Percentage Error vs Percentage Uncertainty: HSC Physics Guide

Your voltmeter reads \(4.8\pm0.2\ \mathrm{V}\). The accepted reference value is \(5.0\ \mathrm{V}\).

The measured value differs from the reference by \(0.2\ \mathrm{V}\), exactly the same amount as its quoted uncertainty. Does that mean the percentage error and percentage uncertainty are both 4%?

Not quite. They use different denominators, because they describe different things.

• Percentage uncertainty compares the absolute uncertainty with the measured value.
• Percentage error compares the difference between the measured value and the accepted value with the accepted value.

For this measurement:

\[ \%\text{ uncertainty} =\frac{0.2}{4.8}\times100 =4.17\% \]

\[ \%\text{ error} =\frac{|4.8-5.0|}{5.0}\times100 =4.00\% \]

Both calculations are correct. The important thing is knowing which question you’re answering.

01Percentage error vs percentage uncertainty: the difference

Let’s establish the two formulas first.

1. Percentage uncertainty

Percentage uncertainty tells you how large the estimated uncertainty is relative to your measured value.

\[ \boxed{ \%\text{ uncertainty} =\frac{\text{absolute uncertainty}}{\text{measured value}} \times100\% } \]

In symbols:

\[ \boxed{ \%\text{ uncertainty} =\frac{\Delta x}{|x|}\times100\% } \]

Where:

• \(x\) is the measured value.
• \(\Delta x\) is its absolute uncertainty, expressed in the same units as \(x\).

For example, if a length is measured as \(20.0\pm0.2\ \mathrm{cm}\):

\[ \%\text{ uncertainty} =\frac{0.2}{20.0}\times100 =1.0\% \]

The measurement has a percentage uncertainty of 1.0%.

Notice that you don’t need an accepted or theoretical value to calculate this. You only need the measurement and its uncertainty.

2. Percentage error

Percentage error tells you how far your experimental result differs from an accepted reference value.

\[ \boxed{ \%\text{ error} =\frac{|\text{measured value}-\text{accepted value}|} {|\text{accepted value}|}\times100\% } \]

In symbols:

\[ \boxed{ \%\text{ error} =\frac{|x_{\mathrm{meas}}-x_{\mathrm{ref}}|} {|x_{\mathrm{ref}}|}\times100\% } \]

Where:

• \(x_{\mathrm{meas}}\) is your experimental measurement.
• \(x_{\mathrm{ref}}\) is the accepted or reference value.

For example, suppose your experiment gives \(19.6\ \mathrm{cm}\), while the accepted value is \(20.0\ \mathrm{cm}\).

\[ \%\text{ error} =\frac{|19.6-20.0|}{20.0}\times100 =2.0\% \]

Your measurement has a percentage error of 2.0%.

The denominator is the accepted value, not your experimental result.

Compare the two calculations

A useful way to remember the distinction:

• Uncertainty asks: How precisely did I measure?
• Error asks: How far was my result from the reference?

These are related, but they are not interchangeable.

02What does measurement uncertainty actually mean?

Before calculating percentage uncertainty, you need to understand what the absolute uncertainty represents.

Suppose you measure a metal rod and obtain:

\[ L=20.0\pm0.2\ \mathrm{cm} \]

This means:

• Your best measured value is \(20.0\ \mathrm{cm}\).
• Your estimated absolute uncertainty is \(0.2\ \mathrm{cm}\).

The corresponding measurement interval is:

\[ 19.8\leq L\leq20.2\ \mathrm{cm} \]

The interval extends \(0.2\ \mathrm{cm}\) below and above the measured value.

This interval is useful when comparing your result with an accepted value.

Consider two possibilities:

Case A: The accepted value is \(20.1\ \mathrm{cm}\).

It lies inside the measurement interval. Your result is consistent with the accepted value within the stated uncertainty.

Case B: The accepted value is \(20.6\ \mathrm{cm}\).

It lies outside the measurement interval. Your result is not consistent with the accepted value under the uncertainty convention you’ve used.

This does not prove that the true value must lie inside your interval. Uncertainty is an estimate based on your measurement method and assumptions, not a guarantee.

