Absolute, Relative & Percentage Errors

A machinist grinding a shaft to fit a bearing misses by one millimetre and the assembly is scrap. A groundskeeper marking out a football pitch misses by the same millimetre and nobody on Earth could detect it. Identical error — $1\ \mathrm{mm}$ — yet one is a catastrophe and the other invisible. The raw size of an error tells you almost nothing; what matters is the error compared to the thing measured, and the moment you make that comparison you have discovered relative error — the most exam-tested number in all of measurement physics. This module builds the complete error toolkit for repeated measurements: how to distil five disagreeing stopwatch readings into one best value, how to attach an honest $\pm$ to it, and how to convert that $\pm$ into the percentage that lets you compare a coin measurement against a planetary orbit. ICSE practicals ask for the full readings-table ritual; JEE-Main and NEET compress it into two-minute numericals ("which student measured best?"); CIE A-Level runs the same machinery through every practical paper it sets. The ideas are three lines of algebra — but the marks live in the conventions ( $n$ or $n-1$ , magnitude or signed, how many digits), and those are exactly where students bleed.

Key takeaways

With repeated readings $a_1, \ldots, a_n$ , the best estimate of the true value is the arithmetic mean $\bar a$ .
The absolute error of a reading is $\Delta a_i = |a_i - \bar a|$ ; the mean absolute error is their average, $\Delta \bar a = \frac{1}{n}\sum |a_i - \bar a|$ — divide by $n$ , not $n-1$ .
Quote the result as $\bar a \pm \Delta \bar a$ : error rounded to 1 (at most 2) significant figures, mean cut at the same decimal place.
Relative error $= \Delta \bar a / \bar a$ is a pure, unit-free number; ×100 gives percentage error — the only fair way to compare measurements of different-sized quantities.
Timing $n$ oscillations instead of one divides the percentage error in the period by $n$ — the classic exam lever for improving a measurement.
Key idea: absolute error tells you how far off you are; relative error tells you how much that matters — and exams almost always pay for the second one.

The core insight

The mean of repeated readings is the best estimate of a measured quantity, the mean absolute error $\Delta \bar a = \frac{1}{n}\sum|a_i-\bar a|$ is its honest uncertainty, and dividing that error by the value itself — the relative error — converts every measurement in physics onto one universal quality scale.

Where this lives today

Every opinion poll you will read this year quotes a "margin of error of $\pm 3\%$ " — a relative-error statement, chosen precisely because it is the unit-free comparison this module built. Your phone's GPS fixes position to about $5\ \mathrm{m}$ against $2 \times 10^{7}\ \mathrm{m}$ satellite orbits — a relative error of $2.5 \times 10^{-7}$ , achieved by the average-many-readings logic of equation (1) running continuously in the receiver. At the frontier, relative error is the headline: chip fabs hold nanometre features across $300\ \mathrm{mm}$ wafers (parts in $10^{8}$ ), GPS caesium clocks drift by parts in $10^{14}$ , and LIGO detects arm-length changes of one part in $10^{21}$ — the best relative measurement humans have ever made, still conceptually a $\Delta \ell/\ell$ from equation (5). When physicists brag, they brag in relative error.

The physical picture

Picture the two measurements side by side. A pencil, about $15\ \mathrm{cm}$ long, measured to $\pm 1\ \mathrm{mm}$ : the band of doubt is $\tfrac{1}{150}$ of the object — visible on any decent diagram. A $100\ \mathrm{m}$ football pitch measured with the same $\pm 1\ \mathrm{mm}$ : the band is $\tfrac{1}{100{,}000}$ of the length — drawn to scale, thinner than the ink of the boundary line. Same absolute error; utterly different meaning.

Figure

One millimetre of error on a pencil and on a football pitch — identical absolute error, wildly different quality

The second picture is what repeated readings actually look like. Time one swing of a pendulum five times and the stopwatch will not repeat itself: reflexes, air currents, the judgement of when the bob "really" passes the mark — all jitter the digits. The readings scatter, and the scatter is not failure but information: its centre is the best value, its width is how much to trust that value. This module is the recipe for extracting those two numbers and expressing the width in the form that makes measurements comparable.

