To find the [tex]$r$[/tex]-value, also known as the Pearson correlation coefficient, between the variables [tex]\(x\)[/tex] and [tex]\(y\)[/tex], we follow these steps:
1. List the data points:
Given data points are:
[tex]\[
\begin{array}{|c|c|}
\hline
x & y \\
\hline
4 & 23 \\
\hline
5 & 12 \\
\hline
8 & 10 \\
\hline
9 & 9 \\
\hline
13 & 2 \\
\hline
\end{array}
\][/tex]
2. Calculate the means of [tex]\(x\)[/tex] and [tex]\(y\)[/tex]:
[tex]\[
\bar{x} = \frac{4 + 5 + 8 + 9 + 13}{5} = \frac{39}{5} = 7.8
\][/tex]
[tex]\[
\bar{y} = \frac{23 + 12 + 10 + 9 + 2}{5} = \frac{56}{5} = 11.2
\][/tex]
3. Compute the deviations from the mean for each variable:
[tex]\[
x_i - \bar{x}: \quad [4 - 7.8, 5 - 7.8, 8 - 7.8, 9 - 7.8, 13 - 7.8] = [-3.8, -2.8, 0.2, 1.2, 5.2]
\][/tex]
[tex]\[
y_i - \bar{y}: \quad [23 - 11.2, 12 - 11.2, 10 - 11.2, 9 - 11.2, 2 - 11.2] = [11.8, 0.8, -1.2, -2.2, -9.2]
\][/tex]
4. Multiply these deviations pairwise and sum them up:
[tex]\[
\sum (x_i - \bar{x})(y_i - \bar{y}) = (-3.8 \times 11.8) + (-2.8 \times 0.8) + (0.2 \times -1.2) + (1.2 \times -2.2) + (5.2 \times -9.2)
\][/tex]
[tex]\[
= -44.84 - 2.24 - 0.24 - 2.64 - 47.84 = -97.8
\][/tex]
5. Calculate the squared deviations and their sums:
[tex]\[
\sum (x_i - \bar{x})^2 = (-3.8)^2 + (-2.8)^2 + (0.2)^2 + (1.2)^2 + (5.2)^2 = 14.44 + 7.84 + 0.04 + 1.44 + 27.04 = 50.8
\][/tex]
[tex]\[
\sum (y_i - \bar{y})^2 = (11.8)^2 + (0.8)^2 + (-1.2)^2 + (-2.2)^2 + (-9.2)^2 = 139.24 + 0.64 + 1.44 + 4.84 + 84.64 = 230.8
\][/tex]
6. Compute the standard deviations:
[tex]\[
s_x = \sqrt{\frac{\sum (x_i - \bar{x})^2}{n-1}} = \sqrt{\frac{50.8}{4}} = \sqrt{12.7} \approx 3.564
\][/tex]
[tex]\[
s_y = \sqrt{\frac{\sum (y_i - \bar{y})^2}{n-1}} = \sqrt{\frac{230.8}{4}} = \sqrt{57.7} \approx 7.594
\][/tex]
7. Calculate the Pearson correlation coefficient:
[tex]\[
r = \frac{\sum (x_i - \bar{x})(y_i - \bar{y})}{(n-1)s_x s_y} = \frac{-97.8}{4 \times 3.564 \times 7.594} \approx \frac{-97.8}{108.218} \approx -0.903
\][/tex]
Therefore, the [tex]$r$[/tex]-value of the data is [tex]\( -0.903 \)[/tex] when rounded to three decimal places, making option [tex]\( \text{A. } -0.903 \)[/tex] the correct answer.