It does, however, give you a useful way to assess the reliability of an experimental result.

03How do you determine absolute uncertainty?

This is where you need to pay attention to the measuring instrument and the experimental method.

You cannot automatically choose an uncertainty without considering how the measurement was obtained.

1. Analogue instruments

For an analogue instrument, a common school convention is to use half the smallest readable scale division.

For example, imagine a ruler with markings every \(1\ \mathrm{mm}\).

Its smallest division is:

\[ 1\ \mathrm{mm} \]

Using the half-division convention:

\[ \Delta L=\pm0.5\ \mathrm{mm} \]

Therefore, a reading of \(85\ \mathrm{mm}\) would be reported as:

\[ L=85.0\pm0.5\ \mathrm{mm} \]

This is a common HSC-style convention, rather than a universal statement about every instrument.

2. Digital instruments

Digital instruments require a little more care.

A display showing \(2.45\ \mathrm{s}\) does not automatically mean the uncertainty is \(\pm0.005\ \mathrm{s}\).

The instrument may have a larger accuracy limitation than its display resolution suggests.

Use the manufacturer’s accuracy specification or the uncertainty convention supplied in the question. If neither is available, state and justify the assumption you make.

3. Repeated measurements

Suppose you measure the same quantity five times:

A common classroom method for a small set of repeated measurements is to estimate absolute uncertainty using half the range.

\[ \boxed{ \Delta x=\frac{x_{\max}-x_{\min}}{2} } \]

Here:

\[ \Delta t=\frac{1.82-1.78}{2} =0.02\ \mathrm{s} \]

The mean time is \(1.80\ \mathrm{s}\), so the result can be reported as:

\[ t=1.80\pm0.02\ \mathrm{s} \]

The half-range estimates the spread of the repeated measurements. It is not automatically the uncertainty of the mean under every statistical method.

What about standard deviation?

Standard deviation is another way of describing how much repeated measurements vary around their mean.

It is a statistical measure of spread, not a synonym for measurement error or absolute uncertainty.

For a sample of repeated measurements, the sample standard deviation is:

\[ s=\sqrt{\frac{\sum(x_i-\bar{x})^2}{n-1}} \]

Where:

• \(x_i\) represents an individual reading.
• \(\bar{x}\) is the mean.
• \(n\) is the number of readings.
• \(s\) is the sample standard deviation.

Standard deviation can be used to estimate measurement uncertainty when an appropriate statistical method is specified. However, it is not automatically interchangeable with half-range uncertainty, instrument accuracy, or uncertainty in the mean.

Here’s the important distinction:

Imagine a stopwatch that consistently reads \(0.10\ \mathrm{s}\) too high.

You could take twenty measurements and obtain almost identical results. The standard deviation might be very small.

However, every measurement would still contain the same systematic bias.

A small standard deviation tells you that your readings are consistent. It does not prove that they are accurate.

05An experimental example: Why measuring more carefully matters

Let’s apply both concepts to an experiment that measures gravitational acceleration using a simple pendulum.

The experimental question is straightforward:

Can we use the length and period of a pendulum to estimate \(g\), the acceleration due to gravity?

The original experimental design

A pendulum consists of a small bob attached to a string. The string hangs from a fixed pivot on a clamp stand.

A metre ruler is placed beside the string so that the length can be measured from the pivot to the centre of the bob.

The bob is pulled slightly to one side and released. A stopwatch is used to measure its oscillation period.

The first idea might be to time one complete oscillation.

But there’s a problem.

Human reaction time affects when you start and stop the stopwatch. If the timing uncertainty is approximately \(0.20\ \mathrm{s}\), that is a substantial fraction of a period of around \(1.8\ \mathrm{s}\).

The percentage uncertainty in the timing would be:

\[ \frac{0.20}{1.80}\times100 =11.1\% \]

That’s a fairly large uncertainty.

Improving the experimental design

Instead of timing one oscillation, time twenty complete oscillations.

Start the stopwatch when the bob passes a chosen position in a particular direction. Stop it when the bob returns to that same position and direction after twenty complete cycles.

Then divide the total time by twenty.

