Here we have data on n = 37 subjects taking the Wechsler Adult Intelligence Test. This test is broken up into four different components:

• Information (Info)
• Similarities (Sim)
• Arithmetic (Arith)
• Picture Completion (Pic)

The data are stored in five different columns. The first column is the ID number of the subjects, followed by the four component tasks in the remaining four columns.

Download the txt file: wechslet.txt

Analysis

We obtain the following sample means.

Variance-Covariance Matrix

\(\textbf{S} = \left(\begin{array}{rrrr}11.474 & 9.086 & 6.383 & 2.071\\ 9.086 & 12.086 & 5.938 & 0.544\\ 6.383 & 5.938 & 11.090 & 1.791\\ 2.071 & 0.544 & 1.791 & 3.694 \end{array}\right)\)

Here, for example, the variance for Information was 11.474. For Similarities it was 12.086. The covariance between Similarities and Information is 9.086. The total variance, which is the sum of the variances comes out to be 38.344, approximately.

The eigenvalues are given below:

\(\lambda_1 = 26.245\),  \(\lambda_2 = 6.255\),  \(\lambda_3 = 3.932\),  \(\lambda_4 = 1.912\)

and finally at the bottom of the table we have the corresponding eigenvectors. They have been listed here below:

\(\mathbf{e_1}=\left(\begin{array}{r}0.606\\0.605\\0.505\\0.110 \end{array}\right)\),  \(\mathbf{e_2}=\left(\begin{array}{r}-0.218\\-0.496\\0.795\\0.274 \end{array}\right)\), \(\mathbf{e_3}=\left(\begin{array}{r}0.461\\-0.320\\-0.335\\0.757 \end{array}\right)\),  \(\mathbf{e_4}=\left(\begin{array}{r}-0.611\\0.535\\-0.035\\0.582 \end{array}\right)\)

For example, the eigenvectors corresponding the the eigenvalue 26.245, those elements are 0.606, 0.605, 0.505, and 0.110.

Now, let's consider the shape of the 95% prediction ellipse formed by the multivariate normal distribution whose variance-covariance matrix is equal to the sample variance-covariance matrix we just obtained.

Recall the formula for the half-lengths of the axis of this ellipse. This is equal to the square root of the eigenvalue times the critical value from a chi-square table. In this case, we need the chi-square with four degrees of freedom because we have four variables.  For a 95% prediction ellipse, the chi-square with four degrees of freedom is equal to 9.49.

For looking at the first and longest axis of a 95% prediction ellipse, we substitute 26.245 for the largest eigenvalue, multiplied by 9.49 and take the square root. We end up with a 95% prediction ellipse with a half-length of 15.782 as shown below:

\begin{align} l_1 &= \sqrt{\lambda_1\chi^2_{4,0.05}}\\ &= \sqrt{26.245 \times 9.49}\\ &= 15.782 \end{align}

The direction of the axis is given by the first eigenvector. Looking at this first eigenvector we can see large positive elements corresponding to the first three variables. In other words, large elements for Information, Similarities, and Arithmetic. This suggests that this particular axis points in the direction specified by \(e_{1}\); that is, increasing values of Information, Similarities, and Arithmetic.

The half-length of the second longest axis can be obtained by substituting 6.255 for the second eigenvalue, multiplying this by 9.49, and taking the square root. We obtain a half-length of about 7.7, or about half the length of the first axis.

\begin{align} l_2 &= \sqrt{\lambda_2\chi^2_{4,0.05}}\\ &= \sqrt{6.255 \times 9.49}\\ &= 7.705 \end{align}

So, if you were to picture this particular ellipse you would see that the second axis is about half the length of the first and longest axis.

Looking at the corresponding eigenvector, \(e_{2}\), we can see that this particular axis is pointed in the direction of points in the direction of increasing values for the third value, or Arithmetic and decreasing value for Similarities, the second variable.

Similar calculations can then be carried out for the third-longest axis of the ellipse as shown below:

\begin{align} l_3 &= \sqrt{\lambda_1\chi^2_{4,0.05}}\\ &= \sqrt{3.931 \times 9.49}\\ &= 6.108 \end{align}

This third axis has a half-length of 6.108, which is not much shorter or smaller than the second axis. It points in the direction of \(e_{3}\)that is, increasing values of Picture Completion and Information, and decreasing values of Similarities and Arithmetic.

The shortest axis has half-length of about 4.260 as show below:

\begin{align} l_4 &= \sqrt{\lambda_4\chi^2_{4,0.05}}\\ &= \sqrt{1.912 \times 9.49}\\ &= 4.260 \end{align}

It points in the direction of \(e_{4}\) that is, increasing values of Similarities and Picture Completion, and decreasing values of Information.

The overall shape of the ellipse can be obtained by comparing the lengths of the various axis. What we have here is basically an ellipse that is the shape of a slightly squashed football.

We can also obtain the volume of the hyper-ellipse using the formula that was given earlier. Again, our critical value from the chi-square, if we are looking at a 95% prediction ellipse, with four degrees of freedom is given at 9.49. Substituting into our expression we have the product of the eigenvalues in the square root. The gamma function is evaluated at 2, and gamma of 2 is simply equal to 1. Carrying out the math we end up with a volume of 15,613.132 as shown below:

\begin{align} \frac{2\pi^{p/2}}{p\Gamma\left(\frac{p}{2}\right)}(\chi^2_{p,\alpha})^{p/2}|\Sigma|^{1/2} &=  \frac{2\pi^{p/2}}{p\Gamma\left(\frac{p}{2}\right)}(\chi^2_{p,\alpha})^{p/2}\sqrt{\prod_{j=1}^{p}\lambda_j} \\[10pt] &= \frac{2\pi^2}{4\Gamma(2)}(9.49)^2\sqrt{26.245 \times 6.255 \times 3.932 \times 1.912}\\[10pt] &= 444.429 \sqrt{1234.17086}\\[10pt] &= 15613.132\end{align}