Answer :
To determine the correlation coefficient for the data in the table, follow these steps:
1. Understand the Data: The data provided is:
[tex]\[ \begin{array}{|c|c|} \hline x & y \\ \hline 0 & 15 \\ \hline 5 & 10 \\ \hline 10 & 5 \\ \hline 15 & 0 \\ \hline \end{array} \][/tex]
2. Calculate the Means: Calculate the mean (average) of [tex]\( x \)[/tex] and [tex]\( y \)[/tex].
[tex]\[ \bar{x} = \frac{0 + 5 + 10 + 15}{4} = \frac{30}{4} = 7.5 \][/tex]
[tex]\[ \bar{y} = \frac{15 + 10 + 5 + 0}{4} = \frac{30}{4} = 7.5 \][/tex]
3. Compute the Numerator:
The numerator of the correlation coefficient is the covariance of [tex]\( x \)[/tex] and [tex]\( y \)[/tex]:
[tex]\[ \text{Cov}(x, y) = \frac{1}{n-1} \sum_{i=1}^{n} (x_i - \bar{x})(y_i - \bar{y}) \][/tex]
[tex]\[ \sum (x_i - \bar{x})(y_i - \bar{y}) = (0-7.5)(15-7.5) + (5-7.5)(10-7.5) + (10-7.5)(5-7.5) + (15-7.5)(0-7.5) \][/tex]
[tex]\[ = (-7.5)(7.5) + (-2.5)(2.5) + (2.5)(-2.5) + (7.5)(-7.5) \][/tex]
[tex]\[ = -56.25 + -6.25 + -6.25 + -56.25 \][/tex]
[tex]\[ = -125 \][/tex]
Since [tex]\( n = 4 \)[/tex],
[tex]\[ \text{Cov}(x, y) = \frac{-125}{4-1} = \frac{-125}{3} \approx -41.67 \][/tex]
4. Compute the Denominator:
This includes the standard deviations of [tex]\( x \)[/tex] and [tex]\( y \)[/tex]:
[tex]\[ \sigma_x = \sqrt{\frac{1}{n-1} \sum (x_i - \bar{x})^2} \][/tex]
[tex]\[ \sum (x_i - \bar{x})^2 = (0-7.5)^2 + (5-7.5)^2 + (10-7.5)^2 + (15-7.5)^2 \][/tex]
[tex]\[ = 56.25 + 6.25 + 6.25 + 56.25 \][/tex]
[tex]\[ = 125 \][/tex]
[tex]\[ \sigma_x = \sqrt{\frac{125}{3}} \approx 6.45 \][/tex]
Similarly, for [tex]\( \sigma_y \)[/tex]:
[tex]\[ \sigma_y = \sqrt{\frac{1}{n-1} \sum (y_i - \bar{y})^2} = \sigma_x = 6.45 \][/tex]
5. Calculate the Correlation Coefficient:
[tex]\[ r = \frac{\text{Cov}(x, y)}{\sigma_x \sigma_y} \][/tex]
[tex]\[ r = \frac{-41.67}{6.45 \times 6.45} \][/tex]
[tex]\[ r \approx -1.0 \][/tex]
Therefore, the correlation coefficient for the given data is [tex]\( -1.0 \)[/tex], indicating a perfect negative linear relationship between [tex]\( x \)[/tex] and [tex]\( y \)[/tex].
1. Understand the Data: The data provided is:
[tex]\[ \begin{array}{|c|c|} \hline x & y \\ \hline 0 & 15 \\ \hline 5 & 10 \\ \hline 10 & 5 \\ \hline 15 & 0 \\ \hline \end{array} \][/tex]
2. Calculate the Means: Calculate the mean (average) of [tex]\( x \)[/tex] and [tex]\( y \)[/tex].
[tex]\[ \bar{x} = \frac{0 + 5 + 10 + 15}{4} = \frac{30}{4} = 7.5 \][/tex]
[tex]\[ \bar{y} = \frac{15 + 10 + 5 + 0}{4} = \frac{30}{4} = 7.5 \][/tex]
3. Compute the Numerator:
The numerator of the correlation coefficient is the covariance of [tex]\( x \)[/tex] and [tex]\( y \)[/tex]:
[tex]\[ \text{Cov}(x, y) = \frac{1}{n-1} \sum_{i=1}^{n} (x_i - \bar{x})(y_i - \bar{y}) \][/tex]
[tex]\[ \sum (x_i - \bar{x})(y_i - \bar{y}) = (0-7.5)(15-7.5) + (5-7.5)(10-7.5) + (10-7.5)(5-7.5) + (15-7.5)(0-7.5) \][/tex]
[tex]\[ = (-7.5)(7.5) + (-2.5)(2.5) + (2.5)(-2.5) + (7.5)(-7.5) \][/tex]
[tex]\[ = -56.25 + -6.25 + -6.25 + -56.25 \][/tex]
[tex]\[ = -125 \][/tex]
Since [tex]\( n = 4 \)[/tex],
[tex]\[ \text{Cov}(x, y) = \frac{-125}{4-1} = \frac{-125}{3} \approx -41.67 \][/tex]
4. Compute the Denominator:
This includes the standard deviations of [tex]\( x \)[/tex] and [tex]\( y \)[/tex]:
[tex]\[ \sigma_x = \sqrt{\frac{1}{n-1} \sum (x_i - \bar{x})^2} \][/tex]
[tex]\[ \sum (x_i - \bar{x})^2 = (0-7.5)^2 + (5-7.5)^2 + (10-7.5)^2 + (15-7.5)^2 \][/tex]
[tex]\[ = 56.25 + 6.25 + 6.25 + 56.25 \][/tex]
[tex]\[ = 125 \][/tex]
[tex]\[ \sigma_x = \sqrt{\frac{125}{3}} \approx 6.45 \][/tex]
Similarly, for [tex]\( \sigma_y \)[/tex]:
[tex]\[ \sigma_y = \sqrt{\frac{1}{n-1} \sum (y_i - \bar{y})^2} = \sigma_x = 6.45 \][/tex]
5. Calculate the Correlation Coefficient:
[tex]\[ r = \frac{\text{Cov}(x, y)}{\sigma_x \sigma_y} \][/tex]
[tex]\[ r = \frac{-41.67}{6.45 \times 6.45} \][/tex]
[tex]\[ r \approx -1.0 \][/tex]
Therefore, the correlation coefficient for the given data is [tex]\( -1.0 \)[/tex], indicating a perfect negative linear relationship between [tex]\( x \)[/tex] and [tex]\( y \)[/tex].