Question 2 of 7: Process Aim and Capability, and Error of Measurement

Question 2: Process Aim and Capability, and Error of Measurement (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.

Given.

Find. Whether the chalk process is aimed at the correct density and what the aim should be, whether the process is capable of holding the 0.6 gm/cc specification band, and a defensible statement about the true length of the part that was measured.

Approach. Aim and spread are separate questions and must be tested with separate quantities: the aim is judged against the specification midpoint using the standard error of the mean, $s/\sqrt{n}$, while capability and the fraction out of specification are judged from the standard deviation of individual pieces, $s$, alone. Part (b) separates the same two ideas one more time: accuracy is a systematic offset that is removed by correction, precision is random scatter that is quantified by an interval.

1. Part (a) — locate the target the process ought to be aimed at. With a two-sided specification and no other information, the best aim is the midpoint of the band, because that maximises the distance to the nearer limit: $$\mu_0 = \frac{\text{LSL}+\text{USL}}{2} = \frac{4.4+5.0}{2} = 4.700\ \text{gm/cc}$$ The observed average sits $4.800 - 4.700 = 0.100$ gm/cc above it, which is half a process standard deviation.
2. Test whether that offset is real, using the standard error of the mean. The precision of the sample average, not of a single piece, decides whether the aim is wrong: $$\sigma_{\bar{x}} = \frac{s}{\sqrt{n}} = \frac{0.2}{\sqrt{100}} = 0.020\ \text{gm/cc}$$ $$z = \frac{\bar{x}-\mu_0}{\sigma_{\bar{x}}} = \frac{4.800-4.700}{0.020} = 5.00$$ A 95 per cent confidence interval for the true process mean is $4.800 \pm 1.96(0.020)$, that is $4.7608$ to $4.8392$ gm/cc, which lies entirely above 4.700. So $$\boxed{\ \text{the process is not aimed correctly; the aim should be moved to } 4.70\ \text{gm/cc}\ }$$ A five-sigma departure could not plausibly arise by chance from a centred process.
3. Now judge capability from the spread of individual pieces. The specification band and the natural tolerance of the process are $$\text{USL}-\text{LSL} = 0.600\ \text{gm/cc}, \qquad 6s = 6(0.2) = 1.200\ \text{gm/cc}$$ so the process spread is exactly twice the width it is allowed to occupy. In index form $$C_p = \frac{\text{USL}-\text{LSL}}{6s} = \frac{0.600}{1.200} = 0.50, \qquad C_{pk} = \frac{\min(\text{USL}-\bar{x},\ \bar{x}-\text{LSL})}{3s} = \frac{0.200}{0.600} = 0.33$$ $$\boxed{\ C_p = 0.50 \ \Rightarrow\ \text{the process is NOT capable}\ }$$ Because $C_p$ is well below 1, re-aiming alone cannot fix the problem; the variability itself is the obstacle.
4. Quantify the cost of the present situation. Assuming density is normally distributed, the tails beyond each limit at the present aim are $$z_U = \frac{5.0-4.8}{0.2} = +1.00 \Rightarrow 15.87\ \text{per cent above USL}$$ $$z_L = \frac{4.4-4.8}{0.2} = -2.00 \Rightarrow 2.28\ \text{per cent below LSL}$$ for a total of 18.14 per cent of production outside specification. Re-aiming to 4.700 makes the two tails symmetric at $z = \pm 1.50$, giving $2(6.68) = 13.36$ per cent. Re-aiming is therefore worth 4.78 percentage points of scrap and costs nothing but a machine setting, yet it still leaves better than one stick in eight unusable.
5. State what a capable process would require. To reach $C_p = 1.00$ the standard deviation must fall to $$\sigma_{\text{req}} = \frac{\text{USL}-\text{LSL}}{6} = \frac{0.600}{6} = 0.100\ \text{gm/cc}$$ a 50 per cent reduction in spread; for the more usual industrial target $C_p = 1.33$ it must fall to 0.0752 gm/cc, a 62 per cent reduction. The recommendation is therefore in two parts: re-centre the aim at 4.70 gm/cc immediately, then attack the variability (raw-material consistency, mixing time, compaction pressure, mould wear) before promising to hold the specification. If the variation cannot be reduced, the alternatives are to renegotiate the tolerance, to screen 100 per cent and grade the output, or to change process or supplier.
6. Part (b) — separate the systematic and random parts of the measurement error. Any observed reading decomposes as $$x_{\text{obs}} = x_{\text{true}} + \text{bias} + e, \qquad e \sim N(0,\sigma_e^2)$$ Accuracy is the bias, here a known $+0.001$ in, and it is removed by subtraction; precision is $\sigma_e = 0.0004$ in and cannot be removed from a single reading, only described.
7. Correct the reading and attach an interval. The best (unbiased) estimate of the true length is $$\hat{x} = 2.638 - 0.001 = 2.637\ \text{in}$$ and with the random error normal, a 95 per cent interval is $\hat{x} \pm 1.96\sigma_e = 2.637 \pm 0.000784$: $$\boxed{\ 2.6362\ \text{in} \le x_{\text{true}} \le 2.6378\ \text{in}\ \ (95\ \text{per cent})\ }$$ On a three-sigma basis the statement widens to $2.6358$ to $2.6382$ in. Note that had the bias been ignored, the interval would have been centred on 2.638 and would have been wrong by two and a half standard deviations of the instrument — the correction matters more here than the width of the band.
8. Say what repeated readings would buy. Averaging $n$ independent readings divides the random part by $\sqrt{n}$, $\sigma_{\bar{x}} = \sigma_e/\sqrt{n}$, so four readings halve the band to $\pm 0.000392$ in and sixteen readings quarter it. Averaging does nothing at all to the bias: only recalibration against a traceable standard removes that. This is the practical conclusion of the question — buy precision with repetition, buy accuracy with calibration.