# Why n-1 instead of n in pooled sample variance

I am currently self-learning hypothesis testing and am looking at the independent samples t-test whose test statistic involves the pooled sample variance (https://libguides.library.kent.edu/spss/independentttest), $$S_p^2 = \frac{(n_1 - 1)S^2_1+(n_2-1)S_2^2}{n_1+n_2-2},$$ where $$n_1, n_2$$ are the sample sizes of the two samples and $$S_1^2, S_2^2$$ their respective sample variance. This test assumes that $$S_1^2 = S_2^2$$.

I understand that the pooled sampled variance is computed as a weighted average with weights $$w_i = n_i -1$$ for $$i=1,2$$. However I am unsure why $$n_i-1$$ is used as a weight instead of $$n_i$$. I understand that the $$n-1$$ is used instead of $$n$$ so that the usual sample variance is an unbiased estimator of the variance (Bessel's correction) but I cannot see why it is necessary for the pooled sample variance since the statistic $$\frac{n_1S^2_1+n_2S^2_2}{n_1+n_2}$$ is also an unbiased estimator.

Can anyone explain this to me? Thanks.

• The factors of $n_i-1$ merely undo the divisions that were made in computing the $S_i^2$ in the first place, thereby producing a sum of squared residuals in the numerator. For this test we don't really care about bias: what matters far more is the distribution of the test statistic. The distribution of your last statistic unfortunately depends on the ratio of sample sizes.
– whuber
May 25, 2022 at 15:41
• @whuber that makes perfect sense - thank you very much! May 25, 2022 at 16:01
• In computing $S_p^2,$ You nave found $S_i^2, i=1,2,$ each of which requires computing a sample mean $\bar X_i, i = 1,2.$ So, $(n_1+n_2 -2)S_p^2/\sigma^2 \sim\mathsf{Chisq}(\nu=n_1+n_2-2).$ May 25, 2022 at 16:02
• Since $\operatorname{Var}S_i^2=...=\frac{\sigma^4}{n_i-1}$, $S_p^2$ can be seen as an inverse variance weighted linear combination of $S_1^2$ and $S_2^2$ and it thus is more efficient than $S_a^2$ as an estimator of $\sigma^2$. May 26, 2022 at 15:14

For a two-sample t test on samples from populations with the same variance $$\sigma^2,$$ you have two proposed variance estimates

$$S_p^2 = \frac{(n_1 - 1)S^2_1+(n_2-1)S_2^2}{n_1+n_2-2},$$

and

$$S_a^2 = \frac{(n_1S^2_1+n_2)S^2_2}{n_1+n_2}.$$

For $$S_p^2,$$ you have found $$S_i^2; i=1,2,$$ each of which requires computing a sample mean $$\bar X_i, 1,2.$$ So,

$$\frac{\nu S_p^2}{\sigma^2} \sim \mathsf{Chisq(\nu)}.$$ where $$\nu = n_1+n_2 - 2.$$

For $$S_a^2,$$ the distribution theory is not so clear. You say something about $$S_a^2$$ being unbiased, but that hardly specifies a distribution. Let's use The same degrees of freedom $$\nu$$ as above for an experiment.

Simulation: Begin by looking at $$m = 10\,000$$ samples x1 of size $$n_1 = 2$$ from $$\mathsf{Norm}(\mu_1 = 100, \sigma_1 = 15)$$ and x2 of size $$n_2=3$$ from $$\mathsf{Norm}(\mu_2 = 110, \sigma_2 = 15).$$
We find the sample variances, the pooled variance estimat and the average variance estimate. Then we look at the corresponding chi-squared random variables.

set.seed(2022)
n1 = 2; m=10^5
M1 = matrix(rnorm(n1*m, 100, 15), nrow=m)
v1 = apply(M1, 1, var)
n2 = 3
M2 = matrix(rnorm(n2*m, 110, 15), nrow=m)

v2 = apply(M2, 1, var)

pool = (v1 + 2*v2)/(n1+n2-2)
q.p = (n1+n2-2)*pool/15^2
avg.v = (v1+v2)/(n1+n2) ####
q.a = (n1+n2)*avg.v/15^2

Then we compare the results with the density functions of the corresponding chi-squared distribution. For the pooled estimate $$S_p^2$$ we get a good match, but for $$S_a^2$$ the fit is not good.

R code for graphs:

par(mfrow=c(1,2))
hist(q.p, prob=T, ylim=c(0,.35), col="skyblue2", main="Pooled")