Matrices need to be symmetric and positive semi-definite.
a.
- \(\begin{pmatrix}2 && 3 \\ 3 && 3 \end{pmatrix}^T = \begin{pmatrix}2 && 3 \\ 3 && 3 \end{pmatrix}\)
- This matrix is symmetric
- \(\begin{vmatrix}2 && 3 \\3 && 3\end{vmatrix} = 2 \times 3 - 3 \times 3 < 0\)
- Determinante is less than 0, so matrix is not positive semi-definite
- Matrix is not symmetric and semi-definite, therefore it is not a covariance matrix
b.
- \(\begin{pmatrix}7 && 6 \\ 5 && 7 \end{pmatrix}^T = \begin{pmatrix}7 && 5 \\ 6 && 7 \end{pmatrix} \)
- This matrix is not symmetric
- Matrix is not symmetric and semi-definite, therefore it is not a covariance matrix
2.
- Distance = \(\sqrt{(x_1-y_1)^2 + (x_2-y_2)^2}\)
| point | distance from mean | Rank (decreasing probability density) |
|---|---|---|
| 2, 5 | 2.0 | 1 |
| -2, 0 | 5.0 | 2 |
| -5, 3 | 7.0 | 3 |
3.
- 4 variables: p, q, r, s
- \(\begin{pmatrix} \sigma_{p}^2 && \sigma_{pq} && \sigma_{pr} && \sigma_{ps} \\ \sigma_{qp}&& \sigma_{q}^2 && \sigma_{qr} && \sigma_{qs} \\ \sigma_{rp} && \sigma_{rq} && \sigma_{r}^2 && \sigma_{rs} \\ \sigma_{sp} && \sigma_{sq} && \sigma_{sr} && \sigma_{s}^2\end{pmatrix}\)
- Covariance is defined as: \(\sigma_{xy} = \frac{\sum(x_i-\mu_x)(y_i-\mu_y)}{n}\)
4.
1.
Diagram 1 has a positive covariance (up from left to right) so the appropriate covariance matrix is: \(\begin{bmatrix}8 && 7 \\ 7 && 7\end{bmatrix}\)
2.
Diagram 2 has a negative covariance (down from left to right) so the appropriate covariance matrix is: \(\begin{bmatrix}7 && -5 \\ -5 && 6 \end{bmatrix}\)
3.
The covariance is 0 (no slope in distribution) and the points are distributed more along the x axis than the y axis, meaning the x variance is higher than the y variance. The appropriate matrix is therefore: \(\begin{bmatrix} 2 && 0 \\ 0 && 1 \end{bmatrix} \)
4.
This could be solved by process of elimination, but the covariance is 0 (no slope in distribution), and the points are distributed more along the y axis than along the x axis, meaning that the y variance is higher than the x variance. The appropriate matrix is therefore: \(\begin{bmatrix}1 && 0 \\ 0 && 4 \end{bmatrix}\)
5.
a.
Red is the mode (the most common value)
b.
| / | < | > | ||
| Red | Black | White | Total | |
| Nissan | 2 | 0 | 0 | 2 |
| Toyota | 1 | 1 | 1 | 3 |
| Honda | 1 | 1 | 2 | 4 |
| Total | 4 | 2 | 3 | 9 |
c.
- \(\rho(\text{Manufacturer}=\text{honda}) = \frac{4}{9}\)
- \(\rho(\text{Color}=\text{white}) = \frac{3}{9} = \frac{1}{3}\)
d.
- Expected number of points (ENP) = total * probability if variables are independent (which is 1/3 in this case)
- ENP(Manufacturer = Honda) = \(9 \cdot \frac{1}{3} = 3\)
- ENP(Color=White) = \(9 \cdot \frac{1}{3} = 3\)
e.
- degrees of freedom = q = \((m_1-1)(m_0-1) = (3-1)^2 = 4\)