Figure

Five stopwatch readings scatter about their mean — the band of half-width 0.11 s is the measurement's honest doubt

One vocabulary note before the algebra. The previous module, Accuracy, Precision & Uncertainty, gave the words: random errors scatter readings both ways, systematic errors shift them all one way. This module gives those words numbers — but only for the random part. A perfect error calculation on a miscalibrated stopwatch still misses the truth; the arithmetic below assumes systematic effects have been hunted down separately.

From readings to a result: the mean and its error

Since the true value $a_{\text{true}}$ is forever out of reach, the working rule of experimental physics is that the best available estimate is the arithmetic mean of all honest readings. Random errors push readings above and below the truth with equal likelihood, so in the sum most of the push cancels:

\bar a = \frac{a_1 + a_2 + \cdots + a_n}{n} = \frac{1}{n}\sum_{i=1}^{n} a_i \qquad (1)

where: $a_i$ — the individual readings; $n$ — how many were taken; $\bar a$ — the mean, adopted as the stand-in for the true value in everything that follows. Read it physically: each reading is truth plus a random nudge; averaging nudges of both signs shrinks their net effect — more readings genuinely buy a better value, not just more paperwork. How far does a typical reading sit from this best value? Take each reading's deviation from the mean and strip its sign — by definition the absolute error of that reading:

\Delta a_i = |a_i - \bar a| \qquad (2)

where: $\Delta a_i$ — the absolute error of the $i$ -th reading, always a magnitude, always in the units of $a$ . The sign is deliberately discarded: signed deviations sum to (nearly) zero around a mean by construction — a useful check on your mean, but averaging deviations with signs would absurdly certify every experiment as perfect. The single number that summarises the scatter is the mean absolute error — the plain average of those magnitudes:

\Delta \bar a = \frac{|a_1-\bar a| + |a_2-\bar a| + \cdots + |a_n-\bar a|}{n} = \frac{1}{n}\sum_{i=1}^{n} |a_i - \bar a| \qquad (3)

where: $\Delta \bar a$ — the typical distance of a reading from the mean, in the units of $a$ . Note the denominator: $n$ , the same $n$ as in the mean. (Statistics courses divide a related quantity by $n-1$ ; that is the sample standard deviation, a different estimator for a different purpose — board physics uses equation (3) exactly as written.) The final result is then reported as

a = \bar a \pm \Delta \bar a \qquad (4)

where: the $\pm$ band declares "the true value very likely lies between $\bar a - \Delta \bar a$ and $\bar a + \Delta \bar a$ ." This one line is the deliverable of an experiment: quoting $2.62 \pm 0.11\ \mathrm{s}$ says more physics than $2.624\ \mathrm{s}$ ever could, because it admits exactly how much of the number is trustworthy.

Derivation AThe pendulum readings table — the full chain from five readings to a quoted result

We want to carry one concrete data set through equations (1)–(4) with no steps skipped, because this exact computation — usually with a pendulum — is the most reliably asked long-form error question in ICSE and CIE practical papers, and its compressed version fuels JEE one-liners. The strategy: compute the mean, round it to the resolution of the data, tabulate magnitude-deviations from that mean, average them, and only then round the error and trim the mean to match. The thing to watch is when rounding happens: once, at the reporting stage — rounding the mean too early or the deviations individually drifts the final digits.

Step 1. The data: five timings of one oscillation, $2.63,\ 2.56,\ 2.42,\ 2.71,\ 2.80\ \mathrm{s}$ (stopwatch reading to $0.01\ \mathrm{s}$ ).

Step 2. The mean, equation (1): the sum is $2.63 + 2.56 + 2.42 + 2.71 + 2.80 = 13.12\ \mathrm{s}$ , so $\bar T = 13.12/5 = 2.624\ \mathrm{s}$ . The data are only good to two decimal places, so the working mean is $\bar T = 2.62\ \mathrm{s}$ . Why round here: the readings carry hundredths as the estimated digit, so a thousandths digit in the mean is bookkeeping, not knowledge; the deviations table is conventionally built from the rounded mean, exactly as NCERT-legacy tables and board mark schemes do.

Step 3. Deviations from $2.62\ \mathrm{s}$ , magnitudes only, equation (2): $|2.63-2.62| = 0.01$ , $|2.56-2.62| = 0.06$ , $|2.42-2.62| = 0.20$ , $|2.71-2.62| = 0.09$ , $|2.80-2.62| = 0.18$ (all in seconds).

