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Accuracy, precision and measurement errors

Key idea: Distinguish accuracy from precision and explain systematic, random, parallax and zero errors using effect-and-improvement exam wording.

  • SEC G3 Physics 2027
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Learning objectives

  • Represent a physical quantity with a numerical magnitude and unit
  • Recall the six prescribed SI base quantities and their units
  • Use the prescribed SI prefixes from nano to tera
  • Compare orders of magnitude from a typical atom to the Earth
  • Select and justify measuring instruments by range and precision
  • Distinguish scalar and vector quantities and give examples
  • Add two vectors graphically to determine a resultant

1. Definitions

Accuracy describes how close a measured value is to an accepted value.

Precision describes how closely repeated readings agree with one another. A precise set can still be inaccurate if every reading has a similar offset.

Accuracy and precision are differentFour target plots compare accurate and precise readings, accurate readings with wider scatter, precise readings offset from the accepted value, and readings that are neither accurate nor precise. Crosses mark accepted values and circles mark readings.Compare centre and spreadAccurate and precise× accepted value · ○ readingsAccurate, less precise× accepted value · ○ readingsPrecise, inaccurate× accepted value · ○ readingsNeither× accepted value · ○ readings
Scroll diagram horizontally to read all labels.
Accuracy compares readings with an accepted value; precision compares their spread. Repeated readings can be precise yet inaccurate when a consistent offset remains.

2. Key Ideas

A systematic error shifts readings in a consistent way. Repeating and averaging do not remove the shift. Check calibration, correct a known zero error or improve a consistently biased method.

A random error causes readings to scatter unpredictably. Repeat measurements, improve the technique and calculate a mean when appropriate.

Avoid treating these as automatic labels without context. Explain the source and its effect on the readings.

Error sourceLikely effectSuitable improvement
instrument has a constant zero offsetreadings shifted in one directionmeasure the zero error and correct every reading
observer views an analogue scale from changing anglesreadings scatterplace the eye perpendicular to the scale each time
hand timing varies between trialsmeasured times scattermeasure a longer interval, repeat and calculate a mean
instrument is miscalibratedreadings may be consistently too high or lowcheck against a standard or use a calibrated instrument
A graph intercept is a clue, not proof

A non-zero intercept may indicate an offset, but it does not by itself prove a systematic error. Consider the physical model, data scatter and whether the relationship is actually expected to pass through the origin.

3. Detailed Explanations

Parallax error

Parallax occurs when an analogue scale and its pointer or object mark are viewed from the wrong angle. The apparent alignment changes with viewing position.

Read with the line of sight perpendicular to the scale. For a rule lying flat, place the eye directly above the mark. For a pointer and scale, use any mirror alignment provided.

If the same wrong viewpoint is used every time, parallax can cause a consistent bias. If the viewpoint changes, it can contribute to scatter.

Zero error and correction

A zero error exists when an instrument gives a non-zero reading when the true input is zero. Check digital calipers and micrometers when closed without an object.

Use the signed correction rule:

corrected reading = observed reading-zero error

This single rule handles positive and negative zero errors.

4. Common Mistakes

  • Calling every spread in repeated readings a zero error. Zero error is an offset seen when the true input should be zero.
  • Adding the signed zero error instead of subtracting it from the observed reading.
  • Claiming that repeats remove a systematic offset; averaging reduces random variation but preserves a consistent bias.

5. Exam Tips

  • Record the zero reading with its sign before taking measurements and write corrected = observed − zero error.
  • For parallax, state the concrete repair: place the eye normal to the scale at the pointer or meniscus.
  • When evaluating data, use the pattern—common offset or scatter—to distinguish systematic from random effects.

6. Worked Examples

Modelled example 1

Precise but inaccurate

Core

Problem

The accepted length is 10.00 cm. A group records 10.49 cm, 10.50 cm and 10.51 cm. Describe the quality of the results.
Study the worked solution

  1. Judge precision from the repeated readings

    Method

    Compare the readings with one another.

    Reason

    Precision concerns the agreement among repeated readings, not their closeness to the accepted value.

    Working

    The range is 10.51-10.49 = 0.02 cm, so the readings have a small spread and are precise.

  2. Judge accuracy against the accepted value

    Method

    Compare the readings with 10.00 cm.

    Reason

    Accuracy concerns closeness to an accepted value.

    Working

    The readings cluster near 10.50 cm, about 0.50 cm above the accepted value, so they are inaccurate.

  3. Identify a plausible next check

    Method

    Check the instrument calibration and zero.

    Reason

    A similar positive offset in every reading is consistent with a systematic effect.

    Working

    The data suggest an offset, but the readings alone do not prove its source.

Guided practice 2

Correct a positive zero error

About 4 min

Problem

A digital caliper reads + 0.03 mm when closed and 12.34 mm around an object. Find the corrected reading.

