Use the given data to Compute the linear correlation coefficient. The linear correlation coefficient for the five pieces of data is enter your response here.
​(Round to three decimal places as​ needed.)
x: 2.2, 3.9, 3, 4.6, 10.5
y: 3.9, 1.5, 3.5, 4.9, 9.3

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?

---

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.

Use the given data to compute the linear correlation coefficient. The linear correlation coefficient for the five pieces of data is (Round to three decimal places as needed.) x: 2.2, 3.9, 3, 4.6, 10.5 y: 3.9, 1.5, 3.5, 4.9, 9.3

Learn how to compute the linear correlation coefficient for the given data points: x = [2.2, 3.9, 3, 4.6, 10.5] and y = [3.9, 1.5, 3.5, 4.9, 9.3]. This statistical calculation uses the linear correlation coefficient formula and is suitable for high school students in Grades 10-12.

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