Question 3 of 6: Variation, Control/Specification/Tolerance Limits, and an Individuals Chart for Tensile Strength

Question 3: Variation, Control/Specification/Tolerance Limits, and an Individuals Chart for Tensile Strength (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) Common-cause vs. special-cause variation; when a process is "in control"; three kinds of limit

Production processes exhibit two fundamentally different types of variation. Common-cause (chance) variation is the inherent, cumulative effect of many small, unavoidable sources — minor material variability, ambient conditions, normal machine and measurement noise — that are always present and, taken together, define the process's natural, stable spread. Special-cause (assignable) variation comes from a specific, identifiable source that is not part of the process's normal operation — a tool wearing out, an operator error, a bad batch of raw material, a machine going out of adjustment — and is, by definition, both unpredictable and (once found) removable. A process is said to be in statistical control when only common-cause variation is present: successive control-chart points fall randomly within the control limits with no trend, cycle, shift, or run pattern, so the process is stable, predictable, and consistent over time even though it is never perfectly constant.

Control limits are statistical boundaries, computed from the process's own data, placed around the centre line of a control chart at a distance (conventionally $\pm3$ standard deviations of the plotted statistic) chosen to make the chance of a false alarm acceptably small; they answer "is the process behaving the way it always has?" and say nothing about customer requirements. Specification limits are engineering/customer requirements on the individual product characteristic itself, set by design intent or contract, independent of how the process actually performs; they answer "is this unit good enough to ship?" Natural tolerance limits describe what the process is actually capable of producing when it is in control — conventionally $\mu\pm3\sigma$ of the individual measurements — and are used to compare the process's inherent spread against the specification limits (process capability, $C_p=(USL-LSL)/6\sigma$) to judge whether an in-control process can even meet the requirement. A process can be perfectly in statistical control while still producing a large fraction nonconforming, if its natural tolerance limits are wider than the specification limits — control and capability are two separate questions.

(b) Individuals and moving-range chart for the 20 tensile-strength readings

Given. Twenty successive single-sheet tensile-strength measurements (N/mm²), one sheet sampled every 30 minutes (so each "sample" is a single reading, $n=1$):

Find. Trial (then, if needed, revised) control limits for the individual measurements and the moving range, and the resulting in-control estimates of the process mean and standard deviation.

Approach. With $n=1$ per sample, the process mean and short-term variation are monitored with an individuals ($I$) chart and a moving-range ($MR$) chart (moving range of span 2), using $\bar\sigma=\overline{MR}/d_2$ with $d_2=1.128$, $D_3=0$, $D_4=3.267$ for $n=2$; any point beyond the trial $I$-chart limits is investigated, removed if assignable, and the limits recomputed.

1. Trial limits. The 20 readings average $\bar x=187.62$ N/mm²; the 19 successive moving ranges average $\overline{MR}=8.1579$, giving $\hat\sigma=\overline{MR}/d_2=8.1579/1.128=7.2322$ N/mm². Trial limits: $$UCL_I=\bar x+3\hat\sigma=187.62+3(7.2322)=\boxed{209.32\ \text{N/mm}^2},\qquad LCL_I=187.62-3(7.2322)=\boxed{165.92\ \text{N/mm}^2}.$$ For the moving-range chart, $UCL_{MR}=D_4\overline{MR}=3.267(8.1579)=\boxed{26.65\ \text{N/mm}^2}$ (no lower limit, $D_3=0$).
2. Revise the trial limits. Comparing all 20 readings to the trial $I$-chart limits, sheet 8 ($214.2$) exceeds $UCL_I=209.32$ — every other reading (180.4–195.1) is well inside the limits. Sheet 8 is treated as an assignable-cause outlier (a genuine excursion, not measurement noise — it sits roughly $3.7\hat\sigma$ above the mean of the other 19 points) and removed; the remaining 19 points and their 18 moving ranges are recomputed: $$\bar x_{rev}=186.22\ \text{N/mm}^2,\qquad \overline{MR}_{rev}=5.3444\ \Rightarrow\ \hat\sigma_{rev}=\overline{MR}_{rev}/d_2=\boxed{4.738\ \text{N/mm}^2}.$$
3. Revised control limits. $$UCL_I=186.22+3(4.738)=\boxed{200.44\ \text{N/mm}^2},\qquad LCL_I=186.22-3(4.738)=\boxed{172.01\ \text{N/mm}^2},$$ $$UCL_{MR}=3.267(5.3444)=\boxed{17.46\ \text{N/mm}^2}.$$ All 19 remaining individuals and all 18 remaining moving ranges now fall inside these revised limits — no further revision is needed.

(c) Expected number of defectives from the shift to the signal

Given. $\hat\mu_0=186.22$, $\hat\sigma_0=4.738$ N/mm² from part (b); production rate 50 sheets/hour; sampling every 30 minutes ($n=1$ per sample, so every sample IS one produced sheet); the mean shifts to $\mu_1=\mu_0+\sigma$ immediately after a sampled measurement.

Find. The expected number of defective (out-of-spec) sheets produced between the shift and the moment the $I$-chart signals.

Approach. Because a sample here IS a single produced sheet, the probability that a given sample signals out-of-control after the shift equals the probability that a given sheet, produced at the shifted mean $\mu_1$, is itself outside $[LSL,USL]$ — the control limits and the (assumed) specification limits are numerically the same boundary. Use the shifted normal distribution to get that common probability $p$, the geometric run-length result $ARL_1=1/p$ for the number of samples (=sheets) to signal, and the known production rate to convert to elapsed sheets/time.

1. Probability a post-shift reading signals / is defective. With $\mu_1=\mu_0+\sigma=186.22+4.738=190.96$ N/mm² and $\sigma=4.738$: $$z_L=\frac{LSL-\mu_1}{\sigma}=\frac{172.01-190.96}{4.738}=-4.00,\qquad z_U=\frac{USL-\mu_1}{\sigma}=\frac{200.44-190.96}{4.738}=2.00,$$ $$p=P(Z<-4.00)+P(Z>2.00)\approx 0.00003+0.02275=\boxed{0.02278}.$$
2. Average run length to signal. Each post-shift sample independently has probability $p$ of signalling, so the number of samples to signal is geometric with $$ARL_1=\frac{1}{p}=\frac{1}{0.02278}=\boxed{43.9\ \text{samples}}\ \big(\approx 43.9\times30\ \text{min}=1{,}317\ \text{min}\approx22.0\ \text{hours to signal}\big).$$
3. Sheets produced per sampling interval. At 50 sheets/hour and a 30-minute sampling interval, $50\times0.5=25$ sheets are produced between consecutive samples.
4. Expected defectives before the signal. Over the (geometrically distributed) $N=ARL_1$ intervals until signal, the expected total sheets produced is $25\,E[N]=25\,ARL_1$, and each of those sheets is independently defective with the same probability $p$ computed in Step 1 (the control limits and the specification limits coincide, so "this sample signalled" and "this sheet is defective" are the identical event for the sampled sheet, and the other 24 sheets each interval share the same shifted distribution). The expected number of defectives is therefore $$E[\text{defectives}]=25\times ARL_1\times p=25\times\frac{1}{p}\times p=\boxed{25\ \text{defective sheets}}.$$ This clean result is a direct consequence of the assumption in the check callout above (control limits $=$ specification limits): it says the expected defectives produced before detection exactly equal one sampling interval's worth of production, regardless of the actual value of $p$, as long as that equality of limits holds.