Design, replicates and sample size · Choose a method
How many biological replicates do you need?
Work out n per group from the smallest difference worth detecting and the variability between animals or cultures. Tables, pilot-study pitfalls, the resource equation and RNA-seq numbers.
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Quick answer. Work n out from two numbers: the smallest difference worth detecting and the standard deviation (SD) between independent animals or cultures. Their ratio is the standardised effect d. For two independent groups of equal size compared once (unpaired, two-sided test at α = 0.05, 80% power), d = 1 needs 17 per group, d = 1.5 needs 9 and d = 2 needs 6 (table below); paired designs, several corrected comparisons and expected losses change these numbers, as shown below. If you have no SD estimate, plan with a published SD or a pilot SD scaled up as shown below, and state which effect your n can detect. “Three per group because that is what the lab does” is a habit rather than a justification.
Terms used below. Standard deviation (SD): the typical spread of values between independent animals or cultures. Standardised effect (d): the difference worth detecting divided by the SD. α (alpha): the false-positive rate you accept, usually 0.05. Two-sided: the test looks for a difference in either direction. Power: the chance that the study detects the difference if it is real; 80% is the usual minimum. Degrees of freedom (df): a count based on the number of independent units, used by the test and in the formulas below. Holm and Dunnett: corrections for making several comparisons (which correction to use). Pilot: a small preliminary experiment.
Start with the two numbers that decide n
The difference worth detecting is a scientific choice: the smallest change that would matter for your question, rather than the difference you hope to see. The SD is the spread between independent units — animals, independent cultures, patients — rather than between wells or cells of one animal. If you are unsure what your unit is, read what counts as n first, because the SD between technical replicates is smaller and gives a falsely small n.
A worked example. You expect a drug to raise grip strength by 4 g, and published data for the same strain and test give an SD of 3 g between mice.
| Step | Result |
|---|---|
| Standardised effect d = 4 ÷ 3 | 1.33 |
| n per group for 80% power (two-sided α = 0.05) | 10 (achieved power 80.5%) |
| Allow for losing 10% of animals (replace with your own expected losses): 10 ÷ 0.9, rounded up | 12 per group — the number to allocate |
| If the SD is really 4 g, n for the same 4 g difference | 17 per group |
| Power of 10 per group if the SD is really 4 g | 56% |
The SD matters as much as the difference: a 33% larger SD raises the required n by 70%. Enter your own numbers in the sample size calculator, which also handles paired designs and unequal group sizes.
What to do: write down the difference, the SD and where each came from before you choose n.
How many per group: the table
Two independent groups, equal group sizes, two-sided α = 0.05, normally distributed outcome with equal SDs (power from the noncentral t distribution; SciPy and R pwr give identical numbers).
| Standardised effect d | n per group, 80% power | n per group, 90% power |
|---|---|---|
| 0.5 | 64 | 86 |
| 0.8 | 26 | 34 |
| 1.0 | 17 | 23 |
| 1.5 | 9 | 11 |
| 2.0 | 6 | 7 |
Three adjustments keep the plan honest:
- You will analyse with Welch’s t-test and expect one group to vary more: plan with the larger SD (Welch vs Student).
- The same animals are measured twice: use the SD of the within-animal changes in the calculator; paired designs need fewer animals when the measurements are correlated.
- Several comparisons will be corrected: do the calculation at the corrected α. Use 0.05 ÷ m, where m is the number of comparisons: it is the strictest level Holm’s method applies, and a safe, slightly generous plan for Dunnett’s test (which correction to use). For vehicle plus three doses, each compared with vehicle, set α to 0.05 ÷ 3 = 0.0167 in the calculator and use the n it gives for every group.
What to do: use the table to find your d, or compute it exactly in the calculator, and add the expected losses.
