Pooled Standard Deviation Calculator
Combine the spread of two or more groups into a single estimate. Enter each group's sample size and standard deviation to get the pooled variance, the pooled standard deviation, and the degrees of freedom — the quantities used by t-tests and Cohen's d.
Before You Calculate
- Use each group's sample standard deviation (the n − 1 version) — that is what the pooling weights expect.
- Pooling assumes the groups share a common population variance (homogeneity of variance); it averages estimates of the same quantity, weighted by information.
- Each group needs n ≥ 2, since a standard deviation from a single observation does not exist.
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Learn More
Standard Error vs Standard Deviation
Standard deviation describes the spread of individual values; standard error describes the precision of an estimate. See how one dataset produces both numbers.
One-Way ANOVA Explained
How ANOVA turns variances into a verdict about means: the full SS/df/MS/F bookkeeping on nine data points, the F-table decision, and what a significant F does not say.
The Pooled Variance Formula
sₚ² = Σ(nᵢ − 1)sᵢ² / Σ(nᵢ − 1)
Two groups: sₚ² = [(n₁ − 1)s₁² + (n₂ − 1)s₂²] / (n₁ + n₂ − 2)
sₚ = √sₚ²
The pooled variance is a weighted average of the group variances, with each group weighted by its degrees of freedom nᵢ − 1 — the amount of information it carries about the common spread. Equivalently, it is the total sum of squares within all groups divided by the total degrees of freedom, since (n − 1)s² recovers each group's SS exactly.
Two details are easy to get wrong by hand: the weighting happens on variances, not standard deviations (square first, pool, then take the root at the end), and the weights are nᵢ − 1, not nᵢ. Simply averaging the SDs of unequal groups gives a biased, formula-incompatible number.
Where the Pooled SD Is Used
- The classic two-sample t-test divides the difference in means by sₚ√(1/n₁ + 1/n₂) and uses df = n₁ + n₂ − 2.
- Cohen's d standardizes a mean difference by dividing it by the pooled SD.
- ANOVA's within-groups mean square (MSW) is exactly the pooled variance of all k groups — this calculator with k rows reproduces it.
- Quality control charts pool subgroup SDs to estimate a process sigma.
The common thread: all these methods assume the groups share one underlying variance, so pooling produces the best single estimate of it. When group variances differ substantially (a common check: one SD more than twice another), Welch's t-test — which does not pool — is the safer choice.
Worked Example: Two Classes, One Spread
Class A has n₁ = 12 students with s₁ = 4.2 points; class B has n₂ = 14 students with s₂ = 5.1 points.
- Square the SDs: 4.2² = 17.64 and 5.1² = 26.01.
- Weight by df: 11 × 17.64 = 194.04 and 13 × 26.01 = 338.13.
- Total: 194.04 + 338.13 = 532.17 over df = 11 + 13 = 24.
- Pooled variance: 532.17 / 24 ≈ 22.1738.
- Pooled SD: √22.1738 ≈ 4.7089.
Note the result (4.71) is not the midpoint of 4.2 and 5.1 (4.65): the larger class pulls the estimate toward its value, and pooling on the variance scale weights big spreads slightly more. This sₚ plugged into a two-sample t-test with df = 24 or into Cohen's d gives exactly the numbers those tools report.
Frequently Asked Questions
What is the pooled standard deviation?
It is a single estimate of spread obtained by combining the sample variances of two or more groups that are assumed to share the same population variance. Each group's variance is weighted by its degrees of freedom (n - 1), the weighted variances are averaged, and the square root of that pooled variance is the pooled SD.
Why not just average the standard deviations?
Two reasons. First, variances - not standard deviations - are the quantities that add and average correctly under the model, so pooling must happen on the squared scale. Second, groups contribute information in proportion to n - 1, so unequal groups need unequal weights. A plain average of SDs matches the pooled SD only when all groups have identical sizes and even then only approximately.
When is pooling appropriate?
When the groups can reasonably be assumed to share one population variance - the homogeneity-of-variance assumption behind the classic two-sample t-test and ANOVA. A common informal check is whether the largest SD is more than about twice the smallest; formal options include Levene's test. If variances clearly differ, use Welch's t-test, which skips pooling entirely.
What degrees of freedom does the pooled estimate have?
The sum of the group degrees of freedom: (n1 - 1) + (n2 - 1) + ... For two groups that is n1 + n2 - 2, which is exactly the df used by the pooled two-sample t-test. The worked example pools 12 and 14 observations into df = 24.
Is this the same as ANOVA's mean square within?
Yes. The within-groups mean square (MSW) in one-way ANOVA is the pooled variance of all k groups: the within-group sums of squares added up and divided by N - k, which equals the sum of the (n - 1) weights. Enter all k groups into this calculator and the pooled variance will match the MSW line of an ANOVA table.
Should I use sample or population standard deviations as inputs?
Sample standard deviations (computed with n - 1). The formula's weights are built to recover each group's sum of squares via (n - 1)s^2, which only works with the sample version. If you have raw data rather than summary statistics, compute each group's sample SD first - the standard deviation calculator on this site shows that working too.