A two-sample statistic

Topic: Estimation and inference for the multivariate normal · more questions on this topic
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Two groups of sizes n_1=n_2=10 in p=2 dimensions have mean difference \bar{\mathbf{x}}^{(1)}-\bar{\mathbf{x}}^{(2)}=(1.5,1.0)^\top and pooled covariance

S=\begin{pmatrix}2&1\\1&2\end{pmatrix}.

Compute T^2 from the two-sample formula, convert it to an F statistic, and decide at the 5\% level given F_{0.05;\,2,\,17}=3.592.

Worked answer (click to show)

Pool the within-group information, form the mean difference and apply the two-sample scaling. The reference distribution requires independent Gaussian samples with a common covariance.

S^{-1}=\frac13\begin{pmatrix}2&-1\\-1&2\end{pmatrix}.

With \mathbf{d}=(1.5,1)^\top,

\mathbf{d}^\top S^{-1}\mathbf{d} =\frac13\bigl(2\cdot 2.25-2\cdot 1.5\cdot 1+2\cdot 1\bigr)=\frac{3.5}{3}=1.1667.

The factor is n_1 n_2/(n_1+n_2)=100/20=5, so T^2=5\times 1.1667=5.8333. Then

F=\frac{n_1+n_2-p-1}{(n_1+n_2-2)p}\,T^2 =\frac{17}{36}\times 5.8333=2.755.

The null reference distribution is F_{2,17}. Since 2.755<3.592, H_0:\boldsymbol{\mu}_1=\boldsymbol{\mu}_2 is not rejected at the 5\% level. The p-value is about 0.09.