How to Find Percentage Uncertainty from Repeated Measurements

Five timing trials for the same trolley produce the following readings:

Which time should you report? The fastest? The slowest? Or the average?

The usual approach for a small set of repeated measurements is to calculate the mean, estimate the absolute uncertainty using half the range, and then convert that uncertainty into a percentage.

The mean is \(0.750\ \mathrm{s}\). The half-range uncertainty is:

\[ \Delta t=\frac{0.752-0.748}{2} =0.002\ \mathrm{s} \]

Therefore, the percentage uncertainty is:

\[ \begin{aligned} \%\text{ uncertainty} &=\frac{\Delta t}{\bar t}\times100\\ &=\frac{0.002}{0.750}\times100\\ &=0.267\% \end{aligned} \]

The final answer is 0.27% percentage uncertainty.

The general method is:

\[ \boxed{ \%\text{ uncertainty} = \frac{\text{half the range}}{\text{mean}} \times100\% } \]

This article explains why that method works, when to use it, and what can go wrong if you apply it without checking your data.

All experimental measurements and datasets below are illustrative.

01Why do we repeat measurements?

Imagine investigating the time taken for a trolley to travel between two points on a track.

You could time the trolley once using a stopwatch. However, your reaction time affects when you start and stop the stopwatch. The trolley might also be released slightly differently each time.

A single measurement cannot tell you how much these variations affect your result.

A first experimental design might use a handheld stopwatch. The student starts the timer when releasing the trolley and stops it when the trolley reaches a marked position.

The problem is that both timing events depend on human reaction time.

A revised design uses two photogates connected to an electronic timer:

• The first photogate starts the timer when the trolley interrupts its beam.
• The second photogate stops the timer when the trolley reaches it.
• The timer displays the elapsed time.

This removes much of the variation caused by human reaction time. It does not eliminate every possible source of uncertainty, such as sensor calibration or inconsistent trolley motion.

What should happen if the experiment is repeated five times under the same conditions?

Ideally, the measurements should cluster around a common value. They won’t necessarily be identical because small variations remain.

That spread is what we use to estimate uncertainty from repeated measurements.

02The half-range method

For repeated measurements, a common school convention is to use half the range as an estimate of absolute uncertainty.

There are three quantities you need to understand.

1. The mean

The mean is the arithmetic average of all your measurements.

\[ \bar{x}=\frac{x_1+x_2+\cdots+x_n}{n} \]

Where:

• \(\bar{x}\) is the mean measurement.
• \(x_1,x_2,\ldots,x_n\) are the individual readings.
• \(n\) is the number of readings.

2. The range

The range is the difference between the largest and smallest measurements.

\[ \text{Range}=x_{\max}-x_{\min} \]

3. The absolute uncertainty

Using the half-range convention:

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

Here, \(\Delta x\) is the estimated absolute uncertainty. It has the same units as the original measurement.

Once you have the mean and absolute uncertainty, calculate percentage uncertainty:

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

The absolute value ensures that the percentage uncertainty is positive.

A visual way to understand half the range

Return to the five trolley measurements:

\[ 0.748,\ 0.752,\ 0.750,\ 0.749,\ 0.751\ \mathrm{s} \]

The smallest reading is \(0.748\ \mathrm{s}\), and the largest is \(0.752\ \mathrm{s}\).

The full range is \(0.004\ \mathrm{s}\).

Half the range is \(0.002\ \mathrm{s}\).

The mean gives you a central estimate, while the half-range gives you an estimate of the spread.

For this dataset, the result is:

\[ t=(0.750\pm0.002)\ \mathrm{s} \]

This notation means that the measured time is \(0.750\ \mathrm{s}\), with an estimated absolute uncertainty of \(0.002\ \mathrm{s}\).

The corresponding symmetric measurement interval is:

\[ 0.748\leq t\leq0.752\ \mathrm{s} \]

Notice that the uncertainty and the mean have different jobs. The mean tells you the central result. The uncertainty tells you how much variation is associated with that result.

03What does the measurement interval actually mean?

When you report a measurement as:

\[ x=\bar{x}\pm\Delta x \]

you are expressing a central estimate and an associated uncertainty.

