The sample correlation coefficient r for both data sets is 0.7071.
To compute r, we can use the following formula:
r = \frac{\sum_{i=1}^n(x_i-\bar{x})(y_i-\bar{y})}{\sqrt{\sum_{i=1}^n(x_i-\bar{x})^2\sum_{i=1}^n(y_i-\bar{y})^2}}
where:
n is the number of data points
xi and yi are the $i$th data points in the x and y data sets, respectively
x and y are the means of the x and y data sets, respectively
For the first data set, we have:
n = 2
\bar{x} = (7+2)/2 = 4.5
\bar{y} = (3+5)/2 = 4
\sum_{i=1}^n(x_i-\bar{x})(y_i-\bar{y}) = (7-4.5)(3-4) + (2-4.5)(5-4) = -5
\sum_{i=1}^n(x_i-\bar{x})^2 = (7-4.5)^2 + (2-4.5)^2 = 25
\sum_{i=1}^n(y_i-\bar{y})^2 = (3-4)^2 + (5-4)^2 = 4
Therefore, the sample correlation coefficient r for the first data set is:
r = \frac{-5}{\sqrt{25 \cdot 4}} = -0.7071
For the second data set, we have:
n = 2
\bar{x} = (3+2)/2 = 2.5
\bar{y} = (7+5)/2 = 6
\sum_{i=1}^n(x_i-\bar{x})(y_i-\bar{y}) = (3-2.5)(7-6) + (2-2.5)(5-6) = 5
\sum_{i=1}^n(x_i-\bar{x})^2 = (3-2.5)^2 + (2-2.5)^2 = 0.25
\sum_{i=1}^n(y_i-\bar{y})^2 = (7-6)^2 + (5-6)^2 = 4
Therefore, the sample correlation coefficient r for the second data set is:
r = \frac{5}{\sqrt{0.25 \cdot 4}} = 0.7071
As we can see, the sample correlation coefficient r is the same for both data sets.