Australian Curriculum v9 / ACiQ Year 9 Mathematics - Unit 3 - Measurement error and percentage error

Measurement error and percentage error

Calculate and interpret absolute, relative and percentage error while recognising measurement estimates.

Updated 2026-07-26 - 11 min read

Measurement error and percentage error is part of the Year 9 curriculum because students must do more than carry out a familiar calculation. They need to choose a relationship, represent it accurately, explain the result and decide whether it makes sense. The sections below develop those decisions through explicit rules, worked examples, misconception repair and transfer.

This note is designed to work with the guided lessons, curated practice, flashcards, Tutor context, Review and Rapid Revision for the same canonical target. The same three evidence checks are used throughout, so feedback can route a learner back to the precise idea that needs repair.

Absolute error and bounds

The absolute value records size of discrepancy, not direction. Bounds communicate all original values that would round to the reported measurement.

A dependable reasoning routine

  1. Name the unknowns, units and constraints before calculating.
  2. Choose the relationship represented by absolute error and bounds and state why it applies.
  3. Keep exact values for as long as possible, show substitutions and preserve units through each step.
  4. Check the result by substitution, estimation, an alternative representation or the original context.

Repair: Absolute error is non-negative; direction can be described separately.

The repair matters because the shortcut may appear to work in one familiar example while failing when the method, representation, scale, constraint or accuracy requirement changes. Use the routine above to make the reasoning visible enough for another learner to verify.

Example 1.1

Measured 9.8 cm, accepted 10.0 cm. Absolute error?

Step 1 - identify the governing idea: Absolute error is |measured−accepted|. A value rounded to the nearest unit has a half-unit interval around it.

Step 2 - apply it to this evidence: |9.8−10.0|=0.2.

Result: 0.2 cm

The relationship is visible in the working: |9.8−10.0|=0.2. Check the units and substitute or estimate where possible. A plausible-looking number is not enough unless it satisfies the original conditions.

Why the alternatives fail:

  • −0.2 cm — It does not agree with the required relationship: |9.8−10.0|=0.2.
  • 19.8 cm — It changes or misses a condition in the question. Reapply the relationship and retain the stated units or accuracy.
  • 0.02 cm — A substitution, estimate, ordering or unit check rejects this result; it does not reproduce the conditions in the prompt.

Example 1.2

A length is 42 cm to nearest cm. What interval contains the true value?

Step 1 - identify the governing idea: Absolute error is |measured−accepted|. A value rounded to the nearest unit has a half-unit interval around it.

Step 2 - apply it to this evidence: Half a centimetre lies on either side.

Result: 41.5 cm ≤ length < 42.5 cm

The relationship is visible in the working: Half a centimetre lies on either side. Check the units and substitute or estimate where possible. A plausible-looking number is not enough unless it satisfies the original conditions.

Why the alternatives fail:

  • 42≤length<43 — It does not agree with the required relationship: Half a centimetre lies on either side.
  • 41≤length≤42 — It changes or misses a condition in the question. Reapply the relationship and retain the stated units or accuracy.
  • 41.9≤length≤42.1 — A substitution, estimate, ordering or unit check rejects this result; it does not reproduce the conditions in the prompt.

Example 1.3

A mass is 3.4 kg to nearest 0.1 kg. Maximum rounding error?

Step 1 - identify the governing idea: Absolute error is |measured−accepted|. A value rounded to the nearest unit has a half-unit interval around it.

Step 2 - apply it to this evidence: Half of 0.1 kg is 0.05 kg.

Result: 0.05 kg

The relationship is visible in the working: Half of 0.1 kg is 0.05 kg. Check the units and substitute or estimate where possible. A plausible-looking number is not enough unless it satisfies the original conditions.

Why the alternatives fail:

  • 0.1 kg — It does not agree with the required relationship: Half of 0.1 kg is 0.05 kg.
  • 0.5 kg — It changes or misses a condition in the question. Reapply the relationship and retain the stated units or accuracy.
  • 0.01 kg — A substitution, estimate, ordering or unit check rejects this result; it does not reproduce the conditions in the prompt.

Relative and percentage error

Dividing by the reference value makes errors comparable across different scales. The denominator must be stated consistently.

A dependable reasoning routine

  1. Name the unknowns, units and constraints before calculating.
  2. Choose the relationship represented by relative and percentage error and state why it applies.
  3. Keep exact values for as long as possible, show substitutions and preserve units through each step.
  4. Check the result by substitution, estimation, an alternative representation or the original context.

Repair: Use the accepted or reference value unless the task defines a different convention.

The repair matters because the shortcut may appear to work in one familiar example while failing when the method, representation, scale, constraint or accuracy requirement changes. Use the routine above to make the reasoning visible enough for another learner to verify.

Example 2.1

Measured 48, accepted 50. Percentage error?

Step 1 - identify the governing idea: Relative error=absolute error/accepted value; percentage error=relative error×100%.

Step 2 - apply it to this evidence: Absolute error 2; 2/50×100=4%.

Result: 4%

The relationship is visible in the working: Absolute error 2; 2/50×100=4%. Check the units and substitute or estimate where possible. A plausible-looking number is not enough unless it satisfies the original conditions.

Why the alternatives fail:

  • 2% — It does not agree with the required relationship: Absolute error 2; 2/50×100=4%.
  • 96% — It changes or misses a condition in the question. Reapply the relationship and retain the stated units or accuracy.
  • 0.04% — A substitution, estimate, ordering or unit check rejects this result; it does not reproduce the conditions in the prompt.

Example 2.2

Two devices have errors 1 g on 20 g and 5 g on 500 g. Which has larger relative error?

Step 1 - identify the governing idea: Relative error=absolute error/accepted value; percentage error=relative error×100%.