The corresponding symmetric interval is:

\[ x_{\min,\text{reported}}=\bar{x}-\Delta x \]

\[ x_{\max,\text{reported}}=\bar{x}+\Delta x \]

For example, a measurement of:

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

gives the interval:

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

This is the interval implied by the stated uncertainty convention.

However, there is a subtle point when dealing with repeated measurements.

The arithmetic mean is not always exactly halfway between the maximum and minimum readings.

Consider these five measurements:

\[ 1.00,\ 1.01,\ 1.01,\ 1.02,\ 1.06\ \mathrm{s} \]

The mean is:

\[ \bar t=\frac{1.00+1.01+1.01+1.02+1.06}{5} =1.02\ \mathrm{s} \]

The half-range is:

\[ \Delta t=\frac{1.06-1.00}{2} =0.03\ \mathrm{s} \]

Therefore, using the mean and half-range convention:

\[ t=(1.02\pm0.03)\ \mathrm{s} \]

The reported symmetric interval is:

\[ 0.99\leq t\leq1.05\ \mathrm{s} \]

But the actual observed range is:

\[ 1.00\leq t\leq1.06\ \mathrm{s} \]

These intervals are not identical.

This happens because the half-range is calculated from the extreme readings, but the reported uncertainty is centred on the mean.

For a typical HSC-style question using the half-range convention, retain the arithmetic mean and use half the range as the uncertainty. Do not silently replace the mean with the midpoint of the extremes.

If a question specifically asks for the observed range, report the minimum and maximum readings instead.

Also remember that a half-range estimate is not a formal confidence interval or a guarantee that the true value lies within the reported interval.

05Uncertainty, error, and standard deviation are not the same

These three terms are related, but they describe different things.

Measurement uncertainty

Uncertainty describes the limitations of your measurement.

For repeated readings, the half-range convention estimates the spread using:

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

You can calculate this without knowing the true value.

Measurement error

Measurement error is the difference between a measured value and the true value.

When an accepted reference value is available, you can calculate the signed difference using:

\[ e=\bar{x}-x_{\mathrm{ref}} \]

Where:

• \(e\) is the signed difference.
• \(\bar{x}\) is the measured mean.
• \(x_{\mathrm{ref}}\) is the accepted reference value.

A positive result means the measurement is higher than the reference. A negative result means it is lower.

The true measurement error is generally unknown because the exact true value is usually unknown.

This is why uncertainty and error cannot be used interchangeably.

For example, you could measure a length five times and obtain almost identical readings. However, if your ruler has an incorrect zero point, all five measurements could be systematically too high.

Your uncertainty might be small, but your measurements could still be inaccurate.

For a more detailed comparison of the two percentage calculations, see Percentage Error vs Percentage Uncertainty: HSC Physics Guide.

Standard deviation

Standard deviation is another way to describe the spread of repeated measurements.

Unlike half-range uncertainty, which only uses the largest and smallest readings, standard deviation considers how every reading differs from the mean.

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

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

Where:

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

A small standard deviation indicates that the readings are closely grouped around their mean.

However, standard deviation is not automatically the same as half-range uncertainty.

It is also important not to confuse standard deviation with standard error. The standard error, commonly calculated as \(s/\sqrt n\), estimates the uncertainty of the mean under appropriate statistical assumptions.

Use the method specified in the question. If it asks for half-range uncertainty, do not substitute standard deviation or standard error.

A comparison: Same mean, different uncertainty

Consider two groups measuring the same physical quantity.

Assume all measurements are in the same units.

Both groups have a mean of 10.0.

However, their half-range uncertainties are different.

For Group A:

\[ \Delta x_A=\frac{10.2-9.8}{2}=0.2 \]

\[ \%\text{ uncertainty}_A =\frac{0.2}{10.0}\times100 =2.0\% \]

For Group B:

\[ \Delta x_B=\frac{10.1-9.9}{2}=0.1 \]

\[ \%\text{ uncertainty}_B =\frac{0.1}{10.0}\times100 =1.0\% \]

Group B has the smaller percentage uncertainty, so its measurements are more repeatable under this convention.

Now suppose the accepted reference value is 9.7.

Both groups have the same signed difference from the reference:

\[ e=10.0-9.7=+0.3 \]

Their percentage uncertainties are different, but their errors are identical.

This is an important distinction: better repeatability does not automatically mean better accuracy.

06A quick decision method for HSC questions

When you encounter a question about repeated measurements, work through this sequence.

