Question 4 of 5: Control Charts for the Mean and the Range

Question 4: Control Charts for the Mean and the Range (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 part (a) the trial control limits on both charts, the estimated standard deviation of individual item weights, and how long the chart takes to signal a shift of the mean to 433 g; for part (b) the three-sigma limits and the probability that a 2 cm shift is caught on the very first subgroup taken after it happens.

Approach. Average the subgroup statistics to obtain the centre lines, apply the Shewhart factors to place the limits, convert $\bar{R}$ to an estimate of the population standard deviation through $d_{2}$, and then treat the shifted process as a normal distribution of subgroup means to obtain the detection probability and the average run length.

1. Part (a) — establish the two centre lines. The grand average and the average range are simply the printed sums divided by the number of 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. Place the limits on the chart for the mean. For $n=3$ the factor is $A_{2}=1.023$, so the half-width is $A_{2}\bar{R}=1.023(4.10)=4.194$ g and $$\text{UCL}_{\bar{X}}=\bar{\bar{X}}+A_{2}\bar{R}=431.0+4.194=435.19\ \text{g}, \qquad \text{LCL}_{\bar{X}}=431.0-4.194=426.81\ \text{g}$$
3. Place the limits on the chart for the range. With $n=3$ the range factors are $D_{4}=2.574$ and $D_{3}=0$, so $$\text{UCL}_{R}=D_{4}\bar{R}=2.574(4.10)=10.55\ \text{g}, \qquad \text{LCL}_{R}=D_{3}\bar{R}=0$$ A zero lower limit is normal for subgroups of five or fewer: with so few observations a range of zero is not improbable, so no lower limit can be set.
4. Estimate the standard deviation of individual weights. The average range estimates the population standard deviation through the factor $d_{2}=1.693$ for $n=3$: $$\hat{\sigma}=\frac{\bar{R}}{d_{2}}=\frac{4.10}{1.693} =\boxed{\hat{\sigma}=2.42\ \text{g}}$$ This is the spread of individual items. The spread of the subgroup mean is narrower by $\sqrt{n}$: $\sigma_{\bar{X}}=\hat{\sigma}/\sqrt{3}=2.4217/1.7321=1.398$ g. As an arithmetic check on the whole chain, $3\sigma_{\bar{X}}=3(1.398)=4.194$ g, exactly the half-width computed from $A_{2}\bar{R}$ in step 2 — which is what “three-sigma limits” means.
5. Part (a), second bullet — find the probability of catching the shifted mean on any one subgroup. If the process average moves to $\mu' = 433$ g, the subgroup mean is still normally distributed with standard deviation 1.398 g but is now centred on 433 rather than 431. A point signals only if it falls outside the limits fixed in step 2, so the standardised distances to the two limits are $$z_{U}=\frac{435.194-433}{1.398}=1.569, \qquad z_{L}=\frac{426.806-433}{1.398}=-4.43$$ The probability of exceeding the upper limit is $1-\Phi(1.569)=0.0583$; the probability of falling below the lower limit is $\Phi(-4.43) \approx 5 \times 10^{-6}$, negligible. Hence the chance of a signal on any given subgroup is $P = 0.0583$, or 5.83 per cent.
6. Convert that probability into a time to detect. Subgroups are independent, so the number taken until the first signal is geometric and its mean is the average run length $$\text{ARL}=\frac{1}{P}=\frac{1}{0.0583} =\boxed{\text{ARL}\approx 17\ \text{subgroups}}$$ On average the shift is caught on about the seventeenth subgroup after it occurs — roughly two shifts' worth of sampling if subgroups are taken hourly, and a long time to be producing off-target parts. Put the other way, the chance the shift is still undetected after ten subgroups is $(1-0.0583)^{10}=0.55$, and after twenty subgroups $(1-0.0583)^{20}=0.30$. The reason the chart is so slow is that the shift is small in the relevant units: $2\ \text{g}/2.42\ \text{g}=0.83$ standard deviations of an individual, which is only 1.43 standard errors of a subgroup of three. Larger subgroups, or a supplementary run rule, or a CUSUM or EWMA chart, would all shorten it.
7. Part (b) — establish the centre lines for the length chart. Now $k=25$ and $n=4$: $$\bar{\bar{X}}=\frac{500}{25}=20.0\ \text{cm}, \qquad \bar{R}=\frac{153.2}{25}=6.128\ \text{cm}$$
8. Compute the three-sigma limits on both charts. For $n=4$, $A_{2}=0.729$, $D_{4}=2.282$, $D_{3}=0$, giving a half-width $A_{2}\bar{R}=0.729(6.128)=4.467$ cm and $$\text{UCL}_{\bar{X}}=20.0+4.467=24.47\ \text{cm}, \qquad \text{LCL}_{\bar{X}}=20.0-4.467=15.53\ \text{cm}$$ $$\text{UCL}_{R}=2.282(6.128)=13.98\ \text{cm}, \qquad \text{LCL}_{R}=0$$ The standard deviation of individual lengths follows from $d_{2}=2.059$: $\hat{\sigma}=6.128/2.059=2.976$ cm, so $\sigma_{\bar{X}}=2.976/2=1.488$ cm and again $3\sigma_{\bar{X}}=4.464$ cm agrees with $A_{2}\bar{R}$.
9. Find the probability of catching a 2 cm shift on the first subgroup. After the shift the subgroup mean is centred on $20.0+2.0=22.0$ cm with the same standard error of 1.488 cm, while the limits stay where step 8 put them: $$z_{U}=\frac{24.467-22.0}{1.488}=1.658, \qquad z_{L}=\frac{15.533-22.0}{1.488}=-4.35$$ $$P=\bigl[1-\Phi(1.658)\bigr]+\Phi(-4.35)=0.0487+0.00001 =\boxed{P \approx 0.0487\ \text{, i.e. }4.87\%}$$ So there is only about a one-in-twenty chance of catching it immediately; the complementary figure, $\beta = 0.951$, is the probability of missing it on that first subgroup, and the average run length is $1/0.0487 \approx 21$ subgroups. The reason is the same as in part (a): a 2 cm shift is only $2/2.976 = 0.67$ standard deviations of an individual part, and a Shewhart chart with three-sigma limits is deliberately insensitive to shifts smaller than about one and a half standard deviations so that false alarms stay rare.

A practical note worth adding on both parts: the range chart must be brought into control before the limits on the mean chart mean anything, because $\hat{\sigma}$ and therefore every limit is derived from $\bar{R}$. The question states that the base-period observations show the process in control, which is what licenses using these as ongoing limits rather than trial limits.