Step 4. Mean absolute error, equation (3): the deviations sum to $0.01+0.06+0.20+0.09+0.18 = 0.54\ \mathrm{s}$ , so $\Delta \bar T = 0.54/5 = 0.108\ \mathrm{s}$ , which rounds to $0.11\ \mathrm{s}$ — errors are quoted to one, at most two, significant figures.

Step 5. Report, equation (4): $T = (2.62 \pm 0.11)\ \mathrm{s}$ . Both numbers end at the hundredths place — the error's last digit and the value's last digit must line up, which is the whole reporting grammar of the significant-figures module in one sentence.

Step 6. The bonus numbers exams love: relative error $= 0.11/2.62 = 0.042$ , i.e. about $4\%$ . A single hand-timed swing is a 4-percent measurement — remember that anchor. Verified: sum $13.12$ and mean $2.624$ recomputed by column addition; deviation set recomputed from $2.62$ ; $0.54/5 = 0.108$ ; using the unrounded mean $2.624$ instead gives deviations summing to $0.536$ and $\Delta\bar T = 0.107 \to 0.11\ \mathrm{s}$ — the same quoted result, confirming the convention is harmless.

Relative and percentage error: the comparison currency

The number $\Delta \bar a$ answers "how wide is the doubt?" — in metres, seconds, kilograms. It cannot answer "is this a good measurement?": the machinist's millimetre and the groundskeeper's millimetre differ only in what they are measured against. So divide:

\text{relative error} = \frac{\Delta \bar a}{\bar a} \qquad (5)

where: $\Delta \bar a$ — the mean absolute error from equation (3); $\bar a$ — the mean value itself. The units cancel, leaving a pure number. That cancellation is the entire point: a pure number can be compared across lengths, masses, times — across anything. A relative error of $0.005$ means "half a per cent of doubt" whether the object is a virus or a galaxy. Multiply by 100 for the exam's favourite form:

\text{percentage error} = \frac{\Delta \bar a}{\bar a} \times 100\% \qquad (6)

where: the same ratio now reads as a percentage — the quality grade of a measurement. Hand-timed pendulum: $\sim 4\%$ . School vernier work: $\sim 0.5\%$ . GPS atomic clocks: $\sim 10^{-14}\%$ . One scale ranks them all. One variant matters for marks: when a question supplies an accepted value (a handbook $g$ , a standard resistance), the comparison is against it, not your mean — measure $g = 9.72\ \mathrm{m/s^2}$ against the accepted $9.81$ and the percentage error is $0.09/9.81 \times 100\% \approx 0.9\%$ . Same formula shape, different reference; read the question to see which reference it hands you.

you want to…	use	why
quote a result from repeated readings	$\bar a \pm \Delta \bar a$	value + honest width, same units
judge or compare measurement quality	relative / percentage error	unit-free, size-free quality grade
compare against a handbook value	$\frac{\|\text{measured} - \text{accepted}\|}{\text{accepted}} \times 100\%$	the reference is the accepted value
feed errors into a formula ( $\rho = m/V$ etc.)	relative errors (next module)	products and quotients add relative errors

Go deeper · Derivation B · 1 minWhy the ratio is the fair judge — scale invariance of relative error

Derivation BWhy the ratio is the fair judge — scale invariance of relative error

We want to prove the claim smuggled into every comparison above: that relative error genuinely measures the technique, independent of the size of the object or the units chosen — so it, and not absolute error, is the legitimate way to rank measurements. The strategy is a two-line thought experiment: transform a measurement by an overall scale factor and watch what survives. Watch how absolute error changes while the ratio does not.

Step 1. Suppose a measurement yields $\bar a \pm \Delta \bar a$ . Now express the same physical result in different units, or measure an object $k$ times larger with the same fractional technique: every reading, hence the mean and every deviation, is multiplied by the same factor $k$ . The quoted result becomes $k\bar a \pm k\Delta \bar a$ .

Step 2. The absolute error has changed: $\Delta \bar a \to k \Delta \bar a$ . Convert centimetres to metres and it shrinks a hundredfold with no experiment performed — absolute error is partly an accident of units and object size.

Step 3. The relative error is untouched: $\dfrac{k \Delta \bar a}{k \bar a} = \dfrac{\Delta \bar a}{\bar a}$ . Nothing about the quality of the work changed, and nothing about the ratio changed either. The ratio tracks technique; the absolute error tracks technique times scale.