There are also a few situations where you need to be more careful.

If the question specifies a different uncertainty convention: Follow the stated method. Half-range is a common classroom convention for a small set of repeated measurements, not a universal rule for every experiment.

If an instrument has a stated accuracy: Do not assume the spread of repeated readings represents the instrument’s entire uncertainty. A calibration limitation may still exist even when readings are closely grouped.

If every reading is identical: Half the range will be zero. This means no variation was observed in the recorded readings. It does not prove that the measurement has zero uncertainty. Instrument resolution and calibration limitations may still matter.

If one reading is unusually different: Do not automatically remove it. Check whether there is evidence of a genuine procedural mistake. A measurement should only be excluded when there is a defensible reason.

If you are calculating a percentage: Keep the original values during your working and round your final result appropriately. A common reporting convention is to give uncertainty to one significant figure and round the measured value to the same decimal place. Follow any specific instructions given in the question.

07Questions and solutions

The following questions progress from straightforward calculations to more demanding experimental reasoning. Use the half-range convention unless another method is explicitly stated.

Question 1

A student measures the extension of a spring under a constant load five times.

The following illustrative readings are obtained:

Calculate:

a) The mean extension.

b) The absolute uncertainty using half the range.

c) The percentage uncertainty.

d) The measurement interval.

Solution 1

The mean extension is \(24.6\ \mathrm{cm}\), the absolute uncertainty is \(0.1\ \mathrm{cm}\), and the percentage uncertainty is \(0.41\%\).

Part a: Mean extension

Add the five readings and divide by five.

\[ \begin{aligned} \bar{x} &=\frac{24.5+24.6+24.7+24.6+24.6}{5}\\ &=\frac{123.0}{5}\\ &=24.6\ \mathrm{cm} \end{aligned} \]

Part b: Absolute uncertainty

The maximum reading is \(24.7\ \mathrm{cm}\), and the minimum reading is \(24.5\ \mathrm{cm}\).

\[ \begin{aligned} \Delta x &=\frac{x_{\max}-x_{\min}}{2}\\ &=\frac{24.7-24.5}{2}\\ &=0.1\ \mathrm{cm} \end{aligned} \]

Part c: Percentage uncertainty

\[ \begin{aligned} \%\text{ uncertainty} &=\frac{\Delta x}{\bar{x}}\times100\\ &=\frac{0.1}{24.6}\times100\\ &=0.4065\%\\ &\approx0.41\% \end{aligned} \]

Part d: Measurement interval

Subtract and add the absolute uncertainty to the mean.

\[ x_{\min}=24.6-0.1=24.5\ \mathrm{cm} \]

\[ x_{\max}=24.6+0.1=24.7\ \mathrm{cm} \]

Therefore:

\[ 24.5\leq x\leq24.7\ \mathrm{cm} \]

The interval is symmetric around the mean in this example because the mean is exactly halfway between the maximum and minimum readings.

Question 2

Two students independently measure the same stable time interval using different timing systems.

Their illustrative measurements are:

Student A

\[ 1.18,\ 1.19,\ 1.20,\ 1.21,\ 1.22\ \mathrm{s} \]

Student B

\[ 1.196,\ 1.200,\ 1.204,\ 1.198,\ 1.202\ \mathrm{s} \]

Using the half-range convention:

a) Calculate the mean and absolute uncertainty for each student.

b) Calculate the percentage uncertainty for each student.

c) Which student obtained the more repeatable measurements?

d) Can you conclude that this student obtained the more accurate result? Explain.

Solution 2

Student B obtained the more repeatable measurements, with a percentage uncertainty of \(0.33\%\), compared with Student A’s \(1.7\%\). However, this does not establish which result is more accurate.

