variables vs. attributes charts, and the demerit chart

Variables charts (e.g. $\bar X$/$R$) use a directly measured continuous characteristic, so they carry far more information per unit inspected (a measurement says how good or bad, an attribute only says good/bad), detect a shift with a much smaller sample than an equivalent attributes chart, and let the analyst separate location ($\bar X$) from spread ($R$) explicitly. Their disadvantage is cost and practicality: precise measurement (gauges, calibration, trained inspectors, time per unit) is more expensive than a simple pass/fail check, and only one characteristic is monitored per chart pair, so several variables charts may be needed to cover a multi-characteristic part. Attributes charts ($p$, $np$, $c$, $u$) need only a go/no-go judgement (or a defect count), so inspection is fast, cheap, and can summarize many different characteristics or an entire unit's overall conformance in a single chart; their disadvantage is that they carry much less information (a unit that barely fails and one that fails badly are recorded identically), so far larger samples are needed to detect a shift of the same practical size, and the underlying dimension that actually drifted is not directly visible from the chart.

A pair of charts is required for a variable because a single number cannot separately convey both location and spread: the $\bar X$ chart alone could look perfectly stable while the process spread silently widens (an $R$-chart signal), and the $R$ chart alone says nothing about whether the process is drifting off target — both must be watched together to fully describe the process's behaviour. An attribute, by contrast, collapses to a single Bernoulli or count outcome per unit or per inspection unit, and one plotted statistic ($\hat p$, count, or $u$) already fully captures everything the chart is designed to track, so a second, independent chart adds nothing.

A demerit chart extends the $c$/$u$-chart idea by weighting nonconformities by severity class (e.g., very serious/serious/moderate/minor, commonly weighted 100/50/10/1) and plotting a single weighted demerit score per unit rather than a raw count, so that a handful of critical defects are not diluted by, nor a large volume of cosmetic defects hidden behind, an unweighted count.

(b) $u$-chart for water-heater workmanship nonconformities

Given. Inspection unit $=1$ water heater; $20$ samples over $20$ days, each sample $n=4$ heaters; total nonconformities across all samples $=40$.

Find. An appropriate control chart, its control limits, and the in-control expected number of defects per heater.

Approach. Because the sample (4 heaters) inspects more than one inspection unit (1 heater) at a time, and the count of interest is nonconformities (not simply nonconforming/conforming units), a $u$-chart (average nonconformities per inspection unit) is the appropriate chart, with a constant sample size of $n=4$ inspection units per sample.

1. Average nonconformities per inspection unit. $$\bar u=\frac{\text{total nonconformities}}{\text{samples}\times n}=\frac{40}{20\times4}=\boxed{0.5\ \text{per water heater}}.$$
2. $u$-chart control limits ($n=4$ constant). $$UCL_u=\bar u+3\sqrt{\frac{\bar u}{n}}=0.5+3\sqrt{\frac{0.5}{4}}=0.5+1.061=\boxed{1.561},$$ $$LCL_u=0.5-1.061=-0.561\ \Rightarrow\ \boxed{0\ (\text{set to }0,\text{ a count cannot be negative})}.$$
3. Expected defects per water heater. With the process assumed in control, the centre line itself is the estimate: $\hat u=\bar u=\boxed{0.5\ \text{nonconformities per heater}}$.

(c) Control limits after redefining the inspection unit to 2 heaters

Given. Same underlying data as (b) (40 nonconformities, 20 samples, 4 heaters/sample); inspection unit re-defined as 2 water heaters; sample size unchanged at 4 heaters, i.e. $2$ (new) inspection units per sample.

Find. The $u$-chart control limits under the redefined inspection unit.

Approach. Re-express the same nonconformity data per the larger (2-heater) inspection unit: the average nonconformities per new unit doubles (twice as many heaters per unit), while the number of new-unit inspection opportunities per sample halves.

1. Average nonconformities per new (2-heater) inspection unit. $4$ heaters per sample $=2$ new inspection units per sample, so $$\bar u_{new}=\frac{40}{20\times2}=\boxed{1.0\ \text{per 2-heater unit}}\quad(\text{equivalently, }2\times0.5\text{ from (b), as expected}).$$
2. Redefined $u$-chart control limits ($n=2$ new units/sample). $$UCL_u=1.0+3\sqrt{\frac{1.0}{2}}=1.0+2.121=\boxed{3.121},$$ $$LCL_u=1.0-2.121=-2.121\ \Rightarrow\ \boxed{0}.$$