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Pooled Standard Deviation

sp=(n1−1)s12+(n2−1)s22n1+n2−2s_p = \sqrt{\frac{(n_1 - 1) s_1^{2} + (n_2 - 1) s_2^{2}}{n_1 + n_2 - 2}}

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When two groups are assumed to share the same underlying variability, you get a better estimate of it by pooling: add the two sums of squared deviations and divide by the total degrees of freedom, n₁ + n₂ − 2. The weighting matters — a group of 30 deserves more say than a group of 5, and the (n − 1) factors deliver exactly that. With s₁ = 6 from 10 observations and s₂ = 4 from 12, the pooled value is √((9 × 36 + 11 × 16)/20) = √25 = 5, closer to 4 than a naive average of 5 would suggest only because the second group is larger.

The trap is pooling when the assumption fails. If one group's variance is more than about four times the other's, the pooled t-test loses its guarantees and you should use the Welch version, which never pools at all — most software now defaults to Welch for exactly this reason. A second, subtler trap: pool the variances, never the standard deviations. Averaging 6 and 4 to get 5 happens to land in the right place here by coincidence; with s₁ = 8 and s₂ = 2 in equal groups the honest pooled value is √((64 + 4)/2) ≈ 5.83, not 5.

Pooled Standard Deviation
sp=(n1−1)s12+(n2−1)s22n1+n2−2s_p = \sqrt{\frac{(n_1 - 1) s_1^{2} + (n_2 - 1) s_2^{2}}{n_1 + n_2 - 2}}
Where
  • sps_p= Pooled standard deviation
  • s1s_1= Standard deviation 1
  • n1n_1= Size of sample 1
  • s2s_2= Standard deviation 2
  • n2n_2= Size of sample 2
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