Part a: Calculate Student A’s mean and absolute uncertainty

\[ \begin{aligned} \bar t_A &=\frac{1.18+1.19+1.20+1.21+1.22}{5}\\ &=1.20\ \mathrm{s} \end{aligned} \]

The half-range is:

\[ \begin{aligned} \Delta t_A &=\frac{1.22-1.18}{2}\\ &=0.02\ \mathrm{s} \end{aligned} \]

Therefore:

\[ t_A=(1.20\pm0.02)\ \mathrm{s} \]

Now calculate Student B’s mean:

\[ \begin{aligned} \bar t_B &=\frac{1.196+1.200+1.204+1.198+1.202}{5}\\ &=1.200\ \mathrm{s} \end{aligned} \]

The absolute uncertainty is:

\[ \begin{aligned} \Delta t_B &=\frac{1.204-1.196}{2}\\ &=0.004\ \mathrm{s} \end{aligned} \]

Therefore:

\[ t_B=(1.200\pm0.004)\ \mathrm{s} \]

Part b: Calculate the percentage uncertainties

For Student A:

\[ \begin{aligned} \%\text{ uncertainty}_A &=\frac{0.02}{1.20}\times100\\ &=1.67\%\\ &\approx1.7\% \end{aligned} \]

For Student B:

\[ \begin{aligned} \%\text{ uncertainty}_B &=\frac{0.004}{1.200}\times100\\ &=0.333\%\\ &\approx0.33\% \end{aligned} \]

Part c: Compare repeatability

Student B has the smaller percentage uncertainty.

The repeated readings are more closely grouped relative to their mean, indicating better repeatability under the chosen convention.

Part d: Compare accuracy

We cannot determine which student is more accurate without knowing the true or accepted reference value.

Student B’s readings are more consistent, but a systematic timing error could shift every reading away from the true value.

A smaller percentage uncertainty is evidence of better repeatability, not automatic proof of greater accuracy.

Question 3

A student investigates the period of a simple pendulum by measuring the time taken for ten complete oscillations.

Five illustrative readings are recorded:

The student’s experimental notes reveal that, during Trial 5, the stopwatch was accidentally stopped after eleven oscillations rather than ten.

a) Should Trial 5 be included in the calculation?

b) Calculate the mean time using the valid measurements.

c) Calculate the absolute uncertainty and percentage uncertainty.

d) Explain why simply removing an unusually large measurement without checking the experimental record would be poor practice.

Solution 3

Trial 5 should be excluded because it measures eleven oscillations rather than ten. Using the four valid measurements, the mean time is \(15.2\ \mathrm{s}\), and the percentage uncertainty is \(0.66\%\).

Part a: Decide whether to include Trial 5

The fifth reading does not represent the same measurement as the other four.

The first four trials measure the time for ten complete oscillations. Trial 5 measures the time for eleven.

It is therefore a documented procedural error, rather than simply an unusually large result.

Exclude it from the calculation and explain why.

Part b: Calculate the mean time

Use the four valid readings:

\[ \begin{aligned} \bar t &=\frac{15.2+15.3+15.1+15.2}{4}\\ &=\frac{60.8}{4}\\ &=15.2\ \mathrm{s} \end{aligned} \]

Part c: Calculate the absolute and percentage uncertainties

The maximum valid reading is \(15.3\ \mathrm{s}\), and the minimum is \(15.1\ \mathrm{s}\).

\[ \begin{aligned} \Delta t &=\frac{15.3-15.1}{2}\\ &=0.1\ \mathrm{s} \end{aligned} \]

Now calculate the percentage uncertainty:

\[ \begin{aligned} \%\text{ uncertainty} &=\frac{0.1}{15.2}\times100\\ &=0.6579\%\\ &\approx0.66\% \end{aligned} \]

The reported result is:

\[ t=(15.2\pm0.1)\ \mathrm{s} \]

The percentage uncertainty is \(0.66\%\).

Part d: Why can’t you simply remove unusual readings?

An unusual reading is not automatically an invalid reading.

Suppose the fifth trial had measured ten oscillations correctly but produced \(16.8\ \mathrm{s}\). That reading would need to remain in the dataset unless there was evidence of a procedural problem.

Removing it merely because it increases the uncertainty would artificially improve the reported result.

The key distinction is between an unusual but valid measurement and a measurement that does not represent the intended experiment.

Question 4

A student measures the time taken for a trolley to complete a particular journey five times.

The illustrative measurements are:

\[ 1.00,\ 1.01,\ 1.01,\ 1.02,\ 1.06\ \mathrm{s} \]

Using the half-range convention:

a) Calculate the mean.

b) Calculate the absolute uncertainty.

c) Calculate the percentage uncertainty.

d) Determine the symmetric interval implied by the mean and half-range.

e) Compare this with the actual observed range. Explain why the two intervals differ and whether this means the half-range calculation is incorrect.