If the start-and-stop timing uncertainty remains approximately \(0.20\ \mathrm{s}\), the uncertainty in the calculated period becomes:

\[ \Delta T=\frac{0.20}{20} =0.010\ \mathrm{s} \]

The percentage timing uncertainty is now:

\[ \frac{0.010}{1.80}\times100 =0.56\% \]

Timing twenty oscillations reduces the fractional effect of reaction time.

It does not eliminate systematic errors, such as an incorrectly calibrated stopwatch or an incorrectly measured pendulum length.

What does the experiment predict?

For a simple pendulum undergoing small-angle oscillations:

\[ T=2\pi\sqrt{\frac{L}{g}} \]

Rearranging:

\[ \boxed{g=\frac{4\pi^2L}{T^2}} \]

Where:

• \(T\) is the period, in seconds.
• \(L\) is the pendulum length, in metres.
• \(g\) is gravitational acceleration, in \(\mathrm{m\,s^{-2}}\).

This model assumes a small oscillation angle, a light string, a compact bob, and negligible air resistance and pivot friction.

If the accepted value is \(9.81\ \mathrm{m\,s^{-2}}\), a pendulum of length \(0.800\ \mathrm{m}\) should have a period of approximately \(1.79\ \mathrm{s}\).

Now we can compare the experimental result with that prediction.

06A quick decision method for HSC questions

When you encounter a question involving percentage error or percentage uncertainty, follow this sequence.

A few additional rules will help you avoid common mistakes.

Common mistakes to avoid

Mistake 1: Using the accepted value as the denominator for percentage uncertainty.

Wrong:

\[ \frac{\Delta x}{x_{\mathrm{ref}}}\times100 \]

Correct:

\[ \frac{\Delta x}{|x_{\mathrm{meas}}|}\times100 \]

Mistake 2: Using the measured value as the denominator for percentage error.

Wrong:

\[ \frac{|x_{\mathrm{meas}}-x_{\mathrm{ref}}|} {|x_{\mathrm{meas}}|}\times100 \]

Correct:

\[ \frac{|x_{\mathrm{meas}}-x_{\mathrm{ref}}|} {|x_{\mathrm{ref}}|}\times100 \]

Mistake 3: Treating measurement error and uncertainty as the same thing.

A result can have a small percentage error but a large percentage uncertainty.

It can also have a small percentage uncertainty but a large percentage error.

Neither percentage automatically determines the other.

Mistake 4: Assuming repeated measurements remove systematic errors.

Repeating an experiment can help reduce random variation and improve the estimate of a mean.

It cannot automatically correct a miscalibrated instrument or a consistently incorrect measurement technique.

Mistake 5: Confusing half-range with standard deviation.

Both describe aspects of variation in repeated measurements, but they are calculated differently and have different statistical meanings.

Always follow the method specified in the question.

07Questions and solutions

Question 1

A student uses a digital force sensor to measure the weight of a hanging object.

The sensor gives:

\[ F=4.90\pm0.05\ \mathrm{N} \]

The accepted reference value is \(5.00\ \mathrm{N}\).

Calculate:

a) The percentage uncertainty.

b) The percentage error.

c) The measurement interval, and whether the accepted value lies inside it.

Solution 1

The percentage uncertainty is \(1.02\%\), the percentage error is \(2.0\%\), and the accepted value lies outside the measurement interval.

Step 1: Calculate percentage uncertainty.

\[ \begin{aligned} \%\text{ uncertainty} &=\frac{0.05}{4.90}\times100\\ &=1.02\% \end{aligned} \]

Step 2: Calculate percentage error.

\[ \begin{aligned} \%\text{ error} &=\frac{|4.90-5.00|}{5.00}\times100\\ &=\frac{0.10}{5.00}\times100\\ &=2.0\% \end{aligned} \]

Step 3: Calculate the measurement interval.

\[ F_{\min}=4.90-0.05=4.85\ \mathrm{N} \]

\[ F_{\max}=4.90+0.05=4.95\ \mathrm{N} \]

Therefore:

\[ 4.85\leq F\leq4.95\ \mathrm{N} \]

The accepted value of \(5.00\ \mathrm{N}\) is outside this interval.

