Question

based on the data shown below, calculate the correlation coefficient (to three decimal places)

x	y
5	12.3
6	12.6
7	10.1
8	14.1

Explanation:

Step1: Identify the formula for correlation coefficient

The Pearson correlation coefficient \( r \) is given by:

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

where \( n \) is the number of data points. Here, \( n = 5 \).

Step2: Calculate the necessary sums

First, we list the values of \( x \), \( y \), \( xy \), \( x^2 \), and \( y^2 \):

\( x \)	\( y \)	\( xy \)	\( x^2 \)	\( y^2 \)
5	12.3	\( 5 \times 12.3 = 61.5 \)	\( 5^2 = 25 \)	\( 12.3^2 = 151.29 \)
6	12.6	\( 6 \times 12.6 = 75.6 \)	\( 6^2 = 36 \)	\( 12.6^2 = 158.76 \)
7	10.1	\( 7 \times 10.1 = 70.7 \)	\( 7^2 = 49 \)	\( 10.1^2 = 102.01 \)
8	14.1	\( 8 \times 14.1 = 112.8 \)	\( 8^2 = 64 \)	\( 14.1^2 = 198.81 \)

Now, calculate the sums:

• \( \sum x = 4 + 5 + 6 + 7 + 8 = 30 \)
• \( \sum y = 9.4 + 12.3 + 12.6 + 10.1 + 14.1 = 58.5 \)
• \( \sum xy = 37.6 + 61.5 + 75.6 + 70.7 + 112.8 = 358.2 \)
• \( \sum x^2 = 16 + 25 + 36 + 49 + 64 = 190 \)
• \( \sum y^2 = 88.36 + 151.29 + 158.76 + 102.01 + 198.81 = 699.23 \)

Step3: Substitute the values into the formula

First, calculate the numerator:

$$ n\sum xy - \sum x \sum y = 5 \times 358.2 - 30 \times 58.5 = 1791 - 1755 = 36 $$

Next, calculate the denominator components:

• \( n\sum x^2 - (\sum x)^2 = 5 \times 190 - 30^2 = 950 - 900 = 50 \)
• \( n\sum y^2 - (\sum y)^2 = 5 \times 699.23 - 58.5^2 = 3496.15 - 3422.25 = 73.9 \)

Then, the denominator is:

$$ \sqrt{50 \times 73.9} = \sqrt{3695} \approx 60.787 $$

Finally, calculate \( r \):

$$ r = \frac{36}{60.787} \approx 0.592 $$

Answer:

\( 0.592 \)