Year 9AC9M9M04

Error in Measurement

Calculate and interpret absolute, relative and percentage errors in measurements, recognising that all measurements are estimates.

Every measurement is an estimate. Absolute error says by how much, relative error says whether that matters.

Method

Absolute error is half the smallest unit

\text{error} = \tfrac{1}{2} \times \text{smallest division}

Measuring to the nearest cm gives 0.5 cm.

Relative error compares error to size

\text{relative} = \frac{\text{absolute error}}{\text{measurement}}

A fraction, not a size.

Percentage error is that as a percentage

\%\text{ error} = \text{relative} \times 100

The same absolute error matters differently

0{,}5 \text{ cm on } 2 \text{ cm vs on } 200 \text{ cm}

25% against 0.25%.

All measurements are estimates

The true value lies in a range, never on a point.

e.g.

\text{A length is measured as } 24 \text{ cm to the nearest centimetre. Find the absolute, relative and percentage error, and the range.}

The smallest division is 1 cm. Halve it. $\text{absolute error} = 0{,}5 \text{ cm}$
Write the range the true value lies in. $23{,}5 \leq \ell < 24{,}5$
Relative error is the absolute error over the measurement. $\frac{0{,}5}{24}$
Work it out. $\approx 0{,}0208$
Multiply by 100 for the percentage. $\approx 2{,}1\%$
Interpret it. About 2% uncertainty is usually acceptable.

Absolute error first, then divide by the measurement.

\text{Absolute error for a measurement to the nearest mm?}

Answer

0{,}5 \text{ mm}

\text{Percentage error in } 5 \text{ cm to the nearest cm?}

Answer

10\%

\text{Percentage error in } 500 \text{ cm to the nearest cm?}

Answer

0{,}1\%

\text{A mass reads } 3{,}4 \text{ kg to the nearest } 0{,}1. \text{ Give the range.}

Answer

3{,}35 \leq m < 3{,}45

\text{Why is relative error more useful than absolute?}

Answer

\text{It shows whether the error matters at that size.}

\text{Two measurements share an absolute error. Same percentage error?}

Answer

\text{Only if the measurements are equal.}