The experimental result is not consistent with the accepted value within the stated uncertainty.

Notice that the absolute difference is \(0.10\ \mathrm{N}\), while the absolute uncertainty is only \(0.05\ \mathrm{N}\). They describe different things.

Question 2

A trolley moves along a straight horizontal track between two photogates.

The distance between the photogates is:

\[ d=2.40\pm0.01\ \mathrm{m} \]

The measured travel time is:

\[ t=1.20\pm0.03\ \mathrm{s} \]

The average speed is calculated using:

\[ v=\frac{d}{t} \]

An independent reference measurement gives an average speed of \(2.05\ \mathrm{m\,s^{-1}}\).

Calculate:

a) The experimental average speed.

b) Its percentage uncertainty.

c) Its absolute uncertainty.

d) Its percentage error.

e) Whether the accepted reference value lies inside the measurement interval.

Use the maximum-uncertainty convention for division.

Solution 2

The experimental speed is \(2.00\pm0.06\ \mathrm{m\,s^{-1}}\), with a percentage uncertainty of \(2.92\%\) and a percentage error of \(2.44\%\). The accepted value lies inside the uncertainty interval.

Step 1: Calculate the experimental speed.

\[ \begin{aligned} v&=\frac{d}{t}\\ &=\frac{2.40}{1.20}\\ &=2.00\ \mathrm{m\,s^{-1}} \end{aligned} \]

Step 2: Calculate the percentage uncertainty in distance.

\[ \%\Delta d=\frac{0.01}{2.40}\times100 \]

\[ \%\Delta d=0.417\% \]

Step 3: Calculate the percentage uncertainty in time.

\[ \%\Delta t=\frac{0.03}{1.20}\times100 \]

\[ \%\Delta t=2.50\% \]

Step 4: Calculate the percentage uncertainty in speed.

Since speed is calculated by division, add the percentage uncertainties:

\[ \begin{aligned} \%\Delta v &=\%\Delta d+\%\Delta t\\ &=0.417+2.50\\ &=2.917\% \end{aligned} \]

Therefore:

\[ \boxed{\%\text{ uncertainty}=2.92\%} \]

Step 5: Calculate absolute uncertainty.

\[ \Delta v=\frac{2.917}{100}\times2.00 \]

\[ \Delta v=0.0583\ \mathrm{m\,s^{-1}} \]

To an appropriate number of significant figures:

\[ \boxed{v=2.00\pm0.06\ \mathrm{m\,s^{-1}}} \]

Step 6: Calculate percentage error.

\[ \begin{aligned} \%\text{ error} &=\frac{|2.00-2.05|}{2.05}\times100\\ &=\frac{0.05}{2.05}\times100\\ &=2.44\% \end{aligned} \]

Step 7: Check the measurement interval.

\[ v_{\min}=2.00-0.06=1.94 \]

\[ v_{\max}=2.00+0.06=2.06 \]

The accepted value of \(2.05\ \mathrm{m\,s^{-1}}\) lies inside this interval.

The result is therefore consistent with the reference value within the estimated uncertainty.

The important trap here is to forget that both distance and time contribute to the uncertainty in speed. Using only the uncertainty in time would underestimate the total uncertainty.

Question 3

A student investigates the density of a metal sample that is approximately cubic.

The measured mass is:

\[ m=71.7\pm0.1\ \mathrm{g} \]

Each side of the cube has a measured length of:

\[ a=2.00\pm0.02\ \mathrm{cm} \]

The density is calculated using:

\[ \rho=\frac{m}{a^3} \]

The accepted reference density supplied for the material is:

\[ \rho_{\mathrm{ref}}=8.96\ \mathrm{g\,cm^{-3}} \]

Calculate:

a) The experimental density.

b) The percentage uncertainty in density.

c) The absolute uncertainty in density.

d) The percentage error.

e) Whether the accepted density lies inside the measurement interval.

Assume the cube has equal side lengths and use the maximum-uncertainty convention.

Solution 3

The experimental density is approximately \(8.96\pm0.28\ \mathrm{g\,cm^{-3}}\), with a percentage uncertainty of \(3.14\%\) and a percentage error of approximately \(0.03\%\). The accepted value lies inside the uncertainty interval.

