When Should an Error Bar Include Zero? HSC Physics Guide

An error bar does not have to include the origin. It should reach zero only when zero lies inside the measurement interval represented by that error bar. On a graph, a vertical error bar that crosses the horizontal axis reaches \(y=0\), but it does not necessarily include the graph origin \((0,0)\).

Consider two measurements of the same extension:

• \(0.30\pm0.40\text{ cm}\) gives an interval from \(-0.10\) cm to \(0.70\) cm. Zero is included.
• \(0.30\pm0.05\text{ cm}\) gives an interval from \(0.25\) cm to \(0.35\) cm. Zero is not included.

The central measurement is identical. The uncertainty determines whether zero is included.

The important rule is simple: draw the error bar according to the stated measurement uncertainty, not according to where you expect the graph to go. If the interval does not reach zero, don’t extend it just to make it touch the origin.

01Start with the measurement interval

An error bar represents the uncertainty associated with a measured value. It shows the range of values allowed by the uncertainty convention being used.

For a measurement \(q\) with a symmetrical absolute uncertainty \(\Delta q\), the interval is:

\[ [q-\Delta q,\ q+\Delta q] \]

Here:

• \(q\) is the measured value.
• \(\Delta q\) is its absolute uncertainty.
• Both quantities must use the same units.

The interval includes zero when:

\[ q-\Delta q\leq 0 \]

and

\[ q+\Delta q\geq 0 \]

For an ordinary positive measurement, the second condition is automatically satisfied. Therefore, the practical test becomes:

\[ \boxed{\Delta q\geq q} \]

In other words, the absolute uncertainty must be at least as large as the measured value.

If the uncertainty is smaller than the measurement, the lower end of the error bar stays above zero.

What if the uncertainty is asymmetric?

Sometimes the uncertainty above and below a measurement is different.

For example, a measurement might be reported as:

\[ q=2.0^{+0.5}_{-1.0} \]

Its interval is:

\[ [1.0,\ 2.5] \]

The lower uncertainty is larger than the upper uncertainty, but zero is still excluded.

The same principle applies: calculate the lower and upper limits, then check whether zero falls between them.

02The graph origin is not the same as the horizontal axis

This is where a lot of confusion starts.

Suppose you plot extension against applied force:

• Horizontal axis: applied force, \(F\), in newtons.
• Vertical axis: extension, \(x\), in centimetres.

A point on this graph might have the coordinates:

\[ (F,x)=(0.50,\ 0.12) \]

Its vertical uncertainty might be \(\pm0.20\text{ cm}\).

The vertical interval is:

\[ 0.12-0.20\leq x\leq0.12+0.20 \]

Therefore:

\[ -0.08\leq x\leq0.32 \]

The vertical error bar crosses the horizontal axis because it includes \(x=0\).

However, the point is still located at \(F=0.50\text{ N}\). It is not at the graph origin.

The graph origin is:

\[ (F,x)=(0,0) \]

To include that point within a two-dimensional uncertainty region, both the horizontal and vertical measurement intervals must include zero.

This distinction is particularly important when interpreting experimental graphs. Saying an error bar reaches zero is not always the same as saying it includes the origin.

03What determines the size of an error bar?

Before deciding whether an error bar includes zero, you need to know what uncertainty it represents.

There is no single universal convention for every experiment. The method used to calculate uncertainty must be clear.

Instrument uncertainty

An instrument has limitations in how precisely it can measure a quantity.

For example, consider a ruler with divisions every \(1\text{ mm}\).

Under the common half-smallest-division convention, its reading uncertainty is:

\[ \Delta x=\pm0.5\text{ mm} \]

A measurement of \(2.0\text{ mm}\) would therefore have an interval of:

\[ [1.5,\ 2.5]\text{ mm} \]

Zero is not included.

Be careful with digital instruments. A display changing in increments of \(0.01\text{ s}\) tells you its resolution, but does not automatically establish its total measurement uncertainty.

Always follow the uncertainty convention specified for the experiment.

