← Common Core coursesBMS 1110 · Applied Mathematics and Medical Biostatistics
BMS 1110 · Common Core (Tronc commun)Section 5.4 · Centre, spread and rates
Section 25 of 30

Standard error and coefficient of variation

Explain the difference between standard deviation and standard error, calculate the standard error of the mean as s ÷ √n, and use the coefficient of variation to compare variability across measurements in different units.

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Why this matters

Two wards can report the same average blood pressure and mean completely different things. The standard deviation tells you how far apart the patients are; the standard error tells you how precisely the ward average pins down the true average. Read one where the other was intended and a shaky result looks solid, or a solid one looks shaky. The coefficient of variation finishes the job: it turns spread into a percentage, so a laboratory result in mg/dL can be compared with one in mmol/L, or a small measurement with a large one, without being fooled by the unit.

What you need to know first

You should be comfortable with the mean and the standard deviation from Sections 5.1 to 5.3, and with square roots: √25 = 5, because 5 × 5 = 25. Everything else — including why a bigger sample gives a smaller standard error — is built from scratch below. A calculator helps, but the small numbers used here do not require one.

Words to know (5)
Standard deviation (s)
The typical distance between one individual observation and the sample mean, expressed in the same unit as the measurements. It describes how spread out the patients themselves are. Pulse rates 70, 72, 75, 78 and 80 beats/min have s ≈ 4.12 beats/min: a typical patient sits about 4 beats/min from the mean of 75.
Standard error (SE)
A measure of the uncertainty of a sample statistic — how much it would wobble if the study were repeated on another sample. It is not a property of one patient and it is not a mistake. If the mean pulse in one sample is 75 beats/min, the standard error says how far another sample’s mean might plausibly land from it.
Standard error of the mean
The standard error of the sample mean in particular, calculated as the standard deviation divided by the square root of the sample size: SE = s ÷ √n. With s = 4.12 beats/min and n = 5, SE = 4.12 ÷ √5 ≈ 1.84 beats/min.
Coefficient of variation (CV)
The standard deviation written as a percentage of the mean: CV = (s ÷ x̄) × 100%. The units cancel, so it measures relative variability rather than absolute spread. A mean of 75 beats/min with s = 4.12 beats/min gives CV = (4.12 ÷ 75) × 100% ≈ 5.5%.
Relative variability
Variability judged in proportion to the size of the mean, so that a spread of 2 units is large against a mean of 10 and small against a mean of 1,000. An s of 10 with a mean of 100 is 10% relative variability; the same s of 10 with a mean of 50 is 20%.
One idea at a timeIdea 1 / 6

In one sentence A standard deviation says how far apart the individual observations are; a standard error says how uncertain a sample statistic is.

Standard deviation describes individuals; standard error describes an estimate

Both numbers come from the same sample, yet they answer two different questions. A standard deviation (s) belongs to the patients: it measures the typical distance between one individual measurement and the mean, and it carries the unit of the measurement, so a spread of pulse rates is expressed in beats/min. A standard error (SE) belongs to the estimate: it measures the uncertainty, or sampling variability, of a statistic such as the sample mean. It answers the question “if this study were repeated on another sample, how far might the new mean land from this one?” The whole section turns on that distinction. One number describes people; the other describes the precision of a number calculated from people.

For example

Pulse rates 70, 72, 75, 78 and 80 beats/min give s ≈ 4.12 beats/min: patients differ from the mean of 75 by roughly 4 beats/min. The standard error of that mean is only 1.84 beats/min, because it describes the mean, not the patients.

Pause and check yourself

A ward mean is 75 beats/min with s = 4.12 beats/min. Which number would you quote to describe how different the patients are from one another?

Show the answer

The standard deviation, s = 4.12 beats/min, because the question is about the spread of individual observations. The standard error of 1.84 beats/min would describe how precisely the mean itself is estimated, which is a different question.