Step 1: Calculate the volume.

The cube has equal side lengths, so:

\[ V=a^3 \]

\[ V=2.00^3=8.00\ \mathrm{cm^3} \]

Step 2: Calculate the density.

\[ \begin{aligned} \rho&=\frac{m}{V}\\ &=\frac{71.7}{8.00}\\ &=8.9625\ \mathrm{g\,cm^{-3}} \end{aligned} \]

Step 3: Calculate the percentage uncertainty in mass.

\[ \%\Delta m=\frac{0.1}{71.7}\times100 \]

\[ \%\Delta m=0.140\% \]

Step 4: Calculate the percentage uncertainty in the side length.

\[ \%\Delta a=\frac{0.02}{2.00}\times100 \]

\[ \%\Delta a=1.0\% \]

Here’s the important part.

The volume depends on the cube of the side length:

\[ V=a^3 \]

Therefore, the percentage uncertainty in volume is three times the percentage uncertainty in the side length.

\[ \%\Delta V=3(1.0)=3.0\% \]

Step 5: Calculate the percentage uncertainty in density.

Density is mass divided by volume, so add their percentage uncertainties.

\[ \begin{aligned} \%\Delta\rho &=\%\Delta m+\%\Delta V\\ &=0.140+3.0\\ &=3.14\% \end{aligned} \]

Step 6: Calculate the absolute uncertainty.

\[ \Delta\rho=\frac{3.14}{100}\times8.9625 \]

\[ \Delta\rho\approx0.28\ \mathrm{g\,cm^{-3}} \]

Therefore:

\[ \boxed{ \rho=8.96\pm0.28\ \mathrm{g\,cm^{-3}} } \]

Step 7: Calculate percentage error.

Use the unrounded experimental density:

\[ \begin{aligned} \%\text{ error} &=\frac{|8.9625-8.96|}{8.96}\times100\\ &=0.0279\% \end{aligned} \]

Therefore:

\[ \boxed{\%\text{ error}\approx0.03\%} \]

Step 8: Check the measurement interval.

\[ \rho_{\min}=8.96-0.28=8.68 \]

\[ \rho_{\max}=8.96+0.28=9.24 \]

The accepted value of \(8.96\ \mathrm{g\,cm^{-3}}\) lies inside this interval.

The result is consistent with the accepted value.

The key difficulty was recognising that the side length is cubed. Using a 1% uncertainty for the volume instead of 3% would substantially underestimate the final uncertainty.

Question 4

Two experimental teams independently measure gravitational acceleration using the same general experimental method.

Both teams perform five trials.

Their results are:

The sample standard deviations are:

• Team A: \(0.007\ \mathrm{m\,s^{-2}}\)
• Team B: \(0.19\ \mathrm{m\,s^{-2}}\)

Use an accepted gravitational acceleration of \(9.81\ \mathrm{m\,s^{-2}}\).

a) Calculate the mean result for each team.

b) Calculate the percentage error for each mean.

c) Which team has more consistent measurements?

d) Does the smaller standard deviation prove that Team A’s result is more accurate? Explain.

Solution 4

Both teams obtain a mean of \(9.71\ \mathrm{m\,s^{-2}}\), giving both teams the same percentage error of approximately \(1.02\%\). Team A has substantially more consistent readings, but its smaller standard deviation does not make its mean more accurate.

Step 1: Calculate Team A’s mean.

\[ \bar g_A= \frac{9.70+9.71+9.71+9.72+9.71}{5} \]

\[ \bar g_A=9.71\ \mathrm{m\,s^{-2}} \]

Step 2: Calculate Team B’s mean.

\[ \bar g_B= \frac{9.40+9.80+9.70+9.90+9.75}{5} \]

\[ \bar g_B=9.71\ \mathrm{m\,s^{-2}} \]

Both teams obtain the same experimental mean.

Step 3: Calculate the percentage error for Team A.

\[ \begin{aligned} \%\text{ error} &=\frac{|9.71-9.81|}{9.81}\times100\\ &=1.02\% \end{aligned} \]

Step 4: Calculate the percentage error for Team B.

