Estimation, errors and uncertainty

Key idea: Good measurement is about the quality of the evidence, not the number of decimal places. Separate scatter from bias, then carry the uncertainty through the calculation.

  • GCE A-Level H1 Physics 2027

H1 Physics 8867 · Lesson 2 of 3

Check your understanding

By the end of this lesson, you should be able to

  • Make defensible order-of-magnitude estimates.
  • Distinguish precision, accuracy, random error and systematic error.
  • Find uncertainty in a derived quantity using absolute, fractional or percentage uncertainties.

Learn the idea

Big question: How do you turn imperfect readings into a result whose reliability is honestly stated?

Estimate before measuring or calculating

An estimate is a physical argument in miniature. Choose a familiar comparison, state one or two assumptions and keep only a sensible order of magnitude. Estimating the mass of classroom air, for example, needs a room volume and an approximate air density—not centimetre-level room dimensions.

A prior estimate also catches calculator slips. If a human walking speed appears as 300 m s⁻¹, the arithmetic or unit conversion deserves another look.

Check your understanding: Would 10⁻³ kg, 1 kg or 10³ kg be the most sensible order of magnitude for a textbook?

1 kg. A milligram-scale book is far too light and a tonne-scale book far too heavy.

Match the uncertainty rule to the operation

Random error produces unpredictable scatter and mainly limits precision; repeated readings help reveal and reduce its effect on a mean. Systematic error shifts readings consistently and mainly limits accuracy. A zero error is one systematic example and must be corrected rather than averaged away.

Absolute uncertainty belongs naturally with addition and subtraction because the quantities share a unit. Fractional or percentage uncertainty belongs with multiplication, division and powers because the result scales with each factor. These rules give a sensible worst-case estimate; they are not a full statistical analysis.

Check your understanding: Why do repeated readings not remove a zero error?

Every reading is shifted in the same direction, so averaging preserves the bias instead of cancelling it.

Random scatter and systematic biasTwo target plots compare scattered readings centred on the accepted value with tightly grouped readings displaced from it.Random scattermean near accepted valuerepeat and averageSystematic biasreadings grouped away from accepted valuecalibrate, correct or redesign
Scroll diagram horizontally to read all labels.
Repeats reveal scatter and improve the estimate of a mean, but they do not remove a systematic offset. Calibration, zero checks or a redesigned method are needed for bias.

Key ideas

  • An estimate should include a sensible power of ten and unit.
  • Precision concerns spread; accuracy concerns closeness to the accepted value.
  • An uncertainty should describe the measurement method, not invented extra precision.

Relationships to know

  • Δ(A ± B) = ΔA + ΔB
  • Δ(AB)/(AB) ≈ ΔA/A + ΔB/B
  • for y = xⁿ, Δy/y ≈ |n|Δx/x

Follow the reasoning

Worked example

Correct a zero error and report uncertainty

Question: A micrometer gives 2.31, 2.35, 2.33 and 2.37 mm for a wire diameter and has a +0.02 mm zero error. Report the corrected diameter with a justified uncertainty.

  1. Step 1: Find the central reading

    Why: The mean reduces the effect of random scatter in repeated readings.

    Working: Mean indication = (2.31 + 2.35 + 2.33 + 2.37)/4 = 2.34 mm.

  2. Step 2: Correct the systematic offset

    Why: A positive zero error makes every indication too large.

    Working: Corrected mean = 2.34 − 0.02 = 2.32 mm.

  3. Step 3: Use the spread to estimate uncertainty

    Why: Half the range represents the observed random spread for this small repeated set.

    Working: Uncertainty = (2.37 − 2.31)/2 = 0.03 mm.

Answer: Diameter = (2.32 ± 0.03) mm.

Check: The value and absolute uncertainty use the same decimal place, and the correction moves the result downward as a positive zero error should.

Now try it with support

Practise with support

A rectangle has length (8.0 ± 0.1) cm and width (5.0 ± 0.1) cm. Find its area and absolute uncertainty.

Hints

  1. Area is a product, so add percentage uncertainties.
  2. Convert the final percentage uncertainty back to cm².
View the guided answer

A = 40.0 cm². Fractional uncertainty = 0.1/8.0 + 0.1/5.0 = 0.0325, so ΔA = 1.3 cm². Report A = (40.0 ± 1.3) cm².

Your turn

Practise independently

Four diameter readings are 2.31, 2.35, 2.33 and 2.37 mm, while the micrometer has a +0.02 mm zero error. Calculate and report the corrected result with a justified uncertainty.

Check your answer

The mean indication is 2.34 mm. A +0.02 mm zero error is subtracted, giving 2.32 mm. The half-range is (2.37 − 2.31)/2 = 0.03 mm, so report (2.32 ± 0.03) mm.

Common mistakes and exam guidance

Watch out for

  • Calling a tightly grouped but biased set accurate.
  • Adding percentage uncertainties when the measured quantities are being added rather than multiplied.

In an exam

  • Name whether an error affects precision or accuracy and explain why.
  • Keep full calculator values while working, then match the result’s decimal place to the absolute uncertainty.

Put the ideas together

Exam-style practice [6 marks]

A block has mass (125.0 ± 0.1) g and dimensions (5.00 ± 0.02) cm by (2.00 ± 0.02) cm by (1.00 ± 0.02) cm. Calculate its density and estimate the absolute uncertainty.

Plan before you answer

  • Calculate volume and density using the central values.
  • Add percentage uncertainties for the product and quotient.
  • Convert the percentage uncertainty into an absolute uncertainty.
View the marking points and model answer

Marking points

  1. Obtains volume 10.0 cm³.
  2. Obtains density 12.5 g cm⁻³.
  3. Finds percentage contributions 0.08%, 0.40%, 1.0% and 2.0%.
  4. Adds them to about 3.5%.
  5. Finds absolute uncertainty about 0.44 g cm⁻³.
  6. Reports (12.5 ± 0.4) g cm⁻³ with sensible precision.

Model answer

Volume = 5.00(2.00)(1.00) = 10.0 cm³, so ρ = 125.0/10.0 = 12.5 g cm⁻³. The percentage uncertainty is 0.1/125.0 × 100% + 0.02/5.00 × 100% + 0.02/2.00 × 100% + 0.02/1.00 × 100% = 3.48%. Thus Δρ = 0.0348(12.5) = 0.44 g cm⁻³, giving ρ = (12.5 ± 0.4) g cm⁻³.

Finish from memory

Three-question recap

  1. Which type of error mainly limits precision?

    Check

    Random error, because it produces scatter between repeated readings.

  2. What happens to percentage uncertainties when quantities are multiplied?

    Check

    Their percentage or fractional uncertainties are added.

  3. Why should an estimate avoid many significant figures?

    Check

    Its assumptions and approximate inputs do not justify false numerical precision.

Try this next

Estimate a familiar quantity, then propagate uncertainty through one sum and one product.

Continue with the next resource in this course.

Course and syllabus information
Course
GCE A-Level H1 Physics
Edition
GCE A-Level H1 Physics 2027