Step 4. Concrete check: pencil $15.0 \pm 0.1\ \mathrm{cm}$ gives $0.1/15.0 \approx 0.7\%$ ; pitch $100.000 \pm 0.001\ \mathrm{m}$ gives $10^{-5} = 0.001\%$ — seven hundred times better as a measurement, exactly as the hero diagram claimed. Verified: $0.1/15.0 = 0.00667$ and $0.001/100.000 = 1.0\times10^{-5}$ recomputed; ratio invariance checked symbolically by cancelling $k$ .

Limiting cases

Case	Result	Meaning
single reading, $n = 1$	$\Delta \bar a$ from (3) is uselessly zero — quote the instrument's least count instead	scatter needs repetition to reveal itself; one reading's doubt is the instrument's resolution
all readings identical	$\Delta \bar a = 0$ , but the doubt is not zero — floor it at the least count	agreement to the last digit means the instrument, not the experiment, is now the limit
one wild outlier	it dominates $\Delta \bar a$ (the $0.20\ \mathrm{s}$ reading contributes over a third of the $0.54$ sum)	check the reading; boards expect you to keep honest data, professionals investigate
$\bar a \to 0$ (e.g. a null result)	relative error blows up	percentage error is meaningless for quantities passing through zero — quote absolute error there
$n$ very large	$\bar a$ steadies; $\Delta \bar a$ approaches the intrinsic scatter of the method	averaging kills random error slowly; it never touches systematic error

The last row is the deep one: no amount of repetition rescues a stopwatch that runs fast — averaging cures randomness only.

Worked examples

Board level — the readings table, start to finish (ICSE and CIE practical pattern)

A student measures the length of a metal rod five times: $5.42,\ 5.46,\ 5.44,\ 5.40,\ 5.48\ \mathrm{cm}$ . Find the mean length, the mean absolute error, and the percentage error, and state the result in the standard form.

Step 1. Mean, equation (1): the readings sum to $5.42+5.46+5.44+5.40+5.48 = 27.20\ \mathrm{cm}$ , so $\bar\ell = 27.20/5 = 5.44\ \mathrm{cm}$ .

Step 2. Magnitude deviations from $5.44$ : $0.02,\ 0.02,\ 0.00,\ 0.04,\ 0.04\ \mathrm{cm}$ . (Quick sanity check: the signed deviations $-0.02, +0.02, 0.00, -0.04, +0.04$ sum to zero — the mean is right.)

Step 3. Mean absolute error, equation (3): the magnitudes sum to $0.12\ \mathrm{cm}$ , so $\Delta\bar\ell = 0.12/5 = 0.024\ \mathrm{cm}$ , quoted as $0.02\ \mathrm{cm}$ (one significant figure).

Step 4. Percentage error, equation (6), computed with the unrounded guard value: $0.024/5.44 \times 100\% = 0.44\% \approx \boxed{0.4\%}$ .

Step 5. Standard form: $\boxed{\ell = (5.44 \pm 0.02)\ \mathrm{cm}}$ — error to one significant figure, mean trimmed to the same decimal place. The sentence "the true length very likely lies between $5.42$ and $5.46\ \mathrm{cm}$ " is the understanding the examiner is paying for.

Competitive level — which student measured the period best? (JEE-Main pattern)

Three students measure the period of the same pendulum. P times 20 oscillations with an analogue stopwatch of least count $0.5\ \mathrm{s}$ and reads $25.0\ \mathrm{s}$ . Q times 40 oscillations with the same stopwatch and reads $50.0\ \mathrm{s}$ . R times 20 oscillations with a digital watch of least count $0.1\ \mathrm{s}$ and reads $25.0\ \mathrm{s}$ . Rank their percentage errors in the period.

Step 1. The trick that unlocks the whole family: the timing error lives in the total time $t$ , but the period is $T = t/n$ with $n$ an exact count. Dividing by an exact number divides the absolute error too: $\Delta T = \Delta t / n$ . Equivalently — and faster — the relative error of $T$ equals the relative error of $t$ , so compute $\Delta t / t$ directly.

Step 2. Student P: $\Delta t/t = 0.5/25.0 = 0.02 \Rightarrow 2\%$ . (The long way agrees: $T = 1.25\ \mathrm{s}$ , $\Delta T = 0.5/20 = 0.025\ \mathrm{s}$ , $0.025/1.25 = 2\%$ .)

