Question 2 of 7: Control Limits, Detection Time and Detection Probability

Question 2: Control Limits, Detection Time and Detection Probability (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.

Find. For each part, the three-sigma limits of the $\bar{X}$ and $R$ charts and the estimated process standard deviation; then, for the stated shift, how quickly the chart reacts — expressed in part (a) as an average run length and in part (b) as the probability of a signal on the very first subgroup.

Approach. Estimate the process centre and spread from the base-period averages, convert the average range into an estimate of the population standard deviation through $d_2$, set the limits with the tabulated factors, and then treat detection as a Bernoulli trial per subgroup whose success probability is the area of the shifted sampling distribution outside the limits.

1. Part (a) — reduce the base period to a centre and a spread. The grand mean and the average range are simple averages over the 30 subgroups, $$\bar{\bar{X}} = \frac{\sum \bar{X}}{k} = \frac{12{,}930}{30} = 431.0\ \text{g}, \qquad \bar{R} = \frac{\sum R}{k} = \frac{123}{30} = 4.10\ \text{g}.$$
2. Set the three-sigma limits on both charts. With $n=3$ the half-width is $A_2\bar{R} = 1.023 \times 4.10 = 4.194$ g, so $$\boxed{UCL_{\bar{X}} = 435.19\ \text{g},\qquad CL = 431.00\ \text{g},\qquad LCL_{\bar{X}} = 426.81\ \text{g},}$$ and on the range chart $UCL_R = D_4\bar{R} = 2.574 \times 4.10 = 10.55$ g with $LCL_R = D_3\bar{R} = 0$ (no lower limit exists for $n \le 6$).
3. Estimate the standard deviation of the individual item weights. The average range estimates the population spread through the tabulated $d_2$ factor: $$\hat{\sigma} = \frac{\bar{R}}{d_2} = \frac{4.10}{1.693} = \boxed{\hat{\sigma} = 2.42\ \text{g per item.}}$$ The corresponding standard error of a subgroup mean is $\sigma_{\bar{X}} = \hat{\sigma}/\sqrt{3} = 1.398$ g, and $3\sigma_{\bar{X}} = 4.19$ g reproduces the half-width found from $A_2$ — the arithmetic proof that the limits really are three standard errors of the mean, not three standard deviations of an item.
4. Part (a) II — find the chance of a signal once the average moves. With the limits frozen and the mean at $\mu' = 433.0$ g, the subgroup mean is still normal with standard error 1.398 g, so $$z_U = \frac{435.19 - 433.0}{1.398} = 1.569,\qquad z_L = \frac{426.81 - 433.0}{1.398} = -4.43.$$ The lower tail is negligible ($< 5 \times 10^{-6}$), leaving $$P(\text{signal on one subgroup}) = 1 - \Phi(1.569) = \boxed{0.0583\ (5.83\ \%),}$$ so the type II risk on any one subgroup is $\beta = 0.9417$.
5. Convert that probability into a detection time. Subgroups are independent, so the number taken until the first signal is geometric and its mean is the reciprocal of the signal probability: $$ARL = \frac{1}{1-\beta} = \frac{1}{0.0583} = \boxed{ARL = 17.2\ \text{subgroups on average.}}$$ Expressed as risk rather than as an average, the shift is still undetected after 10 subgroups with probability $0.9417^{10} = 0.551$ and after 20 with probability $0.304$. If subgroups are taken hourly the shift persists about two shifts on average, which is the operationally useful way to report it.
6. Part (b) — repeat the reduction for the length chart. $$\bar{\bar{X}} = \frac{500}{25} = 20.00\ \text{cm},\qquad \bar{R} = \frac{153.2}{25} = 6.128\ \text{cm}.$$ With $n=4$ the half-width is $A_2\bar{R} = 0.729 \times 6.128 = 4.467$ cm, so $$\boxed{UCL_{\bar{X}} = 24.47\ \text{cm},\qquad LCL_{\bar{X}} = 15.53\ \text{cm},}$$ with $\boxed{UCL_R = 13.98\ \text{cm and } LCL_R = 0.}$ The spread estimate is $\hat{\sigma} = 6.128/2.059 = 2.98$ cm, giving $\sigma_{\bar{X}} = 2.98/\sqrt{4} = 1.488$ cm, and again $3\sigma_{\bar{X}} = 4.46$ cm agrees with $A_2\bar{R}$.
7. Compute the probability of catching a 2 cm shift immediately. The mean moves to 22.00 cm while the limits stay put, so $$z_U = \frac{24.47 - 22.00}{1.488} = 1.658,\qquad z_L = \frac{15.53 - 22.00}{1.488} = -4.35,$$ and the probability of a point outside either limit on the first subgroup after the shift is $$P = \left[1-\Phi(1.658)\right] + \Phi(-4.35) = 0.0487 + 0.00001 = \boxed{P = 0.0487\ (4.87\ \%).}$$ Equivalently $\beta = 0.9513$ on that first subgroup, and the average run length to detection is $1/0.0487 = 20.5$ subgroups.

Both parts make the same point from opposite ends. A three-sigma $\bar{X}$ chart is built to leave the process alone when nothing has changed, and the price of that quietness is that a shift of about 1.3–1.4 standard errors — which is what both of these are (1.43 in part (a), 1.34 in part (b)) — is missed nineteen times out of twenty on any single subgroup. If faster reaction is wanted, the levers are a larger subgroup (which shrinks $\sigma_{\bar{X}}$ as $\sqrt{n}$), a supplementary runs rule, or a cumulative-sum or exponentially-weighted chart, not a tightening of the three-sigma limits, which would flood the operator with false alarms.

Final results.