Remove the signed offset

Unit: mm

Hints

Hint 1: interpret the closed reading
The positive closed reading means the caliper over-reads.
Hint 2: use the signed rule
Subtract the zero error, including its positive sign.
View solution step by step

  1. Write the correction rule

    Method

    Subtract the signed zero error from the observed reading.

    Reason

    The closed reading shows the amount added by the instrument when the true input is zero.

    Working

    corrected reading = observed reading-zero error

  2. Remove the positive offset

    Working

    12.34-(+0.03) = 12.31 mm

Common misconception 3

Negative zero error

Find and correct the mistake

Learner response

An instrument has zero error -0.04 mm and observed reading 2.36 mm. A student calculates 2.36 + (-0.04) = 2.32 mm. Locate the first error and correct the reading.

Keep the zero error's sign

Correct operation

View solution step by step

  1. Locate the first error

    Method

    Reject changing the correction rule from subtraction to addition.

    Reason

    The rule always subtracts the signed zero error; its sign is already part of the value.

    Working

    The incorrect line uses observed + zero error.

  2. Apply the signed correction

    Method

    Subtract -0.04 mm.

    Reason

    A negative zero error means the instrument under-reads, so the corrected value must be larger.

    Working

    2.36-(-0.04) = 2.40 mm

Examiner practice 4

Explain error, effect and improvement

3 marks

Examination question

A student reads a liquid level from above rather than at the level of the meniscus. Explain the error and how to improve the measurement. [3 marks]

Link the error to its effect and repair

View solution step by step

  1. Name the reading error

    1 mark

    Method

    Identify parallax error.

    Reason

    The line of sight is oblique rather than perpendicular to the scale at the meniscus.

    Working

    Error: parallax from viewing above the meniscus.

  2. State its effect

    1 mark

    Method

    Explain that the apparent alignment is displaced.

    Reason

    The meniscus and scale are separated, so changing the viewing direction changes which scale mark appears aligned.

    Working

    The recorded volume may be too high or too low.

  3. Give a specific improvement

    1 mark

    Method

    Place the eye level with the meniscus and view the scale perpendicularly.

    Reason

    This removes the oblique line of sight that causes the apparent displacement.

    Working

    Align eye, meniscus and scale at the correct reading level.

Challenge 5

Correct and evaluate repeated data

Minimal support

Data-evaluation transfer

A digital balance has a zero error of + 0.05 g. A reference mass is labelled 25.40 g.

ReadingObserved mass / g
125.42
225.44
325.43

Calculate first, then judge the pattern

Unit: g
Best description after correction

Hints

Hint 1: separate scatter from offset
Use the spread to judge precision, but correct the known zero offset before comparing with the labelled value.
Hint 2: calculate in the right order
The observed mean is 25.43 g; apply the signed correction to that value.
View solution step by step

  1. Find the observed mean

    Method

    Average the three repeated observations.

    Reason

    The readings show small random variation, so their mean is the appropriate summary.

    Working

    m bar _observed = (25.42 + 25.44 + 25.43)/3 = 25.43 g

  2. Correct the known offset

    Method

    Subtract the positive zero error.

    Reason

    Averaging does not remove a systematic offset.

    Working

    m bar _corrected = 25.43-0.05 = 25.38 g

  3. Evaluate both precision and accuracy

    Method

    Use the spread and the corrected comparison separately.

    Reason

    Repeated agreement and closeness to the reference answer different questions.

    Working

    The range is 0.02 g, and the corrected mean is 0.02 g below the labelled value: precise and close to the labelled value.

Further mistakes to diagnose

  • Claiming that many decimal places guarantee accuracy.
  • Saying averaging removes a systematic offset.
  • Naming an error without explaining its effect on the reading.
  • Assuming every non-zero graph intercept proves zero error.
  • Applying a zero correction without keeping the sign of the error.
  • Describing every parallax error as always systematic or always random.

7. Mind Stretchers

Mind stretcher 1: Correct repeated readingsExtension

A digital caliper has zero error + 0.04 mm. It gives readings 8.36 mm, 8.38 mm and 8.37 mm. Find the corrected mean.

Show answer

First calculate the observed mean:

d bar _observed = (8.36 + 8.38 + 8.37)/3 = 8.37 mm

Then subtract the positive zero error:

d bar _corrected = 8.37-0.04 = 8.33 mm

Averaging reduces the random spread; applying the correction addresses the known systematic offset.

Mind stretcher 2: Decide whether averaging helpsExtension

A student always views a scale from the same position above the correct eye level. Would taking ten readings and calculating a mean remove the parallax error?

Show answer

No. The viewing position produces a consistent bias, so the mean remains biased. The student must place the eye level with the reading and view the scale perpendicularly. Averaging is useful for random variation, not a systematic error that has not been corrected.

Continue with the next resource in this course.

Course and syllabus information
Course
SEC G3 Physics
Edition
SEC G3 Physics 2027