Step 3. Student Q: $\Delta t/t = 0.5/50.0 = 0.01 \Rightarrow 1\%$ . Doubling the oscillation count doubled $t$ while $\Delta t$ stayed fixed — the error halved for free. This is why every practical manual says "time at least 20 oscillations".

Step 4. Student R: $\Delta t/t = 0.1/25.0 = 0.004 \Rightarrow 0.4\%$ . A five-times-finer watch beats a doubled count.

Step 5. Ranking: $\boxed{R\ (0.4\%) < Q\ (1\%) < P\ (2\%)}$ — R measured best. Memorise the general form: % error in $T$ = $\dfrac{\text{least count}}{n \times T} \times 100\%$ — shrink it with a finer watch or more oscillations. Timing note: 90 seconds as an MCQ. The only decision is spotting that $n$ is exact; after that it is three divisions. If you find yourself computing each $T$ and $\Delta T$ separately, you are solving it the 3-minute way.

Exam traps

Dividing by $n-1$ . Statistics teaches the sample standard deviation with $n-1$ ; board physics defines mean absolute error with $n$ , equation (3), full stop. Mixing them turns $0.108$ into $0.135$ and the final digit is wrong.

Quoting the raw error. $\Delta \bar T = 0.108\ \mathrm{s}$ is a working number; the quoted error is $0.11\ \mathrm{s}$ — one or two significant figures, never more — with the mean trimmed to the same decimal place. $2.624 \pm 0.108$ loses the reporting mark even with every digit arithmetically right (the significant-figures module is the grammar book here).

Signed deviations. Averaging $a_i - \bar a$ with signs gives zero every time — that is a property of the mean, not a discovery about your data. Signs are for checking the mean; magnitudes, equation (2), are for the error.

Absolute error used to rank quality. The warning above: convert to relative error before declaring any measurement "better". The bait option in MCQs is always the smaller $\Delta$ .

Mean-based vs accepted-value percentage. "Percentage error of the readings" → divide by your mean. "…given the true value is" → divide by the accepted value. Identify the reference before dividing; ICSE mark schemes list them as different steps.

Forgetting that counts are exact. In $T = t/n$ , the $n$ oscillations carry no error — attach an uncertainty to $n$ and you double the percentage error and miss the answer grid. Counted quantities and defined constants never enter the error budget.

Units on the wrong quantity. Absolute error carries units ( $\pm 0.11\ \mathrm{s}$ ); relative and percentage errors are pure numbers — " $4\%\ \mathrm{s}$ " flags an unsettled concept to the examiner.

Origins · 1 minPierre-Simon Laplace and how this was discovered

Pierre-Simon Laplace

1749–1827 · Paris

Laplace spent a lifetime asking what a pile of disagreeing observations is actually worth. As the leading celestial mechanician of his age he inherited astronomy's embarrassment of riches — centuries of planet and Moon positions, no two of which quite agreed — and he treated the disagreement itself as a mathematical object. Alongside Legendre's least squares and Gauss's error law, Laplace supplied the keystone: his 1810 central limit theorem showed that the accumulation of many small independent errors drives results toward the same bell-shaped clustering around the mean, whatever the individual causes look like. That is the licence behind equation (1) — the reason the mean of scattered readings deserves the title "best estimate" — and behind every $\pm$ quoted in every lab report since. He also weaponised the machinery: his error-weighted analyses of lunar and planetary data pinned down celestial constants to precisions their raw observations never individually possessed.

History

1805

— Legendre publishes the method of least squares for reconciling conflicting astronomical observations.

1809

— Gauss's Theoria Motus derives the normal law of errors and defends the arithmetic mean as the best value.

1810

— Laplace proves the central limit theorem: aggregated random errors cluster universally around the mean.

1898

— Student-era laboratory manuals fix the readings-table ritual (mean, deviations, mean deviation) as standard practical training.

1993

— The ISO Guide to the Expression of Uncertainty in Measurement (GUM) standardises error language for every accredited lab on Earth.

Going further

Texts. ICSE/ISC Class 11 Physics, measurement chapter · Halliday, Resnick & Walker Ch. 1 · CIE 9702 syllabus, Practical Assessment appendix (error-quoting rules) · J. R. Taylor, An Introduction to Error Analysis, Ch. 4 (where $n-1$ properly begins) · legacy NCERT Part 1 error sections (rationalised out of CBSE 2025-26, still the cleanest worked tables in print).

