Hotelling's $T^2$ test on one-sample population

Question

I'm trying to get a grasp with Hotelling's $T^2$ test in the case of a one-sample population. This exercise is a part of a training set; it's not an homework, and I'm not looking for a complete solution, but for help to correct some mistakes.

The Problem

There's a population that can be supposed to follow a bivariate Normal law of mean $\mathbf{\mu}$ and covariance $\mathbf{\Sigma}$. The sample's size is 101, the sample mean is $\mathbf{\bar{x}} = (55.24\quad 34.97)^T$, and the sample covariance is: $$\mathbf{S} = \begin{pmatrix}210.54 & 126.99 \\\ 126.99 & 119.68\end{pmatrix}.$$ Let $\mathbf{\mu_0} = (60\quad 50)^T$; test at $\alpha=0.01$ level of confidence the following hypothesis: $$H_0 : \mathbf{\mu}=\mathbf{\mu_0} \longleftrightarrow H_1 : \mathbf{\mu}\neq\mathbf{\mu_0}.$$ Then, build a $0.99$ confidence ellipsoid for $\mu$.

What I did

Hypothesis testing

Since the true covariance matrix of the population is not known, I have to use the Hotelling's $T^2$ test. Recall that

$$t^2 = (n-1)(\mu_0-\mathbf{\bar{x}})^T \mathbf{S}^{-1}(\mu_0-\mathbf{\bar{x}})$$

follows an Hotelling's $T^2_{p,m}$ distribution of dimension $p$ (with $p=2$ here) and $m$ degrees of liberty (with $m=101$ here). Thus, $$F = \frac{n-p}{(n-1)p}t^2$$ follows an $F(p, n-p)$ distribution of parameters $p, n-p$.

The test is thereby: if $$F_{obs} = \frac{n-p}{p} (\mu_0-\mathbf{\bar{x}})^T \mathbf{S}^{-1}(\mu_0-\mathbf{\bar{x}}) > F_{1-\alpha}(p,n-p)$$ then we reject the null hypothesis $H_0$.

Doing the computations, I get $F_{obs} = 175.182$ (it's huge!) and $F_{0.99}(2,99) = 4.826189$ (computed via the R function qf(0.99, 2, 99)). Obviously, with this value of $F_{obs}$, I'm rejecting the null hypothesis.

Confidence ellipsoid

To compute the confidence ellipsoid, I know that it's given by:

$$\mathscr{E} = \{\mathbf{x}\in\mathbf{R}^2 : (\mathbf{x}-\mathbf{\bar{x}})^T \mathbf{S}^{-1} (\mathbf{x}-\mathbf{\bar{x}}) = F_{0.99}(2,99)\}$$

and is described by the eigenvalues and the eigenvectors of $\mathbf{S}^{-1}$; more precisely, the axes are given by eigenvectors of $\mathbf{S}^{-1}$, and the length of the axes are given by $\sqrt{\frac{\lambda_i}{2}F_{0.99}(2,99)}$ (with $\lambda_i$ being the eigenvalue associated with the eigenvector $\mathbf{v}_i$.)

Here, the eigenvalues are $\lambda_1 = 0.033, \lambda_2 = 0.003$ and the eigenvectors are $$\mathbf{v}_1 = (-0.576\quad 0.818)^T, \mathbf{v}_2 = (-0.818\quad -0.576)^T.$$

What's wrong

I'm feeling like the $F$-statistics is abnormally large compared to $F_{0.99}(2,99)$. Moreover, the confidence ellipsoid should be pretty big (since the confidence level is $0.99$), while the one I computed is tiny.

Can you help me to find where I'm wrong?

Answer 1

The first part of the computations are right: the null hypothesis should be rejected.

However, there are some mistakes when computing the confidence ellipse. Indeed, the confidence ellipse is described as follow:

$$\mathscr{E} = \{\mathbf{x}\in\mathbf{R}^2 : \frac{99}{2} (\mathbf{x}-\mathbf{\bar{x}})^T \mathbf{S}^{-1} (\mathbf{x}-\mathbf{\bar{x}}) \leq F_{0.99}(2, 99)\}$$

One of the mistake was to forget the $\frac{n-p}{p}$ factor (which is equal to $\frac{99}{2}$ here) in the equation. Moreover, the axes' length of the ellipse are not given by $\sqrt{\frac{\lambda_i}{2}F_{0,99}(2,99)}$, but rather by $\sqrt{\lambda_i\frac{2}{99}F_{0.99}(2,99)}$; to see this, recall that $\mathbf{S}$ is a $2\times 2$, symmetric, and positive definite matrix, and such, there exist an orthogonal matrix $\mathbf{V}$ (given by the eigenvectors of $\mathbf{S}$) so that $$\mathbf{S} = \mathbf{V}^T\mathbf{\Delta V},$$ with $\mathbf{\Delta}$ a diagonal matrix formed by the eigenvalues of $\mathbf{S}$. Taking the equation describing $\mathscr{E}$, we get:

$$\begin{align*} &\frac{99}{2} (\mathbf{x}-\mathbf{\bar{x}})^T \mathbf{S}^{-1} (\mathbf{x}-\mathbf{\bar{x}}) \leq F_{0.99}(2, 99) \\ \iff& \frac{99}{2} (\mathbf{x}-\mathbf{\bar{x}})^T \mathbf{R}^T\mathbf{\Delta}^{-1}\mathbf{R} (\mathbf{x}-\mathbf{\bar{x}}) \leq F_{0.99}(2, 99) \\ \iff& \frac{1}{\lambda_1 F_{0.99}(2, 99) \frac{2}{99}} (\mathbf{R}(\mathbf{x}-\mathbf{\bar{x}}))^2_1 + \frac{1}{\lambda_2 F_{0.99}(2, 99) \frac{2}{99}} (\mathbf{R}(\mathbf{x}-\mathbf{\bar{x}}))^2_2 \leq 1 \end{align*}$$ Recall that an ellipse is defined by: $$\frac{x^2}{a^2}+\frac{y^2}{b^2}\leq 1$$ we get that the axes' length are given by $2a = 2\sqrt{\lambda_1 F_{0.99}(2, 99) \frac{2}{99}}$ and $2b = 2\sqrt{\lambda_2 F_{0.99}(2, 99) \frac{2}{99}}$.

Putting everything together, we obtain the axes and axes' length of the confidence ellipse:

$$\begin{align*} \lambda_1 &= 299.982\\ \lambda_2 &= 30.238\\ \mathbf{v_1} &= \begin{pmatrix}-0.818\\ -0.576\end{pmatrix}\\ \mathbf{v_2} &= \begin{pmatrix}0.576\\ -0.818\end{pmatrix}\\ 2a &= 10.816\\ 2b &= 3.434 \end{align*}$$ Which should end the computations and solve the exercise. $\blacksquare$