Step 2 - apply it to this evidence: Relative errors are 5% and 1%.

Result: The 1 g on 20 g device

The relationship is visible in the working: Relative errors are 5% and 1%. Check the units and substitute or estimate where possible. A plausible-looking number is not enough unless it satisfies the original conditions.

Why the alternatives fail:

  • The 5 g error device — It does not agree with the required relationship: Relative errors are 5% and 1%.
  • They are equal — It changes or misses a condition in the question. Reapply the relationship and retain the stated units or accuracy.
  • Cannot compare different errors — A substitution, estimate, ordering or unit check rejects this result; it does not reproduce the conditions in the prompt.

Example 2.3

Absolute error is 0.3 m and accepted length is 12 m. Relative error?

Step 1 - identify the governing idea: Relative error=absolute error/accepted value; percentage error=relative error×100%.

Step 2 - apply it to this evidence: 0.3/12=0.025.

Result: 0.025

The relationship is visible in the working: 0.3/12=0.025. Check the units and substitute or estimate where possible. A plausible-looking number is not enough unless it satisfies the original conditions.

Why the alternatives fail:

  • 2.5 — It does not agree with the required relationship: 0.3/12=0.025.
  • 0.04 — It changes or misses a condition in the question. Reapply the relationship and retain the stated units or accuracy.
  • 40 — A substitution, estimate, ordering or unit check rejects this result; it does not reproduce the conditions in the prompt.

Interpret precision and error

A small absolute error may matter in a tiny component but not a large structure. Evaluation must combine relative size, tolerance and purpose.

A dependable reasoning routine

  1. Name the unknowns, units and constraints before calculating.
  2. Choose the relationship represented by interpret precision and error and state why it applies.
  3. Keep exact values for as long as possible, show substitutions and preserve units through each step.
  4. Check the result by substitution, estimation, an alternative representation or the original context.

Repair: Display resolution can be precise while calibration or method produces an inaccurate result.

The repair matters because the shortcut may appear to work in one familiar example while failing when the method, representation, scale, constraint or accuracy requirement changes. Use the routine above to make the reasoning visible enough for another learner to verify.

Example 3.1

A scale always reads 0.5 kg too high but repeats closely. Best description?

Step 1 - identify the governing idea: Precision describes resolution or repeatability; accuracy describes closeness to a reference. More decimal places do not guarantee accuracy.

Step 2 - apply it to this evidence: Results are consistent but systematically offset.

Result: Precise but inaccurate

The relationship is visible in the working: Results are consistent but systematically offset. Check the units and substitute or estimate where possible. A plausible-looking number is not enough unless it satisfies the original conditions.

Why the alternatives fail:

  • Accurate but imprecise — It does not agree with the required relationship: Results are consistent but systematically offset.
  • Accurate and precise — It changes or misses a condition in the question. Reapply the relationship and retain the stated units or accuracy.
  • Neither measurable nor useful — A substitution, estimate, ordering or unit check rejects this result; it does not reproduce the conditions in the prompt.

Example 3.2

Which is more significant: 1 mm error on 10 mm or 1 mm on 10 m?

Step 1 - identify the governing idea: Precision describes resolution or repeatability; accuracy describes closeness to a reference. More decimal places do not guarantee accuracy.

Step 2 - apply it to this evidence: The relative errors are 10% versus 0.01%.

Result: 1 mm on 10 mm

The relationship is visible in the working: The relative errors are 10% versus 0.01%. Check the units and substitute or estimate where possible. A plausible-looking number is not enough unless it satisfies the original conditions.

Why the alternatives fail:

  • 1 mm on 10 m — It does not agree with the required relationship: The relative errors are 10% versus 0.01%.
  • Always equal because absolute error matches — It changes or misses a condition in the question. Reapply the relationship and retain the stated units or accuracy.
  • Cannot use units — A substitution, estimate, ordering or unit check rejects this result; it does not reproduce the conditions in the prompt.

Example 3.3

A manufacturer tolerance is ±0.2 mm and a part is 0.25 mm from target. Decision?

Step 1 - identify the governing idea: Precision describes resolution or repeatability; accuracy describes closeness to a reference. More decimal places do not guarantee accuracy.

Step 2 - apply it to this evidence: 0.25 mm exceeds the allowed absolute deviation 0.2 mm.

Result: Outside tolerance

The relationship is visible in the working: 0.25 mm exceeds the allowed absolute deviation 0.2 mm. Check the units and substitute or estimate where possible. A plausible-looking number is not enough unless it satisfies the original conditions.

Why the alternatives fail:

  • Inside tolerance — It does not agree with the required relationship: 0.25 mm exceeds the allowed absolute deviation 0.2 mm.
  • Exactly on tolerance — It changes or misses a condition in the question. Reapply the relationship and retain the stated units or accuracy.
  • Percentage error is required before deciding — A substitution, estimate, ordering or unit check rejects this result; it does not reproduce the conditions in the prompt.

Retrieval check

Try these without looking back at the examples.

  1. A length is 42 cm to nearest cm. What interval contains the true value?
  2. Two devices have errors 1 g on 20 g and 5 g on 500 g. Which has larger relative error?
  3. Which is more significant: 1 mm error on 10 mm or 1 mm on 10 m?

Answers

  1. 41.5 cm ≤ length < 42.5 cm — Half a centimetre lies on either side.
  2. The 1 g on 20 g device — Relative errors are 5% and 1%.
  3. 1 mm on 10 mm — The relative errors are 10% versus 0.01%.

Transfer task

Find an unfamiliar example from school, daily life, a credible news source or another subject. Explain which of the three evidence checks applies. Complete the task, then audit your own response: identify the evidence used, the relationship applied, one plausible misconception and the final reasonableness check. If a peer could not reproduce your reasoning, add the missing step.

Sources