Solution 4

The mean is \(1.02\ \mathrm{s}\), the half-range uncertainty is \(0.03\ \mathrm{s}\), and the percentage uncertainty is approximately \(2.9\%\). The reported symmetric interval differs from the observed range because the data are not centred symmetrically around their mean.

Part a: Calculate the mean

\[ \begin{aligned} \bar t &=\frac{1.00+1.01+1.01+1.02+1.06}{5}\\ &=\frac{5.10}{5}\\ &=1.02\ \mathrm{s} \end{aligned} \]

Part b: Calculate the absolute uncertainty

\[ \begin{aligned} \Delta t &=\frac{t_{\max}-t_{\min}}{2}\\ &=\frac{1.06-1.00}{2}\\ &=0.03\ \mathrm{s} \end{aligned} \]

Part c: Calculate percentage uncertainty

\[ \begin{aligned} \%\text{ uncertainty} &=\frac{0.03}{1.02}\times100\\ &=2.941\%\\ &\approx2.9\% \end{aligned} \]

Part d: Calculate the symmetric interval

The lower limit is:

\[ 1.02-0.03=0.99\ \mathrm{s} \]

The upper limit is:

\[ 1.02+0.03=1.05\ \mathrm{s} \]

Therefore, the reported interval is:

\[ 0.99\leq t\leq1.05\ \mathrm{s} \]

Part e: Compare with the observed range

The actual smallest reading is \(1.00\ \mathrm{s}\), and the largest is \(1.06\ \mathrm{s}\).

Therefore, the observed range is:

\[ 1.00\leq t\leq1.06\ \mathrm{s} \]

The midpoint of the observed range is:

\[ \frac{1.00+1.06}{2}=1.03\ \mathrm{s} \]

This is not the same as the arithmetic mean of \(1.02\ \mathrm{s}\).

The half-range calculation is still correct. The difference occurs because the half-range is centred on the mean when using the mean ± half-range reporting convention.

The two intervals describe different things:

• The observed range extends from the smallest to the largest actual reading.
• The reported symmetric interval extends equally above and below the mean.

Do not assume that half-range uncertainty always produces an interval that contains every observed measurement.

Question 5

Two independent electronic timing systems, A and B, measure the same trolley transit time under identical experimental conditions.

Each system records five readings.

The illustrative data are:

System A

\[ 0.994,\ 0.997,\ 1.000,\ 1.003,\ 1.006\ \mathrm{s} \]

System B

\[ 1.029,\ 1.030,\ 1.031,\ 1.032,\ 1.033\ \mathrm{s} \]

An independent calibration check establishes that:

• System A has no detectable fixed timing offset.
• System B adds a constant \(0.030\ \mathrm{s}\) to every recorded time.
• The correction for System B is known with negligible uncertainty for this question.

An independent reference gives the actual transit time as exactly \(1.000\ \mathrm{s}\).

Using the half-range convention:

a) Calculate the mean, absolute uncertainty, and percentage uncertainty for both systems before applying any correction.

b) Correct System B’s readings and calculate its new mean and percentage uncertainty.

c) Compare the precision of the two corrected measurement systems.

d) Determine which system’s central estimate is closer to the reference value.

e) Explain why System B can have a smaller percentage uncertainty than System A while still producing an incorrect uncorrected result.

Solution 5

After correcting System B, it has the smaller percentage uncertainty of approximately \(0.20\%\), compared with System A’s \(0.60\%\). However, System A’s mean is closer to the reference value. System B is more precise, but its original readings were affected by a systematic offset.

This question combines three ideas: repeated-measurement uncertainty, systematic error, and accuracy.

Part a: Calculate the results before correction

For System A, calculate the mean:

\[ \begin{aligned} \bar t_A &=\frac{0.994+0.997+1.000+1.003+1.006}{5}\\ &=1.000\ \mathrm{s} \end{aligned} \]

Calculate the half-range:

\[ \begin{aligned} \Delta t_A &=\frac{1.006-0.994}{2}\\ &=0.006\ \mathrm{s} \end{aligned} \]

Therefore, the percentage uncertainty is:

\[ \begin{aligned} \%\text{ uncertainty}_A &=\frac{0.006}{1.000}\times100\\ &=0.60\% \end{aligned} \]

System A reports:

\[ t_A=(1.000\pm0.006)\ \mathrm{s} \]

