Discussion
If \(T^{2}\) is Hotelling \(T^{2}\)-statistic, then the distribution of \(\dfrac{T^{2}}{n - 1}\,\dfrac{n - p}{p}\) would be :
Answer: A
The keyed answer is the non-central F-distribution.
For \(T^{2} = n(\bar{x} - \mu_{0})'S^{-1}(\bar{x} - \mu_{0})\) from a sample of size \(n\) from \(N_{p}(\mu, \Sigma)\), Anderson's theorem states that \(\dfrac{T^{2}}{n - 1}\dfrac{n - p}{p}\) has a non-central F-distribution with \(p\) and \(n - p\) degrees of freedom and non-centrality \(n(\mu - \mu_{0})'\Sigma^{-1}(\mu - \mu_{0})\).
When the null hypothesis \(\mu = \mu_{0}\) holds the non-centrality is zero and the distribution is the central F, which is the special case used to set critical values; the general statement, with no hypothesis assumed, is the non-central one. A chi-square or t-distribution is not the distribution of this ratio.
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