Provenance. All conventions per ICSE/CIE practical guidance and legacy-NCERT error analysis. A verifier should recompute: pendulum chain $13.12/5 = 2.624 \to 2.62$ ; deviations $0.01, 0.06, 0.20, 0.09, 0.18$ summing to $0.54$ ; $0.54/5 = 0.108 \to 0.11$ ; $0.11/2.62 = 0.042 \to 4\%$ ; rod chain $27.20/5 = 5.44$ ; deviations $0.02, 0.02, 0.00, 0.04, 0.04$ summing to $0.12$ ; $0.12/5 = 0.024 \to 0.02$ ; $0.024/5.44 = 0.44\% \to 0.4\%$ ; stopwatch ranking $0.5/25 = 2\%$ , $0.5/50 = 1\%$ , $0.1/25 = 0.4\%$ ; pencil/pitch $0.1/15.0 = 0.67\%$ vs $0.001/100 = 0.001\%$ ; $g$ example $0.09/9.81 = 0.9\%$ .

Prerequisites. Accuracy, Precision & Uncertainty. Leads to. Propagating Errors: Sums & Differences · Significant Figures, Rounding & Reporting (the reporting grammar) · Error-Question Patterns (the technique drill).

If the true value is unknowable, how can we even define "error"?

By promotion: the mean of many readings is adopted as the working stand-in for the true value, and all errors are measured against it. This is not circular — Laplace's theorem guarantees random scatter clusters symmetrically around the truth, so the centre of the cluster is the best estimate a finite data set can offer. Strictly you compute the uncertainty of your estimate rather than the distance to the hidden truth; school usage says "error" for both. What the convention cannot fix is a systematic shift: a stopwatch running 2% fast makes the mean 2% wrong, and no table of deviations will ever confess it.

Why divide by n and not n−1 like in statistics class?

Two different tools. The physics-lab mean absolute error, equation (3), is the plain average distance of readings from the mean — divide by $n$ , the number of distances you averaged. The statistician's sample standard deviation squares deviations and divides by $n-1$ to correct a subtle bias when estimating a population's spread from a sample. Board physics (ICSE, JEE, NEET, CIE) defines and marks the $n$ version; writing $n-1$ changes your final digits and costs the answer. If a paper wants standard deviation it will say those words — absent them, equation (3) rules.

Can a percentage error be more than 100%?

Yes, and it means your doubt exceeds your value — the measurement barely knows the quantity's size at all. It happens legitimately when the value is tiny compared to the instrument's resolution: timing a single 0.3 s event with a hand stopwatch (reaction time ~0.2 s) is a ~70–100% measurement. It is also why relative error breaks down near zero: as $\bar a \to 0$ the denominator collapses and the percentage explodes even for tiny absolute doubt. For null-type results, professionals quote the absolute band instead — "consistent with zero, $\pm 0.003$ ".

My five readings are all identical. Is my error really zero?

No — your scatter is zero, which only says the instrument cannot see the variation. The doubt never drops below the least count of the instrument: five identical $2.5\ \mathrm{s}$ stopwatch readings mean the truth lies somewhere in the $0.1\ \mathrm{s}$ -wide bin the display rounds into, so quote $\pm 0.1\ \mathrm{s}$ (some boards accept half the least count — follow your board's practical manual). The general rule examiners reward: the quoted error is the larger of the calculated mean absolute error and the instrument's least count. Identical readings are a signal to switch to a finer instrument, not to celebrate perfection.

To how many digits do I write the error and the mean?

Error first: round $\Delta \bar a$ to one significant figure (two are acceptable when the leading digit is 1, e.g. $\pm 0.14$ ). Then trim the mean so its last digit sits at the same decimal place as the error's: $2.624$ with error $0.108$ becomes $2.62 \pm 0.11$ . Never let the value dangle digits beyond the error — a digit the error band has already swallowed is fiction. This is the same one-estimated-digit grammar as the significant-figures module; the error band simply makes the estimated digit's location explicit.

Is "percentage error" the same as the "percentage difference" between two measured values?

No, and papers distinguish them. Percentage error compares a measurement to a reference — your mean, or an accepted value if one is supplied. Percentage difference compares two equally uncertain measurements to each other (their gap divided by their average), used when neither is privileged, e.g. two students' results. Call one by the other's name in a CIE practical and the mark is gone — check which reference the question hands you before dividing.