Question 6 of 6: Shewhart Control Charts and Their Power to Detect a Shift

Question 6: Shewhart Control Charts and Their Power to Detect a Shift (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. Two independent base-period data sets from which Shewhart charts are to be set up.

Find. For each data set: the centre line and 3-sigma control limits of the mean and range charts, the estimated process standard deviation, and the chart's power to detect the stated shift — expressed as an average run length in part (a) and as a single-subgroup detection probability in part (b).

Approach. Estimate the process centre and spread from the base period, build the limits from the tabulated Shewhart constants, then treat detection as a probability problem: with the mean displaced, the sampling distribution of the subgroup mean slides relative to fixed limits, and the tail area beyond them is the chance of a signal on any one subgroup.

Part (a) — item weight, n = 3

1. Establish the centre lines. The two sums given are sums of the 30 subgroup means and of the 30 subgroup ranges, so $$\bar{\bar{X}} = \frac{\sum \bar{X}}{k} = \frac{12{,}930}{30} = 431.0\ \text{g}, \qquad \bar{R} = \frac{\sum R}{k} = \frac{123}{30} = 4.10\ \text{g}$$ That the sums are of subgroup means rather than of the 90 individual readings is settled by the question itself: it later moves the process average to 433 g, which is only meaningful beside a centre line of 431 g.
2. Build the mean chart. With $n = 3$ the Shewhart constant is $A_2 = 1.023$, and the half-width is $$A_2\bar{R} = 1.023 \times 4.10 = 4.194\ \text{g}$$ so that $$\boxed{\mathrm{UCL}_{\bar{X}} = 431.0 + 4.194 = 435.19\ \text{g}, \qquad \mathrm{LCL}_{\bar{X}} = 431.0 - 4.194 = 426.81\ \text{g}}$$
3. Build the range chart. For $n = 3$, $D_3 = 0$ and $D_4 = 2.574$, so $$\mathrm{UCL}_R = D_4\bar{R} = 2.574 \times 4.10 = 10.55\ \text{g}, \qquad \mathrm{LCL}_R = D_3\bar{R} = 0$$ The lower limit is zero for every subgroup size up to six: with so few observations the range simply cannot be small enough to be evidence of anything, so a range chart at $n \le 6$ can only detect an increase in variability.
4. Estimate the process standard deviation. The mean range of a normal process is $d_2\sigma$, so $$\hat{\sigma} = \frac{\bar{R}}{d_2} = \frac{4.10}{1.693} = \boxed{2.42\ \text{g}}$$ As a check that the chart is internally consistent, the standard error of a subgroup mean is $\hat{\sigma}/\sqrt{n} = 2.4217/\sqrt{3} = 1.398\ \text{g}$, and three of those is $4.194\ \text{g}$ — exactly the half-width computed in step 2, which is what "$3\sigma$ limits" means.
5. Set up the detection problem. The limits stay where they are; the process mean moves to 433 g. The subgroup mean is then normal about 433 g with standard error 1.398 g, and the chance of a signal on any one subgroup is the probability that it falls outside the old limits: $$z_{\mathrm{U}} = \frac{435.194 - 433}{1.398} = 1.569, \qquad z_{\mathrm{L}} = \frac{426.806 - 433}{1.398} = -4.430$$
6. Evaluate the detection probability and the run length. From the standard normal tables, $$p = P(z > 1.569) + P(z < -4.430) = 0.0583 + 0.0000 = 0.0583$$ The lower-limit contribution is under five in a million and is rightly ignored. Successive subgroups are independent, so the number of subgroups until the first signal is geometric and its mean is the average run length $$\mathrm{ARL} = \frac{1}{p} = \frac{1}{0.0583} = \boxed{17.2 \approx 17\ \text{subgroups}}$$
7. Interpret that answer. Seventeen subgroups is a poor showing, and the reason is that the shift is small relative to the noise: 2 g against a standard error of 1.398 g is only 1.43 standard errors, and a 3-sigma chart is deliberately insensitive at that scale in exchange for a false alarm rate of about one in 370 when the process is in control. The chance the shift is still undetected after ten subgroups is $(1-0.0583)^{10} = 0.55$, and after twenty it is still 0.30. If a 2 g shift matters commercially, the chart must be changed rather than the arithmetic: enlarge the subgroup (at $n = 9$ the standard error halves and the ARL falls to about 4), add the Western Electric run rules, or replace the Shewhart chart with a CUSUM or EWMA chart, which accumulate evidence across subgroups instead of judging each one alone.

Part (b) — component length, n = 4

1. Establish the centre lines. Over 25 subgroups of four, $$\bar{\bar{X}} = \frac{500}{25} = 20.0\ \text{cm}, \qquad \bar{R} = \frac{153.2}{25} = 6.128\ \text{cm}$$
2. Build both charts from the n = 4 constants. With $A_2 = 0.729$, $D_3 = 0$ and $D_4 = 2.282$, the mean-chart half-width is $A_2\bar{R} = 0.729 \times 6.128 = 4.467\ \text{cm}$, so $$\boxed{\mathrm{UCL}_{\bar{X}} = 24.467\ \text{cm}, \quad \mathrm{LCL}_{\bar{X}} = 15.533\ \text{cm}, \quad \mathrm{UCL}_{R} = 13.98\ \text{cm}, \quad \mathrm{LCL}_{R} = 0}$$
3. Estimate the process standard deviation. With $d_2 = 2.059$ at $n = 4$, $$\hat{\sigma} = \frac{6.128}{2.059} = 2.976\ \text{cm}, \qquad \sigma_{\bar{X}} = \frac{2.976}{\sqrt{4}} = 1.488\ \text{cm}$$ and again $3 \times 1.488 = 4.464 \approx 4.467\ \text{cm}$, confirming the limits are true 3-sigma limits.
4. Slide the mean by 2 cm and take the tail areas. The mean moves from 20.0 to 22.0 cm while the limits stay fixed, so $$z_{\mathrm{U}} = \frac{24.467 - 22.0}{1.488} = 1.658, \qquad z_{\mathrm{L}} = \frac{15.533 - 22.0}{1.488} = -4.346$$
5. Evaluate the probability of detection on the first subgroup. Adding the two tails, $$P(\text{detect}) = P(z > 1.658) + P(z < -4.346) = 0.0487 + 0.0000 = \boxed{0.0487 \equiv 4.87\ \text{per cent}}$$ The complement, $\beta = 0.951$, is the probability the first subgroup after the shift gives no signal at all.
6. Interpret and recommend. Fewer than one chance in twenty of catching the shift immediately is a weak chart, for the same structural reason as in part (a): a 2 cm shift is only $2/1.488 = 1.34$ standard errors, and the corresponding average run length is $1/0.0487 = 20.6$ subgroups. Note that the process spread here is itself very large — a standard deviation of 2.98 cm on a 20 cm nominal length is a coefficient of variation of 15 per cent — so on a newly started component the first priority is not a more sensitive chart at all, but reducing the variation that makes the chart insensitive. Once $\hat{\sigma}$ is brought down, the same 2 cm shift becomes many standard errors and the existing chart detects it almost at once.