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Sagot :
Let's determine the correlation coefficient for the given data set and identify which option is the closest:
Given data:
[tex]\[ \begin{array}{|c|c|} \hline x & y \\ \hline 5 & 19 \\ \hline 7 & 17 \\ \hline 10 & 16 \\ \hline 15 & 12 \\ \hline 19 & 7 \\ \hline \end{array} \][/tex]
Step-by-Step Solution:
1. Define the data points:
[tex]\[ x: [5, 7, 10, 15, 19] \][/tex]
[tex]\[ y: [19, 17, 16, 12, 7] \][/tex]
2. Calculate the means of [tex]\(x\)[/tex] and [tex]\(y\)[/tex]:
[tex]\[ \bar{x} = \frac{5 + 7 + 10 + 15 + 19}{5} = \frac{56}{5} = 11.2 \][/tex]
[tex]\[ \bar{y} = \frac{19 + 17 + 16 + 12 + 7}{5} = \frac{71}{5} = 14.2 \][/tex]
3. Compute the covariance of [tex]\(x\)[/tex] and [tex]\(y\)[/tex]:
[tex]\[ \text{Cov}(x, y) = \frac{\sum (x_i - \bar{x})(y_i - \bar{y})}{n-1} \][/tex]
First, we compute [tex]\((x_i - \bar{x})(y_i - \bar{y})\)[/tex] for each data point:
[tex]\[ (5 - 11.2)(19 - 14.2) = (-6.2)(4.8) = -29.76 \][/tex]
[tex]\[ (7 - 11.2)(17 - 14.2) = (-4.2)(2.8) = -11.76 \][/tex]
[tex]\[ (10 - 11.2)(16 - 14.2) = (-1.2)(1.8) = -2.16 \][/tex]
[tex]\[ (15 - 11.2)(12 - 14.2) = (3.8)(-2.2) = -8.36 \][/tex]
[tex]\[ (19 - 11.2)(7 - 14.2) = (7.8)(-7.2) = -56.16 \][/tex]
Sum these products:
[tex]\[ \sum (x_i - \bar{x})(y_i - \bar{y}) = -29.76 - 11.76 - 2.16 - 8.36 - 56.16 = -108.2 \][/tex]
So then,
[tex]\[ \text{Cov}(x, y) = \frac{-108.2}{4} = -27.05 \][/tex]
4. Compute the standard deviations of [tex]\(x\)[/tex] and [tex]\(y\)[/tex]:
[tex]\[ \sigma_x = \sqrt{\frac{\sum (x_i - \bar{x})^2}{n-1}} \][/tex]
Calculate [tex]\((x_i - \bar{x})^2\)[/tex] for each data point:
[tex]\[ (5 - 11.2)^2 = (-6.2)^2 = 38.44 \][/tex]
[tex]\[ (7 - 11.2)^2 = (-4.2)^2 = 17.64 \][/tex]
[tex]\[ (10 - 11.2)^2 = (-1.2)^2 = 1.44 \][/tex]
[tex]\[ (15 - 11.2)^2 = (3.8)^2 = 14.44 \][/tex]
[tex]\[ (19 - 11.2)^2 = (7.8)^2 = 60.84 \][/tex]
Sum these squares:
[tex]\[ \sum (x_i - \bar{x})^2 = 38.44 + 17.64 + 1.44 + 14.44 + 60.84 = 132.8 \][/tex]
Hence,
[tex]\[ \sigma_x = \sqrt{\frac{132.8}{4}} = \sqrt{33.2} \approx 5.76 \][/tex]
For [tex]\(y\)[/tex]:
[tex]\[ \sigma_y = \sqrt{\frac{\sum (y_i - \bar{y})^2}{n-1}} \][/tex]
Calculate [tex]\((y_i - \bar{y})^2\)[/tex] for each data point:
[tex]\[ (19 - 14.2)^2 = (4.8)^2 = 23.04 \][/tex]
[tex]\[ (17 - 14.2)^2 = (2.8)^2 = 7.84 \][/tex]
[tex]\[ (16 - 14.2)^2 = (1.8)^2 = 3.24 \][/tex]
[tex]\[ (12 - 14.2)^2 = (-2.2)^2 = 4.84 \][/tex]
[tex]\[ (7 - 14.2)^2 = (-7.2)^2 = 51.84 \][/tex]
Sum these squares:
[tex]\[ \sum (y_i - \bar{y})^2 = 23.04 + 7.84 + 3.24 + 4.84 + 51.84 = 90.8 \][/tex]
Hence,
[tex]\[ \sigma_y = \sqrt{\frac{90.8}{4}} = \sqrt{22.7} \approx 4.76 \][/tex]
5. Finally, calculate the correlation coefficient [tex]\(r\)[/tex]:
[tex]\[ r = \frac{\text{Cov}(x, y)}{\sigma_x \sigma_y} \][/tex]
Substitute in the values:
[tex]\[ r = \frac{-27.05}{5.76 \times 4.76} \approx \frac{-27.05}{27.38} \approx -0.99 \][/tex]
Based on the calculations, the correlation coefficient is approximately [tex]\(-0.99\)[/tex].
Comparing the calculated value to the given options:
A. -0.985
B. 0.985
C. 0.971
D. -0.971
The option that is closest to [tex]\(-0.99\)[/tex] is [tex]\(\boxed{-0.985}\)[/tex].