Uncertainty from repeated measurements

Suppose you measure the same quantity several times:

\[ 2.8,\ 3.0,\ 3.1,\ 3.1,\ 3.2\text{ cm} \]

One common school-laboratory method estimates uncertainty using half the range:

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

Therefore:

\[ \Delta x=\frac{3.2-2.8}{2} =0.2\text{ cm} \]

The mean is \(3.04\text{ cm}\).

If the experiment specifies mean ± half-range, the reported result is:

\[ x=3.04\pm0.20\text{ cm} \]

Its interval is:

\[ [2.84,\ 3.24]\text{ cm} \]

Zero is clearly excluded.

This is an estimate based on the spread of the repeated readings, not a universal definition of measurement uncertainty.

Uncertainty is not the same as standard deviation

These terms are related, but they describe different things.

Standard deviation is not automatically the uncertainty of the mean.

For example, if a graph requires error bars representing half the range, you should not replace them with standard deviation simply because you have calculated it.

Likewise, if the instructions specify ±1 standard deviation, use that convention and label it correctly.

Why the distinction between error and uncertainty matters

Imagine an unloaded force sensor gives three readings:

\[ 0.06,\ 0.08,\ 0.07\text{ N} \]

These readings are tightly grouped, so their random variation is relatively small.

However, the sensor is supposed to read zero when no force is applied.

The readings are consistently positive. This suggests a systematic zero offset.

A small uncertainty does not guarantee that a measurement is accurate. An instrument can produce very consistent readings while being systematically wrong.

This is exactly why you should never extend an error bar to zero just because the experiment theoretically predicts zero.

04A zero check in a force-sensor experiment

Consider an illustrative experiment designed to investigate the calibration of a force sensor.

The unanswered question is straightforward:

Does the sensor actually read zero when no force is applied?

The apparatus and prediction

Mount a force sensor vertically on a clamp stand. Connect its output to a data logger.

Attach a mass hanger to the hook underneath the sensor. You can then add known masses and compare the sensor readings with the applied gravitational force.

The applied force is calculated using:

\[ F=mg \]

where \(m\) is the attached mass in kilograms and \(g\) is gravitational field strength, approximately \(9.8\text{ m s}^{-2}\).

For example, a \(0.200\text{ kg}\) mass produces:

\[ F=0.200\times9.8 \]

\[ F=1.96\text{ N} \]

Before adding any masses, however, the sensor should be checked with no applied load.

The prediction is:

\[ F=0\implies \text{sensor reading}=0 \]

What happens if the error bar excludes zero?

Suppose the unloaded sensor produces these illustrative readings:

The mean reading is \(0.07\text{ N}\).

Using the half-range convention, the uncertainty is \(0.01\text{ N}\).

The resulting interval is:

\[ [0.06,\ 0.08]\text{ N} \]

Zero is not included.

This is an important observation. The sensor is registering a positive force even though no external force has been applied.

The error bar should remain between \(0.06\text{ N}\) and \(0.08\text{ N}\). Extending it down to zero would misrepresent the measurements.

What was wrong with the original setup?

The sensor may have had a zero offset because it was not tared before the experiment.

The original design could not distinguish a genuine applied force from an existing sensor offset.

The experiment therefore needs a revision.

Revising the experiment

With the hanger removed, tare the force sensor. Then take several new unloaded readings to check whether the reading remains close to zero.

Suppose the revised readings are:

\[ -0.01,\ 0.00,\ +0.01\text{ N} \]

Their mean is zero, and their half-range is \(0.01\text{ N}\).

The new interval is:

\[ [-0.01,\ +0.01]\text{ N} \]

Zero is now included.

This is consistent with the sensor reading zero when unloaded.

However, taring only addresses the zero offset. It does not establish that the sensor measures larger forces accurately. You still need to test it using known masses.

The experimental conclusion is that an error bar excluding zero can reveal a problem with the measurement system. It should prompt investigation, not an adjustment to the graph.

05Worked examples

06What if the theoretical graph should pass through the origin?

This is a separate question from whether an error bar includes zero.

Consider Hooke’s law:

\[ F=kx \]

Here:

• \(F\) is the applied force in newtons.
• \(k\) is the spring constant in newtons per metre.
• \(x\) is the extension in metres.

Rearranging:

\[ x=\frac{F}{k} \]

If you plot extension against force, the theoretical relationship is a straight line through the origin.

