Question 2 of 6: SPC Tools, X̄/S Control Charts, Fraction Nonconforming, and ARL

Question 2: SPC Tools, X̄/S Control Charts, Fraction Nonconforming, and ARL (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) SPC tools for assignable causes; control limits vs. natural tolerance limits

The "magnificent seven" basic SPC tools used to detect and diagnose assignable causes are: the histogram, the check sheet, the Pareto chart, the cause-and-effect (fishbone/Ishikawa) diagram, the defect-concentration diagram, the scatter diagram, and the control chart. The control chart itself signals that an assignable cause is present (a point outside the control limits or a non-random pattern); the other six are used to investigate which assignable cause is responsible.

Pareto chart: a bar chart of nonconformity/defect categories ranked in descending frequency (or cost), with a cumulative-percentage line overlaid. It applies the 80/20 principle — typically a small fraction of the possible causes account for the large majority of the defects — and is used to prioritize which assignable-cause category to investigate first for the greatest improvement per unit of investigation effort.

Cause-and-effect (fishbone) diagram: a structured brainstorming tool that organizes potential causes of a specific quality problem (the "effect," at the head of the fish) into major branches — classically Man, Machine, Material, Method, Measurement, and Environment (the "6 M's") — with successive sub-branches drilling down to specific, actionable root causes. It is used after a Pareto chart has identified which problem to attack, to systematically generate and organize the candidate assignable causes before they are tested/confirmed with data.

Control limits vs. natural tolerance limits: control limits ($UCL,LCL$, typically $\mu_0\pm3\sigma/\sqrt n$ for an X̄ chart) are computed from the sampling distribution of a statistic (e.g. the subgroup average) and are used to judge whether the process is stable over time (voice of the process, applied to $\bar x$). Natural tolerance limits ($\mu_0\pm3\sigma$) describe the spread of individual measurements from a stable process and estimate what the process is naturally capable of producing; they are not control limits and have no direct relationship to externally imposed specification limits. Because the X̄ chart's limits are narrowed by $1/\sqrt n$ relative to the natural tolerance limits, a process can be in control on its X̄ chart while individual units still fall outside the (much wider) natural tolerance band or specification limits.

(b) X̄-S chart control limits and process-parameter estimates

Given. $m=40$ samples, $n=4$; $\sum_{i=1}^{40}\bar x_i=18{,}200$; $\sum_{i=1}^{40}s_i=380$; chart constants for $n=4$: $A_3=1.628$, $c_4=0.9213$, $B_3=0$, $B_4=2.266$ (Appendix VI).

Find. $UCL/CL/LCL$ for the X̄ and S charts, and the process parameter estimates $\hat\mu_0,\hat\sigma_0$.

Approach. Average the 40 subgroup statistics to get $\bar{\bar x}$ and $\bar s$, then apply the standard 3-sigma chart-constant formulas.

1. Grand average and average standard deviation. $$\bar{\bar x}=\frac{\sum \bar x_i}{m}=\frac{18{,}200}{40}=455.0,\qquad \bar s=\frac{\sum s_i}{m}=\frac{380}{40}=9.5$$
2. X̄ chart 3-sigma limits. $$UCL=\bar{\bar x}+A_3\bar s=455.0+1.628(9.5)=470.47,\qquad LCL=\bar{\bar x}-A_3\bar s=455.0-1.628(9.5)=439.53$$ $$\boxed{\text{X̄ chart: } CL=455.0,\ UCL=470.47,\ LCL=439.53}$$
3. S chart 3-sigma limits. $$UCL=B_4\bar s=2.266(9.5)=21.53,\qquad LCL=B_3\bar s=0(9.5)=0$$ $$\boxed{\text{S chart: } CL=9.5,\ UCL=21.53,\ LCL=0}$$
4. Process parameter estimates. $\hat\mu_0=\bar{\bar x}=455.0$. Since $E[s]=c_4\sigma$, the unbiased estimate of the process standard deviation is $$\hat\sigma_0=\frac{\bar s}{c_4}=\frac{9.5}{0.9213}=10.31$$ $$\boxed{\hat\mu_0=455.0,\ \hat\sigma_0=10.31}$$

Assumptions: the 40 subgroups are independent random samples from a process whose tensile strength is (at least approximately) normally distributed with a constant mean and variance while these baseline samples were taken (i.e. the process was itself in a state of statistical control over the 40 samples used to estimate the limits — the limits describe, they do not yet guarantee, control); and $s_i$ is a valid estimator of the within-subgroup spread (no unusually small subgroup masking real variation).

(c) Fraction nonconforming and the ARL design value $k$

Given. $\hat\mu_0=455.0$, $\hat\sigma_0=10.31$ (from part b); specification $450\pm30\Rightarrow LSL=420,\ USL=480$; $n=4$; 3-sigma X̄ limits ($L=3$).

Find. $P(X\lt LSL)$ and $P(X\gt USL)$ separately, assuming $X\sim N(\hat\mu_0,\hat\sigma_0^2)$; and the smallest $k>0$ such that the out-of-control ARL $\le 5$ when $\mu$ shifts to $\mu_0+k\sigma$.

1. Fraction below LSL. $$z_{LSL}=\frac{420-455.0}{10.31}=-3.394 \quad\Rightarrow\quad P(X\lt LSL)=\Phi(-3.394)=0.000344$$ $$\boxed{P(X\lt LSL)=0.0344\%}$$
2. Fraction above USL. $$z_{USL}=\frac{480-455.0}{10.31}=2.424 \quad\Rightarrow\quad P(X\gt USL)=1-\Phi(2.424)=0.007665$$ $$\boxed{P(X\gt USL)=0.7665\%}$$
3. ARL after a $k\sigma$ shift. With 3-sigma limits and subgroup size $n$, the probability the very next point signals is $$p(k)=\Big[1-\Phi\big(3-k\sqrt n\big)\Big]+\Phi\big(-3-k\sqrt n\big),\qquad ARL(k)=\frac{1}{p(k)}$$ With $n=4$, $\sqrt n=2$. Requiring $ARL(k)\le5$ means $p(k)\ge0.20$. Solving $p(k)=0.20$ numerically (the second, negative-tail term is negligible here) gives $$1-\Phi(3-2k)=0.20\ \Rightarrow\ 3-2k=-0.8416\ \Rightarrow\ k=1.921/2$$
4. Result. Root-finding (bisection) on $p(k)=0.20$ gives $$\boxed{k^*=1.079\ \ (ARL(k^*)=5.00)}$$ For $k>1.079$ the shift is detected faster (ARL < 5); for $k<1.079$ ARL > 5.