To calculate the linear correlation coefficient, $r$, we can use the following formula:

$r = \frac{n \sum xy - \sum x \sum y}{\sqrt{(n \sum x^2 - (\sum x)^2)(n \sum y^2 - (\sum y)^2)}}$

Where:

• $n$ is the number of data points
• $\sum x$ is the sum of all x-values
• $\sum y$ is the sum of all y-values
• $\sum xy$ is the sum of the product of each pair of x and y values
• $\sum x^2$ is the sum of the squares of the x-values
• $\sum y^2$ is the sum of the squares of the y-values

Step 1: Calculate the sums and products needed

Given dataTo calculate the linear correlation coefficient, $r$, we can use the following formula:

$r = \frac{n \sum xy - \sum x \sum y}{\sqrt{(n \sum x^2 - (\sum x)^2)(n \sum y^2 - (\sum y)^2)}}$

Where:

• $n$ is the number of data points
• $\sum x$ is the sum of all x-values
• $\sum y$ is the sum of all y-values
• $\sum xy$ is the sum of the product of each pair of x and y values
• $\sum x^2$ is the sum of the squares of the x-values
• $\sum y^2$ is the sum of the squares of the y-values

Step 1: Calculate the sums and products needed

Given $x = [2.2, 3.9, 3, 4.6, 10.5]$ $y = [3.9, 1.5, 3.5, 4.9, 9.3]$

Now, calculate the required values:

$\sum x = 2.2 + 3.9 + 3 + 4.6 + 10.5 = 24.2$ $\sum y = 3.9 + 1.5 + 3.5 + 4.9 + 9.3 = 23.1$ $\sum x^2 = 2.2^2 + 3.9^2 + 3^2 + 4.6^2 + 10.5^2 = 4.84 + 15.21 + 9 + 21.16 + 110.25 = 160.46$ $\sum y^2 = 3.9^2 + 1.5^2 + 3.5^2 + 4.9^2 + 9.3^2 = 15.21 + 2.25 + 12.25 + 24.01 + 86.49 = 140.21$ $\sum xy = (2.2)(3.9) + (3.9)(1.5) + (3)(3.5) + (4.6)(4.9) + (10.5)(9.3)$ $= 8.58 + 5.85 + 10.5 + 22.54 + 97.65 = 145.12$

Step 2: Substitute into the correlation coefficient formula

[ r = \frac{5(145.12) - (24.2)(23.1)}{\sqrt{[5(160.46) - (24.2)^2][5(140.21) - (23.1)^2]}} ]

First, calculate the numerator:

$5(145.12) = 725.6, \quad 24.2 \times 23.1 = 559.02$ $\text{Numerator} = 725.6 - 559.02 = 166.58$

Now, calculate the denominator:

$5(160.46) = 802.3, \quad (24.2)^2 = 585.64$ $5(140.21) = 701.05, \quad (23.1)^2 = 533.61$ $\text{Denominator} = \sqrt{(802.3 - 585.64)(701.05 - 533.61)} = \sqrt{216.66 \times 167.44} = \sqrt{36270.41} = 190.44$

Step 3: Final calculation of $r$

$r = \frac{166.58}{190.44} = 0.8747$

Rounding to three decimal places:

$r = 0.875$

The linear correlation coefficient is $\mathbf{0.875}$.

Would you like more details or have any questions?

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Here are 5 related questions to consider:

1. How is the linear correlation coefficient interpreted in terms of strength and direction of the relationship?
2. What is the difference between a positive and negative correlation?
3. How would you compute a linear regression line using this data?
4. How does the correlation coefficient change when a data point is added or removed?
5. What assumptions must be true for the correlation coefficient to be valid?

Tip: Always check the scatter plot of the data to visually inspect the relationship before computing the correlation coefficient.