Measurement Error
Theory
No measurement is exact. This Year 9 maths page shows how to find the precision of a reading, its absolute error and the range the true value lies in, then the relative and percentage error, which tell you how accurate a measurement really is. Curriculum: VC2M9M04 / AC9M9M04.
Every measuring instrument has a smallest marked division. This is the size of one unit on the scale, also called the precision. A ruler in millimetres has a unit of \(1\) mm.
The limit of accuracy, or absolute error, is half of one unit. A reading can be out by up to that much either way, so we write it with \(\pm\). For a millimetre ruler it is \(\pm 0.5\) mm.
The true value lies between the lower limit (reading minus the error) and the upper limit (reading plus the error).
The true value can equal the lower limit but stays below the upper limit: \(45\) kg to the nearest kilogram means at least \(44.5\) kg and less than \(45.5\) kg.
The way a number is written also tells you the unit. "To the nearest centimetre" means a unit of \(1\) cm. "Correct to one decimal place" means a unit of \(0.1\). A digital reading of \(7.85\) kg shows hundredths, so its unit is \(0.01\) kg and its absolute error is \(0.005\) kg.
The relative error compares the absolute error with the size of the measurement: absolute error \(\div\) measurement. Written as a percentage, it is the percentage error. The smaller the percentage error, the more accurate the measurement.
| Measurement | One unit | Limit of accuracy | True value lies |
|---|---|---|---|
| \(9\) cm, nearest cm | \(1\) cm | \(\pm 0.5\) cm | \(8.5\) cm to \(9.5\) cm |
| \(3.2\) kg, \(1\) d.p. | \(0.1\) kg | \(\pm 0.05\) kg | \(3.15\) kg to \(3.25\) kg |
| \(70\) g, nearest \(10\) g | \(10\) g | \(\pm 5\) g | \(65\) g to \(75\) g |
How to find the limits of accuracy
- Find one unit: the smallest division on the scale, or the place the value was rounded to.
- Halve it. That is the limit of accuracy, written with \(\pm\).
- Subtract and add it to the reading to get the lower and upper limits.
- For relative error, divide the absolute error by the measurement. For percentage error, multiply that by \(100\%\).
- To compare accuracy, compare percentage errors, not absolute errors.
- For a calculation, use all the lower limits together and all the upper limits together.
(i) the size of one unit on the scale
(ii) the limit of accuracy.
(i) The labels go up by \(10\) kg, and there are \(5\) gaps between them.
| \(\text{one unit}\) | \(=\) | \(10 \div 5\) |
| \(\) | \(=\) | \(\mathbf{2} \text{ kg}\) |
(ii) The limit of accuracy is half of one unit, either way.
| \(\text{limit of accuracy}\) | \(=\) | \(\pm \dfrac{1}{2} \times 2\) |
| \(\) | \(=\) | \(\mathbf{\pm 1} \text{ kg}\) |
Count the gaps, not the tick marks.
(i) A rod measured as \(8\) cm, to the nearest centimetre.
(ii) A mass recorded as \(4.6\) kg, correct to one decimal place.
(i) "To the nearest centimetre" means one unit is \(1\) cm, so the limit is \(\pm 0.5\) cm.
| \(8 - 0.5\) | \(=\) | \(7.5\) |
| \(8 + 0.5\) | \(=\) | \(8.5\) |
The true length is between \(\mathbf{7.5}\) cm and \(\mathbf{8.5}\) cm.
(ii) One decimal place means one unit is \(0.1\) kg, so the limit is \(\pm 0.05\) kg.
| \(4.6 - 0.05\) | \(=\) | \(4.55\) |
| \(4.6 + 0.05\) | \(=\) | \(4.65\) |
The true mass is between \(\mathbf{4.55}\) kg and \(\mathbf{4.65}\) kg.
(i) the absolute error
(ii) the relative error
(iii) the percentage error.
(iv) A \(25\) g letter is weighed to the nearest gram. Which reading is more accurate?
(i) The last digit shows hundredths of a kilogram, so one unit is \(0.01\) kg. Halve it.
| \(\text{absolute error}\) | \(=\) | \(\dfrac{1}{2} \times 0.01 = \mathbf{0.005} \text{ kg}\) |
(ii) Divide by the measurement.
| \(\text{relative error}\) | \(=\) | \(\dfrac{0.005}{1.25} = \mathbf{0.004}\) |
(iii) Write the relative error as a percentage.
| \(\text{percentage error}\) | \(=\) | \(0.004 \times 100\% = \mathbf{0.4\%}\) |
(iv) The letter's absolute error is \(0.5\) g.
| \(\text{percentage error}\) | \(=\) | \(\dfrac{0.5}{25} \times 100\% = 2\%\) |
\(0.4\% < 2\%\), so the flour reading is more accurate, even though its absolute error (\(5\) g) is bigger.
Each side could be up to \(0.5\) m less or more than it was recorded.
| \(\text{length}\) | \(:\) | \(6.5 \text{ to } 7.5 \text{ m}\) |
| \(\text{width}\) | \(:\) | \(2.5 \text{ to } 3.5 \text{ m}\) |
Use the lower limits together, then the upper limits together.
| \(\text{smallest}\) | \(=\) | \(2 \times (6.5 + 2.5) = \mathbf{18} \text{ m}\) |
| \(\text{largest}\) | \(=\) | \(2 \times (7.5 + 3.5) = \mathbf{22} \text{ m}\) |
The recorded perimeter is \(20\) m, but it could really be anywhere from \(18\) m up to \(22\) m.
Common pitfalls
Frequently asked questions
What is the absolute error of a measurement?
It is half of the smallest unit used, also called the limit of accuracy. A ruler marked in millimetres has an absolute error of \(\pm 0.5\) mm.
How do you find the upper and lower limits of a measurement?
Halve the unit the measurement was made to. Subtract it for the lower limit and add it for the upper limit: \(8\) cm to the nearest cm lies between \(7.5\) cm and \(8.5\) cm.
What is relative error?
Relative error is the absolute error divided by the measurement. For \(1.25\) kg with an absolute error of \(0.005\) kg, it is \(0.004\).
How do you calculate percentage error?
Divide the absolute error by the measurement and multiply by \(100\%\). For \(25\) cm to the nearest cm: \(\dfrac{0.5}{25} \times 100\% = 2\%\).
Which measurement is more accurate?
The one with the smaller percentage error. With the same instrument, a bigger measurement has the same absolute error but a smaller percentage error, so it is more accurate.