Pooled T Ratios from Two Normal Samples

Why Pool Two Sample Variances?

When two independent groups are sampled from normal populations with the same variance, each group provides information about that shared variance. Pooling uses both groups’ residual sums of squares to estimate it. The resulting two-sample t ratio has an exact finite-sample distribution, not merely an approximation.

The exact result depends on the model: the samples must be independent, each population must be normal, and the two population variances must be equal. We derive the distribution under those conditions and distinguish the null case from the case of a nonzero difference in population means.

The weights in \(S_p^2\) are the respective residual degrees of freedom, not the sample sizes. Each sample mean uses one degree of freedom, leaving \(n_1-1\) and \(n_2-1\) residual degrees of freedom. Their total is \(\nu=n_1+n_2-2\).

The Exact Distribution

A noncentral t distribution with degrees of freedom \(\nu\) and noncentrality parameter \(\delta\) is the distribution of \((Z+\delta)/\sqrt{U/\nu}\), where \(Z\sim N(0,1)\), \(U\) has a chi-square distribution with \(\nu\) degrees of freedom, and \(Z\) and \(U\) are independent. Thus the theorem requires establishing both the chi-square law of the pooled residual sum of squares and its independence from the sample-mean contrast.

Deriving the Residual and Contrast Laws

Let \(Z\) be the vector formed by standardizing each observation by its population mean and standard deviation: \[ Z=\left(\frac{X_1-\mu_1}{\sigma},\ldots,\frac{X_{n_1}-\mu_1}{\sigma}, \frac{Y_1-\mu_2}{\sigma},\ldots,\frac{Y_{n_2}-\mu_2}{\sigma}\right)^T. \] Its \(n_1+n_2\) entries are independent standard normal variables. Let \(P\) be the block diagonal matrix with blocks \(I_{n_1}-J_{n_1}/n_1\) and \(I_{n_2}-J_{n_2}/n_2\), where \(I_m\) is the \(m\)-dimensional identity matrix and \(J_m\) is the \(m\)-by-\(m\) matrix of ones. Each block removes the sample mean of its group.

To verify the residual distribution, note that \(P\) is symmetric and idempotent: each block \(I_m-J_m/m\) has both properties. Its rank is \((n_1-1)+(n_2-1)=\nu\). The sum of the squared deviations within the two samples, divided by \(\sigma^2\), is \(Z^TPZ\). By choosing an orthonormal basis of eigenvectors for \(P\), its quadratic form becomes the sum of squares of \(\nu\) independent standard normal coordinates: the eigenvalues are \(\nu\) ones and the remaining eigenvalues are zero. Therefore

For independence, define \(W=(\bar X-\bar Y-(\mu_1-\mu_2))/[\sigma\sqrt{1/n_1+1/n_2}]\). This is a linear combination \(c^TZ\) for a unit vector \(c\): its entries are \(1/n_1\) in the first group, \(-1/n_2\) in the second, with all entries divided by \(\sqrt{1/n_1+1/n_2}\). The entries of \(c\) sum to zero within each group, so \(Pc=0\). Consequently \(c\) is orthogonal to the range of \(P\).

Extend \(c\) and an orthonormal basis for the range of \(P\) to an orthonormal basis of the whole space. In these coordinates, the standard normal vector \(Z\) still has independent standard normal coordinates: its density is proportional to \(\exp(-z^Tz/2)\), which is unchanged by an orthogonal transformation and factors into the coordinate densities. The contrast \(W=c^TZ\) is one such coordinate. The quadratic form \(Z^TPZ\) depends only on the \(\nu\) coordinates for the range of \(P\). Those coordinates are independent of \(W\), proving the claimed independence. \(\square\)

A related consequence checks the scale of the pooled estimator. Since a chi-square variable with \(\nu\) degrees of freedom has expectation \(\nu\), the displayed quadratic-form identity gives \(E[\nu S_p^2/\sigma^2]=\nu\), and hence \(E[S_p^2]=\sigma^2\). Thus pooling preserves unbiasedness under the common-variance model.

From the Components to a t Ratio

The sample means are linear combinations of independent normal observations. Their difference is therefore normal, with mean \(\mu_1-\mu_2\) and variance \(\sigma^2(1/n_1+1/n_2)\). It follows that

The preceding result shows that this standardized contrast is independent of \(U=\nu S_p^2/\sigma^2\), where \(U\sim\chi^2_\nu\). Substituting \(S_p^2=\sigma^2 U/\nu\) into the statistic gives

The numerator is normal with mean \(\delta\) and variance one, and is independent of \(U\). This is precisely the defining representation of a noncentral t variable with degrees of freedom \(\nu\) and noncentrality parameter \(\delta\). When \(\mu_1=\mu_2\), \(\delta=0\), giving the central Student t law. \(\square\)

Worked Calculations

Assumptions and a Common Pitfall

The exact t result is useful because it supplies a finite-sample reference law for a standardized difference in means. Its denominator estimates the shared variance using all available within-group residual information, while its numerator measures the difference in sample means. The two components are independent under the stated normal model, which is what makes the exact ratio law possible.

A common mistake is to pool the variances even when the populations have different variances and then claim the same exact t distribution. The proof above uses one common \(\sigma^2\) both to standardize the two samples and to express the pooled residual sum of squares as a single scaled chi-square variable independent of the contrast. If the population variances differ, those steps do not yield the stated ratio representation in general. Independence and normality also matter; the theorem is not a distribution-free guarantee.

Another frequent error is to use \(n_1+n_2\) degrees of freedom. Each sample loses one degree of freedom when its sample mean is estimated, so the residual degrees of freedom add to \((n_1-1)+(n_2-1)=n_1+n_2-2\). The weighted pooled variance uses these same degrees of freedom.