No SD estimate? Three routes, in this order
1. Use a published SD for the same measurement, strain and protocol. Cite it in your protocol.
2. Run a pilot, and scale its SD up before you use it. A small pilot estimates the SD imprecisely, and planning with the raw estimate gave an underpowered study in 53% of the cases in our simulation below (code, output). Kraemer and colleagues argue for pilots on other grounds but “caution against their use for the objective of guiding power calculations” (Kraemer et al. 2006). We simulated it: the true effect is d = 1 (17 per group needed), a pilot of 5 animals per group estimates the SD, and the main study is sized from that estimate.
| Planning rule (pilot of 5 per group, 20,000 simulated pilots) | Studies with power below 80% | Below 50% | Median n per group |
|---|---|---|---|
| Plug in the pilot SD | 53% | 10% | 16 |
| Plug in the upper one-sided 80% confidence limit of the SD (pilot SD × 1.32 for this 5-per-group pilot, df = 8; other pilots need their own multiplier) | 18% | 2% | 27 |
The pilot SD came out below the true SD in 57% of pilots, which is why the plain rule falls short. Browne showed that planning with a 100(1 − γ)% upper confidence bound on the SD gives enough animals for the planned power in at least 100(1 − γ)% of studies (Browne 1995); with γ = 0.2, our simulation reached 80% power in 82% of plans. The price is more animals, so prefer a published SD when one exists. For a pilot with 8 degrees of freedom the multiplier is √(8 ÷ χ²₀.₂₀,₈) = 1.32; for a two-group pilot, df = 2 × (pilot animals per group − 1), so 5 per group gives df = 8. Compute the multiplier for your own pilot with sqrt(df / qchisq(0.2, df)) in R or (df / scipy.stats.chi2.ppf(0.2, df)) ** 0.5 in Python.
3. Turn the question round (sensitivity analysis). If the number of animals is fixed by budget or licence, state the smallest effect that number detects. With 8 per group, a two-sided test at α = 0.05 reaches 80% power from d = 1.51 upwards — a difference of one and a half SDs. Festing recommends exactly this order: make a provisional choice of n, then estimate the effect it can detect, and increase n if that effect is too large (Festing 2018). The n = 3 guide tabulates detectable effects for 3 to 20 per group.
What to do: use route 1 when a published SD exists, route 2 with the scaled-up SD otherwise, and report route 3 whenever n is fixed.
The resource equation: a last resort for exploratory work
When neither an effect size nor an SD can be assumed, the resource equation sets the error degrees of freedom of an ANOVA between 10 and 20; for a one-way design, n per group = DF ÷ k + 1, with the lower end rounded up and the upper end rounded down so the error df stays between 10 and 20 (Arifin & Zahiruddin 2017). For k = 3: 10 ÷ 3 + 1 = 4.33 → 5, and 20 ÷ 3 + 1 = 7.67 → 7, so 5 to 7 per group.
Exploratory use only: the resource equation contains no effect size, so the numbers below carry no power guarantee.
| Groups (k) | Animals per group | Total animals |
|---|---|---|
| 2 | 6 to 11 | 12 to 22 |
| 3 | 5 to 7 | 15 to 21 |
| 4 | 4 to 6 | 16 to 24 |
The authors describe it as “suitable for exploratory studies whenever it is not possible to assume the standard deviation and the effect size”. It contains no effect size, so it says nothing about the power of the study. Use it for exploratory work alone, label the study exploratory in the protocol and the paper, and switch to a power calculation for the confirmatory experiment.
Omics experiments: RNA-seq
Gene-by-gene power depends on each gene’s variability and fold change, so power is worked out per gene rather than from a single d. One large benchmark sequenced 48 biological replicates of yeast (Saccharomyces cerevisiae) in each of two conditions. With three replicates, nine of the 11 tools tested found only 20–40% of the differentially expressed genes identified with the full 42 clean replicates (more than 85% for genes changing more than fourfold); recovering more than 85% of all of them required more than 20 replicates. The authors recommend “at least six biological replicates”, rising to “at least 12 when it is important to identify SDE genes for all fold changes” (SDE: significantly differentially expressed) (Schurch et al. 2016).
What to do: use at least six biological replicates per condition for RNA-seq, twelve if small fold changes matter. These numbers come from a low-variability yeast system (one lab strain, wild type vs Δsnf2); treat six as a floor, and plan more for tissue or human samples. If you are restricted to three, say in the paper that the experiment was designed to detect large changes (above fourfold, where three replicates recovered more than 85% of genes in this benchmark), and treat the gene list as hypothesis-generating rather than as a result.
Journal minimums are floors, not targets
The British Journal of Pharmacology has “set 5 as the minimum ‘n’ required for datasets subjected to statistical analysis” (Curtis et al. 2018). Meeting that floor says nothing about power: 5 per group reaches 80% power from d = 2.02 upwards. Low power is common — Button and colleagues estimated “the median statistical power of studies in the neurosciences is between ∼8% and ∼31%” (Button et al. 2013) — and low-powered studies overestimate the effects they do detect.
