Question 5 of 6: p- vs. np-Charts, a u-Chart for Furniture Scratches, and Chart Design for Fast Detection

Question 5: p- vs. np-Charts, a u-Chart for Furniture Scratches, and Chart Design for Fast Detection (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. np-chart

Both charts monitor DEFECTIVES (units classified pass/fail) from samples with a binomial distribution, and differ only in what is plotted. The p-chart plots the fraction defective $\hat p=D/n$, with mean $E[\hat p]=p$ and standard deviation $\sigma_{\hat p}=\sqrt{p(1-p)/n}$, giving control limits $$UCL,LCL=\bar p\pm3\sqrt{\bar p(1-\bar p)/n}.$$ The np-chart plots the raw COUNT of defectives $D=n\hat p$ itself, with mean $E[D]=np$ and standard deviation $\sigma_D=\sqrt{np(1-p)}$, giving control limits $$UCL,LCL=n\bar p\pm3\sqrt{n\bar p(1-\bar p)}.$$ Because $D=n\hat p$ is just a constant-scaling of $\hat p$, the two charts are mathematically equivalent (identical out-of-control decisions) and are interchangeable ONLY when the sample size $n$ is constant across all samples; a p-chart is the necessary choice whenever $n$ varies from sample to sample (a raw count is not comparable across different sample sizes, but a fraction is), while an np-chart is often preferred for constant-$n$ shop-floor use because operators find a whole-number defective COUNT more intuitive to read and record than a decimal fraction.

(b) u-chart for scratch marks, revised limits, and in-control expected count

Given. 25 samples, each of $n=20$ inspection units (one piece of furniture per unit); number of scratch marks recorded per sample:

Find. Construct and, if needed, revise the appropriate attributes chart; estimate the in-control expected scratches per piece of furniture.

1. Chart choice and trial limits. A piece of furniture can carry more than one scratch, so this is a DEFECTS-per-unit situation (not defectives) — a u-chart on $u_i=c_i/20$, matching directly what part (c) asks for ("expected scratches per inspection unit"). Total scratches over all 25 samples: $\sum c_i=194$ over $25(20)=500$ units, so $$\bar u=\frac{194}{500}=0.388,\qquad UCL,LCL=\bar u\pm3\sqrt{\bar u/n}=0.388\pm3\sqrt{0.388/20}=0.388\pm0.418.$$ $$\boxed{UCL=0.806},\qquad LCL=-0.030\to\boxed{0\ (\text{floored})}.$$
2. Check each sample; revise. Comparing every $u_i=c_i/20$ to the trial limits, samples 3 ($c=18$, $u=0.90$) and 7 ($c=17$, $u=0.85$) both exceed $UCL=0.806$; every other sample is inside. Both are treated as assignable-cause excursions and removed; the remaining 23 samples ($\sum c_i=194-18-17=159$, over $23(20)=460$ units) give revised limits: $$\bar u_{rev}=\frac{159}{460}=0.3457,\qquad UCL_{rev}=0.3457+3\sqrt{0.3457/20}=\boxed{0.7400},\qquad LCL_{rev}=\boxed{0\ (\text{floored})}.$$ The largest remaining point ($u=0.60$, sample 11) is still comfortably below $UCL_{rev}=0.7400$, so no further revision is needed.
3. In-control expected scratches per piece of furniture. $$\hat u_0=\bar u_{rev}=\boxed{0.346\ \text{scratches/unit}}.$$

(c) Chart design for fast detection of a shift to $u_1=0.6$

Given. In-control $\bar u_0=0.3457$ (part b); shifted level $u_1=0.6$; require $P(\text{detect by sample 1 or 2})\ge0.7$ and $LCL>0$.

Find. The minimum sample size $n$ and the resulting control limits.

1. Translate the detection requirement. $P(\text{detect on sample 1 or 2})=1-(1-p)^2\ge0.7$, where $p$ is the single-sample probability of exceeding $UCL(n)$ once the mean has shifted to $0.6$: $$(1-p)^2\le0.3\ \Longrightarrow\ p\ge1-\sqrt{0.3}=0.4523.$$
2. LCL-positivity constraint. $LCL(n)=\bar u_0-3\sqrt{\bar u_0/n}>0\ \Longrightarrow\ n>9\bar u_0/\bar u_0^2\cdot(1/9)$; solving numerically, $LCL(n)>0$ first holds at $\boxed{n\ge27}$.
3. Detection-probability constraint. With $U\sim N(u_1,\ u_1/n)$ after the shift, $p(n)=1-\Phi\!\left[\dfrac{UCL(n)-u_1}{\sqrt{u_1/n}}\right]$ where $UCL(n)=\bar u_0+3\sqrt{\bar u_0/n}$. Evaluating $p(n)$ for successive $n\ge27$: $$n=43:\ UCL=0.6147,\ p=0.451\ (\text{just short of }0.4523),$$ $$n=44:\ UCL=0.6115,\ p=0.461\ (\text{meets the requirement}).$$ So the minimum sample size satisfying BOTH requirements is $\boxed{n=44}$.
4. Control limits at $n=44$. $$UCL=0.3457+3\sqrt{0.3457/44}=\boxed{0.6115},\qquad LCL=0.3457-3\sqrt{0.3457/44}=\boxed{0.0798>0}.$$