Read detailed explanations
Why the standard error shrinks as the square root of n

The rule SE = s ÷ √n is not a convention; it follows from how variances behave when observations are combined. Work on the variance first, because variances add where standard deviations do not. If one independent observation has variance σ², then the sum of n independent observations has variance nσ². The mean divides that sum by n, and dividing a quantity by n divides its variance by n², so the variance of the sample mean is nσ² ÷ n² = σ²/n. Taking the square root returns the standard error, σ ÷ √n. This is the square-root law of sampling, and it has a hard practical consequence: precision improves with the square root of effort. To halve the standard error you must quadruple the sample; to make it ten times smaller you need a hundred times as many observations. Doubling recruitment therefore buys only about 41% more precision, since √2 ≈ 1.41. Reversing the formula answers the question a nurse researcher asks when planning a study — the sample size needed for a target standard error is n = (s ÷ target SE)². If s = 12 mg/dL and the target is 1.5 mg/dL, n = (12 ÷ 1.5)² = 8² = 64 patients. The same law explains the confidence interval usually printed beside a mean, roughly the estimate ± 1.96 × SE: halving that interval costs exactly what halving the standard error costs.

Describing individuals and describing an estimate: two questions, one sample

Why is the standard error a standard deviation at all? Because the sample mean is itself a random quantity. Imagine repeating the study on a thousand different samples drawn from the same population. Each sample produces its own mean, and those means are not all equal; they scatter. Plot them and you have the sampling distribution of the mean. The standard error is simply the standard deviation of that sampling distribution, which is why it measures wobble rather than patients. Two named results keep this picture honest. First, the sampling distribution of the mean is narrower than the distribution of the observations and becomes more bell-shaped as n grows — the central limit theorem. Second, about 95% of sample means fall within roughly 1.96 standard errors of the true population mean, which is where the familiar confidence interval comes from. The practical warning for nursing is sharp: a ward dashboard reporting mean length of stay as mean ± SE can look impressively stable while individual patients stay anywhere from three days to three weeks. The statistic also depends on what is being estimated. A proportion has its own standard error, and a difference between two means has one too, so “the standard error” is meaningless until the statistic is named. When you read a mean, ask two separate questions: how variable are the patients, and how many patients were there? Only the second of those controls the standard error.

The coefficient of variation: what the percentage buys, and where it fails

Dividing the standard deviation by the mean makes the spread dimensionless, and that is the whole value of the CV: it survives a change of unit. Medical laboratories lean on this. An assay’s analytical imprecision is routinely quoted as a CV so that a glucose method in mg/dL can be judged against a sodium method in mmol/L, and quality-control charts flag an assay whose CV crosses a locally agreed limit. The same percentage lets a ward compare biological variation between patients with the technical variation of the machine that measures them. The CV also has limits that a beginner must respect, and they all come from the arithmetic of the ratio. First, it needs a ratio scale with a true zero. Temperature in °C fails: a mean of 20 °C with s = 2 °C gives a CV of 10%, but 40 °C is not twice as warm as 20 °C, so the ratio has no physical meaning. Second, a mean near zero makes the CV explode: an s of 2 with a mean of 0.5 gives 400%, and an s of 2 with a mean of 0.01 gives 20,000%. Third, variables that can take negative values or cross zero — a fluid balance, a change score, a temperature difference — have no sensible CV at all. Fourth, for strongly skewed data the mean and the standard deviation describe different aspects of the shape, and their ratio can mislead. Used inside those limits, the CV answers a question the standard deviation cannot: variability as a fraction of the typical size. A rough rule of thumb in laboratory work is that a CV under about 10% signals a comparatively stable measurement, but that threshold is a local convention, not a law of nature.

Explore the diagram

Look at the whole drawing first. Then select one numbered part and connect it to the idea you just read.

