A small T^2 test

Topic: Estimation and inference for the multivariate normal · more questions on this topic
‹ A determinant identity A two-sample statistic ›

Six subjects give \bar{\mathbf{x}}=(2,1)^\top and

S=\begin{pmatrix}4&1\\1&1\end{pmatrix}

for the changes in two scores. Test H_0:\boldsymbol{\mu}=\mathbf{0} at the 5\% level using T^2, given F_{0.05;\,2,\,4}=6.944. State the exact null distribution you use.

Worked answer (click to show)

The quadratic form measures the observed displacement. Its reference distribution belongs to the random statistic under the null, rather than to the numerical value obtained after observing the sample.

Here n=6 and p=2. Then |S|=4-1=3 and

S^{-1}=\frac13\begin{pmatrix}1&-1\\-1&4\end{pmatrix}.

The quadratic form is

\bar{\mathbf{x}}^\top S^{-1}\bar{\mathbf{x}} =\frac13\bigl(1\cdot 4-2\cdot 2\cdot 1+4\cdot 1\bigr)=\frac43,

so T^2=6\cdot(4/3)=8. Under H_0 and the independent Gaussian assumptions, the random scaled statistic has an F_{2,4} distribution. Its observed value is

F_{\mathrm{obs}}=\frac{n-p}{p(n-1)}T^2=\frac{4}{2\cdot 5}\cdot 8=\frac{4}{10}\cdot 8=3.2.

Since 3.2<6.944, H_0 is not rejected at the 5\% level. The p-value is about 0.15. This non-rejection does not establish the null. Power would require a specified alternative mean and covariance.