When no force is applied, there should be no extension, provided extension is measured relative to the spring’s natural, unloaded length.

However, this is a prediction from the physical model. It does not mean every experimental error bar must touch zero.

Consider two situations.

Situation A: The zero-load error bar excludes zero.

Your measurements suggest that the unloaded spring has a nonzero extension.

Possible explanations include:

• The ruler’s zero was incorrectly positioned.
• The spring was already under tension.
• The sensor has a zero offset.
• The measurement uncertainty has been underestimated.

You should investigate these possibilities before forcing a best-fit line through the origin.

Situation B: The zero-load error bar includes zero.

Your measurements are compatible with zero extension when no force is applied.

This supports the theoretical prediction, but it does not prove that the intercept must be exactly zero.

You still need to examine the other data points and determine whether the overall relationship supports the model.

The key distinction is:

• Error bars describe measurement uncertainty.
• A best-fit line represents a model of the relationship between variables.

They are related, but they answer different questions.

For more on using uncertainty intervals to determine possible gradients, see Error Bars and Maximum-Minimum Gradients: HSC Physics Guide.

07Questions and solutions

The following questions move from straightforward interval calculations to more demanding interpretations of experimental graphs.

Question 1

A displacement sensor records a displacement of:

\[ x=0.24\pm0.05\text{ m} \]

Calculate the upper and lower limits of the measurement interval. Does the error bar include zero?

Solution 1

No, the error bar does not include zero. The entire measurement interval is positive.

Step 1: Calculate the lower limit.

\[ x_{\min}=0.24-0.05 \]

\[ x_{\min}=0.19\text{ m} \]

Step 2: Calculate the upper limit.

\[ x_{\max}=0.24+0.05 \]

\[ x_{\max}=0.29\text{ m} \]

Step 3: Write the interval.

\[ [0.19,\ 0.29]\text{ m} \]

Since the lower limit is greater than zero, the error bar does not reach zero.

The correct interval must be drawn entirely above zero.

Question 2

A student plots extension against applied force. One experimental point has the following measurements:

\[ F=0.60\pm0.04\text{ N} \]

\[ x=0.03\pm0.07\text{ cm} \]

Determine:

a) The horizontal uncertainty interval.

b) The vertical uncertainty interval.

c) Whether the vertical error bar crosses the horizontal axis.

d) Whether the full rectangular uncertainty region contains the graph origin.

Solution 2

The vertical error bar crosses the horizontal axis, but the full uncertainty region does not contain the graph origin.

Step 1: Calculate the horizontal interval.

\[ F_{\min}=0.60-0.04=0.56\text{ N} \]

\[ F_{\max}=0.60+0.04=0.64\text{ N} \]

Therefore:

\[ F\in[0.56,\ 0.64]\text{ N} \]

Step 2: Calculate the vertical interval.

\[ x_{\min}=0.03-0.07=-0.04\text{ cm} \]

\[ x_{\max}=0.03+0.07=0.10\text{ cm} \]

Therefore:

\[ x\in[-0.04,\ 0.10]\text{ cm} \]

Step 3: Check the vertical error bar.

The vertical interval contains zero because its lower limit is negative and its upper limit is positive.

Therefore, the vertical error bar crosses the horizontal axis.

Step 4: Check the graph origin.

The horizontal interval is entirely positive:

\[ 0.56\leq F\leq0.64 \]

Therefore, the uncertainty rectangle cannot contain the graph origin.

The distinction is that zero extension is possible within the measurement uncertainty, but zero applied force is not.

Question 3

A force sensor is tested with no applied force. Five readings are obtained:

\[ 0.02,\ 0.03,\ 0.03,\ 0.04,\ 0.05\text{ N} \]

The experimental instructions specify that uncertainty should be calculated using half the range.

a) Calculate the mean reading.

b) Calculate the uncertainty.

c) Determine whether the resulting error bar includes zero.

d) A student argues that standard deviation should always be used instead of half the range because it is a more accurate measure of uncertainty. Explain why this argument is incorrect.

Solution 3

The specified half-range uncertainty interval does not include zero. Standard deviation is not automatically interchangeable with measurement uncertainty.