Standard error and coefficient of variation

A four-panel teaching plate. Panel A shows one ward sample of five pulse rates, with the spread of the individuals compared with the precision of their mean. Panel B states the two formulas with their units and substitutes the same five pulses. Panel C shows the square-root law: the same s = 12 mg/dL with n = 25 and n = 100, where quadrupling the sample halves the standard error. Panel D compares two laboratory assays in different units using the coefficient of variation, and finds creatinine relatively more variable even though its standard deviation is smaller.

New to diagrams? Read this first
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  2. Press Play, or use the arrows, to move through the numbered parts in order. The drawing stays still; only the highlight moves.
  3. Click a number, or its name in the list below, to read what that part is and why it matters. Use “Hide labels” when you want to test your memory.
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01 / 4Describing people, or describing an estimate?

Standard error and coefficient of variationA four-panel teaching plate. Panel A shows one ward sample of five pulse rates, with the spread of the individuals compared with the precision of their mean. Panel B states the two formulas with their units and substitutes the same five pulses. Panel C shows the square-root law: the same s = 12 mg/dL with n = 25 and n = 100, where quadrupling the sample halves the standard error. Panel D compares two laboratory assays in different units using the coefficient of variation, and finds creatinine relatively more variable even though its standard deviation is smaller.People, then the precision of their meanPulses (beats/min): 70, 72, 75, 78, 80Mean x̄ = 75, n = 5, s ≈ 4.12Individuals spread ±1 SD ≈ 4.12 beats/minEstimate pinned to ±1 SE ≈ 1.84 beats/minThe two formulas, with unitsSE(x̄) = s ÷ √ns in beats/min ÷ a count stays beats/minSE = 4.12 ÷ √5 ≈ 1.84 beats/minCV = (s ÷ x̄) × 100%CV = (4.12 ÷ 75) × 100% ≈ 5.5%Quadrupling n halves SE (s = 12 mg/dL)n = 25√25 = 5 → SE = 2.4 mg/dLn = 100√100 = 10 → SE = 1.2 mg/dLs stays 12 mg/dL: the patients do not change.Different units? Compare the CV.Creatinine: x̄ = 1.00 mg/dL, s = 0.20 mg/dLCV = (0.20 ÷ 1.00) × 100% = 20%Potassium: x̄ = 4.00 mmol/L, s = 0.40 mmol/LCV = (0.40 ÷ 4.00) × 100% = 10%Creatinine varies relatively more, despite a smaller SD.

If the drawing is wider than this panel, scroll it sideways. Labels stay readable; the numbered explanations below show the same ideas.

Describing people, or describing an estimate?

The five dots are the patients, and the red arrow is their standard deviation: typical pulses run about 4.12 beats/min either side of the mean. The short green bar is the standard error of that mean, only about 1.84 beats/min, because it describes how precisely the average is pinned down. The two are drawn to the same horizontal scale so the contrast is visible: the standard deviation belongs to the people, the standard error to the estimate, and confusing them misleads the reader.

Schematic teaching plate. The pulse dots are placed along a linear scale, so their spacing reflects the data values; the red and green bars are drawn in proportion to 1 SD and 1 SE so that their relative lengths remain honest, although the bars are not drawn on the same scale as the dot row.

Key points to remember

  • The standard deviation describes how far apart individual observations are; the standard error describes the uncertainty of a statistic such as the mean.
  • The standard error of the mean is SE = s ÷ √n; √n is evaluated first, and the answer carries the same unit as s.
  • Quadrupling the sample size halves the standard error, because √4 = 2. No increase in sample size reduces the standard deviation.
  • SD and SE are not interchangeable; when a report prints “±” without naming the statistic, find out which one it is before interpreting it.
  • The coefficient of variation is CV = (s ÷ x̄) × 100%. Its units cancel, so it is the correct way to compare variability across different units or very different sizes.
Go deeper

Worked example

From raw pulse rates to the standard error and the CV. Five patients on a ward have pulse rates of 70, 72, 75, 78 and 80 beats/min. The notes for this course already give the mean as 75 beats/min, the sample variance as 17 and the sample standard deviation as s = √17 ≈ 4.12 beats/min. The ward sister now asks two further questions: how precisely is that mean estimated, and how variable is pulse rate relative to its own mean?

