Question 5 of 6: Attributes Charts — p-Chart Design, Revision, and ARL-Based np-Chart Sizing

Question 5: Attributes Charts — p-Chart Design, Revision, and ARL-Based np-Chart Sizing (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) p-chart vs. u-chart

A p-chart monitors the proportion (fraction) of nonconforming units in a sample of fixed (or slowly varying) size $n$, where each inspected unit is classified simply as conforming or nonconforming (a Bernoulli/binomial attribute) — a single unit can only count once against the tally, however many individual defects it might carry. A u-chart monitors the average number of defects PER UNIT, appropriate when a single inspected unit (or a fixed inspection area/length/time) can carry any number of individual defects, modelled as a Poisson count; the u-chart is used precisely when "how many separate flaws does this unit have" is the meaningful quality measure rather than a simple pass/fail classification (e.g. blemishes per painted panel, solder defects per board). Both charts allow the sample/inspection-unit size to vary sample-to-sample, with control limits recalculated (widened for smaller $n$, narrowed for larger $n$) accordingly; a p-chart uses $n_i$ (units inspected) while a u-chart uses $n_i$ inspection units of possibly varying total area/length.

(b) p-chart set-up, revision, and largest $p_1$ for $ARL_{p_1}\ge100$

Given. 10 samples, $n=100$ parts each; nonconforming counts 2, 5, 0, 6, 4, 4, 2, 3, 1, 3.

Find. The p-chart's centre line and trial limits, the in-control $p_0$ (after any needed revision), and the largest shifted proportion $p_1$ for which $ARL_{p_1}\ge100$.

1. Centre line and trial limits. $\bar p=\dfrac{\sum d_i}{10(100)}=\dfrac{30}{1000}=0.03$. $$UCL_p=\bar p+3\sqrt{\frac{\bar p(1-\bar p)}{n}}=0.03+3\sqrt{\frac{0.03(0.97)}{100}}=\boxed{0.0812},\qquad LCL_p=0.03-3\sqrt{\frac{0.03(0.97)}{100}}<0\Rightarrow LCL_p=\boxed{0}.$$
2. Check for revision. The 10 sample proportions ($2\%,5\%,0\%,6\%,4\%,4\%,2\%,3\%,1\%,3\%$) all fall between 0 and $UCL_p=0.0812$ — no point signals out of control, so no revision is needed: $p_0=\bar p=\boxed{0.030}$.
3. Largest $p_1$ keeping $ARL_{p_1}\ge100$. With $LCL_p=0$, a shifted proportion above $p_0$ can only signal via the UPPER limit, so $ARL_{p_1}=1/P(\hat p>UCL_p\mid p=p_1)$. Requiring $ARL_{p_1}\ge100$ means $P(\hat p>UCL_p)\le0.01$, i.e. $\dfrac{UCL_p-p_1}{\sqrt{p_1(1-p_1)/n}}\ge z_{0.01}=2.326$. Solving this (mildly nonlinear in $p_1$) numerically gives $$\boxed{p_1=0.0372\ (3.72\%)}\quad\text{at which }P(\text{signal})=0.0100,\ ARL_{p_1}=100.$$ Any $p_1$ larger than 0.0372 pushes the signal probability above 1%, dropping $ARL_{p_1}$ below 100, so 0.0372 is the largest value satisfying the requirement.

(c) Minimum-$n$ np-chart with a positive LCL, and its ARL at $p_1$

Given. $p_0=0.030$ (from part b). Requirement: an np-chart (plotting the COUNT of nonconforming parts, not the proportion) whose lower control limit is strictly positive.

Find. The minimum sample size $n$, the resulting control limits, and $ARL_{p_1}$ at $p_1=0.0372$ (from part b).

1. Minimum $n$ for $LCL_{np}>0$. $LCL_{np}=np_0-3\sqrt{np_0(1-p_0)}>0\ \Rightarrow\ n>\dfrac{9(1-p_0)}{p_0}=\dfrac{9(0.97)}{0.03}=291.0$, so the minimum integer sample size is $\boxed{n=292}$.
2. Control limits at $n=292$. Centre line $=np_0=292(0.03)=8.76$. $$UCL_{np}=np_0+3\sqrt{np_0(1-p_0)}=8.76+3\sqrt{292(0.03)(0.97)}=\boxed{17.50},\qquad LCL_{np}=8.76-3\sqrt{292(0.03)(0.97)}=\boxed{0.015>0}.$$
3. $ARL_{p_1}$ for this chart. At the shifted proportion $p_1=0.0372$, the count of nonconforming parts in $n=292$ has mean $np_1=10.86$ and standard deviation $\sqrt{np_1(1-p_1)}=3.226$. The (normal-approximation) probability of a point exceeding $UCL_{np}=17.50$ is $$P(\text{signal})=P\!\left(Z>\frac{17.50-10.86}{3.226}\right)=P(Z>2.058)=0.0198,\qquad ARL_{p_1}=\frac{1}{0.0198}=\boxed{50.5}.$$