Measuring Things — Question 3

Question 3

Precision is not accuracy

Two instruments measure a rigid reference whose assigned length is $10.00$ cm; treat the reference uncertainty as negligible here. Both instruments display hundredths of a centimeter. Instrument A reports $10.41,\ 10.42,\ 10.40,\ 10.41,\ 10.41\ \mathrm{cm}$. Instrument B reports $9.94,\ 10.05,\ 10.00,\ 10.03,\ 9.98\ \mathrm{cm}$. For this question, use the range (largest minus smallest reading) to compare repeatability and the difference between the sample mean and the reference to estimate bias.

1. Calculate each instrument’s sample mean, range, and estimated bias.

2. Which set is more repeatable? Which sample mean is closer to the reference? Explain why those answers need not identify the same instrument.

3. A later reading from A is $14.73$ cm. If its bias is a constant additive offset over this range, find the corrected estimate. Explain why the five reference readings alone do not establish that assumption or a complete uncertainty for the corrected result.

Question 3 - Solution

Precision is not accuracy

(a) Calculate the summaries.

$$\overline x=\frac{\sum_{j=1}^{5}x_j}{5},\qquad R=x_{\max}-x_{\min},\qquad b=\overline x-x_{\mathrm{ref}}.$$

$$\overline A=\frac{52.05}{5}=10.410\ \mathrm{cm},\qquad \overline B=\frac{50.00}{5}=10.000\ \mathrm{cm}.$$ $$\begin{aligned}\overline x_A&=10.410\ \mathrm{cm},& R_A&=0.02\ \mathrm{cm},&b_A&=+0.410\ \mathrm{cm},\\\overline x_B&=10.000\ \mathrm{cm},&R_B&=0.11\ \mathrm{cm},&b_B&=0.000\ \mathrm{cm}.\end{aligned}$$ The extra decimal in a calculated mean identifies the arithmetic result; it does not upgrade instrument resolution.

(b) Interpret what the data support.

A has the smaller observed range and is more repeatable in this trial. B’s sample mean coincides with the assigned reference value and is closer to it. A tightly clustered set can still be displaced by a systematic offset. Conversely, a mean close to the reference does not make every individual reading accurate, or prove that future readings are unbiased.

(c) Apply a conditional correction.

Under the stated constant-offset model, $$L_{\mathrm{corrected}}=14.73-0.41=\boxed{14.32\ \mathrm{cm}}.$$ The calibration tests only one reference length. It cannot distinguish an additive offset from a scale-factor error, or establish behavior throughout the range. Repeated readings also do not test future drift. Additional reference lengths, stability checks, and an uncertainty analysis are needed; neither the display resolution nor the observed range alone is a complete uncertainty bound.

Check: A reads too high on the reference, so its additive correction must be subtracted.