The mean is identical, so:

\[ \begin{aligned} \%\text{ error} &=\frac{|9.71-9.81|}{9.81}\times100\\ &=1.02\% \end{aligned} \]

Both teams have the same percentage error.

Step 5: Compare the standard deviations.

Team A has a standard deviation of:

\[ s_A=0.007\ \mathrm{m\,s^{-2}} \]

Team B has a standard deviation of:

\[ s_B=0.19\ \mathrm{m\,s^{-2}} \]

Team A’s readings are much more tightly grouped around their mean.

Therefore, Team A has greater repeatability.

Step 6: Explain why this does not establish greater accuracy.

Both teams have the same mean, and both means are below the accepted value by \(0.10\ \mathrm{m\,s^{-2}}\).

Team A’s smaller standard deviation only tells us that its individual measurements are more tightly grouped.

It does not reveal whether the measuring equipment has a systematic calibration error.

For example, both teams could be using instruments that consistently produce values that are too low.

The repeated readings would still be tightly grouped, but their mean would remain biased.

This is why you must distinguish precision, accuracy, and uncertainty rather than treating them as interchangeable.

Question 5

A student measures a metal rod whose reference length is exactly \(20.00\ \mathrm{cm}\).

The student uses a ruler with \(1\ \mathrm{mm}\) divisions.

However, the ruler has a calibration problem: its physical edge is \(0.20\ \mathrm{cm}\) to the left of the zero mark. The spacing between all the scale divisions is otherwise correct.

Each individual scale reading has an uncertainty of \(\pm0.05\ \mathrm{cm}\).

The student tries two measurement methods.

Method A: The student places the physical edge of the ruler against the beginning of the rod and reads the far end as \(20.20\ \mathrm{cm}\). The student reports:

\[ L_A=20.20\pm0.05\ \mathrm{cm} \]

The student has incorrectly treated the physical edge as the zero mark.

Method B: The student reads both ends of the rod using the scale markings.

The beginning of the rod aligns with the \(0.20\ \mathrm{cm}\) mark, and the far end aligns with the \(20.20\ \mathrm{cm}\) mark.

The student subtracts these readings and reports the difference, using worst-case addition of the two reading uncertainties.

Now consider a second ruler. This one has no zero offset, but every scale interval is \(0.5\%\) longer than it should be.

Answer the following:

a) For Method A, calculate the quoted percentage uncertainty and the actual percentage error.

b) For Method B, calculate the measured length, percentage uncertainty, and percentage error.

c) Which method gives the smaller quoted percentage uncertainty? Does this mean it is the more accurate method?

d) Using the second ruler, determine what length Method B would report for the same rod. Calculate the percentage error.

e) Explain why subtracting two readings removes the first ruler’s zero offset but does not remove the second ruler’s scale calibration error.

f) Would repeating either measurement many times necessarily eliminate the systematic error? Justify your answer.

Solution 5

Method B correctly eliminates the first ruler’s known zero offset, giving \(20.00\pm0.10\ \mathrm{cm}\). However, the second ruler’s scale expansion produces a systematic error that subtraction cannot eliminate.

The crucial distinction is between an additive offset and a multiplicative calibration error.

Step 1: Calculate Method A’s percentage uncertainty.

The student reports:

\[ L_A=20.20\pm0.05\ \mathrm{cm} \]

Using the reported value:

\[ \begin{aligned} \%\text{ uncertainty} &=\frac{0.05}{20.20}\times100\\ &=0.248\% \end{aligned} \]

This is only the percentage uncertainty associated with the scale reading.

It is not a complete uncertainty statement for the uncorrected result because the student has ignored the known zero offset.

Step 2: Calculate Method A’s percentage error.

The actual rod length is \(20.00\ \mathrm{cm}\).

\[ \begin{aligned} \%\text{ error} &=\frac{|20.20-20.00|}{20.00}\times100\\ &=\frac{0.20}{20.00}\times100\\ &=1.0\% \end{aligned} \]

The result is systematically too high by 1.0%.

The known zero offset should have been corrected rather than ignored.

Step 3: Calculate Method B’s measured length.

