Welcome to Westonci.ca, where curiosity meets expertise. Ask any question and receive fast, accurate answers from our knowledgeable community. Get the answers you need quickly and accurately from a dedicated community of experts on our Q&A platform. Get precise and detailed answers to your questions from a knowledgeable community of experts on our Q&A platform.
Sagot :
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].
Thanks for using our service. We aim to provide the most accurate answers for all your queries. Visit us again for more insights. Thank you for choosing our platform. We're dedicated to providing the best answers for all your questions. Visit us again. Thank you for choosing Westonci.ca as your information source. We look forward to your next visit.
