Question 1 of 7: A Control Criterion for the Watch-Gear Dimension

Question 1: A Control Criterion for the Watch-Gear Dimension (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.

Given. A continuing variables inspection of one key gear dimension, summarised by 25 preliminary subgroups of five pieces each.

Find. The statistical criterion that signals loss of control, how that criterion compares with a drawing specification, and the courses of action open to the vendor when the two are not compatible.

Approach. Convert the two summary statistics into a pair of Shewhart charts using the standard subgroup factors, then convert the same $\bar{R}$ into an estimate of the process standard deviation so that the natural tolerance of the process can be laid alongside whatever tolerance the horologist has specified.

1. Part (a) — choose the right criterion. The dimension is a measured variable and the data arrive in rational subgroups of five, so the criterion is a pair of Shewhart charts: an $\bar{X}$ chart to police the location of the process and an $R$ chart to police its spread. A criterion built on one chart alone is incomplete, because a process can drift off aim with a perfectly stable spread, or open up in spread with the mean untouched. For $n = 5$ the factors are $$A_2 = 0.577, \qquad D_3 = 0, \qquad D_4 = 2.114, \qquad d_2 = 2.326$$
2. Set the range-chart limits first. The $R$ chart must be brought into control before the $\bar{X}$ limits mean anything, because $\bar{R}$ is the estimate of spread on which those limits rest. $$\text{UCL}_R = D_4 \bar{R} = 2.114 \times 0.002 = 0.004228 \text{ in}$$ $$\text{LCL}_R = D_3 \bar{R} = 0 \times 0.002 = 0$$ The lower limit vanishes for every subgroup of six or fewer, which is why an $R$ chart at $n = 5$ can only signal an increase in spread.
3. Set the limits on the mean. The $A_2$ factor already contains the conversion from $\bar{R}$ to the standard error of a subgroup mean, so $$\text{UCL}_{\bar{X}},\ \text{LCL}_{\bar{X}} = \bar{\bar{X}} \pm A_2 \bar{R} = 0.125 \pm (0.577)(0.002) = 0.125 \pm 0.001154$$ $$\boxed{\;\bar{X}\text{ chart: } 0.123846 \le \bar{X} \le 0.126154 \text{ in}; \qquad R\text{ chart: } 0 \le R \le 0.004228 \text{ in}\;}$$ The criterion the vendor should set up is therefore: plot the mean and the range of every subgroup of five; declare the process out of control whenever a point falls outside either band, or whenever the points inside the band show a run, trend or other non-random pattern.
4. Convert the same data into a process standard deviation. The control limits describe the behaviour of subgroup averages, which are far less variable than individual gears, so they cannot be compared with a specification directly. The bridge is the $d_2$ factor: $$\hat{\sigma} = \frac{\bar{R}}{d_2} = \frac{0.002}{2.326} = 0.00085985 \text{ in}$$ As an internal check, $3\hat{\sigma}/\sqrt{n} = 3(0.00085985)/\sqrt{5} = 0.001154$ in, which reproduces $A_2\bar{R}$ exactly — the two routes to the $\bar{X}$ limits agree.
5. State the natural tolerance of the process. Individual gears, if the process stays in control and the dimension is approximately normal, spread over $$6\hat{\sigma} = 6(0.00085985) = 0.005159 \text{ in}, \qquad \bar{\bar{X}} \pm 3\hat{\sigma} = 0.122421 \text{ to } 0.127580 \text{ in}$$ $$\boxed{\;\text{natural tolerance} = 0.125 \pm 0.00258 \text{ in}\;}$$ This, and not the control limits, is what must be set against the drawing.

How the criterion should compare with the specification

Statistical control and conformance to specification are two different ideas, and the whole point of the comparison is that a process can be in perfect control and still make scrap. Control limits come from the process; specification limits come from the designer. The quantity that connects them is the capability ratio

For the criterion to be compatible with the specification the process needs $C_p \ge 1$, and industrial practice asks for $C_p \ge 1.33$ so that ordinary drift does not immediately produce non-conforming gears. Because the exam does not print the drawing tolerance, the comparison is best made as a threshold: the horologist's tolerance band must be at least 0.00516 in wide, that is at least plus or minus 0.00258 in about the nominal size. On a plus or minus 0.003 in drawing the vendor has $C_p = 0.006/0.005159 = 1.16$ and the criterion is compatible, if not generous. On a plus or minus 0.002 in drawing $C_p = 0.004/0.005159 = 0.78$, the criterion is incompatible, and about 2.0 per cent of gears fall outside the drawing even when every plotted point sits inside the control limits.

Alternatives when the criterion is not compatible with the specification

Four courses of action are open, and a good answer ranks them rather than listing them. First, centre the process. If $\bar{\bar{X}}$ is off the nominal size, part of the non-conformance is pure aim and is corrected by a tool-setting adjustment at essentially no cost; this is always the first move because it buys capability without touching variability. Second, reduce the variability itself, which is the only permanent cure when the process is already centred: a more rigid fixture, a tighter-running spindle, better material consistency, closer control of tool wear, or a different machine altogether. Third, renegotiate the tolerance. Tolerances are frequently tighter than function requires, and a designed experiment or a functional study that shows the gear works at plus or minus 0.003 in is far cheaper than a new machine; this conversation belongs with the horologist, since it is the customer's drawing. Fourth, and only as an interim measure, screen the output — 100 per cent inspection with sorting or selective assembly — which converts an incapable process into conforming shipments at the price of inspection cost and guaranteed scrap or rework. Screening adds no value and should be treated as a containment action while one of the first three takes effect.