The student reads both endpoints:

\[ L_{\mathrm{start}}=0.20\ \mathrm{cm} \]

\[ L_{\mathrm{end}}=20.20\ \mathrm{cm} \]

The length is the difference:

\[ \begin{aligned} L_B&=20.20-0.20\\ &=20.00\ \mathrm{cm} \end{aligned} \]

The zero offset cancels.

Step 4: Calculate Method B’s absolute uncertainty.

Each endpoint reading has an uncertainty of:

\[ \pm0.05\ \mathrm{cm} \]

Using the worst-case convention for subtraction, add the absolute uncertainties:

\[ \Delta L_B=0.05+0.05 \]

\[ \Delta L_B=0.10\ \mathrm{cm} \]

Therefore:

\[ L_B=20.00\pm0.10\ \mathrm{cm} \]

Step 5: Calculate Method B’s percentage uncertainty.

\[ \begin{aligned} \%\text{ uncertainty} &=\frac{0.10}{20.00}\times100\\ &=0.50\% \end{aligned} \]

Its percentage error is:

\[ \begin{aligned} \%\text{ error} &=\frac{|20.00-20.00|}{20.00}\times100\\ &=0\% \end{aligned} \]

Method B has a larger percentage uncertainty than Method A’s quoted reading uncertainty, but it gives the correct central value.

This is an important result: a smaller quoted percentage uncertainty does not automatically mean a measurement is more accurate.

Method A’s uncertainty statement was incomplete because it ignored a known systematic offset.

Step 6: Investigate the second ruler.

Now suppose the ruler has no zero offset, but its scale intervals are all \(0.5\%\) too long.

This is a different type of calibration problem.

For an actual rod length of \(20.00\ \mathrm{cm}\), the measured scale difference becomes:

\[ L_{\mathrm{meas}}=20.00\times1.005 \]

\[ L_{\mathrm{meas}}=20.10\ \mathrm{cm} \]

Method B would therefore report:

\[ L_B=20.10\ \mathrm{cm} \]

The percentage error is:

\[ \begin{aligned} \%\text{ error} &=\frac{|20.10-20.00|}{20.00}\times100\\ &=0.50\% \end{aligned} \]

The absolute uncertainty from the two endpoint readings remains \(0.10\ \mathrm{cm}\), using the stated worst-case convention.

The percentage uncertainty is:

\[ \begin{aligned} \%\text{ uncertainty} &=\frac{0.10}{20.10}\times100\\ &\approx0.50\% \end{aligned} \]

Step 7: Explain why the two calibration problems behave differently.

This is the most important part of the question.

A zero offset is an additive error. It shifts both position readings by the same amount.

If the measured positions are represented by:

\[ x_1+b \]

and:

\[ x_2+b \]

then subtracting gives:

\[ (x_2+b)-(x_1+b)=x_2-x_1 \]

The shared offset \(b\) disappears.

However, a scale expansion is a multiplicative error.

If every measured position is multiplied by a factor \(k\), the difference becomes:

\[ kx_2-kx_1=k(x_2-x_1) \]

The scale factor remains.

This explains why subtracting two readings can eliminate a shared zero offset but cannot eliminate an incorrect scale calibration.

Step 8: Consider what happens if you repeat the measurements.

Repeating the measurement may reduce random variation and improve the estimate of the mean.

However, every reading from the second ruler will still contain the same \(0.5\%\) scale bias.

The mean will therefore remain systematically too high.

A small standard deviation would not reveal this problem.

The instrument needs calibration or a correction factor to remove the systematic error.

This is the deeper lesson: uncertainty describes the limitations of your measurement, but it does not automatically account for every systematic problem. You must understand how your instrument produces its readings.

08What should you learn next?

Once you can calculate absolute uncertainty, percentage uncertainty, and percentage error, the next step is applying measurement uncertainty to experimental graphs.

In particular, you need to understand how uncertainty intervals become error bars and how maximum and minimum gradient lines help estimate uncertainty in a calculated gradient.

Continue with Error Bars and Maximum-Minimum Gradients: HSC Physics Guide.

That connects the calculations in this article to one of the most useful techniques in HSC Physics experimental data analysis.