Now calculate System B’s mean:

\[ \begin{aligned} \bar t_B &=\frac{1.029+1.030+1.031+1.032+1.033}{5}\\ &=1.031\ \mathrm{s} \end{aligned} \]

Its absolute uncertainty is:

\[ \begin{aligned} \Delta t_B &=\frac{1.033-1.029}{2}\\ &=0.002\ \mathrm{s} \end{aligned} \]

Its percentage uncertainty is:

\[ \begin{aligned} \%\text{ uncertainty}_B &=\frac{0.002}{1.031}\times100\\ &=0.19399\%\\ &\approx0.19\% \end{aligned} \]

System B reports:

\[ t_B=(1.031\pm0.002)\ \mathrm{s} \]

Before correction, System B appears to have much better repeatability.

However, its mean is \(0.031\ \mathrm{s}\) above the reference value.

Part b: Correct System B’s measurements

The calibration check identifies a constant positive offset of \(0.030\ \mathrm{s}\).

Subtract this value from every reading:

\[ \begin{aligned} 1.029-0.030&=0.999\\ 1.030-0.030&=1.000\\ 1.031-0.030&=1.001\\ 1.032-0.030&=1.002\\ 1.033-0.030&=1.003 \end{aligned} \]

The corrected measurements are:

\[ 0.999,\ 1.000,\ 1.001,\ 1.002,\ 1.003\ \mathrm{s} \]

Calculate the corrected mean:

\[ \begin{aligned} \bar t_{B,\text{corrected}} &=\frac{0.999+1.000+1.001+1.002+1.003}{5}\\ &=1.001\ \mathrm{s} \end{aligned} \]

The corrected half-range is:

\[ \begin{aligned} \Delta t_B &=\frac{1.003-0.999}{2}\\ &=0.002\ \mathrm{s} \end{aligned} \]

Notice that the absolute uncertainty has not changed.

Subtracting the same constant from every measurement shifts the entire dataset but leaves the range unchanged.

Calculate the corrected percentage uncertainty:

\[ \begin{aligned} \%\text{ uncertainty}_{B,\text{corrected}} &=\frac{0.002}{1.001}\times100\\ &=0.1998\%\\ &\approx0.20\% \end{aligned} \]

The corrected result is:

\[ t_B=(1.001\pm0.002)\ \mathrm{s} \]

Part c: Compare precision

System A has a percentage uncertainty of \(0.60\%\).

Corrected System B has a percentage uncertainty of \(0.20\%\).

Therefore, System B produces the more precise measurements under the half-range convention.

Its repeated readings are more tightly grouped relative to their mean.

Part d: Compare accuracy

The reference value is:

\[ t_{\mathrm{ref}}=1.000\ \mathrm{s} \]

System A’s mean is:

\[ 1.000\ \mathrm{s} \]

Its central estimate has zero difference from the reference.

Corrected System B’s mean is:

\[ 1.001\ \mathrm{s} \]

Its central estimate is \(0.001\ \mathrm{s}\) above the reference.

Therefore, System A’s central estimate is closer to the reference value.

However, both corrected measurement intervals include the reference:

For System A:

\[ [0.994,\ 1.006]\ \mathrm{s} \]

For corrected System B:

\[ [0.999,\ 1.003]\ \mathrm{s} \]

Both results are consistent with the reference within their stated half-range estimates.

Part e: Explain the systematic offset

This is the most important part of the question.

System B originally produced a smaller percentage uncertainty because all its readings were closely grouped.

However, they were all shifted upwards by the same amount.

Repeating a measurement can reveal random variation, but it cannot automatically identify a fixed systematic offset.

The calibration check was necessary to identify that problem.

The final comparison is:

The surprising result is that the more precise system does not necessarily have the central estimate closest to the true value.

A tightly grouped set of measurements can still be systematically biased.

This is why you should never judge an experiment using percentage uncertainty alone.

08What to learn next

Once you can calculate absolute and percentage uncertainty from repeated measurements, the next step is using that uncertainty when analysing experimental graphs.

For example, you can represent measurement uncertainty using error bars, then use maximum and minimum gradient lines to estimate the uncertainty in a graph’s gradient.

Continue with Error Bars and Maximum-Minimum Gradients: HSC Physics Guide to see how individual measurement uncertainties affect the final experimental result.