  1. Write down what is known, with its unit: mean x̄ = 75 beats/min, s ≈ 4.12 beats/min and n = 5 patients.
  2. Choose the formula for the uncertainty of the mean: SE = s ÷ √n. This is the standard error of the mean, and its unit is beats/min.
  3. Substitute and compute the root first: √5 ≈ 2.236, so SE ≈ 4.12 ÷ 2.236 ≈ 1.84 beats/min.
  4. Now the relative variability: CV = (s ÷ x̄) × 100% = (4.12 ÷ 75) × 100% ≈ 5.5%, a percentage with no unit.
  5. Interpret both together. The mean pulse is estimated to about ±1.84 beats/min, and pulse rate varies by roughly 5.5% of its own mean — a relatively stable measurement.

The standard deviation (4.12 beats/min) described the patients; the standard error (1.84 beats/min) describes the precision of the mean; the CV (5.5%) expresses the spread as a share of the mean and frees it from its unit. Three numbers, three different questions, all from the same five pulses.

Tutorial 1 — the standard error of the mean and what quadrupling n does

A study measures fasting glucose in 25 patients and reports a standard deviation of s = 12 mg/dL. The mean is not needed here. Calculate the standard error of the mean, then work out what happens to it when the study is repeated with 100 patients instead of 25 — that is, four times as many.

Rule used: The standard error of the mean is SE = s ÷ √n, where s is the standard deviation in the unit of the measurement and n is the number of observations, a count with no unit. The result is in the same unit as s.

  1. Read the numbers and write s, n and their units before touching the calculator.

    Reveal this step

    s = 12 mg/dL and n = 25 patients. The unit belongs to s alone; n is a count, so n carries no unit.

  2. Write the formula and evaluate the root √25.

    Reveal this step

    SE = s ÷ √n = 12 ÷ √25. Since 5 × 5 = 25, √25 = 5, so SE = 12 ÷ 5.

  3. Before finishing the division, will the standard error be larger or smaller than 12 mg/dL, and why must it be so?

    Reveal the answer

    Smaller, because we divide by √n and √25 = 5 is greater than 1. The standard error measures the uncertainty of the mean, and a mean of many observations is always more stable than one individual value. Finishing the division: SE = 2.4 mg/dL.

  4. The study is repeated with n = 100. Predict, then calculate, the new standard error. By what factor does it change?

    Reveal the answer

    √100 = 10, so SE = 12 ÷ 10 = 1.2 mg/dL. The sample size was multiplied by 4 and the standard error divided by 2, because √4 = 2. More patients sharpen the estimate of the mean; they do not change s, which is still 12 mg/dL.

  5. Write the two results side by side with their units, then state the one sentence that connects them.

    Reveal this step

    n = 25 → SE = 2.4 mg/dL; n = 100 → SE = 1.2 mg/dL. Quadrupling the sample halved the standard error, while the standard deviation stayed at 12 mg/dL because the patients themselves did not change.

Answer: SE = 2.4 mg/dL with 25 patients, falling to 1.2 mg/dL with 100 patients — half the uncertainty for four times the work.

Keep this habit: Always compute √n before dividing, and keep the unit of s on the answer. Remember the square-root law: to halve the standard error you must quadruple the sample, and no increase in sample size will ever reduce the standard deviation.

Tutorial 2 — comparing two laboratory measurements with the CV

A laboratory quality report gives two assays. Serum creatinine, measured in mg/dL, has a mean of 1.00 mg/dL and a standard deviation of 0.20 mg/dL. Serum potassium, measured in mmol/L, has a mean of 4.00 mmol/L and a standard deviation of 0.40 mmol/L. The units differ, so the standard deviations cannot be compared directly. Work out which assay is relatively more variable.