Step 1: Calculate the mean.

\[ \bar F=\frac{0.02+0.03+0.03+0.04+0.05}{5} \]

\[ \bar F=0.034\text{ N} \]

Step 2: Calculate half the range.

\[ \Delta F=\frac{F_{\max}-F_{\min}}{2} \]

\[ \Delta F=\frac{0.05-0.02}{2} \]

\[ \Delta F=0.015\text{ N} \]

Step 3: Calculate the measurement interval.

\[ F_{\min}=0.034-0.015 \]

\[ F_{\min}=0.019\text{ N} \]

The upper limit is:

\[ F_{\max}=0.034+0.015 \]

\[ F_{\max}=0.049\text{ N} \]

Therefore:

\[ F=0.034\pm0.015\text{ N} \]

The interval is:

\[ [0.019,\ 0.049]\text{ N} \]

Zero is excluded.

Step 4: Consider the standard deviation argument.

The sample standard deviation of these readings is approximately:

\[ s=0.0114\text{ N} \]

Standard deviation describes how spread out the individual readings are around their mean.

It does not automatically describe the uncertainty in the mean, nor does it account for systematic errors such as a sensor zero offset.

The experiment explicitly specifies half the range as its uncertainty convention. Therefore, the correct error bar is \(\pm0.015\text{ N}\), not an alternative value chosen by the student.

The consistently positive readings also suggest that the unloaded sensor may have a zero offset.

Question 4

A student investigates the extension of a spring as the applied force increases.

The following illustrative data are recorded. Assume the uncertainty in the applied force is negligible compared with the vertical uncertainty.

The student makes the following argument:

“The error bar at zero force includes zero extension. Therefore, the best-fit line must pass through the origin.”

Evaluate this argument.

In your response:

a) Calculate the extension interval at zero applied force.

b) Determine the straight-line relationship that passes through the central values.

c) Explain whether the uncertainty bars require the best-fit line to pass through the origin.

Solution 4

The student’s argument is incorrect. The zero-force error bar includes zero, but that does not force the best-fit line to have a zero intercept.

Step 1: Calculate the zero-force interval.

At zero applied force:

\[ x=0.10\pm0.15\text{ cm} \]

The lower limit is:

\[ x_{\min}=0.10-0.15=-0.05\text{ cm} \]

The upper limit is:

\[ x_{\max}=0.10+0.15=0.25\text{ cm} \]

Therefore:

\[ x\in[-0.05,\ 0.25]\text{ cm} \]

Zero is included.

The zero-force measurement is therefore consistent with zero extension within the stated uncertainty.

Step 2: Find the relationship between the central values.

The extension increases by \(1.10\text{ cm}\) for every \(1\text{ N}\) increase in force.

The gradient is:

\[ m=\frac{3.40-0.10}{3-0} \]

\[ m=1.10\text{ cm N}^{-1} \]

The central values therefore follow:

\[ x=1.10F+0.10 \]

The intercept is \(0.10\text{ cm}\), not zero.

Step 3: Consider what the uncertainty allows.

A line with a zero intercept can also be consistent with the measurements.

For example:

\[ x=1.14F \]

At forces of 1 N, 2 N, and 3 N, this line predicts extensions of:

\[ 1.14,\ 2.28,\ 3.42\text{ cm} \]

Each prediction falls within the corresponding measured uncertainty interval.

At zero force, the predicted extension is zero, which also falls within the measured interval.

Therefore, both a nonzero-intercept relationship and a zero-intercept relationship are compatible with the experimental uncertainties.

Step 4: Draw the correct conclusion.

The error bars do not require the line to pass through the origin.

A zero-intercept model may be physically justified if Hooke’s law applies and the extension reference has been correctly established. However, compatibility with zero is not proof that the intercept must be zero.

The complete dataset, experimental calibration, and theoretical model must all be considered.

Question 5

A student investigates a spring using a ruler mounted beside it.

The ruler’s printed zero is not aligned with the spring’s unloaded position.

The following illustrative measurements are obtained:

Assume:

• The ruler readings are independent measurements, each with an uncertainty of \(\pm0.10\text{ cm}\).
• The applied forces are treated as exact.
• When subtracting independent measurements, use the conservative maximum-uncertainty convention: add their absolute uncertainties.
• Extension is defined relative to the spring’s unloaded position.

