Question 2 of 6: Variation, Control vs. Specification vs. Natural Tolerance Limits, and an X̄/R Chart Build

Question 2: Variation, Control vs. Specification vs. Natural Tolerance Limits, and an X̄/R Chart Build (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) Types of variation; statistical control; control vs. specification vs. natural tolerance limits

Production-process variation splits into two types. Common (chance/random) cause variation is the inherent, cumulative effect of many small, unavoidable sources (minor material variation, ambient conditions, normal machine/operator variability) built into the process design itself; it produces a stable, predictable statistical distribution and can only be reduced by changing the process (better equipment, tighter incoming material specs, redesign). Special (assignable) cause variation is a large, identifiable, intermittent source external to the stable system — a tool wearing out, an operator error, a bad batch of raw material — that shifts the mean or inflates the spread and, once identified, can usually be removed without redesigning the process.

A process is in statistical control when only common-cause variation is present, i.e. its output can be described by a stable probability distribution over time: successive control-chart points are randomly scattered within the control limits with no trends, runs, cycles, or out-of-limit points, so the process mean and variance can be treated as constant.

Control limits ($\mu_0\pm3\sigma_{\bar x}=\mu_0\pm3\sigma/\sqrt n$ for an X̄ chart) are computed from the sampling distribution of a subgroup statistic and answer "is the process behaving the same way it always has?" — they are a statistical judgment about process stability, entirely internal to the process. Specification limits (LSL/USL) are externally imposed engineering/customer requirements on individual units and have no statistical relationship to the process's own behaviour — a process can be perfectly in control while still producing parts outside the spec, if its natural spread is wider than the tolerance band. Natural tolerance limits ($\mu_0\pm3\sigma$) describe the spread of individual measurements from a stable process, i.e. what the process is naturally capable of producing; they use the same $\sigma$ as the control limits but are $\sqrt n$ times wider (since they describe individuals, not subgroup averages), and they are compared against the specification limits (not the control limits) to judge process capability.

(b) X̄ and R chart control limits, process-parameter estimates, and fraction nonconforming

Given. $m=20$ samples, $n=5$; $\sum_{i=1}^{20}\bar x_i=5016$; $\sum_{i=1}^{20}R_i=180$; specifications $250\pm10\Rightarrow LSL=240,\ USL=260$; chart constants for $n=5$: $A_2=0.577$, $D_3=0$, $D_4=2.114$, $d_2=2.326$ (Appendix VI).

Find. $UCL/CL/LCL$ for the X̄ and R charts, the process parameter estimates $\hat\mu_0,\hat\sigma_0$, and $P(X\lt LSL)$, $P(X\gt USL)$.

Approach. Average the 20 subgroup statistics to get $\bar{\bar x}$ and $\bar R$, apply the standard 3-sigma chart-constant formulas, then treat the individual tensile-strength measurements as $N(\hat\mu_0,\hat\sigma_0^2)$ to read off the two tail probabilities against the given spec.

1. Grand average and average range. $$\bar{\bar x}=\frac{\sum \bar x_i}{m}=\frac{5016}{20}=250.8,\qquad \bar R=\frac{\sum R_i}{m}=\frac{180}{20}=9.0$$
2. X̄ chart 3-sigma limits. $$UCL=\bar{\bar x}+A_2\bar R=250.8+0.577(9.0)=255.99,\qquad LCL=\bar{\bar x}-A_2\bar R=250.8-0.577(9.0)=245.61$$ $$\boxed{\text{X̄ chart: } CL=250.8,\ UCL=255.99,\ LCL=245.61}$$
3. R chart 3-sigma limits. $$UCL=D_4\bar R=2.114(9.0)=19.03,\qquad LCL=D_3\bar R=0(9.0)=0$$ $$\boxed{\text{R chart: } CL=9.0,\ UCL=19.03,\ LCL=0}$$
4. Process parameter estimates. $\hat\mu_0=\bar{\bar x}=250.8$. Since $E[\bar R]=d_2\sigma$, $$\hat\sigma_0=\frac{\bar R}{d_2}=\frac{9.0}{2.326}=3.869$$ $$\boxed{\hat\mu_0=250.8,\ \hat\sigma_0=3.869}$$
5. Fraction below LSL. $$z_{LSL}=\frac{240-250.8}{3.869}=-2.791\quad\Rightarrow\quad P(X\lt LSL)=\Phi(-2.791)=0.002626$$
6. Fraction above USL. $$z_{USL}=\frac{260-250.8}{3.869}=2.378\quad\Rightarrow\quad P(X\gt USL)=1-\Phi(2.378)=0.008711$$ $$\boxed{P(X\lt LSL)=0.263\%,\quad P(X\gt USL)=0.871\%}$$

Assumptions. The individual tensile-strength readings (not just the subgroup averages) are approximately normally distributed with the SAME $\hat\mu_0,\hat\sigma_0$ estimated from the subgroup statistics; the 20 subgroups were themselves taken while the process was in statistical control (so $\bar{\bar x},\bar R$ estimate fixed, meaningful parameters rather than describing an unstable snapshot); and $\hat\sigma_0=\bar R/d_2$ is a valid unbiased estimator (no unusually restricted-range subgroup masking real variation).

(c) Smallest $k$ for $P(\text{detect by 1st or 2nd sample})\ge0.6$

Given. $\hat\mu_0=250.8,\ \hat\sigma_0=3.869$ (part b); $n=5$; 3-sigma X̄ limits ($L=3$).

Find. Smallest $k>0$ such that the probability of an out-of-limits signal on the first OR second sample after the mean shifts to $\mu_0+k\sigma$ is $\ge0.6$.

Approach. The probability that any single post-shift sample signals is $p(k)=\big[1-\Phi(L-k\sqrt n)\big]+\Phi(-L-k\sqrt n)$; successive samples are independent, so the probability of at least one signal in the first two samples is $1-(1-p)^2$.

1. Single-sample detection probability. With $n=5,\ \sqrt n=2.236,\ L=3$: $$p(k)=\big[1-\Phi(3-2.236k)\big]+\Phi(-3-2.236k)$$
2. Detect-by-2 condition. $$1-(1-p)^2\ge0.6\ \Rightarrow\ (1-p)^2\le0.4\ \Rightarrow\ p\ge1-\sqrt{0.4}=0.3675$$
3. Solve for $k$. The negative-tail term $\Phi(-3-2.236k)$ is negligible for $k>0$, so $$1-\Phi(3-2.236k)=0.3675\ \Rightarrow\ 3-2.236k=\Phi^{-1}(0.6325)=0.3389$$ $$k=\frac{3-0.3389}{2.236}$$
4. Result (numerical root-find on the exact two-term $p(k)$, matching the closed-form approximation above to 3 decimals): $$\boxed{k^*=1.190}$$ Check: $p(k^*)=0.3675$, so $1-(1-p)^2=0.600$.