Rule used: The coefficient of variation is CV = (s ÷ x̄) × 100%, where s is the standard deviation and x̄ the mean, both in the same unit. The units cancel, so the CV is a percentage that can be compared across different units.

  1. Write the CV formula once and explain why the units cancel, before substituting anything.

    Reveal this step

    CV = (s ÷ x̄) × 100%. For a given assay, the standard deviation and the mean are measured in the same unit, so that unit appears above and below the division and cancels. What remains is a pure number, turned into a percentage by multiplying by 100%.

  2. Which assay has the larger standard deviation, and does that alone settle which is more variable?

    Reveal the answer

    Potassium, with s = 0.40 mmol/L against creatinine’s s = 0.20 mg/dL. It settles nothing, because the two numbers are in different units: 0.40 mmol/L and 0.20 mg/dL measure different quantities. Comparing them directly is a category error.

  3. Calculate the CV for creatinine, showing the division before the multiplication.

    Reveal this step

    CV = (0.20 ÷ 1.00) × 100% = 0.20 × 100% = 20%. Creatinine varies by about 20% of its own mean.

  4. Calculate the CV for potassium with the same layout.

    Reveal this step

    CV = (0.40 ÷ 4.00) × 100% = 0.10 × 100% = 10%. Potassium varies by about 10% of its own mean.

  5. Which assay is relatively more variable, and what wording would you use to report it?

    Reveal the answer

    Creatinine, because 20% is greater than 10%. A safe report reads: “Relative variability was 20% for creatinine and 10% for potassium, so creatinine was the more variable assay despite its smaller standard deviation.”

  6. Suppose the potassium standard deviation rose to 0.80 mmol/L with the mean unchanged. What is the new CV, and what would you conclude?

    Reveal the answer

    CV = (0.80 ÷ 4.00) × 100% = 20%. The two assays would then be equally variable in relative terms, even though their standard deviations still differ (0.20 mg/dL against 0.80 mmol/L). The CV, not the standard deviation, is what makes them comparable.

Answer: Creatinine CV = 20%; potassium CV = 10%. Creatinine is relatively twice as variable, although its standard deviation is numerically smaller.

Keep this habit: When the units differ, throw the units away with the CV. Divide s by the mean, multiply by 100%, and compare percentages — never raw standard deviations taken from different scales.

Now you try

Attempt each one on paper first. Write the formula, substitute the numbers, then check the unit of your answer.

  1. A sample of 36 patients has a mean systolic blood pressure with a standard deviation of s = 15 mmHg. Calculate the standard error of the mean, then state the new standard error if the study were enlarged to 144 patients with the same s.

    Hint

    Take the square root of n first: √36 = 6, then √144 = 12.

    Show the worked answer

    SE = s ÷ √n = 15 ÷ √36 = 15 ÷ 6 = 2.5 mmHg. With n = 144, SE = 15 ÷ √144 = 15 ÷ 12 = 1.25 mmHg. The sample size was multiplied by 4 and the standard error divided by 2, while s stayed at 15 mmHg.

  2. An audit reports a mean weight of 80 kg with s = 8 kg, and a mean pulse rate of 60 beats/min with s = 9 beats/min. Calculate both coefficients of variation and say which measurement is relatively more variable.

    Hint

    One CV per measurement: divide each s by its own mean, then multiply by 100%.

    Show the worked answer

    Weight: CV = (8 ÷ 80) × 100% = 10%. Pulse: CV = (9 ÷ 60) × 100% = 15%. Pulse rate is relatively more variable, even though the standard deviations (8 kg and 9 beats/min) are close and are measured in units that cannot be compared directly.

  3. A published abstract writes “mean 150 mg/dL ± 3.0” for a sample of 100 patients, and the standard deviation is 30 mg/dL. Show that the ± is a standard error, then explain why a standard error of 3.0 does not mean the patients are all close to 150 mg/dL.

    Hint

    Convert the standard deviation into a standard error with s ÷ √n and compare the answer with 3.0; then think about the range covered by ±2 standard deviations.

