Question

Question:Repeat Exercise 7 for the data in Exercise 2.

Short answer

The variance of the data by \({y_1}\) obtained as: \(92.8869\% \).

Step-by-step solution

Mean Deviation form and Covariance Matrix

The Mean Deviation form of any \(p \times N\)is given by:

\(B = \left[ {\begin{array}{*{20}{c}}{{{\hat X}_1}}&{{{\hat X}_2}}&{........}&{{{\hat X}_N}}\end{array}} \right]\)

Whose \(p \times p\)covariance matrix is:

\(S = \frac{1}{{N - 1}}B{B^T}\)

The Change in Variance

From exercise 2, the maximum eigenvalue is:

\({\lambda _1} = 21.9213\)

The respective unit vector is:

\({u_1} = \left[ {\begin{array}{*{20}{c}}{0.44013}\\{0.897934}\end{array}} \right]\)

The new variable will be:

\({y_1} = 0.44013{x_1} + 0.897934{x_2}\)

Now, the percentage of change in variance can be obtained as:

\[\begin{array}{c}\Delta = \frac{{{\lambda _1}}}{{tr\left( S \right)}} \times 100\\ = \frac{{21.9213}}{{5.6 + 18}} \times 100\\ = 92.8869\% \end{array}\]

Hence, this is the required answer.