Question 2 of 6: Variation, $\bar X$/$R$ Control Charts, Fraction Nonconforming, and Shift Detection

Question 2: Variation, $\bar X$/$R$ Control Charts, Fraction Nonconforming, and Shift 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) Two types of variation, statistical control, and control limits vs. specification limits

Production processes exhibit two distinct types of variation. Common-cause (chance) variation is the cumulative effect of many small, inherent sources — minor machine vibration, ambient temperature drift, normal material-lot differences — that are always present, individually unidentifiable, and economically impractical to eliminate one by one; a process subject only to common causes behaves predictably and repeatably, even though individual values still scatter. Special-cause (assignable) variation arises from specific, identifiable events — a tool wearing out, an operator error, a bad raw-material lot, a machine going out of adjustment — that are not part of the process's inherent design and can, in principle, be found and removed. A process is said to be in statistical control when only common-cause variation is present: successive plotted points on a control chart fall randomly within the control limits with no trends, runs, cycles, or points beyond the limits, so the process's behaviour is stable and predictable (its future output can be forecast from its past).

Control limits ($UCL$, $LCL$) are computed from the process's own observed variability (typically $\pm3$ standard deviations of the plotted statistic's sampling distribution — the "voice of the process") and are used to judge whether the process is behaving consistently over time. Specification limits ($USL$, $LSL$) are externally imposed engineering or customer requirements on individual units (the "voice of the customer") and have no mathematical connection to the control limits whatsoever. In particular, the control limits on an $\bar X$ chart bound the sampling distribution of the subgroup average (width $\propto\sigma/\sqrt n$), while specification limits bound individual measurements (width $\propto\sigma$); a process can therefore be perfectly in statistical control — every $\bar X$ point inside its (narrow) control limits — while still producing individual units outside the (wider, unrelated) specification limits, exactly the situation quantified in part (c) below.

(b) $\bar X$ and $R$ chart control limits and process-parameter estimates

Given. $m=40$ samples of size $n=5$; $\sum_{i=1}^{40}\bar x_i=18{,}800$; $\sum_{i=1}^{40}R_i=800$.

Find. $UCL/LCL$ for the $\bar X$ and $R$ charts, and the process parameter estimates $\hat\mu$, $\hat\sigma$.

Approach. Average the 40 sample statistics to get $\bar{\bar x}$ and $\bar R$, then apply the standard tabulated control-chart constants for $n=5$ ($A_2=0.577$, $D_3=0$, $D_4=2.114$, $d_2=2.326$).

1. Grand average and average range. $$\bar{\bar x}=\frac{1}{40}\sum_{i=1}^{40}\bar x_i=\frac{18{,}800}{40}=470.0 \qquad \bar R=\frac{1}{40}\sum_{i=1}^{40}R_i=\frac{800}{40}=20.0$$
2. $\bar X$ chart limits. With $A_2=0.577$ for $n=5$, $$UCL_{\bar X}=\bar{\bar x}+A_2\bar R=470.0+0.577(20.0)=\boxed{481.54}\qquad LCL_{\bar X}=470.0-0.577(20.0)=\boxed{458.46}$$
3. $R$ chart limits. With $D_3=0$, $D_4=2.114$ for $n=5$, $$UCL_R=D_4\bar R=2.114(20.0)=\boxed{42.28}\qquad LCL_R=D_3\bar R=0(20.0)=\boxed{0}$$
4. Process parameter estimates. Assuming the process is in control (both charts show no out-of-control signals, which is implicit in being asked to summarize "the" limits from all 40 samples), the grand average estimates the process mean and the average range, scaled by $d_2=2.326$ for $n=5$, estimates the process standard deviation: $$\hat\mu=\bar{\bar x}=\boxed{470.0}\qquad \hat\sigma=\frac{\bar R}{d_2}=\frac{20.0}{2.326}=\boxed{8.60}$$

Assumptions. (i) The 40 subgroups are themselves in statistical control — no special causes were active while the data used to compute $\bar{\bar x}$ and $\bar R$ were collected (otherwise the limits and $\hat\sigma$ would be inflated by the very special causes they are meant to detect). (ii) Tensile strength within a subgroup is approximately normally distributed, which is what justifies $d_2$ and $A_2$ as tabulated. (iii) Subgroups are rational: the five units within a subgroup are drawn close together in time/production sequence so that only common-cause variation acts within a subgroup, while all subgroup-to-subgroup variation is what the chart is meant to monitor. (iv) Successive subgroups are independent.

(c) Fraction nonconforming against specifications $475\pm30$

Given. $USL=475+30=505$, $LSL=475-30=445$; process estimates from (b): $\hat\mu=470.0$, $\hat\sigma=8.60$.

Find. Estimated fraction nonconforming $\hat p$.

Approach. Standardize both specification limits against the estimated (in-control) process distribution and sum the two tail probabilities.

1. Standardize the limits. $$Z_{USL}=\frac{505-470.0}{8.60}=4.07\qquad Z_{LSL}=\frac{445-470.0}{8.60}=-2.91$$
2. Sum the tail probabilities. $$\hat p=\Phi(Z_{LSL})+[1-\Phi(Z_{USL})]=\Phi(-2.91)+[1-\Phi(4.07)]=0.00181+0.00002=\boxed{0.00185}$$ so about 0.185% (roughly 1{,}850 ppm) of production is expected to fall outside the specification, almost entirely on the low side ($Z_{LSL}=-2.91$ is far closer to the limit than $Z_{USL}=4.07$).

(d) Minimum shift $k$ detected with probability $\ge0.6$ on the next sample

Given. $\bar X$ chart from (b): $n=5$, 3-sigma limits at $\hat\mu\pm3\hat\sigma/\sqrt n$. Process mean shifts from $\mu_0$ to $\mu_0+k\sigma$.

Find. The minimum $k>0$ such that the probability the very next sample's $\bar X$ falls outside the (unchanged) control limits is at least 0.6.

Approach. After the shift, $\bar X\sim N(\mu_0+k\sigma,\ \sigma^2/n)$ while the limits stay fixed at $\mu_0\pm3\sigma/\sqrt n$; express both limits in standard-normal units of the shifted distribution and solve for $k$.

1. Detection probability as a function of $k$. Standardizing the (fixed) limits against the shifted sampling distribution ($\text{mean}=\mu_0+k\sigma$, $\text{sd}=\sigma/\sqrt n$): $$P(\text{detect})=P\!\left(\bar X>UCL\right)+P\!\left(\bar X\lt LCL\right)=\Big[1-\Phi\big(3-k\sqrt n\big)\Big]+\Phi\big(-3-k\sqrt n\big)$$ For any $k>0$ of practical size the second term is negligible (the shift moves the mean toward the upper limit, away from the lower one), so the detection probability is governed almost entirely by the first term.
2. Set the target and solve. With $n=5$, require $1-\Phi(3-k\sqrt5)\ge0.6$, i.e. $\Phi(3-k\sqrt5)\le0.4$: $$3-k\sqrt5=\Phi^{-1}(0.4)=-0.2533 \ \Longrightarrow\ k=\frac{3+0.2533}{\sqrt5}=\frac{3.2533}{2.236}=\boxed{1.455}$$
3. Check. At $k=1.455$: $3-k\sqrt5=-0.2533$, $1-\Phi(-0.2533)=0.600$, and the lower-tail term $\Phi(-3-3.254)=\Phi(-6.25)\approx0$ is indeed negligible, confirming the approximation.