Measurement Errors and Accuracy
Theory
No measurement is exact. The absolute error of a measuring instrument is half its smallest division, and the percentage error compares that error with the size of the measurement. This Year 10A maths page revises precision, limits of accuracy and percentage error from Year 9, then looks at the impact of measurement errors on calculated areas, volumes, costs and decisions, including sums, equal shares and significant figures, and at what happens when you round or truncate during a calculation. Curriculum: VC2M10M04, VC2M9M04 (revised) / AC9M10M04, AC9M9M04 (revised).
The precision of a measuring instrument is its smallest division: \(1\) mm on a ruler, \(0.5^\circ\)C on a thermometer marked every \(0.5^\circ\). For a written measurement, the precision is the place value of its last digit: \(5.37\) m is precise to \(0.01\) m.
The absolute error is half the precision. It is also called the greatest possible error (GPE). A reading of \(62\) kg on scales marked in kilograms could really be anything from \(61.5\) kg to \(62.5\) kg, so it is written \(62 \pm 0.5\) kg.
The limits of accuracy are the smallest and largest possible true values: measurement minus and plus the absolute error.
The percentage error is the absolute error as a percentage of the measurement. A smaller percentage error means a more accurate measurement.
Rounding looks at the next digit; truncating just cuts the extra digits off. Truncating \(5.38\) to one decimal place gives \(5.3\), but rounding gives \(5.4\), which is closer.
| Measurement | Precision | Absolute error | Limits |
|---|---|---|---|
| \(62\) kg (nearest kg) | \(1\) kg | \(\pm 0.5\) kg | \(61.5\) to \(62.5\) kg |
| \(1.8\) m (1 d.p.) | \(0.1\) m | \(\pm 0.05\) m | \(1.75\) to \(1.85\) m |
| \(0.36\) L (2 d.p.) | \(0.01\) L | \(\pm 0.005\) L | \(0.355\) to \(0.365\) L |
| \(3700\) people (nearest 100) | \(100\) | \(\pm 50\) | \(3650\) to \(3750\) |
How to find the error in a measurement
- Find the precision: the smallest division on the instrument, the place value of the last digit, or the unit in “correct to the nearest …”.
- Halve it to get the absolute error. Write the answer with \(\pm\) and the unit.
- Add and subtract the absolute error for the limits of accuracy.
- Divide by the measurement and multiply by \(100\%\) for the percentage error. Use the same units on top and bottom.
Errors in a calculation
- Smallest result (area, volume, total): use every lower limit. Largest result: use every upper limit.
- Repeated measurements: multiply the error of one by the number of measurements for the worst case.
- Choosing an instrument: work out the percentage error each instrument would give and pick one that meets the required accuracy.
- Round only at the end. Rounding or truncating a value part-way through makes the final answer less accurate. Compare the two results to see by how much.
- Practical decisions: to be sure there is enough material, use the greatest possible amount needed; to be sure a supply lasts, use the least possible amount available. Then round up or down to whole items.
- Sensible answers: a result calculated from measurements cannot be more accurate than the measurements. Give it to about the same number of significant figures as the least accurate measurement.
(i) the precision
(ii) the absolute error
(iii) the limits of accuracy of the reading.
(i) The smallest division is \(2\) mL, so the precision is \(\mathbf{2}\) mL.
(ii)
| \(\text{absolute error}\) | \(=\) | \(\dfrac{1}{2} \times 2\) |
| \(\) | \(=\) | \(\pm\,\mathbf{1}\text{ mL}\) |
(iii)
| \(\text{lower limit}\) | \(=\) | \(36 - 1 = \mathbf{35}\text{ mL}\) |
| \(\text{upper limit}\) | \(=\) | \(36 + 1 = \mathbf{37}\text{ mL}\) |
The reading is \(36 \pm 1\) mL.
(i) the absolute error
(ii) the percentage error, correct to two decimal places.
Two decimal places means the precision is \(0.01\) kg.
(i)
| \(\text{absolute error}\) | \(=\) | \(\dfrac{1}{2} \times 0.01\) |
| \(\) | \(=\) | \(\pm\,\mathbf{0.005}\text{ kg}\) |
(ii)
| \(\text{percentage error}\) | \(=\) | \(\dfrac{0.005}{3.85} \times 100\%\) |
| \(\) | \(=\) | \(0.1298\ldots\%\) |
| \(\) | \(\approx\) | \(\mathbf{0.13\%}\) |
The true mass lies between \(3.845\) kg and \(3.855\) kg.
Each side could be up to \(0.5\) m less or more. Use the lower limits for the smallest area and the upper limits for the largest.
| \(A_{\min}\) | \(=\) | \(6.5 \times 2.5\) |
| \(\) | \(=\) | \(\mathbf{16.25}\text{ m}^2\) |
| \(A_{\max}\) | \(=\) | \(7.5 \times 3.5\) |
| \(\) | \(=\) | \(\mathbf{26.25}\text{ m}^2\) |
The \(21\text{ m}^2\) found from the rounded sides could be up to \(4.75\text{ m}^2\) too big or \(5.25\text{ m}^2\) too small.
Compare percentage errors. The smaller percentage error is the more accurate measurement.
| \(\text{road}\) | \(:\) | \(\dfrac{0.5}{85} \times 100\% \approx 0.59\%\) |
| \(\text{screw}\) | \(:\) | \(\dfrac{0.05}{4.2} \times 100\% \approx 1.19\%\) |
\(0.59\% < 1.19\%\), so the road measurement is more accurate, even though its absolute error (\(0.5\) km) is far bigger.
Common pitfalls
Frequently asked questions
What is absolute error in maths?
Absolute error is half of the smallest unit of measurement. It is the most a measurement can be wrong by. A length measured to the nearest centimetre has an absolute error of \(\pm 0.5\) cm.
How do you calculate percentage error?
Divide the absolute error by the measurement and multiply by \(100\%\). For \(40\) kg to the nearest kilogram: \(\dfrac{0.5}{40} \times 100\% = 1.25\%\).
What is the precision of a measuring instrument?
The precision is the smallest division marked on the instrument, such as \(1\) mm on a ruler or \(0.5^\circ\)C on a thermometer. The smaller the division, the more precise the instrument.
What are the limits of accuracy?
They are the smallest and largest values the true measurement could be: the measurement minus the absolute error, and the measurement plus the absolute error.
What is the difference between precision and accuracy?
Precision is about the instrument: how small its divisions are. Accuracy is about the measurement: how close it is likely to be to the true value, judged by the percentage error.
What is the difference between rounding and truncating?
Rounding looks at the next digit and rounds up if it is \(5\) or more. Truncating just cuts off the extra digits. Rounding usually gives a value closer to the original.
How do measurement errors affect a calculated area or volume?
Each length could be up to its absolute error too big or too small, so the result has a range. Find the smallest result from the lower limits and the largest from the upper limits; the gap is often much bigger than the error in one length.