What to do: treat a journal’s minimum as the smallest acceptable n, and justify your actual n with the calculation above.
Check what an AI assistant planned
AI assistants can produce a sample size in seconds; check the inputs before you copy it into a protocol.
- Where did d come from? A conventional “medium effect, d = 0.5” gives 64 per group; your system can need far fewer or far more. Replace it with your difference and SD.
- Which SD? It must be between animals or independent cultures, rather than between technical replicates.
- One- or two-sided, and which α? The table uses two-sided α = 0.05; a corrected α needs more animals.
- Was n rounded up, and were losses added?
Reproduce the worked example to check the tool the assistant used:
library(pwr)
pwr.t.test(d = 4/3, sig.level = 0.05, power = 0.80) # n = 9.889 -> 10 per group
from statsmodels.stats.power import TTestIndPower
print(TTestIndPower().solve_power(effect_size=4/3, alpha=0.05, power=0.8)) # 9.889 -> 10 per group
Report it
ARRIVE 2.0 item 2b asks you to “Explain how the sample size was decided. Provide details of any a priori sample size calculation, if done” (Percie du Sert et al. 2020). A template that covers it (replace every number and every bracketed item with yours):
The experimental unit was the individual mouse. We planned to detect a difference in grip strength of 4 g, the smallest change we considered biologically relevant, assuming an SD of 3 g between mice [source]. A two-sided two-sample t-test at α = 0.05 with 80% power required 10 mice per group (R 4.3.3, pwr 1.3.0); allowing for 10% losses, we allocated 12 per group (24 in total).
For cell-culture work:
The experimental unit was the independent culture (separate days and passages). We planned to detect a [difference] in [outcome], assuming an SD of [SD] between independent cultures [source]; a two-sided two-sample t-test at α = 0.05 with 80% power required [n] independent cultures per condition.
Give the error degrees of freedom when the resource equation sized an exploratory study, and the smallest effect detectable with 80% power when n was fixed.
References
- Arifin WN, Zahiruddin WM (2017). Sample size calculation in animal studies using resource equation approach. Malaysian Journal of Medical Sciences 24(5):101–105. doi:10.21315/mjms2017.24.5.11
- Browne RH (1995). On the use of a pilot sample for sample size determination. Statistics in Medicine 14(17):1933–1940. doi:10.1002/sim.4780141709
- Button KS, Ioannidis JPA, Mokrysz C, et al. (2013). Power failure: why small sample size undermines the reliability of neuroscience. Nature Reviews Neuroscience 14:365–376. doi:10.1038/nrn3475
- Curtis MJ, Alexander S, Cirino G, et al. (2018). Experimental design and analysis and their reporting II: updated and simplified guidance for authors and peer reviewers. British Journal of Pharmacology 175:987–993. doi:10.1111/bph.14153
- Festing MFW (2018). On determining sample size in experiments involving laboratory animals. Laboratory Animals 52(4):341–350. doi:10.1177/0023677217738268
- Kraemer HC, Mintz J, Noda A, et al. (2006). Caution regarding the use of pilot studies to guide power calculations for study proposals. Archives of General Psychiatry 63(5):484. doi:10.1001/archpsyc.63.5.484
- Percie du Sert N, Hurst V, Ahluwalia A, et al. (2020). The ARRIVE guidelines 2.0: updated guidelines for reporting animal research. PLOS Biology 18(7):e3000410. doi:10.1371/journal.pbio.3000410
- Schurch NJ, Schofield P, Gierliński M, et al. (2016). How many biological replicates are needed in an RNA-seq experiment and which differential expression tool should you use? RNA 22(6):839–851. doi:10.1261/rna.053959.115
How this page was made
- Written and published by Licklider. Recommendations cite peer-reviewed methods papers and reporting guidelines listed under References; cited figures were checked against the source text, and where only an abstract was accessible the page states the claim at that level.
- The sample-size table, worked example and pilot-study simulation were produced by stored code (Python with SciPy 1.18.1, and statsmodels 0.15.0 for the reproduction snippet; R 4.3.3 with pwr 1.3.0), and the two implementations agree.
- Separate review passes checked the statistics, the citations and readability before publication.