The student makes two claims.

Claim 1: None of the ruler-position error bars includes the ruler’s printed zero, so a graph of extension against force cannot pass through the graph origin.

Claim 2: If the unloaded ruler position is subtracted from every measurement, the resulting graph passes through the origin. This proves that the spring obeys Hooke’s law.

Evaluate both claims. Calculate the corrected extensions and their uncertainties, and explain what the resulting graph actually establishes.

Solution 5

Both claims are incorrect. The ruler’s printed zero is not the same as the graph origin, and subtracting a reference position establishes a coordinate origin without independently proving Hooke’s law.

Step 1: Examine the original ruler measurements.

The unloaded ruler reading is:

\[ r_0=2.40\pm0.10\text{ cm} \]

Its interval is:

\[ [2.30,\ 2.50]\text{ cm} \]

The other ruler readings also have positive intervals.

None of these measurements includes the ruler’s printed zero.

However, this does not mean the graph of extension against force cannot pass through its origin.

The ruler measures an absolute position relative to its own printed scale. The experiment needs the change in position relative to the unloaded spring.

These are different quantities.

Step 2: Calculate the extension at each applied force.

Extension is defined as:

\[ x=r-r_0 \]

For an applied force of 1 N:

\[ x=3.40-2.40 \]

\[ x=1.00\text{ cm} \]

For 2 N:

\[ x=4.40-2.40 \]

\[ x=2.00\text{ cm} \]

For 3 N:

\[ x=5.40-2.40 \]

\[ x=3.00\text{ cm} \]

Step 3: Calculate the uncertainty in each extension.

The extension is found by subtracting two independent ruler readings.

Under the specified maximum-uncertainty convention, their absolute uncertainties are added:

\[ \Delta x=\Delta r+\Delta r_0 \]

\[ \Delta x=0.10+0.10 \]

\[ \Delta x=0.20\text{ cm} \]

The corrected measurements are therefore:

The intervals for the three loaded measurements are:

\[ [0.80,\ 1.20]\text{ cm} \]

\[ [1.80,\ 2.20]\text{ cm} \]

\[ [2.80,\ 3.20]\text{ cm} \]

None of these loaded points has an uncertainty interval that includes zero.

That is entirely reasonable. Each represents a spring under a nonzero applied force.

Step 4: Examine what happens at zero force.

At zero force, the extension is calculated using the same unloaded position as both the initial and final reference:

\[ x_0=r_0-r_0 \]

Therefore:

\[ x_0=0 \]

This zero is established by the definition of extension.

It is not an independent experimental measurement demonstrating that the spring’s extension is zero.

The uncertainty associated with the reference position also affects every corrected extension. Because the same reference reading is reused, its uncertainty shifts the corrected measurements together rather than creating independent random variations in each point.

Step 5: Evaluate whether Hooke’s law has been demonstrated.

The corrected central values follow:

\[ x=1.00F \]

where extension is measured in centimetres and force in newtons.

The three nonzero measurements show a linear relationship over the tested range.

This provides evidence consistent with Hooke’s law.

However, the origin itself was established by the choice of reference position. It cannot independently prove the model.

You would still need to consider whether the spring behaves linearly over the tested range, whether the ruler is correctly calibrated, and whether the results are repeatable.

The important lesson is that changing the coordinate reference can change the graph’s intercept without changing the physical behaviour of the spring.

A ruler’s zero does not have to coincide with a graph’s zero.

08The decision rule to remember

When deciding whether an error bar should include zero, work through these checks:

1. Identify the quantity represented by the error bar.
2. Identify the uncertainty convention used.
3. Calculate the lower and upper measurement limits.
4. Check whether zero lies between those limits.
5. If working with a graph, distinguish crossing an axis from containing the coordinate origin.
6. If the theoretical model predicts zero but the error bar excludes it, investigate possible systematic errors rather than extending the bar.

An error bar is a representation of measurement uncertainty, not a prediction of where the graph should go.

Once your error bars are correctly established, the next useful step is using them to determine maximum and minimum gradients and assess whether an experimental relationship is consistent with its theoretical model.