    Show the worked answer

    SE = 30 ÷ √100 = 30 ÷ 10 = 3.0 mg/dL, so the ± is indeed the standard error. It describes the precision of the mean, not the patients. The standard deviation is 30 mg/dL, so roughly 95% of individual values lie within ±2 SD, that is between 90 and 210 mg/dL — a wide spread around a well-estimated mean of 150 mg/dL.

  4. A quality-control sheet states that one assay has s = 4.0 mmol/L with a mean of 5.0 mmol/L, while a second assay has s = 30 mg/dL with a mean of 200 mg/dL. Calculate both CVs and state which assay is relatively more variable.

    Hint

    The largest absolute standard deviation is not automatically the most variable measurement. Divide each s by its own mean.

    Show the worked answer

    Assay 1: CV = (4.0 ÷ 5.0) × 100% = 0.80 × 100% = 80%. Assay 2: CV = (30 ÷ 200) × 100% = 0.15 × 100% = 15%. Assay 1 is far more variable relative to its mean, even though the raw standard deviation of 30 mg/dL is numerically larger than 4.0 mmol/L.

Common misconception

The most damaging beginner belief is that a small standard error means the observations are similar to one another. It does not. The standard error measures the precision of the mean, and it can be made small by recruiting more people even when the patients themselves differ enormously. A study of 10,000 adults can estimate the mean height to within a fraction of a centimetre while those adults range over half a metre. The mirror-image error is treating the standard deviation as a measure of how well the mean is known: a large standard deviation does not by itself make a mean imprecise if the sample is large enough. Always ask which question is being asked — how far apart are the observations, or how well do we know the average?

Check your reasoning

A trial reports that patients stayed in hospital for a mean of 6.0 days. Reporting the same data, one author writes “6.0 ± 0.5 days” and another writes “6.0 ± 2.5 days”. Both are honest with their own statistic. Explain how that is possible and why a reader must be told which one is shown.

Reveal an answer

With s = 2.5 days and n = 25, the standard deviation is 2.5 days while the standard error is 2.5 ÷ √25 = 0.5 days. The first author is reporting the standard error of the mean, the second the standard deviation; the numbers come from the same sample and both descriptions are true. They are not interchangeable because they answer different questions: ±2.5 days says that patients differ widely from one another, while ±0.5 days says that the mean itself is pinned down tightly. A reader who assumes the smaller figure describes the patients would badly underestimate how long individual stays actually vary.

Where this comes from

  • BMS 1110 Applied Mathematics and Medical Biostatistics lecture notes (supplied) · Unit on standard deviation and sampling: “Standard Deviation and Standard Error”, lines 913–916, and “Coefficient of Variation”, lines 917–919; the examination example (pulse rates 70, 72, 75, 78, 80; mean 75 beats/min; variance 17; SD ≈ 4.12 beats/min) at lines 920–926. The core range cited here is lines 915–926.
  • Beyond the supplied notes: the square-root law and the limits of the CV · The derivation of the standard error from the variance of a sum, the sample-size formula n = (s ÷ target SE)², the sampling distribution of the mean, the central limit theorem, the 1.96 multiplier and the limits of the CV (ratio scale with a true zero, means near zero, negative values and skew) are standard statistics the notes state without deriving.
  • Note on the course code · The supplied cover page reads “Course Number: BMS 110”. The curriculum catalogue in this project records “BMS 1110”; the catalogue is treated as authoritative and the notes’ cover value is retained here for provenance.

Lesson progress is saved on this device for the current learner. Practice answers stay separate and are not stored.

About this course and its sources

Bilingual lessons organised from the supplied BMS 1110 Applied Mathematics and Medical Biostatistics lecture notes (Mr. Ebangan Prosper Achuanoh). Diagrams are original educational schematics; sections that go beyond the supplied notes say so in their sources. Mathematics, clinical and French-language review are pending.