Given data:
[tex]\[ \begin{array}{|c|c|} \hline x & y \\ \hline 5 & 19 \\ \hline 7 & 17 \\ \hline 10 & 16 \\ \hline 15 & 12 \\ \hline 19 & 7 \\ \hline \end{array} \][/tex]
Step-by-Step Solution:
1. Define the data points:
[tex]\[ x: [5, 7, 10, 15, 19] \][/tex]
[tex]\[ y: [19, 17, 16, 12, 7] \][/tex]
2. Calculate the means of [tex]\(x\)[/tex] and [tex]\(y\)[/tex]:
[tex]\[ \bar{x} = \frac{5 + 7 + 10 + 15 + 19}{5} = \frac{56}{5} = 11.2 \][/tex]
[tex]\[ \bar{y} = \frac{19 + 17 + 16 + 12 + 7}{5} = \frac{71}{5} = 14.2 \][/tex]
3. Compute the covariance of [tex]\(x\)[/tex] and [tex]\(y\)[/tex]:
[tex]\[ \text{Cov}(x, y) = \frac{\sum (x_i - \bar{x})(y_i - \bar{y})}{n-1} \][/tex]
First, we compute [tex]\((x_i - \bar{x})(y_i - \bar{y})\)[/tex] for each data point:
[tex]\[ (5 - 11.2)(19 - 14.2) = (-6.2)(4.8) = -29.76 \][/tex]
[tex]\[ (7 - 11.2)(17 - 14.2) = (-4.2)(2.8) = -11.76 \][/tex]
[tex]\[ (10 - 11.2)(16 - 14.2) = (-1.2)(1.8) = -2.16 \][/tex]
[tex]\[ (15 - 11.2)(12 - 14.2) = (3.8)(-2.2) = -8.36 \][/tex]
[tex]\[ (19 - 11.2)(7 - 14.2) = (7.8)(-7.2) = -56.16 \][/tex]
Sum these products:
[tex]\[ \sum (x_i - \bar{x})(y_i - \bar{y}) = -29.76 - 11.76 - 2.16 - 8.36 - 56.16 = -108.2 \][/tex]
So then,
[tex]\[ \text{Cov}(x, y) = \frac{-108.2}{4} = -27.05 \][/tex]
4. Compute the standard deviations of [tex]\(x\)[/tex] and [tex]\(y\)[/tex]:
[tex]\[ \sigma_x = \sqrt{\frac{\sum (x_i - \bar{x})^2}{n-1}} \][/tex]
Calculate [tex]\((x_i - \bar{x})^2\)[/tex] for each data point:
[tex]\[ (5 - 11.2)^2 = (-6.2)^2 = 38.44 \][/tex]
[tex]\[ (7 - 11.2)^2 = (-4.2)^2 = 17.64 \][/tex]
[tex]\[ (10 - 11.2)^2 = (-1.2)^2 = 1.44 \][/tex]
[tex]\[ (15 - 11.2)^2 = (3.8)^2 = 14.44 \][/tex]
[tex]\[ (19 - 11.2)^2 = (7.8)^2 = 60.84 \][/tex]
Sum these squares:
[tex]\[ \sum (x_i - \bar{x})^2 = 38.44 + 17.64 + 1.44 + 14.44 + 60.84 = 132.8 \][/tex]
Hence,
[tex]\[ \sigma_x = \sqrt{\frac{132.8}{4}} = \sqrt{33.2} \approx 5.76 \][/tex]
For [tex]\(y\)[/tex]:
[tex]\[ \sigma_y = \sqrt{\frac{\sum (y_i - \bar{y})^2}{n-1}} \][/tex]
Calculate [tex]\((y_i - \bar{y})^2\)[/tex] for each data point:
[tex]\[ (19 - 14.2)^2 = (4.8)^2 = 23.04 \][/tex]
[tex]\[ (17 - 14.2)^2 = (2.8)^2 = 7.84 \][/tex]
[tex]\[ (16 - 14.2)^2 = (1.8)^2 = 3.24 \][/tex]
[tex]\[ (12 - 14.2)^2 = (-2.2)^2 = 4.84 \][/tex]
[tex]\[ (7 - 14.2)^2 = (-7.2)^2 = 51.84 \][/tex]
Sum these squares:
[tex]\[ \sum (y_i - \bar{y})^2 = 23.04 + 7.84 + 3.24 + 4.84 + 51.84 = 90.8 \][/tex]
Hence,
[tex]\[ \sigma_y = \sqrt{\frac{90.8}{4}} = \sqrt{22.7} \approx 4.76 \][/tex]
5. Finally, calculate the correlation coefficient [tex]\(r\)[/tex]:
[tex]\[ r = \frac{\text{Cov}(x, y)}{\sigma_x \sigma_y} \][/tex]
Substitute in the values:
[tex]\[ r = \frac{-27.05}{5.76 \times 4.76} \approx \frac{-27.05}{27.38} \approx -0.99 \][/tex]
Based on the calculations, the correlation coefficient is approximately [tex]\(-0.99\)[/tex].
Comparing the calculated value to the given options:
A. -0.985
B. 0.985
C. 0.971
D. -0.971
The option that is closest to [tex]\(-0.99\)[/tex] is [tex]\(\boxed{-0.985}\)[/tex].
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