Convert the matrix of observations to mean-deviation form, and construct the sample covariance matrix.

matrix ((27  32  27  32  12  32)(13  43  58  28  53  33))

To convert the given matrix of observations into the mean-deviation form and then construct the sample covariance matrix, follow these steps:

### 1. **Given Matrix:**
\[
\mathbf{X} = \begin{pmatrix}
27 & 32 & 27 & 32 & 12 & 32 \\
13 & 43 & 58 & 28 & 53 & 33
\end{pmatrix}
\]

### 2. **Calculate the Mean of Each Row:**

- For the first row:
  \[
  \mu_1 = \frac{27 + 32 + 27 + 32 + 12 + 32}{6} = \frac{162}{6} = 27
  \]

- For the second row:
  \[
  \mu_2 = \frac{13 + 43 + 58 + 28 + 53 + 33}{6} = \frac{228}{6} = 38
  \]

### 3. **Convert to Mean-Deviation Form:**

Subtract the mean of each row from each element in that row.

- For the first row:
  \[
  \mathbf{X}_1 - \mu_1 = \begin{pmatrix}
  27 - 27 & 32 - 27 & 27 - 27 & 32 - 27 & 12 - 27 & 32 - 27 \\
  \end{pmatrix} = \begin{pmatrix}
  0 & 5 & 0 & 5 & -15 & 5 \\
  \end{pmatrix}
  \]

- For the second row:
  \[
  \mathbf{X}_2 - \mu_2 = \begin{pmatrix}
  13 - 38 & 43 - 38 & 58 - 38 & 28 - 38 & 53 - 38 & 33 - 38 \\
  \end{pmatrix} = \begin{pmatrix}
  -25 & 5 & 20 & -10 & 15 & -5 \\
  \end{pmatrix}
  \]

Thus, the mean-deviation form of the matrix is:
\[
\mathbf{X}_{\text{md}} = \begin{pmatrix}
0 & 5 & 0 & 5 & -15 & 5 \\
-25 & 5 & 20 & -10 & 15 & -5
\end{pmatrix}
\]

### 4. **Construct the Sample Covariance Matrix:**

The sample covariance matrix \(\mathbf{S}\) is calculated using the formula:
\[
\mathbf{S} = \frac{1}{n-1} \mathbf{X}_{\text{md}} \mathbf{X}_{\text{md}}^\top
\]
where \(n\) is the number of observations (columns).

Here, \(n = 6\), so:
\[
\mathbf{S} = \frac{1}{5} \mathbf{X}_{\text{md}} \mathbf{X}_{\text{md}}^\top
\]

First, compute the matrix product:
\[
\mathbf{X}_{\text{md}} \mathbf{X}_{\text{md}}^\top = \begin{pmatrix}
(0)^2 + (5)^2 + (0)^2 + (5)^2 + (-15)^2 + (5)^2 & (0)(-25) + (5)(5) + (0)(20) + (5)(-10) + (-15)(15) + (5)(-5) \\
(-25)(0) + (5)(5) + (20)(0) + (-10)(5) + (15)(-15) + (-5)(5) & (-25)^2 + (5)^2 + (20)^2 + (-10)^2 + (15)^2 + (-5)^2
\end{pmatrix}
\]

This simplifies to:
\[
\mathbf{X}_{\text{md}} \mathbf{X}_{\text{md}}^\top = \begin{pmatrix}
275 & -450 \\
-450 & 1125
\end{pmatrix}
\]

Finally, divide by 5:
\[
\mathbf{S} = \frac{1}{5} \begin{pmatrix}
275 & -450 \\
-450 & 1125
\end{pmatrix} = \begin{pmatrix}
55 & -90 \\
-90 & 225
\end{pmatrix}
\]

### 5. **Result:**
The sample covariance matrix is:
\[
\mathbf{S} = \begin{pmatrix}
55 & -90 \\
-90 & 225
\end{pmatrix}
\]

Would you like more details or have any questions about this process?

Here are five related questions you might find useful:
1. How do you interpret the sample covariance matrix in the context of data analysis?
2. What does a negative covariance value indicate between two variables?
3. How is the covariance matrix related to the correlation matrix?
4. How does the sample size affect the covariance matrix?
5. How would the process differ for a dataset with more than two variables?

**Tip:** The covariance matrix is a key component in principal component analysis (PCA), which is widely used in dimensionality reduction techniques.

Learn how to convert a matrix of observations to mean-deviation form and construct the sample covariance matrix step-by-step. This problem covers statistical concepts, matrix operations, and covariance calculations. Suitable for advanced high school or college students.

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