Question 5 of 6: Accuracy, Precision and the Effect of Repeat Measurements

Question 5: Accuracy, Precision and the Effect of Repeat Measurements (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. One instrument whose systematic error and random scatter have both been measured, one reading taken with it, and a second gage of known precision on which repeat measurements are to be averaged.

Find. (a) a defensible statement about the true length of the part, together with the assumptions it rests on, and (b) the standard deviation of the mean of n repeat readings for each stated multiple, presented as a graph.

Approach. Accuracy and precision are separate defects and are treated separately: the bias is a known systematic error and is subtracted, while the precision is a random error and becomes an interval about the corrected value; in part (b) the standard error of a mean falls as the reciprocal of the square root of the number of readings averaged.

1. Part (a) — correct the reading for the known bias. The instrument reads 0.001 inch high on the average, so the best single estimate of the true length is the reading less the bias: $$\hat{x} = x - b = 2.638 - 0.001 = 2.637\ \text{in}$$ A bias that is known can always be removed by arithmetic; only the part that is not known has to be carried into the answer as an uncertainty.
2. Attach the random error as an interval. A single reading carries the full precision of the instrument, $\sigma = 0.0004$ inch, so a 95 per cent statement is $$\hat{x} \pm 1.96\,\sigma = 2.637 \pm 1.96(0.0004) = 2.637 \pm 0.00078$$ $$\boxed{2.6362\ \text{in} < x_{true} < 2.6378\ \text{in}\ \text{(95 per cent)}}$$ On a three-standard-deviation basis, which is the convention more common in gage work, the corresponding limits are $2.637 \pm 0.0012$, that is 2.6358 to 2.6382 inches.
3. State the assumptions the statement rests on. The interval above is only as good as five assumptions, and the question asks for them explicitly: that the bias of +0.001 inch is known exactly and is stable, so that it can be subtracted rather than being itself a random variable; that the random error is normally distributed about the corrected value; that the standard deviation of 0.0004 inch, established in the error-of-measurement study, still applies to this instrument, this operator and this part; that the reading quoted is a single reading and not already an average; and that the part itself has a well-defined length at the point measured, so that form error and temperature are not adding a further component to the spread.
4. Part (b) — state the law that governs averaging. The mean of n independent readings of the same feature has variance $\sigma^{2}/n$, so its standard deviation is $$\sigma_{\bar{x}} = \frac{\sigma_{1}}{\sqrt{n}} = \frac{0.001}{\sqrt{n}}$$ Note that it is the precision that improves; averaging does nothing whatever to a bias, which repeats identically in every reading.
5. Tabulate and plot the multiples asked for. Evaluating the expression at the seven stated values of n gives 0.000707, 0.000577, 0.000500, 0.000447, 0.000316, 0.000224 and 0.000183 inch respectively, so $$\boxed{\sigma_{\bar{x}} = 0.001/\sqrt{n};\ n=4 \Rightarrow 0.000500\ \text{in};\ n=30 \Rightarrow 0.000183\ \text{in}}$$
6. Read the economics off the curve. Four readings halve the spread and thirty readings reduce it by 81.7 per cent, but the twenty-six extra readings between those two points buy only a further 63 per cent of what four had already achieved, at six and a half times the inspection cost. The square root is unforgiving: each further halving of the spread costs four times as many measurements. In practice, therefore, repeat measurement is used in small multiples — two to five — and a genuinely inadequate gage is replaced rather than averaged.