Please consider the following values for the variables X and Y. Treat each row as a pair of scores for the variables X and Y (with the first row providing the labels "X" and "Y").

X Y
2   7
3   6
4   5
5   4
6   3
7   2

Please calculate Pearson's correlation coefficient (r) for these data and report your answer below. When reporting your answer, please provide three decimal places (if relevant).

Question

Please consider the following values for the variables X and Y.  Treat each row as a pair of scores for the variables X and Y (with the first row providing the labels "X" and "Y").

X   Y
2   7
3   6
4   5
5   4
6   3
7   2

Please calculate Pearson's correlation coefficient (r) for these data and report your answer below.  When reporting your answer, please provide three decimal places (if relevant).

BlackTom AI · Accepted Answer

We start by listing the data pairs for X and Y from the table: (2,7), (3,6), (4,5), (5,4), (6,3), (7,2).
Next, compute the means of X and Y. Since the X values are 2 through 7, the mean of X is (2+3+4+5+6+7)/6 = 27/6 = 4.5. The Y values are 7 through 2 in reverse order, so their sum is also 27 and the mean of Y is 27/6 = 4.5.
Now evaluate the components needed for Pearson's r. For each pair, calculate (Xi - Xmean) and (Yi - Ymean).
Because Yi = 9 - Xi for all data points, Yi - Ymean = (9 - Xi) - 4.5 = 4.5 - Xi. Therefore (Xi - Xmean) equals (Xi - 4.5) and (Yi - Ymean) equals (4.5 - Xi).
The product (Xi - Xmean)(Yi - Ymean) becomes (Xi - 4.5)(4.5 - Xi) = -(Xi - 4.5)^2. Summing across all data points:
- For Xi = 2, (Xi - 4.5) = -2.5, square = 6.25
- Xi = 3, (Xi - 4.5) = -1.5, square = 2.25
- Xi = 4, (Xi - 4.5) = -0.5, square = 0.25
- Xi = 5, (Xi - 4.5) = 0.5, square = 0.25
- Xi = 6, (Xi - 4.5) = 1.5, square = 2.25
- Xi = 7, (Xi - 4.5) = 2.5, square = 6.25
Sum of squares is 6.25 + 2.25 + 0.25 + 0.25 + 2.25 + 6.25 = 17.5. Therefore the sum of (Xi - Xmean)(Yi - Ymean) is -17.5 (since each term is the negative of its squared deviation).
Similarly, sum of (Xi - Xmean)^2 equals 17.5, and sum of (Yi - Ymean)^2 also equals 17.5, due to the symmetric relationship between X and Y in these data.
Pearson's r is defined as the sum of the cross-deviations divided by the square root of the product of the sums of squared deviations: r = Σ[(Xi - Xmean)(Yi - Ymean)] / sqrt[Σ(Xi - Xmean)^2 * Σ(Yi - Ymean)^2].
Plugging in the numbers gives r = (-17.5) / sqrt(17.5 * 17.5) = (-17.5) / 17.5 = -1.0.
Interpreting the result, the data lie on a perfect negative linear relationship, so the correlation is exactly -1.0. In reporting with three decimal places, this would be -1.000.

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